^ IMAGE EVALUATrON TEST TARGET (MT-S) 1.0 I.I iSii^ |ii2,5 ^ lis 1 2.0 1.8 f 1.25 1.4 1.6 « 6" ► 9« # /i / ^ Photpgraphic Sciences Corporation 4? \ iV «'.., 33 WIST MAIN STREET WEBSTER, N.Y. 145S0 (716) 877-4503 \ CIHM/iCMH IVSicrofiche Series. CiHM/ICMH Collection de microfiches. Canadian Institute for Hintorical Micror«productions / Institut cansidien da microraproductions hittoriqua: m m '*t»i i fc ' i'»a m JU^iSm ^r>f*rmr-' Tt'ili ii iiiNi-^rrr'Vir i fcMW iiH-itfr-inJtfii^ -"I"'- ^iv fV- f mathematics are so treated that the student may enter upon the study of Kinetics with clear notions of motion, veloc- ity and acceleration. Part III treats of the KineUcs cf a particle and of rigid Imdiee. tW" !^^!^'S^ -f_ ' t. ' ; Jnu; 1 ^."V g !" ' W^: :;;^:;,;^i----^. -^ PREFACE. - In this arrangement of ti.«« work, with the exception of Kine- matics, I have followed the plan naually adopted, and made the subject of Statics precede that of Kinetics. For the attainment of that grasp aX principles which it is the special aim of the book to impart, numerous examples are given at the ends of the chapters. The jyreater part of thorn will present no serious difficulty to the student, while a few may tax his best efforts. In prepwring this book I have availed myself of the writings of many of the beat authont The chief sources frtm which I have derived assistance are the treatises of Price, Miuchin, Todhunter, Pratt. Routh, Thomson and Tait, Tait and Steele, Weisbach, Venta- roli, Wilson, Browne, Gregory, R»nkine, BoucharUt, Pirie, Lagrange, and La Place, while many valuable hints as well as examples have been obtained from the works of Smith, Wood, Bartlett, Young, Moeek/, Tate, Magnus, tioodeve, Parkinson, Olmsted, Gamett, Benwick, Bot- tomley, Morin, Twisden, Whewell, Oalbraith, Ball, Dana, Byrne, ihe Elusjyclopedia Britannica, and the Mathematical Visitor. 1 have again to thank my old pupil, Mr. B. W. Pi«ntiss, of the Nautical Almanac Office, and formeriy Fellow in Mathematics at the Johns Hopkins University, fbr reading the MS. and for valuable sug. gestiona Several others also of my friisnds have kindly assisted me by correcting proofi^beets and verifying copy aad foimule. E. A. B. BtrroBRS Collbob, i Nbw Bbdhbwick, N. J., June, 1884. ' eptlon of Kine- and made tho which it is tlio ilea are given at will present no J tax his best TABLE OF CONTENTS. ' the writings of I which I have lin, Todhunter, eisbacb, Yentu- 'irie, Lagrange, nples have been oung, Moselc^ , Benwick, Bot. ma, Bjrrne, the r. 'rentiss, of the ematios at the valuable aug. aasisted me vim. E. A. B. PART I. lly CHAPTER I FIRST PBINCIPLE8. m. '^» 1. Definitions— Statics, Kinetics and Kinematies 1 2. Matter * 8. InerJa * 4. Body, Space and Time 8 6. Rest and Motion 8 6. Velocity •• 8 7. Pormube for Velocity * b. Acceleration ■ * 9. Measure of Acceleration S 10. Geometric Representation of Velocitj and Accelerat'on 6 11. -Mass * 12. Momentum "^ 18. Change of Momentum ' 14. Force ® 15. Static Measure of Force 8 16. Action and Reaction ® 17. Method of Comparing Forces • 18. Representation of Forces 1® 19. Measure of Accelerating Forces 10 20. Kinetic Measure of Force H 31. Absolute or Kinetic Unit of Force 18 22. Throe Ways of Measuring Force 14 28. Meaning of fli In Dynamics l** IS'., rr. ,, ".i ' W. ^ g!^. '#■ roirrxyrs. 24. OniTltation Units Df Force and Mam...... ic 25. Gravitation and Absolute Measure 17 Examples 18 ■ STATICS (REST). -^ CHAPTER II. THE COMPOSITION AND BE80LUTI0N OF CONCUBBINO FOROKS — CONDITIONS OF EQUILIBBIUM. 26. Problem of Statics 21 27. Concurring and Conspiring Forces 21 28. Composition oi Conspiiring Forces 22 29. Composition of Velocities 23 30. Composition of Forces 24 81. Triangle of Forces. 25 82. Three Concurring Forces in Equilibrium 26 83. The Polygon of Forces 27 84 Paralleloplped of Forces 28 85. Resolution of Forces 80 86. Magnitude and Direction of Resultant 81 87. Conditions of Equilibrium 32 88. Resultant of Concurring Forces in Space 84 89. Equilibrium of Concurring Forces in Space 85 40. Tenaion of a String 85 41. Equilibrium of Concurring Forces on a Smooth Plane 39 42. Equilibrium of Concurring Forces on a Smooth Surface 41 Examples 45 CHAPTER III. COMPOSITION AND BKSOLCTION OF FOBCES ACTING ON A RIGID BODY. 48. ARlgldBody 57 44. Transmissibility of Force 57 46. Resultant of Two Parallel Forces 58 46. Moment of a Force flo '.*!a3i-Bia»»»»u.im.'. -.v\aiifevj^v.a».£?i^>&":'''■ " 91. Friction 149 92. Laws of Friction 150 93. Magnitudes of Coefflcionts of Friction i( i^ 94. Angle of Friction ^^.^ 95. Reaction of a Rougli Curve or Surface 163 96. Friction on an Inclined Plane 154 97. I'riction or. a Doable Inclined Plane 156 98. Friction on Two Inclined Planes. 159 99. Friction of a Trunnion 159 100. Friction of a Pivot .,,, y-Q Examples 192 CHAPTER VI. THE PRINCIPLE OF VIRTUAL VELOCITIE& 101. Virtual Velocity igg 102. Principle of Virtual Velocities 167 103. Nature of the Displacement 169 104. Equation of Virtual Momenta '. , 169 105. System of Particles Rigidly Connected 170 Examples 172 120 122 123 126 127 130 181 183 136 188 140 149 J50 1/-: ib'a 153 154 156 159 159 ICO 162 coifTEyrs. CHAPTER VII. . "■■ „ MACHINES. „ %' u». - ■■ - ' -"'■•»" 106. Functions of a Machine 177 107. Mechanical / Jvantage 178 108. Simple Machinee , • 1^0 109. Th, i* i i« i i;i. 'i . y.L . ji •n- it'yjr! >f'ft i; Ti^i''viT-r'yKfr'!! mi ' ^.,J V i ' -!tf 'g ^ '! '^r>l !?*P^'7'iy coAnuvTS. PART II. KINEMATICS (N.OTION). CHAPTER I. RECTILINEAB MOTION. ABT. r^ag 184. DefinltlonB— Velocity 281 185. Acceleration 288 188. Relation lKtwe«n Space and Time when Aooeleration = 0.. . 288 187. Relation when the Acceleration is Constant 284 188. Relation when Acceleration Tariea as the Time 235 189. Relation waen Acnolenttion Vfrico ai> the Distance 285 140. Ekiuations of Motion for Falling Bodies. 287 141. Particle Piv^jected Vertically Upwards 280 142. Oompoeitions of Velodtias 242 148. Resolution of Velodtiea. 248 144 Motion on an Inclined Plane 245 146. Times of Descent down <'^ords of a. Circle 247 146. The atraight Line of Quickest Descent 248 Examples. , 249 CHAPTER II CCEVItlNEAE MOTHyN. 147. Remarks on Curvilinear Motion 258 J48. Composition of Unifo.m Velocity and Acceleration 258 140. C3m position and Resolution of Acceleration 259 Examples 2fli 160. Motion «♦ Project llos '•» Vacuo 266 161. The Path of a Particle in Vacuo ?flnition8 321 181. A Particle under the Action of a Central Attraction 821 182. The Sectorial .\rea Swept over by the Radius Vector 825 183. Velocity of Particle at any Point of it« Oruit 826 184. Orbit when Attraction as the Inverse Square of Distance. . . 32 J 186. Suppoee the Orbit to be an Ellipee 833 186. Kepler's L»w8 335 187. Nature of the Force which acta upon the Planetary System. 885 Kxampies .... 338 CHAPTER III. CONSTRAIKED MOTIOK 188. Deflnitions 848 189. Kinetic Energy or Vis Viva— Work 345 190. To Find the Heaetion of the Constraining Curve 848 101 . Point where Particlj will leave Constri ining Curve 849 192. Constrained Motion Under Action of Oraviiy 350 108. Motion on a Circular Arc in a Vertical Plane 850 194. The Simple Pendulum 353 196. Relation of Time, Length, and Force of OravHy 858 196. Height of Mountain Determined with Pendulum ;t64 197. Depth of Mine Determined with Pendulum 866 198. Centripetal and Centrifugal Forces 866 199. The Centrifugal Force at the Equator 058 200. Centrifugal Force at Diffcn>nt Latitudes 869 901. The Conical Pendiilnm— The Uovemor 861 Examples 992 CHAPTER IV, IMPACT. 209. An Impulsive Porce 370 208. Impact i)r Collision 371 204. Direct and Central Im|4u;t 872 206. Elaaticity of Bodies— Coefficient of Restitution 878 ABT. 20(5. ','07 208. 2()9. 210. 211. 212 213. 214. 215. 210. ] ] 217. 1 1 218. I 219. i 230. \ 231. 1 222. > 238. 6 I 224 1 226. 8 236. 237. 238. 239. 230. I 231. J iilif I ^ CONTSNTa, xm ... 821 Ion S21 ector 825 826 Distance. . . 32 J 888 385 aiy System . 885 848 845 848 irvo 849 830 850 , 858 858 , :i54 355 856 aw 859 861 870 871 872 878 ABT. rAOB 30«. Direct Impact of Inelastic Bodies 374 '207. Direct impact of Eliifltic Bodien 875 -m. Loss of Kinetic Energy in Impact of Bodies 878 2()9. Oblique Im|jact of Bodies 880 210. Oblique Impact of Two Smooth Spheres 883 Examples 888 CHAPTER V. WOEK AND ENERGY. Definition and Meaanre of Work 880 General Case of Worli done by a Force l>90 Work on an Inclined Plane 891 Examples 808 Horse Power 895 Work of Raising a System of Weights ?0« Examples 8i»7 Modulus of a Machine 400 Examples 401 Kinetic and Potential Energy— Stored Work 404 Examples 406 Kinetic Energy of a Rigid Body Revolving roond an Axis. . . 403 Force of a Blow 411 Work of a Water Fall 418 The Duty of an Engine 414 Wr.k of a Variable Force 415 Sitiipeon's Rule 415 Exampltw 417 211. 212 213. 214. 215. 210. 217. 218. 219. 230. 231. 223. 233. 234. 225. 226. 227. 238. 229. 230. 231. CHAPTEK VI. MOMENT UF INERTIA. Moment of In»«rtia 480 Moments of Inertia relative to Paral'pl Axes or Planes 438 Radius of Gyration 434 Polar Moment of Inertia 4ii6 Moment of Inertia of a Solid of Revolution 487 Moment of Inertia aliout A: is Perpendicular to Geometric Axis 438 Moment of Inertia of Various Solid Bodies 440 Moment of Inertia of a Lamina with respect to any Axis. . . . 441 ' X-f. ;jy-"'j;i ' v4^; f? g ' ^svj? ! gy ' l^w'^:^M^ l ■fti.^l jija^f[ ^ ^4»^ ,i, \ UAm,v: ' ^> !WSSS^^ 1^ lir CONTMNTB. 232. Principal Axes of E Body.... 448 2S3. Products of Inortia ■■ 446 Examples 447 ^ Hi CHAPTER VII. ROTATORY MOTION. 284. Imprefwed and Effective Forces 461 285. D'Alembert'B Principle 453 286. RoUtion of a Rigid Body about a Fixed Axis 454 287. The Compound Pendulum 457 28M. Length of Second's Pendulum Determined Experimentally. . 462 289. Motion of a Body when Unconstrained 4t?4 240. Centre of Percussion— Axis of Spontaneous Rotation 404 241. Principal Radius of Gyration Determined Practically 467 242. The Ballistic Pendulum 468 248. Motion of a Body about a Horizontel Axle through its Centre 470 244. Motion of a Wheel and Axle 471 245. Motion of a Rigid Body about a Vertical Axis 472 246. Body Rolling down an ..uclined Plane 478 247. Falling Body under an Impulse not through its Centre 475 Examples 477 CHAPTER VIII. MOTION OF A SYSTEM OF RIGID BODIES IN BPAOB. 248. Equations of Motion obtfliner how it is by introduo- f velocity. aerceived by ;ed upon by etrability. ttaphyo'.cian Is . uceive at it u I it is in this iae. jerty of mat- ; or uniform rce. Inertia If change its it if a body otion to rest, il rectilinear temal cause. a portion of uently of a positions of forces. |ry direction, Ly be treated limited por- rhen it con- body is in motion when the body or its parts occupy successively dif- furent positions in space. But we cannot judge of the stute of rest or motion of a body without referring it to the ]K)iiition8 of other bodies ; and hence rest and motion must be considered as necessarily relative. If there were anything which we knew to be absolately fixed in space, we mi^t perceira aiieolate motion by change of place with rcfertince to that object. Bat as we know of no such thing as also- lute rest, it follows that all motion, as measured 1>t ub, must be rolatiye ;%.»., most relate to something which we assume to be fixed. Hence the same thing may often I>e itaid to be at rest and in motion at the same time : for it may be at rest in regard to one thing, anJ. in niution in regard to another. For example, the objects on a vessel may be at rest with reference to each other and to the voseel, while they are in motion with reference to the neighboring shore. So a man, I>unting his barge up the river, by leaning against a pole which rests on the bottom, and walking on the deck, is In motion relative to the barge, and in motion, but in a different manner, relative to the car- rui}t, while he is at rest relative to the earth. Motion is uniform when the body passes over equal spaces iu equal times ; otherwise it is variable. 6. Velocity. — The velocity of a body is its rate of motion. When the velocity ic eonstant, it is measured by the space pas^-ed over in a unit of time. When it is varia- ble, it is measured, at any instant, by the space over which the* body would pass in a unit of time, were it to move, during that unit, with the same velocity that it has at the instant considered. The speed of a railway train is, in general, variable. If we were to Bay, for example, that it was running at the rate of 80 miles an hour, W6 would not mean that it ran 80 miles daring the last hour, nor that it would run 80 miles daring the next hour. We would mean that, if it were to ruu for an hour with the speed which it now has, at the instant considered, it would pass over exactly 80 miles. I In order to have a uniform unit of velocity, it is custom- ary to express it in feet -! VMLOCtTT AND AOCELaBATtOH, i/P rs <; a a HJJ Si" and 2/8 = v«, (5) (6) which determine the velocity and space. 10. Gtoometrio Reprosentatioii of Valoeity and Acceleration.— The velocity of a body may be conveni- ently represented geometrically in magnitude and direction by means of a straight line. Let the line be drawn from the point at which the motion is considered, and in the direction of motion at that point With a convienient scale, let a length of the line be cat off that shall contain as many units of length as there are units in the velocity to be repre- sented. The direction of this line vnll represent the direction of the motion, and its length will represent the velocity. Also an acceleration may be represented geometrically by a straight line drawn in the direction of the velocity generated, and containing as many units of length as there are units of acceleration in the acceleration considered. Also, since an acceleration is measured by the actual increase of velocity in the unit of time, the straight line which represents an acceleration in magnitude and dii*ec- tion will also completely represent the velocity genei-atcd in the unit of time to which the acceleration corresponds. 11. The Mass of a body or particle is the quantity of matter which it contains; and is proportional to the Volume and Density jointly. The Density may therefore be defined as the quantity of m»t>«r in a unit of volume. Let ilf be the mass, p the density, and V the volume, of a homogeneous body. Then we have M^ Vp, (1) if we so take our units that the unit of mass is the mass of the unit volume of a body of unit density. -■i^-VAiia: %. J.- •iv. ▼aloeity and lay be conveui- le and diroction be drawn from red, and in the ionvenient scale, contain as many Mjity to be repre- reprosent the 11 represent the ^metrically by of the velocity lengtli aa there ;ion considered. by the actual he straight line Itude and direc- ity genei-ated iu orrespoud*. the quantity of >rtional to the \y may therefore ►it of volume, the volume, of (1) is the muss of MOMKNTUM. 7 If the density varies from point to point of the body, we liave, by the above formula, and the notation of the Integral Calculus, 3f=/odV = f/fpdx dy dt, (2) wlicre p is supposed to be a known function of x, y, t. In England the unit of mass is the imperial standard l)ound avoirdupois, which is the weight of a certain piece of platinum preserved at the standard office in London. On the continent of Europe the unit of mass is the gramme. This is Icnown as the absolute or kinetic unit of mass. 12. The Quantity of Motion,* or the Momentum of a body moving without rotation is the product of its mass and velocity. A double mass, or a double velocity, would correspond to a double quantity of motion, and so on. Hence, if we take as the unit of momentum the mo- mentum of the unit of mass moving with the unit of velocity, the momentum of a mass M moving with velocity r is Mv. 13. Change of Quantity of Motion, or Change of Momentum, is proportional to the mas^ moving and the change of its velocity jointly. If then the mass remains constant the change of momentum is measured by the product of the moss into the change of velocity ; and the rale of change of momentum, or acceleration of momentum, is measured by the product of the mass moving and the rate of change of velocity, that is, by the product if the mass moving and the acceleration (Art. 9). Thus, ct lling J/ the mass, we have for the measure of the rale of change of momentum, rd*« M Ifl' * This phtwe WM a««d bj Newton in place of Uie more modern term " Momen- mm." 8 STATIC MSAaURS OF FORCE. 14. Force. — Force is any cause which changes, or tends to change, a body's state of rest or motion. A force always tends to produce motion, but may be pre- vented from actually producing it by the counteraction of an equal and opposite force. Several forces may so act on a body as to neutralize each other. When a body remains at rest, though actcu on by forces, it is said to be in equilibriam ; or, in other words, the forces are said to produce equilibrium. What force is, in its nature, we do not know. Forces are known to us only by their effects. In order to measure them we must compare the effects which they produce under the same circumstances. 15. Static Measnre of Force.— 7%e effect of a force depends on: Ist, its magnitude, or intensity ; 2d, its direc- tion; i. e., the direction in which it tends to move the body on which it acts ; and 3d, its point of application .; t. «., the point at which the force is applied. The effect of a force is pressure, and may be expressed by the weight which will counteract it. Every force, statically considered, is a pressure, and hence ha« magnitude, and may be measured. A force may produce motion or not, according as the body on which it acts is or is not free to move. For example, take the case of a body restiug on a table. The ame force which produces pressure on the table would cause the body to fall toward the earth if the table were removed. The cause of this pressure or motion is gravity, or the force of attraction in the earth. In the first case the attrac- tion of the earth produces a pressure; in the second case it produces motion. Now either of these, viz., the pressure which the body exerts when at rest, or the quantity of motion it produces in a unit of time, may be taken as a means of measuring the magnitude of the force of attrac- tion that the earth exerts on the body. The former is S^' V^'/il^^-^t,,^*i":i'«V#i4^,"^^*^.-^-S^:''~<^aSi{^S\ «* hanges, or tends but may be pre- counteraction of 38 may bo act on a body remains is said to be in rces are said to t know. Forcea )rder to measure h thoy produce effect of a force y; 2d, its direc- to move the body ication ,; i. e., tbo y be expressed by Y force, statically magnitude, and B motion or not, i or is not free to >dy resting on a pressure od the the earth if the ~ is graTity, or the ]t case the attrac- e second case it iz., the pressure the quantity of ay be taken as a le force of attrac- The former is MBTffOD OF COMPABiya FOttCKS, called the static method, and tUe forces ftrc called ttatie farces; the latter is called the kinetic method, and the forces are called kinetic forces. Weight is the name given to the pressure which the attraction of the earth causes a l)o(ly to exert Hence, since static forces produce pressure, we may take, as the unit of force, a pressure of one pound (Art. 11). Therefore, the magnitude of a force may be measured ainfically hy the pressure it will produce "t/on some body, ami expressed in pounds. This is called the Static measure of force, and its unit, one pound, is called the Oravitation unit of force. 16. Action and Roactioii are always equal and opposite. — This is a law of nature, and our knowledge cf it comes from experience. If a force act on a body hold by a fixed obstacle, the latter will oppose an equal and con- trary rasistance. If the force act on a body free to move, motion ¥rill ensue ; and, in the act of movitig, the inertia of the body will oppose an equal and contrary resistance. If we press a stone with the hand, the stone presses the liuud in return. If we stiike it, we receive a blow by the act of giving one. If we urge it so as to give it motion, we lose some of the motion which we should give to our limba In the same effort, if the stone did not impede them. In euch of these cases there is a reaction of the same kind as tlic action, and equal ♦^" it. 17. Method of Comparing Forces.— Two forces are orjiial when being applied iu opposite directions to a particle they maintain equilibrium. If we take two ^qual forces, and apply them to a particle in the same direction, wo obtain a force double of either ; if we unite three equal foices we obtain a triple force ; and so on. So that, in general, to compare or measure forces, we hare only to K^iopt the same method as when we compare or measure JO REPRBSENTATIOH OP P0SCX8. any qnantities of the same kind ; that is, we maet take some known force as the unit of force, and then express, in numbers, the relation which the other forces bear to this measuring unit. For example, if one pound be the unit of force (Art. 16), a force of 12 pounds is expressed by 12 ; and so on. 18. Rspresentatioii of Forces by SymbolB and IJneg. — If p. Q. B., etc., represent forces, they are numl)ers expressing the numl)er of times which the concrete unit of force is contained in the given forces. Forces may be represented geomehrically by right lines ; and this mode of reprci^icntation has the advantage of giving the direction, magnitude, and point of application of each force. Thus, draw a line in the direction of the given force ; then, having selected a unit o2 length, such as an inch, a foot, etc., measure on this line as many units of length as the given force contains units of W3ight The marinittide of the force is represented by the measureti length of the line ; its direction by the direction in which the line is drawn; and \t& point of application by the point from which the line is drawn.* Thus, let the force Pact at the point * ^ A, in the direction AB, and let AB ^'■- '• represent us many units of length as P contains units of force; then tlio force P is represented geometrically by the line AB ; for the forti act? in the direction from A to B ; its point of application io at A, and its magnitude iii represented by the length of the line AB. 19. Measure of Acceleratitig Foices. — From our definition of force (Art. 14), it is clear that, when a singk • Forcof", velocities, and accclnratlotiH are direeltd (ptantlUfCy p.nd m majr be roprewntpd by a line, in direction and magnitnde, and may b« coinpoand«d In the Mmu way a* vtduri. If auyUilus bu maRnHude and direction, Uie magoltode and dir«ctlon taken togetlier constitato a r«clor. res. is, we mast take I then express, in orces bear to this nd he the unit of expreasod by 12; Symbols and they are numl)ers J concrete unit of ly by right lines; ivantage of giving pplication of each tiou of the given ength, ftuch as an as many units of . of W3ight The by the measureti ircction in which ation by the point t * « J Fi9. I. ' contains units of goonictrically by direction from A its magnitude Ia xces. — From our lat, when a singlti uantUitr, Rnd ro tuKj b* be compoonded In the tade and diroctton Ukou MBASURS OF AOCKLSSATmO FOSOXS. 