The original of this book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924001080237 HonUoti : C. J. OLAY AND SONS, CAMBRIDGE UNIVERSITY PRESS WAREHOUSE, AVE MARIA LANE. CAMBRIDGE : DEIGHTON, BELL, AND CO. LEIPZIG: F. A. BROCKHAUS. NEW YORK: MACMILLAN AND CO. A TEEATISE ON ANALYTICAL STATICS WITH NUMEKOUS EXAMPLES EDWARD JOHN ROUTH, Sc.D., LL.D, MA., F.R.S., &c, HON. FELLOW OF PETEKHOUSE, OAMBEIDGB ; FELLOW OF THE TJNIVEBSITY OF LONDON. VOLUME I. CAMBRIDGE AT THE UNIVERSITY PRESS 1891 [All Bights reserved] (lamlrcitfge : PRINTED BY 0. J. OLAY, M.A. AND SONS, AT THE UNIVERSITY PRESS. PKEFACE. 1~\URING many years it has been my duty and pleasure t< give courses of lectures on various Mathematical subjects to successive generations of students. The course on Static! has been made the groundwork of the present treatise. It hai however been necessary to make many additions ; for in a treatis< all parts of the subject must be discussed in a connected form while in a series of lectures a suitable choice has to be made. A portion only of the science of Statics has been included ii this volume. It is felt that such subjects as Attractions, Astatics and the Bending of rods could not be adequately treated at th< end of a treatise without either making the volume too bulk] or requiring the other parts to be unduly curtailed. These re maining portions will appear in a second volume. Considerabli progress has already been made and I hope that this additiona volume will soon be ready. In order to learn Statics it is essential to the student to worl numerous examples. Besides some of my own construction, ! have collected a large number from the University and Collegi Examination papers. Some of these are so good as to deserve t< rank among the theorems of the science rather than among th< examples. Solutions have been given to many of the examples sometimes at length and in other cases in the form of hints whei these appeared sufficient. E. s. b VI PREFACE. I have endeavoured to refer each result to its original author. I have however found that it is a very difficult task to effect this with any completeness. The references will show that I have searched many of the older books and memoirs as well as some of those of recent date to discover the first mention of a theorem. When so many analytical results are given it is improbable that there should be no errors. The assistance given to me by Mr E. G. Gallop of Gonville and Caius College in correcting the proof sheets enables me to hope that none of these imperfections are serious. I cannot conclude without expressing how much I am indebted to him for his help and the many valuable suggestions which he has made. EDWARD J. ROUTH. Peteehouse, January 3, 1891. CONTENTS. CHAPTER I. THE PARALLELOGRAM OF FORCES. ABTS. PAGES 1 — 12. Elementary considerations on forces, &c 1 8 13 — 18. Dynamical and statical laws 8 12 19 — 21. Kigid bodies 12 14 22 — 23. Eesultant forces 14 15 24 — 30. Parallelogram of forces 15 — 19 31. Historical summary 19 — 20 CHAPTER II. FORCES ACTING AT A POINT. 32 — 40. Geometrical Method. Triangle, Polygon and Parallelepiped of forces 21—25 41 — 46. Method of analysis. Eesultant of any number of forces 25 — 28 47 — 50. Forces which act normally to the faces of a polyhedron. Applications to a tetrahedron. See also Art. 319 . . 28 — 30 51 — 53. The mean centre. Its use in resolving and compounding forces 30—33 54 — 61. Equilibrium of a particle. Smooth curves and surfaces . 33 — 38 62—69. The principle of Work 38—41 70 — 74. Astatic equilibrium. The astatic triangle of forces . . 41 — 44 75 — 77. Stable and unstable equilibrium. A free body under two and three forces 44 — 46 vm CONTENTS. CHAPTER III. PARALLEL FORCES. AKTS. 78—81. Besultant of two and any number of parallel forces 82 — 84. The centre of parallel forces 85. Conditions of equilibrium . 86. A body suspended from a fixed point 87 — 88. A body resting on a plane . 89—103. Theory of couples PAGES 47—50 50—51 51—52 52—54 54—55 55—65 CHAPTER IV. FORCES IN TWO DIMENSIONS. 104 — 108. Besultant of any number of forces 66—68 109 — 115. Conditions of equilibrium 68 — 70 116—117. Varignon's theorem 70—72 118 — 120. The single resultant force or couple 72—73 121—129. The solution of problems. Constrained rods, discs, spheres, &c. See also note at page 403 . . . 74 — 83 130. Equilibrium of four repelling or attracting particles . . 83—84 131 — 133. Eeactions at joints. Theorems of Euler, Fuss &o. A note on a property of triangles 84 — 91 134 — 140. Polygon of heavy rods. Funicular polygon. See also Arts. 357—358 91—94 141. Examples 94—96 142 147. Eeactions at rigid connections. Bending moment. Beam with weights, diagram of stress. Various-examples . 96 — 102 148—149. Indeterminate problems 102 — 105 150 — 152. Frameworks. Conditions of stiffness .... 105 — 107 153. Equations of equilibrium of a body or framework derived from those of a single particle. See foot-note . . 107 — 108 154 — 155. Sufficiency of the equations to find the reactions when the frame is just stiff. Exceptional cases. See also Arts. 235, 236. Supplementary equations when the frame is overstiff 108—109 156 — 163. Astatics. Astatic resultant. Centre of forces, &o. . . 109 — 113 CHAPTER V. ON FRICTION. 164 — 171. Laws of friction. The friction force and friction couple. The angle of friction and limiting friction . . . 114 — 119 172 — 175. Particle constrained by a rough curve, surface, &c. The cone of friction 119 — 121 CONTENTS. IX ARTS. PAGES 176 — 178. Problems on friction when the direction is known. Dif- ferent methods of solution 121 — 130 179. Wheel and axle with friction 130—131 180 — 187. Friction in unknown directions. Two methods . . 132 — 137 188 — 190. Examples. Eotation of bodies on » finite and an infinite number of supports. String of particles, &o. . . 137 — 145 CHAPTER VI. THE PRINCIPLE OP WORK. 191—195. Proof of the principle 146—149 196 — 198. The forces which do not appear in the equation of work . 150—153 199. One sided constraints 153 — 154 200. Converse of the principle of work 154 — 155 201—202. Initial motion 155 203 — 204. Equations of equilibrium derived from work . . . 155 — 157 205. Examples on the principle 157 — 160 206—212. The work function 160— 16$ 213 — 225. Stable and unstable equilibrium. Analytical method . 164—171 226—228. Attracting or repelling atoms 171 — 173 229 — 234. Determination of stress in a simply stiff frame. Examples 173—180 235—236. Abnormal deformations 180—182 237 — 238. Indeterminate tensions. Theorems of Crofton and Levy. See also Art. 366 182—184 239 — 243. Geometrical method of determining the stability of a body. The circle of stability 184—188 244—246. Booking stones. Examples 188 — 191 247 — 253. Booking stones, spherical and not spherical to a second approximation in two and three dimensions . . . 191—196 254 — 256. Lagrange's proof of the principle of virtual work . . 196 — 197 CHAPTER VII. FORCES IN THREE DIMENSIONS. 257—259. Eesultants of a system of forces. Conditions of equi- librium 198—201 260—262. Components of a force 201—202 263—267. Moment of a force 202—205 268—269. Problems on equilibrium.. Pressures on an axis, rods, spheres, &c 205—209 270. Poinsot's central axis 209 — 210 271 278. The equivalent wrench. Analytical method . . . 210—214 279—283. The Invariants 214—217 284—290. The Cylindroid 217—221 CONTENTS. ARTS. 291—293. 294—297. 298—302. 303—311. 312—314. 315. 316—319. 320—323. 324—335. 336—338. 339. PAGES Work of a wrench 221—222 Reciprocal screws .... ... 222—223 The Nul plane 224—225 Conjugate forces 225—229 Equivalent wrench of two forces. Geometrical method . 229 — 231 Theorems on three forces 231—232 Four forces. The hyperboloid. Forces acting normally to the faces of a tetrahedron. See also Art. 50. Forces normal to a surface 232 — 234 Five forces. The two directors 235 — 236 Six forces. The polar plane. The transversals and other lines. Forces in involution. Eesolution of a force along six arbitrary lines 236 — 242 Theorems on the conditions of equilibrium of a system of forces 242—244 Tetrahedral coordinates 244 — 246 CHAPTER VIII. GRAPHICAL STATICS. 340 — 348. Reciprocal figures. Mode of drawing .... 247 — 252 349_351. Maxwell's theorem 252—254 352. Cremona's theorem 254 — 255 353 — 356. Graphical construction of the resultant force. Conditions of equilibrium . . 255 — 258 357. Parallel forces. The funicular polygon. See also Arts. 134—140 258—259 358. Theorems and examples on forces 259 — 261 359 — 363. Reactions of frameworks 261 — 265 364—365. Method of sections 265—266 366—367. Indeterminate tensions. See also Arts. 237, 238 . . 266 — 267 368—372. Line of pressure 267—269 CHAPTER IX. CENTRE OF GRAVITY. 373 — 379. Definitions and fundamental equations 380 — 382. Working rule. Examples 383 — 386. Area and perimeter of a triangle 387 — 388. Quadrilateral areas 389 — 395. Tetrahedron. Volume, faces, and edges. Pyramid, cone and double tetrahedra. Isosceles tetrahedron 396 — 399. Centres of gravity of arcs. Circle, catenary, cycloid, &c. 400—402. Centres of gravity of circular areas .... 403 — 408. Geometrical and analytical projection of areas . 270—273 273—274 275—277 277—279 279—282 282—284 284—285 285—288 m CONTENTS. XI ABTS. PAGES 409—412. Centre of gravity of any area 288—291 413—417. Theorems of Pappus 291—294 418—419. Areas on the surface of a right cone 295—297 420 — 424. Areas on a sphere 297 — 301 425 — 427. Surfaces and solids of revolution. Moments and products of inertia . 301 — 303 428—430. Ellipsoidal volumes and shells 303 — 305 431—432. Any surfaces and volumes 306 — 307 433 — 434. Heterogeneous bodies. Octant of ellipsoid. Triangular area, &a. . . '. 307 — 310 435 — 438. Lagrange's two theorems 310—311 439 — 441. Applications to pure geometry 312 — 314 CHAPTER X. ON STRINGS. 442 — 445. Catenary. Equations and properties 315 — 319 446 — 448. Problems on free and constrained catenaries . . . 319 — 325 449. Stability of catenaries 326 450 — 453. Heterogeneous chains. Cycloid, parabola, and the catenary of equal strength 326—330 454. String under any forces. General intrinsic equations . 330 — 331 455 — 456. General Cartesian equations 331 — 333 457 — 458. Constrained strings. Light string on a smooth curve. Problem of Bernoulli 333—334 459 — 462. Heavy string on a smooth curve. The anti-centre and statical directrix. Examples 335 — 339 463 — 466. Light string on a rough curve. Bough pegs . . . 339 — 342 467 — 471. Heavy string on a rough curve 342 — 346 472 — 473. Endless strings. Strings which just fit a curve . . . 346 — 348 474 — 477. Central forces. Various laws of force. Kinetic analogy. Two centres 349—354 478 — 480. String on a smooth surface. Cartesian and Intrinsic equations. Various theorems. Case of no forces . . 354 — 357 481 — 482. String on a surface of revolution. Sphere .... 357 — 359 483. String on a cylinder 359—360 484. String on a right cone 360 — 361 485—487. String on a rough surface 361—363 488. Curved string on a rough plane and other examples . . 363 — 364 489—491. Elastic string. Hooke's law 364—366 492. Heavy elastic string suspended by one end .... 366 — 367 493. Work of an elastic string 367—369 494. Heavy elastic string on a smooth curve. The statical directrix 369—370 495. Light elastic string on a rough curve. Bough pegs . . 370—371 496 — 499. Elastic string under any forces 371—373 500—501. Elastic catenary 373—374 Xll CONTENTS. CHAPTER XL THE MACHINES. ARTS. PACES 502—505. Mechanical advantage. Efficiency 375—376 506 — 516. The Lever. Conditions of equilibrium. Pressure on axis. Various kinds of Levers. For lever under any forces see Art. 268 ; rough axis Art. 179 377—380 517. Eoberval's balance 380—381 518—520. The common balance 381—383 521—524. The common and Danish Steelyard 383—386 525—528. The single pulley 386—387 529—530. The system of pulleys with one rope . . . . 387—389 531. Eigidity of cords 389 532 — 536. Systems of pulleys with several ropes ... . 389 — 394 537—539. The inclined plane 394—396 540—543. The wheel and axle. The differential axle . . . 396—397 544—546. Toothed wheels 397—400 547—549. The wedge 400—402 550—553. The screw . . 402—404 Note on two theorems in Conics assumed in Arts. 126, 127. 405—407 EBKATA. Page 24, lines 10 and 11, interchange the letters M and N. Page 36, last line, for W read w. V Page 65, line 11, for direction read same direction. * Page 192, line 18, for non-spherical read unsymmetrical. * / CHAPTER I. THE PARALLELOGRAM OF FORCES. 1. The science of Mechanics treats of the action of forces on bodies. Under the influence of these forces the bodies may either be in motion or remain at rest. That part of mechanics which treats of the motion of bodies is called Dynamics. That part of mechanics in which the bodies are at rest is called Statics. If the determination of the motion of bodies under given forces could be completely and easily solved, there would be no obvious advantage in this division of the subject into two parts. It is clear that statics is only that particular case of dynamics in which the motions of the bodies are equated to zero. But the particular case in which the motion is zero presents itself as a much easier problem than the general one. At the same time this particular case is one of great importance. It is important not merely for the intrinsic value of its own results but because these are found to assist in the solution of the general case in a very remarkable manner. It has therefore been generally found convenient to lead up to the general problem of dynamics by considering first the particular case of statics. 2. Since statics is a particular case of dynamics we may begin by discussing the first principles of the more general science. We should consider how the mass of a body is measured, how the velocity and acceleration of any particle are affected by the action of forces. The general principles having been obtained we may then descend to the particular case by putting these velocities equal to zero. In this way the relationship of the two great branches of mechanics is clearly seen and their results are founded on a common basis. » R. S. 1 2 THE PARALLELOGRAM OF FORCES. [CHAP. I. 3. There is another way of studying statics which has its own advantages. We might begin by assuming some simple axioms relating to the action of forces on bodies without introducing any properties of motion. In this method we introduce no terms or principles but those which are continually used in statics, leaving to dynamics the study of those terms which are peculiar to it. Whether this is an advantageous method of studying statics or not depends on the choice of the fundamental axioms. In the first place they must be simple in character. In the second place they must be easily verified by experiment. For example we might take as an axiom the proposition usually called the parallelo- gram of forces or we might, after Lagrange,, start from the principle of work. But neither of these principles satisfies the conditions just mentioned, for they do not seem sufficiently obvious on first acquaintance to command assent. If we found the two parts of mechanics on a common basis, that basis must be broader than that which is necessary to support merely the principles of statics. We have to assume at once all the experimental results required in mechanics instead of only those required in statics. Now there is an advantage in intro- ducing the fundamental experiments required only as they are wanted. We thus more easily distinguish the special necessity for each, we see more clearly what results are deduced from each experiment. The order of proceeding would be to begin with such elementary axioms about forces as will enable us to study their composition and resolution. Presently other experimental results are introduced as they are required and finally when the general problem of dynamics is reached,, the whole of the funda- mental axioms are summed up and consolidated. In a treatise on statics it is necessary to consider both these methods. We shall examine first how the elementary principles of statics are connected with the axioms required for the more general problem of dynamics, and secondly how they may be made to stand on a base of their own. 4. In mechanics we have to treat of the action of forces on bodies. The term force is defined by Newton in the following terms, ART. 7.] CHARACTERISTICS OF A FORCE. 3 An impressed force is an action exerted on a body in order to change its state either of rest or of uniform motion in a straight line. 5. Characteristics of a Force. When a force acts on a body the action exerted has (1) a point of application, (2) a direction in space, (3) magnitude. Two forces are said to be equal in magnitude when, if applied to the same particle in opposite directions, they balance each other. The magnitudes of forces are measured by taking some one force as a unit, then a force which will balance two unit forces is represented by two units and so on. 6. The simplest appeal to our experience will convince us that many at least of the ordinary forces of nature possess these three characteristics. If force be exerted on a body by pulling a string attached to it, the point of attachment of the string is the point of application, and the direction of the string is the direction of the force. The existence of the third element of a force is shown by the fact that we may exert different pulls on the string. The origins of all the causes which produce or tend to produce, motion in a body a:ce not known. But as they are studied, it is found that they can be analysed into simpler causes, and these simpler causes are seen to have the three characteristics of a force. If there be any causes of motion which cannot be thus analysed, such causes are not considered as forces whose effects are to be discussed in the science of statics. 7. There are other things besides forces which possess these three characteristics. These other things may be used to help us in our arguments about' forces so far as their other properties are common also to forces. The most important of these analogies is that of a finite straight line. Let this finite straight line be A B. One extremity A will represent the point of application. The direction in space of the straight line will represent the direction of the force and the length of the line will represent the magnitude of the force, 1—2 4 THE PARALLELOGRAM OF FORCES. [CHAP. I. Other things besides forces may also be represented graphically by a finite straight line. Thus in dynamics it will be seen that both the velocity and the momentum of a particle have direction and magnitude and may in the same way be represented by a finite straight line. One extremity A is placed at the particle, the direction of the straight line represents the direction of the velocity and the length represents the magnitude. Generally this analogy is useful whenever the things considered obey what we shall presently call the parallelogram law. 8. In order to represent completely the direction of a force by the direction of the straight line AB, it is necessary to have some convention to determine whether the force pulls A in the direction AB or pushes A in the direction BA. This convention is supplied by the use of the terms positive and negative. The positive and negative directions of straight lines being defined by some conven- tion or rule, the forces which act in the positive directions of their lines of action are called positive and those in the opposite direc- tions are called negative. These conventions are often indicated by the conditions of the problem under consideration, but they usually agree with the rules adopted in the differential calculus. Thus the direction of the radius vector drawn from the origin is usually taken as the positive direction, and so on for all other lines. • Sometimes instead of using the term positive, the direction or sense of a force is indicated by the order of the letters, thus a force AB is a force acting in the direction A to B, a force BA is a force acting from B towards A. 9. The third element of a force is its magnitude. This is represented by the length of the representative straight line. A unit of force is represented by a unit of length on any scale we please; a force of n such units of force is then represented by a straight line of n units of length. 10. Measure of a force. We may enquire how we are to measure the magnitude of a force. A force must be measured by its effects. Since a force may produce many effects there are several methods open to us. If we wish the measure of two equal forces acting together to be twice that of a single force equal A ? T - 11] MEi^jjURE OF A FORCE. 5 to either, the effect which is to measure the force must be properly chosen. We may measure a force by the weight of the mass which it will support. Placing two equal masses side by side, they will be supported by equal forces. Joining these together we see that a double force will support a double mass. Thus the effect is proportional to the magnitude of the cause. "We may also measure a force by the motion it will produce in a given body in a given time. If by motion is here meant velocity then it may be shown by the experiments usually quoted to prove the second law of motion that a double force will produce a double velocity. So here also the effect chosen as the measure is propor- tional to the magnitude of the cause. This measure requires some experimental results, necessary for dynamics, but not used after- wards in statics. If we agree to measure a force by the weight it will support the unit will depend on the force of gravity at the place where the experiment is made. Such a unit will therefore present several inconveniences. If also we measure a force by the velocity generated in a unit of mass in a unit of time, it is necessary to discuss how these other units are to be chosen. It is not necessary for as, at this stage of our argument, to decide on the best method of measuring a force. It will be presently seen that our equations are concerned for the most part with the ratios of forces rather than with the forces themselves. The choice of the actual unit is therefore unimportant at present, and we could leave this choice until the proper occasion arrives. The comparative effects of forces will then have been discussed, and the reader will the better understand the reasons why any particular choice is made. When therefore we speak of several forces equal to the weight of one, two or three pounds &c, acting on a body and determine the conditions of equilibrium, we shall find that the same conditions are true for forces equal to the weight of one, two or three oz. &c, and generally of all forces in the same ratio. 11. One system of units is that based on the foot, pound, and second as the three fundamental units of length, mass, and time. The unit force is that force which acting on a pound of matter for one second generates a velocity of one foot per second. This unit of force is called the poundal. The foot and the pound are defined by certain standards kept in a place of security for reference. Thus the imperial yard is the 6 THE PARALLELOGRAM OF FORCES. [CHAP. I. distance between two marks on a certain bar, preserved in the Tower of London, when the whole bar has a temperature of 62° Fah. The unit of time is a certain known fraction of a mean solar day. The units committee of the British Association recommended the general adoption of the centimetre the gramme and the second as the three fundamental units of space, mass and time. These they proposed should be distinguished from absolute units, otherwise 1 derived, by the letters c. G. s. prefixed, these being the initial letters of the names of the three fundamental units. The C. G. s. unit of force is called a dyne. This is the force which acting on a gramme for a second generates the velocity of a centimetre per second. It is found by experiment that a body, say a unit of mass, falling in vacuo for one second acquires very nearly a velocity of 3219 feet per second. This velocity is the same as 981 - 17 centimetres per second. It follows therefore that a poundal is about ^gth part of the weight of one pound, and a dyne is the weight of -g^xth part of a gramme. These numerical relations strictly apply only to the place of observation, for the force of gravity is not the same at all places on the earth. The difference between the greatest and least values of gravity is about ^th of its mean value. The relations of these two systems of units are given in the ordinary tables of units. We have nearly one inch = 2*54 centimetres, one pound = 453 - 59 grammes. It follows from what precedes that one poundal = 13825 dynes. The reader is referred to Prof. Everett's treatise on Units and Physical Constants for minute particulars on the relations of the C. G. s. units. 12. The parallelogram of velocities. This proposition is preliminary to Newton's laws of motion. The velocity of a particle when uniform is measured by the space described in a given time. A straight line whose length is equal to this space will represent the velocity in direction ART. 12.] THE PARALLELOGRAM OF VELOCITIES. \ and magnitude; Art. 8. Suppose a particle to be carriec uniformly in the given time from to G, then OG represents its velocity. This change of place may be effected by moving the particle in the same time from to A along the straight line OA, if while this is being done we move the straight line OA (with the particle sliding on it) parallel to itself from the position OA to the position BG. The uniform motion of the particle from to A is expressed by the statement that it* velocity is represented by OA. The displacement produced bj the uniform motion of the straight line is expressed by the state ment that the particle has a velocity represented in direction anc magnitude by either of the sides OB or A G. It is evident by thi properties of similar figures that the path of the particle in space is the straight line OG. It follows that when a particle move, with two simultaneous velocities represented in direction anc magnitude by the straight lines OA, OB its motion is the sanu as if it were moved with a single velocity represented in direction and magnitude by the diagonal OC of the parallelogram describee on OA, OB as sides. This proposition is usually called the parallel ogram of velocities. Let a particle move with three simultaneous velocities repre sented in direction and magnitude by the three straight line (Mi, OA%, OA s . We may replace the two velocities OA^', OA by the single velocity represented in direction and magnitude by the diagonal 0B y of the parallelogram described on OA lt OA as sides. The particle now moves with the two simultaneous velocities represented by 0B± and OA s . We may again use th< same rule. We replace these two velocities by the single velocit; represented in direction and magnitude by the diagonal OB described on 0B X and OA s as sides. We have thus replacec the three given simultaneous velocities by a single velocity. In the same way any number of simultaneous velocities maj be replaced by a single velocity. If the simultaneous velocities represented by OA^, 0A. 2 &c were all altered in the same ratio, it is evident from the propertie, of similar figures that the resulting single velocity will also be altered in the same ratio. 8 THE PARALLELOGRAM OF FORGES. [CHAP. I. Let the simultaneous velocities 0A lt 0A 2 &c. be such that their resulting velocity is zero. It follows that if all the velocities OA lt 0A 2 &c. are altered in any, the same, ratio the resulting velocity is still zero. 13. Newton's laws of Motion. These are given in the introduction to the Principia. 1. Every body continues in its state of rest or of uniform motion in a straight line, except in so far as it may be compelled by force to change that state. 2. Change of motion is proportional to the force applied and takes place in the direction of the straight line in which the force acts. 3. To every action there is always an equal and contrary reaction ; or the mutual actions of any two bodies are always equal and oppositely directed. The full significance of these laws cannot be understood until the student takes up the subject of dynamics. The experiments which suggest these laws, and their further verification, are best studied in connection with that branch of the science, and are to be found in books on elementary dynamics. The student who has not already read some such treatise is advised to assume the truth of these laws for the present. We shall accordingly not enter into a full discussion of them in this treatise, but we shall confine our remarks to those portions which are required in statical problems. 14. The first law asserts the inertness of matter. A body at rest will continue at rest unless acted on by some external force. At first sight this may appear to be a repetition of the definition of force, since any cause which tends to move a body at rest is called a force. But it is not so. Here we assert as the result of observation or experiment the inertness of each particle of matter. It has no tendency to move itself, it is moved only by the action of causes external to itself. 15. In the second law of motion the independence of forces which act on a particle is asserted. If the effect of a force is always proportional to the force impressed it is clearly meant that each force must produce its own effect in direction and mag- nitude as if it acted singly on the particle placed at rest. ART 15.] NEWTON'S LAWS OF MOTION. 9 Let us consider the meaning of this statement a little more fully. Let a force act on a given particle placed at rest at a point and generate in a given time a velocity which we may represent graphically by the straight line OA. Let a second force act on the same particle again placed at rest at and generate in the same time a velocity which we may represent by OB. If both forces act simultaneously on the particle both these velocities are generated. The actual velocity of the particle is then represented by the diagonal OC of the parallelogram described on OA, OB as sides, Art. 12. In the same way, if any number of forces act simultaneously on a particle at rest, the law directs that we are to determine the velocity generated by each as if it acted alone for a given time. These separate velocities are then to be combined into a single velocity in the manner described in Art. 12. This single velocity is asserted to be the effect of the simultaneous action of the forces. Let a system of forces be such that when they act simul- taneously on a particle placed at rest the resulting velocity of the particle is zero. These forces are then in equilibrium. Let a second system of forces be also such that when they act on the particle placed at rest, the resulting velocity of the particle is again zero. Then this second system of forces is also in equi- librium. Let these two systems act simultaneously, then since the forces do not interfere with each other, the resulting velocity of the particle is still zero. We thus arrive at the following important proposition. Let us suppose that there are two systems of forces each of which when acting alone on a particle would be in equilibrium. It follows that when both systems act simultaneously there will still be equi- librium. This is sometimes called the principle of the superposition of forces in equilibrium. When we are trying to find the conditions of equilibrium of some system of forces, the principle enables us to simplify the problem by adding on or removing any particular forces which by themselves are in equilibrium. Let the forces P 1( P 2 &c. acting on a given particle for a given time generate velocities v u i> 2 &c. respectively. If the same or equal forces were made to act on a different particle the velocities generated in the same time may be different. But since the effect 10 THE PARALLELOGRAM OF FORCES. [CHAP. I. of each force is proportional to its magnitude the velocities gene- rated by the several forces are to each other in the ratios of ^ to « 2 to v s &c. If then a system of forces is in equilibrium when acting on any one particle, that system will also be in equilibrium when applied to any other particle (Art. 12). 16. We notice also that it is the change of motion which is the effect of force. A given force produces the same change of motion in a particle whether that particle is in motion or at rest. In this way we can determine whether a moving particle is acted on by any external force or not. If the velocity is uniform and the path rectilinear there is no force acting on the particle. If either the velocity is not uniform, or the path not rectilinear, there must be some force acting to produce that change. Let two equal forces act one on each of two particles and generate in the same time equal changes of velocity ; these particles are said to have equal mass. If the force acting on one particle must be n times that on the other in order to generate equal changes of velocity in equal times, the mass of the first particle is u times that of the second. It follows that the mass of a particle is proportional to the force required to generate in it a given change of velocity in a given time. Now all bodies falling from rest in a vacuum under the attraction of the earth are found to have the same velocity at the end of the first second of time, Art. 11. We therefore infer that the masses of bodies are proportional to their weights. The units of mass and force are so chosen that the unit of force acting on the unit of mass will generate a unit of velocity in a unit of time. The product of the mass of a particle into its velocity is called its momentum. It follows from what has just been said that the expression "change of motion" means change of momentum produced in a given time. These results are peculiarly important in dynamics, but in statics, where the particles acted on are all initially at rest and remain so, they have not the same significance. 17. In the third law the principle of the transmissibility of force is implied. The principle is more clearly stated in the remarks which Newton added to his laws of motion. The law asserts the equality of action and reaction. If a force acting at a point A pull a body which has some point B held at rest, the reaction at B is asserted to be equal and opposite to the force acting at A. In general, when two forces act at different points of a body there will be equilibrium if the lines of action coincide, the directions of the forces are opposite, and their magnitudes equal. From this we deduce that when a force acts on a body, its effect is the same whatever point of its line of action is taken as the point of application, provided that point is connected with the rest of the body in some invariable manner. ART. 18.] STATICAL AXIOMS. 11 For let a force P act at A and let B be another point in its line of action. We have just seen that the force P acting at A may be balanced by an equal force Q acting at B in the opposite direction. But the force Q acting at B may also be balanced by an equal force F acting at B in the same direction as P (Art. 15). Thus the two equal forces P and P' acting respectively at A and B in the same directions can be balanced by the same force Q. Thus the force P acting at A is equivalent , to an equal force P' acting at B. 18. Statical Axioms. If we wish to found the science of statics on a basis independent of the ideas of motion we require some elementary axioms concerning matter and force. In the first place we assume as before the principle of the inertness of matter. We also require the two principles of the independence and transmissibility of force. The first of these principles is regarded as a matter of common experience. When our attention is called to the fact, we notice that bodies at rest do not begin to move unless urged to do so by some external causes. The other two require some elementary experiments. Let a body be acted on by two forces, each equal to P, and having A, A' for their points of application. We may suppose these to be applied by means of strings attached to the body at A and A' and pulled by forces each of the given magnitude. Let us also suppose the body to be removed from the action of gravity and all other forces. This may be partially effected by trying the experiment on a disc placed on a smooth table or by suspending the body by a string attached at the proper point, or the experiment might be tried on some body floating in a vessel of water. It is a matter of common experience that when the strings are pulled there cannot be equilibrium unless the lines of action of the forces acting at A and A' are on the same straight line. The body acted on will move unless this coincidence of the lines of action is exact. 12 THE PARALLELOGRAM OF FORCES. [CHAP. I. This result is not to be regarded as obvious apart from experiment. In the diagram the points of application A and A' are separated by a space not occupied by the body. The forces have therefore to counterbalance each other by acting, if we may so speak, round the corner E. As the manner in which force is transmitted across a body is not discussed in this part of statics, it is necessary to have an experimental result on which to found our arguments. Let us now suppose that two other forces each equal to Q are applied at B and B' and have their lines of action in the same straight line. These if they acted alone on the body without the forces P, P' would be in equilibrium. Then it will be seen, on trying the experiment, that equilibrium is still maintained when both the systems act. Thus it appears that the introduction of the two forces Q, Q' does not disturb the two forces P, P' so as to destroy the equilibrium. From the results of this experiment we may deduce exactly as in Art. 17, the principle of the transmissibility of force. 19. Rigid bodies. Let two or more bodies act and react on each other and be in equilibrium under the action of any forces. The principle of the transmissibility of force asserts that any one of these forces may be applied at any point of its line of action. If the line of action of any force acting on one of the bodies be produced to cut another, it does not follow that equi- librium will be maintained if the force is transferred from a point on the first body to a point on the second. For example, let two heavy rods AEB, EA' hinged together at E, be suspended from two fixed points M and N by two strings attached to the points A and A'. We may replace the tension of the string A'N by an equal force P. Let the line of action of P, when produced cut the rod AEB in the point B. It is obvious that if P were transferred to act at B, the rod EA' being unsupported would immediately begin to turn round the hinge E. At the same time the rod AEB might also begin to move. It is therefore to be understood that when a force is transferred from any point in its line of action to another the two points are supposed to be rigidly connected together. When the points of application of the forces are connected in some invariable manner, the body acted on is said to be rigid. Such are the bodies we shall in general speak of, though for the sake of brevity we shall often refer to them simply as bodies. ART. 20.] RIGID BODIES. 13 • It appears from this reasoning that when we have to determine the conditions of equilibrium of a system of rigid bodies we should consider each body by itself. "We regard each as acted on, not only by the given external forces, but also by the unknown pressures and reactions of the other bodies in its immediate neighbourhood. Proceeding in this way we shall by help of the elementary axioms of statics be able to form necessary conditions of equilibrium. It will be afterwards seen that these are generally sufficient to determine the positions of equilibrium of the several bodies and their mutual reactions. Briefly stated, the method of proceeding is as follows. Let it be required to find the conditions of equilibrium of a system of bodies which could be moved amongst themselves. We analyse the system into smaller systems (or into single bodies) which are such that the several parts of each are sensibly at invariable distances. These simpler bodies are the bodies called rigid. The conditions of equilibrium of these elementary bodies are first determined. Those of the compound system are afterwards deduced. If we do not obtain in this manner sufficient equations to determine all the unknown quantities, we must either analyse the body into others still simpler or discover some new statical axiom. When the body is made of some flexible material or is fluid, it may be necessary to analyse the body into infinitely small elements. 20. It is sometimes convenient to form the conditions of equilibrium of the whole system (or any part of it) as if it were one body. That this may be done is evident, since the mutual actions and reactions of the several bodies are equal and opposite. But we may also reason thus ; the system being in a position of equilibrium, we may suppose the points of application of the forces to be joined in some invariable manner. This will not disturb the equilibrium. The system being now made rigid we may form the conditions of equilibrium. These are generally necessary and sufficient for the equilibrium of the system regarded as a rigid body, but though necessary they are not generally sufficient for its equilibrium when regarded as a collection of bodies. 14 THE PARALLELOGRAM OF FORCES. [CHAP. I. Suppose, for example, the body acted on were elastic or fluid. After the body has assumed a position of equilibrium we may regard it as rigid and thus obtain some of the conditions of equilibrium. But these are not generally sufficient to express the equilibrium of the body. 21. When a force acts on a rigid body, the principle of the transmissibility of force asserts that the body transmits its action from one point of application to another, but does not itself alter the magnitude of the force. It appears, therefore, that so far as this principle and that of the independence of forces are concerned the conditions of equilibrium depend on the forces and not on the body. If a system of forces be in equilibrium when acting on any body, that system will also be in equilibrium when transferred to act on any other body, provided always the points of application are connected by some kind of invariable relations. It follows that no definition of the body acted on is necessary when the forces in equilibrium are given. The forces must have something to act on, but all we assume here about this something, is that it transmits the force so that the axioms enunciated may be taken as true. For this reason, it is sometimes said that statics is the science which treats of the equilibrium and action of forces apart from the subject matter acted on. 22. Resultant force. When two forces act simultaneously on a particle and are not in equilibrium, they will tend to move the particle. We infer that there is always some one force which will keep the particle at rest. A force equal and opposite to this force is called the resultant of the two forces and is equivalent to the forces. It is obvious that the resultant of two forces acting on a particle must also act on that particle. It is also clear that its line of action is inter- mediate between those of the two forces. Let Pi, P 2 , ... P n be any number of forces acting on the same particle. The two forces P l3 P 2 have a resultant, say ft. We may remove Pi and P 2 and replace them by ft. Again ft and P 3 may be replaced by their resultant ft and so on. We finally have all the forces replaced by a single force. This single force is called their resultant. If the forces of a system do not all act at the same point, it may happen that there is no single force which could balance the system. If so, the system is not equivalent to any single resultant force. 23. To find the resultant of any number of forces acting at a point and having their lines of action in the same straight line. ART. 25.] PARALLELOGRAM OF FORCES. 15 Let be the point of application, and first let all the forces act in the same direction Ox. Since each acts independently of the others, the resultant is clearly the sum of the separate forces and it acts in the direction Ox. If some of the forces act in one direction Ox and others in the opposite direction say Oaf, we sum the forces in these two direc- tions separately. Let X and X' be these separate sums, and let X be the greater. Then by Art. 15 we can remove the force X from both sets of forces. The whole system is therefore equivalent to the single force X — X' acting in the direction of X. By the rule of signs this is also equivalent to a single force represented by the negative quantity X' —X acting in the opposite direction, viz. that of X'. The necessary and sufficient condition that a system of forces acting at a point and having their lines of action in the same straight line should be in equilibrium is that the algebraic sum ol the forces should be zero. 24. Parallelogram of Forces. To find the resultant oi two forces acting at a given point and inclined to each other al any angle. The theorem we are about to prove may be enunciated in th« following manner. Let the two forces act at the point and let them be repre- sented in direction and magnitude by two straight lines OA, OI drawn from the point (Art. 7). Let us now construct a paral- lelogram having OA, OB for two adjacent sides and let OC be thai diagonal which passes through the point 0. Then the residtan of the two forces mil be represented in direction and magnitudi by the diagonal OC. Several proofs of this important theorem have been given As the " parallelogram law " is the foundation of the whole theor; of the composition and resolution of forces, it will be useful ti consider more than one proof, though the student at first reading should confine his attention to one of them. 25. Newton's proof of the parallelogram of forces. Thi ^proof is founded on the dynamical measure of force. Its principL has already been explained in Art. 15. It is repeated here oi 16 THE PARALLELOGRAM OF FORCES. [CHAP. 1. account of its importance. The figure is the same as that used in Art. 12 for the parallelogram of velocities. 26. Suppose two forces to act on the particle placed at in the directions OA, OB. Let the lengths OA, OB be such that they represent the velocities these forces could separately generate in the particle by acting for a given time. Since each force acts independently of the other, it will generate the same velocity whether the other acts or does not act. When both act the particle has at the end of the given time both the velocities represented by OA and OB. These are together equivalent to the single velocity OC. But this is also the measure of the force which would generate that velocity. Thus the two forces measured by OA, OB are together equivalent to the single force measured by OG. 27. Duchayla's proof of the parallelogram of forces. This proof is founded on the principle of the transmissibility of force, Art. 17. It has been shown in Art. 18 that this principle can be made to depend only on statical axioms. To prove the proposition we shall use the inductive proof. We shall assume that the theorem is true for forces of p and m units inclined at any angle, and also for forces of p and n units inclined at the same angle ; we shall then prove that the theorem must be true for forces of p and m + n units inclined at the same angle. Let the forces p and m act at the point and be represented in direction and magnitude by the straight lines OA and OB. On the same scale let BD represent the force n in direction and magnitude. Let BD be in the same straight line with OB, then the length OD will represent the force m + n in direction and magnitude, Art. 23! Let the two parallelograms OBGA, BBFC be constructed and let OC, OF, BF be the diagonals. ART. 28.] DIJgHAYLA'S PROOF. 17 By hypothesis the resultant of the two forces p and m acts along OC. By Art. 18, we transfer the point of application to G. We now replace this resultant force by its two components p and m. These act at C, viz. p along BC produced and m along CF. Transfer the force p to act at B and the force m to act at F. Since BG is equal and parallel to OA, the force p acting at B is represented by BG. The force n may be supposed also to act at B and is represented by BD. Hence by our hypothesis the resultant of these two acts along BF. Transfer the point of application to the point F. The two forces p and m + n are therefore equivalent to two forces acting at F. Their resultant must therefore pass through F, Art. 22. For the same reason the resultant passes through 0, and the forces have but one resultant, Art. 22. Hence the resultant must act along OF. But this is the diagonal of the parallelogram constructed on the sides OA, OB which represent the forces p and m + n. It is clear that the resultant of two equal forces makes equal angles with each of these forces. The resultant of two equal forces therefore acts along the diagonal of the parallelogram constructed on the equal forces in the manner already described. Thus the hypothesis is true for the equal forces p and p. By what has just been proved it is true for the forces p and 2p and therefore for p and 3p and so on. Thus it is true for forces p and rp where r is any integer. Again the hypothesis has just been proved true for forces rp and p ; hence it is true for rp and 2p and so on. Thus the hypothesis is true for forces rp and sp, where r and s are any integers. Thus the proposition so far as the direction of the resultant is concerned is established for any commensurable forces. 28. We have now to find the direction of the resultant when the forces are incommensurable. Let OA, OB represent in direc- tion and magnitude any two incommensurable forces p and q, then if the diagonal OC does not represent the resultant, let 00 be -the direction of the resultant. The straight line OG must lie within the angle AOB and will cut either BG between B and G ox AG between A and G ; Art. 22. Let it cut BG between B and C. Divide OB into a number of equal parts each less than OG and measure off from OA beginning at portions equal to these until E. s. 2 18 THE PARALLELOGRAM OF FORCES. [CHAP. I. we arrive at a point K where AK is less than GG. Draw QH, KL parallel to AG. Since OB and OK are commensurable the B forces represented by these have a resultant which acts along the diagonal OL. Thus the forces p and q acting at are equivalent to two forces, one- of which acts along OL and the other is the force represented by K A. The resultant of these two must act at in a direction lying between OL and OA. But 00 lies outside the angle AOL, hence the assumption that the direction of the resultant is OG is impossible. But OG represents any direction other than OG for then only is it impossible to divide OB into equal parts each less than GG. Thus the resultant force must act along the diagonal whether the forces be commensurable or incommensurable. We have given a separate proof for incommensurable forces. But this is unnecessary. The theorem has been proved for all forces whose ratio can be expressed by a fraction. In the case of incommensurable forces we can still find a fraction which differs from their true ratio by a quantity less than any assigned difference. In the limit the theorem must be true for incom- mensurable forces. 29. To prove that the diagonal represents the magnitude of the resultant as well as its direction. Let OA and OB represent the two forces, and let OG be the diagonal of the parallelogram OACB. Take OD in GO pro- duced of such length as to repre- sent the resultant in magnitude. Then the three forces OA, OB, OD are in equilibrium and each of them is equal and opposite to the resultant of the other two. ART. 31.] HISTORICAL SUMMARY. 19 Construct on OB, OD the parallelogram OBED. Since A is equal and opposite to the resultant of OB and OB, OE is in the same straight line with OA and therefore OE is parallel to GB. By construction 00 is in the same straight line with OD and is therefore parallel to EB. Thus EG is a parallelogram. Hence OG is equal to EB and therefore to DO. Thus the diagonal 00 represents the resultant of the two forces OA, OB in magnitude. 30. Ex. Assuming that the diagonal of the parallelogram constructed on OA , OB represents the magnitude of the resultant, show that it also represents the direction. As before, let OA, OB, OD represent forces in equilibrium. It is given that OA = OE, OG=OD, and it is to be proved that AOE, DOG are straight lines. Since AB and BD are parallelograms, OA=BC, OD=BE. Hence in the quadrilateral EOGB the opposite sides are equal in length. The quadrilateral is therefore a parallelogram. (For the triangles OEB, BGO have their sides equal each to each.) It follows that OE is parallel to BG, and is therefore in the same straight line with OA. 31. Historical Summary. If we enquire on what principles the science of statics has been studied in former times, we find that these may be reduced to three. There is first the principle used by Archimedes, viz., that of the lever. It is assumed as self-evident or as the result of an obvious experiment, (1) that » straight horizontal lever charged at its extremities with equal weights will balance about a support placed at its middle point, (2) that the pressure on the support is the sum of the equal weights. Starting with this elementary principle, and measuring forces by the weights they would support, the conditions of equilibrium of a straight lever acted on by unequal forces were deduced. From this result by the addition of some simple axioms the other proposition of statics may be made to follow. The truth of the first elementary principle named above is perhaps evident from the symmetry of the figure. But Lagrange points out that the second is not equally evident with the first. The second principle which has been used as the foundation of statics is that of the parallelogram of forces. In 1586, Stevinus, in his statics and hydrostatics enunciated the theorem of the triangle of forces. Till this time the science of statics had rested on the theory of the lever, but then a new departure became possible. The simplicity of the principle and the ease with which it may be applied to the problems of mechanics caused it to be generally adopted. The principle finally became the basis of modern statics. For an account of its gradual development we refer the reader to A Short History of Mathematics, by W. W. R. BaU. Many writers have given or attempted to give proofs of this principle which are independent of the idea of motion. One of these, that of Duchayla, has been reproduced above, as that is the one which seems to have been best received. There is another, that of Laplace, which has attracted considerable attention. 2—2 20 THE PARALLELOGRAM OF FORCES. [CHAP. I. This is founded on principles similar to the proofs of Bernoulli and D'Alembert. It is assumed as evident that if two forces be increased in any, the same, ratio the magnitude of their resultant will be increased in the same ratio, but its direction will be unaltered. In comparing these proofs with that founded on the idea of motion, we must admit the force of a remark of Lagrange. He says that, in separating the principle of the composition of forces from the composition of motions, we deprive that principle of its chief advantages. It loses its simplicity and its self-evidence, and it becomes merely a result of some constructions of geometry or analysis. The third fundamental principle which has been used is that of virtual velocities. This principle had been used by the older writers. Lagrange gave, or attempted to give, an elementary proof of the principle and then made it the basis of the whole science of mechanics. This proof has not been generally received as presenting the simplicity and evidence which he had admired in the principle of the composition of forces. CHAPTER II. FORCES ACTING AT A POINT. The triangle of forces. 32. In the last chapter we arrived at a fundamental proposi- tion usually, called the parallelogram of forces. . We now proceed to its application to the problems which occur in statics. The first we shall take is that in which all the forces act at the same point. Experience shows it is not always convenient to draw the parallelogram, for this complicates the figure and makes the solution cumbersome. Several artifices have been invented to enable us to use the principle with facility and quickness. In this chapter we shall consider these in turn. 33. If OA, OB represent two forces P and Q acting at a point 0, we know that their resultant is represented by the diagonal OG of the parallelogram constructed on those sides. Now it is evident that A G will represent the force Q in direction and magnitude as well as OB, though it will not represent the point of application. This however is unimportant if the point of application is otherwise indicated. Thus the triangle OAG may be used instead of the parallelogram OAGB. As the points of application are supposed to be given inde- 22 FOKCES ACTING AT A POINT. [CHAP. II. pendently it is no longer necessary to represent the forces by straight lines passing through 0. Thus we may represent the forces P, Q, R acting at both in direction and magnitude by the sides of a triangle DEF provided these sides are parallel to the forces and proportional to them in magnitude. It is clear that all theorems about the parallelogram of forces may be immediately transferred to the triangle. We therefore infer the following proposition called the triangle of forces. If two forces acting at some point are represented in direction and magnitude by the sides DE, EF of any triangle, the third side DF will represent their resultant. If three forces acting at some point are represented in direction and magnitude by the three sides of any triangle taken in order viz., DE, EF, FD, the three forces are in equilibrium. 34. When three forces in one plane are given and we wish to determine by help of the triangle of forces whether they are in equilibrium or not, we see that there are two conditions to be satisfied. 1. If they are not all parallel two of them must meet in some point 0. The resultant of these two will also pass through the same point. The third force must be equal and opposite to this resultant and must therefore also pass through the same point. Hence the lines of action of the three forces must all meet in one point or be parallel. 2. If the forces are not all parallel, straight lines can be drawn parallel to the forces so as to form a triangle. The magnitudes of the forces must be proportional to the sides of that triangle taken in order. The case in which the forces are all parallel will be considered in the next chapter. 35. We may evidently extend this proposition further. Sup- pose we turn the triangle DEF through a right angle into the position D'E'F', the sides will then be perpendicular instead of parallel to the forces. Also if the forces act in the directions DE, EF, FD they now act all three outwards with regard to the triangle D'E'F. If the forces were reversed they would all act inwards. We have thus a new proposition. ART. 38.] POLJpON OF FORCES. 23 If three forces acting at some point be represented in magnitude by the sides of a triangle, and if the directions of the forces be perpendicular to those sides and if they act all inwards or all outwards, the three forces are in equilibrium. 36. Instead of turning the triangle through a right angle, we might turn it through any acute angle. We thus obtain another theorem. If three forces acting at a point be represented in magnitude by the sides of a triangle and if their directions make the same angle with the sides taken in order, the three forces are in equilibrium. In using this theorem, it is found to be generally inconvenient to sketch the triangle. We therefore put the theorem into another form. The sides of the triangle are proportional to the sines of the opposite angles. This relation must therefore also hold for the forces. Hence we infer the following theorem. Three forces acting on a body in one plane are in equilibrium if (I) their lines of action all meet in one. point, (2) their directions are all towards or all from that point, (3) the magnitude of each is proportional to the sine of the angle between the other two. 37. Polygon of forces. We may further extend the triangle of forces into a polygon of forces. If several forces act at a point we may represent these in magnitude and direction by the sides of an unclosed polygon DE, EF, FG, OH &c. taken in order. The resultant of DE, EF is represented by DF. That of DF and FG is DG and so on. Thus the resultant is represented by the straight line closing the polygon. It is clear that the sides of the polygon need not all be in the same plane. If several forces acting at one point be represented in direction and magnitude by the sides of a closed polygon taken in order, they are in equilibrium. 38. Ex. 1. Forces represented by the numbers 4, 5, 6 are in equilibrium ; find the tangents of the halves of the angles between the forces. By drawing parallels to these forces we construct a triangle of the forces. The angles of this triangle can be found by the ordinary rules of trigonometry. Ex. 2. Forces represented by 6, 8, 10 lbs. are in equilibrium ; find the angle between the two smaller forces. How must the least force be altered that the angle between the other two may be halved ? 24 FORCES ACTING AT A POINT. [CHAP. II. Ex. 3. If OA, OB represent two forces, show that their resultant is represented by twice OM, where M is the middle point of AB. Ex. 4. Two constant equal forces act at the centre of an ellipse parallel to the directions SP and PH, where S and H are the foci and P is any point on the curve. Shew that the extremity of the line which represents their resultant lies on a circle. [Math. Tripos, 1883.] Ex. 5. Forces P, Q act at a point 0, and their resultant is B ; if any transversal cut-the directions of the forces in the points L, M, N respectively, show that _L + A. _ A . [Math. Tripos, 1881.] 0L + OM ON L Clear of fractions and the equation reduces to the statement that the area £0#C is the sum of the areas LOjt, MON. s Ex. 6. A particle is in equilibrium under three forces, viz., a force F given in magnitude, a force F' given in direction, and a force P given in magnitude and direction. Find the lines of action of F by a geometrical construction. If OA represent P, draw AB parallel to F', and describe a circle whose centre is . and whose radius represents F in magnitude. Ex. 7. ABGDEF is a regular hexagon; prove that equal forces acting along AB, CD, EF, AF, CB, ED are in equilibrium. Find also the resultant of 2P along AB, CD, EF, P along ED, CB, IP along AF. [St John's Coll., 1881.] Ex. 8. ABCD is a tetrahedron, P is any point in BC, and Q any point in AD. Prove that a force represented in magnitude, direction, and position by PQ, can be replaced by four components in AB, BD, DC, CA in one and in only one way, and \ find the ratios of these components. [St John's Coll., 1887.] Ex. 9. Lengths BD, GE, AF are drawn from the corners along the sides BC, CA, AB of a triangle ABC; each length being proportional to the side along which it is drawn. If forces represented in magnitude and direction by AD, BE, CF acted on a point, show that they would be in equilibrium. Conversely if the forces AD, BE, CF are in equilibrium, then BD, CE, AF are proportional to the sides. If be any point in the plane of the triangle, show that the forces represented by OD, OE, OF have a resultant which is independent of the magnitude of the ratio BD : BC. ^ Ex. 10. If any number of forces in one plane whose magnitudes are propor- tional to the sides of a closed polygon act perpendicularly to those, sides at their middle points all inwards or all outwards, prove that they are in equilibrium. Let ABCD &c. be the polygon. Join AC, AD, &c. Consider the triangle ABC thus found. The forces across AB, BC meet in the centre of the circumscribing circle, and have therefore for resultant a force proportional to AC acting perpen- dicularly to it at its middle point. Taking the triangles ACD, ADE &o. in turn, the final resultant is obviously zero. 39. Parallelepiped of forces. Three forces acting at a point are represented in direction and magnitude by three straight lines OA, OB, OC not in one plane. To show that the resultant is ART. 41.] PARA]*,ELEPIPED OF FORCES. 25 represented in direction and magnitude by the diagonal of the parallelepiped constructed on the three straight lines as sides: To prove this we consider the parallelogram constructed on OA, OB. The resultant of these two forces is represented by OB. If GE be the parallel diagonal of the opposite face, it is clear fey geometry that OGED will be a parallelogram. The resultant of the forces represented by OG, OB will therefore be OE. But OE is the diagonal of the parallelepiped. Hence the result follows. We may also deduce the theorem from Art. 37. The resultant of the three forces represented by OA, AB, BE is represented by the straight line which closes the polygon AOBEO, i.e. it is OE. "* 40. Ex. If six forces, acting on a particle, be represented in magnitude and direction by the edges of a tetrahedron, the particle cannot be at rest. [Math. Tripos, 1859.] Method of Analysis. 41. We shall now proceed to the second method of treating forces which act at any point of a body. We have seen that any force may be replaced by two others^ called its components, which are inclined at' any angle to each other which may appear suitable. But it is found by experience that when a force has to be resolved it is generally more useful to resolve it into two components which are at right angles. When therefore the component of a force is spoken of it is meant, unless it is otherwise stated, that the other component is at right angles to it. By referring to the figure of Art. 33, we see that the parallelogram OAGB becomes a rectangle. The two components of the force OG are OG. cos CO A and OG. sin CO A. We may put this result into the form of a working rule. // a force R act at in the direction 00, its component in any direction Ox is R cos COx. In the same way its component in the opposite direction Ox' is R cos COx'. In the same way the component of R perpendicular to Ox is R sin COx. It is convenient to have some short name to distinguish the 26 FORCES ACTING AT A POINT. [CHAP. II. rectangular components of a force from its oblique components. The name resolute for the components in the first case has been suggested in Lock's Elementary Statics. Though this name has not yet come into general use it will be occasionally used here when it is necessary to make the distinction. 42. Two forces P 1( P 2 act at a point 0. To find the position and magnitude of their resultant. Let Ox, Oy be any two rectangular axes; and let a u a 2 be the angles the forces P 1; P 2 make with the axis of x. The sums of the components parallel to the axes are X = Pi COS £»! + P 2 COS 0a, Y = Pj sin oti + P 2 sin Oa. If these are the components of a force R whose line of action makes an angle d with the axis of x, we have X = Pcosa, F=Psina. We easily find by adding together the squares of X and Y that P 2 = P* + P 2 2 + 2P 1 P 2 cos 0, where 6 = a 1 — a 2 , so that 6 is the angle between the directions of the forces P 1: P 2 . This result also follows from the parallelogram of forces. For the right-hand side is evidently the square of the diagonal of the parallelogram whose sides are P x and P 2 . The direction of the resultant is also easily found, for we have Y Pi sin t*i + P 2 sin a 2 tan a = t? = -g — „ . X Pi cos ai + P 2 cos a2 43. Ex. 1. Two forces P, Q act at an angle a and have a resultant R. If each force be increased by R, prove that the new resultant makes with R an angle whose tangent is D . (P ~ 9 ' Sm ° . [St John's Coll., 1880.] e P + Q + R + (P+Q,)cosa L J Take the line of action of the resultant R for the axis of x. Ex. 2. Forces each equal to F act at a point parallel to the sides of a triangle ABC. If R be their resultant, prove that R? = F 2 (3 - 2 cos A - 2 cos B - 2 cos C). Ex. 3. The resultant of P and Q is R, if Q be doubled jR is doubled, if Q be reversed, R is also doubled ; shew that P : Q . : R :: *J2 : ^3 : ^2. [St John's Coll., 1881.] Ex. 4. ABCD is a quadrilateral; forces act along the sides AB, BC, CD, DA measured by a, /3, y, 8 times those sides respectively. Show that if there is equilibrium !, cti = 0, o 2 = 12 &c., the equations R cos a = X = %P cos a become R cos 1 = P 1 + P 2 cos 6 l2 + P 3 cos 13 + «fec. In the same way by taking the axis of x along the force P 2 we find R cos 2 = Pj cos 8 12 + P 2 + P 3 cos &2S + . . . and so on. Thus the direction of R has been found. il. Polyhedron of forces. The equations of Art. 44 have a geometrical meaning which is often useful. Let any closed polyhedron be constructed, let A v A 2 &c. be the areas of its faces. Let normals be drawn to these faces, each from a point in the face all outwards or all inwards, and let V 2 &c. be the angles these normals make with any straight line which we may call the axis of z. Let us now project orthogonally all these areas on the plane of xy. The several projections are A x cos 8 1 , A 2 cos 2 &c. Since the polyhedron is closed the total projected area which is positive is equal to the total negative projected area. We therefore have A 1 cos t + A 2 cos 8 2 + ... = 0. Similar results hold for the projection on the other coordinate planes. Thus we obtain three equations which are the same as the equations of equilibrium already ART. 49.] POI/pffiDRON OF FORCES. 29 found, except that we have A lt A 3 &c. written for P lt P 2 &e. We therefore have the following theorem. If forces acting at a point he represented in magnitude by the areas of the faces of a closed polyhedron and if the directions of the forces be perpendicular to those faces respectively, acting all inwards or all outwards, then these forces are in equilibrium. 48. By using the theory of determinants we may put the results of Art. 46 into a more convenient form. Let it be required to find the resultant of any three forces acting at a point. To obtain a symmetrical result we shall reverse the resultant and speak of four forces in equilibrium. Let Pj, P 2 , P 3 , P 4 be four forces in equilibrium. Putting R=0, we have found in Art. 46 four linear equations connecting these. Eliminating the forces, we have the determinantal equation = 0. 1 COS 9 12 COS #13 COS <>14 cos #21 1 COS #23 COS #24 cos *31 COS #32 1 COS #34 cos '« COS #42 COS #43 1 This is the relation connecting the mutual inclinations of any four straight lines in space*. If all these angles except one (say 1S! ) are given, we have a quadratic to find the two possible values which cos 12 could then have. If three of the angles say 12 , 23 , 31 are right angles this determinant reduces to the well-known form cos 2 14 + cos 8 024 + cos 2 9 U = 1. If the angles between the four directions in which the forces act are given, the ratios of the forces are found from any three of the four linear equations above mentioned. It follows that the forces are in the ratio of the minors of the constituents in any row of the determinant. Ex. Show that the squares of the forces are in the ratio of the minors of the constituents in the leading diagonal. For let I„ be the minor of the rth row and sth column, then by a theorem in Salmon's Higher Algebra I 11 I 22 =/ 12 2 . But we have shown above that Pi : J > a=-Tn : ^12 > hence we deduce at once P x 2 : P^=I n : I 2 2- For the sake of reference we state at length the minor of the leading constituent. It is I n = l- COS 2 023 - cos 2 M - cos 2 42 + 2 cos 23 cos 34 cos 42 . . This expression is easily recognized as one which occurs in many formulas in spherical trigonometry. For example, if unit lengths are drawn from any point parallel to the directions of any three of the forces (say P 2 , P 3 , P 4 ) the volume of the tetrahedron so formed is one sixth of the square root of the corresponding minor (viz. / u ). 49. Sometimes it is necessary to refer the forces to oblique .axes. In this case we replace the direction cosines of each force by its direction ratios. Let the direction ratios of P v P 2 &c. be KVi)' i a J>a c s) &0 - Then bv tne same reasoning as * Another proof is given in Salmon's Solid Geometry, Ed. iv., Art. 54, 30 FORCES ACTING AT A POINT. [CHAP. II. before, the sums of the components of the forces parallel to the axes are X=2Pa, F=SP6, Z = XPc. If these are the components of a force R with direction ratios (I, m, n) we have Rl = X, Rm=Y, Rn=Z. The relations between the direction ratios of a straight line and the angles that straight line makes with the axes are given in treatises on solid geometry or on spherical trigonometry. They are not nearly so simple as when the axes of reference are rectangular. For this reason oblique axes are seldom used. 50. Ex. 1. Forces act at the centres of the circles circumscribing the faces of a tetrahedron perpendicular to those faces and proportional to them. Prove that they are in equilibrium if they act either all inwards or all outwards. Ex. 2. Forces act through the angular points of a tetrahedron perpendicular to. the opposite faces and proportional to them. Prove that they are in equilibrium if they act either all inwards or all outwards. [Math. Tripos, 1881.] Let ABCD he the tetrahedron, AK, BL &o. the perpendiculars. Since the product of each perpendicular into the area of the corresponding face is equal to three times the volume of the tetrahedron, it follows that the forces are inversely proportional to the perpendiculars along which they act. Let the forces be nlAK, l*IBL &a. Let us resolve the force iijAK into three components which act along the edges AB, AC, AD. The component F which acts along AB is found by equating the resolutes perpendicular to the plane AGD. This gives where 6 is the angle between the perpendiculars AK and BL. In the same way we resolve the force )i.\BL into components along the edges. The component F' which acts along BA is found from _, AK ix F ■AB = BL e0se - Hence F and F' are equal and opposite forces. In the same way it may be shown that the forces along all the other edges are equal and opposite. The system is therefore in equilibrium. Ex. 3. Three forces act at each corner of a tetrahedron perpendicular to the faces which meet at that corner. If the magnitudes of these twelve forces are proportional to the areas of the perpendicular faces, show that they are in equilibrium. | The resultant force at each corner is found by Art. 47. The result then follows. The mean centre. 51. There is another method of expressing the magnitude and direction of the resultant of any number of forces acting at a point which will be found very useful both in geometrical and analytical reasoning. Let us represent the forces P lt P 2 &c. in direction by the ART. 51.] T^E MEAN CENTRE. 31 straight lines 0A lt 0A 3 &c. To represent their magnitudes we shall take lengths measured along these straight lines, thus the force along OA x is represented hj p 1 .OA 1 , that along (L4 2 by p%. 0A 2 , and so on. The advantage of introducing the numerical multipliers p lt p 2 &c. is that the extremities A u A^ &c. of the straight lines may be chosen so as to suit the figure of the problem under consideration. It is evident that this is equivalent to representing the forces by straight lines on different scales, viz. the scales p u p 2 &c. of force to each unit of length. Taking for origin, let (uc-fl^i), (os 2 y 2 z 2 ) &c. be the coordi- nates of the points A lt A 2 &c. We have already proved that the components of the resultant are X = 2P cos a = %p . 0A t cos a = 2,pso\ Y = %P cos /3 =^py\ (1). Z = 2P cos 7 = l,pz) Let us take a point whose coordinates (xyz) are given by the equations x zp' y ~ %p> z tp {l >- It follows at once that X = x%p, Y = y%p, Z = zlp. These equations imply that the resultant of the forces is repre- sented in direction and magnitude by OG . 2p. This point is known by a variety of names. It is called the centre of gravity, or centroid or mean centre of a system of particles placed at A lt A 2 whose masses or weights are proportional to Pi, Pi &o- The result is, if forces acting at a point be represented in direction by the straight lines OA 1; OA 2 &c. and in magnitude by Pi . OA 1; p 2 . OA 2 &c, then their resultant is represented in direction by OG and in magnitude by 2p . OG, where G is the centre of gravity of masses proportional to pi, p 2 &c. placed at A 1; A 2 &c. This theorem is commonly ascribed to Leibnitz. The utility of this proposition will depend on the ease with which the point G can be found when A 1 , A% {«>, V, *) = 0, y}r (x, y, z) = 0, we have with the usual notation for partial differential coefficients x dx + y dy + z dz = 0, ty x dx + yjr y dy + \jr,,dz = 0. Eliminating the ratios dx : dy : dz, we have the determinant J = = 0. X, 7, Z This determinantal equation joined to the two equations of the curve suffices in general to find the values of x, y, z. There may be several sets of values of these coordinates, and these give all the positions of equilibrium. 57. In some cases it is more convenient to use polar coor- dinates. Suppose the curve to be plane. Let P, Q be the com^ ponents of the forces acting along and perpendicular to the radius vector in the positive direction. We then find by resolving these along the tangent Joining this to the equation to the curve we have two equations to find (r, 6). Suppose the curve to be tortuous. Let Q, T, P be the polar components of the forces, the first two acting perpendicular to the radius vector respectively in and perpendicular to the plane of rz and the last acting along the radius vector. Resolving as before, we have n d0 , „ . Q d we write the angle the direction AB of the A ' D MB A N M A M iV force makes with the direction A 'A opposite to the displacement, the product is called the work done against the force. Let us drop a perpendicular A'M on AB; the work done by the force is also equal to the product P. AM, where AM is to be estimated positive when in the direction of the force. Let P' be the resolved part of P in the direction of the displacement ; the work is also equal to P'. AA'. These expressions for the work of a force are clearly equivalent, and all three are in continual use. 63. The forces which act on a particle generally depend on the position of that particle. Thus if the particle be moved from A to any point A' at a finite distance from A, the force P will not generally remain the same either in direction or magnitude. For this reason it is necessary to suppose the displacement A A' to be so small a quantity that we may regard the force as fixed in direction and magnitude. Taking the phraseology of the dif- ferential calculus this is expressed by saying that the displacement AA' is of the first order of small quantities. We may suppose any finite displacement of the point A to be ART. 66.] m WORK. 39 made along a curve beginning at A and ending at some point 0. Let ds be any element of this curve, and when the particle has reached this element let P be the resolved part of the force along ds in the direction in which s is measured. Then by the above definition JP'ds is the sum of the separate works done by the force P as the particle travels along each element in turn. This sum is defined to be the whole work in any finite displacement. If s be measured from any point on the curve, the limits of this integral will evidently be s = OA and s = OG. 64. The resolved displacement A A' cos cf> is sometimes called the virtual velocity of the point of application. The product P. A A' . cos is called the virtual moment or virtual work of the force. But these terms are restricted to infinitely small displace- ments. When the displacement is finite, the integral of the virtual works is called the work. 65. It is often convenient to construct a proposed displace- ment by several steps. Thus a displacement AA' may be con- structed by moving A first to D and then from D to A' (see figure in Art. 62). Supposing AD and DA' to be infinitely small so that the direction and magnitude of the force P continue constant throughout, it is easy to see that the work due to the whole displace- ment AA' is the sum of the works due to the displacements AD and DA'. For if we drop the perpendiculars DN and A'M on the direction of the force, the separate works with their proper signs will be P. AN and P.NM. The sum of these is P.AM which is the work due to the whole displacement A A'. If the displacement A A' is finite, and the force P remains unaltered in direction and magnitude, the work due to the resultant displacement will be equal to the sum of the works due to the partial displacements AD, DA'. 66. Suppose next that several forces act at the point A ; then as A moves to A' each of these will do work. The sum of the works done by each separately is defined to be the work done by all the forces collectively. We shall now require the following proposition. If any number of forces act at a point A, the sum of the works due to any small displacement AA' is equal to the work done by their resultant. 40 FORCES ACTING AT A POINT. [CHAP. II. The work done by any one force P is equal, by definition, to the product of AA' into the resolved part of P in the direction of AA'. The work done by all the forces is therefore the product AA' into the sum of their resolved parts. By Art. 44 this is equal to AA' into the resolved part of the resultant, i.e. is equal to the work done by the resultant. 67. This theorem leads to another method of stating the conditions of equilibrium of any number of forces P 1; P 2 &c. acting at the same point A. Case 1. If the particle at A is free to move in all directions it is necessary for equilibrium that the resultant force should vanish. The virtual work of the forces P 1; P 2 &c. must therefore be zero in whatever direction the particle is displaced. Conversely, if the virtual work for any displacement AA' is zero it immediately follows that the resolved part of the resultant in that direction is also zero. If then the virtual work of Pj, P 2 &c. is zero for any three different displacements not all in one plane, the three resolved parts of the resultant in those directions are zero. The particle is therefore in equilibrium. 68. Case 2. If the particle is constrained to move on some curve or surface, then besides the forces P 1( P 2 &c. the particle is acted on by a pressure R which is normal to the curve or surface. The forces which maintain equilibrium are therefore P lt P 2 &c. and R. Then by Case 1 their virtual work is zero for all small displacements. If the displacement given to A is along a tangent to the curve or is situated in the tangent plane to the surface, the angle between the reaction R and the displacement is a right angle. The virtual work of that force is therefore zero. It immediately follows that for all such displacements the virtual work of P lt P 2 &c. is zero. Conversely, suppose the particle constrained to move on a curve ; then if the virtual work for a displacement along the tangent is zero the resolved part of the resultant force in that" direction is also zero. The particle is therefore in equilibrium. Next, suppose the particle constrained to move on a surface ; then if the virtual works for any two displacements, not in the ART. 70.] ASTATIC EQUILIBRIUM. 4)1 same straight line, are each zero, the resolved parts of the resultant force in those directions are each zero. The particle is therefore in equilibrium. 69. Ex. 1. Deduce from the principle of virtual velocities the conditions of equilibrium obtained in Art. 56, for a particle constrained to rest on a curve. The forces on the particle are X, Y, Z ; the displacement is ds, the projections of ds on the forces are dx, dy, dz. Multiplying each force by the corresponding projection, we see at once that the condition of equilibrium is Xdx + Ydij + Zdz = 0. Ex. 2. Two small smooth rings of equal weight slide on a fixed elliptical wire, of which the axis major is vertical, and are connected by a string passing over a smooth peg at the upper focus ; prove that the rings will rest in whatever position they may be placed. [Math. Tripos, 1858.] Let P, Q be the two rings, W the weight of either. Let T be the tension of the string, I its length. Let S be the peg, let x, x' be the abscissa of P, Q measured vertically downwards from S; let r = SP, r'=SQ, then r-rr' = l. Since the ring P is in equilibrium, we have by the principle of virtual work Wdx - Tdr=0. The positive sign is given to the first term because x is measured in the direction in which W acts ; the negative sign is given to the second term because T acts in the opposite direction to that in which r is measured. In the same way we find for the other ring Wdx'- Tdr'=0. Since dr=-dr' this gives as the condition of equilibrium Wdx+ Wdx' =0. As yet we have not introduced the condition that the wire has the form of an ellipse. If 2c be its latus rectum and e its eccentricity, we have r = c + ex, r' = c + ex'. It easily follows that dx + dx' = 0, so that the condition of equilibrium is satisfied in whatever position the rings are placed. Ex. 3. A small ring movable along an elliptic wire is attracted towards a given centre of force which varies as the distance : prove that the positions of equilibrium of the ring lie in a hyperbola, the asymptotes of which are parallel to the axes of the ellipse. [Math. Tripos, 1865.] ^ Ex. 4. Two small rings of the same weight attracting one another with a force varying as the distance, slide on a smooth parabolic shaped wire, whose axis is vertical and vertex upwards : show that if they are in equilibrium in any symmetrical position, they are Bo in every one. [Coll. Ex., 1887.] ^ Ex. 5. Two mutually attimoting or repelling particles are placed in a parabolic groove, and connected by a thread which passes through a small ring at the focus ; prove that if the particles be at rest, the line joining the vertex to the focus will be a mean proportional between the abscissae measured from the vertex. | [Math. Tripos, 1852.] Ex. 6. A weight W is drawn up a rough conical hill of height h and slope a and the path cuts all the lines of greatest slope at an angle /S. If the friction be /i times the normal pressure prove that the work done in attaining the summit will be W h (1 + n cot o sec p) . [St John's Coll. 1887.] Astatic Equilibrium. 70. Suppose that three forces P, Q, B acting at a point are in equilibrium. We may clearly turn the forces round that point 42 FORCES ACTING AT A POINT. [CHAP. II. through any angle without disturbing the equilibrium if only the magnitudes of the forces and the angles between them are un- altered. Since a force may be supposed to act at any point of its line of action these three forces may act at any points A, B,G in their respective initial lines of action. If now we turn the forces supposed to act at A, B, G, each round its own point of appli- cation, through the same angle it is clear the equilibrium will be disturbed unless these points are so chosen that the lines of action of the forces continue to intersect in some point, (Art. 34). It is evident that instead of turning the forces round their points of application we may turn the body round any point through any angle. In this case each force preserves its magnitude unaltered, continues to act parallel to its original direction supposed fixed in space, while the point of application remains fixed in the body and moves with it. When equilibrium is undisturbed by this rotation, it is called Astatic. 71. Let A and B be the points of application of the forces P and Q. Let their lines of action intersect in 0. Then as the forces turn round A and B, in the plane A OB, the angle between them is to remain unaltered. Hence n will trace out a circle passing through A and B. The resultant of these two forces / passes through and makes constant / angles with both OA and OB. It there- \y fore will cut the circle in a fixed point 0. A\~ This resultant is equal and opposite to the force called B. If therefore three forces P, Q, R, acting at three points A, B, C, intersect on the circle circumscribing ABC, and be in equilibrium, the equilibrium will not be disturbed by turning the forces round their points of application through any angle in the plane of the forces.'^ ''''■ = Lrw. ~-. <*.?'. J.:t. This proof is given in Moigno's Statics, p. 228. If the forces P and Q are parallel, the point of intersection is at infinity. In this case the circle of construction becomes the straight line AB. The sines of the angles AOO, BOO are now ultimately proportional to A and OB. Hence AG is to GB ART. 74.] ASTATJi TRIANGLE OF FORCES. 43 inversely in the ratio of the forces tending to A and B. If the forces besides being parallel are equal and opposite, the force R acts at a point on the straight line at infinity. 72. When two forces P 1( P 2 act at given points A, B the point at which the resultant acts however the forces are turned round is called the centre of the forces. If a third force P s act at a third given point C, we may combine the resultant of the first two with this force and thus obtain a resultant acting at another fixed point in the body. This is the centre of the three forces. Thus we may proceed through any number of forces. We see that we can obtain a single force acting at a fixed point of the body which is the resultant of any number of given forces acting at any given fixed points in one plane. This single force will continue to be the resultant and to act at the same point when all the forces are turned round their points of application through any angle. This force is called their astatic resultant. 73. Astatic triangle of forces. This proposition leads us to another method of using the triangle of forces. Referring to the figure of Art. 71, we see that the angles ABG, AOG and BAG, BOO being angles in the same segment are equal each to each. If therefore P, Q, R are in equilibrium, they are proportional to the sines of the angles of the triangle ABG. It follows that P, Q, R are also proportional to the sides of the triangle ABG. Thus P : BG=Q : OA = R : AB. The points A, B, C divide the circle into three segments AB, BC, CA. If be taken on any one of the segments, say AB, then the forces whose lines of action pass through A and B must act both to or both from A and B. The third force acts from or to C according as the first two act towards or from A and B. We deduce the following proposition. Let three forces act at the corners of a triangle; they will be in equilibrium if (1) their magnitudes are proportional to the opposite sides, (2) their lines of action meet in any point on the circumscribing circle, (3) their directions obey the rule given above. Also the equilibrium will not be disturbed by turning all the forces round their points of application through any, the same angle, but without altering their magnitudes. The forces are supposed to act in the plane of the triangle. 74. Ex. 1. Any number of forces P, Q, R, S &a. in one plane are in equilibrium, and their lines of action meet in one point 0. Through describe any circle cutting the lines of action of the forces in A, B, 0, D &o. If these points be regarded as the points of application of the forces, prove that the equilibrium is astatic. Ex. 2. If in the figure of Art. 71 GO' be drawn parallel to the opposite side AB to cut the circle in 0', prove that the forces P, Q, R make equal angles with the sides BC', O'A, AB of the triangle BG'A, and thence deduce the conditions of equilibrium, Art. 36. 44 FORCES ACTING AT A POINT. [CHAP. II. Ex. 3. If a, p be the angles the forces P and Q make with their resultant R, prove that the position of the centres of the forces is given by CE _ AE = BE = AB cot j3 cot a cot a + cot /S ' where GED is drawn from C perpendicular to AB. Ex. 4. Let the forces act from a point towards A and B where is on the left or negative side of AB as we look from A towards B. lip, q be the coordinates of A, p', q' of B, prove that the coordinates of the central point of A and B are given by (cot a + cot j3) x =p cot a+p'eot p+(q' -q)\ (cota + cotj8)^ = gcota+s'cot/S-(y-2>)f ' These are obtained by projecting AE, EG on the coordinate axes. Ex. 5. If the forces P and Q are at right angles, prove that the coordinates of their centre are given by (P 2 + Q 2 ) x =pP* +p'Q* + {q'-q)PQ\ (P* + Q?)y=qP> + q'Q*-(p<-p)PQ] ' Stable and Unstable Equilibrium. 75. Let us suppose a body to be in equilibrium in any position, which we may call A, under the action of any forces. If the body be now moved into some neighbouring position B and placed there at rest, it may either remain in equilibrium in its new position (as in Art. 71) or the body may begin to move under the action of the forces. In the first case the position A is called one of neutral equilibrium. In the second case the equili- brium in the position A is called unstable or stable according as the body during its subsequent motion does or does not deviate from the position A beyond certain limits. The magnitude of these limits will depend on the circumstances of the case. Some- times they are very restricted, so that the deviation . permitted must be infinitesimal ; in other cases greater latitude may be admissible. The determination of the stability of a state of equilibrium is a dynamical problem. We must according to this definition examine the whole of the subsequent motion to determine the extent of the deviations of the body from the position of equili- brium. But sometimes we may settle this question from statical considerations. If the conditions of the problem are such that for all displacements of the body from the position A within certain limits, the forces tend to bring the body back to that position, then the position may be regarded as stable for displacements ART. 76.] STABLE AMJ3 UNSTABLE EQUILIBRIUM. 45 within those limits. If on the other hand the forces tend to remove the body further from the position A, that position may be regarded as unstable. This cannot however be strictly proved to be a sufficient condition until we have some dynamical equa- tions at our disposal. Properly we should, for the present, distinguish this as the criterion of statical stability or statical instability. But for the sake of brevity we shall omit this dis- tinction, except when we wish to draw special attention to it. When the body is disturbed, it is of course understood that the forces which act on it are subject to given conditions. Frequently these are that they shall continue to act at the same points of the body, parallel to the same directions in space and remain of the same magnitude. When these are the conditions a state of neutral equilibrium is sometimes called astatic. 76. Two equal given forces P, Q act on a body at two given points A, B, and are in equilibrium. They therefore act along the straight line AB. Let the body be now turned round through any angle less than two right angles and let the forces continue to act at these points in directions fixed in space. It is required to find the condition of stability. Referring to the figure, it is evident that the forces tend to restore the body to its former position if each force acts from the point of application of the other force, while they tend to move the body further from that position if each force acts towards the point of application of the other. In the first case the equilibrium is stable, in the second unstable. If the- body be turned round through two right angles, the forces will again be in equilibrium. The position of stable equili- brium will then be changed into one of unstable equilibrium and conversely. .A B P Q A C- — *- i ;jg 46 FORCES ACTING AT A POINT. [CHAP. II. 77. Ex. 1. Three given forces P, Q, R act on a body in one plane at three given points A, B, C and are in equilibrium. When the body is disturbed, the forces continue to act at these points parallel to directions fixed in space and their magnitudes are unaltered. Find the condition of stability. In the given position of equilibrium the lines of action of the forces must meet in some point 0. If the point lie on the circle circumscribing ABC we have already seen in Art. 71 that the equilibrium will still exist after a displacement of the body. In this case the equilibrium is neutral or astatic. Next let the point lie within the segment of the circumscribing circle opposite to the point C. In order that the resultant of P and Q may pass through G in the position of equilibrium the forces P, Q must act both from towards A, B respectively or both towards from A and B. Let us first suppose them to act both towards A, B while R acts from G towards 0. Describe a circle about OAB cutting OG in C. Then since is within the circumscribing circle, C" is without that circle. By Art. 71, the forces P and Q are astatically equivalent to a force equal and oppo- site to R but acting at C". Thus the whole / qi \ system is equivalent to two equal forces acting / "I^SN* I at G and C" and each tending away from the a\£~ — ^^^ jiji5 point of application of the other. The equi- f\ s\^, — - — ""/ \ librium is therefore stable for all rotatory dis- / J^\"^ / ' placements less than two right angles. In the C'r ^ same way if the forces P, Q act respectively from A and B towards the equilibrium is unstable. If the point lie outside the circumscribing circle, but within the angle ACB, the point C" is within that circle. The conditions of the argument are then just reversed, and therefore if the forces P, Q tend from towards A, B the equilibrium is unstable. If the point lie within the triangle A BG, the forces must all three act from towards their points of application or all three towards from their points of application. By the same reasoning as before we may show that in the former case the equilibrium is stable, in the latter unstable. Summing up, we have the following result. If two at least of the forces in equilibrium act from the common point of intersection towards their points of application A, B, C ; then the equilibrium is stable if lie within the circle circumscribing ABC and unstable if lie outside that circle. If two at least of the forces act from their points of application towards 0, these conditions are reversed. Ex. 2. A smooth circular ring is fixed in a horizontal position, and a small ring sliding upon it is in equilibrium when acted on by two forces in the directions of the chords PA, PB. Prove that, if PC be a diameter of the circle, the forces are in the ratio of BG to AC. If A and B be fixed points and the magnitude of the forces remain the same, show that the equilibrium is unstable. [Math. Tripos, 1854.] CHAPTER III. PARALLEL FORCES. 78. To find the resultant of two parallel forces. Let the two parallel forces be P, Q and let them act at A, B, which of course are any points in their lines of action. In order to obtain a point of intersection of the forces at a finite distance let us impress at A, B in opposite directions two equal forces of any magnitude, each of which we may represent by F, Art. 15. The resultants of P, F and Q, F act respectively along some straight lines AO, BO which intersect in 0. Thus we have replaced the two given forces by two others, each of which may be supposed to act at 0. Draw OG parallel to AP, BQ to cut AB in G. Consider the force acting at along OA. We may resolve this force (as in Duchayla's proof of the parallelogram of forces) into two forces, one equal to P acting along OG and the other equal to F acting parallel to GA. In the same way the other force acting at along OB is equivalent to Q acting along OG and F acting at parallel to GB. The two forces each equal to F balance each other and may be removed. The whole system is therefore reduced to the single force P+Q acting along OG. The sides of the triangle OGA are parallel to P, F and their OG P resultant. Hence prj = p- In the same way oq = q GB F' ™ , , t- AC BG AB We therefore have -tj- = -p- = p.Q • 48 PARALLEL FORCES. [CHAP. III. The resultant of the parallel forces P, Q is P + Q, and its line of action divides every straight line AB which intersects the forces in the inverse ratio of the forces. F A B FC If the forces P, Q act in opposite directions the proof is the same, but the figure is somewhat different. If Q be greater than P, BO will make a smaller angle with the force Q than OA makes with the force P. Hence will lie within the angle QBG. In this case the magnitude of the resultant is Q — P and its line of action divides AB externally in the inverse ratio of P to Q. 79. Conversely any given force R acting at a given point G may be replaced by two parallel forces acting at two arbitrary points A and B, where A, B, G are in one straight line. Let us represent these forces by P and Q. Let CA = a, GB — b, and let these be regarded as positive when measured from G in the same direction. We then find P + Q = R, P = ~R, Q = R. b — a ' a — b~ If A and B lie on the same side of G, a and b are positive ; in this case the force nearer R acts in the same direction as R, the other force acts in the opposite direction and is therefore negative. If G lie between A and B, one of the two distances a, b is negative ; in this case both forces act in the same direction as R. 80. To find the resultant of any number of parallel forces P 1( P 2 &c. acting at any points A 1; A 2 &c. when referred to any axes. Let {x-iy^, (x 2 y z z 2 ) &c. be the Cartesian coordinates of the points A lt A 2 &c. The forces P lt P 2 acting at A u A% are equiva- lent to a single force Pi + P^ acting at a point (7i situated in A,A 2 such that P^A^^P^A.C, (Art. 78). Let (£%&) be ART. 81.] CENTRE # OF PARALLEL FORCES. 49 the coordinates of G x . Since A&, A& are in the ratio of their projections on the axes of coordinates we have i\(ft-0 = P 2 (* 2 -ft) Similar results apply for the other coordinates of C^. The force P a + P 2 acting at Cj and a third force P 3 acting at A 3 are in the same way equivalent to P 1 + P 2 + P 3 acting at a point 2 whose coordinates (fa^fa) are given by (P, + P 2 + P 3 ) ft = (P, + P 2 ) ft + P 3 * 3 = P& + P 2 aj 2 + PsPs with similar expressions for % and f 2 . Proceeding in this way we see that the resultant of all the forces is P 1 + P i +... and if (ffy§) be the coordinates of its point of application, we have (P, + P 2 + &c.) £ = Pjo, + P A + &c. (P x + P 2 + &c.) v = P^ + P 2 2/ a + &c. (P x + P 2 + &c) .? = P A + P 2 s 2 + &c. These equations are usually written SPa SPy SP^ f~ SP' ,_ SP* ? SP" 81. It might be supposed that this proof would either fail or require some modification if any one of the partial resultants P 1 + P 2 , P 1 + P % +P s &e. were zero. The effect of this would be that some of the quantities f x , | 2 &o. would be infinite, and the final result might be thought to fail if 2P=0. But any proposition proved true for general values of the forces must be true for these limiting cases, though its interpretation may not be understood until we come to the theory of couples. We may avoid this apparent difficulty by a slight modification of the proof. Let us separate the forces into two groups. Let the forces in one group act in one direction and in the other group in the opposite direction. Let us suppose the sums of the forces in the two groups are unequal. Let us compound together first all the forces in that group in which the sum is greatest and then join to these one by one the forces of the other group. It is clear that we shall never have any of the partial resultants equal to zero and no point of application of any such partial resultant will be at infinity. If the sums of the forces in the two groups are equal, the centre of parallel forces is infinitely distant. Ex. 1. Parallel forces, each equal to P, act at the corners A, B, C, D of a re-entrant plane quadrilateral and a fifth force equal to - P acts at the intersection // of the diagonals HGA, BHD. If the centre of the five parallel forces coincide with a corner C of the quadrilateral, prove that HG= GA. R. s. 4 50 PARALLEL FORCES. [CHAP. III. Ex. 2. ABC is a triangle; APD, BPE, CPF, the perpendiculars from A, B, G on the opposite sides. Prove that the resultant of six equal parallel forces, acting at the middle points of the sides of the triangle and of the lines PA, PB, PC, passes through the centre of the circle which goes through all of these middle points. [Math. Tripos, 1877.] Ex. 3. ABCD is a quadrilateral whose diagonals intersect in 0. Parallel forces act at the middle points of AB, BC, CD, DA respectively proportional to the areas AOB, BOG, COD, DOA. Prove that the centre of parallel forces is at the fourth angular point, viz. G, of the parallelogram described on OE, OF as adjacent sides where E, F are the middle points of the diagonals AC, BD of the quadrilateral. [Coll. Ex., 1885.] Taking BD as the axis of x we find t\ = \ (p-p 1 ) where p, p' are the per- pendiculars from A and C on BD. It follows that the centre of parallel forces lies on EG. Similarly it lies on FG. 82. These equations determine only one point in the line of action of the resultant. But since the resultant acts parallel to the forces P Jt P 2 &c, we may suppose it to act at any point in the straight line drawn through the point (i-y£) parallel to the given forces. The point (|fo£) determined by the equations of Art. 80 has one important property. Its position is the same whatever be the magnitudes of the angles made by the forces with the coordinate axes. If then the points of application of the given parallel forces viz. A. lt A 2 &c. are regarded as fixed in the body, the point of appli- cation of their resultant is also fixed in the body however the forces are turned round their points of application provided they remain parallel and unaltered in magnitude. This point of application of the resultant is called the " centre of parallel forces." It appears from Art. 51 that the central point of the forces thus determined is the same as the mean centre or centroid of a system of particles placed at A lt A 2 &c. whose masses or weights are proportional to the forces P 1 , P 2 &c. 83. This proposition is really the limiting case of one already considered. If concurrent forces act along OA 1 , OA 2 &o. their resultant may be found by any of the methods considered in the last chapter. By regarding as a point very distant from the points A lt A 2 &o., the forces acting along OA lt OA 2 &o. become parallel and the corresponding theorem follows at once. Thus in Art. 51 it is shown that the resultant of forces proportional to P 1 . OA 2 , P 2 . OA 3 &o. is a force proportional to 2P . OG acting along OG where C is the centre of gravity of particles P x , P 2 &a. placed at A lt A 2 P Y B' 9" Let P, Q be the equal forces, AB the arm of the given couple. Let A'B' be the new arm, P', Q' the new forces. Apply at each of the points A', B' equal and opposite forces, each equal to P'. Then by the conditions of the proposition, P . AB = P' . A'B'. Hence if G be the middle point of both AB and A'B', we have P.AG = P'.A'C. The forces P and P" have a resultant P — P" which by Art. 78 acts at G. In the same way Q and Q" have an equal resultant, also acting at G but in the opposite direction. Removing these two, it follows that the given couple is equivalent to the couple of forces + P' acting at A', B'. ART. 95.] THBORY OF COUPLES. 59 94. It follows from Arts. 91 and 92 that a couple may be transferred without altering its effect from one given position to any other given position in a parallel plane. Thus by Art. 92 we may turn a couple round the middle point of its arm until the forces become parallel to their directions in the second given position. Then by Art. 91, we may move the couple parallel to itself into the required position. It follows from Art. 93 that the forces and the arm may also be changed without altering the effect of the couple, provided its moment is kept the same. Summing up these results, we see that a couple is to be regarded as given when we know, (1) the position of some plane parallel to the plane of the couple, (2) the direction of rotation of the couple in its plane, and (3) the moment of the couple. 95. To find the resultant of any number of couples acting in parallel planes. Let P lt P 2 &c. be the magnitudes of the forces, a x , a 2 &c. the arms of the couples. Let us first suppose the couples all tend to produce rotation in the same direction. By Art. 94 we may move these couples into one plane and turn them about until their arms are in the same straight line. We may then alter the arms and forces of each until they all have a common arm AB whose length is, say, equal to b. The forces of the couples now act at the extremities of AB, and are respectively equal to P&fb, P 2 a 2 /b &c. All these together constitute a single couple each of whose forces is (Pa + P^a-z + &a.)/b and whose arm is b. This single couple is equivalent to any other couple in the same plane with the same direction of rotation whose moment is Pitti + Pattjj 4- &c, i.e. whose moment is the sum of the moments of the separate couples. If some of the couples tend to produce rotation in the opposite direction to the others, we may represent this by regarding the forces of these couples as negative. The same result follows as before. We thus obtain the following theorem ; the resultant of any number of couples whose planes are parallel is a couple whose moment is the algebraic sum of the moments of the separate couples and whose plane is parallel to those of the given couples. 60 PARALLEL FORCES. [CHAP. III. 96. Measure of a couple. We may use the proposition just established to show that the magnitude of a couple regarded as a single element is properly measured by its moment. To prove this we assume as a unit the couple whose force is the unit of force and whose arm is the unit of length. The moment of this couple is unity. By this proposition a couple whose moment is n times as great is equivalent to n such couples and its magnitude is therefore properly represented by the symbol n. 97. Axis of a couple. A couple may tend to produce rotation in one direction or the opposite according to the circum- stances of the couple. One of these is usually called the positive direction and the other the negative. Just as in choosing axes of coordinates sometimes one direction is taken as the positive one and sometimes the other, so in couples the choice of the positive direction is not always the same. In trigonometry the direction of rotation opposite to the hands of a watch is taken as the positive direction. In most treatises on conies the same choice is made. In solid geometry the opposite direction is generally chosen. Having however chosen one of these two directions as the positive one it is usual to indicate the direction of rotation of a given couple in the following manner. From any point G in the plane of the couple draw a straight line CD at right angles to the plane and on one side of it. The straight line is to be so drawn that if an observer stand with his feet at G on the plane and his back along CD, the couple will appear to him to produce rotation in what has been chosen as the positive direction. The straight line CD is called the positive direction of the axis of the couple. To indicate the direction of rotation of a couple it is sufficient to give the direction in space of CD as distinguished from DC. This is effected by the convention usually employed in solid geometry. A finite straight line having one extremity at the origin of coordinates is drawn parallel to CD. The position of this straight line is defined by the angles it makes with the positive directions of the axes of coordinates. The position of the straight line CD, when given, indicates at once the plane of the couple and the direction of rotation. We may also use a length measured along CD to represent the magni- ART. 98.] THMRY OF COUPLES. 61 tude of the moment of the couple, in just the same way as a straight line was used in Art. 7 to represent the magnitude of a force. We therefore infer that all the circumstances of a couple may be properly represented by a finite straight line measured from a fixed point in a direction perpendicular to its plane. This finite straight line is called the axis of the couple. 98. To find the resultant of two couples whose planes are inclined to each other. Let the two couples be moved, each in its own plane, until they have a common arm AB, which of course must lie in the intersec- tion of the two planes. In effecting this change of arm it may have been necessary to alter the forces of the couples, but the moments of the couples must remain unaltered. Let the forces thus altered be P and Q. At the point A we have two forces P and Q; these are equivalent to some resultant R found by the parallelogram of forces. At the point B there are two forces equal and opposite to those at A ; their resultant is equal, parallel and opposite to R. Thus the two couples are equivalent to a single couple, each of whose forces is equal to R, and whose arm is AB. Let the length of AB be b. From any point G (which we may conveniently take in AB) draw Op, Gq in the directions of the axes of the given couples, and measure lengths along them proportional to their moments, viz. to Pb and Qb. These axes are perpendicular to the planes of the couples, and their lengths are also proportional to P and Q. If we compound these two by the parallelogram law we evidently obtain an axis perpendicular to the plane of the forces + R, 62 PARALLEL FORCES. [CHAP. III. whose length is proportional to R. It is evident that the paral- lelogram Cpqr is similar to that contained by the forces PQR, but the sides of one parallelogram are perpendicular to the sides of the other. We therefore infer the following construction for the resultant of any two couples. Draw two finite straight lines from any point C to represent the axes of the couples in direction and magnitude. The resultant of these two obtained by the parallelogram law repre- sents in direction and magnitude the aids of the resultant couple. The rule to compound couples is therefore the same as that already given for compounding forces. It follows that all the theorems for compounding forces deduced from the parallelogram law also apply to couples. The working rule is that if we represent the couples by their axes, we may compound and resolve these as if they were forces. 99. Ex. A system of couples is represented in position and magnitude by the areas of the faces of a polyhedron, and their axes are turned all inwards or all outwards. Show that they are in equilibrium. Mobius. This follows at once from the corresponding rule for forces. 100. A force P acting at any point A may be transferred parallel to itself, to act at any other point B, by introducing a couple whose moment is Pp, where p is the perpendicular distance of B from the line of action AF of P. This couple, acts to turn the body in the direction AFB. P" A Hr B 'P' p P Apply at B two equal and opposite forces P', P", each equal to P. One of these, viz. P', is the force P transferred to act at B. The two forces P" and P then constitute the couple whose moment is Pp. 101. Summing up the various propositions just proved on forces and couples, we find that they fall into three classes. These may be briefly stated thus : 1. Forces may be combined together according to the paral- lelogram law. AET. 103.] THE^BY OF COUPLES. 63 2. Couples may be combined together according to the paral- lelogram law. 3. A force is equivalent to a parallel force together with a couple. The theorems in the subsequent chapters are obtained by continual applications of these three classes of propositions. It is therefore evident that these theorems will apply also to any other vectors for which these three classes of propositions are true. Thus in dynamics we find that the elementary relations of linear and angular velocities are governed by these three sets of proposi- tions. We therefore apply to these, without further proof, all the theorems found to be true for couples and forces. 102. Initial motion of the body. If a single couple act on a body at rest, it is clear that the body will not remain in equilibrium. It is proved in treatises on dynamics that the body will begin to turn about a certain axis. Since a couple can be moved about in its own plane without altering its effect, this axis cannot depend on the position of the couple in its plane. The dynamical results are (1) the initial axis of rotation passes through the centre of gravity of the body, (2) the axis of rotation is not necessarily perpendicular to the plane of the couple, though this may sometimes be the case. The construction to find the axis is somewhat complicated, and its further discussion would be out of place in a treatise on statics. We may show by an elementary experiment that the axis of rotation is independent of the position of the couple in its plane. Let a disc of wood be made to float on the surface of water contained in a box. At any two points A, B attach to the disc two fine threads and hang these over two small pullies, fixed in the sides of the vessel at G and D, with equal weights suspended at the other extremities. Let the strings AG, BD be parallel so that their tensions form a couple. Under the influence of this couple the body will begin to turn round. However eccentrically the points A, B are situated the body begins to turn round its centre of gravity. The body may not continue to turn round this axis for, as the body move.-, the strings cease to be parallel. For this and other reasons the motion of rotation is altered. 103. Ex. 1. Forces P, 2P, iP, IP act along the sides of a square taken in order ; find the magnitude and position of their resultant. [St John's Coll., 1880.] The two forces IP and 2P make a couple which may be turned through a right angle ; Art. 92. In this way two opposite sides of the square become occupied by forces respectively equal to 3P and 6P acting in opposite directions, while the two other sides are left vacant. Since one of these forces is double the other, the position of the resultant is evident, but it may also be found by Art. 78. It acts parallel to the force 4P, at a distance from it equal to the side of the square. Ex. 2. A triangular lamina ABC is moveable in its own plane about a point in itself: forces act on it along and proportional to BG, CA, BA. Prove that if these 64 PARALLEL FORCES. [CHAP. III. do not move the lamina, the point must lie in the straight line which bisects ISC and CA. [Math. Tripos, 1874.] Ex. 3. Forces are represented in magnitude, direction, and position by the sides of a triangle taken in order; prove that they are equivalent to a couple whose moment is twice the area of the triangle. If these sides taken in order represent the axes of three couples, prove that these couples are in equilibrium. Ex. 4. Forces act along the sides of a skew quadrilateral ABGD, proportional to these sides ; show that they are equivalent to a couple, whose moment is represented in magnitude and direction by four times the area of the parallelogram whose vertices are the middle points of the sides. [Coll. Ex., 1890.] The forces at the corner D have a resultant acting at D equal to GA. Those at B have an equal and opposite resultant. The moment of this couple is CA.BD.sia9, where 9 is the angle which CA, BD make with each other. The result follows by similar triangles. Ex. 5. Forces act along the sides AB, BC, CD, DA of a skew quadrilateral proportional to these sides ; show that they are equivalent to a couple, whose axis is parallel to the shortest distance between AC and BD. The proof is similar to that given in Ex. 4. Ex. 6. Forces are represented in magnitude, direction, and position by the sides of a skew polygon taken in order j show that they are equivalent to a couple. If the corners of the skew polygon are projected on any plane, prove that the resolved part of the resultant couple in that plane is represented by twice the area of the projected polygon. Ex. 7. Three couples are represented in position and magnitude by the areas of three faces of a tetrahedron OABC, say the faces OBG, OCA, OAB. If their axes are all turned inwards, prove that their resultant couple is represented in position and magnitude by the area of the fourth face ABC, and that its axis is turned outwards. If the axes of the given couples which act in two of the faces, say OBC, OCA, are turned inwards, while the axis of the couple in the third face OBA is turned outwards, so that the directions of rotation are indicated by the order of the letters as given above, prove that their resultant couple acts in the plane ODE, and is represented by four times the area of the triangle ODE, where D and E are the middle points of BC and CA. The first proposition in the example is a particular case of Mobius' theorem, see Art. 99. We proceed to the second proposition. Let the linear scale be such that the moments of the given couples are represented by twice the areas of the triangles OBG, OCA, OBA. Let one force of the couple AOC be made to act along GA, and let it be represented by GA. Then the other force of this couple acts at in the opposite direction. Treating the other two couples in a similar manner, the whole system is replaced by the forces BC, GA, and BA together with the parallel and opposite forces at 0. The forces BC, CA are equivalent to a force acting at C equal to BA. Hence the three forces BC, CA and BA are equivalent to a force 4DE acting along DE. In the same way the forces acting at parallel and opposite to BC, GA and BA are equivalent to a single force 4ED acting at 0. The whole system is therefore equivalent to four times the couple represented by twice the area ODE. AKT. 103.1 THEORY OF COUPLES. 65 J • Ex. 8. AC, BD are two non-intersecting straight lines of constant length; prove that the effect of forces represented in every respect by AB, BO, CD, DA is the same, so long as AG, BD remain parallel to the same plane, and their projections on that plane are inclined at a constant angle to one another. [Coll. Ex., 1881.] The forces AB, BG are equivalent to a force equal to AG acting at B. The forces CD, DA are equivalent to a force equal to GA acting at D. These make a couple whose plane is parallel to the given plane and whose moment is AG . BD . sin 6, where 6 is the constant inclination of the projections of AG, BD. The result follows by Art. 94. Ex. 9. If two equal lengths Aa, Bb, are marked off in the direction along a given straight line, and two equal lengths Gc, Dd along another given line, prove that forces represented in every respect by AG, ca, GB, be, BD, db, DA, ad are in equilibrium. [Trin. Coll.] Ex. 10. Forces proportional to the sides a lt where c 2 =a 2 -6 2 . Ex. 5. Two wires, bent into the forms of equal catenaries, are placed so as to have a common vertical directrix, and their axes in the same straight line. The extremities of a uniform rod are attached to two small rings which can freely .slide on these catenaries. Show that in equilibrium the rod must be horizontal. Ex. 6. A straight uniform rod has smooth small rings attached to its extre- mities, one of which slides on a fixed vertical wire and the other on a fixed wire in the form of a parabolic arc whose axis coincides with the former wire, and whose latus rectum is twice the length of the rod : prove that in the position of equilibrium the rod will make an angle of 60° with the vertical. [Math. Tripos, 1869.] Ex. 7. AG, BC are two equal uniform rods which are jointed at C, and have rings at the ends A and B, which slide on a smooth parabolic wire, whose axis is vertical and vertex upwards ; prove that in the position of equilibrium the distance of C from AB is one fourth of the latus rectum. [Math. Tripos, 1871.] Ex. 8. Two heavy uniform rods AB, BG whose weights are P and Q are connected by a smooth joint at B. The ends A and G slide by means of smooth rings on two fixed rods each inclined at an angle a to the horizon. If 9 and be the inclinations of the rods to the horizon, show that Pcot0=Qcotd = (P + Q)tana. [Trin. Coll., 1882.] Besolve horizontally and vertically for the two rods regarded as one system; then take moments for each singly about B. 127. Two smooth rods OM, ON, at right angles to each other are fixed in space. A uniform elliptic disc is supported in the same vertical plane by resting on these rods. If OM make an angle a with the vertical, prove that either the axes of the ellipse are parallel to the rods, or the major axis makes an angle B with OM, given by a 2 tan* a -6 s tan 2 d = at-tftania' 80 FORCES IN TWO DIMENSIONS. [CHAP. IV. Let P, Q be the points of contact and let the normals at P, Q meet in I. Let C be the centre, then in equilibrium either C and I must coincide, or GI is vertical. In the former case the tangents OM, ON are parallel to the axes. In the latter case, let D bisect PQ, then OD produced passes through C; but because the tangents are at right angles OPIQ is a rectangle, therefore OD passes through I. Hence OCI is a vertical straight line. These two results follow easily from a " principle to be proved in the chapter on virtual work. As the ellipse is moved round, always remaining in contact with the rods, we know by conies that C describes an arc of a circle, whose centre is 0, and whose radius is v /(a 2 + 6 2 ). Hence when C is vertically over 0, its altitude is a maximum. When the axes are parallel to the rods, C is at one of the extremities of its arc and its altitude is a minimum. It immediately follows from virtual work that the first of these is a position of unstable equilibrium, and that the other two are positions of stable equilibrium. Eesuming the solution, we have now to find when GI is vertical. The perpendicular from C on OM makes with the major axis an angle equal to the complement of 8, hence a? sin 2 8 + 6 2 cos 2 8 = OC 2 sin 2 o = (a 2 + ft 2 ) sin 2 a. The value of tan 2 8 follows immediately. Ex. 2. An elliptic disc touches two rods OM, ON, not necessarily at right angles, and is supported by them in a vertical plane. If (XY) be the coordinates of the intersection of the rods, referred to the axes of the ellipse, prove that the major axis is inclined to the vertical at an angle given by tan 0= - -= .,_ y2 . To prove this we may use a theorem deduced from two given by Salmon in his chapter on Central Conies, Art. 180, Sixth Edition. Let (XY) be a point from which two tangents are drawn to touch a conic at P, Q. The normals at P, Q meet in a point I, whose coordinates (xy) are given by X~ K ; a 2 r 2 +6 2 X 2 ' Y~ v "'a, 1 T>+»X»' The result follows, since CI must be vertical. Ex. 3. An elliptic disc is supported in equilibrium in a vertical plane by resting on two smooth fixed points in a horizontal straight line. Prove that in equilibrium either a principal diameter is vertical, or these points are at the extremities of two conjugate diameters. Let the principal diameters be the axes of coordinates. Let the fixed points P, Q be (xy), (x'y'), and let (£7;) be the intersection I of the normals at these points. In equilibrium IC must be perpendicular to PQ, hence (x - x 1 ) J + (y - y') 7; = 0. By writing down the equations to the normals at P, Q we find |, 77, as is done in Salmon's Conies, Art. 180. This equation then becomes *-*')fc/-y>(5 + fQ = 0. ART. 129.] SOLUTIpN OF PROBLEMS. 81 One of these three factors must therefore vanish. These give the three positions of equilibrium. That there should be equilibrium when P, Q are at the extremities of two conjugate diameters is evident ; for PI, QI are perpendiculars from two of the corners of the triangle CPQ on the opposite sides, hence CI must be perpendicular to the side PQ. This is the condition of equilibrium. That there should be equilibrium when an axis is vertical is evident from symmetry. ^ 128. Ex. 1. A cone has attached to the edge of its base a string equal in length to the diameter of the base, and is suspended by the extremity of this string from a point in a smooth vertical wall, the rim of the base also touching the wall. If a be the semi-angle of the cone, $ the inclination of the string to the vertical, prove that in a position of equilibrium tan a tan 0=xj ■ Assume that the centre of gravity of the cone is in its axis at a distance from the base equal to one quarter of the altitude. Ex. 2. A square rests with its plane perpendicular to a smooth wall, one corner being attached to a point in the wall by a string whose length is equal to a side of the square. Prove that the distances of three of its angular pointB from the wall are as 1, 3 and 4. [Math. Tripos, 1853.] By resolving vertically, and taking moments about the corner of the square which is in contact with the wall, we obtain two equations from which the inclination of any side to the wall and the tension may be found. Ex. 3. AB is a uniform rod of length a ; a string APBG is fastened to the end A of the rod and passes through a smooth ring attached to the other end B ; the end G of the string is fastened to a peg C, and the portion APB is hung over a smooth peg P which is in the same horizontal plane as C at a distance 26 from it (b < a). If AP is vertical, find the angles which the other parts of the string make with the vertical, and show that the string must have one of the lengths %bJS i^-ft 2 ). [King's Coll., 1889.] Ex. 4. Two light elastic strings have their ends tied to a fixed point on the line joining two small smooth pegs which are in the same horizontal plane, so that when they are unstretched their ends just reach the pegs ; they hang over the pegs and have their other ends fastened to the ends of a heavy uniform rod ; show that the inclination of the rod to the horizon is independent of its length, being equal to tan -1 (yj - y^fta, where y 1 and y 2 are the extensions of the strings when they singly support the rod, and a is the distance between the pegs. Show also that the two strings and the rod are inclined to the horizon at angles whose tangents are in arithmetical progression. [Math. Tripos, 1887.] It may be assumed that the tension of each string is proportional to the ratio of its extension to its unstretched length. 129. Ex. 1. A sphere rests on a string fastened at its extremities to two fixed points. Show that if the are of contact of the sphere and plane be not less than 2 ton -1 £f , the sphere may be divided into two equal portions by means of a vertical plane without disturbing the equilibrium. [Math. Tripos, 1840.] It may be assumed that the centre of gravity of a solid hemisphere is on the middle radius at a distance $ ths of that radius from the centre. R. s. 6 82 FORCES IN TWO DIMENSIONS. [CHAP. IV. Consider the equilibrium of the hemisphere ABD and the portion AD of the string in contact with it. The mutual reactions of the string and the hemisphere may now be omitted. This compound body is acted on by (1) the tensions of the string, each equal to T, acting at A and D, (2) the weight W of the hemisphere acting at its centre of gravity G, (3) the mutual reaction R of the two hemispheres. The reaction R is the resul- tant of all the horizontal pressures between the elements of the plane bases and must act at some point within the area of contact. The two bases will separate unless the resultant of the remaining forces also passes inside the area of contact. The arc AD being as small as possible, this separation will take place by the hemispheres opening out at B, for the mutual pressures are then confined to the single point A at the lowest point of the sphere. The hemisphere ABD is then acted on by the three forces, T at D, T- R at A, and W at /(73). [Christ's Coll.] Ex. 3. An elliptic lamina of eccentricity e, divided into two pieces along the minor axis, is placed with its major axis horizontal in a loop of string attached to two fixed points, so that the portions of the strings not in contact with the- — * ellipse are vertical. Show that equilibrium will not exist unless (67re) 2 <(97r-4) (3tt + 4). [Coll. Ex., 1890.] Each semi-ellipse is acted on by two equal tensions along the tangents at the extremities A and B of the axes. These have a resultant inclined at 45° to either , axis. Let it cut the vertical through the centre of gravity G in the point H. The reaction between the semi-ellipses must pass through H. Hence the altitude of H above B must be less than the axis minor. If G be the centre, this gives at once a - CG < 26. Granting that CG = 4a/3?r, this leads to the result. Ex. 4. A circular cylinder rests with its base on a smooth inclined plane ; a string attached to its highest point, passing over a pulley at the top of the inclined plane, hangs vertically and supports a weight ; the portion of the string between the cylinder and the pulley is horizontal : determine the conditions of equilibrium. [Math. Tripos, 1843.] Show that the ratio of the height of the cylinder to the diameter of its base must be less than the cotangent of the inclination of the plane to the horizon. Ex. 5. A uniform bar of length u, rests suspended by two strings of lengths I and V fastened to the ends of the bar and to two fixed points in the same horizontal line at a distance c apart. If the directions of the strings being produced meet at right angles, prove that the ratio of their tensions is al + cV : al'+ el. [Math. Tripos, 1874.] Ex. 6. A smooth vertical wall AB intersects a smooth plane BG so that the line of intersection is horizontal. Within the obtuse angle ABC a smooth sphere ART. 130.] SOLUTION OF PROBLEMS. 83 of weight W is placed and is kept in contact with the wall and plane by the pressure of a uniform rod of length I which is hinged at A, and rests in a vertical plane touching the sphere. Show that the weight of the rod must be greater than Wh cos a cos £a 21 sin \B sin £ (a - 6) cos 2 £ (a - 6) ' where a and B are the acute angles made by the plane and rod with the wall, and h=AB. [Math. Tripos, 1890.] Ex. 7. A set of equal frictionless cylinders, tied together by a fine string in a bundle whose cross section is an equilateral triangle, lies on a horizontal plane. Prove that, if W be the total weight of the bundle, and n the number of cylinders in a side of the triangle, the tension of the string cannot be less than -j—^ [ 1 + - ) 4J3 \ nj W ( 1\ or j-jx (1 1 , according as n is an even or an odd number, and that these values will occur when there are no pressures between the cylinders in any horizontal row above the lowest. [Math. Tripos, 1886.] Ex. 8. A number n of equal smooth spheres, of weight W and radius r, is placed within a hollow vertical cylinder of radius a, less than 2r, open at both ends and resting on a horizontal plane. Prove that the least value of the weight W of the cylinder, in order that it may not be upset by the balls, is given by a W'= (n - 1) (a - r) W or aW'=n (a - r) W, according as n is odd or even. [Math. Tripos, 1884.] Ex. 9. The circumference of a heavy rigid circular ring is attached to another concentric but larger ring in its own plane by n elastic strings ranged symmetrically round the centre along common radii. This second ring is attached to a third in a similar manner by 2rc strings, and this to a fourth by 3n strings and so on. Supposing all the rings to have the same weight, and the strings at first to be without tension, show that, if the last ring be lifted up and held horizontal, all the other rings will be on the surface of a right cone. [Pet. Coll., 1862.] ' Ex. 10. Two spheres of densities p and where [GHY] and [DCX] represent the areas of the triangles GHY and DGX. These areas are equal to iHO .p' and %DX. GX sin X respectively. Also AB . XM is twice the area of the triangle AXB, and is therefore equal to XA . XJB sin X. Again YD _ AD YG BC XA _ AB XB sin A ~~ sin I" sin B ~ sin Y ' sin B ~ sin X ~ sin A ' Substituting we obtain the equation connecting T, T' given in the enunciation. 133. Ex. 1. A series of rods in one plane, jointed together at their extremities, form a closed polygon. Each rod is acted on at its middle point in a direction per- pendicular to its length by a force whose magnitude is proportional to the length of the rod. These forces act all inwards or all outwards. Show that in equilibrium (1) the polygon can be inscribed in a circle, (2) the reactions at the corners act along the tangents to the circle, (3) the reactions are all equal. Let AB, BC, CD, &c. be the rods, L, M, N, &a. their middle points. Let aB/3, flGy, &o. be the lines of action of the reactions at the corners B, G &a. Since each rod is in equilibrium, the forces at the middle points of the rods must pass through * Let D, E, F be three arbitrary points taken on the sides of a triangle ABC. If A, A' be the areas of the triangles ABC, DEF, it may be shown that X_ AF.BD.CE + AE. CD.BF A ~~ abe To form the two products AF . BD . GE and A E . CD . BF, we start from any corner, say A, and travel round the triangle, first one way and then the other, taking on each circuit one length from each side. The sum of the two products so formed, each with its proper sign, is the expression in the numerator. The signs of these factors maybe determined by the following rule. Each length, being drawn from one of the corners of the triangle ABC, along one of the sides, is to be regarded as posi- tive or negative according as it is drawn towards or from the other corner in that side. Thus, AF being drawn from A towards B is therefore positive, BF being drawn from B towards A is also positive. If F were taken on AB produced beyond B, AF would still be positive, but BF would be negative. If F move along the side AB, in the direction AB, the area DEF vanishes when F reaches the transversal ED, and becomes negative when F passes to the other side of it. In the same way, if we draw any three straight lines through the corners of the triangle, say AD, BE, GF, they will enclose an area PQB. If the area of the triangle PQR is A", it may be shown that A"_ (AF.BD.GE-AE.GD. BFf ~~K ~\ab-GE. CD)(bc-AE.AF)(ca-BF.BD) m The author has not met with these expressions for the area of two triangles which often occur. He has therefore placed them here in order that the argument in the text may be more easily understood. 90 FORCES IN TWO DIMENSIONS. [CHAP. IV. a, p, y, &o. respectively. Consider the rod BG ; the triangles BMp, CMp are equal and similar, also the reactions along B/3 and (7/8 balance the force along Jf/S which bisects the angle B/3C. Hence these reactions are equal. It follows that the reactions at all the corners are equal in magnitude. Draw BO, GO perpendicular to the directions of the reactions at B and C. These must intersect in some point on the perpendicular through M to BG. The sides of the triangle OBG are perpendicular to the directions of the three forces which act on the rod BG, and are in equilibrium. Hence GO represents the magnitude of the reaction at C on the same scale that BG represents the force at its middle point. In the same way if GO', DO' be drawn perpendicular to the reactions at G and D, they will meet in some point 0' on the perpendicular through N to CD. Also GO' will measure the reaction at G on the same scale that CD measures the force at its middle point. Hence by the conditions of the question CO= CO', and therefore and 0' coincide. Thus a circle, centre 0, can be drawn to pass through all the angular points of the polygon and to touch the lines of action of all the reactions. Ex. 2. A series of jointed rods form an unclosed polygon. The two extremities of the system are constrained, by means, of two small rings, to slide along a smooth rod fixed in space. If each moveable rod is acted on, as in the last problem, by a force at its middle point perpendicular and proportional to its length, prove that the polygon can be inscribed in a circle having its centre on the fixed rod. Let A and Z be the two extremities. We can attach to A and Z a second system of rods equal and similar to the first, but situated on the opposite side of the fixed rod. We can apply forces to the middle points of these additional rods acting in the same way as in the given system. With this symmetrical arrange- ment the fixed rod becomes unnecessary and may be removed. The results follow at once from those obtained in the last problem. These two problems may be derived from Hydrostatical principles. Let a vessel be formed of plane vertical sides hinged together at their vertical intersections, and let this vessel be placed on a horizontal table. Let the interior be filled with fluid which cannot escape either between the sides and the table or at the vertical joinings. The pressures of the fluid on each face will be proportional to that part of the area of each which is immersed in the fluid, and will act at a point on the median line. These pressures are represented in the two problems by the forces acting on the rods at their middle points. It will follow from a general principle, to be proved in the chapter on virtual work, that the vessel will take such a form that the altitude of the centre of gravity of the fluid above the table is the least possible. Hence the depth of the fluid is a minimum. Since the volume is given, it immediately follows that the area of the base is a maximum. ART. 134.] REACTIONS AT JOINTS. 91 By a known theorem in the differential calculus, the area of a polygon formed of sideB of given length is a maximum when it can be inscribed in either a circle or a semicircle, according as the polygon is closed or unclosed. (De Morgan's Biff, and Int. Calculus 1842, p. 302). The results of the preceding problems follow at once. We may also deduce the results from the principle of virtual work by forming the work function of the forces without the intervention of any hydrostatical principles. We may notioe that both these theorems will still exist if a great many con- secutive sides of the polygon become very short. In the limit these may be regarded as the elementary arcs of a string acted on by normal forces proportional to their lengths. If then a polygon be formed by rods and strings, and be in equilibrium under the action of a uniform normal pressure from within, the sides can be inscribed in a circle, and the strings will form arcs of the same circle. The first of these two problems was solved by N. Fuss in MSmoires de L'Aca- demie ImpSriale des Sciences de St Petersbourg, Tome vm, 1822. His object was to determine the form of a polygonal jointed vessel when surrounded by fluid. 134. Ex. Polygon of heavy rods, n uniform heavy rods A A V A X A 2 & k c, A n _ 1 A n are freely jointed togetlier at A v A 2 <&e. A n _ 1 and the two extremities A and A n are hinged to two points which are fixed in space; it is required to find the conditions of equilibrium. At each of the joints A , A 1 &e. draw a vertical line upwards ; let O , 8 1 &a. be the inclinations of the rods AgA^ A X A 2 &a. to these verticals, the angles being measured round each hinge from the vertical to the rod in the same direction of rotation. Let the weights of these rods be W , W x &a. First Method. The equilibrium will not be disturbed if we replace the weight W of any rod by two vertical forces, each equal to %W, acting at the extremities of the rod. In this way each rod may be regarded as separated into three parts, viz. the two terminal particles, each acted on by half the weight of the rod, and the inter- mediate portion thus rendered weightless. Let us first consider how these several parts act on each other. At any joint the two terminal particles of the adjacent rods are hinged together. Each particle is in equilibrium under the action of the force at the hinge, the half-weight of the rod of which it forms a part, and the reaction between itself and the intermediate portion of that rod. This last reaction is therefore a force. Since the intermediate portion of each rod has been rendered weightless, the reactions on it will act along the rod, Art. 131. Let the reactions along the intermediate portions of the rods A A lt A X A 2 &c. , be T , T 1 &o., and let these be regarded as positive when they pull the terminal particles as if the rods were strings. To avoid introducing the force at a hinge into our equations we shall consider the equilibrium of the two particles adjacent to that hinge as forming one system. This compound particle is acted on by the half-weights of the adjacent rods and the reactions along the intermediate portions of those rods. The result of this argument is, that we may regard all the rods as being without weight, if we suppose them to be hinged to heavy particles placed at the joints instead of to each other, the weight of each particle being equal to half the sum of the weights of the adjacent rods. 92 FORCES IN TWO DIMENSIONS. [CHAP. IV. A system of weights joined, each to the next in order, by weightless rods or strings and suspended from two fixed points is usually called a funicular polygon. Consider the equilibrium of any one of the compound particles, say that at the joint A 2 . Eesolving horizontally and vertically, we have T 1 sin0 1 =!Z', ! sin0 2 ) T 2 cos0 2 -T 1 cos0 1 = i(PF 1 + W 2 )j (!)• We easily find ±!2£±?J =T lS in0, cot 2 - cot e 1 1 1 .(2). .(3). The right-hand side of this equation is the same for all the rods, being equal to the horizontal tension at any joint, we find therefore h(W 1+ W,) ^ j(W 2 +W 3 ) =&c cot0 2 -cot0i cot0 3 -cot0 2 If A r , A, be any two joints we see that each of these fractions is equal to Wr-i+W, + + W,. 1 + jW, cotfl.-cot^.i 135. Second Method. In this method we consider the equilibrium of any two successive rods, say A-^A^, A 2 A 3 , and take moments for each about the extremity remote from the other rod. Let Xj, y 2 be the resolved parts of the reaction at the joint A 2 on the rod A 2 A S . The two equations of moments give - X 2 cos 2 + y 2 sin 2 + J JT 2 sin ft, = 0) - X 2 aoB8 1 + Y 2 sm6 1 -iW 1 sm8 1 =o\ Eliminating Y 2 we find X, (cot 2 -cot 0^=4 (Wi + Jjy (4), which is equivalent to equations (2). 136. Let l , ^ &a. be the lengths of the rods, h, k the horizontal and vertical coordinates of A n referred to A as origin. We then have l COS O +Z 1 COS0 1 + + J B _ 1 cos0 n _ 1 = ftj l sin O + Zj sin X + + £„_! sin n _j = k \ * '' The equations (2) supply n- 2 relations between the angles B , 6 ± &c. and the weights W , W x &o. of the rods. Joining these to (5) we have sufficient equations to find the angles when the weights are known. When the angles and the weights of two of the rods are known, the n - 2 remaining weights may be found from equations (2). 137. It is evident that either of these methods may be used if the rods are not uniform or if other forces besides the weights act on them. The two equations of moments in the second method will be slightly more complicated, but they can be ART. 140.] REACTJpNS AT JOINTS. 93 easily formed. In the first method the transference of the forces parallel to them- selves to act at the joints is also only a little more complicated, see Art. 79. 