?^2»"->-"^E» PRESENTED TO TJBfS! CORNELL UNIVERSITT, 1870, The Hon. William Kelly Of Rhinebeck. Cornell University Library arW3908 A treatise on statics 3 1924 031 364 445 olln.anx The original of tiiis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031364445 A TEEATISE STATICS, CONTAINING THE THEOEY OF THE EQUILIBRIUM OF FORCES; AND ICttrntr^ns (^um^hs ILLUSTRATIVE OF THE GENERAL PRINCIPLES OF THE SCIENCE. BT S.^'EARNSHAW, M.A. 0? ST iohh's COLLXOB, caiibkisos. FOURTH EDITION, REVISED. DEIGHTON", BELL, AND CO. LONDON: BELL AND DALDY. 1858.' CamiriSae : FBINTED BY 0. J. CLAY, M.A. AT THE UHIVERSITY PRESS. PREFACE TO THE THIED EDITION. This edition of the Treatise on Statics differs so little from the last as scarcely to require a separate notice. A few Articles have been added, on the pres- sure which a rigid body is made to exert on a fixed point or axis of support by the action of forces when there is equilibriuni. These will be found useful in those Problems of Dynamics wherein it is reqtiired to find the pressure which a rigid body in motion, under the influence of any forces, exerts on a fixed point or axis. Indeed, it is chiefly with a view to this appli- cation of them that the articles alluded to have been introduced into this edition of the Statics. The collection of Problems for practice given at the end of the Treatise has been considerably enlarged, chiefly by the addition of Examples of an elementary character. In the selection of them care has been taken to choose such as illustrate Statical Principles under every important variation of aspect, without impeding the student's progress through them by analytical and other difficulties foreign to the proper object of this Treatise. Cambkidob, Feb. i, 1845. PREFACE TO THE SECOND EDITION. Though the general plan and arrangement of this edition of the Treatise on Statics are the same as in the former, in the details there will be found, it is hoped, some important improvements. The fundamental proposition of the science, — the Parallelogram of Forces, — I have proved after Du- chayla's method, by reason of its simplicity; but I think it necessary here to inform the reader that, as that method is inapplicable when the' forces act upon a single particle of matter (as a particle of a fluid medium on the hypothesis of finite intervals), on ac- count of its assuming the transmissibility of the forces IV PREFACE. to other points than that on which they act, I have, in an Appendix, given the proof which in the first edition was given in the text. The same objection ^and for the same reason) lies against the proof of the parallelogram of forces from the properties of the lever. This method, though allowable in the infancy of the science, can never be exclusively adopted in a Treatise which professes to take a more philosophical view of the subject; for, were the transmissibility of force not true in fact, the law of the composition of forces acting on a point would «tiU be true; it is evident, therefore, that to make the truth of the former an essential step in the proof of the latter, is erroneous in principle. In the former edition, forces were considered as acting in any directions in space ; a mode of treatment of the subject which necessarily rendered the inves- tigations useless to such readers as had not studied Geometry of Three Dimensions. In the present edi- tion this defect is remedied; and a cha.pter, in which the forces are supposed to act in a plane, is always made to precede the more general investigations. At the end of Chapter IV. several propositions are proved which have hitherto been used in Elementary Books without proof. The fifth Chapter contains a new (and it is hoped a satisfactory) and complete proof of the Principle of Virtual Velocities, and its Converse. The proof given by Lagrange in his MScanique Analytique, page 22, et seq., though highly ingenious, I regard as a fallacy; and, if not fallacious, deficient in generality. In the last Chapter, I have endeavoured to set before the reader such problems as, without involving analytical difficulties, seemed best calculated to make him acquainted with the mode of applying a,ll the most important principles of the science ; and not un- frequently I have added remarks upon important steps with the view of pressing them more particularly upon the reader's attention. St John's Colleoe, CAMsniDaE, March 12, 1842, CONTENTS. INTRODUCTION. ART. PAGE 1 — 22. Definitions and Preliminary Notions 1 CHAPTER I. 23—32. Forces which act in one plane upon a particle, or upon THE same point OF A RIGID BODV. 23, 24, Forces acting in tlie same line on a point 8 26. Parallelogram of forces. ..' 9 27] 28. Three forces acting on a point 10 29—32. Any number of forces acting in a plane on a point 11 CHAPTER II. 33 — 38. Forces which act upon a particle^ or upon the same point of a rigid body, in any direction not in one plane is CHAPTER III. 39 — 68. Forces which act in one plane but not upon the same point op a rigid body. 39—57. The Theory of Couples 19 I 68 — 63. Parallel forces in a plane 29 &i — 68. Non parallel forces in a plane. 32 CHAPTER IV. 69 — 101. Forces not in one plane which act upon different points op a rigid body. 71 — 77. Parallel forces not in a plane 36 78 — 80. Resultant of three couples 41 81 — 95. Any forces acting on a rigid body 42 96, Equilibrium of three forces acting on a rigid body. 63 98 — 101, Conditionsof equilibrium of any forces made general 64 VI CONTENTS. CHAPTER V. ART. FASE 102 — 120. The Principle op Virtual Velocities 57 CHAPTER VI. The Centre op Parallel Forces, and the Centre op Gravity. 121—126. The Centre of Parallel Forces 68 126—142. The Centre of Gravity. 70 143 — 167. Position of Centre of Gravity _ of reotilmear and regular figures 76 168 — 173. General Properties of the Centre of Gravity 83 174 — 183. Application of Integral Calculus in finding the Centre of Gravity of hodies 91 174, 176. Centre of gravity of a curve line — .176. a plane area 95 177 a solid of revolution 97 178 a solid of any form 99 179 a surface of revolution 104 180 a surface of any form .106 181 , a curve of double curvature — 182, 183 hodies of variable density — 184, 186. Guldin's Properties 116 CHAPTER VII. 18G — 230. Mechanical Instruments 118 189—194. The Lever 121 195—207. The Pulley 123 208—211. The Wheel and Axle 128 212—214. The Inclined Plane 129 215. The Screw :.... 130 216. The Wedge 133 217. General Pi-operty of Machines 134 218. White's Pulley 136 219. Hunter's Screw...., ]3(3 220. Compound Wheel and Axle 137 221. The Genou.... 138 223—225. Toothed Wheels 140 .226. The Endless Screw 143 227. The Conimon Balance 144 .228. The Steelyard 146 .229. The Danish Balance I47 230. Roberval's Balance 148 CONTENTS. Vll CHAPTER VIII. ART. PAGB 231—236. Fbiotion 150 CPIAPTER IX. 237—244. Elastic Stbings 155 CHAPTER X. Funicular Polygon, Catenary, Roofs, and Bridges ,,. 159 245—248. The Funicular Polygon — 249. Roofs and Bridges 161 250—267. The Catenaiy 162 CHAPTER XI. Problems 174 Appendix 209 Miscellaneous Problems 216 By the same Author. A TREATISE ON DYNAMICS. Tliird Edition. STATICS. INTRODUCTION. DEFINITIONS AND PEELIMINARY NOTIONS. 1. In the Science of Mechanics of which Statics forms a part, matter is considered as essentially possessing extension, figure and impenetrability. The least conceivable portion of matter is called a particle. 2, We conceive of matter that it can exist either in a state of rest, or mation. If then matter, once at rest, pass into a state of motion, the change, not being essential to the existence or nature of ma,tter, is of necessity ascribed to some agent, which, as to its nature, is essentially independent of the matter influenced. Whether this agent reside in the matter influenced, or in external objects, or in both, are questions which can only be answered after experimental investigation. This agent is called _^ce,' and it will be perceived from this statement, that a force is judged of entirely by the effects which it produces : and hence, if in the same circumstances two forces produce equal effects, we infer that the forces are equal. 3. It is assumed, that the effect of two equal forces acting in concert, is double the effect of one of them ; three, treble ; and so on. The reason of its being necessary to make this an assump- tion is, that in our ignorance of the nature of force, we are com- pelled to judge of it by the change which it produces in the state of rest or motion of matter ; and it is obvious, that we can E. s. 1 2 DEFINITIONS AND PRELIMINARY NOTIONS. no more judge that one such change is twice as great as another, than we can affirm ' that one candle is twice as bright, or one substance twice as sweet, or one noise twice as loud as an- other. 4. A force is considered as having magnitude and direction, and a point of applicaticm. When these three are known, the force is said to be known. From Art. 2, it will be seen that, by the magnitude of a force, we mean the degree of motion which it is capable of prod^icing in matter previously at rest ; and by the direction of a force, we mean the direction in which a particle of matter, under the influence of that force, would begin to move ; and by the point of application of a force, we mean that particular particle of a mass of matter on which the force immediately exerts its influence. 5. If one particle of a rigid* mass of matter be acted upon by a force, it cannot obey the influence of the force without dragging with it the other matter with which it is connected ; the motion therefore which it would receive, if free, is in some manner distributed among the whole mass of which it is a part. It is clear, therefore, that the subject of which we are treating, naturally divides itself into two distinct parts, according as the forces act on a free particle, or on a rigid body. 6. With respect to the motion of a particle of matter, we conceive that it consists in the particle's being found to occupy diiferent parts of space at successive instants, or epochs of time; but with respect to the motion of a rigid body we conceive, (1) That as a whole it may occupy the same portion of space at successive epochs, while some of its parts individually occupy different parts of space in successive instants. This is called rotatory motion. * We define a rigid body to be an assemblage of particles of matter, connected together in such a manner that their relative places never change. DEFINITIONS AND PEELIMINAEY NOTIONS. 3 (2) That as a whole it may occupy different parts of space at successive epochs, without having at the same time any rotatory motion. This is called a motion of translation. (3) That both these kinds of motion may exist together in the same body. This is the most general kind of motion of which we can form a notion. 7. From the preceding articles it will be perceived that we have taken motion as the characteristic effect of force. It will now be necessary to shew, that there exists another effect (and that more convenient for our present purpose) which may be taken as the measure or characteristic of force. If any portion of matter (a stone for instance) be held in the hand, it will be found to exert a pressure ; and if the hand be suddenly removed, will fall. In its fall it may be caught, but the hand will again feel a pressure. This experiment informs us, that that which is the cause of motion, is likewise the cause of pressure. While the stone is held at rest, its continual ten- dency to fall is evidenced by the pressure which is exerted on the hand ; hence, in all cases where motion is prevented, there is pressure. But further, the latter part of the experiment teaches us that, in all cases where motion is retarded, there is pressure. If when the stone is at rest, the hand exert a greater pressure upwards than is necessary to prevent it from falling, the stone will begin to move upwards. Hence we learn that - pressure attends the production as well as the prevention and the destruction of motion. Thus it appears that pressure pro- duces the same results as we have taken to be the characteristic effects of force. We may therefore take pressure as the measure of force, because both pressure and motion are effects of the same cause. 8. The Earth, in some unseen manner, exerts a pressure in a downwards direction upon all matter with which we are 4 DEFINITIONS AND PEELIMINARY NOTIONS. acquainted. This pressure it is which occasions the descent of falling bodies to the ground, and causes all bodies lying on the ground to press against it. More accurate experiments prove that every particle of matter, whether of metal, wood, earth, or of any other substance, is subject to this influence. And it can be shewn that the degree of this pressure exerted upon a given body never changes. Thus, let a spring AB have one end A firmly fixed in an immoveable block. Suspend a proposed sub- stance P fi-om the other end B, then the spring will be bent in the manner represented- in fig. 1, the point B taking a position B'. If the experiment' be again tried with the same body P after any interval of time, it will be found that the spring will be bent exactly as at first ; thus shewing that the Earth exerts an unvarying pressure upon every body. If the experiment be tried with the same spring and sub- stance P at a place in another latitude, or on a hill, or in a pit, the bending of the spring is not found to be the same as before : but at the same place no variation is ever observed in the result. 9. We may easily find other substances P', P", P'"... each of which being suspended from B will bend the spring exactly as P does. By suspending 2, 3, 4, ... of these bodies at a time, and marking the spaces through which the spring is bent in each case, we may form a grxiduated scale, by means of which we can ascertain exactly the degree of pressure which the Earth exerts upon any proposed body whatever, as compared with the pressure which it exerts upon P. If this be done, it is usual to call the pressure on P the unit of pressure ; and the pressure which is exerted upon another body, if it be TF times the pressure on P, is said to be equal to W. 10. The pressure >F which the Earth exerts upon a body, when measured in the manner just described, is called the weight of the body. How great soever be the pressure which any other force exerts upon a body, we can always find (hypo- thetically at least) so many bodies P, P', P", P'"... that the DEFINITIONS AND PEELIMINAEY NOTIONS. 5 Earth shall exert upon them, taken all together, exactly as much pressure as the proposed force exerts upon the proposed body. Hence then, with the assumption in Art. 3, we perceive that every force may be measured, and therefore represented, by a weight. 11. To avoid circumlocution, when a body is prevented by an' obstacle from moving, it is usual to say that the hody exerts a pressure upon the obstacle, and that the obstacle exerts an equal pressure upon the body in the contrary direction. The fact however is, that the body is completely passive ; and the reason why it remains in a state of rest is, that it is under the influence of two equal pressures exerted on it in opposite directions. By the same licence, if a body, which is under the influence of the Earth's action, be suspended by a string, it is often said that the string exerts a force or pressure upon the body ; the fact however in this case is, that the string by being attached to the body, becomes a part of the body ; and the whole remains in a state of rest, for the same reason as before. Hence it will be seen that, in the experiment described in Art. 8, the spring exerts a force equal to that exerted by the Earth upon P, though in the contrary direction. And hence we say, when two bodies are pressed together, that they act and react upon each other with equal forces. 12. It is sufficiently evident, that two equal pressures, acting in opposite directions upon the same point of a body, counteract each other : but it is conceivable that if several pres- sures be applied to a body, even though they be not two and two in opposite directions, nor all applied to the same point of tbe body, they may counteract each other. The science which teaches the relations necessarily existing between the magnitudes of forces, their directions, and their points of application, when they exactly counteract each other, is called Statics. 13. If several forces acting upon a body counteract each other, the body is said to be in equilibrium: and the forces are said to balance each other. 6 DEFINITIONS AND PEELIMINAEY NOTIONS. 14. If sevieral forces acting upon a free particle do not balance each other, the particle will hegin to move in some direction in a certain manner. It may be prevented from so moving by applying a proper force in the opposite direction to that in which there is a tendency to motion. This new force exactly counteracts the whole system of forces : but it might be itself counteracted by a single force equal to itself and acting in a contrary direction. A single force satisfying these conditions would be exactly equivalent to the whole of the original system of forces ; and it is therefore called their resultant. 15. We thus learn that several forces, if they act upon a free particle, may be replaced by one force ; and the converse is evidently true, viz. that one force may be replaced by a system of several forces. When one force is replaced by a system of several forces, they are called its com/ponents. 16. By reference to Art. 6, we see that the motions which a rigid body may take are of two distinct kinds : and therefore the reasoning just stated respecting a free particle does not apply to rigid bodies. We shall hereafter shew that, corresponding to the three cases stated in the Article referred to, a system of forces acting on a rigid body may have (1) A resultant for rotation only, (2) A resultant for translation only, (3) Two resultants, one for the rotation and one for the translation. 17. It is evident from the explanations above given, that a system of forces, acting on a free particle, cannot have more than one resultant : but we have just seen that the same is not neces- sarily true when they act on a rigid body. It is always true, however, that the same force may have different systems of components. 18. If a particle, or a rigid body, be in equilibrium under the action of several forces, we may add to the system, or take away fr-om it, any set of forces which balance each other. DEFINITIONS AND PKELIMINARY NOTIONS. 7 This principle is called the " superposition of equilibrium," and we shall hereafter have frequent instances of its utility. 19. It follows at once from this, that when a body is in equilibrium under the action of a system of forces, they may be all increased, or all diminished in any proportion, without affect- ing the equilibrium. 20. It scarcely needs remarking, that if a set of forces balance each other, any one of them is equal to, and acts in an opposite direction to, the resultant of all the rest. 21.- It is proved by experiment, that when a rigid body is in equilibrium, any point (of the body) in the line of the direc- tion in which a force acts, may be taken for the point of applica- tion of the force, without affecting the equilibrium. 22. If a system of unbalancing forces acts upon the same point of a rigid body, they will have the same resultant as if they acted upon a free particle. CHAPTER I. ON FORCES WHICH ACT IN ONE PLANE UPON A PARTICLE, OR UPON THE SAME POINT OF A RIGID BODY. 23. To find the resultant of several forces acting, in the same line, wpon the same point of a rigid hody. If all the forces act in the same direction along the line, they will produce the same effect as a single force equal to their sum. If some act in one direction and some in the opposite direc- tion, then by the first case the resultant of each set wiU be equal to the sum of the forces of which it is composed : and these two resultants, acting in opposite directions, will be equivalent to a single resultant equal to their difference. Hence then whether the original forces act in the same or in opposite directions, their resultant is equal to their algebraic sum. In forming this sum, we are to account those forces positive which act in one direction, and those negative which act in the opposite direction ; and the algebraic sign of the sum so formed will shew in what direction the resultant acts. 24. Cor. If a number of forces act in the same line upon the same point of a rigid body, they will be in equilibrium when their algebraic sum is equal to zero, for in that case their re- sultant vanishes, and they produce no effect. Hence the condi- tion of equilibrium of any number of forces acting in the same line and upon the same point of a rigid body is that their alge- braic sum shall be equal to zero, 25. Dep. Lines are said to represent forces in magnitude and direction, when they are drawn parallel to the directions in which the forces act, and have their lengths proportional to the magnitudes of the forces. FORCES ACTING IN A PLANE ON A POINT. 9 26. If from a point two lines he drawn representing two forces which act upon a point; and if upon these lines a parallel- ogram he constituted, the diagonal drawn from the same point will represent the resultant of the two forces. This property is usually cited as "the parallelogram of forces." We shall first prove that the diagonal represents the direction of the resultant force. This part of the proposition is evidently true when the two given forces are equal ; let us assume that p, q and r are three forces, such that this is true for p and q ; and also for jj and r. At the point A, fig. 2, apply p in the direction AB: and g-, r, both in the direction AD. Take AB, AC, CD to represent the respective magnitudes of these forces. Complete the parallelograms, and draw the diagonals as in the fig. The resultant of p and q acts in the direction AE, hy hypothesis ; and we may by Art. 21, suppose it to act at E; and we may there resolve it into its original components p and q ; the latter acting in the line EF, and the former in the line GE produced ; this we may suppose hy Art. 21 to act at 0, and the former at F; also the force r may he supposed to act at G. We have now two forces p, r &t C represented hy the lines GE, GD ; their resultant by hypothesis acts in the line GF, and therefore we may suppose at F. The three forces p, q, r, which originally acted at A, are by this process reduced to forces acting at F. F is therefore in the line in which their resultant acts when they are applied at A, (Art. 21). lfow-4Z), AB, represent twoibrces q-\-r and p ; and we have just shewn that their resultant acts in the direction of the diagonal AF. If then our proposition be true for the two forces p, q ; and also for the two forces p,r; it is also true for the forces^ and q-\-r. Now it is true when^, q, f are all equal ; and hence it is true for ^ and 2p : and because it is true for p, p; and also for ^, 2p; therefore it is true for^, 3p;... and by following the same mode of reasoning it is true for p and mp, m being any whole number. Again, because it is true for mp and p, and also for mp and p ; therefore it is true for mp and 2p ; and as before for mp and np, n being an integer. Hence our proposition respecting the direction of the resultant is true for any two commensurahle forces mp, np. E.S. 2 10 FOKCES ACTING I.N A PLANE ON A POINT. If the proposed forces P, Q be incommensurable, by taking p extremely small and the integers m, n correspondingly large, we can make mp differ from P, and np from Q by less than any quantities which can be assigned ; and we may then use mp and np, instead of P and Q, without any sensible error ; and there- fore the proposition is true of P and Q. We shall now prove that the diagonal represents the magni- tude of the resultant force. Let A C, AB (fig. 3) represent the magnitudes and directions of two forces acting on a point. Complete the parallelogram : its diagonal ^^ has been proved to represent the direction of the resultant ; it also represents its magnitude. For in EA produced take ^i^to represent its magnitude; then AB, AG, AF repre- sent three forces which balance each other : wherefore completing the parallelogram AFQB, its diagonal A Q represents the direc- tion of the resultant of AF, AB, and is consequently in the same ■ line with A G (Art, 20). Hence A GBE is a parallelogram, and therefore ^JS^= OB = AF; that is, ^^ represents the magnitude of the resultant oi AB, AG. 27. If three forces acting on a point are represented hy the sides of a triangle taken in order, they will halance each other. And conversely ; If three lines, forming a triangle, he parallel to the directions of three forces which, acting on a point, ~ halance each other, the sides of the triangle taken in order will represent the forces. For let AB, BE, EA (fig. 3) represent the forces P, Q, B which act on a point. Complete the parallelogram BG; then because ^ C is drawn parallel and equal to BE, it also represents the force Q. Hence the resultant of P and Q is represented by AE; which being equal, and in a contrary direction, to EA which represents R, there is equilibrium. Conversely, let the three forces P, Q, B, acting on a point balance each other : and suppose ABE (fig. 4) to be the triangle whose sides are respectively parallel to the directions in which P, Q, B act. Two, at least, {AB, BE suppose) represent the directions of the corresponding forces ; and we are at liberty to FORCES ACTING IN A PLANE ON A POINT. 11 suppose that one of these {AB suppose), represents also the magnitude of its force P: if BH do not represent the magni- tude of the other force Q, take BJE' to represent it, and join AE'. Then (by the former case) B,'Q and a force represented by U'A will balance each other. But F, Q, R balance each other; and therefore R is represented by E'A both in magni- tude and direction; which is impossible (because EA repre- sents R in direction by hypothesis) unless E' coincide with E. Therefore E' does coincide with E, and therefore the forces are represented by AB, BE, EA, which are the sides of the triangle taken in order. 28. If three forces, acting upon a point, balance each other, their directions lie in a plane ; and their magnitudes are respec- ■ tively proportional to the sine of the angle between the directions in which the other two act. Let the forces be P, Q, R (fig. 5) acting in the directions OA, OB, 00. In OA, OB, take points A, B, such that OA, OB represent the magnitudes as well as the directions of P, Q. Complete the parallelogram A OBI), and join OB. Then Q being represented by OB may also be represented by AB; also as the three forces represented by OA, AB, BO, acting on a point will balance each other (Art. 27), therefore P, Q and a force represented hy BO balance each other; but P, Q, R balance each other ; therefore R is represented hj BO: and con- sequently BOG is a straight line, and OA, OB, 0(7 lie in the plane of the triangle OAB. Also P : Q : R :: OA : AB : BO sin OB A : sin BOA : sin OAD sin BOB : sin AOG : sin PAB sin QOR : sin POR : sin POQ, ''' ^^ioR'mSoR^^KTOQ- Therefore, &c. 29. Two forces act upon the same point in directions at right anghs to each other, to find their resultant (fig. 5*). 12 FORCES ACTING IN A PLANE ON A POINT. Let tbe forces be X, Y acting upon the point in the direc- tions Ox, Oy at right angles to each other. Take OM, ON to represent the forces ; complete the rectangle OMPN, and draw the diagonal OP. This line by Art. 26 represents the resultant of X and F. Let B be the resultant, and 6 the angle POM which its direction makes with "the direction of the force X Now OP'=OM' + MP'; .■.P?= X'+Y', which determines the value of B : and then the equation . MP Y ^^^ = -OM^X determines the value of 9. 30. COE. If a force B be given and it be required to resolve it into two components acting in directions at right angles to each other, we must employ the equations X = 5 cos e, and r= ^ sin 6, which are derived from the equations 0M= OP cos 6, and MP= OP. sin 9. 31. Any nwmher of forces act wpom, a point in given directions in a plane; to find their resultant. Let F^, F^, F^...F„ be the forces, and (see fig. of Art. 29) the point upon which they act. In the plane in which are the lines in which the forces act, draw any two lines Ox, Oy at right angles to each other; and denote by a,, a^, ag...a„the angles which the directions oiF^, F^, j?'3...i^„ respectively make with Ox. Then the components of these forces are respectively (by Art. 30), F^ cos «! , F^ cos ttj, jPj cos Kg F„ cos a„ in the direction Ox ; and F^ sin ftj, F^ sin a^, F^ sin ag F^ sin a^ in the direction Oy. Let us replace (see Art. 15) the original forces FORCES ACTING IN A PLANE ON A POINT. 13 by tliese two sets of componentg. These components are re- spectively equivalent to two forces acting in the lines Ox, Oy (Art. 23), and being equal to F^ cos tti + F^ cos a^ + -PI, cos Kg + ... + -F„ cos a„, and F^ sin a, + F^ sin a^ + F^ sin a, + . . . + i^„ sin a„. Let B be the result of the original forces F F F . . F and suppose that 6 is the angle which the line in which it acts makes with Ox. Then since B is equivalent to the original forces, it is also equivalent to the two components of them which have just been found : hence (by Art. 30) B cos6 = F^ cos Kj + i^2 cos a^ + ... + F„ cos a„, and B sm6 = F^ sin a^ + F^ sin a.^+ ... +F„ sin a„. From which two equations both B and 9 may be found. Eemaek. The sum of a number of quantities of the same form is oftMi, for brevity, represented by prefixing the symbol S to a term representing the general form. Upon this principle the above equations may be written thus : B cosd — %{F cos a), ox%.F cos a, and i? sin = X (i^ sin a), or S . i'' sin a ; :. B?={X.FcoaoLY+{t.F sin a)», S.-Fsina and tan 6 = % .Fcosol' 32. To find the conditions of equilihrium of forces acting upon a point in any directions in one plane. Let F^, F^... Fnhe, the forces ; Oj, a, ... a„ the angles which their directions make with a line Ox ; then, proceeding as in the last article, We have the equations there found for the determi- nation of their resultant. But because they balance each other by hypothesis, they have no resultant, and therefore .H = 0, or = (S . i^ cos a)'' ->r{t.F sin a)^ 14 FORCES ACTING IN A PLANE ON A POINT. But as the right-hand member consists of two terms which, being squares, are essentially positive, their sum cannot be equal to unless each be separately equal to ; .•- Q = 't.Fcosa, and = S.i^sina. If the investigation in Art. 31 be examined, it will be seen that the line Ox was taken in any direction in the plane of the forces ; and hence we may in the most general terms state the signification of the two equations just found to be as follows, If any number of forces act in one plane upon a point, that there may be equilibrium. The sums of the components of the forces parallel to any two lines, at right angles to each other, in the plane of the forces must he separately equal to zero. The converse is evidently true also. No other condition is necessary for equilibrium, for if 2.i'^cosa=0, and S.i^sina=0, it follows inevitably that J? = 0, and therefore there is equilibrium. CHAPTEE II. ON FORCES WHICH ACT UPON A PARTICLE, OR UPON THE SAME POINT OF A RIGID BODY, IN ANY DIRECTIONS NOT JN ONE PLANE. 33. If three forces acting upon the same point he respectively represented hy the three edges of a parallelopiped which meet, the diagonal of the parallelepiped drawn from, that point to the opposite corner will represent their resultant. For let OA, OB, OG (fig. 6) be the edges wliicli represent the three forces, and OE the diagonal of the parallelopiped : draw OD, GE. Then because OA, OB represent two forces, OD represents a force which is equivalent to them both (Art. 26): hence the three forces represented by OA, OB, G are equivalent to the two represented by OD, OG, which again are equivalent to the single force represented by OE, for GD is a parallelogram. 