Cornell University Library arV17342 An elementary treatise on kinematics and 3 1924 031 267 382 olln.anx Cornell University Library The original of tliis 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/cu31924031267382 AN ELEMENTARY TREATISE KINEMATICS AND DYNAMICS. AIS^ ELEMENTARY TREATISE ON KINEMATICS and DYNAMICS. JAMES GOEDON MACGREGOR, M.A, D.Sc, FELLOW OF THE ROYAL BOCIETIES OF EDINBURGH AND OF CANADA, MOKEO FSOFEESOB OF PHYSIOS, DALHODSIE COLLEGE, HALIFAX, N.S. ^0nbon: MACMILLAN AND CO, AND NEW YORK. 1887. © All rights reserved. CORNELL UfSilVERSITY ^ UBRARV^' UUIVEBSITY PRESS, GLASGOW I ROBERT MACLEHOSE. PREFACE. This book treats in an elementary manner the whole of what is ordinarily known as Abstract Dynamics, including Kinematics, Kinetics, and Statics, and is designed for use in the higher classes of Schools and the junior classes of Colleges arid Universities. It assumes, therefore, a knowledge of only the. more elementary branches of Mathematical Science — Geometry, Algebra, and Plane Trigonometry. The kinematical portions of the subject are treated by themselves, not only because this course is the more logical, but also because it has been found in my experience to be the better from an educational point of view. The usual division of Dynamics into Kinetics and Statics has not been adopted ; but statical problems are throughout regarded as boundary cases of kinetic prob- lems, the equations of equilibrium being in all cases deduced from the equations of motion. This course also VI PREFACE. has recommended itself to me both by its logical fitness and in my experience as a teacher. A careful analysis of the subject has been made, that the reader may be able to recognise at once the exact re- lation which each department bears to the whole. The scrappiness of treatment which characterizes many of our text-books has thus been avoided. An endeavour has been made to eliminate all un- necessary assumptions, the various so-called " Principles," which have obtained currency in our text-books, being deduced from Newton's three Laws of Motion, which are adopted as the fundamental hypotheses of theoretical Dynamics. It has been found necessary to modify the current definitions of a few important terms, e.g., velocity and acceleration. This is due to the adoption of the distinction, proposed by Prof Tait, between velocity and speed, and the extension of this distinction to acceleration and rate of change of speed. Velocity and acceleration have therefore been defined so as to connote both magni- tude and direction. A large number of illustrative problems have been inserted both in the text and at the end of the volume. These have been drawn, for the most part, from the ex- amination papers of the more important British and American Universities and Colleges; but some of them are original, and some are taken from works mentioned below. PREFACE. vn For problems in Rigid Dynamics, I am especially indebted to Walton's " Problems in Theoretical Mechanics.' ' Readers who wish a larger selection of examples may be referred to Garnett's " Elementary Dynamics," Greaves' " Elemen- tary Statics," and Walton's "Problems in Elementary Mechanics." The traditional chapter on simple machines has been omitted, but the treatment of simple machines has been introduced here and there as illustrative matter. In the preparation of my class lectures, which formed the basis of this book, I derived assistance from a large number of works on Kinematics and Dynamics. As the lectures were prepared without any intention of publica- tion, I am unable now to acknowledge, except in a general way, the assistance thus derived. I am sensible of being directly indebted, however, to a greater or smaller extent, to the following works : — Thomson and Tait's " Treatise on, and Elements of, Natural Philosophy " ; Tait's Article on Mechanics in the "Encyclopaedia Britannica," 9th ed., and his ' Properties of Matter " : Frost's " Newton " ; Clifford's " Elements of Dynamic " ; Maxwell's " Matter, and Motion " ; Parkinson's " Elementary Mechanics " ; Goodeve's " Principles of Mechanics " ; Garnett's "Elemen- tary Dynamics " ; Wormell's " Principles of Dynamics " ; Lodge's "Elementary Mechanics "; Earnshaw's " Statics ''; Minchin's " Treatise on Statics '' ; Routh's " Rigid Dynamics " ; Thomson's Article on Elasticity in the Vlll PEEFACE. "Encyclopsedia Britannica," 9th ed. ; Everett's " Units and Physical Constants"; and "Force, Impulsion and Energy, by John O'Toole." I am indebted for valuable suggestions to my colleagues, Chas. Macdonald, M.A., Professor of Mathematics, and D. A. Murray, B.A., Tutor in Mathematics, and to Professor J. A. Ewing, F.R.S., of University College, Dundee, who have kindly read portions of the proof sheets. To Mr. Murray I am indebted also for the verification of a large number of the examples. I have taken pains to attain as great accuracy as possible ; but errors are inevitable ; and readers will confer a great favour if they will kindly point out to me any they may detect. J. G. MACGREGOR. Dalhousib College, Halifax, N.S., August 19t/i, 1887. CONTENTS. PART I.— KINEMATICS. Chapter I. — Position and Motion. Sec. Subject matter of Kinematics, ... ... 1 Position, .... . 2 Co-ordinates, polar and Cartesian, . . . .3-10 Configuration, .... 11 Measurement, . ... 14-15 Measurement of length, area and volume, . 16-17 Derived units and their dimensions, . . . 18-20 Measurement of angle, .... . 21-22 Motion and rest, . . . 23-24 Time, . . ... 25 Description of instants, , . 26 Measurement of time, . . 27-32 Chapter II. — Translation : Paths. Degrees of freedom of a point, . . .35 Paths, ... 36 Curvature, . . . . . 37-40 Tortuosity, ... . . . 41 Speed of a point moving in a given path, . 42-44 Units of speed and their dimensions, . . 45-49 Change of speed, . . . 51 Kate of change of speed, . . . . . 52-55 Units of rate of change of speed and their dimensions, . 56-58 Motion of a point in its path under given rates of change of speed, 60 The rate of change being zero, . . . . 61 The rate of change being uniform, . . 63-66 CONTENTS. Chapter III. — Translation : Displacements, Velocities, Accelerations. Displacement (linear), Change of the point of reference, Composition of successive displacements, . Composition of simultaneous displacements. Resolution of displacements, Trigduometrioal expression for resultant displacement, Analytical expression for resultant displacement, Velocity (linear), Units of velocity, .... Change of point of reference (relative velocity), . Composition of velocities. Resolution of velocities, Moment of a velocity, . Change of velocity. Acceleration (linear), . Units of acceleration, The hodograph, . . ... Change of point of reference (relative acceleration), Composition and resolution of accelerations. Tangential and normal acceleration. Moment of an acceleration, . Angular displacement of a point. Angular velocity of a point. Units of angular velocity, ... Relation between angular and linear velocity. Moment of linear velocity in terms of angular velocity, Areal velocity, . . . ... Angular acceleration of a point, ^ Units of angular acceleration. Sec. 69-70 71-74 76-77 78 79-84 8.5-89 . 90 92 94 96 98 99-101 . 103-107 109 110 111 113 1].-) .116-118 . 120-121 . 123-124 125 127 128 . 129-131 132 133 135 136 Chapter IV. — Translation : Motion under given Accelerations. Unconstrained motion of a point, . 138 Under zero acceleration, . 138 Under a uniform acceleration, . .... 140 Rectilinear motion (falling bodies, etc. ), . 140 Curvilinear motion (projectiles), .... .142-154 CONTENTS. XI Siso. Under a central acceleration, . . 156 Planetary motion, . . 158-162 Harmonic motion, . . 163-180 Constrained motion of a point, . .181 Motion on an inclined plane, . . 181-183 Motion in a curved path, 185 Motion on a spherical surface (simple pendulum), . 187-190 Motion in a cycloid, . 192-193 Chapter V. — Rotation. Degrees of freedom of a rigid system with one point fixed, . 198 Rotations, .... ... 199 Composition of successive rotations, . . 200-202 Composition of simultaneous rotations, . , 203-206 Resolution of rotations, .... 207 Rotational displacements, ... . 208-211 Angular velocity of a rigid system, ... . 212-215 Composition and resolution of angulat velocities, . . 216-218 Angular acceleration of a rigid system, 219-220 Composition and resolution of angular accelerations, . .221 -223 Motion of a rigid system under given angular accelerations, 224 Under zero angular acceleration, ...... 224 Under a constant angular acceleration, .... 225-227 Geometrical representation of the motion of a rigid system about a fixed point, 22!) Chapter VI. — Motion of Rigid Systems. Degrees of freedom of a free rigid system, . . . . 231 Displacement of a free rigid system, 232-236 Special case of a rigid lamina moveable in its ovpn plane, . 233 Composition of translations and rotations, ..... 238-244 Reduction of any displacement to a translation and a rotation about the direction of translation, . . . 245 Most general possible motion of a rigid body, 246 Composition of linear and angular velocities, . . 247-248 Composition of angular velocities about parallel axes, 249 Composition of linear and angular accelerations, . 250 Motion of a free rigid system under given accelerations, . . 251 Geometrical representation of the motion of a rigid lamina in its own plane, 252 Motion of rigid systems under constraint (Kinematics of machinery), . .' .... 253-254 CONTENTS. Chapter VII. — Strains. Homogeneous strains, Sec. 257 Properties of homogeneous strains, . 258-263 Strain ellipsoid, . 264 Undistorted planes, 265 Kelation of final to initial volume. 266 Pure strains 267 Rotational strains. 268 The shear, . 269 Homogeneity of the shear, . : .... 270 Reduction of the shear to a pure strain and a rotation, 272-275 Relation of amount of shear to principal ratios and elongations, . 276 Torsion, flexure, etc., . 277 Specification of a strain, . 278-283 Heterogeneous strains, 284 PART II.— DYNAMICS. Chapteb I. — The Laws of Motio>'. Forccj . . . . . . 285 First law of motion, . . 286 Second law of motion, . 287-295 Measurement of force and mass, . . 297-298 Gravitational units, . . 298-300 Absolute units, . . . . . 301-303 Density, . . .304 Third law of motion, .... . 307 Hypothetical character and verification of the laws of motion, . 308 Galileo's law,. . . . . . 309 Chapter II. — Dynamics of a Particle. Specification of a force acting on a particle. Composition and resolution of forces acting on a particle, Attractions, Equations of motion of a particle, Equations of motion in terms of impulse, . Impact, . . . 311 312-313 315-316 317-318 319 321 CONTENTS. Xlll Sec. Equilibrium of a particle, ... . 323-326 Friction, Work done. Units of work done, Eate of work done, Units of rate of work done. Work done under given forces, Under a uniform force. Under a central force, 328 330 . 331-332 . 333-334 335 337 337 ..3.38-340 Relation of work done by resultant to work done by com- ponent forces, . . . . . ,342 Energy, . . . 343 Kinetic energy, . 344 Potential energy, . . . 345 The law of energy, . . . ,348 Transformations of energy, . . 349 AppUoatibn of law of energy to kinetic problems, . 351 Application of law of energy to static problems (principle of virtual velocities), . . . . 353 Potential, ... ... . 355-357 Equipotential surfaces, . 358 Lines of force, . . . 359 Tubes of force, . . 360 Gravitational potential, . . ... 361 Integral normal attraction over a surface and applications to calculation of ajitractions, 365-370 Mapping out of field of force, . . . . 371 Chapter III. — Dynamics op Simple Systems of Particles. Internal and external forces . 376 Collision, .377 Systems of particles connected by strings, . 381 Atwood's machine, etc., . . . 382 Chapter IV. — Dynamics of Flexible Inbxtensiblb Strings. Equations of motion and of equilibrium, . . 384-385 Strings under no external forces, . . 387 Strings under forces acting at isolated points, . 389 XIT CONTENTS. Sec Strings under external forces continuously applied, . . 390 The external force being the reaction of a curved surface, smooth or rough, 391-392 The external force being the weight of the string, . . 394-396 Chapter V.— Dynamics or Extended Bodies. Centre of mass • 399 Distance of centre of mass from any plane, . . . 400 Determination of centres of mass in special cases, . . 405-408 Velocity and acceleration of centre of mass in terms of velocities and accelerations of particles and their masses, ... ... . 409-4J2 Acceleration of centre of mass in terms of external forces, . 414 Conservation of linear momentum, .... 416 D'Alembert's principle, . . . 417 Moment of momentum, . . .418-419 Angular momentum, 420 Relation between moment of momentum and angular momentum, 421 Moment of acceleration of momentum, ..... 422 Relation between moment of acceleration of momentum and rate of change of angular momentum, . . . 423 Moment of a force, ... 425-427 Rate of change of angular momentum in terras of moments of exbernal forces, 428 Conservation of angular momentum or of areas, . 429 Equations of motion of extended systems, . . 431 Energy of a system of particles, . . . 432-434 Conservation of energy, . . 435-436 Law of energy, . . . 437-442 Equilibrium of extended systeu,^, . . 444-449 Stability of equilibrium, . . .... 450 Relation between potential energy and stability of equi- librium, . . .... 451-452 Chapter VI. — Dynamics op Rigid Bodies. Application of general equations of motion to rigid bodies, 453 Rotating power of a force on a rigid body, . . . 455 Principle of transmissibility of force, . . . 456 Specification of a force acting on a rigid body, . 457 CONTENTS. XV Sec. Composition of Forces acting on a rigid 1)0(13?, .... 459 Co-planar forces reducible to a single force, 460 Resultant of non-parallel coplanar forces, . 461-463 Resultant of parallel coplanar forces, . . 464-470 Couples, ...'.. . . 467-469 Parallel forces reducible to a single force, . 471 Centre of system of parallel forces, . 472 Centre of gravity, . . ... 473-474 Any system of forces reducible to two forces, ' . . . 476 Condition of reducibility to a single force, .... 477-478 Any system of forces reducible to a single force and a single couple, .... ... 479-481 Poinsot's central axis, . . . 482-484 Moments of inertia, .... 486 Determination by experiment, . 487 Determination by calculation, . . . 488-490 Units of moment of inertia, . . . 491-492 Equations of motion of a rigid body, . 493 Equations of motion in terms of impulse, 494 Conservation of linear and angular momentum, . . 495 Motion of rigid bodies about fixed axes (physical pendulum, etc. ) , 496 Motion of free rigid bodies, . 497 Motion of systems of rigid bodies (Atwood's machine, common balance, direct and oblique impact of spheres, etc.,) . 498 Application of the law of energy to the solution of problems on rigid bodies, ... . . 499 Equilibrium of a rigid body, ... . . 500-506 Equilibrium of the balance, lever, wheel and axle, etc. , . 507 Equilibrium of a system of rigid bodies, ..... 508 Equilibrium of systems of pulleys, levers, bars, etc., . 509 Conditions of equilibrium in terms of work done, . . . 510-516 Equilibrium of systems of pulleys, differential wheel and axle, screw, differential screw, etc 517 Chapter VII. — Dynamics of Elastic Solids and Fluids. Statics of deformable bodies, ... ... 518 Stresses . . .519-521 Homogeneous stress, .... . . 522 Units of stress and their dimensions, . . 523 xvi CONTENTS. Sec. Resultant of stress, . 525-527 Centre of stress, .528-529 Resolution of stress, . .... 530 Specification of stress 531-537 Resolution of a tangential stress into longitudinal stresses, 538 Relation of stress to strain, ■ 5^0 Homogeneous bodies, . ... 541 Isotropic bodies, . . 542 Elasticity, . . . .543-546 Solids and fluids, . . 547 Conditions of elastic isotropy, . 549 Statics of elastic solids, ... . 551 Moduluses of elasticity, 551 Strain due to longitudinal stress, . . . 554 Stress required for longitudinal strain, . . 555 Torsion of a cylinder, flexure of a beam, etc., . 556 Kinetics of elastic solids, . 557 Time of oscillation of body suspended by a twisted wire, . 558 Work done during strain, . . .... 559-564 Resilience, ... . .... 565 Statics of fluids (Hydrostatics), 566 Stresses in fluids, . . . 569 Homogeneity of fluid pressure, . 569 Specification of fluid pressure, . . . 570 Equal transmission of pressure, . 571 Surfaces of equal pressure, . 572 Variation of pressure in fluids acted upon by external forces, 577-578 In the ease of liquids, . . . 579 In the case of gases, . . 580 Archimedes' principle, ... 58 1 Equilibrium of a floating body, . . . . 582 Stability of equilibrium of a floating body, . 583 Kinetics of fluids (Hydrokinetics), . . . • . 584 Stresses in a fluid in motion, . . 585 Equations of motion, . . 586 Steady motion . 587 Equation of energy for steady motion, . . . 588 Flow of liquid through an orifice (Torricelli's theorem), 589 Work done during strain of a fluid, ... . 590 PART I.— KINEMATICS. ERKATA. Page 32, 4th line— /or 0003, read 0-0003. —for PjP and pPg, read pP^ and Pjp. —for these moments, read the moments of OA and OB. — -for are, read area. —forp, read pv. —for KLMN, read HKLM. —for x=x + i, readx^x + ^. —for single read resultant. —for is, read are. from the foot of the page— /oj- ns, read Ans. ? . , read -? . 8. ti I 63, 4th „ 79, 1st ,, 79, 2nd ,, 123, 13th „ 307, 9th „ 320, 19th „ 344, 12th „ 344, 13th „ 370, 2nd „ 375, 3rd „ 383, 7th „ 385, 10th ,, 391, 2nd „ 426, 5th ,, —for X- —for from the foot of the page — the equation should be, r' - w'a cos S = 0. —for s, read is. —the equation should be, *= lE^'^t^ . %;mr-vnir) —fm- hinge-pin, read hinge-pins. PART I.— KINEMATICS. CHAPTEE I. POSITION AND MOTION. 1. Kinematics is that branch of Matliematical Science which investigates motion. It makes no inquiry as to the causes of the changes of motion in bodies, but studies their motion in itself. 2. Position. — We recognize bodies as existing in space and having definite, positions among one another. We recognize them as having positions, however, only by the aid of neighbouring bodies, and we describe the positions of their various points by, reference to chosen points in neighbouring jbodies. Position in space is thus a relative conception. "Absolute position" is a meaning- less phrase. The position of a point P, relative to any other point 0, is completely determined if we have sufficient data to enable us to proceed from' to P. There are various modes of specifying the necessary data. They are called systems of co-ordinates. Of these we may mention two : ,(1) that of Polar Co-ordinates ; (2) that of Cartesian Co-ordinates. KINEMATICS. [ 3 3. Polar Co-ordinates. — If the point P is situated in a given plane, its position relative to 0, a^Qother point in that plane, may be de- scribed by the aid of Olf, a known line in the same plaDe, by a statement of the -" angle NOP and the length OP. Thus, if and P are points in a horizontal plane, and OH the north and south line through 0, the angle NOP (which in that case is called the azimuth of P) and the distance of P from determine the position of P The point is called the pole in this system of co- ordinates, ON is called the initial line, and OP the radius vector. The length o£ OP and the magnitude of the angle NOP are the polar co-ordinates of the point P. They are usually denoted by the symbols r and 6. To describe the position of a point P not in a known plane, let be the pole, ON the initial line, and ONA a known plane containing ON but not P. Let OA be the intersection of the plane ONA with a plane perpendicular to it through OP Then, if the angles NO A and J. OP are given, the direction of OP is known, and if the length of OP is also given, the position of P is completely determined. The length of OP and the angles NO A and AOP are then the polar co-ordinates of P. They are usually denoted by r, OP^==x^ + y^.+z\ Hence OP^ = 0P\cqs\ + cos^^g + cos V), an d cos^a + cos^/3 + cos^y = 1 . If therefore the length of OP, and any two of the angles a, /3, y, be given, the position of P is completely specified. The direction of OP is specified by any two of the angles a, ^, y. The cosine? of these angles are therefore called the direction cosines of OP. 8. It is frequently convenient to be able tt^ express the KINEMATICS. [8 inclination* of two straight lines in terms of their direc- tion cosines. Let OP and OP' be two such lines or lines drawn parallel to them, Ox, Oy, Oz rectangular axes, OL, LM, MP the rectangular co-ordinates of any point P of OP, and a, /8, y the angles of inclination of OP to the axes of x, y, and z respectively. Then OL = OP cos a, LM=OP cos /3, MP = OP cos y. Now the projection! of OP on OP' is equal to the sum of * The inclination of one straight line to another, whether they are in one plane or not, is the angle between two lines drawn parallel to them from any point. t (1) The foot of the perpendicular from a point on a straight line is called the orthogonal projection or simply the projection of the point on the line. (2) The locus of the projections of all the points of any line on a given straight line is called the projection of the former on the latter. (3) The projection of a finite straight line on a straight line is equal in length to the product of the length of the projected line into the cosine of its inclination to the given straight line. Let LM be the projected line, AB the line on which it is projected. In general these lines will not be in the same plane. From L, M, draw 111, Mm, perpendicular to AB. Then Im is the projection of LM. From m draw ml' equal and parallel to ML, and join LV and W. Then LV is parallel to Mm and therefore perpendicular to AB^ Hence the plane LIV and therefore the line IV are perpendicular to A B. Hence Im = I'm cos ImV = LM cos Iml', 9] POSITION AND MOTION. the projections of OL, LM, and MP on the same line; Hence, if Q is the angle between OP and OP', and a, /3', y, the inclinations of OP' to the' axes of x, y, and z respectively, , OP cos d = OL cos a + LM cos 13'+ MP cos y = OPcosa cosa'+ OP cos ^8 cos /3' + OP cos y cos y. Hence cos = cos a cos a + cos j3 cos jS' + cos y cos y. 9. To find the value of sin 6, call cos a, cos j8, and cosy, Z, wi, and n respectively, and cos a', cos^Q', and cos y, r, m', and n' respectively. Then I.e., the projection of LM is equal to the product of LM into the cosine of its inclination to AB. The simpler case in which LM and AB are in one plane may be left to tlie reader. (4) The algebraic sum of the projections of the parts of a broken line ia equal to the projection of the straight line joining its end points. Let OLMP be a broken line, the straight portions of which, OL, LM, MP, are not in one plane. From and P draw Oo and Pp perpendicular to A B. Then ol, Im, and r)ip are the projections of OL, LM, and MP on A B. Also, from the construction, op is the projection of the line OP on AB. And op = ol + lm + mp. Hence the projection of OP on AB is equal to the sum of the projections of OL, LM, and MP on the same line. If the position of L is such that the point I is situated to the left of o, ol being drawn to the left instead of the right must be con- sidered negative, the lines Im and mp being taken as positive. In that case we have op = lm + m.p — ol, i.e., the projection of OP on AB is equal to the algebraic sum of the projections of OL, LM, and MP on the same line. 8 KINEMATICS. [9 sin ={ 1 - {W + mm' + nn'f}i = {(J?j^m''+ n^){l'^ + m'2 + n'^) - {W + mm' + nn'f}i = {{mn -nm'y+{nr -ln'f+{lm' -ml'}^}i. 10. It is frequently convenient also to be able to express the direction cosines (X, jm, v) of the common perpendicular to two lines, in terms of the direction cosines (I, m, n, and I', m', n') of the lines themselves. For this purpose we have (8), since cos ('7r/2) = 0, l\ + mfi. + nv = 0, r\ + m'im+n'v = 0. We have also (7) \^ + fj,^+v^=. 1. From these equations we obtain values of X, fi> v- Writing sin 6 for its value as given above (9) they are mw'— Tim' _nl'—ln'_ _lm'—'ml' " sin ' '^ ~" sin ' ~" sin ■ n. The positions of any two points relative to a third being given, that of either of the two relative to the other can be determined. The positions of P and Q being given relatively to 0, p the lengths and directions of OP and OQ are known. Hence also (8) the *5 angle POQ is known, and consequently all the sides and angles of the triangle OPQ. The direction and length of PQ being thus determined, the position of either of the two P, Q, relative to the other is known. It follows that, if the positions of all the points of a sj'stem relative to any one are known, their positions relative to any other are known also. 12. Configuration. — The arrangement of the points of a system is called its configuration. The configuration 14] POSITION AND MOTION. 9 of a system is thus known if the positions of all points relative to any one are known. 13. Dimensions of Space. — Whatever system of co- ordinates we may adopt, we require, in order to specify the position of a point, to have three quantities given. In the case of rectangular co-ordinates they are distances; in that of polar co-ordinates they consist of two angles and a distance. Hence space is said to be tri-dimen- sional. Similarly, any point in a given surface may be specified by a statement of two; quantities, two distances or a distance and an angle ; and any point in a given line may be specified by the statement of a distance merely. Hence a surface is said to have two dimensions, and a line one dimension. 14. Measurement. — The specification of the position of a point requires therefore that we should be able to measure lengths and angles. The measurement of any quantity is the comparison of its magnitude with the magnitude of a known quantity of the same kind. The known quantity of the same kind is called a standard or unit; and a description of any measurement must include a statement of (1) the unit employed, and (2) the ratio of the magnitude of the quantity to be measured to the magnitude of the unit. This ratio is called the numerical measure or value of the quantity. Prof James Thomson has proposed to shorten these terms to numeric. Any quantity whatever, of the same kind as that to be measured, may be chosen as a standard or unit. But it will be evident that no standard should be employed which is not (1) constant in magnitude, (2) well known, and (3) easily reproduced ; and we shall see farther on, 10 KINEMATICS. [14 that among standards satisfying these conditions, there are reasons for preferring some to others. 15. We have seen that the numerical measure of any quantity in terms of any unit is the ratio of the magnitude of the quantity to that of the unit. It follows that the numerical measure of a given quantity must be inversely proportional to the magnitude of the unit in terms of which the quantity is expressed. Let Q be the numerical value of any quantity, and let [Q] denote the magnitude of the unit in terms of which it is expressed. Then we have Qal/[Q]. 16. Measurement of Length. — The selection of stand- ards of length presents no difficulty. A certain distance in space cannot, it is true, be marked off and kept ; but a body, say a rod, may be selected and carefully preserved, and when it is in a specified physical condition (as to temperature, etc.), its length may be taken as unit of length. The submultiples of the unit thus chosen may then be determined by geometrical methods. For the various methods of comparing the length of a body or the distance between two points in space with the standard length, the reader is referred to works on Laboratory Practice. Different nations have adopted different units of length. The more important are the English and French units. The English unit, the yard, is defined by Act of Parlia- ment to be the distance between the centres of two gold plugs in a certain bronze bar deposited in the Office of the Exchequer in London, the bar having the temperature 62°F. (The specification of the temperature is necessary, because the lengths of bodies vary with temperature.) The foot is one-third of the yard. The inch is one- twelfth of the foot. The statute mile is 1,760 yards. The French unit, the mhtre, is the distance between the 17] POSITION AND MOTION. 11 end planes of a certain platinum bar deposited in Paris, the temperature of the bar being 0°C. The metre was intended to be the ten-millionth part of a quadrantal arc of a meridian on the earth's surface. It is now known to be a somewhat smaller fraction. The decvmetre, centivietre, and Tnillimetre are the tenth, hundredth, and thousandth parts of a metre, respectively. The decametre, hectometre, and kilometre are equal to ten, one hundred, and one thousand metres I'espectively. The decimal division of the metre renders it a much more convenient unit than the yard. The following table shows approximately the relative values of English and French units of length : — 1 centimetre = 0-39370 in. 1 inch = 2-5400 cm. 1 foot =30-4797 cm. 1 yard = 91 -4392 cm. 1 mile= 1-60933 km. do. =0-032809 ft. 1 metre =3-28087 ft. 1 kilometre =0-62138 ml. 17. Measurement of Area and Volume. — "We may notice here, though it is not necessary for our present purpose, the measurement of area and volume. Any arbitrary area may be chosen as unit of surface or area. But the most convenient unit is the area of a square whose side is of unit length. The English units are therefore the square yard, square foot, square inch, etc.; the French units, the square metre, square centi- metre, etc. 1 sq. inch = 6-4516 sq. cm. 1 sq. foot =929-01 sq. cm. 1 sq. yard = •836113 sq.m. 1 sq. mile = 2-59 sq. km. 1 sq. centimetre = 0-1550 sq. in. do. =0-001076 sq.ft. 1 sq. metre =1-196 sq. yd. 1 sq. kilometre =0-3861 sq. ml. Similarly, the most convenient unit of volume is that of a cube ,whose edge is of unit length. The English units are thus the cubic yard, cubic foot, etc.; the French 12 KINEMATICS. [17 units are the cubic metre, cubic decimetre (called the litre), etc. 1 cu. inch = IC'SS? cu. em. 1 cu. foot =283ie' cu. cm. 1 cu. yard= 0'764535 cu. m. 1 cu. cm. =0'06102 cu. in. do. = 3-532 X 10-5 cu. ft. 1 cu. metre = 1 '308 cu. yd. 18. Derived Units. — A unit of a quantity of one kind which is thus defined by reference to the unit of a quan- tity of another kind is called a' derived unit. The magnitude of such a unit will depend upon that of the simple, or arbitrarily chosen, unit, by reference to which it is derived. Thus it is clear that if our unit of length be increased two, three, four, etc., times, our unit of area will be increased four, nine, sixteen, etc., times respec- tively ; or, generally, that the magnitude of the unit of area is directly proportional to the square of the magni- tude of the unit of length. In symbols, if [S] represent the magnitude of the unit of area, and [i] that of the unit of length, [8] a [Lf. A statement of the mode in which the magnitude of a derived unit varies with the magnitudes of the simple units involved in it, is called a statement of the dimen- sions of the unit. The unit of area has thus the dimensions [Z]^. 19. Though this result is suflBciently obvious, we may obtain it by a method which we shall find useful when dealing with more complicated units. Let s be the area of a square whose side is I. Then s=P. Now (15) s a l/[;Sf] and I a 1/[L]. Hence [S] oc [Lf- 20. The reader will find no difficulty in showing in a similar way, that the unit of volume has the dimensions [Lf- 21. Measurement of Angle. — There are two units of plane angle, in ordinary use, the degree and the ladian. 24] POSITION AND MOTION. 13 The degree is the ninetieth part of a right angle ; and its subdivisions are the minute, which is one sixtieth part of a degree, and the second, which is one sixtieth part of a minute. The radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. As the circumference of a circle is 2^ times the radius, the radian is equal to 360°-i-27r, i.e., to 57°"29578... or to 57° 17' 44"-8 nearly. It is subdivided decimally. The numerical measure of an angle in radians is often called its "circular measure." It is obvious that the angle sub- tended at the centre of a circle of radius r, by an arc of length a, is equal to ajr radians, and- that consequently the magnitude of the radian is independent of the mag- nitude of the unit of length. 22. The unit of solid angle is the solid angle sub- tended at the centre of a sphere by a portion of its surface whose area is equal to the square of its radius. It may be called the solid radian. It follows that the solid angle subtended at the centre of a sphere of radius r, by a portion of its surface whose area is A, is Ajr^ solid radians, and that the magnitude of the solid radian is thus also independent of that of th^ unit of length. 23. Motion. — The motion of a point is its change of position in space.- It is therefore completely described by a statement of the changes in the co-ordinates of the point. Motion is thus, like position, a relative con- ception. 24. Rest. — A point which is undergoing no change of position, whose co-ordinates therefore are not varying, is said to be at rest relative to the origin of co-ordinates or point of reference. In any case in which we speak of a body as being simply " at rest," it is assumed that the point of reference is known. A " fixed point " or a " point fixed in space ' is one 14 KINEMATICS. [24 which, during the time under consideration, is at rest relatively to the point which has been chosen as point of reference. A line fixed in space is one containing fixed points, 25. Relation of Motion to Time. — The motion of a body is found to occupy time ; and one important object of Kinematics is to compare the contemporaneous motions of different bodies, and to determine the laws according to which the changes in the co-ordinates of some bodies are related to the contemporaneous changes in the co- ordinates of others. As it is not possible for one observer to make many observations of the positions of bodies at the same instant, it is necessary, for the attainment of this object, to be able to describe instants of time, in order that the observations of different observers may be comparable. 26. Description of Instants of Time. — To describe the times of occurrence of events, it is only necessary that we should fix upon some series .of continually occurring events and keep a record of them. We may choose, for example, the daity passage across the meridian, of a known point in the "heavens, say a "fixed" star. In that case, the time of the occurrence of an event would be described as between the «"■ and the (TC-f-l)* transits of this star. To make the description more definite, we may use a rapidly oscillating pendulum, and describe the event as occurring between the m**" and (m-l-1)*'' oscillations of the pendulum after the n*^ transit of the fixed star. By thus selecting a series of events occurring with sufficient frequency, it is possible to give our de- scriptions of instants of time as great precision as may be desirable. 27. " Measurement " of Time. — As we are thus able to describe instants, it is possible to record the magnitudes of quantities {e.g., distances, angles, etc.) at definite 29] POSITION AND MOTION. 15 instants, and therefore to compare the changes which the positions of bodies may have undergone in any required interval of time. 28. In order to compare the contemporaneous motions of any number of bodies among one another, it is only necessary to compare the motion of each body with that of some one selected as a standard of comparison. In selecting a standard, it will save a great deal of labour if we choose a body whose motion is such that as many as possible of the laws of the motions of other bodies, when expressed in terms of its motion, are (1) simple, and (2) permanent, i.e., independent of the date of their determination. To fix upon such a moving body, it is necessary to make observations of the positions of many bodies at short intervals during long periods of time, and to keep records of them. This has been done by astronomers, whose records extend over 2,500 years. Their observations show, that if the motions of other bodies are compared with the rotation (194) of the earth relative to the "fixed" stars, the laws of their motions take forms which are simpler and more per- manent than if any other motion is taken as the standard. Hence by common consent the motion of the earth about its axis is taken as a standard with which other motions are compared. 29. When the law of the change of the position of a body, with reference to the rotation of the earth about its axis, is determined, we are said to have determined the law of its change of position with reference to time, successive rotations of the earth being assumed to occur in equal intervals of time. Whether they do so or not, we have no means of knowing, as we have no means of measuring time, But this form of speech, which assumes the possibility of measuring time, is conveniently short, and so long as we keep in mind its real meaning, can lead to no error. The period of the earth's rotation with reference to the fixed stars, i.e., the period between sue- 16 KINEMATICS. [29 cessive instants at which a fixed star is on our meridian, is called a sidereal day. When we employ the earth's rotation relative to the fixed stars as a standard motion, we may be said to employ the sidereal day as a unit of time. . 30. Recent discussion of astronomical observations * seems to show that the laws of the motions of heavenly bodies would take simpler forms, and would be more permanent, if the standard motion were that of an ideal earth, rotating so that its rate of rotation would slowly gain on the rate of I'otation of the actual earth. At what rate the ideal earth's rate of rotation should gain on that of the actual earth in order that these laws may take their simplest and most permanent forms, is not known. But the astronomical data are sufficiently definite to show that it is exceedingly small. This result is expressed in the language of time by saying that the sidereal day is increasing at a very slow rate. 31. It is found practically inconvenient to compare the motions of bodies directly with the rotation of the earth relative to the fixed stars. They are usually compared directly with the rotation of the earth relative to the sun ; and the law, according to which the earth rotates relatively to the sun, having been determined in terms of its rotation relative to the fixed stars, they can . thus be indirecjtly compared with the standard motion. In the language of time, it is found more convenient to measure time in terms of the solar day than of the sidereal day. The solar day being a variable period, the mean solar day is chosen as practical unit. It is found to be equal to 1 '002738... sidereal days. 32. It is frequently convenient to compare motions with some periodic motion of much greater frequency than * See Thomson and Tait's " Treatise on Natural Philosophy," pt. IL, § 830. 33] POSITION AND MOTION. 17 the rotation of the earth. In such cases the oscillation of a pendulum is chosen ; for it is found that, if a pen- dulum is kept in a constant physical condition, it will oscillate the same number of times in different sidereal days, and that in I/ti"* of a day (i.e., while the earth is making l/n*^ of a rotation) it will make I/tc*^ of the number of oscillations made in a whole day. The second is the time of oscillation of a pendulum which oscillates 86,400 {i.e., 24 x 60 x 60) times in a mean solar day. The sidereal day contains 86,164 mean solar seconds. A clock is an instrument for maintaining a pendulum in oscilla- tion and for counting its oscillations. 33. Complexity of Motion. — The motions of bodies may be of various degrees of complexity. The simplest form is that in which all points of the body move through equal distances in the same direction. Such a motion is called a translation. If, though the various points of the body maintain the same relative positions during the motion, they do not move through equal distances in the same direction, the motion is partly or wholly a rotation. If, finally, the points of the body do not maintain the same distances from one another during the motion, the motion consists partly of a strain or change of volume or form. We shall see farther on that the action of a force upon a body usually affects the motion of the body in all these ways. It is convenient, however, to study the different kinds of motion separately, assuming bodies to have that kind of motion alone, which, for the time, we may wish to investigate. KINEMATICS. [34 CHAPTER II. TRANSLATION :— PATHS. 34. We have defined translation to be the motion which a hody has when all its points move through equal distances in the same direction. If then the motion of one point is known, the translation of the body is known. Hence the study of the translation of a body is the same as the study of the motion of a point. 35. Degrees of Freedom. — The position of a point, as we have seen, is determined by three numbers, which may be measures of distance or of distance and angle. The motion is determined if the changes in these meas- ures are known. Hence a point is said to have three degrees of freedom to move. If the point be constrained to remain on a given sur- face its position can then be determined by two numbers, and it has therefore two degrees of freedom. One degree of constraint is said to have been introduced. The con- dition of constraint in this case is that the distance of the point from the surface shall be zero. If the point be constrained to remain on each of two surfaces it must remain on their line of intersection. Hence its position and its motion may be determined by one number, the distance or the change of distance from a given point in the line. It has one degree of freedom. Two degrees of 37 ] TRANSLATION : PATHS. 1 9 constraint have been introduced. A third degree of constraint, the condition for instance that the point remain on a third surface, will confine it to the point in which the three surfaces intersect: it has then no freedom. Constraint is of course not necessarily applied in the way mentioned above. Thus the condition that a point shall maintain a given distance from a given fixed point restricts its motion to the surface of a sphere. A second degree of freedom is destroyed by constraining the point to move in a vertical plane, and it can now move only in the curve of intersection of the vertical plane and the sphere.. If now it be so constrained that the line joining it with the fixed point maintains a constant inclination to a fixed line in the given vertical plane, the point has three degrees of constraint and consequently a definite position. 36. Paths. — The path of a moving point is the locus of its successive positions, It must be a continuous line, bnt may have any form whatever. We shall see farther on (295), however, that the path of a material particle (310) can undergo no abrupt changes of direction, unless indeed its motion cease and recommence ; and we shall restrict ourselves to the study of paths which are possible for material particles. The direction of such a path at any point is that of the tangent at that point. 37. Curvature. — The change of direction between any two points of a path lying wholly in one plane is called the integral curvature between these points. It is evidently measured by the angle between the tangents at these points. The mean curvature between two points is the integral curvature between them divided by the length 20 KINEMATICS. [37 of path intercepted by them. Thus, if AB is a portion of the path of a moving point, AC and BD being tan- gents at A and B respectively, inclined at the angle 67. Examples. (1) A railway train is moving with a speed of 20 mis. per hour, and is increasing its speed uniformly at the rate of 10 mls.-per-hour per hour. Find (a) its speed after 1^ hours, and (6) the distance traversed in that time. [For (a)— Data : V(,=20mls. per hour; a=10 mls.-per-hour per hour; t=V5 hours. To be determined : v. And (63) v=v„+at. For (6) — Data: as above. To be determined: s. And (64) «=«o«+|a«2.] Ans. (a) 35 mis. per hour; (6) 41^ mis. (2) A railway train, moving at 50 mis. per hour, has the brakes put on, ami its speed diminishes uniformly for 1 minute, when it is found to have a speed of 20 mis. per hour. Find- (as) its rate of change of speed, and (6) the distance traversed in the time. Ans. (a) - 1,800 ml.-hour units; (6) A ml. (3) A point whose speed is initially 20 m. per sec, and is diminishing at the uniform rate of 50 cm.-per-sec. per sec., moves in its path until its speed is 120 m. per min. Find the length of path between the initial and final positions. [Data: Vq = 20 m. per sec; a= — 50 cm.-per-sec. per sec. = -0'5 m.-per-sec. per sec. ; w=120 m. per min. = 2 m. per sec. And (65) v^=Vg^ + Z as.] Ans. 396 m. (4) A point has a uniform rate of increase of speed of 20 cm.-per- sec per sec. and an initial speed of 30 cm. per sec. Find (a) the speed after 16 sec; (6) the time required to traverse 300 cm.; (c) the change of speed in traversing that distance. Ans. (a) 350 cm. per sec; (6) (•y/T29-3)/2 sec; (c) 10(«yi29-.3) cm. per sec (5) If in Ex. (4) the speed be decreasing instead of increasing, find (a) the distance from the starting point to the turning point, (6) the distance from the starting point after 10 sec. ; (c) the length of path traversed during the time in which the speed changes to 60 cm. per sec; (d) the time required by the moving point to return to the starting point. [To find (a), note that the speed 36 KINEMATICS. [67 Pg, whose other sides PjP^ and P^^ in the one triangle, and Pj) and pPj in the other, represent the displacements A and B respectively. Hence, if two sides of a triangle taken the same way round represent the two successive displacements of a moving point, the third side taken the opposite way round will represent the resultant displacement. Also PjP^ is the diagonal of the parallelogram pP^ through the point of intersection of the adjacent sides P^P^, Pip, which represent the two successive displace- ments A and B. Hence, if two successive displacements of a point be represented by two adjacent sides of a parallelogram, taken opposite ways round, the diagonal of the parallelogram through their point of intersection will represent the resultant displacement. Case II. — More than Tivo Displacements. — Let A,B C represent the successive displacements, Pj being the 76] TRANSLATION : DISPLACEMENTS. 43 initial position of the moving point. Draw P^P^ P^Pp P^Pi equal to and codirectional with A, B, and G respec- tively. Join PJP^. Then P^ being the final position of the moving point, P^P^ is the resultant displacement. The same construction is applicable to any number, of displacements. If A, B, G, etc., are all in one plane, P^P^PJP^... is a plane polygon; if not, it is a gauche polygon. It is clear that the same point P^ is reached in what- ever order the displacements occur. For, if the parallelo- gram PJ^i be completed, and then the parallelograms PjPj, pj^i, and PjPg, it follows from the equality and parallelism of the opposite sides of parallelograms that the line p^P^ will complete the parallelogram p^^, and that the six sets of displacements thus laid down, by which P^ may be reached, are the displacements A, B, G taken in all possible orders. And the same con- struction may be made whatever the number of dis- placements. Hence, if any number of successive displacements of a moving point, in any directions whatever, be represented by 71 — 1 of the sides of a polygon, taken the same way round, the resultant displacement, will be represented by the n*^ side taken the opposite way round. 44 KINEMATICS. [ 77 77. It follows from the last proposition that a given displacement may be resolved into any number of succes- sive displacements, provided these displacements can be represented by w— 1 of the sides of a polygon, taken the same way round, by the to*'' side of which, taken the opposite way round, the given displacement is represented. In the special case in which the successive displacements have directions parallel to the given displacement, it is clear that their algebraic sum must be equal to the given displacement. 78. Composition of Simultaneous Displacements. — A point undergoes simultaneously, given displacements relative to the same point; it is required to deter- mine the resultant displacement. Simultaneous dis- placements of a point are usually called com,ponent displacements. Let A, B, C, etc., be the component displacements, and let each of them (77) be resolved into ii equal successive displacements in its own direction. The magnitudes of these displacements will be A/n, B/n, C/n, etc., respec- tively. Then (76) the same final position will be reached whether the point undergo successively the displacements A, B, C, etc., or undergo, n times, the successive displace- merits A/n, B/n, Gjn, etc. But if n is indefinitely great and therefore A/n, etc., indefinitely small, the successive occurrence of the displacements A/n, B/n, C/n, etc., n times, is the same as the simultaneous occurrence of the dis- placements A,B, C, etc. Hence the same final position is reached when A, B, C, etc., occur simultaneously as when they occur successively. Consequently the pro- positions established in 76 for successive displacements apply also to simultaneous or component displace- ments. These propositions when formulated for simul- taneous displacements are usually called the triangle, the parallelogram, and the polygon, of displace- Tnents. 81] TRANSLATION: DISPLACEMENTS. 45 79. Resolution of Displacements. — A displacement and two straight lines being given, to find two displacements parallel to these lines, of which the given displacement is the resultant. — Let OA be the given displacement, B and the / given lines. From and A / draw lines parallel to B and ' G respectively, meeting in D. Then, by 78, the displacement ^ OA is the resultant of the com- ponent displacements OD and DA, and these displacements are parallel to the given lines, 80. When the components, in given directions, of a given displacement ai"e thus determiiaed, the given dis- placement is said to be resolved into components in those directions. Displacements are frequently resolved in directions which are at right angles to one another, in which case the components are called rectangular components. When we speak of the component of a displacement in a given direction, we mean its rectangular com- ponent in that direction. It is clear that the rectangular component of a displacement in any direction is the (orthogonal) projection of the displacement on any straight line in that direction. 81. The component (rectangular) of a given displacement in a plane parallel to any given plane may also be found. OA being the given displacement, draw from A a line AP perpendicplar to the given plane, and from a line OP perpendicular to AP and meeting it in P. OP is a rectangular component of OA, and it is in a plane parallel to the given plane. It is clearly equal to the projection (orthogonal) of OA on the given plane. 46 KINEMATICS. [82 82. The components of a given displacement, in three directions which are not all in the same plane, may also be found. — Let OA be the given displacement, and OB, 00, OD lines having the given directions. From A draw AE parallel to OB and meeting the plane of OB and 00 in E. Join OE, and through E draw ^i^ parallel to 0(7 and meeting OB in F. Then (78) OE and EA are components of OA; and OF and FE are components of OE. Hence OF, FE, and EA are components of OA ; and they are in the given directions. The special case, in which each of the three directions OB, 00, OD is at right angles to the plane of the other two, is of great importance. In this case the components are adjacent edges of a rectangular parallelopiped of which OA is the diagonal through their intersection. Fig.l 83. The resultant of two given displacements is equal to the algebraic sum of their components in its direction. ^ Let OA, AB be the given dis- placements and OB therefore the resultant displacement. From A draw AG perpendicular to OB. Then 00 and CB are the components in the direction of OB, of OA and AB respectively. In Fig. 1, [the components OC and CB have the same direction, B and we have also 0B= 00+ OB. In Fig. 2, OG and GB have oppo- site directions, and we have also OB = GB—0G. Hence the displacement OB is equal to the algebraic sum of the components OG and GB, in the direction of OB, of the displacements OA and AB. Pg.2 S5] TRANSLATION : -DISPLACEMENTS. 47 84. The component in a given direction, of the result- ant of any number of displacements in any directions whatever, is equal to the algebraic sum of their compon- ents in the same direction. Let OA, AB, BC, GD, DE represent the given displacements. Then (78) OE repre- sents their resultant. Now the projection of OE on any line MP is equal (8) to the algebraic sum of the projections of OA, AB, BC, GD, DE. Hence (80) the above proposi- tion is proved. The proposition of 83 is clearly a special case of the above proposition. 85. Trigonometrical Expression fat the Resultant. — The magnitude and inclination of component displace- ments being giyen, to find expressions for the magnitude and direction of the resultant. First, when there are two given components. — Let A and B be the two components, their magnitudes being d^ and c?jj, and their inclination 6. 6 may be an acute angle (Fig. 1) or an obtuse angle (Fig 2). Let P, be the initial position of the point. Draw P\P^ and P^P^ equal to and codirectional with A and B. Then (78) P^P^ is the resultant. Produce P^P^ to 0. Then OP^P^ is the angle 6. From P, draw P^Q perpendicular to Pfi. 48 KINEMATICS. [85 To determine the magnitude of the resultant, we have from Geometry (Fig. 1) P,P,^ = P,P,^+P,P,^ + 2P,P, . P,Q. (Fig. 2) P^P,^ = P,P,''+P,P,^-2P^P^.P,Q. Now in Fig. 1 P,Q = P,P,coBe, and in Fig. 2 P^Q = PJ'.cos (tt - 0) = - PJP.cos d. Hence in both cases p^p^^ = p,Pi + p^i + 2P,p, . p,P3cos e, and, writing R for PJP^ and d^, d^ for the components, R^ = d^^+d,^+2d,d,cose. To determine the direction of the resultant, we may find its inclination to one or other of the components, either the angle P^ which we may call a, or the angle P^P^P^ which we may call j8. For this purpose we have, from Trigonometry, sinP,:sinP,P,P3 = P,P3: P.P^. Now in both the above cases the angle P^PJP^ is equal to (tt — 0). Hence sin a : sin = (Zj : P, and d^ ■ a sm a = ^ sin d. It 86] TRANSLATION: DISPLACEMENTS. 49 Similarly sin |8 = ^ sin 0. It It follows that the displacement R has in the directions inclined a and /3 to its direction respectively, components whose magnitudes are di = iisinj8/sin(a+/3), d^ = R sin a/sin (a + ^). 86. Of these general results, the following are import- ant special cases.* Case I. — The displacements equal. Let both be called cl. Then R^ = 2^2(1 + cos 0) = Wco&\6l2). Hence R = 2d cos (0/2). Air,« „: d . n cZsin0 . d Also sma = -5sme = jrT- ^^ = sm^. it 2(t cos (0/2) 2 Hence a = 0/2. Similarly ^8 = 0/2. f7asei/. — The displacements equal and their inclina- tion 120°. Then, by Case L, i2 = 2c?cos60° = cZ, and a = /3 = 60°. Case III. — The displacements in the same direction, i.e., = 0. Hence cos = 1. Therefore 'R^ = d:^+d^+2d^d^, and R = d^+d.^. Case IV. — The displacements in opposite directions, i.e., = 180°. Hence cos0=-l. Therefore E" = d^^ + d^ - 2d^d^, and R = d^ — d^. Displacements in opposite directions being considered of opposite sign, Cases III. and IV. may be generalized * The reader should obtain the results of these special cases directly. D 50 KINEMATICS. [86 thus : The resultant of component displacements in the same straight line is their algebraic sum. Case V. — The displacements at right angles to one another, i.e., = 90°. Hence cos 6 = and sin 0= 1. Therefore R^ = d^ + d^, and R={d^ + d^^f. Also sin a = dJR, and sin ^ = d^jR. In this case a + /3 = 90°, and therefore sinjQ = cosa. Hence tan a = djd^. Hence also the component (rect- angular) of a displacement J2 in a direction inclined at the angle a to the direction of .R is equal to iicosa. 87. Secondly, when there are more than two given components. — The magnitudes and directions of three or more displacements being given, expressions may be found for the magnitude and direction of the resultant by finding, first, the resultant of any two, then the re- sultant of this first resultant with a third, then the resultant of this second resultant with a fourth, and so on, until all the component displacements have been compounded. 88. An important special case of 87 is the composition of three displacements, the direction of each of which is perpendicular to the plane of the other two. Let OA, OB, OG be the three displacements, the angles ,405, 90] TRANSLATION: DISPLACEMENTS. 51 AOG, and BOG being right angles. Complete the rect- angle AB and draw the diagonal OD. Then OG, being at right angles to OA and OJ, is also at right angles to OD: Complete the rectangle GD and join OE. Then OD is the resultant of OA and OB, and OE that of OD and OG. OE therefore is the resultant of all three. Now, by Geometry, OE^^OA^+OB^+OC^. Hence, calling the resultant R, and the components d^, d^, d^, we have The direction of OE is known, if either the angles AOD and DOE {<{,' and x), or two of the angles AOE, BOE, GOE (a, |8, y) are known. These angles may be expressed in terms of the magnitudes of the given dis- placements. For we have cos ^ = djOD = dj{d^^ + dj")*, cos X = OD/OE = (rf,2 + d/) Vi2, cos a = dJR, cos ^^d^/R, cos 'Y=djR. 89. It follows that the components d^, d^, dg, into which a given displacement R may be resolved, in three direc- tions which are at right angles to one another and are inclined to the direction of R at the angles a, j8, y, are d^=Rcosa; d^ = Rcos0; d^=Rcosy. 90. Analytical Expression for the Resultant of any number of component displacements. Convenient expres- sions for the magnitude and direction of the resultant of any number of component displacements, may be obtained by resolving the given components in rectangular direc- tions which are the same for all, adding the components in these directions, and finding the resultant by 86 (V.) or 88. 52 KINEMATICS. [90 First, let the given displacements he all in one plane. — Let dj, d^, cZj, etc., be the given displacements. _ Take two lines Ox, Oy at right angles to one another in the plane of d^, d, etc. Let the inclinations of d-^, d^, etc., to Ox be Oj, ag, etc. Then the displacements d^ d^, etc., have, in the direction of the x axis, components d^cos a^, djuosa^, etc., and, in the direction of the y axis, com- ponents djSinoj, d^sina^, etc. Hence, 86 (III.), we have in the direction of the x axis a resultant displacement equal to d^cos aj^ + d2Cos a^+ etc., which may be written 2d cos a ; and, in the direction of the y axis, a resultant displacement equal to djSina^ + d^sma^+ etc., which may be written Ecisina. Ox and Oy being at right angles, the resultant of these resultants is, 86 (V.), B = [(Ed cos af + (Id sin a)^]i. The inclination 6 of this resultant to the x axis is determined by the equation tan 6= (Ed sin a) /(Ed cos a). In adding together the components of the given dis- placements in the x and y axes respectively, we have assumed that the displacements are all in such directions as to give components in the directions of Ox, Oy respec- tively. If the directions of any are such as to give components in the directions xO or yO respectively, they must be (86, IV.) considered as negative in deter- mining the resultant displacements in these axes. Thus Ed cos a, Ed sin a are short expressions for the algebraic ^n TRANSLATION: — DISPLACEMENTS. 53 sums of all components of the form d cos a, d sin a re- spectively. _ Secondly, let the given components have any direc- tions whatever. — Take three rectangular axes, Ox, Oy, Oz, and let the inclinations of the displacements d^, d^, etc., to the X, y, z axes, be a^, ^^, y^, a^, ^^, y^, etc., respectively. Then the components of d^, d^, etc., in the direction of the X axis, are d^cos a^, d^cos a^, etc., and their resultant is Xd cos a. Similarly, the resultant of the component dis- placements in the y and z axes are J.dcos^, Scicosy respectively. Hence (88), if R is the magnitude of the resultant, Ii={(Xd cos a)H (2d cos /3)2+(2c? cos y)^}* Also, if the direction cosines of the resultant with reference to the X, y, z axes, are X, fx, v respectively, we have X = (ScZ cos a)/i2; fi ^ (Id cos ^)/R; v = (Ed cos y)/R. 91. Exa/mples. (1) ABGD is a quadrilateral. Show that, if AC is produced to E, and GE made equal to AG, the resultant of component displace- ments represented by AC, DB, AD, and BC will be represented by AE. (2) ABCD is a parallelogram. E is the middle point of AB. Find the components, in the directions of ABsioA. AD,oia, displace- ment which has the direction and half the magnitude of the resultant of component displacements represented by AC and AD. Ans. .4^ and AD. (3) The resultant of two equal displacements of magnitude, d, and inclined 60°,. is equal to that of a and 2a inclined 120°. (4) Two component displacements are represented by two chords of a circle drawn from a point P in its circumference and perpen- dicular to one another. Show that the resultant is represented by the diameter through the point. 54 KINEMATICS. 1^^^ (5) POPi and QUQj^ are twtf perpendicular chords of a circle, whose centre is C. Show that the resultant of four component displacements represented by OP, OP^, OQ, OQi, has the direction of OC and twice its magnitude. (6) The resultant D of two displacements, d^ and d^, is perpen- dicular to rfj- Fi°d til's resultant of displacements dj2 and c/j, their inclination being the same as that of d^ and c?2. Ans. dJ2 in a direction inclined taxr\Dld^ to that of d^,. (7) A point undergoes two component displacements, 60 ft. W. 30° S., and 30 ft. N. Find the resultant. Ans. 51-9... ft. W. (8) Show that three component displacements whose magnitudes are 1, 2,, 3, and whose directions are represented by the sides of an eqviilateral triangle, taken the same way round, have a resultant whose magnitude is JZ.- (9) A point undergoes three component displacements, 40, yds. N. 60° E., 50 yds. S., and 60 yds. W. 30° N. Find the resultant. Ans. 10 V3 yds. W. (10) A ship is carried by wind 4 mis. N., by her screw 8 mis. N. 15° W., and by a current 3 mis. E. 15° N. Find her resultant displacement in a north-easterly direction. Ans. 9-4265 mis. (11) A boat is headed directly across a river flowing fi'om north to south, and reaches a point from which the starting point is found to bear N. 30° W., and is known to be at a distance of 400 ft. How far has the boat been carried by the current, and what distance would it have made in still water ? Ans. 346'41... and 200 ft. respectively. (12) To an observer in a balloon his starting point bears N. 20° E , and is depressed 30° below the horizontal plane; while a place known to be on the same level as the starting point and 10 mis. from it, is seen to be vertically below him. Find the component . displacements of the balloon in southerly, westerly, and upward directions. Ans. 9"39..., 3"42..., and 5'77... mis. respectively. 92] TEANSLATION: VELOCITIES. 55 92. Velocity. — The mean velocity* of a moving point during a given time is a quantity whose direction is that of the displacement produced during the time, and whose magnitude is the quotient of the magnitude of the dis- placement by . the time. Thus, if . a point move in the path A from P^ to P^ in the time t, its displacement in that time is the straight line P^P^, the direction of its mean velocity is the direction of JPjP^ and the magnitude of its mean velocity is P^Pjt. In general the mean velocity of a point varies with the interval of time to which it applies. Thus, if in a time t' the point moves from P^ to P^ the direction of its mean velocity during t' is that of P-^P^ and the magnitude is PjPg/t. . In the special case in which a point moves so that its mean velocity changes neither in magnitude nor in direction, it is said to move with uniform velocity. In that case its path must be a straight line. For, wherever P^ and Pg may be, PJP^ and P^^ must have the same direction. It must also obviously be moving with uniform speed. It will be seen that a point whose speed is uniform has not necessarily a uniform velocity. The speed is uniform if arc PiP2/i = arc PjPs/f. But the velocity is not uniform, unless the chords PjP.j and P2P3 have the same direction, and their quotients by t and if respectively, the same magnitude. * The term mean velocity is employed by most writers to denote what we have caUed (42) mean speed. 56 KINEMATICS. [^^ 93. The instantavieous velocity at a given instant (usually called velocity simply) is a quantity whose magnitude and direction are the limiting magnitude and direction of the mean velocity between that instant and another when the interval of time between them is made indefinitely small. As bodies are found to require in all cases a finite time to traverse a finite distance, the instantaneous velocity of a body has always a finite value. When the interval of time t (92) is made indefinitely small, Pg is indefinitely near Py and the chord P^P^ coincides with the arc P^P.^- Hence the direction of the instantaneous velocity at a given instant is that of the tangent to the path at the point occupied by the moving point at that instant ; and its magnitude is equal to the instantaneous speed (43) of the point at that instant. Velocity, having both magnitude and direction, is thus, like displacement, a vector (70). 94. MeasureTnent of Velocity. — The specification of a velocity involves specification of both magnitude and direction. The direction may be described , in terms of the unit of plane angle (21). The magnitude, being the quotient of a distance by an interval of time, is a quantity of the same kind as a speed (42), and may therefore be measured in terms of the unit of speed (45). A unit of speed is thus also a unit of velocity; and the results of 47-49 apply to units of velocity as well as to units of speed. 95. Examples. (1) A point (see 91, Ex. 7) moves in a straight line from A to B, 60 ft., W. 30° S., in 10 sec, and thence in a straight line to C, 30 ft. N., in 20 sec. Find (a) the mean speed, and (6) the mean velocity during the whole time. Ans. (a) 3 ft. per sec; (6) 1"73... ft. per sec, "W. 96] TRANSLATION: VELOCITIES. 57 (2) A point moving •with uniform speed in a circular path passes from one end of a diameter to the other in 10 sec. The radius being 30 cm., find (a) the mean speed, (6) the mean velocity, and (c) the instantaneous velocity at any instant. Ans. (a) 9'4... cm. per sec; (6) 6 cm. per sec. in the direction of the given diameter; (c) 9'4... cm. per sec. in the direction of the tangent at the point occupied by the moving point at the chosen instant. (3) A man 6 ft. high is walking at the rate of 4 mis. per hour directly away from a lamp-post 10 ft. high. Find the magnitude of the velocity of the extremity of his shadow. Ans. 10 mis. per hour. 96. Change of the Point of Reference. — Velocity, being defined in terms of displacement, can be specified only by reference to some chosen point, which point of reference it is frequently desirable to change. Since the direction and magnitude of a velocity are the direction and magnitude of a displacement, viz., either one which actually occurs in a unit of time, or one which would occur in that time were the velocity not variable, the propositions established in 71-74 for displacements apply also to velocities. Hence, (1) The velocity of one point relative to another is equal and opposite to the velocity of the second relative to the first. (2) If two sides of a triangle, taken the same way round, represent the velocities of P relative to Q, and of Q relative to respectively, the third side, taken the opposite way round, will represent the velocity of P relative to 0. (3) If two sides of a triangle, taken the same way round, represent, the first the velocity of P relative to 0, and the second a velocity equal and opposite to that 58 KINEMATICS. [96 of Q relative to 0, the third side, taken the opposite way round, will represent the velocity of P relative to Q. In the special cg,se in which the velocities of P and Q relative to have the same direction, the velocity of P relative to Q will be equal to the difference of those of P relative to and of Q relative to 0. 97. Examples. (1) A. is moving with velocity F in a north-easterly direction, B with an equal velocity in a direction 15° east of south. Show that A'a velocity relative to B has a magnitude V iJZ and is in a direc- tion N. 15° E. (2) Two points are moving with equal uniforni speed % the one in a circle of radius r, the other in a tangent to the circle. Both start at the same instant in the same direction from the point of contact of their paths. Find their relative velocity after t units of time. vt Ana. 2« sin — , in a direction inclined to the tangent at an angle K--?)- (3) One railway train is running at 20 mis, per hour in a northerly direction. Another running at half the speed appears to a passenger in the former to be Tunning at 25 mla. per hour. Find the direction of the velocity of the latter, Ans. 71° 47' -4 W. or E. of N". (4) To a person travelling at 8 mis. an hour along a road tending west, the wind appeared to come from the N.W. On his standing still, it seemed to shift 5° to the north. Find its velocity. Ans. 64-905 mis. per hour S. 40° E. (5) A man walks at the rate of 4 mis. per hour in a shower of rain. If the drops fall vertically with a speed of 200 ft. per sec. in what direction will they seem to him to fall ? Ans. Jin a direction inclined 1° 40' "8... to the vertical. 99 ] TEANSLATIOX : VELOCITIES. 5 9 (6) Two candles, A and B, each 1 ft. long and requiring 4 and 6 hours respectively to burn out, stand vertically at a distance of 1 ft. The shadow of B falls on a vertical wall at a distance of 10 ft. from B. Find the speed of the end of the shadow. Ans. 8 inches per hour. (7) Two equal circles touch each other. Two moving points start in opposite directions from the point of contact and move on the circles with equal uniform speeds. Prove that the path of each, relative to the other, will he a circle whose radius is equal to the dianieter of either of the first circles. 98. Composition of Velocities. — A point has two or more component velocities ; it is required to find its resultant velocity. As in 96, it may be shown that the propositions proved in 78 to be applicable to displacements are applicable also to velocities. Hence (1) If two sides of a triangle, taken the same way round, represent two component velocities, the third side, taken the opposite way round, will represent the resul- tant velocity. This proposition is known as the triangle of velocities. (2) If two component velocities be represented by two adjacent sides of a parallelogram taken opposite ways round; the diagonal of the parallelogram through their point of intersection will represent their resultant. This proposition is known as the parallelogram of velocities. (3) If any number of component velocities be repre- sented by n — 1 of the sides of a polygon, taken the same way round, their resultant will be represented by the n*^^ side, taken the opposite way round. This proposition is known as the polygon of velocities. 99. Resolution of Velocities. — It follows also that velocities may be resolved into components in the same manner as displacements (see 79-84). 60 KINEMATICS. [100 100. From the above propositions (98) there may be deduced trigonometrical and analytical expressions for the magnitude and direction of a resultant velocity in terms of the magnitudes and inclinations of the compon- ents, just as in the case of displacements. All the formulae of 85-90 hold if we take d^ and d^ to represent component velocities and R to represent the resultant velocity. 101. In the important case in which the position of a moving point is specified by reference to fixed rectangular axes, Ox, Oy, Oz, the components of the instantaneous velocity of the moving point in the directions of the x, y, and z axes are (93) equal to the rates of change of the X, y, and z co-ordinates. Thej' are thus denoted by x, y, z. 102. Examples. (1) A point has three component velocities, A, B, and C in one plane. Their magnitudes are 4, 5, and 6 respectively, and their directions are such that A is inclined 30° to B, and C 60° to B and 90° to A . Find (a) the resultant of A -and B, (b) the resultant of all three, and (c) the component of the resultant in the direction oiB. Ans. (a) (41 + 20^3)*, inclined to A at sin-' _^|^ ■ (6) (107 + 20V3)* inclined to C at tan-i?J^^ ; (e) 8 + 2 ^3. (2) A boat's crew row 3^ mis. down a river and back again in 1 hour 40 min. If the river have a current of 2 mis. per hour, find the rate at which the crew would row in still water. Ans. 5 mis. per hour. (3) A river 1 ml. broad is running at the rate of 4 mis. per hour ; and a steamer which can make 8 mis. per hour in still water is to go straight across. In what direction must she be steei ed ? Ans. At an angle of 60° to the river bank. (4) A ship has a north-easterly velocity of 12 knots an hour. 103] TEjUs^SLATION : VELOCITIES. 61 Find the magnitude of her velocity (a) in an easterly direction, (6) in a direction 15° "W. of N. Ans. (o) 6 V2, and (6) 6, knots per hour. (5) Trom a ship steaming east at 10 mis. an hour a shot is to be fired so as to strike an object which bears N.E. If the gun, properly elevated, can give the shot a mean horizontal velocity of 88 ft. per sec, towards what point of the compass must it be directed ? Ana. N. 38° 13''9... E. 103. Moment of a Velocity. — The moment of the velocity of a moving point about a given fixed point (24) is the product of the magnitude of the velocity into the perpendicular from the given point on a line through the position of the moving point at the instant under consideration, and in the direction of its motion. — Let P be the position of the moving point at the instant under consideration, A that of the fixed point. Let PC be the direction of the velocity, and v its magnitude. Let p be the length of the perpen- dicular AB from A on PC. Then the moment of v about A is pv. If PC represents the velocity in mag- nitude as well as direction, the magnitude of the moment of the velocity is evidently represented by twice the area of the triangle PAC. If the moving point have a velocity represented by CP instead of PC, the moment of its velocity about A will be of the same magnitude. To distinguish between the equal moments of velocities represented by PC and by CP, they are considered to be of opposite sign. If the motion of the moving point is such that the radius vector AP moves counter-clockwise (i.e., in the opposite direc- tion to that of the hands of a clock), the moment of its velocity is considered to be positive. If its motion is such that the radius vector moves clockwise, the moment 62 KINEMATICS. [103 of its velocity is considered to be negative. Thus the moment of the velocity PC is —fv; that of the velocity GP would be ■\-pv. 104. The moment of the velocity of a moving point about a given line or axis, iixed in space (24), is the moment of the component of the velocity in a plane perpendicular to the given line about the point of intersection of that plane with the given line. If P is the position of the moving point at the instant under consideration, V its velocity, OA the given line, v the component of F in a plane perpendicular to OA, A the inter- section of OA with that plane, and AB (length =p) the perpendicular from A on v, the moment of V about OA is the product pv. The same convention of signs is em- ployed as in 103. 105. The algebraic sum of the moments of two com- ponent velocities about any point in their plane is equal to the moment of their resultant about the same point. — Let OA, OB be two component velocities whose resultant is 00, and P any point in their plane, either (Fig. 1) outside or (Fig. 2) inside the angle between the resultant and either of the components. Then, by a familiar geometrical proposition,* the sum of the triangles OAP * If a point P be tal^en in the plane of a parallelogram OACBi and lines drawn from it to the angular points, the area of the triangle OCP is equal to the sum or the difference of the areas of the triangles OAP, OBP, according as these triangles are on the same side or on opposite sides of OP. For area 0/'C=area OJC+area .4i'C±area OAP; and, since the base OB of the triangle OBP is equal and parallel to the base ^0 of the triangles OAC and APC, and the altitude of UBP equal to the sum of the altitudes of OA C and A PC, area OBP =axea, 0.4 C+ area APC. Hence, area OPC= area 05/" ± area O.-l/'. 106] TRANSLATION ; -VELOCITIES. 63 and OBP in Fig. 1, and their difference in Fig. 2, is equal to the triangle OPC. But these triangles are propor- tional to the moments of the velocities OA, OB, OC Fig.l Frg.2 respectively. And these moments have, in the case of Fig. 1, the same sign, and in that of Fig. 2, opposite signs. Hence the algebraic sum of the moments of OA and OB is equal to the moment of 00. The cases in which the point P is on the line OA or the line OC may be left to the reader. In the former, the moment of the one component is zero, and that of the other is equal to the moment of the resultant. In the latter, the moment of the resultant is zero, and the moment of the one component is equal and opposite to that of the other. This proposition may obviously be extended to any number of component velocities in one plane. 106. If the position of the moving point P is specified by reference to fixed rectangular axes. Ox, Oy, in the same plane with P's velocity, its co-ordinates " being x, y, its component velocities in the directions of the axes are « and y respectively, and their distances from 0, y and x re- spectively. Hence the moments ^ of the components about are (103) —xy and +yx respectively. P\^ *-x The moment of V 64 KINEMATICS. [106 (the velocity of P) about is therefore (105) yx-xy. If V is not in the xy plane, yx — xy is obviously equal to its moment about the axis of z. 107. The algebraic sum of the moments of any number of component velocities about any fixed axis is equal to the moment of their resultant about the same axis. — Let the component velocities of the point P be represented by PA, A B, BC, and its resultant velocity therefore by PG. — Let a, h, c be the feet of perpendiculars from A, B,G on the plane through P perpendicular to the given fixed axis OQ. Then Pa, ah, he, Pc are the components of PA, AB, BC, PG in this plane. Since Pa, ah, he are in a plane perpendicular to the axis, the moment of Pc about the axis is (105) equal to the algebraic sum of the moments of Pa, ah, and he. And since Pa, ah, he, Pe are the components, in the aforementioned plane, oSPA, AB, BG, PG respectively, the moments of the former about the given axis are equal respectively (104) to the moments of the latter. Hence the moment of PG about the given axis is equal to the alge- braic sum of the moments' of PA, AB, and BG. 108. Exam.'ples. (1) AB is a diameter of a circle of which BC is a chord. When is the moment about it of a velocity represented by BC the greatest ? Ans. When angle ABC =45°. (2) A moving point P has two component velocities, one of which is double the other. The moment of the smaller about a point 109] TEANSLATION : VELOCITIES. 65 in their plane is double that of the greater. Find the magnitude and direction of the resultant. Ans. If a and ^ are the inclinations of the greater and smaller components respectively to FO, the resultant is v/5 + 4 cos (/3 - a) times the smaller component, and is inclined to FO at the angle . _i 2sina + sin|8 sin-i— — -!i — \'5+4cos(|8-a)" (3) If the component velocities of a moving point can be irepre- sented by the sides of a plane polygon, taken the same way round, the algebraic sum of their moments about any point in their plane is zero. (4) Show that, if the algebraic sums of the moments of the component velocities of a moving point about two points F and Q be each zero, the algebraic sum of their moments about any point in the line PQ will also be zero. 109. Change of Velocity. — The velocity of a moving point in general changes from instant to instant both in magnitude and in direction. Let P^P^P^ be the path of a point, and let V^ whieh touches the path at P^, repre- sent the velocity of the point at Pj and let V^, F, E 6 6 KINEMATICS. [ ^09 similarly represent the velocities of the point at Pg and Pj respectively. The change in the point's velocity, which has occurred in the time occupied by the point in moving from P-^ to Pg. is that velocity which must be compounded with the initial velocity V^ to produce the final velocity Fj. Take any point ; from it draw OQ^ and OQ^, equal to and codirectional with F, and Fj. Join QjQi- Then (98) the final velocity OQ2 is the resultant of the two components OQ^, the initial velocity, and Q-^Q^- Hence, Q1Q2 represents the change of velocity which the point has experienced between P^ and Pg. The change of velocity must be carefully distinguished from the change of speed. The change of speed in the above figure is Fg— F, and is represented by OQ^—OQ^. 