138. To find the reactions at the joints. If we use the second method, these are easily found from equations (3). But if we use the first method we must transfer the weights J W 1 and \W% back to the extremities of the rods which meet at A % . In the original arrangement of the rods when hinged to each other, let B 2 be the action at the joint A 2 on the rod A%A 3 . The terminal particle of the rod A 2 A 3 is then acted on by the three forces iJ 2 , JW 2 and T 2 . We therefore have B 2 2 =T^ + iW^-W 2 T 2 ao B e 3 (6). The direction of the reaction is easily deduced from equations (2). Suppose that the rods A^A^, A%A 3 are joined by a short rod or string without weight. The position of this rod is clearly the line of action of JJ 2 . Treating this rod as if it were one of the rods of the polygon, we have, if 2 be its inclination to the vertical, W-l _ Wl cot - cot 8 X cot 2 - cot ( W 1 + W.J cot = W 2 cot 0j + W 1 cot 2 . ■(7), 139. The subsidiary polygon. The lines of action of the reactions B v B 2 &o. at the joints will form a new polygon whose corners B v B 2 &o. are vertically under the centres of gravity of the rods A^Ay, A 2 A 3 &c. The weights of the rods may be supposed to act at the corners of this new polygon. Each weight will be in equi- librium with the reactions which act along the adjacent sides of the polygon. If we suppose the corners B lt 2? 2 *°- to be joined by weightless strings or rods we shall have a second funicular polygon. This funicular polygon may be treated in the same way as the former one, except that we have the weights W v W 2 &c. instead of i (W 1 + W 3 ), 4 ( W 2 + W 3 ) &o. 140. Let B B 2 B 2 Sea. be any funicular polygon; W v fT 2 , &c, the weights suspended from the corners B v B 2 &a. From any arbitrary point draw straight lines Ob lt Ob 2 , Ob 3 &o. parallel to the sides B B V B^, BJ3 3 &o. to meet any vertical straight line in the points b v 6 2 , b 3 &o. Since a particle at the point B x is in equilibrium under the action of the weight W x and the tensions R v i? 2 acting along the sides BjB^ BjB^, it follows, by the triangle of forces, that the sides of the triangle Ob^ are proportional to these forces. In the same way, the sides of the triangle Ob 2 b 3 represent on the same scale the weight W 2 and the tensions acting along B^, B 2 B 3 . In general the straight lines 0\, 0\ &c. represent the tensions acting along the sides of the funicular polygon to which they are respectively parallel; while any part of the vertical straight line as b 2 b s represents the sum of the weights at B 2 , JB S and B 4 . 94 FOEGES IN TWO DIMENSIONS. [CHAP. IV. By using this figure we may find geometrically the relations between the tensions and the weights. If 0-, 2 &o. be the inclinations of the sides B B V J9-B- &o. to the vertical, we have ON (cot 0- - eot 2 ) = 6- 6 2 , where ON is a perpendicular drawn from on the vertical straight line. Since ON represents the horizontal tension X at any point of the funicular polygon, this equation gives — = ±Z=— — — --— - — = &c. COt 0-- COt 2 COt 2 -COt In the same way other relations may be established. The use of this diagram is described in Rankine's Applied Mechanics. Such figures are usually called force diagrams. We have here only considered the simple case in which the forces are parallel to each other. In the chapter on Graphics this method of solving statical problems will be again considered and extended to forces which act in any directions. 141. Ex. 1. A chain consisting of a number of equal and in every respect similar uniform heavy rods, freely jointed at their ends, is hung up from two fixed points ; prove that the tangents of the angles the rods make with the horizontal are in arithmetical progression, as are also the tangents of the angles the directions of the stresses at the joints make with the same, the common difference being the same for each series. [Coll. Ex., 1881.] Ex. 2. OA, OB are vertical and horizontal radii of a vertioal circle, A being the lowest point. A string AGDB is fixed to A and B and divided into three equal parts in C and D. Weights W, W being hung on at G and D, it is found that in the position of equilibrium G and D both lie on the circle. Prove that W= W tan 15°. [Trin. Coll., 1881.] Ex. 3. If (2m - 1) particles be connected by equal strings and hang from two fixed points at a horizontal distance 2a, and if the whole series of particles lie on the circumference of a semicircle having the line joining the two fixed points for diameter, then show that the weights of the particles starting from the lowest are connected by the equations w- cos 2 a = w 2 eos acos3a=J» 3 cos 3acos5a=M) 4 cos 5a cos 7a=etc where a = 7r/4«. [St John's Coll., 1886.] Ex. 4. If the weights as we ascend in a funicular polygon are W, -- W, ~» W, ^ 3 W &c, W being attached to one end of the horizontal portion of the string and if the number of weights is very great, the tension at the point of support is approxi- mately S /(T 2 + 4W-), where T is the tension in the horizontal element of the string. [CoU. Ex., 1886.] Ex. 5. Pour equal heavy uniform rods AB, BG, CD, DA are jointed at their extremities so as to form a rhombus, and the corners A and C are joined by a string. If the rhombus is suspended by the corner A, show that the tension of the string is 2 IF and that the reaction at either Zi orDis \W tan \BAD, where IP is the weight of any rod. ART. 141.] REACUONS AT JOINTS. 95 \l Ex. 6. AB, BG, CD are three equal rods freely jointed at B and G. The roda AB, CD rest on two pegs in the same horizontal line so that BG is horizontal. If o be the inclination of AB, and /3 the inclination of the reaction at B to the horizon, prove that 3 tan a tan /3= 1. rSt John's Coll., 1881.1 4 Ex. 7. Three equal uniform rods are freely jointed at their extremities and rest in equilibrium over two smooth pegs, in a horizontal line at a distance apart equal to half the length of one rod. If the lowest side be horizontal, then the resultant action at the upper joint is &J3W and at each of the lower -feJWJW, where W is the aggregate weight of the rods. [Coll. Ex., 1882.] Ex. 8. Three rods, jointed together at their extremities, are laid on a smooth horizontal table ; and forces are applied at the middle points of the sides of the triangle formed by the rods, and respectively perpendicular to them. Show that, if these forces produce equilibrium, the mvMOM at the joints will be equal to one another, and their directions will touch the circle circumscribing the triangle. i [Math. Tripos, 1858.] Ex. 9. Three pieces of wire, of the same kind, and of proper lengths, are bent into the form of the three squares in the diagram of Euclid I., 47, and the angles of the squares which are in contact are hinged together, so that the smaller ones are sup- ported by the larger square in a vertical plane. Show that in every position, into which the figure can be turned, the action, if any, between the angles of the smaller squares will be perpendicular to the hypothenuse of the right-angled triangle. [Math. Tripos, 1867.] Ex. 10. Three uniform rods, whose weights are proportional to their lengths a, b, c, are jointed together so as to form a triangle, which is placed on a smooth horizontal plane on its three sides successively, its plane being vertical : prove that the stresses along the sides a, b, c when horizontal are proportional to (b + c) cosec 2A, (c + a) cosec 2B, (a + 6) cosec 20. Examine the cases in which the triangle is right-angled or obtuse-angled. [Math. Tripos, 1870.] Ex. 11. Three uniform rods AB, BG, CD of lengths 2c, 26, 2c respectively rest symmetrically on a smooth parabolic arc, the axis being vertical and vertex upwards. There are hinges at B and G, and all the rods touch the parabola. If W be the weight of either of the slant rods, show that its pressure against the parabola is equal to W -r~i — 1ST ' w ^ ere ^ a ' s *^ e ^ us rectum of the parabola. [Coll. Ex., 1883.] Ex. 12. ABGD is a quadrilateral formed by four uniform rods of equal weight loosely jointed together. If the system he in equilibrium in a vertical plane with the rod AB supported in a horizontal position, prove that 2 tan 0=tana~tan/3, where a, ft are the angles at A and B, and 8 is the inclination of CD to the horizon ; also find the stresses at C and D, and prove that their directions are inclined to the horizon at the angles tan -1 J (tan /3 - tan 8) and tan -1 J (tan o + tan 8) respectively. [Math. Tripos, 1879.] Ex. 13. Four equal rods AB, BG, CD, DA, jointed at A, B, G, D are placed on a horizontal smooth table to which BG is fixed, the middle points of AD, DC being connected by a string which is tight when the rods form a square. Show that, if a 96 FORCES IN TWO DIMENSIONS. [CHAP. IV. couple act on All and produce a tension T in the string, its moment must be IT.AB-J2. [Coll. Ex., 1888.] Ex. 14. A weightless quadrilateral framework A-^A^A^i rests with its plane vertical and the side A X A% on a horizontal plane. Two weights W, W are placed at the corners A it A s respectively, while a string connecting the two corners A±A 3 prevents the frame from closing up. Show that the tension T of the string is given by nT sin 2 sin 4 = IV cos 0j sin 6 S -W cos 2 sin 4 , where d lt 2 , 3 , 4 are the internal angles of the quadrilateral, and n is the ratio of the side on the horizontal plane to the length of the string. Ex. 15. A pentagon formed of five heavy equal uniform jointed bars is suspended from one corner, and the opposite side is supported by a string attached to its middle point of such length as to make the pentagon regular. Prove that the tension of the string is equal to 4.W cos 2 ^ir, where W is the weight of any rod. Find also the reactions at the corners. Ex. 16. A regular pentagon ABODE, formed of five equal heavy rods jointed together, is suspended from the joint A, and the regular pentagonal form is maintained by a rod without weight joining the middle points K, L of BC and BE. Prove that the stress at K or L is to the weight of a rod in the ratio of 2 cot 18° to unity, and find the stresses at B and C. [Math. Tripos, 1885.] Ex. 17. The twelve edges of a regular octahedron are formed of rods hinged together at the angles, and the opposite angles are connected by elastic strings ; if the tensions of the three strings are X, Y, Z respectively, show that the pressure along any of the rods connecting the extremities of the strings whose tensions are Fand Z is -^[Y + Z-X). [Math. Tripos, 1867.] Ex. 18. Any number of equal uniform heavy rods of length a are hinged together, and rotate with uniform angular velocity w about a vertical axis through one extremity of the system, which is fixed ; if 0„ be the inclination to the vertical of the n tb rod counting from the free end, prove that (2ra + 3) tan $^ - (4n+2) tan n+1 + (2«- 1) tan n + ~ {sin0 m+2 + 4sin0 wH . 1 + sin0 m } =0. [Math. Tripos, 1877.1 Reactions at rigid connections. 142. Let AB be a horizontal rod fixed at the extremity A in a vertical wall, and let it support a weight W at its other extremity B. We may enquire what are the reactions or stresses across a section at any point G, by which the portion GB of the rod is supported. It is evident that the reaction at G cannot consist of a single ART. 143.] REACTIONS ^f RIGID CONNECTIONS. 97 force, for then a force acting at G would balance a force W to which it could not be opposite. It is also clear that the resultant action across the section G (whatever it may be) must be equal and opposite to the force W acting at B. Let us transfer the force W from B to any point of the section G by help of Art. 100. We see that the reaction across the section is equivalent to a force equal to W, together with a couple whose moment is W.BG. If the portion GB of the rod is heavy, we may suppose its weight collected at the middle point of GB. Let W be the weight of this part of the rod. Then we must transfer this weight also to the base of reference G. The whole reaction across the section of the rod will then consist of (1) a force W +W and (2) a couple whose moment is W. BG+^W. BG. Various names have been given to the reaction force and reaction couple at different times. The components of the force along the length of the rod and transverse to it have been called the tension and shear respectively. The former being normal to a perpendicular section of the rod is sometimes called the normal stress. The magnitude of the couple has been called the tendency of the forces to break the rod, or briefly, the tendency to break. It is also called the moment of flexure, or bending stress. See Rankine's Applied Mechanics. In what follows we shall restrict ourselves to the case in which the rod is so thin that we may speak of it as a line in discussing the geometry of the figure. 143. Generalizing this argument, we arrive at the following result : the action across a section at any point C of a rod is equal and opposite to the resultant of all the forces which act on the rod on one side of that point C. The action across C on GB balances the forces on GB. The equal and opposite reaction on A C across the same section balances those on AG. Since the forces on one side of G balance those on the other side when there is equilibrium, it is a matter of indiffer- ence whether we consider the forces on the one side or the other of G provided we keep them distinct. Thus the bending couple at is equal to the sum of the R. s. T 98 FORCES IN TWO DIMENSIONS. [CHAP. IV. moments of all the forces which act on one side of G. So also the shear at G is equal to the sum of the resolved parts of these forces along the normal to the rod at G. If we regard the rod as slightly elastic we may .explain other- wise the origin of the force and couple. The weight W will slightly bend the rod, and thus stretch the upper fibres and com- press the lower ones. The action across the section at G will therefore consist of an infinite number of small tensions across its elements of area. By Art. 104 all these can be reduced to a single force and a single couple at a base of reference at G. 144. Ex. 1. A rod AB, of given length 1, is supported in a horizontal position by two pegs, one at each end. A heavy particle M, whose weight is W, traverses the rod slowly from one end to the other. It is required to find the stresses at any point of the rod. Let AM=£, BM=l-$,. Let R and R' be the pressures of the supports at A and B on the rod. These are evidently given by R'l=W. |, Bl=W{l-Q. Let P be the point at which the stresses are required, and let AP = x. To find these we consider the equilibrium of either the portion AP or the portion BP of the rod. We choose the former, as the simpler of the two, because there iB only one force, viz. R, acting on it. The shear at P is therefore, equal in magnitude to R, and the moment of the stress couple is equal to Rx. Ra 3- "•---... f M W If the point at which the stresses are required is on the other side of M as at P', where AP'=x', it is more convenient to consider the equilibrium of BP'. The shear is here equal to R', and the bending moment to R'(l-x'). As the bending couple is generally more effective in breaking a rod than either the shear or the tension, we shall at present turn our attention to the couple. If at every point P we erect an ordinate PQ proportional to the bending couple at P, the locus of Q will represent to the eye the magnitude of the bending couple at every point of the rod. In our ease the locus of Q is clearly portions of two straight lines, represented in the figure by the dotted line. The maximum ordinate is at the point M, and is represented by either P£ or R' (I - f ), according as we take moments about M for the sides AM or MB of the rod. Substituting for R or R', the bending couple at M becomes W% (I - f)/t This is a maximum when M is at the middle point of AB. This result shows in a general way that, when a rann stands on a stiff plank laid across a stream, the bending couple is greatest at the point of the plank on ART. 145.] REACTIONS iff RIGID CONNECTIONS. 99 which the man stands. Also if he walks slowly along the plank, the bending couple is greatest when he is midway between the two supports. Ex. 2. A uniform heavy rod AB is supported at each end. If w be the weight per unit of length, prove that the bending couple at any point P will be \w . AP . BP. 145. When several forces act on a rod, the diagram by which the distribution of bending stress is exhibited to the eye can be constructed in a similar manner. Let forces JJj , P 2 &c. act at the points A x , A 2 &o. of a rod in the directions indicated by the arrows. Let A 1 A 2 =a 2 , A 1 A 3 =a a and so on. Then the bending moment at any point P, say between A a and A 4 , is obtained by taking the moments of the forces which act at A lt A 2 , A s , these being points on one side of P. Putting A^P^x, the required bending moment is y = R 1 x-B 2 (x-a 2 )+B A {x-a 3 ). Erecting an ordinate PQ to represent y, it is clear that the locus of Q between A s and A t is a straight line. R i | ..--'""' R% r£ ^ V -"31 "W -B A t A 2 A s Ai When the point P moves beyond A 4 we must add to this expression the moment of the force P 4 , i.e. -B i (x -a 4 ). The locus of Q is now a different straight line. It intersects the former at the point x=a it i.e. at the top of the ordinate correspond- ing to the point A 4 , but its inclination to the rod is different. We infer that, when a rod is acted on only by forces at isolated points, the diagram representing the bending couple will consist of a series of finite straight lines. This indicates an easy method of constructing the diagram. Calculate the ordinates representing the bending couples at these isolated points, and join their extremities by straight lines. In this case there can be no maximum ordinate between the isolated points A lt A 2 &c. at which the forces act. Hence the bending couple can be a maximum or minimum only at one of these points. If the rod is heavy, its weight is distributed over the whole rod. The bending couple at P will contain not merely the moments of the forces which act at A lt A% &c, but also that of the weight of the portion A-J? of the rod. If w be the weight per unit of length, the bending couple at P will be y = 2P (x - a) - Jwx 2 , for the weight of A X P will be wx, and it may be collected at the middle point of ^ X P. This is the equation to a parabola. Hence the diagram will consist of a series of arcs of parabolas, each intersecting the next at the extremity of the ordinate along which an isolated foroe acts. All these parabolas have their axes vertical. If the different sections of the rod be of the same weight per unit of length, the latera recta of the parabolas will be equal. This expression gives the bending moment by which the forces on the 1 left or negative side of any point P tend to turn the portion of the rod on the positive side of P in the direction of rotation of the hands of a watch. 7—2 100 FORCES IN TWO DIMENSIONS. [CHAP. IV. Suppose that any portion CD of a rod ACDB has no weight, and that no point of support lies between C and D. The remaining parts of the rod on each side of CD may have any weights and any number of points of support. The bending couple at any point between C and D is always proportional to the ordinate of some straight line. But if y v y v and y are ordinates of any straight line at C, D and P, and if the distances CP and PD are Zj and l 2 , it is easy to see that y(h+k)=yih+ynh- This equation therefore must also connect the bending couples y lt y 2 , and y at the points C, D, and any intermediate point P. Let us next suppose that the portion CD of the rod is heavy. The bending couple at any point of this portion of the rod is now proportional to the ordinate of the parabola y — A-\-Bx-\wx i , where A= -1,11a and B = '2R. If y ly y 2 and y are the ordinates at C, D and any point P, where CP=l lt PD = l 2 , it is easy to prove that y (h+^=yih+y2h+i w hk(h+h)- This equation connects the bending couples at any three points of a heavy rod provided there is no point of support within the length considered. Ex. If y lt y 2 , 2/3 be the bending couples at three consecutive points of support of a heavy horizontal rod whose distances apart are Zj, I 2 , then 2/2 ( ? i + h) =Vik+ yA + i w hk (h + h)~ :R h l 2' where B is the pressure at the middle point of support, and id is the weight of the rod per unit of length. 146. Since the bending couple at any point P is the sum of the moments of the several forces which act on one side of P, it is clear that each force contributes its share to the bending couple as if it acted alone on the rod. In this way it is sometimes convenient to consider the effects of the forces separately. For example, if a heavy rod AB, supported at each end, has a weight W placed at a point M, the bending couple at any point P is the sum of the bending couples found in Art. 144 for the two cases in which (1) the rod is light and (2) there is no weight at M. The bending couple is therefore given by ly = W.BM.AP + \wl.AP.BP. 147. Ex. 1. A heavy rod is supported in a horizontal position on two pegs, one at each end. A heavy particle, whose weight is n times that of the rod, is placed at a point M. If G be the middle point of the rod, show that the bending couple will be greatest either at some point between M and C or at M, according as the distance of M from C is greater or less than «, times its distance from the nearer end of the rod. Ex. 2. A semicircular wire ACB is rotated with uniform angular velocity about a tangent at one extremity A. Show that the bending couple is zero at B, is a maximum at the middle point C, vanishes at some point between C and A, and is again a maximum with the opposite sign at A. Show also that the maximum at A is greater than that at C. It may be assumed that the effect of rotation is represented by supposing the wire to be at rest, and each element to be acted on by a force tending directly from the axis of rotation and proportional to the mass of the element and its distance from the axis. ART. 147.] REACTIONS m RIGID CONNECTIONS. 101' Ex. 3. A horizontal beam AB, without weight, supported but not fixed at both ends A and B, is traversed from end to end by a moving load W distributed equally over a segment of it, of constant length PQ. Show that the bending moment at any point X of the beam, as the load passes over it, is greatest when X divides PQ in the same ratio as that in which it divides AB. Show also that this maximum W bending moment is equal to -j^ . AX . BX {AB - %PQ). [Townsend's Theorem.] Let AX=a, BX=b, AB = a + b, PQ = l. Let R be the shear at X, and y the bending moment. If x be the distance from A of any point between P and X, and AP=x', we have, by taking moments about A for the portion AX of the beam, xdx-y + Ra=0, where W=lw. Similarly, if { be the distance from B of any point between Q and X, and BQ=g, we find for the portion BX ' ( b &-y- J i' -Bb = 0. Eliminating B, we find 2l{a + b) y = W {ab (a + b) -bx'* - a?*}. Making y a maximum with the condition x' + g=a + b-l, the results follow at once. Ex. 4. A uniform horizontal beam, which is to be equally loaded at all points of its length, is supported at one end and at some other point; find where the second support should be placed in order that the greatest possible load may be placed upon the beam without breaking it, and show that it will divide the beam in the ratio 1 to J2 - 1. [Math. Tripos.] Let ABC be the beam supported at A and B. Let wdx be the load placed on dx; wR, wR' the pressures at A, B. Let I be the length of the beam, £=AB, then 2|>J. We easily find R = I - |j., R'= Jr. Let P and Q be two points in CB and BA respectively, x = GP, x'=AQ. By taking moments about P and Q respectively the bending couples y, y' at P and Q are found to be y = - \ixx^, y' = wRx' - %wx' % . The first parabola has its maximum ordinate at B, the second has a maximum ordinate at a point x'=R which must lie between A and B. The bending couples at these points are numerically equal to \w (I - J) 2 and Jio ( I - r- I . If these are unequal, the support B can be moved so as to diminish the greater. The proper position is found by making these equal ; hence ± (I - 1) = I - P/21;. Since £ must be greater than \l, this gives if^/2 = J. Ex. 5. Three beams AB, BG, CA are jointed &tA,B,G,B being an obtuse angle, and are placed with AB vertical, and A fixed to the ground, so as to form the framework of a crane. There is a pulley at C, and the rope is fastened to AB near B and passes along BG and over the pulley. If it support a weight W, large in comparison with the weights of the framework and rope, find the couples which tend to break the crane at A and B. [Math. Tripos.] 102 FORCES IN TWO DIMENSIONS. [CHAP. IV. Ex. 6. A gipsy's tripod consists of three uniform straight sticks freely hinged together at one end. From this common end hangs the kettle. The other ends of the sticks rest on a smooth horizontal plane, and are prevented from slipping by a smooth circular hoop which encloses them and is fixed to the plane. Shew that there cannot be equilibrium unless the sticks be of equal length ; and if the weights of the sticks be given (equal or unequal) the bending moment of each will be greatest at its middle point, will be independent of its length, and will not be increased on increasing the weight of the kettle. [Math. Tripos, 1878.] Ex. 7. A brittle rod AB, attached to smooth hinges at A and B, is attracted towards a centre of force G according to the law of nature. Supposing the absolute force to be indefinitely augmented, prove that the rod will eventually snap at a point E determined by the equation sin \ (o + /3) cos0=sin£ (a-/3), where o, /3 denote the angles BAG, ABC, and 9 the angle AEC. Math. Tripos, 1854. See also the solutions for that year by the Moderators and Examiners. Indeterminate Problems. 148. When a body is placed on a horizontal plane, the pressure exerted by its weight is distributed over the points of support. When there are more than three supports, or more than two in one vertical plane, this distribution appears to be indeterminate. Thus suppose the body to be a table with vertical legs, and let these legs intersect the plane horizontal surface of the table in the points A lt At &c. Let the projection on this plane of the centre of gravity of the body be G. The weight W of the table will then be supported by certain pressures R lt i? 2 &c. acting at A lt -A 2 &c. Let Ox, Oy be any rectangular axes of reference in this plane and let Oz be vertical. Let {x^y^), (x 2 y 2 ) &c. be the coordinates of A 1} A 2 &c. and let (xy) be those of G. Since W is supported by a system of parallel forces we have by Arts. 110 and 111 W = R 1 + R 2 +... Wx = R 1 X 1 + R^OC^ +• • • Wy = R 1 y 1 + R 2 y,+... These three equations suffice to determine R lt R 2 &c. if there are but three of them and these not all in one vertical plane, but if there are more than three, the problem appears to be indeterminate. In this solution we have replaced the supporting power of the floor by forces R 1} R 2 &c. acting upwards along the legs. What we have really proved is that the table could be supported by such ART. 148.] INDETERMINATE PROBLEMS. 103 forces in a variety of different ways. Suppose there were four legs ; we could choose one of these forces to be what we please, the others could then be found from these three equations. It is evident that the problem of finding what forces could support the table must be indeterminate, and accordingly they are found to be so. The actual pressures exerted by the table on the floor are not indeterminate, for in nature things are necessarily determinate. When anything appears to be indeterminate, it must be because we have omitted some of the data of the question, i.e. some property of matter on which the solution depends. We notice that the elementary axioms relating to forces, which have been enunciated in Art. 18, make no reference to the nature of the materials of the body. We have found in the preceding Articles that the equations supplied by these axioms have in general been sufficient to determine all the unknown quantities in our statical problems. In all these problems therefore the magnitudes of the reactions and the positions of equilibrium of the bodies depended, not on the materials of the bodies, but on their geometrical forms and on the magnitudes of the impressed forces. It is evident, however, that these axioms must be insuf- ficient to determine any unknown quantities which depend on the materials of the bodies. In such cases we must have recourse to some new experiments to discover another statical axiom. Thus, when we study the positions of equilibrium of rough bodies, another experimental result, depending on the degree of roughness of the special body considered, is found to be necessary. In the same way the mode of distribution of the pressure over the legs of the table is found to depend on the flexibility of the materials. However slight the flexibility of the substance of the table may be, yet the weight W will produce some deformation however small. The magnitude of this will influence and be influenced by the reactions i^, -R 2 &c. The amount of yielding produced by the acting forces in any body is usually considered in that part of mechanics called the theory of elastic solids. No complete solution of the special problem of the table has yet been found. But when any assumed law of elasticity is given, it is easy to show by some examples, how the problem becomes determinate. Poinsot's tMments de Statique and Poisson's TraiU de Me'ccmique. 104 FOKCES IN TWO DIMENSIONS. [CHAP. IV. .(1). 149. Ex. 1. A rectangular table has the legs at the four corners alike in all respects and slightly compressible. The amount of compression in each leg is supposed to be proportional to the pressure on that leg. Supposing the floor and the top of the table to be rigid, and the table loaded in any given manner, find the pressure on the four legs. Show that when the resultant weight lies in one of four straight lines on the surface of the table, the table is supported by three legs only. [Math. Tripos, 1860, Watson's problem, see also the Solutions for that year.] Let the two sides AB, AD be the axes of x and y. Let the resultant weight W act at a point G whose coordinates are (xy). Let AB = a, AD = b. Since the top of the table is rigid, the surface as altered by the compression of the legs is still plane. Also, since the compression is slight, we shall neglect small quantities of the second order, and suppose the pressures at A, B, G, D to remain vertical. We have the usual statical equations W=R 1 + R 2 + R 3 + R i , 1 Wx={R i + R s )a, Wy = (R s + R i )b\ Because a diagonal of the table remains straight, the middle point descends a space which is the arithmetic mean of the spaces descended by its two ends. It follows that the mean of the compressions of the legs A and C is equal to the mean of the compressions of the legs B and D. But it is given that the pressures are proportional to these compressions. Hence iJ 1 + J R 3 =iJ 2 + i? 4 (2). These four equations determine the pressures. If we put R 3 =0, we easily find that 2a;/a + 2i//6 = l, i.e. the table is supported on the three legs A, B, D when the weight W lies on the straight line joining the middle points of AB, AD. Joining the middle points of the other sides in the same way, we obtain four straight lines represented by the dotted lines. When the weight W lies within this dotted figure all the four legs are compressed ; when without this figure three legs only are compressed. The equations above written are then correct, only if we suppose that some of the reactions are negative. As this cannot in general be possible, we must amend the equations (1) by putting one reaction equal to zero. The equation (2) must then be omitted. Ex. 2. A and G are fixed points or pegs in the same vertical line, about which the straight beams ADB and CD are freely movable. AB is supported in a hori- zontal position by CD and has a weight W suspended at B. Find the pressure at G (1) when there is a hinge joint at D, and (2) when CD forms one piece with AB, the weights of the beams being in each case neglected. [Math. Tripos, 1841.] In the first part of the problem the action at D is a single force, in the second part it is a force and a couple, Art. 142. In both parts of the problem the action at G is a force. In the first part, the actions at G and D are equal and act along CD by Art. 131. Taking moments about A for the rod AD, we easily find that this action is equal to W. ABIAN where AN is a perpen- dicular on GD. ART. 150.] STIW FRAMEWORK. 105 # In the second part there is nothing to determine the direction of the action at G. We only know it balances an unknown force and a couple. If we write down the three equations of equilibrium for the whole body, it will be seen that we cannot find the four components of the two pressures which act at A and G. The problem is therefore indeterminate. Ex. 3. A rigid bar without weight is suspended in a horizontal position by means of three equal vertical and slightly elastic rods to the lower ends of which are attached small rings A, B, and G through which the bar passes. A weight is then attached to the bar at any point G. Show that, on the assumption that the extension or compression of an elastic rod is proportional to the force applied to stretch or compress it, and provided the rods remain vertical, then the rod at B will be compressed if G lie in the direction of the longer of the two arms AB, BG, AB^+BCP and be at a greater distance from B than —= — =yj- . [Math. Tripos, 1883.] AB ~* xjO Ex. 4. ABGD is a square ; six rods AB, BG, CD, DA, AG, BD are hinged together at the angular points, and equal and opposite forces, F, are applied at B and D in the directions DB and BD respectively. The rods are elastic, but the extensions or compressions which occur may be treated as infinitesimal. e± is the ratio of the extension per unit length to the tension (or of the compression to the corresponding force) for the rod AB, and is a constant depending upon the material and the section of the rod. « 2 , e a ...e e are similar constants for the other rods in the order written above. Prove that the tension of the rod BD is (l 2 ^ 2eF forces each equal to F. Let JR 12 , R^ , B ls be the reactions along A^A 2 , A 2 A S , A-^A 3 respectively. Here we have a simply stiff framework and we should therefore find sufficient equations to determine the reactions. The equations of equilibrium for the three corners are however which are evidently insufficient to determine the three reactions in terms of the given forces. The conditions under which these exceptional cases can arise are determined algebraically by the theory of linear equations. The 2re-3 or 3re-6 equations to find the reactions at the corners of the framework are all linear. If a certain determinant is zero, one equation at least can be derived from the others or is contradictory to them. In the latter case some of the reactions are infinite ; this of course is impossible in nature. In the former case one reaction is arbitrary, and all the others can be expressed in terms of it and the given external forces. In a similar manner we can express the condition that two reactions are independently arbitrary. There is however another and a more definite way of expressing these con- ditions. As this part of the theory follows more easily from the principle of virtual work, we shall postpone its consideration until we come to the chapter on that subject. 155. Let us next suppose that the system of n particles has more than the number of bars necessary to stiffen it. In this case there are not enough equations to find the reactions unless something is known about them besides what is given by the equations of statics. The rods connecting the particles are in nature elastic, and the forces acting along them are due to their extensions or compressions. Supposing the law connecting the force and the extension to be known, we have to ART. 156.] •ASTATICS. 109 examine, whether the additional equations thus supplied are sufficient to find the reactions. The framework, being acted on by external forces, will yield, and this yielding will continue to increase until the reactions thus called into play are of sufficient magnitude to keep the frame at rest. For the sake of brevity we shall suppose that the amount of the yielding is very slight. In this case we shall assume, in accordance with Hooke's law, that the reaction along any rod is some known multiple of the ratio of the extension to the original length. This multiple depends on the nature of the material of which the rod is made. Let the framework have m rods, where m exceeds 2n - 3 or 3n - 6 by 7c. Taking the case in which the framework is not acted on by any external reactions, we shall require ft additional equations (Art. 153). By Art. 152 there are k relations between the lengths of these rods. Let any one of these be f{l lt Z 2 , &.)=0 (1), where l v l 2 &o. are the lengths of the rods. Differentiating this we have M 1 dl 1 + M i dl 2 + &o. = (2), where M v M 2 &o. are partial differential coefficients, and dl^, dl 2 &o. are the extensions of the sides. If B v R 2 &o. are the reactions along the sides we may, by Hooke's law, write this equation in the form ^i\h R i + -34VA + &c. = 0, where \ , \ &o. are the reciprocals of the known multiples. It appears therefore that each equation such as (1) supplies one relation between the reactions. Thus the requisite number of additional equations can be deduced from the theory of elasticity. Ex. In the case of the three rods mentioned in Art. 154 show that the three reactions are equal in magnitude if all three rods are made of the same material and are of equal sectional areas. Noticing that the relation corresponding to (1) is l ls + 7 23 - 'i 2 =0, where l n =A 1 A i , &o. the result follows by differentiation. ji Asiatics. 156. Suppose a rigid body is acted on at given points^, A 2 &c. by forces P lt P 2 &c. whose magnitudes and directions in space are given. Let this body be displaced in any manner: it is re- quired to find how the resultant force and couple are altered. Choosing any base of reference and any rectangular axes Ox, Oy fixed in the body, we may imagine the displacement made by two steps. First, we may give the body a linear displacement by moving to its displaced position O u the body moving parallel to itself ; secondly, we may give the body an angular displacement, by turning the body round O x as a fixed point until the axis Ox comes into its displaced position. Then every point of the body will be brought into its proper displaced position, for otherwise the several 110 FORCES IN TWO DIMENSIONS. [CHAP. IV. points of the body would not be at invariable distances from the base and the axis Ox. Since the forces P u P 2 &c. retain unaltered their magnitudes and directions in space, it is clear that the linear displacement does not in any way affect the resolved parts of the forces, or the moment about 0. We may therefore disregard the linear dis- placement and treat and 1 as coincident points. Consider next the angular displacement. It is clear that we are only concerned with the relative positions of the body and forces, for a rotation of both together will only turn the resultant force and couple through the same angle. Instead of turning the body round through any given angle keeping the forces un- altered, we may turn each force round its point of application through an equal angle in the opposite direction, keeping the body unaltered. See Art. 70. 157. We are now in a position to find the changes in the resultant force and couple. Let Ox, Oy be any axes fixed in the body. Let P be any one of the forces P 1; P 2 &c. and let A be its point of application. Let a be the angle its direction makes with the axis of x. Let this force be turned round A through an angle 8 in the positive direction, so that it now acts in the direction indicated in the figure by AP'. Let X, Y, G be the resolved parts of the forces, and the moment about before displacement ; X', Y, 0' the same after displace- ment. Then, as in Art. 106, X' = 2Pcos(a-|-0) = Xcos0-Fsin0, Y' = XP sin (a + 0) = X sin 6 + Ycos 6, G' = 2P {x sin (a + 6) - y cos (a + 6)} = G cos + Fsin 6, where G = 2(xP y -yP sc ), V=X(xP x + yP y ). The symbol G represents the moment of the forces before displacement about the centre of rotation. If the angle of rotation round is a right angle, 6 = \ir and G' = V. Thus the symbol V represents the moment of the forces about after they ]iave been rotated through a right angle'". If it is permitted to alter slightly a name given by Clausius (see Phil. Mag., August 1870), V might be called the Virial of the forces. After a rotation through an angle 6 let V be the new value of the virial, then V = 2P {x cos (a + 6) + y sin (a + 0) } = V cos - G sin 6. * Darboux, Sur VSquilibre astatique, p. 8. ART. 160.] CENJRE OF FORCES. Ill Thus it appears that the moment G is also what the virial becomes (with the sign ohanged) when the forces have been rotated through a right angle. We may find another meaning for the virial V. Let us suppose the components P x , P y to act at 0, and let their point of application be moved to N, where ON=x. The work of P x is xP x , that of P„ is zero. Let the point of application be further moved from N to A, where NA =«/. The additional work of P x is zero, that of P y is yP y . The sum of these two for all the forces is V. Thus V is the work of moving the forces from the base of reference to their respective points of application, the forces being supposed unaltered in direction or magnitude. 158. If the body is in equilibrium before displacement, we have X = 0, 7 = 0, G = 0. Hence after a rotational displacement through an angle 6 we have X' = 0, 7' = 0, G' = Fsin 0. We therefore infer that the only other position in which the body can be in equilibrium is when 6 = nr, i.e. when the position of the body has been reversed in space. If the body is in equilibrium in any two positions which are not reversals of each other, the body must be in equilibrium in all positions. Lastly, the analytical condition that there should be equilibrium in all positions is that V = in some one position of equilibrium. 159. Ex. 1. A body is placed in any position not in equilibrium, and the forces are such that the components X, Y are both zero. Find the angle through which the body must be rotated that it may come into a position of equilibrium. Ex. 2. If a body be in » position of equilibrium under the action of forces whose magnitudes and directions in space are given, show that the equilibrium is stable or unstable according as V is positive or negative in the position of equilibrium. 160. Centre of the forces. It has been shown in Art. 118, that, provided the components of the forces (viz. X and Y) are not both zero, the whole system can be reduced to a single resultant at a finite distance from the base of reference. In any position of the forces, the equation to this single resultant is G'-£7' + v X' = 0, i.e. (G - %Y+ V X) cos 6 + ( V- %X - r, Y) sin 6 = 0. Thus it appears that, as the forces are turned round their points of application, this single resultant always passes through a fixed point in the body, whose coordinates are given by G-ZY+ v X=0, V-%X- V 7=0. This point is called the centre of the forces. The first of these 112 FORCES IN TWO DIMENSIONS. [CHAP. IV. equations represents the line of action of the single resultant when = 0, the second represents its line of action after a rotation through a right angle, i.e. when 6 = ^ir. As every force in this theory has a point of application fixed in the body, it will be found convenient to regard the central point as the point of application of the single resultant. Thus the single resultant, like the other forces, has a fixed magnitude, a fixed direction in space, and a fixed point of application in the body. The centre of forces therefore bears to forces in general the same relation which the centre of parallel forces bears to parallel forces. The centre of the forces may be defined in words similar to those already used in Art. 82. If the points of application of the given forces are fixed in the body, the point of application of their resultant is also fixed in the body, however the body is displaced, pro- vided the given forces retain their magnitudes and directions in space unaltered. This fixed point is called the centre of the forces. Taking any one relative position of the body and forces, and any rectangular axes, the coordinates (£77) of the centre of the forces are given by %R?=VX + GY, V R>=VY-GX, where X, Y, V, G are referred to the origin as base, and R is the resultant of X and Y. 161. Ex. 1. If the forces of a system are reducible to a single resultant couple, show that the centre of the forces is at infinity. Ex. 2. Show that, as the forces are rotated, the value of G/V at any assumed base is always equal to the tangent of the angle which the straight line joining to the centre of the forces G makes with the direction of the resultant force R, while the value of G 2 + F 2 is invariable and equal to iJ' 2 . CO\ Since the system is equivalent to a single force R acting at 0, it is evident that (?=JJ . ON, where ON is a perpendicular on the line of action of R. Turning R through a right angle, we have V=R . GN. The results follow at once. 162. There is another method* of finding the astatic resultant of a given system which is sometimes useful. The body having been placed in any position relative to the forces which may be convenient, let two axes Ox, Oy be chosen so that the resolved parts of the forces in these directions, viz. X and Y, are neither of them zero. Consider first the resolved parts of all the forces parallel to x. By the theory of parallel forces these are equivalent to a single force, viz. X = '2P X , which acts at a point fixed in the body whose coordinates are (x-^y^), where x 1 X = SxP xl yi X=-2yP x . * The method explained in this Article has been used by Darboux, Sur Vequilibre astatique, and by Larmor, Messenger of Matliernaties. ART. 163.] CENT^p OF FORCES. 113 Consider next the resolved parts parallel to y. TheBe also form a system of parallel forces and are equivalent to a single force Y—SP y , which acts at a point fixed in the body whose coordinates are (x^ 2 ), where x a Y=-2xP v , y 2 Y=SyP v . Since the axes of coordinates are arbitrary and need not be at right angles, the forces have thus been reduced to two forces acting at two points fixed in the body in directions arbitrarily chosen but not parallel. The positions of these points depend on the directions chosen. 163. Let the fixed points thus found be called A and I). In any one relative position of the body and forces, let the two forces X and Y intersect in /, and let their resultant act along IF. Let IF intersect the circle described about the triangle ABI in C Then, by the astatic triangle of forces, G is a point fixed in the body, and the resultant of X and Y may be supposed to act at C. The point C is therefore the centre of the forces. Conversely, when the resultant force and the centre of the forces are known, that force may be resolved into two astatic forces by using the triangle of forces in the manner already explained in Art. 73. R. S. CHAPTEK V. FRICTION. 164. When one body slides or rolls on another under pressure, it is found by experience that a force tending to resist motion is called into play. In order to discover the laws which govern the action of this force we begin with experiments on some simple cases of equilibrium, and then endeavour by a generalization to extend these so as to include the most complicated cases. Let us consider the case of a box A resting on a rough table BG. A string DEH attached to the box at D passes over a small pulley E and supports a scale pan II in which weights can be placed. By putting weights into the box A D E q and varying the weight at H, all b q ' i\ cases can be tried. Supposing the ^-^ box loaded, we go on increasing the weight at H by adding sand (which can be afterwards weighed) until the box just begins to move. The result is that the box, whatever load it carries, does not move until the weight at If is a certain multiple of the weight of the box and load. Of course the experiment must be conducted with much greater attention to details than is here described. For example the friction at the pulley E must be allowed for. 165. Laws of friction. The results of this experiment suggest the following laws. 1. The direction of the friction is opposite to the direction in which the body is urged to move. 2. The magnitude of the friction is just sufficient to prevent ART. 165.] LAWSPOF FRICTION. 115 motion. Thus there is no friction between the box and the table until a weight applied at H begins to act on the box, and then the amount of the friction is equal to that weight. 3. No more than a certain amount of friction can be called into play and when more is required to keep the body at rest, motion will ensue. This amount of friction is called limiting friction. 4. The magnitude of limiting friction bears a constant ratio fi to the normal pressure between the body and the plane on which it rests. This constant ratio fi depends on the nature of the materials in contact. It is usually called the coefficient of friction. We do not here assert that the friction actually called into play is in every case equal to fj, times the normal pressure, but only that this is the greatest amount which can be called into play. For smooth bodies fi = 0. For a great many of the bodies we have to discuss //. lies between zero and unity. 5. The amount of friction is independent of the area of that part of the body which presses on the rough plane, provided only the normal pressure is unaltered. 6. When the body is in motion, the friction called into play is found to be independent of the velocity and proportional to the normal pressure. The ratio is not exactly the same as that found for limiting friction when the body is at rest. It is found that the friction which must be overcome to set the box in motion along the table is greater than the friction between the same bodies when in motion under the same pressure. If the box has remained on the table for some time under pressure the friction which must be overcome is greater than if the bodies were merely placed in contact and immediately set in motion under the same pressure by the proper weight in the pan H. In some bodies this distinction between statical and dynamical friction is found to be very slight, in others the difference is considerable. The coefficient of friction /j, for bodies in motion is therefore slightly less than for bodies at rest. It should be noticed that friction is one of those forces which are usually called resistances. This follows from the second of the laws enunciated above. When a body is pressed against a 8—2 116 FRICTION. [CHAP. V. wall, a reaction or resistance is called into play and is of just the magnitude necessary to balance the pressing force. If there is no pressure there is no reaction. In the same way friction acts only to prevent sliding, not to produce it. 166. There is another method of determining the laws of friction by which the use of the pulley and string is avoided and which therefore presents some ad- vantages. Imagine the box A placed symmetrically on an inclined plane BG. Let the inclination of BG to the horizon be 0. If W be the weight of the box we easily find that the normal reaction is R = W cos 0, and the friction ■n F=Wsin0. Hence -= = tan 0. Let us now suppose the inclina- tion of the plane to the horizon to be gradually increased until the box A begins to slide. The friction F is then the limiting friction. It is found by experiment that this inclination is the same, whatever the weight of the box may be. It follows that the ratio of the limiting friction to the normal pressure is indepen- dent of that pressure. This experiment supplies us with an easy method of approxi- mating to the value of fi for any two materials. Place a body A constructed of one of these materials on an inclined plane BG constructed of the other material. Supposing A to be at rest, increase the inclination until A just begins to slide, then /a is slightly less than the value of tan thus found. Next supposing the inclination of the plane to be such that the body A slides, we might decrease it until the box is just stationary, then ji is slightly greater than the value of tan thus found. In this way we have found two nearly equal numerical quantities between which the coefficient of friction, viz. fi, must lie. The value of which makes tan = ft is often called the angle of friction. Ex. Assuming that limiting friction consists of two parts, one proportional to the pressure and the other to the surface in contact, show that if the least forces which can support a rectangular parallelepiped whose edges are a, b, and e on a given inclined plane be P, Q, and B when the faces be, ca, and ab respectively rest on the plane, then (Q-B)bc + {R-P)ca+ (1 J - Q)ab = 0. [Trin. Coll., 1884.] p+p ART. 168.] LAWS £F FRICTION. 117 167. The friction couple. When a wheel rolls on a rough plane the experiment must be conducted in a different manner. Let a cylinder be placed on a rough horizontal plane and let its weight be W. Let two weights P and P+p be suspended by a string passing over the cylinder and hanging down through a — slit in the horizontal plane. Let the plane of the paper represent a section of the cylinder through the string, let G be the centre, A the point of contact with the plane. Imagine p to be at first zero and to be gradually increased until the cylinder just moves. By. resolving vertically the reaction at A is seen to be equal to W + 2P +p. By resolving horizontally we see that there can be no horizontal force at A. Thus the friction force is zero. Taking moments about A we see that there must be a friction couple at A whose magnitude is equal to pr. 168. The explanation of this couple is as follows. The cylinder not being perfectly rigid yields slightly at A and is therefore in contact with the plane over a small area. When the cylinder begins to roll, the elements of area which are behind the direction of motion are on the point of separating and tend to adhere to each other, the elements in front tend to resist compression. The resultant action across both sets of elements may be replaced by a couple and a single force acting at some convenient point of reference. The yielding of the cylinder at A also slightly alters the position of the centre of gravity of the whole mass, but this change is very insignificant and is usually neglected. The cylinder is treated as if the section were a perfect circle touching the plane at a geometrical point A. The whole action is represented by a force acting at A and a couple. The resolved parts of the force along the normal and tangent at A are often called respectively the reaction and the friction force. In our experi- ment the latter is zero. The couple is called the friction couple. The results of experiment show that the magnitude of p when the cylinder just moves is proportional to the normal pressure directly and the radius of the cylinder inversely. We therefore state as another law of friction that the moment of the friction couple is independent of the curvature and proportional to 118 FRICTION. [CHAP. V. the normal pressure. The ratio of the couple to the normal pressure is often called the coefficient of the friction couple. The magnitude of the friction couple is usually very small and its effects are only perceptible when the circumstances of the case make the friction force evanescent. The weight p is commonly spoken of as the friction of cohesion which is then said to vary inversely as the radius of the cylinder. But we have preferred the mode of statement given above. 169. It should be noticed that the laws of friction are only approximations. It is not true that the ratio of the limiting friction to the pressure is absolutely constant for all pressures and under all circumstances. The law is to be regarded as representing in a compendious way the results of a great many experiments and is to be trusted only for weights within the limits of the experiments. These limits are so extended that the truth of the law is generally assumed in mathematical calculations. If we followed the proper order of the argument, we should now enquire how nearly the laws of friction approximate to the truth so that we may be prepared to make the proper allowance when the necessity arises. We ought also to tabulate the approximate values of /i for various substances. But these discussions would occupy too much space and lead us too far away from the theory of the subject. 170. The experimenters on friction are so numerous that only a few names can be mentioned. The earliest is perhaps Amontons in 1699. He was followed by Muschanbroek and Nollet. But the most famous are Coulomb (Savants Strangers Acad, des Sc. de Paris x. 1785); Ximenfes (Teoria e pratica delle resistenne de' solidi ne' loro attriti. Pisa 1782) ; Vinee (Phil. Trans, vol. 75, 1785) and Morin (Savants Strangers Acad, des Sc. de Paris iv. 1833). Besides these there are the experiments of Southern, Bennie, Jenkin and Ewing, Osborne Eeynolds &o. 171. One of the laws of friction requires that the direction of the friction should be opposite to the direction in which the body under consideration is urged to move. When, therefore, the body can begin to move in only one way, the direction of the friction is known and only its magnitude is required. But when the body can move in any one of several ways, if properly urged, both the direction and the magnitude of the friction are unknown. It follows that problems on friction may be roughly divided into two classes. (1) We have those in which the bodies ART. 172.] PARTICLE WON ROUGH CURVE. 119 rest on one or more points of support, at all of which the lines of action of the frictions are known, but not the magnitudes. (2) There are those in which both the direction and magnitude of the friction have to be discovered. We shall begin by using the laws of friction enunciated above to solve some problems of the first class. Afterwards we shall consider how the directions of the friction forces are to be dis- covered when the system is bordering on motion. 172. A particle is placed on a rough curve in two dimensions under the action of any forces. To find the positions of equilibrium. Let X, Y be the resolved forces in any position P of the particle. Let R be the reaction measured inwards of the curve on the particle, F the friction called into play measured in the direction of the arc s. Let i/r be the angle the tangent makes with the axis of x. The particle is supposed to be on the proper side of the curve, so that it is pressed against the curve by the action of the impressed forces. Taking the figure of the next article, we have, by resolving and taking moments, X cos ty + Y sin ■f + F = 0, - X sin i/r + Fcos i|r + R = 0. Now if fi be the coefficient of friction F must be numerically less than fiR. The required positions of equilibrium are therefore those positions at which the expression X cos i|r +Y sin yfr — X sin i|r+ Fcosi/r : is numerically less than fi. This expression is a function of the position of the particle on the curve. Let us represent it by/ (*). The positions of equilibrium in which the particle borders on motion are found by solving the equations/ (x) = ± fi. Since this equation may have several roots, we thus obtain several extreme positions of equilibrium. We must then examine whether equi- librium holds or fails for the intermediate positions, i.e. whether f{x) is < or >/i numerically. We may sometimes determine this last point in the following manner. Suppose an extreme position, say x=x 13 to be determined by solving the equation f(x)=/j.. If equilibrium exist in the positions determined by values of x slightly less than x lt f(x) must be increasing as x increases through the value x=x x . On the contrary if equilibrium fail for these values of x, f(x) must be decreasing. Thus equilibrium 120 FRICTION. [CHAP. V. fails or holds for values of x slightly greater than x x according as f'(x) is positive or negative when x = x 1 . Let us next suppose that an extreme position, say x=x 2 , is determined by solving the equation /(%)=- p. If equilibrium exist in the positions determined by values of x slightly less than s 2 , f(x) must be algebraically decreasing as x increases through the value x = x 2 , and therefore /' (x 2 ) is negative. If therefore any extreme position of equilibrium is determined by the value x=x 1 of the independent variable, equilibrium fails or holds for values of x slightly greater than x x according as /' (x 2 ) has the same sign as n or the opposite. It is clear that this rule may also be used in the case of a rigid body whose position in space is determined by only one independent variable. 173. Cone of friction. There is another method of finding the position of equilibrium which is more convenient when we wish to use geometry. Let e be the angle of friction, so that /u, = tan e. At any point P draw two straight lines each making an angle e with the normal at P, viz. one on each side. Let these be PA, PB. Then the resultant reaction at P (i.e. the resultant of R and F) must act between the two straight lines PA, PB. These lines may be called the extreme or bounding lines of friction. If the forces on P were not restricted to two dimen- sions, we should describe a right cone whose vertex is at P, whose axis is the line of action of the reaction R, and whose semi-angle is tan -1 /i. This cone is called the cone of limiting friction or more briefly the cone of friction. Since the resultant reaction at P is equal and opposite to the resultant of the impressed forces on the particle we have the following rule. The particle is in equilibrium at all points at which the impressed force acts within the cone of friction. In the extreme positions of equilibrium the resultant of the impressed forces acts along the surface of the cone. 174. A particle is placed on a rough curve in three dimensions under the action of any forces. To find the positions of equilibrium. Let X, Y, Z be the resolved parts of the impressed forces. Let R be their resultant, T their resolved part along the tangent to the curve at the point where the particle is placed. Since T must be less than fi times the normal pressure in any position of equilibrium we have T* < fi° (R? - T°). ART. 176.] PARTICLE »N ROUGH SURFACE. 121 If ds be an element of the arc of the curve, this may be put into the form Here X, Y, Z and s are functions of the coordinates x, y, z. The particle will be in. equilibrium at all the points of the curve at which this inequality holds. If we change the inequality into an equality, we have an equation to find the limiting positions of equilibrium. 175. A particle rests on a rough surface under the action of any forces. To find the positions of equilibrium. Let / (%, y,z) = be the surface, let Q be the normal component of the impressed forces at the point where the particle is placed. In equilibrium we must have R^ — Q^< /w. 2 Q 2 . We have therefore (Xf.+ Yfy + ZfiY X>+Y* + Z* Here X, Y, Z and f are functions of the coordinates. If we change the inequality into an equality, we have a surface, which cuts the given surface /= in a curve. This curve is the boundary of the positions of equilibrium of the particle. 176. Ex. 1. A heavy bead of weight W can slide on a rough circular wire fixed in space with its plane vertical. A centre of repulsive force is situated at one extremity of the horizontal diameter, and the force on the bead when at a distance ?• is pr. Find the limiting positions of equilibrium. If 20 be the angle the radius at the bead makes with the horizon, the tangential and normal forces are (JTcos20 -pramd) and (Wsm 28+praos 8). Putting the ratio of the first to the second equal to ± tan e, we find sin [y =p e - 20) = ± cos y sin c, where W=pa tan y and a is the radius. Discuss the positions given by these values of 8. Ex. 2. A heavy particle rests in equilibrium on a rough cycloid placed with its axis vertical and vertex downwards. Show that the altitude of the particle above the vertex is less than 2(^/(1 + /u 2 ), where a is the radius of the generating circle. Ex. 3. A rigid framework in the form of a rhombus of side a and acute angle a rests on a rough peg whose coefficient of friction is ii. Prove that the distance between the two extreme positions which the point of contact of the peg with any side can have is a/* sin a. See Art. 173. [St John's Coll., 1890.] Ex. 4. Two uniform rods AB, BG are rigidly joined at right angles at B and project over the edge of a table with AB in contact. Find the greatest length of AB 122 FRICTION. [CHAP. V. that can project ; and prove that if the coefficient of friction be greater than „„, - the system can hang with only the end A resting on the edge. [Math. Tripos, 1874.] Ex. 5. Three rough particles of masses m 1 , m^, m^, are rigidly connected by light smooth wires meeting in a point 0, such that the particles are at the vertices of an equilateral triangle whose centre is 0. The system is placed on an inclined plane of slope a, to which it is attached by a pivot through ; prove that it will rest in any position if the coefficient of friction for any one of the particles be not less than — ^-^ (mf + mi + m^ - m 2 m 3 - m i m l - m^)*. [Math. Tripos, 1877.] Wl^ T" WLj *T" W^a Ex. 6. A particle rests on the surface xyz=c 3 under the action of a constant force parallel to the axis of z : prove that the curve of intersection of the surface 1 1 u? with the cone - 3 + - 2 = —„ will separate the part of the surface on which equilibrium X" y z is possible from that on which it is impossible ; p being the coefficient of friction. [Math. Tripos, 1870.] Ex. 7. The ellipsoid —„ + £;+-„ = 1 is placed with the axis of x vertical, its a' ly c l surface being rough. Show that a heavy particle will rest on it anywhere above its intersection with the cylinder tt 2 ( 1 + -^^ I + - 2 ( 1 + -j- s 1 = 1, /* being the coeffi- cient of friction. ' ' [Trin. Coll., 1885.] 177. The following problem is regarded from more than one aspect to illustrate some different methods of proceeding. Ex. 1. A ladder is placed with one end ore a rough horizontal floor and the other against a rough vertical wall, the vertical plane containing the ladder being perpen- dicular to the wall. Find the positions of equilibrium. Let AB be the ladder, 21 its length, w its weight acting at its middle point G. Let 8 be its inclination to the horizon. See the figure of Ex. 2. Let B, R' be the reactions at A and B acting along A D, BD respectively ; p, /j.' the coefficients of friction at these points. The frictions at A and B are £R and t)R\ where f, 17 are two quantities which are numerically less than /t and // respectively. In many problems f , n may be either positive or negative. In this case however, since friction is merely a resistance and not an active force, we may assume that the frictions act along AL and LB. We may therefore regard |, n as positive. This limitation will also follow from the equations of equilibrium. By resolving and taking moments we have £R=B' riR' + R=w 2nR' I cos 6 + 2R' I sin $ = wl cos 0. 1 — 6j Eliminating R, R' we find tanfl= . Any positive value of tan 6 given by this '1% equation, where |, n are less than p., p', will indicate a possible position of equilibrium. If the roughness is so slight that pp'<.l, the minimum value of tan is given by tan = — ^- . If the roughness is so great that pp'>l, the ladder will &p rest in equilibrium at all inclinations. ART. 177.] EXAMBftES ON FRICTION. 123 Ex. 2. The ladder being placed at any given inclination 8 to the horizon, find what weight can be placed on a given rung that the ladder may be in equilibrium. Let M be the rung, W the weight placed on it, AM=m. Let ^=tane, /t' = tane'. Geometrical Solution. If we make the angles DAE = e, DBE=c', the resultant reactions at A and B must lie within these angles and must meet in some point which lies within the quadrilateral EF DH. Let Gf be the centre of gravity of the weights W and w. If the vertical line through G pass to the left of E, the weight (W+w) may be supposed to act at some point P within the quadrilateral above mentioned. This weight may then be resolved obliquely into the two directions PA, PB. These may be balanced by two reactions at A and B each lying within its limiting lines. The result is that there will be equilibrium if the vertical through G passes to the left of E. It is evident that this reasoning is of general application. We may use it to find the conditions of equilibrium of a body which can slide with a point on each of two given curves whenever the impressed forces which act on the body can be conveniently reduced to a single force. We draw the limiting lines of friction at the points of contact, and thus form a quadrilateral. The condition of equilibrium is that the resultant impressed force shall pass through the quadrilateral area. The abscissas of the points E and G measured horizontally from A to the right are easily proved to be respectively _ 21 (/j./// cos 8 + n sin 8) ^_(Wm + wl) cos 8 W+w ' If G lie to the right of the vertical through E (i.e. lcos8>x) there cannot be ■ equilibrium unless the given rung lie to the left of that vertical (mcos 8 a;), the weight W placed on it must be sufficiently small not to bring the centre of gravity to the right of that vertical. Lastly, if the vertical through E lie to the right of B (tan -1 in > %ir - 8) there is equilibrium whatever W may be, and on whatever rung it may be placed. Another problem is solved on a similar principle in Jellett's treatise on friction, 1872. Analytical solution. Following the same notation as in Ex. 1 we have by resolving and taking moments ZR=R\ riR' + R=W+w, 2rjR'l cos 8 + 2R'l sin 8 = (Wm + ml) cos 8. 124 FRICTION. [CHAP. V. Eliminating R, R', we find 2i(g?;cos0 + {sin0) _ (Wm + wl)co&0 £ij+T ~~ W+w .(A). y N' N The condition of equilibrium is that it is possible to satisfy this equation with values of f , 17 which are less than /a, p.' respectively. By seeking the maximum value of the left-hand side we may derive from this the geometrical condition that the centre of gravity of W and w must lie to the left of a certain vertical straight line. But our object is to discuss the equation otherwise. Let us regard £, 17 as the coordinates of some point Q referred to any rectangular axes. Then (A) is the equation to a hyperbola, one branch of which is represented in the figure by the dotted line. If this hyperbola pass within the rectangle NN' formed by |=±/i, 7j=±p', the conditions of equilibrium can be satisfied by values of I, 7i less than their limiting values. If the curve does not cut the rectangle, there cannot be equilibrium without the assistance of more than the available friction. The right-hand side of (A) is the quantity already called x. Let it be transferred to the left-hand side and let the equation thus altered be written z=0. We notice that z is negative at the origin. In order that the hyperbola may cut the rectangle it is sufficient and necessary that z should be positive at the point N, i.e. when |=^, »;=/*'. The required condition of equilibrium is therefore that — ^^ ; — ^ - x should be a positive quantity. This is virtually the same result as before and may be similarly interpreted. Ex. 3. Let the ladder AB be placed in a given position leaning against the rough vertical face of a large box which stands on the same floor, as shown in the figure of Ex, 2. Determine the conditions of equilibrium. We have now to take account of the equilibrium of the box BLL'. Let W be its weight. Let R" be the reaction between it and the floor, ffi" the friction. We have then, in addition to the equations of Ex. 1, R"=W' + 7)R', ffl' = R'. Eliminating R" we find {W' + w)l7i£+W'£-wZ=0. We have also by Ex. 1, £tj + 2$ tan 0-1 = (A). Eliminating 7], so as to express both t\ and f in terms of one variable {, we find 2 (W + u>)tan0|f+ie{-(2W + w) f=0 (B). The conditions of equilibrium are that the two equations A and B can be simultaneously satisfied by values of £ , r\, f less than y,, p.', p!' respectively. Regarding f , 1;, f as the coordinates of a representative point Q, these equations represent two cylinders. These cylinders intersect in a curve. If any part of this curve lie within the rectangular solid bounded by £=±/i, 7)= ±/»', f=±,n" the conditions of equilibrium are satisfied. ART. 178.] EXAMBftES ON FRICTION. 125 But instead of using solid geometry we may represent (A) and (B) by two hyperbolas having different ordinates i;, f but the same abscissa f. The frictions being resistances, we shall assume that they act so that f, v , f are all positive, it will therefore be necessary only to draw that portion of the figure which lies in the positive quadrant. Take OM=/i, OM'=n', OJkT" = /i". Let OB and AH represent the hyperbolas (B) and (A). • Then we easily find M>A=* , M "B= (g^ + «)M» M+2tan«' 2{W' + w)/j." ta.n6 + w" The condition of equilibrium is that an ordinate can be found intersecting the two hyperbolas in points Q, Q' each of which lies within the limiting rectangles. The necessary conditions are therefore found by making an ordinate travel across the figure from OM' to N'N". They may be summed up as follows. (1) The hyperbola AH must intersect the area of the rectangle ON'; the condition for this is that M'A/i. If the ladder is so placed that the inequality (2) becomes an equality while (1) is not broken, the frictions 17 and f attain their limiting values while { is not limiting, the ladder will therefore be on the point of slipping at its upper extremity, and the box will be just slipping along the plane. If the ladder is so placed that the inequality (1) becomes an equality while (2) is not broken, { and 17 have their limiting values while f is less than its limit. The box is therefore fixed and the ladder slips at both ends. 178. Ex. 1. A ladder AB rests against a smooth wall at B and on a rough horizontal plane at A. A man whose weight is n times that of the ladder climbs up it. Prove that the frictions at A in the two extreme cases in which the man is at the two ends of the ladder are in the ratio of 2m + 1 to 1. Ex. 2. A boy of weight w stands on a sheet of ice and pushes with his hands against the smooth vertical side of a heavy chair of weight nw. Show that he can incline his body to the horizon at any angle greater than cot -1 2/t or cot -1 2/j.n, according as the chair or the boy is the heavier, the coefficient of friction between the ice and boy or the ice and chair being /*. [Queens' Coll.] Ex. 3. Two hemispheres, of radii a and b, have their bases fixed to a horizontal plane, and a plank rests symmetrically upon them. If p. be the coefficient of friction between the plank and either hemisphere, the other being smooth, prove that, when the plank is on the point of slipping, the distance of its centre from its point of contact with the smooth hemisphere is equal to (a~ &)//*. [St John's Coll., 1885.] Ex. 4. A heavy rod rests with one end on a horizontal plane and the other against a vertical wall. To a point in the rod one end of a string is tied, the other end being fastened to a point in the line of intersection of the plane and 126 FRICTION. [CHAP. V. wall. The string and rod are in a vertical plane perpendicular to the wall. Show that, if the rod make with the horizon an angle o which is less than the complement of 2e, then equilibrium is impossible unless the string make with the horizon an acute angle less than a + e, where e is the angle of friction both with the wall and the plane. [Math. Tripos, 1890.] Ex. 5. A parabolic lamina whose centre of gravity is at its focus rests in a vertical plane upon two rough rods of the same material at right angles and in the same vertical plane ; if be the inclination of the directrix to the horizon in one extreme position of equilibrium, prove that tan 2 (o - 0) tan (a + e - 0) = tan (a - e) ; where e is the angle of friction, a the inclination of one rod to the horizon. [Trin. Coll., 1882.] Ex. 6. Two rods AC, BC with a smooth hinge at G are placed in a given position with their extremities A and B resting on a rough horizontal plane. The plane of the rods being vertical, find the conditions of equilibrium. Let 0, 6' be the inclinations of the rods to the horizon, W and W their weights. Let (B, |iJ), (B', 7]B') be the reactions and frictions at A and B. Eesolving and taking moments in the usual way, we find W+W _ W'+W £~ W7+„« Q' i tour i TJ7'\ +«~ a' V—' W t&a 8' +(2W+W) tan $' ' W tan0 + (2W" + W) tanfl' ' If the value of | thus found is >/* the system will slip at A ; if r/>/i it will slip at B. If the system slip at A only, then £>ij ; this gives 17 tan < W tan 0'. Ex. 7. A groove is cut in the surface of a flat piece of board. Show that the form of the groove may be so chosen as to satisfy this condition, that if the board will just hang in equilibrium upon a rough peg placed at any one point of the groove, it will also just hang in equilibrium when the peg is placed at any other point. [Math. Tripos, 1859.] Ex. 8. A lamina is suspended by three strings from a point ; if the lamina be rough, and the coefficient of friction between it and a particle P placed upon it be constant, show that the boundary of possible positions of equilibrium of the particle on the lamina is a circle. [Math. Tripos, 1880.] Let ON be a perpendicular on the lamina. Let D be the centre of gravity of the lamina, G that of the lamina and particle. Then in equilibrium OG is vertical and NG is the line of greatest slope. The angle NOG is equal to the inclination of the plane to the horizon and is constant because the equilibrium is limiting. The locus of G is a circle, centre N. Since DP : DG is constant the locus of P is also a circle. Ex. 9. Spheres whose weights are W, W rest on different and differently inclined planes. The highest points of the spheres are connected by a horizontal string perpendicular to the common horizontal edge of the two planes and above it. If /x, p! be the coefficients of friction and be such that each sphere is on the point of slipping down, then fiW=fi'W' . [Math. Tripos.] Consider one sphere : the resultant of T and p.B balances that of W and B. By taking moments about the centre T=nB. Hence, by drawing a figure, B=W. Thus T=/xW and the result follows. ART. 178.] EXAMPLES ON FRICTION. 127 Ex. 10. A uniform rod passes over one peg and under another, the coefficient of Motion between each peg and the rod being /i. The distance between the pegs is 6, and the straight line joining them makes an angle /3 with the horizon. Show that equilibrium is not possible unless the length of the rod is >6 {l + (tan/3)//t}. [Coll. Ex.] Ex. 11. A uniform rod AGB, length 2a, is supported against a rough wall by a string attached to its middle point C: show that the rod can rest with G at any point of a circular arc, whose extremities are distant a and a cos e from the wall, where 6 is the angle of friction. Take moments about C. Ex. 12. Two uniform and equal rods of length 2a have their extremities rigidly connected, and are inclined to each other at an angle 2o. These rods rest on a fixed rough cylinder with its axis horizontal, and whose radius is a tan a. Show that in the limiting position of equilibrium the inclination 8 to the vertical of the line through the point of intersection of the rods perpendicular to the axis of the cylinder is given by sin ! a sin 8 = cos (8 - e) sin e, where tan e iB the coefficient of friction. [Coll. Exam.] Ex. 13. Three equal uniform heavy rods AB, BG, CD, hinged at B and C, are suspended by a light string attached to D from a point E, and hang so that the end A is on the point of motion, towards the vertical through E, along a rough horizontal plane (coefficient of friction /u=tan e) : show that cos (a - e) _ cos (/3 - e) _ cos (7 - e) _ p. cos (8 - e) cos a " 3cos/3 ~" 5 cos 7 ~~ 6 cos 8 where a, /S, 7 are the inclinations of the rods to the horizon beginning with the lowest, and 8 that of the string. [Coll. Ex., 1881.] Take moments about B, G, D, E in succession for the rods AB, AB and BG, and so on. Subtracting each equation from the next in order, the results follow at . once. ' Ex. 14. A sphere rests on a rough horizontal plane, and its highest point is joined to a peg fixed in the plane by a tight cord parallel to the plane. Show that, if the plane be gradually tilted about a line in it perpendicular to the direction of the cord, the sphere will not slip until the inclination becomes equal to tan -1 2/*, where /i is the coefficient of friction. [Math. Tripos, 1886.] Ex. 15. A uniform hemisphere, placed with its base resting on a rough inclined plane, is just on the point of sliding down. A light string, attached to the point of the hemisphere farthest from the plane, is then pulled in a direction parallel to and directly up the plane. If the tension of the string be gradually increased until the sphere begins to move, it will slide or tilt according as 13 tan is less or greater than 8, where is the inclination of the plane to the horizon. The centre of gravity of the hemisphere is at a distance from the centre equal to three-eighths of the radius. [Coll. Ex., 1888.] Ex. 16. A circular disc, of radius a, whose centre of gravity is distant c from its centre, is placed on two rough pegs in a horizontal line distant 2a sin a apart. Show that all positions will be possible positions of equilibrium, provided a sin o sin (X x + X 2 ) > c sin (2a =p X ; ± X 2 ) , where \, X 2 are the angles of friction at the two pegs. [St John's Coll., 1880.] 128 FRICTION. [CHAP. V. Ex. 17. A number of equally rough particles are knotted at intervals on a string, one end of which is fixed to a point on an inclined plane. Show that, all the portions of the string being tight, the lowest particle is in its highest possible position, when they are all in a straight line making an angle sin -1 (tan X/ tan a) with the line of greatest slope, X being the angle of friction and a the inclination of the plane to the horizon. Show also that, if any portion of the string make this angle with the line of greatest slope, all the portions below it must do so too. [Math. Tripos, 1886.] Ex. 18. A rough paraboloid of revolution, of latus rectum 4a, and of coefficient of friction cot/3, revolves with uniform angular velocity about its axis which is vertical : prove that for any given angular velocity greater than (#/2a)* cot |/3 or less than {gj2ay tan J/3 a particle can rest anywhere on the surface except within a certain belt, but that for any intermediate angular velocity equilibrium is possible at every point of the surface. [Math. Tripos, 1871.] Let rag be the weight of the particle. It is known by dynamics that we may treat the paraboloid as if it were fixed in space, provided we regard the particle as acted on by a force miA- tending directly from the axis, where r is the distance of the particle from the axis, and w the angular velocity of the paraboloid. We may prove that the ordinates in the limiting positions of equilibrium are given by ^]f-(2a^-g)y + 2ap.g=0. That a belt may exist, the roots of this quadratic must be real. Ex. 19. A rod rests partly within and partly without a box in the shape of a rectangular parallelepiped, presses with one end against the rough vertical side of the box, and rests in contact with the opposite smooth edge. The weight of the box being four times that of the rod, show that, if the rod be about to slip and the box about to tumble at the same instant, the angle the rod makes with the vertical is JX + $cos _1 (JcosX), where X is the angle of friction. [Math. Tripos, 1880.] Ex. 20. A glass rod is balanced partly in and partly out of a cylindrical tumbler with the lower end resting against the vertical side of the tumbler. If a and /3 are the greatest and least angles which the rod can make with the vertical, cir]3 m sin S prove that the angle of friction is i tan -1 -^-= =—^ . suv 5 a cos a + sin 2 /3 cos /3 [Math. Tripos, 1875.] Ex. 21. A heavy rod, of length 21, rests horizontally on the inside rough surface of a hollow circular cone, the axis of which is vertical and the vertex downwards. If la is the vertical angle of the cone, and if the coefficient of friction /i is less than cot a, prove that the greatest height of the rod, when in equilibrium, , ., , ... . , . (l + cos 2 o + sinav/(sin 2 a + 4 / u, 2 ))l above the vertex of the cone is looi a { ^-^ „, ' , — f 2 . I 2 (1 - /j? tan 2 a) ( [Math. Tripos, 1885.] Ex. 22. A heavy uniform rod AB is placed inside a rough curve in the form of a parabola whose focus is S and axis vertical. Prove that, when it is on the point of slipping downwards, the angle of friction is J (SAB - SB A). [Coll. Ex., 1889.] Ex. 23. A rod MN rests with its ends in two fixed straight rough grooves OA, OB, in the same vertical plane, which make angles a and /3 with the horizon : prove that, when the end M is on the point of slipping down AO, the tangent of the inclination of MN to the horizon is ;r—. — J — Ji, — -1 — , . [Math. Tripos, 1876.1 2 sin (/3 + c) sin (a - e) L r ' J ART. 178.] EXAMPI«§S ON FRICTION. 129 Ex. 24. A uniform rectangular board ABCD rests with the corner A against a rough vertical wall and its side BG on a smooth peg, the plane of the board being vertical and perpendicular to that of the wall. Show that, without disturbing the equilibrium, the peg may be moved through a space /* cos a (a cos a + 6 sin a) along the side with which it is in contact, provided the coefficient of friction (/*) lie between certain limits ; a being the angle BG makes with the wall, and a, b the lengths of AB, BG respectively. Also find the extreme values of p. [Math. Tripos, 1880.] Ex. 25. An elliptical cylinder, placed in contact with a vertical wall and a horizontal plane, is just on the point of motion when its major axis is inclined at an angle a to the horizon. Determine the relation between the coefficients of friction of the wall and plane ; and show from your result that, if the wall be smooth, and o be equal to 45°, the coefficient of friction between the plane and cylinder will be equal to Je 3 , where e is the eccentricity of the transverse section of the cylinder. [Math. Tripos, 1883.] Ex. 26. A rough elliptic cylinder rests, with its axis horizontal, upon the ground and against a vertical wall, the ground and the wall being equally rough ; show that the cylinder will be on the point of slipping when its major axis plane is inclined at an angle of ir/4 to the vertical if the square of the eccentricity of its principal section be 2 sin e (sin c + cos e), where e is the angle of friction. [Coll. Ex., 1885.] Ex. 27. Three uniform rods of lengths a, 1>, c are rigidly connected to form a triangle ABC, which is hung over a rough peg so that the side BG may rest in contact with it; find the length of the portion of the rod over which the peg may range, showing that, if ^> — L- r-^cosecC + tan4(C-B), where G>B, the triangle will rest in any position. [Math. Tripos, 1887.] * Ex. 28. A waggon, with four equal wheels on smooth axles whose plane contains the centre of gravity, rests on the rough surface of a fixed horizontal circular cylinder,' the axles being parallel to the axis of the cylinder; investigate the pressures on the wheels, and prove that the inclination to the horizontal of the plane containing the axles is tan -1 {tan o (w-w')lW), where w, w' are the weights on the two axles, W that of the whole waggon, and 2a is the angle between the tangent planes at the points of contact. [Math. Tripos, 1888.] Ex. 29. Three circular cylinders A, B, C, alike in all respects, are placed with their axes horizontal and their centres of gravity in a vertical plane ; A is fixed, B is at the same level, and G at a lower level touches them both, the common tangent planes being inclined at 45° to the vertical. B and C are supported by a perfectly rough endless strap of suitable length passing round the cylinders in the plane containing the centres of gravity. Show that equilibrium can be secured by making the strap tight enough, provided that the coefficient of friction between the cylinders is greater than 1 - 1/^2 ; and find how slipping will first occur if the strap is not quite tight enough. [Math. Tripos, 1888.] Ex. 30. Two uniform rods AB, BC, of equal length, are jointed at B. They are at rest in a vertical plane, equally inclined to the horizon, with their lower ends in contact with a rough horizontal plane. Prove that, if they be on the point of slipping both at A and G, the frictional couple at the joint is Tra(sina-2/icoso), where W is the weight of each rod, a the inclination of each rod to the luiiiniin.Kr t J ' 'la the length of each rod, and /* the coefficient of friction. [St John's Coll., 1890.] R. S. 9 130 FRICTION. [CHAP. V. Ex. 31. Six uniform rods, each of length 2a, are joined end to end by five smooth hinges, and they stand on a rough horizontal plane in equilibrium in the form. of a symmetrical arch, three on each side ; prove that the span cannot be greater than 2a */2 (l + x/i + ^/iV), if * ne coefficient of friction of the rods and plane be i. [Coll. Ex., 1886.] Consider only half the arch. The reaction at the highest point is horizontal, and equal to half the weight of one rod. Take moments (1) for the upper, (2) for the two upper, (3) for all three rods. We find that their inclinations to the vertical are Jtt, tan -1 J, tan -1 $. The result follows easily. 179. Friction between wheel and axle. Ex. 1. A gig is so constructed that when the shafts are horizontal the centre of gravity of the gig and the shafts is over the axle of the wheels. The gig in this position rests on a perfectly rough ground. Find the direction and magnitude of the least force which, acting at the extremity of the shaft, will just move the gig. When an axle is made to fit the nave of a wheel, the relative sizes of the axle and hole are so arranged that the wheel can turn easily round the axle. The axle is therefore just a little smaller than the hole. Thus the two cylinders touch along some generating line and the pressures act at points in this line. Even if the axle were somewhat tightly clasped at first, yet by continued use it would be worn away so that it would become a little smaller than the hole. It is possible that the axle may be so large that it has to be forced into the hole. When this is the case, besides the pressures produced by the weight of the gig, there will be pressures due to the compression of the axle. These last will act on every element of the surface of the axle and their magnitudes will depend on how much the axle has to be compressed to get it into the hole. If the axle and hole are not perfectly circular, these pressures may be unequally distributed over the surface of the axle. When these circumstances of the problem are not given, the pressures on the axle are indeterminate. Let X, Y be the required horizontal and vertical components of the force applied at the extremity S of the shaft. Since it touches a perfectly rough ground Consider the equilibrium of the wheel, at A, the friction at this point cannot be limiting. Let It and F be the reaction and friction. It is evident that the friction F must act to the left, if it is to balance the force X which is taken as acting to the right. The axle will touch the circular hole in which it works at some one point B. At this point there will be a reaction B' and a friction I<", which is limiting when the gig is on the point of motion. Thus F'=ixB'. The resultant of B' and /iB' must balance the resultant of B and F and the weight of the wheel. It therefore follows that the point B is on the left of C, i.e. behind the axle. Let 6 be the angle AGB, let a and 6 be the radii of the wheel and axle. Taking moments about A we have B'a sin 6=/j,B' (a cos 0-b). ART. 179.] EXAMPLES ON FRICTION. 131 Putting yn=tan e, this equation gives sin (e - $) = - sin e. a Since 6 is less than a, this equation shows that 8 is positive and less than e. Consider next the equilibrium of the gig. The forces R' and p.R' act on the gig in directions opposite to those indicated in the figure. Let W be the weight of the gig, then resolving and taking moments about G we have X= —R' sin 8 + iiR' cos 8, Y= - R' cos 6 - ij.R' sin 8+W, Yl=p.R'b, where I is the length of the shaft. These equations give X and Y. Ex. 2. A light string, supporting two weights W and W, is placed over a wheel which can turn round a fixed rough axle. Supposing the string not to slip on the wheel, find the condition that the wheel may be on the point of turning round the axle. If a and b be the radii of the wheel and axle, prove that W-W'a_ /i W+W'b~,J(l + ^)- Ex. 3. A solid body, pierced with a cylindrical cavity, is free to turn about a fixed axle which just fits the cavity, and the whole figure is symmetrical about a certain plane perpendicular to the axle. The axle being rough, and the body acted on by forces in the plane of symmetry, find the least coefficient of friction that the body may be in equilibrium. The circular sections of the cavity and axle are drawn in the figure as if they were of different sizes. This has been done to show that the reaction and friction act at a definite point, but in the geometrical part of the investigation they should be regarded as equal. Let the plane of symmetry be taken as the plane of xy, and let its intersection with the axis be the origin. Let X, Y, G be the components of the forces, and let these urge the body to turn round the axis in a direction opposite to that of the hands of a watch. The axle will touch the cavity along a generating line, let B be its point of intersection with the plane of xy. Let 8 be the angle BOx. Let R and F be the normal reaction and the friction at B ; when the body borders on motion we have F=i>.R. By resolving and taking moments we find R (cos 8 + n sin 8) + X= 0, ■R(sin0-,ucos0) + y=O, -fiRa +0 =0, where a is the radius of the cavity. Putting ^ = tan e, we deduce from these equations tan (8 - e) = Y\X, R 2 = (X s + F 2 ) cos 2 e. These determine the point B and the reaction R. The least value of the coefficient of friction is then given by (-X 2 +r 2 )a 3 sin 2 e=G 2 . 9—2 132 FRICTION. [CHAP. V. 180. Lemma. Before we proceed to extend the laws of friction to a body about to move in any manner, we shall require the following lemma. If a lamina be moved from any one position to any other in its own plane, there is one point rigidly connected to the lamina whose position in space is unchanged. The lamina may therefore be brought from its first to its last position by fixing this point and rotating the lamina about it through the proper angle. Let A, B be any two points in the lamina in its first position, A', B' their positions in the last position. Then if A, B can be brought into the positions A', B' by rotation about some point /, fixed in space, the whole lamina will be brought from its first to its last position. Bisect AA', BB' at right angles by the straight lines LI, MI. Then I A = I A', and IB = IB 1 . Also, since A B is unaltered in length by its motion, the sides of the triangles AIB, A'lB' are equal, each to each. It follows that the angles AIB, A'lB' are equal, and therefore that the angles AIA' and BIB' are equal. If then we turn the lamina round /, as a point fixed in space, through an angle equal to AIA', A will take the position A', and B will take the position B'. Thus the whole body has been transferred from the one position to the other. If the body be simply translated, so that every point moves parallel to a given straight line, the bisecting lines LI, MI are parallel to each other, and therefore the point i" is infinitely distant. If the angle AIA' is indefinitely small, the fixed point I of the lamina is called the instantaneous centre of rotation. 181. Frictions in unknown directions. We are now prepared to make a step towards the generalization of the laws of friction. Let us suppose a heavy body to rest on a rough hori- zontal table on n supports. Let these points be A T , A 2 ,...A n , and let the pressures at these points be P lt P 2 ,...P„. We shall also suppose the body to be acted on by a couple and a force applied at some convenient base of reference, the forces being all parallel to the table. To resist these forces a frictional force is called into ART. 181.] LAWS»OF FRICTION. 133 play at each point of support. The directions and magnitudes of these frictional forces are unknown, except that the magnitude of each is less than the limiting friction, and the direction is opposed to the resultant of all the external and molecular forces which act on that point of support. If the pressures P±,...P n are known, there are thus In unknown quantities, and there are only three equations of equilibrium. The frictions at the points of support are therefore generally indeterminate. By calling the frictions indeterminate we mean that there are different ways of arranging forces at the points of support which could balance the given forces and which might be frictional forces. Which of these is the true arrangement of the frictional forces depends on the manner in which the body, regarded as partially elastic, begins to yield to the forces. Suppose, for example, a force Q to act at a point B of the body, and to be gradually increased in magnitude. The frictions on the points of support nearest to B will at first be sufficient to balance the force, but, as Q gradually increases, the frictions at these points may attain their limiting values. As soon as they begin to yield, the frictions at the neighbouring points will be called into play, and so on throughout the body. When the external forces are insufficient to move the body as a whole, the directions and magnitudes, of the frictions at the points of support depend on the manner in which the body yields, however slight yielding that may be. Even if the external forces were absent, the body could be placed in a state of constraint and might be maintained in that state by the frictions. Thus the frictions depend on the initial state of constraint as well as on the external forces. It is also possible that the body, though apparently at rest, may be performing small oscillations about some position of stable equilibrium. This might cause other changes in the frictions. 182. Limiting Equilibrium. Let us now suppose that the external forces have been gradually increased according to some given law until the whole body is on the point of motion. By this we mean that the least diminution of roughness or the least increase of the forces will cause the body to move. We may enquire what is the condition that these forces may be just great 134 FRICTION. [CHAP. V. enough to move the body, or just small enough not to move the body. When the external forces move the body as a whole, the arrangement of the frictional forces is somewhat different from that described in the -last article. We suppose the body to be so nearly rigid that the distances between the several particles do not sensibly change. Thus their motions are not independent, but are sensibly governed by the law proved in the lemma of Art. 180. The directions of the frictions, being opposite to the directions of the motions, are governed by the same law. In what follows, we shall suppose that the forces just overbalance the resistances, and that the body is beginning to move as a rigid body. It will be seen from what follows that, when a rigid body turns round an instantaneous axis, the friction at every point of support acts in the direction which is most effective to prevent motion. If, therefore, the frictional forces thus arranged are insufficient to prevent motion, there is no other arrangement by which they can effect that result. If the body move on the horizontal plane, no -matter how slightly, it must be turning about some vertical axis ; let this vertical axis intersect the table in the point /. There are then two cases to be considered, (1) the point i" may not coincide with any one of the points of support, and (2) it may coincide with some one of them. Let us take these cases in order. The position of / is un- known ; let its coordinates be £, rj referred to any axes in the plane of the table. The points A 1 ,...A n are all beginning to move each perpendicular to the straight line which joins it to the point /. The frictions at these points will therefore be known when I is known. Their directions are perpendicular to IA lt IA 2 , &c, and they all act the same way round /. Their magnitudes are /*iP,, /i 2 P 2 , &c, if /Uq, ^2, &c. are the coefficients of friction. Since the impressed forces only just overbalance the frictions, we may regard the whole as in equilibrium. Forming then the three equations of equilibrium, we have sufficient equations to find both f, rj and the condition that the body should be on the point of motion. It may be that these equations do not give any available values of £, rj, and in such a case the point / cannot lie away from one of the points of support. ART. 184.] LkVm OF FRICTION. 135 183. Let us consider next the case in which I coincides with one of the points of support, say A v The coordinates £, 97 of / are now known. Just as before the frictions at A 2 ,...A n are all known, their directions are perpendicular to A X A 2 , A^, &c. and their magnitudes are /x 2 P 2 , &c. Since A x does not move, the friction at A x is not necessarily limiting friction. It may be only just sufficient to prevent A x from moving. Let the components of this friction parallel to the axes x and y be F x and F(. Forming as before the three equations of equilibrium, we have sufficient equations to find F lt F^ and the required condition that the body may be on the point of motion. If, however, the values of F lt F x thus found are such that F^ + F^ 2 is greater than fifPf, the friction required to prevent A x from moving is greater than the limiting friction. It is then impossible that the body could begin to turn round A x as an instantaneous centre. We can determine by a similar process whether the body could begin to turn round A 2 , and so on for all the points of support. 184. We shall now form the Cartesian equations from which the coordinates if, i\ and the condition of limiting equilibrium are to be found. These however are rather complicated, and in most cases it will be found more convenient to find the position of I by some geometrical method of expressing the conditions of equi- librium. Let the impressed forces be represented by a couple L together with the components A' and Y acting at the origin. Let the coordinates of A v A 2 &c. be (ic^]), (;*%), &c. Let the coordinates of I be (£ij). Let the distances IA V IA V &e. be r u r 3 &o. Let the direction of rotation of the body be opposite to that of the hands of a watch. Then since the frictions tend to prevent motion, they act in the opposite direction round I. The resolution of these frictions parallel to the axes will be facilitated if we turn each round its point of application through an angle equal to a right angle. We then have the frictions acting along the straight lines IA V IA it &c, all towards or all from the point I. Taking the latter supposition, their resolved parts are to be in equilibrium with X acting along the positive direction of the axis of y and Y along the negative direction of x. We find by resolution Z,«P^ +7=0 ) (1). r / The equation of moments must be formed without changing the directions of the frictions. Taking moments about I, we have 2pPr+Y£-Xri-L = : (2). 136 FRICTION. [CHAP. V. If the instantaneous centre I coincide with A lt the equations are only slightly altered. We write (x^) for (£,), F 1 and - FJ for ^P 1 Vl -^ 1 and ^Pj °^ , and r i r i finally omit the term /ji^P^ in the moment. 185. Another Method. There is another way of discussing these equations which will more clearly explain the connection between the two cases. If the body is just beginning to turn about some instantaneous axis, it would begin to turn about that axis if it were fixed in space. Let then I be any point on the plane of xy and let us enquire whether the body can begin to turn about the vertical through I as an axis fixed in space. Supposing all the friction to be called into play, the moment of the forces round I, measured in the direction in which the frictions act, is u^Z/iPr+Y^-Xn-L. If, in any position of I, u is negative, the moment of the forces is too powerful for that of the frictions, the body will therefore begin to move. If on the other hand u is positive, the moment of the frictions is too powerful for that of the forces, and the body could be kept at rest by less than the limiting frictions. Let us find the position of I which makes u a minimum. If in this position «. is positive or zero, there is no point I about which the body can begin to turn. To make u a minimum we equate to zero the differential coefficients of u with regard to {,17. Since r^=(x-^+(y-r;) 2 , the equations thus formed are exactlythe equations (1) already written down in Art. 184. The statical meaning of these equations is that the pressures on the axis which has been fixed in space are zero when that axis has been so chosen that a is a minimum. If this is not evident, let B x and By be the resolved pressures on the axis. The resolved parts parallel to the axes of the impressed forces and the frictions together with B x and R„ must then be zero. But the equations (1) express the fact that these resolved parts without B x and B y are zero. It evidently follows that both B x and R,, are zero. That this position of I makes u a minimum and not a maximum may be shown analytically by finding the second differential coefficients of u with regard to | and 17. The terms of the second order are then found to be 2/tP {to ~ 2/) <«"- 120° this resultant is >J/uWand is therefore inadmissible. Thus the table cannot turn round an axis at any corner unless the angle at that corner is greater than 120°. If the corner is G, the magnitude of the least couple is llxW(GA + GB). This statical problem might also be solved by finding the position of a point I such that the sum of its distances AI, BI, CI (all multiplied by the constant J|U W) from the corners is an absolute minimum. Ex. 2. Four equal heavy particles A, B, G, D are connected together so as to form a rigid quadrilateral and placed on a rough horizontal plane. Supposing the pressures at the four particles are equal, find the least couple which will move the system. The instantaneous centre I is the intersection of the diagonals or one of the corners according as that intersection lies inside or outside the quadrilateral. Ex. 3. A heavy rod is placed in any manner resting on two points A and B of a rough horizontal curve, and a string attached to the middle point G of the chord is pulled in any direction so that the rod is on the point of motion. Prove that the locus of the intersection of the string with the directions of the frictions at the points of support is an arc of a circle and a part of a straight line. Find also how the force must be applied that its intersection with the frictions may trace out the remainder of the circle. Firstly let the rod be on the point of slipping at both A and B, and let F, F' be the frictions at the two points. Then F, F' are both known, and depend only on the weight and on the position of the centre of gravity of the rod. Supposing the centre of gravity to be nearer B than A, the limiting friction at B will be greater than that at A. Since there is equilibrium, the two frictions and the tension must meet in one point ; let this be P. Then since AC= CB, it is evident that GP is half the diagonal of the parallelogram whose sides are AP, BP. Hence, by the triangle of forces, AP, BP and IPC will represent the forces in those directions. Hence AP : PB :: F: F', and thus the ratio AP : PB is constant for all directions of the string. The ' ^1 ' ^> A J, locus of P is therefore a circle. Let the point G be pulled in the direction PC, so that the line CP in the figure represents the produced direction of the string. The string CP cuts the circle in two points, but the forces can meet in only one of these. It is evident that the rod must be on the point of turning round some one point I. This point is the intersection of the perpendicular drawn to PA, PB at A and B. Now the frictions, in order to balance the tension, must act towards P, and therefore the directions of motion of A and B must be from P. This clearly cannot be the case unless the point I is on the same side of the line AB as P. Therefore the angle PAB is greater than a right angle. Thus the point I cannot lie on the dotted part of the circle. Secondly. Let the rod be on the point of slipping at one point of support only. ART. 188.] EXAMPL* ON FRICTION. 139 Supposing as before that the centre of gravity is nearer B than A, the rod will slip at A and turn round B as a fixed point. Thus the friction acts along QA and the locus of P is the fixed straight line QA. But P cannot lie on the dotted part of the straight line, for if possible let it be at B. Then if AB represent F, BB must be less than F", because there is no slipping at B. But, because B lies within the circle, the ratio AB : RB is less than the ratio AP : PB, i.e. is less than F : F', and therefore BB is greater than F'. But this is contrary to supposition. Thus the string being produced will always cut the arc of the circle and the part of the straight line in one point and one point only. The frictions always tend to that point when the rod is on the point of motion. In order that the locus of P may be the dotted part of the circle it is necessary that the frictions should tend one from P and the other to P and the tension must therefore act in the angle between PA and PB produced. By the triangle of forces APB we see that the tension must act parallel to AB, and be proportional to it. Ex. 4. A lamina rests on three small supports A, B, G placed on a horizontal table ; one of these, viz. G, is smooth and the other two, A and B, are rough. A string attached to any point D, fixed in the lamina, is pulled horizontally so that the lamina is on the point of motion. If the position of the centre of gravity and the coefficients of friction are such that the limiting frictions F and F' at A and B are in the ratio BD : AD, prove that the locus of the intersection P of the string and the frictions F, F' is (1) a portion of the circle circumscribing ABB, (2) a portion of a rectangular hyperbola having its centre at the middle point of AB and also circum- scribing ABB, (3) a portion of two straight lines. Let AD = b,BD = a, then Fb=F"a. Draw LAL', HBH' perpendiculars to AB. If the lamina slip at one point only of the supports A, B, the point P lies on these perpendiculars. If the lamina slip at both A and B, we find, by taking moments about D, that sin. PAD = sin PBD. The angles PAD and PBD are therefore either supplementary or equal. The locus of P is therefore the circle circumscribing the triangle ABD, and a rectangular hyperbola also circumscribing ABD. The first locus follows also from the triangle of astatic forces considered in Art. 71. The second locus may be found by taking AB as axis of x and equating the tangents of the angles PBA and PAB - y, where y is the difference of the angles DAB and DBA. To determine the branches of these two curves which form the true locus of P 140 FRICTION. [CHAP. V. we consider the relative positions of P and the instantaneous centre I. These two points lie at opposite ends of a diameter of a circle drawn round ABP. Hence, if P lie outside the perpendiculars LL', HH', I also must lie outside. The frictions cannot then balance the tension T unless the straight line PD passes inside the angle APB. Similarly, if P lie between the perpendiculars, PD must be outside the angle APB. The straight lines LL', HH', DA, DB divide space into ten compartments. Several of these compartments are excluded from the locus of P by the rules just given. It will be convenient to mark (by shading or otherwise) the compartments in which P can lie. We now sketch the circle and the hyperbola and take only those branches which lie on a marked compartment. The figures are different according as D lies between or outside the lines LL', HH'. Ex. 5. If in the last example the limiting frictions are in any ratio, the locus of the intersection of the string and frictions is a portion of a curve of the fourth degree and of two straight lines. The proper portions, as before, are those branches which lie in the marked compartments. 189. Ex. 1. A uniform straight rod AB is placed on a rough table, and all its elements are equally supported by the table. Find the least force which, acting at one extremity A perpendicular to the rod, will move it. Let I be the length of the rod, w its weight per unit of length. Each element dx of the rod presses on the table with a weight wdx. The limiting friction at this element is therefore nwdx. If I be -4 | I H' B the centre of instantaneous rotation, the friction at each element acts perpendi- cular to the straight line joining it to I, and all these are in equilibrium with the impressed force P at A. The point I must lie in the length of the rod. For suppose it were on one side of the rod, then, rotating (as already explained) the frictions through a right angle so that they all act towards I, these should be in equilibrium- with a force P acting parallel to the rod. But this is impossible unless I lie in the length of the rod. Next, let I be on the rod, and let A 1 = z. The friction at any element H or H' acts perpendicular to the rod in the direction shown in the figure. Resolving and taking moments about A, we have I iiwdx - I /i/wdx=P, I /iwxdx= I /iwxdx. The last equation gives z*J2 = l, and the first shows that P=pW(>J2 - 1), where W is the weight of the rod. Ex. 2. Show that the rod could not begin to turn about a point I on the left of A or on the right of B. Ex. 3. If the pressure of an element on the table vary as its distance from the extremity A of the rod; and P, Q be the forces applied at A, B respectively which will just move the rod, prove that the ratio of P to Q is 2 ( ^2 - 1). Ex. 4. Two uniform equally rough rods AB, BC, smoothly hinged together H ART. 189.] EXAMPLES ON FRICTION. 141 at B, are placed in the same straight line on a rough horizontal table, and the extremity A is acted on by a force P in a direotion perpendicular to the rods. If P is gradually increased until motion begins, show that the rod AB begins to move before BG or both begin to move together according as 2 (^2 - 1) W is greater or less than W, where W, W are the weights of the rods AB, BC respectively. If both rods begin to move together, prove that the instantaneous centre of 2« 2 W rotation of AB is at a distance z from A where ^=- = 1 + 2(^/2-1) -== and I is the l* ™ w length of AB. Ex. 5. A heavy rod AB placed on a rough horizontal table is acted on at some point C in its length by a force P, in a direction making an angle o with the rod, and the force is just sufficient to produce motion. If the instantaneous centre lie in a straight line drawn through B perpendicular to the rod and be a distance from A equal to twice the length AB, prove that tan o=2 (2 - J3)I,J3 log 3. Find the position of G. ' Ex. 6. A hoop is laid upon a rough horizontal plane, and a string fastened to it at any point is pulled in the direction of the tangent line at the point. Prove that the hoop will begin to move about the other end of the diameter through the point. [Math. Tripos, 1873.] Let A be the point, AB the diameter through A. If we rotate each force round its point of application through a right angle the frictional forces will act towards the centre I of rotation Art. 184. The point / is therefore so situated that the resultant of the frictional forces (regarded as acting towards I from the elements of the hoop) is parallel to the diameter AB. It easily follows that I must lie on the diameter AB. Let us next consider the equation of moments. The point I must be so situated in the diameter AB that the moment about A of the frictions at all the elements of the hoop is zero. This condition is satisfied if I is at the end B of the diameter AB, for then the line of action of the friction at every element passes through A. It is, perhaps, unnecessary to prove that no point, other than B, will satisfy this condition. It may however be shown in the following manner. If possible let I lie on AB within the circle. Whatever point P is taken on the hoop the angle IP A is less than a right angle. Since the friction at P acts in a direction at right angles to IP, it will become evident by drawing a figure that the friction at every element acts to produce rotation round A in the same direction. The moment therefore of the frictions about A could not be zero. In the same way we can prove that I cannot lie outside the circle. Ex. 7. A thin homogeneous shell in the shape of a right circular cylinder is placed with its axis vertical on a rough horizontal plane : a string is fastened to a point of the cylinder near the plane and is pulled in a horizontal direction : show that, as the tension is increased, the cylinder will begin to rotate about a vertical axis at a distance x from the axis of the cylinder, given by /; {ccos0 + acos(0-0)} dff=0, where tan<*= a3m , a being the radius of the cylinder, and e the distance r acosB-x between the axis and the direction of the string. [Coll. Ex., 1884.] 142 FRICTION. [CHAP. V. Ex. 8. A uniform semicircular wire, of weight W, rests with its plane horizontal on a rough table,. AB is the diameter joining its ends, and C is the middle point of the arc; a string tied to C is pulled gently in the' direction CA, and the tension increased until the wire begins to move. Show that the tension at this instant is equal to 2 JinWjir. [St John's Coll., 1886.] The instantaneous axis is at B. Ex. 9. A uniform piece of wire, in the form of a portion of an equiangular spiral, rests on a rough horizontal plane ; show that the single force which, applied to a point rigidly connected with it, will cause it to be on the point of moving about the pole as instantaneous centre, is equal to the weight of a straight wire of length equal to the distance between the ends of the spiral, multiplied by the coefficient of friction. Show how to find the point. [Math. Tripos, 1888.] Ex. 10. Three equal weights, occupying the angles A, B, G of an equilateral triangle, are rigidly connected and placed upon a rough inclined plane with the base AB of the triangle along the line of greatest slope, and the highest weight A is attached by a string to a point in the line of the base produced upwards ; if the system be on the point of moving, prove that the tangent of the inclination of the plane is (2+^/3)^/^/3, where /i is the coefficient of friction. [Math. Tripos, 1870.] Suppose I not at a corner, the three frictions are then equal. Since A can only move perpendicular to OA, I must lie in OAB. Unless I lie between A and B and at the foot of the perpendicular from C on AB, the three frictions will have a component perpendicular to A B. Taking moments about I, we find the result given in the question. Next suppose I to be at the corner A. The frictions at B and G when resolved perpendicular to AB are then too great for the limiting friction at A . This supposition is therefore impossible. Ex. 11. A three-legged stool stands on a horizontal plane, the coefficient of friction being the same for the three feet ; a small horizontal force is applied to one of the feet in a given direction, and is gradually increased until the stool begins to move; show that this force will be greatest when its direction intersects the vertical through the centre of gravity of the stool. Show also that if the force when equal to twice the whole friction of the foot on which it acts, applied in a direction whose normal at the foot passes between the two other feet, causes the foot to begin to move in its own direction, the centre of gravity of the stool is vertically above the centre of the circle inscribed in the triangle formed by the feet. [Math. Tripos.] Ex. 12. A flat circular heavy disc lies on a rough inclined plane and can turn about a pin in its circumference; show that it will rest in any position if 32/i>9fl-tani, where i is the inclination of the plane to the horizon. The weight is supposed to be equally distributed over its area. [Pet. Coll., 1857.] Let TV be the weight of the disc. The origin being at the pin the friction at any element rdSdr is pWcosi .rdBdrjira?. Taking moments about the pin the result follows by integration. Ex. 13. A right cone, of weight W and angle 2o, is placed in a circular hole cut in a horizontal table with its vertex downwards. Show that the least couple which will move it is pWr cosec a, where r is the radius of the hole. The pressure Rds on each element ds of the hole acts normally to the surface of ART. 190.] A STRIN# OF PARTICLES. 143 the oone, hence, resolving vertically, fRds sin o= W. The limiting friction on each element is /iJRds, hence, taking moments about the axis of the cone, the result follows. Ex. 14. A heavy particle is placed on a rough inclined plane, whose inclination is equal to the limiting angle of friction ; a thread is attached to the particle and passed through a hole in the plane, which is lower than the particle but not in the line of greatest slope ; show that, if the thread be very slowly drawn through the hole, the particle will describe a straight line and a semicircle in succession. [Maxwell's problem, Math. Tripos, 1866.] Let W be the weight resolved along the line of greatest slope, F the friction, then F= W. As the particle moves very, slowly, the forces F, W and the tension T are always in equilibrium. As long as the hole O is lower than the particle, T is infinitely small and just disturbs the equilibrium. The particle therefore descends along the line of greatest slope. When the particle P passes the horizontal line through 0, T becomes finite. Hence T bisects the angle between F and W. The path is therefore such that the radius vector OP makes the same angle with the tangent (i.e. F) that it makes with the line of greatest slope. This, by a differential equation, obviously gives a semicircle having for one extremity of its horizontal diameter. Ex. 15. If, on a table on which the friction varies inversely as the distance from a straight line on it, a particle is moved from one given point to another, so that the work done is a minimum, the path described is a circle. [Trin. Coll.] By .using the Calculus of Variations this result follows at once from Lagrange's rule. 190. Ex. 1. Two heavy particles A, A', placed on a rough table, are connected by a string without tension and very slightly elastic. The particle A is acted on by a force P in a given direction AG making with A' A produced an angle /3 less than a right angle. As P is gradually increased from zero, will A move first or will both move together ? Let F, F' be the limiting frictions at A, A'. Suppose P to increase from zero : while P is less than F it is entirely balanced by the friction at A. The string, however nearly inelastic it may be, has no tension until A has moved. Let P be a little greater than F ; take AL to represent P and draw LMM parallel to AA' ; with centre A and radius F describe a circle cutting LMM' in M and M', then LM represents the tension of the string. Of the two inter- sections M, M', the nearest to L is chosen, for this makes the friction at A act opposite to P when P=F. As P gradually increases M travels along the arc CH. The equilibrium of the particle A becomes impossible when LMM' does not cut the circle, i.e. when M reaches H. The particle A' borders on motion when the tension LM becomes equal to F'- Now HK=Faotp. Hence the particle A moves alone if Fcot/3<2<" but both move together if .Foot fi>F'. When the limiting frictions F, F' are equal, and /3 is less than half a right angle, both particles move together. One friction acts along AA' and the other makes an angle /3 with the force P. 144 FRICTION. [CHAP. V. In this solution the point M' has been exoluded by the principle of continuity, though statically A would be in equilibrium under the forces represented by AL, LM', M'A. If the string AA' had a proper initial tension, but balanced by frictions at A and A' together with an initial force P along AC, then M ' would be the proper intersection to take. Ex. 2. Two weights A and B are connected by a string and placed on a horizontal table whose coefficient of friction is /*. A force P, which is less than fiA+pB, is applied to A in the direction BA, and its direction is gradually turned round an angle 6 in the horizontal plane. Show that if P be greater than ju Ja* + B 2 , then both A and B will slip when cos0= 2 pp ' but if P tie less than n ^/a^+B 2 and greater than /xA, then A alone will slip when sin 8= -=- . [Math. Tripos.] Ex. 3. The n particles A , A v . . . , A K - V of equal weights, are connected together, each to the next in order, by n - 1 strings of equal length and very slightly elastic. These are placed on a rough horizontal plane with the strings just stretched but without tension, and are arranged along an arc of a circle less than a quadrant. The particle A n _ x is now acted on by a force P in the direction A n _ 1 A n , where A n is an imaginary (m + l)th particle. Supposing P to be gradually increased from zero, find its magnitude when the system begins to move. Let us suppose that any two consecutive particles A m and A^^ both border on motion. Let m be the angle the friction at A m makes with the chord A m+1 A m . Let T m be the tension of the string A m A m+1 . Let jS be the angle between any string and the next in order. Let F be the limiting friction at any particle. Resolving the forces on the particles A m and A m+1 perpendicularly to A m _^A m and A m+1 A m+2 respectively, we find T m sinp=Fs,m( mH .= m +p. Suppose that all the particles from A v to A n ^ 1 border on motion and that t p-i=° 5 we have then 4> p =0, t/>p + . 1 =p, and in general M 1? B N F transference its virtual works are F . AM and F . BN respectively. Since A'B makes with AB an infinitely small angle whose cosine may be regarded as unity, we have MN equal to A'B'. Hence, if the distance between the two points of application remain unaltered, ART. 195.] PROOF 0» THE PRINCIPLE. 149 i.e. AB = A'B', we have BN = AM. It immediately follows that F.AM = F.BN. Thus in all changes of forces into other forces consistent with the principles of statics, the work of the forces due to any given small displacement is unaltered. 195. We may now apply this result to a system of forces Pi, P 2 &c. acting on a free rigid body. All these forces can be reduced to a force R acting at an arbitrary point 0, and a couple G, Art. 105. By what precedes the virtual work of the forces P u P 3 &c. due to any displacement is equal to the virtual work of R and G. If the forces P u P 2 &c. are in equilibrium, both R and G are zero, Art. 109. Hence the virtual work of Pj, P 2 &c. for any displacement is zero. Conversely, if the virtual work of P I; P 2 &c. is zero for all displacements, then the virtual work of R and G is zero. We shall now show that this requires that R and G should each be zero. First let the body be moved parallel to itself through any small space Sr in the direction in which R acts. The virtual work of the force R is RSr. Let AB be the arm of the couple and let the forces act at A and B. Since equal and parallel displacements AA', BB' are given to A and B, while the forces acting at A and B are equal and opposite, it is evident that the works due to the two forces cancel each other. The work of the couple G is therefore zero. Hence the sum of the works of R and G cannot vanish unless R = 0. Next let the body be turned through a small angle 8a> round a perpendicular drawn through to the plane of the couple, and let this rotation be in the direction in which the couple urges the body. Let bisect the arm AB and let the forces of the couple be ± Q. Each of the points A and B receives a displace- ment equal to £ ABSco in the direction of the force acting at that point. The sum of the works due to these two forces is therefore AB . QSa>, i.e. GBca. Since the point of application of R is not displaced, the virtual work of R (even if R were not zero) is zero. Hence the sum of the virtual works of R and G cannot vanish unless G = 0. It immediately follows that the body is in equilibrium. 150 THE PRINCIPLE OF VIRTUAL WORK. [CHAP. VI. 196. On the forces which do not put in an appearance in the equation of virtual work. When the body is not free but can move either under the guidance of fixed constraints or under the action of other rigid bodies it becomes necessary (as explained in Art. 193) to determine what actions and reactions do not appear in the general equation of virtual work. We cannot make an exhaustive list, but we may make one which will include those cases which commonly occur. I. Let two particles A, B of the system act on each other by means of forces along AB, then if the distance A B remain invaria- ble for any displacement, the virtual works of the action and the reaction destroy each other. This follows at once from Art. 194, for the force at A may be transferred to B. The two equal and opposite forces acting at B have then the same displacement. Hence their virtual works are equal and opposite. For example, if the points A, B are connected by an inelastic string, the tension does not appear in the equation of virtual work. II. If any body of the system is constrained to turn round a point or an axis fixed in space, the virtual work of the reaction at this point or axis is zero. This is evidently true, for the displacement of the point of application of the force is zero. III. Let any point A of a body be constrained to slide on a surface fixed in space. If the surface is smooth, the action R on the point A of the body is normal to the surface. Let A move to a neighbouring point A', then A A' is at right angles to the force. The work by Art. 68 is therefore zero. If the surface is rough, let F be the friction. This force acts along A 'A, and its work is — F . A A'. This is not generally zero. IV. If any body of the system roll without sliding on a fixed surface, the work of the reaction is zero. If this ia not evident, it may be proved as follows. In the figure the body DAE rolls on the fixed surface MABN and takes a neighbouring position D'BE'. ART. 197.] FORCES VmiCB. DO NO WORK. 151 The plane of the paper represents a, section of the surfaces drawn through their common normal at A, and contains the elementary arc AB of rolling. In this displacement the point A of the body begins to move along the common normal and arrives at A'. If we replaoe the curves DAE, MAB by their circles of curvature, we know (since the arcs AB, A'B are equal) that AA'-.AB 2 is half the sum of the opposite curvatures. Assuming these curvatures to be finite, it follows that AA' is of the same order of small quantities as AB 2 , i.e. AA' is of the second order of small quantities. Hence, when we retain only terms of the first order, as in the principle of virtual work, we may treat the rolling body as if it were turning round a point A fixed (for the instant) in space. It follows therefore from the result of the last article that, when a body rolls on a fixed surface, which may be either rough or smooth, the virtual work of the reaction is zero. V. If the surface on which the body rolls is another body of the system, the surface is moveable. But we may show that, if both bodies are included in the same equation of virtual work, the mutual action does not appear in that equation. To prove this we notice that we may construct any such displacement of the two bodies (1) by moving the two bodies together until the body MABN assumes its position in the given displacement, and then (2) rolling the body DAE on the body MABN, now considered as fixed, until DAE also reaches its final position. During the first of these displacements the action and reaction at A are equal and opposite, while their common point of application A has the same displacement for each body. Their virtual works are therefore equal and opposite, and their sum is zero. During the second displacement the body DAE rolls on a fixed surface, and the virtual work of its reaction is zero. See Art. 65. 197. Work of a bent elastic string. If the points A, B are connected by an elastic string, it may be necessary to know what the work of the tension is when the length is increased from I to I + dl. We shall show that, whether the string connecting A and B is straight, or bent by passing through smooth rings fixed or moveable or over a smooth surface, the work is - Tdl. For the sake of greater clearness we shall consider the cases separately. (1) Let the string be straight. Referring to the figure of Art. 194, the virtual work of the tension at "A is + T . AM. The positive sign is given because the tension acts at A in the direction AB and the displacement AM is in the same direction, Art. 62. The work of the tension at B is - T.BN. The sum of these two is - T (A'B' - AB) i.e. - Tdl. 152 THE PRINCIPLE OF VIRTUAL WORK. [CHAP. VI If the action between A and B is a push R instead of a pull T, the same argu- ment will apply but we must write - R for T, so that the virtual work is Rdl. If the action between A and B is due to an attractive or repulsive force F the result is still the same ; the virtual works are - Fdl or + Fdl according as the force F is an attraction or a repulsion. (2) Suppose the string joining A and B is bent by passing through any number of small smooth rings G, D &o. fixed in space. Taking two rings only as sufficient for our argument, let these be C and D. Let A, B be displaced to A', B\ and let A'M, B'N be perpendiculars on A C and DB. The whole length I of the string is lengthened by BN and shortened by AM, hence dl=BN- AM. The tension T being the same throughout the string, the work at A is T .AM, that at B is— T .BN. Exactly as before, the whole work is the sum of these two, i.e. - Tdl. (3) Suppose the rings G, D &c, through which the string passes, are attached to other bodies of the system. The rings themselves will now be also moveable. Supposing all these bodies to be included in the same equation of virtual work, the system is acted on by the following forces, viz. T at A along AC, T at C along GA, T at G along CD, T at D along DC and so on. By what has just been proved, the work of the first and second of these taken together is - Td (AC), the work of the third and fourth is - Td (CD) and so on. Hence, if I be the whole length of the string, viz. AC + CD + &C, the whole work is - Tdl. In all these cases we see that, if the length of the string is unaltered by the dis- placement, the tension does not appear in the equation of virtual work. (4) Let the string joining A and B pass over any smooth surface, which either is fixed in space, or is one of the bodies to be included in the equation of virtual work. Each elementary arc of the string may be treated in the manner just explained. The work done by the tension is therefore as before equal to - Tdl. In order not to interrupt the argument, we have assumed that the tension of a string is unaltered by passing over a smooth pulley or surface. To prove this, let us suppose the string to pass over any arc BC of a smooth surface. Any element PP' of the string is in equilibrium under the action of the tensions at P, P' and the normal reaction of the smooth surface. The resolved part of these forces along the tangent at P must therefore be zero. Let T, T' be the tensions at P, P', df the angle between the tangents at these points, and let ds be the length of PP'. Supposing the pressure per unit of length of the string on the surface to be finite and equal to R, the pressure on the arc PP' is Eds. The resolved part of this along the tangent at P is less than lids sin dip, and is therefore of the second order of small quantities. The difference of the resolved parts of the tensions is T- I" coa dip, which, when small quantities of the second order are neglected, reduces to T-T. Since this must be zero, we have T=T'. Taking a series of elements of the string, viz. PP', P'P" &c, it immediately follows that the tensions ART. 199.] ONE SIBED CONSTRAINTS. 153 at P, P', P" &c. are all equal, i.e. the tension of the string is the same throughout its length. If the surface were rough, this result would not follow, for the frictions must then be included in the equation of equilibrium formed by resolving along the tangent. We may also prove the equality of the tensions by applying the principle of virtual work to the string BC. Sliding the string without change of length along the surface, we have T . BB'=T. CC. Hence T= 2". When the surface is a rough circular pulley which can turn freely about a smooth axis, and the string lies in a plane perpendicular to the axis, we can prove the equality of the tensions by taking moments about the axis. Let the string be ABCD and let it touch the cylinder along the arc BG. Let T, T' be the tensions of AB, CD, r the radius of the cylinder. Taking moments about the axis, we have Tr = TV. This gi ves T = T. 198. In the preceding arguments we have tacitly assumed that the pressures which replace the constraints are finite in magnitude. If this were not true it is not clear that the virtual work would be zero. It is not enough to make a product P . dp vanish that one factor viz. dp should be zero, if the other factor P is infinite. Such cases sometimes occur in our examples when we treat the body under consideration as an unyielding rigid mass. But in nature the changes of structure of the body cannot be neglected when the forces acting on it become very great. The displacements are therefore different from those of a rigid body. 199. One sided constraints. In some constraints a point A of the body merely rests against one side of a surface, so that when all possible displacements are taken account of, the point A may be moved off the surface. In such cases the enunciation of the principle, as given in Art. 191, must be modified. Suppose we replace the action of the constraining surface on A by a force R, ; the sum of the virtual works will be zero for all displacements, provided we include the virtual work of R in the equation. Now since the body presses against the surface, the reaction R will act towards that side on which the body is free to leave the surface. For all displacements on that side the virtual work of R is positive. If then we wish to include displacements which take the body off the surface and yet wish to omit the reaction, we must state as the condition of equilibrium, that the virtual work of the impressed forces is zero or negative for all possible displacements. It does not .appear that this mode of enunciating the principle offers many advantages. The chief use of the principle is indeed to find the .conditions of equilibrium free from any unknown reactions, but we are also able so to use it as to find those 154 THE PRINCIPLE OF VIRTUAL WORK. [CHAP. VI. reactions. To effect the former purpose we give the system all necessary displacements which do not separate the point A of the body from the surface on which it rests, and in each case we equate the sum of the virtual moments to zero. In all the equations thus obtained the reaction R does not appear. To effect the latter purpose we displace the system in such a way that the point A of the body is moved off the surface on which it rests. Including the virtual work of the reaction R and still equating the sum of the virtual works to zero, we have an equation to find R. 200. Converse of the principle of virtual work. We shall now prove the converse principle of virtual work for a system of bodies. The system being placed at rest in some position, it is given that the work of the external forces is zero for all small displacements which do not infringe on the constraints. We are to prove that the system is in equilibrium. If the system is not in equilibrium it will begin to move. Let us then examine all the ways in which the system could begin to move from its position of rest. Some one way having been selected, it is clear that by introducing a sufficient number of smooth con- straining curves we can so restrain the system that it cannot move in any other way. Thus if any point of one of the bodies would freely describe a curve in space, we can imagine that point attached to a small ring which can slide along a rigid smooth wire, whose form is the curve which the point would freely describe. The point is thus prevented from moving in any other way. The reaction of this smooth curve has been proved to have no virtual work. It is also clear that these constraining curves in no way alter the work of the external forces during the displacement of the body. In order to prevent the system from moving from its initial position it will now only be necessary to apply some force F to some one point A in a direction opposite to that in which A would move if F did not act. The forces of the system are now in equili- brium with F. Let the system receive an arbitrary virtual dis- placement along the only path open to it. In this displacement let the point A come to A'. Then the work of the forces plus the work of F is zero. But it is given that the work of the forces is zero for every such displacement, hence the work of F is zero, But this work is — F.AA', and since A A' is arbitrary it im- ART. 203.] CONVERSE«OF THE PRINCIPLE. 155 mediately follows that F must be zero. Thus no force is required to prevent the system from moving from its place of rest along any selected path. The system is therefore in equilibrium. Treatise on Natural Philosophy, Thomson and Tait, 1879, Art. 290. 201. Initial motion. Let us imagine a system to be placed at rest, and yet not to be in equilibrium under the action of the given external forces. We shall show that the system will so begin to move * that the work of the forces in the initial displacement is positive. The proof of this is really a repetition of the argument already given in Art. 200. If the system begin to move from the position of rest in any given way, we constrain it to move only in that way. If F be the force acting at A which will prevent motion, we find as before that the work of the forces plus that of F is zero. But F must act opposite to the direction in which A would move if F were not applied, hence its work is negative. It follows that the work of the impressed forces in this displacement is positive. 202. It might be supposed from this result, that it would be sufficient to ensure equilibrium that the work of the forces should be negative instead of zero for all displacements. For then there is no displacement which the system could take from its state of rest. If however the work of the forces is negative for any one displace- ment, it must be positive for an equal and opposite displacement, i.e. one in which the direction of motion of every particle is reversed. To exclude therefore all dis- placements which make the work positive, it is necessary that the work should be zero for all displacements. The case in which one displacement is possible, but its opposite is impossible, arises when a point of a body is constrained to rest on one side of some fixed surface. The point can receive a displacement off the surface on one side but not on the other. But this case has already been considered in Art. 199. 203. To deduce the equations of equilibrium from the principle of work. The equations of equilibrium of a system are really equivalent to two statements, (1) the sum of the resolved parts of the forces in any direction for each body or collection of bodies in the system is zero, (2) the sum of the moments about any or every straight line is zero. The equations of equilibrium of a system in one plane have been obtained in Chap, iv., Arts. 109 — 111. The corresponding equations of a system in space will be given at length in a later chapter. But to avoid repetition they are included in the following reasoning. See also Arts. 105 and 113. * Dynamical proof. When a system starts from a position of rest, it is proved in dynamics that the semi vis viva after a displacement is equal to the work done by the external forces. Now the vis viva cannot be negative, because it is the sum of the masses of the several particles multiplied by the squares of their velocities. It is therefore clear that the system cannot begin to move in any way which makes the virtual work of the forces negative. 156 THE PRINCIPLE OF VIRTUAL WORK. [CHAP. VI. We have now to deduce these two results from the principle of work. As before, let P u P 2 &c. be the forces, A lt A* &c. their points of application, (c^, &, 7^, (a 2 , /3 2 , 7 2 ) &c. their direction angles. Let the body or collection of bodies receive a linear displacement parallel to the axis of x through a small space dx. V Fig. 1. Fig. 2. Then if A be moved to A', AA' = dx, (Fig. 1), and the projection AN on the line of action of P is dx cos a. Hence, by the principle of work, Pj cos ai dx + P 2 cos ck dx + . . . = 0. Dividing by dx, this gives the equation of resolution, viz. P 1 cos <*! 4- P 2 cos a 2 + . . . = 0. In this equation all the reactions on the special body considered due to the other bodies are to be included. To find the sum of the moments of the forces about any straight line, say the axis of z, let us displace the special body considered round that axis through an angle dm. First let the forces act in the plane of xy, and let p lt p 2 &c. be the perpendiculars from the origin on their respective lines of actions. Thus in Fig. 2, OM =p. The displacement AA' of A due to the rotation is 0A . dm. The projection of this on the line of action of P is OA dm sin 0AM, i.e. pdm. Hence by the principle of work Pip 2 dm + P 2 pz dm + . . . = 0. Dividing by dm, we have the equation of moments, viz. P lPl + P 2 p 2 + ...=0. Next, let the forces act in space. We first resolve each force parallel and perpendicular to the axis about which we take moments. The resolved parts of P are respectively P cos 7 and P sin 7. The displacement A A' of its point of application due to a rotation round z is perpendicular to the axis of z. The work of the first of these components is therefore zero. The second component is ART. 205.] EQUATIOH6 OF EQUILIBRIUM. 157 parallel to the plane of xy, and its work is found in exactly the same way as if it acted in the plane of xy. If p be the length of the perpendicular from on the projection on xy of its line of action, the work is P sin ypdto. We therefore find as before P 1 sm.'y 1 p 1 +P 1l siny i p 2 + ... = 0, which is the usual equation of moments. 204. Combination of equations. The equations of equilibrium of each of the bodies forming a system having been found by resolving and taking moments, we can combine these equations at pleasure in any linear manner. For example we might multiply by X an equation obtained by resolving parallel to some straight line x, and multiply by n another equation obtained by taking moments about some straight line &. Adding the results, we get » new equation which may be more suited to our purpose than either of the original ones. We shall now show that this derived equation might be obtained directly from the principle of work by a suitable displacement. Suppose both the equations combined as above to be equations of equilibrium of the same body. Let these be written in the form 2Pcos, ' the angles which the radii CA, CA' make with the horizontal diameter Cx. Let a be the radius of the sphere, 6 the distance between the tubes. If y, y' be the altitudes above Cx of the centres of gravity of the rods, we have by the principle of work, - Wdy-W'dy' = 0. The negative sign is used because the y's are measured upwards opposite to the direction in which the weights are measured. Since y and y ' differ from a sin 0, and a sin $' by constants, viz. half the lengths of the rods, we find But by geometry TPcos 4>d