34. Three forces act upon a point in directions which are at right angles to each other; to find their resultant. Let X, Y, Z be the forces, acting upon the point (fig. 7) in the lines Ox, Oy, Oz which make right angles yOz, xOy, z Ox with each other. From set off OL, OM, ON to represent the forces X, Y, Z respectively. Complete the parallelograms OMQL, OQPN, and join OP; this line, by the last Art., repre- sents the resultant required. Let B denote the resultant, and a, /3, 7 the angles POx, POy, POz which its direction makes with the directions of the given forces. 16 FORCES NOT ACTING IN A PLANE ON A POINT. Then because 0N= OP. cos 7 ; .•. Z= ^ cos 7 1 Similarly F = iZ cos ^ i . . . (A) ; and X=B cos a) .: X'+Y^ + Z' = B' (cos' a + cos' /3 + cos' 7) /OX' ow oir\ ~ [OF"^ Or^ OF') _ ™ OL' + LQ'+QP' _ j^ OQ'+QJP' -^ Oi« ~ OF' OF' 'OF' This equation gives the value of B, and then the three equations marked (A) give the angles a, /8, 7, which fix the line in which B acts. Remark. The reader will observe from the above that cos' a. + cos' /8 + cos' 7 = 1, a formula which will be of frequent use in the following pages. 35. CoE. If a force B be given, and it be required to resolve it into three components, whose directions are at right angles to each other, we must employ the equations marked (A). 36. Any number of forces act in given directions upon a point; to find their resultant. Let (fig. 7) be the point upon which the given forces Fj^, F.^, F^ ... Fn act; from draw three lines Ox, Oy, Oz, arbitrarily taken, at right angles to each other ; and denote by «A7u aA^a) «A7s ••• «A7„, the angles which the directions of the forces make with these three fixed lines. The respective components of the given forces are -F, cos a,, F^ cos a^, ... F^ cos a„, in the direction Ox; POECES NOT IN A PLANE ACTING ON A POINT. 17 F, cos^., F^ cos/3,, ... F„ cos ,S„, in the direction Oy; and F^ cos 7, , F^ cos 7j, . . . F^ cos 7„, in the direction Oz. Replace now the original forces hy these three sets of com- ponents (Art. 15) ; each set is reduced to one force by Art. 23 ; and we then have three component forces 2 . i^ cos a, % . F cos /S, % . F cos 7, acting respectively in the lines Ox, Oy, Oz. Let R be the resultant required, and a', yS', 7' the angles which the line in which it acts makes with Ox, Oy, Oz. Then since R is equivalent to the original forces,^ it is also equivalent to the three components of them which have just been found ; hence R cos a' —%. F cos 0L'\ ^ cos jS' = 2 . i^ cos/si ... (A) ; R cos 7' = 2 . i^cos 7' and therefore since 1 =cosV+ cos'/S' + cosV) (-A^t- 34, Rem.) we find B' ={t.F cos a)" + (2 . ^ cos ^Y +{%.F cos 7)'. This equation gives the value of R ; and then the equations (A) will give a', /3', 7', which fix the direction in which R acts. 37. To find the equations of the line in which the resultant a/its. Suppose OP (fig. 7) the line in which the resultant acts ; and let x, y, z be the co-ordinates of any point P in it. Then if OP be taken to represent R, the co-ordinates will represent the components, and therefore by Art. 25, X _ y _ z 2 . i^cosa~2 . i^cos/3~2 . -FC0S7' which are the equations required. E. s. 3 18 FORCES NOT IN A PLANE ACTING ON A POINT. If the point at whicli the forces act be not the origin of co-ordinates, let its co-ordinates be a,h,c; then since the line whose equations are required passes through this point, x—a _ y—h e—c STFcosa ~ sTTcos^S ~ S . -F cos 7 are the equations of it in this case. 38. To find the conditions that a system of forces acting upon a point in any directions, may balance each other. It is evident that there cannot be equilibrium among the forces F^, F^, F^ F„, unless their resultant be eyanescent, and therefore we must have = (2 . ^ cos a)'' + (S . i^ cos /3)= + (S . i^ cos 7)', which for a reason similar to that assigned in Art. 32 resolves itself into the three independent conditions = S.^cosa, = S.i^cos/3, = 2.^ cos 7. Or, in words, (remembering that the positions of Ox, Oy, Oz were arbitrarily chosen). The sum of the components of the given forces parallel re- spectively to any three lines at right angles to each other must separately he equal to zero. CHAPTEK III. ON FORCES WHICH ACT IN ONE PLANE BUT NOT UPON THE SAME POINT OF A RIGID BODY. THE THEORY OF COUPLES. 39. Remark. It has been stated in Art. 21, that the effect of a force is not altered by supposing it to be transferred from one point of the body in the line of the direction of its action to another: from this it follows that if the directions of the forces which act at different points of a rigid body, all pass through a point, we may fictitiously transfer them to that point, and then by the preceding Chapters find their resultant, which in its turn we may transfer to any convenient point of the rigid body which happens to lie in the line of its direction. It is obvious, that when any two forces in the same plane act upon a rigid body at different points, their directions unless parallel being produced will meet, and therefore after the statement just made it will not be necessary to include the consideration of two non-parallel forces in the present Chapter, we shall there- fore begin with the following. 40. Two forces act in 'parallel directions upon different points of a rigid hody, to find their resultant. Case 1. Let F, F' be the two forces, and let us, first, suppose them to act in the same direction. Let A, B (fig. 8) be any two points of the rigid body in the lines of direction of the respective forces: join A, B; at these points in opposite directions along the line AB apply any two equal forces /, /'. These being in equilibrium produce no effect. 20 THE THEORY OP COUPLES. Now i^ and/ (by Art. 26) and F" and/' will have resultants {rn,, n suppose) acting in certain directions Am, Bn within the angles FAf, and F'Bf: these lines being produced will meet in some point P to which let m, n be transferred : and let them there be resolved into their original components; viz. m into / and F, acting at P in the directions Pf and PR {PR being drawn parallel to AF) ; and n into/' and F' acting at P in the directions Pf and PR, which is also parallel to BF'. The forces /and/' at P being in equilibrium may be removed, and there remain the original forces F, F' both acting at P along the line PR parallel to their direction at A and B. Hence the resultant of F and i^ is a force, equal to their 'sum F-\- F', acting at any point in the line PR ; the position of which we find as follows. Let PR cut AB in Q. Then because m is the resultant of P and/ a force equal to m applied at ul in the direction AP would keep the two forces P,/in equilibrium; and the three being paa-allel to the sides of the triangle APQ taken in order, aie proportional to those sides (Art. 27) ; .-. P : / :: PQ : AQ. Similarly /' : P':: BQ : PQ .: F:F::BQ: AQ; •.•/=/'. Consequently Q divides AB into two parts which are inversely proportional to the forces adjacent to which they lie. 41. Case 2. Let us now suppose the two forces P, P' to act in contrary directions, as in fig. 9, and that they are unequal, P being the greater. Introduce the equal and opposite forces /and/' as before; and let m be the resultant of P and/; and n that of P' and/'. Then since the angle FAf is equal to the angle F'Bf, it will be found, by constructing the parallelograms of force upon FA, Af, and F'B, Bf, that since P is greater than P', the direction of the resultant m lies nearer to AF, than the direction of the resultant n to BF: that is, the angle fAm is. greater than the angle fBn or ABP. Consequently the lines nB, Am being produced will meet on the side towards the greater force P THE THEORY OF COUPLES. 21 as is represented in the figure. From this point proceeding as in the former case we find that the forces F, F' preserving their proper directions, may be removed to the point P. Hence their resultant R is equal to i^— F', the algebraic sum of the forces, and acts in the direction of the greater force. The word sum is used in the statement of this result, because jP being assumed positive, F acting in the contrary direction must be accounted a negative force. (See Art. 23.) The position of the point Q is found as before from the proportion F : F' :: BQ : AQ, and it is to be noticed particularly thfit Q lies in BA produced ; and is situated nearer to A (the point of application of the gi-eater force) than to B. 42. Case 3. Let us lastly sujypose the two forces F, F' acting in contrary directions, to he equal (fig. 9). In this case the angles fAm, f'Bn are equal ; and conse- quently the angles fAm, ABP are cqnal, and the lines Am, nB are parallel, and have no point of concourse. It would appear then, that the former mode of finding the resultant of F and F' fails entirely in this case. The present case may, however, be considered the ultimate state of Case 2, at which we arrive by supposing the magnitude of F' to approach continually nearer to that of F, until at length their difierence becomes less than any assignable quantity. Let us then reconsider Case 2. We have found B = F-F', and F : F' :: BQ : AQ; ■•'BQ-^ • a) or R . BQ = F. AB. (2). Hence, we see from (1) that as F' increases, the point Q moves continually farther from B, and BQ becomes infinite in the ultimate state; and at the same time from (2) we see that the resultant R diminishes in such a manner that the product 22 THE THEORY OF COUPLES. B.BQ never changes; and B becomes zero in the limit. Hence, in the ultimate state, that is, when F' differs from J?' by less than any assignable quantity, we have a resultant zero acting at a point infinitely distant from ^ or -B; yet even then the product B.BQ remains finite, which apart from any other consideration would induce us to conjecture, that some finite effect is due to the action of jFand F' in this case, although not such an one as can be represented by a single force. Upon these grounds we conclude, that a system of two equal forces acting in contrary directions on different points of a rigid hody is not reducible to a single resultant. 43. Deps. a system of two equal forces acting upon a rigid body in opposite directions but not in the same straight line, is denominated A couple. A plane which passes through the two lines in which the forces of a couple act, is called the plane of the couple. When the line AB (fig. 9) is drawn at right angles to the directions of the forces of the couple, it is called the arm of the couple; and the product F . AB (see (2) of Art. 42) is then called the moment of the couple. 44. E.EMAEK. It is obvious from an examination of fig. 9, that one effect of a pair of forces, acting in contrary directions at different points of a rigid body, whether they be equal or unequal, is the communication of a rotatory motion (see Art. 6) to the body on which they act ; what other effect they would produce is not so obvious, nor indeed does it belong to us, in treating of the present subject, to consider what is the effect of unbalanced forces in any case. For the satisfaction of the reader, however, and for convenience in what follows, it may be stated, that it is proved in Dynamics, that the sole effect of a couple is to communicate to the body on which it acts an angular motion about an axis passing through a certain point in the body, called the centre of gravity. 45. Def. If the forces of the couple act .so as to tend to turn the body round in the direction of the motion of the hands of a watch, it is called a right-handed couple, and more fre- quently a positive couple ; but if, as in fig. 9, the forces act so as THE THEORY OP COUPLES. 23 to turn the body in the contrary direction, the couple is styled left-handed, or negative. These terms, to prevent confusion, will be used in this book as here defined ; but the reader will observe that, in Statics as in Algebra, the terms positive and negative are only relative, and may be applied, at discretion, to any two forces acting in contrary directions, or to any two couples which tend to com- municate opposite angular motions to the body on which they act. 46. The reflecting reader will have remarked that a couple, though positive when viewed by a spectator looking at it from one position, appears negative to a spectator looking at it from a position on the other side of its plane. A couple is therefore positive or negative, according to the situation of the spectator, with respect to its plane. It will prevent confusion, if we call that face of the couple's plane the positive face, upon which the spectator looks when the couple appears to him to be positive: the other face of the plane must then be considered negative. 47. Deps. a straight line, in length proportional to the moment of a couple, being drawn perpendicular to the plane of the couple, is called "the axis of the couple" And it is said to be the jpositive, or the negative axis, ac- cording as the perpendicular stands on the positive, or on the negative face of the couple's plane. If the axis of a couple is mentioned without its being stated whether it is positive or negative, we are to understand that the positive axis is alluded to. The angle between the planes of two couples is measured by the angle between their positive axes. 48. The effect of a couple acting on a rigid body is not altered hy twrning its a/rm through any angle in the plane of the couple. Let F and F' be the equal forces of a couple acting at two points in the lines FA, F'B (fig. 10), and having the arm AB. From A, any point in the line in which F acts, draw in an arbitrary direction in the plane of the couple AB' equal to AB; 24 THE THEORY OP COUPLES. and at A, B' in the plane of the couple F, F', and in directions at right angles to AB' apply two pairs of opposite fotces /, g; f, g': each force being equal to JP'or F'. These being in equi- librium, produce no effect. Bg' and F'B^W intersect each other in some point G; join AG. Then because AB==AB', and AG is common to the triangles ABO, AB'G, and the aB= aB'; therefore ^C bisects the angles^ GS', BAB': hence the resultant of the two equal forces F' g" which we may suppose to act at G lies in AC produced ; and that of the equal forces F, g lies in GA pro- duced. But, because AF is parallel to GF' and Ag parallel to Of, therefore the iFAg= iF'Gg'; and consequently the re- sultant of the forces F', g" is equal to thait of the forces F, g : we have just proved also that they act in opposite directions ; therefore, the four forces i?; ^, i^', / balance each other, and may be removed. There is then left only the couple/,/', which is the same as if the arm of the original couple had been turned through the arbitrary angle BAB'. 49. The effect of a cowple acting on a rigid body is not altered hy removing it to any other part of the rigid hody\ pro- vided its new plane he parallel to its original ^Icme (fig. 11). Let F, F' be a couple acting upon a rigid body in the plane HH; let AB be its arm. Let KK be any other plane, in the rigid body, parallel to HH; and in this plane draw the line ah, parallel and equal to AB. At a and h apply pairs of opposite iovcea f g; f, g': each force being equal and parallel to the forces F, F'. These pairs of forces balance each other, and therefore produce no effect on the rigid body. Draw Ah, Ba ; they, being in the plane which contains AB and ah, necessarily intersect in some point 0. In fact, if A, a, and B, h were joined, AahB would be a parallelogram, and therefore Ah, Ba, being its diagonals, bisect each other in 0. Draw PGQ parallel to AF. Then because F=: g', and & C = ^ 0, ' ' .-. F : g' :: hC : AC. Hence Fani g! (by Art. 40) have a resultant F+g', which acts at in the line GP. Similarly it may be shewn, that F' THE THKOKY OF COUPLES. 25 and g have a resultant, F' -\- g acting at C in the line CQ. Now F+g' = F' +g and CP is opposite to CQ, therefore the four forces F, g", F', g balance each other, and may he removed. There remains then only the couple f, f, which is the same as if the original couple had been removed into the new plane KK, retaining its arm 06 parallel to -^5; but we may now (by Art. 48) turn the arm ab through any angle without altering the effect of the couple. And hence the effect of a couple, &c. 50. The effect of a couple acting on a rigid hody is not altered hy removing it to any other part of the rigid hody in its own plane. The demonstration in the last Article will serve for this, using fig. 12 instead of fig. 11. 51. A Gowple acting on a rigid hody may he changed for any other cowple acting upon the same rigid hody, provided the moments of the two couples he equal, their planes parallel, and they he both of the same hind, i. e. hoth positive or hoth negative. Let HE (fig. 11) be the plane of the couple F, F' ; and in any other plane KK oi the rigid body draw, parallel to AB, a line cd> of any proposed length : at a, h apply pairs of equal and opposite forces/, g; f, g'; of such magnitude that F.AB=f.ab, these balance each other, and therefore produce no effect. Now AB and ah being parallel, the lines Ah, Ba lie in one plane, and intersect in some point C: and because AB i^ parallel to ah, the ^ GAB= ^ Cha, and the ^ CBA = ^ Cab ; consequently the two triangles AGB, hca are similar. Now because /=^' .-. F : g' :: ab : AB :: Ch: CA; therefore, by Art. 40, the resultant F+ g' of the two forces F, gf acts at C in the direction GP. In a similar way it may be shewn that F' +g, the resultant of F', g, acts at C in the direc- tion CQ. Now F= F' and g'=g, and therefore F^g'=-F'^g; consequently the four forces F, cf, F', g are in equilibrium, and may be removed; which being done, the original couple is E. s. -4 26 THE THEORY OF COUPLES. replaced Iby the equivalent couple f, f whose arm is ah. This couple/,/' may now he turned through any angle in the plane KK, and thus the proposition is established. 52. Any number of couples act upon a rigid body in the same plane, or in parallel planes ; to find their resultant. Change all the couples into others equivalent to them, and therefore of the same moment, but all having their arms of the length h. Then if F^,F^, F^...F^ be the forces; and a,, a^, a^...a^ the arms of the original couples; and P^, P^, P^,...Pn the forces of the corresponding equivalent couples, we shall have PJ>=F^a„ P,b = F^a^,...PJ = F„a,. Now since the new. couples act in parallel planes and have equal arms, they may be removed into the same plane, and then turned round and transposed so as to make all their equal arms exactly coincide; in which position the system of couples is reduced to one couple, the arm of which is b, and the forces of which are equal to F, + P, + P,+ ...+P^. Hence the moment of the resultant couple = (P, + P, + P3+...+P„).J = F,a, + F^a, + F^a,+ ... + F„a„ = the sum of the moments of the original couples. Whence, the moment of the resultant couple is equal to the sum of the moments of the original couples. The reader will be careful to remark, that if any of the couples are of a negative character, their moments are to be accounted negative in taking this sum. 53. COE. If all the n couples be equal, the moment of their resultant couple is n times the moment of one of them; THE THEORY OP COUPLES. 27 and as the effect of n equal couples must he n times the effect of one of them, it follows that the moment of a couple is a proper measure of its effect in producing or destroying equilibrium. Whenever, therefore, we have occasion to speak of the magnitude of a couple, we shall do so by stating its moment ; thus, the couple G will signify the couple whose moment is G. It will, lead to no inconvenience that the magnitude of the forces which compose the couple is not stated, seeing that the effects of all couples of equal moments acting in the same plane, whatever be the magnitudes and directions of their forces, are the same. It will also be observed, that it is not necessary to state the precise plane in which a couple is situated ; it will be sufficient to know its moment, and the position of some line to which its axis is parallel. 54. It will be observed, that all equivalent couples have their axes equal and parallel. 55. If from a point two straight lines be drawn parallel and equal to the axes of two couples, and upon them a parallelogram be described, the diagonal drawn from the same point will be parallel and equal to the axis of the resultant couple. (This pro- position is usually cited as the parallelogram of couples.) (fig. 13). As the planes {SO A, HOB, suppose) of the couples are not parallel, let them intersect in the line HO, Change the couples into two equivalent couples having their forces FF', ff' all equal; place these new couples so that one extremity of their arms OA, OB shall be at 0, and the forces F, f which act there, shall act in the line OH, as in the figure. Complete the parallelogram OADB, and draw the diagonals OD, AB, bisecting each other in C. Then because F' and /' are equal and act in the same direction, they are equivalent to a resultant F' +f' acting at" G (Art. 40). But such a force at G would likewise be the resultant of the same forces F', f acting at B, 0. We may therefore transpose F' to B, and /' to 0, which being dtone, /' and f s.t balancing may be removed ; and there will only remain Fat and F' at B, forming a couple F, F' whose arm is OB, which is therefore the resultant of the two original couples. 28 THE THEORY OP COUPLES. Now the forces of the two component couples and of their resultant being equal, their axes which are proportional to their moments, are in this case proportional to their arms OA, OB, OD; we may therefore consider OA, OB, OB as being equal to the axes. If therefore from in~ the plane OBA, we draw ■ three lines respectively perpendicular and equal to OA, OB, OB, they will be the axes of the three couples, and will then have the same position as the lines OA, OB, OB would take if the parallelogram OABB were turned through a right angle about the fixed line OE. This figure OABB so turned is the paral- lelogram stated in the enunciation of the proposition to possess the property which we have just proved belongs to it. 56. Two couples act upon a rigid hody in planes which are at right angles to each other ; to find their resultant. From any point 0, draw OA, OB equal and parallel to the axes of the two couples. Complete the rectangle OACB, and draw its. diagonal. OG. By the last article 00 is equal and parallel to the axis of the resultant couple. . Let L, M, G be the moments of the two component couples and of their resultant. 6 = GO A the angle at which the axis of G is inclined to that of L. Then because OA=OG cos 0, and OB=OG sind; .'. L= G cos 6, and M= G sin 6, from which we find G' = L'+M', and tan ^ = -f- ; which equations determine both the magnitude of the resultant couple, and the position of its axis. 57. If it should be required to resolve a given couple whose moment is G into two components acting in planes at right angles to each other, we must use the equations L=Gcoa9, M=Gs[n0. PAEALLEL FORCES IN A PLANE. 29 58. Any number of forces act on a rigid body in parallel directions in one plane at different points of the body; to find their Let F^, F^... F^hQ the forces ; from any point (fig. 14) of the rigid hody in the plane of the forces draw a line cutting their directions perpendicularly in the points A, B ...H; and put OA = x^, OB=x^,...OH=x,. At apply two opposite forces each equal and parallel to F^ ; they do not afiect the system. In the same way apply at a pair of forces for each of the remaining forces F^, F^...F„. By this means we have n forces acting at in the direction OR, respectively equal to F^, F^, ... F^; these are equivalent to a single force B, acting at in the direction OB, and B = F^ + F^+... + F^^t.F. We have, besides, n couples whose arms are a;,, x^...x„, which are (by' Art. 52) equivalent to a single couple, whose moment = F^x^ + F^x^+...+F^x^ = %.Fx. Consequently the given system of forces is equivalent to a single force % . F acting at in the direction OB ; and a negative couple whose moment is S . Fx. 59. The result just obtained is perfectly general, but admits of simplification except in the particular case when 'St .F=0. (1) In the particular case when "t.F^O, there is no re- sultant force acting at 0, and therefore the only resultant is the couple whose moment is "Z . Fx. (2) When "Z .Fia not =0, change the couple whose moment = 't.Fx into an equivalent couple which has its forces B', B" equal to ^ ox't.F, and place it so that its arm OK (fig. 14) shall coincide with the line OH; .-. f^.F).0K^t.Fx (Art. 50.) 30 PARALLEL FORCES IN A PLANE. By the arrangement the force i? at is balanced -by B" one of the forces of our new couple ; and these being removed, there remains only the force B' = % .F&t the point Z" determined by the equation t.Fx 0K-- t.F Consequently, except when S . i^= 0, the resultant is a single force equal Xo%iF acting at the point just found. 60. COE. If the line OH instead of cutting the directions of the forces at right angles, should cut them in an i a, we should have found that (1) "When % . F= 0, the resultant is a couple whose moment is {% . Fx) sin a : and (2) When S . i^ is not = 0, the resultant is a single force S . F acting at the point determined by the same equation as before, viz. t.Fx 0K= t.F 61. Any numh&r of parallel fc/rcm act upon a rigid lody in one plane at different points of the lody ; to find the conditions that they may balance each other. Let the system of forces be that of Art. 58 ; then we have to consider the two cases pointed out in the last Article. In the second case the resultant is the force S . ^acting at K; and there cannot be equilibrium unless this force vanish, or % . F= 0, But if this be the case, the second case coincides with the first ; and the resultant is a couple whose moment = % . Fx : there cannot be equilibrium therefore unless this couple also vanish. Conse- quently the conditions of equilibrium are t.F=0, and.t.Fx = 0; these are both necessary and sufficient for equilibrium. They are necessary, for if the former only be satisfied, there will exist the couple of Case 1 ; and if the latter only be satisfied, there will exist the resultant force acting at 0. And they are NON-PARALLEL FORCES IN A PLANE. 31 sufficient, for they secure that there shall exist neither the re- sultant of Case 1, nor that of Case 2. 62. Def. The products F^x^, F^x^, ... F^x^ are called the moments of the forces F^, F^, ... F^ about the point ; they are also called the moments of the same forces about an axis passing through at right angles to the plane of the forces. Hence remembering that the point was arbitrarily chosen in the plane of the forces, the two conditions of. equilibrium of parallel forces acting on a rigid body in one plane may be thus enunciated in words : — The algebraic sum of the forces, and the sum of the moments of the forces about any point in the plane of the forces or about any axis perpendicular to the plane of the forces, must be each equal to zero. 63. Suppose that there is in the plane of the forces a fixed point, or in the body a fixed axis not parallel to the plane of the forces: to find the conditions of equilibrium. If there be a fixed point In the plane of the forces, let that .point be taken for 0; or if there be a fixed axis it will cut the plane of the forces in a point, which take for ; then the in- vestigations of Art. 58 apply here. The force % . F which acts at 0, can produce no effect since it acts on an immoTeable point ; it is not necessary therefore that % . F should be = 0. But the couple whose moment = S . Fx, if it exist, will turn the body about 0; and therefore that there may be equilibrium, it is necessary and suflScient that t.Fx = 0; hence there is only one condition of equilibrium in this case, which we may thus express in words : — The sum of the moments of the forces about the fixed point, or about that point where the fixed axis cuts the plane of the forces, must be equal to zero. Remark. When there is in the plane of the forces a fixed point, and the forces are in equilibrium, the pressure on the fixed point = 'tF, which is the same as if every force were transposed to that point, without altering the direction in which it acts. 32 NON-PAKALLEL FORCES IN A PLANE. 64. Any number of forces act upon a rigid body in one plane, at different points of the body and in directions not necessarily parallel; to fnd their resultant. Let-i?;, F^,...F„loe the forces, and in the plane in which they act, from a point 0, arbitrarily chosen, draw any two lines Ox, Oy at right angles to each other. To these lines as co-ordinate axes refer the given forces and their points of application (fig 15). Let «!, a^, ... a„ be the inclinations of the lines in which the forces act to Ox; x^y^, x^^, ... x^y^ the co-ordinates of the points of application of the forces; if P be that of jfj, then a!j= OM, y^=PM. From the point draw OQ perpendicular to F^ P; and at apply two opposite forces F', F" each equal and parallel to F^. By this means we have a force F' acting at 0, and a couple {F^F",) whose moment is equal ^la -F^.OQ. Or we may say that the force F^ may be transposed to without altering its direction, if at the same time we also apply to the body a couple whose moment = -F,.OQ = - j; . {ON- QN), since MNis parallel to PQ, = -F^. {x^ cos HON- y^ sin MPQ) = — F^. (iCj sin ttj — y^ cos aj = {F^ cos aj . y^ - {F^ sin aj . x^ ; or =X^y^- Y,x^, if we put X, Fj for the components of F^ parallel to the co-ordinate axes Ox, Oy. The same method being applied in succession to each one of the remaining forces of the system, we shall have transposed all the forces to 0, each preserving its original direction; but there will be acting on the body besides them a number of couples in one plane whose moments are ^1^1 - i^i^^i. -3;^,- Y^x^ ... X„y„ - F„a;„. NON-PARALLEL FORCES IN A PLANE. 33 If Q be the resultant of the couples, and R the resultant of the forces at 0, we shall have = t[Xy-Yx)...Kxi.b% and I^=i^.F cos a)" +{t.F sin a)^ . . Art, 31, = {t.Xf+{t.Yf % Y and tan0 = =-^, 6 being the inclination of the line in which B acts to Ox. 65. The result just obtained is perfectly general, but it can be simplified, being reducible to a single, resultant, except when R = Q, i. e. except when SX= and X Y= 0. For, (1) WhenSX=0 and SY=0, there is no resultant force acting at 0, and the only resultant is the couple whose moment = {Xy — Yx). (2) When the two equations SX= 0, and X Y= are not both satisfied, change the couple whose moment is 2 {Xt/ — Yx) into an equivalent couple which has each of. its forces B' It" equal to B, and place it so that one end of its arm OK (fig. 16) shall be at 0, and one of its forces {B") exactly opposite to B. R and B" balance each other and may be removed ; and there remains only the force J?' acting at the point T such that OT.CQ^e=OK; .: B. OT.cose = B. OK, or, since ^ cos ^ = tX (Art. 31), and B . 0K= t [Xy - Yx), t{Xy-Yx) ^^~ tx • Consequently when the two equations 'tX=(i, 2F=0are not both satisfied, the resultant is a single force B acting at the point just found, or at any point in the line KB'. Eemark. From this it appears that when non-parallel forces, acting in one plane on a rigid body, admit of a single E. s. 5 34 NON-PAEALLEL FORCES IN A PLANE. resultant, there is a certain line to any point of which all the forces admit of heing transposed (each force retaining its original direction) without their effect being in any respect altered. This line is EK, and we shall shew in the following Article how. its equation may be found, 66. When the forces of Art. 64 are reducible to a single resultant force, to find the equation of the line in which it acts. • Let x' y be the co-ordinates of any point in the line UK (fig. 16) in which the resultant acts. Then because this line passes through the point T and, being parallel to OR, makes an angle Q with the axis Ox, its equation is y'-OT=t^-a.e.x, , t{Xy-Yx) _tY , ^^ y ~ tx ~sz"'^' or y'tX-x'tY=t{Xy - Yx). 