11 iorce acts upon a particle, perfectly free to move, it must pioduce motion ; and hence the force may be represented to us by the motion it has produced. But motion is measured in terms of velocity (Art. 7), and consequently the \eiocity communicated to, or impressed upon, a particle, in a given time, may be taken as a measure of the force. That is, if the same particle moves along a right line so that its velocity is inorea/jed at a constant rate, i-"" v/ill be acted upon by a constant force. If a certain uonstani force, acting for a second on a given particle, generate a velocity (if 32.2 feet per second, a douWo force, acting ibr one second on the same particle, would generate a velocity of 'JiA feet per second ; a triple force would generate a V elocity of 96.0 feet pei* second, and so on. If the rate of increase of the velocity, (t. e., the accelera- tion), of the particle is not uniform, the force acting on it if, not nniform, and the magnitude of the force, at any imnt of the particle's path, is measured by the acceleration of the pari.icle at this point Hence, sine* one and the .«imo particle is capable of moving with all possible accelera- tions, all forces may be measured by the velocities they (jennrate in the same or equal particles in the same or equal times. When forces are so measured they are called Accelerating Forces. 20. Kinetic or Abftolute Measnre of Forea.*— Let n equal particles be placed side by side, and let each of them lie acted on uniformly for the same time, by the same force. Each particle, at the end of this time, will have the same velocity. Now if these n separate particles are all united so as to form a body of n times the mass of each particle, and If each one of them is still acted on by the same 'orce as * Arta. M, n, tt, and K, treat of Um Kinetle moarare of force, and may be omitted till Part til la reached ; but It ii e sams velocity that each aeporate partiule had, and will \ i cted on by n times the force which generated this velocity in the particle. Comparing a single particle, then, with tlxe body whose maw is n times the mass of this particle, we see that, to produce the same velocity in two bodies by forces acting on them for the same time, the maf^nitudes of the forces must be proportional to the masses on which they act.* Hence, generally, since force vaiies as the velocity when the mass is constant (Art 10), and varies as the mass when the velocity is constant, we have, by the ordinary law of proportion, when both are changed, force varies as the product of the mass acted npon and the velocity generated in a given time ; that is, it varies OS the quantity of motion (Art 13) it produces in a given mass in a given time. If the force bo variable, the rate of change of velocity is variable (Art 19), and hence the force varies as the product of the cuiss on which it aots and the rtUe of chanyi of velocity, ». e., it varies as the acceleration of the momentum (Art U). Therefore, if any force P act on a mass M, wo have (Art 10) Pn a nnit mass at any latitude A; lolute unit, being id Mass.— If in W of the body, sleration i^ g, (3) m m of the mass. ^tred by the pres- rce is one pound 'My of matter in where the aceel- 1 is oonstant at '; i^e.iiU weight t of matisr in it would weigh at iji Leith Fort it mass of a body ' matter in Ibis place to place, planet its mass OS A vrrATtON MMAStr/fB Of PORCK. 17 would not be altered, but its fimght wonM be very different. Its weight wherever placed would vary directly m the force of gravity ; but the aoeeleratioH afeo would vary diivotly as tlie force of gravity. If placed on the sub, for example, it would weigh about 5J8 tiraca acr macfe m ©b the surface of the earth j but the aeceleration on tht ssn would also b« 28 times as much ae on the surface of th© earth ; that i% the ratio of the weight to^ the acceleration, anywhere in W the universe is constant, and henoe — , which is the 9 numerical value of m (Bq. 2), is oonstuit for tiw same mass at all places. 25. Compsiisoa of Gteavitatioii and Absolnta Maasnra.— The pound weight has been long used for the measurement of force instead of mass, and i»the recognized standard of reference. It came into general use because it afforded the moat ready and simple method of estimating forces. The pressure of steam in a boiler is always reck- oned in pounds per square inch. The tension of a string is estimated in pounds; the force necessary to draw a train of cars, or the pressure of water against a look-gate, is fxnivBsed in pounds. Such expressions as "a force of 10 pounds,^" or " a pressure of steam equal to 50 pounds on the inch," are of every day occurrence. Therefore this method of measuring forces is eminently convenient in practice. For this reason, and because it is the one used by most engineers and writers of mechanics, we shall adopt .it in this work, and adhere to the measurement of force by pounds, and give all our results in the usual gravitation measme. In this measure it is convenient to represent the W mass of a body weighing W pounds by the fraction — (Art 1^4), so that (3) of Ari 20 becomes P = If 0) 18 BXAMFLES. To do so it will only be necessary to assnme that the unit of mass is the quantity of matter in a body weighing // pounds, and changes in weight in the same proportion tbut g changes (Ai-t. 24). Of course, the units of mass and force in (a) of Art. 30 may be either absolute or gravitation units. U absolute, the unit of mass is one pound (Art. 12), and the unit of force is - pounds (Art. 21). If gravitation, the units arc g times as great; i. «., the unit of mass is g pounds (Art. 24), and the unit of foroe is one pound (Art 16). The advantage of the gravitation measure is, it enables ns to express the force in pounds, and furnishes us with a con- stant numerical representative for the same quantity of matter ; that is to say, a mass orepresented by 20 ou the equator would be represented by 20, at the pole or on the sun. Hence, in (1), P is the static measure of any moving force [Art 22, (3)], W is the weight of the body in pounds, g the acceleration of gravity (Art. 24), — the-mass upon which the force acts [(2) of Art 24], and which is frf.e to move under the action of P, the unit of mass being the mass veighing g pounds, and / the acceleration which the force P produces in the mass. EXA.MPLBS * 1. Compare the velocities of two points which move uniformly, one through 5 feet in half a second, and the other through 100 yards in a minute. Ans. As 2 is to 1. 2. Compare the velocities of two points which move uni- formly, one through 720 feet in one minute, and the other through 3Jf yards in three-quarters of a second. Ans. As 6 is to 7. 3. A railway train travels 100 miles in 2 hours ; find the average velocity in feet per second. Ans. 73^. inme that the unit i body weighing // ae proportion tbut B in (3) of Art. 20 mita. If absolute, i), and the unit of Hon, the units are is g pounds (Art. A.rt 16). are is, it enables ns shes us with a con- same quantity of ted by 30 on the at the pole or on static measure of the weight of the gravity (Art. 24), i) of Art. 24], and of P, the unit of unds, and / the in the mass. oints which move a second, and the ns. As 2 is to 1. s which move uni- iite, and the other econd. ns. As 6 is to 7. in 2 hours ; find Ans. 73f ItXAMPLBS. 19 4. One point moves nnjfonnly round the circumference of a circle, while another point moves uniformly along the diameter ; compare th^ir velocities. Ahs. As tr is to 1. 5. Supposing the earth to be a sphere 26000 miles in circumference, and turning round once in a day, deter- mine the velocity of a point at the equator. Ans. 1527| ft. per sec. 6. A body has described 60 feet from rest in 2 second^ with uniform acceleration ; find the velocity acquired. From (1) of Art. 9 we have and from (4) we have /< = v ; .'. V = 60. 7. Find the time it will take the body in the last exam- ple to move over the next 160 feet. From (6) of Art. 9 we have etc Ans. 2 seconds. 8. A body, moving with uniform acceleration, describes 63 feet in the fourth second ; find the acceleration. Ans. 18. 9. A body, with uniform acceleration, described 72 feet while its velocity increases irom 16 to 20 feet per second ; find the whole time of motion, and the acceleration. Ans. 20 seconds ; 1. 10. A body, in passing over 9 feet with uniform accelera- tion, has its velocity increased from 4 to 5 fe *: per second ; find the whole space described from rcsut, and thie accelera- tion. Ans. 26 feet ; |. 20 MXAMPLMa. 11. A body, aniformly accelerated, is found to be mov- ing at the end of 10 seconds with a velocity which, if continued uniformly, would caiJ7 it through 46 mites in the next hour ; find the acceleration. Am. ^. 12. Find the matis of a straight wire or rod, the d^naity of which varies directly as the distance fh>m one end. Take the end of the rod aa origin ; let o = its length ; let the ''aetance of ary point of it from that end = a; ; and let u = the area of its transverse section, and k = the density at the uxut'g distance from the origin. Thea dV=i p. 7, also tl in magni- ralUhgrmn, 'nd direction «st be taken »t the com- ogram fcom It WM «nun- t'^ariguon, the dependent of Ben given by , more or less 1» given by ithon object •nllelogram student who >and in Tod- ftnd in niaoy TMiANOhE or woacsa. other works ; or to UpUoe-s pnxrf, (Sm M^canicine Celeste, Liv. I, chap. 1.) If fl be the angle between the sides of the parallelogram, AJi and AC (Fig. 2), and P and Q represent the two com- ponent forces acting at A, and R represent the resultant, AD, we have from trigonometry, iZ» = i« -f. ^ -H ^PQ cos 9 m an equation which gives the magnitude of the resultant of two forf es in terms of the magnitudes of the two forces and the angle between their directions, the forces being repre- sented by two lines, both dra\ni from the point at which they act. Cob. —If e = 90^ and « and be the angles which the direction of R makes with the directions of P and Q, we have from (1) C0B« = COS/3 = ^; (») (K) from which the magnitude and direction of the resultant are determined. 31. Triangte of Tonm.—If three oonourring forces be repreaenied in magnitude and direoiicn hij the sides of a triangle, taken in order, they will be in equilibrium. Let ABO be the triangle who«" sides, taken in order, represent in magnitude and direction three foroes applied at the point A. Complete % n*>s S6 TBlAJiGLM OF FOBCXS. the pawllelogram ABCD. Then the forces, AB and EC, applied at A, are expressed by AB and AD (since AD is eJ:^«< .ns iMtwosn ThrM Oonewring ForoM in Tic i^■ vm.— Siuce the sides of a plane triangle are as the Biuo' >/ iie opposite angles, wo have (Fig. 8) AB : BC (or AD) : AC : : sin ACB : sin BAC : sin ABC : : sin DAC : sin BAC : sin BAD. Hence, calling P, Q, and R, the forces represented by AB, AD, and AC, and denoting the angles between the direo- lu. mm !8,ABand BC, D (since AD is tant of AB and irefore the three e eqaivaleut to Tom A towards being equal and Therefore t!ie i, acting at the represents the it is not in the ces act at the hree concnrrin/-j sent/nl in m^g- triangle, drawn forces. 8 in magnitude it, and henre to nust be opposed (fore, the three )y AB, BO, and in equilibrium, reeuluuit of the irrinf ForoM me triangle are (Fig. 8) AC: AC sin ABO sin BAD. aseuted by AB, woen the direo- POLTQON OP FORCSa. tions of the forces P and Q, Q and R, and R and P, by AAA PQ> QP> and RP, respectively, we have P ^ _Q R . /^ /\ A em QR sin RP sin PQ 0) There/ore, when three concurring forces are in equilibrium they are respectively in the same proportion as the sines of the angles included between the directions of the other two. 33. The Polygon of Forces.— // amj number of concurring forces be represented in magnitude and. direction by the sides of a closed polygon taken in o: der, they will be in equilibriurrv. Let the forces be represented in It magnitude and direction by the lines AP„ AP„ AP„' AP„ AP,. Take AB to represent AP,, through B draw BC equal and parallel to AP, ; the resultant of the forces AB and BC, or AP, and AP, is represented bj AO (Art. 31). Of courae the resultant acts at A and is parallel to BC. Again through draw CD equal and parallel to AP„ the resultant of AC and CD, or AP,, AP„ and AP, is AD. Also through D draw DE equal and parallel to AP^, the resultant of AD and DE, or AP,, APg, APj, and AP^ is AE. Now if AE is equal and opposite to AP, the system is in equilibrium (Art. 18). Hence the fos-ces represented by AB, BC, CD, DE, EA wiU be in equilibrium. Cob, 1. — Any one side of the polygon represents in magnitude and direction the resultant of all *he forces represented by the remaining sides. Cob. 2.— If the lines representing the forces do not form a closed polygon the forces are not in eqoilibrium ; in this »8 PARALLMI'OPIPED OP P0K0S8. case the last side, AE, taken from A to E, or that which is required to clope up the polygon, represents in magnitude and direction the resultant of the system. 34. Farallelopijied of Porcea.— // three concur- ring forces, not ih. the sattve plane, are represented in magnitude and direction by the three edges of a parallelopiped, then the resultant will be repre- sented in magnitude and direction by the diag- onal; conversely, if the diagonal of a parallel- opipcd represe.its a force, it is equivalent to three forces represented by the edges of the parallci- opiped. Let the three edges AB, AC, AD of the parallolopiped represent the three forces, applied at A. Then the resultant of the forces AB and AC is AE, the diagonal of the face ABCE; and the resultant of the for'^os 4^ and AD is AF, the diagonal of the parallelogram ADFE. Hence AF represents the resultant of the three forces AB, AC, and AD. Conversely, the force, AF, is equivalent to the three components AB, AC, and AD. Lot F, Q, S represent the three forces AB, AC, AD ; R, the resultant ; a, P, y, the angles whicL the direction of B makes with the directions of P, Q, S, and suppose the forces to act at right angles with each other. Then aiuce Sr» = XB* + AC» + AD', we have R^ = P* + (? + 3*; , P alio, cos o = -gi Q cos /3 = ^, 8 (1) m ta, \T that which is I in magnitude Ihree concur- s represented tree edges of vill be repre- by the diag- n parullel- Zent to three 'ihe parallci- represents the D. ; to the three J, AC, AD; R, direction of R kd suppose the Then aiuce (1) (a) KXAMPtiiS. »» from which the magnitude and direction of the resultant are determined. EXAMPLES. 1. Three forces of 5 lbs., 3 lbs., and 2 lbs., respectively, act upon a point in the same direction, and two other forces of 8 lbs. and 9 lbs. act in the opposite direction. What single force will keep the point at rest P Atts. 7'lbs. a. Two forces of 5| lbs. and 3| lbs., applied at a point, urge it in one dire^'^ion ; and a force of 2 lbs., applied at the same point, urges it in the opposite direction. What additional force is necessary to preserve equilibrium ? Ans. t lbs. 3. If a force of 13 lbs. be represented by a line of 6^ inches, what line nWX represent a force of 7^ lbs.? Am. 8f inches. 4. Two forces whose magnitudes are as 3 to 4, acting on a point at right angles to each other, produce a rcuultaut of 20 lbs.; required the component forces. Am. 12 lbs. and 16 lbs. 5. Let ABO be a triangle, and D the middle point of the side BC. If the three forces represented in magnitude and direction by AB, AO, and AD, act upon the point A ; And the direction and magnitude of the rennltant Ans. The direction is in the line AD, and the magni- tude is represented by SAD. 6. When P = Q and fi = 60°, find R. Ana. R = PVS. 7. When P = Q and 8 = 135°, find R. Ans. R- P^'z — V^ 8. When P = ^ and = 120°, find R. Ans. R = P. 30 sssoLurrox or roscss. \ '■ 9. If P =: Q, show that their resultant R — ^P cos ?• 10. If P = 8, and e = 10, and (? = 60°, find R. Ana. R z=2 V^l. 'll. If P = 144, R = 145, and 6 = 90°, find Q. Ans. Q = 17. 12. Two forces of 4 lbs. and 3 V2 lbs. act at an angle of 45°, and a third force of V42 lbs. acts at right angles to their plane at the same point ; find their resultant. Ahs. 10 lbs. 35. Resolntion of Forces.— % the resohition of forces is meant the process of finding tlie components of given forces. We have seen (Art 30) that two concurring forces, P and C = AB and AC, (Pig. 2) are equivalent to a single force 72 = AD ; it is evident then that the single force, R, ucnng along AD, can be replaced by the two forces, P and Q, represented in magnitude and direction by two adjacent sides of a parallelogram, of which AD is the diagonal. Since an infinite nnmber of pamllelogrnnis, of oac'i of which AD is the diagonal, can be constructed, it follows that a single force, R, can be resolved into two other forces iu an infinite number of ways. Also, efl 3h of i-ie forces AB, AC, may be resolved into two others in way similar to that by which ID was resolved .nto twt ; and so on to any extent. Hence, a single force nray be resolved into any number of forces, whose oombinod action is equivalent to the original force. Cor. — The most convenient compo- nent into which a force can be resolved are those whose directions are at right angles to eacli other. Thus, let OX and OF be any two lines at right ujiglea to each other, and P any force acting at in the Fia.a i = 2P cos J 9°, find R. E — 2 V'ai. \ find Q. ns. Q = 17. at an angle of right angles to 111 taut. Atis. 10 lbs. )hition of forces of given forces, l forces, P and a single force force, it, acting rcos, P and Q, ' two adjacent diagonal, ms, of oac'i of :ted, it follows wo other forces e resolved into vhich AD was Hence, a single forces, whose force. a ig at in the MAomrmiS and DinEcrtoir or bssultant. plane XOY. Then completing the rectangle OMPN we find the ocnponenta of P along the axes OX and OF to be Oif and ON, which denote by X and Y. Then we have clearly X = P cos «, ) /jv r = P sin «; ) ^ ' where a is the angle which the direction of P makes with OX. These wmponents X and Y are called the rect- angular components. The i-ectangular component of a force, P, along a right lino is Px cosine of angle between^ line and direction of P. In strictness, when we speak of the component of a given force along a certain line, it is necessary to mention the other line along which the other component acts. In thia work, unless otherwise expressed, the component of a force along any line will be understood to-be its rectangular component; le., the resolution will be made along this line and the line perpendicular to it. 36. To find th« lAagnitode and Direction of the Resultant of any number of Concurring Forces in one Plane.— When there are several concurring forces, the condition of their equilibrium may be expressed as in Art. 33, Cors. 1 and 2. But in practice we obtain much simpler results by using the principle of the Resolutwn of Forces (Art -35), than those given by the principle of Composition of Forces. Let be the point at which all the forces act. Through draw the rectangular axes XX', YY'. I^t P,, P„ P,, etc., be the forces and «,, «„ a J, etc., be the angles which their directions make with the axis of*. • Now resolve each force into its two components along the axes of x and Fig.7 Then the com- 88 MAomruDJi avd otRscnoN of sxauLTAjrr. f L ponenta along the axia of x (aHwmponents) ore (Art 35, Cor.), P, COS a,, P, cos a,, P^ cos «,, etc., and thoae along the axis of y are Pj sin ct|, P, sin «„ P, ain «„ etc. ; and therefore if X and Y denote the algebraic sum of the ^-components and ^-components respectively, we have X= P, cos cr^+P, cos «,4-P, cosa, 4-etc. ) . = 2P coa «, ) ^ ' y = P, sin cci + Pi sin a, + Pj ain a, +etc. =r iP sin a. (2) Let R be the resultant of all the forces acting at 0, and the angle which it makes with the axis of x ; then resolving R into its x- and ^-components, we have i2 COS e = -T = £P coa o, 5 si- 3 3= y =s £P ain a, '} /? = X»+r»; tane = -=. (8) which determines the magnitude and direction of the resultant. ScH. — Regarding OX and OF as positive and OX^ and OY^ as negative as in Anal. Geom., we see that Oa;,, Oy^, Oy, are positive, and Oa;,, Oa;„ Oy, are negative. The forces may always be considered as positive, and hence the signs of the components in (1) and (2) will be the same as those of the trigonometric functions. Thus, since a, is > 90° and < 180° its sine is positive and cosine is negative; since a, is > 180° and < 270° both its sine and cosine are negative. 37. Tha Conditions of Eqnlllbrtam for any nnmbor of ConcTuring ForcM in one Plane.— For the equilibrinm of the forces we must have R = 0. Hence (4) of Art. 36 becomes X» -H r» = 0. (I) Urn VLVAHT, s) ore (Art 3tc., and thoae Kg, P, ain «„ robraic sum of ely, we have 4- eta) (1) +eto. } <^\ g at 0, and 9 then resolving (3) ■', (4) lOtion of the and OX^ and hat Oa;,, Oy„ jgative. The md hence the ) the same as B, since a, is ae is negative; nd cosine are •nynnmbcr tie eqnilibrinm 4) of Art 36 (1) BXAMPLMB. 88 Now (1) cannot be satigfled so long 'as X and F are real quantities unleae X=:0, F.=:a; therefore, X=:£i'co8« = and F=SPsino = 0. (2) Hence these are the two necessary and sufficient conditions for the equilibrium of the forces; that is, the algebraic sum of the rectangular components of the forces, along each of two right lines at right angles to each other, in the plane of the forces, is eqwA to zero. As the conditions of equilibrium must be independent of the system of co-ordinate axes, it follows that, if any number of concurring forces in one plane are in equilibrium, the algebraic sum of the rectan- gular components of the forces along every right line in their plane is zero* EXAMPLES. 1. Given four equal concurring forces whose directions are inclined to the axis of z at angles of 16°, 76°, 136% and 225° ; determine the magnitude and direction of their resultant Let each force be equal * > P ; then X = P ooa 16° + P coe 76 + P cos 136° + P cos 225° .3* -a = p: 2* F = P sin 16° + P sin 76° + P sin 136° + P sin 226" .'. 7? = P(6-2V^)*- 3* tan ds= 8 Err 2 2. Giren two equal concurring forces, P, whose direc- f ions are inclined to the axis of x at angles of 30° and 316°; tlnd their resultant Ans. B = 1,69 P. 2* i piW saa'Wiywwi 34 CONCtmRIlfO FOROSa. i. 3. Given three concurring forces ol 4, 6, and 6 lbs., whose directions are inclined to the axis of « at angles of 0°, 60°, and 136° respectively ; fi nd their resnltant. An$. R= V97 + 16 V 6 - 39 V2. 4. Given three equal concurring forces, P, whose direc- tions are inclined to the axis of x at angles of 30°, 60°, and 165° ; find their resultant. Ans. i2 = 1.67 P. 5, Given three concurring forces, 100, 50, and 200 lbs., whose directions are inclined to the axis of x at angles of 0° 60° and 180°; find the magnitude and direction of their resultant Ans. R = 86.6 lbs. ; = 150°. 38. To find the Magnitnde and Direction of the Resultant of any number of Concnrrlng Forces in Bpace.