'd' - 0. «cos0 + acos0'=6. Differentiating the latter equation, and eliminating d : dtp/, we find Weot &c. their values in the position I., and the upper limit by giving the same coordinates their values in the position II. ART. 212.] THE ^DRK FUNCTION. 163 When the expression (2) is a perfect differential, this integration can be effected without knowing the route by which the system travels from the one position to the other. The integral W is a function of the upper and lower limits, and will thus depend on the initial and final position of the system and not on any intermediate position. It follows that the work due to a displacement from one given position to another is the same, whatever route is taken by the system, provided always none of the geometrical constraints are violated. 211. When the forces are such that the expression %Pdp is a perfect differential, the forces are said to form a conservative system of forces. Suppose we select any one position of the system of bodies as a standard, and let this position be defined by the values of the coordinates = lt = fa, &c. Then taking this standard position as the lower limit of the integral and any general position as the upper limit, we have W=jtPdp = F(0,fa&c.)-F(0 u fa,&c.); when it is not necessary to make an immediate choice of a standard position we write the integral in its indefinite form, viz. W = F(0, fa &c.) + G The function W, particularly when used in the indefinite form, is often called- the force function, or work fwnction. 212. Sometimes the upper limit is made the standard position and the general position the lower limit. Suppose this standard to be determined by the values 6 = 6 it cj>=fa, &c; the integral il P P O TY1 P ci V=F(0 2 ,fa,&c.)-F(0,fa&c). This is usually called the potential energy of the forces with reference to the position defined by 9 = 2 , = fa, &c. If the two standards of reference were identical, we should have W=— V. But both these standards are seldom used in the same problem. ( In every case that standard of reference is generally chosen which is most suitable to the particular problem under discussion. We notice that W+ V is the work of the forces as the system moves along any route from the position {0 X , fa, &c.) to the 11—2 164 THE PRINCIPLE OF VIRTUAL WORK. [CHAP. VI. position (# 2 , <£ 2 , &c), and these being fixed, the sum is constant for all positions of the system of bodies. 213. Maximum and Minimum. Suppose the system to be in a position of equilibrium. We then have dW=0 for every virtual displacement, so that W is a maximum, a minimum, or stationary. The last alternative represents the case in which the evanescence of the first differential coefficients does not indicate a true maximum or minimum. We have therefore another method of finding the positions of equilibrium of a system. We regard the work function as a known . function of the coordinates 8, ,...) + 0. To find the positions of equilibrium we use the rules given in the differential calculus to find the values of 8, <£> &c. which make W a maximum or minimum. If the coordinates 8, &c. are all independent, we make the differential coefficient of W with regard to each of the variables equal to zero. Of course this is equivalent to giving the system the geometrical displacements indicated by varying 8, &c. in turn, and equating the virtual work in each case to zero. But the process is analytical instead of geometrical, and this has sometimes great advantages. If there are several coordinates connected by equations of constraint we may use the method of indeterminate multipliers. More generally we may use any method given in the differential calculus by which we are accustomed to simplify and shorten the process of making a given function a maximum or minimum. 214. Stable and Unstable equilibrium. It should be noticed that it is necessary and sufficient for equilibrium that the work function If be a maximum, a minimum, or stationary. There is however an important distinction between these three cases. Suppose the system is in equilibrium in such a position that W is a true maximum, i.e. W is decreased if the system is moved into any neighbouring position which is consistent with the con- straints. Let the system be actually placed at rest in any one of these neighbouring positions. Not being in equilibrium in this ART. 217.] STABILITY OF EQUILIBRIUM. 165 new position it will begin to move. By Art. 201 it must so move that the initial work of the forces is positive, i.e. it must so move that W increases. The system therefore tends to approach closer , to its original position of equilibrium. The original position is therefore said to be stable. Suppose next the system is in equilibrium in such a position that W is a true minimum, i.e. W is increased if the system is moved into any neighbouring position. Let the system be placed at rest in one of these neighbouring positions, then, by the same reasoning as before, it will begin to move on some path which will take it further off from its original position of equilibrium. The equilibrium is then said to be unstable. Lastly, suppose the system is in equilibrium in such a position that W is neither a true maximum nor a true minimum, i.e. W is decreased when the system is moved into some neighbouring positions and increased when the system is moved into some others. By the same reasoning as in the two preceding cases the equili- brium is stable for some displacements and unstable for others. According to the definition given in Art. 75 this state of equilibrium is to be regarded as on the whole unstable. 215. We have only considered how the system begins to move, and not whether it may afterwards approach or recede from the position of equilibrium. The question of the stability or instability of a position of equilibrium cannot be properly discussed from a statical point of view. The motion of a system about its position of equilibrium is evidently a question of dynamics. It is better to postpone its consideration until the reader takes up that subject. The general result however agrees with what has been proved above. 216. Instead of using the work function we may use the potential energy. Since their sum W + V is constant, the general results are just reversed. When the system is placed at rest in any position other than one of equilibrium, it begins to move so that the potential energy decreases. In a position of equilibrium the potential energy is a maximum, a minimum, or stationary. The equilibrium is stable or unstable according as the potential energy is a true minimum or maximum. 217. We have supposed in what precedes that none of the 166 THE PRINCIPLE OF VIRTUAL WORK. [CHAP. VI. neighbouring positions are also positions of equilibrium. It is of course possible that W should be constant for two consecutive positions of the system of bodies, and yet (say) greater than when the system is moved into any other neighbouring position. In such a case the equilibrium is neutral for the displacement from one of the consecutive positions to the other and stable for all other displacements. Various cases may occur. For example, the equi- librium may be neutral for more than one or for all displacements from a given position of equilibrium ; or again W may be constant for all positions defined by some relations between the coordinates, and yet (say) a maximum for all displacements from this locus. We then have a locus of positions of equilibrium, each of which is stable for all displacements which do not move the system along the locus. In a system with two coordinates 0, , we could regard W as the ordinate of a surface whose x and y coordinates are 6 and <$. Every geometrical peculiarity connected with the maximum and minimum ordinates of such a surface has a corresponding statical peculiarity in the positions of equilibrium of the system. 218. Altitude of the centre of gravity a maximum or minimum. There is one important application of the theorem on virtual work of which much use is made. Let gravity be the onlyi external force acting on the system. Let z lt z 2 &c. be the altitudes! above any fixed horizontal plane of the several heavy particles, and z the altitude of their centre of gravity. If m^ m 2 &c. be the masses of these particles, we have zXm = "Zmz. If g be a constant, so that mg represents the weight of the mass m, the virtual work of the weights is d W = — %mgdz = — gtmdz. The work function is therefore W =— zgXm+G. This is a true maximum or a true minimum, according as z is at the least or greatest height. We deduce the following theorem. Let a system of bodies be under the influence of no forces but their weights, together with such mutual reactions as do not appear in the equation of virtual work, and let it be supported by frictionless reactions with other fixed surfaces, or in some other way by forces which do not appear in the ART. 221.] STABILITY OF EQUILIBRIUM. 167 equation of virtual work ; the possible positions of equilibrium may be found by making the altitude of the centre of gravity of the system above any fixed horizontal plane a maximum, minimum, or stationary. The equilibrium will be stable or unstable according as the altitude of the centre of gravity is or is not a true minimum. 219. Alternation of stable and unstable positions. Suppose the constraints are such that the system moves with one degree of freedom. Then as the system moves through space the centre of gravity will describe some definite curve. The positions in which the ordinate is a true maximum and a true minimum must evidently occur alternately. It follows that the truly stable and truly unstable positions of equilibrium occur alternately. 220. Analytical method of determining the stability of a system. To show how this theorem may be used to find positions of equilibrium in an analytical manner, let us suppose as an example that the system has one degree of freedom. We first choose some convenient quantity by which the position of the system is fixed, and which is therefore called its coordinate. Let this be called 6. Then the value of when the system is in equilibrium is the quantity to be found. Let z be the altitude of the centre of gravity of the system above some fixed horizontal plane. From the geometry of the question we now express z in terms of 8. The required value of 6 is then found by making dzjdd — 0. To deter- mine whether the equilibrium is stable or unstable, we differentiate again and find d^z/dd 1 . If this second differential coefficient is positive, when.0 has the value just found, the equilibrium is stable. If negative, the equilibrium is unstable. If zero we must examine the third and higher differential coefficients of z, following the rules given in the differential calculus to discriminate whether a function of one independent variable is a maximum or minimum. If the coordinate 6 cannot vary from = — oo to = + oo , it may itself have maxima and minima. It must be remembered that these values of 6 may lead to maxima and minima values of z other than those given by the ordinary theory in the differential calculus. r\i/ 221. Examples. Ex. 1. A uniform heavy rod AB rests against a smooth ' vertical wall and over a smooth peg 0. Find the position of equilibrium, and deter- mine whether it is stable or unstable. 168 THE PRINCIPLE OF VIRTUAL WORK. [CHAP. VI. Let the length of the rod be 2a and let the distance of G from the wall be 6. Let the inclination of the rod to the wall be 0. Taking the horizontal through G for the axis of x, we find for the altitude z of the centre of gravity z = a cos - b eot 0, 9= - a sin + b (sin 0) -2 , = - a cos 0-2b (sin 0)~ 3 cos 0. Putting dzjd0 = O, we find that in the position of equilibrium sin 3 0=6/a. Since d 2 2/d0 2 is negative the equilibrium is unstable. Ex. 2. A frustum of a right cone is suspended from a smooth vertical wall by a string, having one extremity attached to a point in its base, and the frustum is in equilibrium with one point of the base in contact with the wall. If the length I of the string is equal to the diameter of the base and the centre of gravity is at a distance hi from the base, show that the tangent of the inclination of the string to the vertical is §£. Is the equilibrium stable ? 222. Ex. A heavy body can move in a vertical plane in such a manner that two of its points, vis. A and B, are con- strained to slide, one on each of two equal and similar smooth curves whose equations are respectively x='f(y) and x= -f (y), y being vertical. The perpendicular on the chord AB drawn from the centre of gravity G bisects AB in E. Show how to find the positions of equilibrium, and determine whether the position in which AB is horizontal is stable or not. Let AB = 2a, GE = h. Let be the inclination of AB to the horizon and (xy) the Cartesian coordinates of G. Then since the points A, B lie on the given curves we find x+h sin + a cos 0=f(y-h cos + a sin 0) x + h sin - a cos 6= -f{y- h cos - Eliminating x, we have 2acos9=/(2/-fccos0 + asin0)+/(2/-7icos0-asin0) (2). Differentiating this and putting dy/d0 = O, we find - 2a sin 0= /' {y - h cos + a sin 0) (ft sin + a cos 0) +f'(y-h cos —a sin 0) (7s sin - a cos 0) Joining this equation to (1) and (2) we have three equations to find x, y, 0. It is clear that (3) is satisfied by = 0, this therefore is one position of equilibrium. To determine if this horizontal position is stable, we differentiate (2) twice to find cPyjdS 2 . We easily find after reduction sin 0) | ■ a sin 0)j ' .(1). (3). .(4). dhj a+ay(y-h) . h d0>- f'(y-h) +n The position of equilibrium is stable or unstable according as the right hand side is negative or positive. ART. 223.] STABILirS»OF EQUILIBRIUM. 169 We may obtain a geometrical interpretation for the equation (4) in the following manner. The straight line AB being in its horizontal position, let n be the length of the normal to the curve at either ior£ intercepted between the curve and the axis of y. Let p be the radius of curvature at A or B, estimated positive when measured from the curve in the direction of n, and let