110. Acceleration. — The integral acceleration during a given time is the change of velocity undergone by the moving point during that time. The mean acceleration during any time is a quantity whose magnitude is the quotient of the integral accelera- tion by the time, and whose direction is that of the integral acceleration. Thus, if t units of time are occupied by the point in moving from Pj to Pg (109), the mean acceleration during that time is in the direc- tion of Q1Q2 and of the magnitude Q-Ji^l^. If t+tf units of time are occupied in moving from Pj to P3, and if OQ3 is drawn equal to F^ and in the same direction, then the mean acceleration during these t+if units of time is in the direction Q^Q^ and of the magnitude Q^QJ(t+t')- Thus the mean acceleration of a point varies in general both in magnitude and direction with the interval of time to which it applies. In the special case in which it varies neither in magnitude nor in direction, the point is said to move with uniform acceleration. 112] TRANSLATION: ACCELERATIONS. 67 _ The instantaneous acceleration of a moving point at a given instant (called usually the acceleration simply) is a quantity whose magnitude and direction are the limit- ing magnitude and direction of the mean acceleration between that instant and another when the interval of. time between them is made indefinitely small. As (295) in the case of a body a finite time is requii-ed for a finite change of velocity, the instantaneous acceleration of a body can never have an infinite value. If the point is moving with uniform acceleration, the instantaneous acceleration at any instant has clearly the same magnitude and direction as the mean acceleration for any interval. Acceleration, having both magnitude and direction, is a vector (70), like displacement and velocity. 111. Measurement of Acceleration. — The specification of an acceleration involves specification both of its mag- nitude and of its direction. Its direction may be described in terms of the uiiit of angle (21). Its magnitude being the quotientof the magnitude of a certain velocity by an interval of time, is a quantity of the same kind as a rate of change of speed (52), and may therefore be measured in terms of the unit of rate of change of speed (56). This unit is thus called also the unit of acceleration ; and the results of 57, 58 hold for units of acceleration. 112. Examples. (1) The initial and final velocities of a moving point during an interval of 2 hours are 8 mis. per hour E. 30° N., and 4 mis. per hour N. Find (a) the integral, and (6) the mean acceleration. Ans. (a) 4^/3 mis. per hour W. ; (6) 2^3 mis.-per-hour per hour W. (2) A point moves in a horizontal circle with uniform speed v, starting from the north point and moving eastwards. Find the 68 KINEMATICS. [112 integral acceleration when it lias moved through (a) a quadrant, (6) a semicircle, (c) three quadrants. Ans. (a) V ^2, S. W. ; (6) 2v, W. ; (c) v ^2, N.W. (3) The- velocity v oi a point moving in a straight line being supposed to vary as the square root of its distance s from a fixed point in the line, show that its instantaneous acceleration in any position is equal to v^l2s. (4) The velocity of a point moving in a straight line varies as the square root of the product of its distances from two fixed points in the line, show that its instantaneous acceleration varies as the mean of its distances from the fixed points. 113. The Hodograph. — The variation of the velocity of a moving particle from one position to another of its path may be studied by means of an auxiliary curve, called the hodograph of the path. The velocity of a particle must have (295) indefinitely nearly the same magnitude and direction at points of its path which are indefinitely near. If therefore (109) the angle between OQ^ and OQ2 is indefinitely small, the length of OQ^ must be indefinitely nearly equal to that of OQ2. Hence the locus of the end Q of the line OQ, which represents the velocity of the moving particle P in its successive positions, must be a curve of continuous curvature. This curve is called the hodograph of the path. The point is called the pole of the hodograph. The hodograph has two important properties which may be proved as follows : — The straight line QjQ^ re- presents in magnitude and direction the integral accele- ration during the time t occupied by P in moving from Pj to P2, and QjQJt represents the magnitude of the mean acceleration during the same time. When Pj is taken indefinitely near P^, the direction of Q^Q^ is the direction, and the magnitude of Q^QJ-r is the magnitude, of the acceleration of P at the instant at which it is at 11^] TRANSLATION: ACCELERATIONS. 69 Pi- But when Pg is taken indefinitely near P^, and therefore Q^ indefinitely near Q^, the direction of Q^Q^ is that of the tangent to the hodograph at Q^, and the ma,gnitude of Q^QJt is that of the velocity at Q^ of the point Q in the hodograph. Hence (1) the direction of the acceleration of the moving point P at the instant at which it occupies a given position in its path is that of the tangent to the hodograph at the corresponding position of Q, and (2) the magnitude of the acceleration of P at the instant at which it occupies a given position in its path is equal to the magnitude of the velocity of Q at the corresponding position in the hodograph. 114. Examples. (1) Show that the hodograph of a point moving with uniform speed in a straight path reduces to a point. (2) A point moves with uniform acceleration, either in a straight or in a curved path. Show that the hodograph of the path is a straight line, and that the point in the hodograph moves with uniform speed. (3) The hodograph of a point which moves with uniform speed in a circle, is a circle, in which the corresponding point moves also with uniform speed. (4) If a point move in either a parabola, an ellipse, or an hyper- bola, so that the moment of its velocity about a focus is constant, the hodograph is a circle. [Note that the locus of the foot of the perpendicular from a focus on a tangent is a circle in the case of either the ellipse or the hyperbola, and in the case of the parabola a straight Une. Note also that the locus of the foot of the perpen- dicular from the vertex of a parabola on a straight line drawn through the focus is a circle.] (5) The hodograph of a point moving in an ellipse so that the moment of its velocity about the centre is constant, is a similar ellipse. [Note that the area of the parallelogram formed by drawing tangents to an ellipse at the extremities of a pair of con- jugate diameters is constant.] 70 KINEMATICS. [115 115. Change of the Point of Reference. — Acceleration being simply the velocity which must be compounded with the velocity of a moving point at a given instant, to produce the velocity which it either has after unit of time, or would have if the acceleration were uniform, the propositions (96) dealing with the change of the point of reference in the case of velocities, apply also in the case of accelerations. 116. Com/position and Resolution of Accelerations. — Similarly the laws of the composition of velocities (98) may be shown to be those according to which accelera- tions also are compounded. We have thus propositions called the triangle, the parallelogram, and the polygon of accelerations identical in form with the corresponding propositions for velocities. Hence also accelerations may be resolved after the same manner as velocities (99) ; and the formulae of 85-90 hold, if c?j, d^, etc., and R denote component and resultant accelerations. 117. The component of an acceleration in any direction is equal to the rate of change of the component in that direction of the velocity. — Let OP and OQ be the initial and final velocities of a moving point *Q during a given time. Then PQ is the integral acceleration. Draw- ing OR, QT at right angles to any line PR, we have PT=RP-Bt. If then T be the time, PT/r=(RP-RT}/T. Now PT, RP, and RT are the components in the line PR of PQ, OP, and OQ respectively. If r is indefinitely short, PT/t is thus the component of the instantaneous acceleration in the direction PR, and {RP — RT)/t is the instantaneous rate of change of the component velocity in the same direction. 119] TRANSLATION: ACCELEEATIONS. 71 118. If the position of a moving point be specified by reference to rectangular axes Ox, Oy, Oz, its component accelerations in their directions will therefore be equal to the rates of change of its component velocities in their directions, namely of x, y, 'z respectively. They are therefore (55) x, y, z. 119. Examples. (1) A ball is let fall in an elevator which is rising with an acceleration of 7 2 kilometres per min. The acceleration of the falling ball relative to the earth is 981 cm.-sec. units. Find its acceleration relative to the elevator. Ans. 1,181 cm.-sec. units towards the floor. (2) Two railway trains are moving in directions inclined 60°. The one A is increasing its speed at the rate of 4 ft.-per-min. per min. The other B has the brakes on and is losing speed at the rate of 8 ft.-per-min. per min. Find the relative acceleration. Ans. 4^/7 ft.-min. units inclined sin-^W^^ to the direction of TT /3 motion of A, and S ~sii^~^\/i7 *° ^^^ of B. (3) The locus of the extremity of the straight line representing either of the two equal components of a given acceleration, is a straight line perpendicular to the straight line representing the given acceleration and through its middle point. (4) A buUet is fired in a direction towards a second bullet which is let fall at the same instant. Prove that the line joining them will move parallel to itself and that the bullets will meet. (5) Find the resultant of four component accelerations, represented by lines drawn from any point P within a parallelogram to the angular points. Ans. If C is the point of intersection of the diagonals, PG repre- sents the direction of the resultant, and 4PC its magnitude. (6) The resultant of two accelerations a and a' at right angles 72 KINEMATICS. ^19 to one another is R If a be increased by 9 units and a' by 5, the magnitude of R becomes increased to three times its former value, and its direction becomes inclined to a at the angle of its former inclination to a'. Find a, a', and R. Ans. 3, 4, arid 5 units respectively. 120. Tangential and Normal Acceleration.— An im- portant special case of the resolution of accelerations is the resolution of the acceleration of a point moving in a plane curve, into components in and normal to the direction of motion at any instant.^ — Let P, Q be points on the path, and PA, QB tangents to the path at P, Q respectively. Let OA, GB represent the velocities of the moving point at P and Q respectively. From GB cut off CD equal to CA. If the point Q be made to approach P, the angle BCA becomes ultimately zero, and the angles GDA and GAD therefore ultimately right angles. Now AB represents the integral acceleration between P and Q, and it may be resolved into AD and DB as components. Hence ultimately AD and DB, divided by the time, re- present the normal and tangential components of the instantaneous acceleration at P. Since CD was made equal to GA, DB is the change of speed. Hence ultimately, when P and Q coincide, DB divided by the time is the rate of change of the speed of the moving point. Hence the tangential component of the acceleration of a moving point is equal to the rate of change of speed. From P, Q draw PO, QO, normals to the path at those points, and let them meet in 0. Then the angle QGA is equal to the angle QOP. Calling these angles d, the velocity at P, V, and the component integral accelera- tion AD, V, we have v = 2 Fsin (0/2) = Fsin 0/cos (6/2). 121] TKANSLATION: ^ACCELERATIONS. 73 If a is the component in the direction AD of the mean acceleration between P and Q, and t the time of motion from P to Q, we have thus _■?;_„ sin 1 Now ultimately the time of motion from P to Q may be put equal to PQ/V. Hence, calling PQ, s, we have a-Y2 sing 1 cos(0/:J) ■ s' Also ultimately PQ may be considered an arc of a circle, and PO, QO become equal to one another and to the radius of curvature (p) of the path at P, in which case s=p6. Hence „a sing 1_ cos (0/2) ■ pff Also, 6 being indefinitely small, sin 0/0 = cos (0/2) = 1. Hence a=V^/p. Now a, being the mean acceleration in the direction AD, becomes ultimately the instantaneous acceleration normal to the path at P. Hence the normal component of the acceleration of a point moving in a curved path is the product of the square of its velocity into the curvature of its path. 121. If the path be a circle, the radius of curvature is the radius of the circle. Also a normal to the circle through any point passes through the centre. Hence the acceleration of a point moving with uniform speed in a circle is directed towards the centre, and is equal to the quotient of the square of the speed by the radius. (The reader should prove this special case directly. Thomson and Tait ("Elements," § 36) prove it by means of the hodograph.) 74 KINEMATICS. [ 121 If T be the time of a complete revolution (the periodic time) of the point in the circle, and if V be the uniform speed and R the radius, F= ^ttRIT. Hence the accelera- tion has the magnitude W^RjT^. (See also 131.) 122. ExaTnples. (1) A circus rider is moving with the uniform speed of a mile in 2 min. 40 sec. round a ring of 100 ft. radius Find liis acceleration towards the centre. Ana. 10'89 ft.-sec. units. (2) Show that a shot fired at the equator with either a westerly- velocity of 8,370'7 metres per second, or an easterly velocity of '7,440'5 m. per sec, will, if unresisted, move horizontally round the earth, completing its circuit in about 1^ or 1^ hours respectively. [Data : The mean radius of the earth is 6,370,900 metres ; the speed of a point on the equator 465'1 m. per sec. ; and the acceleration of a falling body 9'81 m.-sec. units.] (3) A point moving in a circular path, of radius 8 in., has at a given position a speed of 4 in. per sec, which is changing at the rate of 6 in.-per-sec. per sec Find (a) the tangential acceleration ; (6) the normal acceleration ; (c) the resultant acceleration. Ans. («) 6 in. -sec. units ; (b) 2 in.-sec units ; (c) 2 iJlO in.-sec. units, the direction being inclined at tan"' 3 to the normal. (4) If different points be describing different circles with uniform speeds and with accelerations proportional to the radii of their paths, their periodic times will be the same. 123. The moment of an acceleration is defined in exactly the same way as the moment of a velocity. See 103 and 104. Also the propositions of 105 and 107, being deductions from the parallelogram law, apply to accelera- tions as well as to velocities. And it may be shown, as in 106, that, if the position of a moving point be referred to rectangular axes of co-ordinates, the moment of its acceleration about the z axis is equal to yx — xy. 125] TRANSLATION: ANGULAR DISPLACEMENTS. 75 124. The moment of the acceleration of a moving point about a fixed point in the plane of its motion is equal to the rate of change of the moment of its velocity about the same point. For the final velocity in any time is the resultant of the initial velocity and the integral acceleration, and therefore (105) its moment about any point in their plane is equal to the sum of their moments. Hence the moment of the integral acceleration is equal to the difference of the moments of the initial and final velocities; and therefore, dividing by the time and making it indefinitely small, the moment of the instantaneous acceleration is equal to the rate of change of the moment of the velocity. 125. Angular Displacement* of a Point. — The angular displacement of a moving point about a given point in a given time, is the angle between the initial and final positions of the radius vector from the given point. Thus, if the point has moved from P^ to Pj its angular displacement, relative to 0, is PyOP, That an angular displacement ^^ about a given point may be completely specified, the magnitude of the angle must be given through which the radius vector has moved, the direction of the radius vector's motion, and the plane in which its initial and final positions lie. This plane is specified if a line normal to it be given; and the direc- tion of the radius vector's motion is specified if this line is always so drawn from a point in the plane of the dis- placement that, on looking along it towards that plane, * When there is danger of confounding angular displacements with the displacements considered in 69, the latter are called linear displacements. 76 KINEMATICS. [123 the radius vector is seen to move counter-clockwise, i.e., in a direction opposite to that of the hands of a clock. An angular displacement about a point may therefore be completely represented by a line normal to the plane of the displacement, whose direction is determined^ by the above convention, and whose length is proportional to the magnitude of the angular displacement. By the direction of an angular displacement is meant the direc- tion of this line. 126. The angular displacement in a given time of a moving point about a given line or axis, is the inclination of perpendiculars from the initial and final positions of the moving point on the axis. Let OA be the given axis, P and Q the initial and final positions of the moving point, and PR and QS perpendiculars to OA. Then the in- clination of PR to QS is the angular displacement about OA. Complete the rectangle RQ by the lines Rq, Qq. Then (8), since Rq is parallel to 8Q, PRq is the angular displacement. Since the plane of PqR is perpendicular to OA, and Qq being parallel to R8 is perpendicular to that plane, Pq is the projection of PQ on that plane. Hence the angular displacement is the angle subtended by the pro- jection of the linear displacement on a plane perpen- dicular to the axis, at the point of intersection of the axis with that plane. 127. Angular Velocity of a Point* — The mean angular velocity of a moving point about a given point, during a given time, is a quantity whose direction is that of the angular displacement during the time, and * When there is danger of confounding the velocity of 92 with angular velocity, the former is called linear velocity. 129] TRANSLATION: ANGULAR VELOCITY. 77 whose magnitude is the quotient of the angular displace- ment by the time. The mean angular velocity varies in general with the time. In eases in which it does not, the angular velocity is said to be uniform. If the motion of the moving point is confined to a plane, its angular velocity must have one of two opposite directions. In other words, it can vary only as to mag- nitude and sign. The instantaneous angular velocity of a point at a given instant has a magnitude and a direction, which are the limiting magnitude and direction of the mean angular velocity between that instant and another, when the interval of time between them is made indefinitely small. The mean and instantaneous angular velocities about a given axis are defined in a manner similar to that in which they are defined with reference to a given point. 128. Measurement of Angular Velocity. — The measure of an angular velocity being the quotient of the measure of a certain angle by that of a certain time, the most convenient unit of angular velocity will be unit of angle per unit of time. The unit of angle which is usually employed in measuring angular velocities is the radian. Unit of angular velocity in terms of the radian is one radian per unit of time. As the magnitude of the radian is (21) independent of that of the unit of length, the magnitude of the radian per unit of time depends only upon that of the unit of time and is inversely proportional to it. 129. Eelationbetween Angular and Linear Velocity.— Let the moving point P be displaced from P, to P^ in the time t with the mean linear velocity v, and the 78 KINEMATICS. [129 mean angular velocity u> about the point 0. Then w^PfiPJt, and the chord P^P^ = vt. From P-, draw P^N perpendicular to OPi- Then, if the angle P-^P^N be called d, P^N = PjPaSin e = vt sin 0. Hence sin PfiP^ = vt sin djOP^. If now Pg ^^ indefinitely near P^, co and v become instantaneous velocities at Pj, Q becomes the angle between the radius vector and the direction of the linear velocity at P^, and sin P^OPj becomes equal to PfiP^. Hence, if r is the radius vector, &) = ■?; sin 0/r. Hence the angular velocity of a moving point about a given point, expressed in radians, is equal to the component of its linear velocity perpendicular to the radius vector from the given point, divided by the length of the radius vector. 130. If the point be moving in a circle, its linear velocity is at all points perpendicular to the radius vector from the centre. Hence, if r is the radius and to the angular velocity about the centre, co = v/r. 1.31. Hence the normal component of the linear accele- ration of a point moving in a circle, which (121) is equal to v^jr, is, in terms of angular velocity about the centre, equal to wV. 132. Moment of Velocity in Terms of Angular Velocity. — Since P-^PJt (129), when t is small, is the velocity of 136] TEANSLATION: ANGULAE ACCELEKATION. ^79 the moving pointi its moment about is twice the are of the triangle OP^P^ divided by the time. Hence, if p be written for the moment, pv = OP 2 . P^^/t = vr sin Q = wr^. 133. Areal Velocity. — The area swept out by the radius vector of a moving point per unit of time, is sometimes called its areal velocity. It follows from 132 that the areal velocity of the point P (129) is represented by area OP^PJt and ia equal to |cor^. 134. These results (129-133) apply also to angular veloci- ties about axes, provided v stand for the component linear velocity in a plane perpendicular to the given axis, and r for the perpendicular distance of the point from the given axis. 135. Angular Acceleration of a Point. — We might define the angular acceleration of a moving point about a given point, as we did its angular velocity, generally. We restrict ourselves, however, to the useful case of the angular acceleration about a given axis. An angular velocity about a given axis must have one of two opposite directions, and can vary therefore in magnitude and sign only. Hence the integral angular acceleration about a given axis is the diflPerence between the final and initial values of the angular velocity about that axis ;• the mean angular acceleration in a given time is equal to the in- tegral acceleration divided by the time ; and the magni- tude of the instantaneous angular acceleration at a given instant is the limiting value of the mean angular accelera- tion between that instant and another when the interval of time between them is made indefinitely small, pr in other words it is the rate of change of angular velocity. The angular acceleration of a point moving in a plane about a given point in that plane is an angular accelera- tion about a given axis. 80 KINEMATICS. [136 136. Measurement of A ngular Acceleration. — The most convenient unit of angular acceleration is clearly unit of angular velocity per unit of time, e.g., one radian-per-sec. per sec. With the radian as unit of angle its dimensions are [T]-^. 137. Examples. (1) The earth makes a complete rotation in 86,164 mean solar seconds. Assume her radius to be 6,370,900 metres, and find (a) the angular velocity, and (6) the linear velocity of any point on the equator. - Ans. (a) radians per sec. ; (6) 465'1 m. per aec. (2) A wheel of a carriage which is travelling at the rate of 7 mis. per hour is 3 ft. in diameter. Find the angular velocity of any point of the wheel ahout the axle. Ans. 6'8... rad.per sec. (3) Compare the angular velocities of the hour, minute, and second hands of a watch. Ans. As 1 : 12 : 720. (4) Express in terms of the radian per second an angular velocity of 20° per min. Ans. 20-94.... (5) A point is moving with uniform speed via a, circle of radius ■/•. Show that its angular velocity about any point in the circum- ference is »/2n (6) The angular velocity of a point moving with uniform speed in a straight line is inversely proportional to the square of the distance of the point from a fixed point not in the line. (7) Show that the angular velocity of the earth about the sun is proportional to the apparent area of the sun's disc. [Datum : The radius vector from the sun to the earth sweeps over equal areas in equal times.] 137 ] TRANSLATION : ANGULAR VELOCITY. 8 1 (8) If the velocity of a particle be resolved into several com- ponents in one plane, its angular velocity about any fixed point in the plane is the sum of the angular velocities due to the several components. (9) A wheel makes 200 revolutions per hour. Express its angular velocity (a) in radians per sec. ; (6) in degrees per min. Ans. (a)^; (5)1,200. (10) Reduce an angular acceleration of 300 radians-per-min. per min. (a) to revolution-hour units, (6) to degree-second units. . , , 540,000 ,,, 15 Ans. (a) ; (b) — . T 7r (11) A point P moves in a parabola with a constant angular velocity about the focus S. Show that its linear velocity is pro- 8 portional to iSP^- 82 KINEMATICS. [138 CHAPTER IV. TEANSLATION :— MOTION UNDER GIVEN ACCELERATIONS. 138. Unconstrained Motion. — The motion of a point under given accelerations will depend upon the degree of its freedom to move (35). We shall take, first, cases of unconstrained motion, the moving point having all three degrees of freedom. Case I. — The Acceleration being Zero. — If there is no acceleration, there is no change in either the magnitude or the direction of the velocity. The path is therefore a straight line, and the magnitude of the velocity is con- stant. Hence the mean and instantaneous velocities, and therefore also the mean velocity and mean speed, have the same values (93 and 43). The results of 61 are thus at once applicable to this case. 139. Examples. (1) A point moves with a uniform velocity of 2 cm. per sec. Find the distance from the starting point at the end of 1 hour. Ans. 72 metres. (2) Two trains having equal and opposite velocities, and con- sisting each of 12 carriages, of 23 ft. length, are observed to take 9 see. to pass one another. Find the magnitude of their velocities. Ans. 20'91 mis. per hour. 140] TEAKSLATION: FALLING BODIKS. 83 (3) Two points move -with uniform velocities of 8 and 15 ft. per sec. in straight lines inclined 90°. At a given instant their distance is 10 ft., and their relative velocity is inclined 30° to the line joining them. Find (a) their distance when nearest, and (6) the time after the given instant at which their distance will be least. Ans. (a) 5 ft. ; (6) ^ ^3 sec. 140. Case II. — The Acceleration Uniform. — The motion of a point under a uniform acceleration will be different according as the point has or has not at any instant a velocity inclined to the direction of its acceleration. (a) Rectilinear motion. — If at any instant the velocity of the moving point is in the same straight line with the acceleration, the path is a straight line. For (109) OQ, and Qj^Q^ being in the same straight line, so also are OQ^ and OQ^. Hence the velocity does not vary in direction. Also, O'Q.Q^ being a straight line, Q,Q,=OQ,- OQ,. If- then t is the time in which the velocity changes from OQ, to 0Q„ and t being taken indefinitely small, we find that the instantaneous acceleration is equal to the rate of change of the magnitude of the velocity, and therefore (93) to the rate of change of speed. Hence the results of 63-60 are applicable to this case. We have a familiar instance of the motion under con- sideration in the motion of bodies vertically upwards or downwards through short distances at the surface of the earth, except in so far as their velocity is modified by the resistance of the air. For all bodies falling freely near the surface of the earth are found to have a down- ward acceleration of about 32-2 ft.-sec. units, or 981 cm.-sec. units. When represented by. a symbol, the 84 KINEMATICS. [1*0 special symbol g* h usually employed to denote this acceleration. 141. Examples^ (1) A body is projected vertically upwards with a velocity of 300 ft. per sec. Find (a) its velocity after 2 sec. ; (6) its velocity after 15 see. ; (c) the time required for it to reach its greatest height ; (d) the greatest height reached ; («) its displacement at the end of 15 sec; (/) the space traversed by it {i.e., the length of path described) in the first 15 sec. ; (g) its displacement when its Velocity is 200 ft. per sec. upwards ; {h) the time required for it to attain a displacement of 320 ft. [Note that if the upward direction be taken as positive, the acceleration in this case is negative.] Aris. {a) 235"6 ft. per sec. upwards ; (6) 183 ft. per sec. down- wards; (c) 9-3... sec; {d) 1,397-5 ft.; (e) 877-5 ft. upwards; (/) 1,917-5 ft.; (g) 776-3 ft. upwards; (A) 1-13 sec in ascending, 17-5 sec. in descending. (2) A ball is projected vertically upwards from a window half way up a tower 117-72 metres high, with a velocity of 39-24 m. per sec. After what times and with what speeds does it (a) pass the top of the tower ascending ; (6) pass the same point descending ; and (c) reach the foot of the tower ? Ans. (a) 2 sec, 19-62 m. per sec; (6) 6 sec, 19-62 m. per sec; (c) (4 + 2 v'7) sec, 19-62 x ^/7 m. per sec (3) A stone is dropped into a well, and the splash is heard in 3"13 sec Given that sound travels in air with a uniform velocity of 332 metres per sec, find the depth of the well. Ans. About 44-1 m. * The value of g at any place near the earth's surface is given approximately in centimetre-second units by the following formula, in which \ is the latitude of the place, and h its height above sea- g = 980-6056 - 2-5028 cos 2X - -OOOOOSA. t In problems on falling bodies the resistance of the air is not to be taken into account. When the value of g is not given it is to be taken as 32 '2 ft.-sec units or 981 cm.-sec. units. ^41] TEANSLATION: FALLING BODIES. 85 (4) Show that a body, projected vertically upwards, requires twice as long a time to return to its initial position as to reach the highest point of its path, and has, on returning to its initial position, a speed equal to its initial speed. (5) A stone projected vertically upwards returns to its initial position in 6 sec. Find (a) its height at the end of the first seconc), and (6) what additional initial speed would have kept it 1 sec. longer in the air. Ans. (a) 80-5 ft.; (b) ICl ft. per see. (6) A body let fall near the surface of a small planet is found to traverse 204 ft. in the fifth and sixth seconds. Find the accele- ration. Ans. 20'4 ft. -sec. units. (7) A particle describes in the n^ second of its fall from rest a space equal to p times the space traversed in the (n-l)* second. Find the whole space described. Ans. (l-3pyffJiS(l-pYl (8) A body uniformly accelerated, and starting without initial velocity, passes over 6 feet in the first p seconds. Find the time of passing over the next b ft. Ans. ^( ;^2 — 1 ) sec. (9) A ball is dropped from the top of an elevator 4'905 metres high. Find the times in which it will reach the floor, (a) when the elevator is at rest ; (6) when it is moving with a uniform downward acceleration of 9-81 m. per sec. ; (c) when moving with a uniform downward acceleration of 4-905 m. per sec. ; (d) when moving with a uniform upward acceleration of 4-905 m. per sec. Ans. (a) 1 sec. ; (6) oo ; (c) ^^2 sec. ; W /y^ g sec. (10) If «i, «2 a'fe the heights to which a body can be projected with a given initial vertical velocity at two places on the earth's surface at which the accelerations of falling bodies are ff^ and ff^ respectively, show that *i5'i=«s,9'2. (11) A stone A is let fall from the top of a tower 483 ft. high. 86 KINEMATICS., [141 At the same instant another stone B is let fall from a window 161 ft. below the top. How long before A will B reach the ground ? Ans. (^6-2)^5 sec. (12) A ball falling from the top of a tower had descended a ft. when another was let fall at a point h ft. below the top. Show that if they reach the ground together the height of the tower is (a + 6)2/4aft. (13) If two bodies be projected vertically upwards with the same initial velocity V, at an interval of t sec, prove that they will meet at a height |(-^-^). (14) Two stones are falling in the same vertical line. Show that if one can overtake the other it will do so after the same lapse of time, even if gravity cease to act. (15) Bodies are projected vertically downwards from heights li^, hj, A3, with velocities %, v^, v^ respectively, and they all reach the ground at the same moment. Show that (Aj - A2)/(«i - Wj) = (Ag - h^j{V2 - V3) = (A3 - Ai)/(D3 - vj. (16) Two points move in straight lines with uniform accelerations. Show that if at any instant their velocities are proportional to their respective accelerations the path of either relative to the other will be rectilinear. (17) Pa,rticles are projected vertically upwards from different points in a horizontal straight line AX, with velocities respectively proportional to the distances of the points of projection from A. Prove that all the particles when at their highest points will be on a parabola whose vertex is A. 142. (6) Curvilinear motion. — If the moving point has at any instant a velocity inclined to the direction of its acceleration, the direction of the velocity must change with the time, and consequently the path must be 144] TRANSLATION: PEOJEOTILES. 87 a curved line. For if OA is the initial velocity, and AQ, AQ', the integral accelerations after t and if units of time respectively, OQ and OQ' are the velocities after these intervals of time. And since AQ and AQ' have the same direction, OQ and OQ' must have different direc- tions. Nevertheless the component acceleration in any given di- rection being uniform in this case (140), the formulae of 63-66 apply to curvilinear as well as to rectilinear motion, provided we restrict our attention to a component motion in a given direction. Curvilinear motion under uniform acceleration is of interest because it is the motion which projectiles near the earth's surface would have, if they were not resisted by the air, and if their accelerations were rigorously, as they are approximately, the same at all points of their paths. 143. To find the Velocity after any Time. — The moving point has after the time t two component velocities, one the initial velocity V, represented by OA (142), the other the integral acceleration at (if a is the acceleration), represented by AQ. If then the inclination of the two be given, the resultant, represented by OQ, may be found by 100. 144. To find the Displacement of the Poimt after any Time. — The moving point will have two component dis- placements after any time t, one due to the initial velocity V, the other due to the acceleration a ; the one therefore having the value Vt, the other the value ^at^. If their inclination be known, their resultant may be determined by 85. KINEMATICS. [145 145. To find the Displacement in any given Direc- tion* — Let P be the initial position of the moving point, PA the direction of the initial velocity V, and PB that of the acceleration a. ("We draw PB vertical because of the importance of this problem in the study of projectiles.) Draw PJV" perpendicular to PB. Let PA be inclined to PN^ at the angle O.j It is required to dcr i' termine the displacement which the point will reach in the direction of PM, inclined to PN at the angle a. The initial velocity has a component Fsin(0 — a) perpendicular to PM. The acceleration has a component a cos a perpendicular to PM and opposite in direction to the above component of the initial velocity. Hence, if t is the time at the end of which 'the displacement per- pendicular to PM is zero, we have (64) t = 2Vsin(0-a) a cos a Now the point has in the direction PN a velocity Fcos 6 and no acceleration. Hence in the time t its displacement in that direction is 2Fsin(0-a) a cos a X Fcos 0. Let PN represent this displacement. Draw NM perpen- dicular to PN. Then PM is the required displacement * In gunnery the displacement of a projectile in a given direc- tion is called its range on a given plane ; the time required to reach that displacement is called the time of flight. t In gunnery this angle is called the elevation of the projectile. 148] TRANSLATION: PKOJECTILES. 89 in the direction PM. But PM^PNjcosa. Hence, denoting PM by R, P 2F^sin(e-a)cose Expanding sin (0 — a), adding and subtracting Vhmal{a cos^a), and remembering that 2cos^0— l = cos20, we find ^^ F^[sin(2e-a)-sina] a cos^a The required displacement is therefore determined in terms of known quantities. 146. If the given direction be upon the other side of PN, viz., that of the line PM' (the angle M'PN being equal to a), we obtain the result ■^ ^ 7^[sin (2g + g) + sin g] a cos^o If therefore in this case the inclination of the given direction to PN be considered negative, so that angle M'PN= —a, we get the same expression for R as in 145. 147. If 6' is such that 26' -a = 180° - (W - a), we have sin(2©'-g)=sin(20-g). Hence R will have the same value, whether the incli- nation of the initial velocity to PN be 6 or 90°- 0+g. With a given acceleration and an initial velocity of given magnitude, there are therefore two directions of initial velocity, and therefore two paths, by which the point may attain a given displacement in a given direction. 148. The above expression for R involves V, a, 6 and a. If V, a, and g are given, 9 is the only variable. The magnitude of R will therefore depend upon that of 6. 90 KINEMATICS. [148 Now sin (20 — a) has its greatest value when 20 — a = 90°. Hence, with a given acceleration and an initial velocity of given magnitude, the displacement in a given direction has its greatest 'value, viz., 7^(1 — sin a)/(a cos^a), when 9 = 45° + a/2. When 6 has this value, 90' -6 + a- ha,s_ the same value. Hence there is but one direction of initial velocity by which the maximum displacement in the given direction can be attained ; and that direction bisects the angle between the direction opposite to that of the acceleration, and the given direction. 149. In the important special case in which PM co- incides with PJV) we have a = 0. Hence R*=Vhm2e/a, and this displacement is attained whether the inclination of PA to PN have the value d or the value 90° - 6. The greatest possible value of R in this case is F^/a, and it is attained when the inclination of the initial velocity has the value 45°. 150. The important practical problem of determining the direction of an initial velocity of given magnitude, with which the moving point will pass through a given point, may be solved at once by means of the above expression (145) for It. For the point, say M, being given, PM (i.e., iJ) aind a are known, and a and V being given, all the quantities involved in this expressioii except 6 are known. We thus have for determining 6, = Jsm-i[^ y^ — -l-sinaj + 2- As pointed out above, 6 may clearly have either of two values. In practice allowance must of course be made for the resistance of the air. * In gunnery the range on a horizontal plane. — The reader should prove this special case directly. 152] TEANSLATION : PEOJECTILES. 91 151. To Determine the Path of the Point. — The accele- ration and initial velocity being given, the value of R in the above expression (145) will depend upon that of a. If different values be given to a, the displacements of the moving point in known directions, and therefore as many positions as we please of the point ia its path may be determmed. , Thus, as the reader who is familiar with analytical geometry will see, this expression is an equa- tion to the path of the moving point expressed in polar coordinates. 152. The path may be determined also by the follow- ing geometrical method. Let P be the initial position of the moving point, and let PQ represent in direction the direction of the initial velocity and in magnitude the component displacetnent due to the initial velocity in t units of time. Let QR represent the component displace- ment due to the acceleration in t units of time. Then R is the position of the point after t units of time. Now PQ = Vt and QR = ^atl Eliminating t, we find PQ^ = {2V^/a)QR. This relation must hold for all values of t and therefore for all positions, of the moving point. But we know, from the geometry of the parabola, that, if 8 is the focus 92 KINEMATICS. [152 of a parabola whicli touches QF in P and whose axis is parallel to QB, PQ^ = ^SF.QR. Hence the path of the moving point is a parabola which touches PQ in P, has an axis parallel to QR, and has a focus distant Vy2a from P. To find the directrix of the parabola, we know that it must be perpendicular to QR and at a distance from P equal to F8. Hence from P draw FM parallel to QR and make it equal to V^/2a. Then from M draw MM' perpendicular to FM. MM' is the directrix. To find the focus 8 we know that FS and FM must be equally inclined to FQ. Hence from P draw FS, making the angle SFQ equal to MFQ, and make F8 equal to Vy2a. 8 is the focus of the parabola. The directrix and focus being thus known, the parabolic path is known. 153. The acceleration and the magnitude of the initial velocity being given, FM will be constant. The length of F8 will also be constant, but its direction will vary with the direction of the initial velocity. Hence the diflFerent positions occupied by 8, for different directions of the initial velocity, lie on a circle whose centre is P and radius F8. 154. We may apply the geometrical method to deter- mine the displacement in a given direction with given acceleration and initial velocity. Let P be the initial position of the point, PA the direction of the initial 154] TRANSLATION : PfiOJECTILES. 93 velocity, MM' the directrix of the path. Then. the focus of the path must lie ou the circle MAS. If then the angle APS is made equal to the angle MPA, S is the focus. To find at what point the path cuts the given direction Pm, it is necessary to find a point P' in Pm. whose distance from S is equal to its distance from MM', i.e., to find in Pttv the centre of a circle which will pass through S and touch MM' — a simple geometrical problem. Let P' be this point and M'SS' the circle. Then P'M'= P'S. Hence P' is a point on the parabola whose focus is S, and therefore PP' is the displacement of the point in the direction Ptti. As the circles meet in general in two points we have MP=PS', and M'P'=P'8'. Hence P and P' are also points on the parabola whose focus is 8'. Hence there are two paths by which, with a given acceleration and an initial velocity of given magnitude, a given distance in a given direction may be attained. The direction of the initial velocity which gives the second path is that of the line bisecting the angle S'PM. As the angle MPA increases, 8 and 8' approach one •another and PP' increases in length. When MPA = APm, S and 8' coincide, the circles merely touch, and PP' = 94 KINKMATICS. [154 FS+8P'. This is the greatest distance the moving point can attain in the given direction with an initial velocity of given magnitude ; and it can be attained obviously by one path only. The locus 'of P', when PP' is the greatest distance which can be attained in different directions with an initial velocity of given magnitude, is the curve inside which all points can be reached by the moving point with an initial velocity of the given magnitude, outside which no points can be reached. It is evidently a para- Q Q' m' y? bola whose focus is P and vertex M. For if PM be produced to Q, so that PM=MQ, and if QQ' be drawn parallel to MM, and P'M produced to meet it in Q', we have P'P = P'Q'. Hence P' is a point on a parabola whose focus is P and directrix QQ', and whose vertex consequently is M. 155. Examples* (1) A body is projected with an initial velocity of 30 ft. per sec inclined 60° to the horizon. Find the velocity after 20 sec. * In the solution of these problems the resistance of the air is not taken into account. When the value of g is not specified it is' to be taken as 32'2 ft.-sec. units or 981 om.-sec. units. 155] TRANSLATION: PROJECTILES. 95 Ans. 618-2 ft. per sec. inclined 148° 36'-6 to the direction of the initial velocity. (2) Find the direction and magnitude of the velocity of projection in order that the projectile may reach its maximum height at a point whose horizontal and vertical distances from the starting point are 6 and h respectively. Ans. Direction inclined tan-i(2A/6) to the horizon, magnitude [(4A2+6%/2^]i. (3) A particle is projected horizontally with a speed of 32'2 ft. per sec. from a point 128'8 feet from the ground. Find the -direc- tion of its motion when it has fallen half way to the ground. Ans. Inclination to the vertical =tan~i J. (4) A stone is let fall in a railway carriage travelling at the rate of 30 mis. per hour. Find its displacement relative to the road at the end of O'l sec. Ans. 4-4028... ft. inclined 2° 3'-4 to the horizon. (5) A stone is thrown into the air at an angle of 45° to the horizontal plane with a speed of 50 ft. per sec. Find the magnitude of the displacement at the instant at which the stone's velocity is horizontal. Ans. 43-4... ft. (6) A gun is fired horizontally at a height of 144-9 ft. above the surface of a lake and gives the ball an initial speed of 1,000 ft. per sec. Find (a) after what time, and (6) at what horizontal distance, the ball will strike the lake. Ans. (a) 3 sec; (6) 3,000 ft. (7) A stone thrown at an elevation of 1 9° from the top of a tower falls in 5 sec. at a distance of 100 ft. from the base. Find (a) the height of the tower, and (6) the speed of projection. Ans. ia) 368-06... ft.; (6) 21-15... ft. per sec. (8) The elevation of a projectile is that of maximum range on a horizontal plane. Show that the time which elapses before it reaches a point in its path whose horizontal and vertical distances from its starting point are h and k respectively is {^-(h—k)\ . 96 KINEMATICS. [155 (9) Three particles are projected at the same instant from the same point in different directions. Show that the area of the triangle of which they form the angular points varies as the square of the time, and that the plane passing through them remains parallel to itself. (10) The velocities of a projectile at any two points of its path being given, find the difference of their altitudes above a horizontal plane. Ans. (F^in^a- F'%in2o')/25', where V, F' are the magnitudes of the given velocitiesj u, a their inclinations to the horizon. (11) A ball is projected with a velocity of 100 ft. per sec. inclined 75° to the horizon. Find (a) the range on a horizontal plane ; (6) the range on a plane inclined 30° to the horizon ; and (c) what other directions of the initial velocity would give the same respec- tive ranges. Ans. (a) 155-2 ft.; (6) (n/S- 1)207-0... ft.; (c) inclinations 15° and 45° respectively. (12) At what elevation must a body be projected with a speed of 310'8 ft. per sec. that it may reach a balloon 500 ft. from the earth's surface and at a distance of 1,000 ft. from the point of projection ? Ans. Either 39° 17'-7... or 80° 42' -3. (13) On a small planet a stone projected with a speed of 50 ft. per sec. is found to have a maximum range on a horizontal plane of 400 ft. Find the acceleration of falling bodies at the surface of that planet. Ans. 6-25 ft.-sec. units. (14) Show that with a given initial speed the greatest range on a horizontal plane is just half as great as the greatest range down an incline of 30°. (15) The greatest range on a horizontal plane of a projectile with a given initial speed being 500 metres, show that the greatest range on a plane inclined 60° to the horizontal is 2 - ;^3 kilometres. (16) AB being the range of a projectile on a horizontal plane, 155] TEANSLATION: PROJECTILES. 97 show that if t be the time from A to any point P of the trajectory (t.e., the path), and i the time from P to B, the height of P above AB is \gtt'. (17) A particle projected at a given elevation with an initial speed V reaches the top of a tower h ft. high and 2A ft. from the point of projection in t seconds. Find (a) the initial speed of another particle which, being projected at the same elevation from a point distant 4A ft. from the tower, will also reach its summit, and (6) the time it will require. Ans. (a) sl¥gVtl{h+gf^y; {b)[2{h+gt^lg]i. (18) Two stones thrown at the same instant from points 20 yds. apart, with initial velocities inclined 60° and 30° respectively to the horizon, strike a flag-pole at the same point at the same instant. Show that their initial speeds are as 1 : v/3 ; and that the distance of the pole from the nearer point of projection is 10 yds. (19) If a particle, projected with a speed w, strike at right angles a vertical wall whose distance from the point of projec- tion is M^os Jh^ — r^. 156. Case III. — Central Acceleration, the acceleration directed towards a fixed point or centre. (See 138.) If a point move under a central acceleration the moment of its velocity about the centre will be constant. — Since the velocity of the moving point at any instant 157 ] TRANSLATION :-. — CENTRAL ACCELERATION. 99 is the resultant of the velocity at a former instant, and of the integral acceleration during the intervening time, its moment about the centre is (105) equal to the sum of their moments about the same point. But the moment of the acceleration about a point towards which it is directed is zero. Hence the moment of the velocity of the moving point about the centre is constant. It is clear also that the converse proposition holds, that if the moment of the velocity of a moving point about any fixed point be constant, its acceleration must be directed towards the fixed point. ' It follows from 132 that, if to be the angular velocity of the moving point about, and r its distance from, the centre of acceleration, wr^, and therefore ^wr^, are con- stant. Hence (133) the areal velocity of the moving point, or the area described per unit of time by the radius vector from the centre of acceleration, is constant. 157. Examples. (1) Various particles, whose accelerations are all directed to one centre C, are projected from a given point A with equal speeds but in different directions. Show that the areas described in a given time by lines drawn from C to the particles will be proportional to the sines of the inclinations of their initial velocities to the line AC. [The areal velocities are proportional to the moments of the linear velocities, and the perpendiculars on the directions of motion are proportional to the sines of the inclinations.] (2) A point moves in an elliptic path with an acceleration directed to one of the foci. Show that its velocity varies inversely as the square root of its distance from that focus, and directly as the square root of its distance from the other, and has maximum and minimum values when the point is nearest to and farthest from the centre of acceleration respectively. [Note that the product of the perpen- diculars from the foci on a tangent is equal to. the square of the semi-axis minor.] 100 KINEMATICS. [157 (3) A point moves in a parabola under an acceleration directed towards the vertex. Show that the time required to move from any point to the vertex will be found to vary as the cube of the distance of the point from the axis. [If P is a point on a parabola whose vertex is A, and if PM is a perpendicular on the axis of the parabola, the area APM is proportional to the product oi AM into MP.I (4) If an ellipse be described by a point under an acceleration directed towards its centre, the velocity of the point will vary directly as the diameter .conjugate to that which passes through the point. (5) A point moves in an ellipse ABA'B' (ms.]oT a,:Kis, ASS'A' ;. minor axis, BB' ; foci, S and S') with an acceleration directed towards S. Show that the ratio of the times of describing AB and BA' is (ir-2e)l{ir + 2e), where e is the excentricity of the ellipse. (6) A point moves in a circle and is observed to occupy, in passing from a fixed point in the circle to any other point, a time proportional to the sum of the lengths of the arc described and of the perpendicular from one extremity on the diameter through the other. Show that the acceleration of the moving point is directed towards a fixed point. (7) Find the angular velocity of a point moving with a central acceleration, about the centre, in terms of the length of the radius vector (r) and the areal velocity (A). Ans. Sh/A 158. We shall discuss two important cases of central acceleration, viz., that of planetary motion and that of harmonic motion. I. Planetary motion, the acceleration being inversely proportional to the square of the distance of the moving point from the centre of acceleration. This case is of interest, because it is that of the motion of planets about the sun and of satellites about their primaries. 158] TRANSLATION: PLANETARY MOTION. 101 (a) The motion rectilinear, the velocity being in the same straight line as the acceleration at any instant (140). Let s be the distance of the moving point from the centre of acceleration at any instant. Then if a be the acceleration at that distance, and k a constant, a= —Jc/s\ the negative sign being used because the acceleration is towards the point from which the distance s is measured. If v be the speed at the instant under consideration, and v' the speed after an indefinitely short time T, a={v' — v)/t = — Jc/s\ If s'.is the distance after the time t, (s' — s)/t is the mean speed during t ; and as t is indefinitely short, we may consider it equal either to v or to v'. Hence v+v' = 2(s'-s)/t. Hence also , , ,,v'—v Jc s' — s (v+v)^=-2-^.-—- T if T As T is indefinitely small we may consider g' equal to ss'. Hence Let V be the velocity of the point when at a distance 8. Then the space between the positions, whose distances from the centre are s and 8, may be divided into an indefinitely great number- of parts by points whose distances from the centre are Sj, s^, etc., s„. In that case, if %, v^, etc., Vn, are the velocities of the moving point when it is at the above distances respectively, we have F^-.^„=2^(l-A), etc., 102 KINEMATICS. [ l^^ v'-tf = 2k(^---). Vsj s/ Hence by addition we obtain Therefore v^-^=v^--, S s 21c or v^ = ^ (a constant). If s„ is the distance from the centre of acceleration at which the velocity becomes zero (the distance of the starting point, if the moving point start from rest), we have = [-A, and A = So Hence v^ = 2Jc(--~); \s sj and the speed at any given distance from the centre of acceleration is thus expressed in terms of that distance. 159. We may apply the above to the case of the falling of bodies to the earth from great distances. For this purpose we must determine the value of k in this case. Now the acceleration of a falling body at the earth's surface, i.e., at a distance equal to the earth's radius (iJ) from the centre of the earth, is g ; and by Newton's law of gravitation the acceleration of a falling body is in- versely proportional to its distance from the earth's centre. Hence at a distance s we have and therefore in this case k=gR\ Hence, if v is the velocity of a falling body at a distance s from the earth's 160] TRANSLATION: PLANETARY MOTION. 103 centre, its velocity at a distance s„ having been zero, At the earth's surface therefore its velocity will be such that ^2 = 2^iJ2(^-^) = 2^i2(l-^)- _ If the point from which the body has fallen be a short distance h from the earth's surface, Sf^ = R+h, and = 2sri?(l-(l-|+^-etc.)). If now h, be sufficiently small (h/S)^ and higher powers may be neglected. Hence we have v^=2ghi the result obtained in 65. If a body fall to the earth's surface from a very great (practically infinite) distance, we have 1/s = 0, and hence i^ = 2gB. 160. Examples. (1) The acceleration (expressed in ft. -sec. units) of a moving point towards a centre is four times the square of the reciprocal of its distance from the centre. If it start from rest at a distance of 6 ft., find its speed at a distance of 1 ft. Ans. 2'58... ft. per sec. (2) A body falls to the earth from a point 1,000 mis. above its surface. Find its speed on reaching the surface (neglecting resist- ance of air and taking the earth's radius to be 4,000 mis.). Ans. 3'12... mis. per sec. (3) With the data of the last problem find the body's distance ■from the earth's surface'when its speed is 2 mis. per sec. Ans. 535"2... mis. (4) With what speed must a body be projected vertically at the earth's surface that it may never return ? (Assume the earth to have no atmosphere and not to be rotating.) Ans. 6 '98... mis. per sec. 104 KINEMATICS. [160 (5) At what point on a line joining the centres of the earth and moon ■will a body have no acceleration ? (Acceleration of falling bodies at the moon's surface due to moon's attraction=5'5 ft. -sec. units; radius of moon = 1,080 mis.; distance between centres of earth and moon =240,000 mis.) Ans. At a point about 215,900 mis. from the earth's centre. 161. (6) Tlie motion curvilinear, the velocity at any instant being inclined to the acceleration. As (or^ is constant (156), the angular velocity of the moving point P about the centre of acceleration is pro- portional to l/r\ and therefore to its linear acceleration. Now the angular velocity of P is also the angular velocity of the direction of the acceleration, and is therefore (112) equal to the angular velocity of the tangent at the cor- responding point Q of the hodograph. And the linear acceleration of P is equal to the linear velocity of the point Q in the hodograph. If therefore s be the length of the small arc between two points of the hodograph, and ^ the angle between the tangents at these points, (p/s is constant. Now the acceleration of P is in the same plane as its velocity at any instant, and the centre of acceleration, and therefore its path also, is in that plane. Hence the hodograph is a plane curve of constant curvature, i.e. (40), a circle. Let H be the circular hodograph, its pole (which may be either inside or outside or upon the circumference), A its centre, and Q the point in it corresponding to the position P of the moving point in its path. Through draw OM perpendicular to QA or QA produced, and through Q draw QiV perpendicular to OA or OA produced. Since the tangent at Q is in the direction of the acceleration of P, and therefore in that of the radius 162] TRANSLATION: PLANETARY MOTION. 105 vector, OM is the componeat of the velocity of P, in the direction of the radius vector, and is therefore clearly equal to the rate of change of the length of the radius vector. Also QiV is the component perpendicular to ^he fixed line OA of the velocity of P. Hence the ratio of OM to QN is the ratio of a small increment of the radius vector to the simultaneous increment of the distance of the point P from a fixed line in the plane of motion. Now the triangles 0AM and QAN are similar, and the ratio of OM to QH is therefore equal to the ra,tio of OA to AQ, and consequently is constant. P's path is there- fore such that if r and r' are initial and final values of the radius vector in a short time, and if d and d' are corresponding values of the distance of P from a certain fixed line in the plane of laotion, (r' — r)/(d' —d) = k (a constant). Take another fixed line parallel to the given fixed line, and so placed in the plane of motion that, if P's distance from it is S, when the radius vector is r, we may have r/S=k. Also, when the radius vector is r', let S' be the distance of the point from this line. Then d'-d=S^-S. Hence r'-T = l({8'-S). Now T = kS. Therefore t'=W. Hence the ratio of the distance of the moving point from a fixed point to its distance from a fixed line has a constant value, or, in other words, the path must be a conic section. If /c < 1 (and therefore the point inside the circle), the path is an ellipse. If fe=l (the point on the circumference) it is a parabola. If A; > 1 (0 outside the circle) it is an hyperbola.* 162. The astronomical problem is the converse of the above. Kepler generalized from many series of observa- * This proof is due to Prof. Tait. See Encyclopaedia Britaniiica, 9th ed., art. Mechanics. 106 KINEMATICS. [162 tions (1) that the path of each planet is an ellipse, one of whose foci is occupied by the sun ; and (2) that the radius vector of each planet, from the sun, describes equal areas in equal times. These are two of what are known as Kepler's laws. In astronomy, therefore, we have to determine the direction and magnitude of the accelera- tion of a point whose path is an ellipse and whose radius vec- tor from one focus describes equal areas in equal times. Let P be the position of the planet at any instant, V its velocity, APA' its elliptic path, , AA' the axis major of the path, 8 the focus occupied by the sun, and SY a, perpendicular from S on the tan- gent at P- The locus of F is a circle on AA' as diameter. Draw this circle and let YS meet it in Q. By the second of Kepler's laws, V. SYis constant (132), and by a property of the circle SY. 8Q is also constant. Hence 8Q is proportional to V. And it is at right angles to the direction of V. Hence the locus of Q, the circle A YA', turned through a right angle about 8 so that 8Q may become codirectional with V, is the hodograph of P's motion. By a property of the ellipse, GQ is parallel to P8. Hence the tangent QE at Q, whose direction is that of Q's velocity, is perpendicular to P8, and, if the circle be turned through a right angle, will be codirec- tional with P8. But (112) the direction of the velocity of Q is that of the acceleration of P- Hence P's accelera- tion is towards 8. Also the magnitude of P's acceleration is equal to that of Q's velocity. And Q's velocity is proportional to the angular velocity of Q about (7, i.e., since CQ and P8 are parallel, of P about 8. And the areal velocity of P about 1 63 ] TRANSLATION : HARMONIC MOTION. 107 *Si being constant, its angular velocity about S is inversely proportional to the square otPS. Hence the acceleration of P is inversely proportional to its distance from S* 163. II. Harmonic motion, the magnitude of the central acceleration being directly proportional to the distance of the moving point from the centre of accelera- tion. — This case is of interest because it is that of the motion of elastic bodies after compression or distortion. It includes therefore the motion of air and of the lumi- niferous ether in the transmission of sound and light respectively. (a) The motion rectilinear, the velocity at any instant being in the same straight line as the acceleration (140) — simple harm,onic motion. Let a be the acceleration of the moving point when at a distance s from the centre of acceleration. Then, k being a constant, a= — Jcs, the negative sign being used, as in 158. Let the point move to a position at a distance s' from the centre. Then, since the acceleration increases uniformly with the distance, its average value per unit distance during the above displacement must be half the sum of its initial and final values, i.e., —k(s + s')/2. The change of velocity during the displacement is the same as if the point had had an acceleration of this amount during the whole displacement. Hence, if v, v' are the velocities of the moving point at the distances s, s' respec- tively (140, 65), ^'2_^2 = a/ /|(g + s'))(s'-s) = k(s^-s'^). As the point moves away from the centre its velocity diminishes. Let s„ be the distance at which it becomes * This proof is also due to Prof. Tait. See his " Properties of Matter," § 146. 108 KINEMATICS. [ 1 63 zero. Then at any other point distant s its velocity v is such that v^=k(s^^-s^). When the point reaches the centre of acceleration, s = 0, and v^='ks^. Hence its speed on passing through the centre is ^A;.s„. At any point distant — s from the centre its speed is such that and is therefore the same as at a point distant +s. At a point distant — s„its speed is zero. Hence the moving point starting from a distance s^, with zero speed, moves with increasing speed to the centre of acceleration where its speed is i^k . s„ ; thence with decreasing speed to a distance — s,,; and thence back to the starting point, undergoing the same changes of speed in the reverse order ; and so on, its whole motion consisting of a series of such oscillations. Let S be the centre of acceleration, SA the line of motion. From S as centre with a radius equal to Sq describe a circle. From P, whose distance from S is s, draw PM perpendic- ular to SA and meeting the circle Po^ inM. If now the point M move with a uniform speed ^Jk . s„ in the circle, P, the foot of the perpendicular from M on SA, will move in SA with a speed which is the component of M's velocity in the line SA and is therefore Jk . s.cos SMP = Jk . s~ = Jks/^^i If then P's velocity is v, v^ = k(sQ^ — s^). Hence P's velocity, and consequently also its acceleration, at any given distance from S, are the same as the velocity and acceleration respectively of the moving point under consideration when at the same distance from its centre 163] TRANSLATION: HAEMONIC MOTION. 109 of acceleration. Hence the motion of a point moving in a straight line with an acceleration directly proportional to its distance from a centre of acceleration in that line is the resolved part in the direction of that line of the motion of a point moving with uniform speed ^k . «„ in a circle whose centre is the centre of acceleration and whose radius is s„. The time required by the moving point to make a complete oscillation from A to A' and back to A being that required by M to move once round the auxiliary circle is clearly 2xs„ _ 2-7r _ 2 /displacement ^Jh .s^ y/k \ acceleration ' since the magnitude of k is the ratio of the acceleration of the point to its displacement, in any position. The time of a complete oscillation depends therefore only upon the value of k, the constant ratio of the acceleration of the moving point to its displacement from the centre of acceleration. It is independent of the .extent of the oscillation. For this reason such oscillations are said to be isochronous. ■ The time required by the moving point to move from a position P„ to P is, if M„ is the intersection with the circle of a line drawn from P^ perpendicular to SA, 27r &ngle M„SM _ 1 „„„i„ m o jr the angle being measured in radians. The oscillation of a point moving in a straight line about a fixed point in the line towards which its accelera- tion is directed, the acceleration being directly propor- tional to the distance between the points, is called Simple Harmonic Motion.* * Simple Harmonic Motion is thus not only the simplest form of the motion of bodies after release from strain, but is also the apparent motion of bodies moving in circular orbits when observed from a distant point in the plane of the orbit, as, e.g., approximately in the case of the motion of Jupiter's Satellites. 110 KINEMATICS. [ 164 164. It will be obvious that the above results apply also to the case of a point moving in a curved path, pro- vided its rate of change of speed is proportional to its distance (measured along the path) from a fixed point in the path, and is positive, or negative, according as it is moving towards or from the fixed point. 165. The distance of the centre of acceleration or the mean position of the moving point P from its extreme position, 8A in the figure of 163, is called the Amplitude of the simple harmonic motion. The interval of time between two saccessive passages of the moving point through the same position in the same direction is called the Period. The fraction of the period intervening between the instant of the point's occupying its extreme position A in the positive direction, and that at which it occupies any given position is called the Phase. The phase is frequently described by reference to the auxiliary circle. In that case it is defined as the ratio of the angle ASM (P, 163, being the given position of the moving point) to the whole angle (27r radians) through which SM moves in the period. These two modes of defini- tion give clearly the same value of the phase in any particular case. The angle ASM^{P^ being the position of the moving point at zero of time) is called the Epoch of the simple harmonic motion. The epoch is thus equal to the product of the phase at zero of time into 27r. The epoch determines the position of the point at zero of time, the phase its position after any given interval. The epoch has a definite value in any given case of simple harmonic motion, the phase varies with the time, 166. Examples. (1) A point whose motion is simple harmonic has velocities 20 and 25 ft. per sec. at distances 10 and 8 ft. respectively from its centre of acceleration. Find (a) its period, and (6) its acceleration at unit distance from the centre. Ans. (a) --sec. ; (6) 6 "25 ft. -sec. units. 166] TRANSLATION: HARMONIC MOTION. Ill (2) The period of a simple harmonic motion is 20 sec. and the maximum velocity of the moving point is 10 ft. per sec. Find its velocity at a distance of GO/tt ft. from the mean position. Ans. 8 ft. per sec. (3) A point moves from rest towards a fixed point 10 metres dis- tant, its acceleration being everywhere 4 times its distance from the fixed point. At what distance will it have a velocity of 12 metres per sec. ? Ans. 8 metres. (4) Find the mean speed of a point executing a simple harmonic motion, during the time occupied in moving from one to the other extremity of its range, its maximum speed being 5 ft. per sec. Ans. XO/tt ft. per sec. (5) If T be the period and a the amplitude of a simple harmonic motion, and if v be the velocity and « the distance from the centre, of the moving point at a given instant, show that \ 47r2 / (6) A point oscillates about a centre, its acceleration being pro- portional to its distance from the centre. Show that the ratio of its maximum velocity to the square root of the excess of the square of its maximum velocity over the square of the velocity which it has when at a given displacement from the centre, is equal to the ratio of its maximum displacement to the given displacement. (7) A point has a simple harmonic motion whose period is 4 min. 12 sec. Find the time during which its phase changes from ^V to ^ of a period. Ans. 21 sec. (8) A moving point has a velocity of 1 ft. per sec. when at a distance of ^/3 ft. from a fixed point in its line of motion towards which its acceleration is directed, its acceleration being everywhere numerically equal to its distance from that point. After what time will it be at a distance of 1 ft. ? Ans. 7r/12 sec. (9) Show that a point having a simple harmonic motion requires 112 KINEMATICS. [166 ^ of its period to move from a position in which its displacement is a maximum to one in ■which its displacement is one-half the ampli- tude. 167. (b) Curvilinear motion, the velocity of the moving point at any instant being inclined to the acceleration — compov/nd harmonic Tnotion. Let S be the centre of accelera- tion, P the position of the moving point at any instant, and V its velocity at that instant. In the plane of V and 8P take two fixed rectangular axes Sx, 8y. From P draw PM, FN perpendiculars on Sx and Sy respectively. Let the inclination of V to 8x be a. Then the moving point has in the direction of Sx a component velocity Fcos a, and, if s is -the distance of P from 8, a component acceleration - ks cos P8M= -Jcsi^=-k. SM. /5-r Similarly in the direction of Sy, P has a component velocity Fsina and a component acceleration —k.SN. Hence the motion of the moving point is the resultant of two component simple harmonic motions, the one in the direction 8x, the other in the direction Sy. We may therefore determine its motion by determining the laws of the composition of simple harmonic motions. We shall investigate these laws at greater length than is necessary for the mere solution of the above problem, as they are of great importance in the study of sound and light. 168. Composition of Simple Harmonic Motions. — A point has two or more component simple harmonic motions; it is required to determine its resultant motion. 168] TRANSLATION : HAEMONIC MOTION. 113 (1) Two Simple Harmonic Motions in the same line and with the same period. — Let the point P, moving in line BB', have two component simple harmonic th6 motions, of amplitudes OA and GB, and of the same period. Let GP^ and (7Pjj be the component displace- ments due to the respective simple harmonic motions at a given instant. Then the resultant displacement is (86, III.) (7Pj + OPj. Draw the auxiliary . circles, and let ilfj, M^ be the points in these circles corresponding to Pj, Py Complete the parallelogram M^M^, and from 8 draw SR perpendicular to BB'. Since Mfi=SM^&iiA angle GMJP^=^M^SR, GP^==Pfi. Hence GR is equal to the resultant displacement, and ^'s motion is the resultant motion. Since the periods of the motions are the same, the angular velocities of GM^^ and GM^ are the same. Hence the angle MfiM^ is constant, and therefore the length of the diagonal GS of the paral- lelogram MM^,. a,nd its inclination to GM^ or GM^ are constant. 8 therefore moves with uniform speed in a circle. Hence -B's motion is simple harmonic, and there- fore the resultant of two simple harmonic motions in the same line and of the same period is also a simple harmonic motion. As the inclination of G8 toCMj is constant, the period bf the resultant simple harmonic motion is the common H 114 KIKEMATIOS. [168 period of the components. Its amplitude is C8. Its phase , is intermediate between the phases of the com- ponents. If the phases of the two components are the same, the amplitude of the resultant motion is the sum of those of the component motions. If the diflference of phase is tt radians (or [2%+ IJtt radians), the amplitude of the resultant is the difference of those of the com- ponents. 169. As, by taking CM and CM^ of proper lengths, the angles MfiP^ and MpM^ may be made what we please, while GB is kept constant, any given simple harmonic motion may be resolved into two components in the same line, having any desired difference of phase, and one of them having, any desired epoch. 170. (2) Three or more coviponent Simple Harmonic Motions in the same line and of the same period may be compounded two and two, the above process being applied in each case. The resultant motion will evidently be simple harmonic of the period common to the components. 171. (3) Two component Simple Harmonic Motions ■in the same line but of different periods. — If the periods are not the same, the angle MfiM^ (168), and consequently also GS, are variable. At the instants at which the phases of the component motions are the same or differ by 2%-^ radians, GS has its maximum value, viz., GM^ -|- GM^. At the instants at which the phases differ by {tn-^-Vyic radians, GS has its minimum value, viz., GM^—GM^. The angular velocity of GS will also be variable. The direc- tion of GS will oscillate back and forth about that of GM^ their maximum inclination being %\n.~\GMJGM-^. The resultant motion is therefore not simple harmonic but a more complex motion, 172. (4) Component Simple Harmonic Motions in 173] TRANSLATION: — -HARMONIC MOTION. 115 different lines with the same period and phase. — Let the point P have two component simple harmonic motions in the lines \A J.' and BB'. Let CP^ and GP^, and Gp^ and Gp^ be the component displacements of P at times t^ and t^, due to the respective component motions. Then, as periods and phases are the same, GPJGP^ = GpJGp^. Complete the parallelograms pJP^,p^P^. Then GR^ and GR^ are in the same straight line and GPjCP^ = GRJGR^; i.e., the resultant motion is a simple harmonic motion in the line GR^, and is of the same period and phase as the components. The amplitude is the diagonal of the paral- lelogram, whose adjacent sides represent the amplitudes of the components and are inclined at the inclination of the lines in which the simple harmonic motions occur. Hence a simple harmonic motion may be resolved in any two directions into two simple harmonic motions of the same period and phase as the given simple harmonic motion. . It follows that the projection of a simple harmonic motion on any straight line or on any plane is also a simple harmonic motion of the same period and phase as the projected simple harmonic motion. If the component simple harmonic motions are more than two, they may be compounded two by two accord- ing to the above law, and it follows that any number of component simple harmonic motions, in any directions; and of the same period and phase, give as resultant a simple harmonic motion of the same period and phase in a determinate direction and of determinate amplitude., 173. (5) Two component Simple Harmonic Motions in 116 KINEMATICS. [173 different lines with the same period hut with different phases. — We have seen (163) that if a point move uni- formly in a circle, the component of its motion in the direction of a diameter is a simple harmonic motion. Hence the uniform motion of a point in a circle may be resolved into two simple harmonic motions in directions at right angles to one another. These simple harmonic motions will clearly have the same periods and amplitudes. They will differ in phase however by one quarter of a period. For let AA', BB' be perpendicular diameters of the circle ABA'B', in which the point M is moving counter-clockwise. Then the foot P of the perpendicular MP^ will be moving towards U, while P^, the foot of the perpendicular MP^, will be moving towards B. When Pj is at A {i.e., has the phase zero), P^ will be at C, and not until M has moved from A to B will P^ have the phase zero. It follows also that two component simple harmonic motions in perpendicular directions, of the same period, of equal amplitudes, and with phases differing by one quarter of a period, will give as resultant, uniform motion in a circle whose radius is the common amplitude of the components. Now the orthogonal projection of a circle is an ellipse,* the centre of the circle projecting into the centre of the ellipse; the projection of uniform motion in a circle (a motion in which the areal velocity about the centre is constant) is motion in an ellipse with constant areal * If the reader is not familiar with the geometry of projection, he should read the chapter on this subject in Todhimter's " Conic Sections " or some similar work. 1 74 ] TEANSLATION : HARMONIC MOTION. 1 1 7 velocity about the centre; the projections of perpen- dicular diameters of a circle are conjugate diameters of the ellipse, whose inclination and relative length may be made what we please by a proper selection of the plan§ of projection; and the projection of a simple harmonic motion we have seen (172) to be a simple harmonic motion with unchanged period and phase. If, therefore, we project the circle A'BAB', with its perpendicular diameters, on a plane so selected that the projections of the diameters have any des-ired inclination and relative length, the projections of the motions of P^ and P^ will be simple harmonic motions differing in phase by a quarter of a period ; and their resultant motion, the pro- jection of that of M, will be motion in the ellipse which is the projection of A! BAB', the motion being such that the areal velocity of the moving point about the centre of the ellipse is constant. Hence, if a point have two component simple harmonic motions in any dii-ections, of any amplitudes, of the same period, and with phases differing by a quarter of a period, the resultant motion will be motion in an ellipse, with conjugate diameters whose directions are the directions, and whose lengths are the amplitudes, of the component motions, and with constant areal velocity about the centre. Such a motion is called elliptic harmonic motion. 174. If now the two component simple harmonic motions differ in phase by any amount, each of them may (169) be resolved into two in its own direction, which differ in phase by a quarter of a period, and one of which has any desired epoch. Thus we have now two pairs of components, the components of each pair having the same phase, but differing in phase by a quarter of a period from those of the other pair. The components of each pair give as resultant a simple harmonic motion of determinate amplitude and direction, and of their common period and phase. Hence we obtain two simple harmonic motions of determinate amplitude and direction, equal in 118 KI K EMATICS. [ 1 74 period, and differing in phase by one quarter of a period, the resultant of which is determined by 173. Hence the resultant of two component simple harmonic motions of the same period, whatever may be their amplitudes, directions, or phases, is elliptic harmonic motion. 