-Let P^, P„ P„ etc., be the forces, and the whole be referred to a system of rectangular co-ordmates. Let a„ /3t, yj, be the angles which the direction of Pj makes with three rectangular axes drawn through the point of appUcation ; let «„ a„ r„ be the angles which the direc- tion of P, makes with the same axes; aj, Pi, Ts. tne angles which P, makes with the same axes, etc. Resolve these forces along the co-ordinate axes (Art. 34) ; the com- ponents of P, along the axes are Pj cos o„ Pj cos 0,, P, cos y . Resolve each of the other forces in the same way, and let X, Y, Z, be the algebraic snms of the components of the forces along the axes of x, y, and z, respectively ; then we have X = P, cos «i + P, cos «, + Pj cos «, -H etc.) = SP COS a. Y=P. cos /3j + P, COS /3, + Pj COS 0, + etc.( ^^^ = IP cos |3. Z = Pi COS yi + Pg cos y, + P» cos y, + etc.] = SP cos y. 6, and 6 Iba, jf x at angles of Bgultant. V 6 — 39 V2. P, whose direc- of 30°, 60°, and B = 1.C7 P. 50, and 200 lbs., f a; at angles of md direction of M.;e = 150°. Irection of the ring Forces in forces, and the liar co-ordinates. direction of Pi through the point s which the direc- Kj, Ps, Ts. *!»« ses, etc. Resolve rt. 34) ; the com- !„ Pi COS0,, P, in the same way, ! the components a z, respectively ; BOS «j + etc.) cos 0, + etc.^ ^» cosy, + etc.] CONDITIOira OF EQlflUBRIUJl. 3S Let i2 be the resultant of all the forces; and let the angles which its direction makes with the three axes be a, b, c ; then as the resolved parts of B along the three co-or- dinate axes are equal to the sum of the resolved parts of the several components along the same axes, we have ^ cos a = X, iJ COB ft = r, i2 cos c -- Z. (2) Squaring, and adding, we get B? =z X^ + m -^ Z*', X Y cos fl = -p, cos ft = -^, cos c = -p; (3) ('4 which determines the magnitude of the resultant of any system of forces in space and the angles its direction makes with three rectangular axes. 39. The Conditions of Equilibrium for any num- ber of Concnrring Forces in Space.— If the forces are in equilibrium, ^ = ; therefore (3) of Art. 38 becomes 2'» + r«+z» = o. But as every square is essentially positive, this cannot be unless X = 0, F = 0, Z = ; and therefore 2:Pco8a = 0, IP 008/3 = 0, SPco8y = 0; (1) and these are the conditions among the forces that they may be in equilibrium ; that is, the sum of the components of the forces along each of the three co-ordinate axes is equal to zero. 40. Tension of a String. — By the iension of a string is meant the pull along its fibres which, at any point, tends to stretch or break the string. In the application of the preceding principles the string or cord is often used as a ■ 'swviStWMB^.'flKs-wtwsair!'.' ' ■ 36 BXAMPLKa. moiuis of oommnnioating force. A string is said to be per- fectly flexible when any force, however small, which is applied otherwise than along the direction of the string, will change its form. In this work the string will be regarded as perfectly flexible, inextensible, and withont weight If such a string be kept in eqnilibrinm by two forces, one at each end, it is clear that these forces must be equal and act in opposite directions, so that the string assumes the form of a straight line in the direction of the forces. In this case the tension of the string is the same throngh* out, and is measured by the force applied at one end ; and if it passes over a smooth peg, or over any number of smooth surfaces, its tension is the same at all of its points. If the string should be knotted at any of its points to other strings, we must regard its continuity as broken, and the tension, in this case, will not be the same in the two por- tions which stext from the knot BXAMPLES. 1. A and^ B (Fig. 8) are two fixed points in a horizontal line ; at A is fastened a string of length h, with a smooth ring at its other extremity, 0, through which passes another string with one end fastened at B, the other end of which is attached to a given weight W ; determine the position of C. Before setting about the solution of statical problems of this kind, the student will clear the ground before him, and greatly simplify his labor by asking himself the following questions : (1) What lines are tbiue in the flgore whose lengths are already given P (2) What forces are there whose magnitudes are already given, and what are the forces whose magnitudes are yet unknown? (3) What it is required to mid to be per- mall, which ia I of the string, string will be , and without by two forces, must be equal string assumes of the forces. same through* one end ; and my number of 1 of its points, points to other roken, and the in the two por- is required to al problems of i)efore him, and the following le figure whose irces are there what are the nP (3) What mXAMPLKS. 87 variable lines or angles in the figore would, if they were known, determine the required position of P Now in this problem, (1) the linear magnitudes which are given are the lines AB and AC. (2) The forces acting at the point C to keep it at rest are the weight W, a ten- sion in the string CB, and another tension in the string CA. Of these W is given, and so is the tension in OB, which must also be equal to W, since the ring is smooth and the tension therefore of WCB is the sanio throughout and of course equal to W. But as yet there is nothing determined about the magnitude of the tension in CA. And (3) the angle of inclination of tbo string OA tc the horizon would, if known, at once determine the posi- tion of 0. For if this angle is known, we can draw AC of the given length ; then joining to B, the position of the system is completrly known. Let AB = fl, AC = b, CAB = e, CBA = ^, and the tension of the string AC = T. Then, for the equilibrium of the point C under the action of the three forces, W, W, and T, we apply (2) of Art. 37, and resolve the forces horizontally and vertically j and equate those acting towards the right-hand to those acting towards the left ; and those acting upwards to those acting downwards. Then the horizontal and vertical forces are respectively Trsin0 + TsinS = W. Eliminating T we have cos d = sin (9 + 0) ; .-. 25 + = 90°. (1) Also, from trigonometry we have nn (0 -H 0) _ a . sin d ' (2) r 38 EXAMPLES. T\%M from (1) and (3) (? and may be fonn-^ ; and therefore T may be found; and thus all the circumstances of the problem are determined. 2. One end of a string is attached to a fixed point, A, (Fig. 9) ; the string, after passing o?er a smooth peg; B, sustains a given weight, P, at its other extremity, and to a given point, C, in the string is knotted a given weight, W. Find the posi- tion of equilibrium. The entire length of the string, ACBP, is of no conse- quence, since it is clear that, once equilibrium is estab- lished, P might be suspended from a point at any distance whatever from B. The forces acting at the point, 0, are the given weight, W, the tension in the string, CB, which, since the peg is smooth, is /*, and the tension in the string OA, which is unknown. Let AB = a, AC = 4, CAB = », CBA = ^, and the tension of the string, AC = T. Then for the equilibrium of the point Cj we have (Art. 32), (1) P cosfl >r~ sin (e -1- 0) ' also, from the geometry of the figure, we ] ft sin (0 + 0) = a sin 0. From(l) and (2) we get P dcc«( Q W~ a sin^' or rin^ = hW . -p cos tf; («) COS0 = Vo»/'»-ft»Fr» coB« 9 ___ \ c I I I i l i f I jj IWBW id therefore T Btances of the ve P* ■■*>*^p* ■Hfii is of no conse- briam is estab- at any distance be point, C, ore Qg, CB, which, n in the string . = ip, and the ^ the equilibrium (1) (3) >B»» MXAMPLIJS, 99 Expanding sin (fl + ^) in (2), and substitating in it these values of sin ^ and cos 0, and reducing, we have the equation ^ =0, ^,,S!!f^^^±D^», 2W*b from which 6 may be found. (See Minchin's Statics, p. 29.) 3. If, in the last example, the weight, W, instead of being knotted to the string at C, is suspended from a smooth ring which is at liberty to slide along the string, AOB, find the position of equilibrium. Atu. sin = ^. 41. BqniUbrinm of Coneiurrliig Forces on a Smooth Piano. — If a particle be kept at rest on a smooth surface, plane or curved, by the action of any number of forces applied to it, the resultant of these forces must be in the direction of the normal to the surface at the point where the particle is situated, and must be equivalent to the pressure which the surface sustains. For, if the resultant had any other direction it could be resolved into two components, one in the direction of the normal and the other in the direction of a tangent ; the first of these would be opposed by the reaction of the surface ; the second being unopposed, would cause the particle to move. Hence, we may dispense with the t»lane altogether, and regard its normal reaction as one of the forces by which the particle is kept at rest. Therefore if the particle on which the statical forces act b« on a smooth plane surface, the case is the same aa that treated in Art. 39, viz., equilibrium of a particle acted upon by any number of forces ; and in vnnt- ing down the equations of equilibrium, wo merely have to include the normal i action of the plane among all the others. p^ 40 JtXAMPLSa. BXAMPLBS. 1. A heavy particle is placed on a smooth inclined plan», AB, (Pig. 10), and is sustained by a force, P, whidi acts along AB in the vortical plane which is at right angles to AB ; find P, and also the pressure on the in- clined plane. The only eflfect of the inclined plane is to produce a normal reaction, R, on the particle. Hence if we intro- duce this force, we may imagine the plane removed. Let IT be the weight of the particle, and « the inclina- tion of the plane to the horizon. Resolving the forces along, and perpendicular to AB, since the lines along which forces may be resolved are arbitrary (Art 37), we have guooeflsively, P— ITsittft SE 0, or PzszWfAna', and R — W'cos « = 0, or R = IT cos a. If, for example, the weight of the particle is 4 or, and the in hnation of the plane 30°, there will be a normal pressure of 2^3 oz. on the plane, and the force, P, wUl be 2 oz. a. In the previous example, if P act horizontally, find its magnitude, and also that of R. Resolving along AB and perpendicular to it, we have Bucccsdiveiy, P cos a— IT sin a = 0, or PssTTtana; \^ and PBin«+ F oo«« — i? = 0, .•. J? = W oobm* 8 to produce a ce if we intro> amoved. a the inclina- licnlar to AB, >e resolved are )OS a. le ia 4 oz., and I be a normal > force, P, will risontally, find o it, we have Ftan «; V^ coimmoifa op gqpuMnnm. U 3. If the particle is auatained by a ft)roe, P, mining a given angle, 6, with the inclined plane, find the mi^itade of this force, and of the pressure on the plane, ail the forces acting in the same vertical plane. Besolving along and perpendicular to the jdaae succes- sively, we have PcobO— Wnna = 0, and JJ + P sine— ff COB « = 0, from which we obtain P^W^^,; R 008 $' _- |^r Coe(« + ^) COS0 Rem.— The advantage of a judicious selection of direc- tions for the resolution of the forces is evident. By resolv- ing at right angles to one of the unknown forces, we obtain an equation free fh)m that force; whereas if the directions are selected at random, all of the forces will enter each equation, which will make the solution less simple. The student will observe that these values of P and B cculd have been obtained at once, without resolution, by Art. 32. 42. Conditiosui oi SqidUbiiaiii for may number of Ooncnrring ForoM whoa the perticle on which they act ia Oeaatrained to Remain on a Qiven Smooth Bvrfiwe.— If a particle be kept at rest on a smooth sur- face by the action of any number of forces applied to it, the resultant of these forces must be in the direction of the normal to the surface at the point where the particle is situated, and must be equivalent to the pressure which the surface sustains (Art. 40). Hence since the resultant is in the direction of the normal, and is destroyed by the roao- iSt CONDinOtfS OF MqmUBRIUM. tion of the Brrface, we may regard this reaction as an addildonal force directly opposed to the normal force. Let N be the normal reaction of the surface, and a, 0, y, the aiigles which JV makes with the co-ordinate axes of x, y, and t, respectively. Let X, Y, Z, be the sum of the comfoneuts of all the other forces resolved parallel to the three axes respectively. The reaction JVmay i., considered a ne^r force, which, with the other forces, keeps the parti- cle im eqoilibriom. Therefore, resolving N parallel to the three axes, we have (Art 39), X+ JVcosa = 0, F-f-JVcos/S Z+ JVcosy = 0. > (1) Leit u =/(«, y, «) = 0, be the equation of the given Burfa<3e, and x, y, t the co-ordinates of the particle to which the forces are applied. We have (Anal. Geom., Art. :176), a' cos a = oosjS = cosy = y Vo'« -I- »'» + I'i 1 (2) where a' and V are the tangents of the angles which the projections of the normal, N, on the co-ordinate planes xz and yt make with the axis of tt. Since the normal is per- pendicular to the plane tangent to the surface at (x, y, t), t))o projections of the normal are perpendicular to the traces of the plane. Therefore (AnaL Qeom., Art 37, Cor. 1), W6 have and 1 -I- aa' = 0, 1 + W = ; (8) (4) mm/Hmfm^ UM. s reaction as an rmal force. Pace, and a, 0, y, dinate axes of z, the sum of the ed parallel to the nay <., considered keeps the parti- ¥ parallel to the (1) ion of the given [ the particle to e (Anal. Geom., (2) angles which the rdinate planes xz he normal is per- irfaco at {x, y, *), endicular to the Qeom., Art. 37, (8) (4) in Thich a CONDrnOIfB OF EqUIUBRIUK. I ' y dx , daf ,_dy j.__^ (Calculus, Art. 66a.) Substituting in (3) and (4), we have x.S-^ = o; and >-^|-f = <". from which du d^ -§^= ? (OaLArt87) = a', dx du ^ dt (5) du ,„a ^ = -^ = # = *'. <«) M dy du dM Substituting these values of a' and 7/ in (2) and multiply- du , ing both terms of the fraction by ^, we have 008a = 008/3 = dm tSi 3. cosy = (7) I f M 44 CONDITTOm OP SQVIMBMIXrM. \ which give the value of the direction cosines of the aonaai at {x, y, z). Putting the denominator equal to Q, tot shortness, and fiubstitating in (1) and transposing, we have ^ N du r — —— ~ Q' dy' Z — —— ^ ~ Q' dz (8) Pi (10) X T Z du du du dm ^ ST Eliminating N between ihese three equations, we obtain the two independent equations, m which express the conditions that must exist among the applied forces and their directions in order tiiat their resultant may be normal to the surface, t. e., that there may be equilibrium. If these two equations are not satisfied, equilibrium on the surface cannot exist Hence the point on a given surface, at which a given particle under the action of given forces will rest in equilibrium, is the point at which equations (11) are satisfied. Cob. 1.— Squaring equations (8), (9), (10) and adding, we get /rfM\> /rftt\» (du\ \dxf ]dy}_ \dz} SW "^ 1 -m*- immm EXAMPLSa. 47 which in (12) gives =^ + y* = «[^-^^-^} 8. A particle ia placed ingide a smooth spboipe on the con- cave surface, and is acted on by gravity and by a repulsive force which varies inversely as the square of the distance from the lowest point of the sphere; find the position of equilibrium of the particle. , Let the lowest point of the sphere be taken for the origin of co-ordinates, and let the axis of « be vertical, and posi- tive upwards; then the equation of the sphere, whose radius is a, is Lot TT = the weight of the particle, and r = the distance of it from the lowest point; then r« = a^ + ^ + «' = 2o«. Also, let the -epnlsive force at the unit's distance = « ; then at the distuioe r it will be — ^ X = r = Sox' 2as u w -If. ji. i mmmmm^i mmsBss 48 MXAMPLXa. Let iV = the normal pressni-e of the curve ; then (8) and (10) give i.t-^+^'-^-o.. from which we have t = ^a^W^' whence the position of the particle is known for a given weight, and for a given value of u. (See Price's Anal. Mechanics, Vol. I, p. 39.) 4. Two weights, jP and C, are fastened to the ends of a string, (Fig. 11), which passes over a pulley, 0; and ^ hangs Ireely when P rests on a plane curve, ^P, in a vertical plane ; it is required to find the position of equili- brium when the curve is given. The forces which act on P are (1) the tensiou of the string in the line OP, which is equal to the weight of Q, (2) the weight of P acting vertically downwards, (3) the normal reaction of the curve R. Let be the origin of oo-ordinates, and the axis of x vertical and positive down- wards. Let OM — X, MP = If, OP = r, POM =e,OA=a. Then, Flg.ll X= P-Qfmd—R dy T^-Qmxd+R^', then <8) and n for a given Price's Anal. the ends of a jr, 0; and Q ve, AP, in a ion of equili- Fig.ll MXAMPm.9, therefore from (15) we have (P — Q COB 6) dx — Q sin ddy = 0, 4« or But since we have a* + »• = f4, zdx -\- ydy =. rdr\ .'. Pdz—Qdr = 0; (1) which is the oonditidn that most be aatisfed by P, Q, and the equation of the curve. 6. Required the equation of the curve, on ^sill points of which P will rest. Integrating (1) of Ex. 4, we have Px-^Qr= O. (1) But since P is to rest at all points of the curve, this equr tion must be satisfied when P is at A, from which we get a; = r = o ; therefore (1) becomes Pa—Qa=z C; which in (1) gives P ' 1 — ^ cos which is the equation of a conic section, of which the focus is at the pole ; and is an ellipse, parabola, or hyperbola, according ac P <, =, or > Q. 3 IK> SXAMPLBa. EXAMPLES. 1. Two forces of 10 and 20 lbs. act on a particle at an angle of 60° ; find the resuliont. Ans. 26.5 lbs. 2. The resultant of two forces is 10 lbs.; one of the forces is 8 lbs., and the other is inclined to the resultant at an angle of 36°. Find it, and also find the angle between the two forces. (There are two solutions, this being the ambiguous case in the solution of a triangle.) Ans. Force is 2.66 lbs., or 13.52 lbs. Angle is 47° 17' 05", or 132° 42' 55". 3. A point is kept at rest by forces of 6, 8, 11 lbs. Find the angle between the forces 6 and 8. Ana. 77° 21' 52". 4. The directions of "iwc forces acting at a point are inclined to each other (1) at an angle of 60°, (2) at an angle of^ 120°, and the respective resultants are as V? : VS ; compare the magnitude of the forces. Ans. 2 : 1. 5. Three posts are placed in the ground so as to form an equilateral triangle, and an elastic string is stretched round them, the tension of which is 6 lbs. ; find the pressure on each post. ^„«. e Vs. 6. The angle between two unknown forces is 37°, and their resultant divides this angle into 31° and 6° ; find the ratio of the component forces. Ans. 4927 : 1. 7. If two equal rafters support a weight, W, at their upper ends, required the compression on each. Let the length of each lufter be a, and the horizontal distance between their lower ends be & , aW Ans. V4o»-i» particle at an ns. 26.5 lbs. I. 10 one of the . reaaltaut at angle between this being the ingle is 47° 17' ' 6, 8, 11 Iba. 77° 21' 52". at a point are 60°, (2) at an Itants are as rces. Ana. 2 : 1. > as to form an TetcLed round le pressure on Ana. 6 \/3. ses is 37°, and i 6° ; find the ». 4.927 : 1. ;, W, at their och. Let the ontal distance aW V4o» - V ■m ■■• MXAMPUa^ wmmm 51 8. Three forces act at a point, and include angles of 90° and 45°. The first two forces are each equal to 2P, and the resultant of them all is VlOP; find the third fofce. ^ng. P V%. 9. Find the magnitude, R, and direction, 0, of the resultant of the three forces, P, = 30 lbs., P, = 70 lbs., Pj = 50 lbs., the angle included between P, and P, being 56°, and between P, and P, 104°. (It is generally conTenient to take the action line of one of the forces for the axis of a;.) Let the axis of x coincide with the direction of P^ ; then (Art. 36), we have X = 22.16 ; Y = 75.13 ; B = 78.33 ; « = 73° 34'. 10. Three forces of 10 lbs. each act at the same point ; the second makes an angle of 30° with the first, and the third makes an angle of 60° with the second ; find the magnitude of the resultant. Ana. 24 lbs., nearly. 11. If three forces of 99, 100, and 101 units respectively, act on a point at angles of 120°; find the magnitude of their resultant, and its inclination to the force of 100. Ana. VS; 90°. 12. A block of 800 lbs. is so situated that it receives from the water a pressure of 400 lbs. in a south direction, and a pressure from the wind of 100 lbs. in a westerh direction ; required the magnitude of the resultant pres- sure, and its direction with the vertical. Ana. 900 lbs.; 27° 16'. 13. A weight of 40 lbs. is supported by two strings, one of which makes an angle of 30° with the vertical, the other 45° ; find the tension in each string. . Ana. 20 {VQ - Vi) ; 40 ( V3 — 1). tMM 6% BXAMPLSa. 14. Two forces, P and P\ acting along the diagonalfl of a parallelogram, keep it at rest in such a position that one of its sides is horizontal ; show that P sec «' = P' sec o = W cosec (« + «')> where W is the weight of the parallelogram, and a and «' the angles between *;he diagonals and the horizontal side. 16. Two persons pall a heavy weight by ropes inclined to the horizon ut angles of 60° and 30° with forces »jf 160 lbs. and 200 lbs. The angle between the two veitical planes of the ropes is 30" ; find the single horizontal force that would produce the same eflfect Ans. 245.8 lbs. 16. In order to raise vertically a heavy weight by means of a rope passing over a fixed pulley, three workmen pull at the end of the rope with forces -of 40 lbs., 50 lb8.,8tul 100 Ibr. ; the directions of these forces being inclined to the horizon at «n ax^gle of 60". What is the magnitude of the resultant force which tends directly to riise the weight? Ans. 164.64 lbs. 17. Three persons pull a heavy weight by cords inclined to the horizon at an angle of 60°, with forces of 100, 120, atid 140 lbs. The three vertical planes of the cords are inclined to each other at angles of .S0°; find the single horizontal force that would pi-oduoe the same effect Am. 10 '^^146 + 72 Vl lb«. 18. Two forces, P and Q, acting respectively parallel to the base and length of an inclined plane, will each singly sustain on it a particle of weight, W\ to det«rmino the weight of W. Let « = inclination of the plane to the horizon ; then resolving in each case along the plane, so that the normal pressures may not enter into the equations (See Bern., Ex. 3, Art. 41), wo have T !e diagonals of sition that one 1, aod a and «' triiiODtal eide. ' ropes inclined with forces of he two veitical horizontal force nti. 245.8 lbs. eight by means workmen pull at >8., 50 lbs., and iug inclined to e mugnitude of ;J8e the weight ? s. 164.64 lbs. r cords inclined XJ8 of 100, 120, f the cords are find the single le effect + 72 VS lb«. vely parallel to rill each singly i determine tho B horizon ; then that tlie normal See Bern., Ex. 3, MXAMPLBS. St Poofla = fTsina; C=lfBina; jr = PQ {p* W 19. A cord whose length is 21, is faster* • m .■ vnd B, in the same horizontal line, at a distance i iii .*cli other oqual to 2rt ; and a smooth ring upon the cortt sustains a weight W; find the tension of the cord. Ana. T .: --^^==. 