writing a for x and y - h for y we find that dB* apta,n

  • '), where p, q are the perpendiculars drawn from the origin on the tangents, and u, a' are the angles which these perpendiculars make with opposite sides of the axis of x, y being vertical as before. The centre of gravity Gf lies in the bisector of the angle C at a distance h from either of the straight lines GA, CB. Show how to find the incli- nation of CGf to the vertical when the body is in equilibrium, and determine whether the position in which CG is vertical is stable or unstable. Let o be the angle GG makes with either CA or CB, and 9 the inclination of GG to the vertical. Let y be the altitude of G. We first show by geometrical con- siderations that y sin 2o= {p - ft) cos (9 - o) + (q - h) cos (8 + a). Eemembering that p=f{8 + a) and q=f(a-8) we have, by equating dy\d8 to zero, an equation to find 9. In the position in which CG is vertical 8=0, hence p=q. Differentiating a second time, we have sin 2a dhj ( ft -* + «l) o dp . cosa + 2 -^sin a. We may obtain a geometrical interpretation of this value of d i yjd8' i . The body being in the position in which GG is vertical, the straight line CA will touch one of the curves in some point P. Let /> be the radius of curvature of the curve at P, £ the horizontal abscissa of P. We may then show that d 2 u aina-^L = h + p— 2£ sec a. d8 2 r The equilibrium is stable or unstable according as the value of d^y/tiB 1 is positive or negative. If the value is zero, we must differentiate a second time. 226. Examples of atoms. Some good examples of the method of using the work function to determine questions of stability are supplied by Boscovich's theory of atoms. Almost all the following results are enunciated by Sir W. Thomson in an interesting paper contributed to Nature, October 1889. It js enough for our present purpose to say that Boscovich supposed matter to consist of atoms or points between which there is repulsion at the smallest distance, attraction at greater distances, repulsion at still greater distances, and so on, ending with attraction according to the Newtonian law for all distances for which this law has been proved. Boscovich suggested numerous transitions from attraction to repulsion and vice versa, but for the sake of simplicity, we shall here consider problems which involve only one change from repulsion to attraction. Suppose then that the mutual force between two atoms is repulsive when the distance between them is less than p, zero when it is equal to p, and attractive when greater than p. With this supposition we shall consider the stability of the equili- brium of some groups of atoms. 172 THE PRINCIPLE OF VIRTUAL WORK. [CHAP. VI. 227. Ex. 1. Three particles, whose masses are m, to', m" repel each other so that the force between m and m' is F= - mm' (r -p)*- 1 where n is an even integer. The particles are in equilibrium when placed at the corners of an equilateral triangle each of whose sides is equal to p. Show that the equilibrium is stable. The term of the work function W corresponding to F is fFdr= - (r-p) n . When the atoms are displaced, let the three sides of the triangle \>ep+x,p + y,p + z. We have by Art. 208, n ( G - W) = m'm"x u + m"my n + mm' si*. The equilibrium is stable or unstable according as W is a maximum or a minimum, i.e. according as the right hand side is a minimum or a maximum. But, since n is even, the right hand side is a minimum when x, y, z are each zero ; for these values make the right hand side zero and all others make it greater than zero. The equilibrium is therefore stable. We have taken the law of force to be a single power of r -p, but it is clear that the same reasoning will apply if the law of force is expressed by several terms with different odd powers. Even greater generality may be given to the law, for it is sufficient that the lowest power should be odd. In just the same way we may prove that a group of four particles placed at the corners of - regular tetrahedron, each of whose edges is equal to p, is a stable arrangement. Ex,. 2. Three equal atoms A, B, G are placed in equilibrium in a straight line. Supposing the force of repulsion to be F= -fi.(r p)"^ 1 determine if the configuration is stable or unstable. It is clear that in the position of equilibrium the distances AB, BO are each less than the critical distance^, while AG is greater than p. Let AB and BC be each equal to a. As we are only concerned with relative displacements, let A be fixed. Let B', C" be the displaced positions of B, C ; let (xy) be the coordinates of B' referred to B, and (x'y 1 ) those of C referred to G. If r=AB, we have r=)(a + xf + y i Y=a-Yx + ^- + &o. .-. (r-p) n =(a-p) n + n(a-p) n - 1 (*+§-)+'' ^ir («-i>) n ~ 2 a 2 + &c. If we replace (xy) by (x 1 -x, y' - y), this expression gives the value of ()■" -p) n where r"=B'C. If instead we replace (xy) by (x'y') and write la for a, the expression gives the value of (/ -p) n , where r'=AG. Taking all these expressions, we have as before " (G - W) = (r -p) n + (r' -p) n + (f' -p) n =n(a-pr-> |a' + )"- z a:' 2 + &c., where all the constant terms have been absorbed into one constant, viz. G. ART. 229.] ON WUMEWORKS. 173 To find the position of equilibrium, we make W a maximum or a minimum, i.e. WepUt ^ = °' di' =0, 1|= ' % =0 - These give (o-i>)«-i + (2a-p)»-i = 0. Hence, since n - 1 is odd and p lies between a and 2- It appears that, if the length of any rod, not acted on by the external forces, can be increased without undoing the frame the reaction along that rod is determinate. For example, if there are no external forces acting on the frame, the reaction along any such side is zero. 233. If the rod A^A^ is acted on by some of the external forces 176 THE PRINCIPLE OF VIRTUAL WORK. [CHAP. VI. the reactions at the corners A lt -4 2 do not necessarily act along the length of the rod. We may reduce this case to the one already considered in the last article by replacing each of these forces by two parallel forces, one acting at each extremity of the rod. This method has been explained in Art. 134. We may also find the reactions by a more direct process. Let i?!2, 8 12 be the components of the action at the corner A} of the rod A^A^, resolved along and perpendicular to the length of the rod. In the same way i2 21 , $ 21 are the components at the corner -4 2 of the same rod. Let us remove the rod A X A^ and replace its effects on the rest of the frame by applying at its extremities the forces .R^, 8 U and i? 21 , 8 21 . Let R n , Rn be regarded as positive when the rod is in a state of thrust. Let the system be so deformed that the length of the side A t A^ is increased by dl n , while the corner A% and the direction in space of that side are unaltered. The virtual work of the reactions R 21 , 8 n and $i 2 in this displacement is evidently zero. Let dW be the virtual work of the external forces which act on the system, excluding the rod A 1 A a , then R^dlu + dW^O. To find the reaction 8 12 a different displacement must be given to the system. The external forces which act on the rod A^A t having been removed, the remaining external forces are not in equilibrium. Their virtual work for a displacement of the frame as a whole is not necessarily zero. Keeping A 2 as before fixed in space and not altering the length £ 12 , let us turn the frame round an axis perpendicular to the plane containing A 2 and the force S 12 . If d0 be the angle of displacement and dW the work of the forces, we have S 12 d0+dW=O. By giving the frame these two deformations the reactions .R^ and /S 12 at the corner A 1 can be found. If the frame be perfectly free, the deformation necessary to find 8 12 can always be given. The deformation necessary to find R u requires that the length of the rod can be altered. It follows that both these reactions are deter- minate if the length of the rod AjA 2 can be altered without destroying the connections of the frame. If the frame is subject to any external constraints, these may ART. 234.] ON 1RAMEWOKKS. 177 be replaced by pressures at the points of constraint. When the magnitudes of these pressures have been deduced from the general equations of equilibrium, we may regard the frame as perfectly free and acted on by known forces. The reactions at any corner may then be found as if the frame were free. It is not meant that in every case exactly these displacements must be given to the system, for these may not suit the geometrical conditions of the problem. Other displacements may recommend themselves by their symmetry or by the ease with which the virtual work due to those displacements can be found. Any two displacements which introduce only M& and S 12 into the equations of virtual work will supply two equations from which these two components may be found. If the system be in three dimensions, the direction of S u may be unknown as well as its magnitude. In this case the components of /S 12 in two convenient directions may be used instead of S 12 . Three displacements to supply three equations of virtual work will then be necessary. 234. Examples. Ex. 1. Six equal heavy rods, freely hinged at the ends, form a regular hexagon ABGDEF, which when hung up by the point A is kept from altering its shape by two light rods BF, CE, Prove that the thrusts of the rods BF, CE are as 5 to 1, and find their magnitudes. [Math. Tripos, 1874.] Let the length of any side be 2a, and let be the angle which either of the upper sides makes with the vertical. To find the thrust T of BF, we suppose the length of BF to be slightly increased. The inclinations of AB and AF to the vertical are therefore increased by d0. The work of the thrust T is Td (4a Bin 6). The work of the weights of the two upper rods is 2Wd(acosB). The centre of gravity of each of the four other rods is slightly raised, and the work of their weights is 4Wd (2a cos 8) . We have therefore Td (4a sin 8) + 2 Wd (a cos 8) + 4 Wd (2a cos 8) = 0, .-. 2T=5WtB,n0. To find the thrust T of the rod CE, we suppose the length of CE to be slightly altered. No work is done by the weights of the four upper rods. The centres of gravity of the two lower rods are however slightly raised. If 8 be the angle either of the lower rods makes with the vertical, we easily find T'd (4a sin $) + 2Wd (a cos 8) =0, R. s. 12 178 THE PRINCIPLE. OF VIRTUAL WORK. [CHAP. VI. .-. 22" = Wtanfl. The result given in the question follows at once. Ex. 2. A tetrahedron formed of six equal uniform heavy rods, freely jointed at their extremities, is suspended from a fixed point by a string attached to the middle point of one of its edges. It is required to find the reactions at the corners. The tetrahedron is regular, hence the upper and lower rods, viz. AB and CD, are horizontal. Let L and M be their middle points, then LM is vertical ; let LM—z. Let P, P' be the thrusts along the upper and lower rods respectively. Let w be the weight of any rod. Without altering the direction in space of the upper rod, or the position of its middle point, let us increase its length by dr. Since the transverse reactions at its extremities will do no work in this displacement, the equation of virtual work for the other five bars is Pdr + iw . idz + wdz = (1). In the same way, if we increase the length of the lower bar by dr without altering its direction in space or the position of its middle point, the equation of virtual work will be P'dr-4w.ldz-wdz + Tdz=0 (2), where T is the tension of the string. Since T = &w, and the ratio dr : dz is the same for each rod, these two equations give at once P=P'. To find the relation between dr and dz we require some geometrical considera- tions. From the right angled triangles BLG, LCM we have B(P-BL 3 =CL !1 =CM ! < + z i (3). In obtaining equation (1), the half side BL is altered by Jdr, the other lengths CM and BG being unaltered ; we therefore have -BL.dBL=zdz, ;. dr= -IJIdz. In obtaining (2) the opposite half side is altered by \&r, we therefore have as before dr=-2 s /2dz. Substituting these values of. dr in (1) and (2) we find that each of the thrusts P and P' is equal to f^w. We have now to find the other reactions. Since three rods meet at each corner, it is necessary to specify the arrangement of the hinges. We assume that each of the rods which meet at any corner is freely hinged to a weightless particle situated at that corner. Since this particle may afterwards be considered as joined to the extremity of any one of the three rods, we thus include the case in which two of the rods at any corner are hinged to the third. The reaction between a particle and any one of the rods which meet it will be a single force. By taking moments for the rod about a vertical drawn through one end, we may show that the reaction at the other end lies in the vertical plane through the rod. The reaction may therefore be obliquely resolved into a force acting along that rod and a vertical force. Let Q and Z be the components at art. 234.] on Frameworks. 179 A on either of the rods AC, AD, Q being positive when it compresses the rod and Z when acting upwards. In the same way Q' and Z' will represent the components on either of these rods at their lower extremities. Let us now lengthen each of the four inclined rods by dp, keeping the upper rod fixed. The equation of virtual work for the lower bar together with the two par- ticles at each end is then iQ'dp+4,Z'dz+wdz-=0 (4). Since the rod CD has here received simply a vertical displacement, this equation might have been obtained by resolving vertically the forces on the rod and equating the sum to zero, Art. 204. To find the relation between dp and dz we recur to (3). In obtaining the equation (4), BG is altered by dp while BL and CM are unaltered, hence BC . dBC=zdz, .: dz^jidp. We therefore have 2^/2Q' + 4Z' + M7=0 (5). Resolving the forces on the particle at G in the direction CD, we find -P'=2Q'cos60° (6). The value of P' having been already found, we have Q'=-&J2vi, Z' = iw. In the same way, if we lengthen each of the inclined rods by dp keeping the lower rod fixed, the equation of virtual work for the upper rod and the two particles at each end becomes -4Zdz + iQdp-wdz + Tdz=0 (7). Resolving the forces on the particle at A along AB, we have -P=2Qcos60° = Q (8). These give Q = - $J2w, Z = Jw. \ Ex. 3. Two rods GA, OB, freely jointed at G, are placed in a vertical plane, and rest with the points A, B on a smooth horizontal table, A and B being connected by a weightless string AQPB passing through smooth rings at P and Q, the middle points of GA, GB. Prove that the tension T of the string is given by T.AB . \jTp + 2Q + j^j = WcosAeosBcoBecG, where W is the weight of the two rods. [Coll. Exam., 1890.] \ Ex. 4. A frame ABCD is formed of four light rods, each of length a, freely jointed together; it rests with AG vertical and the rods BG, CD in contact with fixed frictionless supports E, F in the same horizontal line at a distance c apart, the joints B, D being kept apart by a light rod of length 6. Show that, when a weight W is placed on the highest joint A, it produces in BD a thrust of magnitude W /2a s c \ (4tf-&P)» V 6 2 )' Examine the case when b-(2a 1 c)'. [Math. Tripos, 1886.] Ex. 5. Four equal rods ABB, CRD, ESB, FSD form with each other a rhombus RBSD ; A and G are fixed hinges at a distance a from R ; R, B, S and D are free hinges, and at E and F forces, each equal to P, are applied perpendicular to the rods. If o be the angle which the reactions at A and C make with AC, 20 the angle ARC, and 6 a side of the rhombus, show that a cot a = 2 (a + b) tan 6 + a cot 0. [Coll. Exam., 1889.] 12—2 180 THE PRINCIPLE OF VIRTUAL WORK. [CHAP. VI. Take AC as axis of x, its middle point as origin. Let X, Y be the reactions at A ; x = a sin 8, y = 2 (a + b ) cos 8 the coordinates of E. Increasing the length of A C without altering its direction in space, or the position of its middle point, we have, by the principle of virtual work, Xd (a sin 8) + P sin 8dy - P cos 8dx = 0. Also by resolution Y+P sin 8 = 0. The result follows at once. Ex. 6. A succession of n rhombus figures of equal sides, each being b, are placed having equal diagonals in a straight line and one angular point common to two successive figures, and the extreme sides of the first and last rhombus are produced through equal lengths a in opposite directions to points A, B, C, D respectively. Consider now all the straight lines in the figure to be rods hinged freely where they intersect and having fixed hinges at C and D. At A and B, the free ends, are applied equal forces perpendicular to the rods ; show that the reactions at C and D make an angle with CD, where a cot

    f s (aoa0 v cos0 2 , l 12 , l w &e.)=0) ■•••■■"■ \ )• * Levy, Statique Graphique. 182 THE PRINCIPLE OF VIRTUAL WORK. [CHAP. VI. Since this particular polygon can have a slight deformation without altering the sides we must have 1^1^=0, f/* 1+ t^=o «. These give (20! =0 and d8 2 =0, unless the special polygon under consideration is such that the determinant J= I d/Jd^ d/j/dSj =0 (5). | dfjde 2 dfjae,, If we vary the lengths of the rods, the corresponding changes of the angles S v 2 are given by f/^£^=- s f^) (6) Multiplying these equations by the minors of the first row of the determinant J, and adding the results, the left hand side will vanish. We thus obtain a relation between the increments of length of the rods of the form P 12 dZ 12 + P 23 1 and BP i =BPIp 2 ; the circle circum- scribing the triangle IP 1 P 2 is the circle of stability. Ex. 2. A body moves in one plane and the instantaneous centre of rotation is known. Show that a straight line attached to the moving body touches its envelope in a point G which is found by drawing a perpendicular IG on the straight line. [Roberval's rule.] Since GI is normal to the locus of G, an element GG' of the path of G lies on the straight line. Thus the straight line intersects its consecutive position in G', i.e. G ' or G is a point on the envelope. Ex. 3. A body moves in one plane and the instantaneous position of the circle of stability is known. Prove the following construction to find the radius of ART. 243.] STABILITY *>F EQUILIBRIUM. 187 curvature of the envelope of a straight line attached to the moving body : draw a perpendicular IQ on the straight line from the instantaneous centre I and let it cut the circle of stability in P r Take IO = IP 1 on QP-J produced if necessary, then is the required centre of curvature. By the last example, 10 is a normal at Q to the envelope. If we now turn the body and the attached straight line round I through an angle d$, and draw from I' a perpendicular I'Q' on the straight line thus displaced, it is clear that Q'T is the consecutive normal to the envelope. Let Q'l' intersect QI in 0, then O is the required centre of curvature. Since 10 and I'O are perpendiculars to two consecutive positions of the same straight line, the angle IOT is equal to dB. Draw I'P' parallel to IP 1 to intersect the circle of stability in P', then as in Art. 239 the angle P'lP^ is also equal to dff. Thus I'O is parallel to P'l and P'O is a parallelogram. Therefore 10 is equal to I'P', and in the limit 10 and IP 1 are equal. Ex. 5. The corners of a triangle ABO are constrained to move along three curves, the normals at A, B, G meet in I and a, |3, y are the angles at I subtended by the sides. If p lt p. 2 , p 3 be the radii of curvature of the curves, prove that APsina Bf 2 sinB CI 2 Bmy ,„. ..... 1 "+ =AIsma+BI&mp + CIsmy. Pi Pa Pa 243. Ex. 1. A homogeneous rod AB, of length 21, rests in a horizontal position inside a bowl formed by a surface of revolution with its axis vertical. Show that the equilibrium is stable or unstable according as Vp is less or greater than n 3 , where p is the radius of curvature at A or B and n is the length of the normal. [See Art. 222.] The normals at A and B meet in a point I on the axis of revolution. Take AL and BM so that each is equal to APjp. The circle described about ILM is the circle of stability. Let the circle drawn through I touching the rod at G out AI in a- point H, then AH. AI=AG\ The equilibrium is unstable if G is within the circle ILM, i.e. if AL is less than AH, i.e. if ri 2 jp is less than P/m. The result follows at once. If the extremities of the rod terminate in small smooth rings which slide on a curve symmetrical about the vertical axis, the position A'B', in which the normals at A'B' meet in a point I below the rod, is also a position of equilibrium. Following the same reasoning the concavity of the path of G is turned towards I when l 2 p < n s . The conditions of stability are therefore reversed, the equilibrium is therefore stable or unstable according as Pp is > or < m 3 . Ex. 2. The extremities of a rod are constrained by small rings to be in contact with a smooth elliptic wire. If the major axis is vertical prove that the lower horizontal position is unstable and the upper stable if the length of the rod is greater than the latus rectum. These conditions are reversed if the length is less than the latus rectum. If the minor axis is vertical the lower horizontal position is stable and the upper unstable. In an ellipse p (6 2 /a) 2 =»i 3 , where la and 26 are respectively the vertical and 188 THE PRINCIPLE OF VIRTUAL WORK. [CHAP. VI. horizontal axes. Using this property, the results follow at once from those of Ex. 1. It has been shown in Art. 126, that when the major axis of the ellipse is vertical the rod is in equilibrium only when it is horizontal or passes through one focus. The condition of stability in the latter case follows easily from the principle that the altitude of the centre of gravity must be a minimum. Let the rod AB be in any position and let S be the lower focus. Let AM, BN be perpendiculars on the lower directrix. The altitude of the centre of gravity above the lower directrix is b(AM+BN) = fr- {SA + SB). Since SA and SB are two sides of the triangle SAB, this altitude is a minimum when S lies on the rod AB. In the same way if S is the upper focus, the depth of the centre of gravity below the upper directrix is represented by the same expression. When therefore the rod passes through tlie lower focus the equilibrium is stable, when it passes through the upper focus the equilibrium is unstable. Ex. 3. The extremities A, B of a rod are constrained by two fine rings to slide one on each of two equal and opposite catenaries having a common vertical directrix and a common horizontal axis. Prove that the lower horizontal position of the rod is stable, see Art. 126, Ex. 5. By drawing a figure it will be seen that the paths of A and B are convex to I. Hence A and B lie inside the circle of stability. Hence G also lies inside the circle and its path also is convex to I. The equilibrium is therefore stable. Ex. 4. A rod rests in a horizontal position with its extremities on a cycloid with its axis vertical. Prove that the equilibrium is stable. Rocking Stones. 244. A perfectly rough heavy body rests in equilibrium on a fixed surface : it is required to determine whether the equilibrium is stable or unstable. We shall first suppose the body to be displaced in a plane of symmetry so that the problem may be considered to be one in two dimensions. The geometrical method explained in Art. 241 supplies in most cases an easy solution. Let / be the point of contact of the two bodies, then / is the centre of instan- taneous rotation. Let CIG be the com- mon normal in the position of equilibrium, 0, C the centres of curvature. We shall suppose these curvatures positive when measured in opposite directions. If the upper body is slightly displaced so that /' becomes the new point of contact, the angle viz. d6 turned round by the body is equal to the angle between the normals ART. 245.] EOCHING STONES. 189 GJ' and G'l', and this is evidently equal to the sum of the angles J' CI, I' CI. We therefore have p p where II' = IJ' = ds as before. See also Salmon's Higher Plane Curves, third edition, Art. 312, or Besant's Roulettes and Glissettes, Art. 33. To construct the circle of stability we measure along the common normal IC in the position of equilibrium a length IS = ds/dd. Writing z for this length, we see that - = - + —. The circle de- Z P f ■scribed on IS as diameter is the circle of stability. Let IG cut this circle in P. If the centre of gravity G lie without this circle, the concavity of its path is turned towards /. Hence the equilibrium is stable or unstable according as G is below or above the point P. If G coincide with P the equilibrium is neutral to a first approximation. The critical altitude IP which separates stability and instability is clearly IP — z cos a. = — 7- , where a is the inclination to the P + P . vertical of the common normal in the position of equilibrium. 245. Ex. 1. A solid hemisphere (radius p) rests on the summit of a fixed sphere (radius p') with the curved surfaces in contact. If the centre of gravity of the hemisphere is at a distance f/a from the centre, prove that the equilibrium is stable or unstable according as p is less or greater than £p'. In this example a = 0, and therefore IG i.e. %p must be less than z if the equi- librium is to be stable. Ex. 2. A solid hemisphere rests on a rough plane inclined to the horizon at an angle p. Find the inclination of the plane base to the horizon and show that the equilibrium is stable. The centre of gravity must lie in the vertical through I, and CG is also perpen- dicular to the base. Hence the required inclina- tion of the base is the supplement of the angle CGI. The vertical through I cannot pass through G if CI sin /3 is greater than CG. Since CG=%p, it is necessary for equilibrium that sin/3 , and therefore z=p. The circle described on IC is therefore the circle of stability, Since the angle CGI is greater than a right angle, it is obvious that G lies inside the circle. The con- cavity of the path of G is therefore upwards, and the equilibrium is stable. 190 THE PRINCIPLE OF VIRTUAL WORK. [CHAP. VI. Ex. 3. A solid homogeneous hemisphere, of radius a and weight W, rests in apparently neutral equilibrium on the top of a fixed sphere of radius 6. Prove that 5o = 36. A weight P is now fastened to a point in the rim of the hemisphere. Prove that, if 55P = 18W, it still can rest in apparently neutral equilibrium on the top of the sphere. [Math. Tripos, 1'869.] Ex. 4. A heavy hemispherical bowl, of radius a, containing water, rests on a rough inclined plane of angle a ; prove that the ratio of the weight of the bowl to that of the water cannot be less than — =—. — , where ira 8 eos 2 d> is the area of sin - 2 sin a the surface of the water. [Math. Tripos, 1877.] When the bowl is displaced the water is supposed to move in the bowl so as to be always in a position of equilibrium. Its statical effect is therefore the same as if it were collected into a particle and placed at the centre of the bowl. The weight of the bowl may be collected at its centre of gravity, i.e. at the middle point of the middle radius. Ex. 5. A parabolical cup, the weight of which is W, standing on a horizontal table, contains a quantity of water, the weight of which is nW: if h be the height of the centre of gravity of the cup and the contained water, the equilibrium will be stable provided the latus rectum of the parabola be > 2 (n + 1) h. [Math. Tripos, 1859.] Let H be the centre of gravity of the water when the axis of the cup is vertical. Let the cup and the contained water be placed at rest in a neighbouring position with the surface of the water horizontal ; Art. 215. It may be shown that the vertical through the centre of gravity H' of the displaced water intersects the axiB of the paraboloid in a point M, where HM is half the latus rectum. The point M is called the metacentre. As in the last example the weight of the fluid may be collected into a particle and placed at the metacentre. The weight of the cup may be collected at the centre of gravity G of the cup. The equilibrium is stable if the altitude of the common centre of gravity of the two weights at M and G satisfies the criterion given in Art. 244. 246. When a cylindrical body rests on a fixed horizontal plane, it easily follows from what precedes that the equilibrium is stable or unstable according as the centre of gravity of the body is below or above the centre of curvature at the point of contact. There is one case however which requires a little further consideration. Let us suppose that the evolute has a cusp which points vertically downwards when the point of contact is at some point A. Let us also suppose that the centre of gravity G of the body is at a very little distance above 0. The position of the body is unstable, but a stable position exists in immediate proximity on each side in which, the tangent from G to the evolute is vertical. That these positions are stable is clear, for since the cusp points down- wards either tangent from G will touch the evolute at a point L or M which is above G when that tangent is vertical. When G moves down to these two flanking stable positions come nearer to the unstable position and finally come up to it. When therefore the centre of gravity is at the cusp of the evolute, the equilibrium is stable. ART. 247.] R0CK4NG STONES. 191 In the same way, if the cusp point upwards and O be situated at a very short distance below 0, the equilibrium is stable with a near position of instability on each side. In the limit when G coincides with 0, the equilibrium becomes unstable. The reader may consult a paper by J. Larmor on Critical Equilibrium in the fourth volume of the Proceedings of the Cambridge Philosophical Society, 1883. 247. Spherical bodies, second approximation. When the equilibrium is neutral it is necessary to examine the higher differ- ential coefficients to settle the stability or instability of the equili- brium. The geometrical method is not very convenient for this purpose. When both surfaces are spherical we can investigate all the conditions of equilibrium by the method of Art. 220. Let the body, as represented in the figure of Art. 244, be dis- placed so that J' comes into the position /'. The position of the body is then represented in the adjoining figure, where J represents that point of the upper body which in equilibrium co- incided with /. ~LetJG = r. Let f = ICT, ■ty = JCT, then p'-yfr' = pty. Let y be the altitude of G above C. The inclinations to the vertical of C'G, GJ and JG are respectively a + ■$•', a + y}r + y]r' and i|r + i/r'. Projecting these three lines on the vertical, we have y = (p + p') cos (a + yfr') - p cos (a + i/r + -f') + r cos {-f + y\r'). We now substitute for i/r its value p'ty'jp and expand the expression in powers of i/r'. The coefficients of i|r', £ip &c. are the successive differential coefficients of y, hence the stability is deter- mined to any degree of approximation by the rule of Art. 220. The coefficient of i/r' is evidently zero. The coefficient of £\p is (z cos a - r) p'*/z 2 , where z has the same meaning as before. The equilibrium is stable or unstable according as this coefficient is positive or negative, i.e. according as r is less or greater than z cos a. If this coefficient also vanish the equilibrium is neutral to a first approximation. To settle this we examine the coefficient of ■yfr' 3 . Unless this also vanishes the equilibrium is stable for dis- placements on one side of the position of equilibrium and unstable for displacements on the other. Supposing however that the coefficient of yjr' 3 does vanish, we examine the terms of the fourth 192 THE PRINCIPLE OF VIRTUAL WORK. [CHAP. VI. order. The equilibrium is then stable or unstable according as the coefficient of yfr'* is positive or negative. 248. Ex. 1. A spherical surface rests on the summit of a fixed spherical surface, the centre of gravity being at such a height above the point of contact that the equilibrium is neutral to a first approximation. If the lower surface is convex upwards as in the diagram, prove that, whether the upper body has its convexity upwards or downwards, the equilibrium is unstable. If the lower surface has its concavity upwards, the equilibrium is stable, unstable or strictly neutral according as the radius of curvature of the lower body is greater, less or equal to twice that of the upper body. The coefficient of \j/' 2 is here zero. The coefficient of ^/' 4 after elimination of r reduces to -p'(p' + 2p) (p' + p)/24p 2 . Since the equilibrium is therefore stable or unstable according as this coefficient is positive or negative, the results follow at once. Ex. 2. A spherical surface rests in apparently neutral equilibrium within a fixed spherical bowl with the point of contact at the lowest point. If the radius of one surface is twice that of the other, show that the equilibrium is really neutral. 249. flnMBBBBarucaies, second approximation. If the boundaries of the bodies in contact are not spherical we may adopt the following method. Suppose the upper body has rolled away from its position of equilibrium into that represented in the figure of Art. 247. Then it is clear that, if G in that figure is to the right of the vertical through I', the body will roll further away from the position of equilibrium, but if G is on the left of the vertical, the body will roll back. Let i be the angle GI' makes with the vertical ; our object will be to find i. Let

    _ cos (p ds ~ r (2)*- * The equation (2) is useful for other purposes besides that of finding the con- ditions of stability. For example it- may be very conveniently used in the differential calculus to find the conic of closest contact at any point I of a curve. If

    is positive when measured behind the normal as P travels along the conic in the direction in which the arc « is measured. Suppose G to be the centre of the conic, then assuming this value of 0, the distance r of the centre of the conic from I is given by the equation (2) in the text. Generally the equation (2) is useful to find the point of contact with its envelope of a straight line IG drawn through each point of a curve making with the normal an angle which is a given function of s. ART. 250.] ROOMING STONES. 193 Lastly, let tp' be the inclination of the normal CC to the vertical, then i=\)/ - and dflds = llp'. Hence by (2) di 1 1 cos0 ,„. ds = -p + i'- — (3)l These three equations supply all the conditions of stability. In the position of equilibrium the centre of gravity is vertically over the point of support. Hence i = 0. In any other position the value of i is given by Taylor's series viz. . di , dH Ss* . l= dl Ss+ d?T^ +&c - If in this series the first differential coefficient which does not vanish is positive and of an odd order, it is clear that the straight line IG will move to the same side of the vertical as that to which the body is moved. The equilibrium will therefore be unstable for displacements on either side of the position of equilibrium. If the coefficient is negative the equilibrium will be stable. On the other hand if the term is of an even order, it will not change sign with 5s, the equilibrium will therefore be stable for a displacement on one side and unstable for a displacement on the other side. The first differential coefficient is given by (3). The second may be found by differentiating (3) and substituting for d^t/ds and drjth from (2) and (1). The third differential coefficient may be found by repeating this process. In this way we may find any differential coefficient which may be required. Firstly. Suppose the body such that di/ds is not zero in the position of equilibrium. The condition of stability is therefore that - + -^- is negative. This leads to the rule already considered in Art. 244. Secondly. Suppose the body such that in the position of equilibrium the centre of gravity lies on the circle of stability. We then have di/ds = 0. Differentiating (3) and substituting for (cos ' we find S-iCW)«-(H)G + ?) '"• Unless this vanishes the equilibrium will be stable for displacements on one side and unstable for displacements on the other side of the position of equilibrium. Thirdly. Suppose the second differential coefficient given by (4) is also zero in the position of equilibrium. We find by differentiating (3) twice and substituting for r as before s-se*j)*e + ?)e + j»-—e + ?)i; -— (H)"(tt)- The equilibrium is stable or unstable according as this expression is negative or positive. 250. Ex. 1. A body rests in neutral equilibrium to a first approximation on the surface of another, and both are symmetrical about the common normal. Show that the equilibrium cannot be stable unless either the point of contact is the summit of the fixed surface or />'= - ip. R. S. 13 194 THE PRINCIPLE OF VIRTUAL WORK. [OHAP. VI. Ex. 2. A body rests in neutral equilibrium to a second approximation on a rough inclined plane. Show that the equilibrium is stable or unstable according as cPpldfi is positive or negative. Ex. 3. A body rests in equilibrium on the surface of another body fixed in space, and the centre of gravity G of the first body is acted on by a central force tending to some point in GI produced and varying as the distance therefrom. If G' be 1 taken on IG so that IG' jfj + jj^t Ae equilibrium is stable or unstable according .(1). as G' lies within or without the circle of stability. 251. Rocking Stones In three dimensions. The upper body being in its position of equilibrium, let the common tangent plane at the point of contact be taken as the plane of xy. Let the equations to the upper and lower bodies be respectively 2z=ax' i + 2bxy + cy i + &o. \ -2z'=a'ifi + 2b'xy + cY + &o.j In the standard case, therefore, the two bodies have their convexities turned towards each other. We shall now suppose the upper body to be displaced from its position of equilibrium by rolling over the lower along the axis of x through a small arc ds. Take OP = OP' = ds. We have first to determine how the upper body must be rotated to bring the tangent plane at P into coincidence with that at P'. Referring to equations (1), we see that the tangents at P and P' to OP and OP' make angles with the plane of xy which are dz\dx=ads and dz'jdx= -a'ds. To make these tangents coincide we must rotate the upper body round Oy through an angle w% = (a + a') ds. Consider next the tangents at P and P' which are perpendicular to OP and OP 1 ; these make angles with the plane of xy which are dz\dy = Ids and dz'ldy = - b'ds. To make these tangents coincide we must rotate the upper body round Ox through an angle w 1= -(b + b')ds. Taking both these rotations either simultaneously or one after the other, the upper body will be rolled along the arc OP=ds. These two rotations wj and w 2 about the axes of x and y are equivalent to a resultant rotation (2 about some axis Oy'. If the angle xOy' = i , we have 11 cos i= u^ and 11 sin i = u 2 . The arc of rolling Ox and the axis of rotation Oy' are not neces- sarily at right angles to each other ; either being given, the other can be found by these relations. ART. 252.] ROClftNG STONES. 195 252. The body being placed at rest in its new position, the centre of gravity G is no longer in the vertical through the point of contact. The weight will therefore make the body begin to move. Let us suppose for example that the system is constrained either to go back to its position of equilibrium by the way it came or to recede further on that course. The equilibrium will then be stable or unstable according as the moment of the weight about a parallel to Oy' through the new point of contact tends to bring the body back to or further from the position of equilibrium. It will be found more convenient to refer the displacement of Gf to the rectangular axes Ox', Oy', Oz instead of the original axes. Let x', y', z be the coordinates of G in the position of equilibrium, let r = OG and let a', /3', y' be the direction angles of 06. Then x' — r cos a',y'= r cos /S', z =r cos y'. If we draw GN a perpendicular on Oy', the point G will be displaced by the rotation 12 along a small arc GG' of a circle whose plane is parallel to x'z, whose centre is N and radius NG. The displacements of G parallel to x' and z' respectively are therefore (Iz and - Qx'. The resolved forces on G parallel to the axes x', y', z are X= - Wcosa', Y= - W cos /3', Z= - WCOS7', where W is the weight of the body. The moment of these about a parallel to Oy' drawn through the new point of contact P is M = (z-Qx') X-(x' + Qz-ds&va.i)Z . = {?-0(cos 2 a'+cos 2 7')-dssin£cos7'} W. To discuss the sign of the moment M let us find the locus of G when M=0. Changing to Cartesian coordinates by writing r cos a' = x' &a. we find for the locus ,„ „ dssini a" + 2 9 -* — jr — =0. This is the equation to a right circular cylinder whose axis, is parallel to Oy" and whose base is the circle described on a diameter measured along the axis of z equal in length to ds (sin i)jii. The left hand side of the last equation represents the product Mr. We see, therefore, that as Gf travels down the vertical GO, M changes sign when G crosses the cylinder at sonle point K remote from 0, while r changes sign when G passes below 0. Thus M is negative for all positions of G below K. If G is inside this cylinder or below the tangent plane the moment M is negative and will tend to turn the body back to its former position, the equilibrium is therefore stable. If G is outside and above the tangent plane the moment is positive and will tend to move the body away from its former position, the equilibrium is therefore unstable. This cylinder may be called the cylinder of stability. Since sin i—u> 2 =(a+a')ds we have a construction for the circular base similar to that for the circle of stability. Referring to equations (1), we notice that a, a! are the reciprocals of the radii of curvature of the sections of the two surfaces made by the plane of xz. We have therefore the following construction. The body being in the position of equilibrium, let Oz be the common normal, Ox the constrained arc of rolling, Oy' the axis of rotation. Let 1/z be the sum of the opposite curvatures of the two surfaces made by a normal plane through Ox. Measure a length z sin 2 xOy' along the common normal and describe a circle on this length as diameter. The cylinder with this circle for base and its axis parallel to Oy' is the cylinder of stability. 13—2 196 THE PRINCIPLE OF VIRTUAL WORK. [CHAP. VI. 253. It will be noticed that the angular rotations &>, and « 2 do not depend on either a, a' or 6, 6' alone, but only on their sums a+a' and 6 + 6'. If therefore we desoribe the surface 2z=(a + a')x i +'i(b + b')xy + (c + c')y i and make this roll on a rough plane, the conditions of stability are unaltered. The indicatrix of this surface viz. (a+a')x* + 2{b + b')xy + (c + c')y !l =N, where N is any constant, may be called the relative indicatrix of the solids given by the equations (1). We notice also that the tangent OP to the arc of rolling and the axis of rotation Oy' are conjugate diameters with regard to the relative indicatrix. To prove this we observe that the equation to the axis of y' is, by Art. 251, w 2 a;-w 1 2/=0, or (a+a')x + (b + b')y = 0. The latter of these is the same as the conjugate of the axis of x. Geometrical proofs of these theorems are given in the Author's Rigid Dynamics, Chap. x. The cylinder of stability is also shown to be useful in finding the time of oscillation when the body is slightly disturbed from a position of stable equilibrium. Lagrange's proof of the principle of virtual work. 254. Let a body ABC be acted on by any commensurable forces P, Q, E &e. at the points A, B, C &o. Let these forces be multiples I, m, n &c. of some force 2K. At the point A of the body let a small smooth pulley be attached, and opposite to it at some point A' fixed in space let an equal pulley be fixed so that AA' is the direction of the force P. Let a fine string be wound round these two pulleys so as to go round each I times. It is clear that, if the tension of this string were K, the force exerted at A would be equal to the given force P and act in the same direction. C1K Imagine similar pulleys to be placed at B, G &o. and opposite to them at B', C" Ac. Let the same string go round the pulleys B, B' m times, and round C, C n times, and so on. Let one extremity of this string be attached to a point fixed in spaoe. Let the other extremity of the string after passing over a smooth pulley D fixed in art. 256.] iagrange's proof. 197 space be attached to a weight K. It is clear that by this arrangement all the forces P, Q, B &o. of the system have been replaced by the pressures due to the tension K of the string. Suppose now the body receives any small displacement so that the pulleys A, B, G & P y , P z ART. 258.] GEMERAL PRINCIPLES. 199 acting at 0, and three couples whose moments are yP z — zP y , zP x — xP z , xP y — yP x , and whose planes are yz, zx, xy respectively. Kepeating this for all the given forces, we see that they may be replaced by three forces X, Y, Z acting along the axes of coordinates, and three couples whose moments are L, M, N, and whose axes are the axes of coordinates, where X = 2P a , L = Z(yP z -zP y ), Y=tP y , M = t(zP x -xP z ), Z=ZP Z , N=%{xP y -yP x ). These are called the six components of the forces. The three components X, Y, Z may be compounded into a single force whose magnitude is R, and whose direction cosines (I, m, n) are given by Rl = X, Rm = Y, Rn = Z, R? = X*+Y*+Z*. The force R is called by Moigno the principal force at the point 0. The three components L, M, N in the same way may be compounded into a single couple whose moment G and the direction cosines (X, /i, v) of whose axis are given by G\ = L, Gfi = M, Gv = N, The couple G is called the principal couple at the point 0. The components L, M, N of the principal couple are also called the moments of the forces about the axes. This is Poinsot's method of finding the resultants of a system of forces. 258. The base of reference to which the forces have been transferred, has been taken as the origin of coordinates. But when it is necessary to distinguish between these points we must modify the expressions for the components. Let some point 0' whose coordinates are £, rj, £ be the base of reference. The expressions for the six components for this new base may be deduced from those for the origin by writing x — fj, y — y, z — f for x, y, z. The expressions for the components of the force R do not contain x, y, z, hence the principal force R is -the same in magnitude and direction whatever base is chosen. 200 FORCES IN THREE DIMENSIONS. [CHAP. VII. The expressions for the components of the couple G become W = Z{{x-?)P y -(y-r l )P x } = N-1;Y+r,X. Thus the magnitude and the axis of the principal couple G are in general different at different bases. 259. Conditions of equilibrium. It has been proved in Art. 105 that the forces on a body can be reduced to a single force R and a single couple G. By the same reasoning as in Art. 109 it is necessary and sufficient for equilibrium that these should separately vanish. We therefore have R = and G = 0. If the axes of reference are at right angles, these lead at once to the six conditions X = 0, Y=0, Z=0, Z=0, M=0, N=0; we may, however, put these results into a more convenient form. In order to make the resultant force R zero, it is necessary and sufficient that the sum of the resolutes of all the forces along each of any three straight lines (not all parallel to the same plane) should be zero. To prove this, let OA, OB, OG be parallel to the three straight lines. If the resolute of R along OA is zero, it is evident that either R is zero, or the direction of R is perpendicular to OA. Hence, if R is not zero, its direction is perpendicular to each of three straight lines meeting in and not all in one plane, which is impossible. In the same way, since couples are resolved according to the same laws as forces, we infer that to make the principal couple G zero, it is necessary and sufficient that the component couple of all the forces about each of any three straight lines intersecting in the base but not all in one plane, should be zero. It will be presently seen that the moment of the component couple for any axis through is also the moment of the forces about that axis, Art. 263. Since a couple may be moved into a parallel plane without altering its effect, it is clear that, when the force R is zero, the moments about all parallel straight lines are equal. It is therefore sufficient for equilibrium that the moment of the forces about each of ART. 261.] COMPONENTS OF A FORCE. 201 any three straight lines {whether intersecting or not) should be zero, but all three must not be parallel to the same plane, and no two must be parallel to each other. The method of finding these moments will be more fully explained a little further on. 260. Components of a force. Usually we suppose a force to be given when we know its magnitude and the equations of its line of action. We see from the results of the proposition in Art. 257 that it will sometimes be more convenient to determine a force P by the values of its six components, viz. P a , P y , P z , and yP z — zP y , zP x — xP z , xP y — yP x . The advantage of this repre- sentation is that the resulting effect of any number of forces is found by adding their several corresponding components. If we wish to represent the line of action of the force apart from the force itself, we may regard the straight line as the seat of some force of given magnitude, and suppose the line itself determined by the six components of this chosen force. Let (I, m, n) be the direction cosines of the straight line, (x, y, z) the coordinates of any point on it. Then, if the force chosen is a unit, the six components or coordinates* of the line are I, m, n,X = yn — zm, /a = zl — am, v = xm — yl, with the obvious relation IX + m/j, + m> = (1 ). If a force P act along this straight line, its six components or coordinates are Pl,Pm,Pn; PX, Pfi, Pp. If we compound several forces together, the six components become X = SPl, Y = tPm, Z = SPm ; L = %PX, M= 2P/*, N=2Pv, but the relation XL+YM+ZN=0 (2) is not necessarily true. 261. We have seen in Art. 257 that all these forces may be joined together so as to make a single force R and a couple G. This combination of a force and a couple is called a dynam e. The six quantities X, Y, Z, L, M, N are the components of the dyname. * The six coordinates of a line are described in Salmon's Solid Geometry (fourth edition, Art. 51) from an analytical point of view. See also Cayley, Quart. Journal, 1860 ; Plucker, Phil. Trans. 1865 and 1866, to whom the nam e dynam e is due ; Cayley, Gamb. Trans. 1867. '^ SL ~ =~* . 202 FORCES IN THREE DIMENSIONS. [CHAP. VII. The three former components are multiples of some unit force, the three latter of some unit couple. It will be shown further on that when the coordinates of the dyname satisfy the condition (2), either the force M or the couple of the dyname is zero. 262. Ex. 1. The six components of a force are 1, 2, 7; 4, 5, -2. Show that the magnitude of the force is ^54, and that the equations to its line of action are (7?-&)/4=(*-7*)/5 = (2s-y)/(-2) = l. Ex. 2. The six components of a dyname are 1, 2, 3 ; 4, 5, 6. Show that the magnitude of the force is ^14, and that its direction cosines are proportional to 1, 2, 3. If this force act at the origin the magnitude of the couple is */77, and the ■ direction cosines of its axis are proportional to 4, 5, 6. 263. Moment of a force. It has already been stated that the expressions for L, M, N in Art. 257 are usually called the moments of the forces about the axes of x, y, z respectively. These expressions are L = % (yP z - zP y ), M=% (zP x - xP z ), N = t (xP y - yP x ). To show how far this definition agrees with that already given in Art. 113, let us examine how the expression for N has been obtained. The force P has been resolved into its components P x , P y , P z \ the two former act in a plane perpendicular to the axis of z, hence by the definition given in Art. 113, the expressions yP x and — xP y are respectively equal to their moments about that axis. The latter P z acts parallel to the axis of z, and if the moment of this component is defined to be zero, the expression N~ will become the moment of the forces about the axis of z. Let Q be the resultant of the two components P x , P z , then the moment of Q about the axis of z is equal to the sum of the moments of P x and P z , Art. 116. Since any straight line may be taken as the axis of z, this explanation applies to all straight lines. It appears therefore that the moment of the component couple for any axis is the same as the moment of all the forces about that axis. We thus arrive at the following definition of the moment of a force about any straight line. Let the straight line be called CD. Resolve the force P into two components, one parallel and the other perpendicular to the straight line CD. The moment of the former is defined to be zero. The moment of the latter is obtained by multiplying its magnitude by the shortest distance between it and the given straight line CD. ART. 266.] MO»ENT OF A FORCE. 203 It is evident that this shortest distance is equal to the shortest distance between the original force P and the straight line CD, each being equal to the distance between CD and the plane of the components. Let r be the length of this shortest distance. Let be the angle between the positive directions of the force P and the line CD, then the resolved part of the force P perpendicular to CD is P sin 0. We therefore find that the moment of the force P about CD is equal to Pr sin 0. When the moments of several forces round the same straight line CD are to be added together, we must take care that these have their proper signs. Any direction of rotation round CD having been chosen as the positive direction, the moment of any force is to be taken as positive when the force acts round CD in the positive direction. 264. It follows from Art. 263 that, if two equal forces act along the positive directions of two straight lines AB, CD, the moment of the former about CD is equal to the moment of the latter about AB. The product r sin is sometimes called the moment of either of the straight lines AB, CD about the other. Let i be the moment of one straight line about the other, and let either line be occupied by a force P. Then the moment of P about the other line is Pi. 265. In some cases it may be necessary to take account of the signs of r and 6. Supposing the positive direction of the common perpendicular to AB and CD to have been already determined, the shortest distance r must be measured in that direction. The angle $ must then be measured in any plane perpendicular to r from the projection of one line to the projection of the other in such a direction that when r and sin are positive, a positive force acting along either line will tend to produce rotation round the other in the positive direction. 266. The volume of a tetrahedron is known* to be equal to * To find the volume of a tetrahedron. Pass a plane through CD and the shortest distance EF between CD and the opposite edge. Then since the tetrahe- dron ABCD is the sum or difference of the tetrahedrons whose vertices are A and B and common base is DEC, its volume is one third the area DEC multiplied by AB . sin 0, where is the angle AB makes with the plane DEC. If a straight line AB cut a plane in E and be at right angles to a straight line EF in that plane, its inclina- tion to the plane is the angle it makes with a straight line drawn in the plane perpendicular to EF. But CD lies in the plane and is perpendicular to EF, hence 6 is equal to the angle between the opposite edges AB, CD. The volume is therefore equal to \AB . CD . EF. sin 0. 204 FORCES IN THREE DIMENSIONS. [QHAP. VII. one-sixth of the continued product of the lengths of two opposite edges, the shortest distance between those edges and the sine of the angle between them. Let then the force P be represented by a length AB situated at any part of its line of action. Then if CD be a unit of length, the moment of P about CD is represented by six times the volume of the tetrahedron whose opposite edges are AB and CD. This result may be put into a more convenient form. Let AB be any length conveniently situated in the line of action of the force P, and let CD be any length in the axis ; the moment of P about 6V CD is equal to P at> r\j\ > where V is the volume of the tetrahedron whose opposite edges are AB and CD. This form of the result follows at once from the former, because the volumes of the two tetrahedra are proportional to the products of their opposite edges. 267. Ex. 1. The moment of a force P acting along either of the straight lines (x~f)\l=(%j-g)lm=(z-h)ln, (x-f) jl' = (y -^jm'^z -%')%', about the other straight line is PA, where A is the deter- /-/' g-g' h-h' minant in the margin and (Imri) (I'm'n') are the actual I m n direction cosines of the two straight lines. I' m' n' This follows at once from the ordinary determinantal expression for the volume of a tetrahedron, the coordinates of whose corners are given. If (/, m, n), {V, to', n') represent the direction cosines of the positive directions of the lines, and if the positive direction of rotation be that usually chosen in solid geometry, viz. clockwise, this expression will give the moment of the force with its proper sign, Art. 97. The order of the terms in the determinant is as follows ; if /, g, h precede /', O x suppose the direction of the pressure to be perfectly Q general. Taking any arbitrary point B on the axis as a base of reference, each pressure may be transferred to act ->r* ■n at B, by introducing a couple whose plane passes through the axis. All the pressures are therefore equivalent to a F y £ resultant pressure which acts at B together with a resultant /() couple whose plane passes through the axis. Let one of the / forces of this couple act at B and let the arm be so altered •' (if necessary) that the other force acts at some other arbitrary point C of the axis. Then compounding the forces which act at B, we see that the pressures on all the elements of length of the axis are equivalent to two pressures which may be made to act at any two arbitrary points B, C of the axis. We may suppose the body attached to its axis at these two points by smooth hinges. Let F x ,. F v , F, and G x , G vt G t be the resolutes of the pressures at B and C re- spectively. Let 6, c be the ordinates of these points. Let X, Y, Z, L, M, N be the 206 FORCES IN THREE DIMENSIONS. [CHAP. VII. six components of the given forces, moments as in Art. 257, Then resolving parallel to the axes and taking -Fyb- F x b + ( F x +G x + X=0\ -F v b-G y c + L=0\ F y + G y +Y=a\, F x b + 0^ + M= 0[. F z +G, + Z = o) N=0) The last equation determines the condition of equilibrium, and shows that the body will turn about the axis unless the moment of the given forces about it is zero. We have therefore five equations to determine the six component pressures on the axis. The pressures F x ,F y , G x , G y are obviously determinate, but only the sum of the components F z , G a can be found. The solution of these equations will be simplified by a proper choice of the arbitrary points B and C. The position of the origin is generally determined by the circumstances of the problem. If we place B at the origin we have 6=0, and the values of G v , G e become evident by inspection. Suppose for example the body to be a heavy door constrained to turn round an axis inclined at an angle a to the vertical. In this case, since the moment of the forces about the axis must be zero, the centre of gravity of the door must lie in the vertical plane through the axis. Let us take this plane as the plane of xz, the axis of the door being as before the axis of z. Let x, 0, z be the coordinates of the centre of gravity, and let W be the weight of the door. To simplify the moments we resolve W parallel to the axes ; we therefore replace W by the two components W sin o and - W cos o acting at the centre of gravity parallel to the axes of x and z. We shall choose the arbitrary point B to be at the origin, while the other G is at a distance c from it. Resolving and taking moments as before, we have F x +G x +W sin a=0\ - GyC=0 j F v +Gy =0>, G x c + Wz Bin a + Wx cos a=0\. F e +G,- Wcosa=0) J It follows from these equations that F y and G v are both zero, so that the resultant pressures act in the vertical plane through the axis. The values of F x , G x and F z + G s may be easily found. Ex. 2. Three equal spheres, whose centres are A, B, C, are placed on a smooth horizontal plane and fastened together by a string which surrounds them in the plane of their centres, and is just not tight. A fourth equal sphere, whose centre is D, is placed on the top of these touching all three. Prove that tlie tension of the string is T=-^W. Let R be the reaction of any one of the lower spheres on the upper, DN a perpendicular from D on the plane ABC, then 3Bcos ADN= W. Consider next the ART. 268.] PROBLEM ON EQUILIBRIUM. 207 sphere whose centre is A ; the other two of the lower spheres exert no pressure on it. The resolved part of R in the direction NA balances the two tensions of the parts of the string parallel to A B and A G. Hence R eos DA N= 2 T cos BAN. The angle BAG =60°, and . .„„ AN „AM „2rsin60» sin ADN= ■r=r=#- r =r=S — s • AD * AD s 2r We now easily find T in terms of W. Ex. 3. Four equal spheres rest in contact at the bottom of a smooth spherical bowl, their centres being in a horizontal plane. Show that, if another equal sphere be placed upon them, the lower spheres will separate if the radius of the bowl be greater than (2^13 + 1) times the radius of a sphere. [Math. Tripos, 1883.] ^ Ex. 4. Six thin uniform rods, of equal length and equal weight W, are connected by smooth hinge joints at their extremities so as to constitute the six edges of a regular tetrahedron; one face of the tetrahedron rests on a smooth horizontal plane. Show that the longitudinal strain of each of the rods of the W lowest face is =-^ . [Coll. Ex.] Ex. 5. A heavy uniform ellipsoid is placed on three smooth pegs in the same horizontal plane, so that the pegs are at the extremities of a system of conjugate diameters. Prove that there will be equilibrium, and that the pressures on the pegs are one to another as the areas of the conjugate central sections. [Coll. Ex.] Ex. 6. Four equal rods are jointed to form a square. One side is held horizontal and the opposite one is acted on by a given couple whose axis is vertical. Show that in a position of equilibrium the lower rod makes an angle 2 sin -1 GjWl with the upper, G being the couple, and W and I the weight and length of a rod. Find the action at either of the lower hinges. [Coll. Ex., 1880.] Ex. 7. An equilateral triangular lamina, weight W, hangs in a horizontal position with its angles suspended from three points by vertical strings each equal in length to the diameter 2a of the circle circumscribing the triangle. Prove that the couple required to keep the lamina at a height 2 (1 - n) a above its initial position is Wa V (1 - n 2 ). [Coll. Ex., 1886.] ^ Ex. 8. A weightless rod, of length 21, rests in a given horizontal position with its ends on the curved surfaces of two horizontal smooth circular cylinders, each of radius a, which have their axes parallel and at a distance 2c. The rod is acted on at its centre by a given force P and a couple. Find the couple when there is equilibrium, and prove that the magnitude of the couple will be least when P acts vertically, provided that c + $aJ1 sec \ is the angle between the rod and the axes of the cylinders. [Math. Tripos, 1889.] ^ Ex. 9. A solid circular cylinder, of height h and radius a, is enclosed in a rigid hollow cylinder which it just fits, and is formed of an infinite number of parallel equally elastic threads, which will together support a weight W when stretched to a length 2ft. The ends of these strings are fastened firmly to two discs, one of which is then turned through an angle « in its own plane : assuming each thread to form a helix, prove that there is a force exerted in the direction of the axis of the cylinder equal to 2 J(f - J s/^^+^) ■ t Math - Tri P° s . 1871 208 FORCES IN THREE DIMENSIONS. [CHAP. VII. Ex. 10. Three equal heavy spheres, of weight W and radius a, are suspended from a fixed point by three equal strings each of length I. A very light smooth spherical shell of radius b is placed symmetrically on the top of them, and water is poured very gently into it. Show that the greater the amount of water poured in the closer must the three lower spheres be to one another in order that equilibrium may be possible, and that equilibrium will be impossible if the weight of the water poured in exceed nW, where n is the positive root of the equation n 2 (J-6)(J+2a + fc) + (2« + 3)(a 2 -6a6-36 2 )=0, it being assumed that b is so small as to admit of the strings being straight. [Math. Tripos, 1890.] 269. Ex. 1. A heavy rod OAB can turn freely about a fixed point 0, and rests over the top CAD of a rough wall. If OC be a perpendicular from on the top of the wall, prove that the angle 8 which the rod makes with OC when the equilibrium is limiting is given by fi=tan /3 sin 0, where /3 is the angle OC makes with the per- pendicular OE drawn from to the vertical face of the wall. To assist the description of the figure, let OAB be called the axis of x. Let z be normal to the plane AOC, and let y be perpen- dicular to x and z. The weight W of the rod acting at G is equivalent to W cos |S parallel to a, and W sin /3 acting parallel to CO. This latter is equivalent to W sin /? cos and W sin j3 sin parallel to x and y respectively. The reaction R at A is perpendicular to both OA and CD, and is therefore parallel to z. The point A of the rod can only move perpendicularly to OA. The friction therefore acts, not along the top of the wall, but opposite to the direction of motion, i.e. parallel to y. Taking moments about y and z respectively, we have W cos/3 . OG=B . OA, W sin/3 sin0 . OG = uE . OA. These give u=t&np sin 0. Ex. 2. Three equal heavy spheres, each of weight W, are placed on a rough ground just not touching each other. A fourth sphere of weight nW is placed on the top touching all three. Show that there is equilibrium if the coefficient of friction between two spheres is greater than tan \a, and that between a sphere and the ground is greater than tan Ja.n/(ji + 3), where a is the inclination to the vertical of the straight line joining the centres of the upper and one lower sphere. Ex. 3. A pole of uniform section and density rests with one end A on the ground (which is sufficiently rough to prevent any motion of that end) and with the other against a rough vertical wall whose coefficient of friction is ft.. If AH be the limiting position of the pole for any position of A, AN the perpendicular from A on the wall, a the angle BAN, and the inclination of BN to the vertical, prove that tan a tan is constant, and find the whole friction exerted at B. Find also the equation to the locus of B on the wall, N being fixed, and prove that the deviation of B from the vertical through N is greatest when a=ff=tan -1 ^/JI. [Coll. Ex., 1886.] \ Ex. 4. A narrow uniform rod of length la rests in an oblique position with one end on a rough horizontal table and the other against a rough vertical wall, the ART. 270.] POINSW'S CENTRAL AXIS. 209 coefficients of friction at the table and wall being ^ and /j.„, and the distance of the foot of the rod from the wall being k ; show that the rod is on the point of slipping at the lower end if the vertical plane in which it lies makes an angle with the wall given by ft^ (*t 2 a sin 2 B - cos 2 ffft = k - 2/^ (4a 2 sin 2 $ - hrf, and that the inclination of the tangential action at the upper end to the horizon is then sec -1 (|U 2 tan 6). [Math. Tripos, 1887.] Ex. 5. A curtain is supported by an anchor ring capable of sliding on a horizontal cylinder by means of a hook fixed at that point of the ring which is lowest when the curtain is hanging. Show (1) that the ring may touch the cylinder at one or two points but not more, (2) that if there be double contact and the weight of the ring can be neglected the ring will not slip along the oylinder however it be pulled unless the coefficient of friction be less than ,-J — rr-h — -p. — =■ , in which b is the r {2a + b)am6-b radius of the generating circle, a that of the circle described by its centre and 8 the inclination of the plane of this latter circle to the axis of the cylinder. [Math. Tripos.] For the sake of the perspective take the axis of the anchor ring as axis of z, and let the plane of the circle whose radius is a be the plane of xy. Let the axis of x pass through the hook. Let B, B' be the two points of contact of the cylinder and ring, B' being nearest the hook. Let (R, /iB) (B', /j,B') be the reactions at these points, then these four forces lie in the plane xz. Taking moments about an axis through the hook and solving, we find (2a + 6)cos0-/>6cos M = (2a + &)sin0-& + p6(l + sin0)' where p is the ratio of B' to B. As long as there is double contact B and JJ' are both positive. But if ft is greater than the value given in the question, this equation shows that p must be negative. The central axis and the invariants. 270. Poinsot's Central Axis. Any base having been chosen, the forces of a system have been reduced to a force R acting at and a couple 0. We shall now examine whether this representation of the forces can be further simplified by a proper choice of the base. Let 6 be the angle between the direction of the force R and the axis of the couple G. We may resolve into two couples, one G cos 8 whose plane is perpendicular to R, and the other 6rsin0 whose plane contains that force. This latter couple together with the force R may be replaced by a single force in its plane equal and parallel to R, but situated at a distance G sin 6/R from 0. We have therefore reduced the system to a force R (acting in a direction parallel to the principal force at any base) together with R. s. 14 210 FORCES IN THREE DIMENSIONS. [CHAP. VII. a couple whose plane is perpendicular to the force. The line of action of this force R is called Poinsot's central axis . To construct geometrically the central axis when the couple and the force R at any base of reference are given, we notice that (1) the central axis is parallel to R, (2) it is at a distance G sin 6/R from R, (3) the perpendicular from on the central axis is at right angles both to R and the axis of G, (4) the perpendicular from must be so drawn that its foot is moved by the couple G sin 6 in the same direction as that in which R acts. 271. Screws and wrenches. A body is said to be screwed along a straight line when it is rotated round this straight line as an axis through any small angle dd, and at the same time trans- lated parallel to the axis through a small distance ds. The ratio ds/dd is called the pitch of the screw. If the pitch is uniform, it may also be denned as the space described along the axis when the angle of rotation is a radian, i.e. a unit of circular measure. The pitch of a screw is therefore a length. For the sake of brevity the axis of the screw is often called the screw. The term wrench, has been applied by Sir R. Ball to denote a force and a couple whose axis coincides with or is parallel to the force. The phrase wrench on a screw denotes a force directed along the axis of the screw and a couple in a plane perpendicular, to the screw, t he moment of the coup l e being equaj to the pro/lnn*. of the force and the pitch of the screw. The force is called the intensity of the wrench. When the pitch of the screw is zero the wrench is simply a force. When the pitch is infinite the wrench reduces to a couple. The phrase wrench on a screw is sometimes abbreviated into the single word, wrench. A wrench is a dyname in which the direction of the force is perpendicular to the plane of the couple. To determine a screw five quantities are necessary. Four are required to determine the position of the axis, for example the coordinates of the points in which it cuts two of the coordinate planes. One more is necessary to determine the pitch. To determine a wrench on a screw a sixth quantity is required, viz. the magnitude of the force. 272. Screws are distinguished as right or left-handed according to the direction in which the body is rotated for the same translation. ART. 273.] THE E ... B n whose weights are proportional to the resolved parts of the forces parallel to GG'. [Coll. Ex., 1889.] Ex. 9. A fixed straight line AB is perpendicular to the central axis of a system of forces. From any point of AB a finite straight line is drawn parallel to the axis of the principal couple at that point and proportional to its moment ; prove that this straight line lies on a certain hyperbolic paraboloid and that its extremity is situated on a fixed straight line. [Caius Coll., 1864.] 279. Invariants of a system. It follows from the third result of Art. 273 that, whatever base is chosen and whatever the directions of the rectangular axes may be, the quantity / = LX + MY -f NZ is invariable and equal to TR. The square of the resultant force, viz. i? s = X 2 + F a + Z 2 is also invariable. These two quantities, viz. I and R 2 , are called the invariants. If the forces of the system are such that the first of these invariants is zero, it follows that either R = or T = 0. The condition that the forces should be equivalent to either a single force or a single couple is therefore 1=0. We may distinguish between these two cases by examining the second invariant. If the forces are to be equivalent to a single force we must have as a second con- dition R not equal to zero. ART. 281.J TUB INVARIANTS. 215 281. When the forces are known to be equivalent to a single force it may be required to find its position and magnitude. But this single resultant is obviously only the equivalent wrench with the pitch of its screw equal to zero. The problem has therefore been already solved in Art. 273. 281. Ex. 1. Two forces Pj, P 3 being given, find tlieir invariants. Referring to the figure of Art. 276, let the line of action of the force P 1 be the axis of 2, let the line of action of P 2 be O'A, and let the shortest distance 00' between these forces be the axis of x. The components of the forces are X=0, L = 0, r=P 2 sin 6, M= - P s r cos 6, Z=P 1 + P 2 cos0, N=P i rsmd. Since the invariants are independent of all axes, we have I=LX+ MY+NZ=P 1 P 2 r sin 9, P? = iV + iV + 2PjP 2 cos e. The expression for I may be written in the form 1= PjN, i.e. the invariant of two forces is equal to either force multiplied by the moment of the other force about the first. Let the positive direction of a straight line be determined by the signs of the direction cosines of the line. The positive direction of rotation round that line is then determined by the rule in Art. 272 or Art. 97. The sign of the invariant of two forces is positive or negative according as the sign of either force and that of the moment of the other are like or unlike. We notice that the invariant of two forces is not altered by changing the directions of both the forces, their lines of action remaining unchanged. The forces P 2 , P 2 being represented by two lengths measured along their respective lines of action, the invariant I is equal to six times the volume of the tetrahedron having these lengths far opposite edges. This tetrahedron is sometimes called the tetrahedron constructed on the two forces. See Art. 266. Ex. 2. Any number of forces Pj , P 2 &e. being given, find the invariant I. Taking any rectangular axes, the six components are given in Art. 257. It follows that J is a quadratic function of P lt P., &c. of the form 1= A U P* + A 2i Pf + 24 12 P I P 2 + &c. where A n &a. are all independent of the magnitudes of the forces. When all the forces except P lt P 2 are put zero this expression should reduce to P^P^r^ sin (P x , P 2 ), where (P x , P 2 ) expresses the angle between the directions of the forces. Hence A n =0, A M =0 ; applying the same reasoning to the other forces, we infer that 1= SP^, sin (P^P.,). It follows that I is half the sum of each force multiplied by the sum of the moments of all the other forces about it, each moment being taken with its proper sign. It also follows that the invariant of any number of forces is the sum of their invariants taken two and two with their proper signs. 216 FORCES IN THREE DIMENSIONS. [CHAP. VII. Show that the invariant of a force P and a couple whose moment is G is PG cos 6, where $ is the angle the direction of the force makes with the axis of the couple. Ex. 3. The invariant I of two wrenches whose forces are P 1 , P 2 and couples i\ , r 2 is r x Pi + r 2 p 2 + (iy^ + iy,) cos e + p^r sin e. Ex. 4. A system of forces in equilibrium being given, we may separate the forces into two portions and determine the two invariants of each portion separately, viz. Ij and Z 2 , R^ and P 2 a . Show that the corresponding invariants for each portion are equal in magnitude viz. -T 1 = I 2 and R 1 *=R 3 a . [Chasles.] If we reverse the forces in one portion, the two sets of forces are equivalent. Their invariants are therefore equal. Ex. 5. If the system is equivalent to the forces X, Y, Z, acting along oblique axes and the couples L, M, N, whose axes coincide with the oblique axes, show that the invariant I is I=LX+MY+NZ + (YN+ZM) cos («/, z) + (ZL + XN) cos(z, x) + (XM+LY) cos (x,y). 282. Invariants of two systems of forces. Given two systems of forces P 1; P 2 &c. and Q lt Q 2 &c, we form the two expressions SPQr sin (P, Q), 2PQ cos (P, Q), where r is the shortest distance between the forces P, Q, and (P, Q) is the angle between these forces, the products being taken with their proper signs. Then each of these expressions is invariable when we change either system into any equivalent system of forces. This theorem is given by Chasles, Liouville's Journal, 1847. To prove this consider both systems as one, then however the forces Q lt Q 2 &c. may be changed, P lt P 2 being unaltered, the invariant / of the united systems remains the same. Hence SJyVasinCPx, P 2 ) + SQxQ/i 2 sin(Q 1 , &) + SPQr sin (P, Q) is invariable. But the two first terms are each invariable. Hence the last term is also invariable. In just the same way by considering the invariant B? we may show that 2PQ cos (P, Q) is also invariable. 283. Examples. Ex.1. Forces la, mb, nc act in three non-intersecting edges of a parallelepiped, where a, b, c are the lengths of those edges. Prove that, if the system be reduced to a wrench, the product of the force and couple of that wrench is (Im + mn + nl) V, where V is the volume of the parallelepiped. [St John's, 1890.] Ex. 2. A system of n given forces is combined with another force P, which is given in magnitude and passes through a fixed point ; prove that, if the n + 1 forces have a single resultant, P must lie on a right circular cone, and that, if their least principal moment be constant, it must lie on a cone of the fourth degree. In the second case, prove that if the n forces reduce to a couple, the central axis of the n + 1 forces lies on a hyperboloid of revolution. [Math. Tripos, 1871.] ART. 284.] THE CYLINDROID. 217 Ex. 3. If a system, consisting of two forces whose lines of action are given and a couple whose plane is given, admit of a single resultant, prove that the direction of this resultant lies upon a certain hyperbolic paraboloid. [Math. Tripos.] Ex. 4. A rigid body is acted upon by three forces 2P tan A, - P tan B, IP tan G along three edges of a cube which do not meet, symmetrically chosen with respect to the axes of coordinates drawn parallel to them through the centre of the cube. Prove that the forces are equivalent to a single force acting along the line whose equations are la cot B - x cot A — 2y cot B + a cot A = - z cot G, where 2A, 'ZB, 2G are the angles of a, triangle whose sides are in arithmetical progression, and 2a is the edge of the cube. [Math. Tripos, 1867.] Ex. 5. If the rectangle under the three pairs of opposite edges of a tetrahedron are equal to each other, show that four equal forces acting along the sides taken in order of the skew quadrilateral formed by leaving out one pair of opposite edges are equivalent to a single resultant force ; and that the lines of action of the three single resultants obtained by leaving out different pairs of opposite edges in succession are the three diagonals of the complete quadrilateral in which the faces of the tetrahedron are cut by a certain plane. [Coll. Ex., 1889.] Consider the system of equal forces which act along the edges AB, BC, CD, DA, the empty edges being AG and BD. The resultant of the forces AB, BC bisects the angle ABC externally, and therefore intersects AG in a point P which divides AG externally in the ratio AB : BC. Since this ratio is equal to AD : DG, the resultant of the forces CD, DA intersects A G in the same point P. Hence the four forces have a single resultant which intersects AC in P. It may be proved in the same way that the single resultant of these four forces intersects BD in a point Q which divides BD externally in the ratio BA : AD or BC-.CD. Taking the given tetrahedron as the tetrahedron of reference, the tetrahedral equation x\lAB.AC.AD + y\/BA.BG~B~D + z*JCA . CB .GD + uslDA . DB .DC=0, represents a plane formed symmetrically with regard to the edges. Putting y = Q u=0, it is evident on inspection that this plane intersects the edge AC in P. In the same way it intersects the edge BD in Q, and therefore contains the single resultant PQ. By symmetry, it will contain all the three single resultants. The plane intersects the faces of the tetrahedron in four straight lines which form the sides of a quadrilateral. The three single resultants pass through the corners of this quadrilateral and yet do not lie on the faces of the tetrahedron. They are therefore the diagonals of the quadrilateral. On Screws and Wrenches. 284. The Cylindroid. This surface has been used by Sir R. Ball for the purpose of resolving and compounding wrenches. Following his line of argument we shall first examine a special case, and thence deduce the general solution. 218 FORCES IN THREE DIMENSIONS. [CHAP. VII. To find the resultant of two wrenches of given, intensities on screws of given pitches which intersect at right angles. Let the axes of these screws be the axes of * and y. Let X, Y be their forces ; p, p' their pitches. Let R be the resul- tant of the forces X, Y, and let OA be its line of action. Let G be the resultant of the couples Xp, Yp' and let OB be its axis. Let the angle A OB = . By resolving G into G cos <£ about OA and G sin