175. It will be obvious that all possible paths of a point having two such component simple harmonic motions, represented by AA' and BB' , must tonch each of the sides of a parallelogram BEFO, whose sides pass through Q F A, A', B, B', and are parallel to AA' and BB'. What the particular path will be, with amplitudes and directions given, will depend upon the difference of the phases of the components. If there is no difference of phase the path is the diagonal GE. If the phases differ by one quarter of a period (that of the simple harmonic motion in AA' being ahead), the point will move in the ellipse ABA'B', and its motion will be counter-clockwise. If they differ by one-half period, the diagonal FD will be the path. If by three-quarters, the point will again move in the ellipse ABA'B', but its motion will be clockwise. For differences of phase of intermediate value the paths will be ellipses in intermediate positions. Thus, for dif- ferences between and i or f and 0, the paths will be such ellipses as HKLM, the motion being counter-clock- wise or clockwise respectively ; and for differences between \ and I or i and | such ellipses as NOPQ traversed counter- clockwise or clockwise respectively. 177 ] TEANSLATION : HAEMONIC MOTION. 119 176. (6) Three or more component Simple HarTnonic Motions in different Unes with the same period but with different phases. — If there be more than two component simple harmonic motions of the same period, but in different lines, and of different amplitudes and phases, each of them may, as in 174, be resolved into two in its own direction, which differ in phase by a quarter of a period, and one of which has any desired epoch. We thus obtain two sets of component motions, all the mem- bers of each set having the same phase, but the members of each set differing in phase from those of the other set by a quarter of a period. The components of each of these sets give, when compounded (172), a simple har- monic motion in a determinate direction, of determinate amplitude, and with the common phase of its components. Hence we obtain two simple harmonic motions in known directions, of known amplitudes, and differing from one another in phase by one quarter of a period. The re- sultant motion is therefore determined by 173. Hence the resultant of any number of component simple har- monic motions of the same period, whatever their ampli- tudes, directions, or phases, is elliptic harmonic motion. 177. (7) Component Svmple HarTnonic Motions differ- ing in period. — If a point ha,ve two or more component simple harmonic motions differing in period, the complete determination of the resultant motion is not possible by elementary mathematical methods. The path of the point may however always be found by determining its positions at a series of instants and drawing a curve through them. For example, let us find the path of a point F which has two component simple harmonic motions in lines at right angles to one another, with periods as 1:2, the simple harmonic motion of longer period having zero epoch, and that of shorter period an epoch 37r/2. Let AA\ BB' be the given lines at right angles to one another. 120 KINEMATICS. [177 GA and CB the given amplitudes of the simple harmonic motions in these lines. Let the simple harmonic motion in AA' be the one of longer period. The component displacement of P from G at zero of time in A A' is GA. Z' ^ ^ ^ 1 H K \ / f \ 1 d " \ /c b a \ \ / \ / \ 3 F / Y \ L M / V ^ V y Since the epoch of the simple harmonic motion in BB' is 3'7r/2, the component displacement of P from G in BE at the same instant must be zero, and its component motion must be from G towards B. Hence the position of , P at zero of time is A. To find other points in the path we may divide AA' and BB' into portions requiring equal times for their description. This may be done by de- scribing semi-circles on A A' and BB' as diameters, dividing the semi-circles into a convenient number (say six) of equal arcs and dropping perpendiculars from the points of section on the respective diameters. Let Ba, a/S, ^G, Gy, etc., be portions of BB', thus determined, requiring equal times for their description, and let Aa, ah, bC, etc., be similar portions of AA'. Through B, a, /3, y, S, B' draw lines parallel to AA', and through A, a, b, c, d, A' draw lines parallel to BB'. Since the simple harmonic motion in AA' is the one of longer period, a component displacement in AA' is accompanied by one in BB' of double the amount. 179 ] TKANSLATION :— HARMONIC MOTION. 121 Hence, while P undergoes the displacement Aa in the line AA', it undergoes the displacement Ca in BB'. Hence P moves from A ;to B. Similarly the component dis- placements ab an4 aB + Ba occur in the same time. Hence P moves from B to K Similarly bC and aC, Co and CS, cd and SB'+B'S, dA' and SO, A'd arid Ga, dc and ' oB + Ba. cC and aC, Cb and CS, ha and ^5'+5'^, and a A and ^(7 are pairs of displacements occurring in the same time. And hence the path passes through the following points in order, via. A, D, E, C, F, G, A', H, K, C, L, M, A, and will be approkimately represented by a smooth curve through these points. The figures on next page represent a few paths of points having two component simple harmonic motions in lines at right angles to one another and differing in period and epoch. The ratio of the periods is indicated at the left of the row of figures to which it refers. The component simple harmonic piotion of shorter period is horizontal. Its epoch is indicated in each case. The epoch of the vertical simple harmonic motion is zero. 178. If the periods of component simple harmonic motions are commensurable, at the end of a period which is their least common multiple the resultant displace- ment of the moving point from the mean position will be the same as at the beginning of the period ; and the path will return into itself, forming a closed curve. If the periods be not ciammensurable, the path will not thus form a closed curve. 179. If the ratio of the periods of two component simple harmonic motions is very nearly a simple ratio, but not exactly, the path very nearly returns into itself ; and it is clearly the same as if the periods were thus simply related, with the difference of epoch slowly in- creasing, the simple haimonic motion with the shorter period gradually gaining in epoch on the othei'. Hence 122 KINEMATICS. [179 |:|(M 181] TRANSLATION: CONSTRAINED MOTION. 123. the point will very nearly pass through all the paths of a point with component simple harmonic motions having periods in the given ratio and of the given amplitudes and directions, with all differences of epoch. Thus, if the amplitudes and the directions are as represented in 175, the periods being very nearly equal, and if at a given instant the phases are the same, the point will first oscil- late in a very elongated ellipse about QE. The ellipse will gradually open out through HKLM to ABA'B', and passing through all such forms as OPQN will gradually come to oscillate in BF. It will then open out again and retrace nearly the same ellipses in the opposite direction, passing through OPQN', ABA'B', and KLMN until it again oscillates in the line GE. Similarly, if the periods be very nearly as 1:2, the path of the moving point will gradually pass through the series of forms- represented on p. 122. 180. Paths similar to those represented on p. 122 are traced out most simply by the aid of Blackburn's pen- dulum, which consists of a bob hung by a Y-shaped arrangement of wires GBEB, the ends G and D being attached at points in a horizontal line. Thus hung, the bob oscillates in a direction perpen- dicula.r to the plane GBEB, about the axis GB. In this plane it oscillates about E. Hence (187) B has two component simple harmonic motions at right angles to one another and of different- periods. The difference of period may be made what we- please by properly adjusting the lengths of the wires.. If the bob be provided with a funnel containing sand or ink, it will leave a tracing of its path. 181. Constrained Motion under given Accelerations^ — We take next certain cases of the motion of a point under conditions of constraint (see 138). 124: KINEMATICS. [ 181 (1) Motion on an Inclined Plane under Uniform Acceleration. — Let a point having a uniform acceleration ■a, whose direction is OA, be constrained to remain iu a plane whose inclination to OA is y. From A draw AB perpendicular to the plane and meeting it in B. Then the angle A OB is y. The effective component of the acceleration in the plane is a cos y in the >B direction OB. For the component normal to the plane cannot affect motion in it. Hence the motion of the point will be rectilinear or parabolic according to the direction of the initial velocity, and will be determined by the equations of 140 and " 142-150, a cosy being, the acceleration in Ihe formulae of those articles instead of a. In the case in which the acceleration is that due to the weight of a body, OA is vertical and the, given plane OB may have any inclination. 182. The speed gained by the point in moviug on the given plane through the distance OB is equal to that which would be gained in moving in the direction of the acceleration OA, through a distance which is the pi'ojec- -tion of OB on OA. To prove this, draw BG from B perpendicular to OA. Then, calling OB I, 00 h, the initial speed V, and the speed at B v, we have t)2 _ y2 = 2al cos y. Had the point moved from to C with the same initial .-speed its speed v' at would have been such that ^'2 —Y^— 2ah = 2al cos y. Hence v' = v. 183. The times required to produce these changes of rspeed are of course different. Thus, if t, t' are the times 184 ] TRANSLATION : CONSTRAINED MOTION. 1 2 5' required by the point to. move from to G and from O to B respectively, we have v=V+at, and v=V+at'cosy. Hence t = t'cos y. 184. Eoca/m/ples. (1) A point having a constant acceleration of 24 ft.-sec. units is. constrained to move in a direction in which its speed changes in 1 min. from 10 to 250 yds. per sec. Find the inclination of its direc- tion of motion to that of the given acceleration. Ans. 60°. (2) A heavy particle (gr=32) is projected* up an inclined plane- whose inclination to the horizon is .30°. Find the distance traversed during a change of speed from 48 to 16 ft. per sec. Ans. 64 ft. (3) A railway carriage has, when 1 mile up an incline of 1 in 50 {i.e., one having an inclination to the horizon of sin~'--s-^, an upward velocity of 30 miles per hou.r. (a) In what time will it come to a standstill ? (6) If it afterwards run back, with what speed will it reach the foot of the incline ? (Take 9 = 32.) Ans. (a) 1 min. 8"75 sec. ; (6) 63'5 miles per hour. (4) A body slides from rest down a smooth sloping roof and then falls to the ground. The length of the slope is 18 ft., its inclination to the horizon 30°, and the height of its lowest point from the ground 40 ft. Find the distance from the foot of the wall to the- point where the body reaches the ground. (Take r. 215. The angular velocity of a system is measured in terms of the same unit as the angular velocity of a point (128). 216. Composition of Angular Velocities. — If a rigid system with one point fixed have any number of com- ponent angular velocities of given magnitudes and direc- tions, we may prove, by reasoning similar to that employed in determining the law of the composition of linear velocities, that their resultant is to be determined accord- ing to the same law. We have therefore propositions . called the parallelogram, the triangle, and the polygon of angular velocities of the same form as the similar pro- positions for linear velocities. The reader can easily construct them for himself 217. All the deductions from these propositions made in the case of linear velocities may also be made in that of angular velocities ; and hence all the formulae of 85-90 apply to angular velocities, d^, d^, etc., being now the mag- nitudes of the component angular velocities and R the magnitude of the resultant. 218. It follows also that angular velocities may be resolved after the same manner as linear velocities or displacements (79-84). 219. Angular Acceleration of a Rigid System. — The angular velocity of a rigid system will in general vary from in.stant to instant both in magnitude and direction. The integral angular acceleration of a rigid system during any time is that angular velocity which must be compounded with the initial angular velocity, in oi'der to produce the final angular velocity. 223] ROTATION. 145 The mean angular acceleration of such a system during any time has & direction which is that of the integral angular acceleration, and a magnitude which is that of the integral angular acceleration divided by the time. In general the mean angular acceleration will be different for different intervals of time. If it is the same, both in magnitude and in direction, whatever be the interval of time, the system is said to be rotating with uniform angular acceleration. The instantaneous angular acceleration of a rigid system at a given instant has a direction and a magnitude which are the limiting direction and the limiting magni- tude of the mean angular acceleration between that instant and another when the interval of time between them is made indefinitely small. The instanta,neous angular acceleration of a rigid body is in all cases finite. 220. The angular acceleration of a system is measured in terms of the same unit as that of a point (1 36). 221. Composition and Resolution of Angular Accele- rations. — The laws of the composition and resolution of angular accelerations are the counterpart of those of linear accelerations. As the latter were deduced from the laws of the composition of linear velocities, so may the former be deduced from the laws of the composition of angular velocities. 222. It follows that the relations between the magni- tudes and inclinations of the components and the mag- nitude and direction of the resultant, as expressed in the formulae of 85-90, hold also for angular accelerations, d^, d^, etc., standing now for the magnitudes of these accelerations. 223. An angular acceleration may, like a linear accele- ration (120), be resolved into tangential and normal K 146 KiNEMAflCS. [223 components, i.e., into components whose respective direc- tions coincide with, and are normal to, the instantaneous axis of rotation ; and in the one case as in the Other, it may be shown that the normal component determines the change of the direction of the angular velocity, i.e., the change of the direction of the instantaneous axis^ while the other component is equal to the rate of change of the magnitude of the angular velocity. If therefore a rotating body have an angular accel- eration which is continually perpendicular to its axis of rotation, its angular velocity will change in direc- tion but not in magnitude. If it have an acceleration which has the same direction as its angular velocity, the direction of the rotation will be constant, and the rate of change of the magnitude of its angular velocity will be equal to its angular acceleration. 224. Motion under given angular accelerations, of a rigid system with one point fixed. We may take a few of the simpler cases of such motion. (1) Angular acceleration zero. — If there is no accelera- tion, the direction and magnitude of the angular velocity must remain constant. Hence, as in 138, if w be the angular velocity and 6 the displacement in a time t, we have 6 = dot; and the axis of the angular displacement is the constant axis of rotation. 225. (2) Angular acceleration constant in magnitude and direction. (a) Direction the same as that of the instantaneous axis at any instant. — The directions of the acceleration and of the instantaneous axis of rotation at a given instant being the same, that of the instantaneous axis is constant (140). Hence, also, the rate of change of the magnitude of the' angular velocity is equal to the angular acceleration. If therefore a be the magnitude of the 226] ROTATION. 147 angular acceleration, w, and Wj the initial and final angular velocities, and the displacement in a time t, we may obtain, as in 63-65, the formulae — 226. (6) Direction any whatever. — Let OA represent in direction the initial angular y, velocity w, OB that of the angular acceleration «, the angle AOB be- tween their directions being (p. In the plane of OA and OB draw 00 perpendicular to OA. The com- ponents of the acceleration in the directions OA and 00 are thus a cos and a sin (f>. ° To find the angular velocity after any time t we know that its components about OA and OG are w-J-a^cos^ and at sin ^ respectively. Hence, if Q is its magnitude, 0= {(ftj+aicos Y}i. Also, if 1^ is the angle made by its direction with 00, ^ at sin To find the angular displacement after any time t, we know that the component displacements about OA and OG respectively are wt+^af cos

. Hence, if 6 is the magnitude of the resultant, e=iiwt+ ^at^cos about a given axis and a linear velocity v in a plane perpendicular 249] MOTIOU" OF RIGID SYSTEMS. 159 to the given axis is an equal angular velocity about a parallel axis distant vfte in a direction perpendicular to that of V. 248. Hence also a given angular velocity w may be re- solved into an equal angular velocity about a parallel axis distant din a. given direction, together with a linear velocity equal to dto perpendicular to the plane of the axes. 249. Composition of Angular Velocities about Parallel Axes. — By the aid of this result, we may determine the resultant of two component angular velocities about parallel axes. Let A, B be the parallel axes, d the distance between them, w^t m.^ ^^^ angular velocities about A, B respectively. Then the angular velocity w^ about A is equivalent to an equal angular velocity about B with a linear velocity perpendicular to the plane of the axes and equal to cZw^. The given angular velocities about A and B are therefore equivalent to an angular velocity equal to their sum about B together with the linear velocity equal to rfwj. Similarly, an angular velocity of Wj+Wj about B is equivalent to a linear velocity — dw^ with an angular velocity w^+w^ about a parallel axis G, distant ^Scoj/Cwi+ftJa) from B in the direction AB, and therefore dwj(a)j^+w^ in the direction BA. Hence, as positive, and negative equal translations destroy one another, the two angular velocities Wj and ca^, about parallel axes A and B respectively, are equivalent to an angular velocity Wj+w^ about a parallel axis through in the same plane as A and B, whose distance from 5 is dwj((i)^+a>^,a,nd from A d — doojicoj^+w^), and which therefore intersects the line BA so that BG:CA= w, : w^. If the component angular velocities about A and B are equal and opposite, Wj + Wa = 0, and doi)J(u)^ + Wg) = o) . The axis of the resultant angular velocity is thus at an infinite distance; in other words, the resultant velocity is a trans- 160 KINEMATICS. [249 lational velocity in a direction perpendicular to the plane of A and B. 250. Composition of Linear and Angular Accelera- tions. — As in 116, it may be shown that the laws of the composition of linear and angular velocities apply also to linear and angular accelerations. 251. Motion of a Rigid System under given Accelera- tions. — -The resultant linear acceleration of any point of the system and the resultant angular acceleration being- known, together with the initial velocities, the displace- ment and the final velocities of the s^'stem may be deter- mined by 138-180 and 224-226. In practical problems the angular accelerations are usually known about axes fixed in the body. Of such cases we must restrict our- selves (227) to those in which the axes fixed in the body have also fixed directions in space. 252. Oeometrical Representation of the Motion of a Rigid Lamina in its own Plane. — The instantaneous centre of such a lamina occupies at any instant a definite position both in the lamina itself and in space ; but from instant to instant its positions, both in the system and in space, change. By 295 the successive positions in the case of a body must be indefinitely near ; and therefore the series of positions of the instantaneous centre in the system forms one curve and the series of positions in space forms another. At each instant these curves must be in contact, the points of the curves in contact being the positions of the instantaneous centre at the given instant. Hence the motion of a rigid plane system in its own plane may be geometrically represented by the roll- ing of a curve fixed in the system on a curve fixed in space. This conception is of great use in the treatment of some of the more difficult problems of the motion of rigid systems. 254] MOTION OF RIGID SYSTEMS. 161 253. Motion of Rigid SysteTns under Constraint.' — If two or more bodies are connected together in any way they form a system, the motion of any one member of which, thus subjected to constraint, depends upon the position and the motions of the others. In such cases it may be required to determine the instantaneous axis of rotation of any one member, to find relations between the velocities of various members of the system, etc. We may illustrate such cases by a few examples. Readers who wish a thorough treatment of the constrained motion of rigid bodies should study works on the Kinematics of Machinery. 254. ExaTnples. (1) The line BE moves, keeping its extremities in two fixed lines ADB, A EC. Find the instantaneous centre and the direction of motion of any point G in DE, when DE occupies any given position. From D and E draw DF and EF perpendicular to AB and AC and meeting in F. F is the instantaneous centre (233); for in- definitely small displacements of D and E have the same directions as AB and AC respectively, and their middle points coincide' ultimately with D and E. Join OF. The line through i2=2v'/r. Similarly w^=Zv'lr, and so on. That the angular velocities may be the same, the radii of the 1st, 2nd... Ji* sheaves must be as the numbers 1, 2... n respectively. (8) AB, BC, CD are three rigid rods jointed to one another at B and C, and to fixed points at A and D, and moveable in one plane. Find the angular velocity of CD when that of 4 i? is w. [The motion of this system is called three-bar motion. The system is one of the " elementary combinations " of machinery.] Produce AB and BC to meet in K Then at any instant the linear velocities of B and C are perpen- dicular to AB and CD respectively. Hence at that instant BC is rotating about E. Now B'a linear velocity is w . AB. Hence the angular velocity of BC about E is aABjBE. Hence also the linear velocity " ^ of C is laAB . ECjBE, and the angular velocity of CD is oiAB . ECj{BE . CD). (9) A disc (radius =/•) rolls without sliding on a plane. Find the relation between its angular velocity w and the linear velocity v of its centre. The point of the disc in contact with the plane has two com- ponent linear velocities, one the translational velocity v which it has in common with the centre, and another equal to wr due to its rotation about the centre. These components are in the same straight line. Hence their resultant is equal to their algebraic sum. But their resultant is zero. For as the disc rolls without sliding the point of the disc in contact with the plane is instantane- ously at rest. Hence v + respectively, the angular velocity of the rod will be V sin^ I. oos(-e) 166 KINEMATICS. [235 CHAPTER VII. STRAINS. 255. We have next to discuss the motion of systems of points whose distances from one another are variable. Any change of configuration of such a system is called a strain. 256. Strains may involve both translation and rotation, i:e., there may be no point of the system which occupies the same position in space in both the initial and the final configurations of the system, and there may be no three intersecting straight lines in the system whose directions in the initial and final configurations are parallel. In considering strains however it is usual to exclude from consideration the translation involved, as occasioning no difficulty. For this purpose one point of the body is assumed to be fixed in space. 257. Homogeneous Strains. — We shall restrict our- selves to the most simple strains to which bodies are subjected, those, viz., which are such that the distances of pairs of points so placed in any pax-t of the unstrained system that the lines joining them have the same direc- tion, are increased or diminished in the same ratio. Such strains are called homogeneous strains. The ratio of the distance of two points after the strain 259] STEAINS. 167 to their distance before the strain is called the ratio of the strain for the direction of the line joining them. The ratio of the increment of the distance of two points to their initial distance is called the elongation of the strain for the direction of the line joining them. The elongations of a strain may be positive or negative. If d and d' are the initial and final distances of two points, a the ratio of the strain, and e its elongation, for the direction of the line joining the points, we have thus a = cZ'M e = {d'-d)/d. Hence a=l+e. 258. Points which lie in straight lines before a homo- geneous strain lie also in straight lines after the strain. Let A, B, C be points lying in a straight line before the strain and let A', B', C be their positions after the strain. Then (257) A'B'IAB = B'G'/BC= A'C'/AC. Hence (A'B'+B'Oy(AB+BG} = {A'B'+B'C')/AG=A'C/AC, and hence A'B'+B'C'=A'C'. B' is therefore a point in the straight line A'C. 259. Since all straight lines remain straight after the strain it is clear that planes must remain planes. 168 KINEMATICS. [260 260. Lines which are parallel in the unstrained state of the system are parallel also after the strain. — Let AB, OD and AG, BD be pairs of intersecting parallel lines in the unstrained system. These lines being in the same plane before the strain must be in the same plane after it. Also AB and CD being equal before the strain must re- main equal. Similarly AG and BD must remain equal. Moreover, as one portion of a material body cannot pass through another portion, GD cannot cut AB or AG cut BD after thestrain. Hence, HA'B'D'G' represent the strained system, A'G' and CD' are equal respectively to D'B' and B'A'; and A'D' being common to the two triangles A'G'D' and D'B! A', these triangles are equal in every respect. The angles B'A'D' and A'D'G' are therefore equal, and likewise the angles FD'A' and UA'G'. Hence A'G' is parallel to B'D' and A'F to G'D'. 261. Parallel straight lines remaining parallel and straight, parallelograms must remain parallelograms, parallel planes must remain parallel, parallelopipeds must remain parallelopipeds, and figures which are similar and similarly situated must remain similar and similarly situated after the strain. 262. Since parallel straight lines must remain parallel and must be increased or diminished in the same ratio, a circle drawn in any part of the system must be strained so that parallel chords remain parallel and become in- creased or diminished in length in a given ratio. Hence (l73) after the strain it will be an enlarged or diminished orthogonal projection of the circle on some plane, i.e., it 263] STRAINS. 169 will be an ellipse, perpendicular diameters of the circle having become conjugate diameters of the ellipse. There is one pair of perpendicular conjugate diameters in every ellipse, viz., the major and minor axes. Hence there is one pair of perpendicular diameters in the circle whose mutual inclination is not changed by the strain. 263. As all plane sections of a sphere are circles, a spherical portion of the unstrained system must after the sti'ain have the shape of a figure whose plane sections are ellipses, i.e., of an ellipsoid. , A cube circumscribing the sphere will become a paral- lelopiped (in general not rectangular) circumscribing the ellipsoid, the points of contact of the cube with the sphere, which are the extremities of three diameters at right angles to one another, becoming the points of con- tact of the parallelopiped with the ellipsoid, which are the extremities of conjugate diameters. Hence perpen- dicular diameters of the sphere become conjugate diameters of the ellipsoid after the strain. There is one set of conjugate diameters of every ellip- soid which are at right angles to one another, viz., the principal axes. One of them is the greatest diameter, another the least, and the third has in general an inter- mediate value. There are thus three perpendicular diameters of the sphere which after the strain become the axes of the ellipsoid. Lines in their directions in the initial configuration have the same mutual inclination in the final configuration, though the inclination of these lines to fixed lines in space or to lines in other directions in the system may have changed. The directions of the axes of the ellipsoid in the strained system and of the corresponding rectangular diameters in the unstrained system are called the prin- cipal axes of the strain. The elongations in these direc- 170 KINEMATICS. [263 tions are called the principal elongations ; the ratios of the strain in these directions, the prvncipal ratios. 264. The ellipsoid into which any spherical portion of the system is strained is called the strain ellipsoid. It has obviously (257) in the case of a homogeneous strain the same form and relative position in whatever part of the system the sphere may be taken. If the principal elongations of a strain are all equal, the strain ellipsoid becomes a sphere, and the ratios of the strain in all directions are the same as the principal ratios. All lines in a system subjected to such a strain, whatever may be their directions, are changed in length in the same ratio. There is no change of form. If two of the principal elongations are equal and the third either greater or less than the other two, the strain ellipsoid is a spheroid, prolate or oblate. If two of the principal elongations are equal to zero, it is also a prolate or oblate spheroid, its equal axes having the same length as the diameter of the sphere. A strain in which two of the principal elongations are •zero is called a si/mple longitudinal strain. 265. There are two sets of parallel planes which remain undistorted after the strain. — Let ABGD be a section of the strain ellipsoid by a plane through the greatest and /< \ ^>\ V . ^' ^^\. J ^< _^^^>f least of the principal axes. Let SOS' and TOT' be dia- meters in this plane equal to the mean principal axis. 268] STEAINS. 171 Then the sections of the ellipsoid by planes through SOS' and TOT' perpendicular to the plane ABGD are ellipses with equal principal axes, i.e., circles. Hence the elonga- tions of all lines in these planes are the same as the mean principal elongation, and these planes therefore, and all planes parallel to them, remain undistorted after a strain though they may be changed in area. The axes AG and BD evidently bisect the angles of inclination of the planes of no distortion. 266. The ratio of the final to the initial volume of a system homogeneously strained is the same as the ratio of the volume of the strain ellipsoid to that of the corre- sponding sphere. Hence, if a sphere of radius r is strained into an ellipsoid whose principal semi-axes are a, h, c, the ratio of the final to the initial volume of the system is M^=?.M=a+.xi+«i+»). if e, f, and g are the respective elongations. If e, f, and g are so small that their products may be neglected, we have Hence, in the case of a small strain, the cubical dilata- tion, or expansion per unit of volume, is equal to the sum of the principal elongations. 267. Pure Strains. — Strains in which the initial and final directions of the principal axes are the same, are called pure strains. They are so called because their characteristic property excludes the possibility of rotation. 268. Rotational Strains. — In general,, however, the initial and final directions of the principal axes of the strain are not the same. In all such cases, since the princi- pal axes maintain their mutual inclinations, they may be 172 KINEMATICS. [268 brought into the positions which they occupy after the strain, by a rotation, and the body thus rotated may then have its final configuration given it by a pure strain.^ It will be obvious also that the same result will be attained if the body be first subjected to the pure strain and then to the rotation. 269. The Shear.— If one plane of a system be held fixed, and if the planes parallel to it be moved in their own planes, without change of form or area, those on the one side of the fixed plane in any one direction, and those on the other side in the opposite direction, and all through distances proportional to their distances from the fixed plane, the system is said to have undergone a shear. The amount of the shear is the relative displacement of any two of the parallel planes divided by the distance between them. The plane of the shear is any plane intersecting the fixed plane normally in a line parallel to the direction of relative motion. The direction of the shear is that of the relative motion of the parallel planes. Similarly, if one line of a plane system be held fixed, and if all lines parallel to it be moved parallel to it in one direction or the other according as they are on one side or the other of the fixed line, and through distances pro- portional to their distances from the fixed line, the plane system will undergo a shear, whose plane is the plane of the system, whose direction is that of the fixed line, and whose amount is the relative displacement of any two lines per unit distance between them. Thus any parallelogram abed may be produced from any other parallelogram ABGD on an equal base {AB=ab) and between the same parallels {aB and Dc) by subjecting it to a shear whose plane is the plane of the parallel lines aB and Dc, whose direction is that of the fixed line EF which is parallel to aB, and whose amount is the quotient of Bd by the perpendicular distance of EF from Be. 270] STRAINS. .16 A familiar approximate illustration of a shear in three dimensions is the change of configuration by which a pack of cards initially forming a rectangular parallelo- piped is made to take the form of a parallelepiped not rectangular. The illustration would be exact if the cards were indefinitely thin. 270. Homogeneity of the Shear. — Let AB and CD be parallel lines having any direction in the unstrained system. From their extremities let fall perpendiculars Aa, Bh, Co, Dd on the fixed plane of the shear to which the system is to be subjected. Let A', B', G', D' be the positions of A, B, C, D after the shear. Then AA', BB', GC, DD' are parallel, and AA'/Aa = BR/Bb = GG'/Gc = DD'/Dd. Let BA and DG produced meet the fixed plane in E and 174 KINEMATICS. [270 F respectively. Then Eah and Fed being the projections of ^J.5 and FOB on the same plane are parallel straight lines. Since Eah is a straight line and Aa is ^parallel to Bh, AE/Aa = BE/Bb. Hence AA'/AE=BB'/BE; and therefore EA'B' is a straight line. Similarly FC'JD' is a straight line. Since EB, Bh, and hE are parallel respec- tively to FD, Bd, and dF, the triangle EBh is similar to the triangle i?'i)d Hence BE/Bh = BF/Bd. Hence also BB'/BE= BB'/BF. Now BB' and 5^ are parallel to BB' and BF respectivelv. Hence the triangle EBB' is similar to FBB'; and therefore A'B/AB^C'B'IOB. Hence the lengths of the parallel lines AB and CB are increased by the shear in the same ratio. The shear is therefore a homogeneous strain. It has consequently principal axes, ratios, and elongations, like all homo- ffeneous strains. 271. It is obvious that as, in a shear, all planes of a body parallel to a given plane are translated in their own planes but not changed in area, there can be no change in the volume of the body. 272. Reduction of the Shear to a Pure Strain and a Rotation. — Let be the centre of a spherical portion of a system subjected to a shear, AGB the intersection of the sphere with the plane of the shear through 0, and AB the intersection of the fixed plane with the same. Let the system be subjected to a shear of amount s, and such that planes parallel to the fixed plane through AB and on the (7- ward side of AB move in the direction AB, parallel planes on the other side oi AB moving in. the opposiite direction. Then (270 and 262) the circle AGB will after the shear have the form of an ellipse ABB whose centre is ; and the sphere intersecting the plane of the shear mAGB will become the ellipsoid intersecting that plane in ABB. Since the distances of points of the 272] STEAINS. 175 system from the plane of the shear through remain constant, this plane must contain the greatest and least principal axes of the ellipsoid. Now, by 265, EE' and FF, the bisectors of the angles AOa and BOa respec- tively, will be the minor and major axes of the ellipse ADB. Hence OE, OF, and a line through 0, perpen- dicular to both and equal to OB, are the least, greatest, and mean principal axes of the ellipsoid. If OC is perpendicular to AB, and CB touches the circle at C, it will also touch the ellipse, and its point of con- tact D will be the position of G after the shear. OB is therefore conjugate to OB. Hence, since 00 is equal to the perpendicular from B on OB, OF/00=OC/OE. Hence the circle may be brought to have the shape ot the ellipse by elongating all chords parallel to one dia- meter in the ratio OF/OC, and shortening all chords perpendicular to that diameter in the ratio of OG/OF. Let a line through- E parallel to OA meet the circle in e. Then E is the position of e after the shear. Hence (262) Oe and a line perpendicular to it in the plane of AGB coincide in direction after the shear with OE and OF. And a line through perpendicular to the plane of AGB remains unchanged in direction. Hence these lines 176 KINEMATICS. [272 are the positions before the shear of the principal axes of the shear. And therefore, if the sphere be first rotated about an axis through perpendicular to the plane of ACB, through the angle eOE, it may then be brought to its final configuration by a pure strain whose axes are OF, OB, and a line perpendicular to both. The ratios of the strain in these axes are OF/ 00, 0E/0G[=1/(0F/0G)] and 1 respectively. If OF/OG be called a, they are a, 1/a, and 1. 273. From the symmetry of the figure it is obvious that a shear of the same plane and amount, but with the plane through ab as fixed plane, is equivalent to the same pure strain as above, together with a rotation of equal amount and about the same axis but in the opposite direction. Hence, rotation being neglected, the same change of configuration is produced in a system by a shear of given plane and amount, whether its direction be one or other of two directions equally inclined to the greatest and least principal axes of the shear. 274. It is obvious from 265 and 272 that planes through AB and ab, normal to the plane of the shear, and all planes parallel to these planes respectively, are both un- distorted and unchanged in area by each (273) of the above shears. Hence in any body subjected to a shear there are two sets of planes which are unchanged in area and form, these sets of planes being equally inclined to the greatest and least principal axes and parallel to the mean principal axis. 275. It is obvious also, with the aid of the above, that Oe may be brought to coincide in direction with OH, its length remaining unchanged, either by a rotation about an axis through perpendicular to the plane J. (75, through, the angle eOE, or by a shear of the amount GD/CO in the 277] STKAINS. 177 plane AGB, and in the direction OD, together with a pure strain whose ratios in the principal axes OF, OE have the values OG/'OF and OC/OE respectively, and in a direction perpendicular to both, the value unity. 276. The amount s of the shear may be expressed in terms of its principal ratios or elongations. By a pro- perty of the ellipse (Fig. of 272) OI)^+OB^ = OF^+OE^ Hence OI)^ = OF^+OE^-OG\ and GI)^=0F^.+0E^-20G\ and (272) GD'' = OF^+ 0E^-20E . OF = {OF-OE)\ Hence GD/OG=s=OF/OG-OE/OG=a-l/a. If e is the greatest principal elongation (257), 8 = 1 + 6-1/(1+6). If the shear be indefinitely small, we have l/(l+e) = l-e, and hence s = 2e. Also, when the shear is indefinitely small, OG, OD and Oa (Fig. of 272) ultimately coincide. Hence Oa is at right angles to OA, and therefore (265) is inclined to OE and OF at angles of 45°. Hence, if a system be subjected to a strain consisting of two indefinitely small elongations, one e in any direction, and the other — e in a perpendicular direction, the resul- ting strain is a shear whose amount is 2e, whose plane is that of the two rectangular directions, and whose direction bisects the angle between them. 277. Examples. (1) Show that, if rotation be left out of account, a small simple elongation e in any direction is equivalent to a uniform cubical M 178 KINEMATICS. [277 dilatation together with two shears, ea,ch having the given direction for one principal axis and lines at right angles to it and to each other for the other axes ; and determine the magnitude of the dila- tation and the amounts of the shears. — Let OA be the direction of ,D the simple elongation and OD a cube of which OA is an edge. The elongation e in the direc- tion OA is equivalent to three elongations in the same direction, each having the magni- tude e/3. As there are no elongations in the directions OB and OC perpendicular to OA and to each other, we may regard the cube as subjected to two elongations in each of these directions, having the magnitudes e/3 and - «/3. Now an elongation e/3 in each of the three rectangular directions OA, OB, and OC is (266) equivalent to a uniform cubical dilatation of the magnitude e. Also, the elongation e/3 in the direc- tion of 0.4 with the elongation -e/3 in the direction of OB are equivalent (276) to a shear whose principal axes are these lines and whose amount is 2e/3 ; and similarly the remaining elongation e/3 in the direction of OA with the remaining elongation —e/3 in the direction of OC are equivalent to a shear whose principal axes are OA and OC, and whose amount is 2e/3. (2) Show that, if a square be subjected to a small shear whose axes are in the directions of its diagonals, it becomes a rhom- bus whose sides are equal to those of the square and whose angles differ from right angles by S radians, B being the amount of the shear. (3) Investigate the strain in the case of a uniform circular cylinder of length I fixed at one end and having its other end twisted through an angle 8. This form of strain is called Torsion. The cylinder being uniform, every noi-mal section of it will rotate about its axis ; and, 9/Z being the amount of the twist per unit length of the cylinder, the amount of the rotation of any section will be the product of Sjl into its distance from the fixed end of the cylinder. Hence, also, any normal section will be twisted relatively to any other normal section distant d from it through an angle edil. Let .4 a be the axis of the cylinder, ABba and ADda planes 277] STRAINS. 179 through Aa, in the unstrained system, inclined at an indefinitely small angle, ABD and abd planes normal to Aa, CE and ce arcs of circles having AG aa radius and A and a as centres respectively, and BD and hd arcs of circles haying A and a respectively as centres, and as radius AB indefinitely nearly equal to AC. Then CE, ce, BD, and hd may be considered to be equal and parallel straight lines, and BcDe a rectangular parallelepiped whose edges Bb, Co, Dd, and Ee are parallel to Aa. After the strain B, C, D, E will have moved relatively to 6, c, d, e to B', C, D', E', BB' and DD' being equal to (e/?)56 . AB, and CC and EE equal to (6ll)Bb .AC. These quantities, when angle BAD and BC are made indefinitely small, are ultimately equal. Hence the small rectangular parallelopiped BcDe becomes after the strain the non-rectangular parallelopiped B'cD'e, on the same base and between the same parallel planes. Hence the parallelopiped BcDe has been subjected to a shear whose plane is BDdb, direction BD, and amount BB'/Bb, i.e., (e/l)AB. Hence at every point distant r from the axis of the cylinder thus subjected to torsion, it undergoes a shear whose plane is parallel to 180 KINEMATICS. [277 the axis and perpendicular to a plane through the point and the axis, whose direction is normal to this plane and whose amount is Brjl. (4) A uniform straight beam is bent so that lines initially longi- tudinal and straight become arcs of circles in parallel planes (called planes of bending), with centres in a line normal to these planes ; transverse sections initially parallel become so inclined that they intersect in this line, and longitudinal lines in a surface, called the neutral surface, normal to planes of bending and initially a plane, are not changed in length. Investigate the strain. Let ABDC be a section of the bent beam by a plane of bending, EF the intersection with ABDC of the neutral surface, ac and hd the intersections with it of two transverse sections of the beam {0 being their inclination), and the centre of curvature oi AB and CD. ' Theii it is obvious that longitudinal lines, such as GIf, between AB and HF are lengthened, and longitudinal lines between EF and CD are shortened, by the strain. The line gh was initially equal to «/. Hence it has undergone an elongation (per unit of its length) equal to {gh-ef)/ef. Now gh=0g.9 and ef=Oe.e. Hence the elongation of gh is {Og- Oe)IOe=gelOe=dlp, if d is the distance of the line GH from EF, and p the radius of curvature of EF. This result applies to all lines parallel to gk and intercepted between the transverse sections ao and bd, d being 278] STRAINS. 181 positive when measured from ef towards ah, and negative when measured from e/ towards cd. Hence at every point of the beam there is a longitudinal strain in the direction of its length, the elongation being equal to d\f. It is positive for all points between the convex surface and the heutral surface, and negative for all points between the neutral and the concave surfaces. It is obvious however that these longitudinal elongations alone would not involve bending, and that in order to bring the beajn into its final configuration longitudinal planes normal to planes of bending, which have thus been elongated, must slide over one another. Hence at each point of the beam there is not only a longitudinal elongation, but also a shear whose plane is the plane of bending and whose direction is longitudinal. By 273 and 276, if this shear is small it is equivalent to another in the same plane, but with a direction transverse to the beam and in the plane of bending, trans- verse slices of the beam sliding over one another in the direction of their intersections with planes of bending. Hence the strain at any point of the beam consists of a longitud- inal elongation equal to c?/p, together with the above shear. 278. Specification of a Strain. — The elongations of a homogeneous strain in any three non-coplanar directions being given, the elongation in any other direction can be found. — Let Ox, Oy, Oz be lines having any three direc- tions and e, f, g the elongations in them respectively. Then any point P whose co-ordinates referred to these lines as axes are x, y, z, has component displacements ex, fy, gz, and the resultant displacement may be determined by 78. The final distance of P from may thus be deter- mined, and hence also the elongation in the direction of OP 182 KINEMATICS. [279 279. Hence a homogeneous strain is completely speci- fied if the elongations in any three non-coplanar directions are given. 280. The specification of three non-coplanar directions requires (7) six numerical data, and that of the elonga- tions in these directions three more. Hence, in general, nine quantities are requisite for the complete specification of a strain. 281. As a pure strain consists simply of elongations in certain rectangular directions, the principal axes, the specification of a pure strain requires only data sufiicient to determine the directions of these axes and the elonga- tions in them. To determine three rectangular directions three numerical data are sufficient. Hence the specifica- tion of a pure strain requires only six numerical data. 282. As any homogeneous strain may be regarded as compounded of a pure strain and a rotation, the nine data necessary for its- specification may consist of the six necessary for the specification of the pure strain and the three necessary (198) for the specification of the rotation. 283. Rectangular Specification of a small Strain. — Let Ox, Oy, Oz be rectangular axes of co-ordinates, OA, OB, OG the principal axes of the pure strain, and Or the axis of the ^ rotation, of which, the given small strain may be regarded as com- -X pounded. The elongations being given for the directions OA, OB, 00, equiva- ^ lent elongations for the directions Ox, Oy, Oz may (278) be determined. The rotation about Or may be resolved into com- ponent rotations about Ox, Oy, Oz. Now the rotation 2S4] STEAINS. 183 about Ox being small may be regarded (275-276) as compounded of a shear whose plane is the yz plane and direction either the y or the z axis, together with a pure strain whose principal axes are Ox, a line bisecting the angle yOz, and a line perpendicular to both these. The elongation in the direction of Ox is zero, and those in the directions of the other principal axes may (278) be con- verted into elongations in the directions of the Oy and Oz axes. Similarly the rotation about Oy may be regarded as compounded of a shear whose plane is the xz plane and direction either the x or the z axis, together with elonga- tions in the x and z axes ; and the rotation about Oz, as compounded of a shear whose plane is the xy plane and direction either the x or the y axis, together with elongations in the x and y axes. Now these various component strains being all small may be applied in any order. The three component elongations in the direction of the x axis are thus equiva- lent to a single elongation in that direction, and similarly for the components in the y and z axes respectively. Hence a small strain may be resolved into three simple elongations e, /, g in the directions of the three rectangular axes Ox, Oy, Oz respectively, and three shears whose amounts may be represented by a, b, c, whose planes are the yz, xz, and xy planes respectively, and whose directions are those of either the y or z axis, either the x or z axis, and either the X or y axis, respectively. Any small strain is there- fore completely specified if the values of e, f, g, a, h, c, are given. 284. Heterogeneous Strains. — The elongations of a homogeneous strain we have seen to have the same values in the same directions- throughout the system. In general however, in the strains to which bodies are subjected, the elongations in a given direction are different at different parts of the system. Such strains are called hetero- geneous strains. If throughout the system the elonga- 184 KINEMATICS. [284 tions at points indefinitely near one another are indefinitely nearly the same, the strain is said to be continuous. The strains of bodies, except in cases of fracture, are usually continuous. The variation of the elongations from point to point being gradual in the continuous strain, they may be con- sidered constant throughout indefinitely small spheres, and the dimensions and position of the ellipsoids into which these spheres are changed may then be determined as in the case of homogeneous strain. The ellipsoids however will in this case be different for different points of the body, and that the strain may be known, the strain ellipsoid, or sufficient data for determining it, must be known for every point of the system. The consideration of strains of this kind requires mathematics of a higher order than readers of this work are supposed to have at command. PART II.— DYNAMICS. PART I L— DYNAMICS. CHAPTER I. THE LAWS OF MOTION. 285. So far we have dealt with the motion of bodies only by means of our mathematical generalizations. We have thus seen how to determine the displacements, velocities, paths, of bodies vl^hen their accelerations are known. Farther mathematics alone does not enable us to go. If we wish to inquire into the way in which bodies come to have accelerations and how they influence one another in their motions, we must obtain additional generalizations on which to build ; and we thus pass from the department of mathematical science into that of physical science. Dynamics is that branch of physical science which treats of the effect of the exertion of force upon bodies. The idea of force is ultimate. It is given us by sense. Like colour, taste, smell, it cannot be described. But we all have the idea, and when any one speaks of exerting force we all know what he means. What the organ of the sense is from which we have this idea physiologists have not definitely settled. It has been supposed to reside in the muscles, and has been consequently called 188 DYNAMICS. [285 the muscular sense. We do not require to know the seat of the sense, and may call it simply the sense of force. A sensation of force is not qualitative merely, but quantitative as well. We recognize ourselves in any case, not only as exerting force on a body, but as exerting a greater or a smaller force in a definite direction. Our power of perceiving the magnitudes of the forces we exert is not naturally strong, but it is susceptible of cultivation ; and it is the education of the sense of force which renders all manual skill attainable. 286. First Law of Motion. — Among our earliest gene- ralizations aie included those with regard to the effects of the exertion of force on bodies. These effects are very different in different circumstances ; but when examined they are found to be in all cases composed of changes of velocity and changes of form or volume. And as a change of the form or volume of a body is a change of the relative positions and therefore of the relative velocities of its constituent parts, we find the effect of the exer- tion of force on bodies to be in all cases change of velocity, or acceleration. Cases of equilibrium (323), i.e., cases in which a body, though acted upon by two or more forces, has an accel- eration zero, apparently form exceptions to this result. But in such cases, if the forces are allowed to act on the body successively, the accelerations produced are f6und to he such as would give a resultant acceleration zero were they to occur simultaneously. Thus, though the forces together produce an acceleration zero, each may be regarded nevertheless as producing its own acceleration. Having exerted forces on all bodies within our reach and found acceleration invariably produced, we are led to ex- pect this effect in all cases whether within the range of our 287] THE LAWS OF MOTION. 189 experiments or not, and to conclude that a force exerted on any body will produce an acceleration in it. We observe also, however, that many bodies move with acceleration when we are exerting no force upon them. Two billiard balls, for example, which impinge upon one another, have their velocities changed. A body which is simply let fall is found to fall with continually increasing speed. One body in short is found to be able to produce acceleration in others, it may be during contact, it may be even without contact. In such cases the effect pro- duced is the same as if we exerted force upon the bodies; and we therefore regard the action between them as of the same kind as our action on them when we are exert- ing force. We are thus led to conclude that the exertion of force on a body is invariably the antecedent of acceleration in it. We may express this result negatively by asserting that a body not acted on by force will experience no acceleration ; and it was in this form that Kepler, and afterwards Newton, enunciated it. Newton called it the first law of motion and expressed it thus — Every body continues in its state of rest or of uniform motion in a straight line except i/n so far as it may he compelled by im,pressed forces to change that state. The necessity of exerting force in order to produce acceleration in a body is said to be due to its inertia. 287. Second Law of Motion. — We have next to ask how the acceleration produced by a force depends upon the magnitude and the direction of the force which pro- duces it. The investigation of this dependence involves the measurement of force. For this purpose we may take as a provisional unit of force that exerted by a given spring when stretched a given amount. We may also prepare several exactly similar springs. Both our educated 190 DYNAMICS. [287 sense of force and our confidence in the " uniformity of nature" assure us that when extended by the given amount they exert the same force. Let us now act by means df these springs on, say, a curling-stone lying on a smooth horizontal surface of ice, taking care so to apply the springs that no appreciable rotation or change of form or volume may be produced. Preliminary trial shows that if, having started the stone, we exert no force upon it, it moves with a nearly uniform speed in a straight line over the surface of the ice. If now we attach one of our stretched springs to the stone and allow it to act on the stone during known intervals of time, keeping the spring stretched to the same extent and in a constant direction as the stone moves, we may, by noting the positions of the stone at a series of instants, determine the direction and magnitude of the acceleration which is produced. The same deter- mination may be made with two or with any number of springs attached and for longer or shorter periods of time. When that is done it is found (1) that in all cases the accelerations produced are uniform : (2) that the direction of the acceleration is always that of the force ; and (3) that the acceleration produced by a force in a given body is proportional to the force, double the force producing double the acceleration, three times the force three times the acceleration, and so on. The same result is obtained, whatever the kind or the condition of the body experi- mented with, whatever its initial velocity, and whatever component accelerations it may have besides that produced by the springs. The rough experiments sketched above apply only to forces whose direction and magnitude are the same during the whole time of their action. As we find, however, that the result does not depend upon the length of time during which the force acts, and as a variable force may be considered to consist of a succession of constant forces of different magnitudes or directions, each acting for a short time, we extend our results to all forces, uniform or 289] THE LAWS OF MOTION. ' 191 variable, and conclude that the magnitude of a variable force is at any instant proportional to the instantaneous acceleration of the body at that instant, and that their directions are the same. 288. If by F we indicate the magnitude of the force exerted on a given body, and by a that of the acceleration thereby produced, the third part of the above result may be expressed in symbols thus: Fcxa. Hence F/a = ei constant, i.e., the ratio of the force acting on a given body to the acceleration thereby produced, in it is constant. The value of this constant ratio will clearly depend upon the magnitudes of the units of force and acceleration. But with given units this ratio will have a iixed value for a given body, whatever its condition (as to tempera- ture, etc.) and whatever the circumstances of its motion. 289. We describe the constancy of the relation between the force acting on a body and the acceleration thereby produced by saying that the mass of the body is constant, the mass of a body being thus defined to be a quantity proportional to the constant ratio of the force acting on the body to the acceleration produced by it. If m denote the mass of the body we have thus : F/a 0:7)1 = km, where k is a constant, whose value for any given body will depend upon the magnitudes of the units of force, accele- ration, and mass which may be employed. The term mass is clearly the scientific equivalent of the popular term m,assiveness. We speak of a body as being massive when we require to exert a great force upon it in order to produce a small change in its velocity. Thus an iron gate is said to be more massive than a wooden gate of the same dimensions, because it takes a greater force to produce in it a given angular acceleration than in the wooden gate, though the friction and other opposing forces may be the same in both cases. The greater'the force required to produce a given acceleration, 192 DYNAMICS. [289 and the smaller the acceleration produced by a given force, in other words, the greater the value of the ratio Fja, the greater do we consider the massiveness to be. 290. The reader should carefully note that the mass of a body is quite a different thing from its weight Its weight is the force with which it is drawn vertically downwards in the neighbourhood of the earth, and will have different values at different parts of, and at different distances from, the earth's surface. Its mass is not a force at all, but, as we have seen, the value of a certain ratio which is the same everywhere. At any one place all bodies fall with the same accelera- tion. Now the acceleration with which a body falls is that produced in it by its weight. Let w and w' be the weights of two bodies, g the acceleration with which they fall at any given place, then their masses are proportional (289) to wig and w'jg respectively. If, then, m and m' are their masses, we have m : mf^wlg : w'lg = w : w'. Hence the masses of bodies are proportional to their weights at the same place, and the ratio of the masses of two bodies is the same as that of their weights. For this reason the term weight is frequently employed not only with its primary signification given above, but also as synonymous with mass. As this double meaning of the term leads to confusion, we shall restrict it to its primary signification. 291. The mass of a body is by many writers defined as the quantity of matter which it contains. As we do not know what matter is, still less how to measure it, this phrase (for it is thus a mere phrase) must then itself be defined ; and such writers define it more or less directly as being proportional to the ratio of the force acting on the body to the acceleration thereby produced. This 294] THE LAWS OF MOTION. 193 mode of definition is clearly the same as that employed above, except that a useless intermediate term is in- troduced. The phrase quantity of inertia has been similarly used as an intermediate term. 292. By 117, the acceleration of a body is equal to the rate of change of its component velocity in the direction of the acceleration; and, by the second part of the above experimental result (287), the acceleration is in the direc- tion of the force. Hence, if v and v' be the initial and final values of the body's component velocity in the direc- tion of the force (F) during a time t (which, if F is- variable, must be small), we have „ , v' — v -.nnv' — viv Ji = Icm — - — = k V 293. The product of the mass of a body into its velocity is called its momentum* The product of its mass into the component of its velocity in a given direction is called its momentum in that direction. Hence the result of 292 may be thus expressed : When a body is acted on by a force its momentum in the direc- tion of the force changes at a rate which is proportional to the force. 294. From the expression of 292 we obtain Ft=k(7nv'—m.v). The product Ft is called the impulse of the force during the time t. Hence we obtain Newton's expression of the second law of motion — * The momentum of a body is often defined as its " quantity of motion," the quantity of motion being then defined as the product of mass into velocity — ^another case of the introduction of a useless intermediate term. N 194 DYNAMICS. [294 Change of momentum, is proportional to the impulse of the imioressed force and takes place in its direction. 295. It follows from the second law of motion that a finite force can produce in a body only a finite change of momentum and therefore a finite acceleration. As we find no infinite forces in nature it follows that the speeds, velocities, and accelerations of bodies cannot have infinite values, and that the directions of their paths, velocities, accelerations cannot undergo abrupt changes. 296. Measurement of Force and Mass. — The second law gives us at once a mode of measuring both force and mass. If forces F, F act on masses* m and m' and produce accelerations a and a' respectively, we have F:=kma and F' — km'a''. To compare two forces, allow them to act successively on the same mass and note the accelerations. We have then F:F' = a:a'. To compare the masses of two bodies, let equal forces act on them and note the accelerations. We then have m : m' = a' : a. 297. Having thus found modes of measurement, we must next choose units. Either both may be chosen arbitrarily, or one being so chosen the other may be derived. If both be chosen arbitrarily, the constant k in the above equation will usually have an inconvenient value. Thus let the weight of the body called a pound be chosen as unit of force, the mass of the pound as unit of mass, and the foot and second as units of length and time. We know that the weight of any body produces in it an acceleration of about 32'2 ft.-per-sec. per sec. Hence our unit of force will produce this acceleration in * This is a shortened expression for : bodies whose masses are m and m'. 299] THE LAAVS OF MOTION. 195 our unit of mass. Substituting these values, F= m = 1 and a = 32-2, in the equation F=Jcma, we find /(; = 1/32-2. 298. If, however, either the unit of force only or that of mass only be chosen arbitrarily, the other may be so chosen as to give k the convenient value unity, in which case the symbolic statement of the second law becomes F=7na. The constant k will have the value unity if F= m = a = l. Hence, if the unit of force be arbitrarily chosen, the unit of mass which will make k=l is that mass in which the unit of force will produce unit of acceleration. And if the unit of mass be arbitrarily chosen, the unit of force which will make A; = 1 is that force which wilL produce in unit of mass unit of acceleration, According as the unit of force or that of mass is chosen arbitrarily do we obtain one or other of two groups of systems of units. (1) Unit of Force chosen Arbitrarily. — We may select as unit of force any force we please ; but practically the weight of some body is always selected. The bodies in general use are the pound, which is a piece of platinum kept in the Standards Office in London, the kilogramme, another piece of platinum, kept in the Palais des Archives in Paris, and multiples or submultiples of these. These units and all units derived from them are called gravitational units because their magnitudes depend upon the attraction of the earth. As the weight of a given body has different values at different points on the earth's surface, gravitational units are not constant. They are sufficiently constant, however, for many non-scientific purposes, and are very extensively used. 299. Corresponding to each unit of force we have a system of gravitational units, as follows : — Foot-pound-second {F.P.S.) Gravitational System. — The unit of force is the weight of the pound. 196 DYNAMICS. [299 The unit of mass is that mass in which a force equal to the weight of a pound will produce an acceleration of 1 ft.-per-sec. per sec. As the weight of the pound produces in the pound an acceleration of g (about 32-2) ffc.-sec. units, it will produce one of 1 ft.-sec. unit in a body whose mass is g lbs. Hence the unit of mass of this system is a mass of about 32'2 lbs. Metre-Icilogramme-second {M.K.8.) GravitationalSystem. — The unit of force is the weight of the kilogramme. The unit of mass is that mass in which a force equal to the weight of 1 kilogramme will produce an acceleration of 1 m.-sec. unit. It may be shown as above that the unit of mass of this system is a mass of 9 'SI kilogrammes. Other gravitational systems, based on other simple units of length, time, and force the reader will readily construct for himself The two given above are those most- generally used and are sufficient for purposes of illustration. 300. Dimensions of Derived Unit of Mass. — The mag- nitude of the unit of mass, derived as above, will depend upon the magnitudes of the simple units of force, length, and time. With the notation of 15 we have (289 and 15) XT r, 1 1 1 [F] [m] [a] Hence [m] a [i'']/[a] ; i.e., the magnitude of the unit of mass is directly pro- portional to the magnitude of the unit of force, and inversely proportional to that of the unit of acceleration. Hence (111 and 57) the dimensions of the derived unit of mass are given by the equation [m]o:[F][L]-\Tf. This equation may be employed in the solution of problems 302] THE LAWS OF MOTION. 197 in the same way as the similar equations in the case of speed and rate of change of speed (47-50, 57-59). 301. (2) Unit of Mass chosen Arbitrarily. — The units ordinarily selected are the mass of the pound and that of the gramme (a body whose mass is 1/lOOOth of that of the kilogramme), with their multiples and submultiples. The English hundredweight is equal to 112 pounds, the American hundredweight to 100 pounds. The ton is equal to 20 cwts. The decagramme and hectogramme are 10 and 100 grammes respectively. The decigramme, centigramme, and milligramme are the tenth, hundredth, and thousandth parts respectively of a gramme. The following are approximately the relative magni- tudes of these units : lib. = 453-59 grm. I 1 grm. = 0-0022046 lb. 1 ton (English) = 1016-05 kgr. I 1 kgr. =0-0009842 ton. As the mass of a body is constant, these units are con- stant ; and the magnitudes of the units derived from them depend therefore only on the magnitudes of the simple units involved in them. Hence they are called absolute units to express their independence of all such varying quantities as terrestrial attraction. 302. Corresponding to each unit of mass selected, we have a system of absolute units. The following are im- portant systems : — F.P.S. Absolute System. — The unit of mass is the mass of the pound. The unit of force is therefore that force which will produce in the pound an acceleration of 1 ft.-sec. unit. This force is called the poundal. As the weight of 1 lb. produces in it an acceleration of 32-2 ft.-sec. units, it is clear that the poundal is equal to the l/32-2th part of 198 DYNAMICS. [302 the weight of a pound, i.e., to about the weight of half an ounce. Gentimetre-gramme-second (G.G.S.) Absolute System. — The unit of mass is the mass of the gramme. The unit of force is therefore that force which will produce in 1 gramme an acceleration of 1 cm.-per-sec. per sec. This force is called the dyne. It will be clear that the dyne is equal to about l/981th of the weight of a gramme. 303. Dimensions of Derived Unit of Force. — From the equations of 300 we obtain at once [F] a [J^][«], i.e., the magnitude of the derived unit of force is directly propor- tional both to the magnitude of the unit of mass and to that of the unit of acceleration. Hence (111 and 57) the dimensions of the derived unit of force are given by the equation [F]o:[M-\[L][T]-^ This equation may be employed in the solution of prob- lems in the same way as the corresponding equations in the case of speed and rate of change of speed (47-50, 57-59). 304 Density. — -The Tnean density of a body is the quotient of its mass by its volume. The density at a given point of a body is the quotient of the mass by the volume of an indefinitely small portion of the body surrounding the given point. If the density of a body is the same at all its points, it is said to be homogeneous or of uniform density. In general the density of a body varies from point to point; the body 'is heterogeneous. The density of a substance in a given state is the quotient of the mass by the volume of any portion of the substance in that state. 304] THE LAWS OF MOTION. 199 If d be the density of a body, m its mass, and v its volume, we have by definition d = mlv. Hence the dimensions of density are given by the expression. [Z»]cx[il/][F]-ia[ilf][i;]-^. The unit of mass of an absolute system of units, in- stead of being arbitrarily selected as in 301, may be defined to be the mass of unit volume of some standard substance whose density in terms of those units is there- fore unity. This amounts to choosing a unit of density arbitrarily and deriving from it the unit of mass. The French unit, the gramme, was intended to be the mass of 1 cubic centimetre of water at its temperature of maxi- mum density (about 4°C.). But though it may for most practical purposes be considered to liave that mass, it has not rigorously ; and thus the gramme must be considered to be an arbitrarily chosen unit. The great advantage of deriving the unit of mass from an arbitrarily chosen unit of density is that the density of any given substance is in that case equal to the ratio of the masses (and therefore (290) of the weights) of equal volumes of the given sub- stance and of the standard substance, or to what is called the specific gravity of the given substance. If the unit of mass is not thus derived, the density of a given sub- stance is obviously equal to the product of its specific gravity into the density of the standard substance (usu- ally water) by reference to which its specific gravity is expressed. The mean linear density of a body whose length is great relatively to its other dimensions is the quotient of its mass by its length. The dimensions of linear density are thus [iif][i]-i. The mean surface density of a thin body is the quo- tient of its mass by the area of one of its surfaces. The dimensions of surface density are thus [ilf][Z]"^. :^00 DYXAMICS. [305 305. Examples. (1) Two forces produce in two masses accelerations of 25 and 30 units respectively. Show that, if the masses are equal, the forces are as 5 to 6, and that, if the forces are equal, the masses are as 6 to 5. (2) Forces of 20 and 30 units acting on two masses produce accel- erations of 40 and 50 units respectively. Show that the masses are as 10 : 12. (3) Show that 1 poundal is equivalent to 13,825 dynes. (4) Prove that the weight of 1 lb. is equal to 4'45 x 10^ dynes approximately. (5) Show that the value of 1 dyne, expressed in terms of the weight of 1 ton,* is 1003 x 10~^^ approximately. (6) Compare the values of the mass of a body as expressed in gravitational units of the ft. -lb. -sec. and yd.-ton-min. systems. Axis. 2,688,000 : 1. (7) The value of a force expressed in dynes has to be expressed in absolute units of the metre-kilogramme-minute system. By what number must it be multiplied ? Ans. 0-036. (8) Reduce 20 poundals to absolute units of the yd.-cwt.-min. system. Ans. 214^. (9) The unit of mass being a mass of 10 lbs., the unit of time 1 min., and the unit of length 1 yd., compare the derived unit of force with the poundal. Ans. As 1 : 120. (10) With 20 lbs. and 40 sec. as units of mass and time respec- tively, find the unit of length that the derived unit of force may be equal to the weight of 1 lb. at a place where gr = 32-2, ft. -sec. units. Ans. 2,576 ft. (11) The unit of acceleration being 6 ft.-per-sec. per sec, find (a) * The ton used in these Examples is the English ton of 2,240 lbs. 306] THE LAWS OF MOTION. 201 the vmit of mass when the derived unit of force is equal to the weight of 20 lbs., and (6) the unit of force when the derived unit of mass is a mass of 20 lbs. Aus. (a) 107^ lbs., (6) 3'7... pounds- weight. (12) Tlie unit of velocity being 20 cm. per sec, the unit of mass 15 grammes, and the derived unit of force the weight of a kilo- gramme, find the unit of time. Ans. 1/3270 sec. (13) The density of water is about 1,000 oz. per cub. ft. Show that it is also about 1687'5 lbs. per cub. yd., and about I'OOl grm. per cub. cm. (14) The masses and radii of two spheres are as 1 : 2. Show that their densities are as 4 : 1. (15) Given that the diameter of the earth is 1*275 x 10^ cm. and its density 5 '67 times as great as that of water, show that its mass is about 6"15 x 10^' grammes. (16) The unit of density being that of water, and the imits of time and mass 1 min. and 1 cwt. respectively, find the magnitude of the derived unit of forces Ans. 0'0378 pouhdals nearly. (17) The nvimber of seconds in the unit of time being equal to the number of feet in the unit of length, the unit of force being the weight of 750 lbs. {c/ = 32 ft. -sec. -units), and a cub. ft. of the stand- ard substance having a mass of 13,500 oz., find the unit of time. Ans. 5^ see. 306. Force is usually exerted upon some portion of the bounding surface of a body and acts therefore across an area. In specifying the magnitude of a force we may do so, as above, without reference to the area across which it acts, or we may divide its total magnitude by this area and thus express its magnitude per unit of area or its intensity. When we do so we usually describe the force as a pressure, a tension, a stress, though these terms have another not inconsistent connotation (307) as well. Thus a force of F poundals which is transmitted by a 202 DYNAMICS. [306 string of s square feet section would be a tension of Fjs poundals per square foot, and the total force exerted through the string would of course be determined by- multiplying this quantity by the area across which the force is acting. It should be noted however that forces are not always measured in this way when they are spoken of as pres- sures, tensions, etc. Unless either it is stated, or the context shows, that they are so measured, they should always be assumed to be measured without reference to the area across which they act. 307. Third Law of Motion. — If we now return to the examination of cases in which bodies are acted on by forces, we find that forces always act between pairs of bodies, never on single bodies alone. I push a body with my hand ; the body is urged forwards ; the forward motion of my hand is lessened. Both the body pushed and the hand are acted on by force. A horse draws a carriage ; the carriage is pulled forwards ; the horse is pulled backwai'ds and does not move forwards so fast as he would otherwise do with the muscular exertion he is putting forth. To investigate this mutual action more thoroughly we may take two of our curling stones and project them, without rotation, on the ice so as to make them collide, noting the direction and magnitude of their velocities before and after collision. Let OA, OB and Oa, Oh be drawn representing the velocities before and after collision, of the respective stones. Then AB, ah °^^\ \ will represent the respective in- tegral accelerations. They will be found in all cases to be parallel and in opposite directions. If the ^B stones used were of equal mass, they will be found equal. If not, it will be found that 308] THE LAWS OF MOTION. 203 if If, m are the masses of the respective stones, then M.AB = in.ab. Now the product of the mass of a body (294 and 117) into its integral acceleration measures the impulse of aforce. Hence the stones during collision have experienced equal impulses in opposite directions. Other simple experiinents give the same result and suggest a third law of motion, which Newton enunciated as follows : — To every action there is always an equal and contrary reaction; or the Tnutual actions of any two bodies are always equal and oppositely directed. The exertion of a force upon one body is thus only a one-sided view of a more complex phenomenon, viz., the simultaneous exertion of equal and opposite forces upon two bodies. When we are thinking of a force as acting not on one body, but between two bodies, we call it a stress. When the stress is such as to make the bodies move towards one another it is called an attraction or a tension ; when its effect is to increase their distance it is called a repulsion or a pressure (see 306). 