2 VP — fit' 20. A heavy particle, whose weight is W, is sustained on ii smooth inclined plane by three forceg applied to it, each equal to ^; one acts vertically upward, another horizon- O tally, and the third along the plane ; find the inclination, «, of the plane. Ans. tan 1 2' 21. A body whose weight ia 10 Ihs. Is supported on a smooth inclined plane by a force of 2 lbs. acting along the plane, and a horizontal force of 5 lbs. Find the inclination of the plane. ^»«. sin-» |. 22. A body is sustained on a smooth inclined plane (in- clination a) by a force, P, acting along the plane, and a horizontal torce, Q. When the inclination is halved, and the forces, P and Q, each halved, the body is still observed to rest; find the ratio of P to ^. ^^^ 23. Two weights, P and Q, (Fig, 12), rest on a sipooth doxihle-inolined plane, and are attiiched to the extremitios of a string which ijaases over a smooth peg, 0, at a point vertically over the intersection of the Hues, the jieg and the weights being in a Fi|.a ■PBiWiP M MXAMPLSa, vertical plane. Find the position of equilibrium, if Z = the length of the string and h = CO. Ans. The position of equilibrium is given by the equa- tions p 8in_a _ -, sin /3 ► cos * ~ ^ cos ^ * cos « oos_^ _ I sin sin ~ ^* 24. Two weights, P and Q, connected by a string, length /, rest on the convex side of a smooth vertical circle, radius a. Find the position of equilibrium, and show that the heavier weight will be higher up on the cirele than the lighter, the radius of the circle drawn to P making an angle with the vertical diameter. Ana. P sin 6 = Q sin ( eh 25. Two weights, P and Q, connected directly by a string of given length, rest on the convex side of a smooth vertical circle, the string forming a chord of tii«? circle ; find the position of equilibrium. Ans. If 2a is the angle subtended at the centre of the circle by the string, the inclination 0, of the string to the vortical is given by the equation P cot fl = p- ^^ tan o. 26. Two weights, P and Q, (Fi^. 13), Teat on the concave Hide of a pari< )la whose axis is horizontal, and are con- nected by a string, length I, which ]iasse8 over a smooth peg at the focus, F. Find the {wsition of equilibrium. A»8. Let — the angle which FP Fi|.,ll turn, if 2 r= the n by the equa- by a string, nooth vertical lilibriuni, and ler up on the le drawn to P wmm- »n G'-»)- directly by a le of a smooth of tii'5 circle ; centre of the 3 string to the 'IfcU BXAMPLSa. 65 makes with the aria, and 4m = the latus reotnm of the parabola, then e Py/l- 2m _ 27. A particle is placed on the convex side of a smooth ellipse, and is acted upon by two forces, F and F', towards the foci, and a force, F"y towards the centre. Find the position of equilibrium. ^^ ,. _ __A__, where r = the distance of the par- Vl — «• tide ftom the centre of the eUipse ; b = semi-minor was, F-F' and n = —^ — 28. Let the curve, (Fig. 11), be a circle in which the origin and pulley are at a distance, c, above the centre of the circle : to determine the position of equilibrium. Q Ans. r = -pa. 29. Let the curve, (Fig. 11), be a hyperbola in which the origin and pulley are at the centre, 0, the transverse axis being vertical ; to determine the position of equilibrium. 30. A particle, P. is acted upoL oy two forces towards two fixed points, 8 and H, these forces being ^ and ^p respectively ; prove that P will rest at all points inside a smooth tube in the form of a curve whose equation is 8P. PH = ]^,k being a constant 31. Two weights, P and Q, connected by a string, r«8t on the convex side of a smooth cycloid. Find the position of equilibrium. 56 aXAMPUUL Am. If 2 s the length of the string, and « s radios of generating circle, the position of eqoilibriam is defined by the equation .0 Q I "" a = pTq ' 45' where B is the angle between the vertioal and the radius to the point on the generating circle which corresponds to P. 32. Two weights, P and Q, rest on the convex side of a smooth vertical circle, and are connected by a string which passes over a smooth peg vertically over the centre of the circle ; find the position of equilibrinm. Ana. Let h =. the distance between the peg, B, and the centre of the circle ; and ^ = the angles made with the vertical by the radii to P and Q, respectively ; a and j3 = the angles made with the tangents to the circle at P and Q by the portions PB and QB of the string ; / =b length of the string; then COS a ' _ nsin^ - ^C08/3' \cos a Bin *\ cos /)/ - *» ft cos (fl -f a) =. a cos a. A COS (^ + /3) = a cos j3. « =: radios of u defined by . the radias to esponds to P. nvex ride of a a sti-ing which centre of the Bg, B, and the made with the ; a and j3 = irole at P and b; ; 2 =B length CHAPTER III. COMPOSITION AND RESOLUTION OF FORCES ACTING ON A RIGID BODY. 43. A Xtigid Body. — In the last chapter wo considered the action of forces which have a common point of applica- tion. We shall now consider the action of forces which are applied at different points of a rigid body. A rigid body is one in which the particles retain invari- able positions with respect to one anotbeis so that no pxternal force can alter them. Now, as a matter of fact, there is no such thing in nature as a body that is perfectly rigid ; every body yields more or less to the forces which act on it. If, then, in any case, the body is altered or com- pressed appreciably, we shall suppose tiuit it has assumed its figure of equilibrium, and then consider the points of application of the forces as a system of invariable form. The term body in this work means rigid body. 44. TransmisBibility of Force.— When & force acts at a definite point of a body and along a definite line, the iffect of the force will be unchanged at whatever point of its direction we suppose it applied, ])rovided this point be I'ither one of the points of the body, or be invariably con- iioctcd with the body. This principle is called the trans- missibilily of a force, to any point in its line of action. Now two Ciiual forces acting on a particle in the same line and in opposite directions neutralize each other (Art. 10); BO by this principle two equal forces acting in the same line and in opposite directions at any points of a rigid body in that line neutralize each other. Hence it is dear that when many forces are acting on a rigid body, any twc, which aie equal and huvo the same line of action ■k$'f % ip!ii)UWU I l ( «WI I UPM. I lllM I WMIlMi l » l l.l i HMM 68 SBSXTLTANT OF PARALLEL FOBCKS. rHhM and act in opposite directions, may be omitted, and also that two equsJ forces along the same line of action and in opposite directions, may be introduced without changing the circumstancQi of the qrstem. 46. Rorahast of Two PanlM Force*.*— (1) Let P and Q, (Fig. 14), be the two parallel forces acting at the points A and B, in the same direction, on a rigid body. It is re- qaired to find the resultant of P and Q. At A and B introduce two equal and opposite forces, F. Th^ introduction of these forces win not disturb the action of P and Q (Art. 44). P and F at A are equivalent to a single force, R, and Q and /* at B are equivalent to a single force, 8. Then let R and S be snpposed to act at 0, the point of intersection of their lines of action. At this point let them be resolved into their components, P, F, and Q, F, respectively. The two forces, F, at 0, neutralise each other, while the components, P and Qy act in the line OG, parallel to their lines of action at A and B. Hence the magnitude of the resultatti is P-\- Qt (Art S8). To find the point, G, in which its line of action outs AB, let the eztremitief of P and R (acting at A) be joined, and complete the parallelogram. Then the triangle PAR is evidently nmikr to GOA ; therefore, P GO . ., , f the arm and »wn from any le couple, and udr of the mo- le direction in ad not actital, position of it ; of the couple, itch placed in m let the axes tation in the COUfLKB. 6S direction of the motion of the hands be drawn upward through the face of the watch, and the axes of those which tend to produce the contrary rotation be drawn tlowntmrd through the back of the watch. Thus each couple is oom- plotoly represented by its axis, which is drawn upward or downward according as the moment of the couple is posi- tive or negative ; and couples are to be resolved and compounded by the same geometric constructions performed with reference to their axes as forces or velocities, witii reference to the lines which diredly represent them. Wo shall now give three propositions showing that the effect of a couple is not altered when cu-tain changes are made with respect to the couple. 51. Tlie Effect of a Couple on a Rigid Body is not altered if the arm- be turned through an;/ angle about one extremity in the plane of the Couple. Let the plane of the paper be the plane of the couple, AB the arm of the original couple, AB' its new posi- tion, and P, P, the forces. At A and B' respectively introduce two forces each equal to P, with their action lines perpendicular to the arm AB', and opposite in direction to each other. The effect of the given couple is, of course, unaltered by the introduction of those forces. Let BAB' = W ; then the resultant of P acting at B, and of P acting at B', whose lines of action meet at Q, is 2F sin 6, acting along the bisectw A^ ; and the result- ant of P acting at A peipendicular to AB and of P per- pendicular to AB', is %P sin 9, acting along the bisector AQ in a direction opposite to the former resultant Hence these two resultants nentjalire each other; and there remains the couple whose arm is AB', and whose forces aro P, P. Hence the effect flf the couple is not altered. ,.--m':f.x»mum\uimm ummmmmmi:'. 64 COUPLES. 4 \ ^B Fig. 19 52. The Effect of a Couple on a Rigid Body is not altered if we transfer tlie Cojtple to any other Parallel Plane, the Arm remaining parallel to itself. Let AB be the arm, antl P, P, the forces of the given couple; let A'B' be the new position of the arm par- allel to AB, At A' and B' apply two eqaal and opposite forces each equal to P, acting pi^rpendicular to A'B', and in a plane parallel to the plane of the original couple. This will not alter the effect of the given couple. Join AB', A'B, bisecting each other at ; then P at A and P at B', acting in parallel lines, and in the same direction, are equivalent to 2P acting at ; also P at B and P at A', acting in parallel lines and in the same direction, are equivalent to 2/* acting at 0. At O therefore these two resultants, being equal and opposite, neutralize each other ; and there remains the couple whose arm is A'B', and whose forces are each P, acting in the same directions as those of the original couple. Hence the effect of the couple is not altered. 53. The Effect of a Couple on a Rigid Body is not altered if we replace it by another Couple of which the Moment is the same ; the Plane remain- ing the same and the Arms being in the same straight line and having a common extremity. Let AB be the arm, and P, P, the forces of the given couple, and sup- pose P=Q+R. Produce AB to C Bo that AB : AC :: Q : P{=Q + B), and therefore AB : BO :: Q : B', tP=Q+R F{gM ♦ P=Q+R (1) igid Body is to any other parallel to \/ 7' Fifl.l9 I effect of the !i other at O ; 1 lines, and in ]g at ; also es and in the : at O. At O and opposite, couple whose acting in the Hence the ^,id Body is r Couple of ne rcniain- the same mmm .30 Q P=Q+R (1) (2) FOSCS AND A COUPLB. 66 at introduce opposite forces each equal to Q and parallel to P ; this will not alter the effect of the couple. Now E &t A and Q at C will balance Q + li at B from (3) and (Art. 45); hence there remain the forces, Q, Q, acting on the arm, AC, which form a couple whose moment is equal to that of P, P, with arm, AB, since by (1) we have P X AB = Q X AC. Hence the effect of the couple is not altered. Rem. — From the last three articles it appears that we may change a couple into another couple of equal moment, and transfer it to auy position, either in its own plane or in a plane parallel to its own, without altering the effect of the couple. The couple must lomain unchanged so far as concerns the direction oj rotation which its forces would tend to give the arm, i. e., the axis of the couple may be removed parallel to itself, to any position within the body acted on by the couple, while the direction of the axis from the plane of the couple is unaltered (Art 50). 54. A Force and a Couple acting in the same Plane on a Rigid Body arc equivalent to a Single Force. Let the force be /'and the couple {P, a), that is, P is the magnitude of each force in the couple whose arm is a. Then (Art 53) the couple (P, o) = the couple \F, ^. Let this latter couple be moved till one of its forces acts in the same line as the given force, F, but in the opposite direction. The given force, F, will then be destroyed, and there will remain a force, F, acting in the same direction as the given one and at a perpendicular distance from it aP = -F' S • i v8 RBSULTANT OF COUPLXa. Cob.-— J force and a couple acting on a rigid h}dy cannot produce equilibrium. A couple can be in equilibrium only with an equivalent couple. Equivalent couples are thotie whose moments are equal.* The resultant of several couples is one which will produce tk'j safne effect singly as tfie component couples. 55. To find the Resultant of any number of Couples acting on a Body, the Planes of the Couples being parallel to each other. Let P, Q, R, etc^ be the forces, and a, b, c, etc., their arms respoctively. Suppose all the coaples transferred to the same piano (Art. 52) ; next, let them all be transferred sa as to have their arms in the same straight line, and one extremity common (Art. 51) ; lastly, let them bo replaced by other couples having the same arm (Art. 53). Let » be the common arm, and P,, Qi, R^, etc., the new forces, 80 that P^a ::=: Pa, Qia = Qb, Uia — Re, etc., then P| = P-, Qi = e-, Rt=R*-, etc., i.e., the new forces are P , Q , R-, etc., actlug on the common arm a. Hence their resultant will be a ooaple of which each force equals p? 4. ^* 4- i?*: ^. etc., a a a •nd the arm ~ n, or tho moment cquaU Pa+ Qb + Re + etc If one of tho oonplcs, as Q, act in a direction opposite to • The nioDteaU nroqiUmlent cou|iit« iiiajr >wt(i like or uullke ilgiM. ^?«?? ' id body cannot mlibrium only pUa are thoKC h will produce number of ines of the », c, etc., their transferred to } transferred sa ', line, and one ;m bt) replaced •3). Let a be 10 new forces, c, etc., actliig on the le a couple of 11 m 1 opponito tt iliko llgM. aSSULTANT OF TWO COUPLES. •Y the other couples its sign will be negative, and the foroe at each extremity of the nrm of the resultant couple jv ill be p?_^* + i?- + etc. Hence the moment of the resultant couple is equal to the algebraic sum of the momentfi of the component couples. S6. To Find the Resultant of two Couplet not acting in the same Plane,* Let the planes of the con pies b^ inclined to each other at an angle y ; let the couples be trejis- ferred in their pianos so as to have the same arm lying along the line of intersection of the two planes ; and let the forces of the couples thus traiisfbn-ed Ixi P and Q. Let AB hk\ the com- mon arm. I^et R be the rosnltant of the forces J"' and Q at A acting in the ^i! rection AJS ; and of P and (? at B acting in the direction Jtit. Then einoe P and Q at A are parallel to P and Q at B respectively, tlierefore £ at A is parallel to i2 at B. Hence the two couples are equivalent to the single couple R, R, acting on the arm AB ; and since PA.Q =. y, V.0 LiTO i? = P« + g» + iPQ cofi Y (Art 30). (1) Draw A«, Bi perpendicular to the planes of the conples /•, /', and Q, Q, respectively, and proportional in length to thflir moments. Draw Ac perpendicular to the plane of R, R, and in the same proportion to Aa, Bb, that the moment of the couple, R, R, is t<^) those of P, P, and Q, Q, respeotively. Then Art, Kb, Ac, nuiy hi' Inkoii bb the axes of P, P ; Q, Q; and * tMhoDlar't autlM, p. «. AtM PraU't UmIuuiIo*, p. K ■HMMMRMP G8 RESntTAXT OF TWO COUPLES. R, R, respectively (Art. 50). Now the three strai^^^ht linos. Art, Ar, Kb, make the same angles with each other that AiF^ A^, A^ make with each other; also they are in the euine proportion in which or in which AB • P, AB . i2, AB . ^ are, P, /?, ^ are. But R is the resultant of P and Q ; therefore Ac is tho diagonal of the parallelogram on Ao, Ai (Art. 30). Hence if two straight lines, having a common extrnmitif, represent the axes of two couples, that diagonal of the parallelogram descril/ed on these straight lines as adjacent sides which passes through their comn*on extremity repre- sents the axis of the resultant couple. Con. — Since R • AB is the axis or moment of tho reeui'- ant couple, we have from (1) /p. All' = /". Al?+ Q»- A^+iP- AB- Q- ABcos y. (2) If X and M represent the axes or moments of the com- ponent couples and G, that of the resultant couple, (2) Dtcones CP = L* + M' + 2L • M COB y. (3) Pen. 1. — \{ L, M, N, are the axes of three component couples which act in planes at right angles to one a.iother, and G tho axis of the resultant couple, it may easily be shown that C*= D + M*^ N\ (4) If A, n, V be the angles which the axis of tho resultant makes with those of the componont8, we have 009 A = ^' M cos /* = ^-, y COS V = ^. vs. B straight linos, ach other that ;hoy are in the ;fore Ac \a tho rt. 30). mon extremity, iagonal of the Ines as adjacent xtremity repre- it of tho resuit- AB-cosy. (3) ta of the com- iit couple, (2) (3) ree component o one aiiothcr, may easily bo (*) tho rosnltanf o' ^ggggggm*«*- 1 1 , 111 , 1 1 iiiii« || p | i i i. ii . VARroifoy's theorem of moments. iBil"i»SBS.r 6tf Sen. 2. — Hence, conversely any couple may be replaced by three coupIos acting in planes at right aiigloa to ono another ; their moments being G cos k, G cos /u, G cos v ; where G is the moment of the given couple, and A, p,, v tho angles its axis makes witli the axes of the three couples. Thus the composition and rcaolution of covplcs follow laws similar to those which apply U) forces, tho ••xis of the couple corresponding to tho direction of the force, and the moment of the couple to the magnitude of tho force. 57. Varignon's Theorem of Moments.— TTta mo- rri'CMt of the resultant of two conifjonen!, forces with respect to any point in their pic %e is equal to the algebraic sum of the mnmenis of the two components with respect to the same point. Let A P and .4 Q represent two com- ponent forces ; complete the parallelo- gram and draw the diagonal, Ali, representing the resultant force. lict O be tho origin of moments (Art. 46). Join OA, OP, OQ, OR, and draw PC and QB parallel to OA, and let p = tho perpendicular let fall from to All. Now tho moment of A P about is the product of A P and the per|)endicular let fali on it from (Art. 40), which IS double the area of the triangle, AOP (Art. 48). But the area of the triangle, AOP, = the area of tlie triangle, AOC, since these triangles have tho same base, AO, and are between tho same parallels, AO and CP. Hen^o the moment of AP about = the moment of vlC about O = AC -p. Also the moment of AQ v^^.cni is dooblo tho area of tho triangle, A OQ, = double the area of tho triangle, A OB, since the two triangles hovo the same baso, AO, end are between thn same parallels, AO and QB. Hence tlie moment of AQ about — Uu' moment of AU Fig.Zl 70 VASIOjrON'S TBBOREX OF MOMEKTS. about = AB • p. Therefore the sura of the moments of AP and AQ about = tlie sum of the moments ot AC and AB about = {AC + AB)p, = {A3 + £i?)p, {auci AC = fift firom the equal triangles ^PCand QBR) = AR ' p = the moment o£ the resulttmt. If the origin of momente fall between AP and AQ, the forces will tend to produce rotation in opposite directions, and hence their moments will have contrary signs (Art 4t). In this caae the moment of the resultant = the dif- ference of the moments of the components, as the student will find no difficulty in showing. Hence in either case the moment of the resultant is equal to the algebraic sum of the moments of the components. Con. 1.— If there are any number of component forces, wo may compound them in order, taking any two of them first, then finding the resultant of these two and a third, and so on ; and it follows that the sum of their momenta (with their proper signs), is equal to the moment of the redultojit. OoR. 2.— If the origiv of moments be on the line of action of the reroltant, p = 0, and therefore the moment of the resultant = ; hence the sum of the moments of the componeuta is ecjual to isero. In this case the moments of the forces in one direction balance those in the opposite dirpction ; t. e,, the forces that tend to produce rotation in one direction Mf^ counteracted by the forces that tend to produce rotation in the opposite direction, and there is no tenden'jy to rotation. Oo». 3.— If all the forces are in eqailibriam the resultant J? = 0, and therefore the moment oi R — Q; henoe the sum of the momontfl of the conponents is equal to aewj, and there is no teudeacty to motion either of tnuiulatiou or rottktion. VlfTS. the moments of loments ot AC {A3 + BB)p, IPC and QBJi) iP and AQ, the >aite directions, rary signs (Ai-t. [«nt = the dif- I as the student i in either case 3 alj/ebratc sum nponent forces, ,ny two of them ,\vo and a third, their moments moment of the on the line of re the moment he moments of se the moments in the opposite uoe rotation in [>s that tend to md there is no im the resultant = 0; hence the equal to serri, trauulatiou or p R Q 1 ( 1 B VARIGNOIf'a TBEORMM FOR PAKALLSL FORCES. 71 CoK. 4. — Therefore when the moment of the resultant = 0, we conclude either that the resultant — (Cor. 3), or that it passes through the point taken aa the origin of moments (Cor. 2). 58. Varignon'B Tlieoram of MomamtB for Parallel Forcea. — The aum of the rnamients of two parallel forces abotit any point is equal to the moment of their resultant about the point. Let P and Q be two paraliel forces acting at A and B, and R their result- ant acting at 6, and let be the point about which moments are to be taken. Then (Art. 45) we have P X AG = Q X BG, .•. P(OG - OA) = C (OB - OG), .-. (P + C) OG = P X OA + C X OB, /Z xOG = PxOA + G xOB; that is, the earn of the moments = the moment of the resultant OoB.