67. Any number of forces act on a rigid body in one plane at different points of the body; to find the conditions that they may balance each other. Let the system of forces be that of Art. 64, then we have to consider the two cases of Art. 65. In the second of these cases the resultant is the force R' (= R) acting at T; there cannot be equilibrium unless this force vanish, or ^ = 0. But if this be the case, the second case coincides with the first ; and the re- sultant is a couple whose moment = % {Xy — Yx)'] there cannot be equilibrium unless this couple also vanish. Consequently the conditions of equilibrium are XX=0, tY=0, S(Xy-Fa;)=0. These three conditions are both necessary and sufficient. By referring to Art. 64 we perceive that X,y^ — F,a5, is equal to the moment of i^, about the point 0, consequently ;S (Xy— Yx) is equal to the sum of the moments of all the forces about 0. If then we remember that the point 0, and the directions of the axes Ox, Oy, were arbitrarily chosen in the plane of the forces, we may enunciate the conditions of equilibrium as follows : NON-PARALLEL FORCES IN A PLANE. 35 The algebraic sums of the components of the forces parallel to any two lines at right angles to each other in the plane of the forces must he each equal to zero ; and the sum of the moments of all the forces about any point in the plane of the forces, or about any axis at right angles to the plane of the forces^ must also be equal to zero. 68. Suppose that there is in the plane of the forces a fixed point, or perpendicular to the plane of the forces a fixed axis ; to find the conditions of equilibrium. Let the fixed point, or the point where the fixed axis cuts the plane of the forces, be taken for the point in the investi- gation of Art. 64. Then the force R which acts at can pro- duce no effect since it acts on an immoveable point, it is not necessary then that R should "he equal to zero. But the couple whose moment is % i^y ~ ^a;), if it exist, will turn the body about 0, and therefore that there may be equilibrium it is neces- sary and sufficient that t{Xy-Yx)=0. There is therefore in this case only one necessary condition of equilibrium, viz ; — TTiat the sum of the moments of all the forces about the fixed point or axis should be equal to zero. Eemaek. When there is equilibrium, that is, when the above condition is satisfied, the pressure on the fixed point is due entirely to the force R which acts directly upon it. Hence the pressure on the fixed point is the same as if all the forces which act on the body were transposed to the fixed point with- out altering their directions. CHAPTEE IV. on forces, not in one plane, which act upon d;iffbebnt points of a rigid body. 69. If the directions of the forces all pass through a point we may transfer them to that point, and find their resultant by Chapter I. or II. 70. In the present Chapter we shall meet with couples of which the planes are not parallel. We can however always reduce them to other couples in the planes of rectangular co- ordinates. It is necessary therefore only to observe that when a couple acts in a co-ordinate plane, it wiU be considered a posi- tive couple when its axis stands on the positive side of that plane. Thus a positive couple in the plane yz has its axis coinciding with + Ox, xz + Oy, xy + Oz. 71. Parallel forces not in one plane act, on different points of a rigid hody; to find their resultant. Take any point (fig. 17) in the rigid body, fi-om which draw Oz parallel to the direction of the proposed forces, which take for axis of z. Draw Ox, Oy in any directions at right angles to each other and to Oz, and take them for the axes of x and y. Let Z^, Z^...Z^ be the forces ; P the point where the line in which Z^ acts cuts the plane xy. OM=x^, MP=y^ the co-ordinates of P. Complete the parallelogram OMPN and join OP. At the point apply two pairs of opposite forces Z', Z" each equal and parallel to Z^ ; these do not afiect the system. PARALLEL FORCES NOT IN A PLANE. 37 Now it has been shewn in Art. 55, that when two equal parallel forces act in the same direction at the extremities of one of the diagonals of a parallelogram, they may be transposed to the extremities of the other diagonal. Let us on this principle transpose Z^ to M, and Z' to N. We have then, one, force Z' acting at in the direction Oa, and two couples in the plane xz, yz whose arms are OM, ON, the former couple being nega- tive. By this means we have transposed the force Z^ to 0, retaining its proper direction, and have introduced the couples — Z^x^, +Z^yi in the planes xz, yz respectively. Proceeding in the same manner with the remaining forces ^, ^...^, we shall have, instead of the original system, the forces Z^, Z^,...Z^ acting in the line Os, which (by Art. 23) have a resultant R^t.Z.:. (1); and, in the plane xz, a set of negative couples, which (by Art. 52) are equivalent to a single couple in that plane whose moment =^-t{Zx); and, in the plane yz, a set of positive couples which are equi- valent to one whose moment = t{Zy). If C be the moment of the resultant of these two couples, and 9 the angle which its arm makes with Ox, we shall have from Art. 55, G.cose = t{Zx), and O . %va.e = t {Zy) ; .-. CP={t.Zxy+{t.ZyY (2), andtan^ = i-^ (3), Equation (1) gives the resultant force acting parallel to the original forces at the origin of co-ordinates ; and equations (2) (3) give the magnitude of the resultant couple, and the position of its plane. 72. We have determined the position of the arm of the resultant couple. That result supposes, as in fig. 17, a negative force acting at that extremity of the arm which is at 0, and a 38 PARALLEL FORCES NOT IN A PLANE. positiye force at the other extremity. It will sometimes be more convenient to know the position of the positive axis of the couple. Let a be the inclinations of this axis to the axis of x. Then t.Zy , . t.Zx cos a = — Yr~ > ^^^ sm a = ^y— . In which equations G is to be accounted positive. 73. The results obtained in Art. 71 are perfectly general, but they admit of reduction to a single force except when M or %Z=0. (1) When 'tZ= 0, there is no force acting at 0, and the only resultant is the couple whose moment is G. (2) When "tZ is not equal to zero, change the couple whose moment is G into an equivalent couple which has each of its forces B'B" equal to i? or %Z; its arm will be equal to ,-^; place this couple as in fig.. 18, so that one of its forces B" balances the resultant B. By this mode the whole are reduced to a single force B' {=%Z) acting at a point P whose co-ordinates x'y are known from the equation x' = OPcos e, and y' = OP sin 6 G „ G . a t.Zx _ t.Zy ~ %Z ~^ZZ~' These equations are free from ambiguity. 74. To find the conditions that the system of forces in Art. 71 may balance each other. We must consider the two cases mentioned in the last Arti- cle. In the second of these cases the resultant is the force "ZZ acting at a point whose co-ordinates are S. Z x , % .\Z/ PARALLEL FOllCES NOT IN A PLANE. 39 There can be no equilibrium as long as this force exists ; we must therefore have %Z= 0. But if this be the case the 2nd case coincides with the first; so that the resultant is a couple whose moment is G. There cannot be equilibrium there- fore unless (r = ; an equation which is equivalent to S . ^a; = and 'Z .Zy = Q. Hence the conditions of equilibrium are tZ=0, %.Zx = 0, %.Zy = 0; which three conditions are both necessary and sufficient. 75. General definition of "the moment of a force about a line^ If the direction of the force be perpendicular to the given line, the moment is equal to the product of the force into the length of a line which is perpendicular hoth to the line in which the force acts and the line about which the moment is required. If the direction of the force be not perpendicular to the given line, it must be resolved into two components, one perpendicular _ and the other parallel to the given line; the moment of the former will be found by the definition just given, and that of the latter will be zero. 76. According to this definition Z^y^ and Z^x^ are the moments of Z^ about the axes of x and y respectively; and hence we may state the three conditions of equilibrium of parallel forces acting on a rigid body as follows : The sum of all the forces must he equal to zero ; and the sums of their moments about any two lines at right angles to each other in a plane which is perpendicular to the direction of the forces must he respectively equal to zero. 77. To find the conditions of equilibrium of the forces in Art. 71, when there is in the hody a fixed point; or a fixed line at right angles to the direction of the forces. (1) When there is a fixed point. Let it be taken for the point in Art. 71 ; then, as by this an-angement R acts upon an immoveable poiijt, it is not 40 PARALLEL FORCES NOT IN A PLANE. necessary that %Z should be = ; but as the couple O would turn the body round it is necessary and sufficient for equili- brium that G — Q, or that %{Zx)=Q, and 2(%)=0. That is ; the sums of the moments of the forces about any two lines drawn from the fixed ^oint at right angles to each other, in a ^lane perpendicular to the direction of the forces, must be separafeli/ equal to zero. ^ Remark. In this case, i. e. when there is equilibrium, the pressure on the fixed point is S^ acting directly upon it; i. e. it is the same as if all the forces were transposed to that point without altering the direction in which they act. (2) When there is a fixed line in the body, at right angles to the direction in which the forces act. Let it be taken for the axis Oy in Art. 71, being any point in it. Then since the force B acts upon a fixed point it is not necessary for equilibrium that it should be = 0. Also the couple O is equivalent to the two S {Zx), ^ {Zy) : the latter of which being in the plane yz can be so placed that its forces shall both act upon points in the line Oy, which being immove- able, it Is not necessary that this couple should be equal to zero. The remaining couple % (Zx) tends to turn the body about the ^ fixed line Oy, so that there cannot be equilibrium as long as it exists. Wherefore the only condition which in this case is necessary and sufficient for equilibrium is t{Zx) = 0, that is, the sum of the moments of all the forces about the fixed line must be equal to zero. Eemark. In this case the pressure on the fixed axis is equivalent to the force XZ at 0, and the couple S {Zy) ; as in Art. 59 (2) these are equivalent to a single force "tiZ acting at a point K in Oy such that 0K= %z RESULTANT OF THREE COUPLES. 41 The force %Z acting at this point represents the pressure on the axis. But the pressure may be otherwise represented, for a com- parison of the equation 0K==-^1 with the result found in Art. 59 shews that the pressure on the axis is just the same as if the forces Z^Z^Z^ . . . were transposed to the fixed axis and applied without changing their directions to points in the axis at the respective distances i/^i/^, ... from 0: that is, if through every force we draw planes at right angles to the fixed axis we may transpose each force without altering its direction to the point where the corresponding plane cuts the axis. The forces thus transposed produce the same pressure on the axis as the given system. 78. To find the resultant of three couples which act upon a rigid hody in different planes, no two of which are parallel. From any point (fig. 6) draw three lines OA, OB, OG to represent the axes of the couples, the moments of which are L, M, N; complete the parallelepiped, and join OD, OF. Then the couple whose axis is OD is equivalent to L, M whose axes are OA, OB: and OE is the axis of a couple which is equi- valent to 00, OD; i. e. to the three couples L, M, K 79. To find the resultant of three couples L, M, N whose planes are mutually at right angles. From any point (fig. 7) take QL, OM, ON to represent the axes of the given couples. Then as before we may shew that OP represents the axis of the resultant couple O. Let a, j8, 7 be the angles POL, POM, POiV between the axis of G and the axes of L, M, N. Then since OL = OPcos a, 0M= OP cos ^, 0N= OP cos 7 ; .•.L=G cos a, M=G cos /3, N=G cosy, and .'.L' + M''+N'=G\ From which four equations the magnitude and position of the resultant couple are known. E. s. 6 42 ANY FORCES ACTING ON A EIGID BODT. 80. By means of the four equations just given we may- resolve a couple into three components acting in planes at right angles to ei>ch other. 81. To find the resultants of any forces, acting on different points of a rigid body, in lines which are neither parallel nor in one plane. Take any point (fig. 17) of the rigid hody as origin, and from it draw any three lines perpendicular to each other for axes of co-ordinates. Let x^y^s^, x^y^s^, ...be the co-ordinates of the points- at which the forces act; resolve each force into three components parallel to Ox, Oy, Oz. Denote the components parallel to a; by X^, X^, X^... 2^ by Y„ Y„ F,... ■ « by Z„ Z„ Z,... The resultants of the last set of forces we have already found (in Art. 71) to be a force %Z acting at in the line Oz, a couple S {Zy) acting in the plane yz, and a couple — S {Zx) acting in the plane xz. The forces FjFgFj ...form a system of parallel forces, of which the resultants may be deduced from those of ^^^... by writing Y, x, z for Z, y, x respectively : they are therefore equi- valent to a force % Y acting at in the line Oy, a couple S ( Yx) acting in the plane xy, and a couple — % ( Yz) acting in the plane zy. And in these, writing X, z, y for Y, x, z we find the forqes Xj, Xjj, Xj ... equivalent to a force 1X acting at in the line Ox, a couple % [Xz) acting in the plane zx, and a couple — 2 {Xy) acting in the plane yx. ANY FORCES ACTING ON A RIGID BODY. 43 Collecting these results it appears that the original forces are equivalent to tX, tY, SZ acting at 0; and the three couples t{Yx)-t {Xy) =^{Yx-Xy) in the plane xy, t {Xz) - t (Zx) =t{Xz- Zx) in the plane zx, and t{Zy)-t{Yz)=t {Zy - Yz) in the plane yz. Now if R he the resultant of the forces acting at 0, and a, /3, 7 the angles which the line in which it acts makes with Ox, Oy, Oz ; and if G he the resultant of the couples, and a', /S', 7' the angles which its axis makes with Ox, Oy, Oz, we have hy Arts. 34, 79, i?cosa = SX, Bcos^ = tY, Ecoay = tZ, R={tXy+{tY)' + {tZ)': and G cos a' = S {Zy — Yz) = L suppose aco5p' = t{Xz-^Zx)=M Gcosy' = %{Yx-Xy) = N G^ = L^ + M' + Jsr. These eight equations give both the magnitude and direction of the resultant force which acts at the origin of co-ordinates ; and the magnitude and position of the axis of the resultant couple. These results are quite general, but we shall now shew that under certain conditions the original forces admit of a single resultant. 82. To jind the condition that the forces in Art. 81 may admit of a single resultant, and to find the magnitude and position of it. If (? be = 0, no reduction is necessary ; but if not, change the couple G into an equivalent couple, whose forces R'R" are each equal to R; place this couple so that one of its forces (as R') shall act at 0, and if possible in a direction opposite to R ; in this case R and R' balance each other and may be re- moved ; there will then be left only the force R", which is the same as if the force R had been transposed to R", and the couple 44 ANY FORCES ACTING ON A EIGID BODY. taken away. It appears, then, that a couple and a force are reduced to a single force (where the problem is possible) by ' taking away the couple and transposing the force to some other point. The possibility of being able to do this, depends on its being possible to place the forces R and E in the same line. The student will perceive that this can be done only when the force B acts in a line which is perpendicular to the axis of the couple Q, the analytical condition of which is = cos a cos a' + cos /3 cos j8' + cos 7 cos 7' _tX L tYM tZN ~ R -a'^ B 'G^ B •(?• Hence the conditions required are (1) if must not be = ; and (2) LtX-ir Mt Y+ NtZ must = 0. We have yet to find the line in which the resultant force B" acts. We have remarked above, that B" is the force B transposed without altering its magnitude or direction. If we had begun the investigations of Art. 81 by taking a point in B" for the origin of co-ordinates, we should have found B" acting at that origin, and no resultant couple ; that is, denoting the co-ordi- nates referred to this origin by x", y", z" we should have found = S (X»" - Zsl'), and = S(ra;"-Z/); these are in fact the conditions that the origin may be a point in the single resultant force. Let a?', y, z'- be the co-ordinates of this origin referred to the original origin, x, y, z being the same as before ; then x'= x - x', y" ~y- y', z" = z- z', which being substituted in the above equations give ANY FOllCES ACTING ON A RIGID BODY. 45 y'%Z- z't Y=t{Zy- Yz) == L, z'tX- x'tZ = t{Xz~ Zx) = M, x'tY-y'tX=t{Yx-Xy)=N, x, y', z are the co-ordinates of any point in the line in which i2" acts. There being three, equations between these quantities, it would seem as if there existed only a single point at which ^".can be applied, which is contrary to ^rt. 21: but if we multiply these equations by %X, % Y, "ZZ and add the results we shall find = LtX + MtY+NtZ, which being satisfied by hypothesis, the three equations are not independent, but any one is derivable from the other two. Con- sequently any two of these are the equations of the line in which the single resultant acts. 83. When the farces in Art. 81 do not admit of being reduced to a single force, they can be reduced to a force and a couple the axis of which is parallel to the force. For let <^ be the angle between the axis of the couple G in Art. 81, and the force B. Kesolve G into two components G cos ^, G sin ^ whose axes are respectively parallel and per- pendicular to R. The latter of these, being compounded with B as in the last Article, will be destroyed, ■ and R will be trans- posed to some other point of the rigid body, without altering its direction ; it is therefore still parallel to the axis of the couple whose moment = G cos <\) = G (cos a. cos a' -H cos /3 cos /3' + cos 7 cos 7') LtX + MtY+NtZ B This appears to be the simplest form to which the forces in Art, 81 are in general reducible. They may however be pre- sented in another simple form as in the following Article. 84. The forces in Art. 81 can he reduced to two forces acting in two Unes which in general do not meet; and to find the shortest distance between these lines, 46 ANY FORCES ACTING ON A EIGID BODY. Let them be reduced as in the last Article to a force E and the couple Gf cos (f). Let Q (fig. 19) be the point at which R acts ; and let the couple G CO? be placed so that one of its forces K acts at Q, PQ being its arm. Then QR being parallel to the axis of the couple is perpendicular to QK; hence li H\>& the resultant of R and K, and -4^ be the angle HQK, the forces are now reduced to K at P, and ^at Q, such that „ „ (tCOS^ H sin yjr = R ; , , , R.PQ and tan vr = -79 ; . ^ - 6r cos 9 Now PQ being at right angles both to QH and Pff" is the minimum distance between them. It appears from the above equations that PQ is arbitrary ; but when it is of given length then both K and JI are known, and their relative position from the last- equation. Q is known by the preceding Article. 85. To find the equations of the line in which E, acts, and of the plane in which G acts, in Art. 81. Since R passes through the origin its equations are t t I X y ^ z cos a cos /3 cos 7 ' ^ _ y' _ 2' Again, we may suppose the plane of G to pass through the origin. And since a', /3', 7' are the angles which a perpendicular upon it makes with the co-ordinate axes, its equation is X cos a' + y' cos /8' + «' cos 7=0; ANY POECES ACTING ON A EIGID BODY. 47 .-. Lx +My' + Nz'=(i is the equation to the plane in which G acts. 86. The conditions that the plane of O may be perpen- dicular to the line in which R acts uic ^r^= =rT^==r^. 87. To find the equations of the line in which the resultant force acts when the resultant couple acts in a plane at right angles to it. (Art. 83). Let 0' he any point in the line in which the resultant acts in this case ; x, y, z' its co-ordinates referred to the origin used in Art. 81. If with the origin 0' we were to proceed as in Art. 81, we should find a resultant R acting at 0', and a couple G', the plane of which would he found to be perpendicular to the direction of R ; and therefore L' M' N' where L', M', N' represent the quantities t {Zy" - Yz"), X iXz" - Zx"), t ( Yx" - Xy"), and x", y", z" are the co-ordinates of a point referred to the origin 0': hence x" = x — x', y" = y—y', z" = z — z', as in Art. 82 ; t{Z{y-y')-Y{z-z')} _ t{X{z-z')-Z{x-x')} '• XT' 2F t\Y{x-d)-X{y-y')] - %Z ' or bringing x, y', z' outside of the symbol S, and writing L, M, Niox their equals, the equations required are z ,tY ,%Z^ L _,tZ ,tX M_ %x~y %x^^X~ sr"'' XY'^tY ,%X .tY N 48 ANT FOECES ACTING ON A RIGID BODY. 88. In Art. 83 we were able, by transposing B to destroy the couple G sin ^. If afterwards we transpose B to some other point, we shall thereby introduce a new couple, the axis of which being at right angles to the axis of the couple G cos would be compounded with it, and make a resultant couple greater than G cos ^. ' Hence to whatever point B be transposed, the resultant couple will always be greater than in Art. 83. Consequently the resultant couple is a minimum when its axis is parallel to the resultant force. This is sometimes called the principal couple. 89. Def. The line in which B acts when the resultant couple is a minimum, is called the central axis. Its equations are found above in Art. 87. 90. If B be transposed from the central axis to a distance a from it, a couple is thereby introduced whose arm is a and moment Ba ; consequently the resultant couple for this position of B is ViiV + G^ cos^ 0, which is constant as long as a is con- stant. Hence if we construct a cylindrical surface having the central axis for its axis, the surface of this cylinder will be the locus of the origins, which will give equal resultant couples. 91. To find the conditions that the forces in Art. 81 inay balance each other when the iody upon which they act is free. (1) Suppose the direction of B to be not parallel to the plane in which G acts ; then since E and G cannot in this case be reduced to a single force, they must be separately equal to zero ; which are equivalent to the six following : = tX, O^tY, = -ZZ, = L, = M, = K ANY FORCES ACTING ON A RIGID BODY. 49 (2) Suppose the direction of R to te parallel to the plane in which Q acts; then R and G can be reduced to a single force, the effect of which reduction is to transpose R and de- stroy G. There can therefore he no equilibrium unless R = 0; it is necessary therefore that R should be = 0. But if J? = 0, R and G cannot be reduced to a single force ; that is, G cannot be destroyed by transposing iJ; it is therefore also necessary that G should separately be = 0. Hence the conditions of equi- librium are the same in this as the preceding case; and, ob- serving that i, M, N are the moments of the forces about the lines Ox, Oy, Oz, we may thus state them in words : The sums of the resolved parts of the forces parallel to any three lines at right angles to each other must he separately equal to zero ; and the sums of the moments of the forces, about any three lines at right angles to each other and passing through a point, must be separately equal to zero. 92. To find the conditions that the forces in Art. 81 may balance each other when one point of the rigid body is fixed. Let this point be taken for the point in Art. 81. Then since by this arrangement R acts upon a fixed point, it is not necessary for equilibrium that R should vanish ; but as the couple G would turn the body about this point, it is necessary and sufficient for equilibrium that G' be = ; that is, that L = 0, M= 0, N= 0. Or, in words : The sums of the moments of the forces, about any three lines at right angles to each other passing through the fiooed point, must be separately equal to zero. Eemark. The pressure on the fixed point is represented by R acting directly upon it : i. e. it is the same as if all the forces of the system were transposed to the fixed point without changing their directions. 93. To find the conditions that the forces in Art. 81 may balance each other vjhen there is in the body a fixed axis. E. s. 7 50 ANT FORCES ACTING ON A RIGID BODY. Let the fixed axis be taken as the axis of z in Art. 81, and any point in it as the point ; then since B, acts upon a fixed line it is not necessary for equilibrium that B, should be equal to zero ; also the couples L, M, acting in the planes yz, xz, can be turned round and so placed that their forces shall all act upon the fixed line Oz; but the couple N acting in the plane xy cannot be so placed, a&d therefore as long as it exists it will turn the body round the line Oz ; consequently it is necessary and suflScient for equilibrium that N= ; — or in words,. ' The sum of the moments of the forces dbawt the fixed axis must he equal to zero. Remark. The pressure on the fixed axis is represented by the force JB at the origin and the forces of the two couples L, M applied directly to the axis. But R is equivalent to the three 2X, "ZY, "tZ; of which %X can be compounded with the couple M (the forces of which are in the same plane with it) as in Art. 59 (2) ; the result of this compounding is a single force %X acting at a point in the axis the abscissa of which is yY . The force S Y may in like manner be compounded with the couple M; and the result in this case is S F acting at the M point — =-y.. Hence then the pressures on the axis are repre- sented by {^), "ZX fat the point g^j parallel to x, ZyI ~Yy) P^'*^^®^ *° ^> ZZ (at any point of the axis) parallel to z. The last of these {tZ) may be compounded with either of the others ; and thus in the most general case the pressure on a fixed axis may be represented by two forces. COE. The pressure X^ urges the axis in the direction of its length, the other two pressures tX, %Y can be reduced to a ANT FORCES ACTING ON A EIGID BODY. 51 force and a couple, the plane of the couple being perpendicular to the direction of the force. As this reduction is useful in cer- tain cases, we shall shew how it may he effected. Through the fixed axis draw a plane so inclined to the two forces "tX, 't Y that the resolved parts of %X, S Y along this plane may be equal and in contrary directions. Let a normal to this plane make an angle 6 with %X, and therefore an angle 90° — 6 with X Y; then the components ^ f "ZX. cos parallel to the normal, I "tX . sin along the plane ; ^ ('%Y.sm0 parallel to the normal, I'tY. cos along the plane ; of which four components SX.sin^ is equal to 'ZY.cos0 by hypothesis ; and therefore these two form a couple, and fix the value of ; for since XX. sin = tY. cos 0, %Y tan0 = ^, the arm of this couple is %X'^%Y' ^™™ ^^^' and therefore its moment is ■.^.■ZX^sm0 + ^.-ZY.cos0 LtY+mx -{{txy+i%Yy}i' And the positive axis of this couple is inclined to the axis of x at the angle 0, given above. Of the four components mentioned above, the two not yet reduced are tX.cos0, tY.sin0, acting in one plane on the points ^. -^- They are therefore (Art. 40 or 59) equi- valent to a single force = tX. co30 + t Y. sin = {{tXy + {t Yff, 52 ANY FORCES ACTING ON A KIGID BODY. acting at an inclination 6 to the axis of x, upon a point in the ■fixed axis the distance of which from the origin is (by Art. 59) ^X.cosg.^-:SF.sing.^ ^ ^^^_^y SX. cos ^ + Sr. sine I^Xf + (S Yf ' 94. To find the, conditions that the forces in Art. 81 may balance each other, when there is in the body a line moveable lengthwise but in no other direction. Let this line he taken as the axis of z in Art. 81, and any point in it as the point ; then B acts upon this line, and being resolved into its components XX, ^Y, "ZZ, the first two acting in directions in which the line cannot move produce no eifect ; but "ZZ acting in the direction in which the line can move must be equal to^ zero if there be equilibrium. Also the couples L, M, being turned round and so placed that their forces shall act upon the line Oz, produce no effect because they urge it in directions in which by hypothesis it cannot move: but the couple N cannot be so placed, and therefore as long as it exists it -^111 tm-n the body about the line Oz ; it is therefore necessary that N should be equal to zero. Hence the conditions necessary and suflScient for equilibrium in this case are, The sum of the resolved parts of the forces parallel to the given line must be equal to zero; and the sum of the mament^s of the forces about the same line must also be equal to zero. This Art. will be applied when we come to investigate the power of a screw, 95. The preceding Articles have been enunciated for rigid bodies only : but since when a flexible body or a body that has joints is in equilibrium it may be supposed to become rigid without affecting its equilibrium, all the conditions of equilibrium before investigated must be satisfied by a flexible body or a body that has joints. But it is to be noticed particularly that all these conditions may be fulfilled and yet such a body not be in equilibrium, for some of its parts may not be in equilibrium. As a simple instance take the following. A straight rod placed in a horizontal position with its ends on two props will THREE FORCES ACTING ON A RIGID BOBY. 53 be in equilibrium ; but a chain, or a rod with a joint in the middle, so placed would fall. Hence then in equilibrium flexible and jointed bodies satisfy all the conditions which rigid bodies satisfy; and besides them such other conditions as are necessary to secure the equilibrium of every part into which they are divided by joints : the actions at each joint are, though unknown generally, equal and opposite upon the two pai'ts joined there. 96. If three forces acting upon a rigid hody ialance each other, the lines in which they act must he in one plane, and either he parallel or pass through a point. When a rigid body is in equilibrium, we may suppose any line or point in it to become fixed without affecting the equili- brium: upon this principle let an axis, not parallel to any of the forces and intersecting the lines in which two of the given forces act, become fixed; then these two forces acting upon fixed points may be removed; which being done the body having a fixed axis is kept in equilibrium by the remaining force, which is impossible unless the line in which this force acts either intersect the fixed axis, or be parallel to it. But it is not parallel to it by hypothesis, therefore it intersects it. It appears then that any axis, not parallel to one of the forces, and intersecting two of them, must meet the directions of all the forces, consequently they are all in one plane. Again,, since they are all in one plane they must either be all parallel, or some two of them must intersect ; in the latter case, the point of intersection may be supposed to become fixed^ and the cor- responding forces removed ; and then the rigid body having a fixed point is kept in equilibrium by a single force, which is impossible unless its direction pass through tlie fixed point; consequently, the directions of all the forces either are parallel or pass through a point. 97. The student will have remarked that when forces (as in Chaps. I. II.) act on a point, it is not a necessary condition of equilibrium that their moments about an axis should be equated to zero. The same is true of every system which is 54 EQUILIBRIUM OF A ETGID BOD'X'. capable of being reduced to forces acting on a point. Also if in any case of equilibrium we know that the forces are capable of being reduced to three forces not parallel, since these by the last Art. must act in lines passing through a point, the same is true. 98. In such of the preceding Articles as relate to the conditions of equilibrium of a rigid body under the action of a system of forces, the lines parallel to which the forces are to be resolved, or about which the moments are to be taken, and equated to zero, have been spoken of as necessarily perpen- dicular to each other. This necessity, however, has entirely arisen from the mode in which we have conducted our investi- gations ; from our having, in fact, assumed the co-ordinate axes to be rectangular. We shall shew that it may be dispensed with; and that it is sufficient if the forces be resolved in directions of, and the moments taken about, any three lines, provided no two of them are parallel, and all three not in the same plane. For this purpose it will be necessary to prove the following propositions. 