308. The experiments which we have sketched above as leading up to the laws of motion are of necessity rough, and are quite insufficient to demonstrate the truth of these laws. They merely serve to suggest them. They apply moreover only to bodies of so large size that experi- ments may be made with them. Now, in studying the motion of bodies, we are forced to regard them as consist- ing of indefinitely small parts called particles, and the extension of the above laws to indefinitely small bodies we cannot prove to be warranted. Hence the laws of motion as employed in Dynamics are simply hypotheses suggested by rough experiments, and their accuracy must be tested by the agreement of deductions made from them with observed fact. The body of deductions from these hypotheses constitutes the theoretical portion of Dyna- 204 DYxN^AMICS. [308 mics. Many of the deductions which will be made in subsequent chapters may be tested by experiment. But for the most part we shall have to do with ideal bodies and our deductions will be only approximately true of real bodies. The most satisfactory tests of the laws of motion are furnished by astronomical calculations. These laws are assumed in the determination of the positions of the moon and other heavenly bodies at given times, of the times of occurrence of eclipses, of the dates of the return of comets, etc., and the precision with which such predictions are fulfilled is well known. The assumption of the truth of these laws has even led to the discovery of heavenly bodies not previously known to exist. In short, they have stood such rigorous tests that not the slightest doubt is now entertained of their truth. And we may make deductions from them, even in cases in which veri- fication by experiment is impossible, with full confidence that, if our mode of deduction is correct, the result will be true. 309.' The three laws of motion adopted by Newton as the fundamental hypotheses of Theoretical Dynamics have not been universally adopted. Some authors sub- stitute for Newton's second law one first enunciated by Galileo, and therefore bearing his name, which has been expressed by Thomson and Tait in, the following words : — When any forces whatever act on a body, then whether the body be originally at rest or moving with any velocity in any direction, each force produces in the body the exact change of motion vjhich it would have produced if it had acted singly on the body originally at rest. As Newton's second law is perfectly general it includes Galileo's law. Those who make Galileo's law the second law of motion must deduce Newton's law from it. This deduction is made as follows : — Let two forces each equal to F act in the same direction on a particle. Then if a is the acceleration which each would produce if it acted 309] THE LAWS OF MOTION. 205 singly, 2a is by Galileo's law the acceleration produced when they act together. Similarly 3a is that which would be produced by three forces each of the magnitude F and in the same direction; na that which would be produced by n such forces. And hence the acceleration produced in the particle is proportional to the force. It will be noticed that the assumption is here made that n. equal forces in the same direction are equivalent to a force of n times the magnitude, a special case of the Law of the Composition of Forces (313 and 86, iii). We made the same assumption in discussing the rough ex- periments used to suggest the fundamental hypotheses. But such an assumption made after the choice of three fundamental hypotheses is equivalent to the introduction of a fourth. " For D'Alembert's "Principle," which is extensively employed instead of Newton's second and third laws in the solution of problems on the motion of extended bodies, see 417. 206 DYNAMICS [310 CHAPTER II. DYNAMICS OF A PARTICLE. 310. We shall first restrict ourselves to the considera- tion of force as affecting the translation of bodies. Now the translation of an indefinitely small body difiers in no respect from that of a body of finite size, while rotation is possible only for bodies of finite size. Hence in con- sidering the effect of force on the translation of bodies, in order to exclude the possibility of its having rotational effects, we imagine the bodies acted upon to be indefin- itely small. Such bodies are called material points or particles, or, if they form parts of a continuous body, elements. 311. A force which we imagine as acting on a particle is of course one whose place of application is a point. The lines of action of forces which act on the same par- ticle must intersect in the position of the particle. A force is completely specified if its place of application, its direction, and its magnitude are given. If it act on a particle, its place of application is the position of the particle itself In that case therefore it is completely specified if its direction and magnitude are given. It may therefore be completely represented by any straight line of the proper length and direction. 312. Composition and Resolution of Forces. — Forces 314] OF A PARTICLE. 207 which act simultaneous!}' on a particle are called com- ponent forces. The resultant of any number of compon- ent forces is that single force by which' the same resultant acceleration would be produced. 313. Let OA and OB represent two component forces. Since these forces act upon the same particle, OA and OB represent also the accelerations they would produce act- ing singly. Now OA and OB representing the compon- ent accelerations, OG the diagonal '^ of the parallelogram AB represents the resultant acceleration (116). And OA, OB, OG representing acr celerations of the same particle are proportional to the forces which would produce them. Hence OG represents the resultant force. Forces acting on a particle therefore are to be com- pounded according to the parallelogram law after the manner of the displacements, or velocities, or accelerations of a point. We have therefore propositions called the parallelogram, the triangle, and the polygon of forces, the same in form as those enunciated under velocities (98). Hence forces are to be resolved in the same manner as displacements, or velocities, or accelerations. Hence all the consequences of the parallelogram law, as deduced in the case of the displacements of a point, apply also to forces acting on a particle, and the formulae of 85-90 are applicable to component forces, if the symbols representing displacements are taken to represent forces. 314. Examples. (1) The resultant of forces of 7, 1, 1, 3 units represented in direction by lines drawn from one angle of a regular pentagon towards the other angles, taken in order, is ;^7l. (2) P and Q are two component forces whose resultant is R. S 208 DYNAMICS f'^^* is the resultant of R and P. Show that if P ana Q be inclined to each other at a right angle, and if Q=2P, then S=2P ^/2. (3) Component forces P, Q, R are represented in direction by the sides of an equilateral triangle taken the same way rovmd. Find the magnitude of their resultant. Ans. (P^+Q^ + R^-QR-PR- PQ)'-. (4) Three component forces are represented by lines drawn from the angular points of a triangle to the points of bisection of the opposite sides ; show that their resultant is zero. (5) Three component forces are represented in direction by lines drawn from the angular points A,B, G oi a, triangle to the points of bisection of the opposite sides, and have magnitu.des equal to the cosines oi A, B, and C respectively. Prove that their resultant is equal to (1 — 8 cos A cos B cos C)*. (6) The centre of the circumscribed circle of a triangle ABC is 0, and the intersection of the perpendiculars from angular points ou opposite sides is P. Prove that the resultant of forces represented in magnitude and direction by OA, OB, 00 will be represented by OP. (7) Three forces are represented by the sides AB, AC, BC of a triangle. Show that the resultant has the direction AG and is represented in magnitude by 24C. (8) ABCD is a parallelogram. Prom AB, AE is cut off equal to one-third of AB. Prove that the resultant of forces represented by AG and 24/) is equal to three times the resultant of forces repre- sented ^yy AD and AE. (9) li AB represent the resultant of two forces AC and AD, and if the angle CAD be given, show that the extremities of the lines representing the two forces {AC and AD) will lie on two circles, which, if the given angle be a right angle, will be coincident. Also show that, if the given angle be obtuse, each force has its maximum value when the other is perpendicular to the resultant. (10) A particle is acted upon by two forces represented by the lines joining the particle to two given points. Show that, if the 3151 OF A PAETICLE. 209 particle be made to describe any plane curve, the end of the straight line representing the resultant of the above forces will de- scribe an equal and similar curve. (11) Give a geometrical construction for resolving the force represented by the diagonal BB of a square ABC3 into three forces, each represented in magnitude by a side of the square and one represented by DC in direction. Ans. Upon EG describe an equilateral triangle BGE. The required components are represented by DC, CE, EB. 315. Attractions. — An important case of the composi- tion of forces is the determination of the attractive force exerted on a particle by an extended body, the law of the attraction being that of gravitation, viz., that the force exerted between two particles is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In such cases the attraction on the particle is the resultant of compon- ent attractions exerted on it by the elements into which the attracting body may be divided. Its determination requires usually the application of the Integi-al Calculus. But in a few important cases it may be found by elemen- tary methods. If m, mf are the masses of two particles, d their dis- tance, and F their mutual attraction, the law of gravita- tional attraction ia expressed by the equation P mm' _ inm^ where Ic is a constant. The value of k, when units of force, mass, and distance already chosen are employed, may readily be determined from our knowledge of the dimensions and density of the earth and of the value of g. We may give it the more convenient value unity, however, by choosing a new unit of either mass or force ; for example, by taking as unit of mass a mass which attracts an equal mass at unit distance with unit force o 210 DYNAMICS [315 This is called fJie astronomical unit of mass. We shall use it in the following examples. 316. Examples. (1) Find the attraction of a uniform thin circtdar disc on a particle placed at any point on a line through its centre and perpendicular to its plane. Let A£ he the disc, its centre, OP a line through perpen- dicular to AB. Let P be the position of the particle, and m its mass. Consider first the component attraction exerted by the element (i.e., small portion) of the disc surrounding any point D, the line DP having the length r, and its inclination to CP being 6 radians. Let the element at I) subtend at P the small solid angle a (solid radians, 22). DP's inclination to CP being 0, the surface of the element at D is inclined to a surface normal to DP at the same angle. The element at D being indefinitely small, the cone of which it is a section is one of indefinitely small angle. Hence the ortho- gonal section of this cone at D is the projection of the element at D on a plane inclined B to the plane of the element. If therefore A is the area of the element, A cos S is the area of the orthogonal ', section. But a being the solid angle subtended at P by this section, its area must be ur\ Hence the area of the element at I) 316] OF A PARTICLE. 211 is ar^/cos0. Let p be the surface density of the disc. Then the mass of the element is ar^p/coa 8. Hence the force exerted by the element on the particle at P is in the direction of PD and of the magnitude — ^x»i cos t> up r' cosS This force has one component in the direction PC of the magnitude — i-jjmx cos » = worn, cos 8 "^ ' and another in the direction CD of the magnitude apm tan 8. If DO be produced to D', and CD' made equal to CD, the element of area A at D' will exert on the particle at P a force whose components in the directions PC, CD' are of the same magnitudes as the components above determined. The components CD and CD' therefore neutralize each other, and hence the only effective component of the attraction of the element at D is that perpen- dicular to the disc, whose magnitude is apm. Now the same is true of all the elements into which the disc may be divided. Hence the resultant attraction will be perpendicular to the disc, and equal to the sum of the effective components of magnitude upm, for all the elements of the disc, i.e., since pm is con- stant, to the product oipm into the solid angle subtended at P by the whole disc. If a is the radius of the disc and h the distance of P from it, the area of the segment of the sphere whose centre is P and radius PA or ^IW+ofi is 2ir >Jh? + a\ xJh^ +a^-hy Hence the solid angle subtended at P by the disc is 2T(l-h'/ s/h^ + a^); and therefore the attraction of the disc on the particle at /' is 27rp«i(l-A/\/A2+a2). If the disc be of indefinitely great extent (a=oo), or if the particle be indefinitely near it (A=0), the attraction becomes 2Trpm. (2) Find the attraction of a thin circular ring of gravitating matter of uniform linear density p and radius a on a particle of unit mass on its axis, at a distance h from its centre. Ans. 2Trpah/(a^+h^)i. (3) All parallel slices of equal thickness of a homogeneous cone 212 BYNAMICS [316 of gravitating matter exert the same attraction on a particle at its vertex. [First prove for a cone of indefinitely small angle and then extend to one of finite angle.] (4) A right cone of gravitating matter of semi-vertical angle o, length I, and uniform density p, attracts a particle of unit mass at its vertex with a force 2irp?(l — cos a). (5) Show that the attraction of a thin spherical shell of uniform thickness and density on a particle inside it is zero. Let P be the position of the particle, and A any point in the spherical surface. Join AP and produce it to meet the surface iii A . Consider a small element of the shell at A . If, from points in its boundary, lines be drawn through P, their end points will mark off a corresponding element about A'. These corresponding elements are both sections of the same cone, and as they coincide with the tangent planes at A and A', we know from the geometry of the sphere that they are equally inclined to the line AA'. Hence their areas, and therefore their masses, are directly proportional to the squares of their distances from the vertex P. But their attractions on a particle at P are directly proportional to their masses and inversely proportional to the squares of their distances from P. Hence their attractions have the same magnitude. And they have opposite directions. Hence the pair of elements about A and A' exert no resultant attraction on the particle at P. But the whole shell may be divided into such pairs of elements. Hence the resultant attraction of the shell on a particle at P is zero. 316] OF A PAETICLE. 213 Clearly the same would told for a shell of any thickness, provided it is either uniform in thickness and density, or uniform in thick- ness and symmetrical about the centre as to density. (6) The attraction of a uniform thin spherical shell on a particle placed outside it is the same as if the whole mass were condensed at the centre. Let P be the position of the particle and C the centre of the spherical shell. Join GP, meeting the shell in D, and divide it ./I P at B, so that CB : GD=GB : CP. Take any point A in the shell. Join AB and produce it to meet the shell in .4'. Join GA, GA', PA, PA'.* Since GB ■:,GA = GA : GP, the triangles GAB and CPA are similar, and the angle GAB equal to the angle GPA. Similarly, the angle GA'B is equal to the angle GPA'. Hence also the angle GPA is equal to the angle GPA'. If straight lines be drawn from the boundary of a small element surrounding A, through B, their end points will mark out a corre- sponding element about A'. These elements are sections of a cone whose vertex is B and solid angle u (solid radians); and their common inclination to an orthogonal section of the cone, is the angle GAB. Hence, as in Ex. (1), their attractions on a particle of mass m at P are respectively in the directions PA and PA', and of the magnitudes mpia . AB^ , mpa . A'& AP^. cos GAB ^^°- A'PKcosGAB' p being the surface density of the shell. Now AB_GA_GA' _A'B AP GP GP A'P * PA and PA' are not tangents, as would appear from the figure. 214 _ DYNAMICS [316 Hence the magnitudes of the above attractions are equal, and they are equally inclined to PC. Hence the direction of their resultant is PC, and its magnitude is 2mpa . GA^jCPK Now the whole sphere may be divided by lines through B into pairs of corresponding elements similar to the above, the resultant attraction of each pair being in the direction PC, and equal to the product of its solid angle into the constant 2«ip . CA^IGP\ Hence the resultant attrac- tion of the spherical shell is in the direction PC, and is equal to the product of this constant into the sum of the solid angles of all the pairs of elements into which the sphere may be divided, which is clearly the solid angle subtended at its centre by a hemisphere. Its magnitude is therefore ^mnp . CA^jCP^ which is equal to the product of. the masses of the particle and shell divided by the square of the distance of the particle from the centre of the shell. Hence the shell attracts the particle as if its mass were condensed at its centre. Hence also a spherical shell of any thickness, and a sphere also, attract particles outside them as if their masses were condensed at their centres, provided their density is symmetrically distributed about their centres. (7) Show that the attraction of a homogeneous sphere on a particle of unit mass inside its bounding surface is directly pro- portional to its distance from the centre. (8) Assuming the earth to be a homogeneous sphere, compare its attraction on a given mass at a distance from its centre equal to one- half its radius with the attraction when the given mass is at a distance equal to twice the radius. Ans. As 1 : 8. (9) rind in dynes the attraction of two homogeneous spheres, each of 100 kgr. mass, with their centres 1 metre apart. [Data. — Quadrant of earth, assumed spherical = 10' cm.; mean density of ■ earth=5'67 grms. per cu. cm.; ^'=981 cm. -sec. units.] Ans. 0-0649 nearly. (10) A pendulum beating seconds at the surface of the earth is taken (a) up a mountain 1,400 ft. high, and (6) down a mine of 317] OF A PARTICLE. 215 equal depth. Find its loss or gain per day in each case, assuming the earth to be a uniform sphere of 21,000,000 ft. radius. Ans. (a) loss of 5-76 sec. ; (6) loss of 2-88 sec. (11) Show that, if a pendulum oscillates in the same time at the top of a hill as at the bottom of a mine, the depth of the mine is very nearly twice the height of the hill. (12) Show that the astronomical unit of mass of the C.G.S. system is 3,928 grammes (mass of earth=6"14x 10^' grms. ; radius of earth = e'37 x 10^ cm. ; 5^=981 cm. -sec. units). (13) Find in C.G.S. units the value of h in the formula F=h-j^- Ans. e-48 X 10-8. (14) Compare with the dyne the unit of force employed when it is stated that the attraction between two masses of m and m' grms. at a distance d cm. has the value mrti'ld!^. Ans. Unit employed = 6'48x 10~8 dynes. 317. Equations of Motion. — The second law of motion provides us with an equation, F—7na, by means of which any one of the three quantities, force acting, mass of particle acted upon, and acceleration produced, may be determined, if the other two are given. These two being expressed in the units of a derived system, the third determined by the above equation will be expressed in terms of the unit of the same system. The acceleration of a particle being determined, the character of its motion is known from Kinematics. Hence the above equation is called the equation of motion of a particle. If a particle is given as acted upon ,by several forces, the resultant force may be found as in 313, or, the com- ponent accelerations having been found by the equation of motion, the resultant acceleration may be determined by 116. It follows that if F^, F^, etc., are the components in a 216 DYNAMICS [317 given direction of the forces acting on a particle, and a its component acceleration in that direction, 2i''= ma. 318. It is in many cases found convenient in describing the forces acting on particles, to specify not their mag- nitudes and directions, but the magnitudes of their components in three given rectangular directions. Ex- pressions for the component accelerations in these three directions may be at once written down. Thus, if X^, X^, etc., Fj, Fj, etc., Z^, Z^, etc., be the components in the x, y, z axes of the forces acting on the particle, if ax, ay, a« be the component accelerations of the particle in these direc- tions, and X, y, z the co-ordinates of the particle at the instant under consideration, we have (317 and 118) ax=x=^(2X)lm, ay=y = (27)/m, a^=z = {I,Z)/m. 319. It is frequently convenient to express the equation of motion in terms of the impulse of the force rather than of the force itself. If the force (F) is constant, its impulse ($) during a time t is (294) Ft, and we have from 294 and 298 ^=Ft=mv'—7nv, where v' and v are the final and initial values of the component velocity in the direction of the impulse, and m is the mass of the particle acted upon. If the force is variable, it may be considered constant during indefinitely short intervals of time. Let t be divided into n such short intervals t^, t^, etc., t^. Let the components, in any given direction, of the force (supposed constant) during these intervals be F^, F^, etc., Fn ; and let the components, in the given direction, of the initial velocity and of the velocities at the ends of the above intervals be v, v^, v^, etc., v' respectively. Then 320] OF A PAKTICLE. 217 etc., FJn = '>nv' — mvn - \. The impulse ($) of the force, in the given direction, is the sum of the impulses Ff^, F^t^, etc. Hence $ = 'EFt = mv' — mv. This form of the equation of motion is especially con- venient when the force is one whose magnitude is great and time of action small, as in cases of impact, collision, explosion, etc. Such forces are therefore frequently called impulsive forces. It will be obvious however that the above form of the equation of motion is applic- able generally, and that the restriction of the term impulsive force to one whose time of action is short is merely a matter of convenience.* 320. Examples. (1) A constant force of 20 poundals acts on a mass of 10 lbs. Find (a) the acceleration, (6) the displacement in 5 sec, the initial velocity having been 4 ft. per sec. in the same direction as the acceleration ; (c) the velocity at the end of the sa«ie time, the initial velocity of 4 ft. per sec. having been inclined 60° to the direc- tion of the force. Ans. (a) 2 ft. per sec. per sec, (6) 45 ft. in the direction of the initial velocity, (c) 2 ,/39 ft. per sec, inclined sin-\5/2 >/13) to the direction of the initial velocity. * The term impulse is unfortunately sometimes applied to these short-lived forces. But it should be restricted to the sense in which it is used above. Otherwise it becomes necessary to speak of the impulse of an impulse. The term impulsive force is some- times used to denote the impulse of a short-lived force. But this use of the term leads to confusion and should be avoided. 218 DYNAMICS [320 (2) An unknown force produces in a body of 50 lbs. mass an acceleration of 12-5 f t.-sec. units. Express the force (a) in poundals, (6) in terms of the weight of a pound. Ans. (a) 625, (6)19-4.... (3) A uniform force of 200 dynes changes the velocity of a body moving in a straight line from 250 to 300 metres per sec. in 1 minute. Find the mass of the body. Ans. 2 '4 grammes. (4) What acceleration will be produced in a mass of 20 lbs. by a force equal to the weight of 50 lbs. ? Ans. bgl% (5) How long must a force of 14 lbs. -weight act on a mass of 1,000 tons to move it from rest through 1 inch ? Ans. 28'8 sees, nearly. (6) A spring balance (an instrument for measuring force, being a spring provided with a scale to show the amount of its elongation) is graduated for a place where 5' = 32'2 and indicates 1 '6 pounds- weight at a place where 5' =32. Find the correct value of the force thus measured. Ans. 1'61 pounds- weight. (7) Find the force which must be exerted by a man in an elevator on a body of 1 lb. mass which he holds in his hand, to prevent its moving relatively to the elevator when the elevator is moving (a) with uniform^ speed, (6) with an upward acceleration of 8 ft. per sec, (c) with a downward acceleration of 8 ft. per sec, {d) with a down- ward acceleration of 33 ft. per sec. Ans. {a) 32'2 poundals upwards, (6) 40'2 poundals upwards, (0) 24'2 poundals upwards, {d) 0'8 poundals downwards. (8) Show that in any motion of a particle the tangential compon- ent of the force acting on it may be measured by the rate per sec. at which momentum is increased. (9) Prove that if W lbs. be acted upon by a uniform force of P pounds-weight for t sec, the velocity acquired will be PgtjW, and the distance traversed Pgfi/(2 W). 320] OF A PARTICLE. 219 (10) A body of 10 lbs. mass and with an initial velocity of 20 ft. per sec. in a northerly direction is acted upon by two forces, one of 100 poundals in a north-easterly direction and the other of the same magnitude in a north-westerly direction. Find its velocity after 1 min. Ans. 868 '5... ft. per sec. in a northerly direction. (11) Find the impulse necessary to produce in 20 lbs. a speed of 25 ft. per sec. Ans. 500 absolute ft. -lb. -sec. units. (12) Two particles, each of mass m, are at rest side by side when one is struck a blow of impulse * in a given direction, while a con- stant force F begins at the same instant to act upon the other in the same direction. Prove that if after travelling a distance s in the time t, they are again side by side, 2# = i^< and %^^=mFs. (13) A particle of mass m is moving in an easterly direction with a velocity v. Mnd" the impulse necessary to make it move in a northerly direction with an equal velocity. Ana , mv ^2 in a north-westerly direction. (14) A particle of mass m moves with uniform speed « in a circle of radius r. Find the force acting upon it. The particle has an acceleration equal to v^/r directed towards the centre of the circle (121). Hence the force must be iu the same direction and equal to mv^/r. [A body moving in a curved path was formerly thought to have what was called centrifugal force, which required to be neutralized by a force applied to the body (through a string or by other means) towards the centre of curvature (and called therefore centripetal force), in order that the body might be kept in the curvTed path. Thus a body moving with uniform speed in a circle was considered to be in equilibrium {i.e., to have no acceleration) under equal and opposite forces, the supposed centrifugal force and the actually applied centripetal force. The necessary centripetal force being known to have the magnitude mv^lr, the centrifugal force was supposed to have that magnitude also. According to our modem conception of force, a body cannot be said to home a force. More- 220 DYNAMICS [320 over, we now know that if no force be applied to a body it will move with uniform speed in a straight line, and that, if it is to be made to move in a circle, the resultant force on it must be centripetal. Though the old notion of centrifugal force has been abandoned, the term is still used, being applied by different writers in different ways. It is applied (1) in its original sense, some writers finding it still convenient in some cases to imagine a body moving uniformly • in a circle as acted on by a force equal and opposite to the actual centripetal force under which it moves ; (2) to the actual centri- petal force under which the body moves ; (3) to the reaction of the moving body on the body by which the centripetal force is exerted, the centrifugal and centripetal forces being thus opposite aspects of the same stress ; (4) to the acceleration of the moving body. Such varying usage leads to great confusion. The old term should be laid aside with the old hypothesis on which it was based.] (15) Find the horizontal force which must be exerted on an engine of 20 tons which is to go round a curve of 600 yds. radius at the uniform rate of 30 mis. an hour. Ans. 0*67 ton-weight nearly. (16) A stone of 4 kgr., attached to a fixed point by a weightless inextensible string 3 metres long, moves uniformly in a circle in the horizontal plane through the fixed point. Find (a) the tension in the string when the speed of the stone is 20 cm. per sec, and (6) the time of revolution when the tension of the string is equal to the weight of 12 kgr. [We shall investigate farther on (383) the action of forces on bodies through strings. Meantime, we may consider the above string to be a means of keeping the particle at a constant distance from the fixed point and of exerting on it a force, usually called a tension, directed towards the fixed point.] Ans. (a) 5,333J^ dynes ; (&) 2 sec. approximately. (17) A man, standing at one of the poles of a rotating planet, whirls a body of 20 lbs. mass on a smooth horizontal plane by a string 1 yd. long at the rate of 100 turns per minute. He finds that the difference of the forces which he has to exert- according as 320] OF A PARTICLE. 221 he whirls the body one way or the opposite is O'Ol pound-weight. Find the period of rotation of the planet. Ans. 13 h. 37 min. 21-6 sec. (18) A railway carriage is going round a curve of 500 ft. radius at the rate of 30 mis. per hour. Find how much a plummet hung from the roof by a thread will be deflected from the vertical. Ans. 6° 51'-4.... (19) A particle of mass m is attached by a massless string of length Z to a fixed point, and moves with uniform speed i; in a circular path about a vertical axis through the fixed point. Find the tension in the string and the time of a revolution, when the string has a given inclination 8 to the axis. [This arrangement is called the conical pendvlum. The distance A of the fixed point from the plane of the particle's motion is called the height of the pendulum.] The particle is acted upon by two forces, its weight mg vertically downwards, and the tension in the string T directed towards the fixed point. Its resultant acceleration is v^jil sin B) and is directed towards the centre of its path. The sum of the components of the acting forces in this direction is ^sin 8, Hence Ta.mB = mv''j(l sin 8). The particle has no acceleration in a vertical direction, and the components of the forces in that direction are mg downwards and TcosS upwards. Hence Tco&e-mg=0. From either of these equations T may be found. Eliminating T, we obtain «2=?g' sin Stan 6. Hence, if r is the radius of the circular path, i^=r'^glh. If therefore u is the angular velocity of the particle about the centre of its path, w= i^gjh, and if t is the time of revolution, «=27r/w=27r\/%, 222 DYNAMICS [320 I. which also (187) is the time of oscillation of a simple pendulum of length h. If d is indefinitely small, k and I are ultimately equal, and hence t = %Tr>Jllg ultimately. Compare this result with that of 190, which shows that in this case the motion is the resultant of two simple harmonic motions whose common period is 2jr iJLjg. (20) A particle of mass m, attached by an inextensible string (length =Z) to a fixed point, moves in a vertical plane through the fixed point in a circle of radius I. Tind the tension T of the string in any position. Let V be the speed at the highest point A of the path, v the speed at any point P, the angle subtended at the fixed point by the arc A P. The normal component of the particle's acceleration when at P is v^jl. Since the vertical distance through which it has fallen from A is then 1(1 — cos 6), we have (185) 2,2= 172+2^^(1 -cose). Hence the normal acceleration v'^ll=V^ll+'iig{\-cose). The forces acting on the particle at P are the tension T towards the fixed point and the weight of the particle mg downwards. The sum of the components towards the fixed point is T+mgcosB. Hence T+mg cos$ = m{ V^/l+2g{l - cos 6)}, by which equation T is determined. Show that the least and greatest values of ?'are m(V^/l—g) and m{V^/l + 5g) respectively, and that 2' has these values at the highest and lowest points of the path respectively. Show also that the least value of V with which a 'circle will be described is s/lg, and that, when V has this value, the greatest value of T is equal to six times the weight of the particle. (21) A particle moving in a straight line is acted upon by a force directed towards a fixed point in the line and proportional to the distance of the particle from it. Show that the particle's motion is simple harmonic, and that, if / is the force on the particle when 320] OF A PAKTIOLE. 223 at unit distance from the fixed point, the period of its simple harmonic motion is 2ir 'Jmjf, m, being its mass. (22) A mass of 7 lbs., hung from a fixed support by a massless spiral spring, and set to vibrate in a vertical line, makes 80 com- plete vibrations per minute. What force will the spring exert when extended 2 inches ? [The force exerted by a compressed or ex- tended spiral spring is proportional to the amount of the compres- sion or extension.] Ans. 81 '9 poundals approximately. (23) A particle moves in' an ellipse under a force directed towards one of the foci. Show that the force is inversely proportional to the square of the distance of the particle from the focus. (24) A particle of mass m slides down a smooth inclined plane (inclination = e), its motion being opposed by a force F, inclined to the plane at the angle (p. Find (a) the acceleration, and (6) the reaction of the plane. [A smooth body is one which reacts upon another body in contact with it in a direction normal to its surface at the point of contact. Smooth bodies are of course purely ideal. The stresses between actual bodies in contact are not in general normal to the surface. We shall see farther on (328) how their directions are determined.] Ans. (as) i^cos 0/m -g sin B, up the plane ; (6) mg cos 8 - Fsin . (25) A particle slides down a smooth curve in a vertical plane, starting from rest at a given point. If the curve have such a form that at every point the resultant force on the particle is equal to its weight, the radius of curvature at anyipoint will be twice the intercept of the normal to the curve at that point between the curve and the horizontal line through the starting point. (26) A particle (mass = m) slides in a vertical plane down the edge of a smooth circular disc (radius =r) whose axis is horizontal. Show that if it start from rest at the highest point, it will quit the disc after describing an arc subtending at its centre an angle cos~^|. Let V be the speed of the particle after describing an arc sub- 224 DYNAMICS [320 tending at the centre an angle B, then (Ex. 20) the normal accele- ration is i>2/?-=2p'(l-cose). The forces acting on the particle are the reaction R of the disc, normally outwards, and its weight mg ; and the sum of their normal components is mg cos 6 — R. Hence ingco&6-R== 2mg(l - cos d) ; and R = mg(3 cos 9 - 2). Hence, for 9=cos-if, R=0 ; and for e>cos-i|, R is negative, i.e., the disc must attract the particle if they are to remain in contact. (27) A particle slides down a smooth cycloid placed in a vertical plane, with its vertex upwards and hase horizontal. It starts from rest at the vertex. Show that it will leave the curve at the point where the horizontal line drawn through the centre of the generat- ing circle cuts the curve. [If from a point /" of a cycloid a normal be drawn meeting the base in the point S, the radius of curvature at P is equal to 2FS.] 321. Impact. — When two bodies in relative motion come into contact, they are said to impinge upon one another or to undergo impact. The consequence of the impact is a change in their velocities. Hence during the impact a stress must have acted between the bodies; and in applying the equation of motion it is often necessary that we should have some means of deter- mining the stress. In actual bodies the stress is usually of very short duration, and it is thus more convenient to determine the impulse of the stress than the stress itseK In all cases it affects only the component velocities of the im- pinging bodies in its own direction. In some cases it is of sufficient magnitude only to equalize these component- velocities; in others its magnitude is such as to make the bodies recoil, or move away from one another, after impact. Whether or not particles would behave, on im- 322] OF A PARTICLE. 225 pinging, like actual bodies, we have no means of knowing. For the purpose of illustrating the subject by problems, we may assume that they would. At our present stage we have to consider only the special case of a particle impinging upon a smooth surface of a fixed body. In that case the direction of the stress is normal to the surface. If u and u' are the components normal to the surface, of the particle's velocity just before and just after impact, u is called the velocity of approach and u' that of recoil. Now the stress must be sufficient to change a velocity of approach u into a velocity of recoil ft', i.e., if m is the mass of the particle, to produce a change of momentum equal to mu +17111,'. Hence, if $ is the impulse, ^='mu+mu'. If e is the ratio of the velocity of recoil to that of approach, e = u'/u. Hence ^ = 'mu(l + e). If is the stress which is just sufficient to destroy the velocity of approach, and produces no recoil, we have ^=mu. Hence * = 0(l+e). Newton found by experiment (379) that with given impinging bodies the ratio of the velocity of recoil to that of approach was constant. The ratio e is therefore called the coefficient of restitution for the given bodies. 322. Examples. (1) A body of 4 lbs. mass, moving -with a velocity of 10 ft. per sec. in a direction inclined 60° to a smooth surface, impinges upon and is reflected by that surface, the coefficient of restitution being 0'5. Find the impulse of the stress. Ans. 30 ^JZ absolute ft.-lb.-sec. units. P 226 DYNAMICS [322 (2) A particle impinges on a smooth plane, the coefficient of restitution being e, the angle between the direction of the particle's motion before impact and the normal to the plane (called the angle of incidence) being a, and the angle between the direction of its motion after impact and the normal (called the angle of reiiection) being 9. Show that tanS/tana = l/e. [Let u, V be the components of the particle's velocity normal and parallel respectively to the given plane before impact ; u', v the same quantities after impact. Then vlv,=twxa and v/u' = tajid. And u' = eu.1 (3) A particle of mass m is let fall from a height h upon a smooth horizontal plane and rebounds to a height h'. Find (a) the impulse of the stress, and (6) the coefficient of restitution. Ans. (a) mg v'2( Jh+ ^h')y (b) sfhTih. (4) Prove that the velocity of a particle moving on a smooth horizontal plane is reversed in direction after impinging successively , on two fixed smooth vertical planes at right angles to one another, the coefficients of restitution being the same for both planes. (5) A particle is projected from a point A in the circumference of a circle and after impinging at three other points in the circum- ference returns to A. Show that the tangents of the four angles of incidence are e^, «z, e~4, and e~i, e being the coefficient of restitution. (6) A ball falls vertically from rest for 1 sec. and then strikes a smooth plane inclined 45° to the horizon, the coefficient of restitu- tion being 1. Show that it will again strike the plane in 2 sec. (7) A particle, after sliding from rest for 4/ ,^3 sec. down a smooth plane inclined 60° to the horizon, strikes a horizontal plane (coefficient of restitution = J) and rebounds. At what distance will it again strike this plane ? Ans. 37-18... ft. (8) A ball is projected at an elevation a towards a smooth vertical wall (coefficient of restitution =«) from a point whose 322] OF A PARTICLE. 227 distance from the wall is a. What must the velocity of projection be that the ball may return after its rebound to the point of projection ? Ans. [3'a(l + e)/(esin2a)]i. (9) A particle is projected from a point on an inclined plane of inclination it, with a velocity v, inclined j3 to the plane. Find the time between the n^ and the {n + Vf^ rebounds, the coefficient of restitution being e. Ans. 2e''« sin |8/(gr cos a). (10) A ball is projected with a velocity of given magnitude from a given point in a smooth plane inclined a to the horizon (coefficient of restitution =e). Find the direction of the velocity that the ball may cease to hop just as it returns to the point of projection. [The ball ceases to hop after an infinite series of hops. Express the time in which the ball returns to the point of projection (1) by the aid of the last example, noting that 1/(1 -e) = H-e+eH etc. ; and (2) by considering the component motion parallel to the plane ; and equate these expressions.] Ans. The inclination of the velocity to the inclined plane is cot-^[tana/(l-e)]. (11) A ball is projected from a pdnt in a plane of inclination a (coefficient of restitution = «), with a velocity Fat right angles to the plane. Find its distance from the point of projection when it ceases to rebound. Ans. 2 F^sin o/[.g'(l - e)^cos^a]. (12) A stream of particles, each of mass m grm., moving in the same direction with a velocity u em. per sec, impinge successively on a fixed plane (coefficient of restitution = e) inclined a to the direction of their velocity. If n particles reach the plane per sec, find the force exerted by the plane. [The plane exerts on the particles a series of impulses. The force exerted is the sum of all the impulses occurring in 1 sec] Ans. mnvil + e) sin a dynes. 228 DYNAMICS [322 (13) A uniform chain (linear density = p lbs. per foot) is held in the hand by one end, its other end being in contact with a horizon- tal table (coefficient of restitution =0). At a given instant it is let go. Show that the force exerted by the table ou the chain after t sec. is three times the weight of the portion of the chain then lying coiled on the table. [The various links of the chain having all at any given instant the same velocity fall as though they were unconnected particles. After t sec. ^gf lbs. of chain lie on the table, and the force exerted by the table is also destroying the momentum of pgt lbs. of chain per sec] 323. Equilibrium. — A particle is said to be in equi- librium or in a static condition when the forces acting on it produce in it a resultant acceleration equal to zero. The acting forces are also said to be in equilibrium in this case. A particle in equilibrium must therefore either be at rest or be moving with uniform speed in a straight line. The subject of equilibrium, together with all those portions of Dynamics which are necessary for its discus- sion are frequently treated separately under the title Statics, the other department of Dynamics, which treats of forces as producing acceleration, being then called Kinetics. Some writers employ the term Dynamics as synonymous with Kinetics, and apply the term Mechanics to what we have called Dynamics. 324. Condition of Equilibrium. — That the resultant acceleration of a particle may be zero, the resultant force actiiig on it must (317) be zero also. And if the result- ant force be zero, the resultant acceleration will be zero also. Hence the vanishing of the resultant force is the necessary and sufficient condition of equilibrium. This condition may be otherwise expressed, viz., that any one of the forces acting on a particle must be equal and opposite to the resultant of all the rest. 325] OF A PARTICLE. 229 325. Expressions for Condition of Mquilibriuin in Special Cases. — In special cases the following different modes of expressing completely or partially the condition of equilibrium are found convenient in the solution of statical problems. (a) If two forces only act on a particle, they must be equal and opposite. (6) If three forces act on a particle, they must all lie in one plane. For the resultant of any two must be in their plane and must be opposite to the third. (c) If three component forces caij be represented by the sides of a triangle taken the same way round, the resultant is zero. This is an immediate inference from the triangle of forces (313). (d) Conversely, if three forces are in equilibrium and if they can be represented in direction by the sides of a triangle taken the same way round, they will also be represented by them in magnitude. —Let AB, BG, CA represent in direction the three „ component forces P, Q, R re- j ~7 ^~ spectively^ which are in equi- librium. Let AB represent P in magnitude also. If BC does not represent Q in magnitude, cut off BD from it of the proper length to do so. Then the resultant AI) oi AB and BD must be in equilibrium with the third force whose direction is CA. That is, two forces whose directions are not in the same straight line may be in equilibrium, which is impossible. Hence the assumption that BC does not represent Q in magnitude was wrong. Similarly it may be shown that CA repre- sents R in magnitude. (e) If three forces P, Q, R are in equilibrium, they are proportional to the sines of the angles between the 230 DYNAMICS [325 directions of Q and E, P and R, and P and Q respectively. OA and OB representing P and Q, R must be represented by GO the diagonal of the parallelogram AB. Now OA:AG: CO = sm OCA : sin COA : sin OAC. If the angle between the directions of P and Q be written PQ, we have sin OCA = sin (180° - QB.) = sin QR, sin 00 A = sin (180° - PR) = sin PR, sin J. C = sin (1 80° - PQ) = sin PQ. Hence P : Q : P = sin QiJ : sin Pi? : sin PQ. (/) If more than three forces act, it may be shown from the polygon of forces that, if any number of com- ponent forces can be represented by the sides of a polygon taken the same way round, their resultant must be zero. But the converse proposition similar to that of (d) does not hold. 326. Analytical Expression for Condition of Equi- librium. — We may express the condition of equilibrium of a particle in a way applicable to all cases by employ- ing the analytical expression for the resultant of any number of forces. If Pj, F^, etc., are the magnitudes of component forces, a^, /8,, y^, a^, /Sj, y^, etc., the angles made by their directions with three fixed rectangular axes, and if R is their resultant, we have (313 and 90) 327] OF A PAKTICLE. 231 E={(2F COS aY + (2F cos /3)H (E,F cos y)^}*. If there is equilibrium, R = 0. Hence in that case (2F cos a)2 + (E^cos jS)2 + (I,F cos y)^ = 0. But if the sum of three essentially positive quantities is zero, the quantities themselves must each be zero. Hence 2i?'cos a = S^cos /3 = Si?' cos y = 0, i.e., the condition that the algebraic sum of the compon- ents of the acting forces in each of any three rectangular directions must be equal to zero is a necessary condition of equilibrium. It is evident that it is also sufficient. If the forces are coplanar, the angles y become right angles and the angles ^ become the complements of the angles a. Hence the equations expressing the condition of equilibrium become Si' cos a = 2^ sin a = 0. 327. Flxamples. (1) A particle of weight W rests upon a smooth inclined plane of inclination a (to the horizontal plane) under a force F acting up the plane (i.e., in the direction of a line of greatest slope). Find the magnitude of ^ and of the reaction Ji of the plane. Let ABhe the line of greatest slope of the inclined plane, AC a, horizontal line in the vertical plane through AB. Then the particle at P is in equilibrium under the three forces, TF acting vertically downwards, R acting at right angles to ji 5 in a vertical plane, and F in the direction AB. If a perpendicular be let fall from Z) (the point in which a vertical line ^'' through F meets AC) on the direction of R and meet it in JE, we form a tri- angle PBH whose sides FD, BE, and EF taken the same way round have the same directions as the forces W, F, R respectively. Hence (325, rf) W : F: R=PD:I)E: EF. 232 DYNAMICS [327 Now A PE being a right angle, the angle DPE is equal to o. Hence DE= PD sin a and EP = PD cos o. Therefore W:F:R=\ :sina:cosa. And hence i?'= PFsina, and iJ= PTcoso. Otherwise thus :— The angle RF is ir/2 radians, the angle RW (tt - o) radians, and the angle i?* If (tt/S + a) radians. Hence (325, e) W : F : £=sm{r/2) : sin(7r-a) : sin(7r/2 + a) = 1 : sin a : cos a. Otherwise thus :— Choose any two directions at right angles to one another and put the algebraic sum of the components of the forces in each of these directions equal to zero (326). To simplify the equations it is well to choose the directions so that they may coincide with those of as many of the forces as possible. In the direction AB 'we have F- If sina=0. In the direction perpendicular to AB we have R- Fcosa = 0. When the inclined plane is used as a simple machine the ratio of W the weight of the body kept in equilibrium on it to F the force which must be applied to the body for this purpose is called its mechanical advantage. Hence in the case in which the force F acts up the plane the mechanical advantage is cosec a. If from B a perpendicular BG be drawn to AC, BO is called the height of the inclined plane, whose length is AB and base AC. The letters h, I, b are frequently used to denote these lines. Hence in the present case W/F=l/h. (2) (a) rind the mechanical advantage of a smooth inclined plane (length = ?, height = A, base =6) when the applied force acts in a horizontal direction, and (6) express the reaction (R) of the plane on a particle in equilibrium on it in terms of the weight ( W) of the particle. Ans. {a)b/h,{b)R=Wl/b. (3) A particle is in equilibrium on a smooth inclined plane (in- clination = a) under the action of a force F whose inclination to the 327] OF A PARTICLE. 233 inclined plane is 6 and to the horizon (o+fl). Find (a) F, and (6) the reaction of the plane, in terms of the weight W of the particle. Ans. (a) TT sin a/cos 9,(6) frcos(a + e)/cos0. (4) A body is kept at rest on a smooth inclined plane by a force acting up the plane and equal to half the weight of the body. Find the inclination of the plane. Ans. 30°. (5) A body is in equilibrium on a smooth inclined plane, and the applied force and the reaction of the plane are each equal to the weight of the body. Determine (a) the inclination of the plane, and (6) the direction of the applied force. Ans. (a) 60°, (6) inclined 30° to both inclined and horizontal planes. (6) A body is supported on a smooth inclined plane by a force equal to its weight. Show that the reaction of the plane is double what it would be if the body were supported by the least possible force. (7) P is the value of the force which, acting up a smooth inclined plane, keeps a body on it in equilibrium. Q is the magnitude of the force necessary to support the body when its direction is such that it is equal to the reaction of the plane. Show that P acting up the plane could just support a body of weight § on a plane of twice the inclination. (8) A heavy body of 12 lbs. mass is kept in equilibrium by two applied forces, one horizontal and the other inclined 30° to the horizon. Find the forces. Ans. Horizontal force=772-8 pdls., the other =669 '2... pdls. (9) Forces of 2J, 6, and 6^ poundals keep a particle in equilibrium. Show that two of them are at right angles, and find the angle be- tween the greatest and least. Ans. cos-^ - A). (10) A heavy bead (weight = W) capable of sliding on a smooth circular wire in a vertical plane is held at a distance equal to the radius of the circle from its highest point by a force directed to 234 DYNAMICS [327 that point. Find (a) the force, and (6) the reaction of the wire on the bead. Ans. (a) W, (6) W. (11) R is the smallest and R' the greatest force which, along with J' and §, can keep a particle in equilibrium. Show that, if P, Q, and a force (R + R')/2 keep a particle in equilibrium, two of these forces are equal ; and that, if P, Q, and a force s/RR' do so, two of them must be at right angles. (12) Two equal particles, each attracting with a force varying directly as the distance, are situated at the opposite extremities of a diameter of a horizontal circular wire on which a small smooth ring is capable of sliding. Prove that the ring will be kept at rest in any position under the attraction of the particles. (13) Show that there is but one point in a triangle at which a particle would be in equilibrium if acted upon by forces represented by the lines drawn from it to the angular points of the triangle. (14) Show that a particle is in equilibrium if acted upon by three forces represented in direction by the perpendiculars from the angular points of a triangle on the opposite sides, and in magnitude by the reciprocals of the lengths of those perpendiculars. (15) On a smooth inclined plane of inclination cos~^J a particle is in equilibrium under the action of a certain force up the plane, rind the direction in which an equal force must be applied, that it, along with a horizontal force of the same magnitude, may also keep the particle in equilibrium. Ans. Inclination to inclined plane=cos~-^f. (16) Show that a particle is in equilibrium under the following forces :— 4, JST. ; 2, N. 30° H. ; 4, X ; 2 J3, E. 30° S. ; 4 sl% S. W. ; 2 ^3, W. 30° S. ; and 2, N. 30° W. (17) From two points lines are drawn to the angular points of a triangle. Find the condition that a particle acted upon by forces represented by these six lines may be in equilibrium. Ans. The given points must be on a straight line through the point of intersection of the straight lines drawn from the angular 327] OF A PARTICLE. 235 points of the triangle to the middle points of the opposite sides, and must be at equal distances from this point. (18) A string whose ends are fixed at two points A and B in the same horizontal line has, knotted at C, anotjier string carrying a hea-vy body. Compare the tensions in CA and GB, when they are of such length that A CB is a right angle, the whole system being in equilibrium. [We shall prove farther on (389) that when strings are knotted together, the stresses or tensions in them are in general different. In such cases, if there is equilibrium, the knot must be considered to be in- equilibrium under the action of the stresses in the strings.] Ans. As CBiCA. (19) A string has its ends fixed at A and B. Another string is knotted to it at C and supports a body of weight W. The inclina^ tions of CA and CB to the horizon are 6 and ^ respectively. Find the tensions in CA and CB when there is equilibrium. Ans. W cos ^/sin {B + ) and W cos 9/sin {B + (p) respectively. (20) Three strings have one end each knotted together at C. Two of them are attached to fixed points at A and B, and the tensions in them are T and T' respectively. The third supports a particle whose weight is W. Find the inclinations 6 and of C^.and CB to the horizon when there is equilibrium. An. ^=sin-^^^^^. ^-in-^::^-^^^. (21) A string whose length is 10 feet has its ends fastened at two points in a horizontal line 6 feet apart. A small smooth massless ring slides qn the string carrying a body weighing 10 lbs. Find the tension in the string when there is equilibrium. ["We shall prove farther on (391) that when the direction of a flexible string is changed by its being bent round a smooth body the stress through- out the string is the same. In this problem the portion of the string in contact with the ring is in equilibrium under the action of a force equal to and codirectional with the weight of the body which the ring carries and of the equal tensions in the two portions of the string.] Ans. 201 J poundals. 236 DYNAMICS [327 (22) A fixed smootli hemispherical bowl whose rim is horizontal has, resting inside it, a particle of weight W attached by a string which passes over the rim to another particle of weight W which hangs freely. Find the position of the particle in the bowl. Ans. If S is the angle subtended at the centre of the bowl by the portion of the string within it ''=2<=o« ^^^^ ^• 328. Friction. — We are now able to understand the experimental determination of the direction of the stress between two actual rough bodies in contact with one fji another. The figure shows 'g the apparatus employed, AB is a horizontal table, G a Y flat-bottomed box upon it. lAl) w r^ To (7 a string is attached which passes over a pulley at B and supports a pan B. Weights (i.e., standards of mass) are placed in G and B. Before the string is attached C remains motionless on the table. The only forces acting on it are its weight and the reaction of the table. Hence this reaction must be vertical and therefore normal to the surface of contact between G and the table. If now the string be attached and weights added gradually to B, G remains motionless until they reach a certain amount. For aU loads in B less than this amount, G is in equi- librium* under three forces — its weight W, the reaction of the table R, and the force F exerted by the string (equal to the weight of B and in the direction of the string). Hence R must be in the plane of F and W, and so inclined to W, which is normal to the surface of contact, that it has a component F equal and opposite to F. * The motion of the box is (454) the same as that of a particle acted upon by the same forces, provided the box undergo transla- tion only. 328] OF A PAETICLE. 237 This component F resists the sliding of the box over the surface of the table, and is called ^he friction between the box and the table. It increases with F until the box is just on the point of moving, when it has its greatest value and is called the limAtvng static friction. If we increase F still more the box begins to move with an acceleration ; and the greater we make F, the greater is the acceleration. If the acceleration be observed, the resultant horizontal force may be determined, and the difference between this force and F is the value of F in this case. The value of F' when the box is in motion is called the kinetic friction. It is found (by more re- fined experimental methods than the above) to be usually slightly less than the limiting static friction and to be (at any rate very nearly) independent of the velocity of the box. If weights of different amounts are put into the box C it is found that the friction (whether limiting static or kinetic) is, within limits, proportional to the weight of the box and its contents, and therefore to the normal component of the reaction. If boxes of the same sub- stance and weight, but with bottoms of different areas, are used, the friction is found to be independent of the area of the surface of contact. If the substance of the bottom of the box and that of the table, or their state of surface, be changed, the friction is found to change also. If F is the value of the friction (whether limiting static or kinetic), and JR,' the normal component of the reaction R, we have thus F = fiR', where /jt is a constant for two bodies of given substances with their surfaces of contact in given states. It may be determined by such experiments as the above, and it has different values according as relative motion of the one body over the surface of the other is on the point of occurring or is actually occurring, being usually slightly greater in the former case than in the latter. In the former case, fx is 238 DYNAMICS [328 called the coeflBcient of static friction ; in the latter, that of kinetic friction. The inclination to the normal of the reaction R of the bodies in contact may be expressed in terms of the co- efficient of friction. B' and F being the normal and frictional components of R, we have, if e is the inclination of J? to the normal, ta,n e = F'jR' = fi, and e = tan-V- If (U is the coefficient of static friction the value of e thus determined is the greatest possible inclination of the reaction to the normal. It is called the angle of repose. As R is in the plane containing the normal to the surface of contact, and the direction in which the acting forces tend to produce sliding, and as this direction may be any whatever in the tangent plane at the point of contact of the bodies, the direction of the reaction when sliding is on the point of occurring may be any line on the surface of a cone whose axis is the normal at the point of contact, and whose semi-vertical angle is the angle of repose. The direction of the reaction under all circumstances must be included in this cone. The ideal perfectly rough body is one over whose surface sliding is impossible. In the case of such a body the reaction is supposed to have any direction and mag- nitude that may be necessary to prevent sliding. The above statement of the laws, of friction is sufficient for our purpose. For a more detailed statement of our knowledge of this subject, see recent works on en- gineering. 329. Examples. (1) A particle of mass m is moving up an inclined plane (co- efficient of friction = a) of inclination a, being acted upon by a force F up tlie plane. Find its acceleration and the reaction of the plane. — Let R be the normal component of the reaction of the plane. Then 11.R is the component in the plane, and as the point is moving 329] OF A PARTICLE. 239 up the plane, ii,R is directed down it. Hence, if a is the acceleration up the plane, a={F-iiR- Wama)/m. As there is no acceleration normal to the plane, = iJ- Ifcosa. Hence a = [P- ■pr(/iC0sa4-sina)]/w. Also the resultant reaction is (313 and 86, V) R Vl+^= W n/iTm^ cos a. In the above formulae /t is the coefficient of kinetic friction. (2) A body of 100 lbs. mass, moving on a horizontal surface with a speed of 10 ft. per sec, comes to rest in 2 sec. Find the coefficient of kinetic friction (supposed independent of the velocity). Ans. 0-15.... (3) A mass of 100 lbs. is moved along a horizontal plane by a constant horizontal force of 20 Ibs.-weight. Determine the dis- placement in 10 sec, the coefficient of kinetic friction being 0"17. Ans. 48-3 ft. (4) A force equal to the weight of 28 lbs. is required to draw a mass of 30 lbs. up a plane inclined 30° to the horizon. Find (cs) the coefficient of friction ; (6) the force that would be necessary if the inclination were 45°. Ans. (ffl)0-5...; (6) 45/ ;v/2 Ibs.-weight approximately. (5) A train is going up an incline of 1 in 70, at the rate of 10 mis. per hour, the friction being equivalent to a force of 8 lbs.- 240 DYNAMICS [329 weight per ton of the train's mass. The incline is 500 ft. in length, and when the train is half way up, a coupling-chain breaks. Find (a) how far the train will go up the incline, and (6) its speed at the foot of the incline. Ans. (a) 187-05... ft.; (6) 17-36... ft. per sec. (6) A particle impinges on a fixed rough plane (coefficient of friction =/i, that of restitution = e) with a velocity v inclined a° to the normal. Find (a) the magnitude, and (b) the inclination to the normal, of the velocity after impact. [The frictional impulse is equal to fi times the normal impulse.] Ans. (a) { e Vcos^a + \v sin a — fi.i>{ 1 + e) cos aP } ' ; (6) tan-i[- tan a - ^ (1 + e)]. (7) A particle is in equilibrium on a rough inclined plane of inclination u, being just prevented from moving do-wn by a force F acting up the plane. The coefficient of static friction being fi, find F and the reaction of the plane. This problem may be solved by means of the result of Ex. 1. That the particle may be in equilibrium on the plane we must have a = 0. Hence F- W{ii. cos a. + sma) = 0. In the formula of Ex. 1, /* was the coefficient of kinetic friction. When we make a=0, it becomes the coefficient of static friction. Also that formula was obtained on the assumption that /iR acts down the plane, and therefore that the iparticle moves, or tends to move, up the plane. If we make f- negative and thus obtain F - T'F(sin a - /i cos a) = 0, we reverse the direction of fi-R, i.e., we get a formula applicable to the case in which nR acts up the plane, and the particle therefore is prevented by friction from moving down the plane. Otherwise thus : — The particle is acted upon by three forces, its weight W, F, and the reaction of the plane R'. R' is inclined to the normal FN at the angle of repose (e), the angle being measured towards PB, because the body is on the point of moving down the plane. Since 329] OF A PARTICLE. 241 Tra'=180°+e-a, Ji'F=QO°-e, and WF=9(f + a, we have and hence and F : W : R'=siD.(a-e) : cose : coso ; F= FT sin (a - e)/cos e, R'= TT cos a/cose. Otherwise : — Eeplaeing R' by its normal component R and its frictional component iiR (up the plane), and resolving in and per- pendicular to the direction of AB, we have and Hence and F+iiR- Trsino=0, R- Fcoso=0. F= Tf (sin a — /t cos a), ij'=ij\/r+^= pfcosctVrtv. Recollecting that c=tan~V, it is easy to show the consistency of the above results. The same equations may be obtained in other ways. [See 327 (1)]. (8) A body is in equilibrium on a rough inclined plane of incli- nation u, under a force F, inclined at the angle S to the inclined plane. Find the ratio of the weight of the body to the force F (a) when the body is on the point of moving up the plane ; (J) when it is on the point of moving down. Ans. (a) cosS+iJ,aia6 sina-l- Oleosa ' (6) cos S—fi sin sin a — fi cos a 242 DYNAMICS [329 (9) Prove that the horizontal force which will just sustain a heavy particle on a rough inclined plane will sustain the particle on a smooth inclined plane provided its inclination is less than that of the rough plane by the angle of repose. (10) Show that the least coefficient of friction that will allow of a heavy body's being just kept from sliding down an inclined plane of inclination a, the body (weight = W) being sustained by a given horizontal force F, is ( PFtan a - F)/{F ta,ii a + W). (11) A heavy body is kept at rest on a given inclined plane by a force making a given angle with the plane. . Show that the reaction of the plane when it is smooth is a harmonic mean between the normal components of the greatest and least reactions when it is rough. (12) A bead, capable of sliding on a rough circular wire (radius =r, coefficient of friction =/i) in a vertical plane, is in equilibrium in the highest position (not being the highest point of the wire) in which equilibrium is possible. Find its position. Ans. Its distance from the lowest point of the circle measured along the circumference is rtan~V. (13) Show that it is easier to lift a body a given height than to drag it up an inclined plane of that height by a rope parallel to the pla,ne, if the coefficient of friction is greater than the ratio of the difference between the length and height of the plane to its base. 330. WorJc done. — Work is said to be done by a force on a body when its place of application has a component displacement in the direction of the force, and by a body against a force when the place of application of the force has a component displacement in the direction opposite to that of the force. Work may in both cases be said to be done by the force if in the one case it is considered as positive and in the other as negative. If the force doing the work is uniform, the work done is measured by the product of the force into the compon- ent in its direction of the displacement of its point of application. If W, F, s, a represent the work done, the 331] OF A PABTIOLK. 243 force acting, the displacement, and the inclination of the directions of the displacement and the force, we have thus, by definition, W (xFs cos a. The work done is therefore measured also by the product of the displacement into the component in its direction of the acting force. If the force doing the work is variable, the motion of the body may be supposed to be broken up into a large number of small displacements during each of which the force may be considered constant, and the work done is the sum of all the quantities of work done during these small displacements. 331. Measuremsnt of Work done. — If we write I for the component of the displacement in the direction of the force, we have W—hFL, where fc is a constant whose value will depend upon the units involved in the other quantities. We have already selected units of force and length. We can give h the convenient value unity there- fore only by properly selecting the unit of work. If 'W = F=l = l, k will be equal to 1. Hence we take as unit of work the work done when under the action of unit force a particle has a component displacement of 1 unit in the direction of the force. This derived unit of work will of course vary with the units chosen as simple units. F. P. S. Gravitational System. — The weight of the pound being the unit of force, the unit of work is the work done when a body under this force moves through a distance of 1 foot in its direction. This unit is very largely used in Engineering. It is called the foot-pound, and is usually defined as the work done in lifting one pound one foot vertically. if. K. S. Oravitatiorial System. — The weight of the 244 DYNAMICS [331 kilogramme being the unit of force, the unit of work is that done when under this force a body moves through 1 metre in its direction. It is largely used by French, engineers, and is called the kilogramme-metre. The kilogramme-metre is equivalent to 7'2331 foot-pounds. F. P. 8. Absolute System.— The unit of work is the work done when under a force of 1 poundal, a body moves through 1 foot in the direction of the force. This unit is called the foot-poundal. It is clear that, as the weight of 1 lb. is gr times the poundal, the foot-pound must be g times the foot-poundal. C. G. S. Absolute System. — The unit of work is the work done when under a force of 1 dyne a body moves through 1 centimetre in the direction of the force. It is called the erg. The joiile is 10,000,000 ergs, and is equi- valent to nearly f of a foot-pound. 332. Dimensions of Unit of Work. — From the equa- tion W X Fl, we deduce, as in 300 and 303, [W] x [F][L] and [W] oc [i(/][i]^[T]"^. The former expression applies to gravitational units in which the unit of force is a fundamental unit. The latter applies to absolute units in which the unit of mass is chosen arbitrarily. A know- ledge of the dimensions of units of work is applied in the solution of problems in exactly the same way as in the case of units of speed and rates of change of speed. 333. Bate of Work — The mecwi rate at which a force does work in a given time is the quotient of the work done in the time by the time. In general the mean rate varies with the interval of time to which it applies. In any case in which it does not, the rate of doing work is said to be uniform. The i/nstantaneous rate at a given instant is the mean rate between that instant and another when the interval 335] OF A PAETICLK 245 of time between them is made indefinitely small. It has (295) in all cases a finite value. The rate at which an agent {e.g., a steam engine) can do work is called its Power or Activity. 334. Let W be the work done by a force F on & particle of mass m in a short time t, and R the rate at which the work is done. Then R= W/t. If s is the distance traversed by the body in the direction of F during t, R = Fsjt = Fv, where v is the component of the instantaneous velocity of the particle in the direction of F. If a is the instantaneous acceleration produced in the particle by F, we have F=ma, and therefore iJ=mau, whence a = RI(rav), i.e., the acceleration produced in a particle by a force working at the rate R, is equal to the quotient of this rate by the momentum of the particle in the direction of F. If the work is done against a force F', which has a direction opposite to that of F, and produces an accelera- tion a', the resultant acceleration isa — a' = R/{mv) — F'/m. As V increases a decreases. When a=a' there is no resultant acceleration and v becomes uniform and has its greatest value. Hence the greatest velocity which a force working at the rate R can produce against an opposing force F' is equal to R/F'. 335. Measurement of Rate of Work. — We have by definition R = W/t. When Tf = t = l, R=l. Hence unit rate of work is unit of work per unit of time. The fol- lowing are therefore the units of rate of work in the various systems. F. P. 8. Gravitational System — One foot-pound per second. — The unit employed by English engineers is a multiple of this, viz., 550 ft. -pounds per sec, or 33,000 ft.-pounds per min., which is called the horse-power. M. K. 8. Gravitational System — One kilogramme- 246 . DYNAMICS [3^^ metre per second. — The unit practically employed by French engineers is 75 kilogramme-metres per sec. (equivalent to .542-486 ft.-pounds per sec), which is called the force de cheval. F. P. S. Absolute System — One ft.-poundal per sec. G. G. S. Absolute System — One erg per sec. — A mul- tiple of this unit, viz., 10,000,000 ergs per sec. (equivalent, to nearly | ft.-pound per sec.) is extensively employed in electrical work. It is called the watt. The dimensions of units of rate of work can be readily shown from the formulae of 334 and 332 to be [F}[L][T]-^ or [M][Lf[T]-^ 336. Eaumples. (1) Reduce 50 ergs to kilogramme-metres. Ans. 5'09 X 10~' approx. (2) Reduce 20 foot-pounds to ergs. Ans. 2'712 X 10* approx. (3) Show that 1 foot-poundal = 421,390 ergs. (4) Find the multiplier by which ergs are reduced to foot-pounds. Ans. 7-3'7xlO-8. (5) The second and the foot being the units of time and of length respectively, determine the unit of mass that the derived unit of work may be equal to the foot-pound. Ans. 32-2 lbs. (6) The units of mass, work, and length being taken as funda- mental units, find the dimensions of the derived unit of time. Ans. [X][J/]*[W]-i. (7) A man weighing 168 lbs. climbs a mountain 11,000 feet high in 7 hours, the difiiculties of the way being equivalent to the carry- ing of an additional weight of 42 lbs. Show that he has worked at ^ horse-power. 337] OF A PAETICLE. 247 (8) A boy drags a body of 50 lbs. mass on a smooth horizontal plane, doing work upon it at the rate of yS, horse-power. Find its acceleration when its speed is 1 mile per hour. Ans. 24"15 ft.-sec. units in the direction of motion. (9) An engine is employed in lifting vertically a bale of goods weighing 1 cwt. (a) If the engine is working at 5 H.-P. and the bale has a speed of 5 ft. per sec, find its acceleration. (5) At what H.-P. must the engine work to lift the bale with a uniform speed of 1 ft. per sec. Ans. (a) 158'1 ft.-sec. units upwards ; (b) i H.-P. approx. (10) A train weighing 75 tons ascends an incline of 1 in 800 with a uniform speed of 40 miles per hour. Assuming the friction io be equivalent to a force of 6 pounds-weight per ton of the train's mass, find the rate at which the engine is working. Ans. 70'4 H.-P. approx. (11) Find the greatest speed an engine of 100 H.-P. can give a train of 70 tons mass on an incline of 1 in 100, friction being equiva- lent to a force of 8 pounds- weight per ton. Ans. 17"62... miles per hour. (12) Reduce 20 horse-power to ergs per second. Ans. 1-492x10". (13) In 1 force de cheval how many ergs per second? Ans. 7-36x109. (14) If the acceleration of a falling body be taken as unit of acceleration, 1 ton as unit of mass, 1 horse-power as unit rate of work, and 1 minute as unit of time, find the derived unit of length. Ans. 14-7... feet. 337. Determination of Work done under given Forces. (1) Under a JJmiform Force. — If a particle undergo an}' motion under a uniform force, no difficulty arises in determining the work done. It is simply the product of the magnitude of the force into the component displace ment in its direction. 248 DYNAMICS [338 338. (2) Under a Central Force, i.e., a force directed towards a centre and varying with the distance from the centre.— Let be the centre of force, AB any path of a particle from A to B, and PQ any indefinitely small portion of the path. Join OA, OP, OQ, OB, and from as centre describe arcs of circles Aa, Pp, Qq, M being the point of intersection ofPp withOQ. 0,^::^ — -^ '' PQ being small, the force on the particle between P and Q may be considered constant. Let i'be its magnitude. QO may be considered its direc- tion. Hence the work which must be done in moving the particle from P to Q is F. PQ cos MQP, which, since PM and PQ may be considered straight lines and PM is at right angles to OQ, is equal to F . MQ. Now p and q being at the same distance from as P and Q respectively, the force on the particle, if taken from p to q, would be F also, and MQ =pq. Hence the work which must be done in moving the particle from P to Q is the same as that necessary to move it from p to q. We may treat every element of the path in the same way. Hence, by sum- mation, the work necessary to move the particle from A to B is equal to that necessary to move it from a to 5 in a straight line. Hence also the work done in moving a particle from J. to £ is independent of the path, and depends only on the initial and final distances of the points. 339. (a) The Force directly proportional to the Distance of the Particle from the Centre. — Let / be its value at unit distance. Then its values at a and B are /. Oa and /. OB respectively. Hence its mean value per unit dis- tance between a and B is ^f{Oa+OB} ; and consequently the work which must be done in moving the particle from a to B, and therefore from A to B, is 341] OF A PAKTICLE. 249 y{Oa-\- OB)(OB- Oa) = 1/(05^- Oa^) = IfiO^-OA^). If r and It are the initial and final distances respectively, the work done is ^/{R^—r^). 340. (6) The Force inversely proportioTial to the Square of the Distance of the Particle from the Gevdre. — Let / be its value at unit distance. Then its values at p and q a,vef/Op^ and f/Oq^ respectively. Since pq is indefinitely small, the value of the force between p and q may be put equal to either or to the intermediate value f/{Op . Oq). Hence 'the work which must be done in moving the particle from ^ to g' is Let the line aB be divided into the indefinitely small portions (or elements) ap^, p^^ etc., Pn-iB. Then, adding together the amounts of woi-k done throughout all the elements of aB, the work done in moving the particle from a to B, and therefore from A to B, is ^Aoa~OB) ^AoA~ Ob) ^Ar'B.)' if r and R are the initial and final distances respectively. Hence also the work done in moving the particle from a point at distance r to a point at an infinite distance from the centre is f/r. 341. Eoaamples. (1) Find the work done by the weight of a body of 20 lbs. mass during the first three seconds of its fall from rest. Ans. 93,315-6... ft.-poundals. (2) A body of 80 lbs. mass is projected along a rough horizontal 250 DYNAMICS ■ [341 plane (coeiRcient of friction =0-25) with a speed of 50 ft. per sec. Find the work done against friction in 1 sec. Ans. 919'5 ft. -pounds. (3) Show that the work done in drawing a heavy body up a rough inclined plane is the same as if the body were drawn along the equally rough base arid then lifted through the vertical height. (4) The distance from J!" to Yia 105 miles, and there are 27 inter- mediate stations. Train A stops at all stations. Train B runs through without stopping. The average resistances to A and B with the brakes off are equal to ,^1,^ and -^ of their respective weights. With the brakes on, the resistances are in both cases equal to ^V of the respective weights. Suppose the brakes to be always applied when the speed has been reduced to 30 miles per hour and not before, find which train is the more expensive and by how much per cent. Ans. Traill A, by 9'4. . . per cent. (5) A particle of mass m moves in a circular path of radius r, (a) with uniform speed, (6) with uniform rate of change of speed /. Find the work done in both cases during the motion of the particle through a semicircle. Ans. (a) none, (6) irmrf. (6) Show that in the case of a particle which is oscillating with a simple harmonic motion, the work done during its motion from its extreme position to its mean position is twice that done during its motion from a distance equal to f of its amplitude to a distance equal to J of its amplitude. (7) Aparticle weighing ^ oz. has a simple harmonic motion of 0"5 sec. period. Find the work done during the motion from a distance of 3 inches to a distance of 1 inch. Ans. 0'0274 foot-poundals. (8) Find the work done by the sun's attraction during the motion of the earth from Aphelion to Perihelion. (Mass of earth=6'14xl02' grammes, mass of sun = 324,000 times that of 342] OF A PARTICLE. 251 earth,' distance at Aphelion = l'512x 10'^ cm,, distance at Peri- helion= 1-462 x 10" cm. See 316, Ex. 12.) Ans. 1-79 X 10^9 ergs. (9) At the three corners A, B, C, of a square A BCD (side =100 metres) are material particles of 3,928, 7,856, and 11,784 grammes respectively. Find the work done against the gravitational attrac- tion of the particles in moving 1 gramme from the centre to the fourth comer. Ans. 7-82 x lO"* erg approx. 342. Relation of Work done by Component Forces to that done by Resultant. — The work done by a force during any displacement of a particle is equal to the sum of the quantities of work done by its components. — Let 00 be the force, OA, AB, EG, its components, whose directions may be any whatever. Let OD be the displacement. By 8 (foot- note), 6: a, ^, y being the in- clinations of OC, OA, AB, and ° BC respectively to OJD, OC cos 6 = OA cos a+AB cos /3-|-5Ccos y. Multiplying by OD we obtain OC.OBcose=OA.ODcosa+AB.ODcos^+BG.ODcosy, by which the proposition is proved. If F^, F^ etc., R denote the component and resultant forces respectively, d^, d^, etc.