— It follows that the algebraic sum of the momenta of any number of parallel forces in one plane, with respect to a point in their plane, is equal to the moment of their resultant with respect to the point 69. Centre of Parallel Forcea. -To iind ttte mag- nitude, direction, and point of application of the resultant of any number of parnllel forces acting on a rigid body in one plane. kHH iT'lMMBtmill 72 CBNTRV OF PARALLEL FORCES. Let P., P,, P., etc., denote the ZM fonjM, if,, Jf,, M3, etc., their points ^ -. >■''• of ufplicution. .Take any point in the plane of the forces as origin and draw the rectangular axes OX, OV. L«t («i. yi). (««» 1/%)f etc., be the points of application, if,, i/,, etc. Join M^M^; and take the point if on M^M,, so that ^. /• 6 «& Fig.24 if, if (1) then the rcsnitant of P, and P, is P, + Pg, and it acts through M parallel to P, (Art 46). Draw M^a, Mb, M^c parallel, and Mie perpendicular to the axis of y. Then wo have p Mb-yy= -jr—j- iy, - Vi) '. , , Mb- p^^p^ , (2) which gives the ordinate of the point of application of the rcEultant of P, and Pg. Now since the resultsat of P, and P,, which is P, 4- P,, acts at M, the resultant of P, + P, at M, and P, at M^, is P, + P, -f- P, at g, and substituting in (8) P, + /'„ P„ ifi, and y, for P,, P„ y,, and y, rcgpec- tiywiy, we have mm^ mm mmt m. 7m ^ i^ g-^U, a c » b b Fig.24 r,, BO that 0) Pg, and it acts erpendicalar to « -yi); y». (2) ieation of the Pg, which is I\ at if, and jtnting in (3) md yg ro§pec- ItJ- P^Vi . '• + ^3 ;(3) CBHTRS or PARALLSL P0BCE8, T3 and this process may be extended to any number of parallel forces. Let R denote the resultant forCe and y the ordi- nate of the point of applicjation ; then we have i? = Pi + P, + ^3 + etc. SP. 7, - P^Vx + ^«y« + ^sy._+_.etc. _ I.Py y - p, + p, + P3 + etc. ^P IP Similiirly, if x bo the abscissa of the point of application of the resultant, we have X = XP' The values of zl y are indeponth^nl of the angles which the directions of the forces muko wllh the axcM. Hence t' these directions be turned about the points of application (f the forces, their parallelism being preserved, tlie point of ii|)pliciition of the resultant will not move. For this reason tlie point (i, y) is called the centre of parallel forces. We shall hereafter have many applications in which its position is of great importance. ScH. 1. — The moment of a forcf. toifh respect to a plane to which it is parallel, is the product of the force into ♦ho perpendicular distance of its point of application from the plane. Thui, Pi^i is the moment of the force P,, iu reference to the plane through OX perpendicular to OV. This must be carefully distinguished from the moment of a force with respect to a point. Hence the Aquations for determining the posit'on of the centre of pfprailel forcet show that the sum of tlie mmufnts of the parallel fofoen icitti respect to any plane parallel to tfifun, i« eq„ etc., mr.y be treated in like man- ner. We thus obtain a stt o1 forces, P^, P„ P„ etc., acting at along OY, and a sot of couplr.., P^r^, P,x,, /'jij, etc., in the plane of the forces tending to turn the ^fcbdy from die axis of x to the axis of ;/. These forces are etjuivalcnt to a single resultant force /'^ f\ + /*, + etc., and the couples iirc equivalent to a s^gji i^ciilt&n! rioaj^ Pjjr, + P,x, + P,«, + etc. {Art 6^. Hence denoting the resultant force by 72, and tlie mumout of the resultAiil couple by 0, we have il = P, -I- P, 4- /», -H eta =r SP; = P,*, + P,a!, + P,x, 4- eta .-= S.PTi that is, a system of parallel forces can be reduced to a single fiitj* Hiiij a couple, which (Art. fl4. Cor.) (annot pioduoe equilibrium. Heuoo, for equilibrium, the force and the couple must vanish ; or XP = 0, and LPx = a TM. ^n the line and B Rigid Body 18 Plane.— Let ♦P. turb the equili- it along OF, e., P,a;,. The ied in hko man- ' >» ■* S' 6tc., rig 10 turn (he Phese forces are /% + /*, + etc., ■oopte, uU the momt)ut IP; reduced to n Cor.) runnnf uin, the force V 1 ^1 >R H '^ ,yi -A ' M' F i8.a« coifDmoira ojc squiLiBBiuM. 75 Hence the ooinditions of eqailibriam of a system of pur- iillel forces acting on a rigid body in one pkue are : The sum of the forces must = 0. The suin of the momenta of the forces about every point in their plane mttst = 0. 61. Conditioiift of EqaililMiiiin of a Rigid Body acted on by Forces in any direction in one Plane.— Let Pj, Pg, Pj, etc, be the forces acting at the points (^i» yi)> (a^8. y«)« i^i' ys)> etc., in the plane xy. Resolve the force P, into t>Yo components, X,, F,, parallel to OJT and OY respectively. Let the direc- tion of Kj meet OX at M, and the direction of X, meet OFat iV. Apply at two opix)sing forces each equal and parallel to Xj, and also two opposing forces each equal and parallel to F, . Hence Fj at .4,, or M, is equivalent to Y^ at O, and a louplo whoso moment is F, • OM; and X^ at J,, or iV, is equivalent to Xj at 0, and a couple whose moment is V, ov. Ifenc. Fj is replaced by F, at 0, and the couple F,.t, ; and A", is voplanul by Xj at (), and the couple X|//i (Art. 47). Therefore the force P, may bo replaced by the com- ponents Xj, F, acting at 0, and the couple whose moment is and which equals the moment of P, about (Art. 67). By a similar resolution of all the forces we shall have them replaced by the forces (X,, F,), (Xj, Fj), etc., acting at along the axes, and the couples TtX, — X,y„ Fa^s — Xj^,, etc Adding together the couples or moments of P,, P,, etc., ■i; 76 SQVILIOBr'M VNDEH THREE FORCES. and denoting by G the moment of tho resultant conple, wc get the total moment If the sum of the cumponents of the forces along OX is denoted by SX, and the sum of the components along OV by £F, the resultant of the forces acting at is given by the equation i? = (SX)» -h (£F)«. If a be the angle which R makes with the axis of X, wc have R cos a = IX, Ji Bin a = I.V; tan a = iX' Therefore, any system of forces acting in any direction in one plane on a rigid botly may be reduced to a single force, li, and a single couple whose moment is O, which (Art, 64, Cor.) cannot produce equilibrium. Hence for equilibrium we must have R = 0, and G' = 0, which requires that tx=o, ir=o, ^irx — Xy) = 0. Hence the conditions of equilibrium for a system of forces acting in any direction in one plane on a rigid body arc : Tlfte sum of the components of the forces parallel to each of two rectangular axes ..ivM = 0. The sum of the mofir its of the forces round ever tf point in their plane must = 0. rt*i FORCES. ultant coaple, wc EXAMPLES. 77 roes along OX is inents along OV at is givon by ;he axis of X, we in any direction iced to a single ent is O, wliich Hence for O — 0, wliich m 3r a system of m a rigid body rallel to each of i every point in Cor. — Convoraely, if the forces are in equilibrinni tho sum of the com{K>nents of the forces parallel to any diroc- tioii will = l>, and also the sum of the moments of the forces about any point will = 0. 62. Condition of Equilibrium of a Body under the Action of Three Forces in one Plane -// thn forces jtiaintain a hi'dy in equilibrium,, their directions must meet in- a point, or be parallel. Suppose the directions of two of the forces, P and Q, to meet at a point, and take moments round this point ; then tlio moment of each of these two forces = 0; thcrefon> ilio moment of the third force li = (Art. 61, Cor.), which requires either that Ji — 0, or that it pass through the point of intersection of P and Q. If R is not = 0, it must jtiiss through this jmint. Hence if any two of the forces meet, tiie third must pass through their point of intersec- tion, an keep it at rest, and each force must be equal and opposite to the resultant of the other two. If the angles ])etween them in pairs I c p, q, r, tho forces must satisfy the conditions r : Q '. R = «inp : sin & : sin r (Art. 32). If tro of the forces are parallel, the third must bo parallel o them, aud equal md directly opposed to their resultant. EXAMPLES. . 1. Suppose six parallel forces proportional to the numbers 1, 2, 3, 4, 5, 6 to act at points (—2, —1), (—1, 0), (0, 1), (1, 2), (2, 3), (3, 4) ; find the resultant, R, and the centre of parallel forces. By Art 69 we have J2 = SP = l + 2 + ...6 = 21; i .v^«ari«W«MaH*ea«^^j^j,^j^. ^^Wis^^^l^^^j^i i=i5te*S*iS6ii6'&i^.^^;.o , ■^ J ^%. ,^'t*'„ w. \^>1' * V ^v IMAGE EVALUATION TEST TARGET {MT-3) 1.0 U 22 i.25 1.4 I 1.6 t/. •\ Photographic Sciences Corporation 33 WfST MAIN STRUT WMSTIR.N.Y MSW (716) •73-4503 i\ V [V . 'w ^^^ f CIHM/ICMH Microfiche Series. CIHfVI/iCIVIH Collection de microfiches. Canadian Institute for Historical Micror«productions / Insti'iut Canadian da microraproductions historlquaa ^ . .^aea'iifcii (»i, y»), («„ Jf,) be the vertices, and let a, i, c be the sides opposite to them; then ;_ aa;,+te,+cr , -_ ay, +» »-,+ gy, a+4+c ' » a+i+c 3. If two parallel foroeu, P and ^, act in the same direc- tion at A and 5, (Pig. ji), and make an angle, B, with ^5, find the moment o/ each about the point of epplicur tion of their resnltant The moment of F with respect to (7 is P-^(?8iu(!»(Art.46). But from (1) of Art 4ft, we have P+ Q .'. AG = 4B AG* P + Q aB, which in P - AG an gives for the moment of P which also e<]uals the moment of (j. lOfflHP^:- ( = 49. y 49 21* ■nrallel forces are J to the opposite Bse forces, iiidoe^ and let a, k the Mune direc- u ao^, 0, with point ol epplicur Qoment of (j. MXAMPtML n I Fit-V 4. Two paiallel foroet, aotiBg in ike nm» direction; liave their magnitndes 5 and 13, and th«r points of appUcar tion, J and B, 6 feet tipmti. Find the nu^iind« of their resultant, and the point of application, 0. Ant, B = 18, J.0 rts 4|, BG = If 6. On a Ftraif^ht rod, AF, there are snspended a weights of 6, 15, 7, 6, -uid 9 poands respectively at the points A, B, D, E, F; AB = 3 feet, BD = Q feet, DE = & feet, EF = 4 feet l^'ind the magnitnde of the resultant, and the distance of its point of application, O, from A. Ana. Ji = i2 pounds. AG = 6f feet 6. A heavy nniform beam, AB, rests in a vertical phuae, with one end. A, on a emooth horizontal plane and the other end, B, against a smooth vertical wall ; the end. A, is prevented from sliding by a horizontal string of given length ii£- tened. to the end of the beam and to the vail ; determine the tension of the string and the preesurss against the horizontal plane and the wall Let ito =: the length of the beam, and let IF be its weight, which as the beam i> uniform, wn may suppose to act at its middle point, G. Let B be the vertical pressure of the hori«>ntal plauvj against the beam ; '\nd R the horizoot'U pressure of the vertical wall, and T the tension of the hor- izontal string, AC ; let BAG = o, a known angle, since the lengths of the boaa and th« string are given. Then (Art 61), we have tot horiaontal forces, T=s R; for vertioal forces, 17 = JK ; for moments abont A (Art 47), 2/2*^ dn « = TTa cos •> ; 80 XXAMPLSa. i 7. A heavy beam, AB = 2a, rests on two givea amooth planes which are inclined at angles, a and /3, to the horizon ; required tho angle which the beam makes with the horizontal plane, and the presfiores on the planes. Let a and b be the segmeats, AG and BO, of the beam, made by its centre of gravity, G; let E and R' be tho pressures on the planes, AC and BC, the lines of action of which are perpendicular to the planes since they are smooth, ard let W be the weight of the beam. Then we have for horizontal forces, Rsin a = R' mnP; \ (1) for yertical f oroM, R coa a + R" coa p = W; (8) for moments about G, ^ cos (a—0)=:R'b cos {li+0)\ [ (3) pi.riding (3) by (1), we have aoot a -{- atanO = bcotp — hta,n6; a cut a — 6 cot i? therefore, tan9 = a + b and from (1) and (2) wo have Wain (3 R = sin {a + 13)' and R' IT sing sin (a 4-0)' Otherwise thus : since the beam is in equilibrium under the ftotion of only three forces, they must meet in a point 0, (Art. 62), and therefore we obtain immediately from the geometry of the figure^ P w sing siu (c+/3)' « = Wainfi mi (a + 0)' IG, of the beam, t and R' be the lines of action of they are smooth, ;n we have ramU; \ (1) ip= W; {%) Aoo8(/3-fe), (3) tan $; Bin a lilibrium under Bet in a point O, lately fh)m the sin /3 « + /3)* ■^ipBHMK^'^ SXAMPLSa. 81 ana -n-, = R W sin a R' = IT Bin a sin (a + (iy • " — gin („ ^ py Also since the angles, GOA and OOB, ai-e equal to a and (i, respectively, and AGO = 5 — (?, we have « (a + ft) cot AGO = o cot GAO — b cot GOB; a cot a — ft cot /3 therefore, tan0 = a + b Hence, if x = K^r-gi the beam will rest in a horizontal position. 8. A heavy uniform beam, AB, rests with one end, A, against a smooth vertical wall, and the other end, B, is fastened by a string, BO, of given length to a point, C, in the wall ; the beam and the string are in a vertical l)lane ; it is required to determine the pressure against the wall, the tension of the string, and the position of the beam and the string. Let AG = GB = a, AC = x, BC = ft. F)|.M weight of beam = W, tension of string = T, prousare of waU = Ji, BA£ = 0, BOA = t. Then wo have for horiiontttl forces, R =: Tain ^; (1) for vertical forooB, If = 7* cos ^; (2) for momenta about A, Ifa sin 9 = T- AD = Tz sin ^; (%) . ' . a sin 9 = a; tan 9 ; (4) 89 SXAMPLSa, and by the goometr/ of the figure sin B 2a sin ^ ' z _ sin (O—ji) Ha" sin ^ (8) (6) Solving (4), (6), and (6), we get . . 1 ri6o» - ft»"]* nn from which ^ and T become known. (Price's Anal. Moch'a, Vol. I, p. 69). To determine M tho anknown qoantitles many problems in Stat cs require eqoatlona to be formed bj geometric relations as well as ttatic relations. Thus (1). (S), (8) are stntic equations, and (5) is a geometric equation. 0. A uniform heavy beam, AB = 2a, rests with one end, A, against the inter- nal surface of a smooth hemispherical bowl, radius = r, while it is supported at some point in its length by the edge of the bowl ; find the position of equili- brium. The beam is kept in equilibrium by three forces, vias., the reaction, R, at A perpendicular to the surface of contact, (Ari 42) and therefore perpendicular to the oowl, the reaction, R', at C which, for the 8an)e reason, is perpen- dicular to the beam, and the weight IK acting at 0. rtg.so ti w 01 (1 k .S( li t1 E (5) (6) (Price's Anal. problems in Stat cs ins M well as tttUic id (6) is a ge(HU«tric na.30 forccB, via!., the irface of contact, the bowl, the Mon, is perpon- ng at O. MXAMPLSa. 83 Let = the inclination of tho beam to the horizon = ■{• d. (4) From (3) and (4) a value of 6 can be obtained, and hence tlie position of equilibrium. Other^'ise thus : since the beam is in equilibrium under tiie action of only three forces they must meet in a point, 0. Geometry then gives us 2 cot 0GB = cot AOG — cot GOB = cot AOG, or 2 tan 9 = tan ^ which is the same as (3). 63. Centre of Parallel Forces in Different Planes. — To find the magnitude, direction, and point of application of the resultant of any number of parallel forces dieting on a rigid body. The theorem of Art. 59 is evidently true also in the cose in which neither the parallel forces nor their fixed points of application lie in the same plane, hence, calling $ the third co-ordinate of the point of application of the resultant, we have for the distance of the centre of parallel forces from tlie planes yz, zx, and xy, m = :lPx - _ I;Py . _ 1P« iF y = ii* £P' Hence (Art. 59, Sch.) the equations for determining the position of the centre of parallel forces show that the sum of the f I' omenta of the parallel forces with respect to any plane paraiUi to litem is equal to the mommt of their reanUani. li'i »^^ ii i iiJMia i iili i iiui i ^^ •^ V Fis*31 86 squtLTBRum of parallsl forcbb m spacxs. 64. Conditioiu of Equilibrinm of a Sjrstem of Parallel Forces Acting upon a Rigid Body in Space. — Let P,, P„ P,, etc., denote the forceF, and let them be referred to three rectangular axes, OX, or, OZ; the last parallel to the forces ; let (a;,, y„ «,), (z„ y,, «,), etc., be the points of application of the forces, /•,, Pj, etc. Ijet the direction of P^ meet the plane, ay, at Af,. Draw if,i\r, perpendicular to the axis of X meeting it at JV,. Apply at O, and also at JV,, two opposing forces each equal and parallel to P^. Then the force Pj at Mi is replaced by (1) P, at along OZ; (2) a couple formed of P, at Jf, and P, at JV, ; (3) a couple formed of P, at JV, and P, at 0. The moment of the first couple is Pijft, and this couple may be transferred to the plane ye, which is parallel to its original plane, without altering its effect (Art. 52). The moment of the second couple is P,a;,, and the couple is in the plane xz. Replacing each force in this manner, the whole ejstem will be equivalent to a force ■Pi +P, + Ps +etc., or SP at along OZ, together with the couple Ayi +-P»yt +^»y8 +eto., or I,Py, in the plane yz, and the couple P,a:i + Pt^g+P^x, +etc., or SPx in the plane xz. The first couple tends to turn the body from the axis of y to that of z round the axis of x, and the second couple 8 IN BPACMa. r a System of Xigid Body in be forces, and let Plfl<32 1 also at JVj, two » P,. Then the »,atJV,; 1 at 0. and this conplc is parallel to its (Art. 62). The the couple is in he whole system ong OZ, the plane yt^ he plane xz. am the axis of y second couple MQVTLIBRIPS Of PARALLEL FORVBa IN SVACB. 87 tends to turn the body from the avis of a; to that of t round the axis of y. It is customary to consider those couples as positive which tend to turn the body in tlie direction indicated by the natural order of the letters, i. «., positive from x to y, round the «-axi8 ; from y to « round the z-axis ; and from z to x round the y-axis ; and negative in the contrery direction. Hence the moment of the first couple is +^Py, and therefore OX is its axis (Art. 60) ; and the moment of the second couple is —I,Px, and therefore OV \» its axis. The resnltant of these two couples is a single couple whose axis is found (Art. 66) by drawing OL (in the positive direction of the axis of x) — ^Py, and OM (in the nega- tive direction of the axis of y) = ZPx, and completing the parallelogram OLOM. If 00, the diagonal, is denoted by 0, we have and i2 = 2:P; R being the resultant force. Now since this single force, R, and this single couple, O, cannot produce equilibrium (Art 64, Cor.), we must have R = 0, and = 0, and O cannot be = unless I,Px = and £Py = ; the conditions therefore of equilibrium are R = 0, IPx = 0, SPy 0. Hence, the conditions of equilibrium of parallel forces in space are: 7%0 sum of the forces must = 0. I%e sum of the moments of the forces mth respect to every plane parallel to them must = 0. wm 88 SqUrLtBRWM OF FORCXa. i KM. Fig. 33 65. ConditioiiB of EqtdUbriiim of a System of Forces acting in any Direction on a Rigid Body Sn Space. — Let P,, P,, Pj, etc., denote the forces, and let them be referred to three rectangular axes, OX, OY, 02; let (a;,, yi» «i), (*i» y». 2|). etc., be the pointa of applica- tion of P,, Pf, etc. Let J, be the point of application of Pjj resolre Pj into components X,, J",, Zj, parallel to the co-ordinate axes. Lot the direction of Z^ meet the plane xy at if J, and dr»»w if,iVi perpendicu- ^ lar to OX Apply at iV\ and also at two opposing forces each equal aud par- allel to Z,. Hence Z^ at J, or Jf, io equivalent to Z, at O, and two couples of which the former has its moment = Z, X iV",iIf, r= Z,yj, and may be supposed to act in tho plane yz, and the latter has its moment = Z, x ONi = — Z^x^ aud act* in the plane zar. Hence Z, is replaced by Z, at 0, i. couple Zi^j in the plane yz, and a couple — Z,a;j (Art. 64) in the plane zx. Similarly X, may be replaced by X, at 0, a couple X,«, in the plane zx, and a couple — X,y, in the plane xy. And F, may be replaced by T, at 0, a couple JT^a;, in the plane xy, and a couple — Y^Zi in the piano yz. Therefore the force P, may be replaced by X,, F,, Z,, acting at 6>, and three couples, of which the moments are, (Art. 54), Z,y, — Fi«, in the plane yz, around tho axis of a?, X,«, — ZiX^ in the plane zx, around the axis of y, I'jo;, — X^y, in the plane xy, around the axis of «r. By a similar resolution of all the forces we shall Lave them replaced by the forces sx, sr, iz, acting at along the axes, and the couples a SyBtein of Rigid Body ^ >e forcoa, and ht , OX, or, OZ; tointfl of applica- ,f 4 *z. r^tT' !J-X ivalent to Z, at KB ita moment = ed to act in the Z, X ON^ = iple Ziff^ in the n tho plane zx. a conple X,«, a the plane xy. pie J",*, in the yz. Therefore ,, acting at 0, ■e, (Art. 54), tho axis of x, the axis of y, tho axis of z. !s we shall Lave tiimuMm ■MMmn)'^ BqviLiBntmt of forces. 89 £ (Zy — F«) = L, suppose, in tho plane yt, S (Xz — Zx) = M, suppose, in the plane tz, S ( Fa; — Xy) = N, suppose, in the plane xx. Let U be the resultant of the forces which act at 0; a, b, c, the angles its direction makes with the axes ; then (Art. 38), R* = (1X)» + (2F)» + (^Zy» sx , sr xz coc a = -jT-> cos — -g- » cos c = -pT • Let G be the moment of the couple ^"rhich is the result- ant of the three couples, L,M,N; A, ^<, v, the angles its axis makes with the co-ordinate axes ; then (Art. 5G, Sch.), <]h = D + M^ + IP, , L M cos A = -^, cos /i = -g, COS V = N Therefore any system of forces acting in any direction on a rigid body in space may always be reduced to a single force, R, and a single couple, 0, and cannot therefore pro- duce equilibrium (Art 64, Oor.). Hence for equilibrium we must have ^ = and Q = Q; therefore ijLXf -H (sr)» + {-LZf = 0, and Z» -H iP + ^ = 0. These lead to the six conditions, SX=0, 2r=0, SZ=0, £ (% - r«) = 0, 1 (Xz — Zx) =r 0, l,{Tx^Xy) = 0. fel- 90 KXAMPLXa. EXAMPLES. 1. If the weights, 1, 2, 3, 4, 6 lbs., act pevpendicalarly to a stniight line at the reep^ctive distances of 1, 2, 3, 4, 5 feet from one extremity, find the resultant, and tho dis- tance of its point of application from the first extremity. Ans. i? = 15 lbs., x — ^ feet. 2. Four weights of 4, —7, 8, —8 lbs., act perpendicularly to a straight line at the points A, B, C, D, so that AB = 6 feet, BC = 4 feet, CD = 2 feet ; find the resultant and its point of application, G. Ana. i? = 2 lbs., AG = 2 feet. 3. Two parallel forces of 23 and 42 lbs,, act at the points A aud B, 14 inches apart; find GB to three places of decimals. -Ans. 4.954 ins. 4. Two weights of 3 cwts. 2 qrs. 15 lbs., and 1 cwt, 3 qis. 25 lbs. are su; ported at the points A and B of a straight line, the length AB - 3 feet 7 inches ; find AG to tlirfco places of decimals offset Ans. 1.268 ft. 0. A bar of iron 15 inches long, weighing 12 lbs., and of uniform thickness, has a weight of 10 lbs. suspended from one extremity ; at what point must the bar be supported that it may jnst balance. Tbe weight of the bar acts at its centre. An«. 4-^ in. from the weight. 6. A bar of uniform thickness weighs 10 lbs., and is 6 feet long ; weights of rf lbs. and 5 lbs. are suspended from its extrumitios ; on wlut point wiU it balance ? Am. b in. from the centre of the bar. 7. A beam 30 feof long balances itself on a point at one- third of its length fcom the thioker end ; but when a weight of 10 lbs. is suspemtod from tbe smaller end, tho prap must Mte pei'pendicalarly »8 of 1, 2, 3, 4, ut, and tho dis- ■st extremity. « = 3| feet. perpendicularly ), 80 that AB = e resultant and A.G = 2 feet. ict at- the points three places of ru. 4.954 ins. nd 1 cwt. 3 qid. B of a straight id AG to thrfco Ins. 1.268 ft. 12 lbs., and of suspended from ix be supported n the weight. 10 lbs., and is mspendod from 0? re of tho bar. a point at one- when a weight tho prap must n XIXAMPLSa. n be moved two feet towards it, in order to maintaiu the equilibrium. Find tho weight of the beam. Am. 90 lbs. 8. A uniform bar, 4 feet long, weighs 10 lbs., and weights oi 30 lbs. and 40 lbs. are appended to its two extremities ; where must the flilcrum* be placed to produce equilibrium P An». 