99. If from a point there he drawn three, lines not in one plane, and the sums of the components, parallel to them, of all the forces he separately equal to zero; and also the sums of the moments of all the forces about them he separately equal to zero; there will he equilihrium. For from the proposed point let there be drawn a system of three rectangular co-ordinate axes Ox, Oy, Oz ; and let one of the proposed lines make angles ^i, Vu S with them. Then the sum of the components of the forces in the direction of this line is % (Xcos f,) + 1 (FcosT?.) + 1 (^cos rj, which is equal to tX . cos fi + S r. cos i7j + tZ. cos §;: and therefore by hypothesis = SX.cosfj + SF.cos')7, + S^.cos?,. EQUILIBRIUM OP A RIGID BODY, 55 Or, if B be tlie resultant of ^X, % Y, %Z; and a, /9, 7 the angles which its direction makes with the co-ordinate axes ; = jB cos a . cos ^, + J? cos /S cos ■t]^ + R cos 7 cos Jfj = R cos a. Similarly, = J? cos J, and = i? cos c ; where a, h, c, are the angles which the direction of i? mates with the three proposed lines. Now these three equations require either that "i? should be = 0, or that cos a, cos h, cos c should each be = ; but this last supposition is impossible, because the given lines are not all in one plane by hypothesis ; Again, the couples L, M, N have their axes parallel to Ox, Oy, Oz respectively : hence resolving them each into two com- ponents, one of which has its axis parallel to the line ^1, 171, ?i, and the other has its axisL perpendicular to it, we have the sum of the former = L cos ^^ + -M"cos rj^ + Ncos, ^, this, being the couple which tends to turn the body about the line under consideration, is the moment of all the forces about that line, and therefore by hypothesis = i cos ^i + ilf cos T/i + -W cos 5'j = (? . cos a' cos ^1+ Gr cos /8' cos 1^^+ G cos 7' cos ^^ = Qcosa. Similarly 0= Gcos h', and = (r cos c, a', V, c being the angles which the axis of G the resultant of L, M, N makes with the three proposed lines. From these three equations it follows as before, that G = 0; and we have already shewn that R = 0; consequently there is equilibrium. 100. CoK. If there be drawn three lines not in one plane, no two of which are parallel, and the sums of the components, parallel to them, of all the forces be equal to zero; then the resultant R is equal to zero. For the first part of the preceding 56 EQUILIBEIUM OF A RIGID BODY. demonstration applies here, since it does not depend upon the positions of the lines, but only' on their directions. 101. If ih&re he three lines not in one plane, no two of which are parallel, and the sums of the components, parallel to them, of all the forces he separately equal to zero; and if there he three lines {not necessarily the same as the former) not in one plane, no two of which are parallel, and the sums of the momsnts of all the forces about them he separately equal to zero; there will he equilihrium. The demonstration contained in the former part of Art. 99, does not depend at all upon the three lines being drawn from a point as required in the proposition, and therefore it will apply here; consequently ^ = 0. From this it follows, that if our present system of forces be not in equilibrium, their resul- tant is a couple, Q suppose. Let a', V, c' be the angles which the axis of G makes with the three, lines mentioned in the latter part of our proposition ; then resolving G into two com- ponents, the axis of one being parallel, and that of the other perpendicular to the first of the three lines, we have the moment of all the forces about that line (which is equal to the former component couple) = G cos a', which by hypothesis is equal to zero. Hence = (r cos a'. Similarly = (? cos h', and = G* cos c'. Consequently (? = ; and we have already shewn that B = Q; therefore there is equilibrium. CHAPTER V. ON THE PRINCIPLE OF VIRTUAL VELOCITIES. 102. Def. If the parts of a rigid body, or of a system of rigid bodies, in equilibrium, be geometrically transferred through a very small space in any manner, the space moved over by any particle is called, in Statics, the velocity of that particle. The path described by any particle is supposed to be so small, that it may in every case be taken as a straight line, on the principle that an arc of a curve ultimately coincides with its chord. The velocity of a point, estimated in the direction of the line in which the force acted upon the point when the body was in its position of equilibrium, is called the mrtual velocity of the point. 103. Having given the velocity of a point, to estimate its velocity in any proposed direction in the plane of motion. Let AB (fig. 20) be the velocity of a point, -EFthe direction in which it is required to estimate it. Draw EG perpendicular to EF; Aa, Bb parallel to EF; and AG parallel to EQ. Then every line perpendicular to EQ in the plane FEQ is parallel to, and therefore in the same direction as EF. Hence, to find the velocity in the direction EF, is the same as to find the space through which the point has receded from the line EQ. Now at A the distance from EQ is Aa, and at B the distance is Bh, consequently the velocity estimated in the direction EF is E. s. 8 58 PRINCIPLE OP VIRTUAL VELOCITIES. Bh-Aa = BG=ABcosABG. And ABC is equal to the angle which the velocity AB makes with the proposed direc- tion. Hence we can estimate a velocity in a proposed direction, by multiplying the velocity into the cosine of the angle at which it is inclined to the proposed direction. 104. From this it will be seen, that when a particle, which is acted on by a force, is displaced, the virtual velocity of that particle will be found as follows; — drop a perpendicular from the new position of the particle upon the line in which the force acted before displacement, and the line intercepted between the foot of this perpendicular and the first position of the point, is the virtual velocity required. Thus, in fig. 21, let the force F act upon the point A in the direction AF, and let A be moved to A'; draw A! a perpendicular to AF, then Aa is the virtual velocity of A. If A were moved to A" so that FAA" is a right angle, the virtual velocity of A would be zero. If A were moved to A" so that the perpendicular A"'a" falls on FA pro- duced, the virtual velocity Ad" of A is said to be negative. 105. If a rigid hody he displaced in any manner, the velo- cities of any two of its particles, estimated in the direction of the line which joins them, are equal. Let A, B (fig. 22) be two particles of a rigid body, and let AA', BB' be their velocities. Then, because the body is rigid, A'B'= AB. Through A draw a plane at right angles to AB, and upon it drop the perpendiculars A'a, B'h. It will be easily seen, that the estimated velocity of A is A a; and that of B, B'h — BA ; and we are to prove these equal. Join a&, and draw A' parallel to it. The angles at G are right angles, and therefore B'b -BA = A'a + B'G-BA = A'a + A'B' cos A'B' G-BA = A'a -BA {I- cos A'B'G) A'B'n = A'a -2BA. sin' ^^J^. PRINCIPLE OF VIRTUAL VELOCITIES. 59 But the last term, containing the square of the very -4 '5' small quantity sin — - — as a factor, must be omitted in con- formity with our definition in Art. -102. Hence Bh-BA=A'a. This proposition is true, if ^, 5 be two particles of different bodies connected by a rigid rod, or inextensible string ; for in the preceding demonstration nothing more is assumed than that AB' is equal to AB^ which is satisfied in these cases. 106. If the reader should have any doubt respecting the propriety of omitting the last term, we would recommend him to reconsider the consequences of the definition in Art. 102, where it is stated that the displacement of every particle is so small, that curve lines may be considered as coinciding with their chords; this requires us to consider the deflection of an arc from its tangent as evanescent in comparison of the arc itself, which arc is the velocity with which we J^ave to deal. Hence BA{l-cosAB'G), being the versed sine (or deflection from the tangent) of the arc which represents a quantity less than the velocity, may be a fortiori neglected. 107. If the displacements of the two points in Art. 102 be in parallel straight lines through finite spaces, the propo- sition of Art. 105 will then also be accurately true; and our definitions in Art. 102, and the property in Art. 103, will also strictly hold, how large soever be the spaces through which the particles are displaced, 108. If the particles A, B, in Art. 105, are urged by two equal forces T, T' in opposite directions along the line AB, the TOtual velocities hs, hs of A, B for those forces will be equal, but of contrary signs: and consequently the quantity Ths + T'hs is equal to zero. Now ii A, Bha two particles of the same rigid body (or of two different bodies connected in such a manner by a rod or cord AB that the distance between 60 PEINCIPLE OP VIRTUAL VELOCITIES. them does not change), their influence upon each other is ex- erted along the line AB, and is called tension. This tension is the same for both, but acts upon them in opposite directions, viz. either to draw them towards each other, or to push them asunder. Hence it follows, that for the tensions acting between A and B, TSs + T'Bs =0. The same maybe proved for any and every two points in a whole system of bodies, provided they are connected in such a manner, that the distance between the points of connection is not changed by the displacement. It is obvious, that the tensions we are now considering, occur in pairs. Hence it follows, that if the forces of tension throughout a whole system of bodies in equilibrium be respectively multiplied by the virtual velocities of the points on which those tensions are supposed to act, the sum will be equal to zero. 109. If a body rest against a smooth fixed point, there will be a pressure of the point against the body in the direction of a normal to the surface of the body. This pressure is one of the forces which teep the body in equilibrium. Let A (fig. 23) be the fixed point, FA the surface of the body resting against it, AB a normal at A, and let the body be displaced without lifting it off the point, so that A comes to some point A' suppose. Then the virtual velocity is ^^' cos ^^^' = ^^' sin P^^' ; but PAA' is an indefinitely small angle, and therefore AA' sin PAA' is indefinitely smaller than AA', and may be neglected. Hence, if B be multiplied into its virtual velocity, the product may be neglected. 110. If a body rest against a smooth fixed curve line or surface, there will be a pressure of the curve or surface against the body, in the direction of a normal at the point of contact. Let PA (fig. 24) be the body resting against the curve line or surface QA ; and let the body, by sliding and rolling, come into the position FBA', B being now the point of contact, arid A' the new position of A. The virtual velocity of .4 = BA' sin ABA'; which, for the same reason as before, may be neglected. PRINCIPLE OF VIRTUAL VELOCITIES. 61 111. If two smooth bodies of a system rest against each other, there will be a mutual pressure, which will act upon them at the point of contact in opposite directions, coinciding with the common normal at that point. If they are disturbed without heing separated, the distance between their centres of curvature, at the point of contact, will remain unchanged ; and, therefore, the virtual velocities will be exactly equal, but of contrary signs for the two bodies. If, then, B, E be the equal pressures exerted by each against the other, and Sr, S/ the virtual velocities, Rh- + R'h-'=0. 112. From the last three Articles, it appears that in any system of bodies kept in equilibrium by the action of external forces, and by tensions, reactions of smooth fixed obstacles, and mutual pressures of smooth bodies of the system, the sum of the products of each tension, reaction, and pressure, into the corresponding virtual velocity of the point on which it acts, is equal to zera. The student will remark, that the Articles referred to, are only true when the displacement of the system is so small as to agree with the definition of a velocity given in Art. 102 : also, in the case of pressures, the surfaces must be smooth, and the con- tact must not be broken; and in the case of tensions, the connecting line must be of unaltered length. 113. Let there he any number of connected bodies of a system kept in equilibrium by the action of external forces, and also by the tensions of connecting rods, cords, '^o'^ similar quantities for T^, T^,... T^ and jB,,5j,...i2„. Then, because the particle is in equilibrium under the action of these forces, therefore (Art. 38) = S (-Fcos a)+t {Teas a)+t{fi cos a'), = t (i^cos/3) + t-(!rcos b)+%{B cos 6'), = S (i^cos7) + 1 {Tcos c)+t{B cos c'). Let now the system be displaced, the Telocity of the particle under consideration being B8^, and ^f the angles which S8j^ makes with the co-ordinate axes. Then, if Bs^ be the virtual ' velocity for the force F^, jPjSsj = F^ (cos Hj cos f + cos /3, cos j] + cos 7^ cos f ) 88^ . Similar expressions are true for the other external forces which act on the point, and therefore t{FSs) = {t{Fcosa.)eos^+tiFcos^)cosv+'^{Fcosi)cos^}SS^. Similarly, if Bt^ and Br^ be the virtual velocities corresponding to 2; and B^, t (TBt) = {t (Tcosa) cosl+'Z{Tc()sb)cosri+t {Tcosc) cos^\B8,; and % {BBr)={t (-Bcosa')cosf +S {Rcosb') cosij+S (-Bcos c) cos ^} B8,. Hence by adding the last three equations we obtain t{FBs) + t{TBt) + t{BBr) = (1) in which the symbol % extends to all the forces, tensions-, and reactions which act upon the point under consideration, but has no reference to the other particles of the system. We may form equations similar to (1) for every other point in the whole system upon which forces of any kind whatsoever act. If all these be added together, the terms belonging to the PRINCIPLE OF VIRTUAL VELOCITIES. 63 tensions along the lines which join points of the same body, and also those which act along rods and cords connecting two points of. separate bodies of the system ; and likewise the reactions of fixed points, and surfaces, and the mutual pressures of two bodies of the system, will all disappear, by Art. 112,' in forming the sum. But these, together with the external forces, are all the forces which act on the system ; consequently, there remains only the equation where %' extends to all the points of the systenf upon which external forces act, S' and % together denote that the sum of the products of all the external forces which act upon all the points of the system into their respective virtual velocities is to be taken, and the equation shews that this sum is equal to zero ; which is the principle of virtual velocities. It is not necessary to employ both S and X', if we suppose X to extend over the whole system, the equation may be written % (FSs) = 0, which is called the equation of Virtual Velocities. 114. The great advantage of the equation of virtual veloci- ties consists in this, that it furnishes at once a relation among the external forces which act upon a system, free, from tensions and pressures. Since the bodies are rigid, and supposed to be connected by strings or rods of unchangeable length, it is obvious that, in general, when one part is arbitrarily disturbed, the disturbance of the other parts will depend upon it by geo- metrical relations. In this case, Ss^ being given, Zs^, Ss, ... will be determinable in terms of Ss, ; and these values being written in the equation % (FSs) = 0, will give only one relation among the forces, and will not therefore enable us to find the forces themselves, if their number exceed two. It will, however, sometimes be possible to disturb one part of the system without affecting other parts ; or the system may consist of several parts, each one of which it may be possible to disturb in such a. manner as not to affect the other parts. In this case it is manifest, that the equation of virtual velocities 64 PRINCIPLE OF VIRTUAL VELOCITIES. will furnish us as many equations between the forces, as there are parts of the system which can he independently disturbed. Now two points can be independently disturbed when no geometrical relation exists between their virtual velocities. Wherefore, in using the equation % {FBs) = 0, we must find, from the geometrical properties of the system, as many of the quan- tities Ssj, Ssj, Ssj... in terms of the others as possible, and sub- stitute them in the equation; the virtual velocities which are stiU left in it are independent, because no geometrical relation exists among them; and, therefore, the corresponding parts of the system admit of independent disturbance; we must conse- quently equate the coefficients of each of these terms to zero. The resulting equations are the conditions of equilibrium. To illustrate what is here meant, we will solve the two fol- lowing problems by the principle of virtual velocities. 115. A particle rests upon a plane curve line, heing acted on. hy two forces X, Y parallel to the co-ordinate axes : to find the conditions of equilibrium. Let y=f{x) be the equation of the curve, x, y being the co-ordinates of the position of equilibrium of the particle. Then since after the disturbance the particle still remains upon the curve, if 2^ + By, and x + hx'he, the co-ordinates of its new posi- tion they must satisfy the equation of the curve ; ••• y + By =f{xJrhx) =y + d^ .Bx; .-. Sy = d^y . Sx. Now Bx, By are the virtual velocities of the particle for the two forces X, Y; .*. XBx + YBy = by the principle ; .■. XBx + Yd^yBx = for all values of Bus,- and .-. X+ Yd^=0, which is the condition of equilibrium. 116. A particle rests upon a smooth curve surface acted on by three forces X, Y, Z parallel to the co-ordinate axes: to find the conditions of equilibrium. PRINCIPLE OP VIRTUAL VELOCITIES. 65 Let s =f{so, y) be the equation of the curve surface, x, y, z being the co-ordinates of the position of equilibrium of the particle. Then if as + Sas, y^ By, s + Bs be ^he co-ordinates of the position of the particle after disturbance, Bx, By, Bz are the virtual velocities of the particle for the forces X, Y, Z respec- tively ; and therefore by the principle of virtual velocities, X$,x-\-YBy + ZBz = f). But because x -f«Sa;, y-\-By, z + Bz are the co-ordinates of a point in the curve, z + Bz=f[x + Bse, y + By) — z + d^z.Bse + dyZ.By; .". Bz = d^z . Bx + dyZ . By. By substituting this value of Bx, we have (X+ Zd,z) Bai+{Y+ Zd,z) By = 0. There is no geometrical relation existing between By and Bx ; consequently, the equations of equilibrium are X+Zd^z = Q, Y+ZdyZ = 0. 117. ^ two forces P, P' whose virtual velocities are Sp, Sp', act upon a rigid body at different points, and he such that the equation PSp + P'Sp' = is true for all arhiiyrary displacements of the hody, then P and P' are equal and act in the same line in opposite directions. The equation shews that Bjy and Bp are always zero together. Now disturb the body in such a way that the point at wTiich P acts may remain stationary ; then since the body is rigid, the point on which P' acts must have described a circular arc about the stationary point ; and as Byp = 0, that arc must be perpen- dicular to the direction in which P acts, therefore P' acts in the direction of a normal to the arc, i. e. in a line passing through the point on which P acts. In the same way it may be shewn that P acts in a line passing through the point at which P' acts ; hence they both act in the same line: it will therefore be possible to disturb the body so that Bp and Bp/ may be equal E. s. ' 9 66 PRINCIPLE OP VIETUAL VELOCITIES. in magnitude; and they must have different algebraic signs (•.■ PSp + P'Sp' = 0), which can only happen, since the body is rigid, by reason of P and F acting in opposite directions ; and therefore P and P' are likewise equal. 118. If the eguation % (FSs) = &e true for all arUtrary displacements of a rigid hody under the action of external forces F, , Fjj . . . there is equilibrium. For if not, there will be at most two restfttants (Art. 84) ; apply forces P, P' equal to these resultants and in the contrary directions to them, and then the body is in equilibrium under the action of the forces F^, F^ ... P, P; consequently by the Principle of Virtual Velocities, S(PSs)+PSp + PS/ = 0. But % {FSs) = by hypothesis, and therefore PBp + P'Bp' = : and hence it follows from the last article that P and P are equal and act in opposite directions ; consequently they destroy each other ; they may therefore be removed without affecting the equilibrium ; hence the body is in equilibrium when Pj, Pj, P^ ... are the only external forces which act on the body. 119. When a system of connected bodies is in equilibrium under the action of external forces, pressures, &c., the equilibrium would not be affected if the connecting joints, cords, &c. were all to become rigid : and hence any force may be transmitted to any point of the system in the line of its action (Art. 21), pro- vided the original point and the new point of application are not situated in independent parts of the system. 120. If the equation t (FSs) =0 be true for all arbitrary displacements of a system of connected rigid bodies, there is equilibrium. If the system consist of independent parts, let one of those parts alone be displaced, then for that part % {FBs) = by hypo- thesis. If that part is not in equilibrium we may apply forces to each body of it which shall keep each of them in equilibrium : these forces (Art. 119) may be transmitted and reduced to two. PRINCIPLE OF VIRTUAL VELOCITIES. 67 P, P, acting upon the part under consideration. Hence reasoning as in Art. 118, we find Pand P equal and opposite, and therefore they may be removed without disturbing the equilibrium of the part. The same may be proved of each of the independent parts ; and, consequently, the whole system is in equilibrium. Remark. We have seen that the principle of virtual velocities is true only when the displacements are so small as to allow us to consider an arc as coincident with its chord or tangent. Now the reader who is familiar with the Differential Calculus will know, that an arc and its tangent coincide ana- lytically only as far as the second term of Taylor's theorem inclusive : hence the principle of virtual velocities embraces only quantities of the first order of smallness. The second term of Taylor's theorem has been called the differential of the first term ; wherefore, in applying the principle of virtual velocities, we ought always to use ds instead of hs. The equation of virtual velocities in its proper form is % (Fds) = 0. Also because this equation involves only differentials of the first order, it is a matter of indifference whether a body rest upon a curve or its tangent, a surface or its tangent plane ; or on any other curve or surface having the same tangent or tangent plane at the point on which it rests. CHAPTER VI. ON THE CENTRE OF PARALLEL FORCES, AND ON THE CENTRE OF GRAVITY. THE CENTRE OF PARALLEL FORCES. 121. If a rigid hody he acted on at dAfferent 'points hy forces in parallel directions, there is a certain point through which their resultant passes, whatever he the position of the hody with respect to the direction in which the forces act. Let F^,F,...F„ act on the points A, B ... K (fig. 25) of a rigid body. From any point in the body draw the rectan- gular co-ordinate axes Ox, Oy, Oz. Join A, B; and let the resultant of F^, F^, pass through G. Draw Aa, Bh, Cc parallel to Os ; join a, b passing through c. Let aj^yjZj, x^y^z^ ... x„y^z^ be the co-ordinates of the points on which the forces act; xy'z those of C; and let Q be the in- clination of AB to ah. Then ^ — z^=Gc — Aa = AG sin Q, and z^-z' = Bh- Gc=BGwa.e; z.-z BG F.,, . ^ ,^v whence we find {F^ +F^ z' = F^z^-\-F^z^. AgAin, take away the forces jP^, F^ and replace them by their resultant F^ + F^ acting at G; then if we put x"y"z" for the co- ordinates of the point through which the resultant of i''^, F^, F^, or, which is the same, of the two {F^-\- F^ audita passes, we hare as before {F, + F, + F,)z'' = {F, + F,)z- + F,z, = F z +F z +F z . CENTRE OF PARALLEL FOECES. 69 In this manner, introducing successively one force at a time, until all have been taken in, and denoting hj x^s the co-ordinates of the point at which the final resultant acts, we shall at length obtain, {F, + F, + F,+ ...+F,)^ = F^z^ + F,z^+ F^z^+ ... + F^z,, or, more concisely, SF. i = S {Fz). By similar reasoning we shall obtain tF.^ = X{Fy), uni tF.x=t{Fx). The last three equations determine the values oixy'z; and since those values do not contain any terms depending on the inclina- tions (to the co-ordinate axes) of the lines in which the forces act, those forces may be turned about the points on which they act without affecting the position of the point whose co-ordinates are xy^. On this account this point is called the centre of parallel forces. 122. Def. The product of a force into the distance of the point on which it acts from a plane, is called the moment of the force with reject to the plane. Hence % {Fx), ^ [Fy), % {Fz) are the sums of the moments of the forces with respect to the planes of yz, xz, xy : and ^F . x, SF. y, %F. z, are the moments of their resultant with respect to the same planes. Hence, remem- bering that the co-ordinate planes were taken in any position, it follows, that the sum of the moments of any parallel forces with' respect to a plane is equal to the moment of their resultant with respect to the same plane. 123. If the proposed plane be drawn through the centre of parallel forces, the moment of the resultant with respect to it will be zero ; consequently, the sum of the moments of any parallel farces with respect to any plane passing through their centre is equal to zero. 124. If %F be equal to zero, there is then no centre of parallel forces, as we likewise know from Art. 73. 125. The formulae of (121) are true if the co-ordinates are oblique: and in that case t [Fx), t {Fy), t {Fz) are called the 70 CENTRE OF GRAVITY. obliqiie moments of the forces with respect to the co-ordinate planes of yz, zx, xy. THE CENTRE OF GRAVITY. 126. It has been found by experiment, that under the exhausted receiver of an air-pump bodies of unequal magnitudes, and differing altogether in their nature and form (such as a piece of lead, a shilling, a feather, &c.) fall from the top to the bottom of the receiver exactly in the same time: from which it has been inferred, that the Earth exerts an equal force on all equal portions of matter ; and that the weight of a body at a given place, measured according to the principles laid down in Arts. 7 — 10, is proportional to the quantity of matter in the body; that is, if M be the quantity of matter in a body whose weight is Wat a given place, then But we have stated in Art. 8, that the weight of a body, measured by a standard spring, is not the same at all places of the Earth's surface ; it is in fact (as is shewn in Dynamics) proportional to the accelerating force of gravity, at the respective places. This force is generally denoted by g ; and hence . we have for a given body W^g. Consequently, for different bodies at different places WccMg, For reasons stated in Dynamics we assume that W=Mg. 127. The size or bulk of a body is called its volume and is denoted by V: but it is necessary to explain, both with regard to V and M, that they are expressed in numbers on the following principle. A known body, composed of matter uniformly dif- fused through all its parts, is taken as a standard to which all others are referred. The volume and mass of this body are called the units of volume and of mass. If a body be V times tlie size, and contain M times the quantity of matter, of the standard body ; V and M are taken as the measures of the volume and mass of that body. Also, supposing the matter of the second CENTRE OF GRAVITY. 71 body to be uniformly diffused through its parts, if a portion of it of the same size as the unit of volume contains p times as much matter, p is called the density of the body; and it is evident that M=pV. 128. The direction in which a body descends when let fall is called the vertical direction ; it may be discovered by suspend- ing a heavy body by a thread, or by drawing a line perpendicular to the surface of still water. A plane at right angles to the vertical is called a horizontal plane; and it is evident, since the Earth is spherical, that the horizontal plane changes its position in passing from place to place: but since the distances of the bodies of systems usually treated of in Statics are exceedingly small compared with the radius of the Earth (4,000 miles, nearly) we may consider the sujface of still water as a horizontal plane to a small extent, and consequently the verticals as parallel. 129. Hence it appears, and from Art. 121, that in every body, and in every rigid system of bodies, there is a certain point through which the resultant of the forces which the Earth exerts on the different parts always passes in every position of the body or system. This point is called the centre of gravity of the body or system: it is sometimes also called the centre of mass. 130. One property of the centre of gravity, particularly worthy of remark, is, that it does not depend at all upon the intensity of the force of gravity. For divide the whole system into very small equal molecules, the quantity of matter in each being m, and their number n, and denote the force exerted upon a unit of matter by g ; then the force exerted on each molecule = mg. And if x^y^z^, x^^z^,... be the co-ordinates of the molecules, and xyz those of the centre of gravity, we have, by Art. 121, _ mgx, + mgx^ + mgx^ + . . . to w terms *~ mg-\-mg + mg + ...ion terms _ X, + X.;, + X, + . n 72 CENTRE OF GRAVITY. Similarly, ^ = ^dJ^i+i^a+inii' ^ andJ = "' + "- + "- + It appears then, that the co-ordinates of the centre of gravity are the means* of the co-ordinates of the equal molecules, and consequently its position is independent of the intensity of gravity. Hence the centre of gravity of any body is a certain point within it, the place of which depends only on the relative disposition of its equal molecules. The investigation of its place is therefore purely geometrical, and may be applied to any body whatever; and for this reason we often speak of the centre of gravity of bodies far removed from the influence of the Earth, and when, in fact, no reference is intended to be made either to the Earth or to gravity; the point alluded to being no other than the one determined from the geometrical principles just laid down, viz. — that its co-ordinates are the respective means of the co-ordinates of all the equal molecules of which the body is composed. 131. When a body is acted on by no other force than gravity, since the resultant of the forces which act on the particles of the body passes through its centre of gravity, if that point' be supported the body will be in equilibrium in every position. For instead of the forces themselves, we may substitute their resultant, which will be counteracted by the point of support, and as this will be the case if the body be turned round that point into any position whatsoever, it follows that there will be equilibrium in any position whatever. 132. And since the resultant may be applied at any point in the line of its direction (Art. 21), if the point of support be not in the centre of gravity, but in any point of a vertical passing through it, the body will be in equilibrium. And con- versely, if a body be suspended from any point in it, it will not be at rest till the centre of gravity and the point of suspension are situated in the same vertical. them. Hence the centre of gravity of two equal "bodies is ihe middle point between CENTRE OP GRAVITY. 