^ r the component displace- ments in the directions of the forces respectively, the above may be written F^d^ + F^d^ -I- etc. = "LFd = Rr, care being taken that, where F and d have opposite directions, the product must have the negative sign. 252 DYNAMICS [343 343. Energy. — We have seen that work is, said to be done by a body against a force which is acting on it, when it undergoes a displacement having a component in a direction opposite to that of the force. When a particle is thus able to do work it is said to possess work-power or energy. Energy being power of doing work is measured in terms of the unit of work. 344. Kinetic Energy. — A particle which has a velocity is able to do work against a force which has a component in a direction opposite to its velocity. It is therefore said to possess kinetic energy. Kinetic energy is thus work-power due to the possession of a velocity. To lind the kinetic energy of a particle we determine the work done by it against any force during a given diminution of its velocity. — Let the particle of mass m have an initial velocity F, and let it do work against a constant force F. It will undergo a displacement hav- ing a component in a direc- ^ tion opposite to that of F. ^ "^ "> ' Let that component be d and let the velocity of the particle be reduced to v. Let the inclination of V to F'& line of action be a. Then the particle has in that line a component initial velocity Fcos a and an acceleration —F/m, and at right angles to it an initial velocity Fsina and no acceleration. Hence, after the displacement, its component velocity u in F's line of action is such that u^ - V^cos^a = - 2Fd/m. Its resultant velocity v is such that 345] OF A PARTICLE. 253 Hence i;2 _ 72 = _ 2Fdlm, and Fd^^mV^ — \mv^. If the force against which the work is done be variable, let the path of the particle be divided into a large num- ber {n) of small displacements, so small that the force may be considered constant during each. Let F^, F^, etc., be the magnitudes of the force during these small displacements, d^, d^, etc., the components of the displacements in the lines of action which the force has during the displace- ments, and v^, Vjj, etc., the velocities of the particle after the successive displacements. Then F^d^ = imV^-^mv^^, F^d^ = lmv^^-lmv^\ etc., Fndn = hmv^ _ 1 — Jmi)^. Hence, if IT be the whole work done, we have by sum- mation W=\mY^-lmv\ Hence the work which can be done by a particle during a given reduction of its velocity is equal to the change produced in the product of half its mass into the square of its velocity. If its final velocity is zero, its work-power due to its velocity is exhausted. In that case W=^mV^. Hence the kinetic energy of a particle is equal to half the pro- duct of its mass into the square of its velocity. 345. Potential Energy. — A particle which is acted upon by such a force as that of gravitational attraction and is in a position from which it can move in the direc- tion of the force, can in virtue of its being so acted upon do work against a second force having a component in a direction opposite to that of the first. It therefore pos- sesses work-power or energy. Thus a heavy body in a position from which it can fall can do work against a 254 DYNAMICS [345 force acting on it in an upward direction, water from a mill-pond, e.g., againstihe reaction of the buckets of a mill- wheel. So also the string of a bent bow can do work against the reaction of the arrow in contact with it. This form of energy has been called energy of 'position for an obvious reason, and static einergy to indicate its independence of the particle's possessing a velocity. The latter term however is defective as seeming to imply that the particle possessing this form of energy must be in equilibrium (323). A particle acted on by a force and in a position firom which it can move in the direction of the force may also be recognized as possessing energy, if we note that, even if no other force be supposed to act on it, it must move in the direction of the force, gaining velocitj'' and therefore work-power. For this reason energy of position has appeared to some writers to be simply a potential form of kinetic energy and it has been named for this reason potential energy. We have seen however that a particle acted on by a force and in a position from 'which it can move in the direction of the force, may do work without first acquiring kinetic energy ; and energy of position must therefore rank as an independent form of energy. The term potential energy should not therefore be em- ployed in the sense in which it was first proposed. We shall see however (356), that this form of energy has a very simple relation to a quantity called the potential ; and to indicate this relation- the term potential energy is employed. In speaking of a particle as possessing potential energy we are taking a narrow view of the phenomenon and neglecting the third law of motion, which states that a force acting on a particle is but one aspect of *a stress which acts between it and another. What we have said of the one particle applies obviously equally to the other. Hence the potential energy belongs not to either of the 347] OF A PARTICLE. 255 particles alone but to the pair, and it is due to the stress between them. When we speak of the potential energy as possessed by the one, we are imagining the other for the moment to be immoveable, or, in other words, we are taking the position of the other as our point of reference. 346. It follows from 345 that all forces do not confer potential energy on the particles on which they act, but those only in whose directions the particles can move. If, e.g., a particle be in motion in contact with a rough surface, it will be acted upon by friction. But the direction of this force must always be opposite to that of the particle's velocity, and the particle therefore cannot move in its direction. Hence friction cannot confer potential energy on a particle. Now all natural forces may be divided into two classes, those of the one class (including such as gravitational attraction) depending only on the position of the particle acted upon, those of the other class (including such as friction) depending upon its velocity and having in all cases directions opposite to that of its velocity. Potential energy is thus conferred on a particle only by forces of the former class whose action depends upon the position of the particle only, and is independent of its velocity. 347. The particle will possess potential energy at whatever point it may be placed of the region through- out which the force acts, but the farther it is displaced in the direction of the force the less it will have. The excess of the value of its potential energy at any point P over its value at another point Q is equal to the work it can do in moving from P to Q. Now, even if it have no initial velocity, it can move from P to Q though acted on by a force F' opposite and indefinitely nearly equal to the force F, to which its potential energy is due. And the work done against F' during this displacement is the product of F' into the component d of PQ in its direc- tion. But since F' is indefinitely nearly equal to F, we 256 DYNAMICS [347 may put F'd = Fd. Hence the excess of the potential energy at P over that at Q, and therefore the diminution of potential energy during the motion from P to Q, is equal to the work done on the particle, during the motion from P to Q, by the force- to which the potential energy is due. Hence also the increase of the potential energy of a particle during any displacement is equal to the work done against this force. Thus the potential energy of a par- ticle of massm at a heightA,'is greater than that of a particle of the same mass at the smaller height h by the amount mg(h' — h). For the work done by the weight (mg) of m during its motion from height h' to height h would be mg(h' — h). Hence also the increase of potential energy of a mass m in moving from a height h' to a height h is mg{h—h'). 348. The Law of Energy. — Let a particle having any initial velocity V undergo displacement when under the action of any number of forces F^, F^, etc. Let R be their resultant and r the component in its direction of the displacement b of the particle; then, the acceleration of the particle being It/m, it may be shown as in 344 that Rr = Imv^ — |m V^. Now, if d^, d^, etc., ' 2 are the components of the displacement -f, in the directions- of the forces F^, F^, etc., we have (342) Rr = 1,Fd. Hence '2Fd = ^mv^—^mV^ ; i.e., the algebraic sum of the quantities of work done by the acting forces is equal to the increase of the kinetic energy. If the displacement be finite, it may be shown as in 344 that the same result holds. We may write this result ^mv^ - Jm F2 + (2f i?W) = 0. The quantity ^^{ — Fd) is (330) the algebraic sura of all the 349] OF A PAETICLE. 257 work done against the acting forces. Hence, in any dis- placement of a particle, the increase of kinetic energy together with the work done against the acting forces is zero. If now the acting forces are all independent of the velocity of the particle, the work done against them is equal to the increase of the potential energy of the particle. Hence, in any displacement of a particle acted on by forces independent of its velocity, the sum of the increments of the kinetic and potential energies is zero, or, in other words, the sum of the potential and kinetic energies is constant. This result is the Law of the Conservation of Energy as applied to a single particle. Forces which depend only on the position of the particle acted upon are usually called conservative forces, as being subject to the above law of conservation of energy. Those which depend on the velocity are called non-conservative forces. If any of the acting forces are dependent upon the velocity of the particle, the work done against them does not result in the production of an equivalent amount of potential energy. In such case, therefore, the sum of the increments of the kinetic and potential energies, and of the work done against such forces, is equal to zero. This result is the Law of Energy as applied to a single particle. The law of the conservation of energy is obviously a special case of the more general law of energy. 849. If a particle acted on by forces be in motion, its energy at any instant consists partly of kinetic,, partly of potential energy. During the motion the relative amounts of these energies will in general change. In such a case its energy is said to be undergoing trcmsformation. Thus the energy of a pendulum at the extremity of its swing 258 DYNAMICS [349 is wholly potential energy. In its mean position (if it be supposed that the string cannot be cut, and that the bob therefore cannot fall lower than the mean position) the energy is wholly kinetic; at intermediate positions it possesses energy of both kinds. The transformations of a particle's energy are always subject to the law of energy. Thus the sum of the kinetic and potential energies of the pendulum at any instant, together with the work done since any former instant against non- conservative forces, must be equal to the energy of the pendulum at that former instant. If the forces acting are all of the conservative class, the sum of the kinetic and potential energies of the pendulum must be the same at all instants. 350. Examples. (1) Compare the amounts of the momentum and kinetic energy in (a) a mass of 20 lbs. having a speed of 16 ft. per sec, and (6) a mass of 1 oz. moving at 5,120 ft. per sec. Ans. Momenta the same, kinetic energy of (6) 320 times that of (a). (2) A cannon ball of 5,000 grammes is discharged with a speed of 500 metres per sec. Find the kinetic energy in (a) ergs, and (6) foot-pounds. Ans. {a) 6-25 x 10^2, (6) 4-61 x 10^, approx. (3) A bale of goods weighing 1 cwt. is lifted 20 ft. Tind the increment of its potential energy. Ans. 2,240 ft.-pounds. (4) A bow 1 yard long is straight when the string is just tight, but when bent has the form of a circular arc of 1 ft. 6 in. radius. The mean force exerted by the hand in bending, per unit distance through which it has moved, is equal to the weight of 10 lbs. Find the potential energy of the bow. Ans. 483 ft.-poundals. (5) A body is projected either (a) vertically upwards, or (6) in 352] OF A PAKTICLE. 259 any direction. Show, by calculating its kinetic and potential energies after any time, that in both cases the energy of the body is the same at all points of its path. [Neglect the resistance of the air and assume g to have the same value at all points of the path.] (6) A meteorite falls in a straight line towards the earth from a great distance, no other heavenly body being supposed near. Show, by calculating the changes produced in its kinetic and potential energies between any two points of its path, that there is no change produced in its energy. (7) A particle weighing 1 lb. has a simple harmonic motion with a period of 20 sec. and an amplitude of 1 ft. Find (a) its kinetic energy in its mean position, (5) its potential energy in either ex- treme position, (c) its kinetic energy and potential energy and their sum when at a distance of 8 inches from the mean position. Ans. (a) 7r2/200 ft.-poundals, (6) the same, (e) kinetic energy = 7r7360 ft.-poundals, potential energy = 7r2/450 ft.-poundals, their sum=7r2/200 ft.-poundals. 351. Application of the Law of Energy to Kinetic Problems. — The law of energy being a generalized form of the laws of motion may be applied at once to the solution of kinetic problems such as those of 320. If the forces acting are all conservative, the law of the conser- vation of energy is applicable. If some of the forces are non-conservative, and if the work done against them cannot be determined, the law of energy cannot be applied. 352. Examples. (1) What speed will the bow of 350, Ex. 4, communicate to an arrow weighing 2 oz. [Assume no work done against non-conserv- ative forces.] Ans. 87'9 ft. per sec. (2) A ball weighing 5 oz. and moving with a speed of 1,000 ft. per sec. strikes a shield 2 inches thick and after piercing it moves on with a speed of 400 ft. per sec. Find the force (supposed 260 DYNAMICS [352 uniform) with which the shield resisted the ball. [Assume as above.] Ans. 787500 poundals. (3) Find the height (h) to which a body weighing 2 grammes and projected vertically upwards with a speed of 20 metres per sec. will have risen before its speed is reduced to 5 metres per sec, assuming the mean resistance of the air to the motion of the body per unit of distance travelled to be 10 dynes. Loss of kinetic energy = 3,750,000 ergs, gain of potential energy = 1,962A ergs, work done against resistance = 10A ergs. Hence A = 1,091 -6 cm. (4) A body of mass m is projected with speed V up an inclined plane of inclination a, the coefficient of kinetic friction being /i. Find the space s traversed before the body comes to rest. Loss of kinetic energy =\m,V^, gain of potential energy = mgs sin a, work done against friction =/im5'« cos a. Hence mgs sin a + langa cos o — Jm V^ = 0, and s = V^/l^g (sin o + ;ti cos a)]. Hence also the acceleration is constant and equal to ,9(sin o +ai cos a). (5) Find the speed v of the bob (mass = m) of a simple pendulum (length = Z) which has swung from its extreme position through a given angle, neglecting the resistance of the air. Let ;3 be the angle made with the vertical in the extreme position, e the angle made with the vertical in the position in which the speed of the bob is to be determined. The kinetic energy gained is ^v\ The vertical height through which the bob has fallen is I cos S — I COS p, and therefore the potential energy lost is mgl(cos e — cos p). The stress in the string has done no work because the bob has had no displacement in its direction. Hence ^nv^ - mgl{cos 6 - cos /3) = 0, and i)2 = 2^Z(cose-cosi3). The reader should solve some of the Examples of 320 and 329 by the application of the law of energy. Those of 322 cannot be solved 354] OF A PARTICLE. 261 by this method, because we as yet know too little of the forces which come into play during impact. We cannot tell whether or not they are conservative forces, nor can we calculate the work done against them. 35.3. Application of the Law of Energy to Static ProhlemiS. — The law of energy may also be employed to obtain an expression for the conditions of equilibrium of a particle. A particle in equilibrium must either be at rest or be moving uniformly. In any indefinitely small displacement of a particle, therefore, from a position in which it is in equilibrium, whether or not it be one which the particle actually undergoes, there can be no change of velocity, and hence no change of kinetic energy. But in any displacement the sum of the increments of potential and kinetic energies, together with the work done against non-conservative forces, must be zero. Hence in any indefinitely small displacement from a position of equilibrium the increment of the potential energy, to- gether with the amount of work done against non- conservative forces, or, in other words, the work done against (and therefore by) all the forces acting on the particle, must be zero. With the symbols of 342, "LFd = 0. This equation might have been deduced at once from that of 342, viz., 1,Fd = Br. For, since for equilibrium E = 0, wehave2FcZ = 0. A small displacement which a particle in equilibrium may be supposed to undergo is often called its virtual displacement or virtual velocity, and its product into the component of any acting force in its direction the virtual work or the virtual moment of the force. The condition of equilibrium as obtained above is then called the Principle of Virtual Work or of Virtual Velocities. 354. Example. A particle of weight W is on the point of moving up an inclined 262 DYNAMICS . [354 plane of inclination a under a force F inclined 6 to the plane, the coefficient of friction being n. Find F in terms of W. The inclination of the reaction E of the plane to the normal FNise= tan" V As we wish to find F in terms of W, we select a displacement FA perpen- dicular to B. If then AB, AC be drawn perpendicular to the directions of F and W respectively, the work done by F, W, and R during the dis- placement are F. PB, - W. PC, and zero respectively. Hence F.PB-W.PC=0, and consequently F.oosAPB- WcosAPO=0. Now the angle APB is equal to O-e, and the angle AFC to 90°-n-e. Hence Fcoa(e-e)-Wsia{a + e) = Q, and substituting for e its value tan~V, E-_p|7 sin g -Ha cos g " cos S -I- A sin S The reader should apply this method to some of the Examples of 327 and 329. 355. Potential. — The region surrounding one or more centres of force (an attracting mass, for example) is called a field of force. If a particle be moved from any one to any other point in such a field, work is in general done either by or against the resultant force of the field, and the amount .of work so done we have seen to be inde- pendent of the path (338). If therefore some convenient point of reference be chosen, the work done in bringing a given particle, say a particle. of unit mass, from any other point to the chosen point has a definite value for every point of the field. So also will the work done in carrying the given particle from the chosen point to all other points of the field. This definite value, when the given particle is one whose mass is unity, is called the potential of the point. The magnitude of the potential of a point will depend upon the position of the point of 356] OF A PARTICLE. 263 reference, and its sign will vary according as we give the name potential to the work done during motion to or from the point of reference and by or against the force of the field. The choice of the point of reference and of the exact mode of defining potential are matters of conven- ience and vary with the kind of field of force under consideration. 356. The importance of the potential depends upon the following proposition : — The rate of change of the potential per unit distance in any direction at any point of afield of force is equal to the coTnponent force in that direction with which a particle of unit mass would be acted upon if placed at that point. — Let A, £ he two points in the field of force and G the chosen point of reference. Since the work done during any displacement is independent of the path of the particle, the work done in carrying unit mass from A to £ is equal to the difference of the amounts of work done in carrying it from AioC and from S to C. Hence, if F^ and Vb are the potentials of A and B, the differ- ence Va ~ Vb is equal to the work done in carrying the unit mass from A to B or from B to A. If now F is the component in AB of the mean force per unit distance acting on the particle between A and B, the work done between A and B is F . AB. Hence F.AB^Va-Vb, and ^ F=(rA^rs)/AB. If now B be indefinitely near A, F becomes the compon- ent force at A in the line AB, and (F4 - Vb)/AB the rate of change of the potential at A per unit distance in the line AB. Hence the above proposition is proved. As the value of a central force at any point of the region through which it acts is equal to the rate of change of the potential at that point, such forces are said to be derivable from a potential. 264 DYNAMICS [356 It follows from 347 that F .AB is the difference be- tween the potential energies of unit mass at A and at B. This difference is thus equal to the difference in the values of the potential for these points. Hence the appropriate- ness of the term potential energy. 357. If at any point, F= 0, there also the rate of chaoge of potential must be zero. Hence, e.g. (316, Ex. 5), at all points inside a uniform spherical shell the gravitational potential is the same. 358. Fqwipotential Surfaces. — A surface, at every point of which the potential has the same value, is called an equipotential surface. The attraction on a particle placed at any point of such a surface will be normal to the surface. For in no direction tangential to the surface is there a rate of change of potential or, consequently, a component force. We may imagine equipotential surfaces drawn in any field of force for any values of the potential. If they be drawn for values increasing by equal amounts, which are also small, the resultant force acting at any point will be inversely proportional to the distance between successive equipotential surfaces in the neighbourhood of the point. For, if A and B are the successive equipotential sur- faces, and AB the distance between them at any point, y^A ~ "^B is constant, and hence (356) F oc 1/AB. 359. Lines of Force. — ^A line so drawn in a field of force, that its direction at any point is also the direction of the resultant force at that point, is called a line of force. As the resultant force at a point has no component in the tangent plane of the equipotential surface passing through the point, lines of force must be normal to the equi- potential surfaces they may meet. 360. Tubes of Force. — If from points in the boundary of any portion of an equipotential surface lines of 362] OF A PARTICLE. 265 force be drawn, the space thus marked off is called a tube of force. 361. Oravitational Potential.— We may consider, as of special importance, the potential in a field of force due to gravitational attraction. If such a field is due to the attraction of a single particle of mass m, the force on unit mass at unit distance (the astronomical unit of mass (315) being employed) is m. Hence (340) the work done in moving unit mass from one point at a distance r to another at a distance JR is m(l/r— 1/i?). If R is in- finitely great, the work done is equal to m/r. Hence if the chosen point of reference be a point at an infinite distance from the attracting particle, the potential of a point at a distance r has the magnitude m/r. If the field is due to any number of particles of masses, m^, m^, etc., the magnitude of the potential will be 'Si{m/r). It is convenient to have the potential for aU points of a gravitational field positive. Now gravitational force being in all cases attractive, the work done by the force of a field in moving a particle from a greater to a smaller distance from the attracting mass is always positive. Hence in this case we define the potential of a point as the work done by the force of the field in moving unit mass from a point at an infinite distance from the attracting mass, to the given point. 362. It follows that the component force on unit mass at a given point of a gravitational field in a given direc- tion is equal to the rate of increase of the potential per unit distance in the same direction. It follows also (347) that with the above convention, if Pa and Pb are the potential energies of unit mass at A and B, and Va and Fg the potentials of these points respectively, Va-Vb=-(Pa-Pb); 266 DYNAMICS [ 362 and that therefore the rate of increase of the potential with distance in a given direction is equal to the rate of decrease of the potential energy of unit mass in the same direction. 363. Calculation of the Potential. — The value of the quantity 2(m/r) for a given point may, in simple cases, be determined by elementary mathematical methods. Usually, however, the Integral Calculus is necessary to effect the summation. 364. Examples. (1) Show that the potential at a given point due to particles of masses m^, m^, etc., situated on either a circle or a, sphere whose radius is r and centre the given point, is equal to (2m)/r. (2) Particles of masses 3'928, 39"28, and 392'8 kilogrammes are situated at three of the corners of a square whose side is 1 metre. Find the potential at the fourth corner. Ans. 1-0807 C. G. S. units. (3) Find the potential (a) at the centre of a thin circular wire of linear density p, and (5) at a point on a line through its centre per- pendicular to its plane, distant I from all points of the wire. Ans. (a) 27r/), (6) ^icprjl. (4) Find the potential at the centre of a circular plate of radius r and surface density p. Ans. 'ivpr. (5) Find the potential at the centre due to a sector of the plate of Ex. 4, of angle 6 radians. Ans. r0p. (6) Find the potential (a) at any point inside a uniform spherical shell of mass m and radius r, and (6) outside it at a distance d from its centre. [See 316, Exs. 5 and 6.] Ans. (a) mir, (b) m/d. 365] OF A PARTICLE. 267 (7) A, a point uear the earth's surface, is h feet above another such point B. Find the excess of the potential of A over that of A Ans. -gh. 365. Integral Normal Attraction over a Surface. — If any closed surface in a field of force be divided into indefinitely small portions, the sum of the products of the areas of these portions into the normal components out- wards (or inwards) of the forces exerted at them on unit mass is called the integral normal attraction over the surface (in the language of the Higher Mathematics, the surface integral of normal attraction). The integral normal attraction over any closed surface in a gravitational field of force is equal to 4x times the mass enclosed by the surface. — Let m be the mass of any Figi Figs particle, at 0, of the attracting mass. Let a cone of indefinitely small solid angle meet the closed surface 8 at Pj, Pj, P3, P^, etc., marking out on it areas A^, A^, A^, A^, etc., inclined to orthogonal sections of the cone at the angles 6^, 6^ 6^, 6^, etc., radians. The resultant force due to this particle at P^, P^, P^, P^, etc., is towards and inversely proportional to OP,^ OP^, OP^, OP^, etc. The normal components at these points are therefore pro- portional to cos ejOP^, cos ejOP^, etc. The orthogonal sections of the cone at P^, P^, etc., have areas proportional to OP^, OP^, OP^, etc. Hence the sections inclined to them at the angles d^, 6^, etc., at the same points, have, since the cone is of indefinitely small angle, areas pro- 268 DYNAMICS [365 portional to OP^^/cos 6,, 0P,7cos d^, etc. The products of the areas A^, A^, etc., into the respective normal components of the force over them are therefore con- stant. If now the point be outside the surface, the force at Pj is outwards, that at P^ inwards, that at P^ out- wards, that at P^ inwards, and so on ; and the cone must meet the surface an even number of times. Hence, if the forces at Pj, P^, etc., be reckoned either all outwards or all inwards, as many of the above equal products are positive as negative, and their algebraic sum is con- sequently zero. But the whole surface may be divided into indefinitely small areas by such cones. Hence the integral normal attraction over the surface is equal to zero when the attracting mass is a particle outside it. If the point be inside the surface, the cone whose vertex is will cut the surface in whatever direction it be drawn an odd number of times. Hence the sum of the products of the areas intercepted by the cone into the normal components of the attractions at them is equal to the value of the product at any one section. At P the normal component of the attraction is m cos dJOP/- If w is the 'solid angle of the cone (in solid radians), the area of the section at Pj is w. OP^/cos 0^. Hence the value of the above product at Pj is wm, and consequently the value of the integral normal attraction, is the product of m into the sum of the solid angles of all the cones with (inside the surface) as vertex, by which the surface may be divided into small elements, which is 4^. Hence the value of the integral normal attraction when and therefore m are inside is 47rm. Hence its value when any mass M is inside is 366. In a tube of force whose ends are indefinitely 368] OF A PARTICLE. 269 small portions of equipotential surfaces, the force perpen- dicular to the tuhular portion of the surface vanishes, and at the end surfaces the resultant force is normal. — Let jF\ and F^ be the resultant forces at the ends, and let Sj, Sj be the indefinitely small areas of the ends. If then the tube contain no attracting mass, one of the two, F^, F^, is outwards, the other inwards, and we have, by 365, F^s^—F^8^ = 0, i.e., if F be the force at any point of a small tube of force and s its normal section at that point, ^s = constant, ov F oc 1/s. Hence the resultant force at any point of a small tube of force is inversely proportional to its transverse section at that point. 367. If the attracting mass is a uniform spherical shell or a sphere with density proportional to distance from the centre, the lines of force are clearly straight lines radiating from the centre, and the tubes offeree are cones, right sections of which are directly proportional to the squares of their distances from the vertex of the cone or centre of the sphere. Hence the attraction exerted by a sphere such as that specified above, at external points, is inversely proportional to the squares of their distances from its centre. (Compare 316, Ex. 6.) 368. If the attracting mass is a cylinder of circular section and infinite length and with density proportional to distance from the axis, the lines of force are clearly straight lines perpendicular to the axis of the cylinder, and the tubes of force are therefore wedges, the areas of right sections of which are directly proportional to their distances from the axis. Hence the forces at external points are inversely proportional to their distances from the axis. 270 DYNAMICS [369 369. If the attracting mass is a plate of uniform thick- ness and infinite extent, and with the density at its various points proportional to their distances from either bounding surface, the lines of force are clearly straight lines normal to either bounding surface, and the tubes of force therefore are cylinders of constant section. Hence the forces exerted at all external points are the same. 370. If a tube of force cut through a plate of attracting matter of surface density p, and if the area of the plate inside the tube he or, we have (365 and 366) ^A--f'2S2-47r/)cr. If the plate be indefinitely thin and the ends of the tube indefinitely near the surfaces of the plate, s^ = (t = s^. Hence F^-F^^4>Trp, i.e., the attractions on unit mass on opposite sides of the plate at points indefinitely near it difier by inrp. As they are clearly equal in magnitude and opposite in direc- tion, the attraction at either side indefinitely near the plate is thus 2x/). (Compare 316, Ex. 1.) 371. The potential cannot have a maximum or a mini- mum value at a point in free space. For, if it could, it must increase or diminish respectively from point to point in all directions outwards from the given point, and hence the force at all points of a small surface enclosing the given point must be outwards or inwards respectively, / and must consequently have a finite value, though the surface encloses no attracting mass. Hence, if the potential is constant over a closed surface containing no attracting mass, it must be constant throughout the whole enclosed space. For otherwise there must be somewhere in it a point of maximum or minimum potential. 373] OF A PAKTICLK. 271 372. A field of force whose law is that of gravitation may be so mapped out by lines of force that they may indicate not only the direction, but also the magnitude of the forces acting at different parts of the field. For let Sj, Sj be normal sections of any tube of force not enclosing any attracting mai^s, and F^, F^ the resultant forces on unit mass at these sections. Then these sections are cut through by the same number of lines of force. Let the number (n) be such that F.^ = n/Sj^. Then, since FjS^ = F^8^, we have also F^ = n/s^. Hence, if the lines of force in a tube of force are so drawn that at any one point the quotient of their number by the normal .section of the tube is equal to the force at that point, the same will be true for any other point. If therefore the lines of force of a field are so drawn that over any equipotential surface the number of lines of force per unit of area at every point is equal to the force at that point, then throughout the field the number of lines of force per unit of area normal to them at any point will be equal to the force at that point. 373. A uniform field of force is one at all points of which the resultant force has the same magnitude and direction. The tubes of force must therefore be cylinders, and the lines of force must be parallel straight lines, equal numbers of which pass through equal areas normal to their direction. 2 72 DYNAMICS [374 CHAPTER III. DYNAMICS OF SIMPLE SYSTEMS OF PAETICLES. 374. For the discussion of the motion of a single particle we have found the first two laws of motion to be sufficient. If we wish, however, to discuss the motions of even only two particles which act upon one another, we have to deal with both aspects of the stress between them and must know how the stress affects both particles. The third law tells us that, it affects them equally in opposite directions, producing in them equal and opposite changes of momentum in the same time. "With the aid of the third law it is often possible to pass from particle to particle of a simple system, applying to each particle the equations of motion or the conditions of equilibrium of a single particle, and thus determining the motion of the whole system. 375. It is hardly necessary to point out that the law of energy also may be applied to a system of particles. For since, if the system is in motion, the increment of the potential and kinetic energies, together with the work done against non-conservativfe forces, during any displacement, is for each particle equal to zero, it must be equal to zero also for the whole system. And since, if the particles of the system are in equilibrium, the sum of the quantities of work done by the forces acting on each particle during any small displacement is zero, it is zero also for all together. 377] ~ OF SIMPLE SYSTEMS OF PARTICLES. 273 376. The forces acting on a system of particles may be divided into two classes, those acting between the particles of the system and bodies exteriial to the system, called external forces, and those acting between the particles of the system themselves, called internal forces. The internal forces may be mutual attractions such as gravitational attraction, explosive forces, reactions exerted during collision, or the stresses or tensions in connecting strings. Some of these cases may be dealt with without further comment. 377. Examples. (1) Two particles of masses, 20 lbs., and 1 lb. respectively, initially at rest on a smooth horizontal table attract one another. After a time the greater mass has a velocity of 10 ft. per sec. Find the velocity of the other. Ans. 200 ft. per sec. in the opposite direction. (2) Two attracting particles initially at rest on a smooth hori- zontal table are observed at a given instant to be approaching one another with a speed w, the speed of each particle being measured relatively to the other. If m and M are their masses, find their speeds v and V respectively relative to a fixed point in the table. Ans. v=Mu/{m+M), V=mu/{m + M). (3) A body having a velocity of 10 ft. per see. in a given direction is divided by an explosion into two portions whose masses are 2 lbs. and 1 lb. respectively. Both portions move, after the explosion, in the original line of motion, and the portion of smaller mass has a velocity of 25 ft. per sec. in the original direction of motion, (a) Find the velocity of the other portion. (6) Find what it would have been had the velocity of the smaller portion been 50 instead of 25 ft. per sec. (c) Find the value of the explosive impulse in the latter case. Ans. (a) 2J ft. per sec. in the given direction, (6) 10 ft. per sec. in the opposite direction, (c) 40 absolute ft.-lb.-sec. units of impulse. (4) Two particles of masses m and m' astronomical units, moving on a smooth horizontal table, attract one another according to the S 274 DYNAMICS [377 gravitational law. Find the acceleration of either relative to the other when they are at a distance d. Each is acted on by a force mm'IcP. Hence their accelerations are m'l(P and ml

, they must be equal. Hence m(u— V) = = n(V—v). 7nu+nv Hence also V= m+n -. mniu — v) and d> = ^ \ ^ m+n 379. If there be recoil, the impulse of the stress $ may be determined in terms of the value it would have if there were none. For u' and v' being the component 379] OF SIMPLE SYSTEMS OF PARTICLES. 275 velocities of iii and n respectively after collision, in the direction of the stress, we have Hence m{u — xd) = ^ = n{v'— v). # Vj — u' v'—v

m') are connected by a massless inextensible string which passes over a smooth horizontal cylinder (or peg, or pulley). Knd their accelerations and the ten- sion in the string. As the tension T is the same throughout the string, each particle is acted upon by two forces, T vertically upwards and its weight down- wards. As the string is inextensible and m is greater than m', m will move downwards with the same acceleration a with which m' moves upwards. The resultant force downwards on m is ^'' mg-T. Hence (317) a={mg-T)lm. The result- ant force upwards on w! is T—m'g. Hence a=(T-m'g)lm,'. Equating these values of a we obtain {mg- T)/m=(T-m'g)/m'. Hence T=^^^g; m + m' and substituting this value of J' in either of the above expressions tor a, we have a= a. m + m'" Otherwise, by applying the law of energy, thus : Let m move down and therefore m' up through a distance s, and let the initial and /•■T mg 280 DYNAMICS [382 final velocities be V and v respectively. Then the gain of kinetic energy is ^{m + m'Xv^— V^). Equal amounts of work are done by and against T. The work done against the weight of m' is m'gs, that done by the weight of m is mc/s. Hence the total gain of potential energy is {m' — m)gs. If we neglect the resistance of the air there are no othe* forces acting. Hence and »«- V^ = ^~'^',qs, m+m' from which it follows that the particles are moving with constant acceleration of magnitude (m — m')g/(m+m'). If m and m' are known and if a be observed, g may be determined. But, as no smooth bodies exist in nature and the conditions of the above ideal problem cannot therefore be realized, this mode of determining 5' is of no value. Atwood's Motchine, a piece of appar- atus of historic interest, is an attempt to realize as nearly as possible the above ideal arrangement. The string passes round a pulley so rough that the string does not slip on it. The axis of the pulley is mounted on " friction wheels " which diminish the friction of the axis very greatly. When the particles move the pulley rotates, and the kinetic energy produced exists partially in the rotating pulley. Work is also done against friction and the resist- ance of the air. The complete discussion of this apparatus is therefore too complicated for us at our present stage. (2) At the extremity of a string which passes over a frictionless pulley moving in a vertical plane are masses of m and 3 lbs. Initially the masses are at rest at the same height, and 3 seconds later the mass m is 72 feet below the other. Show that ra = 5 lbs. (3) Bodies of p and q grms. {p > q) respectively are attached to the ends of a string which passes over a pulley. At the end of each second after motion begins, 1 grm. is taken from p and added to q without jerking. Show that the motion will be reversed after^ — g"-!-! seconds. (4) Two particles of masses m and m' move on two rough inclined planes (inclinations a and a', coefficients of friction m and mO in a 382] OF SIMPLE SYSTEMS OF PARTICLES. 281 vertical plane normal to the intersection of the inclined planes. They are connected by a strjng which passes over a smooth peg at the common summit of the inclined planes. Find their common acceleration, assuming m to move down its plane. Ans. {m(sin a - /t cos a) - ?n'(sin a' +/co8 a')]gl{m + m'). (5) A body weighing 19 lbs. is placed at the centre of a smooth round table 6'44 ft. in diameter. It is moved by a body weighing 1 lb. at the end of a cord passing over the edge of the table. How long will it be before it reach the edge of the table ? Ans. 2 sees. (6) A mass of 6 oz. slides down a smooth inclined plane whose height is half its length, and draws another mass by means of a string along a smooth horizontal table which is level with the top of the inclined plane over which the string passes. In 5 sees, from rest it moves through 3 feet. Find the mass on the table. Ans. 396-5 oz. (7) A string having at one end an unknown mass M, and at the other a small smooth massless ring, hangs over a smooth horizontal cylinder. Through the ring a second string passes, having at its ends masses m and m! {mf ■>m). Find (a) what value M must have in order that m', if initially at rest, may remain at rest during the motion of the system, and (6) the acceleration of the ring. Ans. (a) ;, (6) ^ a. (8) Three particles A, B,G of masses mj, mj, m^ {m^>m2) are connected by strings, .4 to i? and B to C. The string between A and B passes over a smooth horizontal cylinder. C lies on a table vertically below B and the string joining B and C is slack. At a given instant A and B begin to move from rest, and after t sec. the string between B and C tightens. Determine the subsequent motion. The acceleration with which A and B move while BC is slack is, Ex. 1, (jKi - m^glim^ + m^. Hence their velocity at the instant at which BC becomes tight is {m,i-m^gtl{m^ + m,^. Call this v. Let u be the common speed oi A, B, and C immediately after the tight- ening of BC. Then C"s momentum upwards has suddenly increased 282 DYNAMICS [382 by TO3M. It has therefore been acted upon by a short-lived stress of impulse m^u upwards. A'& momentum in an upward direction has changed from —m^ to — m^M. Hence it has been acted upon by a stress of impulse m,i{v-u) upwards. Both these stresses have acted on 5— the latter upwards, the former downwards. Hence the impulse of the resultant upward stress on B is m^{v-u)-miU. Now ^'s momentum upwards has changed from m^i) to m^. Hence (319) and mj(i> - It) — m^ii = TTi^iu -v), The acceleration with which A, B, and C move after BC becomes tight may be shown (as in Ex. 1) to be m^+m^ + m^ Hence the subsequent motion is determined. If m^+in^^ m,^, the acceleration is negative, the velocity will gradually diminish from u to zero, and the direction of motion will then be reversed. "We cannot apply the law of energy to a problem such as the above, because we do not know what non-conservative forces may be acting, and cannot therefore determine the work done against them. (9) Three bodies A, B, C oi equal mass are connected by strings A to B and B to C. A and B are placed close together on a smooth horizontal table, and C hangs over the edge. The string AB is 3 ft. long, and B is 3'5 ft. from the edge. Find the velocity of A (a) when it begins to move, and (5) when B arrives at the edge. Ans. (a) Z'Jg/S, (b) Jbgfs. (10) Two particles A and B of equal mass are connected by a string which passes over a smooth horizontal cylinder. While moving with a speed v (A moving downwards) a third particle, C, of the same mass and at rest, is suddenly attached to the string between A and the cylinder. Find {a) the common speed of A and 382] OF SIMPLE SYSTEMS OF PARTICLES. 283 C immediately after C's attachment, (6) the time after which the string BC again becomes tight, (c) the common speed of A and C and the speed of B just before the string BC tightens, and {d) the common speed of all three just after the string tightens. Ana. (a) »/2, (6) v\<^g), (c) v, ?)/2 ; (rf) 5«)/6. (11) In Ex. 4, find the ratio of the masses m and m' that there may be equilibrium with m on the point of moving down its plane. Equating to zero the value of the acceleration found above (Ex. 4), we have m(sin a - M cos a) - m'(sin a ' + /I'cos a') = 0, m _ siu a' + yn'cos a' ml sin a — At cos a ' Otherwise thus : The acting forces are represented in the figure. and which will be understood without explanation. Resolving forces parallel to the planes we obtain T—imgwsxa + ii,R=Q, T -m'g sina - ij! R' =0. Resolving in directions perpendicular to the planes we obtain R-mgco&a=Oj R' -m'gcosa'—O. Henee, substituting for R and R' in the first pair of equations their values as given in the second pair and eliminating T, we obtain m _ sin a + /i'cos a' m! sin a — /* cos o ' 284 DYNAMICS [382 Otherwise thus : Let the particles move a short distance s, m down the plane CA and m' up the plane CB. Then, equating to zero the algebraic sum of the amounts of work done by the various forces on both particles during the displacement, we have mgs sin o - m'ffs sin a' — fiRs — /i'-R'n = ; and substituting for R and R' their values mg cos a and m'g cos a' respectively, we obtain the same result as above. (12) Two particles A and B of weights W and W are connected by a string. A rests on a rough inclined plane (inclination = o, co- efficient of friction=Ai) over whose smooth summit the string passes ; and B hangs freely. Find the ratio of W' to W that there may be equilibrium with A on the point of moving up the plane. Ans. sin a + M cos a. (13) A string fastened by one end at a fixed point A passes through a fixed smooth ring at B, AB being horizontal, and is pulled by a force at its other end. Between A and B a body of weight W is hung by a smooth ring moveable on the string. How near to AB will it be possible to raise this ring by pulling at the string, if the string can bear a tension equal to 2 PT only. Ans. ABjQQK (14) A rough parabolic wire is placed with vertex upwards and axis vertical. A small ring of weight W moving on the wire is supported at one extremity of the latus rectum by a body of weight W attached to a string passing over a smooth peg at the focus. Find the coefficient of friction. Ans. {W-W')I{W+W'). (15) Two small smooth rings sliding on a circular wire in a vertical plane are connected by a string which passes through a small smooth fixed ring at the highest point of the circle. Show that if the masses of the moveable rings are inversely proportional to the adjacent segments of the string, there will be equilibrium. (16) Two particles A and B (masses m, and m') rest upon smooth inclined planes of inclinations o and ^ respectively. They are con- 382] OF SIMPLE SYSTEMS OF PARTICLES. 285 nected by a string (length =i) which passes over a smooth peg P, vertically over the common summit G of the two planes, and distant h feet from it. Find the inclinations e and

. Hence m'g . Bb . sin ;8/cos ^ -mg .Aa. sia a/cos 9 = 0, and We have moreover and Hence m m sin a 'cos 6 COS0 sin|8' AP : Persia AGP : sin PAG = coao : sine, ; PC= cos ^ : sin0. BP . ap+pb=pg(^-^+?^) Vsinff sin0/ V sin 6 sin 0/ 286 DYNAMICS [382 This equation with that obtained above are sufficient to determine and (p when there is equilibrium. The reader should solve the problem also by applying the condi- tions of equilibrium in other ways. (17) Two particles, of masses m and m', connected by a string, rest upon the edge of a smooth vertical circular disc. Find the position of equilibrium and the tension T in the string. Ans. If a is the angle subtended at the centre by the string and ;3 the angle subtended by the portion between m and the highest point, ^=tan-^ m'sina y^ T'^/^? u- m + m cos a {m^ + m^ + 2mm cos o)' (18) Three smooth tacks A, B, C are driven into a vertical wall, B and C being on the same level. A string, to whose ends bodies of equal weight to are attached, is hung over the three tacks. Find the forces exerted by them on the string when there is equilibrium. Ans. v&(i. + cos A)\ w2*(l - sin 5)*, ■w2*(l - sin C)* respectively. (19) Two rough bodies rest on an inclined plane and are con- nected by a string which is parallel to the plane, W and W being the weights of the bodies, and |U, / their coefficients of friction, and the rougher body having the higher position. Find the greatest inclination of the plane consistent with equilibrium. Ans. ta.u-\(^,W+,,'W')/(W+W')l (20) Two bodies weighing A and B lbs. respectively are connected by a string and placed on a rough horizontal table (coefficient of friction =/*). A force P, which is less than n(A+B) but greater than /I, s/a^ + B^, is applied to A in the direction BA, and its direc- tion is gradually turned round through an angle in a horizontal plane. Show that both A and B will slip when 6 = cos .tjj:\E^-A^)+P'^ 'ift.BP Show also that, if P is less than y. iJa^+B'' but greater than liA, it will cause A only to slip, and that -4 will slip when S=sin-^(M/^)- (21) Particles A, B, etc., n in number, of weights iVi, w^, etc., are connected to one another and to two fixed points P and Q whose 382] OF SIMPLE SYSTEMS OF PARTICLES. 287 horizontal distance is d and vertical distance h, by weightless strings, F being connected to A, A to B, and so on, and the last to Q. The string oounecting P and A has the length a^, that connect- ing A and B the length aji ^^^ so on. Find the tensions Ti, T^, etc., in these strings and their inclinations Oj, Oj, etc., to the horizon, when the particles are in equilibrium. Each particle is acted upon by three forces, its weight and the tensions in the strings attached to it. Since Wj, Wj, etc., are all verti- cal, and since T^ and T^ are in the same plane as Wj, 'f^ and T^ in the same plane as w^, and so on, the whole system must be in the verti- cal plane through P, Q. A is in equilibrium under the forces 1\, 7*2, and Wj. Hence, resolving horizontally and vertically, we get ^icos oi - 2'2Cos ttj =0, and 7\sin oj - T^sin og - it'i = 0. Similar equations may be obtained for each particle — in all in equations. Moreover and d=^a^cosai+a20osa2+e,tc. H-a,„+iCos a„+i, h = ajsin a^ + cjjsin a,^ + etc. + a„+isin a„+ j. We have therefore 2«-l-2 equations involving 2w-l-2 unknown quantities, viz., m -1-1 tensions and n+\ inclinations. The inclina- tions of the strings being determined and their lengths given, the positions of the particles are known. 288 DYNAMICS [382 (22) Particles A, B, C, etc., m in number, are connected by- weightless strings A to B, B to (7, etc., and the n^ to a fixed point Q. A force of given magnitude T^ is applied to A through the string PA. Knd the weights of the particles that the strings PA, AB, etc., may have given inclinations a^, a^, etc., to the horizon. P. From any point draw OP^, OP 2, etc., with inclinations bj, a^, etc., to the horizon. These lines have therefore the same directions as the strings PA, AB, et-a. Draw a vertical line meeting OP-^, OP 2, etc., in Py, P^, etc. If T2, T^, etc., are the tensions in AB, EC, etc., the particle A is acted upon by three forces, T-^, T^, and Wj. These are represented in direction by the lines OPj, P^O, and PxPi respectively. Hence (325, c?) they are also represented by these lines in magnitude. Similarly the forces acting on B, viz., T^, 1\ and w^, are represented in direction and thei'efore also in magnitude "Pn.i ijy Qp^^ p^Q^ and P2-P3 respectively. Thus it may be shown that P1-P2, P2P3, PsPt, etc., P^Pn+i represent the weights %, W2, etc., on the same scale as that on which OP^ repre- sents 7\. Hence the values of w^, w^, etc., may be determined by carefully drawing the diagram (called a force diagram) and measuring the lengths of PiP^, Pji^s, etc. For this reason the above method is called a graphic method. It is of great practical value for the rapid solution of engineering problems. (23) Particles A, B, C, etc., are connected together and to two fixed points, as in Ex. 21, and are in equilibrium, their masses m^, m2, etc., and the inclinations of the strings Oj, Ug, etc., being known. Any one of the strings is cut, say BC. Find the tensions t-^, (2 in PA, AB respectively immediately afterwards. [These tensions are called initial tensions, because they are the tensions when G, B begin to move.] A moves, after the cutting of the string, in a circle about P. At the beginning of its motion its speed is zero, and hence the com- ponent of its acceleration normal to its path {i.e., in the direction ^P) is zero. Its acceleration is therefore initially wholly tangential 382] OF SIMPLE SYSTEMS OF PARTICLES. 289 to its path (i.e., in a direction perpendicular to AF), A'a accelera- tion in the direction AP being zero, the sum of the components in that direction of the forces acting on A is zero also. Hence ti — *2cos(oi — Og) — Mifli sin Oj = 0. A'a acceleration in a direction perpendicular to AP is the quo- tient by its mass of the sum of the component forces acting on it in this direction, and is therefore [m^g cos Oj — i^sinC"! — "2)]/™!- B also after the cutting of the string moves in a circle about A , and as above it may be shown that it has no acceleration relative to A in the direction BA. Hence its acceleration in this direction is equal to ^'s component acceleration in the same direction, and is therefore equal to But it is also equal to (i2 — ^iff ^™ "2) Ma- Hence -^ - Cf sin a2=g cos ajsin (a^ — Og) ^ sin2(ai — 02)- We have therefore two equations containing no other unknown quantities than t■^ and t^, which therefore may be determined. The instantaneous changes of tension on cutting BC are of course 2\ — t-^ and T^-t2, where 2\ and T^ are the tensions before cutting as determined in Ex. 21. (24) A particle is connected by two equal strings to two points in the same horizontal line and is in equilibrium. Show that, accord- ing as the inclination of the strings is less or greater than a right angle, will the tension of either string be instantaneously increased or diminished by cutting the other. 290 DYNAMICS [383 CHAPTER IV. DYNAMICS OF FLEXIBLE INEXTENSIBLE STRINGS. 383. A string or cord or chain may be considered to be a series or row of particles or elements placed end to end. It may thus be regarded as a system of particles less simple than those of Chapter III., but more simple than those of subsequent chapters. A perfectly flexible string is one which is capable of being bent without the exertion of any finite force. An inex- tensible string is one whose length is constant. Flexible and inextensible strings are ideal. Real strings all re- quire force to bend them and can be elongated. In many cases however the forces required to bend real strings are so slight and the elongations under the acting forces so small that they may be considered to be practically per- fectly flexible and inextensible. Since such a string may be bent at any point without the exertion of a,nj finite force, the internal forces acting at that point can have no component normal to the direc- tion of the string. For, otherwise, this component would have to be overcome in bending the string and a finite force would be necessary. Hence the stress in a flexible string has at any point the direction of the string at the point. We restrict our attention to the simple case in which 384] OF FLEXIBLE INEXTENSIBLE STRINGS. 291 the string itself and the external forces acting on ,it are in the same plane. 384. — Equations of Motion. — Let AB be a tense string of which PP' is any element. Let the stresses in the string at P, P' be T, T'. Then the element PP' is acted upon at its end-points by forces T, T tangential to the string at P, P' respectively. Let it also be acted upon by some external force whose magnitude we may indicate by the product F\ where A is the length of the element PP' and F consequently the magnitude of the external force acting on the string per unit length of the string. Let the lines of action of T and T' be inclined at the angle Q, those of T and F\ at the angle (p. Also, let a be the linear density of the string at PP', and a« and «„ the components of the acceleration of the element in directions tangential and normal to the string at P. Then, resolving tangentially and normally, we have, as the equations of motion of the element (317), T'cos e-T+F\coB4, = a-Xat, T'sin 6-FXsm^ = aXan- 292 DYNAMICS [385 385. Conditions of EquiUbrium. — Putting at = an = in the equations of motion we obtain those of equilibrium, viz., T'cos e-T+ FXcos = 0, T'sind-F\sm^ = 0. 386. The above equations hold for every element of the string. The results which may be deduced from them will vary with the nature of the external forces. 387. (1) No External Forces. — If there are no external forces, F= 0. Hence the equations of motion become T'coaO-T^aXat, T'siu 6 = crXan. Ultimately, when P' is very near F, 6 is indefinitely small, and consequently cos = 1 and sin 6 = 6. Also, ultimately T' is indefinitely nearly equal to T, and X/6 = p the radius of curvature at P. Hence the above equations become {T'-T)/\ = =^0. 294 DYNAMICS [390 Hence, when there is equilibrium, the rate of change of the stress in the string at any given point, with respect to its distance measured along the string from a fixed point in the string, is equal to the tangential component of the external force per unit of length, at the given point ; and the curvature of the string at any point is equal to the ratio of the normal component of the external force (per unit length of the string) to the stress at that point. As instances of external forces continuously applied, we may take the reactions of continuously curved sur- faces on strings wound round them, and the weights of heavy strings. 391. (a) The External Force being the Reaction of a Continuously Curved Surface. — First, let the surface be a smooth one over which the string is stretched. Then, as we are supposing the string to have no weight (and in many cases the weight is so small relatively to the stress that it may be neglected), each element of the string is acted upon by three forces only, viz., the reaction of the surface, FX, normal to the surface, and the tensions T, T', whose directions are those of consecutive tangents to the string. Hence in the special case to which we restrict ourselves (383) the osculating plane (41) of the string at any point is normal to the curved surface, and the form of the string is that of what is called a geodetic line on the surface. Since F\ is normal to the surface and therefore to the string,

, Eoi a, regular hexagon ABCDEF. Ans. On the line joining the centre of the circumscribing circle ■with C, and at a distance from equal to OC/5. (6) At the corners A, B, G, D, E, F, O, H of n cube of 1 ft. edge, particles are placed of 1, 2, 3, 4, 5, 6, 7, 8 lbs. respectively. Find the centre of mass. Ans. Distance from face ABCD, i% ft. ; from ABOF, f ft. ; from AD£F, f ft. (7) A piece of uniform wire is bent twice at right angles so as to form three sides of a square of side a. Show that the distance of the centre of mass from the centre is a/6. (8) Find the centre of mass of a uniform wire, bent into the form of a scalene triangle. Ans. It is at the centre of the circle inscribed in the triangle formed by joining the middle points of the sides of the scalene triangle. (9) Find the centre of mass of a uniform wire, bent so as to have the shape of n of the sides of a regular polygon. Let ABCDEF be the wire. Then the centres of mass of the portions AB, BC, etc., are at their middle points a, h, c, d, e. Let be the centre of the inscribed circle and let its radius be r. Let ae subtend at an angle 2a, and let ab, be, etc., subtend each the angle $. Then 2a={n — l)ff. Take Oa as axis of x and a line perpendiciilar to it as axis of ,y. Then the distances of a, b, c, etc., from Oi/ are r, r cos 0, r cos 2ff, etc., and their distances from Ox are 0, rsinfl, rsmie, etc., respectively. Hence, 310 DYNAMICS [408 if X, y are the distances from Oy, Ox respectively of the centre of mass of the wire, !=-[! + cos 9 + cos 29 + etc. + cos {n - 1 )9] ^r cos (w - 1/2) 9 si n (w/2)g ^ n' sin (9/2) _r cos a sin {njn — 1 )a '1 sin(o/n — 1) y = -[sin 9 + sin 29 + etc. + sin (n - 1 )9] n _^r sin (h - 1/2)9 sin (m/2)9 ^ _r sinasin(ra/ji — l)a ™' sin(o/»r^l) Hence y/S=tana, /=2-r=2 '• 8in(m/m-l) a and 'Jx' + v' = r,- . , , — - -> " n sm(a/m-l) i.e., the centre of mass is on a line through whose inclination to the X axis is a, and is at a known distance from 0. (10) rind the centre of mass of a uniform wire in the form of a circular arc. [If n of Ex. 9 be made indefinitely great, and the sides of the polygon mdefinitely short, ABCDEF becomes a circular arc subtending at its centre an angle 2a.] Ans. Distance from centre=?'sino/a. (11) The distance of the centre of mass of a uniform semi-circular wire of radius r from its centre is %r\Tr. (12) Find the centre of mass of a uniformly thin homogeneous triangular plate. — Let ABC be such a triangular plate, and let it be divided by lines parallel to BC into an indefinitely great number of indefinitely narrow strips. Then the centre of mass of ^ach strip is its middle point. Now the middle points of all these strips lie on the line AD drawn from il to Z) the middle point of BC. Hence * See Todhunter's " Plane Trigonometry," chapter on Summation of Series. 408] OF EXTENDED BODIES. 311 the centre of mass of the plate is in the line AD. Similarly if E is the point of bisection of AB, the centre of mass lies in EC. Hence it is the point F in which AD and EC intersect. Since E and D are the middle points of AB and BC, ED is parallel to AC. Hence the triangles AFCsxA DFE&re similar, and DF:FA=DE:AC=\ : 2. Hence the centre of mass is on the line DA, and at a distance from D equal to DAjZ. (13) Show that the centre of mass of the triangle* formed by join- ing the middle points of the sides of a triangle has the same posi- tion as that of the latter triangle. (14) ABC is a triangle and D a fixed point in BC. A triangle BPC is cut away, whose vertex P is in AD. Show that whatever be the position of P, the centre of mass of the remainder lies on a fixed straight line. (15) Given the base and perimeter of a triangle, show that the locus of its centre of mass is an ellipse. (16) Prove that the centre of mass of the trapezoid formed by joining the middle points of two sides of a triangle is on the line joining their point of intersection to the middle point of the third side, at a point which is 2/9 of this line's length from the middle point of the third side. (17) P is the point of intersection of the diagonals of a quadri- lateral, Q the point which bisects the line joining the middle points of * By the centre of mass of a surface is meant that of a uniformly thin homogeneous plate having the form of the surface. 312 DYNAMICS [408 these diagonals, and R a point in PQ produced, such that QR=FQj3. Prove that R is the centre of mass of the quadrilateral. (18) ABC is an isosceles right-angled triangle, right-angled at B. Squares are described on its three sides. Show that the distance of the centre of mass of a uniform thin plate of this form is at a dis- tance from B equal to 13'J2ABI27. (19) One circle touches another internally. The diameter of the latter is d, that of the former ^d. Find the distance from the point of contact, of the centre of mass of the crescent or lune thus formed. Ans. i^d. (20) Find the centre of mass of a sector of a circle of angle 2a and radius r. If the curved portion of the boundary of the sector be divided into an indefinitely large number of equal arcs, the sector may be regarded as consisting of an indefinitely large number of equal isosceles triangles whose bases are the elements of the circular arc and whose equal sides are radii. The centre of mass of each of these triangles is at a distance fr from the centre of the circle. Hence the centre of mass of the sector is the same as that of a cir- cular arc of the angle 2a and the radius fr, and is therefore (Ex. 10) at a distance from the centre equal to frsin a/o. (21) The centre of mass of a uniform thin semi-circular plate of radius »• is at a distance from the centre equal to 4r/37r. (22) The centre of mass of a uniform thin conical shell is on the axis, and at a distance from the vertex equal to f of the height of the cone. (23) Find the centre of mass of a homogeneous triangular pyra- mid. Let the triangular pyramid A BCD be divided by planes parallel to ABC into an indefinitely great number of indefinitely thin triangular plates of which abc is any one. Let ^ be the centre of mass of the plate ABC, and let the plane jBZJiP' intersect ABC, abc and ABC in BS, he, and DE respectively, and let DF and he, which are in the plane BDF, intersect in f. Since F is the centre 408] OF EXTENDED BODIES. 313 of mass of ABC, Eis the middle point of AC. Since ao is parallel to AC, these lines being the intersections of parallel planes aha, ABC with the plane ADC, ae : AE=De : DE=ee : EC. Hence e is the middle point of ao. Since eb is parallel to EB, these lines being intersections of the parallel planes ahc, A BC, with the plane BDE, ef: EF=Df: DF=fb : FB. Hence ef=^eb, and /is therefore the centre of mass of ahc. Hence the centres of mass of all the triangular plates into which the pyramid is divided, and therefore the centre of mass of the pyra- mid, lie on the line DF. Similarly, if EO be equal to ^ED, O will be the centre of mass of, ACD, and the centre of mass of the pyra- mid will lie on the line OB. Now OB and DF are in the plane DBE, and intersect in B. Hence H is the centre of mass of the pyramid. Since EO : OD=EF:FB, OFia parallel to DB, and the triangle OHF similar to the triangle BHD. Hence FB : HD=FO : BD=EO : ED= 1 : 3. Hence the centre of mass of the pyramid is on the line drawn from 314 DYNAMICS [408 the centre of mass of the base to the vertex, and at a distance from the base equal to J of the height of the pyramid. (24) The centre of mass of a homogeneous pyramid with poly- gonal base, or of a cone -with plane base, is on the line joining the centre of mass of the base to the apex, and at a distance from the base equal to J of the height. (25) The centre of mass of a homogeneous triangular pyramid coincides with that of four homogeneous spheres of equal mass, whose centres are at the four angles of the pyramid. (26) The centre of mass of a homogeneous wedge, bounded by two similar, equal, and parallel triangular faces and by three rectangular faces, coincides with that of six equal particles placed at its angular points. (27) A cube (edge = l) is truncated at one angle by a plane which bisects three adjacent edges. Show that the distance of the centre of mass of the remainder from the angle opposite to that which is cut off is ill x/3. (28) From a right cone standing on a circular base another right cone is cut, standing on the same base, and the centre of mass of the remainder is at the vertex of the smaller cone. Show that this smaller cone is 3^ of the whole body. (29) A cylindrical vessel of radius «, stands vertically and con- tains water to a height h. A heavy sphere of radius a/2 is dropped into the water, and lies at the bottom of the vessel. Prove that the new centre of mass of the water lies somewhere within a circle whose radius is a^jl^h, and whose centre is at a distance ajQ - 5a^/'72h from the old centre of mass of the water. 409. Velocity of the Centre of Mass. — The component velocity, in a given direction, of the centre of mass of a system of particles is equal to the sum of the products of the masses of the particles into their component velocities in the same direction, divided by the sum of the masses of the particles. Let m^, TOj, etc., be the masses of the particles of the 411] OF EXTENDED BODIES. 315 system, Sj, s^, etc., s the distances from a plane perpendi- cular to the given direction, of the particles and their centre of mass respectively, at a given instant, s/, s/, etc., n', their respective distances after a short time t. Then -_ mjg,-|-m^s, + etc. . -,_ 'mX + mX + etc. mj+mj+etc. ' mj+mg+etc. Hence, subtracting, and dividing by t, ^ ^' mi+m^+etc. or (44 and 101), y being the velocity of the centre of mass in the given direction, 1,ms v = s = 2m' This result may be otherwise expressed thus: — The velocity of the centre of mass in a given direction is equal to the momentum of the system {i.e., to the alge- braic sum of the momenta of the various particles) in the given direction, divided by the mass of the system. 410. It follows that the momentum of the system in any given direction relative to the centre of mass is zero. For from 409 we have "^iTTis — 's'Lvfi = 2m(s — s) = ; and from 96 (3) it is obvious that 2m(s - s) is the momentum in the given direction relative to the centre of mass. , 411. Acceleration of Centre of Mass. — The component acceleration, in a given direction, of the centre of mass of a system of particles is equal to the sum of the products of the masses of the particles into their component accelerations in the same direction, divided by the sum of the masses of the particles. 316 DYNAMICS [411 The proof may be left to the reader. It is similar to that of 409, s, s^, etc., being replaced by s, «j, etc. Hence (118), a being the acceleration of the centre of mass in the given direction, - 1,ms 2m 412. It follows, as in 410, that the sum of the pro- ducts of the masses of the particles of a system into their component accelerations, in any given direction, relative to the centre of mass, is zero. 413. Examples. (1) If two particles move with unifonn speed in straight lines, their centre of mass will either be at rest or will move with uniform speed in a straight line also. (2) A number of particles of masses, m^, m^, etc., are projected at the same instant vertically upwards from given positions with given speeds, »i, «2i etc-j respectively. Find (a) how long, and (b) how far, their centre of mass will rise. (3) Two particles connected by a string are placed on two smooth inclined planes, the string passing over a smooth peg at the common summit of the planes. Show that the path of their centre of mass is the straight line which joins them when they are in such a posi- tion that the parts of the string on the two planes are to one another as the masses of the particles at their extremities, and that that particle will descend which in this position is the lower of the two. (4) Of three equal particles which start from the highest point of a vertical circle, one drops down the vertical diameter, and the others slide down chords of 60° and 120° respectively, on the same side of the diameter. Show that the centre of mass slides down a chord of cos-'( - H), and that its rate of change of speed is |./19. 415] OF EXTENDED BODIES. 317 414. Acceleration of Centre of Mass vn terms of Ex- ternal Forces. — The component acceleration, in any given direction, of the centre of mass of a system of particles is the same as the acceleration of a particle of mass equal to the mass of the system, acted on by a force in the given direction equal to the sum of the components in that direction of the external forces acting on the particles of the system. Let ^^, F^, etc., be the components, in the given direc- tion, of the resultants of all the external forces acting on the particles (masses = m^, m^, etc.); and let F^, F^, etc., be the components, in the same direction, of the resultants of all the internal forces acting on m^, m^, etc., respectively. Then s^, s^, etc., being the distances from a plane perpen- dicular to the given direction, we have (317 and 318) F^ + F^ = m;s^; F^+F^' = m,s,; etc. Hence I,F+'EF' =1,ms Now by the third law of motion, the internal forces consist of pairs of equal and opposite forces, whose sum is there- fore zero. Hence XF' =0, and I,F='Ems. Now (411) s = 1,ms fETn. Hence, calling a the acceleration, in the given direction, of the centre of mass, _ - j:f And it follows from 317 that a, as determined by this for- mula, is the acceleration that a particle of mass 2m would have, if acted on by a force equal to 2^. If therefore the external forces acting on the system and the mass of the system are known, the acceleration of the centre of mass may be determined. 415. If the components of the external forces in three rectangular directions', the axes of x, y, z, are Xj, X^, etc., 318 DYNAMICS [415 Fj, Fj, etc., Zj, Z^, etc., respectively, the component acce- lerations of the centre of mass in these directions are ^_2Z ^_SF^_ 2^ and the component accelerations being known, the magni- tude and direction of the resultant acceleration may be determined. 416. In the special case in which a system of particles is acted upon by external forces, the sum of whose com- ponents in any given direction is zero, the acceleration of the centre of mass is zero. For "LF being zero, so also is s. It follows also from 411 that if s is zero, "^ms is zero also, and 2ms may easily be shown to be the rate of change with time of the momentum of the system, in the given direction. Hence if there are no external forces the momentum of the system is constant. This result is often spoken of as the "Principle" of the Conservation of Linear Moinentum. 417. {D'Alemhert's Principle. — In 1742 D'Alembert proposed as a law of motion what is called his "principle." It is usually enunciated in the following form, though this is not the form in which it was enunciated by D'Alembert himself : — The impressed forces, with the reversed effective forces, of a system of material particles, constitute together a system of forces in equilibrium. By the term " impressed force " is meant an external force acting on the system. The " effective force " on a particle was the name given to the product of its mass into its acceleration, and this hypothetical force was sup- posed to act in the direction of the acceleration. A re- versed effective force would thus act in the opposite direc- 418] OF EXTENDED BODIES. 319 tion. If F is the component in a given direction of the resultant of all the external forces acting on a particle, m its mass and a the component in the given direction of its acceleration, 2-F is the sum of all such components of external forces, 2ma the sum of the components, in the given direction, of the effective forces, and — Zma there- fore the sum of the components in the given direction of all the reversed effective forces. Since the sum of the components in a given direction of all the forces of a system of forces in equilibrium (323 and 326) is zero, D'Alembert's principle may be expressed by the equation "EF— 'Ema = 0. This equation is obviously that obtained in 414 ; and thus D'Alembert's principle may be deduced immediately from Newton's second and third laws, which were formulated in 1687. By D'Alembert's principle every kinetic problem was re- duced to one of equilibrium between actual and fictitious forces. It was thus of great practical importance, as enabling the equations of motion to be written down for any system for which the conditions of equilibrium had been investigated.] 418. The Moment of Momentum of a particle about a given line is the product of its mass into the moment of its velocity (104) about the line. If v be the component velocity of the particle of mass m, in a plane perpendicular to the given line, and p the distance of this component velocity from the line, the moment of momentum of the particle is Tnvp, the sign being determined according to the convention of 103. The algebraic sum of the moments of momentum of ail the particles of a system about a given line, is the moment of momentum of the system about that line. If the given line be taken as axis of z, the analytical expression for the moment of momentum of the particle 320 DYNAMICS [418 will be (106), 'm{yx — xy), and for that of a system of particles, '2vi{yx — xy). 419. The moment of momentum of a system of particles about a given axis is equal to its value about a parallel axis through the centre of mass, together with the moment of momentum, about the given axis, of a particle of mass equal to the mass of the system, and situated at, and hav- ing the acceleration of, the centre of mass. y Take the given axis as axis of z, and Ox, Oy, as the other rectangular axes of co-ordi- nates. Let X, y, z, be the co- ordinates of anj'' particle of mass m,, at P, and x, y, 2, those of the centre of mass C. Let P's distances from G in the directions of Ox, Oy, Oz, be ^, 17, ^. Then c i 9 X s / / / / / Also (96) x = oc+i,y = y+r,. '^=^+i,y==y+i -2/Sm^ Hence the moment of momentum of the system I,m{yx-xy) =I,m{{y + ^)(x+i)-(sc+^){y+,,)} = (yJ- - xyyXm + 'Em{^^- ^r,) + xlmii + fEm^— xEmri - = (yx - xy)'Em + 2m(^^- ^r,), since (404 and 410) 'Emi] = Im^ = I,mi] = 2m^= 0. ^Mni~i^) ^s clearly the moment of momentum of the system about an axis, parallel to the given axis, through the centre of mass, and (^x-xy)'Em is equal to the mo- ment of momentum of a particle of mass 2m, whose co- 423] OF EXTENDED BODIES. 321 ordinates and component velocities are those of the centre of mass. 420. The Angular Momentum of a particle about a given line is the product of its mass (m), into its angular velocity (w) about the given line, into the square of its distance (r) from the given line — in symbols mwr^. The algebraic sum of the angular momenta of all the particles of a system about a given line is the angular momentum of the system about that line. 421. It follows from 132 that mvp = m,octr^. Hence the moment of momentum about any given axis is equal to the angular momentum about the same axis of either a single particle or a system of particles. Hence also the rate of change of angular momentum about any given axis is equal to the rate of change of moment of momen- tum about the same axis. 422. The Moment of the Acceleration of Momentum, of a particle about a given line is the product of its mass into the moment of its acceleration about the given line. The algebraic sum of all such products for all the particles of a system is the moment of the acceleration of momen- tum for the system. If a be the component acceleration of a particle in a plane perpendicular to the given line, and if j? be the dis- tance of a from it, the moment of the acceleration of momentum for the particle is map and for the system Swiop. The analytical expression for it will be (123, and 106) 1,m{yx—x'y). 423. It follows from 124 that the moment of the acce- leration of momentum of a particle about a given axis is equal to the product of its mass into the rate of change of the moment of its velocity, and therefore to the rate of change of its moment of momentum, and therefore to the 322 DYNAMICS [*23 rate of change of its angular momentum about the same axis. In symbols 7rhap = 7n (^). Hence also for a system of particles Smap = 2m {^). 424. It may be shown by the method employed in 419 that the moment of the acceleration of momentum of a system about a given axis is equal to its value about a parallel axis through the centre of mass, together with the moment of the acceleration of momentum about the given axis, of a particle, having a mass equal to the mass of the system, and situated at, and having the acceleration of, the centre of mass. With the symbols of 419, 'Em(yx — xy) = (yx — Sy)Sm+ 2m('j f — ^rj). 425. The Moment of a Force about a line or axis is the product of the component of the force in a plane perpen- dicular to the axis, into the distance from the axis of the line of action of the component. If F is the magni- tude of the component, and p its distance from the given axis, Fp is the magnitude of the moment of the force about the axis. Its sign is determined by a convention similar to that of 103. 