8 in. from the centre of the bar. 9. A bar of iron, of uniform thickness, 10 ft long, and weighing 1^ cwt, is supported at its extremities in a hori- zontal position, and carries a weight of 4 cwt suspended from a point distant 3 ft from one ex^^^mity. Find the pressures on the points of support ^ns. 3.55 cwt, and 1.05 owt l'^ \ bar, each foot in length of which weighs 7 lbs., rests upon a fulcrum distjint 3 feet ttoTa one extremity ; whttt must be its length, that a weight of 71^ 'Hs. sus- pended from ihat extremity may just be balai i?ed by 20 lbs. suspended from the other r Ans. 9 ft 11. Five equal parallel forces act at 6 angles of a regular hexagon, whose diagonal is a ; find the point of application of Lheir resultant Ana. On the diagonal passing through the sixth angle, at a distance from it of \a. 12. A body, /*, suspended from ona end of a lever with- out weight, is balanced by a weight of 1 lb. at the other end of the ie^ er ; and when tho fulcrum is removed through naif the length of the lever it requiiea 10 lbs. to balance P ; find the weight of P. Ana. 5 lbs. or 2 lbs. 13. A carriage wheel, whoM weigbi is W and radius r, rests upon a level road ; show tha*; the force, F, necessary to draw the wheel ever an obstacle, of height A, is „V2rA-A» F=z W -T" * nto ■wpo't oil whieh U N^s. 0S BXAMPLEa. 14. A beam of nniform thickoesB, 5 feet long, weighing 10 lbs., ia Bupported on two props at the ends of the beura ; iind where a weight of 30 lbs. must be placed, so that the pressures on the two props may be 15 lbs. aud ib lbs. Ana, 10 ins. from the centre. 15. Forces of 3, 4, 5, lbs. act at distances of 3 ins., 4 ins., 5 ii s. 6 ins., from the end of a rod ; at what distance from the same end does the resultant act ? Ans. 4| inches. 16. Four vertical forces of 4, 6, 7, 9 lbs. act at the four corners of a squai'e ; find the point of application of the resultant. Ana. -f^oi middle line from one of the sides. 17. A flat board 12 ins. square is suspended in a hori- zontal position by strings attached to its four comers, A, B, C, D, and a weight equal to the weight of the board is Inid upon it at a point 3 ins. distant from the side AB and 4 ins. from AD ; find the relative tensions in the four strings. An$. As | : | : | : ^. 18. A rod, AB, moves freely about the end, B, as on a hinge. Its weight, W, acts at its middle point, and it is kept horizontal by a string, AG, that makes an angle of 45" with t. Fina the tension in the stiinr. . W " Ana. — — • 19. A rod 10 inches long can turn freely about one of its ends ; a weight of 4 lbs. is slung to a point 3 ins. from this end^ and the rod is held by a string attached to its free end and inclined to it at an angle uf 120° ; find the tension in the striug when the rod is horizontal. Ana. \ v/S lbs. 20. Two forces of 3 lbs. and 4 lbs. act at the extremities cf a straight lever 12 ins. long, and inclined to it at angles of 120° aud 136° respectively ; find the position of the fulcrum. j^na. (8-3 V6) x 9.6 ins. from one end. Ji long, weighing i of the beuni ; id, 80 tliat the J 25 lbs. 1 the (cntre. ices of 3 ins., : what distance >. 4| inches. ct at the foar lication of the of the sides. ied in a hori- ur corners, A, f the board is side AB and I in the four id, B, as on a )int, and it is angle of 45" Ans. — — • V2 abont one of nt 3 ins. from led to ita free 0° ; find the al. t \/3 lbs. le extremities } it at angles sition of the m one end. MXAMPLKS. ^ 21. Find the true weight of a body which is found to weigh 8 ozs. and 9 ozs. when placed in each of the scale- pans of a false balance. ^,„. eVaozs. 22. A beam 3 ft. long, the weight of which is 10 lbs., and acts at its middle point, rests on a rail, with 4 lbs. hang- ing from one end and 13 lbs. from the other ; tind the point at which the beam is supported ; and if the weights at thu two ends change plaoee, what weight meat be added to thn lighter to preserve equilibrium ? Ans. 12 ins. from one end ; 27 Ibe. 23. Two forces of 4 lbs. and 8 lbs. act at the ends of a bar 18 ins. long and make angles ot 120° and 00° with it; find the point in the bar at which the resultant acts. An*. H (* — VS) ins. from the 4 lbs. end. 24. A weight of 24 lbs. is suspended by two flexible strings, one of which is horizontal, and the other is inclined at an angle of 30° to the vertical. What is the tension in each string ? ^^s. 8 V3 lbs. ; 16 ^3 lbs. 25. A pole 12 ft long, weighing 25 lbs., rests with one end against the foot of a wall, and frotn a point 2 ft. from the other end a cord mns horizontrtlly to a point in the wall 8 ft. from the ground ; find the tension of the cord and the pressure of the lower end of the pole. Ans. 11.25 lbs.; 27.4 lbs. 20. A body weighing 6 lbs. is placed on a smooth piano which is inclined at 30° to the horizon ; find the two direo- tionn in which a force equal to the body may act to produce equilibrium. Also find vha*. is the pressure on the plant in each case. Atu. A force at 60° with the pUne, or vertically upwards ; R = ey/3, or 0. 27. A rod, AB, 5 ft. long, without weight, is hnng fn>m a point, 0, by two strings which are attached to its ends i 94 SXAMPLSa. and to the point ; the atring, AG, is 3 ft^ and BO is 4 ft in length, and a weight of 2 lbs. is hang from A, and a weight of 3 lbs. from B ; find the tensions of the strings. Ana, V5lbs.; 2 V5 lbs. 28. Find the height of a cylinder, which can just rest on an inclined plane, the angle of which is 60°, the diameter of the cylinder being 6 ins. and its weight acting at the middle point of its axis. Am. 3.46 ins. 29. Two equal weights, P, Q, are connected by a string which passes over two smooth pegs, A, B, situated ii> a horizontal line, and supports a weight, W, which hangs from a smooth ring through which the string pusses ; find the position of equilibrium. Atu. The depth of the ring below the line W AB = 2 ^4/^ - W AB. 80. The resultant of two forces, P, Q, acting at an angle, $, is = {2m + 1) \//* + Q^; when they act at an angle, a — 6, it is = (2f» — 1) VP*+ Q*; show that tan e = « tn — l m + 1 81. A uniform heavy beam, AB = 2a, rests on a smooth peg, P, and against a smooth vertical wall, AD ; the horizontal distance of the peg from the wall l)cing h ; find the inclination, 0, of the beam to the vertical, and the pressnrcs, R and S, on the wall and peg. Ans. e = s.n-. (*)*; S = w(^f; R = ff^^^. 83. Two equal smooth cylinders rest in oortaot on two smooth planes inclined at angles, a and (i, to the horizon ; Pig.M ■■- 9. mtr s fmmvmt 1 BO is 4 ft in i, and a weight ings. 1.; 2 V5 1b8. an just rest on , the diameter : acting at the ns. 3.46 ins. ;d by a string situated in a which hangs g passes; find g at an angle, at an angle, that tan 6 = rig.J4 i wall and peg. A* ntaot on two tlie horizon; MXAMPLHa. 90 find the inclination, 9, to the horizon of the line joining their centres. Ans. tan =- 1 (cot a — cot j3). 33. A beam, 6 ft. long, weighing 5 lbs., rests on a ver- tical prop, CD = ^ ft. ; the lower end, A, is on a hori- zontal plane, an-l is prevented from sliding by a string, AD = 3^ f t. ; find the tension of the string. Ans. T—\ lbs. 34. A uniform beam, AB, is placed with one end, A, inside a smooth hemispherical bowl, with a point, P, rest- ing on the edge of the bowL If AB = 3 times the radius R, find AP. Ans. AP = 1.838 R. 35. A body, weight W, is suspended by a cord, length I, from the point A, in a horizontal plane, and is thrust out of its vertical position by a rod without weight, acting at another point, B, in the horizontal plane, such that AB = rf, and making the angle, 6, with the plane ; find the tension, T, of the cord. Am. r = IF ^ cot <». 36. Two heavy uniform bars, AB and CD, movable in a vertical plane about their extremities. A, D, which rest on a horizontal plane and are prevented from sliding on it; find their position of equilibrium when leaning against each other. Let the bars rest against each other at B, and let AD = o, AB = h, CD = «?, BD = a;, W and }\\ = the weighte of AB and CD, respectively acting at thoir middle points ; then we have 2«« »r (a» -h 4» - a^) = c FT, (a« -H *• - «») (i« + a^ - ««), which is an equation of the fifth degree, and hence always has one real root, the value of which may be determined when numbers are pat for and AFP = ft ♦w rrs.M Awi. = 3co8-»(|)* 38. Find the form of the curve in a vertical plane such that .1 heavy bar resting ou its concave s'de and on a peg at a given point, say the origin, may be at rest in all positions. Ans. r = y + k eeo e, in which I = the length of the bar, k an arbitrary constant, and d the inclination of the bar to the vertical. It is the equation of the conchoid of Nicomedes. 39. A rod whose centre of gravity is not its middle point is hung from a smooth peg by means of a string attached to its extremities ; find the position of equilibrium. A)U. There are two positions in which the rod hanga vertically, and there is a third thus defined :— Let F be the extremity of the rod remote {torn the centre of gravity, k the distance of the centre of gravity from the middle point of the rod, 2a the length of the string, and 2c the length of the rod ; then measure on the string a length FP from F equal to o (l + - 1, and place the point P over the peg. This will define a third poutiou of equilibrium. 3 co«r« (^f. ical plane such jnd on a peg ut it rest in all length of the ination of the le conchoid of middle point itring attached brium. he rod hangs -Let F be the of gravity, k middle point the length of FP from F )Ter the peg. a. MXAMPLMa. W 40. A smooth hemisphere is flxdd on a horizontal plane, with its convex side turned upwards and its base lying in the plane. A heavy uniform beam, AB, rests against the hemisphere, its extremity A being just out of contact with the horizontal plane. Supposing that A in attached to a rope which, passing over a smooth pulley placed vertically over the centre t^t the hemisphere, sustains a weight, tind the position of tqnilibrium of the beam, and the requisite magnitude of the snspended weight Ans. Let W be the weight of the beam, 2o its length, P the suspended weight, r the radius of the hemisphere, h the height of the pulley above the plane, 6 and ^ the inclinations of the beam and rope to the horizon ; then the position of equilibrium is defined by the equations. r oosec = A cot ^, r cosec* = a (tan ^ -f- cot 6), which give the single equation for 6, r (r — a sin 9 cos 6) = ah sin* B. Also W sin 9 cos {p — (?) = W a sin» e Vr* + A» sin* (9 (1) («) (8) (4) 41. If, in the last example, the position and magnitude of the beam be given, find the locus of the pulley. Ans. A right line joining A to the point of intersection of the reaction of the hemisphere and W. 42. If, in the same example, the extremity. A, of the beam rest against the plane, state how the nature of the problem is modified, and find the position of equilibrium. Ans. The suspended weight must be given, instead of being a result of calculation. Equation (1) still holds, but BXAMPLSa, li ■' not (2) ; and tho position of equilibrium is defined by the equation Ph* co8^ = War Bin» 0. 43. If the fixed hemisphere be replaced by a fixed sphere or cylinder resting on the plane, and the extremity of the beam rest on the ground, find the position of equilibrium. Ans. If A denote the vertical height of the pulley above the point of contact of the sphere or cylinder with the plane, we have r cot 5 = A cot ^, Pr (1 + cot 5 cot 0) cos = FTa cos 0. 44. One end, A, of a heavy uniform beam rests against a smooth horizontal plane, and the other end, B, rests against a smooth inclined plane ; a rope attached to B passes over a smooth pulley situated in the inclined plane, and sustains a given weight; find the position of equilibrium. Let 6 be the inclination of the beam to the horizon, a the inclination of the inclined plane, W the weight of the beam, and P the suspended weight ; then the position of equili- brium is defined by the equation cos « ( ITsin « - 2P) = 0. (1) Hence we draw two conclusions: — (o) If the given quantities satisfy the equation IF sin a — 2P = 0, the beam will rest in all positions. (*) There is one position of equilibrium, namely, that in which the beam is vertical. This position requires that both planes be conceived as prolonged through their line of intersection. 45. A uniform beam, AB, movable in a vertical plane about a smooth horizontal axis fixed at one extremity, A, is li •taatV fttatrar^ i::tl, defined by the r a fixed sphere tremity of the t equilibrium, iie pulley above uder with the cos 0. rests against a B, rests against > B passes oyer le, and sustains um. horizon, a the \t of the beam, ition of equili- (1) nation TFsin « IB. lamely, that in )e conceived aa vertical plane xtremity. A, is MXAMPLES. 99 attached by means of a rope BC, whose weight is negligible, to a fixed point in the horizontal line through A, such that AB =: AC; find the presau.^ on the axis. Am. If d = there will be le the reaction weight down- ody will rotate -According to ought to rest that its cent;^ i]gh the point iy 80 situated ition of equili- onld not snb- )8ition by the ; further from This kind of possible, are stable. Thus 'e equilibrJum, tion of stable generally as f slightly dis- )n, the equili- er away from its orifiutf J^sition, its equilibrium is unstable. When it reinaim in its new position, its equilibrium i . neutral. A sphere or o^ndrical roller, resting on a horizontal surface, {equilibrium. In stahk equilihrium the centre IS m of grwntp occupies the lowest possible position; and in unstable it occupies the highest position. Wc shall first give a few elementary examples. 71. Oiven the Centres of Qravlty of two Mauea, Ml and M„ to find the Centre of Qravity of the two Masses as one System.— Let gi, denote the centre of gravity of the mass if,, and g, the centre of gravity of the mass Mg. Join g^ g, and divide it at the point, O, so that Og.- M\' ma6ses as one system (Art 46). then G is the centre of gravity of the two 72. Oiven tbe Centre of Gravily of a Body of Mass, M, and also the Centre of Ghnvity of a part of the Body of Mass, m, to find the Centre of Qravity of the remainder. — Let denote the centre of gravity of the mass, M, and jr, the centre of gravity of the mass, nt|. Join Og^ and produce it through O tog,, so tliat ^^ =r ^_* — , then g, is the centre of gravity of the remainder (Art 45). 73. Centre of Gravity of a Triangnlar Figure of Uniform Thickness and Density. — Let ABO be the triangle; bisect BC in D, and joiu AD; draw any line bdc parallel to BC ; then it is evident that this line will be bisected by AD in d, and will therefore have its centre of gravity at d ; similarly every line in the triangle parallel to BO will have its centre of gn^vity in AD, and therefore the centre of gravity of the triangle must be somewhere in AD. Frg.37 ■ ?>*/■' 104 CXXTBB OF ORAVnr or il TRZAJteLM. In like )> .annor the pentre of gravity must lie on the line BE which joiuB B to the middle point of AC. It is there- fore at t le intersection, G, of AD and BE. Join DE, whih wiU be parallel to Afi ; then the triangle^ ABO, DEG', are similar : therefore AG GD AB_ BO DE~DO i' or GD = iAG = ^AD. Hence, to find ih« centre of gravity of a triangle, bisect any . side, join the point of bisection with the opposite angle, the centre of gravity Ivss one third the way up this bisection. Cor. 1.— If three equal particles be placed at the vertices of the triangle ABC their centre of gravity will coincide wich that of the triangle. For, the centre of gravity of the two equal particles at B and is the middle point of BO, and the centre of gravity of the three lies on the line joininjj this point to A. Similarly, it lief> on the line joining B to the middle of AC. Therefore, etc. Cob. 2.— The centre of gravity of any plane polygon may be found by dividing it into triangles, finding the centre of gravity of each triangle, and then by Art 69 deducing the cenb'c of gravity of the whole figure. Let the coK)rdinate8 of A, referred to any axes, «. ; those of B!, x,, y„ «, ; and those of 0, ar„ *t » Cob. 8. be a;,, yj, .. , y„ «, ; then (Art 69), the co-ordinates, 5, y, t, of the centre of gravity of three equal particles placed at A B, 0, respec- tively, are • = 3 * ~ 3 + y. ; = 'jL±_«ijtii; M.. i«ta AjreLK. t lie on the line 0. It is there- en the trianglcSy %ngle, Usect any . poaite angle, the 'tis bisection. d at the Tcrtices ty will coincide 1 particles at B entre of gravity point to A. middle of AC. 18 ms polygon may ig the centre of 9 dedacing the >ed to any axes, i> those of C, X •, of the centre A. B, 0, respec- + y. CJWVTBi OF ORAVTTr OF A PTRAXID, which aro also the co-ordinates of the<»ntre of gravity «f the triangle ABO (Oor. 1). 74. Cemtre of Qncvity of a Trlangnlar Pyramid of Unifonn Density.— Let D-ABO be a triangular pyramid; bisect AO at E; join BE, DE; take EP = ^EB, tLan P is the centre of gravity of ABC (Art 73). Join FD ; draw ab, h;-, ea parallel to AB, BC, CA respectively, and let DF meet the plane, abc, at /; join bf and produce it to meet DE at e. Then ^^ since in the triangle ADC, ac is parallel ^*''* ° to AO, and DE bisects AC, e ip tne middle point of «c; also but 'herefore K-W - sL. BF ~ DF ~ EF • EF = IBF, therefore /is the centre of gravity of the triangle a Jc (Ari 73). Now if we suppope the pyrnmid to be divided by planes parallel to ABC into an indefinitely great number of triangular laminae, each of these laminae has its centre of fjravity in DF. Henoe the centre of gravity of the pyramid if^ in DF. Again, take EH r= JED ; join HB cutting DF at G. Then, as before the centre of the pyramid must be on BH. It is therefore at the intersection, G, of the liues DF and BH. Join FH ; then PH is parallel to DB. Also, EP = pB, therefore FH = JDB ; and in the similar triangles, FGH and BGD, we nave therefore PG _ FH _ 1 M ~ D B "~ 8 ' FG = JDG = iDF. 106 CMSTRM OF QRA. VrrT OF A CONS. \.s' 3 u m Eli' I nonce, the centre of gravity of the pyramid ia one-fourth of the way up ike line joining the centre of gravity of the base with the vertex. (Todhuni a Statics, p. 108. Also Pratt's Mechanics, p. 63.) Cor. 1.— The centre of gravity of four equal particles placed at the vortjices oi the pyramid coincideo with the centre of gravity of the pyramid. Cob. 2.— Let (x,, yj, «i) be one of the vertices ; («„ y„ «,) a second vertex, and so on ; let (i, y, i) bto the centre of gravity of the pyramid ; then (Art 59) • = J (*, + «, + ;r, + «4). • » = i (yi + yi + y» + ^t). ; = J («, + «,+«, + «4). Cor. 3.— The perpendicular distance oi' t'le centre of gravity of a triangular pyramid from the base is 64031 to { of the height of the pyramid. 75. 0«ntn of Qnvitj cX a Cone of Uniform Denaity havjog mnf Plana Baaa.— Consider a pyramid whose base ij a polygon of any number of sides. Divide the base into triangles ; join the vertex of the pyramid with the vertices of all the triangles ; ihen we may consider the pyramid as composed of a number of triangular pyramids. Now the centre of gravity of each of these triangular jiyramids lies in a plane whose distance fr&m the base is one-fourth of the height of the pyramid (Art 74, Cor. 3) ; tf ,refore the centre of gravity of the whole pyramid lies in this plane, ». u, i*« perpendicular distance from the base is one-fourth of the height of the pyramid. Again, if we suppose the pyramid to be divided into an i?»deflnitely great yiumber of laminu. as in Art 74, each of these lamina!) has its centre of gravity on the right UiM nid is om'fourih of gravity of the :8, p. 108. Also r equal particles incidef) with the tice«;(a;„y„«,) to the centrt) of )i' the centre of Nisc is e4a3l to \ M of Uniform laidcr a pyramid of sides. Divide he pyramid with nay consider tho ovular pyramidfi. Iicse triangnlar f^m tho base is Art. 74, Cor. 3) ; pyramid lies io from the base is divided iato an Art 74, each of the right liiM csNTRs OF Qturmr. 107 joining the vertex to the centre of gNTity of the base ; and hence the centre of gravity of the whole pyramid lies on tills Une, and hence it mast bo one-fourth tiie way up this line. There is no limit to the number of sides of the poly- gun which forms the base of the pyramid, and henoe they may form a continnous curve. Therefoie, the centre of gravity of a cone whose base is any plane curve whatever is found by joining the centre of gravity of the base to the vertex, and taking a point one- fourth of the tvay up this line. 76. Contre of Ckavity of tho Fnutom of a Pyra- mid. — Let ALC-abc (Fig. 38) be the frustum, formed by ilie removal of the pyramid, D-abc, from the whole pyramid, D-ABO ; let A, and ff be the perpendicalar heights of these ])yramids, respectively; let m and if denote their masses; and let Xi, «,, denote the prpendicular distances of the centres of gravity of the pyramids D-ABO, and D-abe, and tho frustum, from the base \ then we have (Art 60, Soh. 1) Jf«j = i ( J/" — m) + ««, ; or Rat mz M-m ». = -r; i. (1) 4' = (ir-»,) + V = ^-**»- Also, the maasea of the pyramids are to each other an the!. M)lumos* by (1) of Art \%, and therefore as the oubsa of tlieir heights. Henoe (1) becomes * irtiM bodiM Mw iMMMRenmut, tlw roiniiiM or Ike weigtato m« proportkwal to I 111- oMwaM, and m»f be mibi>tlwia4 kn tlHM. 108 SXAMPLSa. H- H* + iTA, + A, * (2) Instead of the heights we may use any two corresponding lines in the lower and npper baaes, to which the heights are proportional, as for example AB and m its lower end = 3 a-\'b 2. If out of any cone a similar cone ic cut so that their axes are in the same line and their bases in the same plane. And the height of the centre of gravity of tha remainder above the base. Take momenta with refefenoe to the VLV +JV (2) '0 oorresponding I the heights are Denoting these ostum hy A, (2) (3) on any base, a ds, and hence it I any plane base. oid in t«rm8 of d b, and of the lol tide. sides and at a t so that their the same plane, thd remainder llfTSOBATIOJf FORMULA. 109 Ana, J . TTZTTt* "^^^^ *' " *^® height of the"^)^iginal cone, and h', the height of that which is cut out of it. 3. If out of any cone another cone is cut having the same base and their axes in the same line, find the height of the centre of gravity of the remain«ier above the base. Ans. J(A + Ai), where h and A, are the respective heights of the original cone and the one that is cut out uf it. 4. If out of any right cylinder a cone is cut of the same base and height, find the centre of gravity of the remainder. Ans. Iths of the height above the base. 77. iDvestigKtions Iiivolvivf Integration.