    )/R. The force along the axis is equal to R and the couple round it is equal to G cos (f>. Since G cos $ and G sin are the moments about OA and a perpendicular to OA, we see that, if be the angle xOA, GcoS(f> = Xp cos + Yp' sin0 = R (p cos 2 8 +p' sin 2 0) G sin

    be the angle between their re- spective axes and let h be the shortest distance between them. We suppose that in the standard case, when and h are positive, the positive direction of each axis is such that a force acting along it would produce rotation about the other axis in the positive direction ; see Art. 26o. Let R be the force of the wrench. 222 FORCES IN THREE DIMENSIONS. [CHAP. VII. Take the axis of the screw as the axis of z and the shortest distance OH as the axis of x. Let H G and HB be drawn parallel to the axes of z and y respectively. The force R may be resolved into R cos , R sin along HG and HB. When the body is translated a space pdO parallel to the axis of s and rotated an angle dO about it, the work of the former force is R cos $ . pdd ; the work of the latter is R sin . hd6. The couple Rp' of the wrench may be resolved into two couples Rp' cos and Rp' sin whose axes are HG and HB. The work of the former is Rp' cos + h sin $}. We notice that this is a symmetrical function of p and p , so that if the two screws are interchanged the work is unaltered. 293. Ex. If I be the invariant of two wrenches whose forces are P, P' and couples T, I", and dW, dW be the works done by these wrenches when each is twisted about the screw of the other through the same angle dd, prove that PdW'ld8 = P'dWjd8=l - PT-P'T'. See Art. 281. 294. Reciprocal screws.* Two screws are said to be reci- procal when a wrench acting on either does no work as the body is twisted about the other. The analytical condition that two screws are reciprocal is therefore (p +p') cosd + h sin = 0. For example, two intersecting screws are reciprocal when either they are at right angles or their pitches are equal and opposite. It follows from the principle of virtual work that a body free to move only oh a screw a is in equilibrium if acted on by a wrench on any screw reciprocal to a. 295. If a screw a is reciprocal to each of two given screws, say a and /3, it is also reciprocal to every screw on the cylindroid containing a and /3. For a wrench on any third screw 7 on this cylindroid may be replaced by two wrenches on the screws a and * The theory of reciprocal screws is due to Sir R. Ball and the substance of Arts. 294 to 297 is taken from his book on Screws. To this work the reader is referred for further development. ART. 297.] RECIPROCAL SCREWS. 223 /3, if the forces on a. and /3 are the components of the force on 7 (Art. 288). Since the virtual work of each of these when twisted along a- is zero, the screws 7 and a are reciprocal. We may say for brevity that the screw z' and 0>0'; Art. 292. It will be seen that a force acting along the positive direction of the axis of either screw would tend to produce rotation round the axis of the other in the negative direction. We therefore put h=z-z',

    , ' with these two planes such that the given ratio n is equal to Fp sin : F'p' sin 0'. Here p and p' are the perpendicular distances of from the given straight lines. Ex. 6. The moments of the forces about the sides of a triangle ABC are respectively M 1 ,M i ,M 3 , and Z is the resolved force perpendicular to the plane of the triangle. Prove (1) that the trilinear coordinates of the nul point of the plane referred to the triangle ABO are MJZ, MJZ, MJZ ; and (2) that the nul planes of the three corners A, B, C intersect the plane of the triangle in AO, BO, GO respectively. 312. To find the equivalent wrench of two conjugate forces F, F'. Let the shortest distance between the forces be taken as the axis of x, and let them intersect this axis in A and A'. Since Poinsot's central axis is parallel to the resul- tant It of the forces transferred to any base, that axis must be per- pendicular to x. Let us then choose as the axis of z a straight line paral- / lei to this resultant, and let the origin be at some point yet to be found. Let a be the angle between the forces F, F', and let 7, 7' be the angles they make with their resultant. Then = 7 + 7' an( i +--£-- J- a). sin 7 sin 7 sin a w Let OA = a, OA' = a' measured positively from in opposite directions. The forces F, F' may be transferred to by introducing the couples Fa, F'a in the planes containing Ox and the forces F, F' respectively. The resultant of these will be a couple Y in the plane xy if each moment is proportional to the sine of the angle between the axes of the other two, i.e. if Fa = F'a = T sin (i7r — 7') sin (\ir — 7) sin a ^ 230 FORCES IN THREE DIMENSIONS. [CHAP. VH. If the origin be placed so that this relation holds, the two given forces have been reduced to a force R along the axis of z and a couple Y in the plane xy. The axis of z is therefore the central axis. We easily infer by division a tan y = a' tan 7 = F/R (3). It follows from this transformation that the central axis of two forces intersects at right angles their shortest distance, and divides - that distance in the ratio of the tangents of the angles those forces make with their resultant. Since the central axis is perpendicular to the shortest distance, it is obvious beforehand that the central axis must also intersect that shortest distance. For the moment of the two forces about their shortest distance is zero, but no line perpendicular to the axis of the equivalent wrench can have zero moment unless it intersects that axis. 313. When the forces B\ F' are at right angles the con- struction is somewhat simplified. In this case 7 and 7' are complementary and a is a right angle. We easily find by multiplying (1) and (2) that F 2 a = F'W, so that the central axis" divides the shortest distance in the inverse ratio of the squares of the forces. We also find from (3) that ad = (r/Rf, so that the product of the segments of the shortest distance is the same for all sets of conjugate forces at right angles. It also follows that R? = F 2 + F* and TR = FF'r, where r is the length of the shortest distance between F and F'. 314. Ex. I. A given straight line making an angle 7 with the central axis cannot be a characteristic of any plane unless its shortest distance from the central axis is (r tan y)jE. This follows from Art. 312. Ex. 2. If a given system of forces be reduced to twp forces, each equal to F, show that their lines of action must be opposite generators of the hyperboloid p 1 z 2 =a! i {x i + y i -w i ), where the axis of 2 is the central axis of the given system, p is the pitch of the equivalent wrench, and a 2 is such that 2Fa / (a 2 + p 2 )* is the intensity of the wrench. Ex. 3. If a given system of forces be reduced to two conjugate forces of given magnitudes F and F', prove that the first of these is a generator of the hyperboloid {(o 2 +p 2 )TO 2 -p a }(a; 2 +2/ 2 -a !! )=2)% !! , where F=nF' and a is a constant depending on the magnitudes of F, F' and the given system. ART. 315.] CONJUGATE FORCES. 231 Ex. 4. Prove that the axis of the resultant of two given wrenches (P tI l\) and (P 2 , r„), the axes of which are inclined to each other at an angle 8, intersects the shortest distance (2c) between the axes at a point the distance of which from the middle point is P? p* '+ p tf ^S "" ' ' [Math " Trip0S - 1887>] Ex. 5. One polyhedron being given, another is derived from it by taking as its faces the nul planes of the corners of the first. Show (1) that the first polyhedron may be derived from the second by the same construction, (2) that each face of one has as many sides as the corresponding corner of the other has edges, (3) each edge of one is conjugate to an edge of the other, (4) that conjugate edges project into parallels on a plane perpendicular to the central axis. [Cremona.] Ex. 6. The locus of the axes of the principal couples at all bases situated on a given straight line is a hyperbolic paraboloid. This paraboloid is a plane when the given straight line can be a characteristic, and in this case the envelope of the axes of the principal couples is a parabola whose focus is the pole of the plane. [Chasles.] Let AB be the straight line, CD its conjugate. The axis of the principal couple at any point on AB is perpendicular to the plane OGD, Art. 303. If the straight line AB were turned round CD as an axis of rotation through any small angle d0, each point on AB would move a small space perpendicular to the plane OGD, i.e. it would move a small space along the axis of the principal couple. Hence these axes all intersect two straight lines, viz. AB and its consecutive position, and are all parallel to a plane which is perpendicular to CD. The locus is therefore a hyperbolic paraboloid. Theorems on forces. 315. " Three forces. If three forces are in equilibrium, they must lie in one plane. Let A and B be any two points on two of the forces. Since the moment about the straight line AB is zero, this straight line must intersect the third force in some point G Let A be fixed and let B move along the second line ; the straight line AB will describe a plane, and the second and third forces must lie in this plane. If we fix G and let B move as before, we see that the first force must also lie in the same plane. Ex. 1. The forces of a system can be reduced to three forces which act along the sides of an arbitrary triangle ABC together with, three other forces which act at the corners A, B, C at right angles to the plane of the triangle. Resolve each force of the system into two, one in the plane ABC and the other perpendicular to that plane. See Art. 88, Ex. 1, and Art. 120, Ex. 3. Ex. 2. The forces of a system may be reduced to three forces which act at the corners of an arbitrary triangle, ABC, in an infinite number of ways. Show also 232 FORCES IN THREE DIMENSIONS. [CHAP. VII. that one of these forces, say that at A, may be made to act perpendicular to the plane of the triangle, and a second, say that at B, may be made to act so that the plane containing the forces at A and B is perpendicular to the plane of the triangle. This follows from the last example by making the force along the side All act at /J, and those along BG, GA act at G. 316. Four forces. If four forces are in equilibrium, show that they must be generators of the same hyperboloid. Mcebius Lehrbuch der Statik. An infinite number of straight lines can be drawn cutting three of the forces. Each of these must intersect the fourth, for other- wise the moment of the forces about that transversal could not be zero. Taking any three of these transversals as directors, the four forces must act along generators of the same system of a hyperboloid of one sheet. 317. Ex. 1. If four forces P lt P 2 , P 3 , P 4 are in equilibrium, prove that the volume of the tetrahedron constructed on any two as P lt P 2 is equal to the volume of the tetrahedron constructed on the remaining two P 3 , P 4 . [Chasles.] Reversing the directions of the two latter forces, the forces P lt P 2 become equivalent to the forces P 3 , P 4 . Their invariants are therefore equal, and hence the volumes of the two tetrahedra are equal, Art. 231. Ex. 2. Four forces acting along the straight lines a, b, c, d are in equilibrium. If the symbol ab represent the product of the shortest distance between a, b into the sine of the angle between them, show that the forces acting along these lines are proportional to (6c . cd . db)i, (cd . da . acfi, {da . ab . bdfi, {ab . be . ca)i. [Oayley, Comptes Rendus, 1865.] We have by Chasles' theorem PjP 2 . ab = P 3 P 4 . cd and P^ . ac=P i P i . bd. Multiplying these together we have the ratio of P a s : P/. The result agrees with that given above. 318. Four forces act along generators of the same system of a hyperboloid. Their magnitudes are such that if transferred parallel to themselves to act at one point they would be in equilibrium. Prove that they are in equilibrium when acting along the gene- rators. Refer the system to the axes of the hyperboloid as coordinate axes. Let a, b, cV — 1 be these axes. Let any one of the gene- rators be (x - a cos 6)1 a sm 6 ={y-b sin 6)j -bcosd = s/c. ART. 318.] THEOHEMS ON FORCES. 233 Let P be the force along this generator. Let X, Y, Z be its resolved parts ; L, M, N its moments about the axes. We easily see that X = -Zsm6, Y=- h -Zcos0, c c lJZx, m=™y, n=-^z. a b ' c Taking all four forces the three equations of resolutions are therefore XZ sin 6 = 0, XZ cos 0=0, 1Z=0. The three equations of moments are evidently the same as these. Ex. 1. Four forces in equilibrium act along four generators of a hyperboloid and intersect the plane of the real axes in A v A 2 , A 3 , A v Show that the resolved parts of the forces parallel to the imaginary axis are proportional to the areas of the triangles 4 2 4 3 4 4 , A g A t A 1 &o., the forces at adjacent corners of the quadrilateral A^AgAi having opposite signs. The ratios of Z 1 ...Z i may be found by solving the three equations of resolution given above. The determinants which express these ratios are proportional to the areas of the triangles. Ex. 2. Any number of forces act along the generators of the same system of a hyperboloid whose semi-axes are a, b and c^J - 1. If Z be the resolved part parallel to the imaginary axis of the force along the generator whose parameter is 6, the invariant of the forces is I=— 2{l -cos (6- 6')} ZZ'. Ex. 3. Forces act upon a rigid body along a number of rectilinear generators of the same system of an hyperboloid of one sheet. Prove that the necessary and sufficient condition of their being reducible to a single resultant is that their central axis should be parallel to one of the generating lines of the asymptotic cone. [Math. Tripos, 1877.] Ex. 4. Any number of forces act along generators of the same system of a hyperboloid' whose axes are a, b and c,/ - 1. Show that the central axis is a generator of a system of concyelic hyperboloids whose equation referred to the principal axes of the given hyperboloid is where p is the pitch of the system of forces. Show also that the pitch p must lie between - ab/c and the greater of the quantities beja and cajb. be If X, Y, Z, L, M, N be the six components of the given forces, we find L — — X, a M= — Y, N= Z. The equation to the central axis is b ' c (L-riZ + tY)IX=W-tX+&)IY={N-SY+riX)IZ. 234 FORCES IN THREE DIMENSIONS. [CHAP. VII. Equate each of these to p and substitute for L, M, N. We thus obtain three linear equations from which we may eliminate X, Y, Z. The result is the hyper- boloid required. Ex. 5. A system of forces have their directions along any non-intersecting generators of a hyperboloid of one sheet ; Bhow that the resultant couple at the centre of the hyperboloid lies in the diametral plane of the resultant force, and the least principal moment is ,. _ , — 2 _ 2 ; D 1 and -D 2 being the semi-axes of the section of the hyperboloid by the plane of the couple, and a, b, c the semi-axes of the surface, and B the resultant force. Explain the difficulty in the geometrical interpretation of these results for a single force. [Math. Tripos, 1880.] 319. Ex. 1. Forces act at the centres of gravity of the four faces of a tetrahedron perpendicularly to those faces and proportional to them in magnitude. Show that they are in equilibrium. A force equal to P acting at the centre of gravity of any face may be replaced by two parallel forces, viz. one equal to fP acting at the centre of gravity of the volume, and the other equal to £P acting in the opposite direction at the corner of the tetra- hedron opposite to the given face. The four forces at the centre of gravity of the volume are in equilibrium because they act at a point and are both proportional to and perpendicular to the faces of a tetrahedron : Art. 47. The forces at the four corners are in equilibrium by Art. 50, Ex. 2, Chapter II. Ex. 2. Forces act at every element of the surface of a solid figure perpendicularly to those elements and proportional to them in magnitude. Prove that they are in equilibrium. Taking any rectangular axes, we see by Art. 47, that their resolved parts in the directions of the axes are zero. Consider next their moment round the axis of x. If the force at any element d) da, where (p, , z) are the cylindrical coordinates of da, prove that the moment about the axis of 2 is zero. Ex. 5. One-eighth of an ellipsoid is cut off by the principal planes, and along the normal at any point a force acts proportional to the element of surface at that point. Shew that all these forces are equivalent to a single force acting along the line where 2a, 26, 2c are the principal axes of the ellipsoid. [June Exam.] ART. 321.] THEOREMS OX FORCES. 235 v 320. Five forces. Five forces acting on a rigid body are in equilibrium. Show that they must intersect two straight lines, which may be coincident or imaginary. Mcebius, Statik. Describe the hyperboloid which contains three of the forces as generators of one system. The fourth force will intersect this hyperboloid in two points. One generator of the other system passes through each of these points and intersects all the four forces. Each of these must intersect the fifth force, for otherwise the moments about them could not be zero. These two straight lines may be called the directors of the five forces. 321. Let the shortest distance between two straight lines be taken as axis of z. Let any five forces intersect these straight lines at distances (r x r/) (r 3 r 2 ') &c. from, that axis, and let Z it Z^ disc, be the z resohites of these forces respectively. Prove that the condi- tions of equilibrium are %Z=0, XZr=0, ZZr' = 0, 2Zrr' = 0. Let the origin bisect the shortest distance between the two directors of the forces, and let this shortest distance be 2c. Let 2a be the angle between the directors, and let the axes of x and y be its bisectors. The equation to any force may then be written (x — r cos a)/(r — r') cos a = (y — r sin a)/(r + r') sin a = (z — c)/2c. Writing l//* 2 = (r - r') 2 cos 2 a + (r + r'f sin 2 a + 4c 2 , and representing the forces by P x . . . P 5 , the equations of equilibrium formed by resolving along the axes are 2Pfj.(r -r) cos a = 0, 2P/i (r + r') sin a = 0, SP/ttc = 0. The equations of moments are %(yZ-zY)= SP/i (r -r')c sin a = 0, % (zX - xZ) = - XPfJ, (r + r') c cos a = 0, 2 (x Y — yX) = 22P/* rr sin a cos a = 0. When c and sin 2a are not zero, these six equations reduce to the four given above. These four equations determine the ratios of the five forces P l ...P 5 when the intersections of their lines of action with the directors are known. 236 FORCES IN THREE DIMENSIONS. [CHAP. VII. 322. Let the two directors be moved so that either their mutual inclination 2o or their distance apart 2c is altered, but let them continue to intersect the axis of z at right angles. It follows from these results that equilibrium will continue to exist provided (1) the forces always intersect the directors at the same distances from the axis of z, and (2) the z component of each is unchanged. When five forces in equilibrium are given in one plane, which besides the three conditions of equilibrium also satisfy the condition 2Zrr' = 0, we may by this theorem construct five forces in space which are also in equilibrium. 323. Ex. Show that, if the algebraic sums of the moments of a system of forces about (1) three, (2) four, (3) five straight lines are zero, the central axis of the system (1) lies along one of the generators of a system of concyclic hyperboloids, (2) intersects a fixed straight line at right angles, (3) is fixed. [Math. Tripos, 1888.] Replace the system by two conjugate forces, one of which cuts the three given straight lines. Then the other force also cuts the same three lines. They are therefore rectilinear generators of a fixed hyperboloid. The first result follows at once by Art. 318, Ex. 4. Choose one of the conjugates to cut the four given straight lines as in Art. 320. The other also cuts the same four lines. Both these forces are therefore fixed in position. By Art. 312 the central axis cuts the shortest distance between these at right angles. If the moments about five straight lines are zero, we can by taking two sets of four forces obtain two straight lines each of which is cut at right angles by the central axis. The central axis is therefore fixed. y. \ja 324. Six forces*. Forces acting along six straight lines are ^\ in equilibrium. Show that, five of these lines and a point on the ^ sixth being given, the sixth straight line must lie in a certain plane. Let Pi,... Pi be four of the forces whose lines of action are given. Let us represent these lines by the symbols 1, 2, 3, 4. Draw the two transversals (Art. 320) which intersect these four straight lines. We shall call these the transversals a and b. We notice that the positions of these two transversals are independent of the magnitudes of the four forces P^.^P^ * The theorem that the locus of the sixth force is » plane is due to Mcebius, Lehrbueh der Statik, 1837. But he omitted to give a construction for the plane. This defect was supplied by Sylvester " sur Vinvolution des lignes droites dans Vespace considered comme des axes de rotation." Comptes Bendus, 1861. He gives several theorems on the relative positions of the fifth and sixth lines. The terms " involution " and " polar plane " are due to him. In a second paper in the same volume he states as the criterion for the involution of six lines the determinant given in Art. 333, the moments (12) &c. being replaced by secondary determinants when the equations of the straight lines are given in their most general form. He mentions that Cayley had found a determinant which is the square root of that given by himself and which would do as well to define involution. A proof of this is given by Spottiswoode, Comptes Rendus, 1868. See also Scott's Theory of Determinants. Analytical and statical investigations connected with involution are given by Cayley, "On the six coordinates of a line," Cambridge Transactions, 1867. The extension of the determinant of Art. 333 to six Wrenches is given by Sir R. Ball, Theory of Screws, 1876. 0* ART. 325.] THEOHBMS ON FORCES. 237 Since the six forces P 1 ...P e are in equilibrium, the moment of P 5 and P 6 about each of the transversals is zero. Hence P 5 (5a) + P 6 (6a) = 0, P 6 (5&) +• P 6 (66) = (1) where the symbol (5a) represents the moment about either of the lines 5 or a of a unit force acting along the other. We therefore have (56) (6a) -(5a) (66) = (2). Thus the sixth line is so situated that the moment about it of two forces proportional to (5b) and — (5 a) acting along a and b respec- tively is zero. We notice that the ratio of the two forces which have thus been applied to the transversals a and b is independent of the magnitudes of the four forces P x . . . P 4 . It follows from this reasoning that the sixth line is a nul line of these two forces. Hence, if any point in the line of action of P 6 is given, that force must act along a straight line which lies in the nul plane of taken with regard to these two forces. The positions of the transversals a and b depend on the positions of the four lines 1, 2, 3, 4. The ratio of the forces which are applied to these transversals depends on the position of the line 5 relatively to a and b. Let these forces be called P a and Pa respectively. The transversals (a, 6) and the lines (5, 6) are so related that a, b are nul lines of the forces P 5 , P 6 and 5, 6 are nul lines of the forces P a ,P . 325. Conversely, the five lines 1, 2, 3, 4, 5 being given, and any line drawn through the given point in the appropriate plane being chosen as the sixth line, it is required to prove that forces can be found which, when acting along these six lines, would be in equilibrium. Since the equation (2) is satisfied by this choice of the sixth line, the two equations (1) determine the ratio P 6 : P 6 . Since a and b are nul lines for these two forces, they may be replaced by two finite conjugate forces (say P 7 , P 8 ), each of which intersects the transversals a and b. See Art. 311, Ex. 4. When the positions of the forces P 7 and P 8 have been thus chosen, the ratio of their magnitudes may be found by taking moments about their lines of action. The moments of P 8 about P 7 and P 7 about P 8 are respectively equal to those of the system P 5 and P 6 about the same axes. The ratio of P B to P 6 being known, that of P, to P 8 may be deduced. We have now the forces P 1 ..P 4 , P 7 , P 8 intersecting a and 6. Proceeding as in Art. 321, the six necessary and sufficient equations of equilibrium are reduced to four linear equations. From these we can find the four ratios of P 1 . . P 4 , P 7 , though in special cases the magnitudes of some of these forces may be zero. 238 FORCES IN THREE DIMENSIONS. [CHAP. VII. 326. Besides arranging the possible positions of the sixth line in a series of planes, we may distribute them in other ways. The positions of the sixth line are distinguished as being the nul lines of two forces made to act along the transversals a and b. Assuming, for example, any two points A and B, their nul planes with regard to these two forces will intersect in some straight line CD. It is easy to see that the straight lines AB and CD will be the seats of two conjugate forces equivalent to the two forces along a and b (Art. 310). Any straight line intersecting these conjugate forces will be a nul line, and may therefore be a possible position of the sixth force. 327. When six straight lines are so placed that forces can be found to act along them and be in equilibrium, the six lines are said to be in involution, The plane passing through the point which is the locus of the sixth line is called the polar plane of with regard to the five given lines. When the force along one line is zero, the five remaining lines are said to be in involution. When six lines are in involution, any force acting along one of them can be replaced by five finite components acting along the remaining five of proper magnitudes, provided these five lines alone are not in involution. 328. The sixth line will remain in involution with the five given straight lines 1 ...5 as it revolves round in the polar plane of 0. The ratios of the forces P!...P S will however change as the line rotates. Let the straight line joining to the intersection of its polar plane with the transversal a be taken as the sixth line. Then since the sixth line is a nul line of the forces which act along the transversals, it will also intersect the transversal b. Thus the polar plane of intersects the transversals a and b in two points which lie in the same straight line with 0. The position in space of this straight line may be constructed when the four straight lines 1, 2, 3, 4 and the point are known. Let it be called the line c of the point with regard to the four lines 1, 2, 3, 4. To construct this line, we first find the two trans- versals a and b, we then pass a plane through and each of these transversals. The intersection of these planes is the line c. ART. 330.] THEOREMS ON FORCES. 239 If we had begun by finding the two transversals a', b' of some other four of the five given lines say 1, 2, 3, 5, we must have arrived at the same plane as the polar plane of 0. Thus by combining the forces in sets of four, we may arrive at five such lines as c. All these lie in the polar plane of 0, and any two will determine that plane. Ex. When the four liaes 1, 2, 3, 4 and the point are given, the fifth line being arbitrary, show that the polar plane of passes through a fixed straight line. This is the straight line c. 329. By equations (1) of Art. 324 we have P 6 (5a)+P 6 (6a) = 0, P 5 (56) + P„(66)=0 (1). When the sixth line is in the position c, the moment of, the sixth force about each of the transversals a and 6 is zero. When the sixth line has revolved in the polar plane of from this position through an angle 0, the moment of the sixth force may be found by resolving P 6 into two forces, one along the line c and the other along a line d drawn perpendicular to c in the polar plane of 0. The moment of the first is zero, that of the second is P, sin . {da) or P 6 sin 6 . (db). It follows from either of the equations (1) that the ratio P 6 : P 6 is proportional to sin and is there- fore greatest when the sixth line is perpendicular to c. We have assumed that the moments (5a) and (56) are not both zero, i.e. that the five given straight lines are not so placed that they all intersect the same two straight lines ; see Art. 320. When this happens the lines 1, 2, 3, 4, 5 alone are in involution. The equations (1) then show that the force P s is zero when its line of action does not intersect the same directors. 330. Ex. 1. If A, B, C, D, E, F be six lines in involution, the polar plane of with regard to A, B, G, D, E is the same as the polar plane of with regard to A, B, G, D, F, the forces along E, F not being zero. For let M be any straight line through in the first polar plane, then a force acting along M can be replaced by five forces along A, B, C, D, E. But the force along E can be replaced by forces along A, B, C, D, F, hence the force along M is equivalent to forces along A, B, G, D, F, i.e. M lies in the second polar plane. The two polar planes therefore coincide. Ex. 2. Supposing two transversals, say a and 6, to be known, we may take with regard to these the convenient system of coordinates used in Art. 321. Let 2c be the shortest distance between the transversals, 2o the angles between their directions. Let (l + /i)/(l-/n) be equal to the known ratio (5a) : (56), i.e. to the ratio of the moments of the fifth force about the transversals a and 6 (Art. 324). Show that the polar plane of is x sin a (h + /ic) + y cos a {fih + c) - z {fain a + /j.g oob a)=c (/if sin a + g cos a). This is obtained by substituting in (2) of Art. 324 the Cartesian expression for a moment given in Art. 267. 240 FORCES IN THREE DIMENSIONS. [CHAP. VII. 331. Analytical condition that six lines are in involution. Let a force P be given by its six components PI, Pm, Pn ; P\, P/j., Pv, Art. 260. I{(fgh) be any point on its line of action, then X = gn — hm, /i = hl —fn, v =fm — gl. Let us suppose that each of the six forces P l . . . P 6 is given in this way, so that ft, »»], %, Xi, /*i, v^) (l 2 , &c.) &c. may be regarded as the coordinates of their several lines of action. Since the six forces are in equilibrium, they must satisfy the six necessary and sufficient equations given in Art. 259. We have therefore 2PZ = 0, 2Pto = 0, 2P« = 0; 2P\=0, 5P/* = 0, IPv = 0. These six equations will in general require that each of the forces P 1 ...P 6 should be zero. But if we eliminate the ratios of these forces we obtain a determinantal equation which is the condition that the six lines should be in involution. This determinant has for its six rows the six coordinates of the six given straight lines, viz. k, «i, n u #!% - hinh, hj, x -/i«i, fw-gJt 1%, &c. Let us suppose that five of the lines are given and that the sixth is to pass through a given point (f e , g 6 , h t ). Let (so, y, z) be the current coordinates of the sixth line, then writing for (Z 6 m„ n s ) in the last row their ratios oc—fe, y — g e , z — h e this determinantal eqxiation becomes the equation to the locus of the sixth line. It is clearly of the first degree and this proves in an analytical way that the locus of the sixth line is a plane. This therefore is the equation to the polar plane of the given point (/ 6 g s A 6 .) 332. If the six straight lines are the seats of six wrenches of given pitches, instead of six forces, we may by an extension of this determinant form the condition that these wrenches may be in equilibrium. Let P be the force of any wrench, p the pitch of its screw. Let (I, m, n, A, ft, v) be the six coordinates of its axis. Then, resolving parallel to the axes of coordinates and taking moments as before, we have 2PZ = 0, SPto = 0, XPn=0. 2P(\+pl) = 0, 2P(/u+pro) = 0, 2,P(v+pn) = = 0. ART. 335.] «BIX FORCES. 241 Eliminating the forces, we have the following six-rowed deter- minantal equation in which the first line only is written down. k,rrh, «i. \+Pik, fh+Piini, v 1 +p 1 n 1 =Q The other lines are repetitions of the first with different suffixes. This determinant has been called the seoeiant by Ball. By giving to the pitches px.-.p^ of these screws values either zero or infinity we can express the condition that m forces and n couples (m + n = 6) connected with six given straight lines should be in equilibrium. 333. If we take moments in turn for the six forces Pi-.-Pn about their lines of action, we obtain six equations of the form P 1 .0 + P 1 (12) + P I (13)+P«(14) + P I (15)+P,(16) = 0. Eliminating the six forces, we obtain a determinant of six rows equated to zero. This is the necessary condition that the six lines should be in involution. Taking any five of these equations, we can find the ratios of the six forces. Thus, if I 12 represent the minor of the constituent in the first row and second column, we have P 1 // u = P 3 // ia = P s // 13 = &c. Since by Salmon's higher algebra I u I a = I\, we may deduce the more symmetrical ratios PiY/ii = Pflln = PsV^as = &C This symmetrical form for the ratios of the forces is given by Spottiswoode in the Comptes Rendus for 1868. 334. We have thus two determinants to define involution. One expresses the condition in terms of the coordinates of the six lines, the other in terms of their mutual moments. These are not independent, for one determinant is the square of the other. This may be shown by squaring the first and remem- bering the expression for the mutual moment of two lines given in Ex. 2 of Art. 267. 335. Let A, B, C, D, E, F be six lines not in involution, then any given force B may be replaced by six components acting along these six lines. R. S. 16 242 FORCES IN THREE DIMENSIONS. [CHAP. VII. Let be any point on the force R, let a and /3 he the polar planes of with regard to two sets of five lines, say A,B,C,B,E and A, B,G,D,F. These two polar planes intersect along the line c, Art. 328. If the planes a and /3 are different, the force R at may be resolved into two forces OA, OB one in each plane. Each of these is in involution with one of the two sets of five forces and may therefore be replaced by its components along the five of its own set. Thus R has been replaced by six forces acting along the six lines. If the force R acts along c, it can be replaced by four forces acting along A,B,G, B, the forces along E and F being zero. The planes a and /3 could not be coincident, for then two equal and opposite forces acting at along some line other than c could be replaced by two sets of five forces acting along the two sets of five lines. Thus six forces acting along the six lines would be in equilibrium, and therefore the six lines would be in involution. Ex. 1. The force R may be replaced by the two forces along OA, OB in an infinite number of ways. Show that all these ways lead only to the same distri- bution of force along the six given lines. To find the distribution due to the force OA or OB resolve it into two forces, one along c and the other perpendicular in its own plane. These are fixed lines and given forces along them have fixed distributions. It may then be shown that whatever angle OA may make with the line c, the components of OA, OB along these fixed lines are the same and equal to the components of R in the same directions. Ex. 2. Show that in general there is only one way of reducing a system of forces to six forces which act along six given straight lines. If the lines of action of five of the forces be given and the magnitude and point of application of the sixth, prove that the line of action of the sixth will lie on a certain right circular cone. [Coll. Exam., 1887.] 336. If the moments of a system of forces about six straight lines not in involution are all zero, the forces are in equilibriit/m. If they are not in equilibrium let (r, R) be their equivalent wrench. Let the axis of this wrench be taken as the axis of z, and let the six lines make angles {8 X , i, fa), (8 2 , <£ 2 , fa), &c. with the axes of z, as, y. Let (r 1; r/, r"), (r 2 , r{, r 3 ") &c. be the shortest distances between the six lines and the axes of z, x, y. Since each of the six lines must be a nul line with regard to the wrench, we have for each T cos 8 + Rr sin 8 = 0. We shall now prove that, if these six equations can be satisfied by values of T and R other than zero, the six lines are in involution. ART. 337.] GENIAL THEOREMS. 243 If forces Pi.^Pa can be found acting along these six lines in equilibrium, they must satisfy the six necessary and sufficient equations of equilibrium. These are 2Pcos0 = O, 2Pcos<£ = 0, 2Pcos^ = 0, tPr sin = 0, 2Pr' sin = 0, %Pr" sin yfr = 0. These six equations in general require that each of the forces Pj . . . P e should be zero. But when the six conditions given above are satisfied the two equations 5p cos 8 = and %Pr sin 6 = follow one from the other. There are therefore only five necessary and sufficient equations connecting the six forces. The ratios of the forces can be found. Hence the lines must be in involution. Hence, if the lines are not in involution, they cannot all six be nul lines of a Wrench, i.e. T and R must both be zero. It follows that six equations of moments about six straight lines are insufficient to express the conditions of equilibrium of a system if those six lines are in involution. 337. It has been shown in Art. 336 that, if the moment of a system about each of six lines is zero, the forces are in equilibrium, provided the six lines are such that no forces (which are not all zero) acting along the lines can be in equilibrium. By placing some of these lines at infinity we may deduce the following theorem. If a system of forces is such that its moment about each of m lines is zero and its resolute along each of n lines is also zero, where m + n = 6, the system is in equilibrium provided the six lines are such that forces acting along the m lines and couples having their axes placed along the n lines cannot be in equilibrium. The forces and couples are not to be all zero. The expression for the moment of a system of forces about a straight line drawn in the plane of xz parallel to x and at a distance I from it is by Art. 258, L' = L + IY. If I be very great the condition Z' = leads to Y=0. It follows that to equate to zero the resolved part of the forces along y is the same thing as to equate to zero their moment about a straight line perpendicular to y but very distant from it. Now a zero force along such a line at infinity is equivalent to a couple round the axis of y. Since the axis of y is any straight line, it follows that, if a system be such that its moments about m lines are each zero and its resolutes along n lines are also each zero, where m + n= 6, 16—2 244 FORCES IN THREE DIMENSIONS. [CHAP. VII. then the system will be in equilibrium provided the six lines are such that m forces along the m lines and n couples round the n line's cannot be found which are in equilibrium. It is evident that, if m is less than 3, the m forces and the n couples could be chosen so as to be in equilibrium whatever positions the six lines may have. It follows that m equations of moments and n equations of resolution are insufficient to express the conditions of equilibrium if m is less than 3. 338. We may obtain another proof of this theorem by using the same kind of reasoning as that in Art. 336. To simplify the proof, let us suppose that the moments of the system about each of the four lines 1, 2, 3, 4 is zero, and that the resolute along each of the lines 5 and 6 is zero. If the system is not in equilibrium, let (r, R) be the equivalent wrench. Let the axes of coordinates and the notation be the same as in Art. 336. We thus have given the four equations rcos0 1 + Pr 1 sin0 1 =O, V cos 2 + Rr 2 sin 2 = 0, &c. = 0, and the two resolutions R cos 6 = 0. R cos 6 = 0. These six equations may be called the equations (A). Let four forces P^.^ act along the four lines 1....4 and let two couples M s , M e have their axes placed along the lines 5, 6. If these can be in equilibrium, they must satisfy the equations P 1 cos0 1 +...+P 4 cos0 4 =O, P^ sin 0j + . . . + P 4 r 4 sin 4 + M s cos 6 + M e cos e = 0, with four other similar equations obtained by writing and \ji for 8. These six equations may be called the equations (B). The equations (B) in general require that the four forces Pj.-.P, and the two couples M 6 , M e should be zero. But if the equations (A) can be satisfied by values of T and R which are not both zero, the six equations (B) are not independent. If we multiply the first by r and the second by R and add the products together the sum is evidently an identity by virtue of equations (A). The equations (B) are therefore equivalent to not more than five equations, and thus forces P 1 ...P i and couples M s , M e , not all zero, may be found to satisfy them. It follows that, if the six lines are such that the forces P 1 ...P t and the couples M s , M 6 cannot be in equilibrium, the values of r and R given by equations (A) must be zero, i.e. the given system is in equilibrium. Tetrahedral Coordinates. 339. Show that the forces of any system can be reduced to six forces which act along the edges of any tetrahedron of finite volume. Let A BCD be the tetrahedron, let any one force of the system intersect the face opposite B in the point D'. Resolve the force into oblique components, one along DB' and the other in the plane ART. 339.] TETRAHHKRAL COORDINATES. 245 ABC. The former can be transferred to D and then resolved along the edges which meet at D. The second can by Art. 120 be resolved into components which act along the sides of A BG. We shall suppose that the positive directions of the edges are AB, BG, CA, AD, BD, CD ; the order of the letters being such that a positive force acting along any edge tends to produce rotation about the opposite edge in the same standard direction. See Art. 97. We shall represent the forces which act along these sides by the symbols .F 12 , F at . F a , F u , F u , F u . The directions of the forces, when positive, are indicated by the order of the suffixes. When we wish to measure the forces in the opposite directions, the suffixes are to be reversed, so that F- 1S = — F S1 . The volume of the tetrahedron will be represented by V. Ex. 1. Show that the six straight lines forming the edges of a tetrahedron are not in involution. If the forces were in equilibrium we see, by taking moments about the edges, that each would be zero. Ex. 2. A force P acts along the straight line joining the points H, K, whose tetrahedral coordinates are (x, y, z, u) (x', y', z', u') in the direction H to K. If this force is obliquely resolved into six components along the edges of the tetrahedron ABGD, show that the component F 12 acting in the direction AB is P ■==* . , , where the terms in the leading diagonal follow the order indicated by the directions HK, AB, of the forces. To prove this we equate the moments of F u and P about the edge CD. The result follows from the expression for the moment given in Art. 267, Ex. 4. Ex. 3. Show that the mutual moment in the standard direction of the straight &V \ Z U t line HK and any edge as AB is -7-= — ====. ,' , . Here z precedes u in the AB . HK \ z , u I determinant because CD is the positive direction of the opposite edge, and unaccented letters precede accented ones if HK is the positive direction of the line. Art. 267. Ex. 4. Two unit forces act along the straight lines HK, LM in the directions H to K and L to M. If the tetrahedral coordinates of H, K, L, M are respectively {x, y, z, u), (x' &c), (a, /3, 7, 8), (a', &c), prove that the moment of either about the other in the standard direction is == — ^^-r where A is HK . MN the determinant in the margin. The order of the rows is determined by the directions HK, LM in which the forces act ; the order of the columns by the positive directions of the edges. Ex. 5. Show that the nul plane of the point {x, y, z, u) with regard to the six forces .F 12 , F a &o. is F^U.n I+F23 \x,u l + Pjj \y ,u \+F u \y , z | + i^ 4 \z,x \ + F^\x,y 1=0, AB\z',u'\ BC\x',u'\ CA\y',u'\ AD\y',z'\ JBX> ( *', as' | CD\x',y'\ where the accented letters are the running coordinates. Ex. 6. Show that the nul plane of any corner, as D, is xFJB C + yF^ICA + zF^AB = 0, where the suffixes are arranged in cyclical order. X, y, «. u x' ,y' 2', u' a, P, 7> S a' P V S' 246 ' FORCES IN THREE DIMENSIONS. [CHAP. VII. Ex. 7. Show that the areal coordinates of the nul point of the face ABC are proportional to F^AD, F 2 JBD, F-^JGD. If the six forces are equivalent to a single force, show that every plane through this point is a nul plane of the point. Ex. 8. Find the equation to the conjugate line of the edge AB, see Art. 310. Show that any edge is a nul line if the force along the opposite edge is zero, Art. 306. Ex. 9. Forces F 12 , F 23 , F sl , F u , F u , F si act along the six edges AB, BC, GA, AD, BD, CD of a tetrahedron. Prove that they will have a single resultant if 3r^ + SfS + ^=°- r Math - Tr * os ' 1870 -i Ex. 10. Six forces acting along the edges of a tetrahedron have a single resultant force. Show that its line of action intersects each face in its nul point, see Ex. 7. Show also that, if (x, y, z, u), (x' t y', z', u') are the coordinates of two points H , K in the resultant B . x'-x_y'-y_z'-z_u'-u_ HK S 1 S 2 Sa S i R Ex. 11. Six forces F r , &a. act along the edges of a tetrahedron. Find their equivalent wrench. The tetrahedron being given, the cosine of the angle between any two sides is known. Proceeding as in Art. 46, the resolved parts of the forces F 13 &c. along the edges may be found. Let these be denoted by X 12 , X^, &c. The direction angles and magnitude of the resultant B may now be found by Art. 46. If 13 , i 3 ROCAL FIGURES. 253 Considering both the given figure and the reciprocal, we have the relations E=B', C=F', F=C, G + F=E + 2, G' + F'=E' + 2. If the given figure is such that C = F, we have E — 1G-1, E'=2C'-2. In this case the number of corners in either figure is equal to the number of polygons, and each figure has one side more than is necessary to stiffen it. That either figure may be possible, a condition for each must exist connecting the directions of the lines. When the given figure has two faces or polygons only on each side, it may be regarded as the projection of a polyhedron ; it then follows from the theorem proved in Art. 350 that a reciprocal figure can be drawn. The conditions connecting the directions of the lines must therefore be satisfied. If G2C-2, E'<2<7'-2 ; on the same supposition the reciprocal figure is indeterminate. If C>F we have £<20-2, £'>2C-2; in this case the construction of the reciprocal figure is impossible unless G-F conditions are satisfied. A figure therefore that has just the number of sides required to stiffen it, cannot, in general, have a reciprocal figure unless the sides are such as to satisfy some geometrical condition. This result agrees with what has been proved in Art. 232. In a framework which is freely dilatable there can be no tensions apart from the external forces, unless the polygon satisfies such a geometrical condition that' it has an abnormal deformation. Accordingly we find that a freely dilatable figure has no reciprocal unless its sides satisfy some geometrical condition. 350. Maxwell's Theorem. It is clear that each of the figures (1) and (2) of Art. 342 may be regarded as the orthogonal projection of a tetrahedron. Maxwell has generalized this into a theorem by which we can construct reciprocal figures with the aid of some solid geometry. See Edin. Trans., Vol. 26, 1872. Let "one polyhedron be given, and let a second be derived from it by the following rule. If z=Ax+By + C be the equation to any face of the first, the corresponding corner of the second is at the point whose coordinates are £=hA, ?j = 7iB, f= - G, where h is any arbitrary length which is the same for all the faces. Geometrically, these polyhedra are so related that each face of either is the polar plane of the corresponding corner of the other with regard to the paraboloid of revolution x 8 + j/ 2 = 27k. To prove this we notice that the equation to the face may be written in the form h (z + f) = a:{ + yy. But this is known to be the equation to the polar plane of the point (£ijf ) with regard to the paraboloid. To each face of one polyhedron there corresponds a corner of the other. The intersection of two faces is an edge of one polyhedron, and the straight line joining the poles of these faces is an edge of the other. These edges are said to correspond to each other. Consider the edges which meet at a corner A of one polyhedron ; the corresponding edges of the second polyhedron lie in the polar plane of A and are the sides of the face which corresponds to that corner. Thus for every corner in one polyhedron there corresponds a face with as many sides as the corner has edges. unity, G and E remaining unchanged, we therefore have G+F=E + 2 for closed polyhedrons. The limiting case of a polyhedron, all whose corners are in one plane, is a rectilineal figure having two faces only on each side. In such a figure Euler's relation must be true, 254 GRAPHICAL STATICS. [CHAP. VIII. We shall now prove that the projection of each edge of one polyhedron is at right angles to the projection of the corresponding edge of the other, the projections being made orthogonally on a plane perpendicular to the axis of the paraboloid. To prove this we write down the equations to the faces of one polyhedron which are the polar planes of the two corners (ifr;f), (|Vf ) °f trle other. These are Eliminating z, we have the equation to the projection of an edge of the first polyhedron, viz. 7«(f-f )=x (£- %') + y (n-n 1 ). Tne equation to the projection of the edge joining the two corners is (y - rj) (£-£') - (a -J) {n- V)=0. These two projections are evidently at right angles. The two tetrahedra have therefore the property that their projections on the plane of xy are reciprocal figures, with their corresponding sides at right angles. We also infer that any figure which is the projection of a plane-sided polyhedron on any plane can have a reciprocal figure. 351. The following examples indicate a geometrical method of arriving at. these results. Ex. 1. A point H is taken on the axis of a parabola at a distance from the vertex equal to half the latus rectum ; prove that a perpendicular drawn from H on any chord intersects the conjugate diameter of that chord in the tangent at the vertex. Ex. 2. A point H is taken on the axis of a paraboloid of revolution at a distance from the vertex equal to half the latus rectum ; prove that the projection II on the tangent plane at the vertex of any point P lies on the perpendicular drawn from II on the polar plane of P. 3. Let P, P' be two corners of one polyhedron, n, II' their projections, then II II, HIT are perpendicular to the corresponding faces of the other polyhedron. It is evident that the plane HIM' is at right angles to the intersection of these two faces ; it easily follows that nil' is perpendicular to the projection of that inter- section. 352. Cremona's Theorem. Cremona has given another construction for reciprocal figures which may also be used. Let one polyhedron be given, and let a second be derived from it by joining the- poles of each face of the first. The pole of a given plane in this construction lies on the plane itself, and is found by the following rule. If z=Ax + By + C be the equation to any face, the coordinates of its pole are |= - hB, n = hA, f=C, where h is any arbitrary length. If the edges of these two polyhedra be orthogonally projected on the plane of xy, their projections will be reciprocal figures, with their corresponding sides parallel. Cremona arrives at results equivalent to these by considerations on the theory of forces in three dimensions. Le figure reciproche nella statica grafica 1872, see Art. 302, Ex. 1 and Art. 314, Ex. 5. But we may easily deduce his construction from that of Maxwell. If we turn Maxwell's reciprocal figure round the axis of z through a right angle, the coordinates of the pole used by him become |= - hB, n = hA, f = - C. If we also change the sign of f, the coordinates become the same as those of the pole used ART. 353.] CONSTRUCTION «BF THE RESULTANT FORCE. 255 in Cremona's construction. The effect of the rotation is that the corresponding lines in the projections of the two polyhedra become parallel, instead of per- pendicular. The effect of the change of sign in f is that we replace the reciprocal polyhedron by its image formed by reflexion at the plane of xy as by a looking- glass. Since this last change does not affect the orthogonal projections on the plane of xy, it follows that the two constructions lead to the same reciprocal figures, except that the corresponding lines are in one case perpendicular to each other, in the other parallel. Ex. A straight line, called the axis of z, intersects a given plane in a point A and makes an angle 6 with the plane. Through the point A a straight line is drawn in the plane perpendicular to the axis of z. Prove that the pole of the plane in Cremona's construction lies in this perpendicular at a distance h cot 8 from the axis of z. See Art. 302, Ex. 2. Graphical construction of the Resultant Force. 353. Having given the lines of action and the magnitudes of the forces Pj, P 2 , &c. P 5 , it is required to find their resultant by a graphic construction*. We find the magnitude and direction of the resultant by con- structing as in fig. (2) the force polygon, Art. 342. As before, only the suffixes of the forces are marked in the figure for the sake of brevity. The arrows indicate the directions in which the forces act. The figure has been drawn on the supposition that five forces are given, but it may evidently be extended to any number. If the extremity of the force 5 does not coincide with that of the force 1, the polygon is unclosed. The required result- ant is a force equal and opposite to the force closing the polygon, represented in the figure by 6. On the other hand if the ex- tremity of the force 5 does coincide with that of the force 1, the polygon is closed. The given forces Pj.-.Pg are then either in equilibrium or equivalent to a couple. We have now to find the line of action of the resultant. From any point taken arbitrarily in the force polygon we draw radii vectores to the corners. These radii vectores divide the figure into a series of triangles, the sides of which are used to resolve the forces Pj &c. in convenient directions by. the use of the triangle of forces. The side joining to any corner occurs in two triangles, and therefore represents two forces acting in opposite directions. * A much more full account of this method of finding the resultant than that given in the following brief sketch may be found in the Statique Graphique of M, Levy. 256 GRAPHICAL STATICS. [CHAP. VIII. No arrow has therefore been placed on that side. The arbitrary point is usually called the pole of the polygon. The corners, as before, are represented by two figures ; thus the intersection of the sides 1 and 2 is called the corner 12 and the straight line joining to this corner is called the polar radius 12. We are now in a position to construct the funicular polygon. Taking any arbitrary point L as the point of departure, we draw a straight line LA 1 parallel to the polar radius 61 to meet the line of action of P 1 in Aj. From A t we draw A t A 3 parallel to the Fig. l. Fig. 2. polar radius 12 to meet P 2 'm A 2 ; then A 2 A 3 is drawn parallel -to the polar radius 23 to meet P 3 in A s ; then A s A t and A t A 5 are drawn parallel to the polar radii 34 and 45. Finally A s A e is drawn parallel to 56 to meet A X L (produced if necessary) in A e . Then A 6 is the required point of application of the resultant force. To understand this, we notice that the force Pj at A 1 is re- solved by one of the triangles of the force polygon into two forces acting along LA 1 and A 2 Ai respectively. The latter combined with P 2 is equivalent to a force acting along A S A 2 . This combined with P 3 is equivalent to one along -A 4 4 3 , and so on. We thus see that all the forces P lt &c. P 5 are equivalent to two, one along LA t and the other along A 6 A 5 , These two must therefore intersect in a point on the resultant force. 354. If we take some point, other than L, as a point of departure we obtain a different funicular polygon having all its sides parallel to those of A^A^^A^ In this way by drawing ART. 356.] CONSTRUCTION ^>F THE RESULTANT FORCE. 257 two funicular polygons we can obtain (if desired) two points on the line of action of the resultant. If we take some point other than as the pole in the force diagram, but keep the point of departure L unchanged, we obtain another funicular polygon whose sides are not parallel to those of A^.^Ag. A few of these sides are represented by the dotted lines. But the resulting point A e must still lie on the resultant. We thus arrive at a geometrical theorem, that for all poles with the same force diagram the locus of A e is a straight line. 355. Conditions of equilibrium. In this way we see that, whenever the force polygon is not closed, the given system of forces admits of a resultant whose position can be found by drawing any one funicular polygon. When the force polygon is closed the result is different. In order to use the same two figures as before let us suppose that the six forces Pi.-.P,) form the given system. Taking any arbitrary point L, we begin as before by drawing LA X parallel to the polar radius 61. Continuing the construction for the funicular polygon, we arrive at a point A s on the now given force P 6 . To conclude the construction we have to draw a straight line from A s parallel to the same polar 61 with which we began. This last straight line may be either coincident with, or parallel to, the straight line LA r with which we began the construction. The whole system of forces has thus been reduced to two equal, opposite, and parallel forces, one along AJj and the other along its parallel drawn from A s . If these two straight lines coincide, the equal and opposite forces along them cancel each other: The system is therefore in equilibrium. In this case the funicular polygon drawn (and there- fore every funicular polygon which can be drawn) is a closed polygon. If these two straight lines are parallel, the forces have been reduced to two equal, parallel, and opposite forces. The system is therefore equivalent to a couple. In this case the funicular polygon is unclosed. The moment of this resultant couple is the product of either force into the distance between them. 356. We notice that any set of forces acting at consecutive corners of the funicular polygon (such as P 4 , P 6 , P 6 ) are statically R. S. 17 258 GRAPHICAL STATICS. [CHAP. VIII. equivalent to the tensions or reactions along the straight lines at the extreme corners (such as A 3 A t and A^). These sides must therefore intersect in the resultant of the set of forces chosen. Hence, whatever pole is chosen and whatever point of departure L is taken, the locus of the intersection of any two corresponding sides of the funicular polygon (such as A 8 A 4 and A a A 6 ) is a straight line. In a closed funicular polygon this straight line is the line of action of the resultant of either of the two sets of forces separated by the sides chosen. Thus the sides A 3 A t) A X A Z meet in the resultant either of P 4 , P 6 , P„ or of P 3 , P 2 , P 2 . 357. Parallel forces. When the forces are parallel, both the force diagram and the funicular polygon are simplified, see Art. 140. Thus let A A U A x A it A 2 A S , A 3 A t be bars hinged together at A u A%, A 3 , and attached to fixed points at A , A t by hinges. Also let the weights P 1( P 2 , P 3 act at A lt A it A 3 . Here the force diagram is a straight line ah divided into seg- ments representing the forces P 1; P s , P 3 . If Oa, Ob be parallel to the extreme bars A^A^, A 3 A it then these lengths represent the tensions of these bars, and the lengths drawn from to the corners 12, 23 represent the tensions of the intervening bars. We may also suppose the polygon inverted so as to rest in the form of an arch with its extreme points A and A t fixed in space. It is evident however that this position of equilibrium is unstable. To find the resultant of three given forces P^P^, P 8 we assume any arbitrary pole in the force diagram and draw the corre- sponding funicular polygon A^A^.-.A,,. The extreme sides A^, A^A 3 produced meet in a point on the line of action of the ART. 358.] CONSTRUCTION %F THE RESULTANT FORCE. 259 resultant. The magnitude is obviously the sum of the given forces, and its direction is parallel to those forces. 358. Ex. 1. Prove the following construction to resolve a given force P 2 acting at a given point A 2 into two forces, each parallel to P 2 and acting at two other given points A lt A 3 . Let a length ac represent P 2 in direction and magnitude on any given scale. Draw aO, cO parallel to A.,A 3 , A^A^ respectively, and from their intersection draw 06 parallel to A X A 3 to intersect ac in 6. Then ab and be represent the required components at A :j and A 1 . Another construction. Produce P 2 to cut A±A 3 in N. Then AjN and NA 3 represent the forces at A 3 and A 1 respectively on the same scale that A X A 3 represents the given force P 2 . These would have to be reduced to the given scale by the method used in Euclid vi. 10. Ex. 2. If the forces are not in one plane, show that in general there is no funicular polygon. To show this, let the resultant of P lt P 2 &c. , P n be required, and if possible let AyA 2 &c, A n be a funicular polygon. Then this polygon must satisfy two conditions ; (1) since any one force P can be resolved into two components acting along the adjoining sides, each force and the two adjoining sides must lie in one plane, (2) the components of two consecutive forces along the side joining their points of applica- tion must be equal and opposite. When the forces lie in one plane, the first condition is satisfied already and the second condition alone has to be attended to, and this one condition suffices to find all the possible polygons. If any one side A X A,, of the polygon is chosen, the first condition in general determines all the other sides. To show this we notice that the plane through AjA a and P 2 must cut P 3 in A 3 ; thus A SS A 3 is determined and so on round the polygon. Thus there are not sufficient constants left to satisfy the second condition, though of course in some special cases all the conditions might be satisfied together. Ex. 3. If we remove any set of consecutive forces from a funicular polygon, and replace them by other forces statically equivalent to them, show that the sides bounding this set of forces remain fixed in position and direction though not in length. Suppose we replace P 4 , P 5 by their resultant, then in the force diagram we replace the sides 4, 5 by the straight line joining 34 to 56. The polar radii 34 and 56 are therefore unaltered. But the bounding sides A 3 A 4 , A S A 6 are drawn parallel to these bounding radii from fixed points A s , A e , hence they are unaltered in position and direction. See fig. of Art. 353. Ex. 4. With the same force polygon let the pole move from any given position along any straight line 00'. Prove (1) that each side of the funicular polygon (with the same point of departure L) turns round a fixed point, and (2) that all these fixed points lie in a straight line, which is parallel to the straight line 00'. Referring to the figure of Art. 353, let L, M, N &e. be the points of intersection of corresponding sides of two polygons constructed with 0, 0' respectively as poles. Let (B gl , P 21 ) (P' 6 i, P' 2X ) be the reactions along the sides which meet on the force P 1 on the two polygons. Since these have a common resultant P v the four forces B a , R' w , P 21 and P' 12 are in equilibrium. Hence the resultant of P 61 , P' 16 acting at L must balance the resultant of R iV B' n acting at M. Each of these resultants 17—2 260 GRAPHICAL STATICS. [CHAP. VIII. must therefore act along LM. But looking at the force polygon, the forces R ev E' 61 are represented by the polar radii drawn from 0, 0' to the corner 61. Hence the resultant of P 61 , P/ 16 is parallel to 00'. Similarly MN is parallel to 00'. Hence LMN is a straight line. [Levy, Statique Graphique.] We may also deduce this theorem from the ordinary polar properties of Maxwell's reciprocal polyhedrons, see Art. 350. The force polygon in Art. 353 is clearly the orthogonal projection of two polyhedra having the hexagon 1...6 for a plane base and two vertices whose projections are 0, 0'. Let the same drawings represent the polyhedra in perspective. The poles of the two faces passing through the polar radius 12 are respectively A x and A v Hence the pole of every plane through this polar radius (and therefore that of the plane 0'0 12 ) lies on the straight line A y A v Similarly the pole of the plane 0'0 a lies on BjB 2 . Hence A X A 2 and B X B 2 intersect in a point M which is the pole of the plane 00' w . Similarly the intersection L of A S A 1 and .BgBj is the pole of the plane 00' el ; and so on all round fig. 1. Since all these planes pass through 00', their poles L, M, &a. lie on a straight line, and that straight line is the edge corresponding to 00'. By Art. 350 the projections of corresponding edges are at right angles. The two figures in Art. 353 have however been drawn with their corresponding lines parallel instead of perpendicular. It follows therefore that in these figures the straight lines LM &c. and 00' are parallel. Ex. 5. Find the locus of the pole of a given force polygon that the corre- sponding funicular polygon starting from one given point M may pass through another given point N. The locus, by Ex. 4, is known to be a straight line parallel to MN : the object is to construct the straight line. Case 1. If the given points M, N lie between any two consecutive forces (say Pj, P 2 ), we may take MN as the initial side A X A V The pole must therefore lie on the straight line drawn through the corner 12 of the given force polygon parallel to the given line A X A 2 (see Art. 353). Case 2. Let the point M lie between any two forces (say P lt P 2 ) and N between any other two (say P 3 , P 4 ). By Ex. 1 we can remove the intervening force P 2 , and replace it by two forces acting at M and N each parallel to P 2 ; let these be Q 2 , Q 2 . Similarly we can replace the other intervening force P 3 by two forces, each parallel to P 8 , acting also at M and N; let these be Q s , Q s '. If we now adapt the given force polygon to these changes, the sides 2 and 3 only have to be altered. We have to draw forces parallel to Q 2 , Q 3 , Q 2 ', Q 3 ', beginning at the terminal extremity of the force 1 and ending (necessarily) at the initial extremity of the force 4. The points M, N now lie between the two consecutive forces Q s Q 2 ', hence by Case 1 the locus of is the straight line drawn parallel to MN through the intersection of these forces in the force diagram. [Levy, Statique Graphique.] Ex. 6. With given forces, show how to describe a funicular polygon to pass through any three given points L, M, N. By Ex. 5 we find the locus of the pole when the funicular polygon has to pass through L and M, and also the locus when it has to pass through L and N. The intersection is the required point. Ex. 7. With given forces show how to describe a funicular polygon so that one side may be perpendicular to a given straight line. ART. 359.] HRAMEWORKS. 261 Suppose the side A X A 3 is to be perpendicular to a given straight line, then the polar radius 12 is also perpendicular to that line, Art. 353. Hence the pole must lie on the straight line drawn through the corner 12 of the force polygon per- pendicular to the given straight line. Ex. 8. Show that a given force P can be resolved in only one way into three forces which act along three given straight lines, the force and the given straight lines being in one plane. Prove also the following construction. Let the given straight lines form the triangle ABC, and let the given force P intersect the sides in L, M, N. To find the force S which acts along any side AB, take Np to represent the force P in direction and magnitude, draw ps parallel to GN to intersect AB in 8, then Ns represents the required force S. See Art. 120, Ex. 3. Let Q, R, S be the forces which act along the sides. The sum of their moments about G must be equal to that of P. The moment of S about G is therefore equal to that of P. If then Np and Ns represent P and S, the areas CNp and GNs are equal, i.e. ps is parallel to GN. Ex. 9. Show how to resolve a couple by graphic methods into three forces whioh shall act along three given straight lines in a plane parallel to that of the couple. Prove also the following construction. Move 1;he couple parallel to itself until one of its forces passes through the corner C of the given triangle, and let the other force intersect AB in N. Take Np to represent this second force, and draw ps parallel to GN to meet AB in A, then the required force along the side AB is represented by Ns. Ex. 10. Referring to the figure of Art. 353, let the force polygon be divided into two sets, each of six triangles, by taking two poles 0, 0'. Also let two funicular polygons be described, one for each pole, forming six quadrilaterals and two hexagons. Show that the figures formed by the force polygon and these two funiculars are reciprocal. Ex. 11. Prove that, if the resultant of two of the forces is at right angles to the resultant of one of these and a third force of the system, a funicular polygon can be drawn with three right angles. [Coll. Ex., 1887.] 359. Frameworks. To show how the reactions along the bars of a framework may be found by graphical methods, the external forces being supposed to act at the corners. There are two methods of finding the magnitudes of these reactions. We may either construct a force diagram or use the method of sections. This last is sometimes called Culmann's method. These will be best understood by considering an example. Let the given framework consist of a combination of three triangles, such as frequently occurs in iron roofs*- Let any forces * Rankine's Applied Mechanics, 1885, Art. 155. Major Clarke's Graphics, 1888, Art. 28. Graham's Graphic and Analytic Statics, 1887, Art. 6. 262 GRAPHICAL STATICS. [CHAP. VIII. Pi, P 5 act at the corners A lt A it A 3) A the whole be in equilibrium. A e , and let If these forces were parallel three 34 of them might represent weights placed at the joints, while the structure is supported on its two extremities A lt A s . The five forces are in equilibrium, hence the five lines 1...5 which represent them in the force diagram form a closed pentagon. Taking any arbitrary pole 0, we may draw the corresponding funicular polygon. If this be Oj a 2 ...a s , it also will form a closed figure. To find the tensions of the bars we must complete our force diagram by including these in our figure. We must therefore draw a reciprocal figure. It is however clear that the figure A^...A S does not admit of a reciprocal. We shall extend the figure by drawing a polygon cutting the lines of action of the forces in the points a!...a 5 . Looking at this extended figure and testing it by the rule in Art. 343, we see (1) that at every corner three sides at least meet, (2) that every side forms a part of two polygons and only two. Thus A X A% is a side of the triangle AiA^As and also of the quadrilateral A^A^a^a^ and so on round the figure. This polygon therefore satisfies that test and may be expected to have a reciprocal. It is clear that this extended figure can be regarded as the orthogonal projection of a plane-sided polyhedron, hence by Maxwell's rule it must have a reciprocal, Art. 350. We shall now proceed to complete the sketch of the reciprocal figure, and it will appear from the result that there is one and only one such figure. ART. 359.] FBAMEWOBKS. 263 The side A X A S forms part of a quadrilateral A^^ a^. This quadrilateral corresponds to four lines in the reciprocal figure which meet in a point. Hence the reciprocal of the straight line A X A 5 is a straight line drawn through the intersection of the consecutive forces 1, 5 parallel to A^A^. The same argument applies to every bar of the frame A^A^.-.A^; each is represented in the reciprocal by a straight line which passes through the junction of the consecutive forces at its extremities. This easy rule enables us to draw the reciprocal figure without difficulty. Thus the reciprocal of the side A^ is a straight line drawn parallel to A^A^ through the point of junction of the consecutive forces marked 1 and 2. These straight lines are marked in the force diagram with the suffixes of the straight lines to which they correspond in the framework. The triangle representing the forces at A x having now been constructed, we turn our attention to those at the next corner A s . These will be represented by a quadrilateral. Following the rule, we draw 45 parallel to A t A 5 through the point of junction of the consecutive forces 4, 5. Thus three sides of the quadrilateral are known, viz. 5, 15, 45. Through the known intersection of 12 and 15 we draw a parallel to A 2 A S completing the quadrilateral. The sides are 5, 15, 25, 45. Turning our attention to the corner A t , we draw 34 by the rule and again we know three sides of the corresponding quadri- lateral, viz. 34, 4 and 45. The fourth side is completed by drawing 24 through the known intersection of 45 and 25. The four sides are 4, 45, 24, 34. The triangle corresponding to the corner A s is completed by joining the known intersection of 34 and 24 to the point of junction of the consecutive forces 2, 3. By the rule this line should be parallel to the side A%A 3 . This serves as a partial verification of the correctness of the drawing. Lastly the forces at the corner A a must be represented by a pentagon, but looking at the figure we find that all the sides of this pentagon, viz. 2, 23, 24, 25, 12, have been already drawn. The magnitudes of the reactions along the bars of the given frame may now all be found by measuring the lengths of the different lines in the diagram. 264 GRAPHICAL STATICS. [CHAP. VIII. 360. The directions of the reactions along the bars of the framework are not usually marked by arrows, because two equal and opposite forces act along each bar. It is more convenient to mark them as bars in tension or in thrust. Different marks have been used by different authors to distinguish these two states. Prof. R. H. Smith in his recent work on Graphics suggests + for tension and — for compression. The actual state of tension or thrust is easily settled when sketching the figure. Thus at the corner A r the bars are parallel to the sides of the triangle 1, 12 and 15. The direction of the force being known, those of 12 and 15 follow the usual rule for the triangle of forces. Hence at the point A x the forces act in the directions 15, 21. Therefore A^ s is in a state of tension and should be marked +, A t A s is in a state of compression and should be marked — . 361. The reciprocal figure is not yet completed, for we have still to draw the lines which are the reciprocals of the sides of the polygon a 1 a 2 ...o 5 . If this be a closed funicular polygon of the forces P lt P 2 ,...P 6 , the required reciprocals are the dotted straight lines drawn from the corresponding pole to the points of junction of the forces. It is evident that these lines are practically separate from the rest of the figure. Unless therefore we wish to assure ourselves that the forces P v .. P 5 are really in equilibrium, it is unnecessary to draw either the funicular polygon a v ...a s or its reciprocal. It is usual to omit this part of tlie figure. If the polygon a 1 a i ...a s were any closed polygon instead of a funicular of the given polygon, a reciprocal figure could still be constructed by following the rule in Art. 348 instead of that in Art. 359. But it would not be necessarily true that the forces or reactions acting along the sides A^, &a., A s a B are the given forces. What they really are would be discovered when the reciprocal had been drawn, for their magnitudes would be proportional to the corresponding sides of that diagram. But it is easy to see that their magnitudes would be such that the closed polygon o^...a 6 is one of their funiculars. It is therefore necessary to the argument that the polygon a^.-.a^ should be a funicular polygon of the given forces. It may be remarked that the process described in Art. 348 for drawing a recipro- cal figure is not in all respects the same as that adopted in Art. 359. The reason however is not far to seek. In the former case the whole figure whose reciprocal is required is given geometrically. In the latter case only part of the figure, viz. A 1 A 2 .. .A 5 , is given geometrically, and the remaining portion of the figure has to be adapted so that the stresses along A^, ... A 5 a s are the given forces P^^Py 362. The figure A V ..A S has 5 corners and 7 sides ; hence, by the rule of Art. 151, viz. £=2(7-3, it has just the number of sides necessary to render it stiff. The figure' has been extended by adding two new sides and one new corner for each force. Hence the relation E = 1G - 3 holds for the extended figure as well as for the original frame. It follows from Art. 349 that the extended figure will not in general have a reciprocal unless some geometrical condition is satisfied. ART. 364.] KRAMEWORKS. 265 This geometrical condition is that the line 23 formed by joining two points previously determined in the figure should be parallel to A%A 3 . That the condition is satisfied is clear, for otherwise the forces on the corner A s would not be in equi- librium. 363. Ex. Nine weightless rods are jointed together at their ends ; six of them form the perimeter of a regular hexagon, and the other three each join one angular point to the opposite one ; to each joint a weight W is attached, and the frame is hung in a vertical plane by strings attached to adjacent angles A, B, so that AB is horizontal, and the strings bisect the hexagon angles externally. Find or show by a diagram the forces in all the rods. [Coll. Ex., 1887.] 364. Method of sections. We shall now show how the reactions are found by the method of sections. Let it be required to find the reactions along the rods A^A lt A 2 A S , A 5 Ai. Let these reactions be called Q, R, 8 respectively. Draw a section cutting the frame along these rods, and let the points of intersection be B, G, B. If we imagine the whole structure on one side of this section to be removed, the remainder will stand if we apply the forces Q, R, 8 to the points B, G, B along the three rods respectively. Let us remove the structure on the right hand as being the more complicated, we have now to deduce the forces Q, R, S from the conditions of equilibrium of the structure on the left hand side. In our example not more than three bars were cut by the section. Since there are only three forces the problem is determinate. By Art. 358, Ex. 8, each force of any system can be replaced by three forces acting along three given straight lines, and this resolution can be effected by a graphical construction. These reactions may also be easily found by the ordinary rules of analytical statics, as in Art. 120, where this problem is solved by taking moments about the intersections of these straight lines. When the figure is so little complicated as the one we have just considered, either the method of the force diagram or the method of sections may be used indifferently. In general each has its own advantages. In the first we find all the reactions by constructing one figure with the help of the parallel ruler, but if there be a large number of bars the diagram may be very compli- 266 GRAPHICAL STATICS. [CHAP. VIII. cated. In the method of sections when only three reactions are required we find these without troubling ourselves about the others, provided these three and no others lie on one section. 365. In these frameworks, each rod, when its own weight can be neglected, is in equilibrium under the action of two forces, one at each extremity. These forces therefore act along the length of the rod, and thus the rods are only stretched or compressed. This is sometimes a matter of importance, for a rod can resist, without breaking, a tensional or compressing force when it would yield to an equal transverse force. The structure is therefore stronger than when rigidity at the joints is relied on to produce stiffness. In actual structures some of the external forces may not act at a corner, for instance, the weight of any rod is distributed over its length. In such cases the resultant force on any bar must be found either by drawing a funicular polygon or by the ordinary processes of statics. This resultant is to be resolved into two parallel components acting one at each of the two joints to which the rod is attached. This transformation of the forces which act on a bar cannot affect the distri- bution of stress over the rest of the structure, so that when these components are combined with the other forces which act at those joints the whole effect on the rest of the structure has been taken account of. So far as the beam itself is concerned, it is supposed to be able to support, without sensible bending, its own weight or any other forces which may act on it at points intermediate between its extremities. 366. Indeterminate Tensions. Let P 1 , P 2 ,...P n be a system of forces in equilibrium. hetA 1 ...A n , A-[...A,l be two funicular polygons of this system. Let the corresponding corners A x , A-l ; A it A% Sec. be joined by rods. Let us also suppose that the external polygon is formed of rods in a state of tension and the internal polygon of rods in thrust. It is clear from the properties of a funicular polygon that the framework thus constructed will be in equilibrium. It is also evident that the thrusts along the cross rods A^A-l &o. will be equal respectively to the original forces P v P 2 , . . . P n . In this way a frame has been constructed with tensions along the rods apart from all external forces. See Art. 237. Prom the property of funicular polygons proved in Art. 356 the corresponding sides of this frame intersect in points all of which lie in a straight line. If there are only three forces the polygons become triangles. Since the forces P v P 3 , P 3 are in equilibrium the three straight lines A-^A-y, A^A^, A. A A^ which join the corresponding angular points must meet in a point. Such triangles are called co-polar. We see therefore that co-polar triangles admit of indeterminate tensions. 367. Levy's theorem, given in Art. 238, follows also from this proposition. Taking only six forces, because the figure has been drawn for a hexagon, let (I\, Pi), (P 2 , P 5 ), (P 3 , P 6 ) be three sets of equal and opposite balancing forces. Let AET. 368.] THE %JNE OF PRESSURE. 267 A 1 ...A 6 be any funicular polygon, but let the second funicular polygon be constructed so that Aj 1 coincides with A v and let the pole be so chosen that A 3 ' and A.J coincide with A s and A 6 , Art. 358, Ex. 6. It then follows that the second funicular coincides throughout with the first. The cross bars A X A±, A 2 A S , A 3 A e become the diagonals of the hexagon. Thus a frame of any even number of sides has been constructed in which the diagonals are in a state of thrust and the sides in tension. The line of pressure. 368. Let us suppose a series of connected bodies, such as the four represented in the figure, to be in equilibrium under the action of any forces, say the three P, Q, R. "We suppose these bodies to be symmetrical about a plane which in the figure is taken to be the plane of the paper. The first body is hinged to some fixed support at A and also hinged at B to the body BGG'. This second body presses along its smooth plane surface GC' against a third body GC'D. This third body is hinged to a fourth body at D, and this last is hinged at E to a fixed point of support. The pressure at A acts along some line Ap and intersects the force P at p. The resultant of these two must balance the action at the hinge B, and must therefore pass through B. This force acting at B intersects the force Q at q, and their resultant must balance the pressure at CC. This resultant must therefore cut GC at right angles in some point M. Also the point M must lie within the area of contact, and the resultant must tend to press the surfaces at CC together. This pressure on the third body acts along qMD and intersects R at D. Finally the resultant of these two must pass through E. 268 GRAPHICAL STATICS. [CHAP. VIII. It is evident that the line ApqDE is a funicular polygon of the forces P, Q, R. When therefore such a series of bodies as we have here described rests in equilibrium with its extremities supported it is sufficient and necessary for equilibrium that some one funicular polygon can be drawn which passes through all the hinges and cuts at right angles the surface of pressure. This particular funicular polygon is called the line of pressure. 369. Let us take an ideal section, such as xy, which separates the whole system into two parts, and let it be required to find the resultant action across this section. This action is really the resultant of the forces across each element of the sectional area. But since each portion of the system must act on the other portion in such a way as to keep that portion in equilibrium, we may also find the resultant from the general principle that it balances all the external forces which act on either of the two portions of the system : see also Art. 143. It immediately follows that the resultant action across xy is the force already described which acts along pq. Similar remarks apply to every section ; we therefore infer that the resultant action across any section is the force which acts along the corre- sponding side of the line of pressure. 370. If we move the section xy from one end A of the system to the other B, there may be some difficulty in determining which is the "corresponding side of the line of pressure " when the section passes the point of application of a force. Suppose for example a to be the point of application of P. If a section as x'y' is ever so little to the left of a, the corresponding side is Ap, but when the section is ever so little on the right of a, the corresponding side is pq. If the section is parallel to the force P, the side corresponding to any section is the side of the line of pressure intersected by that section. When therefore the forces are all vertical it will be found more convenient to consider the actions across vertical sections than across those inclined. 371. We notice also that the resultant action across any section such as x'y' does not necessarily pass within the area of that section. The reason is that this action is the resultant of all the small forces across all the elements of area. As some of these elementary forces across the same sectional area may be tensions and some pressures, the line of action of the resultant may lie outside the area. If the forces all act in the same direction like those across the section CG' (where two bodies press against each other), the resultant must pass within the boundary of the section. Sometimes it is more useful to move the resultant parallel to itself and apply it at any convenient point within the boundary ; we must then of course introduce a couple. This is often done when the body AB is a thin rod. See Art. 142. ART. 372.] THE SINE OF PRESSURE. 269 372. When the todies are heavy we may find the action at any hinge or boundary between two bodies by the same rule. The weight of each body is to be collected at its centre of gravity and included in the list of external forces. The resultant action at any boundary is the force along the corresponding side of the funicular polygon. But if the action across some section as xy is required, this partial funicular polygon will not suffice. We must now consider the body BCG' to be equivalent to two bodies separated by the plane xy. The weights of each of these portions may be collected at its own centre of gravity, and a funicular polygon may be drawn to suit this case. Thus, if Q is the weight of the body BGC acting at its centre of gravity |3, we remove Q and replace it by two weights acting at the respective centres of gravity of the portions Bxy and xyGC. The funicular polygon will therefore have one more side than before. It also loses the corner on the force Q and gains two new corners which lie on the lines of action of these new weights. But since the action at B must still balance the external forces whose points of application are on the left of B, and the action at M must still balance the forces on the right of CC, it is clear that the sides pB and MB of the funicular polygon are not altered. Therefore the two corners of the new funicular polygon must lie respectively on Bq and qD. Thus the new polygon is inscribed in the former partial funicular polygon. If we continue this process of separating the bodies into parts, we go on increasing the number of sides in the funicular polygon, but the side which passes through any real section is unchanged in position. Finally, when the bodies are subdivided into elements, the line of pressure becomes a curve. This curve will touch all the partial polygons of pressure at each hinge and at each real surface of separation. CHAPTER IX. CENTRE OF GRAVITY. 373. The centre of parallel forces. It has been proved in Art. 82 that the resultant of any number of parallel forces P u P 2 , &c, acting at definite points A u A 2 , &c, rigidly connected together, is a force 2P. Let the rigid system of points be moved about in any manner in space ; let the forces P u P 2 , &c. continue to act at these points, and let them retain unchanged their magnitudes and directions in space. It has also been proved that the line of action of the resultant always passes through a point fixed relatively to the points A lt A 2 , &c. This point is therefore regarded as the point of application of the resultant. It is called the centre of the parallel forces. The chief property of this point is its fixity relative to the system of points A lt A u &c. When the forces P lt P 2 , &c. are the weights of the particles of a body, the centre of parallel forces is called the centre of gravity. Thus the centre of gravity is a particular case of the centre of parallel forces. This will be made clear in the following defi- nition. 374. Definition of the centre of gravity. We take as a system of parallel forces the weights of the several particles of a body. Each particle is supposed to be acted on by a force which is parallel to the vertical. This force is called gravity. The resultant of all these forces is the weight of the body. We infer from the theory of parallel forces that there is a certain point fixed in each body (or rigid system of bodies) such that ART. 377.] THE FTJN1&A.MENTAL EQUATIONS. 271 in every position the line of action of the weight passes through that point. This point is called the centre of gravity*. It is evident from this definition that if the centre of gravity of a body is supported the body will balance about it in all positions. 375. A body has but one centre of gravity. This is evident from the demonstra- tion in the article already quoted. The following is an independent proof. If possible let there be two such points, say A and B. As we turn the system into all positions, the resultant keeps its direction in space unaltered. Place the body so that the straight line AB is perpendicular to the direction of the resultant force. Then the line of action of that force cannot pass through both A and B. 376. Let (*!, y u z^), (a? 2 , y 2 , ^2) &c. be the coordinates of the points of application of the parallel forces P lt P 2 , &c. respectively. Let these coordinates be referred to any axes, rectangular or oblique, but fixed in the system. By what has been already proved in Art. 80, the coordinates of the centre of parallel forces are __%Px -_2Py __2Pz X ~ 2P ' y ~^ZP' Z ~tP' It is important to notice that, if all the forces were altered in the same ratio, the magnitude of the resultant would also be altered in the same ratio, but the coordinates of its point of application would not be changed. 377. When the weight of any two equal volumes of a substance are the same, the substance is said to be homogeneous or of uniform density. In such bodies the weights of different volumes are proportional to the volumes. The weight of any elementary volume dv may therefore be measured by the volume. Hence by Art. 376 we have r- f dv - 0B Tl - i dv -y ^_f dv - z jdv ' y ~ \dv ' ~jdV We have here replaced the 2 by an integral, because the parallel forces we are considering are the weights of the elements of the body. * The first idea of the centre of gravity is due to Archimedes, who flourished about 250 B.C. In his work on Centres of gravity or aequiponderants he determined the position of the centre of gravity of the parallelogram, the triangle, the ordinary rectilinear trapezium, the area of the parabola, the parabolic trapezium, &c. See the edition, of his works in folio printed at the Clarendon Press, Oxford, 1792. 272 CENTRE OF GRAVITY. [CHAP. IX. From these equations all trace of weight has disappeared. We might therefore call the point thus determined the centre of volume. When the body is not homogeneous the weights of the elements are not proportional to their volumes. Let us represent the weight of a volume dv of the substance by pdv. Here p will be different for each element of the body, and will be known as a function of the coordinates of the element when the structure of the body is given. For our present purpose the body is given when we know p as a function of x, y, z. We therefore have -_jpdv.x __Jpdv.y __ fpdv.z Jpdv ' y ~ Jpdv ' '" Jpdv In these equations we may replace p by icp, where « is any quantity which is the same for all the elements of the body. All that is necessary is that pdv should be proportional to the weight of dv. We may therefore define p to be the limiting ratio of the weight of a small volume (enclosing the point (xyz)) to the weight of an equal volume of some standard homogeneous substance. For the sake of brevity we shall speak of p as the density of the body. If the body is homogeneous the product of the density into the volume is called the mass. If heterogeneous, then pdv is the mass of the elementary volume dv, and Jpdv is the mass of the whole body. If we write dm = pdv, the equations become __\dm.x __jdm.y __fdm.z fdm ' y fdm ' Jdm When we wish to regard the mass of an element as a quality of the body apart from its weight, we may speak of the point determined by these equations as the centre of mass. 378. Equations similar to these occur in other investigations besides those which relate to parallel forces. In such cases the quantity here denoted by P or m has some other meaning. Accordingly the point defined by these coordinates has had other names given to it, depending on the train of reasoning by which the equation has been reached. This may appear to complicate matters, but it has the advantage that the special name adopted in any case helps the reader to understand the particular property of the point to which attention is called. We here arrive at the point as that particular case of the centre of parallel forces in which the forces are due to gravity. There may therefore be some propriety in using the term centre of gravity. There are also obvious advantages in using the short and colourless term of centroid. Another name, much used, is the centre of inertia. This expresses a dynamical property of the point which cannot be properly discussed in a treatise on statics. ART. 381. J THE FUlS^AMENTAL EQUATION. 273 379. The positions of the centres of gravity of many bodies are evident by inspection. Thus the centre of gravity of two equal particles is the middle point of the straight line which joins them. The centre of gravity of a uniform thin straight rod is at its middle point. The centre of gravity of a thin uniform circular disc is at its centre. Generally, if a body is symmetrical about a point, that point is the centre of gravity. If the body is symmetrical about an axis, the centre of gravity lies in that axis, and so on. 380. Working rule. To find the centre of gravity of any body or system of bodies, we proceed in the following manner. We divide the body or system into portions which may be either finite in size or elementary. But they must be such that we know both the mass and position of the centre of gravity of each. Let mi, wia, &c. be the masses of these portions, and let the coordinates of their respective centres of gravity be (xi, y lt z^), (# 2 , y 2 , s 2 ), &c. The weight of each portion is the resultant of the weights of the elementary particles, and may be supposed to act at the centre of gravity of that portion (Art. 82). We may therefore regard the whole body as acted on by a system of parallel forces whose magnitudes are proportional to rfii, m 2 , &c, and whose points of application are the centres of gravity of mj, m 2 , &c The position of the centre of gravity of the whole system is therefore found by substituting in the formulae _ _ Sma; _ _ %my _ _ %mz %m ' " 2m ' 2m 381. In using this rule it is important to notice that some of the masses may be negative. Thus suppose one of the bodies is such that its mass and centre of gravity would be known if only a certain vacant space were filled up. We regard such a body as the difference of two bodies, one filling the whole volume of the body (including the vacant space) whose particles are acted on by gravity in the usual manner, the other filling the vacant space but such that its particles are acted on by forces equal and opposite to that of gravity. To represent this reversal of the direction of gravity it is sufficient to regard the mass of the latter body as negative. Since in the theory of parallel forces the forces may have any signs, E. S. 18 274 CENTRE OF GRAVITY. [CHAP. IX. it is clear that we may use the same formulae to find the centre of gravity of this new system. 382. Ex. 1. A painter's palette is formed by cutting a small circle of radius 6 from a circular disc of radius a. It is required to find the distance of the centre of gravity of the remainder from the centre of the larger circle. Let O and C be the centres of the larger and smaller circles respectively. Let OG=c. We take as the origin and OC as the axis of x. The masses of the two circles are proportional to their areas ; we therefore put m 1 = ira i , m 3 = - tt6 2 . The latter is regarded as negative because its material has been removed from the larger circle. The centres of gravity of the two circles are at their centres, hence x 1 = 0, x s , = c. We have therefore ~Z.mx ira? . - irb 2 . c -& 2 c x = Tim ira 2 -7r& 2 a 2 -/ The negative sign in the result implies that the centre of gravity of the palette is on the side of opposite to C. Ex. 2. If any number of bodies have their centres of gravity on the same straight line, the centre of gravity of the whole of them lies on that straight line. Take the straight line as the axis of x, then the y and z of each centre of gravity are both zero. Hence by Art. 380 y = 0, and z=0. Ex. 3. Two particles of masses %, m 2 are placed at A, B respectively. Prove that their centre of gravity G divides the distance AB inversely in the ratio of the masses. Let G be taken as the origin and AGB as the axis of x. Then x = 0, hence m 1 %+m a a; 2 =0. But x-±= - GA and x 2 =GB. The theorem follows at once. Ex. 4. Three particles are placed at the corners of a triangle ; if their weights, W], u 2 ,w 3 , vary so that they satisfy the linear equation !w 1 + mw 2 + noi s =0, show that the locus of their centre of gravity is a straight line. What is the areal equation to the straight line? * Ex. 5. Four weights are placed at four given points in space, the sum of two of the weights is given, and also the sum of the other two : prove that their centre of gravity lies on a fixed plane. [Math. Tripos, 1869.] * Ex. 6. Water is poured gently into a cylindrical cup of uniform thickness and density ; prove that the locus of the centre of gravity of the water, the cup, and its handle, is a hyperbola. [Math. Tripos, 1859.] Ex. 7. Water is gently poured into a vessel of any form ; prove that, when so much water has been poured in that the centre of gravity of the vessel and water is in the lowest possible position, it will be in the surface of the water. [Math. Tripos, 1859.] Ex. 8. In the figure of Euclid, Book I. Prop. 47, if the perimeters of the squares be regarded as physical lines uniform throughout, prove that the figure will balance about the middle point of the hypothenuse with that line horizontal, the lines of construction having no weight. [Math. Tripos, I860.] If we take the hypothenuse as the axis of x and its middle point as origin, it follows immediately that x = 0. ART. 384.] TRI JUGULAR AREAS. 275 383. Area of a triangle. To find the centre of gravity of a imiform triangular area ABC. Let us divide the area of the triangle into elementary portions or strips by drawing straight lines parallel to one side of the triangle. Bisect BG in D and join AD, and let AD intersect any straight line PNQ drawn parallel to BO in N. Then by similar triangles PN : NQ = BD : DC; but BD=DG, hence PNQ is bisected in N. Thus every straight line drawn parallel to BG is bisected at its intersection with AD. Since we can make each strip as narrow as we please, it follows that the centre of grayity of each (like that of a thin rod, Art. 379) is at its middle point. The centre of gravity of each strip therefore lies in AD. Hence the centre of gravity of the whole triangle lies in AD ; see Art. 382, Ex. 2. In the same way, if we draw BE from B to bisect AG in E, the centre of gravity lies in BE. The centre of gravity of the triangle is therefore at the intersection G of BE and AD. Since D and E are the middle points of GB and GA, the triangle GED is similar to the triangle GAB. Hence ED is parallel to AB and is equal to one half of it. The triangles DEG, ABG are therefore also similar, and DQ : GA = ED : AB. Thus DG is one half of AG, and therefore DG is one third of AD. This proof requires that the triangle should be divided into elements. It can however present no difficulty to anyone who has even only a moderate acquaintance with the Integral Calculus. Poinsot in his Statique has given another demonstra- tion which though long does not depend on the use of elements. 384. We have thus obtained two rules to find the centre of gravity of a uniform triangle. 18—2 276 CENTRE OF GRAVITY. [CHAP. IX. (1) We may draw two median straight lines from any two angular points to bisect the opposite sides. The centre of gravity lies at their intersection. (2) We may draw one median line from any one angular point, say A, to bisect the opposite side in D. The centre of gravity G lies in AD so that AG = %AD. It will be found useful to observe that the centre of gravity of the area of the triangle is the same as that of three equal particles placed one at each angular point of the triangle. Let the mass of each particle be to. The centre of gravity of the particles at B and C is the point D. The centre of gravity of all three is the same as that of 2to at D and mat A ; 'it therefore divides AD in the ratio 1 : 2 (Art. 382). But the point thus found is the centre of gravity of the triangle. If the mass of each of these three particles is equal to one- third of the mass of the triangle, the resultant weight of the three particles is equal to the resultant weight of the triangle. And these two resultants have just been shown to have a common point of application. Hence these three particles are equivalent to the triangle so far as all resolutions and moments of weights are concerned. Also, when we use the method of Art. 380 to find the centre of gravity of any figure composed of triangles, we may replace each of the triangles by three equivalent particles whose united mass is equal to that of the triangle. The centre of gravity of the whole figure may then be found by applying the rule to this collection of particles. 385. Ex. 1. The centre of gravity of the area of a triangle is the same as the centre of gravity of three equal particles placed one at each of the middle points of the sides. Ex. 2. Lengths AP, BQ, CR are measured from the angular points of a triangle along the sides taken in order so that each length is proportional to the side along which it is measured. Show that the centre of gravity of three equal particles placed one at each of the points P, Q, ft is the same as that of the triangle. Ex. 3. Prove that the centre of gravity of a wedge, bounded by two similar equal and parallel triangular faces and three rectangular faces, coincides with that of six equal particles placed at its angular points. Ex. 4. The perpendiculars from the angles A,B, C meet the sides of a triangle in P, Q, R: prove that the centre of gravity of six particles proportional respec- ART. 387.] QUADRILATERAL AREAS. 277 tively to sin 2 A, sin 3 B, sin 2 C, cos 2 A, oos 2 JB, cos 2 C, placed at A, B, G, P, Q, E, coincides with that of the triangle PQB. [Math. Tripos, 1872.] Ex. 5. On the sides of any triangle similar regular polygons are described, and equal masses are placed at all the corners : prove that the centre of gravity of the masses coincides with that of the triangle. [St John's, 1879.] 386. Perimeter of a -triangle. Ex. 1. A triangle ABC is formed by three thin rods whose lengths are a, b, c. If H be the centre of gravity, prove that the areal coordinates of H are proportional to b + c, c + a, a + b. Ex. 2. The centre of gravity of the perimeter of a triangle ABC is the centre of the circle inscribed in the triangle DEF, where D, E, F are the middle points of the sides of the triangle ABC. [Lock's Statics.] Ex. 3. If H be the centre of gravity of the perimeter of a triangle, G the centre of gravity of the area, I the centre of the inscribed circle, prove that H, G, I are in one straight line, and that GH is one half of 1G. If be the centre of the circumscribing circle, and P the orthocentre, show that the triangles IGP, HGO are similar. 387. Quadrilateral areas. To find the centre of gravity of any quadrilateral area ABCD. Using the rule in Art. 380, we replace the triangle ADC by three particles situated at A, D, C respectively, each equal to one- third of the mass of ADC In the same way we replace the triangle ABG by three masses at A, B, C, each one-third of the mass of ABG. In this way the masses at A and G are each ^M, if M be the mass of the whole quadrilateral. Consider next the masses at B and D ; call these m x and m 2 . Their united mass is also ^M, but this total mass is unequally divided between the particles in the ratio of the triangles ABC : ADC, i.e. in the ratio BE : ED. To obtain a more convenient distribution, let us replace these two masses by three others placed at B, D, and E. If the masses placed at B and D are each \M and the mass placed at E is — %M, the sum of the masses is the same as before. It is also clear that their centre of gravity 278 CENTRE OF GRAVITY. [CHAP. IX. is the same as that of the masses m^ and m a . For by Art. 380 the distance of their centre of gravity from E is given by _ Jtmx _ \M . BE -\M . BE + pf . ~ 2m \M But the distance of the centre of gravity of the masses m 1; m 2 from E is given by _ _ m^.BE-m^I)E _ BE*-DE* X ~ nh + mz ~ BE+DE ' which is the same as before. The centre of gravity of the area of the quadrilateral is therefore the same as that of four equal particles, placed one at each angular point of the quadrilateral, together with a fifth particle of equal but negative mass, placed at the intersection of the diagonals. We may put the result of this rule into an analytical form. Let (*,, 2/i), (x 2 , y 2 ), &c. be the coordinates of the four angular points and of the intersection of the diagonals, then clearly x = %(x l +x i + x s +x i — x s ), with a similar expression for y. See the Quarterly Journal of Mathematics, vol. XL 1871, p. 109. The reader is advised to use the rule of equivalent points partly because the analytical result follows at once, and partly because these equivalent points are used in rigid dynamics to enable us to write down the moments and products of inertia of a quadrilateral. 388. Ex. 1. Prove the following geometrical construction for the centre of gravity of a quadrilateral area. Let P, Q be points in BD, AC such that QA, PB are equal respectively to EC, ED ; the centre of gravity of the quadrilateral coincides with that of the triangls EPQ. This construction is given in the Quarterly Jownal of Mathematics, vol. vi. 1864. Ex. 2. A quadrilateral is divided into two triangles by one diagonal BD, and the centres of gravity of these triangles are M and N. Let MN cut BD in I, from the greater NI take NG equal to MI the lesser. Prove that G is the centre of gravity of the area of the quadrilateral. [Guldin. ] Ex. 3. A trapezium has the two sides AB = a and CD = b parallel. Prove that the centre of gravity G of the quadrilateral area lies in the straight line joining the middle points M and N of AB and CD. Prove also that G divides MN so that MG : GN=a+2b : 2a + b. [Archimedes and Guldin.] To prove this, produce AD, BC to meet in O. Then regard the trapezium as the difference of the triangles OAB, OCD, and apply the rule of Art. 380. Take OMN as the axis of x. ART. 389.] TETRJilEDRAL VOLUMES. 279 Notice that the ratio MG : GN does not depend on the height of the trapezium but only on the lengths of the parallel sides. [Poinsot.] A geometrical construction, also given by Poinsot, is as follows. Produce AB to H so that BH=CD and produce DC in the opposite direction to K so that DK=AB. The straight line which joins HK will cut MN in the centre of gravity of the trapezium. 389. Tetrahedron. hedron ABCD. To find the centre of gravity of a tetra- Let us divide the tetrahedron into elementary slices by drawing planes parallel to one face. Let abc be one of these planes. Bisect BO in E and join DE, then, exactly as in the case of the triangle, DE will bisect all straight lines such as be which are parallel to BG. Join AE and ae, then these are parallel to each other. Take AF — %AE, then F is the centre of gravity of the base ABC. Join DF and let it cut ae in /, then by similar triangles of : AF=Da : BA=ae : AE. Hence af= § ae, that is / is the centre of gravity of the triangle abc. It therefore follows that the centre of gravity of every elementary slice lies in DF. Hence the centre of gravity of the whole tetrahedron lies in DF. Thus the centre of gravity of a tetrahedron lies in the straight line which joins any angular point to the centre of gravity of the opposite face. Let K be the centre of gravity of the face BCD ; join AK. The centre of gravity also lies in AK. It is therefore at their in- tersection G. Exactly as in the corresponding theorem for a triangle, we have FK parallel to AD and = %AD. Hence from the similar triangles AGD, KGF, we see that FG = %GD. Thus DG = \DF. To find the centre of gravity of a tetrahedron we join any corner (as D) to the centre of gravity (as IT) of the opposite face. The centre of gravity G lies in DF so that DG = |DF. a/ \\b "Ye \ \ \ \ *^" / 280 CENTRE OF GRAVITY. [CHAP. IX. As in the case of a triangle, we may fix the position of the centre of gravity of a tetrahedron by means of some equivalent points. The ventre of gravity of a tetrahedron is tlie same as that of four equal particles placed one at each angular point. The proof is exactly similar to that for a triangle. 390. Pyramid and Cone. To find the centre of gravity of the volume of a pyramid on a plane rectilinear base. Proceeding as in the case of the tetrahedron, we divide the pyramid into elementary slices by drawing planes parallel to the base. These sections are all similar to the base. The centre of gravity of each slice, and therefore that of the whole pyramid, lies in the straight line joining the vertex of the pyramid to the centre of gravity of the base. Next, we may divide the base into triangles. By joining the angular points of these triangles to the vertex, we divide the whole pyramid into tetrahedra having a common vertex. The centre of each tetrahedron, and therefore that of the pyramid, lies in a plane parallel to the base such that its distance from the vertex is f of the distance of the base. Joining these two results together, we have the following rule to find the centre of gravity of a pyramid. Join the vertex V to the centre of gravity F of the base and measure along VF from the vertex a length VG equal to three quarters of VF. Then G is the centre of gravity of the pyramid. To find the centre of gravity of the volume of a pyramid on a plane curvilinear base. We regard the base as the limit of a polygon with an infinite number of elementary sides. The rule is therefore the same as for a pyramid on a rectilinear base. To find the centre of gravity of the volume of a cone on a circular or on an elliptic base; join the vertex V to the centre F of the base, and measure along VF from the vertex a length VG equal to three quarters of VF, then G is the centre of gravity of the cone. ' 391. Ex. 1. A cone whose semivertioal angle is tan -1 — - is enclosed in the V 2 circumscribing spherical surface ; show that it will rest in any position. [Math. Tripos, 1851.] ART. 393.] TETRASEDRAL VOLUMES. 281 Ex. 2. A pyramid, of which the base is a square, and the other faces equal isosceles triangles, is placed in the circumscribing spherical surface ; prove that it will rest in any position if the cosine of the vertical angle of each of the triangular faces be f. [Math. Tripos, 1859.] Ex. 3. A frustum of a tetrahedron is bounded by parallel faces ABC, A'B'C. Prove that its centre of gravity G lies in the straight line jqjning the centres of gravity E, B' of the faces ABC, A'B'C and ia such that EG E'G EE' l + 2n + 3ri> m 2 +2»+3 4{l + n + n?)' where n is the ratio of any side of the triangle A'B'C to the corresponding side of the triangle ABC. [Poinsot.] Ex. 4. A frustum of a tetrahedron ABGD is bounded by faces ABC, A'B'C not necessarily parallel. Find its centre of gravity. Let DA, DB, DC be regarded as a system of oblique axes, let the distances of A, B, C, A', B', C from D be a, b, c, a', V, c'. Then __ a?bc - a'Wc' _ aWc - a'b'%' __ abc 2 - a'b'c' 2 I_ * abc-a'b'c' ' V ~* abe-a'b'c' ' "~* abc-a'b'c' ' To prove these results, we regard the tetrahedra as the difference of two tetrahedra whose volumes are as abc : a'b'c'. Ex. 5. The top of a right cone, semivertical angle a, cut off by a plane making an angle /3 with the axis, is placed on a perfectly rough inclined plane with the major axis of the base along a line of greatest slope of the plane ; in this position the cone is on the point of toppling over : prove that the tangent of the inclination of the plane to the horizon has one of the values = =r^ • cos 2a - cos 2/3 [Math. Tripos, 1876.] 392. Faces and edges of a tetrahedron. Ex. 1. The centre of gravity of the four faces of a tetrahedron is the centre of the sphere inscribed in a tetrahedron whose corners are the centres of gravity of the faces of the original tetrahedron. If we take any face as the plane of xy, we find a symmetrical expression for Jp-2, where p is the perpendicular from the opposite corner oh the plane of xy. Ex. 2. If H be the centre of gravity of the faces of a tetrahedron, G the centre of gravity of the volume, I the centre of the inscribed sphere, then H, G, I are in one straight line and HG is equal to one third of GI. Ex. 3. If o, /3, 7, S be the distances of the centre of gravity of the six edges of a tetrahedron from the faces, then (2a - Pl ) r 1 = (2/3 -p 2 ) r 2 = (2 7 -p 3 ) r s = (28 -pj r t , where ft,y 8 ,^3,^4 are the perpendiculars from the corners on the opposite faces, and r lt r 2 , r s , r 4 are the radii of the circles inscribed in the triangular faces. 393. The isosceles tetrahedron. An isosceles tetrahedron is one whose opposite ed0s are equal. It follows from this definition that the sides of any two faces are equal each to each. 282 CENTRE OP GRAVITY. [CHAP. IX. Ex. 1. Show that the following five points are coincident, viz. (1) the centre of gravity of the volume, (2) the centre of gravity of the six edges, (3) the centre of gravity of the four faces, (4) the centre of the circumscribing sphere, (5) the centre of the inscribed sphere. Let this point be called G. Ex. 2. Show that the perpendicular from Gf on any face intersects that face in the centre of the circumscribing circle. Ex. 3. The straight lines which join the middle points of opposite edges of a tetrahedron are called the median lines. Show that the medians pass through G, axe bisected by it and are perpendicular to their corresponding edges. Show also that the three medians are at right angles and form a. system of three rectangular axes. See Casey's Spherical Trigonometry, 1889, Art. 127. 394. Double tetrahedra. To find the centre of gravity of the solid bounded by six triangular faces, i.e. contained by two tetraliedra having a common face. Let the common base be ABC and D, D' the vertices. Join CD', and let it cut the base in E. We replace the tetrahedron ABCD by four particles, each one-fourth its mass situated at the points A, B, G, D. Treating the other tetrahedron in the same way, we have at each of the points A, B, C a particle whose mass is equal to one-fourth of the solid, and at D, D' two particles whose united mass makes up the remaining fourth of the solid, and whose separate masses are in the ratio of the tetrahedra, i.e. in the ratio BE : ED'. Following exactly the steps of the reasoning in the case of a quadrilateral, it is easy to see that we can replace these two masses by two other masses situated at D and D', and each one-fourth that of the whole solid, together with a third particle situated at E of the same mass but taken negatively. The centre of gravity of the whole solid is the same as that of five equal particles placed at A, B, C, D, D' together with a sixth particle equal and opposite to any of the five placed at the intersection o/DD' with the common face ABC. 395. Ex. The centre of gravity of a pyramid on a plane quadrilateral base is the same as that of five equal particles placed at the five apices, and a sixth equal but negative particle placed at the intersection of the diagonals of the base. If we draw a plane through the vertex and a diagonal of the base, the solid becomes two tetrahedra joined together at a common face. The result follows at once. Centre of gravity of an arc. 396. To find the centre of gravity of an arc of a circle. Let A CB be the arc, its centre. Let the radius OG bisect the arc, let OG-a, and the angle A 0B = 2a. ART. 399.] CENTRE OF GRAVITY OF AN ARC. •283 Let PQ be any element of the arc, and let the angle POC = 6. Then in the fundamental formula of Art. 380 m = add, x = a cos 6. If x be the distance of the centre of gravity of the arc from 0, _ 'Zmx x= -= — = 2,m jadO . a cos 6 '' jade sin a since the limits of are = — a and 9 = + a. As this result is frequently used, it will be convenient to put it into a form which will be convenient for reference. Distance of G. G. of arc from centre sin (half angle) , chord . : — ^-^ P — - . rad. = . rad. halt angle arc This result was given by Wallis. 397. Ex. A series of 2n straight Hues are inscribed in a circular arc, each straight line subtending an angle 20 at the centre. Prove that the distance of their centre of gravity from the centre is r cos sin 2n0/2nsin 0. Thence deduce the centre of gravity of a circular arc of any angle. [Guldin's Problem.] 398. Centre of gravity of any arc. The coordinates of the centre of gravity of the arc of any uniform plane curve are given by the formulfe _ _ %mx _ fads _ _ fyds Jte' y ~~]ds' Sm where we write for the elementary arc ds its value given in the differential calculus. Thus we have ds- ■{i + $MV- \dx) j or ds- -©?»■ according as the equation to the curve is given in the Cartesian form y =f(x) or the polar form r = F (0). If the curve be in three dimensions we have an expression for z similar to those written above. The corresponding expressions for ds are given in works on the differential calculus. 399. The process of finding the centre of gravity of an arc is merely that of substituting for ds from the given equation to the curve and then integrating. It seems unnecessary to give at length examples of what is merely integration we shall therefore state only the results in a few cases likely to be useful. 284 CENTRE OF GRAVITY. [CHAP. IX. Ex. 1. The coordinates of the centre of gravity of an arc of the catenary ) from x = to x = x are x=x — — -, S = i(y-\ — )■ l/=|( c "° + « These admit of a geometrical interpretation. Let PQ be any arc of the catenary. Let the tangents at P and Q meet in T and the normals at P and Q meet in N. If x, y be the coordinates of the centre of gravity of the arc PQ, then x = abscissa of T, and #=half the ordinate of N. Ex. 2. Find the centre of gravity of the arc OP of a cycloid between the vertex where 0=0 and the point P, the equations to the curve being x = 2a