426. It foUows from 313 and 107 that the moment of a force is equal to the algebraic sum of the moments of its components about any fixed axis. 427. If the given line be taken as axis of z, and other lines perpendicular to it and to one another (as in 419) as axes of X and y, and if x, y, and z be the co-ordinates of the particle on which the force acts, and X, Y, Z the components, in the directions of the axes, of the given force, then X and Y are rectangular components of F in the plane perpendicular to the given line, and y and X their respective distances from the given line, and therefore Yx and —Xy their respective moments about it. Hence (426) Fp=Yx — Xy. This is the analytical expression for the moment of a force. 429] OF EXTENDED- BODIES. 323 428. The sum of the moments, about an axis fixed in space, of the external forces acting on a system of particles is equal to the rate of change of the angular momentum of the system about the given axis. R being the component, in a plane perpendicular to the given axis, of the I'esultant force acting on the particle, a its component acceleration in the same plane, and m its mass, we have R = ma. If p is the common distance of R and a from the given axis, Rp = map. If F and F are the components in the same plane of the resultants of the external and internal forces respectively, acting on 7n, F and F' are components of R. If therefore P and P' are their respective distances from the given axis (426), Rp = FP+F'F' = map. Hence, for the system (423), 'EFP + XFP' = Imap = m2(^^). Now the internal forces consist of pairs of equal and opposite forces equidistant from the axis. Hence ■2F'P' = 0, and 'EFP = I,m(^). The analytical expression of this result is (427 and 422) 2( Tx — Xy) = 2m(2/a; - xy). 429. In the special case in which the sum of the moments of the external forces about the given axis is zero, the angular momentum of the system about the given axis is constant. For we have 2jPP = 2m(^) = 0. Hence limwT^ is constant. This result is called the "principle " of the conservation ofamgular momentum. It follows that Sma)r72 = constant. Now (133) u^l2 is the area swept over per unit of time by the radius vec- 324 DYNAMICS [429 tor of the particle of mass m. Hence the above result is also called the " principle " of the conservation of areas. 430. From 428 and 424 it follows that 2{7x- Xy) = {yx - xy)I,m + 2m(^"^- ^l?). This applies, as we have seen, to an axis fixed in space. If we choose the axis so that at the instant under con- sideration it is passing through the centre of mass of the system, we have x = y = 0', x=^, and y = i], and therefore Now 2m(^"^— ^) is the rate of change of angular momen- tum about an axis parallel to the given axis through the centre of mass; and 1,(Y^—X^) is (427 and 428) the value the rate of change of angular momentum would have if the centre of mass were fiked. Hence the rate of change of angular momentum about the centre of mass, produced in a system of particles by the forces to which it is subjected, is the same as that which would be pro- duced if the centre of mass were fixed. 431. Equations of Motion. — We have now obtained two important results; the first, that of 414, by which the acceleration of the centre of mass of a system is expressed in terms of the external forces acting on it, and its mass ; and the second, that of 428, by which the rate of change of angular momentum of a system about a fixed axis (or, 430, about an axis through the centre of mass) is expressed also in terms of the external forces acting on it. These equations tell us all that we can know of the motion of a system of particles without data as to the internal forces. They are therefore called the equations of motion of a system of particles or of an extended body. The principles of the conservation of linear and angular momentum are special cases of these equations of motion. 434] OF EXTENDED BODIES. 325 432. Energy of a System, of Particles. — The energy of a system of particles is the sum of the amounts of energy possessed by its particles. The amounts of potential energy possessed by its particles depend upon their mutual attractions, and on the action of external forces, and vary as their distances from one another vary, and as the position of the system varies with respect to the bodies exerting force on it from outside. The potential energy therefore depends only upon the configuration and position of the system. If the system is isolated, so that no external forces act upon it, its potential energy depends upon its configuration only. 433. If the forces of the system are independent of the velocity of the particles, the change of potential energy during any change of configuration will be independent of the paths in which the particles have moved, and therefore of the series of configurations through which the system has passed, and will be equal to the work done against the forces of the system during the change of configuration. If we choose some convenient configu- ration of the system as the configuration of zero potential energy, the potential energy of the system in any other configuration will be the work done against the forces of the system during the passage from the zero configuration to the other. 434. The work done by the forces acting on the par- ticles of a system, during any change of configuration, is equal to the change produced in the kinetic energy of the system. Let X, T, Z be the components, in three rectangular directions Ox, Oy, Oz, of the resultant of the external and internal forces acting on a particle of mass m at a point P, whose co-ordinates are OL, MP, LM, or x, y, z. Then (414) taking all the particles of the system into considera- tion, we have 326 DYNAMICS [434 and therefore S(Z-mic) = 0, 1.{Y-my) = 0, I,{Z-mz) = 0. y -s-x Let the particle undergo any indefinitely small displace- ment, say to a point P', whose co-ordinates are OL', M'P', L'M', or x', y', 0'. The displacement being indefinitely small, the resultant force acting on the particle may be considered uniform. The components of the displacement in the axes of x, y, and z will be x'—x, y'—y, ^—z. Multiplying by these components, we have S(Z-m^)(a;'-a;) = 0, 2(F-mi/)(2/'-2/) = 0, 1.(Z-'mz){z-z) = Q. Hence, also, S(Z-m«)(a;'-aj) -|-2( 7-m2/)(2/'-2/) ■^'L{Z-m,z){z!-z) = K), and 'l.{X{p^ -x)^-Y(:y -y)+Z{z' -z)-m\x{x -x)Vy{y' -y) +0(2'-^)]}=O. The component velocities of the particle at P and P' will 434] OF EXTENDED BODIES. 327 be X, y, z and x', y', z' respectively ; and, as the time t of the displacement is small, the mean component velocities may be taken to be (x-\-x')l% {yfy')l2, (z+!sf)/2. The component accelerations are {x—x)/t, (y' — y)/t, {z' -z)lt; and, as t is small, these may be taken to be the same as X, y, z. Hence j:\^X{x'-x)+7(y'-y)+Z(z-z)-m(^^ . ^t I y'-y y'+y ^ , ^ -^ l+i+ and t • 2'*+V-^^)}=«' j:{X{x' -x) + 7(y' -y)+Z(z' -z)-'^[x'^+y'^+6'-' -(x^ + y^+z^)]}=0. If W denote the work done on the particle m, v the velocity of the particle at P, and v its velocity at P', we have (342, 98 and 88), W=X(x.-x)+ Y{y-y)+Z{z'-z), v'^ = x'^+y'^+z^, v^ = a? + y^ + z^. Hence 2{Tr-^(t;'2-'i;2)}=0. Hence the sum of all the quantities of work done by the forces acting on the particles of a system, during any indefinitely smalL change of configuration, is equal to the sum of the quantities of kinetic energy gained by the particles. Any finite change of the configuration of a system may be broken up into an indefinitely large number of inde- finitely small changes, to each of which an equation 328 DYNAMICS [434 similar to the above applies. Adding them, we have for a finite change of configuration, 2S{F-|^K-'y2)} = 0, or the work done by the acting forces during any finite change of configuration of a system of particles is equal to the increment of the kinetic energy of the system. 435. Conservation of Energy. — If all the forces (ex- ternal and internal) are dependent only on the positions of the particles (346) on which they act, the work done against them results (347) in the increment of the poten- tial energy of the particles. The amount of the potential energy produced is equal to the work thus done against the forces, and is therefore equal to minus the work done by them. If therefore P is the increment of potential energy of a particle in any small displacement, P= — W; and hence EE{-P-Jm('y'2-'y2)}=0; and if K denote the increment of kinetic energy, SE(P-fir) = 0. Hence the sum of the potential and kinetic energies of a system of particles is constant, provided all the acting forces are dependent only on the positions of the particles on which they act. This result is called the law of the Conservation of Energy. A system of particles to which it applies is called a conservative system. 436. A cycle of transformations of a system is a series of changes of configuration by which the particles are brought finally to their initial positions. If the system is conservative and isolated, it is clear that the initial and final potential and kinetic energies must be the same. If therefore a conservative system of particles be so arranged that when set in motion it undergoes a cyclical 437] OF EXTENDED BODIES. 329 transformation, the cycles of transformations will go on for ever. If, for example, heavenly bodies moving in space met with no resistance to their motion, of the nature of friction, the solar system would form a system of this, kind and the planets must continue to move round the sun for ever. If we had materials of perfect smoothness and with other properties excluding the pos- sibility of the action of non-conservative forces, it would be possible to make a machine which, once started, would run for ever without work being done upon it, provided work were not done by it. An isolated conservative system thus undergoing cycles of transformation can never, however, increase the total quantity of its energy. If therefore natural forces are of the conservative kind, it will be impossible to devise a machine which, when set in motion and left to itself, will both run itself and do external work — in other words, the "perpetual motion" will be an impossibility. The universal failure of efforts to discover the perpetual motion have placed it in the same category as the philo- sopher's stone and the elixir vitae. Many writers accept the impossibility of the perpetual motion as having been proved by experience and make it a fundamental law of motion (usually without saying so), deducing the law of the conservation of energy imme- diately from it. It will be clear that such a course is unphilosophical if Newton's three laws have already been chosen as fundamental laws of motion, because the conservation of energy and the impossibility of the perpetual motion may be deduced from these laws. If the impossibility of the perpetual motion be chosen as a law of motion, one or more of Newton's laws should be obtained as deductions from it. 437. Law of Energy. — If any of the forces acting on the particles of the system are of the nature of resistances which depend upon the velocity of a particle, not on its 330 DYNAMICS [437 position merely, the work done against them does not result in the production of potential energy. In such systems therefore, which are at any rate apparently non- conservative, the work done against the acting forces is equal to the sum of the increment of potential energy and of the work done against such resistances. If this work be denoted by w, we have therefore P-^-w^ — W, and hence 22(P-|-w + ^) = 0. Hence the kinetic and potential energy of a material system, together with the energy expended in overcoming friction and other forms of non-conservative force, is a constant quantity. This result is the general law of energy, of which the law of the conservation of energy is a special case. 438. If therefore a non-conservative system of particles be so arranged that, when set in, motion, it undergoes cyclical transformations, its energy will gradually dimin- ish, and its cyclical transformations cannot therefore go on for ever. It is probable that the planets move in a resisting medium, whose resistance they expend energy in overcoming. If so, they must be moving in spiral paths and getting gradually nearer the sun. No machine can be constructed whose parts in their relative motions do not meet with frictional resistance and other forms of (apparently) non-conservative force. Hence no machine can be constructed which will run itself even if no ex- ternal work be done. 439. When work is done against forces, such as friction, which are, apparently at least, non-conservative, there seems at first sight to be no return in the form of energy ; and until recently energy thus expended was believed to be lost. Experiment, however, has shown that when energy is thus expended heat is always produced, that heat is a form of energy, and that the amount of thermal 441] OF EXTENDED BODIES. 331 energy produced when work is done against friction or other such forces, is the exact equivalent of the work so done. Hence the law of energy in the case of material systems which are apparently non-conservative may be thus expressed : The energy of the system, including kinetic, potential, and thermal energy, is a constant quantity, if the system is isolated, so that it can neither give energy to, nor re- ceive energy from, outside bodies. 440. The frictional and other non-conservative forces which we find acting on bodies in their relative motions, and to whose action is due the apparent non-conservative character of material systems, are observed to act between bodies of finite size. It is possible therefore that, if we could observe all the motions of the particles or small parts of bodies, their apparent non-conservative character might disappear. When work is done against friction, fpr ex- ample, it may be that the relative motions of the particles of the bodies in contact are increased, so that though the rubbing bodies do not gain potential energy their particles gain kinetic energy. Thermal energy is generally be- lieved, though not yet proved, to be the kinetic energy of the particles of a body due to their motion among one another. If so, the laws of Thermodynamics should be capable of deduction from the laws of motion. At present however we do not know enough about the relative motions of the particles of a body, or how they are afiected when the body meets with frictional or other such resistances, to make this deduction. 441. In applying the law of energy, obtained above, to the solution of problems on the motion of material systems, it is important to notice that forces acting on fixed portions of the system, stresses between particles whose distances are invariable, and forces acting on par- ticles whose motion is normal to the direction of the force. 332 DYNAMICS [441 ■do no work and therefore do not appear in the equation of energy. 442. In the solution of such problems the following proposition will be of use to facilitate the calculation of the kinetic energy of the system : The kinetic energy of a system of particles is equal to the sum of the kinetic energies of the particles of the system moving with velocities equal to their velocities relative to the centre of mass, together with that of a particle having a mass equal to the mass of the system and a velocity equal to the velocity of the centre of mass. Ox, Oy, Oz being rectangular axes, let the co-ordinates of a particle of mass m be x, y, z. Then its component velocities are x, y, z and its kinetic energy ^m(a;^+2/^ + i^). The kinetic energy of the system is thus ^^mlx^ + y^ + z^). Let X, y, z be the co-ordinates of the centre of mass and ^, rj, f the distances of a particle from it in the directions of Ox, Oy, Oz respectively. Then (419) x = x + ^, y = y+v, and z = z + ^. Also (96) x = x+^, y = y + rj, and z = z + ^. Hence the kinetic enei'gy of the system, ^^m(x^ + y^ + z^) = -Ehn{{k + if + {y + n)' + {i+m = 2im(S2 +y^+ P) + Sim(f +^^+ f ) since (410) 2m£=2m»^ = 2m^=0. And 1,im{^^ + rf+^^} is the sum of the kinetic energies of the particles mov- ing with velocities equal to their velocities relative to the centre of mass, and ^^mi^^ + y^ + z^) is the kinetic ■energy of a particle having a mass equal to the mass of the system and a velocity equal to the velocity of the centre of mass. 443] OF EXTENDED BODIES. 333 443. Examples. (1) Two particles of masses m and M moving in a straight line, with velocities v and V respectively {y > V) come into collision, the stress between them during collision being in the line of motion, and the co-efficient of restitution being e. Find the loss of kinetic energy. Let U be the velocity of the centre of mass, which (416) is the same after as before the collision. Then their velocities relative to- the centre of mass before the collision are v— U and V- irrespec- tively ; and if v' and V are the velocities of m and M respectively after the collision, v'— U and V —U are their respective velocities, relative to the centre of mass after the collision. Hence (442) the kinetic energy is, before the collision, and, after the collision, ^{m,+M)U^+\m{^- Uf+\M{V'-Uf Hence the loss of kinetic energy is Now, as both particles have at the instant of collision the velocity V, we have (416), (m+M) U='mv + MV, and (380, Ex. 1), j_ mv + MY-eM{v- V) m+M ' y,_ mv+MV— em{ V- v) m + M Hence t/-tr=-!^P m+M =e(U-v). And similarly V- U=e(U- F). Hence the loss of kinetic energy is i{l-e^)[m{v- Uf+M{ V- Uf]. If therefore e=l, there is no loss of energy. If e=0, the loss of energy is equal to the energy due to the motion of m and if relative to their centre of mass before the collision. 334 DYNAMICS [443 (2) In any displacement of a system of heavy particles, the work done against the weights of the particles is equal to the product of the weight of the system into the vertical displacement of the centre of mass of the system. Let mj, m^, etc., be the masses of the particles, c^j, d^, etc., the vertical components of their displacements, .ri, X2, etc., their initial distances from a horizontal plane. Then the amounts of work done on the various particles are m^gd^, m^gd^, etc. Hence the whole work done is g'Simd. Now the vertical displacement of the centre of mass is 'Smijc + d)_ Smx _ Smd Sot Sot Sot Hence the product of the weight of the system into this vertical displacement, ^Smx^=^S«j^, which is the whole work done. (3) Find the work done in raising from the ground the materials (cubical blocks of stone of 1 foot edge and of density 1 cwt. per cubic ft.) in building a uniform column 66 ft. high and 20 ft. square. Ans. 42,800 foot-tons. (4) A right pyramid on a square base of 16 ft. side, has an alti- tude of 24 ft., and stands on a horizontal plane. "Find the work necessary to turn it round one of its edges, its density being 100 lbs. per cubic ft. Ans. 819,200 ft. -pounds. (5) A chain whose mass is 100 lbs. and length 50 ft. hangs freely by the upper end, which is attached to a drum, upon which the chain can be wound, the diameter of the drum being so small relatively to the length of the chain that it may be neglected. Find the work done against the weight of the chain in winding up one half of it. Ans. 1875 ft. -pounds. 446] OF EXTENDED BODIES. 335 (6) The cylindrical shaft of a mine, whose section is 50 sq. ft., contains water (density = 1000 oz. per cubic ft.) to within 90 ft. of the surface. How much wiU the surface of the water be lowered by an engine working at 10 horse power for 1 hour. Ans. 54-1 ft. (7) Find the initial speed of a shot of 1000 lb. mass, discharged from a 100-ton gun, the energy of the charge being 300,000 ft.- potinds, and 1 per cent, being lost in heat, light, etc. Ans. 24'3... ft. per sec. 444. Equilibrium, of Extended Systems. — By the equilibrium of a system of particles may be denoted either of two states of motion : (1) a state in which the centre of mass of the system has no linear acceleration, and the system a constant angular momentum about the centre of mass, (2) a state in which the particles of the system are all without linear acceleration. The former may be called a state of molar equilibrium or equilibrium of the system as a whole, the latter a state of molecular equilibrium or equilibrium of the individual particles or molecules of the system. 445. The necessary and sufficient conditions of molar equilibrium may be obtained at once from the equations of 431, viz., d=j:F/Im, and I,FP=Xm{;^). For in order that the acceleration of the centre of mass may be zero, and the angular momentum constant, we must have 2^= and liFP = ; and if these conditions are fulfilled, we have a = and 2m(^) = 0. Hence the necessary and sufficient conditions of molar equilibrium are (1) that the algebraic sum of the components, in any given direction, of the external forces must be zero, and (2) that the algebraic sum of the moments of the same forces about any axis must be zero also. 446. An expression of the condition of molecular 336 DYNAMICS [446 equilibrium may be obtained from the equation of energy (437), which may be written (434) W is here the work done by all the forces acting on m in any small displacement whose components are x'—x, y' — y,z' — z. 2Tr is therefore the work done by all the forces of the system during its corresponding change of configuration. Dividing by t, the time of the small dis- placement of m, we get 2{|- f (^*(a='+*)4-^^(2/'+^)4:=^V+^))} = 0. If now the given change of configuration be from one of equilibrium to one indefinitely near it, the component accelerations {x' — x)jt, etc., may be put equal to zero. Hence we have 2( WH) = 0, i.e., the rate at which the forces acting oti the system do work is zero. If the work done by external forces be denoted by w, and that done by internal forces by w'; we have Sir=2w+Sic;', and therefore 'E{w/t) = 1,( — w'/t). Hence if a material system in any given configuration be in molecular equilibrium, the rate at which the external forces do work during any small motion through that configuration is equal to the rate at which work is done against the internal forces ; or, if the system is conservative, to the rate of increase of the potential energy of the system due to internal forces. 447. Conversely, if in any small motion of a material system through a given configuration, the rate at which the forces of the system do work is zero, the given con- figuration is one of molecular equilibrium. For if not, some of the particles of the system must in that configuration have accelerations. Let them be re- 430] OE EXTENDED BODIES. 337 duced to equilibrium by the action of forces F^, F^, etc., equal to the products of their masses into their accelera- tions and in directions opposite to these accelerations. Let the system now undergo an indefinitely small change of configuration, such that the particles having accelera- tions move in the directions of their accelerations. Then work will be done against F^, F^, etc., and the rate at which work is done by these forces will in all cases be negative. But the rate at which work is done by all the forces of the system together with F^, F^, etc., is zero (446), since the system is now in molecular equilibrium. Hence the rate at which work is done by the forces of the system alone is positive, and cannot be zero. Hence none of the particles of the system can, in the given con- figuration, have accelerations, and that configuration is therefore one of molecular equilibrium. 448. Hence the necessary and sufficient condition of the molecular equilibrium of a material system in any given configuration is that in any small motion through that configuration the rate at which the external forces do work shall be equal to the rate at which work is done against the internal forces, or, if the system is conserva- tive, to the rate of increase of the potential energy of the system due to internal forces. 449. Hence also the necessary and sufficient condition of the molecular equilibrium of a material system in any given configuration is that in any small motion through that configuration the work done by the external forces shall be equal to that done against the internal forces, or, in other words, that the algebraic sum of the amounts of work done by all the forces shall be zero. In symbols, if F^, F^, etc., be the forces acting on the particles of the system, and d^, d,, etc., their component displacements in the directions of I'j, F^, etc., respectively, 2-P'd = 0. 450. Stability of Equilibrium. — If a system of particles Y 338 DYNAMICS [450 which has undergone any indefinitely small change of configuration from that of equilibrium, returns, when left to itself, to the configuration of equilibrium, its equilib- rium is said to be stable for a change of configuration of that kind. If, when left to itself, the system deviates still more from the configuration of equilibrium, its equi- librium is said to be unstable. If the new configuration is also a configuration of equilibrium, the equilibrium of the system is said to be neutral. Thus the position of equilibrium of the bob of a pendulum is the lowest point of its swing. If it be slightly displaced from that position and left to itself it will return to that position. Hence its equilibrium is stable. A symmetrical egg may be made to stand on one end ; and this position is thus one of equilibrium. But if it be displaced from this position ever so slightly and left to itself, the displacement increases with the time and it falls over on its side. Hence an egg standing on one end is in a position of unstable equilib- rium. If a uniform sphere, resting in equilibrium on a horizontal plane, be slightly displaced and left to itself, it will still remain in equilibrium; and thus a uniform sphere on a horizontal plane is in neutral equilibrium. A configuration of equilibrium of a system may be such that for different small changes of configuration the stability of its equilibrium may be different. Thus a sphere resting on a horizontal cylinder is in neutral equilibrium for small displacements, which are rotations about an axis through the point of contact and perpen- dicular to the axis of the cylinder, while, for rotations about all other axes through the same point, its equilib- rium is unstable. The equilibrium of a sphere resting in a cylindrical trough is stable for some displacements and neutral for others; and that of a sphere resting on a saddle-back, or col, is stable for some displacements and unstable for others. The equilibrium of a system which is stable, unstable, or neutral, as the case, may be, for all possible small displacements of the system is said to be 452] OF EXTENDED BODIES. 339 wholly or absolutely stable, unstable, or neutral. The equilibrium of a system which is unstable for any small change of configuration, though it may be stable or neutral for others, is said to be practically unstable. 451. There is a simple relation between the potential energy of a conservative system in its configuration of equilibrium and the stability of its equilibrium. If, after a small displacement from a configuration of equi- librium, the system, when left to itself, returns to the configuration of equilibrium, the forces of the system on the whole do work on the particles of the system in bringing it back to the configuration of equilibrium. Hence, in the configuration of equilibrium, the potential energy of the system is less than in the other configura- tion. If therefore a system has a configuration in which it is in wholly stable equilibrium, that configuration is one of minimum potential energy. If, after a small dis- placement from a configuration of equilibrium, the system, when left to itself, deviates still more from the configura- tion of equilibrium, the forces of the system on the whol& do work during the given small displacement, and hence the potential energy of the system is less after the dis- placement than in the configuration of equilibrium. If therefore a system in a given configuration is in wholly unstable equilibrium, the given configuration is one of ma;ximum potential energy. If, finally, after a small displacement from a configuration of equilibrium a system of particles is still in equilibrium, the forces of the system have neither done work nor had work done against them during the displacement, and hence the potential energy after the displacement is the same as before it. 452. If the potential energy of a system of particles depends wholly upon their weights, the increase of potential energy in any change of configuration (443, Ex. 2) is the product of the weight of the system into the height through which its centre of mass has been 340 DYNAMICS [452 raised. A configuration in which the value of the poten- tial energy is a maximum or minimum, therefore, is one in which the centre of mass has a maximum or minimum height respectively. Hence a configuration of wholly stable or wholly unstable equilibrium is one in which the centre of mass has a lower position or a higher position respectively than in any other configuration into which the system may be brought by an indefinitely small change of configuration. Thus a rod, one end of which is fixed, is in stable equilibrium if the other end, and therefore the centre of mass, is vertically below the fixed end, and is in unstable equilibrium if the other end, and therefore the centre of mass, is above the fixed end. 453] OF RIGID BODIES. 341 CHAPTER VI. DYNAMICS OF EIGID BODIES. 453. A rigid body or system of particles is one whose configuration is invariable, the particles maintaining con- stant relative positions. Such bodies are purely ideal. But in many cases solid bodies are so slightly deformed by the forces acting on them that for many purposes they may be considered rigid. It follows from the constancy of the configuration of a rigid body that, if it is rotating about an axis fixed in itself, all its particles must have the same angular velocity, and consequently the same angular acceleration, about that axis; and that the distance of any particle from the axis must be constant. Hence (420) the angular momentum about the given axis, viz., 'Lynm^, may be written wSmr^ and ^2 becomes ior^. Hence (225) the rate of change of angular momentum if a denote the angular acceleration about the given axis. We found (428) that about an axis fixed in space, 2^P = 2m(^rtS). Hence, if a is the angular acceleration about any axis fixed both in space and in the body, 'LFP = aS,m,i^. 342 DYNAMICS [453 By 430 the same formula applies if a be the angular acceleration about an axis fixed in the body and passing through its centre of mass, whether or not it be fixed also in space. 454. We have thus two equations (414 and 453), 2m' 2mr^' expressing, the one the linear acceleration of the centre of mass, the other the angular acceleration about that point. Hence (251) these equations completely deter- mine the motion of the body. 455. From the second of these equations it follows (2/m,r^ being constant) that the rotating power of a force, or of several forces, about a given axis is proportional to its moment, or to the algebraic sum of their moments respectively, about that axis. This result is frequently assumed by writers on elementary statics. 456. From the two equations of 454 it follows that a force produces in a rigid body the same kinetic effect at whatever point of its line of action it may be applied. For a has the same value, provided the magnitudes and directions of the applied forces are the same ; and a has the same value, provided the magnitudes of the applied forces and the distances from the axis of their lines of action are the same. This result is usually called the "principle of the transmissibility of force," and is usually made a fundamental hypothesis by writers on Statics. 457. It follows, from the result of 453, that for the complete specification of a force which is acting on a rigid body, it is necessary to know not only its magnitude and direction, as in the case of a particle, but its line of action or some point in its line of action as well. 460] OF KIGIl) BODIES. 343 458. It follows, from the second equation of 454, that if a free rigid body be acted upon by a force whose line of action passes through the centre of mass, it produces in the body no angular acceleration about the centre of mass, and therefore (as is evident from 244) no angular acceleration whatever. 459. Composition, of Forces.— It is often convenient in investigating the motion of a rigid body to replace the forces acting on it by a simpler set of forces, which would produce the same kinetic effect. Before applying the equations of 454 to the solution of problems, there- fore, we shall investigate the composition of forces acting on a rigid body, i.e., the reduction of such forces to simpler equivalent systems. The resultant of the forces acting on a rigid body is the single force or the simplest system of forces which will produce in it the same accelerations as are produced by the given forces. 460. Any eoplanar forces acting on a rigid body are reducible to a single force. — A force F, whose components in rectangular directions in the plane of the forces are Fx, Fy, will produce the same linear acceleration of the centre of mass as the acting forces (components Xy Fj, X^, Fj, etc.), provided F^='EX and Fy='27 ; and it will produce the same angular acceleration about any point in the plane of the component forces if its line of action is at such a distance (p) from the point that Fp is equal to the algebraic sum (N) of the moments of the forces about it,* if therefore Fp=N. Hence, as F=(Fx^+Fy^)i, the forces are reducible to a single force if p[C2Xf+(2Y)^i=N. * The moment of a force about a point is its moment about an axis through the point perpendicular to the plane containing the point and the line of action of the force. 344 DYNAMICS [*60 A finite value of p can always be found to satisfy this equation, provided [(2Z)H (2 7)^]* is not equal to zero. Hence, except in this case, a single resultant can always be found. 461. Determination of the Single Resultant of Co- planar Systems of Forces. — The forces may or may not be parallel. Case I. — Non-parallel Coplanar Forces — Analytical Determination.— It follows, from 460, that the magnitude of the single resultant is [(SX)2 + (SF)2]* its direction cosines (I,T)/F, {I,Y)/F, and its distance from the point NjF. The magnitude and direction of the single force is thus the same as if forces of the same magnitude and direction as the given forces acted upon a particle (313 and 90). 462. Qeometrical Determ,ination. — The magnitude and line of action of the resultant may also be found by the aid of 456. — Let coplanar forces F^, F^, F^ act on a rigid body at A, B, G respectively. Produce jPj and F^ till they meet in D. Let both forces (456) act at D instead of aX A and B, and let R^, their resultant, be determined by 313. Produce B^ to meet F^ in E. Let now R^ and F^ act at E instead of C and D, and let R^, their resultant, be determined. Then R^ is the resultant of the given forces. 463] OF RIGID BODIES. 345 By thus applying the parallelogram law the resultant may be determined either by calculation or graphically (382, Ex. 22). 463. The following is a more elegant graphical method:; — Let forces F-^, F^i F^, F^ act as represented in the diagram at the points A, B, C, D. From any point E draw EG, from G draw GH, from H HK, and from K KL, repre- senting in magnitude and direction the forces F^, F^, F^, F^ respectively. Then (461) EL represents their resul- tant in magnitude and direction. To find a point in its line of action, take any point and join it to E, G, H, K, L. From any point in F^, say a, draw a line parallel to OG and meeting F^ in b. From b draw a line parallel to OB, meeting F^ in c. From c draw a line parallel to OK, meeting F^ in d. From d draw a line parallel to OL, and from a a line parallel to OE, and let them meet in M. A force represented by EG may be resolved into two represented by EO and OG. Hence F^ is equivalent to forces proportional to EO and OG, with lines of action Ma and ba. Similarly F^ may be resolved into forces proportional to GO and OH, with lines of action ab and cb, Fg into forces proportional to HO and OK, with lines. 346 DYNAMICS [463 of action be and dc, and F^ into forces proportional to KO and OZ, and with lines of action cd and dM. Hence the given system of forces is equivalent to single forces in the lines dM and Ma, and pairs of equal and opposite forces in each of the lines ab, be, cd. The resultant of this system is clearly a force through M. Hence the re- quired resultant is a force repi'esented by EL and acting atilf. 464. Case II. Parallel Coplanar Forces. — If the given forces are parallel, the constructions of 462 and 463 fail. In any such case, however, a system equivalent to the given system may be obtained by introducing two, equal and opposite forces in the same line, and with direc- tions inclined to those of the given parallel forces ; and to this equivalent system the above constructions may be applied. 465. We may find the resultant of parallel forces more readily, however, as follows : — First, let there be two such component forces. These may be either codirectional or opposite in direction. (a) The Forces Codir'ecUonal. — Let P and Q be forces A acting in the same direction dn a rigid body of mass m. Then, that the resultant iJ may produce in the centre of mass of the body the same acceleration as P and Q, its 466] OF RIGID BODIES. 347 line of action must be parallel to theirs, and we must have a = {P+Q)lm=Blm, and hence R = P+Q. Also that R may produce about any point 0, the same angular acceleration, as P and Q, its moment about must be equal to the algebraic sum of their moments. From draw OBA perpendicular to P and Q, and therefore to R, and meeting P, Q, and R in A, B, and C respectively. Then P . AO+Q . BO=R . CO = {P+Q)GO. It follows that CO is intermediate in length between AO and BO, and that C is therefore between A and B. Sub- stituting for AO and BO their values we have P{AG+ CO) + Q{CO - CB) = {P+Q)CO. Hence • P . AC= Q . CB, i.e., R's line of action cuts the line AB (and, therefore, any line intersecting P and Q), so that the products of the forces into the segments adjacent to them are equal. 466. (6) The Forces Opposite in direction and Unequal. — Let P and Q be the given forces. Then, as above, if P be greater than Q, R=P—Q and is codirectional with P, and rp P . AO-Q. BO = R . CO = (P-Q)CO. Now BO is less than A 0. Hence (P- Q)AO < P. AO-Q. BO; 348 DYNAMICS [466 and therefore AO is less than CO, and is a point in BA produced. Substituting in the above equation the values oi AO and BO, we have P{GO - CA) - Q{GO - OB) = (P - Q)CO, and P .CA = Q.CB; i.e., R's line of action cuts the line BA produced, so that the products of the component forces into the segments adjacent to them are equal. 467. (c) The Forces Opposite and Equal. — A system of two forces, equal and opposite, but not in the same straight line, is called a couple. In this case B, = 0, and P .AO-Q.BO=-P .AB=0xG0. Now P . AB has a finite value. Hence GO must be in- finitely great. The single resultant of two equal and opposite parallel forces is therefore a force zero at an in- finite distance. In other words, a couple can produce rotational, but not translational, acceleration in the body on which it acts. As P . AB has the same value for all positions of 0, the moment of a couple about all points in its plane, and therefore about all axes perpendicular to its plane, is the same, and is equal to the product of either force into the distance between their lines of action. This distance is called the arm of the couple. A couple is therefore completely specified if its moment and the direction of a line perpendicular to its plane are given. It may therefore be represented by a straight line, whose length is proportional to the magnitude of the moment of the couple, whose direction is normal to the plane of the couple, and which is so drawn, according to a convention similar to that of 103, as to indicate 469] OF RIGID BODIES. 349 the sign of the moment. Such a line is usually called the axis of the couple. 468. It follows from the above that all couples which have equal moments of the same sign, and are in the same or in parallel planes, produce the same kinetic effect, or are equivalent, whatever may be their length of arm or the magnitudes or lines of action of their forces. 469. Composition of Couples. — The resultant of any number of component couples is a couple, and is to be determined by the parallelogram law. Let the planes of two component couples intersect in the line AB. At A and B let equal and opposite forces, F, act in the plane of one of the component couples at right angles to AB, and of such magnitude and direction that the couple F . AB has the same moment and' sign as the component couple in its plane. At A and B let equal and opposite forces F" act in the plane of the second component couple, and at right angles to AB, F' being of such magnitude and direction that the couple ¥ . AB has the same moment and sign as the second component couple. Then the couples F . AB and F" . AB are equiva- lent to the two component couples. Let AF and BF, AF' and BF, represent the forces F and F. Then if the parallelograms AFRF, BFRF, be completed, the diago- 350 DYNAMICS [469 nals AR and BR will represent the resultants of F and F at A and B respectively. Since the angles FAF and FBF' are equal, the parallelograms FF are similar. Hence the angles FAR and FBR are equal, and therefore the equal resultants, R, are in the same plane. Since in each case R is in the same plane as F and F', AR and BR are perpendicular to AB. Hence the two component couples are equivalent to the couple R . AB. From B draw Bf, Bf, and Br, the axes of the couples F . AB, F' . AB, and R . AB respectively. Bf, Bf, and Br are thus perpendicular to the planes FABF, FABF, and RABR respectively ; and consequently the angles /Br, f'Br are equal to FBR, F'BR respectively. Also, since the couples represented by Bf Bf, and Br have the same arm, we have Bf: BF=Bf' : BF = Br : BR. Hence, if r be joined to /' and /, rfBf will be a parallelo- gram ; and consequently the axis of the resultant couple is to be determined from the axes of the component couples by the parallelogram law (78). If there are more than two component couples, the resultant of any two may be compounded with a third, their resultant with a fourth, and so on until the resul- tant of all has been found. It follows that the laws of the resolution of couples are the same as in the case of displacements, velocities, etc. 470. Secondly (465), let there be any number of com- ponent parallel forces. In that case the resultant of any two may first be determined, then the resultant of their resultant, and a third, and so on, until the resultant of all has been found. 471. Any system of parallel forces, whether coplanar or not, may be reduced to a single force. — For, as any 474] OF RIGID BODIES.. 351 two parallel lines are necessarily in the same plane, the resultant of any two of the given forces is coplanar with a third, that of any three with a fourth, and so on. Thus the single resultant of a non-coplanar system may be de- termined as in 470. 472. From 471, 465, and 899 it is clear that the above process is exactly that by which the centre of mass of a system of particles was determined, the magnitudes of the parallel forces taking the place of the masses of the particles, and the positions of their points of application that of the positions of the particles. Hence, as in 400, it may be shown that if Fj^, F^, etc., are the magnitudes of the parallel forces, and d^, d^, etc., the distances of their points of application from any given plane, the distance from it of the point of application of their resultant is l!,Fd/'2F. The point of application of the resultant is called the centre of the system of parallel forces. 473. In the special case in which all the particles of a body are acted upon by parallel forces proportional to their masses, the centre of parallel forces is an important point. If F^, F^, etc., are the parallel forces, and m^, m^, etc., the masses of the particles on which they act, F^ = h7n^, F^ = hm^, etc., where Ic is a constant. Hence the distance of the centre of the system of parallel forces from any plane from which the distances of the particles are d^, d^, etc., is IJcmd _ k1,md _ 2mci! And this is the distance of the centre of mass. Hence the centre of the above system of parallel forces coincides with the centre of mass. 474. If a body be sufficiently small relatively to the earth, the weights of its particles may be considered to be parallel forces; and they are proportional to the masses 352 DYNAMICS [474 of the particles, for they produce in the particles the same acceleration, g. Hence the weights of the particles of a sufficiently small body are reducible to a single force equal to g times the mass of the body and acting verti- cally downwards through the centre of mass, whatever the position of the body may be. For this reason the centre of mass is often called the centre of gravity. The term centre of gravity has also the following signification to which it should be restricted : If a body attracts and is attracted by all external bodies, whatever their distance and relative position, as though its mass were concentrated in a point fixed relatively to it, that point is called its centre of gravity, and the body is said to be centrobaric or barycentric. In general, bodies are not centrobaric. We have seen (316, Ex. 6) that a uniform sphere or spherical shell has this property. If a body has a centre of gravity it necessarily coincides with the centre of mass. For, as we have seen (473), the resultant attraction of an infinitely distant body, whose attractions on its particles would be parallel forces, would pass through the centre of mass whatever the position of the body. 475. Examples. (1) Three forces act at the middle points of the sides of a i-igid triangular plate, in its plane, each force being perpendicular and proportional to the side at which it acts. If the forces are all inwards or all outwards, the resultant is zero. (2) If a rigid plane quadrilateral ABCD be acted upon by four forces, represented in magnitude, direction, and line of actioii by AB, CB, AD, CD respectively, the line of action of the resultant will be the line joining the middle points of the diagonals ; and its magnitude will be represented by four times the length of that line. (3) A system of any number of coplanar forces being represented 475] OF RIGID BODIES. 353 by the several sides of a closed polygon, as described by the con- tinued motion of a point in a plane, show that the sum of their moments round any point in the plane is independent of the position of the point. (4) If six forces acting on a rigid body be completely represented, three by the sides of a triangle taken the same way round, and three by the sides of the triangle formed by joining the middle points of the sides of the original triangle, and if the parallel forces act in the same direction, and the scale on which the first three forces are represented be four times as large as that on -which the last three are represented, the given six forces produce neither translational nor rotational acceleration. (5) Forces of 10, 20, 30, and 40 poundals act on a rigid body at A, B, 0, D, the comers of a square whose side is 2 feet, and in its plane. Their inclinations to AB, BC, CD, 1)A are 45°, 90°, 30°, 60° respectively. Show that their resultant is a force of35"65...poun dais , and that its line of action is distant 3'03... ft. from C. (6) Parallel forces in the same direction, and of the magnitudes 10, 15, 20, 25, act at points A, B, C, J) respectively of a straight rod, the distances AB, BC, CD being 2, 3, and 4 respectively. Find the distance of the point of application of the resultant from A. Ans. 5-07.... (7) Two parallel forces in opposite directions, and of magnitudes 20 and 5, act at points A and B respectively of a rigid body 4 feet apart. Find the distances from A and B of the point in which tlieir resultant line of action cuts AB. Ans. 1^ and 5j ft. (8) At each end of each side of a uniform triangular plate a force acts parallel and proportional to the line drawn from the opposite vertex to bisect that side. Show that the resultant of the six forces passes through the centre of mass of the triangle. (9) A triangular lamina ABC at rest is moveable in its own plane about a point in itself. Forces act on it along and propor- tional to BC, CA, BA. Show that if they do not move the lamina, Z 354 DYNAMICS [475 the point must lie in the straight line bisecting BC and CA, and that the reaction at the point is proportional to 'iAB. (10) Two parallel forces in the same direction and proportional to two of the sides of a triangle, act at the angles of the triangle, opposite the sides to. which they are proportional respectively. Show that their resultant passes through the point in the third side iu which it is cut by the line bisecting the opposite angle. (11) The numerical measures of the magnitude of a force which acts at a point in a given direction, and of the distances of the point from two straight lines at right angles to one another in the same plane with it, are denoted by a, 6, c ; but it is not known which is which. Find the centre of all the forces which may be represented. . „. , , , ,. ab + bc+ca Ans. Distance from each line = - a+b + e (12) Forces I, -3, -5,1 act on a rigid rod at points A, B, C, D, whose distances are such that AB=3, BC=2, CD=% Find the magnitude of the resultant couple. Ans. 15. (13) Three equal and codirectional forces {P) act at three corners of a square (side = a) perpendicularly to the square. Find (as) the magnitude of the force which, applied at the other corner of the square, would with the given forces constitute a couple, and (6) the moment of the couple. , Ans. (a) 3^; (6)i^a2^/2. (14) ABC is a triangle right-angled at B. At A a force F is applied in the plane of the triangle perpendicular to AC ; at C a force 2F in the same direction ; and at .B a force ZP in the opposite direction. Find the moment of the resulting couple. Ans.P{AE'-'2,BC^)IAC. (15) The resultant of three forces represented by the sides of a triangle taken the same way round is a couple whose moment is proportional to the area of the triangle. 476. Any forces whatever acting on a rigid body are reducible to a system of two forces. 476] OF RIGID BODIES. 355 A force, F, whose components in the directions of rec- tangular axes are F,., Fy, F^, acting at any chosen point whose distances from the centre of mass of the body in the directions of the components are ^, rj, f, will produce the angular acceleration about the centre of mass pro- duced by the acting forces, provided (427) ^ F,r,-Fyt=L,F4-F4=M,Fyi-F^ = N, where L, M, N are the algebraic sums of the moments of the acting forces about axes through the centre of mass, parallel to the x, y, z axes respectively. We may reduce these equations to one by multiplying the first by F^, the second by Fy, and the third by Ft, and adding, by which process we find that LF^-VMFy+l^F^ = ^ is the condition which must be satisfied that the force F may produce the required angular acceleration. It is obvious that values of F^i^, Fy, F^ can always be found to satisfy this equation. These values will be different for different chosen points of application. The force F with another force F" , whose components are F^, Fy, Fz, and which acts at the centre of mass, will produce the linear acceleration produced by the acting forces, provided F,+F^ = SZ, Fy+Fy' ^ 2F, F.+F^ = ^Z, where 2X, SF, '2Z are the sums of the components of the acting forces in the directions of the x, y, z axes respec- tively. As F acts at the centre of mass it has (458) no efiect on the body's angular acceleration. Now, whatever may be the values of F^, Fy, F^ which satisfy the first condition, values of F^, Fy, Fl may be found to satisfy the last three equations. Hence any forces acting on a rigid body are reducible to two forces. As the point ^, r\, f chosen above was any point what- 356 DYNAMICS 476 ever, the forces acting on a rigid body may be reduced to any one of an infinite number of pairs of forces. 477. To determine the condition of the reducibility of a system of forces acting on a rigid body to a single force. As this force must produce both the linear and the angular accelerations produced by the acting forces, we ha,ve, if F^, Fy, F^ are its rectangular components, and- ^, ri, f the co-ordinates, relative to the centre of mass, of its point of application, and if L, M, N are the moments of the acting forces about axes, parallel to the axes of co- ordinates, through the centre of mass, F„ = l^X,Fy = l:Y,F,= ^Z; and Fzri - Fy^= L, F^^- F4= M, Fyi- F^r, = N. ■ These six equations may, as in 476) be reduced to the single equation LI,X+MI,7+NI,Z=0, which therefore is the condition which must be fulfilled that the resultant of the given forces may be a single force. 478. The magnitude of this resultant force is clearly B = s/(EXf + (2¥f + (ZZf. Its direction cosines are liXjR, HiT/R, JIZ/R. If we put ^=0 in the equations of 477, we obtain ^=M/Fx, t]= —N/Fx. These therefore are the co-ordinates of the point in which the line of action of the force cuts the jjf plane. 479. Any forces acting on a rigid body may be reduced to a single force and a single couple. 480] OF KIGID BODIES. 357 If F^ be any one of the forces acting on a rigid body, there may be introduced at any point 0, without any change of the motion of the body, a pair of equal and opposite forces, F^, parallel to the original F^-, and for every force acting on the body we thus obtain an equal force in the same direction acting at 0, and a couple (called the couple of transference). The forces at give a resultant force at 0, and the couples compound into a resultant couple (469). Whatever point may be chosen, the direction and magnitude of the resultant force will clearly be the same. The resultant couple will however be different for diffe- rent positions of 0. 480. To determine the resultant force and couple for any given system of forces and for any given position of 0. Let Xj, Fj, Z^, X^, Fj, Z^, etc., be the components of the forces of the system in the directions of rectangular axes through 0, and let x^ y^, z^, x^, y^, z^, etc., be the co-or- dinates of their respective points of application. Then as the resultant force R is the same for all positions of 0, it must be the same as the force which at the centre of mass would produce the linear acceleration produced by the system of forces. Hence n = V(2Z)M-T2 7)2 + i^Zf ; and its direction cosines are 2X/i2, 1,Y/R, '2Z/R. As the component couples must produce about the chosen axes the same angular accelerations as the forces of the system, they must be equal to the moments of the forces about these axes. Hence if L, M, N are the component couples whose axes have the directions of the x, y, z axes respectively, 358 DYNAMICS [480 L = -E{Zy- Yz),.M='2(Xz-Zx), N='2,{Yx-Xy). Hence (469 and 88) the resultant couple = J\^{Zy - Yz)f + [-LiXz - Zx)J + [S( Yx - Xy)f, and its direction cosines are LjG, MjG, NjG. Hence also (8) the inclination of the axis of the resultant couple to the resultant force is ^°^ \R ■ 0^ E ■ G^ R- GJ- 481. The resultant force being given, and the resultant couple for a given point of application of the resultant force, to find the resultant couple for any other point of application. Let OR and OG represent the resultant force and couple , when the resultant force acts at 0. I -^ Let 0' be the other point of applica- tion. At 0' introduce two opposite forces R equal and parallel to the force R at 0. They will not affect the motion of the body. Now the forces i2 at and 0' constitute a couple, whose axis ON is perpendicular to the plane of ROO'R, and is proportional to the product of R into the perpen- dicular distance of OR from O'R. The two couples OG and ON give (469) a resultant couple OG' . Hence the given system of forces is equivalent to a force R acting at 0' and a couple OG'. If 0' is in the line of action of OR, it is evident that ON is zero and that OG' is the same as OG. If 0' is any- where else, ON will have a value and OG' will differ from OG. It is obvious that any other line of action of the resultant force than that through 0' must either be at a 482] OF RIGID BODIES. 359 greater or smaller distance from OB than O'R, or be so placed that the plane through it and OR is inclined to the plane ROO' R, and that therefore ON, and con- sequently also OQ', can have a given magnitude and direction only for one line of action of the resultant force. Hence in a given body, acted on by given forces, there is but one line, such that, if the resultant force acts in it, the resultant couple will have a given magnitude and direction. 482. Any forces acting on a rigid body may be reduced to a single force and a couple whose axis is parallel to the line of action of the force. Let R be the force {OR being its line of action) and the couple {OQ being its axis), which together form the resultant of the acting forces according to 479. The couple OG may be resolved into two, whose axes are OH and OJ, in and perpendicular to the direction of OR respectively. The couple OJ is in the plane through OR perpendicular to the plane of OR and OQ. Let 00', " drawn perpendicular to the plane of OR and OQ, be the length of arm of the component couple OJ when its forces are made equal to R, and let the forces R of this couple act at and 0' in directions perpendicular to 00'. Then we have two forces, R, acting at in opposite directions. Hence the original force R, together with the component couple OJ, are equivalent to a force R at 0', having the same direction as the original R. Hence the given system, viz., the force R acting at and the couple OQ, is reduced to the force R acting at 0' and a couple whose axis OH is parallel to R. When this reduction is made, the line of action of the force is called the central aa>is of the system of forces, and as this theorem is due to Poinsot, it is usually called 360 DYNAMICS [482 Poinsot's central axis. It follows from 481 that a given system of forces can have hut one central axis. Sir E. S. Ball has given the name wrench to the re- sultant force and couple to which a given system of forces may be reduced when the line of action of the resultant force is the central axis. 483. If the angle ROQ (482) is 6, the component couples OH and OJ are cos Q and Q sin Q respectively. Hence 00' = {Q sin 6)IR If the direction cosines of OR and OG are I, m, and n, X, fi, and v respectively, those of 00' will be (10) (mv — n/ji.)! sia. 6, {n\ — lv)/sin 6, (l/n. — mX)/sin 6. Now the products of 00' into its direction cosines are the co- ordinates of 0' relative to 0. Hence, employing the values of I, m, n, X, fi, v found in 480, we obtain as co- ordinates of 0' {N^Y-M^Z)IR\ {LI,Z-W:2X)lR\ (MI,X -LI,T)IR\ The direction of the central axis being thus known, and the position of one point in it, the axis is completely determined. 484. The magnitude of the resultant couple is less when its direction is that of the central axis than when it has any other direction. Let OA be the central axis of a given system of forces, O'A' any parallel line, and 00' a line perpendicular to both. Let R, acting at 0, be the resultant force, and OH the resultant couple. At 0' in- troduce two opposite forces, equal and parallel to R. Then the system is equi- valent to a force R acting at 0' and the couples OH and ON, ON being perpendicular to the plane 485] OF RIGID BODIES. 361 ROO'R and therefore to OH, and proportional to the product of R into 00'. The resultant of these couples is one represented by 00, which is necessarily greater than OR. 485. Examples. (1) The magnitudes, directions, and lines of action of four forces acting on a rigid body are represented by four sides of a skew quadrilateral taken the same way round. Show that the system is equivalent to a couple whose axis is perpendicular to both diagonals. (2) Show, by using the result of 477, that the resultant of any system of parallel forces is a single force. (3) ABCD is a tetrahedron, the angles BAG, CAD, DAB being right angles. At the centres of mass of the faces BA G, GAD, DAB forces act (all inwards or all outwards), with directions perpendicular to the faces, and magnitudes proportional to the areas of the faces. Show that their resultant is a single force. (4) When a force is transferred to any point 0, the resolved part of the couple of transference in any direction OZ is equal to the moment of the given force about OZ. (5) OA, OB, OG are conterminous edges of a cube and GD, EF are edges parallel to OB and OG respectively. Find the distance from of the central axis of a system of three equal forces com- pletely represented by OA , GD, and EF. Ans. AGIZ. (6) OA, OB, OG are conterminous edges of a rectangular parallelo- piped, so related that a positive rotation of 90° about OA as axis would bring OB to the initial position of OG. Forces proportional to OA, OB, OG (whose lengths are a, b, c respectively) act at B, G, and A in the directions OA, OB, OG respectively. Find the central axis. Ans. Its direction is that of the diagonal through 0, and it passes through a point whose distances from the planes BG, AG, and AB are respectively ae^ — a}fi c?h — bc^ b^c — o?c a2+6= + c2' a2 + j2+c2' aa+ja+^a- 362 DYNAMICS [485 (7) A system of forces can always be reduced to two forces whose lines of action are at right angles to one another. [Through a point P in the central axis draw a straight line, OPO', perpendicular to the central axis, bisected in P, and of such length that if F is the magnitude of the resultant force when its line of action is the central axis, F . OP is equal to the resultant couple. Resolve the resultant force, F, into two equal and parallel forces, F/Z, at and 0', and let the forces of the resultant couple act at and 0' also.] (8) The volume of the tetrahedron, opposite edges of which repre- sent the forces of any one of the infinite nunjber of pairs of forces to which a given system of forces may be reduced, is constant. Let F and F' be any such pair of forces, and let DF, OF' repre- sent them, DFCF' being therefore the tetrahedron referred to in the problem. Let AB be a line perpendicular to both DF and CF' At A introduce two opposite forces F' equal and parallel to GF'. Let R be the resultant of F and the force F' which acts at A and is codirectional with CF'. Draw AG perpendicular to the plane of the forces F' and representing the couple whose arm is AB and whose forces are the forces F' acting at C and A. Then the force AR and the couple AG form the resultant of the given system. DF, AF', and AG being all perpendicular to AB are in the same plane and AG is perpendicular to AF'. The resultant couple H whose direction is that of the central axis is (482, 483) such that 487] OF EIGID BODIES. 363 H=AO cos RAG ==CF'.AB.amEAF' =CF'. AB.sm FAF'. DFjR. Hence H. R^CF'. DF.AB. sin FAF'. Now (481) H.R is constant. And AB.CF' being equal to twice the area of the triangle ACF', and DF sin FAF' being the length .of the projection of DF on a line perpendicular to the plane of the same triangle, CF' . DF .AB .sin FAF' is equal to six times the volume of the tetrahedron FDCF'. Hence the volume of this tetrahedron is constant. 486. Moments of Inertia. — The quantity 'Lmr^ (453), the sum of the products of the masses of the particles of a rigid body into the squares of their distances from a fixed axis in it, is called the moment of inertia of the body about the given axis. If M is the mass of the body, a quantity h can always be found such that M¥ = 'Lm7^. The quantity k thus found is called the radius of gyration of the body about the given axis. Moments of inertia may be determined either by ex- periment or by calculation. 487. Determimation by Experiment. — Let the body whose moment of inertia I about a given axis is to be determined be so mounted that the given axis is fixed. Let it then be acted upon by a known force -F at a known distance p from the axis, and in a plane perpendicular to the axis, and let the angular acceleration a be observed. We have then (453) aI=Fp and I=Fp/a. It is practically impossible to apply a known force at a known distance from a given axis in it and to observe the angular acceleration. But it is generally easy to apply the same force or set of forces at the same distance or distances in successive experiments. Hence a moment of inertia is more readily determined by two experiments 364 DYNAMICS [487 than by one. First, let the angular acceleration of the body under investigation be observed when under the action of forces whose moment "LFp is constant from experiment to experiment. We have as above aI=^Fp. Next, let a body, whose moment of inertia (/') about a given axis is known, be rigidly attached to the given body so that the axis about which its moment of inertia is known is in the same straight line with the fixed axis of the given body ; and let the same forces be applied in the same way as before. If the angular acceleration is now found to be a, we have a'(I+I')=^'2Fp. Hence al=a'(l+r), a.nd. ■* /• a — a It is practically impossible to observe the angular acceler- ation. But the forces employed may readily be so applied (see 588) that the sum of their moments may be directly proportional to the angular displacement (6) of the body, and that they may tend always to bring the body to a position in which its angular displacement is zero. In that case the body will oscillate, the ratio of its angular acceleration to its angular displacement will be independent of its angular displacement, and every point of the body will therefore execute simple harmonic motions. Hence (163) the time of oscillation will be t = 2Tr>/6ia, and will be independent of the extent of the oscillation. For any given value of 6 therefore aocl/i^. Hence, the times of oscillation in the above experiments having been observed to be t and t' respectively, we have The best methods of applying the force and of observing the times of oscillation will be found described in books on Laboratory Practice. 488. Determination by Calculation. — To effect the 488] OF RIGID BODIES. 365 summation indicated by the formula 2mr^ the Integral Calculus is in general necessary. But in the case of bodies of simple geometrical form and of uniform density the summation may be effected by elementary mathe- matical methods. In the determination by calculation the following pro- positions will be found useful : (1) The moment of inertia of a body about a given axis is equal to the moment of inertia of the body about a parallel axis through the centre of mass, together with the product of the mass of the body into the square of the distance between the two axes. Let P be the position of any particle of the body of mass m. Let the plane of the diagram intersect the given axis and the parallel axis through the centre of mass, nor- mally in A and respectively. Let d be the distance between the axes, s the distance from the axis G of the foot M of the perpendicular PM from P on CA or GA produced, i.e., the distance of P from a plane through the centre of mass and perpendicular to CA. Let the length of PM be p. The moment of inertia of the body about A, Im.AP^ = i:m[{s - dY +p'], = 'Lm{s'^ + V^) + ^^2m - 2d1,ms, = I,m.GP'+d^I,m; since (403) 2ms = 0. Now Im . GP^ is the moment of inertia of the given body about an axis parallel to the given axis, through the centre of mass ; and d^Hm is the product of the mass of the body into the square of the distance of the axes. If M is the mass of the body, and Ic its radius of gyra- 366 DYNAMICS [488 tion about an axis in the given direction through the centre of mass, the moment of inertia about the given axis is M{1 (10.) Find the moment of inertia of a uniform thin plate (mass = M) of a regular polygonal shape, about an axis through its centre of figure and normal to its plane. If there are n sides, each having the length a, and each distant r from the centre of figure, as the polygon may be divided into n isosceles triangles with vertices at the centre of figure, of base a, height r, and mass M/n, \24 2/ (11) Find the moment of inertia of a uniform thin circular plate 2a 370 DYNAMICS [490 (mass=i^, radius =»•) about an axis through its centre and normal to its plane. As the circle may be considered to be a polygon with an inde- finitely great number of indefinitely short sides, each distant r from the centre of figure, we have I=Mr'^l'2i. (12) Find the moment of inertia of a uniform thin circular plate (mass = J/, radius =r) about (a) a diameter, (6) a tangent. Ans. (a) Mr'^14. ; (b) bMr^i. (13) Fiiid the moment of inertia of a uniform circular cylinder (mass= J/", radius=>') (a) about its axis, (6) about a generating line. Ans. (a) Mr^l2 ; (6) 3ify2/2. (14) The moment of inertia of a uniform sphere (mass = J/, radius =?•) about a diameter being 2Mr^l5, find its radius of gyration about a tangent line. Ans. rv7/5. 491. MeasureTTient of Moment of Inertia. — The unit of moment of inertia is that of a particle of unit mass at unit distance from the axis of rotation. In specifying moments of inertia, no mention is usually made of the unit, but they are described as of such and such a value when expressed in such and such units of mass and length. The dimensions of the unit of moment of inertia are clearly [M] [Lf. 492. Examples. (1) Express in oz.-in. units a moment of inertia of 20 ft.-lb. iinits. Ans. 46,080. (2) A moment of inertia has the value 500 when expressed in terms of the centimetre and the gramme. Find its value in terms of the metre and the kilogramme. ns. 0-00005. (3) An author speaks of a rectangular parallelepiped (edges 1, 2, 493] OF EIGID BODIES. .371 and 12 cm. respectively and density 4 grm. per cub. cm.) as having a moment of inertia equal to 1 "207 about an axis through its middle point and perpendicular to the face of greatest area. He is known to have employed the cm. as unit of length, and to have worked where g has the value 980'94 cm.-sec. units. What must have been his unit of mass ? Ans. The unit of mass of the C.G.S. system of gravitational units. (4) By what nximber must we multiply the value of a moment of inertia expressed in the units of the ft.-lb.-sec. absolute system in order to determine its value in the Unit of the C.G.S. absolute system ? Ans. 421390-7.... (5) At the end of a thin rod, of length 2 ft. and hnear density 1 oz. per in., are particles of masses 1 and 2 lbs. respectively. Ex- press its moment of inertia about an axis perpendicvdar to the rod through a point distant 3 in. from the particle of smaller mass, in the units of (a) the absolute, and (i) the gravitational in.-oz.-sec. systems (.9 = 32 ft.-sec. units). Ans. (a) 17,352 ; (6) 45-18.... 493. Equations of Motion. — The moment of inertia being thus a quantity capable of determination, we can apply at once, to cases of motion, the equations of 454. If the motion be about an axis fixed both in the body and in space, the angular acceleration about it a='EFP/'2mr^, where XFP is the algebraic sum of the moments of the external forces, and 2mr^ the moment of inertia, about the fixed axis. If the body be quite free to move, the linear accelera- tion of the centre of mass is given by the equation a = 2172m ; and the angular acceleration about any axis fixed in the body through the centre of mass, by the equation a = 2-P'P/2mr^, where 2-FP is the algebraic sum 372 DYNAMICS [493 of the moments of the external forces and 2m?'^ the moment of inertia, about this axis. These accelerations being determined and the initial velocities being given, the final velocities and the dis- placement may be found (see 251). The above equations are therefore called the equations of motion of a rigid body. We can apply them only in simple cases of the motion of rigid bodies (227), in cases, viz., in which one line in the body has a fixed direction in space. More complex cases require higher mathematical treatment than the readers of this work are supposed able to apply. 494. In many cases, especially when the forces act only for a very short time, it is convenient to have the equa- tions of motion expressed in terms of the impulses of the acting forces rather than of the forces themselves. Let V, w and v', w' be the initial and final values of the com- ponent linear velocity of the centre of mass in the direc- tion of the impulse, and of the angular velocity of the body about the fixed axis, respectively. Then (117, 225, and 319), -, - ^Ft 2* 2m zm and w —w = ^ — 9 = v» — 9> where $ is the impulse of the force F. 495. The laws of the conservation of linear and of angular momentum, deduced from the equations of motion of extended systems (416 and 429) apply necessarily to rigid systems. The expression of the latter becomes somewhat modified however. In the case of extended systems, it is expressed by the equation Smwr^ = con- stant. In the case of a rigid body, either about an axis 496] OF RIGID BODIES. 373 fixed both in it and in space, or about an axis fixed in it through the centre of mass, it is expressed by the equa- tion w^rrir^ = constant. 496. Motion about Fixed Axes. — We shall now discuss some examples of the application of the equations of motion to the determination of the motion of rigid bodies. We take first cases in which the axis of rotation is fixed both in the body and in space. Examples. (1) Find the angular acceleration of a uniform circular ddac, moveable about an axis through its centre, perpendicular to its plane, under a force applied in the plane of the disc by means of a string fixed at a point of the rim of the disc and wrapped round the rim.— Let the disc have a mass M and a radius r, and let its radius of I gyration about the given, axis be k. The force F acts tangentially to the disc. If it acts as in the diagram, its moment about the fixed axis is (425) - Fr. Hence (493) Fr The angular acceleration is therefore constant. Hence, if the initial angular velocity be given, the final angular velocity and the displacement after any time may be determined as in 225. (2) A rigid rod, 12 ft. long, whose mass may be neglected, has at one end a particle of 10 lbs. mass and at the other a particle of 5 lbs. mafis. It is free to rotate in a horizontal plane about an axis through the centre of mass of the system. Find the force which must be applied to the smaller particle perpendicularly to the rod that unit angular velocity may be produced in the rod in 1 sec. Ans. 60 poundals. 374 DYNAMICS [496 (3) Find the time of oscillation of a heavy body capable of rota- tion about a fixed horizontal axis, which does not pass through its centre of mass. [Such a body is .called a compound or physical pendulum.'] — Let the plane of the diagram be a plane through the centre of mass perpendicular to the fixed axis. Let S be the point in which that plane intersects the axis, C the centre of mass, and SN a vertical line. The external forces are the weight mff (if m is the mass of the body) and the forces by which the axis is fixed. The latter have no moment about the axis. The moment of the weight which (474) acts at the centre of mass is, if we denote CS by A and the angle CSN by 6, -mgh sin 0. If k is the radius of gyration of the body about an axis parallel to the fixed axis through the centre of mass, the moment of inertia of the body about *S is (488) ?n(F+A2). Hence, if a is the angular acceleration about S, _ mgh sin "~~^', be the components vertically upwards of the velocities of the centre of mass before and after impact respectively. Let w and u be the angular velocities about the centre of mass before and after impact respectively (the positive direction of rotation being counter- clockwise). Then the integral linear and angular accelerations are v'—v and a' — u respectively. Hence and w' >— w = — *a cos 6/(mk^. (a) 'If there is no recoil, A remains in contact with CI) after the impact. Hence the sum of its component velocities after impact must be zero. These are «' upwards and u'acosS downwards. Hence "' —u'acoa8 = 0. These three equations are sufficient to determine ¥, u', and * in terms of v, u, and $. (b) If there is recoil, let the coefficient of restitution be e. Then if *' be the impulse on impact, and * the value the impulse would have were there no recoil, we have (379) *'=(l+e)*. 384 DYNAMICS [497 ' is therefore known in terms of v, u, $, and e. If its value be sub- stituted for * in the first two equations, v' and o>' may then be determined for this case also. (15) A beam which is moving without rotation in a horizontal plane impinges without recoil on a fixed rod at right angles to the plane. Find (a) the impulse of the reaction of the rod, and (6) the angular velocity of the beam immediately after impact. Ans. (a) mi%aino/(c^+A^); (6) CMsino/(c2+P) ; where m= mass of beam, i= radius of gyration about normal axis through centre of mass, M= velocity of centre of mass before impact, a=inclination of direction of u to the beam, c= distance of centre of mass of beam from fixed rod at instant of impact. 498. Motion of Systems of Rigid Bodies. — If the motion is to be determined of several bodies which act upon one another, the equations of motion must be applied to each of them. The following cases will serve as illustrations : Examples. (1) Two particles of masses m, and m' are connected by an inex- tensible string which hangs over a pulley moveable about a fixed horizontal axis. The axle of the pulley is smooth, its rim so rough that the string does not slip. Find the acceleration of the particles. (A twood's Machine. See 382, Ex. 1). Let yand T be the tensions in the portions of the string attached to m, and m' respectively. The moments of T and T about the axis of the pulley are, if r is the radius of the pulley, Tr and — Tr respectively. Hence, if a is the angular acceleration of the pulley, Tr- Tr "" MW ' where Jf is the mass of the pulley and h its radius .J,' >^ of gyration about its axis. As the string is in- extensible, the acceleration a of m' is the same as that of m; and we have as in 382, Ex. (1), a = {m'g-T')lm; = {T-'mg)lm. 498] OP RIGID BODIES. 385 We have also, since the string does not slip, Hence we have four equations involving four unknown quantities, a, a, T, T. Eliminating o, T, and T, we find _ m! — m which differs from theresult of 382, Ex. 1, in which the string hangs over a smooth peg by the introduction of the quantity MFjrK (2) A uniform cylinder weighing 100 lbs. tarns without friction on its axis which is horizontal. Motion s communicated by a body of 10 lbs. mass, attached to an inextensible string without weight, which is coiled round the surface of the cylinder. Find the distance through which the body will descend from rest in 10 sec. Ans. \Qg. (3) To the string coiled round the wheel of the simple machine called the Wheel and Axle (254, Ex. 5) a mass of 10 lbs. is attached ; to the string around the axle a mass of 100 lbs. Given that the radii are 3 ft. and 3 in. respectively, that the moment of inertia about the axis expressed in terms of the pound and foot is 2400, and that the machine is frictionless, find the number of revolutions made in 1 minute from rest, taking g to be 32 ft. -sec. units. Ans. 60/7r. (4) Find the time of a small oscillation of a Balance with nearly equal masses in its pans. [The Balance consists of a' practically rigid body called the beam, moveable about a horizontal axis fixed in it, and symmetrical about a plane through this axis and the centre of mass of the beam. It carries pans or scales to contain, one the body to be " weighed," and the other standards of mass. The pans are moveable about axes fixed in the beam, which are parallel to the axis of the beam, and are equidistant both from it and from the centre of mass. A plane through the centre of mass of the beam, perpendicular to the three axes, intersects them in three points, which are called the points of suspension of the beam and pans respectively. The distances of the points of suspension of the pans from that of the beam are called the arms of the balance. The 2b 386 DYNAMICS [498 centre of mass of the beam is usually below its fixed axis, the beam being provided with an adjustment by which the position of that point may be varied. The line joining the points of suspension of thte pans passes in general below the axis of the beam. In some in- struments it is made to pass as nearly as possible through the axis, in others a little above it, when the pans are unloaded.] Let OBC be the beam, being its point of suspension; (? its centre of mass, and B and C the points of suspension of the pans P and Q. Let A be the point in which OG produced cuts CB. It is obvious that OA is at right angles to BC. In the diagram the angle OCA is for clearness made large. It is usually small. The beam is acted upon by three forces, its weight and the resultants T, T' of the tensions in the strings or rods supporting P and Q respectively. The motion of the beam is usually slow and through small angles. Hence, though P and Q will oscillate about G and B, we may for an approximate result assume T and T' to be vertical. If, then, M be the mass of the beam and h its radius of gyration about 0, m and m! the masses of the pans P and Q respec- tively with their contents, ;8 the angle OCA , and 9 the inclination of BG to the horizon at any instant, we have for the angular ac- celeration (a) of the beam _ T{AGcose-OAsme)-T\ABao&e+0Aaine)-Mg.0a.sm e. For the reader will have no difficulty in proving by the aid of the 498] OF EIGID BODIES. 