73 This property may sometimes be employed in finding the centre of gravity in a practical manner. For if the body be successively suspended from two points in it, and the correspond- ing verticals be drawn upon or through the body, their common point of intersection will be the centre of gravity. 133. It follows at once, from Art. 131, that if all the par- ticles which are situated in a line passing through the centre of gravity be supported, the body will rest in equilibrium on that line in all positions. And the converse is true, viz. — that if a body rest in equilibrium, in all positions, on a fixed line, the centre of gravity must be in that line ; for, unless the centre of gravity were in that line, a position might be found in which the vertical through the centre of gravity did not pass through a point of support, and consequently the body would not be in equilibrium in all positions, which is contrary to the hypothesis. Hence, if we can find two lines on which a body will rest in all positions, the centre of gravity will be in their common point of intersection. 134. Since the resultant of all the forces of gravity, which act on the particles of a body, may be supposed to act at the centre of gravity, and is equal to their sum (Art. 121), we may, in any investigation in which this resultant is required, suppose the whole mass united at the centre of gravity ; and hence it becomes important to know the situation of this point in bodies of different figures. 135. It is not always convenient to divide a proposed body into equal molecules, as was done in Art. 1 30, it therefore be- comes necessary, in that case, to use other formulae for the determination of the centre of gravity. Let >»,, m^, wij, be very small masses into which the body may conveniently be supposed to be divided ; a;, y^ z^ , x^y^z^, x^y^Zg... their co-ordinates. Then the forces which urge them are ffm^, gm^, gm^, respectively; and therefore, substituting in Art. 121, we obtain E. 8. 10 74 CENTRE OF GEAVITY, gm, + gm^ + gm,+ ... m^ + m^ + m^ + ... _ % imx) and, similarly, S (my) _ 2 (mz) y = '- " — ' — z ■■ %m ' Xm 136. Since, whatever be the position of the plane yz, we always have x.'$m = 'Z (mx), it appears that the moment, with respect to any plane, of the whole mass collected at its centre of gravity, is equal to the sum of the moments of all the molecules, with respect to the same plane. • 137. If the origin of co-ordinates he in the centre of gravity, then X {mx) = 0, S {my) = 0, and 2 (wis) = ; for x, y, and z are, in that case, each equal to zero. 138. Since the mass of a body of uniform density is mea- sured by the product of its volume into its density (Art. 127) ; if Pj, Pj, Pj, ... be the densities, and V^, V^, Fg, ... the volumes of the molecules m,, m^,m^, ... we shall have w,=p,Fj, m, = p,F„ jw,=/o,F3,... the molecules being so small, that every part of each one may be considered of uniform density. Hence, by substitution in the formulae of Art. 135, we have P^yt + P.K + PsK+- _ t{pVx) CENTRE OP GEAVITY. 75 and «-iMM S(pF)- 139. If the density of the whole system be the same in every part, then p^ = p^ = p^ ... and these formulae are simplified by dividing out p, thus, ™-?iM r,-li]M 7-li]^ «- ^Y ' y- iv ' %v ■ But it is to be carefully observed, that these formula are only to be applied to such bodies as are of homogeneous mate- rials. 140. The general application of these formulae depends on the Integral Calculus, but there are a few cases which can be made to depend upon the more simple principles of Art. 133, and with them we shall accordingly commence our series of examples on the subject of finding the position of the centre of gravity in bodies of various forms. All bodies will be supposed homogeneous, or of uniform density, unless the contrary is mentioned. 141. If through any figure a plane can he drawn, so that the figure shall be symmetrical with regard to it; that is, so that the two parts of the figure which are situated on opposite sides of that plane are perfectly similar and equal; the centre of gravity is in that plane. For the volume of the body being similarly disposed on the two sides of this plane, the moment of the volume on one side is exactly equal to the moment of that on the other side, with respect to that plane, and these moments will have contrary signs, and therefore their sum will be equal to zero. But this sum (Art, 139) is equal to the moment of the whole volume, collected at its centre of gravity, with respect to the same plane ; which cannot be the case unless the centre of gravity be in that plane. 76 CEBTTEE OF GRAVITY. 142. Hence, if we can find two such planes differently situated, the centre of gravity will be in the line of their inter- section ; and if we can find a third plane, the centre of gravity will be that point where it cuts the line of intersection of the other two ; in other words, it will be the common point of inter- section of any three planes, by which the figure can be sym- metrically divided, 143. It follows, from these properties,-^ (1) That the centre of gravity of a sphere, or of a spheroid, or of a cube, is its centre. (2) That the centre of gravity of a parallelopiped is the middle point of one of its diagonals; and of a cylinder the middle point of its axis. (3) That the centre of gravity of any figure of revolution is some point in the axis. 144. When we speak of the centre of gravity of a line, or of a plane figure, it is to be understood that the line consists of material particles, and the plane figure of a single lamina of particles, or else, that the thickness is everywhere the same, and inconsiderable. 145. Hence the centre of gravity of a straight line is its middle point ; of a circle, or, ellipse, or square, its centre ; and it will follow, from reasoning precisely similar to that of Art. 141, that if we can draw two straight lines in a plane, by each of which the figure is divided into two equal and symmetrical parts, the centre of gravity is the point of their intersection. This property will enable us to determine at once, by inspection, the centre of gravity gf almost all regular plane figures. 146. To find the centre, of gravity of a plane triangle, ' Let ABC (fig. 26) be the triangle, bisect one of the sides as 50 in D, and join AD. Then we may suppose the triangle made up of material particles, arranged in lines parallel to BC; let Ic be any one of them. Then, by the similar triangles BAD, bAd, BD : DA :: hd : dA, CENTRE OP GKAVITY. 77 and, similarly, BA : DG :: dA : dc, .-. BD '. DG :: Id : dc. But BD =DG, therefore bd = dc; and consequently, ,+m3+ ).{Goy or, t [m ( Omf] = t{7H{GmY]+%ni.{ (^Of. From this equation it appears, that the sum of the products of each particle into the Square of its distance from the point 0, is greater than X{in{Ghny} by the quantity "tm . GO'; and since S [m ( GmY} does not depend at all upon the position of the point 0, the sum will be the least possible when G0 = 0, that is, when the point is in the centre of gravity of the system. 159. Cor. 1. So long as the distance of from G remains the same the quantity X {m {GrmY} +'tm . GC retains the same value; if, therefore, be fixed in space, and the body be made to turn round its centre of gravity, the sum of the products of GENERAL PKOPEETIES OP THE CENTKE OF GEATITr. 85 each particle of the system into the square of its distance from remains unaltered. 160. Cob. 2. The last two articles are equally true if 7«j, OTj, m, ... he large bodies instead of single particles, observ- ing, in that case, that x^ y^ a, ,x^y^e^,x^y^z^... wiU be the co- ordinates of their respective centres of gravity. 161. CoE. 3. Suppose the bodies each equal to m, and let tlierr number be n, then S{TO(Omn=OT,.(Om,)= + m,.(OOT,)' + »»,.(Om,)'' + = m {( OmJ* + ( Ow/ + ( Om.)= + ......} = m.t{OmY; and, similarly, %{m{ Gmf] = m . S ( Omf ; also S»i = m^ + m^ + m^ + =m + m + m-\- to«terms = «m; consequently, by substituting in the equation of Art. 158, we obtain m . t {OrrCf = m.%. {Grrif + nm . {G-Of; . .-. t {OnCf = t . {Qm") +« . {G0)\ It appears then, that in a system of to. equal bodies, the svm of the squares of the distances of their centres of gravity from a given point, is greater than the sum of the squares of the corresponding distances from the centre cf gravity of the system, by n times the square of the distance of this latter from, the given point. 162. CoE. 4. Hence, if ABG be a triangle, G its centre of gravity, and a point situated either in the plane of the triangle or not, we have A(y + 30" ^ CO" = AG^ ^BGI^^ CG-" -vz . G0\ And in a triangular pyramid whose angular points are A, B, C, D, and centre of gravity G, A(J-vBQ-^G(J^B(y = AG'+BG'+0G^ + DG' + i.GO'. For, by Art. 147, the centre of gravity of the triangle coin- cides with that of three equal bodies placed at its angular points ; 86 GENERAL PROPERTIES OF THE CENTRE OP GRAVITY. and the centre of gravity of the pyramid with that of four equal bodies at its angular points, (Art. 151). 163. If each particle of a system he ^multipUed, as in Art. 158, hy the square of its distance from a given point, the sum of the proiMds will he greater than it would he if the whole system were collected at its centre of gravity, hy a qvxintity which is found hy multiplying the products of the hodies taken two and two respectively, hy the squares of their mutual distances, and dividing the sum of these products hy the sum of all the hodies. For let be the given point, G the centre of gravity of the system of particles or bodies m^, m^, m^... Take for the origin, and let x, y, z be the co-ordinates of O; x^y^e^, x^y^z^, x^y^z^,... those of m„ m^, »»,...; also, let [m^m^, {m^m^, (tojOTj),... be used to denote the distances between j»i and m^, m^ and m^, m^ and m^ Then, by Art. 135, 5 . 2»ra = »»ia;i + mjjOjjj + jMja;, + ... 'y-tm = m^y^-^ m^^-\-m^^+ ... s .%m = m^z^ + m^^ + m^^ + ... squaring each of these equations and adding the results we obtain OG\{lmy.^m^.{Om^' + m^. {Om^' + m^' . {Om^' + ... ■ +2m,m^ . {x,x^ + y^y^-¥z^z^ + + 2OT,»re, . {x^x^ + y^^ + z^z^ + + 2m,w,. Ka33+y^, + a^,) + + by writing 0(P, (Om^\ (Om,)', {Om^\.,. for their equals S'+Z + P, x,'+y,' + z,\ x,' + y^' + z,\ x.^ + y^' + z^ re- spectively. But (mjjw,) being the distance between two points whose co-ordinates are x^y^ z^ , x^y^ s, , we have GENERAL PROPERTIES OF THE CENTRE OF GRAVITY. 87 {ni,m;)' = K - x,y + {y^ - yj» + {z, - z,y = ^1 + yi + a.' + a;/ + y/ + «/ - 2 {x,x, + y^^ + z^z^ = ( OmJ' + ( Om^'' - 2 {x^x^ + y,y, + v,) 5 = m^m, {( Omi)» + ( 0>w J - K»»/}. Smilarly, 2}«j»M, (a;,a;, + 2^j, + 3iS3) = m^m, {( 0»»/ + ( Om^' - Km,)"]. Consequently, by substitution, OG' (tmf=^m,\{Om,y + m^. {Om^^ + m^ . {Om,y + ... + 7«,7«, {( OmJ" + ( Om,)= - Km/} + »Wi'»s {( Om;)'' + ( Omj)^' - Km,)^} + m,m3 {( OmJH {Om^Y - {m,m,y} + = (»»i+»ij + OT3+ )mj (Ooti)'' + (mi + »nj + m3+-. )m^{Om^Y + {m^+m^+m^ + )m^{Om^Y+ - m^m^ . {m^m^Y- m^m^ . {m^m^''- m^m^ . {m^m^Y- ... - %n. . S {«i ( Om^] — S \m^^ . {;m^^^\ ; the term % {»»i»»2 • K"?!)"} feeing understood to represent the sum of the products of the particles, taken two and two, into the squares of their mutual distances. Hence dividing by %m and transposing, S{m(Omr} = (Sm). 0(?» + ^i»;|?^^, which expresses the property to be proved. 88 GENERAL PEOPEETIES OP THE CENTRE OP GRAVITY. 164. Cor, 1. From Art, 158, we hare t {m {OmY} = t{m [GhaY] + tm . {QOf; which, substituted in the equation above obtained, gives Am, A result which might have been obtained at once without the aid of Art. 158 by supposing to coincide with G. 165. Cor. 2. If now, as in Art. 161, we suppose all the bodies equal and n in number, the last equation becomes .•. % ('Wi'mJ =71.% { Qmf. Hence, in any system i}f n equal bodies, the sum of the squair^ of the lines joining their centres of gravity, two and two, is equal to n times the sum of the squares of the distances of those points from the centre of gravity of the system,. 166. Cor, 3. Consequently, in the case of the triangle (Art. 147), BG^ ^ AG^ ^AR = Z .{AG^ ^-BG^ + CGf). Hence the sum »»s5'-" ^^^ tlie forces acting upon the particles of the system. Let now the system be disturbed in a manner subject to the same restrictions as were pointed out in the Chapter on virtual velocities, (i. e. rods must not be bent, cords must be kept of invariable length, contacts must not be broken, «S;c.) and let (?a,, dz^, x8s; and it is less than if PQ were all collected at Q ; .■. Su< {x + Bx) Bs. 92 GENERAL PROPERTIES OF THE CENTRE OP GRAVITY. Hence j- always lies between x and x + Bx, consequently the limit of -«- = a; ; but by the principles of the Differential Calculus the limit of -=r- = t" 5 OS as du _ '■•■^"'"' /, u — Jxds, the integral to be taken from x=OC to x= OD. But if xy be the co-ordinates of the centre of gravity of AP, we have by Art. 1S9, xs = u=jxds', - Jxds Similarly, y=^^—. 175. These formulae will suffice for the determination of the point required in any given example: but it may be remarked with respect to these, and other formulsa, which will be investi- gated for finding the centres of gravity of areas and volumes, that they are not always of convenient application. It is, gene- rally speaking, more easy to work out an example by taking an element 8m of the figure, and then applying the equations _ _ X {xZm) -_'%{yhn) If this method be applied to the case investigated in the last Article, we have 8»» = 8s; .*. ^ (Sm)=SSs=/tfe = s; and % {xSm) = S {xSa) =fx8sj ; and .'. x = — , the same result as before GENEEAI. PEOPEKTIES OF THE CENTRE OP GEAVITT. 93 Ex. 1. To find the centre of gravity of the arc of a semi- cycUnd. Let BG (fig. 34) be the base, AB the axis, and ^C-the arc of the semi-cycloid; x = AM, y = MP, 8 = AP, 2a = AB; then the equation of the cycloid is If = {2ax — ar')' + a vers"* - ; and ds= ( — J dx; .'. s = 2 V2aa;. Also, xds = '^2ax.dx', .'. Jxds = -x^f2ax^, o .*• 5=1 .(1). Again, fyd8=ys-Jsdy = ^5 - J]2 V2aS f — - 1 j cfe = ys - 2 V2a/(2a - a;)* dx = 2^« + |V2a(2a-a!)'+a Now this integral ought to vanish when x = 0; ••• C^=-T"' and fyd8=ys + -^ '/Za {2a - x)' - — a' ; . 5:^„ ,H(2lZ^ 8^ (2). ••^^^3 V5 3V2^ 94 GENERAL PROPERTIES OF THE CENTRE OP GRAVITY. Tte equations (1), (2) give tlie co-ordinates o£ the centre of gravity of any arc AP: and if we write in them 2a for x, we find |aiid(^-|)a, for the co-ordinates of the centre of gravity of tlce arc A G. Ex, 2. To find the centre of gravity of an arc of a circle. Let AB (fig. 35) be the given arc, its centre, G its middle point; join OA, OB, OG: and letP^ be a very small element of the arc. Draw Oy perpendicular to OG. a= OA, a =-4 0(7, 6 = GOP, h6 = POQ : the centre of gravity oiAB is manifestly in the line OG, let x be its distance fi'om measured along OG. Then the element PQ = aSd, its moment about Oy = aSd .acos0; .: moment of the arc ^5= a" /cos Odd feom 0=: — ato0 = + a = a" sin 6, irom 6 = — atx.hV; and it is less than it would be if SFwere all collected in the circular plane generated by QN, that is, Sm < (a; + Sa;) . S V. Hence ,c- always lies between x . k— and x.^- + S F. Whence, as in Art. 174, dx ' dx' .-. u^J{xdV). But X, y being the co-ordinates of the centre of gravity of F, x.V=u = S{xdV). Now dV= nry^dx, by the Differential Calculus ; and, therefore, V='7rji/'dx; consequently X Ji^dx = J xy'dx ; - _Jxi/^dx " /y<^x • From A'rt. 143, it is manifest that y = o. Ex. 1. To find the centre of gravity of a hemisphere. A hemisphere is generated by the revolution of a quadrant whose equation is y = 2aa;' — x ; •■• Sy'dx = ax'--x^, which gives, for the whole hemisphere, by writing a for x, the 2 3* 2 quantity -a'. CENTRE OF GRAVITY OP A SOLID 0¥ ANY FORM. 99 Again, Jxy'dx = / {2aas' - x') dx, 2 3 1 1 which, by writing a for x, becomes — a* ; 1^ x = 5 5 = o a = - of the radius, o o Ex. 2. Given the altitude (c) and the radii (a, 5) of the ends of a parabolic frustum, to find its centre of gravity ; * = 3-^^T6^' ^^^2^ = = a; being measm-ed along the axis from the smaller end whose radius is a. Ex. 3. In a cone, generated by the revolution of a right- angled triangle about one of its sides, 5 = f of that side. Ex. 4. In the solid formed by the revolution of a semi- cycloid about its axis, __« 637r^ - 64 '*'~6" 97^-16 ■ X being measured from the base along the axis. Ex. 5. In the paraboloid, formed by the revolution of the parabola, whose equation is y"''*" = a^a;". _ m + Sn X m + 2n' 2 ' 178. To find, the centre of gravity of a solid of any form. Let Ox, Oy, Os (fig. 36) be the rectangular co-ordinate axes to which the solid is referred by its equation. Let ASPG be a portion of the surface of the solid, comprehended between the 100 CENTRE OP GEAVITY OP A SOLID OP ANY POEM. co-ordinate planes xOz, yOz, and the planes PpNG, PpMB re- spectively parallel to them. Through the point S very near to P draw planes 8snc, Ssmh parallel to the former. Let xyz be the co-ordinates of P, and x-'r^x, y + By, z + Se those of 8. Then, denoting the volume of the parallelepiped Ps hy A, its moment about the axis Ox is greater than if it were all collected in the plane Pq, and less than if collected in the plane Us ; that is, the moment of A is greater than yA, and less than {y + Sy) A. But now if u be the moment of the solid PO about Ox, the moment of BBmPn about Ox will be (by Taylor's theorem ap- plied to two variables x, y) djui . Bx -\-\d^u . {SxY + ... dyU . hy + d^dyU . SxSy + ... + ^d,'u.{Syy+... + ... and by the same theorem, applied to the variable x, the moment of the solid BmP. ahout Ox is d^u . Sx + ^d^u . {SxY + ... and, similarly, the moment of the solid CnP, is d^u . Sy + \dyU . {SyY + ... Subtracting both these from the former, we find the moment of the parallelopiped Ps to be equal to d^d^u.SxSy+ ...; conse- quently, this quantity always lies between yA and {y + Sy) A ; and, therefore, dji^u + ... always lies between ■^°^ SxSy ^®^^® ^° ^ ^^ ^*^ ^™^*' ^^^ consequently the two quantities 2,. g^ and 3/. g|-+S3,.^tend to equality with ys; and d^d^u+ ... which always lies between them, tends to CENTRE OF GRAVITY OP AN ELLIPSOID. 101 d^d^u as its limit ; the three limits are therefore equal ; conse- quently, d^u^yz; •■• u = Uy[yz). Now the volume of PO is equal to jj^z, and its moment about Ox is wherefore, by Art. 139, y-U^=Uv{y.^) (2). By a similar investigation, we should find «-/J»«=/J»M- (!)• And observing that the centre of gravity of the parallelepiped A is ultimately in its middle point, we should find ^•U« = 4U(^') (3). Remark. It is evident, that by taking an elementary parallelepiped, at right angles to the plane xOz, we might also obtain ^•/J«2/=/J«(«y); and if the elementary parallelopiped were at right angles to the plane y Oe, we should find y-!y!>«i = !yh{«>y)> « •/!,/««= /J«M- These formulae are in fact, often more convenient than those first given; and which are the most convenient in a given example is to be determined by the form of the body and its situation with respect to the co-ordinate planes ; the choice must, however, be left to the skill of the reader, as no general rule can be laid down. In every case, the greatest care is requisite to take the integrals between proper limits. 102 CENTRE OP GRAVITY OF AN ELLIPSOID. All the three sets of formulje are comprehended in the following : — which may be readily investigated after the manner of Art. 175. Ex. 1. To find the centre of gravity of the eighth part of an The equation of the surface of the ellipsoid is _2 "r 12 "r _a '■• + (7 This integral is to be taken from y = 0, to that value of y ich mi therefore which makes » = ; or from y = 0, to y = - Va* — a;'* ; and CENTRE OF GRAVITY OF A SURFACE OP REVOLUTION. 103 This integral is to be taken from x = 0, to a; = a ; and therefore Again, to find the value of J^Jy {xz) we observe that ly{xz)=xf^z bcir . i 2- ••• /J. (fl'«)=:5 •/.(«'«; -a;') which, taken between the same limits as before, viz. x = 0, and x = a, gives iTifbc hL (aJ^) = ■ Hence x 16 ■ 16 ' _ 3 a; = -o. — 3 Similarly, y = -^b; and z =xc. o Ex. 2, To find the centre of gravity of a portion of a paraboloid, comprehended between two planes passing through its axis at right angles to each other. If a be its length, and h the radius of its base, the co-ordi- nates of its centre of gravity will be _ 2 _ _ 16& 104 CENTRE OF GRAVITY OF A SUEFACEOP KEVOLUTION. 179. To find the centre of gravity of a surface of revo- Employing the notation and figure of Art. 177, let u be the moment of the surface generated by the arc AP, and therefore Bu the moment of- that generated by PQ; let 8 denote the former, and 88 the latter of these surfaces so generated. Then the moment of 88 about Oy is greater than if it were all col- lected in the circumference of the circle described by P, and less than if collected in the circumference of that described by Q, that is, Bu is greater than x . B8, and less than (x + Bx) . B8; .•. c^ lies between x ^- and x ■^- + BS. bx ox ox. Equating the limits, as before, we have du dx d8_ dx . •. u = 2Tr J{xyds). But u = the moment of 8 about Oy = x8= X . 2irf{yds) ; .'. X. ^TrJiyds) = 27r J{xyds) ; ,', ■xfiyds) = = J{xyds). And it is evident, from the symmetrical form of the surface, that ^ = 0. Ex. 1. To find the centre of gravity of the surface of a cone. If a be the altitude and h the radius of the base of the cone, the equation of the line by which the surface is generated is y=-^ hx ' = (i + -s) <^ ; CENTRE OP GRAVITY OF A SURFACE OF REVOX,UTION. 105 ■which, taken between the limits x=0, and x = a, gives Also, 3a V bx' /, by „ which, between the same limits, gives J{xyds)=iab^^+¥; .: x.^b'/J+¥ = iab'/^ni'', Ex. 2, To find the centre of gravity of the surface generated by the revolution of an arc of a circle about a diameter. The centre of gravity bisects the axis of the zone. Ex. 3. To find the centre of gravity of the surface generated by the revolution of a semi-cycloid about its axis, Ex. 4. To find the centre of gravity of the surface of a paraboloid. Taking the focus as origin of polar co-ordinates, we find the distance of the centre of gravity from the directrix Q sec' - - 1 3m 2 sec'--! E. S. 14 106 CENTKE OF GRAVITY OF A SUEPACE OP ANT POEM. Ex. 5. To find tlie centre of gravity of the surface generated by the revolution of a node of the Lemniscate about its axis. a 1 - cos' ie a 2 V2 - 1 aj = 6 ■ 1 - cos 12 ■ V2 - 1 180. To find the centre of gravity of a surface of any form. If, in Art. 178, we use A to denote the elementary surface P8 instead of the prism Ps, we shall have the limit of ~- = \/TT{d;^f+Wf ; and by proceeding exactly as in that Article, we shall find X . u ^^1 + {d^y + id^f=Uy {« ^^1 + {d^r + iA^)\ z.Uy >/T+Wf+Wf=Uy {« '^i + {dj,r+{d,zy}. 181. To find the centre of gravity of a curve of dovMe curvature. If we use 8 for the length of the curve line, and hS for the length of a very small portion of it, we shall have the so limit of s- = d^8 = Vl + [d^yf + {d^Y, and it will be found that x8=jjx>^/\ + {d^yy+{d,zy, y8=J.y'Jl+Jd~yy+Wr, z8 = !^^/l + {d^yy+{d^zy. 182. We shall now add a few examples of finding the centre of gravity when the density is variable. Questions of this kind depend upon the formulae of Art. 138, viz. — 53 tjpVx) , - tjpVy) , -_{tpVzl ''-%{pV)^ ^-t{pV)' '-t{pV)- CENTRE OP GEAVITY OP A LINE. 107 183. To find the centre of gravity of a physical line, the density of which at any point varies as the n* power of its distance from a given point in the line produced. Let AB be the given line, and C j j-y- the given point ; n = the density at a '^ •* ^ point in AB, whose distance from C=l; a= CA, b= CB, X = CP, hx = PQ. Since a physical line is of uniform thickness throughout, we may take the length of any portion of it as the measure of the volume of that portion ; hence hx = the volume oi PQ, and as the density varies as (distance from 0)"; .'. r : a;" :: /* : fix". Wherefore the density at P is /ao;", and PQ is ultimately of uniform density, therefore the mass of PQ is = fix^hx ; .-. the mass of AB= S {fuxfSx) = fi% (af'Bx) = fi,Jx''dx = li = '*• n + l ' between the limits x = a and x = b. Again, the moment of the mass of PQ about C = lj,x"-'^Sx; .: moment of AB about C = fijaf*^dx x"^' + G n+l &"«- ««« = fi. ^ +G n + 2 "''*■ w + 2 between the same limits as before. 108 A TRIANGLE OF VARIABLE DENSITY. Wherefore x being the distance of the centre of gravity of the line from 0, we have -_ t{pVx) _ n+l h'^-a"'^ Eemakk. When m = — 1, M/X = /*/j = fi . logjo; + G And ,ijjc'"-^ = ,j.{b-a); - h — a .'. x = Again, when m = — 2, and fji,Jx"*'^dx = fj, I dx X 1 ^ = /*log.-; a5 , 6 Ex. 2. To ^^jm:? and ^ + S^ respectively; then the portion comprehended between them will be equal to the volume generated by Pq, in revolving through an angle S^, and therefore is = r sin 9 .B<}). rB6 , &• = r'Sr . sin 689 . S^. . And the density of this element is /lir", and therefore its mass is fir^'^'Sr. sin 989. S4>, and its distance from the plane ABO is r-sin^.sin^, as is evi- dent from the construction J and therefore its moment with jespect to the plane ay = /ir"+»S»- . sin" 9S9 . sin ) = ^!B{-ain'6cos6+C) n+4 = — 4/«(l-cos;8)sin'^, taken from ^ = to ^ = /8. Now /» sin" = J/« (1 - cos 20) = \{9-^sin26)+ G _ a sin 2a ~2 ^' taken fi-om ^=0 io 9 = ol; SPHERICAL -WEDGE OP VAEIABLE DENSITY. 115 therefore the moment of the solid with respect to the plane m/ n + 4 sin" -x (a — sin a. cos a) ; ■ ,/8 , „ sm"^ , - _ w + 3 a 2 a — sm a cos a _ ■■^~w + 4"2' . ,a is ' — — B and therefore w = s cot — ^ 2 B . ^ , „ cos - sm ^ _n + S a 2 2 a — sm a cos a ~w + 4"2* . jK ■ is ■ sm"- Ex. 5. Find the centre of gravity of a cone, the density at every point of which varies as the square of its distance from a plane through the vertex parallel to the base. 5 = - of the cone's axis. 6 Ex. 6. Find the centre of gravity of the eighth part of a sphere, the density at any point, whose distance from the centre is r, being proportional to a . irr -sm— -, 7- 2a' where a denotes the radius of the sphere. 5 = y = i = a(Jl_y. 116 guldin's peoperties. guldin's properties. 184. The surface generated hy a plane curve line, which re- volves about a fixed axis, is equal to the product of the length of the curve line by the length of the path described by its centre of gravity. For let AB (fig. 33) be the curve line, and Ox the line about which it revolves through an angle 0; then using the same notation as in Art. 174, the point F describes an arc =yd, con-^ sequently the arc FQ describes a zone, of which the length is y9 ultimately, and the breadth = Ss ; hence the area of the zone is ultimately = 6yBs ; and therefore the area of the whole surface generated is = -Z{dySs) = 0Jyds; the integral being taken between the limits corresponding to x=OC, x=OI). But if y be the distance of the centre of gravity of the arc AB from the axis Ox, we have shewn in Art. 174, that y . (arc AB) = J yds, between the same limits ; hence the surface generated = 0y . (arc AB). Now 0y is the length of the path described by the centre of gravity, consequently the last equation expresses the property to be proved. 185. The volume generated by a plane area, revolving ahout a fixed axis in its own plane, is equal to the product of the area into the length of the path described by its centre of gravity. Let A be the revolving area; hA a portion of it so small that it may be all considered to be at the same distane y from the axis. Then if ^ be the' angle through which the area re- volves, BA will describe a volume which may be considered to be a thin cylinder bent into the form of a portion of a ring. The area of the base of this cylinder is hA, and its length is y0, consequently the volume generated by hA = 0yhA; guldin's properties. 117 and therefore the whole volume generated But if y be the distance of the centre of gravity of the area A from the fixed axis, we have from the nature of the centre of gravity X{BA).y = t{yBA), or Ay = %{yhA); hence the whole volume generated = eyA: an equation which expresses the property which was to be proved. Remark. If the curve line in Art. 184, or the plane area in Art. 185, does not revolve about a fixed axis during its whole motion but moves in any such manner that it may at any moment be assumed to be revolving for an instant about a fixed axis in its plane; then the propositions in those articles will be true for each instant; and consequently, by adding these results together, those articles mil he true for the whole motion whatever be the nature of the path of the centre of gravity. But it is necessary to notice that when the instantaneous axis, about which the generating curve or area is supposed to be revolving, is in such a position that the instantaneous axis divides the curve or area into two portions, the part generated by one of those portions during that instant is to be considered positive, and that gene- rated by the other negative, and the propositions fail in this case. As long therefore as the line of instantaneous revolution lies en- tirely out of the limits of the generating curve or area, the pro- positions in Arts. 184, 185 hold true, viz. : The surface generated hy a plane curve line which moves in any manner (subject to the limitations just named), is equal to the product of the length of the curve line hy the length of the path described hy its centre of gravity. And The ■volume generated hy a plane area, which moves in any wiawwer (subject to the same limitations), is equal to the product of the area into the length of the path described hy its centre of gravity. CHAPTEE VII. ON MECHANICAL INSTRUMENTS. 186. Every machine, how complicated soever its con- stmction, is found to be reducible to a set of simple ones, called the Mechanical Powers. These, though authors differ considerably on the subject, are generally said to be six in number, viz.: 1. The Lever;' 2. The Pulley; 3. The Wheel and Axle; 4. The Inclined Plane; 5. The Screw; 6. The Wedge. These are not the most simple machines ; for, rods used in pushing, and cords used in pulling, are much more simple ; in fact, every machine will be found to be a combination of levers, cords, and inclined planes, and these might consequently be called the simple Mechanical Powers, with much greater pro- priety than the six before mentioned. As, however, these are not very complicated in construction and application, and as levers, cords, and inclined planes do always, in actual practice, present themselves in machinery, in one or more of these six combinations, it will very much facilitate our enquiries into the properties of any proposed machine, to be acquainted with their forms and the advantages to be expected from their use. In speaking of any machine, the force which is applied to work it is called the Working Power, or simply, the Power; the weight to be raised, or resistance to be overcome, is called the Weight; the point where the machine is applied to produce its effect is called the Worhing Point; and the fraction Weight Power MECHANICAL INSTEDMENTS. 119 is called the Mechanical Advantage (by some authors the Power, but this creates confusion by confounding it with the former definition of power) of the machine. 187. Every machine is useless until put in motion, and therefore its parts ought to be so arranged and adapted that the given power may be able to overcome the proposed weight, and move it with the requisite degree of celerity ; but, in dis- cussing the theory of the Mechanical Powers, it will be suffi- cient to determine the ratio of the weight to the power when they balance each other, for then the slightest addition made to the power will cause it to preponderate and put the machine in motion. 