    - sm(p , and y=\ Ex. 3. If G be the centre of gravity of any arc AP of the lemniscate f'=a?co&2B, prove that Off bisects the angle AOP. One case of this is given in Walton's Problems on Theoretical Mechanics. Ex. 4. The centre of gravity of any arc PQ of the curve r 3 sin 30= a 3 lies in the straight line joining the origin to the intersection of the tangents .at P and Q. Ex. 5. If the density at any point of the arc vary as r n ~ 3 , prove that the centre of gravity of any arc PQ of the curve r n smn6 = a n lies in the straight line joining the origin to the intersection of the tangents at P and Q. Ex. 6. The locus of the centre of gravity of an arc of given length of the lemniscate r 2 =a 8 cos 29 is a curve which is the inverse of a concentric ellipse. [R. A. Robert's theorem.] Centres of gravity of Areas. 400. Sectors of circles. To find the centre of gravity of a sector of a circle. Let A GB be the arc of the sector, its centre. As in Art. 396 let the radius OG bisect the arc, OG = a and the angle A OB = 2a. We divide the sector into elemen- tary triangles of equal area. Let OPQ be any one of these triangles ; following the rule of Art. 380 we collect its mass into its centre of gravity, i.e. into a point p where 0p = %0P. Repeating this process for every triangle, we have a series of particles of equal mass arranged at equal distances along an arc ab of a circle. These are represented in the figure by the row of dots. In the limit when ART. 403.] METHOI»OF PROJECTIONS. 285 the triangles are infinitely small this becomes a homogeneous arc of a circle. .The distance of the centre of gravity of the sector from is therefore given by the result in Art. 396, viz. x = la = % -r^- . radius 0G. a 6 3 arc AB This result was given by Wallis. 401. Ex. To find the coordinates of the centre of gravity of the area of a quadrant of a circle AOB. This is a particular case of the last article, viz. when a=Jir. If x, y be the coordinates of G referred to OA, OB as axes, we have x = OGoos a = z- , y= 5- . 402. Ex. 1. The distance of the centre of gravity of the area of a segment of a circle measured from the centre is 4 : . where a is the semianele o - sin o cos a of the segment. [Guldin.] 403. Projection of areas. If any plane area is orthogo- nally projected on any other plane, the centre of gravity of tJie projection is the projection of the centre of gravity of the primitive area. Let the plane on which the projection is made be the plane of xy, and let a be the inclination of the two planes. Let dS be any element of the area of the primitive, dH the area of its projection. Then by a known theorem in conies dU. = dS cos a. We also notice that the x and y coordinates of dS and dU. are the same because the projection is orthogonal. The coordinates of the centre of gravity of either area are known from _ _ Imx _ _ %my Xm u Zm where the m for one area is dli. and for the other is dS. Since these are in a constant ratio, the values of x and y are the same for each area. In order to use effectively the method of projections we join to it the two following well known theorems which are proved in books on conies : (1) the projections of parallel straight lines are parallel, (2) the ratio of the lengths of two parallel straight lines is unaltered by projection. Suppose we had any geometrical relation between the lengths of lines in the primitive figure, and that we require the correspond- ing relation in the projected figure. We first express the given 286 CENTRE OF GRAVITY. [CHAP. IX. relation in the form of ratios of lengths of parallel straight lines. To do this it may be necessary to draw parallels to some of the lines in the primitive if there are no parallels to them mentioned in the given relation. Having put the geometrical relation into the form of ratios, the same relation is true for the projected figure. 404. Elliptic areas. Since an elliptic area is well known to be the orthogonal projection of a circle, we can deduce the centres of gravity of the various parts of an ellipse from those of the corresponding parts of a circle. The circle used for this purpose is sometimes called in conies the auxiliary circle. 405. To find the centre of gravity of an elliptic area. The coordinates of the centre of gravity of a quadrant AOB of a circle, referred to A, OB as axes, may be written in the form 0A OB Sir K) since 0A, OB are both radii. But x and OA, are parallel straight lines, and so also are y and OB. Hence these relations hold in the projected figure also. If then OA, OB are the major and minor semiaxes of an ellipse, the coordinates of the centre of gravity of the area of the quadrant are given by (1). If we make the plane on which we project intersect the quadrant of the circle in any straight line not one of the bounding radii the circular quadrant projects into an elliptic quadrant bounded by two conjugate diameters. If then OA, OB are any two semiconjugates of an ellipse, the coordinates of the centre of gravity of the contained area are given by equations (1). The position of the centre of gravity of a semi-ellipse was first found by Guldin. 406. Ex. 1. A chord PQ of an ellipse, centre C, passes always through a fixed point O. Prove that the locus of the centre of gravity of the triangle GPQ is a similar ellipse. [Coll. Exam.] Ex. 2. The centre of gravity O of any elliptic sector bounded by the semi- diameters OP, OP' lies in the diameter OA' bisecting the chord PP', and is such that yr-r, = § — g- . where sin is the ratio of half the chord PP' to the semicon jugate of OA'. X TP~~ ART. 407.] METH9D OF PROJECTIONS. 287 Ex. 3. The area A of any elliptic sector POP' is A = \ab(' -0), and the coordinates of the centre of gravity referred to the principal diameters, are a; sin 0' - sin y _ t cos 0- cos ' a~* 0'-0 " b~ ls '-

    a. Put k= k's/ - 1 and use the formula in analytical trigonometry, *J — 10= log (cos0+\/-lsin0), where 0=sin _1 fc; we find for the centre of gravity of a hyperbolic sector * = | *' , where ftUfgV-lf 4 . 409. Centre of gravity of any area. After having obtained the fundamental formulae of Art. 380 the discovery of the centres of gravity of any area is reduced to two processes. (1) We have to make a judicious choice of the element m, and (2) we have to effect the necessary integrations. The latter process is fully dis- cussed in treatises on the integral calculus, in fact it is a part of that science rather than of statics. It will thus be unnecessary to do more here than make a few remarks on the choice of m with special reference to centres of gravity. If the centre of gravity of the area bounded by two ordinates Aa, Bb be required, we put the equation of the curve into the form y=f(x). We choose as our element the strip PQM. Here PM=y and m=ydx. The coordinates of the centre of gravity of m are x and \y. Hence, Art. 380, the formulae to be used are a _ Zmx _ jydx . x ~_ jydx.jy y= 2m jydx ' jydx If the centre of gravity of the sectorial area A OB is wanted, we put the equation into the form r=/(0). We choose as our element the triangular strip POQ. Here OP=r, and m= %r 2 dd. The Cartesian coordinates of the centre of gravity of m are f r cos and \ r sin 0. The formulae to be used are J&rW.frflinfl -_ fyr*d8.%r cos \\rH9 '' f=i farUS a M b ART. 411.] AREi*# BY INTEGRATION. 289 Sometimes the equation to the curve is given with an auxiliary variable t, thus x = if>(t), y = \j/[t). It is in this form for example that the equation to the cycloid is generally given. See Ex. 2, Art. 399. In this case when the polar area is required we quote the formula from the differential calculus r*d$=xdy-ydx. Substituting half of this for m in the standard expressions for 3! and f, we have a convenient formula to find the centre of gravity. 410. If the figure whose centre of gravity is required is a triangle or quadri- lateral whose sides are curvilinear, the proper choice for the element m will depend on the form of the curves. If we join the angular points to the origin we have three or four sectors whose areas and centres of gravity may be separately found and thence, by Art. 380, the centre of gravity of the figure. Sometimes the bounding curves are of the same species so that when the process has been gone through for one sector the results for the other sectors may be inferred. In such cases the method is very advanta- geous. For example, we have already seen how the area and centre of gravity of a quadrilateral bounded by four elliptic arcs could be immediately deduced from the area and centre of gravity of an elliptic sector. See Ex. 6, Art. 406. Putting this in an analytical form, we have for a curvilinear triangle whose sides arer=A(0), r'=f 2 (6'), r"=f 3 (B"), Swra =4 J r s cos ede+ij y r" cos B'dd' + J J r" 3 cos 6" 6,8" where o, fi, y are the inclinations of the radii vectores of the angular points to the axis of x. In forming these integrals we travel round the triangular figure taking the sides in order. It might appear at first sight that we are adding together all the three sectors instead of adding some together and subtracting the others. But it will be clear after a little consideration that in those sectors which should be subtracted from the others the dB is made negative by taking the limits in the same order as we travel round the triangle. Instead of joining the angular points to the origin we might draw perpendiculars on the axis of x. We then have fb fe fa 2mx= I xydx+ I x'y'dx + I x y dx , J a J b J e where a, b, c are the abscissae of the angular points. As before, in taking the limits we travel round the sides in order. 411. Sometimes we may use double integration. Suppose we can express the equations to both the opposite sides of a curvilinear quadrilateral in one form by using an auxiliary quantity u. That is, let the one equation represent one boundary when u=a, and let the same equation represent the opposite boundary when m=6. Let this one equation be (x, y, u)=0. It is always possible to do this, for let f x (x, y) = 0, f 2 (x, y)=0 be the boundaries, then . In the same way let \ji (x, y, v) = represents the other two boundaries when v = e and v =/. When this has heen accomplished we have only to follow the rules of the integral calculus. By giving u and v all values between u=a and u=b, v — e and v=f, we obtain a double series of curves dividing the space into elements. Let m be the area of one of these elements and J the Jacobian determinant of x, y with regard to u, v, then m=Jdudv. Hence __jjJdudv . x \\Jdudv.y x ~ jjjdudv ' V~ tfJdudv " To find the Jacobian it may be necessary to solve the equations 0=0, i]/=0, so as to express x, y in terms of u, v. We then have J=-^- -r- - — ~ . Unless we have r " du dv dv du been able in the first instance to express tj> and \j/ so conveniently that this Jacobian takes a simple form when expressed in terms of u, v, this method may lead to com- plicated analysis. The advantage of the method is that the limits of integration u=atob,v = e to /are constants, so that the integrations may be performed in any order or simultaneously. 412. Ex. 1. An area is cut off from a parabola by a diameter ON and its ordinate PN: prove that ~x = \x, f=%y. Ex. 2. Two tangents TP, TP' are drawn to a parabola: show that the co- ordinates of the centre of gravity of the area between the curve and the tangents are x = \ TP, y = \TP' referred to TP, TP' as axes. See Art. 406, Ex. 5. [Walton.] Regard the area as the difference between a triangle and a parabolic segment. Ex. 3. The centre of gravity of half the area of the cycloid x=a(l-coBd), y = a(d + Bia$) is given by x = \ a, V = \\^-^)- [Wallis.] Ex. 4. Find the centre of gravity of the half of either loop of the lemniscate r 2 =a 2 cos 20 bounded by the axis. The result is __ ™ _ 31og(V2 + l)-V2 X -£j2' y 6V2 a - Ex. 5. Four parabolas whose equations are y 2 =a s x, y*=lfix, x^ — e s y, x 2 =fh/ intersect and form a quadrilateral space. Find the abscissa of its centre of gravity. We take as the equations to the opposite sides y 2 =u 3 x and x % =v 3 y. Solving, we find x=uv 2 , y=v?v and J=3v?v 2 . This gives by substitution x -™(b»-a?)(f*-e*)- Ex. 6. The centre of gravity of the space bounded by two ellipses and two hyperbolas all confocal lies in the straight line y _ ("2 - «i) (« a ' - %') («a a + US' THEOREMS. 291 where the unaccented letters denote the semiaxes of the ellipse and the accented letters those of the hyperbola. We take as the equation to the opposite sides *» s/ a , a? y* , u u-h v v-h where u > ft and v < ft. As shown in Salmon's conies, these give ftx 2 =w, - hy 2 =(u - h) (v - ft). The result then follows easily enough. Ex. 7. If the density at any point of a circular disc whose radius is a vary directly as the distance from the centre, and a circle described on a radius as diameter be cut out, prove that the centre of inertia of the remainder will be at a distance ^ T7 - from the centre. [Math. Tripos, 1875.] 107T — 10 \ Ex. 8. A circular disc of radius r, whose density is proportional to the distance from the centre, has a hole cut in it bounded by a circle of diameter a which passes through the centre. Show that the distance from the centre of the disc of the g a 4 centre of gravity of the remaining portion is -= — 8 3 . [Coll. Ex., 1888.] \ Ex. 9. The curve for which the ordinate and abscissa of the centre of gravity of the area included between the ordinates x=a and x=x are in the same ratio as the a 3 6 s bounding ordinate y and abscissa x is given by the equation -g — j = l- [Math. Tripos, 1871.] Pappus' Theorems. 413. Before treating of the centres of gravity of surfaces or volumes it seems proper to discuss a method by which the centres of gravity of the arcs and areas already found may be used to find the surface or volume of a solid of revolution. The two following theorems were first given by Pappus at the end of the preface of his seventh book of Mathematical Collections. Let any plane area revolve through any angle about an axis in its own plane, then (1) The area of the surface generated by its perimeter is equal to the product of the perimeter into the length of the path described by the centre of gravity of the perimeter. (2) The volume of the solid generated by the area is equal to the product of the area into the length of the path described by the centre of gravity of the area. 19—2 292 CENTRE OF GRAVITY. [CHAP. IX. In both these theorems the axis is supposed not to intersect the perimeter or area. 414. Let AB be an arc of the curve, and let it lie in the plane xz. Let it revolve about the axis of z through any elementary angle dd. Any element PQ = ds of the perimeter is thus brought into the position P'Q', and the area traced out by PQ is ds . PP' = ds . xdd. The whole area or surface traced out by the finite arc A B is ddjxds. But this is dd.xs, if s be the arc AB and * the distance of its centre of gravity from the axis of z. If the arc now revolve again about Oz through a second elementary angle dd, an equal surface is again traced out. Hence, when the angle of rotation is d, the area is s . xd. But xd is the length of the path traced out by the centre of gravity of the arc. The first proposition is therefore proved. Next, let any closed curve in the plane of xz revolve as before about the axis of z through an angle dd. By this rotation any elementary area dA at R will describe a volume which may be regarded as an elementary cylinder. The base is dA, the altitude xdd, the volume is therefore dA . xdd. The volume traced out by the whole area of the closed curve is ddfxdA. But this is dd . xA, if A be the area of the curve and x the distance of its centre of gravity from the axis of revolution. Integrating again for any finite value of 6, we find that the volume generated is A . xd. This as before proves the theorem. In both these proofs we have assumed that the whole of the curve lies on the same side of the axis of rotation. For suppose ART. 416.] PARTIS' THEOREMS. 293 P x and P. 2 were two points on the curve on opposite sides of the axis of z, then their abscissae a^ and sc 2 would have opposite signs. Thus the elementary surfaces or volumes (having the factor xdd) would also have opposite signs. The integral gives the sum of these elementary surfaces or volumes taken with their proper signs. It follows that, when the axis cuts the curve, Pappus' two rules give the difference of the surfaces or volumes traced out by the two parts of the curve on opposite sides of the axis of revolution. ' 415. Ex. 1. Find the surface and volume of a tore or anchor ring. This solid may be regarded as generated by a complete revolution of a circle about an axis in its own plane. Let a be the distance of the centre from the axis, b the radius of the generating circle. Then a > b if all the elements are to be regarded as positive. The arc of the generating circle is iirb, the length of the path described by its oentre of gravity is lira. The surface is therefore iifiab. The area of the circle is tt& 2 , the length of the path described by its centre of gravity is lira. The volume is therefore 27r 2 a& 3 . ' Ex. 2. Find the volume of a solid sector of a sphere with a circular rim and also the area of its curved surface. This solid may be regarded as generated by a complete revolution of a sector of a circle about one of the extreme radii. Let 2a be the angle of the sector, O its centre. The arc of the sector is 2aa. The length of the path described by its centre of gravity G is 2r . OG sin a, where OG=(a sin o)/a. The spherical surface is there- fore 47ra 2 sin 2 a. The area of the sector is a?a. The length of the path of its centre of gravity G' is 2tt . Off' sin a, where OG'=% OG. The volume is therefore ^7ra 3 sin 2 a. It appears that both the surface and the volume vary as the versine of the sector. * Ex. 3. A solid is generated by the revolution of a triangle ABC about the side AB: prove that the surface is w(a+b)p and the volume is iircp*, where p is the perpendicular from C on AB. 416. It should be noticed that for any elementary angle dd the axis of rotation need only be an instantaneous axis. Suppose the plane area to move so as always to be normal to the curve described by the centre of gravity of the area. Then as the centre of gravity describes the arc ds, the area A may be regarded as turning round an axis through the centre of curvature of the path. Hence the elementary volume is Ads, and the volume described is the product of the area into the leiigth of the path described by the centre of gravity of the area. 294 CENTRE OF GRAVITY. [CHAP. IX. In the same way, if the area move so as always to be normal to the path described by the centre of gravity of the perimeter, the surface of the solid is the product of the arc into the length of the path of the centre of gravity of the perimeter. 417. When the axis of rotation does not lie in the plane of the curve, we can use a modification of Pappus' rule to find the volume generated by the motion of any area. Let us suppose that the axis of rotation is parallel to the plane of the curve. Referring to the figure of Art. 414, let GL be the axis, and let BL be a perpendicular to it from any point R within the closed curve. The elementary area dA at B will now describe a portion of a thin ring whose centre is at L. The length of this portion is . BL. The area of the normal section of this ring is dA cos 0, where is the angle the normal BL to the ring makes with the area dA. The volume traced out is therefore BL . cos

    bea perpendicular on the axis of z, then OL = z. Let a cylinder circumscribe the sphere and touch it along the circle of which AB is a quadrant. Any point P on the sphere is 298 CENTRE OF GRAVITY. [CHAP. IX. projected on the cylinder by producing LP to meet the cylinder in F. According to this definition any point P and its projection F are so related that their z's and <£'s are the same. The area of any element PQR on the sphere is PQ . QB, and this is equal to a sin 0d(j>. add. The area of the projection on the cylinder, viz. P'Q'R' is FQ' . Q'R', and this is adcj) . dz 1 , where z' = CL = a - a cos 6. Substituting for z, we see that these two areas are equal. Hence any elementary area on the sphere and its projection on the cylinder are equal*. It follows from this result that the area of any finite portion of the spherical surface is equal to the area of its projection on any circumscribing cylinder. This rule enables us to find many areas on the sphere which are useful to us. Thus the area cut off from the sphere by any two parallel planes whose distance apart is h is equal to the area of a band on the cylinder whose breadth is h. The area on the sphere is therefore 2irah. We notice that this result is independent of the position of the planes, except that they must be parallel. Thus the area of a segment of a sphere whose versed sine is h is Zirah. 421. This important theorem is used also in the construction of maps. The places on a terrestrial globe are projected in the manner just described on a circumscribing cylinder. The cylinder is then unrolled on a plane. In this way the whole earth may be represented on a map of a rectangular form. The advantage of this con- * The relation of the sphere to the cylinder in regard to their measurement was first discovered by Archimedes. He wrote two books on this subject. He investi- gated both their surfaces and volumes, whether entire or cut by planes perpendicular to their common axis. He was so pleased with these discoveries that he direoted a cylinder enclosing a sphere to be engraved on his tombstone in commemoration of them. ART. 423.] SPHERAL SURFACES. 299 struction is that any equal areas on the globe are represented by equal areas on the map. This is true for large or small areas in whatever part of the globe they may be situated. The disadvantage of the construction is that any small figure on the map is not similar to the corresponding figure on the globe. If the figure is situated near the curve of contact of the cylinder, the similarity is sufficiently close for practical purposes, but if the figure is situated nearer the pole of this curve of contact, the dissimilarity is more striking. Thus a small circle very near the pole is represented by an elongated oval. In some other systems of making maps, as for example Mercator's, any small figure on the map is made similar to the cor- responding figure on the globe, but in that case equal areas on the map do not correspond to equal areas on the globe. 422. The altitude of the centre of gravity of any portion of the sphere above the plane of contact is equal to the altitude of the centre of gravity of its projection on the circumscribing cylinder. To prove this it is sufficient to quote the formula z = Xmzj'Zm, and to remark that for the surface and its projection the m's and z's are equal, each to each. From this we infer that the centre of gravity of the band on the sphere between any two parallel planes is the same as that for the corresponding band on the cylinder, and is therefore half way between the parallel planes, and lies on the radius perpendicular to them. In the same way the centre of gravity of a hollow thin hemi- sphere of uniform thickness bisects the middle radius. 423. Ex. 1. A segment of a sphere of height h rests on a plane base: show that the centre of gravity of the surface including the plane base is at a distance j r from the base, where a is the radius of the sphere. Ex. 2. The distance of the centre of gravity of the surface of a lune from the axis is -; , where 2a is the angle of the lune. 4 a Ex. 3* A bowl of uniform thin material in the form of a segment of a sphere is closed by a circular lid of the same material and thickness, which is hinged across a diameter. If it be placed on a smooth horizontal plane with one half of the lid turned back over the other half, show that the plane of the lid will make with the horizontal plane an angle given by Sir tan 0=4 tan Jo; a being the angle any radius of the lid subtends at the centre of the sphere of which the bowl is part. [Math. Tripos, 1881.] Ex. 4. An ellipsoidal surface is enveloped by a cylinder, and the density at every point of the two surfaces is proportional to the perpendicular from the centre on the tangent plane. Show that the mass of any portion of the ellipsoid is equal to the mass of the corresponding portion of the cylinder. Points are said to corre- spond when the straight line joining them is parallel to the plane of contact of h 300 CENTRE OF GRAVITY. [CHAP. IX. the ellipsoid and cylinder and intersects the conjugate diameter of this plane. This is an extension of Archimedes' theorem to ellipsoids. Thence show that the centre of gravity of the band on this ellipsoid between two planes parallel to a principal plane is half way between the planes. 424. To find the centre of gravity of any spherical triangle. ^ Let us begin by projecting any portion of the surface of the sphere on a central plane. Let this be the plane of xy. Let dS be any element of area, = area ANB : area ABC. Also (area ANB) = (area AOB) - (area A OC) cos A - (area BOG) cos B =i/3 2 (c-b cos A -a cos B). If E be the spherical excess of the triangle, i.e. if E = A + B + C -ir, we know by Spherical Trigonometry that the area ABG=f?E. Hence z_, c-baosA-aaosB This formula gives the distance of the centre of gravity from the plane AOB containing any side AB of the triangle. The distances from the planes BOG, GOA containing the other sides are expressed by similar formulae. Ex. 1. If p, q, r be the perpendicular arcs from the angular points A, B, G on the opposite sides, and G the centre of gravity of the spherical triangle, prove that cos AOG _ cos BOG _ cos GOG _ 1 aeiap ~ 6 sing — csinr ~ IE' This is equivalent to the result given in Moigno's Statique. * We have here followed the method proposed by Prof. Giulio, chiefly because the lemma on which it depends is of general application and may be useful in other cases. His memoir was published in the fourth volume of Liouville's Journal de Mathematiques. An English version is also given in Walton's Mechanical Problems, p. 36. ART. 425.] SURFACES AND*SOLIDS OF REVOLUTION. 301 Ex. 2. A surface is generated by the revolution of the catenary about its axis. Let this be the axis of z and let the plane generated by the directrix be that of xy. Any portion S of its surface is projected orthogonally on the plane xy, and V is the volume of the cylindrical solid formed by the perpendiculars from the perimeter of S. Prove that the x~ and y of 8 and V are equal each to each, but the z of the first is double that of the secoud. [Giulio, also Walton.] 425. Any surfaces and solids of revolution. A known plane curve revolves round an axis in its own plane which we shall take as the axis of z, and the angle of revolution is 2a. It is required to find the centres of gravity of the surface and volume thus generated. It is clear that every point describes an arc of a circle whose centre is in the axis of z. Thus the whole solid is symmetrical about a plane passing through z and bisecting all these arcs. Let this be the plane of xz. The centres of gravity lie in this plane. Let PP' be half the arc described by P, the other half being behind the plane xz and not drawn in the figure. Let PQ = ds be any arc of the generating curve, then the area of the elementary band described by ds is m = 2xads by Pappus' theorem. Its centre of gravity lies in MP at a distance from M equal to (x sin a)/a. Hence the coordinates of the centre of gravity of the surface are x = "Zmx _ Ja?ds sin a. _ _ Jxzds %m fxds, ' a ' Jxds In the same way the coordinates of the centre of gravity of the volume are _ _ "Zrnx _ Jafda sin a _ fxzda Sto fxda a ' ~ jxda ' where da is any element of the area of the given curve. We may write for da either dxdz or rdddr according as we choose to use Cartesian or Polar coordinates, replacing the single integral sign by that for double integration. 302 CENTRE OF GRAVITY. [CHAP. IX. It is evident that these integrals are those used in the higher Mathematics for the moments and products of inertia of the arcs and areas. When therefore we have once learnt the rules to find these moments of inertia, we seldom have to perform any integra- tion. We simply quote the results as being well known. These rules are usually studied in connection with rigid dynamics, as a knowledge of them is essential for that science. But they are now given in some of the treatises on the integral calculus, for example in that by Prof. Williamson. Ex. I. A portion of an anchor ring is generated by the complete revolution of a quadrant of a circle (radius a) about an axis parallel to one of the extreme radii and distant 6 from it. Prove that the distances of the centres of gravity of the curved surface and volume from the plane described by the other extreme radius are a(26±o) a(86±3a) 7r6±2a 2(3ir6±4o) " The axis of revolution is supposed not to cut the quadrant. Ex. 2. A semi-ellipse revolves through one right angle about the bounding diameter. Show that the distance from the axis of the centre of gravity of the volume generated is 3abjiJ2r, where 2r is the length of the diameter. Ex. 3. A triangular area makes a revolution through two right angles about an axis in its own plane. Prove that the distance of the centre of gravity of the volume from the axis is £- — — , where o, 8, y are the distances of the middle points of w a + p + y the sides from the axis. Ex. 4. The portion of a parabola y 1 =px bounded by the axis and any ordinate revolves round the axis through a right angle. Show that the distance of the centre of gravity of the volume from the tangent plane at the vertex is %x, and its distance from the axis is 16 x /22//15ir. [Walton.] Ex. 5. A circular area of radius a revolves about a line in its plane at a distance c from the centre, where c is greater than a. If 2a be the angle through which it revolves, find the volume of the solid generated and prove that the centre of gravity of the solid is at a distance from the line equal to (4c 2 + a 2 ) sin a/ica. [Coll. Exam., 1887.] 426. To find the centre of gravity of a solid sector of a sphere with a circular rim. Referring to the figure of Art. 400, let OG be the middle radius of the solid sector, N the centre of the rim, the centre of gravity of the sector, V its volume, V the volume of the whole sphere, a the radius, then „ 0N+ OG F F W Otr=rf 2 > K = K -^- . [Walhs.] ART. 428.] ELLIPSOIDAL VOLUMES. 303 To prove this we follow the same method as that adopted to find the centre of gravity of a sector of a circle. Let PQ be an elementary area of the surface, then OPQ is a tetrahedron whose centre of gravity is at^> where Op = f OP. Hence, if G' be the centre of gravity of the surface, OG = %OG'. ' But OG' = %(ON + 00) by Art. 422. Hence the result follows. The volume V has been already found in Art. 415. The centre of gravity of a solid hemisphere follows immediately from this result. Putting ON = 0, we see that the centre of gravity of a solid hemisphere lies on the middle radius and is at a distance f of that radius from the centre. The centre of gravity of a solid octant also follows at once. There are four octants on one side of any central plane and the centre of gravity of each of these is at the same distance from that plane. Hence the centre of gravity of all four must be also at the same distance, and this has just been proved to be fa. Hence, for any octant, the distance of the centre of gravity from any one of the three plane faces is § of the radius. 427. Ex. 1. The centre of gravity and volume of a solid segment of a sphere bounded by a plane distant z from the centre are given by 0G =^' V=l { a-, n sa +z) . Ex. 2. Prove that in a sphere, whose density varies inversely as the distance from a point in the surface, the distance of the centre of gravity from that point bears to the diameter the ratio 2:5. - [Math. Tripos, 1867.] Ex. 3. Prove that the centre of gravity of a solid sphere, whose density varies inversely as the fifth power of the distance from an external point, is at the centre of the section of the sphere by the polar plane of the external point. [Math. Tripos, 1872.] 428. Centres of gravity of volumes connected with the ellipsoid. In order to deduce the centre of gravity of any portion of an ellipsoid from that of the corresponding portion of a sphere, we shall use an extension of that method of projections by which we passed from the areas of circles to those of ellipses. One point (xyz) is said to be projected into another (oc'y'z) when we write x = ax', y = by, z = cz'. The points are then said to correspond. Volumes V, V correspond when their boundaries are traced out by corresponding points. If (xyz), (x y" s!) be the centres of gravity of V, V we have V=JJJdx dy dz = abc fjdx' dy' dz 1 = abcV. In the same way x = ax,y=bf, z = cz'. 304 CENTRE OF GRAVITY. [CHAP. IX. It appears from these equations that any corresponding volumes have a constant ratio, and the centre of gravity of one corresponds to the centre of gravity of the other. We may also show* that (1) parallel straight lines correspond to parallels, and (2) the ratio of the lengths of parallel straight lines is unaltered by projection. Thus the rule already explained in Art. 403 for areas is true also for solids. We may apply these principles to an ellipsoidal solid. The equation to an ellipsoid of semi-axes a, b, c is changed into that of a concentric sphere by writing x => ax', y = by', z = cz 1 . It follows that all projective theorems may be transferred from one to the other. 429. Ex. 1. Find the centre of gravity of a solid 'sector of an ellipsoid with an elliptic rim. Let and N be the centres of the ellipsoid and of the rim. Then ON is the conjugate diameter of the plane of the rim. Let it cut the ellipsoid in C. The corresponding theorem for a spherical sector is given in Art. 426. Since the values of Off and V there given depend on the ratios of parallel lengths, we may transfer them to the ellipsoid. The centre of gravity ff of the ellipsoidal sector therefore lies in ON, and we have og _ s on+oo F __^L F Ex. 2. The coordinates of a solid octant of an ellipsoid bounded by three conjugate planes are Jc=fa, j7=|6, z=%c. Ex. 3. The centre of gravity and volume of any solid segment of an ellipsoid are given by ff=s^±£>- 2 F _( C -«)'(ac+«) * 2c + z • v ~ Idbc "' where 2c is the conjugate diameter of the plane of the segment, z its ordinate measured along c, and V the volume of the whole ellipsoid. 430. Let us construct two concentric and coaxial ellipsoids forming between them a thin solid shell. Let (a, b, c), (a + da, &c.) be the semi-axes of these ellipsoids, p and p + dp the perpendiculars on two parallel tangent planes. Then t=dp is the thickness of the shell at any point. Let da- be an element of the surface of one ellipsoid, dli. its projection on the plane of xy, then dn = d, D' be the densities of these elements. Since D-fn.li 3 . D'—p.jr' 3 , we find vDr=v'DY, i.e. the centre of gravity of two elements is at 0. It easily follows that the centre of gravity of the whole thin shell is at 0. Joining many thin shells together, it also follows that the centre of gravity of a thick shell is at 0. Next, let be an external point, and let the elementary cone whose vertex is at intersect the polar plane of in an element whose distance from is p. Since p is the harmonic mean of »• and r', we easily find vDr+v'DY = (vD + v'D')p, i.e. the centre of gravity of the two elementary volumes v and v' lies in the polar plane of 0. It follows that the centre of gravity of the shell lies in the polar plane of 0. Lastly, let any number of particles %, m 2 , &c, attract the origin according to the Newtonian law, and let the resultant attraction be a force X acting along the axis of x. If the coordinates of the particles be (x± y x z^} , y = r sin sin , z = r cos 6, and replace 2 by the sign of triple integration. These relations are proved in treatises on the integral calculus. We find - _ III* 3 sia - s eoos drd9d