387 second figure, (in which the dotted lines are all either horizontal or vertical) that the numerator of this expression for a is the algebraic sum of the moments of the forces about 0. The pans P and Q are acted upon by two forces each, viz., their weights and the resultants T and T' of the tensions in the strings or rods supporting them. Hence the vertical linear acceleration of P is {mg — T)jm ; and it is equal to the vertical component of the linear acceleration of C, which is a . OC . cos (e+jS). Hence m.g~T= maOG cos (0 + ;8). Similarly 7"-TO'gr=m'oOCcos(9-|3). Now OCcos(e±^) = OC(cQa « cos /S + sin e sin |3) ==ACcoBe+OA sine. Hence aMk^ = 'mg(A C cos e - OA ain B) - m'g{A O cos e+OA sine) - MgOa sin e - moOC^ 008^(0 + /3) - m'a.OC^ cos2(e - A ) For small values of e therefore aMlc^=mg(AC- OA.0)- m'g{AG+ OA .e)-Mg . OG .6 - (m + m')aOC^ cos^^ Hence, noting that AC= OC cos j8, _ m(AC- OA.e)- m'{A C+ OA .' e) -M.OG.e " Mk^ + {m + m')AC^ ^' . If the masses of P and Q are the s^-me, viz. m, we have __ 2m.0A+M.0O „ " Mk^+2m.AG'^ ^ ■ Hence (496, Ex. 3) the time of a small oscillation is / Mk^ + im.AC-'' \/ {2m.OA+M.0O}ff' If B, 0, and C are in the same straight line, AC= OC and OA =0. Hence «=2x^ _^^_^^__ . Quickness of motion is a desirable characteristic of a balance, and it should therefore be so constructed that i may be as small as 388 DYNAMICS [498 possible. Hence for this purpose the mass of the beam, its radius of gyration, and the distance between the points of suspension of the pans, should be as small as possible, and the distance from the axis both of the centre of mass and of the line joining the points of suspension of the pans should be as great as possible. For conditions of sensitiveness, see 507; Ex. 11. (5) A uniform sphere S rolls -without sliding down AB a line of greatest slope of the inclined plane surface of a wedge AG, which lies upon a smooth horizontal table Ox, the centres of mass of the sphere and wedge and the given line AB being in the same vertical plane. It is required to determine the motion of the sphere and wedge. Let the vertical plane containing the line AB and the centres of mass of sphere and wedge, Z) and O, intersect the table in the line Ox ; and let Oy be drawn in this plane, and perpendicular to Ox. Let m be the mass of the sphere, r its radius, k its radius of gyra- tion about any axis through its centre of mass, m' the mass of the OB C wedge, B the normal component, and J'' the frictional component of the reaction between the sphere and wedge, and 4> the angle ABO of the wedge. As it is the frictional component of the reaction of the wedge on the sphere which causes the sphere to roll, it must be directed up the plane. Hence the same component of the reaction of the sphere on the wedge is directed down the plane. (The forces acting on the wedge are indicated by double arrow-heads.) The equations determining the motion of the sphere are as fol- lows. If a^c is the linear acceleration of the centre D of the sphere in the direction Ox, ■*98] OF EIGID BODIES. 389 ««=( - R sin +F cos )lm. If dn is its linear acceleration in the direction (?y, a„ —{Rco&4> + F sin - mg)lm. If a is the angular acceleration of the sphere about an axis through D perpendicular to the plane of the diagram a.=Frl{mk^). The wedge obviously moves so that BC remains in contact with the table. Hence the weight of the wedge and the normal reaction of the smooth table do not affect its motion, and the acceleration a'x of the centre of mass of the wedge is therefore a'x=(R sin - i^cos )/m'. In the above four equations we have six unknown quantities, R, F, a, a^, Uy^ a'x. But two more equations may be obtained by a statement of the kinematic conditions of the problem. First, since S rolls down the inclined plane, the change produced in any time in i^s vertical distance from any point B in the wedge divided by the change in its horizontal distance from the same point is equal to tan 0. Hence at every instant the ratio of Ifs vertical velocity relative to B to its horizontal velocity relative to B has this value ; and hence also the ratio of the vertical acceleration of D to its horizontal acceleration relative to B has the value tan 0. Now, the vertical accelerations of D relative to B and to are the same, for B has no vertical acceleration relative to 0. And the horizontal acceleration of D relative to is equal to that of B relative to 0, together with that of D relative to B ; and therefore the horizontal acceleration of D relative to B is equal to that of D relative to 0, minus that of B relative to 0. Hence the first kinematic condition may be expressed by the equation «»/(«! - a'l) = tan 0. Secondly, since there is no sliding of the sphere on the inclined plane, the linear velocity in any direction of that point of the sphere which is in contact with the wedge must at every instant be equal to the velocity of the wedge in the same direction. Now, this point of the sphere has in the direction Ox two component 390 DYNAMICS [49S linear velocities, one v^ the horizontal velocity of 3, and another ria cos 4' due to the angular velocity w of the sphere about D. Hence if v^' is the horizontal velocity of the wedge, v^'^Vx + rucos. As this holds for every instant, it follows that afx=ax+ra cos ' + Tnv' cos x- As no forces act on either sphere in the- direction perpendicular to CD M sin 0=1*' sin ^', and 'wsinx=»'sinx'. If there is no recoil the component velocities in the direction of CD are simply equalized by the impact. Hence w'cos0' = i/cosx'. 498] OF EIGID BODIES. 393 These four equations are sufficient to determine «', «', 0', x in terms of u, V, tp, and x- If there is recoil, and if e is the coefficient of restitution, we have (379) vf cos x' - u' cos <)>' — e(u cos - w cos x) ; and this equation, with the first three of those obtained above, are sufficient to determine the four unknown quantities. The impact of two spheres, whose centres at the instant of impact are not moving in the same straight line, is said to be oblique. (12) A smooth ball A , weighing 20 grm., strikes another ball E which is at rest, the direction of A's motion being inclined 30° to the line joining the centres of A and B at the instant of impact, and glances off in a direction perpendicular to that of its motion before impact. Find the mass of B, the coefficient of restitution being 0'4. Ans. 400 grm. (13) A billiard ball A (mass=m) impinges upon another B (mass =«i') which is at rest, the direction of A'a motion before impact being inclined 45° to the line joining the centres at the instant of impact, rind the direction of A'a motion after impact, assuming the coefficient of restitution equal to unity. Ans. Inclination to the line joining the centres =tan~*^ — — — . m — m (14) Two billiard balls A and B are lying in contact on a table. Find the direction in which B must be struck by a third ball C so as to be driven off in a direction inclined at a given angle d to the line joining the centres of A and B, all three balls being smooth and of equal volume and mass. Show that the result is the same whatever be the value of the coefficient of restitution. Ans. The line joining the centres of C and B at the instant of impact must be inclined to the line joining the centres of A and B at the angle tan-\| tan ff). (15) Two straight rods ACBaxiA CD, whose thickness and density are equal, and whose coefficient of restitution is unity, lie on a smooth horizontal plane at right angles to each other, the end C ■">94 DYNAMICS [498 of the latter being in contact with the former. Determine the point at which ACB may be struck without consequent rotation. Ans. If AC=a, CB=h, and CD = c, and if a>h, the required a2 - 62 point is in AC, and its distance from C is y:-, ; .. 2(a + 6 + c) (16) A ball (mass=«i, radius=?', radius of gyration about its centre = ^) sliding without rotation along a smooth horizontal plane, with velocity u, strikes against a perfectly rough vertical plane, its direction of motion before impact being inclined at the angle 6 to the vertical plane. Show (1) that if there is no recoil the impulse, during impact, of the frictional component of the reaction of the vertical plane is 7nJc\ cos f + Jc' ' and (2) that if the coefficient of restitution is e, the ball's direction of motion after impact is inclined to the vertical plane at the angle 499. The Law of Energy. — The general law of energy, deduced (437) from the equations of motion for extended systems, including the law of the conservation of energy (435), applies of course to those extended systems which are rigid. Its application to the solution of problems is simplified in the case of rigid bodies for two reasons. First, as the particles of the system are at invariable distances from one another, the internal forces do no work in any displacement, and therefore the external forces only appear in the equation. Secondly, the ex- pression for the kinetic energy relative to the centre of mass is very simple. If m is the angular velocity about an axis fixed in the body, and r the distance of a particle from the axis, cor is its linear velocity relative to points in that axis. Hence the kinetic energy of' the system rela- tive to points in the given axis is 2 499} OF EIGID BODIES. 39& if M and h are the mass of the body and its radius of gyration about the given axis respectively. If the given axis pass through the centre of mass, the above is the expression for the kinetic energy relative to that point. We shall illustrate the application of the law of energy to the solution of problems by a few examples : Exarri'ples. (1) A uniform rod moves in a vertical plane within a fixed smooth hemisphere. To determine its angular velocity in any position, its initial position being one of instantaneous rest. — Let ABB' be the hemisphere, its centre, Oa; and i o o ff Oy horizontal and vertical lines respectively in the plane of the rod's motion. Let A'B' be the initial position of the rod, h being the distance (ffJ)') of its centre of mass (C) from Ox. /^ Let AB \)& its position at the instant under consideration, OB and DC, or x and ^, being the co-ordinates of its centre of mass. The component velocities of C will be x and y. Hence if m is the angular velocity of the rod about C, m its mass, and h its radius of gyration about a normal axis through C, the kinetic energy of the rod in the position AB and therefore the increase of kinetic energy during the motion from the position A'B' to the position AB is (442) \m{a? +y^+ Pf^. The external forces acting on the rod are the reactions of the smooth sphere and the weight of the rod. As the ends of the rod move in all positions in directions perpendicular to the reactions exerted on them, no work is done by or against these reactions. Work has been done by the weight of the rod, and, as the centre of mass has fallen through the distance y — h verti- cally, the potential energy has diminished by the amount mg(^ — h). Hence, by the law of conservation of energy, Now the instantaneous centre of the motion (233) of ^^ is 0. Hence 396 DYNAMICS [499 the linear velocity of C is perpendicular to OG and has the magni- tude OC. w, or, if c be the distance of the rod from the centre, cu. Hence its components in the directions Oip, Oy are, if the angle COy be written S, x= —cbi cos 6, y = cuamd. Also _ y = c cos 0, and A=ccos^, if is the initial value of 6. Hence, substituting in the above equation these values of x, y, y, and k, we obtain {c'+k^)a^=2cg{cos - cos + Qsm9 (10) The radii of the wheel (If) and the axle {A) of the simple machine called the Wheel and Axle (254, Ex. 5) are R and r respectively. Find the force P exerted through a string coiled round the wheel which will balance a force Q exerted through a string coiled round the axle (the axle being smooth). Since the sum of the moments about the fixed axis must be zero, PR -Qr=0 a,ndP=QrlR. (11) Find the conditions that must be fulfilled that a Balance maybe stable and sensitive. (See 498, Ex. 4.) The beam without the pans will be in stable equilibrium (450), if G (fig. of 498, Ex. 4) be vertically below 0, in which case BG will be horizontal. If the centre of mass of the beam be at 0, the beam without the pans will be in neutral equilibrium. With pans of equal mass the beam will be in stable equilibrium, with BC horizontal, provided (1) that G and BC are both below 0, or (2) that if O be above 0, BC be sufficiently far below it, or (3) that if BC be above 0, G be sufficiently far below it. When the balance is in equilibrium, T and T' are vertical and equal to mg and m'g respectively. As the sum of the moments of the forces acting on the beam about must be zero, we have, if m and m' are the masses of the pans, and 8 the inclination of BC to the horizon, mg[AC cos e-OAaiai)- m'g{AB cos 9 + 0^ sin e) - Mg . 00 . sin 9=0. The greater the angle 8 for a given value of m — m' the greater is the sensitiveness of the balance. Hence for sensitiveness the mass 507] OF RIGID BODIES. 407 of the beam, the load (i.e., the total mass in both pans) and the dis- tance of the axis of the beam both from its centre of mass and from the line joining the points of suspension of the pans must be as small as possible, and the distance between the points of suspension of the pans must be as great as possible. Except with regard to the mass of the beam, the conditions for sensitiveness and for quickness of motion (498, Ex. 4) are antagonistic. Hence in all balances the mass of the beam is made as small as is consistent with sufficient rigidity, and a compromise is struck between the demands of sensitiveness and of quickness of motion with regard to length of arm, etc. If the line BC pass through 0, OA = 0. Hence tane = (^,-;^-^. •M.OG In this case, therefore, the sensitiveness is independent of the load. (12) nil, ™2i ^'"® t^^ apparent values of the mass of a body when weighed successively in both pans of a balance which has its three suspension points in a straight line, (a) If its pans are equal and its arms unequal, show that the real mass of the body is \fm^n^. (6) If its arms are equal but the pans unequal, show that the difference of the masses of the pans is ^(m^^m^i. (13) The beam of a false balance (i.e., one having unequal arms) is 3 ft. long. If a certain body is placed in one scale it weighs 4 lbs., if in the other 6 lbs. 4 oz. Find (a) the real mass of the body, and (6) the lengths of the arms. Ans. (a) 5 lbs.; (6) 1 ft. 4 in. and 1 ft. 8 in. (14) The shaft of a steam engine carries a strong wheel (radius Jn. = r) with a flat I'im. An iron strap lined with blocks of wood is 408 DYNAMICS [507 fitted round it, and presses the blocks against the flat rim of the ■wheel. A rod is attached to the strap, and carries at its end (dis- tant I from- centre of shaft) a pan for standards of mass. When the shaft is making n revolutions per second, and doing no work except that of overcoming the friction of the strap, the rod is main- tained in a horizontal position by putting standards, of ■weight W, into the pan. Find the rate at ■which the engine is working. [This arrangement is called the Friction Brake Dynamometer. It should be called an ergometer, as it is not force that ■we measure by it, but rate of ■work done.] The rigid system, consisting of strap, rod, and wooden blocks, is in equilibrium under its own weight, the weight of the standards, and the friction of the rotating wheel. The rod is always counter- poised, so that its own. weight passes through the shaft of the engine, which is the fixed axis of the rigid system under considera- tion. Hence if F is the friction of any small element of the surface of the wheel, we have -ZFr- Wl=Q. Hence Wl^JiFr^r-ZF. The work done against friction during n revolutions of the shaft at each element of the surface of contact is iTrmF. Hence the whole work done against friction per second is ^vnrJiF='2,Tn'Wl. Hence the rate at which the engine is working is expressed in terms of n and W. (15) A uniform thin triangular plate is supported in a horizontal position by three props at its angular points. Show that the pres- sures on the props will be equal. [As there is equilibrium, the moment of the force exerted by any prop about an axis through the points at which the other props touch the plate must be equal to the moment of the weight about the same axis.] (16) A triangular plate is hung with its plane horizontal by three vertical chains, attached to the middle points of its sides. 507] OF EIGIP BODIES. 409 What must its mass be that a ton may be placed anywhere on it without tilting it ? Ans. 3 tons. (17) A rod AB hinged at one end so that it can move in a ver- tical plane rests with the other end on a smooth inclined plane, whose line of intersection with the horizontal plane is perpendicular to the plane of AB's motion. Find the force exerted on it by the hinge and the reaction of the plane. Let ABhe the rod hinged at A and resting with the end B on the inclined plane BC. Let AC J) be the horizontal plane, and let the angles BCD, BA C, be e and ip respectively. The rod is acted upon by three forces — its weight W acting vertically through G its centre of mass, the normal reaction B of the inclined plane, and the force F exerted by the hinge. Hence (506) the force F must pass through the intersection ^ of TT and R. The direction of F is known if the angle EAB(\li) is known. Now, AG sm.AEQ _ e.m.{\^r~-^|/) _ cos( + f) ~(iJf~ sin f sin ^ sin ^ ' GB sin GEB_ sin g _ sin g and Hence and Inf sin GBE sin (^x + ^ - e) cos (fi - 0) ■<^g_ cos(9-^) cos(0 + \{') . ^B sing ' sin^ t-ifjiG sing ^=cot" \GB cos cos ^)+*-4 410 DYNAMICS [507 For the magnitude of F we have (506) Hence W:F=smFR:si3x WR =sinAEB:sia6FB = sin{^Tr-4>-i' + 9) : sin 9 = cos(^ + ^ - S) : sin 6. coa{

gainst a rough vertical wall (angle of repose = e), and a cord tied to the other extremity B is fastened at a point in the wall above A, the vertical plane through the rod being perpendicular to the wall. Show that, if the rod is to be horizontal, the length of the cord must be - Vw^ + tan^e. n (28) A uniform heavy rod, 2 ft. long, is hung up to a peg by means of two strings tied to its ends, the lengths of, the strings 414 DYNAMICS [507 being 1 ft. and ^3 ft- respectively. Show that, when the rod is in equilibrium, it will make an angle of 30° with the horizon, and the tension of the shorter string will be equal to half the weight of the rod. (29) A uniform heavy rectangular trap-door is moveable about one edge as a hinge-line. To the middle point A of the opposite edge is attached a string which passes over a smooth pulley at the point occupied by A when the door is horizontal, and sustains a body of weight w. If W be the weight oif the door, shotv that the inclination of the door to the horizon is given by the equation cos^^ff - -™ cos ^9 — ^ = 0. (30) A carriage wheel (weight = W, radius =»•) rests upon a level road. Show that the force necessary to draw, it over an obstacle of height h is W'Jh{2r-h)/{r - A). (31) A heavy uniform sphere hangs from a peg by a string, the length of which is equal to the radius, and rests against another peg, vertically below the former, the distance between the two being equal to the diameter. Show that the tension of the string is equal to the weight, and the reaction of the peg to half the weight, of the sphere. (32) A beam or lever is moveable about a fixed rough cylindrical R. axle (radius =r, angle of repose = e), which very nearly fills the hole in the beam through which it passes. Find the relation d07] OF RIGID BODIES. 415 between two forces P and Q acting on the beam at given points A and B and in given directions in a plane perpendicular to the axle, when the beam is on the point of moving. — Let 8 be the inclination of P and Q, and let p, q be their distances from C the centre of the axle. Produce P and Q to meet in 0. Then R, the reaction of the axle, must pass through 0. Since the axle only nearly fills the socket there is contact, at any instant, only along a single line. If this line is represented in the diagram by the point D, BO will be the direction of R and will be inclined to CD produced at the angle e. Hence the distance of R from C is rsine. For equilibrium therefore we have Qq = Pp + Rr sin e. Hence Qq = Pp + rava. e >jF'JrQ'+'2,PQ cos e, which is the required relation between P and Q. (33) A heavy homogeneous cubical block rests on a rough hori- zontal plane, and a force is exerted on it by means of a string attached to the middle point of one of the upper edges, the string and the centre of mass being in the same vertical plane. The force being gradually increased, find the nature of the initial motion of the block. — Let ABO be the plane in which all the forces act, and let F'& line of action be above the centre of mass D. Then the initial motion of the block will clearly be either a sliding in the direction of AB or a turning about the edge B. For F=0, the re- action of the plane is normal to AB ; but, as F is gradually in- creased, the reaction (328) becomes gradually more and more inclined to the normal, passing, since there is equilibrium, through the inter- section oi F and W. If the cube turn about B, the reaction must pass through B. If therefore it is on the point of turning about B, the line of the reaction must be BO. Hence, if the fric- tion is such that the angle CBO is less than the angle of repose, the initial motion will be a turning about B. If however CBO is 416 DYNAMICS [507 greater than the angle of repose, slipping will be the initial motion ; for the block will begin to slip when the reaction is inclined to the normal at the angle of repose. Let d be the inclination of F to A B, e the angle of repose, and a the edge of the cube, then the condition for initial turning is tan e > tan CBO >|/(a-|tane) > 1/(2- tan e). Hence also the condition for initial sliding is tane < 1/(2- tan «). If F's line of action is below />, the possible initial motions are sliding in ihe direction AB and turning about the edge A. Show . that the condition of initial turning about A is tarie>l/(tane-2). (34) A homogeneous right cone (vertical angle = 29) is placed with its base on a rough inclined plane (coefficient of friction =/i), whose inclination is gradually increased. Show that, if /n > 4 tan B, the initial motion of the cone will be tumbling, and if ii < 4tan 9, its initial motion will be sliding. (35) A rectangular block is placed with one of its edges horizontal on a rough inclined plane. Show that, if a is the length of the edge of the block which is perpendicular to the plane, and b the length of the other non-horizontal edge, and if /a is the coefficient of friction, the initial motion will be one of tumbling, provided ,11 > hja, and of sliding, provided /i< hja. (36) A rectangular block, weighing 20 lbs., with a square base 8 inches in side, is set up on a level table, and it is found that a horizontal force equal to the weight of 5 lbs., if applied below a certain point, is just able to make it slide, while, if it is applied above that point, the block topples over. Find (a) the position of this critical point, and (6) the coefficient of friction between the block and the table. Ans. {a) 16 in. from the base ; (6) 0'25, 5091 OF EIGID BODIES. 417 508. Equilibrium of a System of Bigid Bodies. — If two or more rigid bodies be connected by strings, rods, joints, etc., the system is said to be in equilibrium pro- vided (1) the system behave as a rigid body and be in molar equilibrium, or (2) each body of the system be in molar equilibrium. (1) If the system behave as a rigid body, its parts not moving relatively to one another, the necessary and sufficient condition of equilibrium is (500) the satisfac- tion of the equations 2^=0, ^FP = 0, the forces in- volved being those external to the system only. (2) The necessary and sufficient condition for, the equilibrium of each body of the system is the satisfaction for each body of the equations 2-F=0, 2iT = 0, the forces involved in the equations including forces external to the body to which they apply, and therefore in general some forces which are internal to the system. 509. Examples. (1) A body is to be supported by means of the system of smooth pulleys represented in Fig. 1, p. 418. The weight of the body being W, and that of the block (254, Ex. B) w, find the force F ■which must be applied at the end of the cord. The pulleys being smooth, the stress throughout the whole string is F (391). Hence, if there are n sheaves in each block, . the lower block is acted upon by 2>i + 2 forces, 2n having each the magnitude F and an upward direction, and two the magnitudes W and w respectively and downward directions. If the directions of all are taken to be vertical, we have therefore 2nF=W+w. The ratio W/F is called the mechanical advantage of the system of pulleys. If M)=0, it has in this system the value 2n. (2) rind the mechanical advantage of the system of smooth 2d 418 DYNAMICS [609 weightless pulleys represented in Fig. 2, there being n moveable pulleys. Ans. 2". (3) Find the mechanical advantage of the system of smooth weightless pulleys represented in Fig. 3, there being n pulleys, and the ropes being so long that they may all be considered vertical. Ans. 2»-l. 509] OF KIGID BODIES. 419 (4) A system of smooth weightless pulleys, like that of Ex. 1, but with only one moveable pulley, is in equilibrium. Show that if the body supported by the moveable pulley have its mass doubled, and the other its mass halved, the tension in the string will be unaltered. (5) Two smooth spheres rest on two smooth inclined planes and press against each other. Determine their position and the magni- tudes of the reactions. — Let A and B be the spheres, C and C" their centres, DE and EF the inclined planes of inclinations ^ and e A respectively. Each sphere is acted upon by three forces — its weight (W, W), the normal reactions of the planes (R, R'), and the equal normal reactions of the spheres on one another {S, S'). As each sphere is acted on by three forces only, these three must in each case be in the same plane, but as the lines of action of S and S' coincide with the line CC, W, W, IS and S' are in the same plane. Hence all six forces are in the same plane, which is consequently a vertical plane and perpendicular to both inclined planes. Let that be the plane of the diagram. The positions of the spheres are determined by the angle ^, the inclination of CC to the horizon. For the equilibrium of A we have, resolving in the direction otDE Wsin tp - Scos {(p - f)=0, and resolving in a perpendicular direction, R- Wcos-Ssm{-R' cos — ^'sine = 0. Also equating to zero the sum of the moments of external forces about 0, we have, W cos xp-R' cos {8 + \l')=0. We have thus three equations for the determination of the three unknown quantities R, R', and ^. (6) Two smooth spheres of equal radius r and weight W are placed inside a uniform thin hollow cylinder (radius =?•'< 2r) which is open at both ends and rests with one end on a horizontal table. What must the weight of the cylinder be that it may not upset ? Ans. 2 W(/ -?•)//. (7) A smooth sphere (weight = W) rests upon two equally in- clined planes (inclination = a) which are placed on a smooth hori- zontal table, and are prevented from sliding apart by a horizontal string which binds them together. Find the tension in the string. Ans. ^PTtana. (8) Of four equal smooth spheres (weight of each= W) three rest in contact on a smooth horizontal plane, and the fourth is placed upon them. Find the horizontal force which must be applied to each of the three to preserve equilibrium. Ans. ir/3v/2. (9) A heavy uniform smooth beam (weight = w, length = 2a) is moveable in a vertical plane about a smooth hinge at one end. 509] OF EIGID BODIES. 421 A heavy smooth sphere (weight= W, radius=r) is attached to the hinge by a string (length = Z), and the two bodies rest in contact. Obtain equations for determining the inclination 8 of the string to the vertical, the inclination (p of the beam to the vertical, the reaction S of the hinge on the beam, and the stress R bet-ween the beam and the sphere. An s. W{1 + r) sin e = wa sin , {l+r) sin. (e + )=r, Roos{e+)= Waine, (S^- w^) cos=(fl + cos(e + ). (10) A uniform heavy rod (weight = W, length = 2Z) connects the centres of two equal heavy wheels (radius =r), which rest on a roiigh inclined plane (coefficient of friction =/i) in a vertical plane, which is a plane of greatest slope of the inclined plane, the lower wheel being locked. Find the greatest inclination of the plane that will admit of equilibrium. id Ans. tan~'- -fir (11) Three horizontal weightless levers, ABB, BFC, CGD, the fulcrums of which are at E, F, G, act upon one another perpendi- cularly, the first and second at B and the second and third at C. They are kept in equilibrium by bodies hanging from the points A, D, and weighing W and 2 Jf respectively. AE, EB, BC, CG, GD are 1, 2, 7, 2, 3 ft. respectively. Find (a) the position of F, and (6) the reaction of the fulcrum at F. Ans. {a) FG=\ ft.; (6) V Tr/2. (12) Two beams whose weights are proportional to their lengths (9 and 7 ft.) rest with their lower ends in contact on a smooth horizontal plane, and their upper ends leaning against two smooth vertical and parallel walls 10 ft. apart. Show that if B and 9' are the respective inclinations of the beams to the horizon, 7tane = 9tane', and 9 cos 9 + 7 cos e' = 10. (13) Two uniform straight rods of equal length rest with their 422 DYNAMICS [509 lower ends on a rough horizontal plane (coefficient of friction =jn) and their upper ends in contact, and are on the point of slipping. Find the common inclination to the horizon. Ans. tan-Xl/S/t). (14) Two bars which are connected by a smooth hinge or joint are in equilibrium. Investigate generally its reactions on the bars. (flf) If the hinge pin is rigidly connected with one of the bars, the reactions between the bars are obviously equal and opposite, their magnitude and direction depending upon the external forces acting upon the bars. (6) If the hinge pin is distinct from both bars, and if no external forces act on the pin (which condition requires either that the weight of the pin should be negligible, or that it should be neutralized by an equal and opposite external force), its reactions on the bars must be equal and opposite. For the pin is in equili- brium under the two forces exerted upon it by the bars at the points or rather lines of contact, and as these forces must therefore be equal and opposite, the reactions of the pin on the bars must also be equal and opposite. If however the pin is acted upon by an external force, the forces exerted upon ^it by the bars will not have the same line of action, and its reactions on the bars will therefore also have different lines of action. (15) In a system of jointed thin bars, in which the hinge-pins are distinct from the bars, if the external forces act only on the hinge-pins (this condition implies that the weights of the bars are negligible), the reactions of the pins on the bars wiU be in the directions of the bars. For in that case any bar is acted upon by two forces only, the reactions of the hinges at its ends. These forces must therefore be equal and opposite, and their lines of action must consequently be the direction of the bar. (16) Two equal uniform rods, equally inclined to the horizon, and connected by a smooth hinge at their higher ends, pass through two small iixed rings in a horizontal line. Find the inclination of either rod, when the system is in equilibrium, and the reactions of the hinge on the rods. 509] OF EIGID BODIES. 423 Let AB, AC be the two rods hinged &t A ; D and E the small rings. The rods are acted upon by their weights ( W, W) and the reactions of the rings {R, R') and of the hinge {S, S'). The reac- tions at A must be in the same straight line, must pass through the intersection of the lines of action of the weight and of the reaction of the ring in the case of each rod, and must therefore be horizontal. Hence . the centres of mass of the rods must be be- tween the rings and their lower end points. Let I be the length of each rod, d the distance between the two rings, and 6 the inclination of each rod to the horizon. Resolving the forces acting on AC in the direction of AC, we have, Wame-Scose=0, and, taking moments about £!, we have W cos e{l- d/cos 0)-Sdta,ne=O. From these equations we may obtain both S and e. (17) Three rods jointed together at their extremities, are laid on a smooth horizontal table, and horizontal forces are applied at their middle points perpendicularly to them. Show that if these forces produce equilibrium, the stresses at the joints will be equal and their directions will touch the circle circumscribing the triangle (See 475, Ex. 1.) (18) Two uniform rods A C, BC (weights= w and W) are connected by a smooth hinge at C, their other ends A and B being fastened to 424 DYNAMICS [509 fixed hinges in a vertical line. Find the reactions of the hinges on the bars. Let Xj, J'j and Xj, T^ be the horizontal and vertical components of the reactions oa. AC and BG at A and B respectively. (It does not matter in what direction, up or down, we draw Y^, T^ or whether we draw X^, X^ to the right or left. If the actual re- actions have components in directions opposite to those assumed in the diagram, the values of X-^, Zj, etc., as the case may be, will be found negative.) Let X3, Y^ be the components of the reaction of the hinge at C on the rod AC. Then, since the hinge-pin is not acted on by external forces, its weight being negligible, its reaction on -BC will have components — X3, — I3. As the four forces shown in the diagram as acting at Cact two on j4Cand two on BC, it is often advisable to draw a special diagram for each bar. Such diagrams are shown above. The equations of equilibrium may be vmtten down by their aid without danger of inserting BC& forces in AC'& equation. Thus for the equilibrium of .4 C we have Xi-1-X3=0, Fi-l-r3-J»=0, and taking AC=%a., and calling its inclination to AB, a. 509] OF RIGID BODIES. 425 2a( A'l cos a+ Yj sin a) — wa sin a—0. And for the equilibrium of BC, calling its length 26, and its inclina- tion to AH, |3, we have We have thus six equations -which are sufficient to determine the six unknown components of the reactions. (19) Two uniform beams J Cand CB (weights = W and W respec- tively) connected at C by a smooth joint are placed in a vertical plane, their extremities A, B being connected by a string and rest- ing on a smooth horizontal plane (inclinations of AC, CB to the horizon = a and a respectively). Find (a) the tension in the string, and (6) the reaction at the joint. Ans. (a) 2(tan a + tan a') ' • (6) magnitudes -^^'^ ^T^^^T"; "" ''""''' ' ^ * 2(tana-f-tano') line of action inclined to horizon at tan~^( .■?"„,. ^^'^ \ (20) Two rods AB, CD are connected by smooth hinges a,t A, D to two fixed points in the same horizontal line, and at B, C, also by smooth hinges to the ends of a rod BC. Show that if all three rods are of equal length, and if either AB or CD is inclined a° to the horizon, the inclination to the horizon of the reaction of the joint at its lower end will be tan-i( J tan o). (21) A plane polygonal frame, composed of a system of rigid bars, moveable freely round their jointed extremities, is in stable equilibrium imder the action of a system of forces proportional to, and bisecting perpendicularly, its several sides. Show that its several vertices lie in a circle. (22) If a system of .thin jointed bars, in which external forces act on the bars, be in equilibrium, and if the external forces acting on each bar be resolved into components acting at the joints, the 426 , DYNAMICS [509 stress in each bar is the resultant of the reaction of the joint on the bar and the ccvmponents at the joint of the external forces acting on the bar. Let AS be the bar, A and B being the points of contact with the hinge-pin. Let Ji^ and R^ ^^ ^^^ reactions of the hinge-pins on the bar, and F the external force. Resolve F into two components Fi and F2 acting at A and B. Then R-^, R^, F^, F^ are equi- valent to ifj, R^, and F. Now ^1 and Fy, and R^ and F^ give in each case a single resultant 8^ and S^ respectively ; and as the bar is in equilibrium under these two resultants they must be equal and opposite. It will be noticed that S-^ and S^ are not the reactions of the hinge-pins, but the resultant forces acting at the end points A and B which are in contact with the pins. These forces, as repre- sented in the diagram, tend to shorten the distance AB. In actual bars a compression brings elastic forces into operation, and S^ and 8^, having changed the length of the bar somewhat, will thus be equilibrated by the elastic stress in the rod. In dealing with rigid rods, we imagine the stress produced in the rod, though the change of length is supposed indefinitely small. The particles at A and B are thus in equilibrium under the forces Sy, 8^ respectively, and the stress in the bar. Hence that stress has the same value as 8^ or 8^, 509] OF KIGID BODIES. 427 and its directions at ^4 and B are opposite to the directions of S{ and S2 at A and JS respectively. It may be mentioned that in frame work a bar which is short- ened by the forces acting on it is called a strut ; one which is lengthened is called a tie. (23) If a system of thin jointed bars in which external forces act on the bars be in equilibrium, the hinge-pins, supposed weightless, may be considered as being in equilibrium under the stresses in the bars they connect and the components at the joints of the external forces acting on the bars. Since Si (fig. of Ex. 22) is the resultant of R^ and F^, a force equal and opposite to 5^ would equilibrate M^ and F^. Hence a force equal and opposite to S^ together with F^ would give as resultant a force equal and opposite to Ri. Now, if a hinge-pin connect two or more bars, it is in equilibrium under any external forces acting on it, together with forces equal and opposite to the reactions it exerts on the bars. Hence it may be considered as being in equilibrium under the stresses in the bars and the components at the hinge of the external forces ,acting on the bars, with the external forces acting on itself. This result is of importance in engineering as enabling us to determine the stresses in framework subjected to given external forces. (24) Three weightless bars AB, BG, CA, jointed at their ex- tremities, are kept in equilibrium by three forces acting at the joints — F acting a,t A, Q a,t B, and R at C. Show that if Sa, /Sj, Sc are the stresses in BC, GA, AB respectively, whose lengths are a, b, e respectively, and if is the point in which P, Q, and R intersect, „ ^ e « • 04 f>OB c.OG (25) Four heavy bars AB, BG, CD, DE (weights =iOi, w^ Wg, w^ respectively) are jointed each to the next at B, G, D and to fixed points at A and E. The rod BG being horizontal, and 6 being the inclination of CD to the horizon, show that the inclination of DE is tan-if^Vh^^VtBtanO- 428 DYNAMICS [509 (26) Three beams, AB, BC, VA are connected at their ends by smooth hinge-pius, so £is to form a triangle. The ends of the beam €A rest upon pillars of equal height. The other two are in a vertical plane ; and at the joint which connects them hangs a body whose weight W is so great that the weights of the beams may be neglected. The lengths of the beams being given, show how to de- termine the stresses in the beams and the reactions of the pillars. Take any point 0, and from it draw Op vertically downwards, making its length nuraierically equal to W. From draw Oq parallel to BC, from p,pq parallel to AB, and from q, qr in a horizontal direction. By Ex. 23 the pin at B is in equilibrium under forces whose directions are those of AB, CB, and W, or of pq, qO, and Op ; and as Op represents W in magnitude it follows that pq represents the stress ia AB and qO that in CB. Similarly it may be shown that qr represents th.d stress in CA and that pr and rO represent R^ and R^, the reactions of the pillars, respectively. The diagram Opq constructed as above is called a Force Diagram. It may be used to solve the problem in two ways. (1) By its aid we may obtain formulae by which the stresses may be calculated. Thus the sides of the triangle ABC being known, we may express the angles CBD and ABD and therefore the angles qOp and qpO, and therefore also the angle Oqp in terms of them. Hence also since OqlOp = sin Opqjwa. Oqp, Oq/Op may be expressed in terms of them. But if S is the stress in BC, SI W= Oq/Op. Hence the stress in BC may be expressed in terms of W and the lengths of the beams. And expressions for the other stresses may be obtained in the same way. (2) The lengths of the beams being given, exact values of the angles ABD, CBD may be obtained, and the force diagram may be care- fully drawn to scale. Then Op having been drawn with a length numerically equal to W, a careful measurement of the lengths of 510] OF RIGID BODIES. 429 Op, pq, and qr determines the stresses. Tke stresses thus deter- mined are said to be determined graphically ; and in complicated framework the labour of calculation is much reduced by the graphic method. (27) A Warren girder consists of 19 rods AB, BC, etc., of equal length, jointed together as in the diagram. Bodies of equal weight, and so heavy that the weights of the rods may be neglected, are hung at the joints B, C, D, E, and the girder is supported on piers of equal height at A and F. Show that there is no stress in KG or KD. 510. Conditions of Equilibrium in terms of Work Done. — If a rigid body is in equilibrium {i.e., molar equilibrium) the algebraic sum of the amounts of work done by the external forces during any indefinitely small displacement consistent with rigidity is equal to zero. Any such displacement may (245) be resolved into a translation, and a rotation about the direction of trans- lation. Any point P of the body therefore will undergo a linear displacement compounded of one S' in the direc- tion of the translation, and another (5 in a plane perpen- dicular to that direction, and due to the rotation about the direction of the translation. Resolve the force F, acting at this point, into two components, one F' in the direction of translation, the other F" in the perpendicular plane. Since the body is in equilibrium, 1,F = 0. Multiplying by S', which is the same for all points of the body, we have 8"2F' = 'LF'S'=d- 430 DYNAMICS [510 i.e., the algebraic sum of the amounts of work done by the components of the external forces in the direction of the translation is zero. Let the plane of the diagram be the plane perpendicular to the axis of the rotation, the intersection of the two being at 0. Let PA perpendicular to OP be the small linear dis- placement S of P due to the rotation w about 0. From draw OM (length =p) at right angles to F", the component of F in the plane of the diagram. Since there is equilibrium, 1iF"p = 0. If 6 is the inclination of S to F", p = OP cos 6 ■ S „ S" ■■ - cos y = — 0) w if 8" denote the component of S in the direction of F". Hence (0 U) Hence also Si^'X = 0, i.e., the work done by the com- ponents of the external forces perpendicular to the direc- tion of the translation is zero. If d is the component of the resultant linear displace- ment of P in the direction of F, Fd is the work done by F during 'the displacement. Hence (342) Fd=F'S'+F"S" and i:Fd = i:F'S'+^F"S!' = 0. And "EFd is the algebraic sum of the amounts of work done by the external forces during the indefinitely small displacement selected. 511. Conversely, if during any indefinitely small dis- 514] OF RIGID BODIES. 431 placement of a rigid body, consistent with its rigidity, the algebraic sum of the amounts of work done by the external forces be zero, the body will be in equilibrium {i.e., molar equilibrium). For it may be shown by the steps of 510 in the reverse order that I.Fd = I,F'S' + I,F"S" = S'XF + a,I,F"p = 0. •Now S' and m are arbitrary and unrelated, the displace- ment being any displacement whatever. Hence 'ZF' = 0, and 1,F"p = 0, i.e., the body is in equilibrium. 512. Hence it is a necessary and sufficient condition of the equilibrium (molar) of a rigid body that the algebraic sum of the amounts of work done by the external forces during any indefinitely small displacement consistent with rigidity, be equal to zero. 513. It follows from 449 that the necessary and suffi- cient condition of the molecular equilibrium (444) of a rigid body, which is obviously consistent with translation but not with rotation of the body, is that the algebraic sum of the amounts of work done by all forces, external and internal, during any indefinitely small displacement, be equal to zero. If the displacement be one consistent with the rigidity of the body, the internal forces (499) do no work. Hence the conditions of molar and of molecular equilibrium for a rigid body are identical. Thus the same conditions must be fulfilled, that a rigid body acted upon by external forces may spin without angular acceleration, as that it may move without angular velocity about an axis fixed in itself. 514. In the case of a system of rigid bodies rigidly con- nected [508 (1)], the necessary and sufficient condition of equilibrium is obviously that expressed in 512, the ex- ternal forces involved being these external to the system. 432 DYNAMICS [515 515. In the case of a system of rigid bodies, not rigidly connected [508 (2)], since the necessary and sufficient condition of the equilibrium of each body is expressed by the equation 'EFd = (510-512), the forces involved being forces external to the body, that of the equilibrium of the system is expressed by the same equation, the forces appearing in it being both those external to the system and such internal forces as stresses in strings and rods and reactions of surfaces. If these internal forces do no work during the small displacement to which the equation applies, as will be the case if they are tensions in inextensible strings, stresses in rigid rods, or reactions of smooth surfaces, the equation "SFd = involves only forces external to the system. 516. If we wish to determine one of the internal forces, of a system of rigid bodies connected by rigid rods or in- extensible strings, we may imagine a small displacement in which the parts, between which the required force acts, so move that the required force does work, in which case the equation 'EFd = involves the external forces and the required internal force. 517. Fxamples. (1) A beam (yf eight =W, length = Z, distance of centre of mass from lower en2, etc., are the intensities of the stress on Sj, s^, etc., respeC' tively, we have p^=hh^, p,^=kh^, etc., where i is a constant. Hence the forces acting across s^, s^, etc., are M^Si, Ich^^ etc. Hence the integral stress is Tkhs = hMs= kU'E/is/lis, for S = 2s. Now 2hs/'2s is obviously (400) the distance from the given plane of the centre of mass of a uniform thin material lamina of the same form and area as the given surface, and of sur- face density unity (304), or, as it is called for shortness, the centre of mass of the surface. Hence the integral stress is equal to the product of the constant k, into the area of the surface, into the distance of its centre of mass from the given plane. (3) Find the resultant of a normal stress on a plane surface of rectangular form (sides = a and 6), the stress at any point being pro- portional to its distance from a given plane parallel to the sides of length a and inclined to the sides of length b at the angle 8, and that side of length a which is nearest the given plane being at a distance h from it. (Use result of Ex. 2.) Ans. kab (6 sin e + 2/0/2. (4) Find the integral stress over a spherical surface of radius r, the stress at any point being proportional to its distance from the tangent plane at the highest point of the sphere and the stress at a point at unit distance being L Ans. 4T^r^. (5) Find the integral stress over the curved surface of a right cone of height h and semi-vertical angle 8, the stress at any point of it being numerically equal to p times the distance of the point from the base. Ans. irphhin 9/3 cos^fl. ' 528. Centre of Stress. — If a single force can be found which is equivalent to a given stress on a given surface, its point of application is called the centre of the stress. 446 DYNAMICS [528 In the case in which the stresses at all points of the surface are parallel, the centre of the stress is the centre of the system of parallel forces of which the stress may be regarded as consisting. Hence, in this special case, if the surface consist of small portions of areas s^, s,, etc., at which the intensities of the stress are p^, f^, etc., if the distances of s-^, s^, etc., from any plane are A^, \, etc., and if the distance from the same plane of the point of ap- plication of the resultant (2ps, by 526) is H, we have by 472 If the distances of the centre of stress be determined from any three intersecting planes, its position is com- pletely specified. 529. Examples. (1) Show that the centre of stress for any plane surface subjected to a uniform stress is the centre of mass of the surface. 5=2p«A/Sp«=^SsA/joS3 = S«A/2s. [See 527, Ex. 2.] (2) Find the centre of stress for a plane triangle, the stresses at all points being uniform in direction and varying as the distances of the points from a plane through one of the sides. If the triangle be divided into narrow strips of equal width par- allel to this side, the stress will be uniform over each strip. Hence the centre of stress for each strip is its middle point, and that of the whole triangle is on the line drawn from the middle point of the above-mentioned side to the opposite angle. The resultant stresses on strips equidistant from the middle point of this line may easily be shown to be equal. Hence the middle point of this line is the centre of stress for the triangle. (3) Find the centre of stress on a parallelogram ABCD, the stress at all points being uniform in direction and varying as their dis- tance from a plane through AB. If the parallelogram be divided into narrow strips of equal width parallel to AB, the resultant stress on each will act at its middle point and be proportional to its distance from the given plane, and 529] OF ELASTIC SOLIDS AND FLUIDS. 447 therefore to its distance from AB. Hence the resultant stresses on the strips are proportional to the lengths of the portions of the strips intercepted between straight lines drawn from G and D to H, the middle point of AB ; and hence the centre of stress of the parallelo- gram coincides with the centre of mass of the triangle BCD. (4) Find the centre of stress for any plane surface, the stresses at its various points being parallel and proportional to their distances from any given plane. With the symbols of 528 we^ave ^j = Mj, p^ = kh^, etc. Hence Sp«A/Sp« = ■S.sh^l'S.gh. The determination of the value of ^sh'/Ssh in special cases requires in general the application of the Integral Calculus. (5) Find the centre of stress on a triangular plane ABC, the stresses at all points being uniform in direction and proportional to the distances of the points from a plane through C parallel to AB. Let the triangle be divided into n narrow strips of equal width parallel to AB. These may be treated as rectangles if n be very great. If AB have the length a, and if 6 be the distance of C from it, the areas of these rectangles in the order in which they occur from C towards AB are ab/v?, Zab/rv', 3ab/n% etc. As they are very narrow the distances of their centres of mass from the given plane, if h is the distance of AB from it, may be taken to be hin, 2A/j7, Zh/n, etc. Hence the distance of the centre of stress from the given plane is (Ex. 4) ab/i'ln* + 2^abk^/n*' + etc. + n^abh?/ni^ _^ . (lH2S + etc. + w3)/ji« abhln? + Z^abAjn' + etc. + m'abh/n.^ ' (1^ + 2^ + etc. + n')ln/> =3A/4, since n is indefinitely great. And it is obviously in the line joining with the middle point of AB. (6) Find the distance from a given plane of the centre of stress on a triangle ABO, the point A being in the given plane and the points B and C at distances h^ and h^ from it, the stress at any point being normal and proportional to the distance of the point from the given plane. [Let BO meet the given plane in B. Then the resultant stresses on AGD and ABD may be determined in terms of the length ot AD and the inclination of the plane of ABO 448 DYNAMICS [529 to the given plane, and their centres of stress may be determined by Ex. 2. Then the resultant stresses and the centres of stress on the whole ABD and the part ACIi being known, the centre of stress of the part ABC may be readily determined.] Ans. (V+^^2+V)/2(^i + ^2)- (7) Find the centre of stress on a parallelogram A BCD, the stress at any point being normal and proportional to its distance from a given plane which is parallel to the sides AB and CD, and distant Aj and h^ from them respectively. Ans. ■i{h^ + kJi^+h^lZ{k^ + h^. 530. Resolution of Stress. — A stress being in general oblique to the surface across which it acts, may be re- solved into tangential and normal components. For each of the forces acting at points, of which it may be con- sidered to consist, may be so resolved. A stress which is normal to the surface across which it acts is often called a longitudinal stress. One which has the inclination zero is called a tangential or shearing stress. 531. Specification of Stress. — The magnitudes and direc- tions of the stresses at a point across any three plane sur- faces through the point being given, the stress across any other plane through the point can be determined. First, let the stress throughout the body be homoge- neous, and let there be no external forces. Let be the given point, and Ox, Oy, and Oz the intersec- tions of the three planes through 0; and let any fourth plane in- tersect these planes in AB, BG, GA. Then the tetrahedron OABG being in equilibrium under the re- sultant stresses on its four faces, and those on the three faces OAB, OBG, OGA being known, the magnitude and direction of that on ABG may 532] OF ELASTIC SOLIDS AND FLUIDS. 449 be determined by 500 ; and the area ABC being known, the stress at any point of ABG, and consequently the stress at 0, across a plane parallel to ABC becomes known. 532. It is usually convenient to take rectangular planes as planes of reference. Let OABG be a tetrahedron whose faces OAB, OBG, OCA are at right angles to one another ; and let the normal to the plane ABC have the direction cosines I, m, n relative to the x, y, z axes respectively. Let the stress at across OAB (the xy planle) have components T, 8, R in the directions of Ox, Oy, Oz respectively, that across OBC (the yz plane) components P, U', T', and that across OAC (the xz plane) components U, Q, 8', in the same directions respectively. Also let Fx, Fy, Fz be the com- ponents in these directions of the stress F eAj across a plane parallel to ABG, and therefore across ABC. Then the tetrahedron is in equilibrium under forces equal to the products of these various stresses into the areas of the faces across which they act, and acting (529, Ex. 1) at the centres of mass of the faces. Hence (500) F^ . ABC=P . OBC+U. OAG+T . OAB, 2f 450 DYNAMICS [532 ABC, OBG, etc., standing for the areas of the faces. Now OBG, 0^(7, and OAB are the projections of ABC on the yz, xz, and xy planes respectively. Hence (see 173) OBG=ABV.l; OAC=ABC.m; and OAB = ABG . n. Hence Similarly and F„=Pl+Um+Tn. Fy=U'l+Qm+8n, F^=T'l+S'm+Rn. 533. It is also necessary for equilibrium (500) that the sum of the moments of the acting forces about Ox, Oy, Oz should be equal to zero. The relations between the com- ponents of the stresses, which are obtained by applying this condition, however, may be more easily obtained by considering the equilibrium of a cube of which Ox, Oy, Oz aj-e adjacent edges. Let OD, OE, OF be three faces of such a cube. The component stresses at all points of these faces are the same as at all points of the correspond- ing faces of the tetrahedron ; and the component stresses at all points of the faces opposite to OD, OE, OF are equal and opposite to those on OB, OE, OF respectively. Let the component stress equal and opposite to P on the 536] OF ELASTIC SOLIDS AND FLUIDS. 451 face opposite to OE be called p, and let the stresses simi- larly related to Q, R, etc., be called q, r, etc. If the cube be one of unit edge, the components of the resultant stresses on its faces are P, Q, R, etc., p, q, r, etc., and the points of application of these component forces are the centres of the faces. Hence, equating to zero the sum of the moments about Ox of all the forces acting on the cube, and noting that P, T, U, p, t, u, which are parallel to Ox, and 8 and >S' which intersect it, have no moments about it, that R and r, Q and q, T and If, and U' and u' have equal and opposite moments about Ox, and that s and s' are equidistant from it and have moments of opposite sign about it, we obtain s = s', and therefore Similarly we jBnd T=T', and V=U'- 534. Substituting these values of S', T, U', in the ex- pressions of 532 for Fx, Fy, F^, we have F^=m-\-'Um-\-Tn, Fy=Ul+Qm+8n, F,=Tl+8m+Rn. 535. Hence if P, Q, 5, 8, T, U are known, the stress at across any surface through is known. The com- plete specification of the stress at a point requires then only these six numerical data. P, Q, and R are the com- ponent stresses at 0, normal to the yz, xz, and xy planes respectively. 8 is the tangential or shearing stress either on the xy plane parallel to the y axis, or on the xz plane parallel to the z axis ; T, that on the xy plane parallel to the X axis, or on the zy -plane parallel to the 2: axis; TJ, that on the xz plane parallel to the x axis, or on the yz plane parallel to the y axis. 536. Secondly (531), if the stress is not homogeneous the same result may be obtained, provided the tetrahedron. 452 DYNAMICS [536 and cube above be taken indefinitely small. For in that case the stresses at across the planes of the faces may- be taken to be the stresses at all points of the faces. 537. The above conclusions (534-5) hold also if the body is acted upon by external forces. Such forces must either be forces acting on the outer surface of the body, or forces, such as gravitational attraction, acting throughout the mass. Forces acting at the outer surface of the body act only on tetrahedra or cubes having faces in the bound- ing surface, and they constitute the stresses on those faces. Forces acting throughout the mass of the body are proportional to the mass acted upon. Hence such of these forces as act on the tetrahedron or cube are pro- portional to its volume. The stresses on its faces are proportional to the areas of these faces. The former are therefore proportional to the cubes, and the latter to the squares of any edge. Hence, if the tetrahedron or cube be gradually diminished, the external forces diminish more rapidly than the stresses ; and if it be made inde- finitely small, the external forces become indefinitely small relatively to the stresses, and may therefore be neglected. 538. Resolution of a Tangential Stress into Longi- tudinal Stresses. — Let a body be subjected to a tension P in a given direction, and a pressure of the same intensity in a perpendicular direction, the state of stress being homo- geneous. And let ABDG be a section of a cube of unit edge, with its faces normal to the directions of the tension and the pressure, through their cen- tral points. Then the resultants of these stresses on the faces of the cube may be considered as acting at the middle points p i » B — ^ / P ( c \ ) 541] OF ELASTIC SOLIDS AND FLUIDS. 453 of the sides of ABDG and as having the magnitudes P. The triangle AGB, or rather the triangular prism of which AGB is a section, being in equilibrium under the two forces P, and the resultant stress on GB, this resul- tant stress must be equal and opposite to the resultant of the two forces P, on AB and AG. Now in a direction normal to GB, these forces have equal and opposite com- ponents, and in the direction of GB each has a component P cos 45°. Hence the resultant stress on GB must be in the direction BG, and of the magnitude IP cos 45°. Now the area of the section of the cube through GB perpen- dicular to ABBG, which is the surface on which this stress acts, has the area 1/cos 45°. Hence the intensity of the stress on GB is 2P cos^ 45° or P. Hence a tension parallel to one line, and an equal pres- sure parallel to any line at right angles to it, are together equivalent to a shearing stress of the same value on planes cutting these directions at angles of 45°. (Compare 276.) The directions of the pressure and tension may be called the axes of the shearing stress. 539. It follows that since a stress at any point of a body may be completely specified in terms of longitudinal and shearing stresses, it may also be completely specified in terms of longitudinal stresses alone. 540. Relation of Stress to Strain. — In considering the detei-mination of the strain produced in a body when sub- jected to given stresses, we must restrict ourselves to the simple case in which the body is homogeneous, isotropic, and perfectly elastic. 541. A body is said to be ho7nogeneous provided any two equal, similar and similarly situated parts of it are not distinguishable from one another by any difference in quality. Probably no bodies perfectly fulfil this con- dition without limit as to the smallness of the parts. But 454 DYNAMICS [541 many bodies are so nearly homogeneous that their hetero- geneity eludes observation. 542. A homogeneous body is said to be isotropic, when any two equal and similar portions of it, whether simi- larly situated or not, are not distinguishable from one another, or, in other words, when it has the same qualities in all directions. A body which exhibits diflferences of quality in different directions is said to be ceolotropic. A body may be isotropic with respect to some qualities, and geolotropic with respect to others. We have to do with isotropy only with respect to the relations of stress to strain. 543. A body is said to be elastic, provided (1) the ap- plication of force is required to produce a change in its shape or its bulk; and (2) a continued application of force is necessary to maintain the change, in which case it will return towards its initial shape or bulk when the applied force is removed. A body is said to be perfectly elastic for a strain of a given kind, provided the, same application of force is re- quisite to maintain the given strain as to produce it, in which case it will obviously return to its initial configu- ration when the stress is removed. 544. Probably no natural bodies fulfil this condition of perfect elasticity, unless in producing strains in them care be taken to keep them at constant temperature. For in all bodies the stress required to maintain a given strain is found to vary with temperature ; and we know from Thermodynamics that consequently a change of configu- ration must be accompanied by a change of temperature. 545. In all bodies it is found that the amount by which the stress required to produce a strain exceeds that re- 548] OF ELASTIC SOLIDS AND FLUIDS. 455 quired to maintain it, is greater than the amount due merely to this change of temperature ; and the difference between these amounts is found to depend upon the rapidity with which the change of configuration is pro- duced. Thus the relative motion of the parts of a body are resisted in the same way as the relative motion of different bodies in contact ; and bodies are therefore said to exhibit molecular friction, or as it is called viscosity. Even a perfectly elastic body will not therefore appear to be perfectly elastic unless its changes of configuration are carried out with infinite slowness. 546. For most bodies, and for most kinds of strain, there are limiting values of the stress by which a strain of a given kind is produced, within which the elasticity for that kind of strain is perfect, and beyond which the elasticity is imperfect. Such limiting value of the stress is called the limit of perfect elasticity for that kind of strain. 547. All bodies exhibit some degree of elasticity of volume. If a body possess any degree of elasticity of shape, it is called a solid. If a body possess no degree of elasticity of shape, it is called a flv/id. 548. That a body may be elastically isotropic, i.e., isotropic so far as the relation of stress to strain is con- cerned, it must obviously satisfy two conditions: — (1) Any spherical portion of it must, if subjected to a uniform normal pressure or tension over its whole surface, undergo no deformation, the compression or dilatation produced being the same in all directions ; (2) Any cubical portion of it, subjected to shearing stresses on the planes of its faces, must undergo distortion or shear ; and the amount of the shear must be the same to whatever side of any face the shearing stress is parallel. 456 DYNAMICS [549 549. Hence the relation of stress to strain in a perfectly elastic homogeneous isotropic body is completely defined if we know (1) the ratio of the intensity of the stress, uniform in all directions, to the dilatation or condensation (266) which is produced by it ; and (2) the ratio of the inten- sity of the shearing stress to the amount of the shear pro- duced by it. The former of these ratios is called the resistance to compression or the elasticity of volume, the latter the rigidity or the elasticity of figure or form. The former may be denoted by the symbol k, the latter by the symbol n. 550. Did we know the laws of the forces with which the particles of bodies act upon one another when in close proximity, and the distribution of the particles in the body, it would be possible, by the aid of the laws of motion, to determine the values of the elasticities of figure and volume for strains of different magnitudes, in the case of different bodies, and in the case of the same body in different physical states. In our ignorance of the laws of these forces, however, we find it necessary to have re- course to experiment. 551. Statics of Elastic Solids. — Hooke's Law gives us the necessary experimental basis for the study of the strains of elastic solids. Hooke expressed the law as follows : " [ft tensio sic vis ; That is. The Power of any Spring is in the same proportion with the tension there- of: That is, if one power stretch or bend it one space, two will bend it two, and three will bend it three, and so forward." In modern phraseology it takes the follow- ing form : Strain is proportional to stress. This law has been subjected to the most minutely accurate experi- mental tests, and the simple proportionality of stress to strain is found to hold in the case of all solids for sufllcieutly small strains, and in the case of metals and hard solids {i.e., solids in which the stress applied, if maintained, 552] OF ELASTIC SOLIDS AND FLUIDS. 457 does not produce a continually increasing strain) for all strains within the limits of perfect elasticity. The strains, by the investigation of which Hooke's law has been established, i.viz., the stretching of wires by appended weights, the compression of rods, the flexure of beams, the extension of spiral springs, the torsion of wires, etc., are all more or less complex strains, involving in most cases both change of volume and change of form. The constancy of the ratio of stress to strain, within the limits of perfect elasticity, in strains involving both change of form and change of volume, warrants us in holding that within the same limits the elasticity of figure and the elasticity of volume must be constant also. 552. Moduluses of Elasticity. — A modulus of elasticity is the ratio of the intensity of a stress to- the magnitude of the strain which it produces. Thus the elasticity of figure (n) and the elasticity of volume (k) are moduluses of elasticity. The elasticity of figure is often called therefore the mo(hblus of rigidity (or of siTnple rigidity), and the elasticity of volume the modulus of bulk elas- ticity. The reciprocal of the latter is called the com- pressibility of the body. Young's modulus, or the modulus of simple longi- tudinal stress, is the ratio of the intensity of the stress applied at the end of a wire or rod in the direction of its length to the increase or diminution which each unit of its length undergoes, the strain being one within the limits of perfect elasticity. The extension of a wire or rod by longitudinal stress involves change of both volume and form. Hence Young's modulus may be expressed in terms of k and n. A modulus of elasticity, being the ratio of a stress to a strain, has the same dimensions as a stress ; for a strain is the ratio of two quantities of the same kind, two lengths, for example, or two volumes, and has therefore 458 DYNAMICS [552 no dimensions. The dimensions of a modulus of elasticity are thus [M][L]-XT]-\ The value of such a modulus expressed in any one system of units may thus re?idily be reduced to any other system of units. Moduluses are usually expressed in gravitational measure, in pounds (i.e., pounds-weight) per square inch, e.g., or in grammes (i.e., grammes-weight) per square centimetre. In the measurement of moduluses however a special unit of force is frequently employed, viz., the weight of unit of volume of the substance to which the modulus applies. The value of the modulus thus expressed is to be obtained from its value expressed as above in ordinary units of stress by dividing by the weight of unit volume of the substance, i.e. (304), by the product of the specific gravity of the, substance into the weight of the unit volume of water at the standard temperature. Thus, if a modulus be expressed in pounds per square inch, its value in terms of the special unit of force is obtained bj' dividing by the product of the specific gravity of the substance into the weight of a cubic inch of water, which in gravitational units is equal to the density of water in pounds per cubic inch. If the modulus be expressed in grammes per square centimetre, its value has to be divided only by the specific gravity of the substance, for the density of water in grammes per cubic centimetre may be taken to be unity. The dimensions of " weight of unit volume " being [-F'][F]-i (where [F] and [F] are the magnitudes of the units of force and volume respectively), and therefore [ilf][ii]'"^[2^"^, those of moduluse's expressed in terms of the weight of unit volume as unit of force are [M][L]--'[T]-y[M][L]-^[T]-^ or [L]. The modulus thus expressed is therefore a length, and its value is there- fore usually called the "length of the modulus." Thus the value of a modulus obtained by dividing its value in pounds per square inch by the product of the specific 553] OF ELASTIC SOLIDS AND FLUIDS. 459 gravity of the substance into the density of water in pounds per cubic inch, is the length of the modulus in inches. The term modulus is also applied to the following ratios, though they are not the ratios of stresses to strains : — The modulus of torsion of a rod or wire is the ratio of the couple applied at one end (the other end being fixed) to the torsion produced per unit length of the wire. The modulus of flexural rigidity, in any plane, of a rod or beam, slightly bent in that plane, is the ratio of the couple producing the curvature to the curvature thereby produced. The dimensions of the modulus of torsion are obviously \M\{Lf{Ty^; those of the modulus of flexural rigidity the same. 553. Examples. (1) The modulus of rigidity of a piece of glass is 245 x 10' grammes per sq. cm. Express it (a) in kilogrammes per sq. mm. ; (6) in absolute C.G.S. units ; and (c) in pounds per sq. in. Ans. (a) 2,450 ; (J) 240 x W ; (c) 3-48 x 10«. (2) The modulus of bulk-elasticity for steel is 1,841 x 10' dynes per sq. cm. Show that its value in grammes per sq. cm. is 1,876 X 10', and in poundals per sq. ft. 1,237 x 10^. (3) Young's modulus for lead (specific gravity = 11 "215) being 177 X 10' grammes per sq. cm., show that the length of the modulus is 15'78 x 10' cm. (4) The length of Yoimg's modulus for iron (specific gravity = 7"5) being 9x10' feet, show that its value in grammes per sq. cm. is 2,057 x 10', and in pounds per sq. ft. 4,218 x 10'. (A cubic foot of water weighs 1000 oz. approximately.)' (5) The modulus of torsion of a certain wire has the value 460 DYNAMICS [553 12 X 10* in the gravitational C.G.S. system, the absolute ft.-lb.-sec. system. Ans. 9166. Knd its value in 554. Strain due to Longitudinal Stress. — ^As the stress at any point of a body may (539) be completely specified in terms of simple longitudinal stresses, the determination of the strain produced by any given stress requires only that we should determine the strain produced by a simple longitudinal stress.— Let ^G be a unit cube of a body P subjected to a simple longi- tudinal stress, of intensity P, normal to the faces ABGD and EFOH. We may obviously apply to each of the other faces two equal and opposite normal stresses of the inten- sity P/3. (Each arrow-head in the figure denotes a stress of the intensity P/8.) Then it is evident that the simple longitudinal stress P is equi- valent to a uniform dilating tension P/3, together with two distorting stresses (538), each equal to P/3 and having one axis in the direction of the simple longitudinal stress, their other axes being at right angles to it and to one another. Hence (549) the efi'ect of the simple longitudinal stress P will be a uniform cubical dilatation of the amount (per unit of volume) PjZh, together with two shears, each of the amount P/Sri and having one axis in the direction of P, their other axes being perpendicular to it and to one another. Each of these shears, if small, is (276) equivalent to a positive elongation equal to P/Qn in the direction of P and a negative elongation of the same magnitude in the direction of the other axis. Also the cubical dilatation 555] OF ELASTIC SOLIDS AND FLUIDS. 461 P/S/c is (266) equivalent to an elongation the same in all directions and equal to P/9h. Hence the effect produced by P is a positive elongation in its own direction equal to P/91c+P/Sn or P(2k+n}/9hn, and a positive elongation equal to P/9k-P/8n or P{2n-3k)/18kn, in each of two perpendicular directions at right angles to one another, and therefore in all directions at right angles to that of P. 555. Stress required for Longitudinal Strain. — Simi- larly, as any strain may (279) be specified in terms of simple longitudinal strains, the determination of the stress required to produce a given strain requires only that we should determine the stress required to produce a simple longitudinal strain. By 277 (Ex. 1) a small simple elongation e is equivalent to a cubical dilatation e (due to elongations e/3 uniform in all directions), together with two shears, each of the amount 2e/3, having the direction of the given simple elongation as major axis or axis of positive elongation, and having as other axes lines perpendicular to the direction of the elongation and to one another. For the production of the cubical dilatation e a tension he, uniform in all directions, is necessary. For the production of each of the shears (538) a tension in the direction of the elongation, and of the intensity 2e7i/3,. together with a pressure of the same intensity in a perpendicular direction is necessary, the pressures required for the two shears being perpendicular to one another. Hence the elonga- tion e requires altogether a tension in the direction of the elongation of the intensity (k+4!n/3)e, and tensions of the intensity {k—2n/S)e in two directions perpen- dicular to that of the elongation and to one another. 462 DYNAMICS [555 and therefore in all directions perpendicular to that of the elongation. 556. The above results are sufficient to enable us to solve a few important problems on the strains produced in elastic solids when subjected to given stresses, and on the stresses required to produce or maintain in them given strains. Exairnples. (1) A rod, bar, or wire is subjected to equal and opposite forces acting at its ends in the direction of its length. Find the ratio (called Poisson's ratio) of the linear contraction it undergoes laterally to the elongation produced in the direction of its length. Ans. Obviously from 554, (3^-2?i)/2(3A + n). (2) Find in Ex. (1) the diminution, per unit area, of the cross section of the rod, P being the intensity of the stress applied at the ends. Ans. P{3k-2n)l9hi. (3) Show that in Ex. (1) the dilatation per unit volume is FjSk, P being the intensity of the stresses at the ends of the rod. (4) Express Young's modulus in terms of the moduluses of bulk- elasticity and of rigidity. The stress P applied at the end of a rod or wire in the direction of its length will (554) produce an elongation per unit of length of P{3k+n)l9i:n. Hence Young's modulus, the ratio of this stress to the elongation produced, is equal to 9knl{3i:+n). (5) Show that in the extension of a band of India-rubber, for which ^ is large in comparison with n, the area of the cross-section is diminished in nearly the same proportion as that in which the band is lengthened, and that there is therefore but little change of volume. (6) Find (a) the stress produced at any point in a circular cylinder of length I, one end of which is fixed while the other is twisted 558] OF ELASTIC SOLIDS AND FLUIDS. 463 through an angle 6, and (6) the moment of the couple which must be applied at the free end of the cylinder to maintain the torsion. (a) By 277,. Ex. (3), the cylinder is, at every point distant r from the axis, subjected to a shear whose plane is perpendicular to a plane through the point and the axis, and is parallel to the axis, whose direction is normal to the plane containing the point and the axis, and whose amount is 6rjl. Hence the stress at any point is a shearing stress of the intensity nBrjl, on a plane normal to the axis and in a direction perpendicular to the plane through the axis and the given point. (6) If the normal section at the end of the cylinder be divided into an indefinitely large number of indefinitely small portions of areas «i, s^, etc., distant r^, r^ etc., from the axis, the resultant shearing stresses on them will be nBr^SiJl, nBr^^jl, etc. The moments of these resultants about the axis will be nBr-^SiJl, n0r^S2ll, etc. Hence, if T is the moment of the couple which must be applied at the free end to maintain the given torsion, T= n0ri\jl + nBr^Sijl + etc. = ^ndrhjl = {nejDSsr'. Now Ssr' is (486) the moment of inertia of a uniform thin lamina of the shape and size of the section of the cylinder and (304) of sur- face density unity (called for shortness the moment of inertia of the section), about an axis through its centre perpendicular to its plane, and (490, Ex. 11) if a is the radius of the cylinder, is equal to n-a*/2. Hence T=nSira,*l'il. Hence also the torsion produced in a wire is directly proportional to the twisting couple and to the length of the wire, and inversely proportional to the rigidity and to the fourth power of the radius. The proportionality of the angle of torsion to the twisting couple was discovered experimentally by Coulomb, and is called Coulomb's law. (7) Express the modulus of torsion of a wire (552) in terms of its dimensions and its rigidity. Ans. nwa*i2, a being the radius of the wire. (8) A uniform straight beam, with one end fixed, is slightly bent by a force F applied at the other end normally to its length and in 464 DYNAMICS [556 the plane of bending, F being so great that the weight of the beam may be neglected. Find the flexural rigidity (552) of the beam in the plane of bending. Since the beam is uniform, and is but slightly bent, the strain produced may be taken to be that of 277, Ex. 4. — Let AECE' be any transverse section of the beam. Then the part of the beam between this section and the free end is in equilibrium under the force F, the normal stress over AECE', due to the longitudinal strain, and the shearing stress over A ECE', due to the shearing strain, to which the beam is subjected. Let EM be the inter- section with AECE' of the neutral surface. Then at any point Q, distant d from EE, there is a longitudinal strain in a direction normal to AECE', the elongation being dip, where p is the radius of curvature of longitudinal lines in the neutral surface and therefore, since the bending is slight, of all longitudinal lines. Hence, if S is the intensity of the longitudinal stress at O, and M is Young's, modulus for the beam (552), M=S\{dlf), and therefore S=Mdlp. If s is an indefinitely small area surrounding O, the resultant stress on this area is Msdjp. The moment of this resultant stress about EE' is therefore Msd^jp. Now the whole area AECE' may be divided into an indefinitely large number of indefinitely small portions. Hence the moment about EE' of the normal stress over the whole surface AECE' is 2(Msd^/p) = {Mlp)'Ssd^, the summation applying to all the small areas into which AECE' is' divided. Now 2rf jg (486 and 556, Ex. 6) the moment of inertia of the surface AECE' about EE'. Calling this /, we find the moment about EE' of the normal stress on AECE' equal to MI/ p. The shearing stress on ^^C^' being tangential has no moment about EE'. If the distance from AECE of the free end of the beam be S, the moment of i^ about EE' is Fs. The portion of the beam between A ECE' and the free end is 556] OF ELASTIC SOLIDS AND FLUIDS. 