— The general formula for the co-ordinates of the centre of gravity vary according as we consider a material line, an area or thin lamina, or a solid ; and assume different forms accord- ing to the manner in which the matter is supposed to be divided into mfinitesimal elements. In either case the principle is the same ; the quantity of matter is divided irto an infinite number of inficitesimal elements, the mass of the element being dm ; multiplying the element by its co-ordinate, x, for example, we get X ' dm, which is the moment of the element* with respect to the plane yt (Art. 63) ; and /x • dm is the sum of the moments of all the elements with respect to the plane yx, and which corresponds to SPz of Art 63. Also, /dm is the sum of the nuases of all the elements which correspond to £P of the come Article. Hence, dividing the former by the latter we have * The momsnt of th« ft>TC* Mtlng on elemAnt dm U Mriotlj dm -gic, bnt cinM the eooittnt g appean In botb terma of expreeilnn ror co-ordinate* of centre of Kravlty, it may be omitted and It becomet more conrenlent to ipaak of the ptomtnt ot the timtmt, mtaniag by It the product of the maM of the al«awF>t dm, and it« arni.z. Tha momant of an ato a a i it maaa nw a lt« «g» a dateraUUng the poaltfcm of the oantrt of gmTUj. 110 CKIfTKS OF QBA VITT OF A LiytU m fx • dm /dm Similarly i = ^ dm t = /dm * /t'dm ^ ~/dm'* (1) (2) (3) the limits of integration being determined by the form of the body ; the sign, /, is used as a general symbol of sum- mation, to be replaced by the symbols of single, doable, or triple integration, according as dm denotes the mass of an elementary length or surface or solid. Hence, the co-or- dinate of the centre of gravity referred to any plane is i tal to the gum of the moments of the elements of the mass referred to the same plane divided by the sum of the elements, or the whole mass. If the body has a plane of symmetry (Art. 67), we may take it to be the plane xy, and only (1) and (2) are necessary. If it has an axis of symmetry we may '^ke it to bo the axis of z, and only (1) is necessary. 79. Centn of Qn^ity of the Are of a Conro.— If the body whose centre of gravity we want is a material line in the form of the arc of any curve, dm denotes the mass of an elementary length of the carve. Let ds =■ the length of an element of the curve ; let h = the area of a normal section of the curve at the point ix, y, z), and let p = the density of the matter at this point Then (Art. 11), we have dm = kpds, which is the mass of the element ; multiplying this mass by its co-or- dinate, X, for example, we have the moment of the element, (kftxds), with respect to the plane, ? ». Hence, substituting for dm in (1), (2), (8), of Art. 77, the linear element, kpds, we obtain, for the position of the centre of gravity of a b«jdy in the form of any curve, the oquutioni u, (2) (3) 1 by the form of I symbol of Bum- single, doable, or I the mass of an Hence, the co-or- \ny plane is t lal ents of the mass m of the eUmenis, Hie of symmetry ty, and only (1) of symmetry we ) is necessary. 9f aCnnr*.— If s a material line lotea the mass of ' the curve; let rte at the point matter at this ds, which is the ass by its co-or- of the element, (8), of Art. 77, position of the any cnrre, the MXAkPLJSH. fkpxdi C := -_/kpyds ^ - J'kpds ' • = /kptd$ fkpds' (1) (8) (8) The quantities h and p mast be given as functions of the position of the point {x, y, «) before the integrations can be performed. If the curve is of doable corvature all three equations required. If it is a plane curve, we may take it to be are in the plane xy, and (1) and (2) are sufficient to determine the centre of gravity, since i = 0. If the curve has an axis of symmetry, the axis of * may be made to coincide with it, and (1) is sufficient « KXAMPL.ES. 1. T find the centre of gravity of a circular 4ro of uni- form thickness and density. Let BC be the arc, A its middle point, and the centre of the circle. Theti as the arc is symmetrical with respect to OA its centre of gravity must lie on this line. Take the origin at 0, and OA as axis of x. Then, since k and p are constant, (1) be- comes fxds m X being the co-ordinate of any point, P, in the arc. Let 9 be the angle POA, and a the radius of the circle, and let =: the angle BOA. Then II m 113 and Hence m = /!- BXAMPLSa. 2 = a COS 9, da = ade. J COB Od$ COB e dO = a f*a do PdO = a Bin a Therefore, the distance of the centre of gravity of the arc of a circle from tJte centre is the product of the radius and the chord of the arc divided by the length of the arc. Cob.— The distance of the centre of gravity of a semi- circle* from the centre is 2a 3. Find tb« centre of gravity of the quadrant, AT/, (Fig. 89), referred to the co-ordinate axes OZ, OF. The equation of the circle is a* + y« = a». . ^ aements of the first order (Arts. 79 and 80). Suppose that the density of the hunina AOB (Fig. 41), is not nniform. If we divide it into triangular eletnents, POQ, the element of mass will be no longer proportional to the element of area, POQ = ^i*d9', nor will the centre of gravity of the triangle, POC* ^ ^ distant from 0. Let a series of circles be described with as a centre, the distance between any two successive circles being dr. These circles will divide the triangle, POQ, into an infinite number of rectangular elements, abed = rdBdr. If k is the thickness and p is the density of the lamina at this ele- ment, the elemeii' of mass will hedtn — kprdSdr; and the coH>rdinates of its centre of gravity will b '' ri?9 6 and r sin 0, Hence, from (1) and (3) of Art '/7. f n'* m = and r fk pr done rdBdr f fkpr* cop & n'J - r ffhr de dr ff^ ^ *''* V — /A'* Bin e de dr //^ d»dr (1) (•) In each of these integrals the values of k and p are to be snbstitnted in terms of r and 6, and the intQgrstiona taken between proper limits. 3al.,Art.l57). nnlaB.— When I point, it may Hocond order ts of the first OB (Fig. 41), ular eletaientfl, >roportional to the centre of >mO. as a centre, ^les being dr. tto an infinite ddr. If k is oa at this ele- prdBdr; and > K ' fi?» 8 and 6 rrj r.r - , (1) )dr (2) d p are to be rations taken MXAMPLM. EXAMPLB. m Find the centre of gravity of the area of a cardioid in which the density at a point increases directly as its distance from the cusp. Let ft = the density at the unit's distance from the cusp, then p = (ir, is the density at the distance r from the cusp. As the axis of the curve h an axis of symmetry (Art 67), y = 0, and the abscissa ol the whole curve is the same as for the halfabove the axis ; then (1) becomes w = = } f /f*cmedddr r rt*dedr Jr*0Med6 fr*dB by performing the reintegration. The equation of the curve is d r s a (1 + COB 9) = 2a oo*» 5 Substituting this value for r, and putting ^ = 0, we have 2 s: la J 00^ ^ (2 cos* ^ — 1) «^ / oo^^d^ = H«. in RXCTJLNaVLAB FORMULAE. 82. DonUe Integntioii.— Reotugiilar Fommte.— Let » aeries of consecative straight lines be drawn parallel to the axea of x and y respectively, diTidiog the area, ABOD, (Fig. 40), into an infinite nnmber o<' rectangular elements of the second order. Then the area of each element, as abed, — d^dy; and if k and p are the thickness and density of the lamina at this element, the element of man will be dm s= kpdxdy, and the co-ordinates of its centre of gravity will be X and y. Uenoe from (1) and (2) of Art 77, we have • = 9 — J jk (Kcdxdif yykpdxdy JJhpydxdy J I lepdxdy the integrations being taken between proper limita (1) (») EXAMPLE Find the centre of gravity of tVa area of a cycloid the density of which varies as the nth power of the distance from the base. Take the base as the axis of x and the starting point as the ori^!ii. Then the equation of the curve is • s a vers-*^ - (Say - y»)* ; dx- V2ay- r Fonnnlie.— drawn parallel le area, ABOD, ^lar elements ich element, as ess and density f man will be sntre of gravity of Art 77, we (1) (2) limits. a cycloid tbe the diatance rting point as aUBFACa OP RSVOLUnON. 123 Let p = ^y" = density at the distanoe y from the base. It is evident that tbe centre of gravity will be in the axis of the cycloid ; therefore i = rra ; and as it is constant (3) becomes y = J^J^trdydx n + I t/p * _n + l ^o V2ay — y « «-»-2* n + 3 " 'v/2ffy — y» rtf*** dy V2ffy — y«. r V2«y — p y = n_-f_l 2n + 5 n + 2" n + 3 ' 83. Centre of OniTitjr of • Bnrfiuse of Revola- tion. — Let a surface be generated by tbe revolution of the curve, aB (Fig. 40), round the axis of x. Then the elementary arc, PQ, (= ds), generates an element of the nirf«o« whose aree = Stry di (Gal., Art 1.93). It k ia the tbicknoB8 and p the density of the lamina or shell in this elementary zone, the element of mass will be dm = %nkpy ds. Also the centre of gravity of this zone is in the axis ul x at 134 SXAMPLSa. the point M whose absoiaaa is x and ordinate 0. Hen<» (1) of Art. 77 becomes, after cancelling *in, (1) Jkpxy da X := — — __ Jkpydt (.he integrations being taken between proper limits. BXAMPLSS. 1. Find the centre of gravity of the surface formed by the revolution of a semi-cycloid round its base. The equation of the generating curve is a » =z a vers-* * — \^^y — y*; d» or V3a-y which in (1) gives, after cancelling 's/'ia kp. 2. Find the centre of gravity of the sarteoe formed by the revolution of a semi- cycloid round its axis. It is clear that the centre of gravity lies on the axis of the curve ; hence y = 0. ' 0. Hence (1) (1) [imits. ace formed b; e. e formed by D the axis of MXAMPLM8. The equation of the generating carve is 135 y = a yerr-> - + V-Sox — a». a Here which in (1) gires ds = VZax'^dx, J^ifxidx m = J^y7r\ix [lya;* — \fx's/%a — x dxT \%yx^ — 2 /V2a — xifoT* _ |ff^ - fy **'•"/* r <^dy * J J {a*-^ r* - y«)* I BnifAoe.— ! ;hth of the 30i«/D or aUVOLOTION. 137 First perform the y-inljegn- tion, X being constant, firom y ^0 to y = -W = Sfi = \/«* — a^ ; the effect will be to sum up all the elements similar to pq from H to I. The eflbct of » subsequent a;-integration will be to sum all these elemental strips that are comprised in the surface of which OAB is the projec- tion, and the limits of this iutegratiou are x = and X = OA =z a. Hence t Fig. ♦?. a = Jo Jn (a' xdxdy »»)* P» /f' dxdy Jn Jq (a» - a» - y»)* J^rxdz id. Similarly P = fi, i =r idz (1) the int^rations being extended over the whole area, CABD, of the bounding curve. If the density varies, the element of mass may require to be taken differently. If the density varies with z alone, t. c, if it is uniform all over the rectangular strip, PQNM, the volume may be divided up as already done, and the element of mass = Ttfyj^ dx. Hence, we shall have v\ this case, m = Jpj/*xdx — — — ■ — ■ ' ■ - — - • fpfdx (2) If the density varies as y alone, we may take a rectangular element of area of the second order, dx dy, at the point (x, y) ; this area will generate an element of volume = 2ny dx dy ; therefore the element of mass = 2irpy dx dy, and we have » = JJpxydx :y fj'pydxdy (8) the y-integrations being performed first, from to y, the ordinate of a point P, on the bounding curve ; and then the x-integrations from OC to OD. mm' 8). Henoe if the »r the position of I the axis of x), (1) the whole area, 18 may require to with « alone, t. 0., rip, PQIf^M, the and the element in this case, («) ke a rectangular iff, at the point lent of volame = 2iTpy dx dy, (8) ■om to y, the iirve ; and then BXAMPLSa. SXAMPLE8. 189 - 1. Find the centre of gravity of the hemisphere generated by the revolution of the quadrant, AD, (Fig. 39), round OA (taken as axis of x), (1) when the density is uniform ; (2) when it is constant over a section perpendicular to OA and varies as the distance of this section from OD ; (3) when it is constant at the same distance from CA And varies as this distance. (1) From (1) we hare X = jj/hidx J y*dx Putting 35 = r cos 6, and y = r sin e, where r is the i-adius of the circle and integrating hetween = and e 2' we have (2) Since p = fia;, we have from (2) faN/»dx w = which gives (3) Since p » =: ~m - tV- Hy, we have from (3) r Ixj^dxdy Cxf^dx ffy*dxdy fi^dx' 180 axAMPLm. ■•, If aatl the previons Bnbstituv^ons for x and y give 16r u = ISir 2. Find the centre of gravity of a paraboloid of revolu- tion, the length of whose uxis is A. Ant. i = \h. 3. Find the centre of gravity (1) of a portion of a prolate spheroid, the length of whose axis measured from the vertex is c, and (2) of a heini-spheroid. An». (1) i = I -g— ~; (2) * = |a. 66. Polar Fonnnlie.— a solid be generated by the revelation of AB, (Fig. 41, 1 the axis of x. Then the elementary rectangle, ahcd, whose mass = pr dd dr, (Art 81), the thickness being omitted, generni'ts a ring which is an element of the solid whose volume = %nr sin pr dd dr ; and the abscissa of the centre of gravity of the ring is r cos 0. Hence (1) of Art. 77 becomes r jp^ sin cos (/d dr ~7P- sin eMdr (1) in which p mnst be a function of r and 6 in order that the integrations may be effected. If the density depends only on the distance from a fixed point in the axis of revolution, this point may be taken as origin, and p will be a function of r ; if the density depends only on the distance from the axis of revolution, p will be a function of r sin 6. IXAMPLK. The vertex of a right circular cone is in the sur&ce of a sphere, the axis of the cone coinciding with a diameter :>f ygire rsboloid of revolu- Ang. i = |A. wrtion of a prolate rod from the vertex I generated bj the 18 of X. Then the = pr dO dr, (Art 5^8 a ring which is Inr sin 6 pr dS dr ; ty of the ring is dr (1) in order that the nee from a fixed may be taken as density depends evolution, p will the surface of a th a diameter c»f OENTRK OF ORAVtTT Of AITT 80UO. 131 the sphere, the base of the cone being a portion of the sur- face of the sphere. Find the distance of the centre of gravity of the cone from its vertex, 2a being its vertical angle, and a, the radius of the sphere. Here the r-limits are and 2a cos ; the 0-limits are and « ; p is constant ; hence from (1) wo have a = *'0 «^0 y sin 6 cos dS dr r Ht* sin e dd dr = 1 r(2fl (,08 (?) * ein cos d$ = |« r(2a cos 6)* sin d9 /"co8^ e sin 9 d9 /■ cos* sin d do 1 — coal* a 1 — cos* a. 87. Centre of Gkavity of any Solid.— Let (x, y, z) and {x •{■ dx, y ■{■ dy, z + dz) be two consecutive points E und F, (Fig. 42), within the solid whose centre of gravity is to be found. Through E, pass three planes parallel t. .'. »=s^a. Similarly y = fft, ■ = |«. 2. Find the centre of gravity of the solid bonnded by th ' planes » =r (ix, n = yx, and the cylinder y» = 2a« — i*. 88. Polar El«m«iit» of MMm.— Lei Fig. 43 repre- sent the portion of the volume of a solid included between its bounding surfaov; and three rectangular oo-ordinato planon, 'Ul 134 POLAR MLEMSSTB OF MASS. mi (1) Throngh the axis of z draw a aeries of consecutive plaues, divid- ing the soUd into wedgenshc^d slices such as OOBA. (2) Round the axis of t describe a series of right cones with their vertices at O, thus dividing each slice into elementary pyramids like 0-PQST. (3) With as a centre describe a series of consecutive spheres; thus the solid is divided into elementary rectangular par- allelopipeds similar to abpt, whose volume — ap-ps' at. Let XOA = , COP = e. Op =2 r, AOB = dit, rOQ =:de, pa- dr. Then pq is the arc of fe circle whose radius is r, and the angle is de ; therefore pq = rd9. Also pa is the furc of a circle in which the angle is d^ sin 6 dr dd dp ; and if p is the density of the solid at p, the element of mass is pr* »iu e dr de dp. Also the oo>ordiDates of the centre of gravity of this element are r sio ooe ^, r ein 9 aiu , 'm p on OZ, or >araIlelopiped =r the element of rwvity of thii rco8ff; mmm HXAMPLSa. 135 hence for the centre of gravity of the whole solid we have / / fpf* sin* Ocoifclrddd^ J'fjp** Bin edrdddit a = fffpr* »in* e iia ift dr de d^ Jjffpf* sin edrded^ r r ipr* sin COB dr dO d^ ti = fffpf* sin Bdrded^ the limit! of integration being determined by the figure of the solid considered. The angles, 9 and ^, are sometimes called the eo4at%tud»t and longitude, respectively. BXAMPLB8. 1. Find the centre of gravity of a hemisphere whose density varies as the nth power of the distance from the centre. Take the axis of f perpendicnlar to the plane bMC of the hemisphere. Let a = the radios of the sphere, and p = fir*, where n is the density at the nnits distance from the centre. First integrate with respect to r from to a, and we obtain the infinitesimal pyramid 0-PQST. Then integrate with respect to from to ^n, and we obtain the sum of all the pyramids which form the elemental slice, CODA. Then integrating with respect to ^ from to arr, we obtain the sum of all the slices included in the hemi- sphere. Heuce, / / Bin (9 COS rfd d^ y / sin rf0 0?^ i = P = 0. 2. Find the centre of gravity of a portion < f a solid sphere conlaired in a right cono whose vertex is the centre of the sphere, the density of the solid varying as the «th power of the distance from the centre, the vertical angle of the cone being = 2«, and the radius = a. Take the axia of the cone as that of t, and any plane through it as that from which longitude is nieaaured. ^*^- • = ^ ^ ^(1 + cos e), and i = y = 0. 89. 8p«oi«l Metfiods.— In the preceding Articles we have given the usual forraulw for finding the centres of gravity of bodies, bnt particular cases may occur which may be most conveniently treat U by special methods. BXAMPLBS. 1. A circle revolves round a tangent line through an angle of 180° find the centre of gravity of the solid generated. drd$dift iedi» i6d TTxdx JT'y nx dx „ / 3^ V2ax — afidx A I'D ir /**« ~ / X V a«w — s^dx 6a 2^* 3. Find the centre of gravity of a right pyramid of uni- form density, whose base is any regular plane figure. Let the vertex of the pyramid be the origin, and the axis of the pyramid the axis of x; divide the pyramid into slices of the thickness dx by planes perpendicular to the axis. Then as the areas of Ihnse sections are as the squares of their homologous sides, and as the sides are as their dis- tances from the vortex, so wiU the areas of the sections be as the squares of their distances from the vertex, and therefore the masses of the slices arc as ^ho squares of their distances' from the vertex. Now imi^ne each slice to be condensed into its centre of gravity, which point is on the a^ic of x. Then the problem is reduced to finding the centre of grav> 1S8 TBXORKMS OF PAPPUS. Jty of a material line io which the density varies as the square of the distance from one end, and which may be found as in Ex. 6, (Art 78). Calling a the altitude of the pyramid, we have a^dz X ■■= _ fa, which if the same as in Art. 75. ■I I 90. TlMormiui of Pappus.*— (1) If a plans curve revolve rourd any axis in its plane, the area of the surface generated is equal to the length of the revolving curve multiplied by the length of the path described by its centre of gravity. Let « denote the length of the carve, x, y, the co-ordinates of one of its points, i, y, the co-ordinates of the centre of gravity of the curve; then, if the curve is of constant thickness and decity, we have from (3) of Art. 78, y- —71—; J" in pa = Zn / yds; (1) the second member of which is the area of the surface generated by the revolution of the curve whose length is 8 about the axis of x, (GaL, Art. 193) ; and thf first member is the length of the rovolviug curve, s, multiplied by the length of the path described by its centre of gravity, 2ffy. • tJnkllr oafled Onldin'i Tbeormm, bnt origlDally •■randated Ity n^vpns. 0m Watton** MeduBloai Protdnaa, p. 48, Sd W.) lity varios aa the d which may be le altitade of the a plane eurve the area of the length of the length of the 'ty. the co-ordinatea of the centre of e is of constant Art. 78, (1) of the surface lose length is a u first member iltiplied by the gravity, 2»ry. Med Iqr Fipptu. ifim wmm. wm wammm TBSORBMB OF PAPPUS. 139 (2) // a plane area revolve round any axis in its plane, the volume generated is equal to the area of the revolving figure multiplied by the length of the path described by its centre of gravity. Let A denote the plane area, and let it be of constant thickness and density, then {%) of Art 82 becomea V = jfy dx djf ffdxd/ or 2?ry y fdA ■= %ir J jy dx dy, (substituting dK for dx dy), (») the integral being taken for every point in the perimeter of the area; but the second member is the volume of the solid generated by the revolution of the area (Oal., Art. 208) ; and the first member is the area of the revolving figure, A, multiplied by the length of the path described by its centre of gravity, 2ny. Cob. — If the carve or anaa revolve throngh any angle, 0, instead of 2it, (1) and (2) become and 6i8 = efyd$, 9yK = ^fi/»dx. itnd the theorem* are still true. SoH. — If the axis cuts the revolving curve or area, the theorems still apply with the convention that the surface or volume generated by the portioni of the curve or area on opposite sides of the axis are affected with opposite rignc. i*fl3! m 140 JSXAMPLMS. EXAMPLES. 1. A circle of radios, a, revolves round ua axis in its own plane at a distance, c, from its centre; find the surface of the ring generated by it. The length (circumference) of the revolving carve = 2na; the length of the path described by its centre of gravity = inc; .'. the area of the surface of the ring = in*ac. 2. An ellipse revolves round an axis in its own plane, the perpendicular distance of which from the centre is c ; find the volume of the ring generated daring a comph t€ revolution. Let a and b be the semi-axes of the ellipse ; then the revolving area = nab ; the length of the path described by itf( centre of gravity = 2nc ; . • . the volume of the ring = in'abc. Observe that the volome is the same for any positioii of the axes of tlie ellipse with respect to th« axis of revolution, provided the per- pendicular diiitance from that axis to the centre of the ellipse is the same. 3. The surface of a sphere, of radius a, = inefi; the length of a somi-circumference = na ; find the length of the ordinate to the centre of gravity of the arc of a semi- circle. ^ - 2a Ans. y = __. 4. The volume of a sphere, of radius a, = fTro* ; the area of a semicircle = rra* ; find the distance of the centre of gravity of the semicircle from the diameter. Aru. y = 40 37r* 6. A circular tower, the d''uneter of which is 20 ft., is being bnilt, and for every foot it rises it inclines 1 in. from I axis in its own i the Bur&ce of )lving carve = ly its centre of its own plane, the centre is c ; ing a complete lipse; then the th described by idtion of the axes provided the per- the elllpoe ia the , = 4rro*; the the length of arc of a ecmi- - 2a ns. y = — . It = ^ira' ; the of the centre 4tf "*• ^ = rn h is 20 ft., is nes 1 in. from JSXAMPLSa. 141 the vei:tical ; find the greatest height it can reach without falling. Ans. 240 ft. 6. A circular table weighs 20 lbs. and rests on four tegs in its circumference forming a square ; find the least ver- tical pressure that must be applied at its edge to overturn it. Ans. 20 {V2 + 1) = 48.28 lbs. 7. If the sides of a triangle be 3, 4, and 5 feet, find the distance of the centre of gravity from each side. Ans. I, 1, f ft. 8. An equilateral triangle stands vertically on a rough plane ; find the ratio of the height to the base of the plane when the triangle is on the point of overturning. Ans. V3 : 1. 