188. It is veiy important to remark, that when a power is employed in working a machine, a very considerable portion of it is found not to reach the working point, being spent in over- coming the stiffness of the cords and the roughness of surfaces whichjnib against each other. Much power is also lost through the imperfection of workmanship, the bending of rods, beams and other materials, which are intended to be rigid, the resist- ance of the air, &c. ; but the introduction of the consideration of these things, though very important in a practical point of view, would only tend to embarrass the student by rendering our investigations tedious and perplexing. We shall therefore at first suppose cords to be perfectly flexible, surfaces quite smooth, workmanship geometrically exact, rods and beams per- fectly rigid, the air to offer no resistance ; &c. " It is scarcely necessary to state, that, all these suppositions being false, none of the consequences deduced from them can be true. Nevertheless, as it is the business of Art to bring machines as near to this state of ideal perfection as possible, the conclusions which are thus obtained, though false in a strict sense, yet deviate from the truth in but a small degree. Like the first outline of a picture, they resemble in their general features that truth, to which, after many subsequent corrections, they must finally approximate. " After a first approximation has been made on the several suppositions which have been mentioned, various effects, which 120 MECHANICAL INSTRUMENTS, have been previously neglected, are successively taken into account. Eoughness, rigidity, imperfect flexibility, the resist- ance of air and other fluids, the effects of the weight and inertia of the machine, are severally examined, and their laws and pro- perties detected. The modifications and corrections thus sug- gested, as necessary to be introduced into our former conclusions, are applied, and a second approximation, but still only an ap- proximation to truth is made, For, in investigating the laws which regulate the several effects just mentioned, we are com- pelled to proceed upon a new group of false suppositions. To determine the laws which regulate the friction of surfaces, it is necessary to assume that every part of the surfaces of contact is uniformly rough; that the solid parts which are imperfectly rigid, and the cords which are imperfectly flexible, are con- stituted throughout their entire dimensions of a uniform material ; so that the imperfection does not prevail more in one part than another. Thus all irregularity is left out of account, and a general average of the effects taken. It is obvious therefore, that by these means we have still failed in obtaining a? result exactly conformable to the real state of things : but it is equally obvious, that we have obtained one much more conformable to that state than had been previously accomplished, and suffi- ciently near it for most practical purposes. " This apparent imperfection in our instruments and powers of investigation, is not peculiar to Mechanics ; it pervades all departments of natural science. In Astronomy, the motions of the celestial bodies, and their various changes and appearances, as developed by theory, assisted by observation and experience, are only approximations to the real motions and appearances which take place in nature. It is true that these approximations are susceptible of almost unlimited accuracy ; but still they are, and ever will continue to be, only approximations. Optics, and all other branches of natural science, are liable to the same observations*." • Captain Kater's Treatise on Machines. THE LEVEE. 121 I. On the, Lever, 189. Def. a Lei)&r is a rigid rod straight or bent, moveable in a certain plane about one of its points, which is fixed and called its ful6rum.' 190. In a lever iohen there is equilibrium the power and weight are to each other inversely as the perj^endiculars from the fulorum upon the directions in which they act. (Both the power and weight are supposed to act in the plane in which the lever is moveable, which is technically called the plane of the leVer). Let AB (figs. 40, 42) or AO (fig. 41) Or BO (fig. 43), be a lever whose fulcrum is 0; A, B the points at which the power P and Weight TF'act; €Y, C^ perpendiculars from upon their directions. Then the equilibrium will not be dis- turbed by applying at two for6es F, F" parallel and equal to P, and two others W', W" parallel and equal to W: We haVe thus, six forces acting on the lever, of which (P, P") and ( W, W") form two couples, and the two remaining forces P', W being counterbalanced by the reaction of the fulcrum, may be retnoved. Hence the couple (P, P") whose arm is OY, balances the couple {W, W") whdse arm is G^j consequently their moments must be equal ; .-. P. 6T= W. OZ. 191. To find the pressure on the falcrum C. We liave shewn that P and W are equivalent to two forces P, W acting at 0, and two equal couples (P, P"), {W, W"); these couples may be removed because they are equal and opposite and therefore balance each other. It appears, then, that Pand PTare equivalent to P' and W' acting at 0. Con- sequently the pTessure on the fulcrum is the same as if the powet and Weight Were both transposed tb it parallel to themselves. 192. We have considered the weight of the lever incon- siderable when compared with P and W, but if this should not E. s. 16 122 .THE LEVEE. be the case, let w be its weight, G its centre of gravity. Then we may suppose the whole force w, which gravity exerts upon the lever,. to be applied at O; this force may be converted into a couple whose moment is w . GG, and as^ there is equilibrium between the three couples, the sum of the moments of the two which act in one direction [i.e. positive or negative) must be equal that of the third ; .-. P.GY + w.CG=W.GZ, is the equation of equilibrium in this case. 193. Exaniples of levers of the same kind as the one in fig. 40, are the common balance, steelyards, pokers, &c. ; and scissors, pincers, &c. are instances of two such levers having a common fulcrum. Examples of levers of the same kind, as those in figs. 41j 43, are the oars and rudders of boats, cutting-knives moveable about one end, &c.'; and tongs, sheep-shears, &c. are instances of the combination of two such levers with a common fulcrum. Examples of the bent lever, in fig. 42, are gavelocks, jemmies, bones of all animals, &c. 194. We have defined a lever to be a rigid rod, but we may consider any rigid body having a fixed axis as a compound lever, whose fulcrum is the axis ; and if powers Pj, Pjj, Pg...P„, act upon this lever, and balance the weights W„W„W, Tn,, then P,Pi + P,p^+ +P„Pn= W^w^+ W^w^+ + W^w^, or t{Pp) = t{Ww); the powers and weights being supposed to act in planes at right angles to the axis, andpj, p^...p„; w^, w^...w^ being the respective perpendiculars from the axis upon the directions in which the powers and weights act. This may be proved as before, by converting the powers and weights into couples, and then transposing them into one plane ; and it will also appear, that the pressure on the axis or fulcrum is the same as it would be if all the forces were transported in their own planes parallel to themselves to the axis. THE PULLEY. 123 II. On the Pulley, 195. Def. a Pulley is a wheel of wood or metal, turning on an axis through its centre at right angles to its plane, and usually enclosed in a frame or case, called its hhck, which ad- mits a rope to pass freely over the circumference of the pulley, in which there is usually a groove to receive it and prevent its slipping out. The pulley is said to be fixed or moveaile, accord- ing as its axis is stationary or not. An assemblage of several pulleys is called a system of pulleys. 196. It will be necessary before investigating the properties of the pulley to premise, that if a cord be stretched by two equal forces applied at its extremities in contrary directions, there will be a tendency to break ; the force which the rope, in consequence of the cohesion of its particles, exerts to resist this tendency, must be equal and opposite to that which causes the tendency ; it is called the tension of the rope. Hence tension is a force which is exerted equally in every part, tends fr-om the ex- tremities of a cord towards the middle, and is always equal to either of the equal forces, by which the cord is stretched. If one end of the cord, instead of being acted on by a force, be fastened to a fixed point, the tension will not be altered; for the fixed point will, by its reaction, exactly supply' the place of the force. 1 97. In the single fixed pulley when there is equilibrium the power and weight are equal.. Let ABK (fig. 44) be the pulley, C its centre, CN its block ; P and W the power and weight acting at the extremities of the cord passing over the pulley, and having the part AB'ia contact with it. Then we may consider the pulley ABK as a lever whose fulcrum is C; and therefore drawing the radii OA, CB to the points A and B, we have P.GA= W.CB; .-. P= W. 124 THE PULLEY. Hence it appears that no mechanical advantage is gained by the use of this pulley ; the only purpose for which it is used is to change the direction in which a force is transmitted. 198. To det$rmvne the jaressure on the fulcrum C. Transpose the forces P and W to that point, and put 9 for the angle at which APaaA. BW &t& inclined to each other, and let R be the pressure, which is, of course, the resultant of these transposed forces, and bisects the aiigle between liiem ; hence resolving these forces io the direction of B, we find 6 Ii = P cos - + Pcos- - 2Pcos - . This pressure is transmitted to N, the fixed point to which the block is attached. 199. In the single nwveable pulley when there is eguilibrmm the power is to the weight :: 1 : 2 x cosine of half the angle be- tween the strings. Let the power P act at the extremity P of the cord PABD (fig. 45), which passing under the pulley has the part AB in contact with it; and its other extremity fastened at D. The weight W hangs from the block at N. Exactly as in the last case, we find the pressure on the centre C to be 2Pcos|, 6 being the angle between the strings AP, BD ; this force is transmitted through the block in the direction C^, bisecting the angle d ; wherefore the action of W must be equal to it and in the opposite direction, otherwise there cannot be an equiUbrinm ; .-. PF=2Pcos|, and consequently P : TF :: 1 : 2 cos - . THE PULLKY. 125 200. No mechanical adeantage can be gained by the use of this pulley, unless d 2 cos - > 1, Q and .'. cos - > ^ > cos 60°; that is, unless the strings are inclined to each other at a less angle than 120°. The greatest possible advantage will be gained when the Q strings are parallel, for then ^ = 0, and cos - = 1, and therefore W = 2P. 201. If the weight of the pulley and its block be con- siderable, it must be considered as an additional weight, and added to W in the above expressions. 202. To find the conditions of equilibrium in a system of pulleys, where each pulley hangs hy a separate string, the strings being all parallel. Let A^, A^, As, ••• (fig- 46) be the pulleys; M^, M^, M^ ... the points where the strings are fastened to an immoyeable block. Then P is the tension of the string passing under A^. The two strings A^P, A^M^ have to support the tension of N^Aj^; so N^A^ and M^^ support that of N^A^, and so on; therefore, (P=) tension of A^P : tension of N^A^ :: 1 : 2, tension of N^A^ : tension of N^^ "1:2, tension of ^y4, : tension of iV^gTr(== PF) :: 1 : 2; .-. P : W :: 1 x 1 x 1 x : 2x2x2.. If n be the number of moveable pulleys, then P ; W:: r : 2"; ,-. W=^TP. 126 THE PULLEY. 203. If the weights of the pulleys and blocks are con- siderable, let A^, A^, A^... represent the weights of the' pulleys and blocks denoted by those letters in the figure; and let Tj, Tj ... be the tensions of the strings N^A^, N^A^ .... Then, as before, the weights of the pulleys must be added to the tensions of the respective cords which they support ; .-. P: T^+A^ :: 1 : 2; .-. T^ = 2P-A^. Similarly, T^ = '2T^-A^ =^2'P~2A,-A„ = 2»P-2Mj-2^,-^„ and so on, the law being manifest ; then, since the tension of the last string = W, we have W= 2"P- 2"-' A, - r-^A, - 2"-M, - -A,. It appears from this expression, that the weights of the pulleys diminish the advantage of this system. 204. If all the pulleys are equal, then W= TP~ A^ (2"-' +2"-"+ + 1) 9"— 1 = 2"P-(2"-l)^, = 2»(P-^.)+A; .'. Tr-^i=2"(P-^,). Hence, if we suppose both the power and weight diminished by the weight of a pulley, we may then neglect the consideration of the heaviness of the pulleys. 205. In the system (fig. 47) where each string is attached to the weight, let T^, J'^-.-be the tensions of the first, second ...strings; then if the Weights of the pulleys are inconsiderable, we have THE PULLEY. 127 2;=22; = 2'P, y,= 22; = 2'P; and if there be n sepaxate strings, Now W is supported by the tensions of the n strings fastened to the block B, and .-. W=T,+ T^+ + T„ = P(l+2 + 2''+...2"-') 2-1 = P(2''-1). 206. In the system (fig. 48), let T^, T^... be the tensions of the first, second... strings; then Tj^ = F; and T^ has to support three tensions equal to P; therefore T, = P, 2; = 3?;=3'P; and if there be («) difierent strings, the tension of the last is ?;=3''-'P. Now the weight W is supported by two strings whose tensions are T^, two of which the tensions are T^, &c.; .-. W=2T^'+2T^+ + 2y„ = 2P. (1-1-3 + 3'+ 3"-') = 2P.^ 3-1 = P.(3»-1). Remark. If the weights of the pulleys and blocks are not ■inconsiderable, they may be taken into account, in this and every 128 THE WHEEL AND AXLE. other system, by adding each to the tension of that string which supports it, as in Art. 203. 207. In the system, fig. 49, the weight W is supported by the tensions of all the strings at the lower block, and as it is the same string which passes round all the pulleys, the tension of every part =vP; wherefore, if there be n pulleys in the lower block, there are 2ra strings supporting the weight, and therefore W= 2nP. III. On the Wheel and Axle. 208. The wheel and axle consists of a cylinder and a wheel firmly attached to each other, and being moveable about a fixed axis coinciding with the axis of the cylinder, and passing through the centre of the wheel at right angles to its plane, as in fig. 50. The power P acts by means of a cord wrapped round the cir- cumference of tte wheel O, and the weight W is fastened to a cord which is wound upon the cylinder AS as P turns the machine round its axis ; and thus Wia raised. 209. To find the condition of equilibrium on the wheel and axle. We may consider P and W as forces acting upon a rigid body with a fixed axis, and therefore their moments about that axis must be equal ; .". Px (perpendicular upon its direction firom the axis), = W. (perpendicular upon its direction from the axis). Now these perpendiculars are respectively the radii of the wheel and of the cylinder ; .•. P. (radius of the wheel) = W. (radius of the axle). 210. If the thickness of the rope be considerable, it must be taken into account. We may suppose the actions of P and TF to be trans- mitted along the middle or axis of the rope, and then the per- THE INCLINED PLANE. 129 pendiculars upon the directions of F and W will be respectively equal to radius of wheel + radius of rope, and radius of axle + radius of rope, and the condition of equilihrium is P . (rad. wheel + rad. of rope) = W (rad. axle + rad. of rope) . This diminishes the advantage of the machine. 211. The pressure on the axis of this machine may be found by transposing P and W in their own planes, parallel to them- selves, to the axis. IV. On the Inclined Plane. 212. This machine is nothing more than a plane inclined to the horizon. The condition of equilibrium may be thus found. Let A^ (fig. 51) be the plane ; A parallel and BC perpen- dicular to the horizon ; W the weight, P the power. Draw TFS perpendicular to the plane, WG perpendicular to the horizon. P is supposed to act in the plane R WB. The weight W is kept at rest by three forces, viz. P in the direction WP: gravity (= W) in the direction WG, and reaction R of the plane in the direc- tion WR. Denote the angle PWB by 6, and the inclination BA G of the plane to the horizon by i; and resolve the three forces, acting on the point W, in a direction parallel to the planes; the sum will be PcobPWB- W cos AWG + B cos RWB = Pcos 9 — TF. sin i. But since there is an equilibrium, this sum must be equal to zero, .•. Pcos 5= Wsini, which is the condition of equilibrium. E.8. 17 130 THE SCREW. 213. If P's direction should happen to be parallel to the plane, 6 = and cos ^ = 1 ; .-. F= Wsinl But if P's direction should happen to be parallel to the horizon, d = — i and cos (— i) = cos i; .". Pcosz= TFsint; /. P= Wta.ni. 214. To Jind the reaction of the plane. Kesolve the forces in a direction at right angles to that in which P acts : .-. B sin BWF- Wain G WF= 0, or Bcoad-Wam{90 + {+6)=0; cos a V. On the Screw.' 215. This mechanical power is a combination of the lever and inclined plane; it may be conceived to be thus gene- rated. Let ABCD (fig. 52) be a cylinder; BEFO a rectangle whose base BE is equal to the circumference of the cylinder. Divide this rectangle into any convenient number of equal rectangles QFJ, IH, CK; and draw their diagonals BH, OK, IF. Then, if this rectangle CE be wrapped upon the cylinder, so that BE coincides with the circumference of the base, E, H, K, F will respectively fall upon the points B, G, I, of the cylinder, and the lines BH, GK, IF will trace out upon its •surface a continuous spiral thread BLGMINO winding uni- formly up the cylinder. The cylinder is usually made pro- tuberant where the spiral line BL GMINC ialh upon it so that the thread becomes a winding inclined plane, projecting from the cylinder as in fig. 53, and differing from the inclined plane THE SCREW. 131 BH* in nothing but its winding course. This is the external screw. The internal screw is formed by applying the paral- lelogram BEFO to a hollow cylinder, equal to the former, and making a groove where the thread falls to fit the protuberant thread of the external screw. This internal screw is often called a nut, and the other the screw. When the two screws are thus adapted to each other, the external or the internal screw, as the case requii'es, may be moved by means of a lever about their common axis, as in figs. 54, 55. The force being applied to the lever at right angles to jt, in a plane parallel to the base of the cylinder. The screw and nut thus applied to each other, resemble two inclined planes, such as BHQ and HBE, one of which is laid upon, and slides down the other ; and as the planes wind round the cylinder a rotatory motion ensues. When the machine is worked, the weight is laid upon the nut, and thus causes its inclined plane to press upon that of the screw in the direction of gravity. The consequence would be, that the nut and weight with it would begin to slide down the thread of the screw and descend, but this is prevented by confining the nut so that it cannot have a rotatory motion, but only one of ascent or descent. The screw is then turned round by means of a lever passing through its head, and thus its inclined thread sliding under that of the nut, forces the nut and the weight upon it to ascend, just as by pushing the inclined plane EBH in the direction EB, the • The following illustration renders this very clear : — *' When a road directly ascends the side of a hill, it is to be considered as an in- clined plane ; but it will not lose this mechanical character, if, instead of directly ascending towards the top of the hill, it winds successively round it, and gradually ascends so as after revolutions to reach the top. In the same manner a path may be conceived to surround a pillar by which the ascent may be facilitated upon the prin- ciple of the inclined plane. Winding stairs constructed in the interior of great columns partake of this charaeler; for although the ascent be produced by successive steps, yet if a floor could be made sufficiently roush to prevent the feet from slipping, the ascent would be accomplished with equal facility. In such a case the winding path would be equivalent to an inclined plane, hent into such a form as to accom- modate it to the peculiar circumstances in which it would be required to be used. It will not be difficult to trace the resemblance between such an adaptation of the inclined plane and the appearances presented by the thread of the screw; and it may hence be easily understood that a screw is nothing more than an inclined plane, constrncted upon the surface of a cylinder."— Captain Kateb's Machines. 132 THE SCREW. plane QBH would be made to ascend. One turn oif the screw raises the weight through an altitude equal to the distance be- tween two threads. Sometimes, however, the nut is firmly fixed so as to admit of no motion whatever (as. in fig. 54) ; and then the thread of the screw, in sliding under that of the nut, forces the screw to descend and press violently against any obstacle which may be opposed to it. In some cases the weight is not applied to the nut, but to the screw ; but as the two inclined planes are perfectly equal and similar, it will require the same force to support a weight on one ^s on the other, and for this reason one investigation will serve for both. As before observed, the screw is worked by applying a power P at the end of a lever ; and the moment of P to turn the screw round = P X length of the lever, and therefore P is equivalent to a force P X length of the lever rad. cylinder acting immediately at the thread of the screw in a horizontal direction parallel to that in which P acts. Now the inclined plane on which W rests, by means of the nut, is only BH wrapped round the cylinder; its inclination to the horizon or base of the cylinder is therefore HBE. Hence we have ^^ length of lever ^^,^^^ (Art. 213) rad. 01 cylmder ^ BE _ T^r distance between two threads ~ ' circumf. of cylinder But the radii of circles are proportional to their circum- ferences ; length of lever _ circumf. described by power rad. of cylinder circumf. of cylinder ' THE WEDGE. • 133 p circumf. by power _ tt?. dist, between two threads _ ' circumf. of cylinder ~" ' circumf. of cylinder ' „ „7- dist. between two threads circumf. described by power ' As the distance between two successive threads can be made very small, and the circumference described by the power as large as we please, the advantage of this machine is very great ; and it is remarkable, that it does not depend upon the thickness of the screw. VI. On the Wedge. 216. A wedge is the solid figure defined by Euclid (Book XII. Def. 4) as a. triangular prism. Its two ends are equal and similar triangles, and its three sides rectangular parallelograms (see fig. 56). It is principally used in splitting timber, and separating bodies which are very strongly united, and in raising very heavy weights through a small altitude, for the purpose of introducing a lever, or some other more convenient machine. AB is called its edge, GDEF its head, GABD and FABE its faces. When used, its edge is introduced into a small cleft pre- pared to receive it, and then by violent blows with a hammer on its head its body is driven between the substances, which are thus separated by an interval equal to the breadth of the head. After this, a larger wedge may be introduced, if neces- sary, and treated as before, until the requisite degree of sepa- ration is effected. As the wedge is di-iven in by violent blows, if its sides were perfectly smooth it would start back by the pressure of the obstacles upon them in the interval between the strokes; and thus we should fail in effecting and maintaining the requi- site degree of separation, and the machine would be rendered useless. In practice, however, the friction in this machine is always so great as to prevent any recoil, and forms, in fact, the principal resistance to be overcome in driving the wedge. The mode of working this machine will at once present itself to the 134 - GENERAL PROPERTY OF MACHINES. reader as Ibeing totally different in principle from that of all the other machines we have described. These are made to work by the constant and steady exertion of a power, uniformly press- ing upon that point of the machine at which it is applied, and gradually producing motion in the weight ; but in this machine motion is accumulated in a hammer, by suffering it to descend fi'om an altitude, and is suddenly by an impulse transferred to the wedge. In this case it must evidently be a useless labour to attempt to calculate the ratio of P to W, when they act by pressures, as in the other mechanical powers, and are in equi- librium. It is true, when we know this ratio, a slight increase* of P will gradually produce a motion in W, and thus separate the obstacles; but this mode of working the machine is so widely different from that actually practised, that it would be a waste of time and labour to attempt an explication on Statical principles. A slight^stroke with a hammer is found to be far more effective than several tons of pressure. The only theo- retical property of the wedge which agrees with practice is that its advantage is increased by diminishing its angle DBE. All cutting instruments, such as knives, swords, hatchets, chisels, planes used by carpenters, nails, pins, needles, &c. are modifications of the wedge. Of these, knives, planes, pins and needles, are usually worked by pressure, but swords, hatchets, chisels, nails, &c. are worked by percussion. GENEEAL PROPERTY OF MACHINES. 217. If the nature of a machine be such, that when the power and weight balance each other in one position of the machine they will balance in every position of it, a very re- markable property appertains to it, deducible from the principle of virtual velocities, which we may state as follows : The power is to the weight as the space moved through iy the weight when the machine is put in motion is to the space moved * This, however, supposes the sides to be perfectly smooth, for otherwise the friction itself, without the assistance of any power at all, would preserve the equili- brium. white's pulley. 135 through hy the power in the same time ; the spaces being measured respectively in the directions in which the power and weight act. Let tlie whole space (measured thus) through which the power P moves be divided into a very large number of spaces Sj, Sj..., and let s\, s',... be the corresponding spaces described by the weight W; then S=Si + «j+ t 1,1, s=s, + s^+ But because P and W are always in a position of equilibrium ; «i, s\, are their virtual velocities for the first position; .-. Ps^+Ws\ = 0, Similarly Ps^ +'Ws'.^ = 0, for the 2nd position Ps^ + Ws'^ = 0, 3rd .-. Ps+Ws' = 0; P __£ •'■ W s' This equation expresses the property enunciated. The negative sign points to the fact, that the direction of the action of one of the two forces P, W is opposed to the direction in which the point moves on which it acts. Mechanical powers possessing this property are ; — (1) The straight lever supporting weights. (2) Air the pulleys in which the strings are parallel. ' (3) The Wheel and Axle. (4) The Screw. (5) The Inclined Plane, only when the Power haaigs by a string passing over the top of the plane. WHITE'S PULLEY. 218. In the common systems of pulleys each pulley has its own independent centre of motion; and consequently as they 136 hunter's screw. air move with different velocities and with different degrees of- pressure, some of them will be liable to greater wear than others, which will very much tend to increase the friction and other inequalities and resistances ; and will greatly diminish the efficiency of the machine. To obviate these difficulties, Mr James White invented a system of pulleys (fig. 57), consisting of two blocks A, B, into which grooves were cut, the radii of those in the upper block being as the numbers 1, 3, 5... and the radii of those in the lower block being as the numbers 2, 4, 6... Now, suppose the lower block to be raised through one inch, then each of its strings will be shortened one inch, and therefore the circumference of the pulley BB^ describes one inch ; that of AA^^, two inches; that of BB^, three inches, and so on; which numbers being proportional to the radii of the respective pulleys, they will all move, with the same angular velocity; and, con- sequently, each block instead of being composed of separate pulleys may consist of one solid piece of wood or metal, contain- ing the grooves before mentioned. The disadvantage of this system is, that if the cord be at all elastic it cannot be kept stretched in every part on account of the tension not being the same throughout, so that the smaller grooves are rather a hin- drance to the motion than a help. HUNTER'S SCREW. 219. We have seen (Art. 215) that the advantage of a screw increases in proportion as the distance between the threads dimi- nishes, and as the length of the lever at which the power acts increases ; therefore, by making the threads of the screw suffi- ciently fine, we may increase the advantage as much as we please ; but there is a limit to the fineness of the threads ; for as all the weight is borne upon them, if they are too fine they will not be sufficiently strong to bear the load. If we, on the other hand, increase the length of the arm of the lever, with the view of increasing the advantage of the screw, the power will have to describe an inconveniently large circle. To obviate these natural defects, and yet increase the advantage to any degree, Mr Hunter invented the screw in fig. 58 ; A and B are two common screws, of which A is also a hollow screw to admit B, which is fastened hunter's screw. 137 to the moveable plate D of wpod or metal. If D, d be the dis- tances between two threads of the screws. A, B respectively; then, while the power describes one circumference, A descends through Z>, and B ascends in A through d, and the space de- scended by the plane D\^D—d; for when A descends it carries B along with it, though B is at the same time ascending in A. Wherefore, by Art. 217, P. (circumf. described by P) = PF. (-D -^ " PP — =sm«, (Art. 213) W 1 — - — =-!-., (Art. 214) pressure cos ^ ^ ' friction sin i . = tan^; pressure cos i .'. friction = (pressure) . tan i. 234. The fraction , is usually called the coefficient pressure of friction, and is taken as its measure. It appears then, that in the last experiment the coefficient of friction is equal to the tangent of the inclination of the plane. 235. It being granted that the friction is proportional to the pressure when the surfaces are given, then, whatever be the magnitude of the surfaces in contact, the friction will remain the same, so long as the pressure is the same. Let the body W (fig. 51) have faces, whose areas are C and P square inches ; then when the first face is in contact with the plane, the whole pressm'e is supported on square inches, and therefore the pressure on each square inch, is equal to pressure and therefore the friction upon each square inch of surface pressure ^ = *- — ry — .tant. Consequently the friction upon the whole surface pressure ^ . i r ■ i. = ^ — ^ — . tan t X number oi square inches pressure . . „ = ~ — py — .tan IX u = (pressure) . tan i. In the same way it may be shewn that the friction upon the second surface = (pressure) . tan *, FRICTION. 153 and therefore the friction of a body is the same whether the surface on which it rests be large or small. When the surface is very small in proportion to the weight, the pressure on each square inch becomes very large, aijd then the friction, as ob- served in Art. 233, becomes somewhat less in proportion to the pressure; and therefore the friction is less, in a slight degree, when the body rests upon a small surface than a larger. 