465 thus in equilibrium under the two moments Fi and MIlo. Hence (500) FS=MIIp, and FSp=MI. Now 1/p is the curvature of the beam, and therefore Fdp is the flexural rigidity of the beam in the plane of bending. Hence the flexural rigidity of a beam in the plane of bending is the product of Young's modulus for the beam into the moment of inertia of a transverse section about the line in which this section .intersects the neutral surface. We must therefore determine the position in the beam of the neutral surface. We have seen that, s being any small portion of a transverse section, the resultant stress on it noi'mal to the trans- verse section has the magnitude J^sd/p. Hence the resultant normal stress over the whole section is :2Msdlp={Mlp)I,sd. Now the bending being slight, the direction of this resultant longi- tudinal stress is perpendicular to the directions of the other acting forc.es. Hence for equilibrium this resultant stress must be zero, and therefore 2sd=0. Hence (403) the line FE", distant d from the little area s, passes through a point which is the centre of mass of the section AECE' (527, Ex. 2), and therefore the neutral surface is the surface passing through the centres of mass of the transverse sections of the beam, and normal to the plane of bending. (That line of the neutral surface which passes through the centres of mass of the sections of the beam is called the elastic central line.) We can now calculate the flexural rigidity of a beam of given section. Thus let the transverse section be rectangular, its sides being a and b. Then (490, Ex. 4) the moment of inertia of a trans- verse section about an axis parallel to the sides a, in its plane, and through its centre of mass, is ah^j\% Hence the flexural rigidity in a plane normal to the sides a is Mab^/lZ, where M is Young's Modulus for the beam. (9) A uniform straight horizontal beam of length Z has one end fixed, and is slightly bent in a vertical plane by the weight F oi a. 2g 466 DYNAMICS [556 body attached to the other end. Find the distance through which the free end "will be lowered. Let the unstrained beam be divided into an indefinitely large number of transverse slices of thickness t, and let abda be one of these slices in the ^ strained state. The bending being slight the transverse sections will in- tersect one another in a horizontal line vertically below the fixed end of the beam. Let this line intersect the plane of the diagram in 0, and let 6 be the inclination of ac to bd. Let ak and bl, tangents at a and b respectively, inter- sect a vertical line through Bini: and I respectively. Then M is the lowering of B due to the strain of abdo. The whole lowering of B will be the sum of the amounts of the lowering due to the strains of the various slices. Hence, if M be denoted by X, the total lowering of B will be SX. Now the angle between ai: and bl is e. Hence, since the bending is slight, if the distance of the slice abdc from the free end be denoted by S, we have \=de. Now (277, Ex. 4) t=pe. Hence \=Stlp. Also (556, Ex. 8) MIIp=F8. Hence \=^FtS^jMl. Hence also the total lowering of the free end SX= I^iFtS'lMI) = (FlMI)-ZtS\ the summation extending to all the slices of thickness t into which the beam of length L is divided. Now (486) S/2. 562. If the body be perfectly elastic, and if the strain be conducted so slowly that no change of temperature results, and no effect of viscosity is appreciable, then the stresses called into play depend only on the configuration of the body, and it thus constitutes a conservative system. Hence the potential energy of the body in its final con- figuration is equal to the work done in producing it. 563. If the body be perfectly elastic, and if the strain 472 DYIfAMICS [563 be not effected with infinite slowness, the stresses at the various stages of the strain are not dependent wholly upon the configuration, but depend also upon the varying temperature and upon the viscosity. Hence in this case the body does not behave as a conservative system, and the final potential energy is less than the work done in producing the change of configuration, the difference being the amount expended in the production of heat. 564. If the body be not perfectly elastic, then, even if the change of configuration be effected with infinite slow- ness, the stress required to produce a strain is not equal to that required to maintain it. Hence in this case also the body does not behave as a conservative system, and the final potential energy is less than the work done. 565. The potential energy of a body strained to the extreme limit of perfect elasticity is called the resilience of the body for that -kind of strain. It is usually mea- sured in gravitational units, and expressed per unit mass of the body. It is obvious that the resilience of a body thus expressed is equal to the .height to which the body would be lifted if an amount of work equal to the resi- lience were done in lifting it. The term resilience is also used by some writers as synonymous with elasticity. 566. Statics of FVwids {Hydrostatics). — A fluid is a body which possesses no degree of elasticity of shape, i.e., its shape may be changed by a stress of any magnitude however small, and no stress is required to maintain the strain thus produced, the body exhibiting no tendency to return to its initial shape when the distorting stress is removed. In consequence of the viscosity of fluids how- ever, a flnite stress is necessary to produce a change of shape, if the change is to be effected with finite rapidity. 567. All fluids are perfectly elastic for condensation strains. But they (fiffer greatly in compressibility. 569] OF ELASTIC SOLIDS AND FLUIDS. 473 Liquids are fluids whose compressibility is small ; gases, fluids whose compressibility is great. The compressibility of most liquids is so small that the properties of the ideal liquid, a liquid of constant density, are approximately those of many real liquids. Hooke's law applies to the condensation of liquids up to the highest pressures to which they have been subjected. In discussing liquids, however, we shall assume their density to be invariable. The relation of the pressure to the volume of a given mass of gas kept at constant temperature is approximately expressed in Boyle's law, which states that the pressure is inversely proportional to the volume, and therefore directly proportional to the density. All gases at suffi- ciently high temperatures follow Boyle's law with con- siderable accuracy through extensive ranges of pressure. But the lower their temperature the greater their devia- tion from it. We may take as the ideal gas one which follows this law, and in dealing with gases we assume it to hold. 568. The distinctive property of fluids, that the main- tenance of a shearing strain requires no stress, may ob- viously be expressed thus : — Provided the parts of a fluid body are not moving relatively to one another, the shear- ing stresses at all points of the fluid are zero, or the stresses at all points on all surfaces through the points are normal. 569. Stresses in Fluids. — The stresses of fluid bodies are usually pressures, though in certain cases they may be tensions. The centre of stress in the case of a fluid is thus usually spoken of as a centre of pressure. The stress throughout a fluid, which is in equilibrium and is not acted upon by external forces throughout its mass, is homogeneous (522). For (1) any hemispherical por- 474 DYNAMICS [569 tion of it is in equilibrium ; and the pressures on the small portions into which, its curved surface may be divided being all normal to these portions, and therefore passing through the centre of the sphere, their resultant also passes through that point. Hence also the resultant of the pressure on the plane surface passes through its centre ; and the pressure over it is therefore uniform. Also (2) any cylindrical portion, with ends normal to the axis of the cylinder, is in equilibrium, and the pressures on the curved portion of its surface being normal to the axis, the pressures on its ends must be equal and opposite. Hence the pressures on parallel surfaces are equal. 570. Specification of Fluid Fressv/re. — The stress throughout a fluid in equilibrium and not acted on by external forces being homogeneous, the results of 531-535 apply to the case of a fluid in this state. In the case of a fluid however the equations of 534 are much sim- plified by the absence of shearing stresses (568), and thus become F^ = Fl; Fy=Qm; F, = Rn, whence F={FW+Q^m^+E^n'^)K Now in the special case in which Z = m = «, we have since F is now a fluid pressure and therefore a normal stress, Ftc — Fy = Fz; and therefore in all cases, P = Q=R. Hence (7) F==P = Q=R. K therefore a fluid be in equilibrium and be not acted upon by external forces, the pressures at all points across all surfaces through these points are the same. If it be acted upon by external forces (537) the pressures at any one point across all surfaces through that point are 574] OF ELASTIC SOLIDS AND FLUIDS. 475 the same, or, as it is usually put, the pressure at any point is the same in all directions. The pressure at any point of a fluid in equilibrium is therefore specified by one numerical datum. 571. Uqual Transmission of Pressure. — If P and P' be the pressures on the ends (normal to the axis) of a cylinder of unit section, and of any length and in any direction, and if F be the sum of the components in the direction of the axis of the external forces acting on the cylinder, then for equilibrium P'-P = F. Hence, if P be increased by any amount, P' becomes increased by the same amount. This result is often called the " Principle " of the equal transmission of pressure. 572. Surfaces of Equal Pressure in a fluid acted upon by external forces and in equilibrium are surfaces at all, points of which the pressure is the same. Lines of force in a fluid acted upon by external forces are lines whose directions at all points coincide with the directions of the resultant external force at those points. 573. Surfaces of equal pressure are at all points normal to lines of force. For the resultant external force on a small cylinder of the fluid with ends normal to its axis, and so placed that the pressures on its ends are equal, can have no component in the direction of the axis. 574. If the external forces are central forces (338), and the various points of the fluid have therefore potentials (355-6), the resultant force at a point must be normal to the equipotential surface through the point (359). Hence surfaces of equal pressure coincide with equipotential surfaces. 476 DYNAMICS [575 575. In that case also (356) the resultant external force on unit mass of the fluid at any point is equal to the rate of change of potential per unit of distance in its direction. Now, if the fluid between two surfaces of equal pressure, indefinitely near one another, be divided by lines of force into columns of equal section, the differ- ences of pressure on the ends being the same for all, and all being in equilibrium, the resultant external forces acting on all must be the same. Let F he the resultant external force on any column, in its mass, I its length and Fand F'the potentials of its ends. Then^/m=(F'- F)/Z. Hence, the difference of potential between the ends being the same for all, the ratio of the mass to the length and therefore of the mass to the volume must be the same for all. And therefore surfaces of equal pressure are also surfaces of equal density. 576. In the case of heavy fluids, the attraction of the earth is the external force. Hence in that case level or horizontal surfaces are surfaces of equal pressure. The free surface of a heavy liquid in equilibrium, being exposed to the pressure of the atmosphere, is therefore a horizontal surface throughout the region in which the pressure of the atmosphere has the same value. 577. Variation of the Pressure of Fluids acted upon by External Forces. — Let F be the resultant external force acting on each unit of volume of the fluid, in one of the columns of 575, s being the area of either of its ends, I its length, and P and P' the intensities of the pressures on its ends. Then {P'-P)s = Fls, and (P'-P)ll = F. Hence the resultant force on unit volume of the fluid is equal to the rate of change of pressure in its direction per unit of distance. 579] OF ELASTIC SOLIDS AND FLUIDS. 477 578. If the external forces are derivable from a potential, we have also (356) V and Fbeing the potentials at the ends of the column at which the pressures are P' and P respectively, and p being the density. Hence (P'-P) = (F~ "F)p. If gravitational attraction is the only external force, we have therefore, with the convention of 361, since the external force is directed from the end of smaller to the end of greater pressure, F'-F = p{J'-Y). Now in this case Hence P'-P=pgl; and therefore the rate at which pressure increases per unit of distance in a direction normal to surfaces of equal pressure in a heavy fluid is equal to pg. 579. In the case of liquids p is a constant. Let P„, P^, Pj, etc., P be the pressures at a series of surfaces of equal pressure indefinitely near, let l^, l^, etc., be the inter- cepts between these surfaces of a line of force, and let the surfaces whose pressures are P„ and P be so near that g may be considered constant, then P,-P„ = p5rZ^,P,-P, = /.flrZ,.etc. Hence, if L be the length of the line of force extending from any point of the surface whose pressure is P to that of which the pressure is P„, we have by addition P-Po = pg{h+h+^^) = P9L. Gravitational attraction being the only external force acting thi'oughout the mass of the fluid, the surfaces of equal pressure are horizontal surfaces and the lines of force are vertical lines. Hence the difference of pressure 478 DYNAMICS [579 between two points of a heavy liquid is equal to their difference of level multiplied by pg, and therefoi^e to the ■wTeight of a column .of the liquid whose length is the difference of level and .whose section is unity. If n is the pressure of the atmosphere at the free, sur- face of a heavy liquid, the pressure at any point at depth L is thus TL+pgL, which may be written pg{L+L'), pro- vided L' = JXfpg, i.e., provided L' is the length of a. column of the liquid of unit section whose weight is equal to 11. The determination of the resultants and centres of the pressures on the surfaces of bodies immersed in heavy liquids is of great practical importance. The reader will find, on looking back to 527, Exs. 2-5, and 529, Eks. 2-7, that examples of such determinations have already been given in considering resultants and centres of stress. 580. In the case of gases kept at a constant tempera- ture we have (567) p = kP, where A is a constant and p the density of a gas at the point at which its pressure is P. Hence P'-P=icP(V'-V), and P'=P[l+k{W-V)]. Let P„, Pj, Pg, etc., P be the pressures at a series of n+1 surfaces of equal pressure indefinitely near, and so chosen that the differences of potential of neighbouring surfaces are the same, and let F, V„ be the potentials of the sur- faces whose pressures are P, P„. Then P=Pll+k{V-r,)/nl P,=Pll+kir-V,)/nl = Pll + k(r-V,)/nf, etc., P=Pll+k(V-V,)lnf.. 581] OF ELASTIC SOLIDS AND FLUIDS. 479 Hence, if n be made indefinitely great, we have (392) where e is the base of Napier's Logarithms. Gravitational attraction being the only external force acting throughout the mass of the gas, and the volume of the gas under consideration being so small that g may be considered constant, we have r-r,= -gh, where hia the height of the point whose potential is F above that whose potential is Vq. Hence It is obvious that since P = p/k, 1/k is equal to g times the length of a c'olumn of the given gas of uniform density p and of section unity, whose weight is equal to P. It is therefore equal to g times the height which an atmosphere of the gas would have if its density were the same throughout its whole extent as at the earth's surface. This height is consequently often called the "height of a homogeneous atmosphere" or the "pressure-height" of the given gas for the temperature to which the given value of k applies. If this height be denoted by H, since 1/k = gH, we have P=P,e-i- The value of H for any gas depends only on its nature and temperature and on the value of g. For dry atmospheric air at 0° 0. in the latitude of Paris it is 7-990 X 10^ cm. 581. Archimedes' Princi/ple.—li. a body be wholly or partially immersed in a heavy fluid, the resultant of the pressure over its surface is a single force acting vertically upwards through its centre 6f mass and equal to the weight of the fluid displaced. For a portion of the fluid having the same position, shape, and size as the given 480 DYNAMICS f581 body or the part of it which displaces fluid, would be in equilibrium under its own weight and the resultant, pressure on its surface, which, since the pressure at a point of a heavy fluid varies only with its depth beneath or height above any chosen level surface, must be the same as the resultant pressure on the body. 582. Equilibrium of a Floating Body. — It follows, from 581, that a body floating at the surface of a heavy liquid will be in equilibrium provided (1) the centres of mass of the body and of the displaced liquid are in a vertical line, and (2) the weight of the body is equal to that of the displaced liquid. 583. Stability of the Equilibriwin of a Floating Body. — The general discussion of the stability of" the equilib- rium of a floating body is beyond the scope of this book. But in the important special case of a homogeneous rigid cylinder, of any section, for angular displacements about its axis, the condition of stability admits of simple ex- pression. — Let ABO be a transverse section of such a cylinder, through its centre of mass G; and let jE" be the centre of mass of the portion beneath the surface 8S' of the liquid, and therefore of the displaced liquid, in the position of equilibrium, in which the line GE is obviously vertical. Also let F be the centre of mass of jthe sub- merged portion when the cylinder has been rotated through a small angle about a longitudinal axis, M being 585], OF ELASTIC. SOLIDS AND FLUIDS. 481 the point in which a vertical line through F intersects QE. Then the cylinder is acted upon by equal and opposite vertical forces through G and /; and it is obvious that if the point M be above Q these forces will tend to diminish the angular displacement and to bring the cylinder back to the positibn of equilibrium ; where- as, if M be below Q, they will tend to increase the dis- placement. In the former case therefore the equilibrium is stable, in the latter unstable. The point M is called the metacentre. The equilibrium is therefore stable, provided the metacentre be above the centre of mass^ This result applies to the rolling of a ship so built and laden that 0, E, and F are in the same plane. 584. Kinetics of Fluids (Hydrokinetics). — When the parts of a fluid move relatively to one another, shearing stresses make themselves manifest. If, e.g., a cylindrical vessel, with its axis vertical, and containing a liquid, be made to rotate uniformly about its axis, the liquid will be found after a time to be rotating with the vessel, and if th^ vessel be now brought to rest the motion of the liquid gradually subsides. Hence any cylinder of the liquid coaxial with the vessel is acted upon by stresses having tangential components when the liquid outside it is in motion. For otherwise that cylinder must remain at rest or in uniform motion. In many important practical cases however the effect of these shearing stresses is small and may be neglected ; and as the consideration of the motion of fluids exhibiting tangential stresses is attended with great difficulty, we restrict our attention to these cases. 585. If the stresses at a point of a moving fluid on all planes through the point are normal, they have also the same intensity. For if we consider a tetrahedron, such as that of 532,' 2h 482 DYNAMICS [585 we have, as the equation of its motion in the direction of Ox, using the symbols of 532, P . OBG-Fl . ABC+X = md, where X is the component of the resultant external force, m the mass, and a the component acceleration of the centre of mass in the x axis. Now X is proportional to the mass, and therefore to the volume, of the tetrahedron, If therefore (537) the tetrahedron be indefinitely small, both X and m may be neglected relatively to P and F. Also we have OBG=ABC. I. Hence P=F. 586, Equations of Motion. — The motion of a fluid under given forces may be determined by applying the general equations of the motion of extended bodies and expressing in equations the conditions imposed by the distinctive peculiarities of fluids. Of these equations there are two. The first expresses the relation which holds between the pressure and the density of the fluid. In the case of a gas at constant temperature it is p = JcP; and in that of a liquid, yo = const. The second is the equation of continvAty which expresses in mathematical language the general law that a fluid in motion is always a continuous mass. The employment of these equations however in the solution of problems is beyond the scope of this book. 587. Steady Motion. — In general the velocity of the fluid particles passing through a given point in space varies with time. When at each point in space through which fluid is passing the velocity of the fluid is constant both in magnitude and direction, the motion is said to be steady. The paths of the particles of a fluid which is moving steadily, are lines of motion, i.e., lines whose directions at all points are the directions of the motion 6i the fluid at those points. They are therefore called streavn lines. S88] OF ELASTIC SOLIDS AND FLUIDS. 483 588. Equation of Energy. — We may obtain, as being simple and important, the equation of energy applicable to cases of the steady motion of liquids under external forces which have a potential. Consider a tube whose curved surface is bounded by stream lines, and whose ends A and B are small and normal to the stream lines. Let p be the pressure, v the speed of the liquid, Fthe potential due to external forces, and 8 the area of the section, at A ; and let p', v', V, s' be the values of the same quantities at B, The masses of liquid entering the tube at A and leaving it at B in unit of time are pvs and pv's' respectively, and since the liquid moves as a continuous mass and does not vary in density, we have (the equation of continuity for this case) pvs = pv's' ; and hence V8=v's'. Unit mass entering the tube at A has the kinetic energy 1^12, the same mass leaving it at B the kinetic energy v'^1'2,. Hence the excess of the amount of the kinetic energy entering the tube at A over that leaving it at B iu unit of time is pvs(v^ — v'^)/2. The potential energy of unit mass at A is greater than that of unit mass at B by V — V, if we adopt the definition of potential given in 361. Hence the excess of the amount of the poten- tial energy entering the tube at A over that leaving it at B in unit of time is pvs{V'—V). Also, the work done by the pressure at A on the liquid entering in unit of time is pvs, and that done by the liquid leaving at B in unit time against the pressure at B is p'v's'. Hence the energy gained by the tube in unit of time on account of the work done by the pressures at the ends is pvs —p'v's' = (p —p')vs. Now as the motion is steady the energy of the tube is constant. Hence 484 . DYNAMICS [588. f)Vs(v^-'V'^)l2+pvs(y'- V) + (p-p')vs = 0; and p + p(vy2 -V) = p' +p(v'y2 - V) ; or p+p(vy2-V) = C, where G is a constant for the same stream line. 589. We may apply the above result to one important hydrokinetic problem. Problems on the motion of fluids, ev.en of liquids, in general require higher mathemati- cal attainments than readers of this book are supposed to JSccmnple. A vessel is kept filled to a constant level with liquid, which escapes through a small orifice in its wall. Knd the speed of efflux. In this case the flow of liquid soon becomes steady. Since the upper surface is large relatively to the orifice, the speed of the moving liquid there is small, and, if the orifice be suflioiently small, may be neglected. The pressure at the upper surface is that of the atmosphere II. If that surface be taken as the level of zero poten- tial, V—0. Hence the above equation of energy for a point at the upper surface on any chosen stream line reduces to Let P be the pressure of the liquid at the orifice and v its speed. Also, let the depth of the orifice below the upper surface of the liquid be k. Then, at the orifice, V=gh. Hence, for a point of the orifice on the above stream line If the pressure at this point be taken to be that of the atmosphere, we have therefore ^ .. v^='ig!i. And if the pressures at all points of the orifice be taken to be equal to that of the atmosphere, this equation would give us the speed 590] OF ELASTIC SOLIDS AND FLUIDS. 485 with. which all liquid particles leave the orifice. This result is called Torricdli's Theorem. Torricelli's Theorem cannot be applied to the calculation of the rate of efflux, i.e., the amount of liquid escaping per unit, of time, for two reasons : (1) The stream lines at the orifice are for the most part not normal to it, for the jet diminishes in diameter from the orifice outwards. Hence the v of the above formula is not the value of the normal velocity of the liquid particles. (2) The assumption that the pressure at all points of the orifice is equal to that of the atmosphere is not well grounded. For, as the jet con- tracts from within outwards, the speed of the liquid particles must be increasing, and therefore the pressure of the liquid must be diminishing, in that direction. The pressure in the interior of the jet at the orifice must therefore be somewhat .greater than that of the atmosphere. At a short distance from the orifice the contraction of the jet ceases, the section of the jet at which it ceases being called the Vena Contrwita. Here the stream lines are normal to the trans- verse section of the jet, and the pressure may thus be taken to be the same at all points of the section and therefore to be that of the atmosphere. Hence the speed of the liquid particles in passing through the Vena Contracta wiU be (2ghy, where h' is the depth of the Vena Contracta beneath the upper surface. Also, the liquid particles are here moving normally to the Vena Contracta. Hence, if (S is the area of the Vena Contracta and p the density of the liquid, the rate of efflux is pS(2ffhy. 590. Work done dunng Strain. — As tangential stresses exist in a fluid during the relative motion of its parts, the expressions obtained (559) for the work done in an elastic solid during a change of configuration apply also to fluids. Since work is done during the straining of a fluid, in overcoming its viscosity, a fluid, like a solid, will behave during a strain as a conservative system only if the strain be effected with sufiicient slowness. 486 DYNAMICS. [590 Since a fluid in equilibrium exhibits no she9,ring stresses, the work done against shearing stresses during a strain has no result in the form of production of potential energy. In the case of liquids, on account of their incompressi- bility, a strain involves no change of volume. Hence the work done in producing a strain in their case has no result in the form of potential energy. It is wholly ex- pended in overcoming molecular friction and results only in the production of heat. Hence Joule employed the agitation of water as a means of determining the me- chanical equivalent of heat, the water employed having, after its agitation, the same potential energy as it had before. MISCELLANEOUS EXAMPLES. (1) A point moves in a plane curve so that its distance, s feet measured along the curve from its starting point, is represented by the formula s=25+6t\ where t is the time in seconds reckoned from the instant of starting. Knd (a) the mean speed between the beginning of the 10th and the end of the 12th second ; (6) the instantaneous speed at the end of the 10th second ; (c) the mean rate of change of speed between the instant of starting and the end of the 10th second ; (d) the instantaneous rate of change of speed after any time. Ans. (a) 126 ft. per sec; (6) 120 ft. per sec; (c) 12 ft.-per-sec. per sec; (d) 12 ft.-per-sec per sec. (2) The breadth between the rails of a certain railway is 4 ft. 8 in. Show that in a curve of 500 yds. radius the outer rail ought to be raised about 2^ inches for trains travelling 30 mis. an hour, that there may be no horizontal pressure on the rails. (3) The velocity of a point moving in a given elliptic orbit is the same at a certain point, whether it describe the orbit in a time t when its acceleration is directed towards one focus, or in a time f when its acceleration is directed towards the other focus. Show that, if 2a is the length of the major axis, the focal distances will be 2afl{l+t') and 2at/{t+t^. (4) A large number of equal particles are fastened at unequal intervals to a fine string and then collected into a heap at the edge of a smooth horizontal table with the extreme one just hanging 488 KINEMATICS AND DYNAMICS. over the edge. The intervals are such that the times between successive particles being carried over the edge are equal. Prove that if c„ be the interval between the «."• and (re + l)"" particles, and v„ the velocity just after the (n+1)*^ particle is carried over, c„/Ci=w„/»i = ra. (5). Reduce 20 cm. per sec. to yards per hour. Ans. 787-38. (6) If a particle move on any smooth curve under the action of any force, and if, at any point, F be the component of this force normal to the curve and towards the concavity of the curve, the reaction of the curve on the particle towards the concavity is equal to mvyp-F, when p is the radius of curvature of the curve ^ind v the speed of the particle. (7) A uniform rod hangs horizontally supported by two equal vertical strings, of length I, attached to its ends. It is twisted horizontally through a very small angle so that its centre of mass remains in the same vertical line, and is then let go. Find the time of a complete (double) oscillation, neglecting the inertia of the strings. Ans. Sir ijljig. (8) A point is moving with a simple harmonic motion of ampli- tude a and period T. Show that, if d is its displacement from its mean position after a time t, the epoch being $, d=acos{ZirtlT+e). (9) A straight staircase consists of stairs each 1 ft. wide and 6 in. high, A smooth particle is projected from a point on one of the stairs near its edge and in the vertical plane perpendicular to the edge of each stair. Find the velocity of projection that the particle may strike the diiTerent stairs in succession at the same distance from the edge, the coefficient of restitution being 0-5. Ans. iJ^gjZ feet per second, inclined 45° to the horizon. (10) The imit of rate of change of speed being a rate of change of speed of 100 cm. -sec. units and the unit of time 1 min., show that the unit of length is a length of 36 x 10* cm. MISCELLANEOUS EXAMPLES. 489 (11) If a ball impiiige successively against two adjacent sides of a rectangle, its velocity will be diminished in the ration of l:e,e being the coefficient of restitution. (12) Two uniform solid cylinders, of weights w and w', descend from rest directly down the two faces of two smooth inclined planes, of inclinations a and t' respectively, over the common summit of which passes a thin inextensible string which goes under and round the central transverse sections of the cylinders, to which the ends of the strings are fastened. Find (a) the tension of the ^tring, and (b) how much it will have slid along the planes at the end of any time t. Ans. (a) wu/ (siua + sina')/3(w + 2»'); (6) gt^w sin a-iv' sin a')/2(w + w'). (13) A particle weighing xV lb. moves backwards and forwards in a straight line 3 inches long with simple harmonic motion, 25 times per second. Find the force acting on it (a) at the end of the range, and (6) at a point at one half the maximum distance from the centre. Ans. (a) 616-8... p lis.; (6) 308-4... pdls. (14) A particle of mass m is suspended from two points in the same horizontal line by two strings of equal length I (inclination = o). One of the strings is suddenly cut. Find the initial change of tension of the other string. Ans. mg{2 cos^a — 1)/(2 cos a). (15) A heavy smooth tetrahedron rests with three of its faces against three fixed pegs and the fourth face horizontal. Prove that the reactions of the pegs are as the areas of the faces on which they are exerted. (16) A point is moving with a uniform rate of change of speed of 2 ft. -sec. units. Show that, if its initial speed is 3 ft. per sec, the ratio of its final to its initial speed during the time required to traverse 4 feet of its path is 5/3. (IV) If particles are dropped from given heights upon a fixed horizontal plane, the heights being inversely as the squares of the coefficients of restitution, they all rise to the same height after reflection. 490 KINEMATICS AND DYNAMICS. (18) A uniform lever ACB, whose arms AC and BG are at right angles to each other, is in equilibrium when ^10 is inclined at an angle j3 to the horizon. If ^<7 be raised to a horizontal position, C being fixed, find the angle through which it will fall. Ans. 2/3. (19) A particle of O'l grm. mass executes 512 simple harmonic oscillations per second, the amplitude of the oscillations being 0*25 cm. Find the maximum value of the force exerted upon it. Ans. 258,736-1... dynes. (20) A rope hanging over a rough horizontal cylinder carries two bodies. The mass of one is 20 lbs.; that of the other is m lbs. («i > 20). But the rope does not slip off the. cylinder, on account of friction. If the coefficient of friction, when the rope is just on the point of slipping, is 0'4, what is the value of m ? Ans. 70-269 lbs. (21) §'s displacement relative to P is n times as great as i"s relative to 0, and they are inclined at an angle 6. Show that if e < 7r/2, §'s displacement relative to increases with n, and that, if 9>7r/'2, it decreases as n increases until m=-cosS, increasing with n for greater values of n. (22) A particle is projected from a point on an inclined plane and after n rebounds returns to its point of projection. Prove that, if o is the inclination of the plane, ;8 the angle between the direc- tion of projection and the plane, and e the coefficient of restitution, cotacotj3= , l^e (23) The time of descent of a heavy particle sliding freely from rest down a smooth inclined plane of given height varies as the cosecant of the inclination. (24) A chain, whose weight per unit length is equal to that of 1 lb., is to be stretched between two points in a horizontal line 800 ft. apart, so that the tension at the lowest point may be equal to the weight of 1,600 lbs. Find (os) the length of chain required, and (6) the depth of its lowest point below the points of suspen- sion. Ans. (a) 808-32 ft ; (6) 50-24 ft. MISCELLANEOUS EXAMPLES. 491 (25) A point undergoes component displacements represented by straight lines drawn from a point within a triangle to the angular points. Show that its resultant displacement is the same as if it had iindergone component displacements represented by lines drawn from the same point to the points of bisection of the sides. (26) On the sides of a right-angled triangle squares are described, the square BODE on the hypothenuse being on the same side of BC as the triangle, the squares CAFO, ABHK on GA, CB on the- opposite side of each to the triangle. Prove that if forces repre- sented by AB, BC, CA,BH, HK, KA, CD, BE, EB, AF, FG, GC, act on a particle, it will be in equilibrium. (27) A particle slides from rest down the whole length of a smooth inclined plane. Prove that the distance between the foot of the inclined plane and the focus of the parabola which the particle describes after leaving the plane is equal to the height of the plane. (28) Trucks containing each a ton of ballast are sustained upon a smooth plane of iiiclination a by an equal number of empty trucks upon a smooth plane of inclination /3. Find the mass of a truck. Ans. sin o/(sin ;8 - sin a) tons. (29) A right cylinder whose weight is to the diameter of its base as 3 : 4, stands on a perfectly rough inclined plane whose inclination is 45°. From the lowest point of its uppermost circular section a body is suspended whose weight is a little greater than one-sixth of the weight of the cylinder. Prove that it will overturn the cylinder. (30) A ship sails from A to B, ^3 miles N. 30° W., in 15 minutes; from jB to C, 1 mile N. 60° E., in 7 minutes; from G to B, 4 miles N. 45° W., in 20 minutes ; and from Dto £,4 miles N. 45° E., in 18 minutes. Show that her mean speed has been 9+^3 miles per hour, and that her mean velocity has been 2 -f- 4 ^2 miles per hour, N. (31) ABCD and A'B'C'iy are two parallelograms. Show that if a particle be acted upon by forces represented by AA', B'B, GG', and lyO, it will be in equilibrium. 492 KINEMATICS AND DYNAMICS. (32) A uniform straight plank (length = 2a) rests with its middle point upon a rough horizontal cylinder (radius =>•), their directions being perpendicular to each other. Supposing the plank to be slightly displaced so as to remain always in contact with the cylinder without sliding, determine the period of an oscillation. Ans. %ital>Jzgr. (33) Two circles lie in the same plane, the lowest point of the one being in contact with the highest point of the other. Show that the time of descent from any point of the former to a point in the latter down the chord passing through the point of contact, is constant. (34) Four pegs are fixed in a wall at the four highest vertices of a regular hexagon, the two lowest being in a horizontal line. Over the pegs a loop is thrown supporting a body of weight W, the loop having such a length that the angles formed by it at the lowest pegs are right angles. Find (a) the tension in the string, (6) .the reactions of the two highest pegs, and (c) those of the two lowest pegs. Ans. (a) F; (6) W, inclined 60° to the horizontal; (c) 'W sl% inclined 15° to the horizontal. (35) Two points, P and §, move in straight lines (inclination = 9) with uniform accelerations a and a', and at a given instant have velocities v and v' respectively. Show that their relative velocity will be perpendicular to §'s line of motion after a time {v cos e — «;')/(«' - a cos 9), and will have its least value after a time {av' + ffl'«)cos e-av- aV a^+a'^ — '2,aa' cos B Show also that if v/v' = a/a', the least value of their relative velocity will be zero. (36) A particle of weight W is supported on a smooth inclined plane of inclination o, by means of two strings attached to fixed points in the plane and inclined at angles 8 and 8' to a line of MISCELLANEOUS EXAMPLES. 493 greatest slope. Piud (a) the tensions in the strings, and (6) the reaction of the plane. Ans. (a) .^7^' and -J^-, (i) Wcosa. (37) A given circle and a given straight line which does not cut the circle are in the same vertical plane. Show that if a tangent be drawn to the circle at its lowest point P, meeting the given line in A, and if from the given line ^§ be cut olF equal to AP, and if P^ intersect the circle in if, §iJ is the straight line of quickest descent from the given straight line to the given circle. (38) Two equal heavy particles slide along the arc of an ellipse whose plane and major axis are vertical. They are connected by a string passing through a smooth ring at the focus. Prove that the particles will be in equilibrium in all positions. (39) A point has three component coplanar velocities, »i, v^, v^, the angles between v^ and % Vg and «j, v^ and v-^ being o, ^, y re- spectively. Show that its resultant velocity is («i^ + v^ + v^ + 2«2»3 cos o + ^-^^ cos ^ + aiJ^Vj cos 7)i (40) If the height of a rough inclined plane be to the length as OS is to sld'+V, and a body of k-Ja^+b^ lbs. mass can just be supported by friction alone, required the least force acting along - the plane which will draw the body up the plane. Ans. 2kag. (41) Two bodies pf equal weight w are tied to the ends of a fine string which passes over two puUies without mass in a horizontal line (distance = a). Supposing a body of weight W(W>2w) to be fixed to the middle point of the horizontal portion of the string, determine how far it will descend. Ans. 2wWa/(4w^- W). (42) A pendulum which -would oscillate seconds at the equator would gain 5 minutes a day at the pole. Show that the ratio of the value of g at the equator to its value at the pole is 144 : 145. (43) If there are n particles in a straight line, of masses m, 2m, Zm, etc., and at distances a, a/2, a/3, etc., respectively from a point in the line, the distance of the centre of mass from jt is 2a/(M+l). 494 KINEMATICS AND DYNAMICS. (44) A square board is hung flat against a wall by means of a string attached to the extremities of its upper edge and passing round a smooth nail. Prove that if the length of the string is less than the diagonal of the board, there will be three positions of equilibrium. (45) A point moves in a circle of radius r ft. with a uniform speed of 7rr/6 ft. per sec. Show that its mean acceleration during 6 seconds is 7rr/18 ft. -sec. units in a direction opposite to the initial direction of the velocity, and that the mean acceleration is 2/ir times the magnitude of the uniform instantaneous acceleration. (46) A particle is just supported by a rough inclined plane of variable inclination when its inclination is i. Find its acceleration up the plane when moving upwards on a line of greatest slope under the action of a force equal to twice its weight acting up the plane. Ans, g'[2 — tan i(3 cos^i - sin^i)]. (47) At a given instant a pendulum begins to oscillate in a vertical plane at a place of latitude 60°. Find after what time it will be apparently oscillating in a plane perpendicular to the former. Ans. 1/2 ^3 day. (48) ABCD is a square from which a corner AEF is cut oif by a straight line drawn parallel to BD and at a distance from A equal to | of the diagonal. Show that the distance of the centre of mass of AEFirovn ^ is J of the diameter. (49) A and B are points in a horizontal line. A uniform and smooth rod AC (weight = W) is fastened to a hinge at A and can swing in a vertical plane through AB. A string passes over a pulley at B, supporting at one end a body of weight P, and at the other being attached to a small smooth ring which slides on the rod. Prove that there will be equilibrium in any position if W.AC^-iP.AB. (50) If a conic section be described under the action of a force tending to a f ocijs, the hodograph will be a circle. MISCELLANEOUS EXAMPLES. 495 (51) Show that 1 foot-grain is equivalent to 1"975 gramme- centimetres. [1 grain =0'064799 gramme]. (52) A rod (length = Z) is fixed at one end about which it can move freely in any direction. When it is inclined to the horizon without motion at the angle a, a horizontal velocity V is com- municated to 'its other end. Determine the velocity of the free, end at the instant at which the rod becomes horizontal. Ans. (V+Slgsma^ inclined to the vertical at the angle tan-^[ Fcos a/( K^sin^a + Zlg sin o)i]. (53) A three-legged stool stands on the floor of an elevator sliding in its frame-work with perfect freedom. Show that it has four degrees of freedom. (54) The distance of the centre of mass of haK a' hexagon 2^ inscribed in a circle from the centre is equal to where }• is the 3 ^3 radius. (55) Two uniform beams of given weight are in equilibrium in a vertical plane, the lower end of each beam resting on a horizontal ' floor and the upper ends being in contact. Show that the friction between either beam and the floor varies inversely as the sum of the tangents of the angles which the beams make with the floor. (56) A particle moves in a parabola under the action of a constant force parallel to the axis. Show that the hodograph of its path is a straight line parallel to the axis. (57) Show that one horse-power is equivalent to about 746 watts. (58) A cone is revolving round its axis with a given angular velocity when the length of the axis begins to be diminished uniformly, and the vertical angle to be increased so that the volume of the cone remains unchanged. Show that if u is the initial angular velocity of the cone, and h the initial length and r the rate of decrease of its axis, its angular velocity after any time t will be a)(l-rt/A). (59) Show that a body has two degrees of freedom, when two of its points are constrained to remain in given curves. 496 KINEMATICS AND DYNAMICS. (60) A body consists of two portions and one of them is moved into a new position. Show that the line joining the two positions of the centre of mass of the whole is parallel to, and bears a fixed ratio to, the line joining the two positions of the centre of mass of the part moved. (61) A regular hexagon is formed of rods jointed at their ex- tremities. Strings are stretched between every pair of alternate angles of the hexagon so as to form two equilateral triangles. Show that the tension of any string is equal to f of the sum of the tensions of the strings which cross it, minus ^ of the tension of the string which is parallel to it. (62) The kinetic energy of a particle, which is constrained to move in a circular path of radius r, varies as the square of its distance s, measured along the path from a fixed point in the path. Show that its tangential acceleration in any position is to its normal acceleration as r : «. (63) If an agent working at the rate of one horse-power, perform the unit of work in the unit of time, and the acceleration of a falling body be unit of acceleration, a pound being the unit of mass, find the unit of (a) time and (6) length. [j7=32 ft. -sec. units.] Anfi. (a) 17A see. ; (6) 9453^ ft. (64) A rod is kept in a vertical position by means of two small rings and its lower end is supported on an inclined plane (inclination =i) which is freely moveable on a horizontal plane. Show that if ■» is the velocity of the rod and «' that of the inclined plane, !;=■!;' tan i. (65) Show that if Q be the centre of mass of the triangle ABC Z{GA-'-V GB"' + GC^) =AE'+BC"'+ CAK (66) Two equal and similar rods AB, BO are freely hinged at B, and rest in a plane of greatest slope of a rough inclined plane, in a position of limiting equUibrium, with the end A hinged at a point in the plane, and the end C resting on the plane. If o, 0, e are respectively the angle of inclination of the plane to the horizon, the angle of inclination of the rods to the plane, and the angle of friction, show that 3 cos(0 -1- e)cos(^ - ") = cos(0 - c)cos(0 -1- a). MISCELLANEOUS EXAMPLES. 497 (67) A point is moving in a straight line with an acceleration varying as its distance from a point in that line. Prove that the corresponding point in the hodograph moves with a similar ac- celeration. (68) The mass of a railway train is 150 tons and the resistances to its motion (from air, friction, etc.) amount to 16 pounds- weight per ton. Find (a) the horse-power of the engine which can just keep it going at 60 miles an hour on a level plane, and (6) the greatest speed which an engine working at 200 horse-power can give it on a level plane. Ans. (a) 384, (6) 31J miles per hour. (69) A uniform rod (length = 2c) moves in a vertical plane within a hemisphere with angular velocity a. Show that if $ be the inclination of the rod to the horizon at any instant the horizontal and vertical velocities of its middle point have the magnitudes cw cos 8 and cw sin S. (70) The corners of a pyramid are cut off by planes parallel to the opposite faces. Show that if the portions cut off be of equal mass, the centre of mass of the remainder will coincide with that of the pyramid. (71) Two uniform rods AB, AG of lengths a, b respectively, are of the same material and thickness and are smoothly jointed at A. A rigid weightless rod of length I is jointed at £ to AB, and its other end B is fastened to a smooth ring sliding on AC. The system is hung over a smooth peg at A. Show that AG makes with the vertical an angle ta,n~\all(b^+a s/a'-P)']. (72) If each unit involved in the measurement of g become m times its former value, show that the new value of ff will be m times its former value also. (73) A particle of 10 lbs. mass, whose motion is simple harmonic, has velocities 20 and 25 ft. per sec. at distances 10 and 8 ft. per sec. respectively from the centre of fdrce. Find the work done during the motion from the distance 10 to the distance 8 feet. Ans. 112'5 foot-poundals. 2l 498 KINEMATICS AND DYNAMICS. (74) A uniform rod in falling strikes, when in a. horizontal position, -with one end against a stone. Show that the impulse of the blow it receives is half that of the impulse of each of the blows which it would have received had both ends struck simultaneously against two stones, the blows being in all cases supposed to be at right angles to the rod. (75) Show that a force of 100 dynes is equivalent to the weight of 1-019 X IQ-* kilogrammes. (76) Show that in the direct impact of elastic balls of masses m and M and initial velocities v and F, and with coefficient of restitu- tion e, an amount of kinetic energy equal to (1 — ef gpn-r — \ ( F- »)^ is lost. (77) How much water will be pumped from a vertical cylindrical shaft of 10 feet diameter by an engine working for 6 hours at 200 horse-power, the water being discharged at a point 10 feet above the mouth of the shaft, and the surface of the water being initially 20 feet below the mouth of the shaft. [Density of water = 1,000 oz. per cub. ft.] Ans. 2,157'1... tons. (78) Determine the unit of time in order that with the foot as unit of length g (32 ft.-per-sec. per sec.) may have the value unity. Ans. 1/4 ,y/2 second. (79) Find the work done on a body of 12 lbs. mass in falling to •the earth's surface from a point 1,000 miles above it. [Earth's radius = 4,000 miles ; 5' = 32 ft.-sec. units.] Ans. 22,628-5... ft.-tons. (80) A ball rolling on a horizontal plane strikes obliquely an .equal ball at rest. The direction of motion of each ball after irtipaot makes the same angle 6 with that of the striking ball before impact. Show that the coefficient of restitution is equal to tan^ff. (81) If the weight of one ounce be the unit of force, one second the unit of time^ and 162 the density in pounds per cubic foot of the standard substance, find the unit of length, g being taken to be 32 ft.-sec. units. Ans. 4 inches. MISCELLANEOUS EXAMPLES. 499 (82) If from any point in the plane of a polygon perpendiculars be drawn to its sides, and if forces act along these perpendiculars, either all inwards or all outwards, each force being proportional to the side to which it is perpendicular, the system is in equilibrium. (83) A rough heavy body bounded by a curved surface rests upon two others, which themselves rest upon a rough horizontal plane. Show that the three centres of mass and the four points of contact lie in one plane. (84) Two points move in concentric circles of radii r and /. When their radii vectores from the common centre are inclined 9 radians, their angular velocities about the centre are w and u' respectively. Find the magnitude of their relative velocity. Ans. ((<)r'' + ()IR + (b sin )IG, and being the angles made by the given straight lines with the central axis, a and b the shortest distances between these lines and the central axis, R the resultant force and G the resultant couple. (101) Show that a stress of 40 grammes-weight per square centimetre is equivalent to one of 0-5689... pound's-weight per sq. inch. (102) If s and s' are the spaces traversed by a point moving with uniform acceleration in a straight line in the times t and t' respec- tively, reckoned from the same instant, show that the acceleration and the initial velocity are, respectively, 2^7-sOa„df?!lzf^' tt'{i-t) tiii-t)' 502 KINEMATICS AND DYNAMICS. (103) A gun is suspended freely by two equal parallel cords and a shot is fired from it. Prove that the range on a horizontal plane is, for a given gun and shot, directly proportional both to the height through which the gun rises in the recoil and to the tangent of its initial inclination to the horizon. (104) Find the force exerted by two equal uniform discs (radius = a, distance =c, surface density = p) placed perpendicularly to the line joining their centres, on a particle of unit mass in that line at a distance h from the nearer disc, it being given that the one disc attracts, while the other repels, according to the gravitational law. Ans. 27rp[(6 + c)/ Va^ + (6 + of - b/ is/^^+F]. (105) Find the moment of inertia of a fly-wheel (mass=if) formed by cutting from a circular plate of radius r^ a circular portion (concentric with the plate) of radius j-j. Ans. ^M(r-i^+r^^). (106) A hollow vessel has the form of a pyramid, four of whose five faces are equilateral triangles (side = a). It is placed with its square face on a horizontal plane and filled with a liquid of density P through a small aperture in the vertex. Find the integral stress on the four triangular faces. Ans. a^pffs/^IS. (lOV) Two inclined planes intersect in a horizontal line, their inclinations to the horizon being a and /3. If a particle be pro- jected at right angles to the former from a point in it so as to strike the other at right angles, the velocity of projection must be sin /3[2gra/(sin a - sin p cos (o -I- /S))]}, a being the distance of the point of projection from the intersection of the planes. (108) Two particles, of masses 9,820 and 1,964 grammes respec- tively, attract one another. Find the acceleration of either relative to the other, when the distance between them is 4 cm. Ans. 0'75 cm. -sec. units. (109) Find the acceleration produced by a mass of 1 kilogramme in a particle at a distance of 1 metre. [Earth's mass=6'14x 10^' grammes; earth's radius =6 '37x10* cm. ; ^' = 981 cm. -sec. units.] Ans. 6-48x10-8. MISCELLANEOUS EXAMPLES. 503 (110) A moment of inertia is expressed in terins of the units of the ft.-lb.-sec. gravitational system. By what number must its value be multiplied that it may be expressed in terms of the metre- kilogramme-second gravitational system. Ans. 0-138.... (111) Two particles are projected from two given points in the same vertical line with the same v.elocities. Prove that lines touching the path of the lower will cut off from the path of the upper, arcs described in equal times. (112) One bullet is fired towards another bullet which is let fall at the same instant. Prove that if, on meeting (see 119, Ex. 4), they coalesce, the latus rectum of their joint path will be one-fourth of that of the original path of the first bullet. (113) A uniform bar of length a rests suspended by two strings of lengths I and V fastened to the ends of • the bar and to two fixed points in the same horizontal line at a distance c apart. Prove that if the directions of the strings are perpendicular the ratio of their tensions is al+d' : al'+cl. (114) In the expression for' the attraction of two particles, F=kmm' jcP, how does the value of h depend upon the units of mass, length, and time. Ans. Its dimensions are [M]-\L]\Ty'. (115) Show that the radius of gyration of a uniform square disc (side = a) about one of its diagonals is aj \/l2. (116) A particle describes an ellipse under a force directed towards its centre. Show that the time between the extremities of conjugate diameters will be constant. (117) A particle is dropped from a point A and a second equal particle is simultaneously projected vertically upwards from a point B so that the balls impinge, the stress during impact being in a line inclined tan-^ V« (« being the coefficient of restitution) to the verti- cal. Prove that both balls will strike the horizontal plane through B simultaneously, and that if the velocity of projection at B be 504 KINEMATICS AND DYNAMICS. that due to AB (i.e., that which the particle would have had had it fallen through the vertical distance AB), their distance there will heABs/Se. (118) If a body, attached at its centre of mass to one end of a string of length r, the other end being attached to a fixed point in a smooth horizontal plane, make n revolutions in one unit of time, prove that the ratio of the tension in the string to the force exerted on the plane is 4irVr : g. (119) Find the time of a small double oscillation under gravity of a uniform one foot cube suspended by one edge as horizontal axis. Ans. l"-07.... (120) A particle describes a parabolic orbit under a force directed towards the focus. Show that the sum of the squares of the velocities at the extremities of a focal chord is constant. (121) A body of mass P pulls one of mass § over a smooth pulley, and § in ascending, as it passes a certain point A, catches and car- ries with it a third body B, which in its descent is again deposited at A. Supposing no jerk to occur when B is caught up and that § oscillates through equal distances above and below A, prove that the mass of B is (P^ _ q^iq. (122) A uniform triangular lamina suspended from a fixed point by three cords attached to its three vertices is in equilibrium. Show that the tensions in the cords are proportional to their lengths. (123) Three inches of rain fell in a certain district (5^=32 ft. -sec. units) in 12 hours. Assuming that the drops fell from a height of a quarter of a mile and neglecting the resistance of the air, find the pressure on the ground due to the rain during the storm. [The mass of a cubic foot of rain water = 1,000 oz.] Ans. 0'105... poundals per sq. foot. (124) Show that in the case of a right-angled isosceles triangular plate the times of small oscillations are the same about horizontal axes perpendicular to its plane through its vertex and through the middle point of its base. INDEX. {The Numbers refer to Sections.) Absciasa, 4. Absolute units, 301. Acceleration, 110; angular, see An- gular acceleration ; central, 156 ; change of point of reference, 115 ; planetary, hodograph and path of point moving under, 161 ; moment of, 123-4; motion under uniform, 140 ; normal, 120 ; of falling bodies at the earth's surface, formula for, 140 (footnote) ; of momentum, mo- ment of, 422; of point moving subject to Kepler's Laws, 162; of point moving uniformly in a circle, 121 ; relative, 115 ; tangential, 120; units of, 111. Accelerations, composition and reso- lution of, 116. Activity, 333. -Slolotropic bodies, 542. Amount of shear, 269. Amplitude of simple harmonic motion, 165. Angle, plane, units of, 21. Angle, solid, unit of, 22. Angular acceleration, of a point, 135 ; of a rigid system, 219 ; units of, 136. Angular accelerations, composition and resolution of, 221. Angular and linear velocity, relation between, 129. Angulardisplaoementof apoint, 125-6. Angular momentum, 420. _ Angular velocities, composition of, 216. Angular velocity, of a point, 127; of a rigid system, 212 ; of rigid system, and linear velocity of one of its points, relation between, 214; mo- ment of velocity, in terms of, 132 ; units of, 128. Approach, velocity of, 321. Archimedes' Principle, 581. Area, units of, 17-19. Areas, conservation of, 429. Areal velocity, 133. Arm of couple, 467. Atmosphere, homogeneous, 580. Attraction, integral normal, over a surface, 365; of infinite uniform plate, 369 ; of thin circular disc, 316 (1) ; of thin uniform spherical shell, 316 (5 and 6), 367 ; of uniform circular cylinder, 368. Attractions, 315; difference of, on opposite sides of attracting plate, 370. Atwood's machine, 382 (1), 498 (1). Axes of co-ordinates, 4. Axes, principal, of strain, 263. Axis, of couple, 467 ; of rotational dis- placement, 208; of rotation, in- stantaneous, 213 ; Foinsot's central, 482. Azimuth and Altitude, 3. Balance, common, conditions of sta- bility and sensitiveness, 507 (11) ; quickness of motion, 498 (4) ; time of oscillation, 498 (4). Balance, spring, 320 (6). Ballistic pendulum, 499 (8). Baryoentrio bodies, 474. Bending of a beam, 277 (4), 556 (8-10). Blackburn's pendulum, 180. Boyle's law, 567. ■ Bulk-elasticity, modulus of, 552. Cartesian co-ordinates, 4. Catenary, common, 396. Central acceleration, 156. 506 INDEX. Central axis, 482. Centre, instantaneous, 233 ; of gravity, 474 ; of inertia, 399 ; of mass, 899; do., acceleration of, 411 ; do., acceleration of, in terms of external forces, 414 ; do., distance of, from any plane, 400-401 ; do., of a surface, 408 (13) ; do., of circular arc, 408 (10) ; do., of com- posite bodies, 407; do., of homo- geneous symmetrical bodies, 406; do., of lune, 408 (19) ; do., of sector of circle, 408 (20) ; do., of triangle, 408(12) ; do., of triangular pyramid, 408 (23); do., velocity of, 409; of oscillation of physical pendulum, 496 (3) ; of percussion, 496 (9) ; of pressure, 579; of stress, 528; of suspension of physical pendulum, 496 (3) ; of system of parallel forces, 472. Centrifugal force, 320 (14). Centripetal force, 320 (14), 338. Centrobaric bodies, 474. Chain, see String. Clock, 32. Coefficient, of elasticity, 321 ; of limit- ing static friction, 328; of kinetic friction, 328 ; of restitution, 321, 378. Collision, of particles, 378; of spheres, 380 (1), 498 (10 and 11). Component accelerations, etc., see composition of accelerations, etc. Components, 80-82 ; resultant equal to algebraic sum of components of, in its direction, 83, 84 ; trigonometrical expressions for, 85, 86, 89. Composition, of accelerations, 116 ; of angular accelerations, 221 ; of an- gular velocities, 216; of angular velocities about parallel axes, 249 ; of couples, 469 ; of forces acting on a particle, 312 ; of forces acting on rigid body, 459; of linear and an- gi]dar accelerations, 250; of linear and angular velocities, 247; of simple harmonic motions, 168; of simultaneous displacements, 78 ; of simultaneous rotations, 203 ; of suc- cessive displacements, 76; of suc- cessive rotations, 200 ; of transla- tions and rotations, 238 ; of veloci- ties, 98. Compound harmonic motion, 167. Compound pendulum, 496 (3). Compressibility, 552. Configuration, 12. Conical pendulum, 190, 320 (19). Conservation, of angular momentum, principle of, 429, 495 ; of areas, 429 ; of energy, law of, for single particle, 348 ; of energy, law of, for extended bodies, 435 ; of linear momentum, pi'inciple of, 416, 495. Conservative forces, 348. Conservative system, 435. Constraint, of a point, 35 ; motion of points under, 181; motion of rigid systems under, 253 ; one degree of, of most general kind, 246. Continuity, equation of, 586. Continuous stress, 522 ; strain, 284. Co-ordinates, 2 ; Cartesian, 4 ; polar, 3; rectangular, 4-5. Coplanar forces on rigid body, single resultant of, 460-463. Cord, see String. Coulomb's law, .556 (6). Couples, 467 ; composition of, 469. Cubical dilatation, 266. Curvature, centre, chord, circle, and diameter of, 39 ; of a circle, 38 ; of any path, 37 ; radius of, 39. Curved path, motion in, under uniform acceleration, 185. Cycle of transformations, 436. Cycloid, motion in, 192. Cycloidal pendulum, 193. D'Alembert's Principle, 417. Day, sidereal, 29 ; sidereal, variation of, 30 ; solar, 31. Degree, 21. Degrees of freedom, see Freedom. Density, 304 ; surfaces of equal, 675. Derived units, 18. Differential screw, 517 (7) ; wheel and axle, 517 (5). • Dilatation, cubical, 266. Dimensions of space, 13 ; of units, 18. Direction cosines, 7; inclination of two straight lines in terms of, 8; of common perpendicular to two given lines, 10 ; sum of squares of, equal to unity, 7. Direction of shear, 269. Displacement of free rigid systems, 232-233. 1 Displacements, 69; angular, see An- INDEX. 507 gular displacement ; change of point of reference, 71-74; composition of simultaneous, 78; composition of successive, 76; resolution of, 79; rotational, 208. Dynamics, subject matter of, 285, 323. Dynamometer, friction brake, 507(14). Dyne, 302. Efficiency of a machine, ,517 (8). Efflux of liquids, 589. Elastic central line, 556 (8). Elastic isotropy, conditions of, 548; Elasticity, 543 ; of figure, 549 ; of vol- ume, 549; perfect, 543; perfect, limit of, 546. Element, 310. Elevation of a projectile, 145. Elliptic harmonic motion, 173. Elongation of strain, 257. Energy, 343; equation of, for fluids, 588 ; law of, application to kinetic problems, 441, 499 ; law of conser- vation of, for extended bodies, 435; law of, for extended bodies, 4.37, 439 ; law of, for single particle, 348 ; law of, for single particle, applica- tion to kinetic problems, 351 ; law of, for single particle, application to static problems, 353; loss of, on impact, 443 (1 ) ; of a system of par- ticles, 432 ; of position, 345. Epoch of simple harmonic motion, 165. Equation of continuity, 586. Equations of motion, of a particle, 317; of extended bodies, 431; of fluids, 586 ; of rigid bodies, 493 ; of strings, 384, 387, 390, 391, 392, 394. Equilibrium, of elastic solids, 551 ; of extended systems, 444; of floating body, 582; of floating body, stabi- lity of, 583 ; of fluids, 569 ; of a par- ticle, 323 ; of a particle, analytical expression for condition of, 326 ; of a particle, condition of, 324 ; of a particle, expressions for condition of, 325 ; of a rigid body, 500 ; of a rigid body, analytical conditions of, .501 ; of a rigid body, conditions of, in terms of work done, 510; of a rigid body, expressions for condi- tions of, 504 ; lof a system of rigid bodies, 508; of strings, 385, 387, 390, 391, 392, 394 ; stability of, 450. Equipotential surfaces, 358. Erg, 331. Ergometer, 507 (14). Extended bodies, 398. External forces, 376, 398. Falling bodies, 140-141, 159 ; value of acceleration of, 140 (footnote). Field of force, 355 ; mapped out by lines of force, 372 ; uniform, 373. First Law of Motion, 286. Fixed point, 24. Flexural rigidity, modulus of, 552 ; of bent beam, 556 (8). Flexure of a beam, 277 (4), 556 (8-10). Floating body, equilibrium of, 582; stability of equilibrium of, 583. Flow of liquids, 589. Fluent, 44. Fluid, 547. Fluid pressure, 569; centre of, 569; resultant, 569 ; specification of, 570. Fluids, shearing stresses, in, 584. Flux, 44. Fluxion, 44. Foot-pound, 331. Foot-poundal, 331. Force, acting on particle, specification of, 311 ; acting on rigid body, speci- fication of, 457 ; centrifugal, 320 (14) ; centripetal, 320 (14), 338 ; de cheval, 335'; diagram, 382 (22), 509 (26) ; dimensions of derived unit of, 308; in terms of potential, 356; line of, 359, 572 ; moment of, 425 ; origin of idea bf, 285; rotating power proportional to moment, 455; tube of, 360; units of, 297- .302, 552. Forces, conservative, .348; internal and external, 376, 398 ; non-conser- vative, 348; on particle, composi- tion and resolution of, 312 ; on rigid body, composition of, 459 ; on rigid body, condition of reducibility to single force, 477 ; on rigid body re- ducible to a force and a couple, 479 ; on rigid body reducible to two forces, 476. Foucault's pendulum, 228 (5). _ Freedom, degrees of, of a point, 35 ; of a rigid system with one point fixed, 198, 211 ; of free rigid system, 231, 236. 508 INDEX. Friction, kinetic and limiting static, 328; molecular, 645. Friction brake dynamometer, n07 (14). Gkilileo's Law of Motion, 309. Gas, real and ideaJ, 567. Geometrical representation of motion of rigid systems, 229, 252. Graphic methods, 382 (22), 509 (26). Gravitation, law of, 315. Gravitational potential, 361. Gravitational units, 298. Gravity, force of, 140 (foot-note) ; centre of, 474. Gyration, radius of, 486. Harmonic motion, 163; compound, 167 ; elliptic, ;173 ; simple, 163. Heterogeneous strains, 284. Heterogeneous stress, 522. Hinge, reactions of, on bars, 509 (14 and 15). Hodograph, 113, 161. Homogeneous atmosphere, 580 ; bodies, 541 ; strains, 257 ; stress, 522. Hooke's Law, 551, 567. Horse-power, 335. Hydrostatics, 566. Hydrokinetics, 584, Ideal gas, 567. Impact, direct, of spheres, 380 (1), 498 (10) ; oblique, of spheres, 498 (11) ; of particle on smooth surface, 821. Impulse, 294 ; equations of motion of a particle in terms of, 319 ; equa- tions of motion of rigid body in terms of, 494. Impulsive forces, 319. Inclination of two lines, 8 (footnote) ; in terms of direction cosines, 8-9. Inclined plane, equilibrium of body on, 327 (1), 329 (7), 354 ; motion on, under uniform acceleration, 181, 329 (1), 352 (4). Inertia, 286 ; centre of, 399 ; moment of, see Moment of inertia ; quantity of, 291. Initial line of polar co-ordinates, 3. Initial tensions, 382 (23). Internal forces, 376, 398. Instantaneous axis of rotation, 213. Instantaneous centre, 233. Intensity of stress, 523. Isotropic bodies, 542. Isotropy, elastic, conditions of, 548. Joule's experiments on mechanical equivalent of heat, 590. Joule, the, 331. Kater's pendulum, 496 (3). Kepler's Laws, 162. Kilogramme-metre, 331. Kinematics of machinery, 253. Kinematics, subject matter of, 1. Kinetic energy, 344, 432; change of, equivalent to work done, 344, 434 ; loss of, on impact, 443 (1). Kinetics, subject mattei: of, 323; of elastic solids, 5.57 ; of fluids, 584. Law of energy, see Energy. Laws of motion, 285. Length, units of, 16. Level surface, 576. Lever, 507 (9); rough, 507 (32) ; rough, efficiency of, 517 (9). Limit of perfect elasticity, .546. Linear and angular velocity, relation between, 129. Linear density, 304. Linear displacement, 125. Linear velocity, 127 . Lines, of force, 359, 572 ; of motion (of fluids), 587 ; of quickest descent, 184 (11). Liquid, 567. Longitudinal stress, 530. Machinery, kinematics of, 2.53. Machines, simple, see Simple ma- chines. Mass, 289 ; astronomical unit of, 815 ; centre of, see Centre of mass ; di- mensions of derived unit of, 300 ; to be distinguished from weight, 290 ; units of, 297-802, 304, 315. Material point, 310. Matter, quantity of, 291. Measurement, 14 ; of length, area, etc., see Length, Area, etc. Mechanical advantage of simple ma- chine, 327 (1 and 2). Mechanical powers, see Simple ma- chines. Mechanics, 323. Metacentre, 583. INDEX. 509 Moduluses of elasticity, 552 ; dimen- sions of, 552. Modulus, length of the, 552 ; of hulk elasticity, 552 ; of flexural rigidity, 552 ; of rigidity, 552 ; of simple longitudinal stress, 552 ; of torsion, 552 ; of torsion, in terms of rigidity, 556 (7) ; Young's, 552, 556 (4). Molar equilibrium of extended sys- tems, 444. Molecular equilibrium of extended systems, 444. Molecular friction, 545. Moment, of acceleration of momentum, 422 ; of a force, 425 ; of a force, ana- lytical expression for, 427 ; of an acceleration, 123-4; of a velocity, 103 ; of a velocity, analytical ex- pression for, 106 ; of a velocity in terms of angular velocity, 132 ; of inertia, 486 ; of inertia, dimensions of units of, 491 ; of inertia, deter- mination by calculation, 488, by experiment, 487 ; of inertia of an area, 556 (6) ;of inertia, units of, 491 ; of momentum, 418. Moments of inertia, 490. Momentum, 293 ; angular, 420 ; con- servation of linear and angular, see Conservation ; moment of, 418; mo- ment of acceleration of, 422. Motion, in cycloidal path, 192 ; of free rigid bodies, 497; of particle, equations of, 317 ; of particle, under given rates of change of speed, 60 ; of particle under uniform accelera- tion, 140 ; of particle, under uniform velocity, 138 ; of rigid body about fixed axis, 496 ; of rigid system under constraint, 253 ; of rigid system under given accelerations, 251 ; of rigid system under given angular accelerations, 224 ; of sys- tems of rigid bodies, 498 ; rela- tion of, to time, 25. Neutral equilibrium, 450. Neutral surface, 277 (4). Newton's experiments on collision of spheres, 378, 380 (12). Newton's Laws of Motion, 285. Non-conservative forces, 348- Non-conservative system, 435. Normal acceleration, 120. Numeric, 14. Numerical measure or value, 14; in- versely proportional to magnitude of unit, 15. Oblique impact of spheres, 498 (11). Ordinate, 4. Origin of co-ordinates, 4. Orthogonal projection, 8. Oscillation, centre of, of physical pendulum, 496 (3). Osculating plane, 41. Parallel forces, resultant of, 464-467, 470. Parallelogram of accelerations, 116 ; of displacements, 78 ; of forces, 313 ; of velocities, 98. Particle, 310. Particles, systems of, 374, 398. Path, 36; of point ujider planetary acceleration, 161 ; of point under zero acceleration, 138 ; of point with harmonic motion, 163, 164, 168, 170-180; of a projectile, 1.51-2. Pendulum, ballistic, 499 (8); Black- burn's,180; Captain Kater's,496 (3) ; compound or physical, 496 (3) ; conical, 190, 320 (19); cycloidal, ,193; Foucault's, 228 (5); mathe- matical or simple, 187, 352 (5). Percussion, centre of, 496 (9). Perfect elasticity, 543 ; limit of, 546. Perfectly rough body, 328. Period of simple harmonic m otion, 165. Perpetual motion, 436; impossibility of, as law of motion, 436. Phase of simple harmonic motion, 165. Physical pendulum, 496 (3). Pitch of screw, 245, 254 (4). Plane of shear, 269. Planetary motion, 158. Poinsot's central axis, 482. Point of reference of displacements, change of, 71-74. Poisson's ratio, 556 (1). Polar co-ordinates, 3. Pole of polar co-ordinates, 3; of the hodograph, 113. Polygon, of accelerations, 116 ; of dis- placements, 78; of forces, 313; of velocities, 98. Position, 2, 11. Potential, 355; calculation of, 363; central forces derivable from, 356 ; 510 INDEX. force in terms of, 356 ; no maximum or minimum value of, in free space, 371. Potential energy, 345-347, 356, 432. Poundal, 302. Power, 333. Pressure, 306 ; centre of, 579 ; differ- ence of, between two points of a heavy fluid, 579, 580; equality -of, in all directions in fluids, 570, 585 ; fluid, 569 ; of fluids acted on by external forces, 577 ; resultant, 579 ; surfaces of equal, 572. Pressure-height, 580. Projectiles, 142 ; displacement of, after given time, 144 ; displacement of, in given direction, 145 ; elevation of, 145 ; path of, 151-2 ; range on given plane, 145 ; range on hori- zontal plane, 149 ; velocity of, after given time, 143. Projection, orthogonal, elementary propositions on, 8 (foot-note); of simple harmonic motion, 172. PuUeys, 254 (6-7), 509 (1. 2, 3), 517 (2). Pure strains, 267. Quickening, 54. Quickest descent, lines of, 184 (11). Radian, 21 ; soUd, 22. Radius, of curvature, 39 ; of gyration, 486. Radius vector, 3. Range, of a projectile, 145, 149. Rate of work, 333 ; dimensions of units of, 335 ; units of, .335. Ratio of strain, 257. Recoil, velocity of, 321. Rectangular components of a displace- ment, 81. Rectangular co-ordinates, 4. Relative acceleration, 11-5. Relative velocity, 96. Repose, angle of, 328. Resilience, 565. Resistance to compression, 549. Resolution, of accelerations, 116; of angular accelerations, 221 ; of dis- placements, 79 ; of forces, 312 ; of rotation into translation and rota- tion, 241 ; of rotations, 207 ; of stress, 530 ; of velocities, 99. Rest, 24. Restitution, coefficient of, 321, 378. Resultant displacement, 76; analy- tical expression for, 90 ; trigonome- trical expression for, 85. Resultant pressure, 579. Resultant stress, 525. Rigid bodies, 194. Rigid bodies, motion about fixed axes, 496 ; motion of free, 497 ; motion of systems of, 498. Rigid dynamics, 453. Rigidity, ,549. Rigidity, flexural, modulus of, 552. Rigidity, modulus of, 552. Rotation, 33, 194. Rotational displacements, 208. Rotational strains, 268. Rotations, 199 ; composition of simul- taneous, 203; composition of suc- cessive, 200 ; resolution of, 207. Rough bodies, 328. Scalar quantity, 42. Screw, 245, 2.54 (4), 517 (6) ; differ- ential, 517 (7) ; rough, efficiency of, 517 (11) ; rough, mechanical advan- tage of, 517 (10). Second, 32. Second Law of Motion, 287. Shear, 269 ; amount of, 269 ; direction of, 269 ; homogeneity of, 270 ; plane of, 269 ; reduction of, to a pure strain and a rotation, 272. Shearing stress, 530. Sidereal day, 29 ; variation of, 30. Signs, convention of, for co-ordinates, 4, 5 ; for moments of forces, 425 ; for moment of velocity, 103. Simple harmonic motion, 163 ; ampli- tude of, 165 ; epoch of, 165 ; period of, 165; phase of, 165; projection of, 172. Simple harmonic motions, composition of, 168. Simple longitudinal strain, 264. Simple machines, see Pulley, Inclined Plane, Wheel and Axle, Lever, Screw. Simple pendulum, 187, 352 (5). Simple rigidity, see Rigidity. Smooth body, 320 (24). Solar day, 31. Solid, 547. Solid radian, 22. Space, dimensions of, 1 Specific gravity, 304. INDEX. 511 Speed, 42 ; change of, 51 ; dimensions of units of, 47 ; motion of a point with uniform, 61 ; motion of a point under given rates of change of, 60; motion of a point under uniform rate of change of, 63 ; rate of change of, 52, 53 ; rate of change of, dimensions of, 57; rate of change of, units of, 56 ; units of, 45. Spring balance, 320 (6). Stabifity of equilibrium, 450; of floating body, .583 ; relation of potential energy to, 451. Stable equilibrium, 450. Standard substance, 804. Standards, 14; of length, etc., see Length, etc. Static energy, 345. Statics, subject matter of, 323. Steady motion, 587. Strain, 33, 255 ; continuous, 284 ; due to longitudinal stre,s8, 554 ; ellipsoid, 264 ; elongations of, 257 ; heterogeneous, 284; homogeneous, 257 ; principal axes of, 263 ; princi- pal elongations of, 263 ; principal ratios of, 263 ; pure, 267 ; ratio of, 257 ; rectangular specification of a small, 283 ; relation of final to initial volume, 266; relation of stress to, 540 ; rotational, 268 ; simple longitudinal, 264 ; specifica- tion of, 278. Stream lines, .587. Strings, flexible and inextensible, 383. Stress, 306 ; centre of, 528 ; con- tinuous, 522; dimensions of imits of, 523 ; homogeneous, 522 ; hetero- geneous, 522 ; longitudinal, 530 ; relation of, to strain, 540 ; required for longitudinal strain, 555 ; resolu- tion of, 530 ; resultant of, .525 ; specification of, 531; tailgential or shearing, 530; tangential, resolu- tion of, into longitudinal stresses, 538 ; units of, 523. Stresses, 519; in bars of framework, 509 (22 and 23). Surface density, 304. Surface integral of normal attraction, 365. Surfaces of equal density, 575 ; of equal pressure, 572. Suspension, centre of, of physical pendulum, 496 (3). Systems of particles connected by strings, 381. Tangential acceleration, 120. Tangential stress, 530 ; resolution of, into longitudinal stresses, 538. Tautochrone, the cycloid a, 192. Tension, 306. Thermal energy, 440. Third Law of Motion, 307. Three-bar motion, 254 (8). Time, description of instants, 26 ; measurement of, 27; relation of motion to, 25 ; units of, 29-32. Toggle-joint, 517 (13). Torrioelli's Thrtirem, 589. Torsion, 277 (3) ; modulus of, 5.52 ; stress required to maintain, 556 (6) ; time of oscillation of a body sus- pended by a twisted wire, 558. Tortuosity, 41. Transformations of energy, 349. Translation, 33, 34. Transmissibility of force, principle of, 456. Transmission of pressure, principle of the equal, 571. Triangle of accelerations, 116 ; of dis- placements, 78 ; of forces, 313 ; of velocities, 98. Tubes of force, 360. Twist, 245. Uniform acceleration, motion in curved path under, 185 ; motion on in- clined plane under, J8 1, 329 (1), 352 (4) ; motion under, 140. Uniform velocity, motion under, 138. Units, absolute, 301 ; derived, 19 ; dimensions of, 18 ; gravitational, 298; systems of, 46 ; of length, etc, see Length, etc. Unstable equilibrium, 450. Vectors, 70. Velocities, composition of, 98 ; reso- lution of, 99 ; virtual, 353. Velocity, 43, 92 ; angular, see Angular velocity ; areal, see Areal velocity ; change of, 109 ; change of point of reference, 96 ; instantaneous, 93 ; mean, 92 ; moment of, about a point; 103, about a line, 104; motion under uniform, 138 ; relative, 96 ; units of, 94. 512 INDEX. Fcjia contraeta, 589. Virtual displacement, 353; moment, 353; velocities or work, principle of, 353. Viscosity, 545. Volume, dimensions of units of, 20; units of, 17. Warren girder, 509 (27). "Watt, the, 335. "Weight, 290 ; to be distinguished from mass, 270 ; proportional to mass, 290. Weights of particles of small body reducible to single force acting at centre of mass, 474. Wheel and Axle, 254 (5), 498 (3), 507 (10); differential, 517 (5). Work done, 330 ; by component forces and resultant, rek,tion between, 342 ; dimensions of units of, 332 ; dimensions of units of rate of, 335 ; during strain, .559, 590 ; under cen- tral forces, 337; under uniform force, 337 ; rate of, 333 ; units of, 331 ; units of rate of, 335. Young's modulus, 552; in terms of moduluses of elasticity, 556 (4). <^p END. PEIHTED AT THE DHIVEKSITY PRESS, ET ROBEKT HACLEHOSE, OLASOOW.