    drd8d(f> - _ /.Ifr 3 sin $ eos 8 drddd(p jjjr* sin 8 drd0d ' V ~ \tfr* sin 8 drd0d(p ' tfjr* sin 8drd8d " For cylindrical coordinates we have m = pdcj> . dp . dz, and x = p cos , y = p sin dpdz ' z ~ jjfpdipdpdz ' Or again, if x, y, z be given functions of three auxiliary variables u, v, w, we can use the Jacobian form corresponding to that of Art. 411. We have then m = Jdudvdw. 432. To find the centre of gravity of the swrface of a solid we find the value of m suitable to the coordinates we wish to use. If the equation to the surface is given in the Cartesian form z =f( a! > y)> we project the element of surface on the plane of xy. The -area of the projection is dxdy. If (a/87) be the direction angles of the normal to the element, the area of the element must be secy dxdy. This therefore is our value of m. We find -_ JJsecydxdy.x ___ JJsecydxdy.y „ JJ sec 7 dxdy ' y //sec 7 dxdy C ' ART. 434.] ANY SUflCACE AND SOLID. 307 Taking the equation to the normal, we find In a similar way, if the equation to the surface is given in cylindrical coordinates z =f(p, ), we find {/dz\* / dz V " If the surface is given in polar coordinates r =/(0, ), we have ™=^#{(|) 2 +- s n*j+^Bin^} i . Substituting these values of m, we have the equations to find x, y,z. 433. In some cases it is more advantageous to divide the solid into larger elements. We should especially try to choose as our element some thin lamina or shell whose volume and centre of gravity have been already found. Suppose, for example, we wish to find x for some solid. We take as the element a thin slice of the solid bounded by two planes perpendicular to x. If the boundary be a portion of an ellipse, triangle, or some other figure whose area A is known, we can use the formula _ _/ Adxx ~ JAdx In this method we have only a single instead of a triple sign of integration. If the centre of gravity of A is known as well as its area, we can find y and z by using the same element. To take another example, suppose the solid heterogeneous. Then instead of the thin slice just mentioned we might take as the element a thin stratum of homogeneous substance. If the mass and centre of gravity of this stratum be known, a single integration will suffice to find the centre of gravity of the whole solid. This method will be found useful whenever the boundary of the whole solid is a stratum of uniform density, for in that case the limits of the integral will be usually constants. 434. Ex. 1. Find the centre of gravity of an octant of the solid ©"♦«)"♦©■- 20—2 308 CENTRE OF GRAVITY. [CHAP. IX. From the symmetry of the case it will be sufficient to find z. It will also evidently simplify matters if we clear the equation of the quantities a, b, c ; we therefore put x=ax', y = by', z = cz', Art. 428. If we take as our element a slice formed by planes parallel to xy, we shall require the area A of the section PMQ. This area is 1 A=jy'dx'=$(l-z"'-x' n )' i dx', where the limits of integration are to (1 - z' n ) n . If we write x' n =(l - z' n ) f, this reduces to A = (l-z' n ) n ij(l-f) ,l f n d%=(l-z' n ) n B, where the limits of the integral have been made to 1, so that B can be expressed in gamma functions if required. 2 „ . z Udz'.z' [(l-z'^dz'.z' \z' = to We have now, -= ' =1 5 , { , = 1 • C ]AdZ J(l-*"r^ l If we put z' n = | and write m for 1/n, this reduces to 2 _ /(l-lf"! 8 " 1 " 1 ^ _ r(2m + l)T(2m) T(3m + 1) _ ~c~ f(l - 1)** f ^ _ r(4m + l) r(2m + l)r(m) ; using the equation T (x + l)=xY(x), this reduces to - J =j ^' r |M , where „=*. c * T (m) T (4m) n Ex. 2. Find the centre of gravity of a hemisphere, the density at any point varying as the nth power of the distance from the centre. Here we notice that any stratum of uniform density is a thin hemispherical shell, whose volume and centre of gravity are both known. We therefore take this stratum as the element. We have the further advantage that the limits are constants, because the external boundary of the solid is homogeneous. Let the axis of z be along the middle radius, let (r, r + dr) be the radii of any shell, and let the density X>=/nr™. Then m=2iri 3 dr .jur", also the ordinate of its centre of gravity is \r, see Art. 422. Hence __ [IvfidTia^^ r TO + 3 a"+*-b a+i "~ jim'dria* ~*m + 4 a n+3 - &**+»■ The limits of 'the integral have been taken from r=b to r=a, so that we have found the centre of gravity of a shell whose internal and external radii are b and a. For a hemisphere we put 6 = 0. If n + 3 is positive, we then have z = ^ ^i-y . In other 2 ra + 4 cases we find 2 = 0. If either tj + 3 or ti + 4 is zero the integrals lead to logarithmic forms, but we still find 2 = 0. Ex. 3. Find the centre of gravity of the octant of an ellipsoid when the density at any point is D =nx?y m z n . To effect this we shall have to find the values of 2mz and 2to, which are both integrals of the form jjjx^^dxdydz ART. 434.] HETEROGENEOUS BODIES. 309 for all elements within the solid. To simplify matters, we write (x/a) 2 =£, &c. The limits of the integral are now fixed by the plane f + ij + f=l. But these are the integrals known as Dirichlet's integrals, and are to be found in treatises on the Integral Calculus. The result is usually quoted in the form JJJS ' s * ' * T(l+m + n+l) though Liouville's extensions to ellipsoids and other surfaces are also given. Here r (p + l) = l . 2 .3.. p when p is integral, and in all cases in which p is positive r(j>+i)=j>r( P ). Also r(4) =N /x. The result now follows from substitution ; we find z _ Fj (n + 2) .T% (l+m+n + 5) c~ Ti(n+l).Ti(l+m + n+&)' When I, m, n are positive integers there is no difficulty in deducing the values of these gamma functions from the theorems just quoted. In this way we can find Smz and 2m and thence 2 whenever the density D is a function which can be expanded in a finite series of powers of x, y, *. If the density at any point of an octant of an ellipsoid is D=nxyz, show that I=16c/35. Ex. 4. If the density at any point of an octant of an ellipsoid vary as the 5c a 2 + 6 2 + 2c 2 square of the distance from the centre, show that z = jg . . 8 . . Ex. 5. To find the centre of gravity of a triangular area whose density at any point is D=/jix'y m . To determine x and y we have to find 2m, 2mx and 2my. All these are integrals of the form jjx'y m dxdy. If y v y 2 , y s are the ordinates of the corners of the triangle and A the area, it may be shown that 2A jjyn dxdy=( —^—^^y^ + y i n-ly 2 + y 1 ^ ys+ ...} (1), where the right hand side, after division by A, is the arithmetic mean of the homo- geneous products of y v y 2 , y s . Thus when the density is D=/u/ n the ordinate y may be found by a simple substitution. If we take y + kx = as a new axis of x, the equation (1) may be written in the form 2A tf{y + kx)*dxdy = , +1)(w + 2) {(yi + ^ ] )" + (2/i + ^i) n ~ 1 (y2 + fa !! ) + -}- Equating the coefficient of k on each side, we find 2A ftnxr- 1 dxdy = (w+1)(w + a) WA" -1 + (« - 1) Vi^V^x + &0 - 1 In general, if H n be the arithmetic mean of the homogeneous products of Vi,Vi, 2/ 3 > wehave d p I d d d \ „ UxP^dxdy = A (^ +s 2 ^ +x s ^J H n . One corner of a triangle is at the origin ; if the density vary as the cube of the distance from the axis of. x, show that y = = V ±f~h ■ Also write down the value of s. 310 CENTRE OF GRAVITY. [CHAP. IX. The same method may be used to find the centre of gravity of a quadrilateral, a tetrahedron- or a double tetrahedron, when the density is D=iL3fy m z n . See a paper by the author in the Quarterly Journal of Mathematics, 1886. Lagrange's two theorems. 435. Def. If the mass of a particle be multiplied by the square of its distance from a given point 0, the product is called the moment of inertia of the particle about, or with regard to, the point 0. The moment of inertia of a system of particles is the sum of the moments of inertia of the several particles. 436. Lagrange's first Theorem. The moment of inertia of a system of particles about any point is equal to their moment of inertia about their centre of gravity together with what would be the moment of inertia about of the whole mass if it were collected at its centre of gravity. Let the particles m,, m 2 &c. be situated at the points A lt A 2 &c. Let (x^iZ^), (x^y^), &c. be the coordinates of A lt A 2 &c. referred to as origin. Let x, y, z be the coordinates of the centre of gravity G. Also let x = x + x', y = y + y', &c. Now 2 (m . OA*) = 2m {(x + xj + (y + yj + (z + a') 2 } = 2m . OG* + IxZmaf + Zytmy + 2z%mz' + 2 (mGA*). Since the origin of the accented coordinates is the centre of gravity, we have 2m*' = 0, 2m/ = 0, 2m^' = 0. Hence putting M = 2m, we have 2(m.04 2 ) = i!f.0(? 2 + 2(m.G ! ^ 2 ) (A). This equation expresses Lagrange's theorem in an analytical form. We notice that the moment of inertia of the body about any point is least when that point is at the centre of gravity. An important extension of this theorem is required in rigid dynamics. It is shown that, if f(x, y, z) be any quadratic function of the coordinates of a particle, then 2m/ («, y, z) = Mf(x, y, z) + %mf(x', y', z 1 ). 437. Lagrange's second theorem. If m, m' be the masses of any two particles, A A' the distance between them, then the theorem may be analytically stated thus 1(jnm'.AA'*)=MZ(m.GA*) (B). AET. 438.] LAGBjfcNGE'S THEOREMS. 311 The sum of the products of the masses taken two together into the square of the distance between them is equal to the product of the whole mass into the moment of inertia about the centre of gravity. This may be easily deduced from Lagrange's first theorem. We have by (A) •Em a OA a s = M.OG* + Xm a GA a \ where 2 implies summation for all values of «. Putting the arbitrary point successively at A 1} A 2 &c. we have 2m tt A 1 A C ? = M. A.G 3 + 2m a GA a \ 2m a A 2 A* = M. A 2 G* + 1m a GA*, &c. = &c. Multiplying these respectively by ntj, m 2 &c. and adding the products together, we have tm a mp A p A a * = M ZmpApG* + tm^ . tmjGA a \ The 2 on the left hand side implies summation for all values of both a and j8. Each term will therefore appear twice over, once in the form m^m^ . A$ A a a , and a second time with a and yS inter- changed. If we wish to take each term once only, we must take half the right hand side. But the terms on the right hand side are the same. Hence Snviip . A a A,? = M %m a . GA a *. 438. Ex. Let the symbol [ABC] represent the area of the triangle formed by joining the three points A, B, G. Let [ABGD] represent the volume of the tetrahedron formed by joining the four points in space A,B, C, D. We may extend the analytical expression for the area and volume to any number of points by the same notation. We then have the following extensions of Lagrange's two theorems Zm a OA a s =M . 0<3 2 + 2m a GA a 2 Znynp [OA a A^f=MSm a [OGA J' + Hmjn,, [QA a ApJ> Sm„m p ra v [OA a A^A y f=M^m a m fi [OGA^f + Zm^m^GA^ApAyf &c.=&c. Hm a m^A a A p a = M 2m a GA a * Sm a mpm y [A a A fl A y f=M2m a mp [GA a A ? f 2m a mpm y m s lA a ApA y A s f=M2m a m ? m y [GA a A p A y ¥ &c. = &c. The first of each of these sets of equations is of course a repetition of Lagrange's equations. The remaining equations are due to Franklin. [American Journal of Mathematics, Vol. x., 1888.] 312 CENTKE OF GRAVITY. [CHAP. IX. Application to pure geometry. 439. The property that every body has but one centre of gravity* may be used to assist us in discovering new geometrical theorems. The general method may be described in a few words. We place weights of the proper magnitudes at certain points in the figure. By combining these in several different orders we find different constructions for the centre of gravity. All these must give the same point. The following are a few examples. Ex. 1. The three straight lines which join the middle points of the opposite edges of a tetrahedron are called the median lines. Prove that each median passes through the centre of gravity of the volume of the tetrahedron and is bisected by it. Place particles of equal weight at the corners A, B, C, D of the tetrahedron. The centre of gravity of the particles A, B is at the middle point M of the edge AB. The centre of gravity of the particles G, D is at the middle point N of the edge CD. The common centre of gravity of all four is at the middle point of MN. In what- ever order we combine the four particles, we must always arrive at the same centre of gravity. But the centre of gravity of the four equal particles is the centre of gravity of the volume. The result therefore follows at once. See Art. 393. Ex. 2. The two straight lines which join the middle points of the opposite sides of a quadrilateral and the straight line which joins the middle points of the two diagonals, intersect in one point and are bisected at that point. [Coll. Exam.] Ex. 3. The centre of gravity of four particles of equal weight in the same plane is the centre of the conic which bisects the lines joining each pair of points. [Caius Coll.] In a conic only one chord is bisected at a given point, unless that point is the centre. Since, by the last example, three chords are bisected at the same point, that point is the centre. Ex. 4. Through each edge of a tetrahedron a plane is drawn bisecting the angle between the planes that meet in that edge and intersecting the opposite edge i prove that the three lines joining the points so determined on opposite edges meet in a point. [St John's Coll. 1879.] Place weights at the corners proportional to the areas of the opposite faces. The centre of gravity of these four weights lies in each of the three straight lines. 440. The theorems on the centre of gravity are also useful in helping us to remember the relations of certain points, much used in our geometrical figures, to the other points and lines in the construction. For instance, when the results of Ex. 1 have been noticed, the distance of the centre of the inscribed conic from any straight line can be written down at once by taking moments about that lioe. * In Milne's Companion to the weekly problem papers 1888, a number of ex- amples will be found of the application of the "centroid" and of "force" to geometry. ART. 441.] APPLICATION TO PURE GEOMETRY. 313 Ex. 1. The areal equation to the conic inscribed in the triangle of reference is ijlx + Jmy + sjnz=0; show that the centre of the conic is the centre of gravity of three particles placed at the middle points of the sides, whose weights are proportional to I, m, n. It is also the centre of gravity of three particles whose weights are proportional to m + n, n + l, l + m, placed either at the points of contact or at the corners of the triangle. See Art. 79, or Art. 382, Ex. 3. Thence show that the conic is a parabola if l + m + n=0. If this condition is satisfied, show that the far focus is the centre of gravity of three particles whose Weights are proportional to I, m, n placed at the corners, and that the near focus is the centre of gravity of three particles of weights cpjl, 6 2 /m, c 2 /» placed at the same points. Ex. 2. The areal equation to the conic circumscribed about the triangle of reference is - -\ h - = 0. Show that its centre is the centre of gravity of six x y z particles, three placed at the corners whose weights are proportional to Z 2 , m 2 , n 2 , and three at the middle points of the sides whose weights are - 2mn, - 2nl, - 2lm. Ex. 3. Three particles of equal weight are placed at the corners of a triangle, and a fourth particle of negative weight is placed at the centre of the circumscribing circle. Show that the centre of gravity of all four is the centre of the nine-points circle or the orthocentre, according as the weight of the fourth particle is numeri- cally equal to or double that of any one of the particles at the corners. Ex. 4. The general equation to a conic in tangential coordinates is Af + Bq* + Cr* + 2Dqr + SErp + 2Fpq = ; show that the centre of the conic is the centre of gravity of three weights proportional to A + E + F, B + F+D, G + D + E placed at the corners. For other theorems see a paper by the author in the Quarterly Journal, Vol. vm. 1866. 441. Theorems concerning the resolution and composition of forces may be used, as well as those relating to the centre of gravity, to prove geometrical properties. Ex.1. A straight line is drawn from the corner D of a tetrahedron making equal angles with the edges DA, DB, DO. Show that this straight line intersects the plane ABG in a point E such that AE/AD, BE/BD, GEjGD are proportional to the sines of the angles BEG, GEA, AED. Show also that ; + iTF, + . 3cos0 AD ' BD ' CD ED ' where S is the angle DE makes with any edge at D. Ex. 2. ABGD is a quadrilateral, whose opposite sides meet in X and Y. Show that the bisectors of the angles X, Y, the bisectors of the angles B, D and the bisectors of the angles A , C intersect on a straight line, certain restrictions being made as to which pairs of bisectors are taken. See figure in Art. 132. [Math. Tripos, 1882.] Apply four equal forces to act along the sides of the quadrilateral, and find their resultant by combining them in different orders. 314 CENTRE OF GRAVITY. [CHAP. IX. Ex. 3. Prove, by mechanical considerations, that the locus of the centres of all ellipses inscribed in the same quadrilateral is the straight line joining the middle points of any two diagonals. [Coll. Exam.] Let A, B, C, D be the corners taken in order. Apply forces along AB, AD, CB, CD proportional to these lengths. The tangents measured from each corner to the adjacent points of contact represent forces whose resultant passes through the centre. Again the resultant of AB, AD and also that of GB, CD bisect the diagonal BD. Similarly the resultant force bisects the other diagonal. Ex. 4. If X, Y are the intersections of the opposite sides of a quadrilateral ABCD, prove that the ratio of the perpendiculars drawn from X and Y on the diagonal AC is equal to the ratio of the perpendiculars on the diagonal BD. Show also that each of these ratios is equal to the ratio of AB . CD sin Y to AD . BC sin X. See figure of Art. 132. CHAPTER X. ON STRINGS. The Catenary. 442. The strings considered in this chapter are supposed to be perfectly flexible. By this we mean that the resultant action across any section of the string consists of a single force whose line of action is along a tangent to the length of the string. Any normal section is considered to be so small that the string may be regarded as a curved line, so that we may speak of its tangent, or its osculating plane. The resultant action across any section of the string is called its tension, and in what follows will be represented by the letter T. This force may theoretically be positive or negative, but it is obvious that an actual string can only pull. The positive sign is given to the tension when it exerts a pull on any object instead of a push. The weight of an element of length ds is represented by wds. In a uniform string w is the weight of a unit of length. If the string is not uniform, w is the weight of a unit of length of an imaginary string, such that any element of it (whose length is ds) is similar and equal to the particular element ds of the actual string. 44'J. The Catenary. A heavy uniform string is suspended from two given points A, B, and is in equilibrium in a vertical plane. It is required to find the equation to the curve in which it hangs. This curve is called the common Catenary*. * The following short account of the history of the problem known under the name of the " Chalnette " is abridged from Montucla, Vol. ii., p. 468. The problem 316 INEXTENSIBLE STRINGS. [CHAP. X. Let G be the lowest point of the catenary, i.e. the point at which the tangent is horizontal. Take some horizontal straight line Ox as the axis of x, whose distance from G we may afterwards choose at pleasure. Draw GO perpendicular to it, and let be the origin. Let ty be the angle the tangent at any point P makes with Ox. Let T and T be the tensions at G and P, and let GP = a. The length GP of the string is in equilibrium under three forces, viz. the tensions T and T acting at G and P in the direc- tions of the arrows, and its weight ws acting at the centre of gravity G of the arc GP. .A y T N If * Resolving horizontally, we have T cos yjr = T (1). Resolving vertically, we have T sin ty = ws (2). Dividing one of these equations by the other, of finding the form of a heavy chain suspended from two fixed points was proposed by James Bernoulli as a question to the other geometers of that day. Pour mathematicians, viz. James Bernoulli and his brother, Leibnitz and Huyghens had the honour of solving it. They published their solutions in the Actes de Leipsick (Act. Erud. 1691) but without the analysis, apparently wishing to leave some laurels to be gathered by those who followed. David Gregory published a solution some years after in the Phil. Trans. 1697. It is the custom of geometers to rise from one difficulty to another, and even to make new ones in order to have the pleasure of surmounting them. Bernoulli was no sooner in possession of the solution of his problem of the chainette considered in its simplest case, than he proceeded to more difficult ones. He supposed next that the string was heterogeneous and enquired what should be the law of density that the curve should be of any given form, and what would be the curve if the string were extensible. He soon after published his solution, but reserved his analysis. Finally he proposed the problem, what would be the form of the string if it were acted on by a central force. The solutions of all these problems were afterwards given by John Bernoulli in his Opera Omnia. Montucla remarks that the problem of the chainette had excited the curiosity of Galileo, who had decided that the curve is a parabola. But this accusation is stated by Venturoli to be without foundation. Galileo had merely noticed the similarity between the two curves. See Venturoli, Elements of Mechanics, translated by Gresswell, p. 69, where the problem of the chainette is discussed. ART. 443.] JHE CATENARY. 317 If the string is uniform w is constant, and it is then con- venient to write T = wc. To find the curve we must integrate the differential equation (3). We have .-. dy=± VO 2 + c 2 ) ' .-. y + A = + VO 2 + c 2 ). We must take the upper sign, for it is clear from (3) that, when x and s increase, y must also increase. When s = 0, y + A = c. Hence, if the axis of x is chosen to be at a distance c below the lowest point G of the string, we shall have A = 0. The equation now takes the form 2/ 2 = s 2 + c 2 (4). Substituting this value of y in (3), we find cds _ , where the radical is to have the positive sign. Integrating, c log {s + V(s 2 + c 2 )} = x + B. But x and s vanish together, hence B = c log c. From this equation we find aJ(s* + c 2 ) + s = ce° . Inverting this and rationalizing the denominator in the usual _x manner, we have V( s * + c i ) — s = ce °- Adding and subtracting we deduce by (4) y= -(ec +e e y s =_^ e c_ e «j ( 5 ). The first of these is the Cartesian equation to the common catenary. The straight lines which have here been taken as the axes of x and y are called respectively the directrix and the axis of the catenary. The point G is called the vertex. 318 INEXTENSIBLE STRINGS. [CHAP. X. Adding the squares of (1) and (2), we have by help of (4) T 2 = v? (s 2 + c 2 ) = wy ; .-. T = wy (6). The equations (1) and (2) give us two important properties of the curve, viz. (1) the horizontal tension at every point of the curve is the same and equal to wc ; (2) the vertical tension at any point P is equal to ws, where s is the arc measured from the lowest point. To these we join a third result embodied in (6), viz. (3) the resultant tension at any point is equal to wy, where y is the ordinate measured from the directrix. 444. Referring to the figure, let PN be the ordinate of P, then T = w . PN. Draw NL perpendicular to the tangent at P, then the angle PNL = yfr. Hence PL = PN . sin sfr = s by (2), NL = PN.cos^ = c by (1). These two geometrical properties of the curve may also be deduced from its Cartesian equation (5). By differentiating (3) a j 1 dilr 1 C /tT . we find — =— y -j- = -, .'. p = — r-r- (7). ' COS 2 l/r as C COS 1 -\jr v ' We easily deduce from the right-angled triangle PNH, that the length of the normal, viz. PH., between the curve and the directrix is equal to the radius of curvature, viz. p, at P. It will be noticed that these equations contain only one undetermined constant, viz. c ; and when this is given the form of the curve is absolutely determined. Its position in space depends on the positions of the straight lines called its directrix and axis. This constant c is called the parameter of the catenary. Two arcs of catenaries which have their parameters equal are said to be arcs of equal catenaries. Since p cos 2 yjr = c, it is clear that c is large or small according as the curve is flat or much curved near its vertex. Thus if the string is suspended from two points A, B in the same horizontal line, then c is very large or very small compared with the distance between A and B according as the string is tight or loose. The relations between the quantities y, s, c, p, f and T in the common catenary may be easily remembered by referring to the rectilineal figure PLNH. We have PN=y, PL=s, NL=c, PH=p, T=w.PN and the angles LNP, NPH are each equal to f. Thus the important relations (1), (2), (3), (4), and (7) follow from the ordinary properties of a right angled triangle. ART. 447.] TgE CATENARY. 319 445. Since the three forces, viz., the tensions at A and B and the weight are in equilibrium, it follows that their lines of action must meet in a point. Hence the centre of gravity G of the arc must lie vertically over the intersection of the tangents at the extremities of the arc. This is a statical proof of one part of the more general theorem given in Art. 399, Ex. 1, where both coordinates of the centre of gravity have been found and expressed in a convenient geometrical form. 446. Ex. 1. Show that it is impossible to pull a heavy string by forces at its extremities so as to make it quite straight unless the string is vertical. If it be straight let if/ be the inclination to the horizon, W its weight. Then, resolving perpendicular to its length, W cos ^=0, which gives \p equal to a right angle. This proof does not require the string to be uniform. Ex. 2. If a string be suspended from any two points A and B not in the same vertical, and be nearly straight, show that c is very large. Let f, \j/ be the inclinations at A and B, and I the length of the string. Then l=s-s'=c (tan^-tan^'). Since f and \j/ are nearly equal, c is large compared with I. Ex. 3. A heavy uniform string AB of length I is suspended from u fixed point A, while the other extremity B is pulled horizontally by a given force F=wa. Show that the horizontal and vertical distances between A and B are a log — — and sJ{P+a 2 ) - a respectively. — Ex. 4. The extremities A and B of a heavy string of length 21 are attached to two small rings which can slide on a fixed horizontal wire. Each of these rings is acted on by a horizontal force F=wl. Show that the distance apart of the rings is2Zlog(HV2). Ex. 5. If the inclination \j/ of the tangent at any point P of the catenary is taken as the independent variable, prove that a; = clogtan ( 7 + i ), y=- -, s=c tan f, p= — 5—. \4 2/' ' cosf r cos 2 ^ If x, y be the coordinates of the centre of gravity of the arc measured from the vertex up to the point P, prove also that a;=a;-ctan^ , * 2 \cos \p r J 447. If the position in space of the points A and B of suspension and the length of the string or chain are given, we may obtain sufficient equations to find the parameter c of the catenary, and the positions in space of its directrix and axis. Let the given point A be taken as an origin of coordinates, and let the axes be horizontal and vertical. Let (h, k) be the coordinates of B referred to A, and let I be the length of the string AB. These three quantities are therefore given. Let (x, y), (x + h, y + k) be the coordinates of A, B referred to the directrix and axis of the catenary. Then x, y, c are the three quantities to be found. We have by (5) of Art. 443, x x g-ffa x+h y = i(e~°+e~\ y + k= c -(e ° +e' * ) (A). 320 INEXTENSIBLE STRINGS. [CHAP. X. Also by (5), since I is the algebraic difference of the arcs CA, GB, x±h x+h x x l=li*° ") .(B). If G lie between A and B, x will be negative. These three equations are sufficient to determine x, y and u. They cannot however be solved in finite terms. We may eliminate x, y in the following manner. .(C). Writing u = e c ,v=e?,-m find from (A) and (B) '-IH;)*'- 1 )) We notice that v contains only c and the known quantity h. Hence, subtracting the squares of these equations in order tp eliminate u, we find h _ h ±J(V-lc i ) = c(<*»-e *) (D). This agrees with the equation given by Poisson in his Traite de Mecanique. The value of c has to be found from this equation. It gives two real finite values of c, one positive and the other negative but numerically equal. A negative value for c would make y negative and would therefore correspond to a catenary with its concavity downwards. It is therefore clear that the positive value of c is to be taken. To analyse the equation (D), we let = 1/7, aa d arrange the terms of the equation in the form z = e m t -e~ m V -ay=0 (E), so that a and m are both positive. We have a? = P-k 2 , and 2m=h. Since the length I of the string must be longer than the straight line joining the points of suspension, it is clear that a must be greater than 2m. By differentiation, rlz ~=m(e m y + e- mv )-a. Thus dzjdy is negative when 7 = 0, so that, as 7 increases from zero, a is at first zero, then becomes negative and finally becomes positive for large values of 7. There is therefore some one value of 7, say y—i, at which z = 0. If there could ' be another, say 7=1', then dzjdy must vanish twice, once between 7=0 and 7 = i,and again between7=i and y=i'. We shall now show that this is impossible. By differentiating twice we have ^=m?(e m y-e- m y); thus dh/dy 2 is positive when 7 is greater than zero. Hence dz/dy continually in- creases with 7 from its initial value 2m -a when 7=0. It therefore cannot vanish twice when 7 is positive. It appears from this reasoning that the equation gives only one positive value of e. The solitary positive value of c having been found from (D), we can form a simple equation to find u by adding one of the equations (C) to the other. In this way we find one real value of x. The value of y is then found from the first of the equations (A). Thus it appears that, when a uniform string is suspended from two fixed points of support, there is only one position of equilibrium. ART. 448.] THE CATENARY. 321 The equation (D) can be solved by approximation when hjc is so small that we can expand the exponentials and retain only the first powers of hjc which do not disappear of themselves. This occurs when c is large, i.e. when the string is nearly tight. In such cases, however, it will be found more convenient to resume the problem from the beginning rather than to quote the equations (D) or (E). 448. Ex. 1. A uniform string of length I is suspended from two points A and B in the same horizontal line, whose distance apart is ft. If ft and I are nearly equal, find the parameter of the catenary. Referring to the figure of Art. 443, we see that s=JJ, x=\h. Hence using one of the equations (5) of that article, we have l=c(e*>- e *°). Whatever the given values of ft and I may be, the value of c must be found from this equation. When ft and I are nearly equal, we know by Art. 446, Ex. 2, that ft/c is small. Hence, expanding the exponentials and retaining only the lowest powers of hjc which do not disappear, we have ft 3 "24 (I- ft)" Since the string considered in this problem is nearly horizontal, the tension of every element is nearly the same. If the string be slightly extensible, so that the extension of any element is some function of the tension, the stretched string will still be homogeneous. The form will therefore be a catenary, and its parameter will be given by the same formula, provided I represents its stretched length. In order to use this formula, the length I of the string and the distance ft between A and B must be measured. But measurements cannot be made without error. To use any formula correctly it is necessary to estimate the effects of such errors. Taking the logarithmic differential we have 2Se_3Sft_ ±5l±5h c ~ h l-h Here 5ft and 51 are the errors of ft and I due to measurement. We see that the error in c might be a large proportion of c if either ft or l-h were small. In our case I - ft is small. Hence to find c we must so make our measurements that the error of I - ft is small compared with the small quantity I - ft, while the length ft need be measured only so truly that its error is within the same fraction of the larger quantity ft. Thus greater care must be taken in measuring I - ft than ft. Suppose, for example, that ft =30 feet and Z=31 feet, with possible errors of measurement either way of only one thousandth part of the thing measured. The value of c given by the formula is 33-5 feet, but its possible error is as much as one thirtieth part of itself. Ex. 2. A uniform measuring chain of length I is tightly stretched over a river, the middle point just touching the surface of the water, while each of the ex- tremities has an elevation k above the surface. Show that the difference between 8 ft 2 the length of the measuring chain and the breadth of the river is nearly 5 — . A I R. s. 21 322 INEXTENSIBLE STRINGS. [CHAP. X. Ex. 3. A heavy string of length 21 is suspended from two fixed points A, B in the same horizontal line at a distance apart equal to 2a. A ring of weight W can slide freely on the string, and is in equilibrium at the lowest point. Find A E B the parameter of the catenary and the position of the weight. Let D be the position of the heavy ring, then BD and AD are equal por- tions of a catenary. Produce BD to its vertex C, and let Ox, OG be the directrix and axis of the catenary DB. Let x be the abscissa of D. Then since I is the difference of the arcs CB, CD, we have «-5<« ")-l^ 1- .(1). Also, since the weight of the ring is supported by the two vertical tensions of the string, W=2w-(e c -e °).. a .(2). The equations (1) and (2) determine x and c. Thence the ordinates of D and B may be found, and therefore the depth of D below AB. If the weight of the ring is much greater than the weight of the string, each string is nearly tight. Thus a\c is small, but xjc is not necessarily small, for the vertex G may be at a considerable distance from D. If we expand the terms con- taining the exponent ajc and eliminate those containing xjc, we find c=Wafiw,J(P-a?) nearly. The contrary holds if the weight of the ring is much smaller than the weight of the string. If W were zero the two catenaries BD and DA would be continuous, and the vertex would be at D. Hence when W is very small, the vertex will be near D and therefore x/a will be small. But a/c is not necessarily small. Ex- panding the terms with small exponentials, we find from (2) that x= Wj2w. Then (1) gives i »5('-o+s«<' + '" , - 1 >- If the weight W were absent this equation would reduce to the one already dis- cussed above. If y be the change produced in the value of c there found by adding the weight W, we find, by writing c + y for c in the first term on the right hand side, W, that H) 7+^(fc-c) = 0, where ft is the ordinate of B before the addition of W. If the weight W had been attached to any point D of the string not its middle point, AD, BD would still form catenaries, whose positions could be found in a similar manner. We may notice that, however different the two strings may appear to be, the catenaries have equal parameters. For consider the equilibrium of the weight W; we see by resolving horizontally that the wc of each catenary must be the same. If the string be passed through a fine smooth ring fixed in space through which it could slide freely, the two strings on eaoh side must have their tensions equal. ART. 448.] TljE CATENARY. 323 Hence the two catenaries have the same directrix. The parameters are not necessarily equal, for the difference between the horizontal tensions of the two catenaries is equal to the horizontal pressure on the ring, which need not be zero. Ex. 4. A heavy string of length I is suspended from two points A, A' in the same horizontal line, and passes through a smooth ring D fixed in space. If DN be a perpendicular from D on AA' and NA=h, NA'=h', DN=k, prove that the parameters c, c' may be obtained from 4#= P jcosh |J cosech ( * + *)} - ft* (coseeh i) , and a similar equation with the accented and unaccented letters interchanged. Ex. 5. A portion AC of a uniform heavy chain rests extended in the form of a straight line on a rough horizontal plane, while the other portion CB hangs in the form of a catenary from a given point B above the plane. The whole chain is on the point of motion towards the vertical throughB. If I be the length of the whole chain and h be the altitude of B above the plane, show that the parameter c of the catenary is equal to /i (l + ixh)- n,J{(ii? + l)h! i +2iM}. ' Ex. 6. A heavy string hangs over two small smooth fixed pegs. The two ends of the string are free, and the central portion hangs in a catenary. Show that the free ends are on the directrix of the catenary. If the two pegs are on the same level and distant 2a apart, show that equilibrium is impossible unless the length of the string is equal to or greater than 2a e. [Coll. Exam. ] ' Ex. 7. A heavy uniform chain is suspended from two fixed points A and B in the same horizontal line, and the tangent at A makes an angle 45° with the horizon. Prove that the depth of the lowest point of the chain below AB is to the length of the chain as ^/(2) -1:2. Ex. 8. A uniform heavy chain is fastened at its extremities to two rings of equal weight, which slide on smooth rods intersecting in a vertical plane, and inclined at the same angle o to the vertical : find the condition that the tension at the lowest point may be equal to half the weight of the chain ; and, in that case, show that the vertical distance of the rings from the point of intersection of the rods is I cot a log (J2 + 1), where 2t is the length of the chain. [Math. Tripos, 1856.] Ex. 9. A heavy string of uniform density and thickness is suspended from two given points in the same horizontal plane. A weight, n times that of the string, is attached to its lowest point ; show that, if 0,

    = ( 1 + - j tan B. [Math. Tripos, 1858.] Ex. 10. If o, jS be the angles which a string of length I makes with the vertical at the points of support, show that the height of one point above the other is I cos % (a + |3)/cos i (a - /3). [Pet. Coll., 1855.] Ex. 11. A heavy endless string passes over two small smooth fixed pegs in the same horizontal line, and a small smooth ring without weight binds together the upper and lower portions of the string : prove that the ratio of the cosines of the angles which the portions of the string at either peg make with the horizon, is equal to that of the tangents of the angles which the portions of the string at the ring make with the vertical. [Math. Tripos, 1872.] 21—2 324 INEXTENSIBLE STRINGS. [CHAP. X. Ex. 12. A and B are two smooth pegs in the same horizontal line, and C is a third smooth peg vertically below the middle point of AB ; an endless string hangs upon them forming three catenaries AB, BO, and GA : if the lowest point of the catenary AB coincides with 0, prove that the pegs AB divide the whole string into two parts in the ratio of 2w + w' to 2w-w', where w and w' are the vertical com- ponents of the pressures on A and G respectively. [Math. Tripos, 1870. ] Ex. 13. An endless uniform chain is hung over two small smooth pegs in the same horizontal line. Show that, when it is in a position of equilibrium, the ratio of the distance between the vertices of the two catenaries to half the length of the chain is the tangent of half the angle of inclination of the portions near the pegs. [Math. Tripos, 1855.] Ex. 14. A heavy uniform string of length 41 passes through two small smooth rings resting on a fixed horizontal bar. Prove that, if one of the rings be kept stationary, the other being held at any other point of the bar, the locus of the position of equilibrium of that end of the string which is the further from the stationary ring may be represented by the equation x=2J (ly) log -. [June Exam.] Ex. 15. A heavy uniform string is suspended from two points A and B in the same horizontal line, and to any point P of the string a heavy particle is attached. Prove that the two portions of the string are parts of equal Catenaries. Prove also that the portion of the tangent at A intercepted between the verticals through P and the centre of gravity of the string is divided by the tangent at B in a ratio independent of the position of P. If 0, be the angles the tangents at P make with the horizon, a and |3 those made by the tangents at A and B, show that 7 - — ? is constant for all posi- tana + tan/3 r tions of P. [St John's Coll.] Ex. 16. A heavy uniform string hangs over two smooth pegs in the same horizontal line. If the length of each portion which hangs freely be equal to the length between the pegs, prove that the whole length of the string is to the distance between the pegs as \/(3) to log V (3). Compare also the pressures on each peg with the weight of the string. Ex. 17. A string of length 2 has one end attached to a wall and the other to a ring, which slides on a smooth vertical rod at a distance m from the wall. If t, t' be the lengths of string whose weights are respectively equal to the tensions of the string at its junctions with the ring and wall, w the length of string whose weight is equal to that of the ring, show that t*-t*=(l + w)*-w*, {t+w)e"):=2 (l + w). [Caius Coll.] Ex. 18. An endless inextensible string hangs in two festoons over two small pegs in the same horizontal line. Prove that, if be the inclination to the vertical of one branch of the string at its highest point, the inclination of the other branch at the same point must be either or , where has only one value and is a function of only. If cot \0 = e aece , then = 0. [June Ex.] ART. 448.] T»E CATENARY. 325 Ex. 19. Four smooth pegs are placed in a vertical plane so as to form a square, the diagonals being one vertical and one horizontal. Round the pegs an endless chain is passed so as to pass over the three upper and under the lower one. If the directions of the strings make with the vertical angles equal to a at the upper peg, £ and 7 at each of the middle and 8 at the lower peg, prove the following relations : log cot Ja tan J/3 _ log cot %y tan £8 sin 7 sin /3 ' sin /3 sin 8 + sin a sin 7 = 2 sin a sin 5. [Caius Coll.] Ex. 20.. A bar of length 2a has its ends fastened to those of a heavy string of length 21, by which it is hung symmetrically over a peg. The weight of the bar is n times, and the horizontal tension Jm times the weight of the string. Show that m a + rfi = \(n + 1) cosech — ■ - n coth — I " . [Coll. Ex. 1889.] Ex. 21. One end of a heavy chain is attached to the extremity of a fixed rod, the other end is fastened to a small smooth ring which slides on the rod : prove that in the position of equilibrium log {cot \9 cot {\v - J^)} = cot 8 (sec \\i - cosec 6), being the inclination of the rod to the horizon, and i// that of the chain at its highest point. [Coll. Ex.] Ex. 22. An endless uniform heavy chain is passed round two rough pegs in the same horizontal line, being partly supported by a smooth peg situated midway in the line between the other pegs, so that the chain hangs in three festoons. If a, /3 are the angles which the tangents at one of the rough pegs make with the vertical, and /t is the coefficient of friction, prove that the limiting values of a and £ are given by the *c(»-»+ffl „ sin o log cot io , , , . ,. . _, , . . ., equation e =2 - — -r-p- 2 — -*- , and obtain an equation for determining the * sin £ log cot |/3 * B length of the chain in terms of a, |3 and the distance between the pegs. [Math. Tripos, 1879.] Ex. 23. A string of length ira is fastened to two points at a distance apart equal to 2a, and is repelled by a force perpendicular to the line joining the points and varying inversely as the square of the distance from it. Show that the form of the string is a semi-circle. [Coll. Ex. 1882.] Ex. 24. A chain, of length 21 and weight 2W, hangs with one end A attached to a fixed point in a smooth horizontal wire, and the other end B attached to a smooth ring which slides along the wire. Initially A and B are together. Show that the work done in drawing the ring along the wire till the chain at A is inclined at an angle of 45° to the vertical is W I (1 - V 2 + lo g l + \7 2 )- [° o11 - Ex - 1883 -] Ex. 25. Determine if the catenary is the only curve such that, if AB be. any arc whose centre of gravity is G, and AT, BT tangents at A and B, then GT is always parallel to a fixed line in space. Ex. 26. A uniform heavy chain of length 2a is suspended from two points in the same horizontal line ; if one of these points be moveable, find the equation of the locus of the vertex of the catenary formed by the string ; and show that the area cut off from this locus by a horizontal line through the fixed point iB 4a 2 (7r a -4). [Math. Tripos, 1867.] 326 INEXTENSIBLE STRINGS. [CHAP. X. 449. Stability of equilibrium. Some problems on the equilibrium of heavy strings may be conveniently solved by using the principle that the depth of the centre of gravity below some fixed straight line is a maximum or minimum, Art. 218. If the curve of the string be varied from its form as a catenary, the use of this principle will require the calculus of variations. But if we restrict the arbitrary displacements to be such that the string retains its form as a catenary, though the parameter c may be varied, the problem may be solved by the ordinary processes of the differential calculus. This method presents some advantages when we desire to know whether the equilibrium is stable or not. We know, by Art. 218, that the equilibrium will be stable or unstable according as the depth of the centre of gravity below some fixed horizontal plane is a true maximum or minimum. Ex. 1. A string of length 21 hangs over two smooth pegs which are in the same horizontal plane and at a distance 2a apart. The two ends of the string are free, and its central portion hangs in a catenary. Show that equilibrium is impossible unless I be at least equal to ae ; and that, if I > ae, the catenary in the position of stable a equilibrium for symmetrical displacements will be defined by that root of ce° = l which is greater than a. [Math. Tripos, 1878.] Let 2s be the length of the string between the pegs. Taking the horizontal line joining the pegs for the axis of x, we easily find (Art. 399) that the depth y of the centre of gravity of the catenary and the two parts hanging over the pegs is given by 2ly=sy-ca+(l-sf. Substituting for y and s their values in terms of c, we find I \ p 2 (c - a) - (c + a) '*-(-;)' 21- c where p stands for e c . It is easy to see that the second factor on the right hand side is negative for all positive values of c. Equating dyjdc to zero, we find that the possible positions of equilibrium are given by l=cp. To find the least value of I given by this equation we put dljde=Q ; this gives c=a, so that I must be equal to or greater than ae. For any value of I greater than ae there are two possible values of c, one greater and the other less than a. To determine which of these two catenaries is stable, we examine the sign of the second differential coefficient, Art. 220. We easily find, r. , o, <^3 i \ p a (c - a) - (c + a) when l=cp, 21 -j£ = (c - a) r —± ' v — '- . In order that the equilibrium may be stable, this expression must, be negative. This requires that e should be greater than a. Ex. 2. A heavy string of given length has one extremity attached to a fixed point A, and hangs over a small smooth peg B on the same level with A, the other extremity of the string being free. Show that, if the length of the string exceed a certain value, there are two positions of equilibrium, and that the one in which the catenary has the greater parameter is stable. 450. Heterogeneous chain. A heavy heterogeneous chain is suspended from two given points A and B. Find the equation to the catenary. ART. 451.] ^HE CATENARY. 327 This problem may be solved in a manner similar to that used in Art. 443 for a homogeneous chain. Since the equations (1) and (2) of that article are obtained by simple resolutions, they will be true with some slight modifications when the string is not uniform. In our case the weight of the string measured from the lowest point is jwds between the limits s = 0, s = s, Art. 442. We have therefore by the same resolutions Tcoaf = T (1), Tsia^=Jwds (2). Dividing one of these by the other as before, we find Jwds = T tan yJr = T ^ (3), d?y _ 'da? ' dot? ds j 1 [dy\ m d^y dec ° da? ... Hiy When the form of the curve is given by its Cartesian equation, the equation (4) determines by a simple differentiation the law of density of the chain. If the curve is given by the equation T P = f( y l r )> th e equation (3) gives by differentiation w = \j . Conversely, when the law of density is known, say w =/(«), the equation (3) gives a relation between s and dy/dx which we may write in the form dy/dx =/i (s). We easily deduce from this *=/{i +(/i WW"**. y =/{i +(/x wrViW*. whence x and y can be expressed in terms of an auxiliary variable which has a geometrical meaning. 451. Cyoloidal chain. Ex. A heterogeneous chain hangs in the form of a cycloid under the action of gravity : find the law of density. In a cycloid we have p=4a cos <]/ and s=4asin \p, where a is the radius of the rolling circle. Substituting, we find 2o = r -"sec s if'= *-= ia (16o 2 -s 2 )*" It appears from this result that all the lower part of the chain is of nearly uniform density ; thus the density at a point whose distance from the vertex measured along the arc is equal to the radius of the rolling circle is about nine tenths of the density at the vertex. The density increases rapidly higher up the chain and is infinite at the cusp. If then the chain when suspended from two points in the same horizontal line is not very curved, the chain may be regarded as nearly uniform. 328 INEXTENSIBLE STRINGS. [CHAP. X. The chief interest connected with this chain is that, when slightly disturbed from its position of equilibrium, it makes small oscillations whose periods and amplitudes can be investigated. 452. Parabolic chain. Ex. 1. A heavy chain A OB is suspended from another chain DGE by vertical strings, which are so numerous that every element of A OB is attached to the corresponding element of DCE. If the weights of DGE and of the vertical strings are inconsider- able compared with that of A OB, find the form of the chain DCE that the chain AOB may be horizontal in the position of equilibrium. The tensions at 0, M of the chain AOB being equal and horizontal, the weight of the length OM is supported by the tensions at C and P of the chain DCE. Thus DCE may be regarded as a heterogeneous heavy chain, such that the weight of any length PC is wx. Eesolving horizontally and vertically for this portion of the chain, we have Tcos^=T , Taia\l/=wx. Dividing one of these by the other, wx = T tan \p = T^dyjdx .: bwa?=T (y-c). The form of the chain DCE is therefore a parabola. One point of interest connected with this result is that the chain AOB might be replaced by a uniform heavy bar to represent the roadway of a bridge. The tensions of the chains due to the weight of the bridge would not then tend to break or bend the roadway. It is only necessary that the roadway should be strong enough to bear without bending the additional weights due to carriages. But this would not be true if the light chain DGE were not in the form of a parabola. The results are more complicated if the weight of the chain DCE is taken into account, and if the chains of support, instead of being vertical, are arranged in some other way. Ex. 2. Referring to the figure of Ex. 1, we notice that, since the tensions at G and P support the weight of the roadway OM, the tangents at G and P must intersect in a point vertically over the centre of gravity of OM. Thence deduce that the curve CP is a parabola. This problem was first discussed by Nicolas Fuss, Nova Acta Petropolitame, Tom. 12, 1794. It was proposed to erect a bridge across the Neva suspended by vertical chains from four chains stretched across the river. He decided that the chains of his day could not support the necessary tension without breaking. Ex. 3. If the weight of any element ds of the string DGPE is represented by w (ds+ndx), show that the form of DGPE is given by x= I jj= 2 * , where z is the tangent of the inclination of the tangent to the horizon, and e is a constant. [Puss.] Ex. 4. Prove that the form of the curve of the chain of a suspension bridge when the weight of the rods is taken into account, but the weight of the rest of the ART. 453.] ME CATENARY. 329 bridge neglected, is the orthogonal projection of a catenary, the rods being supposed vertical and equidistant. [Math. Tripos, 1880.] l^j 453. The Catenary of equal strength. Ex. 1. A heavy chain, suspended from two fixed points, is such that the area of its section is proportional to the tension. Find the form of the chain. If wds be the weight of an element ds, the conditions of the question require that T=cw, where c is some constant. The equations (1) and (2) of Art. 450 now become T cos f= T , T &xa.\p=-\Tds. Substituting in the second equation the value of T given by the first, we have ctan^=J sec \j/ds. Differentiating, we find p cos ip=c. Substituting for p and \p their Cartesian values, MS'r'S4 ■■■- *2-:«- If the origin be taken at the lowest point, the constant A is zero. We then find 2/=clogsec -. Tracing this curve, we see that the ordinate y increases from zero as x increases from zero positively or negatively, and that there are two vertical asymptotes given by x = ±%irc. When x lies between Jn-c and fare, the ordinate is imaginary; when x lies between f7i-c and fire, the curve is the same as that between x= ±J«. For greater values of x, the ordinate is again imaginary and so on. The curve therefore consists of an infinite number of branches all equal and similar to that between x= ±Jjrc. This is therefore the only part of the curve which it is necessary to consider. This curve was called the catenary of equal strength by Davies Gilbert, who invented it on the occasion of the erection of the suspension bridge across the Menai Straits. See Phil. Trans. 1826, part iii., page 202. In the first volume of Liouville's Journal, 1836, there is a note by G. Coriolis on the " chainette " of equal resistance. Coriolis does not appear to have been aware that this form of chain had already been discussed several years before. Ex. 2. Prove the following properties of the catenary of uniform strength. (1) x = vp, (2) s=c log tan i(7r + 2f), (3) the altitude of the centre of curvature at any point above that point is constant, (4) the tension at any point is proportional to the radius of curvature at that point, (5) since the ordinate of the bridge must be finite, the values of x are restricted to lie between ±Jttc. The span therefore cannot be so great as ire. Ex. 3. The distance between the points of support of a catenary of uniform strength is a, and the length of the chain is I. Show that the parameter c must be found from tanh -r = tan -j- . Show also that this equation gives a positive value 4c 4c of c greater than a/ir. 330 INEXTENSIBLE STRINGS. [CHAP. X. Ex. 4. Show that the horizontal projection of the span is in every case less than ir times the greatest length of uniform chain of the same material that can he hung by one end. Assume the strength of any part of the chain to be proportional to the mass per unit of length. [Sir W. Thomson, Math. Tripos, 1874.] If L be the length of uniform chain spoken of, the tension at the point of support is its weight, i.e. wL. Again, the tension at any point of the heterogeneous chain is cw, hence c must be less than L. Hence by the fifth property mentioned in Ex. (2) the span must be less than irL. String under any Forces. 454. The intrinsic equations. To form the general in- trinsic equations of equilibrium of a string under the action of any forces. Let A be any fixed point of reference on the string, AP = s, AQ = s + ds. Let T be the tension at P; then, since T is a function of s, T + dT is the tension at Q*. Let the impressed forces on the element PQ be resolved along the tangent, radius of curvature, and binormal at P. Thus Fds is the force on ds resolved along the tangent in the direction in which s is measured; Gds is the force on ds resolved along the radius of curvature p in the direction in which p is measured, i.e. inwards ; Rds is the force on ds resolved perpendicular to the plane of the curve at P, and estimated positive in either direction of the binormal. These three directions are called the principal directions or principal axes of the curve at P. Let dyjr be the angle between the tangents at P and Q. Hence also the angle PGQ = dyjr. The element ds is in equilibrium under the forces T, T + dT acting along the tangents at P, Q and the forces Fds, Gds, Hds. Kesolving along the tangent at P, (T+ dT) cos cty - T+ Fds = 0, which reduces to dT+Fds = (1). * It should be noticed that, if s were measured from B towards A, so that BQ=s then T would be the tension at Q, T + dT that at P. ART. 455.] STRIN# UNDER ANY FORCES. 331 Resolving along the radius of curvature at P, we have (T + dT)smdy}r+Gds = 0, dv :.T- + Gds = (2). P We have now to resolve perpendicular to the osculating plane at P of the curve. Since two consecutive tangents to a curve lie in the- osculating plane, the tensions have no component perpendicular to this plane. We have therefore Hds = (3). The three equations (1), (2), (3) are the general intrinsic equations of equilibrium. The density of the string is supposed to be included in the expressions Fds, Gds, Hds for the forces on the element. The equations of equilibrium therefore apply, whether the string is uniform, or whether its density varies from point to point. From these equations we infer that the tensions T and T + dT, acting at the extremities of any element, are equivalent to two ds other forces, viz. dT and T — , acting respectively along the tangent to, and the radius of curvature of, the curve at either extremity of the element. In problems on strings it is often convenient to replace the tensions by these two forces. The advantage of this change is that the direction cosines of the tangent and of the radius of curvature are known by the differential calculus. When therefore we form the equations of statics, we can more easily resolve these two forces and the given impressed forces in any directions we may find convenient. Ex. Show that the form of the string is such that at every point the resultant of the applied forces lies in the osculating plane, and makes with the principal normal to the string an angle tan -1 — jf— ■ 455. The Cartesian equations. To form the general Cartesian equations of equilibrium of a string*. * The equations of equilibrium of a string under the action of any forces in two dimensions were given in a Cartesian form by Nicolas Fuss, Nova Acta Petropolitance, 1796. He gives two solutions, one by moments, and another by considering the tension. In this second solution, after resolving parallel to the axes, he deduces algebraically equations equivalent to those obtained by resolving along the tangent and normal. He goes on to apply his equations to the chainette and other similar problems. 332 INEXTENSIBLE STRINGS. [CHAP. X. Let ds be the length of any element PQ of the string. Let the forces on this element when resolved parallel to the positive directions of the axes be Xds, Yds, Zds. The element is in equilibrium under the action of the tensions at P and Q and these three impressed forces. Let us resolve all these parallel to the axis of x. The resolved tension at P is T -5- , and pulls the element PQ towards the left hand. At Q, s has become s + ds, the hori- zontal tension at Q is therefore ('s)+£( r s)* and this pulls the element PQ to- wards the right hand side. Taking both these and the force Xds, we have is( T Ws) ds+Xds=0 - Treating the other components in the same way, we find dx\ ds) + X = hi*®*'- 456. Ex. 1. Show that the polar equations of equilibrium of a string in one plane under forces Pds, Qds, acting along and perpendicular to the radius vector, are |(I*cos0)-?sin^ + P = O | d T (' T-(Tsin0) + - sin0cos0 + Q = O where cos = drjds and sin = rd Thence deduce the equations of equilibrium of a string in space of three dimen- sions, referred to cylindrical coordinates. Ex. 2. A string is in equilibrium in the form of a helix, and the tension is constant throughout the string. Show that the force on any element tends directly from the axis of the helix. Ex. 3. The extremities of a string of given length are attached to two given points, and each element ds of the string is acted on by a repulsive force tending ART. 457.] CONSTRAINED STRINGS. 333 directly from the axis of 2 and equal to 2/irds. If (r6z) be the cylindrical coordinates of any point, prove that T=A - /*)- 2 , d£_B dz~ r 2 ' Otf- •('-!")' Show how the five arbitrary constants are determined. Explain how the helix is, in certain cases, the solution. Ex. 4. A heavy chain is suspended from two points, and hangs partly immersed in a fluid. Show that the curvatures of the portions just inside and just outside the surface of the fluid are as D - D' to D, where D and D' are the densities of the chain and fluid. [St John's Coll.] The weights of the elements just above and just below the surface of the fluid are proportional to Dds and (D - D') ds. If T be the tension, the resolved parts of these weights along the normal must be Tdsjp and Tdsjp'. Hence D/(D - D') =p'/p. Ex. 5. A heavy string is suspended from two fixed points A and B, and the density is such that the form of the string is an equiangular spiral. Show that the density at any point P is inversely proportional to r cos 2 ^, where r is the distance of P from the pole, and \p is the angle which the tangent at P makes with the horizon. [Trin. Coll., 1881.] Ex. 6. A heavy string, which is not uniform, is suspended from two fixed points. Prove that the catenary formed of a given uniform string which touches at any point the curve in which the string hangs and has the same tension at that point will be of invariable dimensions. Constrained Strings. 457. A string rests on a curve of any form in one plane, and is acted on by forces at its extremities. It is required to find the conditions of equilibrium and the tension at any point. There are four cases of this proposition which are of con- siderable importance ; we shall consider these in order. Let us first suppose that the weight of the string is so slight that it may be neglected compared with the forces applied at the two extremities of the string. Let us also suppose that the curve is perfectly smooth. The forces on an element ds are merely the tensions at its ends and the reaction or pressure of the curve. Let Rds be this pressure, then R is the pressure per unit of length of the string. For the sake of brevity this is usually expressed by saying that B is the pressure at the element. It is usual to estimate the pressure of the curve on the string as positive when it acts in the direction opposite to that in which the radius of curvature is measured. 334 INEXTENSIBLE STRINGS. [CHAP. X. Kesolving along the tangent and normal to the string, we have by Art. 454, dT = 0, T - - Rds = 0. P We infer from these equations that, when a light string rests on a smooth curve, the tension is constant, and the pressure at any point varies as the curvature. 458. This theorem has a wider range than would perhaps appear at first sight. Since the curve may be of any form, the result includes the case of a string in equilibrium under any forces which are at every point normal to the curve. Supposing the normal forces given, the form of the curve can be found from the result just proved, viz. that at every point the curvature is proportional to the normal force. As an example we may consider Bernoulli's problem ; to find the form of a rectangular sail, two opposite sides of which are fixed so as to be parallel to each other and perpendicular to the direction of the wind. The weight of the sail is neglected compared with the pressure produced by the wind. Let us enquire what iB the curve formed by a plane section of the sail drawn perpendicular to the fixed sides. Two answers may be given to this question according as the wind after acting on the Bail immediately finds an issue, or remains to press on the sail like a gas in equilibrium. On the former hypothesis we assume as the law of resistance, that the pressure of the wind on any element of the sail acts along the normal to the element and is proportional to the square of the resolved velocity of the wind. We have therefore B=w cos a \j/, where ^ is the angle the normal to the section of the sail makes with the direction of the wind, and w is a constant. This gives c/p=cos 2 ^. By Art. 444 we infer that the curve is a catenary, whose axis is in the direction of the wind, and whose directrix is vertical. If the air presses on the sail like a gas in equilibrium, the pressure on each side of the sail is equal in all directions by the laws of hydrostatics, but the pressure is greater on one side than on the other. We have therefore E equal to this constant difference, hence also p is constant, and the required curve is a circle. Ex. 1. A " square sail " of a ship is fastened to the mast by two yard-arms, and is such that when filled with wind every section by a horizontal plane is a straight line parallel to the yards. Show that, assuming the ordinary law of resistance, it will have the greatest effect in propelling the ship when 3 sin (a - 20) - sin a = 0, where a is the angle between the direction from which the wind comes and the ship's keel, and is the angle between the yard and the ship's keel. [Caius Coll.] Ex. 2. A light string has one end fixed at the vertex of a smooth cycloid; prove that as the string, while taut, is wound on the curve, the line of action of the resultant pressure on the cycloid envelopes another cycloid of double parameter. [Coll. Ex. 1890.] The resultant pressure of the curve on an arc of the string balances the tensions at the extremities of the arc. It therefore passes through the intersection of the tangents at those extremities and bisects the angle between them. AttT. 460.] HEAV» SMOOTH STRING. 335 2/ cv 2k ^3 IV M \S K N G 459. Heavy smooth string. Let us next suppose that the weight of the string cannot be neglected. Let wds be the weight of the element ds. Let i/r be the angle the tangent PK at P makes with the horizontal. The element PQ is in equilibrium under the action of wds along the ordinate PN, Eds along the normal PG, and the tensions at P and Q. Resolving along the tan- gent and normal at P, we have dT - wds sm^ = 0\ (1), T— - wds cos iJr-Rds = 0|. ,,...(2). Since sin i/r = dy/ds, the first equation gives by integration T=wy+G (3). Hence, if 2\, T 2 be the tensions at two points whose ordinates are y u y iy we have Ti-T x = w{y,-y^ This important result may be stated thus, If a heavy string rest on a smooth curve, the difference of the tensions at any two points is equal to the weight of a string whose length is the vertical distance between the points. 460. It may be remarked that this result has been obtained solely by resolving along the tangent to the string, and is alto- gether independent of the truth of the second equation. If then the whole length of the string does not lie on the curve, but if part of it be free and stretch across to and over some other curve, the theorem is still true. Thus if the string ABGD stretch round 336 INEXTENSIBLE STRINGS. [CHAP. X. the smooth curves L, M, N, as indicated in the figure, the tension at any point B or G exceeds that at A by the weight of a string whose length is the vertical distance of B or C above A. Since the tensions at A and B are zero, it follows that the free extremities of a heavy chain are in the same horizontal line. In the same way the tension is a maximum at the highest point. Also no point of the string, such as G or G', can be beneath the horizontal line joining the free extremities. To determine the pressure at any point P (see fig. of Art. 459) we write the equation (2) in the form Rp = T — wp cos yjr, where the pressure R of the curve on the string, when positive, acts outwards, i.e. in the direction opposite to that in which the radius of curvature p is measured, Art. 457. If 2\ be the tension at any fixed point A, and z the altitude of any point P above A, we have by (3) T= 2^ + wz. It therefore follows that Rp = T 1 + w(z — p cos l|r). If we measure a length PS = p along the normal at P out- wards, the point S may be called the anti-centre. It is clear that z — p cos i|r is the altitude of S above A. Hence, if a heavy string rest on a smooth curve, the value of Up at any point P exceeds the tension at A by the weight of a string whose length is the altitude of the anti-centre of P above A. If the extremity A be free, as in the figure of this article, then Rp at any point B is equal to w multiplied by the altitude of the anti-centre of B above A. If part of the string is free, as at G and C, the pressure R is zero. Hence the anti-centres of curva- ture all lie in the straight line joining the free extremities A and D. This is the common directrix of all the catenaries. In these equations Rds is the pressure outwards of the curve on the string. It is clear that, if R were negative and the string on the convex side, the string would leave the curve and equilibrium could not exist. At any such point as B, the anti-centre is' above B and R is clearly positive. But at such a point as E the anti- centre is below E, and if it were also below the straight line AD, the pressure at E would be negative. If the string rest on the concave side of the curve, these conditions are reversed. In ART. 461.] HEAVY STRffcG ON A SMOOTH CURVE. 337 general, it is necessary for equilibrium that Rp should be positive or negative according as the string is on the convex or concave side of the curve. Summing up the results arrived at in this article, we see that a horizontal straight line can be drawn such that the tension at each point P of the string is wy, where y is the altitude of P above the straight line. This straight line may be called the statical directrix of the string. No part of the string can be below the statical directrix, and the free ends, if there are any, must lie on it. If R be the outward pressure of the curve on the string, Rp is equal to wy', where y' is the altitude of the anti-centre of P above the directrix. It is therefore necessary that at every point of the string the anti-centre should be above or below the directrix according as the string is on the convex or concave side of the curve. Ex. 1. Show that the locus of the anti-centre of a circle is another circle. Ex. 2. Show that the coordinates of the anti-centre at any point P of an ellipse referred to its axes are given by ax = 2a 2 cos - c 2 cos 3 , where c 2 = a 2 - 6 2 , and

    'la, the string cannot rest in contact with the circle. Secondly, let us suppose that a portion of the string hangs freely in the form of a catenary. Let P' be one of the points of contact of the catenary with the circle. Let P be any point on the catenary, drawn in the figure merely to show the triangle PLN, Art. 444. Let the angle P'OD = \j/, so that t// is the inclination of the tangent at P' to the horizon. Let x, y be the coordinates of P', s= GP'. By examining the triangle PLN, we see that y = caeef, s = c tan \j/. Since a; = asin^, we have by (5) of Art. 443 sec f + tan

    ), we find that sun//=(e 8 -l)/(e 3 +l). As dl/df changes sign from - to + as sin \p increases, we see that I is a minimum. Effecting the numerical calculations, we have ^="86, and l-fira=(e-ij/)a, which reduces to (1*85) a. For any given value of I, greater than this minimum, there are two positions of equilibrium. In one a portion of the string hangs freely in the form of a catenary ; in the other the string fits closely to the cylinder or hangs free according as the given value of I - f ira is greater or less than 2a. Ex. 2. A uniform chain, having its ends fastened together, is hung round the circumference of a vertical circle. If a be the radius of the circle, 2ay the arc which the string touches, and I the whole length, prove {l-2ay) {log (-cos 7)-log(l + sin7)}=2asin 2 7SeC7. [May Exam.] Ex. 3. A uniform inextensible string of given length hangs freely from two fixed points. It is then enclosed in a fine fixed tube which touches no part of the string, and is cut through at a point where the tangent makes an angle y with the horizon. Prove that at a point where the tangent makes an angle f with the horizon the ratio of the pressure on the tube to the weight of the string per unit of length becomes cos a

    (» + 2)J(oos ^) TC+1 e'^di/,. Putting V v+1 for the last integral, we have {jf+n?) V n -n(n-l) 7 n _ 2 = (cos ^) n ~ 1 e~^ (n sin^ - p. cos f). This formula of reduction may be obtained by integrating by parts, or by differen- tiating the right-hand side of the equation. There is a similar formula for (sin tj/) w e ~ . See Gregory's Examples on the Differential and Integral Calculus, 1846. ^~ 469. Ex. 1. A heavy string occupies a quadrant of the upper half of a rough vertical circle in a state bordering on motion. Prove that the radius through the lower extremity makes an angle a with the vertical given by tan (o - 2e) = e - ' 1 " 1 , where /x=tane. Ex. 2. A heavy string, resting on a rough vertical circle with one extremity at the highest point, is on the point of motion. If the length of the string is equal to a quadrant, prove that \ir tan e=log tan 2e, where tan e is the coefficient of friction. [Coll. Ex. 1881.] Ex. 3. A single moveable pulley, of weight W, is just supported by a power P, which is applied at one end of a cord which goes under the pulley and is then fastened to a fixed point ; show that, if