236. These are the chief properties of statical friction; it does not belong to us to investigate those of dynamical friction ; but to make the subject complete we shall annex the following summary of results which have been obtained by various experi- mentalists. (1) Dynamical friction is a uniformly retarding force : and it diminishes as the pressm^e increases. This is only true when the surfaces in contact are hard ; for in experiments made with bodies covered with cloth, woollen, &c. the friction was found to increase with the velocity. (2) In the same body Statical friction is greater than Dy- namical friction ; i. e. it requires a greater force to put a body at rest in motion, than is requisite to .preserve the motion un- diminished when once it is produced. This was thought by Professor Vince to arise from the cohesion of the body to the plane when it is at rest, which does not happen when the body is in motion. (3) When a body of wood is first laid upon another, the Statical friction increases for a few minutes, when it attains its maximum, and no further alteration takes place. In making experiments, therefore, it is necessary to wait some time before the body is put in motion. (4) Friction between substances of the same kind is greater than when they are of different kinds. (5) The velocity has very little, if any, influence except when one body is composed of wood and the other of metal, in which case the resistance increases with the velocity. E. S. 20 154 FKICTION. (6) It is also found that friction is diminished: by oiling and polishing the surfaces in contact. There is a limit however to the latter, for if they be very highly polished, the resistance increases. (7) The friction of cylinders rolling on planes, is propor- tional to their pressures directly and their radii inversely. It is remarkable, that friction of this kind, unlike that between two planes, is not diminished by greasing or oiling the surface of the planes and cylinder. This kind of friction is much less than that produced by rubbing. CHAPTEE IX, ON ELASTIC STEINGS. 237. Strings made of certain substances are found to be elastic ; that is, they admit of being lengthened by the^ appli- cation of forces to their extremities, and regain their original, dimensions, or nearly so, when the forces are removed. Spiral springs composed of steel wire, such as the one exhibited in fig. 71, are found to possess the same property in a remarkable degree. The connection between the force which stretckes a string, or a spring of the kind here mentioned, and the increase of length cannot be investigated from mathematical considera- tions, but is to be determined entirely by experiments. Let MN (fig. 72) be a very smooth horizontal table ; AB an elastic string or spring laid upon it and fastened at A ; W a weight stretching the string by means of a thread passing over the pulley G, whose position is such that ABG coincides with the table. Then, if W stretches the string to h, and another weight W stretches it still farther to V, it is found that Bh : BV :: W : W; that is, the excess of a given elastic string or spiral spring above its natural length is proportional to the weight which stretches it. 238, This excess is, in different springs of the same make and materials, proportional' to their lengths. For the tension of a string being the same in every part, if we divide the string into any number of equal parts, the increase of length in each part will be the same, and therefore the increase of the whole string will be proportional to the number of these equal parts which it contains : that is, to its length. 156 ELASTIC STRINGS. 239. Consequently, upon the whole, the increase of length of a string is proportional to (its length) x (weight which stretches it). Wherefore if L lie the natural length of a string, and I its length when stretched by a weight W, l-LcxL.W=C.LW; where G denotes a constant dependent on the material, thickness and make of the string. 240. Suppose the string AB (fig. 73), whose ncctural length is &, to he suspended vertically from one end A, and stretched by its own weight w only ; to determine the increase of its length. In AB take any points F, Q very near to each other, and when the string is stretched let b, p, q, a be the points cor- responding to £, F, Q, A; x = BF, Sx = FQ, y=bp,By =pq. Then Sx is stretched into S^ by the weight of bp or BF ■,. 1 wx which = — ; a .'. By — Bx= C.Bx " wx J a therefore, dividing by Sx, and taking the limits. wx d^-l=0.^; a .'. y — x = — . — , by integration ; .', ab — AB = — . = 4 Cwa. 2 a 2 Hence the increase is one half of what it would be, ifKQ were stretched upon a horisontal table by i weight equal to its own weight. 2.41. A weight W is now suspended Jrom, b, to determine the further increase of length. The weight which stretches jjg" is, in that case, a ELASTIC STRINGS. 157 .-. al-AB= CWa + ^Cwa. 242. Of this increase the part ^Cwa we have seen is due to the weight of the string, and therefore GWa, the part due to the weight W, is the same as if the string had no weight. Hence when a string is stretched by several forces, each one produces the same increase of length as it would do if the other forces did not act. By way of illustration we shall add the following ex- amples. 243, Two weights P, Q (fig. 74) resting .on two incUned planes AB, AC, are connected hy a given elastic string; to find the position of equilibrium. Let a, yS be the inclinations of AB, A G, and 6 that of PQ to the horizon; a=the natural length oiPQ; r= its tension. Then P is kept in equilibrium on the plane AB by the force T acting in the direction PQ ; .-. T cos APQ = P sin a, (Art. 212) . But^P^=a-^; „_ P sin g cos (a — ^) ' a- -1 1 m Qsin/3 Similarly, T^ ^^^^^y , p cos (j8 + g) _ ^ cos (g-g) . cos g sin /8 cos sin g ' .-. P (cot ^-i&ne)=Q (cot a + tan 6) ; .-. tan = p , ^ , which gives ; SiniPQ = a+C.a.T =''1 +cos(«-g)r 158 ELASTIC STRINGS. From which PQ is known and thence AP and AQhj means of the triangle APQ, whose angles are all known. 244. Two egual weights P, Q (fig. 75) are connected hy an elastic string, whose natural length is BC ; to find the nature of the curves BP, CQ, on which they will always rest in equilibrium with the string parallel to the horizon/ the plane of the curves being vertical. It is manifest, since the weights are equal, that the curves must also be equal. Bisect BO in A, and draw AM vertical ; AB = AO = a, AM= x, MP=:MQ= y, T= the tension of P^ ; .-. PQ-BG=G.BG.T, or 2y-2a== C'la. T; .'. y — a= CaT. But P being sustained upon the curve BP by its gravity P and -the force T, we have by Art. 213, T=Pd^; .'. y — a= GaPd^ ; .*. {y — a)' = 2 GaPx, by integration, which is the equation of a parabola. Hence BP, GQ are two semi-parabolas, whose vertices are B, G. CHAPTEE X. ON THE FUNICULAR POLYGON, ON THE OATENAEY, ON ROOFS AND BRIDGES. ON THE FUNICULAR POLYGON. 245. ABCDEF (fig. 76) is a cord, supposed devoid of weight, suspended, from two points A., Y in a horizontal line; at the knots B, C, D, E weights, W,, W,, W,, W^......are hung; to determine the proportions of these weights that it may hang in a given form. From A draw Ac, Ad, Ae, Af respectively parallel to the portions BG, CD, BE, EF of the cord ; and denote the re- spective inclinations of jdJ?, 5 C, GB to the horizontal line AF hj a, )3, 7 ; draw MB vertical. Then B is kept at rest hj the tensions of AB, BG and the weight W^, which forces are respectively parallel to the sides BA, Ac, cB of the triangle ABc, and are therefore proportional to them. Therefore W^ is proportional to Be. In the same manner W^ is proportional to cd; and they are on the same scale, for in both Ac represents the tension of BG. W^_Bc^ BM-cM ''■ W~ cd" cM-dM AMi&Ta. a - AMixa ^ ~ ^ilf tan p — ^ilf tan 7 tan a — tan ^ ~tan)8 — tan 7' „. ., 1 PFj tan^-tan7 Similarly, ^ = ^-— ^-^ 160 THE FUNICULAR POLYGON. It appears, therefore, that any one of the weights is propor- tional to the difference of the tangents of the angles at which the two sides of the polygon, which form the angle at which it is suspended, are inclined to the horizon. The angles MAe, Maf, which are situated above the line AF, are to be accounted negative. 246. The horizontal tension of any part of the string is repre- sented by AM, for it is the resolved part of the lines AB, Ac, Ad which represent the whole tensions ; and this horizontal tension : any weight ( W^ suppose) :: AM : cd :: \ : tan ^8 — tan 7. Cob. The tension of any string BG : the horizontal tension :: Ac : AM :: ^If sec /3 : AM :: sec /8 : 1. 247. liAB,BC, GD in the preceding figure, instead of being lines devoid of weight, be heavy beams of wood, or bars of metal, connected at the joints A, B, G, D by hinges, we must consider each beam as exerting by means of its weight vertical forces at its extremities. Thus, if Wj, w^, w^ be the weights of AB, BG, OD we may consider BG as exerting equal pressures ^w^at B and C in a vertical direction, the centre of gravity of the beam being supposed at its middle point ; in like manner AB exerts a vertical pressure equal to Jwj at B, and therefore we may consider W^ + ^{w^+w^) as the whole weight suspended at B. Similarly, the weights to be considered as suspended at G, D are respectively and these weights are to be used instea'd Of those given in the preceding articles. These considerations are intimately connected with the con- struction of suspension bridges. 248. If TFj, W^, Tfg are evanescent, then the weights to be considered as suspended are ^(Wi + Wj), \{w^ + w^ and if the beams are all equal, each of these become equal to w^. ROOFS AND BRIDGES. 161 ON ROOFS AND BRIDGES. 249. If the whole figure of Art. 245, be inverted or turned round the horizontal line ^i^ through an angle of 180°, as in fig. 77, we shall find the same relations between the weights as be- fore ; jt will also appear, from the same reasoning as in Art, 247, that the weights to be considered as hanging from B, G, D are the same as there investigated. In this state the problem contains the whole theory of roofs, arches, and bridges. If ABGBEFhe considered as a roof, of which AB, BG. are the beams, then the horizontal thrust at A and F tending to push out the walls on which the roof is erected, is represented by AM, on the same scale as that wherein Be represents the weight to be suspended from B; it is therefore equal to TF. + IK + w,) ^ tan a — tan yS This thrust is usually prevented from taking effect upon the walls by inserting the ends. A, F of the beams AB, FE into another AF called the tie-leam, which is thus made to sustain the whole thrust ; at other times the walls are prevented from bulging by buttresses, or shores, built against them. If it were required to construct a roof of given span with given beams, which has to support given weights, we must take an equal number of smaller proportional beams, and connect them by strings or pins at the joints, so as to allow them "to move freely, and load them with proportional weights. Then if this model be suspended from its extremities at a propor- tional distance, as in Art. 245, it will assume the required form, which we have merely to turn round AF through an angle of 180°, and it will be a perfect model of the required roof; and will possess the property of being in equilibrium in every part. In such a roof there will be no unnecessary strain on any part of the materials of which it is constructed, and con- sequently no part will require to be unnecessarily strong. In this simple manner we may also obtain the model of a bridge of given span, by taking a great number of very short beams to represent the arch stones, and connecting them as before. If E. s. 21 162 THE CATENARY. when we suspend this model-string of arch stones loaded with weights proportional to what (in the place they occupy in the bridge) they will have to sustain, we find that the bridge would be too lofty, we niust remoye the points of suspension farther apart, until we have obtained the proper altitude. This method- will give us a bridge, in perfect equilibrium in every part, and in which there is, therefore, no injurious strain, no useless strength, nor dangerous weakness in any part. ON THE CATENARY. A Catenary is the curve assumed by a fine chain or flexible string when suspended from its extremities. 250. To investigate the equation of the catenary. Let AOF (fig. 78) be the catenary; A, F being the points from which the chain is suspended, and being the lowest point of it. Through draw BOD vertical, which take for the axis of x, being the origin. From P any point of the chain draw PM perpendicular to OD ; and draw PT a tangent at P. x= OM, y = MP, s = OP. Since there is equilibrium we may suppose the part OP to become rigid ; then since it is kept in equilibrium by the action of three forces (its own weight and the tensions at Pand 0), which act upon it in the directions of the sides of the triangle MTP taken in order, we have tension at PM T>m\r j weight of OP MT "^ But if the chain be uniform, the weight of OPmaj be repre- sented by its length s, and the tension at by the length of a piece of the same chain of the length a ; .-. 1 'Jd+r x+C='U+?. THE CATENARY. 163 But when x = 0, s = 0, and therefore 0=a; .-. x + a = '/^T7, and .'. 03° + 2ax = s", which is the relation between any arc and its abscissa. 251. To find the equation of the catenary in terms of the rectangular co-ordinates x, j. The equation required is expressed in its most simple form hj taking for the origin of co-ordinates the point £, which is such that OB=a. Let then BM=x; then from the last article, \la?—a^ = i ■d^s IX —a=s; X .-. y = a log, {x + 'Ja?-a^) + C. But when x — a, y = 0, consequently 0= — a log, a ; , x + 's/a^-a' •'• 3/ = «log. • This is the equation required. 252. The relation between x and y, and that between s and y may be expressed in very simple exponential forms as follows. From the last equation we have a \a I a \a J 164 THE CATENARY. -=6" +6- (1. a Again, - = 2^-,-. :e — e (2). 253. Def. If through 5 we draw BG horizontal, it is called the directrix oi the catenary. 254. The tension of the chain at any point P is eqyt/xl to the weight of apiece of the same chain of the length BM. J, tension at P 2!P 1 tension at PM sin PTM yi + id^yf a X a _ weight of lengtii x ~ weighf of length a ' But the denominators of these fractions are equal ; .•. tension at P= weight of chain of length x. 255. We have supposed the chain to be uniform; if it should be of variable density or of variable thickness, let p be such a quantity that phs may represent the mass of a small element (S« =) PQ of the chain. Then the weight of OP is S {gpZs) =gjpds = gjypd^; and representing the tension at by ga, we have by proceeding as in Art. 250, - tension at weight of OP ga THE CATENARY. 165 .'. pd^ = adyX (1). When jO is given, this equation being integrated will give the form of the catenary. 256. To find the law of density that the catenary may he of a given form. In this case the relation between x and y is given to find p. From the last Article we have Now the quantity which is multiplied into p is the radius of curvature of the curve at P, and d^ is the secant of the inclina- tion of the tangent at P to the horizon ; wherefore _ a . sec^ of the inclination '^ radius of curvature 257. To find the form of the catenary when the chain is acted on hy a force tending to a fixed centre. Let BAG (fig. 79) be the catenary, suspended fi"om B, 0. 8 the centre of force, A that point of the chain which is nearest to 8; therefore 8A is a* normal at A. Let P be any point, and PQ a small element of the curve, a = 8A, s = AP, Ss = PQ, r = 8P, r- + Sr = 8Q, t = tension of the chain at P, t + Bt= the tension at Q, pBs = the mass of PQ, and F= the force which acts at P towards 8. Then the weight of PQ = FpBs, which we may suppose to act ultimately in the direction Q8. Hence resolving the forces, which act upon PQ, in the direction of the. tangent at Q or P, we have t + FpSs cos PQ8='t + Bt.- But if we draw PP' perpendicular to 8Q, we have Bs cos PQ8 = Br ; .-. FpBr = Bt; .:JFpdr = t (1). 166 THE CATENAEY. The integral is to be taken from r= 8A to r= 8P. Again, the chain AP is kept in equilibrium by the tensions at A and P, and by the weight of each particle of it in a direc- tion passing through 8. Hence taking the moments of all these forces about 8, we have a . (tension at A) =pt, where p is the perpendicular from 8 upon the tangent at P. The left hand member of this equation is constant, and there- fore representing it by C, we have «=- (2). P JFpdr = ^; Hence, combining equations (1) and (2), we have i' ,.Fp = -^. P When jO is given, this equation being integrated will give the form of the catenary. 258, To find the law of density that the chain may hang in a given form when acted on hy a given central force. In this case the relations between F, r and p are given to find p. From the last Article we have Cd,p 259/ Cor. Since QF = hr, FpSr = the weight of a piece of the given chain of the length QP' and density p ; hence if the density of the chain be the same throughout, the equation (1) taken between its proper limits gives tension at P— tension at A = weight of chain of the length P8 — weight of chain of the length SA. This result corresponds to that obtained in Art-. 254. THE CATENARY. 167 260. To find the form of the catenary when the chain is acted on by any forces in its own plane. Let AB (fig. 80) be the curve, in the plane of which take two lines Ox, Oy perpendicular to each other for co-ordinate axes. Let X, Y be the components of the accelerating force which acts at P, parallel to Ox, Oy respectively. . Let FQ be a very small arc; FT, QT tangents at P and Q meeting in T. From T draw TG a normal to the curve. Let X = OM, y = MP; s = AP, Ss = PQ; x + Bx, y+ Sy the co-ordinates of Q; and pSs the mass of the arc PQ. We suppose Bs so small that the accelerating forces X, Y may be considered the same for every point of it ; consequently XpSs, YpSs are the weights of PQ estimated parallel to Ox, Oy re- spectively. Now PQ is kept at rest by three forces, the tension t at P, the tension t + St at Q, and the resultant of XpSs, . YpBs ; consequently, as PQ may be considered rigid without disturbing the equilibrium, these three forces all pass through the point T; they therefore satisfy the conditions of equilibrium of forces acting on a point. Eesolve the forces parallel and perpendicular to the normal GT; .: = < cos PTG + {t + Bt) cos QTG- XpSs . d,y + YpBs . dfc, and = « sin PTG -{t + Bt) sin QTG- XpBs . dpe - YpBs . d.y. Now cos PTG = -^^— = hBs 'iidfW+WW ' rad. curv. ^ ^ ' '^ "" ' and sin PTG = 1 ultimately ; hence by substitution and dividing by Ss, we have = t '/{dWWW - ^p^.y + '^9^?'^ and = d}, -\-Xpdfja + Ypd,y. By eliminating t between these equations, we obtain the dif- ferential equation of the required curve. [The remaining Articles of this Chapter are from the pen of the Eev. J, A. Coombe, Fellow of St John's College : by whose permission they are here inserted.] 168 THE CATENARY. 261. Peop. To find the fiyrm of eqiiilibrmm of a unifirm tnextensible string on a surface and acted on hy any forces. Let M = be the equation to the surface, xyz the rectangular co-ordinates of any point in the string, and therefore of a point in the surface; s the length of a portion of the string inter- cepted between a fixed point in the string and the point {xyz) ; XYZih& resolved parts of the forces at the point {xyz) parallel to X, y, z: R the normal reaction at the point {xyz), making angles aySy with the axis of co-ordinates ; T the tension of the string at the point {xyz), one extremity of an element Zs of the string, and acting in the tangent at that point. Hence Td^ will be the resolved part in x, and Tdpi+d,{Tdp;).Zs will ultimately be the resolved part in x of the tension at the other extremity. Hence d, {Tdp;) Ss will be the difference of the resolved parts in X. The other forces acting on Ss parallel to x are XSs and liSs cos a, supposing the forces to act equally on every point of the very small element Ss. Now Ss being at rest under the action of these forces, we may suppose it to become rigid and apply the equations of equilibrium. Hence we have (dividing by Ss), d,{Tdp:) + X+Rcosa=0 (1). Similarly, d,{Td^)+ F+^cos/3 = (2), and d,{Td,z) +Z + Bcosy = (3). The equations of moments may be dispensed with for the following reasons. Consider three adjacent points P, Q, B, in the curve, Q being in the middle, and the tangents FT, BT, meeting in T. Then the plane containing these points and the tangents PT, BTvnR be the osculating plane. Now the forces X and B cos a being supposed to act equally on every point of PB will have a resultant through Q, and so will the like forces parallel to y and z. THE CATENARY. 169 Hence the whole of the forces acting on PR being reducible to three, lying in the osculating plane, will pass through a single point T in that plane; and the equations of moments taken about this point will be identical. Recurring to the above equations, our object will be to elimi- nate T and R between them. Now (1). d,x+ (2). d,y + (3). d,z = gives d,T-\- Xd^x + Yd,y + Zd^ + R (cos adjc + cos ^d.y + cos^c?/) = 0, since {dp-f + {d,yf + {A/i' = l; and .■ . dfis. d^x + d,y. d^y + d,z . d^z = 0. Also because the tangent to the curve is perpendicular to the normal to the surface, we have cos a . dp: + cos /3 . d,y + cos 7 . Y = (c - 2a cos 0)'' {4 + (cot + cot a)'}. From this equation must be foimd by approximation, and' then <]) will be known from (2). 13. Three rods AD, AE, BC are connected hy hinges at A, B, C; AE is vertical and fixed ai E, and AD horizontal. At D a given weight W is suspended. Find the strain upon the hinges. (Fig. 91.) Since the rod BG has hinges at both ends, it is incapable of exerting any action except in direction of its length ; let T be the pushing force which it exerts upon the hinge B in the direc-. tion GB, and upon G in the direction BG. Let the strain upon the hinge A be resolved into the forces X, F in the directions BA,AG. Let d^^iABG, a = BC, h=^BD. Then the rod AD is kept at rest by X, Y, T and W; resolve them vertically and horizontally, and take their moments about B; .: 0=Tr-7'sina+F, = Tcos a.-X, and = Wi — Ya cos a. From these three equations we find, Wb Y= T=W. a cos a a cos a + b a cos a sm a a cos a + h asm a The first and last of these determine the magnitude and direction of the strain upon the hinge A ; and the second equa- PROBLEMS. 185 tion gives" the magnitude of the strain upon B or C; the direc- tion of this strain has heen stated already to he CB for B, and 5(7 for C. If the joints at A, B, were rigid, the action of BC not heing necessarily in the direction of its length would be inde- terminate: BG might even be removed without affecting the equilibrium. 14. AB is a heavy uniform rod, moveable in a vertical plane about a hinge A; a given weight P sustains the rod by means of a string BCP passing over a smooth pin C situated in a vertical through A and at a distance AC=AB. Fimd the position of eguilibrium of the rod by the principle of virtual velocities. (Fig. 92.) Let W be the weight of the rod, a = AB its length, Q any point in it ; draw QM perpendicular to AG. x=AQ, Q=i. BA C. Then the virtual velocity of P = d. GF= d {BGP- BG) = d (bGP- 2a sin | 6 = — d.2asm-= — acoa-.d6. ' ' The weight of a small element of the rod at Q, the length of which is equal to Bx=W — ; and the virtual velocity of this = d. GM=d{a — a; cos ^) = aj sin ( Hence for the whole rod the value o{'%{Fds) (Art. 113) = t{W-.xsmede) W ^ = --amed0.t{xBx) W = — sin dda Jxdx, from » = to x = a = ^Wa sin dd0. E. s. 24 186 PROBLEMS. Hence, by the principle of virtual velocities, . e p Remark. The preceding solution is more difficult than is absolutely necessary; we preferred giving it however as an illustration of the meaning of the symbol % in Art. 113. The following is more simple. We may consider the weight of the beam as being collected at its centre of gravity. Let Q be this point. Then by the principle of virtual velocities, (i = P.d.CP+W.d.OM = P(- a cos^dd) + W.d{a-^ cos 6) = - Pa cos 5 «?^ + PF^ sin Odd, the same result as before. 15. Two heavy particles P, Q are connected by an injlexible rod; and one of them (P) rests upon a given smooth inclined plane. Required the nature of the curve on which the other (Q) must rest, that there may he equilihrium in all positions. Since there is equilibrium in all positions the common centre of gravity of the bodies neither ascends nor descends (Art. 169), it is therefore always at the same height above a given hori- zontal plane. Let the equation of the given inclined plane be y' = mx' (1), the axis of x being horizontal, and that oiy vertical. Let x'y' be the co-ordinates of P, and xy those of Q. Then denoting the altitude of the common centre of gravity of P and Q above the axis of x by b, and the length of the rod by a, we have {P+Q)b = I),'+Qy (2), a'^^ix-xy+iy-yT (3). PROBLEMS. 187 From (1) and (2) we find <3^z, Q. which being substituted in (3), giye the following equation of the required curve : The values of the coeflScients of the first three terms shew that the required curve is an ellipse. 16. A rod AB is placed in a smooth hemispherical howl, so as to lean against the edge of the howl at P, with one end A within it. Find the position of eguilihrium of the rod. (Fig. 93.) Let C be the centre of the bowl, Q the centre of gravity of the rod. The rod is kept in equiKbrium by the reaction of the bowl at A, the direction of which passes through C; by the reaction of the edge of the bowl at P, which will be in a line PQ at right angles to AB; and by its own weight acting vertically at G. There being but three forces, their directions all paas through a point (Art. 96) ; hence QG \s, ver- tical, and AQ _ smAGQ AG~BinAQG' Let AG = a, and r = the radius of the bowl; then because APQ is a right angle,. ^^ is a diameter of the sphere, and therefore = 2r ; also if ^ = GPA, the inclination of the rod to the horizon, AGQ='rr-PGQ = 'rr-(^-e^='^ + e, 188 PROBLEMS. and AQG = PGQ-PAG^(^-ej-e^'^-2e; 2r ^™U'"y cosg a • /T „/i\ cos 25 sm cos 5 2cos»5-l^ .-. cos''5--^cos5 = i: from which equation is known. 17. ^ smooth heavy rod AB moveable in a vertical plane about a hinge at A, leans against a heavy prop CD also moveable in the same plane about a hinge at C. Find the position of equi- librium. (Fig. 94.) Let G, G' be the centses of gravity of l3ie two rods, at which points we iaaj suppose their weights W, W to "be applied. Let M' be the pressure which the rod AB exerts against the prop; i? the reaction of the prop against AB; these forces will be equal and opposite, and act in a line perpendicular to AB. The rod AB is a lever whose fidcrum is A, kept in equi- librium by B and W; hence putting AG = a, GD = b, AG=c, and the angles BA C, A GD =6, ^ respectively, we have by taking the moments of B and W about A, W.acosO^B.AD = i2J.^ (1). Similarly we perceive that the prop GD is a lever whose ful- crum is G, kept in equilibrium by B' and W, hence taking the moments of these forces about G, we have PEOBLKMS. 189 W'.lcos==B'bsinODB' = m cos CD A = - Bb cos (0+^). Hence eliminating B by means of {1), we find O = 2Wacos0 sin 6 cos {9 + (f>) + W'b cos ^ sin ^. Also from the geometrical property of the figure, c_sin(0+^) b sinO The last two equations will give the values of and . 18. Two rods AB, AC rest against each other upon the horizontal plane ED atK, and against two smooth vertical parallel walls af B, C ; given the lengths and weights of the rods, to find the distance of the walls when the angle between the rods is a right angle. (Fig. 95.) Let a, b be the lengths, and W, W be the weights of the rods AB, A G, which we may suppose acting at their centres of gravity. Let B, B' perpendicular to UB, DO he the reactions of the vertical walls. = BAK Then the rod AB is kept in equilibrium by W, B, and the forces which act upon it at A. To avoid these last take the moments of all the forces which act on AB aboui A ; .-. O = W,^cos0-Basm0. a Similarly for the beam A G, = W. I sin 0-B'b cos 0. To obtain an equation between B and E, not involving the forces at A, let us suppose the rods to become rigidly joined at A, which will not disturb the equilibrium, nor afiect B and B' ; BAG being now a rigid body kept at rest by B, B' acting 190 PROBLEMS. horizontally, and its weight and the reaction of ED at A acting in a vertical direction, we hare, taking the horizontal forces, = R-E. Hence, eliminating R and R between the three equations now found, we obtain and .•. ED = a cos ^ + J sin ^ _ a + J tan 6 ~VT+t^? _ a'JW+l4W ^/W'+W 19. Two given rods connected hy a hinge are laid across a smooth horizontal cylinder of given radius. Determine the posi- tion of equilibrium and the strain ujpon the hinge. Let AC, BO (fig. 96) be the rods, resting upon the circle whose centre is at the points P, Q. Let G, G' be their centres of graviiy. Join OG, draw OS vertical, and upon it drop the perpendiculars GM, G'M'. Let OOH=e, AOO = BOO = tt>, GO=a, G'0=b, the radius of the cylinder = r, W, W the weights of the rods. Then if i be the altitude of the common centre of gravity of the rods above a horizontal plane passing through the point 0, {W+W')s=W. 0M+ W. OM'. Now if O^cut AO'va. H, and BO pf&duced in H', 0M= 00 cose- OG . cos OHH' rcos.d , >,, = — r — ; a cos (

+0) +W'h sin (0- ^) ; .-. {W+W) rcoa&={Wa-W'h) coi + {Wa +W'h) cot0, and ( W+ W')r cosec> ={Wa- W'h) tan^ + ( TFa + PF'S) tan 0. From which we find, by subtraction, tm_20 _ Wbj-Wa ,. tan20~ W'b + Wa ^ '' and by eliminating 0, 4 TFTr'aJ sin*0 - ( W+ W) ( Wa+ W'h) r tan^ + ( Pr+ Tr')V= ; firom which ^ being found by approximation, will be known from (1). To find the strain upon the hinge. Let T be the strain, exerted upon AO in the direction OT, and upon OB in the opposite direction. To avoid the reaction at P, which is unknown, resolve the forces, which keep AG &t rest, parallel to AC, and take their moments about P; .: 0=Tco3 {it- A OT) - Wcoa AEO, and 0=T. PO sin A OT- W. PG . sin AHO. 192 PROBLEMS. From these equations we find iwh cos' ASO + ^, . Bm' AHO PG = cos' { + e) + (^ tan ^ - l) ^in" (^ + 6) from which jP is known; and ta.nAGT=-^. tan AHO = (l--.tan(^)tan(^ + ^), gives the direction in which T acts. 20. A given weight W is sustained on a given inclined plane, partly hy friction and partly hy a power P acting in a given di- rection. Find the greatest and least values ofF. Let (fig. 97) be the body placed on the inclined plane AB; let S be the reaction of the plane in a normal direction, and /jl the coefficient of friction between the bodj and the plane : then if the tendency of is to slide down the plane, P having its least value, the friction /iB will act in the direction CB to pre- vent the motion ; and therefore resolving the forces parallel and perpendicular to AB, = fiR + Pcos6- Wsin i, and = ^ + Psin^-}Fcosz, i representing the inclination of the plane to the horizon, and the iPOB. Hence eliminating B, -. TT7- sin i — iM cos * P=W. ^ — . a - cos —fismff This is the least value of P; i.e. if Pbe less than this, (7 will begin to slide down the plane. If P have its greatest value, G will be on the point of moving up the plane, and therefore the friction fiB will act down the PEOBLEMS. 193 plane ; this will be taken account of by writing — /i for /* in the preceding result ; consequently the greatest limit of P _ ™ sin « + /* cos i ' cos d + fi sind' Any value of P between these two limits will maintain equi- librium. CoE. The limiting values of P found above may be put under the forms ■ffr sin(t-tan-V) ^^^ ^ sin (t + tan"^!^) _ ' cos {9 + tan"' fj.) ' ' cos (0— tan"'/*) ' and, from comparing which with Art. 212, we perceive that the least and greatest values of Pare such as would balance TFif the inclined plane were smooth and its inclination diminished or in- creased respectively by the angle tan~'/n. 21. To fi/nd the limiting values of "2 in the common screw when friction acts. Let TF be the weight sustained, « = the inclination of the thread of the screw to the horizon ; R = the reaction perpen- dicular to the thread, fiB = the friction along the thread : and suppose that P has its least value. Let r = the radius of the screw, and p the length of the arm by which P acts ; then resolving all the forces vertically, and taking their moments about the axis of the screw {fiB acts up the thread), since the axis of the screw is only moveable lengthwise, by Art. 94 we have for equilibrium = Iico3i+ fiR sin i — W, 0= {Rsini — fiB cost) r — Pp. By eliminating B between these equations, we find p 1 + /x. tan I r E. S. W. - . tan (i — tan"». p 25 194 PEOBLEMS. Hence the least value of P is the same as in a screw without friction, the thread of which is inclined to the horizon at the angle ^— tan"*/*. By writing -ytt for /i, we find that the greatest value of F is the same as in a screw without friction, the thread of which is in- clined to the horizon at the angle i + tan-*/A, 22. Let AC he a curve line m a vertical plane; P, Q given weights attacked to the extremities of a string which passes over a smooth pin at B/ to shew how to find the position of equili- hrium. (Fig. 98.) Take the vertical line Bx for the axis of x ; and any fixed point ^ in it for the origin of co-ordinates : draw QM perpen- dicular to Bx; and put AM=x, QM=y, Bp = x'; then if P descend through a small space dx', the corresponding space descended by Q will be dx ; and as P and Q are acted on by no other forces than gravity, except the tension of the string and re- action of the curve line, the virtual velocities of P and Q are dx' and dx; and consequently, by Art. 113, PdK^ + Qdae = ; this is the only mechanical condition of equilibrium. The geo- metrical nature of the machine is expressed by the equation of the curve y = F{x), and the equation, (6 being the length of the string, and a de- noting AB) h = x+ '^{a + xf + f. Ex. Let AC he a parabola, and B the point where the axis intersects the directrix. In this case y' = iax ; .'. J = a;' + V (a + a;)' + iax ; .'. = dx'+r-i — :; ^a-dx (o' + 6ax + a?)* p-'^+{a' + Gax + x'f^- PEOBLEMS. • 195 Hence dividing by dx, and reducing the result, we find 2a Va x = — Sa + p2\i" (-g 23. The weight P in Prob. 21 instead of hanging perpen- dicularly, rests upon a given curve line AD/ to find the position of equilibrium. (Fig. 99.) If x, y' be the co-ordinates of P, and x, y those of Q, both measured from B as origin, the virtual velocities of P and Q will be respectively dx and dx ; consequently Pdx + Qdx = 0. To this we must join the equations of the two curves y' = F{x'), &ndiy = F{x), and the equation b^'Jx'^+y'^ + V^f+f. Ex. Let AD be a circle whose centre is in BA produced; and AC a parabola, the directrix of which passes through B. Then the equation of AC is y = 4a (a; — a), and that of AD is y==2c(a;'-a)-(a;'-a)', c being the radius of the circle ; .-. b = '^x" + y" + '/^^Tf = V2£b' {c + a)- 2ac -d'+ 'J a^ + iax - ia' ; (c + a) dx' {x + 2a) dx 'J2x' {c + a)-'2ac — d? slx^ + ^ax—^a^ _ {c + a) dx (a;+2a) dx ~ BP ^ BQ ' Q _ doc _ x + 2a BP _x + 2a /_5_ _ \ ~ c + a '\BQ J' 196 PROBLEMS. from which equation, BQ being known in terms of x, x may be found. 24. Two given weights P, Q are connected by a string PAQ which is laid across a horizontal cylinder / to find the position and nature of the equilibrium. (Fig. 100.) It is evident the string will lie in a vertical plane perpen- dicular to the axis of the cylinder. Let G be the centre of the circular section of the cylinder by this plane. Draw CA vertical, and BGD, PM, QN horizontal : join CP, OQ. Then since the length of the string and the radius of the cylinder are given, the angle PCQ is known ; let it be denoted by 2a : and let (x + 0, a — 6 represent the angles QCA, PC A; and a= CA. Then if i be the altitude of the common centre of gravity of P and Q above BD, we have {P+Q)z = P.CM+Q.CN = Pa cos {a— 6)+ Qacos (a + 0) ; .-. (P+ Q)des = Pa sin {a.-ff)- Qa sin (a + 6), and (P+ Q) diz = — Pa cos (a - 5) - Qa cos (a + 6) = -{P+Q)z (1). Now in the position of equilibrium dgs = 0, and therefore Pa sin (a — 5) = Qa sin {ix + 6), from which we find /, P- Q tan = p „ . tan a, which gives the position of equilibrium, which is unstable, be- cause equation (1) shews that s is then a maximum. 25. A hollow paraboloid is placed with its vertex downwards and axis vertical; a given rod rests within it, leaning against a pin at the focus, and having its lower end upon the parabolic surface. Find the position of equilibriv/m. (Fig. 101.) Let PQ be the rod, G its centre of gravity, 8 the focus of the paraboloid, AS its axis, BA G a section of it by a vertical plane PROBLEMS. 197 passing through the rod; a = AS, b = PG, r = P8, 6 = ASF; through 8 draw L8 horizontal, and draw MQ vertical; let i = MG. Then by the nature of the parabola whence cos = 1 . 1 + cos r Also l=GM=80cose = {r — b) cos 6 =('-»)C^->) 2ab , = 2a-r h b ; r and d;s= — ^ (1). In the position of equilibrium d^ = 0, and therefore r = V2aS ; from which the position of the rod is known. Equation (1) shews that the altitude of G is then a minimum; and therefore the position is one of stable equilibrium. 26. A paraboloid, formed by the, revolution of a given para- bolic area about its axis, is placed with its convex surface upon a horizontal plane; to find the position in which it will rest. (Fig. 102.) Let .4C be the axis of the parabola, inclined at an ^ ^ to the horizontal plane : P the point on which it rests ; draw PN vertical: then since there is equilibrium the centre of gravity must be in the line PN (Art. 132), but it is also in AG, the axis of the parabola, consequently it is at N. Draw PM perpendicular to AG; let a = ^ C, 6 = BO ; then the latus rectum = — , a and.-.iW=g; 198 PEOBLEMS. 2a ' AM= — T^r- = — . cor : ftr\ 4a .-. AN=AM+MN = £(cof^ + 2). 2 But because N is tlie centre of gravity, AN= - a. (Ex. 5, Art. 177) ; cof^ = g-2, from which the position is known. CoE. The least value of cot 6 is when ^ = ^ : hence when 8aV 1 = or < 3^is=or<2, or, when a is = or < , the solid can only rest in equilibrium with its axis vertical. 27. Two heavy rods AC, CB connected hy a hinge at C rest on two smooth points D, E, situated in a horizontal line : find the position of equilibrium. (Fig. 103.) Let G, g be the centres of gravity; and W, W the weights of the rods AC, BC; B, B' the reactions of the points B, E which will be at right angles to the rods, because the points on which they rest are smooth. Join DE, and let 6, ^ denote the angles CDE, OED; and put Oa = a, Cg = a', DE=l. The rod -40 is kept in equilibrium by three forces, viz. its own PROBLEMS. 199 weight at G, the reaction R at D, and the tension of the hinge C; to avoid the last, (the magnitude and direction of which are unknown, and are not required,) let us take the moment of these forces about G; .: R.DO- W.aQ,os6 = (1). Proceeding in a similar manner with the beam OB, we find E.EG- TF'.a'cos<^ = (2), Again, when the equilibrium is once established, we may suppose the hinge G to become rigid; under this hypothesis- the rigid body AGB is kept in equilibrium by four forces, viz. R, R', W and W. Hence resolving them vertically and horizontally, we find Rcose + R'cos^- W-W' = (3), andi?sin0-^'sin^ = O (4). These four are all the independent equations which can be derived from the mechanical properties of the machine; there are however two others, which express its geometrical properties, derived from the triangle J) GE, viz. (5) DG= . ,„ ^,. , and^C=-r— T3 — ^. (6). From (1) and (5), we find „_ Wa cos sin {0+) ~ 6 sin ^ ' and from (2) and (6) ,_ W'a cos ^ sin (g + <^) ^~ bsind which being substituted in (3) and (4) give ^ ^' \ bsno.(ji sm p and = Wa sixi' cos0- W'a sin" ^ cos <^, which two equations are sufiicient for the determination of and <^. 200 PROBLEMS. CoE. If the rods are equal in all respects, these two equations become 2^ • la , j.\ /cos'^ cos''(f>^ ,„, and = sin'' ^ cos 5 - sin" ^ cos ^ ; the last of which gives ^ = <^ (A or 1 = cos" + cos 5 cos ^ + cos^ {B). Let us consider these two results separately, and (1) When 6 = ^ equation (7) gives COS0 ~ (2 J ' whence the position of the rods is known. This position is symmetrical with regard to a vertical line through C. (2) The equations [B) and (7) shew that 6 and ^ are in- terchangeable, and consequently there are, besides the symme- trical position just found, two unsymmetrical positions of equili- brium, similarly situated on each side of the first found position. They may be found by means of (7) and {B). 28. A solid of any form is placed with its convex surface upon a horizontal plane ; to find the position of equilibriu/m. Let z =/(«, y) be the equation of the surface, referred to three rectangular axes in the body: and let xyz be the co-ordinates of the point in contact with the horizontal plane, and xyz those of the centre of gravity referred to the same axes. Then the plane on which the body stands being a tangent plane, if a/Sy be the inclinations of the co-ordinate axes to the horizon, -5 — a, -5 — /8, — — 7 will be the inclina- Ji Z A tions of the vertical through the point of contact to the co-ordi- nate axes ; this vertical line is a normal, and therefore PEOBLEMS. 201 sma = sin/S = -dji sm7 = 1 •(!)• But since the solid rests upon a point, the vertical through that point must pass through the centre of gravity of the solid, i. e. the normal at the point of contact passes through the centre of gravity of the solid ; hence the equations of the normal give 0=^x-x + d^.{z-~s)\ ^=y-y+d^- (s-i)3 These two equations, together with the equation 2 =fip^ y) will enable us to find x, y, z ; and thence a, /3, 7 from (1). Ex. Sw^ose the solid to he the eighth part of a sphere com-- pnihended hetween three rectangular planes: to find the position in which it will rest with its convex surface on a horizontal plane. Let its equation be ••• 3,^=--^, and J^ = -|. Also from Ex. 1, Art. 177, we have O Q Q hence making substitution in equations (2) we obtain x = y = z; these two equations joined with.the equation of the surface of the sphere give a a a pass through G' the body is still in equilibrium ; but if G' lie to the right of i^ (as in the figure), the body when left to itself will roll back into its original position ; and lastly, if G' lie to the left of i^, the body will roU farther from its first position. Hence the first position is one of stable, unstable, or neuter equilibrium according as A'j>is>< ox = AG. COK. 1. If the surface on which the body rests be concave, ■we must account p negative in the above result. COE. 2. If the surface be a plane, we must make p infinite, and then since -^--^ = P, P+P i+fi. P in that case, the equilibrium will be stable, unstable, or neuter ac- cording as p' is >< or = ^ G'. CoE. 3. If the lower surface of the body be a plane, we must make p infinite, and then the result is p is >< or = AG. Ex. Find what segment of a paraboloid will rest in a posi- tion of neuter equilibrium upon a spherical surface of given radius. Let X be the length of the axis of the paraboloid, im its latus rectum ; and a the radius of the spherical surface. Then from Ex. 5, p. 101, we have and by the Differential Calculus p = 2m, T 2« 2ma and .'. -7r = : 3 2«i + a 5ma 2m + a' 204, PEOBLEMS. 30. A string is stretched along a smooth curve line of any form hy two equal forces, required the unit of pressure exerted hy it upon the cylinder at any point. (Fig. 105.) Let AHB be the curve line along which the string is stretched by the two equal forces P, Q. Let EH' be a very small arc, and at H, H' draw tangents meeting in K, and nor- mals HO, H' 0. Join KO, and put ^ HOH' ^ Sd. The portion of string HH' is kept in equilibrium by the tensions at H, H, each of which is equal to P or Q; and by the reactions of the curve line HH', which being smooth, the reaction at every point will be in the direction of a normal. Hence the re- sultant of all the reactions on HH' will pass through 0, and as it must also pass through K, it acts in the line OK. Hence by Art. 28, resultant reaction on HH : P :: sin HKH' : sin OKH :: sin HOH : cos KOH :: Z9 : 1 ultimately ; .'. resultant reaction on HH' = P. W. Now when the arc HH' is diminished indefinitely, the pres- sures upon it may be all considered parallel, and therefore their resultant is equal to their sum. Consequently the pressure upon the indefinitely small arc HH is equal to PS^ or to p a-rc ' rad. curv. ' and the unit of pressure (or pressure on an arc of the length unity) _ tension of string ~ rad. curvature Cor. Let C, D be the points where the string leaves the arc ; and let p be the whole pressure upon CH; and let 6 = the angle between the normal at G and that at H; then Sp = the pressure upon HH, and by what has just been proved hp^ne-, .-. _p + 0= pe. PROBLEMS. 205 But when ^ = 0, j) = 0, and therefore C-0; .•.p=pe. If a be the angle included between the normals at C and D, and p the pressure upon the whole arc OHD, It is remarkable that this result is independent of the form of the curve, provided it be in every part convex towards the string in contact with it. 31. A string is stretched along a rough curve line of any form iy two such forces that the string is on the point of moving. Having given the coefficient of friction, find the proportion of the forces. (Fig. 105.) Let Q be the larger force ; <, J + S< the tensions of the string at H, H' ; /jl the coefficient of friction ; 6, SO, a as before. Then the pressure on MS' = tSd, and therefore the friction on SH' = fjLtBd; but the arc MH' being pulled in opposite directions by the forces t, t + St, the latter is prevented from producing motion only by the friction on HH' ; .-. t + St - t = /jLtSe ; det ■■-1=1'' .'. log,* =^fi,e+o, when 6 = 0, t = P, and when 6 = a, t= Q; wherefore loge<9-log,P=/ia; .-. Q = P^. This result is independent of the form of the curve. 32. A unifcyrm heavy chain is laid upon a smooth arc of a quadrant of a circle, and coincides with it; one of the hounding radii of the quadrant heing horizontal, and the other vertical. Find the force necessary to prevent the chain from sliding down the arc : and compare the pressure upon the curve with the weight of the chain. (Fig. 106.) 206 PROBLEMS. Let F te the force whicli, acting horizontally at B, juat prevents the chain from sliding down the quadrant. Let F, Q be two points very near to each other ; a = AO, —A OP, Sd = POQ, t and t+ St the tensions of the chain at P and Q; J) = the pressure upon the arc AP, and Sp = that upon PQ ; p = the weight of a piece of the chain of the length 1. Then the elementary portion of -chain PQ is kept in equilibrium by the tensions t, t + St, its own weight paSd, and the reactions of the curve PQ ; to avoid the latter, take the moments of these forces about the point ; .•. = {t + St)a — ta — paS6 .acosd; .'. dgt = pa cos ; .\ t=pasia6+ C^ But at ^, t = 0, ^ = 0, and.-. (7 = 0; .'. t = pa sin 0. And at 5, < = ^ and 0=^; .-. F=pa = the weight of a piece of the chain, the length of which is equal to the radius. Again, to find the pressure upon the quadrant. The pressure of the elementary portion PQ is due to two causes, viz.- its own weight, and the tensions t and t + St. The former part = paS0 sin 0, and the latter part = tS0, by Prob. 29 j .-. Sp =paSd sin d+tS0; .-. dep = pa sin + pa sin 0', .-. p = — 2pa cos 0+0. At 2,'0 = O and ^ = 0; .•.C=2pa; .'. ^s=2joa(l — cos^), PEOBLEMS. 207 and at J5, ^ = x and p = the whole pressure on the quadrant, = 2pa; press, on quad. _ 2pa _ 4 weight of chain tt tt " 33. Supposing the quadrant to be rough, to find the least value of F which can prevent the chain from sliding off; having given the coefficient of friction (=/*). In this case the chain is on the point of moving towards the point A, consequently friction acts up the quadrant. The forces which keep PQ in equilibrium are t, t-\- Si, /iSp, paW, and the normal reactions; to avoid the last, take the moments of the forces about the point ; .'. Q = {t + Zi)a — ta + fJbZpa — paZd . a cos d ; .". det+ fidgp=pacos6 (1). ■Also as before Sp = paZO . sin 6 + tW ; .'. dgp = pa sin 6-\-t. Hence substituting this value of d^p in (1), we obtain del + /i {pa sin 6 + t) = pa cos ; .'. det + fit = pa (cos — fj, soi. 9) . Multiplying this equation by ^ and integrating, we find . 2w cos 0+(l — u^ sin ..a , n 1 + /A Now when 0=0, t = 0; And when = J, t = F; 20S PROBLEMS. l + jjr '^ l + fir Cor. If the pressure be required, it may be found by inte- grating equation (1) ; .". t + /i^ = pa sin 6, no constant being added, because t, p, 6 vanisli together ; pa . „ t .'. p='— . smo . /* ^ Hence when = — , we have the pressure on the quadrant ^pa_F " /* /* _pa l—fj? pa 2pa -^ 2pa , , -^ l + /i' APPENDIX. ON THE COMPOSITION OF TWO FORCES ACTINO ON A POINT. 1. Since two forces which are in equilihrium must neces- sarily be equal and opposite, two forces F^ and F.^ which do not act in opposite directions, .must necessarily have a resultant, the position of which we shall proceed to determine. (1) The resultant of two forces Fj Mnd F^ acting on a point P, is situated in the plane F^PF^. For if it be not in that plane, it must be either above or below. But it cannot be above ; for, any reason which would assign it such a position might be used to assign it a similar position below ; for these two positions are similarly situated with regard, to the forces F^ and F^ ; there would consequently be two resultants, which is impossible. The resultant then cannot be situated above the plane of the forces ; and in a similar way we may shew that it cannot be situated below, and therefore it must be in the plane. (2) It lies within the angle FjPF^. For the tendency of F^ is to draw the particle P in the direc- tion Pjfj, while that of .F^ is to draw it in the direction PF^, and hence the real motion, which is the result of these united ten- dencies, will not be in the direction of either, but intermediate to both ; and therefore within the angle F^PF^ : consequently the resultant, which. is a single force that would produce the sanae motion, must be situated within the angle FJPF^. 2. Since F^ and F^ do in part hinder each other from pro- ducing their whole effects, it appears that their resultant must be less than their sum ; for their resultant Can only be equal to their E. s. 27 210 APPENDIX. sum when neither interferes with the other, which is not the case unless they act in the same direction; consequently 3. If the forces Fj avd F^ are egual, their resultant R will bisect the angle FjPF,. For if there be a reason why FB should lie nearer to FF^^ than to PFj, there must he a similar reason why it should lie nearer to Pi^^ than to FF^, since the forces are equal; and hence there would he two resultants, which is impossible ; consequently PF bisects the angle F^FF^. 4, Having thus determined the direction of the Tesultant of two equal forces, we proceed to find its magnitude. Let F^,f (fig. 107) be two equal forces acting on the particle P, and i? their resultant bisecting the angle F^Ff. Since R is less than the sum of the two forces F^ and f it is clear that R R ■ fc, J , or its equal -^^ , is always less than 1 ; and, conse- quently, an angle 6 inay be found such that 2^=cos0, or P=2PjCOs^. The angle 6 is .unknown at present, but from Art. 19 we learn that so long as the angle F^Pf^ remains the same, 6 con- tinues unchanged ; that is, if we have two sets of forces inclined at the same angle with each other respectively, we shall have B = 2F^ cos 6, and R' = 2i?"j cos 9, and therefore R: R' :: F^: F,' (A), that is, the resultants are proportional to the components. Let now F^, fhe two other eqttal forces acting on P whose resultant is also equal to R, the angles F^FF^, flf^ being each equal to RFF^ or RPf^. Now at P apply four forces, each equal to X, two of them respectively in the directions FF,, Pf^, and the •other two in the direction PR ; and let them be of such magni- APPENDli. 211 tude, that F^ may be the resultant of the one in the direction PF^ and one in the direction PB. Then, since these two contain the same angle as F^ and f^, and F^ is their resultant, F^ = Ix cos Q. Also, if we substitute instead of F^ and f^, their components, we may consider R as the resultant of the forces x, x, x, and x ; of which two act in the same direction as M ; and, consequently, P—2x is the resultant of the two x, x, which act in the direc- tions PF^, Pf^ ; and since, by hypothesis, B is the resultant of F^ and f^, which act in the same directions as x, x, .: B : B-^x :'. F^: x, from (A) ; • ^-^ 2 But i2= 2i?; cos 5= 2 . 2a;cos^ . cos 6 = lx cos'^; .-. 4=4cos^^-2 = 2(2cos=e-l) = 2 cos 2^; .-. jB=2i^jCos2^. It appears then, that in the formula B = 2^1 cos e, if we double the angle at which the forces are inclined, we must also double 0. We will now suppose, that when the angle at which the forces act is a multiple n, or any inferior multiple, of FJPf^, it is true that in the same formula the corresponding equimultiple of is to be taken ; so that B = 2i^„ cos n0 = 2F^^ cos (w - 1) ^ = . . . = 2F^ cos 6. . Apply (fig. 108) at P, as before, four forces in the direc- tions PF^, PF^_^, Pf^^, and Pf„_^ respectively, each of such a 212 APPENDIX. magnitude x that F may Ije the resultant ,of the two in the direc- tions -Pi^„+,, Pi^„_i, and/„ of the other two ; .\ F„ = 2a; cos d* But if, instead of the forces F^, /„, we substitute their four components, we may consider B as the resultant of the forces 03, X, X, and x, of which two acting in the directions PF^^ , Pf„_i will have 2a; cos (w — 1) for their resultant in the direction FB, and consequently B — 2x cos {n — l)d is the resultant of the other two which act in the same directions as F^^ and_^j; consequently, from (A), B : ^-2a;cos(M-l)^ :: F^i : x; B -H „ / ,x /I ••• ^^ = --2cos(«-l)5 = 2 COS cos «0 — 2 sin sin w0. a; But B = 2i^„ cos n0 = 4cc cos cos n0 ; .". -p— = 4 cos 5 cos w^ — 2 cos cos «^ — 2 sin sin n0 = 2 (cos cos «0 — sin sin w^) = 2 cos (w + 1) ; .•.'5 = 2i?'^,cps(»i+l)5. Hence the formula is true for a multiple {n + 1) if it he true for n and all inferior multiples: but it has been shewn to be true for 2 and 1, and consequently it is true for multiples 3, 4, 5, 6, ... and generally, by induction, for any multiple whatever. It appears then, that as we mcrease the angle at which two equal forces {F, f) act, we must increase the angle in the same proportion, and then, that the formula B = 2i^cos For the lF.^.t PFn-i = /Fj Pf„ (fig. 107). APPENDIX. 213 still holds good. This, however, supposes the angle between the forces to be a multiple oi F^Pf^ (fig. 107), which may not happen to be the case ; but by taking the original angle Fj^Pf^ exceedingly small, we may find a multiple of it which shall differ from FPfsi proposed angle by less than any assignable quantity. It is evident then, that FFf and have an invariable ratio to each other, so that if FPf— 2^, then 6 — = constant quantity = o suppose ; .-. i2 = 2^cosc^. To determine the value of c, we observe that if i^ and /act at an angle tt, or are opposite to each other, (in which case ^ = -k") they have no resultant ; .-. = 2Fcos^, .'. cos — = 0. Now none but angles which are odd multiples of — have their cosines = ; .'. c = an odd integer, = 1 as we shall shew. TT For if c is not = 1, let the angle FPf be such that ^ = ^ > which is therefore less than a right angle, and then R = 2ii'cos cj> = 2i^cos ^ = 0. But since the angle FPf is, in this case, = - , and therefore less than tt, the resultant cannot be == 0, which is absurd, and consequently c = I. We arrive therefore at the general result, that if ii^,/be two equal forces acting on a particle, and inclined to each other at the angle 2<^, their resultant R is inclined to ^14 APPENDIX. each of them at the angle ^, and its magnitude is determined by the equation jB = 2^cos^. 5. It will be immediately obvious that, since the forces F and / are perfectly equal and similarly situated with respect to PR, they contribute equally to the resultant R ; and, conse- quently the efficiency of each in the direction PR is equal to ^R, or Fcos = a + j8 ; /. cos (^ = cos a cos |8 — sin a sip /&; and, consequently, E' = F' + 2Ffcosb — 2a tan and < b, where 25 is the altitude of the cylinder, a the radius of the base, and d the angle which the string makes with the vertical. ■'o 40. A flat board in the form of a square is supported upon two props with its plane vertical; determine its positions of equilibrium, friction being neglected, and the distance between the props being equal to half a side of the square. 41. Determine the position of equilibrium of a uniform rod, one end of which rests against a plane "perpendicular to the horizon, and the other on the interior surface of a given hemi- sphere. 42. If the sides of a triangle ABG be bisected in the points D, E, F; then the centre of the circle inscribed in the triangle DEF is the centre of gravity of the perimeter of the triangle ABG. 43. Three equal rods, loosely connected together by one extremity of each, have their other extremities placed upon a MISCELLANEOUS PEOBLEMS. 221 rough horizontal plane at the angular points of a given equi- lateral triangle. A smooth heavy ring ia then placed on the rods ; find tlie coeflficieut of friction between the rods and the plane that the machine may just he on the point of falling. 44. A cone and sphere of given weights support each other between two given inclined planes, the cone resting on its base on one of the planes ; determine what must be the vertical angle of the cone, that the equilibrium may subsist. 45. A given cylinder with its axis horizontal is held at rest on a given rough inclined plane by a string coiled round its middle and then fastened to the top of the plane ; find the position of equilibrium. 46. A given weight is placed upon a rough horizontal plane ; required the magnitude and direction of the least force which will be able to move it. 47. The resultant and sum of two forces are given, and also the angle which one of them makes with the resultant; it is required to determine the forces and the angle at which they act. 48. A circular hoop is supported in a horizontal position, and three weights of 4, 5, and 6 pounds respectively are sus- pended over its circumference by three strings fastened together in a knot. Shew that the knot must be in the centre of the hoop, and find what must be the positions of the strings so that they may sustain one another. 49. Four beams, AB, BG, CD, DA (fig. 27) connected by hinge joints, have the opposite comers connected by two elastic strings A C, BD. Shew that AE.EG BE. ED tension oi AG : tension of BD AG ' BD 50. A uniform straight rod rests with its middle point upon a rough horizontal cylinder, their directions being horizontal and perpendicular to each other. Find the greatest weight which may be attached to one end of the rod without causing it to slide oif the cylinder. 222 MISCELLANEOUS PROBLEMS. 51. Two equal uniform beams connected by a binge joint ^are laid across a smooth horizontal cylinder of given radius. Find their inclination to each other when in equilibrium, 52. A particle is placed on the surface of an ellipsoid, and is attracted towards the principal planes by forces which are respectively proportional to its distance from them; find the conditions of equilibrium. 53. Prove the equality of the power and weight in Rober- val's balance by couples ; and find the strains upon the joints and pins. 54. A particle is placed on the arc of a given parabola, and is acted on by gravity parallel to the axis, and a force perpen- dicular to it which is proportional to the distance of the particle from the axis ; find the position of equilibrium. 55. If three parallel forces acting at the angular points A, B, C of a plane triangle are respectively proportional to the opposite sides a, b, c ; prove that the distance of the centre of parallel forces from A 2bc A ' = , — cos—. a+h+c 2 56. A ladder rests with its foot on a horizontal plane, and its upper extremity against a vertical wall; having given its length, the place of its centre of gravity, and the ratios of the friction to the pressure both on the plane and on the wall ; find its position when in a state bordering upon motion. 57. If a lever, kept at rest by weights P, Q, suspended from its arms a, b, so that they make angles a, ^,, with the horizon, be turned about its fulcrum through an angle 20, prove that the vertical spaces desci-ibed by P and Q, are to one an- other as a cos (a 4- ^) : b cos {fi—d); and thence deduce the equation of virtual velocities. 58. If /S and D represent respectively the semi-sum and semi-difference of the greatest and least angles, which the direc- tion of a power supporting a weight on a rough inclined plane MISCELLANEOUS PROBLEMS. 223 may make with the plane, and ^ be the least elevation of the plane when a body would slide down it; prove that the cosine of the angle, at which the same power being inclined to a smooth plane of the same elevation would support the same weight, COS 8 ,„ ,, cos ^ ^ ^' 59. A roof A CB consisting of beams which form an iso- sceles triangle with its base AB horizontal, supports a given weight at C; find the horizontal force at A. Why must a pointed arch carry a heavy weight at its vertex ? 60. Four equal uniform beams connected by joints are sym- metrically placed in a vertical plane, in the form of a roof; shew that if the extremities be in a horizontal line, and 6, ^ be the inclinations of the beams to the horizon, tan = 3 tan ^. 61. A beam ^B, capable of motion in every direction about a fixed ball and socket joint at A, rests with its end B against a rough vertical plane ; determine the extreme positions of equi- librium. 62. In the last question suppose the end B rests against a rough inclined plane; determine the extreme positions of equilibrium. 63. Three weights are suspended from the angles of an isosceles triangle, whose plane is vertical and which is supported by a horizontal axis, passing through its centre of gravity, about which it is capable of revolving. Determine its positions of equilibrium, the two weights suspended from the extremities of the base being equal, and each greater than the third: and shew in each case whether the equilibrium will be stable, un- stable, or neutral. 64. A uniform rod, whose length is a, moveable freely in a vertical plane about a hinge at one extremity, is. attracted by a force varying as D^, and acting from a centre at a height a directly above the hinge ; find the position in which it will rest, and the nature of the equilibrium, supposing that the attractive force on the hinge is \g. 224 MISCELLANEOUS PROBLEMS. 65. A hollow cylinder stands upon a smooth horizontal plane, and a light rod of given length, being in the same vertical plane with the axis of the cylinder, passes over the upper edge and rests against the interior surface. A given weight is attached to the other extremity of the rod, and the cylinder is just on the point of turning over. Determine its weight. 66. A cylinder is laid upon two equal cylinders all in parallel positions, and the lower ones resting in contact with each other upon a rough horizontal plane ; find the relation between the coefiicients of friction between the cylinders, and the co- efficient of friction between a cylinder and the plane, that all the points of contact may begin to slip at the same instant. 67. Determine the conditions of equilibrium of a material point situated in an indefinitely thin bent tube of any form and acted upon by any number of forces. 68. A chain of uniform density is suspended at its extre- mities by means of two tacks in the same horizontal line at a given distance from each other ; find the length of the chain so that the stress upon either tack may be equal to the chain's weight. 69. A uniform chain is suspended from two tacks in the same horizontal line at a given distance from each other. Find the length of the chain that the stress on the tacks may be the least possible. 70. A cylinder rests with the centre of its base in contact with the highest point of a fixed sphere, and four times the altitude" of the cylinder is equal to a great circle of the sphere ; supposing the surfaces in contact to be rough enough to prevent sliding in all cases, shew that the cylinder may be made to rock through an angle 90", but not more, without falling. ^ 71. A man runs round in the circumference of a given circle with a given velocity ; determine the inclination of his body to the horizon. MISCELLANEOUS PROBLEMS. 225 72. One end of a heavy rod can turn in every direction about a fixed point. The other end rests on the upper surface of a rough plane, (coefBciei^t of friction fi) which is inclined to the horizon at an angle a. If /3 be the angle which the beam makes with the plane, prove that the rod will not rest in every position, unless cofa be not less than —^ + tan^/S. 73. A chain suspended at its extremities from two tacks in the same horizontal line forms itself into a cycloid; prove that the density at any point cc sec' {^0), and the weight of the corresponding arc . '^7' :iz o ^/M I- / / / -^ p v~ 1 „ (JSi siz TUtAe2. ^ a — yA f r / u /^ .d '-^ a c ~~ & I^la£e f3 . (4<^) (4-9) (^O) jZ^^ (6ffi Tlate'S. C:=^*te^^S*^ Tj^^2j&^: ^^mmm^ftx^^ ,*VX;ja^'^ ^^m^^ «L