    where tan e is the tan e + (1 - 3 cos 2 e) tan n coefficient of friction. Ex. 6. A heavy string rests on a rough cycloid with its base horizontal and plane vertical. The normals at the extremities of the string make with the vertical angles each equal to a, which is also the angle of friction between string and cycloid. If, when the cycloid is tilted about one end till the base makes an angle o with the horizontal, the string is on the point of motion, show that 2a tan a cos 2 a e 3 cos 2 o - 2 ' [It is assumed that none of the string hangs freely.] [Coll. Ex. 1883.] Ex. 7. A heavy uniform flexible string rests on a smooth complete cycloid, the axis of which is vertical and vertex upwards, the whole length of the string exactly coinciding with the whole arc of the cycloid ; prove that the pressure at any point of the cycloid varies inversely as the curvature. [Math. Tripos, 1865.] Ex. 8. A heavy string AB is laid on a rough convex curve in a vertical plane, and the friction at every point acts in the same direction along the curve. Show that it will rest if the inclination of the chord AB to the horizon be less than tan -1 ,u, where /* is the coefficient of friction. [June Ex. 1878.] 470. The following proposition will be found to include a number of problems which lead to known integrals. Let the form be known in which a heterogeneous unconstrained string, supported at each end, rests in equilibrium in one plane under the action of any forces. Let this known curve be y=f(x). Let us now suppose this string to be placed in the same position on a rough curve fixed in space whose equation is also y =f (x). If the extremities of the string be acted on by forces such that the string is on the point of slipping, then {T+Gp)e~>" l '=C, Rpe'^^C (1), where C is constant throughout the length of the string. Here, as in Art. 454, Gds is the resolved normal force inwards on the element rfs. The standard case is the same as that taken in Art. 467. The string is just slipping in that direction along the curve in which the \j/ of any point of the string increases. Also the pressure R of the curve on the string, when positive, acts outwards. If either of these assumptions is reversed, the sign of /* must be changed. In order that the string may not leave the curve, the sign of G should be such that R acts from the curve towards that side on which the string lies. To prove these results, we refer to equations (1) and (2) Art. 454. Introducing the pressure B into these equations, we have Tils dT+Fds-nRds = 0, —+Gds~ Bds=0 (2). P Eliminating R, we find as in Art. 467 Te-^^-KF-^pe-^dip+C (3). 346 INEXTENS1BLE STRINGS. [CHAP. X. When the string is hanging freely, E = ; by eliminating T between the equa- tions (2) we find that Fp = — (Gp) is *™e along the curve. When the string is constrained to lie on a curve which possesses this property, we can substitute this value of Fp in the equation (3). We then find Te~^= -e'^Gp+C. The first result to be proved follows immediately, the second is obtained by substituting this value of T in the second of equations (2). 471. Ex. 1. A uniform heavy string AB is placed on the upper side of a rough curve whose form is a catenary with its directrix horizontal. If the lower extremity is at the vertex, find the least force F which, acting at the upper extremity, will just move the string. At the upper end of the string we have T=F, G= - g cos = 0, where $ is the radial angle, i.e. the angle OP A. Since cos tf> = dr/ds, dT this reduces to -^- +F = (1). We might obtain a second equation by resolving the same forces along the normal at P, but the result is more easily found by taking the moment of the forces which act on the finite portion of string AP. The portion A P is in equilibrium under the action of the tensions T a , T and the central force tending from on each element. Taking moments about 0, these latter disappear; we have therefore Tp = A (2), where p is the perpendicular from on the tangent at P, and A is the moment about of the tension T . These two equations are sufficient for our purpose. Two cases have now to be considered. First. Suppose the form of the string to be given, and let the force be required. By known theorems in the differential calculus 350 INEXTENSIBLE STRINGS. [CHAP. X. we can express the equation to the curve in the form p = tJt (r). The equations (1) and (2) then give T = J- F = A ^ (3). The constant A remains indeterminate, for it is evident that the equilibrium would not be affected if the magnitude of the central force were increased in any given ratio. The tension at any point of the string and the pressures on the fixed points of suspension would be increased in the same ratio. 475. Secondly. Suppose that the force is given, and that the form of the curve is required. Eliminating T between (1) and (2), we find j = B-JFdr (4). This differential equation has now to be solved. Put u = 1/r and JFdr =/(w); we find by a theorem in the differential calculus ^ 2 +(sy}=cB-» 2 ( s > \dd. . Separating the variables, we have r + Adu ]{(B-f u y-Aw} i = 6 + G (6) - When this integration has been effected the polar equation to the curve has been found. There are three undetermined constants, viz. A, B, G, in this equation. To discover their values we have given the polar coordinates (u 6 ), (uj0i) of the points of suspension. After inte- grating (6) we substitute in turn for (ud) these two terminal values, and thus obtain two equations connecting the three con- stants. . We have also given the length of the string. To use this datum we must find the length of the arc. We easily find (dsy = (dry + (rdey = - t {(duy + (udd)% Substituting from (5), we have f (B — fu) du S= )v?{{B-fu)*-A*u^ <*)• ART. 476.] •CENTRAL FORCES. 351 Taking this between the given limits of u, and equating the result to the given length of the string, we have a third equation to find the three constants. - The equation (6) agrees with that given by John Bernoulli, Opera Omnia, Tamus Quartus, p. 238. He applies the • equation to the case in which the force varies inversely as the nth power of the distance, and briefly discusses the curves when n=0 andrc=2. 476. Ex. 1. A string is in equilibrium under the action of a central force. If F be the force at any point per unit of length, prove that the tension at that point =Fx, where x i g * ne semi-chord of curvature through the centre of T force. Show also that F=:A -=- , where A is a constant. p*p Ex. 2. A uniform string is in equilibrium in the form of an arc of a circle under the influence of a centre of force situated at any point 0. Find the law of force. Let C be the centre, OG=c, CP=a. Then 2op=r 2 + a 2 -c s , ... F= -A% -=4aA- ' drp (f + a?- 1. Show that the form of the string between A and B is r n-2 =&»- 2 cos (n - 2) 0. If m=2 the curve is an equiangular spiral. Ex. 5. A closed string surrounds a centre of force =im n , where n>l and <2. Show that, as the length of the string is indefinitely increased so that one apse becomes infinitely distant from the centre of force, the equilibrium form of the string tends to become r" -2 = 6 n_a cos (n - 2) 8. If n = f the form of the curve is a parabola. 352 INEXTENSIBLE STRINGS. [CHAP. X. Ex. 6. A uniform string of length 21 is attached to two fixed points A, B at equal distances from a centre of repulsive force =/*u 2 . If OA = OB = b and the angle A OB = 2/3, prove that the equation to the string is M _ 1 cos (8 sin o) r ~~ cos o ' where the real and imaginary values of M and a are determined from the equations M , cos (S sin a) . , 6 . ,„ . . — =1 + 5C ' sma= ±-rSin(Ssina). 6 cos a I The equations (1) and (2) of Art. 474 become here dT=/j.du, Tp = A. f Ada Proceeding as explained in Art. 475, we find ± I , = 9 + C. J {(B + tmf-AWy This integral is one of the standards in the integral calculus, and assumes different forms according as A* — i& is positive, negative or zero. Taking the first assumption, we have after a slight reduction A ^u= li ±Acos(l-^ (8 + C). This formula really includes all cases, for when A* - /t 2 is negative we may write for the sine of the imaginary angle on the right-hand side its exponential value. Proceeding to find the arc in the manner already explained, we easily arrive at Bs = ± { (Br + n¥ - 4 2 }* + D, where the radical must have opposite signs on opposite sides of an apse. The conditions of the question require that the string should be symmetrical about the straight line determined by 8 = 0. We have therefore (7=0 and D = 0. Putting A=fi8eca, the equation to the curve reduces to /itan 2 a 1_ 1 cos (8 sin a) ... B r~ coso We also have B 2 J 2 =(B6 + / n) 2 -/i 2 sec 2 o (2). Eliminating B between these equations, we find I sin o = ± b sin (/3 sin o). We now put M for the coefficient of 1/r in (1) and include the double sign in the value of a. Since r=b when 8= ±/3 the three results given above have been obtained. Ex. 7. A string is in equilibrium in the form of a closed curve about a centre of repulsive force =/i« 2 . Show that the form of the curve is a circle. Eeferring to the last example, we notice that, since r is unaltered when 8 is increased by 2ir, r must be a trigonometrical function of 8. Hence sin o=l or 0. Putting M ooaa=M', the first makes M'lr=aos8, which is not a closed curve, the Becond gives M=r, which is a circle. Ex. 8. If the curve be a parabola, and the centre of force at the focus, and if the equilibrium be maintained by fixing two points of the string, find the law of force, and prove that the tension at any point P is 2/r, where r=SP and / is the force at P per unit of length. [St John's Coll. 1883.] Ex. 9. An infinite string passes through two small smooth rings, and is acted on by a force tending from a given fixed point and varying inversely as the cube of the distance from that point. Show that the part of the string between the rings assumes the form of an arc of a circle. [Coll. Ex. 1884.] ART. 477.] • CENTRAL FORCES. 353 Ex. 10. If a string, the particles of which repel each other with a force varying as the distance, be in equilibrium when fastened to two fixed points, prove that the tension at any point varies as the square root of the radius of curvature. [Math. Tripos, I860.] Ex. 11. A string is placed on a smooth plane curve under the action of a central force F, tending to a point in the same plane ; prove that, if the curve be such that a particle could freely describe it under the action of that force, the pressure of the string on the curve referred to a unit of length will be equal to — =-^ + - , l p where is the angle which the radius vector from the centre of force makes with the tangent, p is the radius of curvature, and c is an arbitrary constant. If the curve be an equiangular spiral with the centre of force in the pole, and if one end of the string rest freely on the spiral at a distance a from the pole, then the pressure is equal to ^JELr ( _ + _ J . [Math. Tripos, I860.] Ex. 12. A free uniform string, in equilibrium under the action of a repulsive central force F, has a form such that a particle could freely describe it under a central force F' tending to the same centre. Show that F=kpF', where k ib a constant. If v be the velocity of the particle and T the tension of the string, show also that T= kpv 2 . See Art. 476, Ex. 1. Ex. 13. It is known that a particle can describe a rectangular hyperbola about a repulsive central force which varies as the distance and tends from the centre of the curve. Thence show that a string can be in equilibrium in the form of a rectangular hyperbola under an attractive central force which is constant in magnitude and tends to the centre of the curve. Show also that the tension varies as the distance from the centre. For a comparison of the free equilibrium of a uniform string with the free motion of a particle under the action of a central force, see a paper by Prof. Townsend in the Quarterly Journal of Mathematics, vol. xin., 1873. 477. When there are two centres of force the equations of equilibrium are best found by resolving along the tangent and normal. Let r, r' be the distances of any point P of the string from the centres of force ; F, F' the central forces, which are to be regarded as functions of r, r' respectively. Let p, p' be the perpendiculars from the centres of force on the tangent at P. We then have dT+Fdr + F'dr' = \ (1), T- F l- F .^=0 | (2). p r r' ) The first equation gives T=B- jFdr-jF'dr 1 (3). We may suppose the lower limits of these integrals to correspond to any given point P on the string. If this be done B will be the tension at P . Substituting the value of T thus obtained from (1) in (2) and remembering that p=rdr\dp, we have ±(pSFdr) + ±(p'!F'dr') = B (4); R. S. 23 354 INEXTENSIBLE STRINGS. [CHAP. X. on the other hand, if we find T from (2) and substitute in (1), we find after reduction H^h^m-' <* Thus of the four elements, viz. (1) the force F, (2) the force F', (3) the tension T, (4) the equation to the curve, if any two are given, sufficient equations have now been found to discover the other two. Ex. A string can be in equilibrium in the form of a given curve under the action of each of two different centres of force. Show that it is in equilibrium under the joint action of both centres of force, and that the tension at any point is equal to the sum of the tensions due to the forces acting separately. String on a surface. 478. A string rests on a smooth surface under the action of any forces. To find the position of equilibrium. Let the equation to the surface be f(x, y, z) = 0. Let Rds be the outward pressure of the surface on the string. Let (I, m, n) be the direction cosines of the inward direction of the normal. By known theorems in solid geometry, I, m, n are proportional to the partial differential coefficients of f (x, y, z) with regard to x, y, z respectively. If the equations are required to be in Cartesian coordinates, we deduce them at once from those given in Art. 455 by including R among the impressed forces. We thus have £(r*) + x-*-o> >. We have here one more unknown quantity, viz. R, than we had in Art. 455, but we have also one more equation, viz. the given equation to the surface. 479. Let us next find the intrinsic equations to the string. Let PQ be any element of the string, PA a tangent at P. Let APB be a tangent plane to the surface, PB being at right angles to PA. Let PN be the normal to the surface. Let PC ART. 479.J STRING»ON A SMOOTH SURFACE. 355 be the radius of curvature of the string, then PC lies in the plane BPN. Let x De tne angle GPN, then % is also the angle the osculating plane OP A of the string makes with the normal PN to the surface. The element PQ is in equilibrium under the action of (1) the forces Xds, Yds, Zds acting parallel to the axes of coordinates, which are not drawn in the figure, (2) the reaction Rds along NP, (3) the tensions at P and Q, which have been proved in Art. 454 to be equivalent to dT along PQ and Tds/p along PG. Resolving these forces along the tangent PA, we have d T + Xds $ + Yds $ + Zds ^ = 0, ds ds ds :. T + f(Xdcc+ Ydij + Zdz) = A (1). The forces are said to be conservative, when their components X, Y, Z are respectively partial differential coefficients with regard to x, y, z, of some function W which may be called the work function, Art. 209. Assuming this to be the case, the integral in (1) is equal to the work of the forces. It follows from this equation that the tension of the string plus the work of the forces is the same at all points of the string. Taking the integral between limits for any two points P, P' of the string, we see that the difference of the tensions at two points P, P' is independent of the length or form of the string joining those points and is equal to the difference of the works at the points P', P taken in reverse order. We shall suppose that, while p is measured inwards along PC, the pressure R of the surface on the string is measured outwards along NP, Art. 457. We shall also suppose that (I, in, n) are the direction cosines of the normal PN measured inwards. 23—2 356 INEXTENSIBLE STRINGS. [CHAP. X. With this understanding we now resolve the forces along the normal PN to the surface ; we find Tds — cos y + Xds I + Yds m + Zds n — Rds = 0. P * By a theorem in solid geometry, if p be the radius of curva- ture of the section of the surface made by the plane NPA, i.e. by a plane containing the normal to the surface and the tangent to the string, then p' cos x = P- We therefore have -, + Xl+Ym-t-Zn = R (2). P It follows from this equation that the resultant pressure on the surface is equal to the normal pressure due to the tension plus the pressure due to the resolved part of the forces. The tension at any point P having been found by (1), the pressure on the surface follows by (2), provided we know the direction of the tangent PA to the string. This last is necessary in order to find the value of p'. Lastly, let us resolve the forces along the tangent PB to the surface. Let \, /j,, v be the direction cosines of PB. Since PB is at right angles to both PN and PA, these direction cosines may be found from the two equations *+/./.+*=<>, 4>g + 4:=o. We then have by the resolution T - smx + X\+Yfi + Zv = (3). Ex. 1. A uniform heavy string rests on a smooth surface, and at each point P a length PC is measured along the normal to the surface outwards, equal to the radius of curvature p' of a normal section of the surface drawn through the tangent to the string. The point U thus constructed is the anti-centre of P. If T be the tension at P, R the outward pressure of the surface on the string, x the angle the radius of curvature of the string makes with the normal to the surface, show that the equations (1), (2), (3) are equivalent to T=wz, Rp'=wz', z tanx=/>'sin 0, where z, z' are the altitudes of P and U above a certain horizontal plane, and is the angle the vertical makes with the plane containing the normal to the surface and the tangent to the string. See Art. 460. ART. 481.] STRING«6N A SMOOTH SURFACE. 357 Ex. 2. When the applied forces form a conservative system, the three equations of equilibrium may be expressed in terms of the work instead of the resolved forces. If dp, dq be elements of length measured from P along the normal PN and PB, show that the equations (1), (2), (3) become T+W=A, ^cos x + ^=.R, -sinx + ^=0. p dp p "• dq Ex. 3. Show, by eliminating R and dT/ds from the equations of Art. 478, that = 0, I + X, x\ I m Y, y', m n z, z't n where accents denote differential coefficients with regard to 8. Show also that this result is the analytical equivalent of equation (3) of this article. Substituting for T from (1) and joining the result to the equation of the surface, we have the differential equations of the curve in which the string rests. Ex. 4. An endless uniform string is lying along a central circular section of a smooth ellipsoidal surface, of mean semi-axis b. The force per unit length at any point of the string resolved perpendicular to the string in the tangent plane to the surface is F, and the tension at that point is T. Prove that T=b 2 F (b 2 -p 2 )~K where p is the perpendicular from the centre to the tangent plane to the surface at the point. [Trin. Coll. 1890.] 480. Case of No Forces. If any portion of the string is not acted on by external forces, we have for that portion X = 0, Y=0, Z = 0. The equation (1) then shows that the tension of the string is constant. The equation (2) shows that the pressure at any point is proportional to the curvature of the surface along the string. The equation (3) (assuming the string not to be a straight line) shows that x = 0, 1,e - a * ever y point the osculating plane of the curve contains the normal to the surface. Such a curve is called a geodesic in solid geometry. Conversely, if the string rest on the surface in the form of a geodesic under the action of some forces, we see by (3) that the forces must be such that at every point of the string their resolved part perpendicular to the osculating plane of the string must be zero. 481. A string on a surface of revolution. When the surface on which the string rests is one of revolution, we can replace the rather complicated equation (3) of Art. 479 by a much simpler one obtained by taking moments about the axis of figure. If also the resultant force on each element is either parallel to or 358 INEXTENSIBLE STRINGS. [CHAP. X. intersects the axis of figure, there is a further simplification. This includes the useful case in which the only force on the string is its^weight, and the axis of figure of the surface is vertical. Let the axis of figure be the axis of z, and let (r', &, z) be the cylindrical coordinates of any point on the string, so that in the figure r'^ON, s = PN, and 8' = the angle NOx. Then from the equation to the surface we have z =f(r'). Let the forces on the element ds be Pds, Qds, Zds when resolved respectively parallel to r', r'dO', and z. We shall now take moments about the axis of figure. The moment of R is clearly zero. To find the moment of T, we resolve it perpendicular to the axis and multiply the result by the arm r' . In this way we find that the moment is Tr' sin , where <£ is the angle the tangent to the string makes with the tangent to the generating curve of the surface, i.e. is the curvilinear angle OP A. The equation of moments is therefore d(Tr' sin _ f dji It- ( . The equations (1) become Tp=B', T+\Fdr=C (2). These are the equations of equilibrium of a string in one plane under the action of a central force, and the constants of integration are determined by the same conditions in each case. We may therefore transfer the results obtained in Art. 474 to the string on the cone. In transferring these results we notice that the point (r, 6) on the plane corresponds to (r'S'z) on the cone, where r 1 = r sin o, 8' sin a = 8, 2 = r cos a. By referring to equation (2) of Art. 479, we may show that the pressure R is , „ T sin0 i?cosa 1 cos 2 d> sin 2 , „ , , ,, given by R = -. = — ~ • -=— s — > since -. = + -. — — by Euler's theorem on p r 2 sin'a p oo )■ sec a curvature. Ex. 2. The two extremities of a string, whose length is 21, are attached to the same point A on the surface of a right cone. The equation to the projection of the string on a plane perpendicular to the axis is 7rr' = Zcos(0'sina), the point A being given by 8'= v. Show that the string will rest in equilibrium under the influence of a centre of force in the vertex varying inversely as the cube of the distance. Ex. 3. A heavy uniform string has its ends fastened to two points on the surface of a right circular cone whose axis is vertical and vertex upwards, the string lying on the surface of the cone. Prove that, if the cone be developed into a plane, the curve on which the string lay is given by p(a + br) = l, the origin being the vertex, p the perpendicular on the tangent, and a, b constants. [Coll. Ex. 1890.] 485. String on a rough surface. A string rests on a rough surface under the action of any forces, and every element borders on motion ; to find the conditions of equilibrium. The required conditions may be deduced from the equations for a smooth surface by introducing the limiting friction. The pressure of the surface on the element ds being Rds, the limiting friction will be fiRds. This friction acts in some direction PS lying in the tangent plane to the surface. See figure of Art. 479. 362 INEXTENSIBLE STRINGS. [CHAP. X. Let sjr be the angle SPA. Resolving along the principal axes at any point of the string exactly as in Art. 479, we have dT+Xdx+Ydy + Zdz + pRds cos i|r = \ -+Xl+Ym + Zn-R = T 7 tan x + -3^- + Yfi + Zv + /j,R sin i/r = These three equations express the conditions of equilibrium. 486. The simplest case is that in which the applied forces can be neglected compared with the tension. We then have, putting zero for X, Y, Z, dT D -j- + fiK cos y}r T P T = = B >■ 7 tan % + /iR sin i|r = It easily follows from these equations that tan % + fi sin i/r = 0. This requires that tan ^ should be less than /n ; thus equilibrium is impossible if the string be placed on the surface so that its osculating plane at any point makes an angle with the normal greater than tan -1 /*. Eliminating y and R from these equations, we have dT T i •. log T=C- ft (^ - tan* x) Thus, when the string is laid on the surface in a given form and is bordering on motion, the tension at any point can be found. 487. It also follows from the equations of Art. 486 that, if X = 0, then y = 0. If therefore the string is placed along a geodesic line on the surface, the friction must act along a tangent to the string. Putting y(r = 0, we have from the two first equations logT=C-f,j^. ART. 488.] STRINW ON A ROUGH SURFACE. 363 Since along a geodesic p = p, we may deduce from this equation the following extension of the theorem in Art. 463. If a light string rest on a rough surface in a state bordering on motion, and the form of the string be a geodesic, then (1) the friction at any point acts along the tangent to the string, and (2) the ratio of the tensions at any two points is equal to e to the power of ±fi times the sum of the infinitesimal angles turned through by a tangent which moves from one point to the other. The conditions of equilibrium of a string on a rough surface are given in Jellett's Theory of Friction. He deduces from these the equations obtained in Arts. 486 and 487. 488. Ex. 1. A fine string of inconsiderable weight is wound round a right circular cylinder in the form of a helix, and is acted on by two forces F, F' at its F' cos 2 a extremities. Show that, when the string borders on motion, log ■= = ± ft s, where s is the length of the string in contact with the cylinder, o the angle of the helix and a the radius of the cylinder. Since the helix is a geodesic, this result follows from the equations of Art. 487 by writing for 1/p' its value cos 2 a . ja given by Euler's theorem on curvature. Ex. 2. A heavy string AB, initially without tension, rests on a rough hori- zontal plane in the form of a circular arc. Find the least force F which, applied along a tangent at one extremity B, will just move the string. Let be the centre of the arc, let the angle AOP=6, the arc AP=s. Let the element PQ of the string begin to move in some direction PP', where P'PQ ■= a sin a. It is evident that this result does not hold if the length of the string exceed a quadrant, for then where E is some constant, which is independent of the natural length of the string and of the force by which it is stretched. It is clear that, if two similar and equal strings are placed side by side, they will together require twice the force to produce the same extension that each string alone would require. It follows that the force required to produce a given extension is proportional to the area of the section of the unstretched string. The coefficient E is therefore proportional to the area of the section of the string when unstretched. The value of E when referred to a sectional area equal to the unit of area is called Young's modulus. To find the meaning of the constant E, let us suppose that the string can be stretched to twice its natural length without violat- ing Hooke's law. We then have l = 2l t , and therefore E=T. Thus E is a force, it is the force which would theoretically stretch the string to twice its natural length. 490. This law governs the extension of other substances besides elastic strings. It applies also to the compression and elongation of ejastic rods. It is the basis of the mathematical theory of elastic solids. But at present we are not concerned with its application except to strings, wires, and such like bodies. The law is true only when the extension does not exceed certain limits, called the limits of elasticity. When the stretching is too great the body either breaks or receives such a permanent change of structure that it does not return to its original length when the stretching force is removed. In all that follows, we shall suppose this limit not to be passed. The reader will find tables of the values of Young's modulus and the limits of elasticity for various substances given in the article Elasticity, written by Sir W. Thomson for the Encyclopaedia Britannica. 366 ELASTIC STRINGS. [CHAP. X. 491. Ex. 1. A uniform rod AB, suspended by two equal vertical elastic strings, rests in a horizontal line ; a fly alights on the rod at C, its middle point, and the rod is thereupon depressed a distance h ; if the fly walk along the rod, then when he arrives at P, the depression of P below its original level is 2h(AP 1 + BF i )IAB 2 , and the depression of Q, any other point of the rod, is 2k(AP. AQ + BP .BQ)jAB 2 . [St John's Coll. 1887.] Ex. 2. A heavy lamina is supported by three slightly extensible threads, whose unstretched lengths are equal, tied to three points forming a triangle ABC. Show that when it assumes its position of equilibrium the plane of the lamina will meet what would be its position in case the threads were inelastic in the line whose tv 7/7/ ZZ areal equation is -=^ + ^ + -^ = 0, where E, F, G are the moduli, and x , «„, « the areal coordinates of the centre of gravity of the lamina referred to the triangle ABC. [St John's Coll. 1885.] 492. A uniform heavy elastic string is suspended by one extremity and has a weight W attached to the other extremity. Find the position of equilibrium and the tension at any point. Let 0A 1 be the unstretched string, P^, any element of its length. Let OA be the stretched string, PQ the corresponding position of P^. Let w be the weight of a unit of length of unstretched string, l 1 =OA 1 , x 1 =0P 1 ; o q l = 0A,x=0P. The tension T at P clearly supports the weight of PA and W. Hence Pi I- -P Q T=w(l 1 -x J )+W (1). If PA were equally stretched throughout we A i could apply Hooke's law to the finite length PA. But as this is not the case we must apply the law to an elementary length PQ. We have therefore dx — dx z = dXi eT (2), where e has been written for the reciprocal of E. From these equations we find ^ i = l+e{w(l 1 -x 1 )+W}. Integrating, we have x = #i -r e [w (1& — %xf) + Wx-i\ + C. The constant G introduced in the integration is clearly zero, since «! and x must vanish together. Putting x 1 = l ] , we find l-l^ie.wtf + eWk. ART. 493.] HEAVY ST§tING ON INCLINED PLANE. 367 If the string had no weight, the extension due to W would be eWk- If there were no weight W at the lower end, the extension due to the weight of the string would be %ewl?. This is the same as that due to a weight W equal to half the weight of the string attached to the lowest point. We also see that the extension due to the weight of the string and the attached weight is the sum of the extensions due to each of these treated separately. Ex. 1. A heavy elastic string OA placed on a rough inclined plane along the line of greatest slope is attached by one extremity to a fixed point, and has a weight W fastened to the other extremity A. Find the greatest length of the stretched string consistent with equilibrium. When the string is as much stretched as possible, the friction on every element acts down the plane and has its limiting value. Let a be the inclination of the plane to the horizon. Let /*, p' be the coefficients of friction between the plane and the string and between the plane and the weight respectively. If /= sin a + /i cos a, then fw replaces w in Art. 492. We therefore find for the whole elongation V--l = yfwl* + <:f'Wl, where/' is what /becomes when p' is written for /*. Ex. 2. A heavy elastic string AA' is placed on a rough inclined plane along the line of greatest slope. Supposing the inclination of the plane to be less than tan -1 /*, find the greatest length to which the string could be stretched consistent with equilibrium. Compare also the amount of stretching of the different elements of the string. The frictions near the lower end A of the string will act down the plane, while those near the upper end A' will act up the plane. There is some point separating the string into two portions OA,OA' in which the frictions act in opposite directions. Each of these portions may be treated separately by the method used in the last example. An additional equation, necessary to find the unstretched length z of OA , is obtained by equating the tensions at due to the two portions. The results are L / tano\ , , , , a /, tan 2 o\ g = 2\ ~T~)' l-h=i^weosal 1 ^l-—^j. Ex. 3. A series of elastic strings of unstretched lengths l lt l it l 3 ... are fastened together in order, and suspended from a point, ^ being the lowest. Show that the total extension is i[e 1 w 1 l 1 i + e 2 w 2 l^+...)+w 1 l 1 (e 2 l i + e s l s +...) + w ss l !i (e 3 l s + ..,) + &a., where w 1 , w 2 , &o. are the weights per unit of length of unstretched string, t lt e 2 , &c. the reciprocals of the moduli of elasticity. [Coll. Exam. 1888.] 493. Work of an elastic string. If the length of a light elastic string be altered by the action of an external force, the work done by the tension is the product of the compression of the string and the arithmetic mean of the initial and final tensions. 368 ELASTIC STRINGS. [CHAP. X. Iln the standard case let the length be increased from a to a', then a — a' is the shortening or compression of the string. As before, let ^ be the unstretched or natural length. By referring to Art. 197, we see that the work required is - STdl = - JE 1 -^ dl = - z ia'-W-ia-W , the limits of the integral being from I = a to I = a'. This result may be put into the form \ (T x + T 2 ) (a - a'), where T y and T 2 represent the values of T when a and a' are written for I. The rule follows immediately. See the author's Rigid Dynamics 1877, or any later edition. This rule is of considerable use in dynamics where the length of the string may undergo many changes in the course of the motion. It is important to notice that the rule holds even if the string becomes slack in the interval, provided it is tight in the initial and final states. If the string is slack in either terminal state, we may still use the same rule provided we suppose the string to have its natural or unstretched length in that terminal state. Ex. 1. Show that the depth below the point of suspension of the centre of I 1 8 W\ gravity of the elastic string considered in Art. 492 is $l t + — I - +-y I , where S is the weight of the string. Show also that the work done by gravity as the string and weight are moved from the unstretched position OA 1 to the stretched position OA, is Ex. 2. Let one end of an elastic string be fixed to the rim of a wheel sufficiently rough to prevent sliding, and let the other be attached to a mass resting on the ground, so that when the string (of length a) is just taut it shall be vertical. Show that the work which must be spent in turning the wheel so as just to lift the mass off the ground is Mga + EalogEI(E + Mg), where E is the tension which would double the length of the string, neglecting the weight of the string. [Math. Tripos.] Ex. 3. A disc of radius r is connected by n parallel equal elastic strings, of natural length Zj , to an equal fixed disc ; the wrench necessary to maintain the discs at a distance x apart with the moveable one turned through an angle 9 about the common axis, consists of a force X and a couple L given by X=nEx (j - ~\ , L=2nEr* sin 8 Q - ^ V where ?=x i +ir i sin 2 \B. [Coll. Exam., 1885.] One disc being moved to a distance x from the other and turned round through an angle 6, we first show that the length of each string is changed from 2j to {. Using the rule above, the work function is ART. 494.] HEAVY WRING ON SMOOTH CURVE. 369 By Art. 208 we have Xdx + Ld$ = ~ dx + ^Z de. ax d6 Effecting the differentiations X^dW/dx, L = dWjdO, we obtain the results given. 494. Heavy elastic string on a smooth curve. Ex. 1. A heavy elastic string is stretched over a smooth curve in a vertical plane : show that the difference between the values of T+ lrs at any two points of the string is equal to the weight 2E of a portion of the string whose unstretched length is the vertical distance between the points. It follows from this theorem that any two points at which the tensions are equal are on the same level. If ds 1 is the unstretched length of any element ds of the string, we have by Hooke's law ds^dsEKT+E). If then w is the weight per unit of unstretched length, the weight of any element ds of the stretched string is equal to to'ds, where w'=wEI(T+E). Let us now form the equations of equilibrium, using the same figure and reasoning as in Art. 459, where a similar problem is discussed for an inextensible string. We evidently arrive at the same equations (1) and (2) with so' written for w. Substituting for w' and integrating, we find that (1) leads to whence the theorem follows. Ex. 2. A heavy elastic string is stretched on a smooth curve in a vertical plane: show that a horizontal straight line' can he drawn so that where T is the tension at any point P, B the outward pressure of the curve on the string per unit of length of unstretched string, y and y' the altitudes of P and its anti-centre above the horizontal line, and w the weight of a unit of length of unstretched string. This straight line may be called the statical directrix of the string, Art. 460. Show also that no part of the string can be below the directrix, and that the free ends, if there are any, must lie on it. Ex. 3. Suppose the string has a free extremity A 1 at which the tension T 1 is zero, and let the equation to the curve be given in the form y=f{s), where y and s are measured from A v Prove that the unstretched length of any arc s is given by Eds Hi (E*+2Ewy) i ' where the limits are s =0 to 8=s. Ex. 4. A heavy elastic string rests in equilibrium on a smooth cycloid with its cusps upwards. If one extremity is attached to a point on the curve while the free extremity is at the vertex, prove that the stretched length of any unstretched arc s x measured from the vertex is given by 7«=sinh ys 1 , where iaEy*=w, and a is the radius of the generating circle. R. S. 24 370 ELASTIC STRINGS. [CHAP. X. Ex. 5. An elastic string rests on a smooth curve whose plane is vertical with its ends hanging freely. Show that the natural length a may be found from the equation I — J = , where y is the vertical height above the free extremities, and 6 the natural length of a portion of the string whose weight is the coefficient of elasticity. If the natural length of each vertical portion be I, and if (Z + 6) 2 =2a&, and if the curve be a circle of radius a, prove that the natural length of the portion in contact with the curve is 2 J (ab) log (V2 + 1). [June Exam., 1877.] Ex. 6. If the impressed forces per unit mass, acting on a stretched string in the plane of xy, be Ax parallel to the axis of x, and By paraEel to the axis of y, A and B being constants ; and if the string, fixed at the point x = a, y = 0, rest along the whole quadrant of the circle x" + y 2 =a?, for which the coordinates of any point are both positive : prove that the unstretched length of the string is equal to nearly, when Aa? and Ba? are both small compared with X, the modulus of elasticity of the string. [Coll. Exam., 1880.] Ex. 7. An elastic string, uriiform when unstretched, lies at rest in a smooth circular tube under the action of an attracting force (jj.r) tending to a centre on the circumference of the tube diametrically opposite to the middle point of the string. If the string when in equilibrium just occupies a semicircle, prove that the greatest tension is {\(\ + 2npa?)}*-\, where X is the modulus of elasticity, a the radius of the tube, p the mass of a unit of length of the unstretched string. [Trinity Coll., 1878.] Ex. 8. An infinite elastic string, whose weight per unit of length when un- stretched is m, and which requires a tension ma to stretch any part of it to double its length (when on a smooth table), is placed on a rough table (coefficient p.) in a straight line perpendicular to its edge. The string just reaches the edge, which is smooth. A weight fymap. is attached to the end and let hang over the edge. If the weight takes up its position of rest quietly, so that no part of the string re-contracts after having been once stretched, show that the distance of the weight below the edge of the table is \ap.(Zp. + i), and that beyond a distance Aa(/u + 2) from the edge of the table the string is unstretched. [Trinity Coll.] 495. Light elastic string on a rough curve. Ex. 1. An elastic string is stretched over a rough curve so that all the elements border on motion. If no external forces act on the string except the tensions F, F' at its extremities, then F' */# - = e , where \p is the angle between the normals to the curve at its extremities. This follows by the same reasoning as in Art. 463. Ex. 2. A light uniform elastic string, placed on the surface of a rough sphere along a line of greatest slope, has one extremity fixed at the summit of the sphere. A weight W attached to the other extremity stretches the string so that it occupies a quadrant of a circle. Show that the unstretched length of the string lies between the two values of (ajp.) log (W+E)I(W+Ep), where logi/= ± \jm. ART. 496.] STRK* UNDER ANY FORCES. 371 Ex. 3. An elastic string (modulus X) is stretched round a rough circular arc so that every element of it is just on the point of slipping ; if T, T are the tensions at its extremities, the ratio of the stretched to the unstretched length is log % : log j ff ^ • [St John'* Coll., 1884.] Ex. 4. An endless cord, such as a cord of a window blind, is just long enough to pass over two very small fixed pulleys, the parts of the cord between the pulleys being parallel. The cord is twisted, the amount of twisting or torsion being different in the two parts, and the portions in contact with the pulleys being unable to untwist. If the pulleys be made to turn slowly through a complete revolution of the string, show that the quotient of the difference by the sum of the torsions is decreased in the ratio e 4 : 1. [Math. Tripos, 1853.] Ex. 5. An elastic band, whose unstretched length = 2a, is placed round four rough pegs A, B, C, D, which constitute the angular points of a square of side=a. If it be taken hold of at a point P between A and B, and pulled in the direc- tion AB, show that it will begin to slip round both A and B at the same time if AP=aj{e ilx " + l). [May Exam.] Ex. 6. An endless slightly extensible strap is stretched over two equal pulleys : prove that the maximum couple which the strap can exert on either pulley is t=-4 l—r T, where a is the radius of either pulley, c the distance of their e cothJ|U7r + 2a//4 ■" centres, /* the coefficient of friction, and T the tension with which the strap is put on. [Math. Tripos, 1879.] Ex. 7. A rough circular cylinder (radius a) is placed with its axis horizontal, and a string, whose natural length is I, is fastened to a point Q on the highest generator of the cylinder and to an external point P at a distance I from Q, PQ being horizontal and perpendicular to the axis of the cylinder ; the cylinder is then slowly turned upon its fixed axis in the direction away from P ; show that the string will slip continually along the whole of the length in contact with the cylinder until S (the natural length of the part wound up)=a//i, when all slipping will cease, and that up to this stage the relation between S and 8 (the angle turned through by the cylinder) is le** =(l- a, where S = a. [Coll. Exam. , 1880.] . 496. Elastic string, any forces. To form the equations of equilibrium of am elastic string under the action of any forces. Let dsx be the unstretched length of any element ds of the string. Then by Hooke's law ds = ds 1 (T+E)/E. The forces on the element, due to the attraction of other bodies, will be pro- portional to the unstretched length. Let then the resolved parts of these forces along the principal axes of the string be Fds 1 , Gds^, Hdsx, as in Art. 454. The equations of equilibrium (1), (2), and (3) of that article are obtained by equating to zero the resolved parts of the forces along the principal axes of the curve ; these equations will therefore apply to the elastic string if we replace 24—2 372 ELASTIC STRINGS. [CHAP. X. Fds, Gds, Hds, by Fds u Gds lt Hds^ The equations of equilibrium for the elastic string may therefore be derived from those for an inelastic string by treating the forces as E- E E Fds yr, n, GdS yr. y. , Hds T+E' T+E' ~""T+E' i.e. reducing all the impressed forces in the ratio E-.T + E. 497. Suppose, for example, the string rests on any smooth surface. The resolution along the tangent to the string (as in Art. 479) gives fl + ^\dT + Xdx + Ydy + Zdz = 0. 712 ■'• T +^+tt Xdx + Yd y+ zdz )= G - It follows that T + T 2 /2.E + the work function of the forces is constant along the whole length of the string, Art. 479. Ex. 1. When gravity is the only force acting, show that the equations of equilibrium of an elastic string corresponding to (1), (2), (3) of Art. 479 may be written in the simple forms jt2 j>2 / 2 12 \ T + 2E = WZ ' Bp '~M =WZ '' (wz + —)ta,nx=wp'sm0, where T is the tension at any point P, It the outward pressure of the surface on the string per unit of unstretched length, x the angle the radius of curvature of the string makes with the normal to the surface, z and 2' the altitudes of P and the anti-centre S above a certain horizontal plane, 6 the angle the vertical makes with the plane containing the normal to the surface and the tangent to the string, and w the weight of a unit of unstretched length. If PS be a length measured outwards along the normal to the surface equal to the radius of curvature of a normal section of the surface drawn through the tangent at P to the string, S is the anti-centre of P. See Ex. 1, Art. 479. Ex. 2. A heavy elastic string rests on a surface of revolution with its axis vertical ; show that the equations of equilibrium are JT2 Ji2 T + 2B =WZ ' Bp ' ~ VET WZ '' 2Vsintf>=P, where r' is the distance of P from the axis of the surface,

    V, where w is the weight per unit of length of the unstretched string, /i the coefficient of friction and p the radius of curvature. [Math. Tripos, 1881.] Ex. 5. An elastic string has its two ends fastened to points on the surface of a smooth circular cylinder of which the axis is vertical ; show that in the position of equilibrium of the string on the surface the density of the string at any point varies as the tangent of the angle which the osculating plane at that point makes with a normal section of the cylinder through the direction of the string. [Math. Tripos, 1886.] 500. A heavy elastic string is suspended from two fixed .points and is in equilibrium in a vertical plane. To find its equation. We may here use the same method as that employed in Art. 443 to determine the form of equilibrium of an inelastic string. Referring to the figure of that article, let the unstretched length of GP (i.e. the arc measured from the lowest point up to any point P) be s u and let the rest of the notation be the same as before. Consider the equilibrium of the finite portion GP ; we have Tcos^ = T (1) Tsmf = ws 1 (2), dy , ws-, s. 374 ELASTIC STRINGS. [CHAP. X. From these equations we may deduce expressions for x and y in terms of some subsidiary variable. Since Sj = c tan ^ by (3), it will be convenient to choose either s 1 or ijr as this new variable. Adding the squares of (1) and (2), we have 2 T2 = w 2 (c 2 + s 1 2 ) (4). Since dxjds = cos i/r and dy/ds = sin yjr, we have by (1) and (2) [ T , [wet. T\, wo , , g I + V(c 2 + g 1 s ) where the constants of integration have been chosen to make x = and y = c + c 2 w/2E at the lowest point of the elastic catenary. The axis of x is then the statical directrix, Art. 494, Ex. 2. 501. Ex. 1. Prove the following geometrical properties of the elastic catenary (1) wy = T+^, (2) p= ^!±£l jl + |v{e 2 + Sl 2 )}, (3) s^ 1+ ^\ h ^wn^^sp^, all of which reduce to known properties of the common catenary when E is made infinite. Ex. 2. Let M, M! be two points taken on the ordinate PN so that MM' is bisected in N by the statical directrix and let each half be equal to T 2 /2Sw. If M be above the directrix draw ML perpendicular to the tangent at P. Show that T=w.PM, «!=Pi, c = ML, w.MN=T 2 I2E and that M' is the projection of the anti-centre on the ordinate. Ex. 3. An elastic string, uniform when unstretched, is hung up by two points. Prove that the intrinsic equation of the catenary in which it will hang under gravity is s=e tan^+gr- -ttan ^secf + logtan ( - + ¥ U , where c is the natural length of the string whose weight is equal to the tension at the lowest point, from which s is measured, and X is the natural length of the string whose weight is equal to the modulus of elasticity. [Coll. Exam., 1880.] Ex. i. An elastic string, uniform when stretched, hangs in the form of a parabola of latus rectum 4a, with its vertex downwards, the density being unity ; the forces per unit length on every particle are its weight, and a force downwards •s/m along the tangent equal to g ./ ( — — I , r being the focal distance : prove that the tension at each point is 2gr, and if the modulus of elasticity is equal to the tension at the lowest point, find the unstretched length of the string between the extremities of the latus rectum. [Trinity Coll., 1880.] CHAPTER XI. THE MACHINES. 502. It is usual to regard the complex machines as constructed of certain simple combinations of cords, rods and planes. These combinations are called the mechanical powers. Though given variously by different authors, they are generally said to be six in number, viz. the lever, the pulley, the wheel and axle, the inclined plane, the wedge and the screw*. Mechanical advantage. In the simplest cases they are usually considered as acted on by two forces. One of these, viz. the force applied to work the machine, is usually called the power. The other, viz. the force to be overcome, or the weight to be raised, is called the weight. The ratio of the weight to the power is called the mechanical advantage of the machine. 503. As a first approximation, we suppose that the several parts of the machine are smooth, the cords used perfectly flexible, the solid parts of the machine rigid, and so on. In some of the machines these suppositions are nearly true, but in others they are far from correct. It is therefore necessary, as a second approximation, to modify these suppositions. We take such account as we can of the roughness of the surfaces in contact, the rigidity of the cords and the flexibility of the materials. After these corrections have been made, our result is still only an approximation to the truth, for the corrections cannot be accurately made. For example, in making allowance for friction we assume that the bodies in contact are equally rough throughout, and that the coefficient of friction is properly known. The results however thus obtained are much nearer the real state of things than our first approximation. * In the descriptions of the machines given in this chapter, the author has derived much assistance from Oapt. Kater's Treatise on Mechanics in Lardner's Cyclopeedia, 1830, Pratt's Mechanical Philosophy, 1842, Willis' Principles of Mechanism, 1870, and other books. 376 THE MACHINES. [CHAP. XI. 504. Efficiency. Suppose a machine to be constructed of a combination of levers, pulleys, &c, each acting on the next in order. Let a force P acting at one extremity of the combination produce a force at the other extremity such that it could be balanced by a force Q acting at the same point. Then, for this machine, P may be regarded as the power and Q as the weight. Let the machine be made to work, so that its several parts receive small displacements consistent with their geometrical relations. Such a displacement is called an actual displacement of the machine. Taking this as a virtual displacement, the work of the force P is equal to that of the force Q together with the work of the resistances of the machine. These resistances are friction &c, in overcoming which some of the work done by the power is said to be wasted or lost. The work done by the force Q is called the useful work of the machine. The efficiency of a machine is the ratio of the useful work to that done by the 'power when the machine receives any, small actual displacement. It appears that the efficiency of a machine would be unity if all its parts were perfectly smooth, the solid parts perfectly rigid, and so on. In all existing machines however the efficiency is neces- sarily less than unity. 505. Ex. In any machine for raising a weight show that, if the weight remains suspended by friction when the machine is left free, the efficiency is less than one half. If however a force P be required to raise the weight, and a force P' be required P + P' to prevent it from descending, show that the efficiency will be — jrp~ . supposing the machine to be itself accurately balanced. [St John's Coll., 1884.] When the force P just raises a weight Q, the friction acts in opposition to the power P ; on the contrary it assists P' in supporting Q. The frictions in the two cases are evidently the same in magnitude, being the extreme amounts which can be called into play. Let x, y be the virtual displacements of the points of appli- cation of P, Q when the machine is worked, and let the same small displacement be given in each case. Let U be the work of the frictions. Then Px = Qy + U, P'x = Qy~U. The efficiency of the machine is measured by the ratio -p^ . Eliminating U, we easily obtain the result given in the example. If any of the resistances, other than friction, have no superior limit, but continually increase with the increase of the power, it is easy to see by the same reasoning that the efficiency will be less than the value found above. ART. 508.] • THE LEVER. 377 506. The lever. A lever is a rigid rod, straight or bent, moveable about a fixed axis. The fixed axis is usually called the fulcrum. The portions of the lever between the fulcrum and the points of application of the power and the weight are called the arms of the lever. The forces which act on the lever are usually supposed to act in a plane which is perpendicular to the fixed axis. When the forces act in any directions at any points of the body, the problem is one in three dimensions, the solution of which is given in Art. 268. In what follows we shall also neglect the friction at the axis, as that case has already been considered in Art. 179. 507. To find the conditions of equilibrium of two forces acting on a lever in a plane perpendicular to its axis. The axis of the lever is regarded in the first approximation as a straight line ; let G be its intersection with the plane of the forces. D Let the forces be P and Q. Let them act at A and B on the arms GA, GB in the directions DA, DB. When the lever is in its position of equilibrium, the forces P, Q and the reaction at the fulcrum must form a system of forces in equilibrium. Hence the resultant of P and Q must act along DG, and be balanced by the pressure on the fulcrum. The conditions of equilibrium follow at once from the principles stated in Art. 111. Let CM, GN be perpendiculars drawn from G on the lines of action of the forces. Taking moments about G, we have P . CM — Q . GN = 0. It follows that in a lever, the power and the weight are to each other inversely as the perpendiculars drawn from the fulcrum on their lines of action. 508. To find the pressure on the fulcrum, we find the resultant of the two forces P, Q by any one of the various methods usually employed to compound forces. For example, if the position of D be known, let be the angle ADB ; we then have P 2 =P 8 + Q 2 + 2PQ cos0, where R is the required pressure. "I or 378 THE MACHINES. [CHAP. XI. Let CA=a, GB = b, and let a, £ be the angles the directions of the forces P, Q make with the arms GA, CB. Let 7 be the angle AGB. If these quantities, are known, we may find the pressure by another method. Let $ be the angle the line of action of R makes with the arm CA, so that the angle DC A is v-6. Then, resolving the forces along and perpendicular to CA, we have Pcos0=Pcosa+Qcos (7-/?)) R sin B=P sin a + Q Bin (7 - /3)j whence tan 6 and B can be easily found. Other relations between P, Q and R may be found by taking moments about A, B or some other point suggested by the data of the question. In the same way other resolutions will sometimes be more convenient than those given above as specimens. 509. When several forces act on the lever, we find the condition of equilibrium by equating to zero the sum of their moments about the fulcrum, each moment being taken with its proper sign. The moments are taken about the fulcrum to avoid introducing into the equation the reaction at the axis. To find the pressure on the fulcrum we transfer each force parallel to itself, in the plane perpendicular to the axis, to act at the fulcrum. We thus obtain a system of forces acting at a single point, viz. the intersection of the axis with the plane of the forces. The resultant of these is the pressure on the axis. 510. In the investigation the weight of the lever itself has been supposed to be inconsiderable compared with the forces P and Q. If this cannot be neglected, let W be the weight of the lever. There are now three forces acting on the body instead of two. These are P, Q acting at A and B, and W acting at the centre of gravity G of the lever. Let the fulcrum be horizontal, and let CL be the per- pendicular distance between the fulcrum and the vertical through G. Let us also suppose that in the standard figure the weight W and the force P tend to turn the lever round the fulcrum in the same direction. The equation of moments now becomes P . CM- Q . CN+ W . CL = 0. The pressure on the fulcrum is found by compounding the forces P, Q,W. 511. Levers are usually divided into three kinds according to the relative positions of the power, the weight, and the fulcrum. In the first kind, the fulcrum is between the power and the weight. In the second kind the weight acts between the fulcrum and the power, and in the third kind the power acts between the fulcrum and the weight. The investigation in Art. 507 applies to all three kinds, the only distinction being in the signs given to the forces and the arms, in resolving and taking moments. 512. The mechanical advantage of the lever is measured by the ratio Q : P. This ratio has been proved to be equal to CN : CM. By applying the power so that its perpendicular distance from the fulcrum is greater than that of the weight, a small power may be made to balance a large weight. Thus a crow bar when used to move a body is a lever of the second kind. The ground is the fulcrum, the weight acts near the fulcrum, and the power is applied at the extreme end of the bar. 513. If the lever be slightly displaced by turning it round its fulcrum through a small angle, the points of application A, B of ART. 515.] • THE LEVEll. 379 the forces P, Q are moved through small arcs AA', BB', whose centres are on the fulcrum. Thus the actual displacements of the points of application of the power and the weight are proportional to their distances from the fulcrum. It is however the resolved part of the displacement AA' in the direction of the force P which measures the speed of working. For example, if the force P were applied by pulling a rope attached to the point A, the amount of rope to be pulled in would be measured by the resolved part of AA' in the direction of the length of the rope. The resolved parts of AA', BB' in the direction of the forces P, Q are evidently A A '.sin a, BB'.sin/3. These are proportional to CA sin a, CB sin /3, i.e. to CM, CN. (See fig. of Art. 516.) These resolved displacements are clearly the same as. the virtual displacements of the points of application ; Art. 64 If then mechanical advantage is gained by arranging the lever so that the weight is greater than the power, the displacement of the weight is less, in the same ratio, than that of the power, each displacement being resolved in the direction of its own force. It follows that what is gained in power is lost in speed. 514. The reader may easily call to mind numerous instances in which levers are used. As examples of levers of the first kind we may mention the common balance, pokers, &c. Wheelbarrows, nutcrackers, &c. are examples of levers of the second kind. In these the weight is greater than the power. They are used when we wish to multiply the force at our disposal. In levers of the third kind the weight is less than the power, but the virtual displacement of the weight is greater than that of the power. Such levers therefore are used when economy of force is a consideration subordinate to the speed of working. 515. The most striking example of levers of the third kind is found in the animal economy. The limbs of animals are generally levers of this description. The socket of the bone is the fulcrum; a strong muscle attached to the bone near the socket is the power ; and the weight of the limb, together with what- ever resistance is opposed to its motion, is the weight. A slight contraction of the muscle in this case gives a considerable motion to the limb : this effect is particularly conspicuous in the motion of the arms and legs in the human body ; a very inconsiderable contraction of the muscles at the shoulders and hips giving the sweep to the limbs from which the body derives so much activity. The treddle of the turning lathe is a lever of the third kind. The hinge which attaches it to the floor is the fulcrum, the foot applied to it near the hinge is the power, and the crank upon the axis of the fly wheel, with which its extremity is connected, is the weight. 380 THE MACHINES. [CHAP. XI. Tongs are levers of this kind, as also the shears used in shearing sheep. In these oases the power is the hand placed immediately below the fulcrum or point where the two levers are connected. Capt. Kater's Mechanics. 516. The principle of virtual work may be conveniently used to investigate the conditions of equilibrium in the lever. Let P, Q be two forces acting at A and B, and let G be the fulcrum. If the lever be displaced round C through a small angle 80, so that A, B come into the positions A', B', we have P.AA'sma-Q.BB'smP = 0, where a, /3 have the same meanings as in Art. 507. This im- mediately leads to the result P . GM=Q . ON. 517. Roberval's Balance. This machine supplies an excellent example of the principle of virtual work. In this balance the four rods A A', A'B', B'B, BA are hinged at their extremities and form a parallelogram. The sides AB, A'B' are also hinged at the points C, C" to a fixed vertical rod 00 C. The line GC must be parallel to AA' and BB', but need not necessarily be equidistant from them. Two more rods MM', NN' are rigidly attached to AA', BB' so as to be at right angles to them. These support the weights P and Q suspended in scale pans from any two points H and K. As the combination turns smoothly round the supports C, C, the rods AA', BB' remain always vertical, and MM', NN' are always horizontal. The peculiarity of the machine is that, if the weights P, Q balance in any one position, the equilibrium is not disturbed by moving either of the weights along the supporting rods MM', NN'. It may also be remarked that, if the machine be turned round its two supports C, C" so that one of the rods MM', NN' descends and the other ascends, the two weights continue to balance each other. ART. 518.] THE«COMMON BALANCE. 381 To show this, let the equal lengths CA, G'A' be denoted by a, and the equal lengths CB, G'B' by 6. Let the inclination to the horizon of the parallel rods AB, A'B' be 0. If the machine is displaced so that the angle 8 is increased by dB, the rod AA' desoends a vertical space a cos BdB, and the rod BB' ascends a space b cos BdB. When the weights of all the parts of the machine are neglected in comparison with P and Q, we have by the principle of virtual work Pa cos 0dB = Qb cos BdB. This gives Pa= Qb ; thus the condition of equilibrium is independent of the positions H, K at which P and Q act on the supporting rods, and is also independent of the inclination 8 of the rods AB, A'B' to the horizon. If the balance is so constructed that the weights P, Q are equal, when in equili- brium, we can detect whether any difference in weight exists between two given bodies by simply attaching them to any points of the supporting rods. The advantage of the balance is that no special care is necessary to place them at equal distances from the fulcrum. Ex. 1. If the weights of the rods AB, A'B' are w, w' and the weights of the bodies AA'M', BB'N' are W, W, prove that the condition of equilibrium is {P+W)a-(Q+W')b + b(w + w')(a-b) = 0. Thence show that, if the weights P, Q balance in one position, they will as before balance in all positions. Find also the point of application of the resultant pressure of the stand EF on the supporting table. Ex. 2. A common form of letter-balance consists of a jointed parallelogram two of whose sides are vertical, and the middle points of the others fixed. The scales are rigidly connected with the vertical sides. What is the use of this arrangement, and how does it act ? If the balance be at rest and horizontal, prove that the horizontal pressure on either support bears to either weight the ratio of the difference of the horizontal distances of the centres of gravity of the weights from the central plane of the balance to the distance between the supports. [Math. Tripos, 1874.] 518. The Common balance. In the common balance two equal scale-pans E, F are suspended by equal fine strings from the extremities A, B of a straight rod or beam. The rod AB can turn freely about a fulcrum 0, with which it is connected by a short rod OC which bisects AB at right an'gles. The oentre of gravity Gf of the beam AOB lies in the rod OC, and therefore, when the beam and the empty scales are in equilibrium, the straight line AB is horizontal. The bodies to be weighed are placed in the Boale-pans, and if their weights are unequal, the horizontality of the beam AB is disturbed. The centre of gravity G of the beam is now no longer under the point of support, and in the new position 382 THE MACHINES. [CHAP. XI. of equilibrium the inclination of the rod AB to the horizon is such that the moment of the weight of the beam about the fulcrum is equal to that of the weight of the bodies and the scale-pans. It is therefore evident that the fulcrum should not coincide with the centre of gravity of the beam. Let P, Q be the weights in the scales E and F, w the weight of either scale, let W be the weight of the beam AOB. Let OG = h, OC=c, AB = 2a. Let be the inclination of All to the horizon when the system is in equilibrium. Taking moments about 0, we have (P+w) (a cos + c sin $) - {Q + w) (a cos - c sin 0) + Wh sin = 0. The coefficient of P + w in this equation is the length of the perpendicular from on the vertical AE, and is easily found by projecting the broken line OC, GA on the horizontal. The other coefficients are found in the same way. We therefore have tan 0= lv ^~ F \ a m . \P+Q + 2w)c+Wh 519. A good balance has three requisites. The first is that when loaded with equal weights in the pans the rod AB should be horizontal. This is secured by making the arms AG, GB equal. To determine when the beam is horizontal, a small rod called the tongue is attached to it at right angles at its middle point. The beam is usually suspended from a point above 0, and when the beam is horizontal the direction of the tongue should pass through the point of suspension. The second requisite is sensibility. When the weights P, Q differ by a small quantity, the angle should be so large that it can be easily observed. For a given difference Q-P the sensibility increases as tan $ increases. We may therefore measure the sensibility by the ratio -= — =, = -p= — -= — =— : ==r- . The 3 J Q-P (P+Q + 2w)c+Wh sensibility is therefore secured by so constructing the balance that the expression on the right hand side of this equation is as large as possible. The sensibility is therefore increased (1) by increasing the length of the rod AB, (2) by diminishing the length of the rod OC, (3) by diminishing the weight of the beam. If the balance is so constructed that h and c have opposite signs, the sensibility can be greatly increased. This requires that the fulcrum should lie between Gf and G. The third requisite of a balance is usually called stability. When the balance is disturbed, it should return readily to its horizontal position. The beam oscillates about its position of equilibrium, and the quicker the oscillation the sooner can it be determined by the eye whether the mean position of the beam is or is not horizontal. The balance should be so constructed that the times of oscillation are as short as possible. The discovery of the nature of the oscillations is a problem in dynamics, and cannot properly be discussed from a statical point of view. 520. Ex. 1. If one arm of a common balance, whose weight can be neglected, is longer than the other, prove that the true weight of a body is the geometrical mean of the apparent weights when weighed first in one scale and then in the other. [Coll. Exam.] Ex. 2. A balance has its arms unequal in length and weight. A certain article appears to weigh Q x or Q 2 according as it is put in the one scale or the other. Similarly another article appears to weigh P x or P 2 . Find the true ART. 521.] TBE STEELYARDS. 383 weights of these articles ; and show that if an article appears to weigh the f) 7? — Q 7? same in whichever scale it is put, its weight is Vl 2 — „ a * . Vi _ Y2 - *h + -"a [Ooll. Exam., 1886.] Ex. 3. In a false balance a weight P appears to weigh Q, and a weight P' to weigh Q' : prove that the real weight X of what appears to weigh Y is given by X(Q-Q') = Y(P-P') + FQ-PQ'. [Math. Tripos, 1870.] Ex. 4. A true balance is in equilibrium with unequal weights P, Q in its scales. If a small weight be added to P, the consequent vertical displacement of Q is equal to that which would be the vertical displacement of P were the same small weight to be added to Q instead of to P. [Math. Tripos, 1878.] Looking at the expression for tan 8 in Art. 518, we notice that the changes produced in 6 by altering either P or Q by the same small quantity are equal with opposite Bigns. The effect of increasing P or Q is therefore to turn the • balance the one way or the other through the same small angle. The vertical displacements of the weights are therefore equal in the two cases. Ex. 5. If the tongue of the balance be very slightly out of adjustment, prove that the true weight, of a body is nearly the arithmetic mean of its apparent weights, when weighed in the opposite scales. [Coll. Exam.] Ex. 6. If the fulcrum can be moved on the line OG, let c be its distance from the straight line AB when the sensibility is infinite with weights P , Q in the scales. If c be its distance when the weights are- P and Q, prove that the sensibility s is given by als = (P+Q+ W+ '&w) c~(P + Q +W+ 2w) e . Ex. 7. A delicate balance, whose beam was originally suspended by a knife- edged portion of itself (higher than its centre of gravity) resting upon a horizontal agate plate, has its knife-edge worn down a distance e so that it becomes curved (curvature = 1/r), and has a corresponding hollow made in the agate plate (curvature =l//>). If slightly different weights P and Q be placed in the scales (whose weights may be neglected), show that the reciprocal of the sensibility is increased by (P + Q + W) fe + ?2-\ - . [Ooll. Ex., 1890.] 521. The Steelyards. The common steelyard is a lever AGB with unequal arms AG, GB, the fulcrum being situated at a point a little above G. The body Q to be weighed is suspended from the extremity B of -the shorter arm, and a given 5 Ha A\ weight P is moved along the longer arm GA to some point H such that the system balances. Let be the centre of gravity of the beam, w its weight. The three weights, P acting at H, w at G, and Q at B are in equilibrium. Taking moments about G, we have P . HG+w , GC=Q.GB (1). 384 THE MACHINES. [CHAP. XI. Let D be a point on the shorter arm CB, such that w . GC=P. CD ; the equation (1) then becomes P.HD = Q. CB (2). Thus the weight of Q is determined by measuring the distance HD. To effect this easily, we measure from D towards A a series of lengths DE X , E-JS^, E 2 E 3 , &o. each equal to CB. The weight of the body Q is therefore equal to P, 2P, 3P, &c. according as the weight P is placed at the points E l , JS 2 , E s , &o. when the system is in equilibrium. The intervals E^E^, EJ£ 3 , &a. are usually graduated into smaller divisions, so that the length HD can be easily read. The points E lt E 2 , &o. are marked 1, 2, &o. in the figure. An instrument of this form was used by the Romans and is therefore often called the Eoman steelyard. 522. In the Danish steelyard the weights P and Q act at fixed points of the lever, but the fulcrum or point of support C is made to slide along the rod AB J until the system balances. The weight P, being fixed, can be conveniently joined to that of the lever. Let, then, P' be the weight of the instrument, so that P'=P + m>, and let G be the centre of gravity. Taking moments about C, we have P' BG P'.GC=Q.CB, ,.BC = ^^. This expression enables us to calculate the values of BG when Q=P', 2P', 3P', &c. Marking these points of the rod AB with the figures 1, 2, 3, 1859.] "n+2 n n+l "n n„ r. s. 25 386 THE MACHINES. [CHAP. XI. Ex.9. An old Danish steelyard, originally of weight W lbs., and accurately graduated, is found coated with rust. In consequence of the rust, the apparent weights of two known weights of X lbs. and Y lbs. are found when weighed by the steelyard to be (X - .t) lbs. , {Y-y) lbs. respectively. Prove that the centre of gravity of the rust divides the graduated arm in the ratio W (x - y) : Yx - Xy ; and that its W+ Y W-v X weight is, to a first approximation, - — - x+ — -^ ij. [Math. Tripos, 1885.] A. — 1 ± — A Ex. 10. A brass figure ABDC, of uniform thickness, bounded by a circular arc BDG (greater than a semicircle) and two tangents AB, AC inclined at an angle 2a, is used as a letter-weigher as follows. The centre of the circle, 0, is a fixed point about which the machine can turn freely, and a weight Pis attached to the point A, the weight of the machine itself being w. The letter to be weighed is suspended from a clasp (whose weight may be neglected) at D on the rim of the circle, OD being perpendicular to OA. The circle is graduated, and is read by a pointer which hangs vertically from : when there is no letter attached, the point A is vertically below and the pointer indicates zero. Obtain a formula for the graduation of the circle, and show that, if P=^w sin 2 o, the reading of the machine will be \w when i ..wi. t - i i ii * [(it + 2 ■ e * ( (ir + 2a) sin 3 a + 2 cos a ) [Math. Tripos, 1878.] 525. The Pulley. The common pulley consists of a wheel which can turn freely on its axis. A rope or cord runs in a groove formed on the edge of the wheel, and is acted on by two forces P and P' one at each end. If the pulley is smooth and the weight of the string infinitesimal, the tension is necessarily the same throughout the arc of contact. It follows that the forces P, P' acting at the extremities of the string are equal to each other and to the tension. See fig. 1 of Art. 527. The same thing is true if the pulley is rough and circular, but can turn freely about a smooth axis ; Art. 197. 526. When the axis of the pulley is fixed one of the forces P, Q is the power and the other is the weight. Thus a fixed pulley has no mechanical advantage in the technical sense. A machine, however, which enables us to give the most advantageous direction to the moving power is as useful as one which enables a small power to support a large weight. 527. A moveable pulley can however be used to obtain mechanical advantage. Suppose a perfectly flexible string to be fixed at A, pass under a pulley C of weight Q, and to be acted on at B by a force P; see fig. 2. In the position of equilibrium ART. 529.] «ffHE PULLEY. 387 the strings on each side of the pulley meet in the line of action of the force Q (Art. 34), and must therefore make equal angles with the vertical (Art. 27). Let a be the inclination of either string to the vertical, then 2Pcosa=Q. Fig. l. Fig. 2. The mechanical advantage is therefore 2 cos a. Unless a is less than 60° the mechanical advantage is less than unity. When the strings are parallel, we have 2P = Q. 528. Ex. 1. In the single moveable pulley with parallel strings a weight W is supported by another weight P attached to the free end of the string and hanging over a fixed pulley. Show that, in whatever position the weights hang, the position of their centre of gravity is the same. [Math. Tripos, 1854.] Ex. 2. A string is attached to the centre of a heavy circular pulley of radius r and is then passed over a fixed peg, then under the pulley, and afterwards passes over a second fixed peg vertically over the point where the string leaves the pulley and has a weight W attached to its extremity. The second peg is in the same horizontal line as the first peg and at a distance 4}r from it. If there is equilibrium, prove that the weight of the pulley is | W, and find the distance between the first peg and the centre of the pulley. [Coll. Ex., 1886.] Ex. 3. An endless string without weight hangs at rest over two pegs in the same horizontal plane, with a heavy pulley in each festoon of the string ; if the weight of one pulley be double that of the other, show that the angle between the portions of the upper festoon must be greater than 120°. [Math. Tripos, 1857.] 529. Systems of pulleys may be divided into two classes, (1) those in which a single rope is used ; and (2) those in which there are several distinct ropes. We begin with the first of these systems. 25—2 388 THE MACHINES. [CHAP. XI. D Two blocks are placed opposite each other, containing the same number of pulleys in each. Three are repre- | sented in each block in the figure. The string ; passes over the pulleys in the order ADBEGF, and has one extremity attached to one of the blocks. The power P acts at the other extremity of the string, while the weight Q acts on a block. Let n be the number of pulleys in either block, W the weight of the lower block; we then have Q + W supported by 2m tensions. Since the tension of the string is the same throughout, and equal to P, we have by resolving vertically 2nP = Q + W. If the pulleys were all of the same size, and exactly under each other, some difficulty might arise in their arrangement so that the eords should not interfere with each other. For this, and other reasons, the parts of the string not in contact with the pulleys cannot be strictly parallel. Except when the two blocks are very close to each other the error arising from treating the strings as parallel is very slight, and may evidently be neglected when we take no account of the other imperfections of the " x , w 2 w n tne weights of the moveable pulleys, there will be no mechanical advantage unless W-w n +2(W-w n _ 1 ) + 2*(W-w n _ ;s ) + +2»-i(r-u> 1 ) be positive. [Math. Tripos, 1869.] Ex. 6. In the system of n heavy pulleys in which each hangs by a separate string, P is the power (acting upwards), Q the weight, and B the stress on the beam from which the pulleys hang : show that R is greater than Q (1 - 2 -n ) and less than (2 n - 1) P. [Math. Tripos, 1880.] Ex. 7. If there be two pulleys, without weight, which hang by separate strings, the fixed ends only of the string being parallel, and the power horizontal, prove that the mechanical advantage is JS. [St John's Coll., 1883.] Ex. 8. In that system of pulleys, in which all the strings are attached to the weight, if the power be made to descend through one inch, through what distance will the weight rise ? Illustrate by reference to this system of pulleys the principle which is expressed by the words, " In machines, what is gained in power is lost in time." [Math. Tripos, 1859.] Ex. 9. In the system of pulleys in which all the strings are attached to the weight Q, prove that, if the pulleys be small compared with the lengths of the strings, the necessary correction for the weight of the strings is the addition to Q, w lt w 2 ...io n _ 1 respectively, of the weights of lengths 7i 1 + ft a +... + ft K _i + fi > 2(7^-/1), 2(\-\), ...2(ft B _i-'*n-2) of string ; where U x , h^, h 3 ...h n are the heights of the n pulleys (whose weights are w lt w 1 .,.w ll respectively) above the line of attachment, supposed horizontal, of the strings to the weight Q, and h the height of the point of attachment of the power above the same line. [Math. Tripos, 1877.] •394 THE MACHINES. [CHAP. XI. Ex. 10. In that system of pulleys in which the strings are all parallel, and the weights of the pulleys assist the power, show that, if there are n pulleys, each of diameter 2a and weight w, the distance of the point of suspension of the weight from the line of action of the power is equal to 2"+ 1 Q + [(re-3)2"+TO + 3]w " 2(2»-l)Q a ' where Q is the weight. [Math. Tripos, 1883.] Ex. 11. In a system of four pulleys, arranged so that each string is attached to a bar carrying the weight, the string which usually carries the power is attached to one end of the same bar, and the fourth string to the other end. The weight and diameter of each pulley are respectively double of those of the pulley below it, and the strings are all parallel. The weight being 33 times that of the lowest pulley, find at what point of the bar it is hung. [Trin. Coll., 1885.] Ex. 12. In the system of pulleys, in which each pulley hangs by a separate string with one end attached to a fixed beam, there are n moveable pulleys of equal weight w. The rth string, counting from the string round the highest pulley, cannot bear a greater tension than T. Prove that the greatest weight which can be sustained by the system is 2" _r+1 T - (2 n_r+1 - 1) w. [Trin. Coll., 1890.] Ex. 13. It is found that any force P being applied to the extremity of a string passing over a pulley can just raise a weight P (1 - 0). In the system of pulleys in which each hangs by a separate string a weight Q is just supported, the weight of each pulley being a.Q. If a and 8 are small quantities, whose squares and products may be neglected, show that an additional power equal to n8Qj2 n can be applied without affecting the equilibrium. [Coll. Exam., 1888.] 537. The Inclined Plane. To find the relation between the power and the weight in the inclined plane. Let AB be the inclined plane, G any particle situated on it. Let GN be a normal to the plane and GV vertical ; let a be the inclination of the plane to the hori- zon, then the angle NGV = a. Let Q be the weight of G, P a force acting on G in the direction GK, where the angle NGK = . It is supposed that GK lies in the verti- cal plane VGN. Fig. l- If the plane is smooth the reaction R of the plane on the particle acts along the normal CN. We therefore have by Art. 36 JL = Q= . * . (1). sin a sm

    ~a.) ' It is necessary for equilibrium that B should be positive, for otherwise the particle would leave the plane. It follows from ART. 538.] TH» INCLINED PLANE. 395 these equations that must be greater than a. This result follows also from an examination of fig. (1), for Q acting along VC and R along CN cannot be balanced by a force P unless its direction lies within the angle formed by GV and NG produced. If P act up the plane, = \tt and P = Q sin a, R = Q cos a. If P act horizontally, = \tt + a., and P = Q tan a, R = Q sec a. 538. If the plane is rough, let /*=tan e be the coefficient of friction. With the normal CN as axis describe a right cone whose semi-angle is a ; this is the cone of friction, Art. 173. The resultant action R' of the plane on the particle lies within this cone; let CH be its line of action and let the angle NCH=i; then i lies between ±e. Let the standard case be that in which o is greater than e, and greater than either ; this is represented in fig. (2). We therefore have Fig. 2. Fig. 3. Q •(2); sin(o-i) sin(0-i) sin(0-a) when the force P is so great that the particle is on the point of ascending the plane, the reaction R' acts along GE, and i= - e. Let P 1 be this value of P, then Q R' sin(a + e) sin(0 + e) sin(^-o) *■■(•»)• When the force P is so small that the particle is only just sustained, the reaction JS' acts along CD, and % = e. Let P 2 be the value of P, then Q sin (o - e) sin {

    e as in fig. (2), it is clear that the particle will slide down the plane if not supported by some force P, Art. 166. When the particle is just supported the reaction R' acts along CD and Q along VG ; it is clear that these forces could not be balanced by any force P unless its direction lay within the angle made by GV and DC produced. Accordingly we see from (4) that R' is negative unless

    a. In the same way it is impossible to pull the particle up the plane (without pulling it off) by any force whose direction does not lie between GV and EC produced. Assuming

    a, the least force required to keep the particle at rest is given by (4), and the greatest by (3). 396 THE MACHINES. [CHAP. XI- If e > a as in fig. (3), the particle will rest on the plane unless disturbed by some force P. To just pull the particle up the plane the force must act within the angle formed by GV and EG produced, and its magnitude is given by (3). In order that the particle may be just descending the plane the force must act within the angle formed by GV and DG produced, and its magnitude is given by (4). 539. Ex. 1. If a power P acting parallel to a smooth inclined plane and sup- porting a weight Q produce on the plane a pressure M, then the same power acting horizontally and supporting a weight E will produce on the plane a pressure Q. [Coll. Exam., 1881.] Ex. 2. Find the direction and magnitude of the least force which will pull a particle up a rough inclined plane. By (3) we see that P 1 is least when + e = |;r, i.e. when the force makes an angle with the inclined plane equal to the angle of friction. Ex. 3. Find the direction and magnitude of the least force which will just support a particle on a rough inclined plane. Ex. 4. A given particle G rests on a given smooth inclined plane and is supported by a force acting in a given direction. If the inclined plane is without weight and has its side AL moveable on a smooth horizontal table, find the force which when acting horizontally on the vertical face BL will prevent motion. Find also the point of application of the resultant pressure on the table. Ex. 5. A heavy body is kept at rest on a given inclined plane by a force making a given angle with the plane ; show that the reaction of the plane, when it is smooth, is a harmonic mean between the greatest and least reactions, when it is rough. [Math. Tripos, 1858.] Ex. 6. A heavy particle is attached to » point in a, rough inclined plane by a fine rigid wire without weight, and rests on the plane with the wire inclined at an angle 6 to a horizontal line in the plane. Determine the limits of 6, the angle of inclination of the plane being tan -1 (/* sec |8). [Coll. Exam.] Ex. 7. Two equal particles on two inclined planes are connected by a string which lies wholly in a vertical plane perpendicular to the line of junction of the planes, and passes over a smooth peg vertically above this line of junction. If, when the particles are on the point of motion, the portions of the string make equal angles with the vertical, show that the difference between the inclinations of the planes must be twice the angle of friction. [Math. Tripos, 1878.] 540. Wheel and Axle. To find the relation between the, power and the weight in the wheel and axle. Let a be the radius of the axle AB, c that of the wheel. The power P acts by means of a string which passes round the wheel several times and is attached to a point on the circumference. The weight Q acts by a string which passes similarly round the axle. Taking moments round the central line of the. axle, we have Pc = Qa. The mechanical advantage is therefore equal to cja. ART. 544.] flHEEL AND AXLE. 397 b-B Fig. 1. Fig. 2. If p, q be the spaces which the power and weight pass over while the wheel turns through any angle, we have p/q = c/a=Q/P. 541. When a great mechanical advantage is required we must either make the radius of the wheel large or that of the axle small. If we adopt the former course the machine becomes unwieldy, if the latter the axle may become too weak to bear the strain put on it. In suoh a case we may adopt the plan represented in fig. (2). The two parts of the axle are made of different thicknesses, and the rope carried round both. As the power P descends, the rope which supports the weight is coiled on the thicker part of the axle and uncoiled from the thinner. Let a, b be the radii of these two portions of the axis. If Q be the weight attached to the pulley, the tension of the string is JQ. Taking moments about the central line of the axis, we have Pc = ^Q (a - 6). The mechanical advantage is therefore equal to the radius of the wheel divided by half the difference of the radii of the axle. By making the radii of the two portions of the axis as nearly equal as we please, we can increase the mechanical advantage without decreasing the strength of the machine. This arrangement is called the differential axle. 542. Ex. 1. A rope passes round a pulley, and its ends are coiled opposite ways round two drums of different radii on the same horizontal axis. A person pulls vertically upon one part of the rope with a force P. What weight attached to the pulley can he raise, supposing the parts of the rope parallel ? [Coll. Exam.] Ex. 2. In the differential axle if the ends of the chain, instead of being fastened to the axles, are joined together so as to form another loop in which another pulley and weight are suspended, find the least force which must be applied along the chain in order to raise the greater weight, the different parts of the chain being all vertical. [Math. Tripos.] 543. When both the power and the weight act on the circumference of wheels there are various methods of connecting the two wheels besides that of putting them on a common axis. Sometimes, when the wheels are at a distance from each other, they are connected by a strap passing over their circumferences. In some other cases one wheel works on the other by means of teeth placed on their rims. 544. Toothed Wheels. To obtain the relation between the power and the weight in a pair of toothed wheels. 398 THE MACHINES. [CHAP. XI. Let A, B be the centres of two wheels which act on each other by means of teeth, the teeth on the axis of one wheel working into those on the circumference of the other at the point G. Let a u a 2 be the radii of the axles, b u b 2 those of the wheels. Let p, q be the virtual velocities of the power P and weight Q, then Pp = Qq. If the teeth are small the average velocities of the E points near G on the two wheels are equal, and the common direc- tion is perpendicular to the straight line AB. If then 6 lt 2 are the angles turned through by the wheels when the power P receives a small displacement, we have a 1 6 1 = b i 6 i . But p^b^, q = a i 6 i . It follows that S = —- We have here omitted the work lost in overcoming the friction at the teeth in contact and at the points of support. 545. Let a tooth on one wheel touch the corresponding tooth on the other in some point D, and let EDF be a common normal to the two surfaces in contact at D. The point D is not marked in the figure because the teeth are not fully drawn, but it is necessarily situated near G. The actual velocities of the points of the teeth in contactat D when resolved in the direction EDF are equal. If, then, h and k are the perpendiculars drawn from A , B on EDF, it is clear that ^ft = 2 ft. As the wheels turn, if the lengths h and k alter, and the ratio hjk is not constant, there is more or less irregularity in the working of the machine. To correct this defect, the teeth are sometimes cut so that the normal at every point of the boundary of a tooth is a tangent to the circle to which the tooth is attached. When this is done, the line EDF is always a common tangent to the two circles. The ratio hjk is therefore constant throughout the motion and equal to the ratio of the radii of the circles. One cause of irregularity will thus be removed and the motion will be made more uniform. If the normal at every point of the surface of a tooth is a tangent to a circle, each of the two halves of that tooth is bounded by an arc of an involute of the circle. The two involutes are unwrapped from the circle in opposite directions and portions of each form the sides of the tooth. ART. 547.] TdBTHED WHEELS. 399 When the centres of the toothed wheels are given, and the ratio of the angular velocities at which they are to work, we may determine their radii in the following manner. Let A, B be the given centres ; divide AB in G so that AC . 1 =BC . 2 . Through C draw a straight line ECF, which should not deviate very much from a perpendicular to AB. With A and B as centres describe two circles touching the straight line ECF. The sides of the teeth are to be involutes of these circles. By this construction the common normal to two teeth pressing against each other at D is the straight line ECF. As the wheels turn round, and the teeth move with them, the point o'f contact D travels along the fixed straight line ECF. The perpen- diculars h and k are equal to the radii of these circles and are constant during the motion. Their ratio also is evidently equal to the ratio of AC to BG, i.e. of 2 to r It has already been shown that Pp = Qq, and p = b 1 6 1 , g = a 2 2 . Since 1 /! = 2 fr, we find as before ^ = — 2 . We may notice that, if the distance between the centres A and B is slightly altered, the pair of wheels will continue to work without irregularity and the ratio of the angular velocities will be the same as before. To prove this, we observe that the common normal to two teeth pressing against each other is still a common tangent to the two circles, though in their displaced positions. Thus, though the inclination to AB of the straight line ECF is altered, the lengths of the perpen- diculars h and k are the same as before. That the teeth should be made of the proper form is a matter of importance to the even working of the machine. Many other considerations enter into the theory besides that mentioned above. But the subject is too large to be treated of in a division of a chapter. The reader who is interested in this matter is referred to books on the principles of mechanism. Another defect arises from the wearing of the teeth if the pressure be very great at the point of contact. There may also be jolts and jars when the teeth meet or separate. 546. Ex. 1. In a train of n wheels, the teeth on the axle of each wheel work on those on the circumference of the next in order. Show that the power and weight are connected by the relation -^ = 1 2 '" " , where a lt u 2 &c. are the radii of the axles and b lt 6 2 &c. those of the wheels. Ex. 2. In a pair of toothed wheels show that, if the ratio of the power and weight is to be" approximately constant, the height and breadth of the teeth must both be small relatively to the radius of each wheel. Two equal and similar wheels, with straight narrow radial teeth, are started with a tooth of each in contact and in the same straight line ; show that they will work together without locking, provided that the distance of their centres be greater than 2a cos — and less than 2a cos - , where a is the radius of either wheel ° n n measured to the summit of a tooth, and n the number of teeth. [Math. Tripos, 1872.] Ex. 3. Investigate the relation Q/P=6 1 6 2 /a 1 a 2 for a pair of toothed wheels without using the principle of virtual work. 400 THE MACHINES. [CHAP. XL The reaction B between two teeth acts along the straight line EDF. Taking moments in turn about A and B, we have Pb^Bh, Qa 2 =Bh. As before, we have when the teeth are small ft/ft = a 1 /6 2 . The result follows at once. 547. The Wedge. To find the relation between the power and the weight in the wedge. Let M, N be two obstacles which it is intended to separate by inserting a wedge ABC between them. For the sake of distinctness these obstacles are represented in the figure by two equal boxes placed on the floor, but it is ob- vious they may be of any kind. We shall suppose that the wedge used is isosceles, and that it has its median line GIV vertical. Let the angle ACB be 2a. Let D, E be the points of contact with the obstacles, R, R the normal reactions at these points, F, F the frictions. When the wedge is on the point of motion we have F = R tan e, where tan e is the coefficient of friction. Let P be a force acting vertically at N urging the wedge downwards. Supposing P to prevail, the frictions on the wedge act along CA, CB ; we therefore find by resolving vertically P = 2R (sin a + tan e cos a) = 2R sin (a + e) sec e. The resultant reaction R' at _D is then found by compounding R and fiR. If the obstacle M can only move horizontally, the whole of the reaction R' is not effective in producing motion. The horizontal component of R' tends to move M, but the vertical component presses the box on the floor and possibly tends to increase the limiting friction between the box and the floor. Let X be the horizontal component of R'; we find X = R cos a — R tan e . sin a = R cos (a + e) sec e. The mechanical advantage X/P is therefore equal to \ cot (a + e). ART. 549.] *HE WEDGE. 401 i f 548. It may be noticed that the mechanical advantage of the wedge is increased by making the angle a more and more acute. There is of course a practical limit to the acuteness of this angle, for that degree of sharpness only can be given to the wedge which is consistent with the strength required for the purpose to which it is to be applied. As examples of wedges we may mention knives, hatchets, chisels, nails, pins, &c. Generally speaking, wedges are used when a large power can be exerted through a small space. This force is usually applied in the form of an impulse. It has not been considered necessary to consider separately the case in which the wedge is smooth, as the resultB obtained on so erroneous a supposition have no practical bearing. 549. If the force is applied in the form of a blow so that the wedge is driven forwards between the obstacles, the problem to determine its motion is properly one in dynamics. Our object here is merely to find the conditions of equilibrium of a triangular body inserted between two rough obstacles and acted on by a force P. When a series of blows is applied to the wedge, we may however enquire what happens in the interval between two impulses. The wedge may either stick fast, held by the friction, or begin to return to its original position, being pressed back by the elasticity of the materials. Assuming that these forces of restitution may be represented by two equal pressures R, R, acting on the sides of the wedge, let P, be the force necessary to hold the wedge in position. The friction now acts to assist the power. To determine P, we write — e for e in the equations of equilibrium. We therefore have Pj = 2R sin (a — e) sec e. If a is greater than e, P t is positive and therefore some force is necessary to hold the wedge in position. If a is less than e, P t - is negative, thus the friction is more than sufficient to hold the wedge fast. A force equal to this value of P, with the sign changed, is necessary to pull the wedge out. The result is that the wedge will stick fast or come out according as the angle AGB is less or greater than twice the angle of friction. Ex. 1. Referring to the figure of Art. 547, show that if either of the equal angles A or B of the wedge is less than the angle of friction, no force P however great could separate the obstacles M, N. If the angle A is less than 6, we find that a + e is greater than a right angle, and therefore that X is negative. It is easy also to see that, if the angle A is equal to e, the resultant reaction between one side of the wedge and an obstacle is vertical. The wedge therefore merely presses the obstacle against the floor-. r. s. 26 402 THE MACHINES. [CHAP. XI. Ex. 2. If the obstacles M, N are not of the same altitude and are unequally rough, the position of the wedge when in equilibrium is such that the force P 1 and the resultant actions RJ, R 2 ' across the faces meet in a point. Supposing the force Pj to act perpendicularly to the face AB of the wedge and to be just sufficient to hold the wedge at rest, show that A Mn iV sin (2a - e 1 - e 2 ) cos (a - Cj) cos (a - e s ) ' assuming the obstacles to be of such form that the wedge must slip at both simul- taneously. Show also that, if the wedge be such that the angle C is less than the sum of the angles ej + a,, the wedge can be held fast by the frictions without the application of any force. Ex. 3. Deduce from the principle of virtual work the relation between the force X and the power P in a smooth isosceles wedge as represented in the figure of Art. 547. Discuss the two cases in which (1) one obstacle is immovable and (2) both move equally when the wedge makes an actual displacement. 550. The Screw. To find the relation between the power and the weight in the screw. Let AB be a circular cylinder with a uniform projecting ridge running round its surface, the tangents to the directions of the ridges making a constant angle a with a plane perpendicular to the axis of the cylinder. The screw thus formed fits into a hollow cylinder with a corre- sponding groove on its internal surface, in which the ridge works. The grooves on the hollow cy- ■ linder have not been sketched, but are included in the beam EF. The position of the ridge on the cylinder is easily understood by the following construction. Let a sheet of paper be cut into the form of a right angled triangle LMN, such that the altitude MN is equal to the altitude of the cylinder AB and the angle the base LM makes with the hypothenuse LN is equal to o. Let this sheet of paper be wrapped round the cylinder AB ; if the base LM is long enough to go several times round the base of the cylinder, the hypothenuse will appear to wind gradually round the cylinder. The line thus traced by the hypothenuse is the curve along which the ridge lies. Let P be the power applied perpendicularly at the end of a lever CD. Let AG = a, and let b be the radius of the cylinder. art. 551.] •the scrkw. 403 Supposing the body EF in which the screw works to be fixed in space, the end B of the cylinder will be gradually moved as C describes a circle round AB. Let Q be the force acting at B. Let a be any small length of the screw which is in contact with an equal length of the groove. Let Ra be the normal reaction between these small arcs, fiRa the friction. In some screws the ridge is rectangular, so that it may be regarded as generated by the motion of a small rectangle moving round the cylinder with one side in contact with the surface and its plane passing through the axis. When the ridge has this form, the line of action of R lies in the tangent- plane to the cylinder and its direction makes with the axis of the cylinder an angle equal to a. In other screws the section of the ridge has some other form such, for example, as a triangle. In such cases the line of action of R makes some angle 8 with the tangent plane to the cylinder. We therefore resolve R into two components, one intersecting at right angles the axis of the cylinder and the other lying in the tangent plane. The magnitude of the latter is R cos 8, and its direction makes with the axis of the cylinder an angle equal to a. Since the ridge is uniform the angle 8 will be the same throughout the length of the screw. Let us suppose that the power P is about to prevail, then the friction acts so as to oppose the power. Resolving parallel to the axis of the cylinder and taking moments about it, we have Q = %Ra . cos 9 cos a — 2-Ro- . (i sin a, Pa = 2i?o- . b cos 8 sin a + %R, 0' be the eccentric angles of two points P, Q on the conic. Taking the principal axes of the curve as the axes of coordinates, the equations of the normals at these points are cos sin cos sin ' The ordinate i\ of their intersection is therefore given by frn sin 4 (0+0') . -T-T2 = lil I? sm Bm (!)■ a'- 6 s ... cos J (0-0) •••vi The ordinate of the middle point of the chord PQ is f =£& (sin + sin 0') = 6 sin J (0 + 0') cos \ (0 - 0'), ■ 6" i?_ -sin0sin0' _ cosH(ft + 0') " a B -6 2 y eos 2 4(0-0') cob 2 £(0-0') ( ^" Again, the equation to the chord PQ is -cos4(0 + 0') + ~sinJ(0 + 0')-cos£(0-0')=O (3J . If p, j)' and g are the perpendiculars on the chord from the foci and the centre, we have by the usual formula for the length of a perpendicular pp^_ {cosi(0-0')-ecos^(0 + 0')} {cos&(0-0') + ecos^(0 + 0')} t cos 2 J (0-0') It follows by an easy reduction that fl AbJ tf \y J «■' i* ' It is explained in the text that the corresponding form for £ is an inconvenient one because the foci on the minor axis are imaginary. If the chord cut the axes in L and M, we find, from the equation to the chord PQ given above, that GL _ cos j (0 - 0') CM _ cos j (0 - 0') a - cosi(0 + 0')' 6 sinj(0 + 0')" (8). 406 NOTE ON CONICS. We have immediately from (2) J! 2 (l A _ CZi'-a' + ft' a 2 /£_ \ _ _CM=-^+a 2 _ « 8 W / CX 2 ' & 2 \* / CM 2 l '" The second follows from the first by changing the letters. These are the formula used in Art. 126, Ex. 3. By introducing CM into the right hand side of (1) we find "Ja^P^" 11 ^ 8 " 10 '' TfZ^ 00 ^ 008 *' -( 6 )- When the points P, Q coincide, |, t\ become the coordinates of the centre of curvature at P. We then deduce from (1) the well known formula -A 2=Biu3 *' A- 0083 * < 7 >- The coordinates x, y of the middle point G of the chord being given, the chord itself is determinate. The equation to the chord is {£-x)x , (r)-y)f _ n a 2 + ft 2 We then readily find the intercepts GL, CM. We deduce from (2) or (5) j V y } pi ypg, lo a -6 a f" t ' f V t '& 2 / ~o 2 f « a i , ,1 1? , n s _r ' I a 3 -i 2 S J lo 2 ft 2 / ~A 2 J Let .ST, F be the coordinates of the intersection T of the tangents at P, Q, then X_Y ?*£?_! S ~ % ' o 2 + 6 2 ' because (r is the intersection of the straight line joining the origin to T with the polar line of T. We easily find x, y in terms of X, Y, and the equations (7) then become r, _ (a 2 -a 2 )(Z 2 -a 2 ) £ (a 3 - 6 2 ) (r 2 -S a ) . . r~ a 2 7 2 + 6 2 .X 2 ' X~ a 2 y 2 + & 2 .X 2 W ' which are the equations used in Art. 127. Ex. 1. A uniform rod, whose ends are constrained to remain on a smooth elliptic wire, is in equilibrium under the action of a centre of force situated in the centre C and varying as the distance, see Art. 51. Show that the centre of gravity G must be either in one of the axes or at a distance from the centre equal to CR 2 /(a 2 +6 2 ) , where GR is the semidiameter drawn through G. Show that in the latter case half the length of the rod is equal to CD 2 /(a 2 + ft 2 ) , where CD is conjugate to GR. Show also that the tangents at the extremities of the rod are at right angles. Find the lengths of the shortest and longest rods which could be in equilibrium. Ex. 2. One extremity of a string is tied to the middle point of a rod whose extremities are constrained to lie on a smooth elliptic wire. If the string is pulled in a direction perpendicular to the rod, show that there cannot be equilibrium unless the rod is parallel to an axis of the curve. N«TE ON CONICS. 407 Ex. 3. When the conic is »• parabola, show that the equations (5), (8), (9) take the simpler forms, „_ AR 237 / STA 2 »7=237. — =-J-[x- J - )=--XY, !72 QV2 $=2