9. A heavy bar 14 feet long is bent into a rigbc angle so that the lengths of the portions which meet at t.ie angle are 8 feet and 6 feet respectively ; show that the distance of the centre of gravity of the l»r so bent fh)m the point of the bar which was the centre of dravity when the bar 9 V^ was straight, is — ^ — feet. 10. An equilateral triangle rests on a sqnare, and the base of the triangle is equal to a side of the square ; find the centre of gravity of the fligure thus formed. Ans. At a distance from the base of thn triaugle equal to 7= of the base. 8 + 2^3 11. Find the inclination of a rough plane on which half p. i^gnkiT hexagon can just rest in a vertical position with- out overturning, with the shorter of its parallel sides in contact with the plane. Ans. 3 y/Z : 5. 12. A cylinder, the diameter of which is 10 ft, and height 60 ft, rests on another cylinder the diameter of which is ■i*ii 143 EXAMPLMS, 18 fL, and height 6 ft ; and their axes coincide ; find their common centre of gravity. Ans, 273|f ft. from the base. 13. Into a hollow cylindrical vessel 11 ins. high, and weighing 10 lbs., the centre of gravity of which is 5 ins. from the bai>e, a uaiform sohd cylinder C ins. long and weighing 20 lbs., is just fitted ; find their common centre of gravity. Ans. 3f ins. from base. 14. The middle points of two adjacent sides of n square are joined and the triangle formed by this straight line and the edgep is cut ofF; find the centre of gravity of the remainder of the square. Ans. ^ot diagonal from centre. 15. A trapezoid, whose parallel sides are 4 and 12 ft. long, and the other sides each equal to 5 ft., is placed with its plane vertical, and with its shortest side on an inclined plane ; find the relation between the height and base of the plane when the trapezoid is on the point of falling over. Ans. 8 : 7. 16. A regular hexagonal prism is placed on an inclined plane w:th its end faces vertical ; find the inclination of the plarie so that the prism may just tumble down the plane. Ans. 30". 17. A regular polygon just tumbles down an inclined plane whose inclination is 10° ; how many sides has the polygon ? Ans. 18. 18. Prom a sphere of radius R is removed a sphere of radius r, the distance between their centres being c; find the centre of gravity of the remainder. Ans. It is on the line joining their centres, and at a dis- cr* tance /2«-r» from the centre. 19. A rod of uniform thickness is made np of equal lengths of three sabstauoes, the densities of which taken in icide; find their from the base. 1118. high, and which is 5 iiis. 6 ins. long and foimon centre of na. from base. des of a square traight line and gravity of the from centre. e 4 and 12 ft. , is placed with on an inclined and base of the falling over. J««. 8 : 7. on an inclined i inclination of iown the plane. Ans. 30°. vn an inclined sides has the Ans. 18. 'ed a sphere uf i being e ; find , and at a dis- • tip of equal rbich taken in tXAMPLWa. 143 order are in the proportion of 1, %, and 3 ; find the position of the centre of gravity of the rod. Ans. At -^ of the whole length from the end of the densest piu*t. 20. A heavy triangle is to bo suspended by a string pass- ing through a point on one side ; determine the position of the point so that the triangle may rest with one side vertical. Ans. The distance of the point from one end of the side =r twice its distance from the other end. 21. The sides of a heavy triangle are 3, 4, 5, respectively ; if it be suspended- from the cetitre of the inscribed circle show that it will rest with the shortest side horizontal. 22. The altitude of a right cone is h, and a diameter of the base is i ; a string is fastened to the vertex and to a point on the circumference of the circular base, and is then put over a smooth peg ; show that if the cone rests with its axis horizontal the length of the string is ^/{Ifl + ¥). 23. Find the centre of gravity of the helix whose equa- tions are X =■ a cos ^ ; y = a sin ; « = ha^. Ans. X , y . J a—x - % ka^; y = ka -— -; z = ^y Z z a 24. Find the distance of the centre of gravity of the catenary (Cal., Art 177), from the axis of x, the curve being divided into two equal portions by the axis of y. Ans. If 2Z is the length of the curve and (/*, k) ig the extremity, the centre of gravity is on the axis of y at a distance — ^ — trom the axis of x. KXAMPLBS. 25. Find the centre of gravity of the area included between the arc of the parabola, y» = 4aa:, and the straight line y = kx Arts, i = ?1 y = 2rt 26. Find the centre of gravity of the area bounded by the cissoid and its asymptote, the equation of the cissoid being y» = --^--. ^^. ^ ^ j^, 27. Find the centre of gravity of the area of the witch of Agnesi. Afis. At a distance from the asymptote equal to ^ of the diameter of the base circle. 28. Find the centre of gravity of the area included be- tween the arc of a semi-cycloid, the circumference of the generating circle, and the base of the cycloid, the common tangent to the circle and cycloid at the vertex of the latter being taken as axis of z, the vertex bein^r origin, and a the radius of the generating circle. . .37r» — 8 _ Ans. x= —^--— a; y == in. 39. Find the centre of gravity of the area contained be- tween the curves y' = ax and f = 2ax — a^, which is above the axis of a;. , . 15 n- — 44 y = Atis. i = o a IStt — 40' 3n - 8 30. Find the centre of gravi>;^ of the area included by the curves y* = ax and a^ = ly. Ana. i: = ^ijl ; y — ^ibK 31. Find tno distance of the centre of gravity of the area of the circular sector, BOCA, (Fig. 39), from the centre. Let 20 = the angle included by the bounding radii .„„ - .sin © Ans. « = f* -^-. SXAMPLSa. 146 area inclndod ind tho straight 8a _ _ 2rt ea bonnded by I of tho cissoid ins. i = \a. a of the witch [ual to I of the [. included be- ference of the , tho common of the latter ^n, and a the y — {a. contained be- ■ «*, which is — ^ included by = A«***. _, of the area :he centre. g radiL , sin 9 32. Find the distance of the centre of gravity of the circular segment, BCA, (Fig. 39), from the 'centre. . - , a sin' B Ana. « = f BC' » — sin «• cos e ~ 12 urea of ABC 33. Find the centre of gravity of the area bounded by the cardioid r = o (1 + cos 6). Ana. i = ^a. 34. Find the centre of gravity of the area included by a loop of the curve r == a cos 20. _ 28a Vi Ans. X = 105n- 35. Find the centre of gravity of the area included by a loop of the curve r = a cos 3 . _ gja y^ Ana. te = 80rr 36. Find the centre of gravity of the area of the sector in Ex. 31, if the density varies directly as the dis- tance from the centre. Ana. X = 3a sin 6 e 37. Find the centre of gravity of the area of a circular sector in which the density varies as the «th power of the distance from the centre. Ana. — ^t_ . _, where a is the radius of the circle, I the fi + o I length of the arc, and c tue length of the chord, of the sector. 88. Find the centre of gravity of the area of a circle in which the density at any point varies as the nth power-of the distance from a given point on the circumference. Ana. It is on tho diameter passing through the given 2(n + point at a distance from this point equal to a being the radius. « + 4 ^^^^ the vertex is the origin and a the altitude. 46. Find the distance of the centre of gravity of a henii- aphere from the centre, the radius being a, Ans. i = |fl. 47. Find the centre of gravity of the solid generated by the revolution of the cycloid, y = V2ax — a» + a vers"*?, a (1) round the axis of z, and (2) round the axis of y. A /i\- (637r»_64)« .... /16 Tr»\ 2a Am. (1) . = V(9^;ii)-5 (2) » - (y + i) IT- 48. Find.the centre of gravity of the volume formed by the revolution round the axis of x of the area of the curve f — axf + X* — 0. ^ Ant. i 3an 32 ' 49. Find the centre of gravity of the volume generated by the revolution of the area in Ex. 2U round the axis of y. . - 8a ^'"- >' = aTr5inri4)- 60. Find the centre of gravity of a hemisphere when the density varies as the square of the distance from the ooutro. . . 5a ''12' Ant. 51. Find the centre of gravity of the solid generated by a scmi*pai-abola bounded by the latns reotum, rsvolviug round the lutus reotum. Ant. Distance from focus = /^ of latus reotum. ^ . (I'fc'fi 148 SXAMPLBS. 62. The vertex of a right circular cone is at the centre of a sphere ; find the centre of gr&vity of a body of uniform density contained within the couo and the sphere. Ans. The distance of the centre of gravity from tue ver- tex of the cone = ~ (1 + cos «), where a = the semi- o vertical angle of the cone and a = the radius of the sphere. 53. Find the distance firom the origin to the centre of gravity of the solid generated by the revolution of the cardioid round its prime radius, its equation being r = a (1 + cos 6). Ana. X = fa. 54. Find by Art. 90 (1) the surface and (2) the volume of the solid formed by the revolution of n cycloid round the taigent at its vertex. Ans. Surface = ^na'; Volume = rrV. 55. Find (1) the surface and (2) the volume of the solid formed by the revolution of a cycloid round its liaso. Am. (1) *^na* ; (2) orV. 56. Au equilateral triangle revolves round its base, whose length is a ; find (1) the area of tho surface, and (i) the volume of the figure described. Ana. (1) na* a/3 ; (2) -^. 67. Find (1) the surface and (2) the volnmo of a ring with a circular section whose intomal diameter is 12 ins., and thickness 3 ins. Am. (1) 444.1 sq. in.; (2) 833.1 cub. in. at the centre of •ody of uoifonu phere. y from tue ver- a = the Bemi- radiua of tbe the centre of relation of the 1 heing ins. X = |a. (2) the volume cycloid round ume = tV. me of the Rolid its liasc. »» ; (2) 5rW. und its base, tho surface, \/3 ; (2) iTrt» nniu of a ring iieter is 12 ins., 33.1 cnb. in. CHAPTER V. FRICTION. 91. Friction. — Friction is that force which acts between two bodies at their surface of contact, and in the direction of a tangent to that surface, so as to resist their sliding on each other. It depends on the force with which the bodies are pressed together. All the curves and surfaces which we have hitherto considered were supposed to be smooth, and, as such, to offer no resistance to the motion of a body in contact with them in any other than a rormal dircctiou. tSuch curves and surfaces, however, are rot to be tbund in nature. Every surf two is capable of destroying a certain amount of force in its tangent plane, i.e., it possesses a certain degree of roughness, in virtue of which it resists tho sliding of other surfaces upon it. This resistance is called friction, and is of two kinds, viz., sliding and rolling friction. The rst is that of a heavy body dragged on a plane or other surface, an vie turning in a fixed box, or a vertical shaft turning on a narizo'zU! plate. Friction of the Be<,ond kind is that of a v;beel rolling along a plane. Both kinds of friction are governed by the same laws ; the former is much Teater than the latter under the same circumstances, and < I he only one that we shall consider. .'\. smooth surface i i one which opposes no resistance to the motion of a body upon it A rough surface is one which does oppose a resistance to the motion of a body u])ou it. TIiP fliirfkcM of all bodiM eonslBt of rerj Rintll elevations and deproMlons, ho that if thejr are prMsed againut vacli (ith«r, the elevatlona of ooo fit, mom or lesa, into the deprpwions of thv other, and tbe Burfaoea iitt«rp«netrate each otkor ; and the- mutual penetra- \ht 160 LAWS OF FRIcnON. tion is of ooane greater, if the presBing force is greater. Henoe, when a force is applied bo as to cause one liody to move on another with which it is iu contact, it is necessary, lieforo motion can take place, either to break off the elevations or compress them, or force tlio iKxliea to separate far enough to allow them to pass each other. Much of this roughneM may be removed by polishing ; and the effect of much of it may be destroyed by lubrication. Friction always vets along a tangent to the surfiuje at the point of contact ; and its direction is opposite to that of the lice of motion ; it presents itself in the motion of a body as a passive force or resistance,* since it can only hinder motion, but can never produce or aid it. In investigations in mechnnius it can be considered as a force acting in opposition to erery motion whose direction lies in the plane of contact of the two bodies. Whatever may be the direction in which we move a body resting upon a boriaontal or inclined plane, the friction will always act in the opposite direction to that of the motion, i. «., when we slide a body down an inclined plane, it will appear as a force up the plane. A surface may also resist sliding motion by means of tha adhetion > ^ween its substance and that of another body in contact with it.f Tho friction of a body on a surface is measured by the leaxt force which will put the body in motion along the surface. 92. Laws of FriotioB. — In onr ignorance of the ooudtitutiun of bodies, the laws of friction must be deduced from experiment. Experiments made by Coulomb and Morin have established the following laws of friction : (1) The friction varies as the normal pressure when the materials of the surf..^js in contact remain the same. Subse- quent experiments have, however, considerably modified this law, and shown that it can be regarded only as an approximation to the truth. When the pressure is very great it is found that tho friction is less than this law would give. * WalHbwh, p. «». t 8e« Haaklue'i ApplM MsetMDks, p. au». I greater. Henoe, movA on another 3 motion can take them, or force tlie paw each other, ag i and the efl'cxrt aoe at the point of line of motion ; it brce or resistance,* duee or aid it. In ts a force acting in lie plane of contact in which we move le, the friction will motion, i.e., when >pear as a force up a by means of tha ler body in contact leasured by the )tion along the loranoe of the ust be deduced Coulomb and frictiou : |e«jiur(? wJien the same, Subsc- Irsbly modified only as nn assure is very than this law ijiwa or rRicnoN. (2) Tht frictioH is independent of the extent of the aur- facee in contact so hng as the normal pressure remains the same. When the surfacec iu contact are very small, as for instance a cylinder resting on a surface, this law gives the friction much too great. These two laws are true when the body is on the point of moying, and also when it is actually in motion ; but in the case of motion the magnitude of the friction is not always the same as when the body is beginning to move ; when there is a difference, the fHction is greater in the state bordering on motion than In actual motion. (3) The friction is independent of the velocity when the body is in motion. It follows from those laws that, if .R be the normal pressure between the bodies, F the force of friction, and ft the constant ratio of the latter to the former when slipping is about to ensite, we have F= fiR. (1) The fraction n is called the co-efficient qf friction ; and if the first law were true, fi would be strictly constant for the ■ame pair of bodies, whatever the magnitude of tho normal pressure between them might bo. This, however, is not tho case. When the normal pressure is nearly equal to that which would crush either of the surfaces in contact, the force of friction increases more rapidly than the normal pressure. Equation (1) is nevertheless very nearly true when the differences of normal pressure are not very jfrrinienM wert; made at Mi-ts by Morin, 1881-S4, by direction of the French military authorities, the raialt of I f . 153 AlfOLK OF FRICTION. i which has been to oonfirm, with slight exeeptions, all the results of Coulomb, and to determine witli conBldersble precision tlie numerical values of the coefficients of friction, for all the substances usually employed in the construction of machiaes. (8ee Oalbraith's Me- chanics, p. 68, Tiwisdeu's Practical Mechanics, p. 188, and Weisbach's Mecbauica. VoL I, p. 817.) 93. Magnitndes of CoeffloientB of Friction.— Prac- tically there is no observed coeflRcient much greater than 1. Most of the ordinary coefficients are less than ^. The fol- lowing results, selected from a table of coefficients,* will afford an idea of the amount of friction as determined by experiment : these results apply to the friction of motion. For iron on stone i* varies between .3 and .7. For timber on timber " " " .2 and .5. For timber on metals " " " .2 and .6, For metals on metals " " '* .15 and .25. For full particulars on this subject the student is referred to Rankine's Applied Mechanics, p. 209, and Mosilcy's Engineering, p. 124, also to the treatise of M. Morin, where he will find the subject investigated in all its completeness. 94. Angle of Friction. — The angle at which a rough plane or surface may be inclined so that a body, when acted upon by the force of gravity only may just rest upon it with- out sliding, is called the Angle of Friction, f Let a bo the angle of inclination of the piano AB just as the weight is on the point of slipping down; W the weight of the body ; R the normal pres- sure on the plane ; F the force of fric- tion acting along the plane = fiR (Art. 92). Then, resolving the forces along and perpendicular to the plane we have for equilibrium * Ranklne'B AppHod Mechautcs, p. m. t SoiuetimM called " Ibe aogto of ropoae; " al»o caUad " the Umltlng asglo of resUtaiMe." na-45 all the rasolts of ion tlie namerical ubatauces nsnalljr Galbnith's Me- B, and Weisbach's fiction.— Prac- greater than 1. an f The fol- efficiente,* will determined by on of motion. .3 and .7. .2 and .5. .2 and .6. 15 and .25. dent is referred and Mosiley'fl Morin, where ) completeness. which a rough ody, tvhen acted upon it with' erpendicnlar to the limiting angle of BK ACTION or A SOUOH CVBVK. uB = ffaina; R — ffoosa; tan a = /», a) which gives the limiting valae of the inclination of the plane for which equilibrium is possible. The body will rest on the plane when the angle of inclination is less than the angle of friction, and will slide if the angle of inclination exceeds that angle ; and this will be the case however great W may be ; the reason being that in whatever manner we increase W, in the same proportion we increase the friction upon the plane, which serves to prevent IT from sliding. From (1) we see that the tangent of the angle of friction is equal to the coefficient of friction. 95. ReactiOB of a Rough Cnrvo or Surface. — Let AB be a rough cuive or surface ; P the position of a particle on it ; and suppose the forces acting on P to be confined to the plane of the paper. Let R^ = the normal resistance of the surface, acting in the normal, PN, and F = the force of friction, acting along the tangent, PT. The resultant of if, and F, called the Totai Resistance* of the surface, is represented in magnitude and direction by the line PR = R, which is the diagonal of the parallelo- gram determined by R^ and F. We have seen that the total resistance of a smooth surface is normal (Art. 41) ; but this limitation does not apply to a rough surface. Let ^ denote the angle between R and the normal £, ; then ^ is given by the equation * tan = ^• * MlBcitia'p Btatici, p. (M. ris.4e 154 WBiCTION ON Air INCLiySD PLAKIl, Hence, ^ will be a maximum when the force of friction, F, bears the greatest ratio to the normal pressnre B^. But this greatest ratio is attained when the body is just on the point of slippin/; along the surface, and is what we called the coefficient of friction (Art 92), that is Tr = ^; . • . t«m ^ = ^. Therefore the greatest angle by which the Total Reaietanc^ of a rough curve or surface can deviate from the normal is the angle whose tangent is the coefficient of friction for the bodies in contact ; and this deviation is attained when slip- ping is about to commence. Cob.— By (1) of Art 94, tan « = f» ; ft. hence, the direction of the total resistance, R, is inclined at an angle a to the normal ; i.e., the greatest angle that the Total Resistance of a rough curve or surface can make with the normal is equal to the angle of friction, correspmiding to the two bodies in contact. 96. Friction on an Inclined Plane. — A body rests on a rough inclined plane, and is acted on by a given force, P, in a vertical plane which is perpendicular to the inclined plane ; find the limits of the force, and the angle at which the least force capable of drawing the particle up the plane must act Let t = the inclination of the plane to the horiison ; 6 = the angle between the inclined plane and the line of action of P; (I =■ the coefficient of friction ; aud let us first snp- pose that the body is on the point of moving down the force of friction, )Tes8are B^. But »dy is just on the is ivhat we called i Total Betistance' om the normal is '' friction for the tained when slip- R, is inclined at st angle that the ce can make with m, correspotuling A body rests on a given force, P, to the inclined angle at which cle up the plane le horizon ; = le line of action let us first snp- loving dowti the mmmmmmmBm^':. FRICTION ON AN INCLINMD PLANS. 156 plane, so that friction is a force acting up the plane^ then resolving along, and perpendicular to, the plane, we have F + Pco&e =z W fan if B + Pmie= Wooai, F^fiB; . sin t — ^ cos t P=W cobO — fianO (1) And if P is increased so that motion up the plane is just beginning, F acu in an opposite direction, and therefore the sign of /u must be changed and we have p — j|r »^P*'4-^C0St COS $ + n Bind' (2) Hence, there will be equilibrium if the body be acted on by a force, the magnitude of which lies between the values of P in (1) and (2). Substituting tan ^ for /« (Art. 95) ; (2) becomes n_ nr wn(t-f ») COB («) To . determine 6 in (2) so that P shall be a minimum we must put the first derivative of P with respect to fl = 0, therefore dP nr / • • ■ ^ and — fi COB 6 ,*. tan 9=:^; that is, the' force P necessary to draw the body up the plane will be the least possible when =. the angle of friction. 166 DODBhE-UrCLINMD PLANE. Henoe we infer that a given force acts to the greatest advantage in dragging a weight up a hill, if the angle ut which its line of action is inclined to the hill is equal to the angle of friction of the hilL Similarly, a force acts to the greatest advantage in dragging a weight along a hori- zoatal plane if its line of action is inclined to the plane at the angle of friction of the plane. We may also deter- mine from this the angle at which the traces of a drawing horse should be inclined to the plane of traction. These results are those which are to be expected, becanse some part, of the force ought to be expended in lifting the ight from the plane, so that friction may be diminished. V oe Price's Anal. Mech's, Vol. I, p. 160.) 97. Friction on a Doubla-Inclined Plane.— Two bodies, whose weights are P and Q, rest on a rough double- inclined plane, and are connected by a string which passes over a smooth peg at a point. A, vertically over the intersec- tion, fi, of the two planes. Find the position of equili- brium. Let « and /3 be the inclinations of the two planer ; let 2 = the length of the string, and h = AB; and let and 0' be the angles the portions of the string make with the planes. Suppose P is on the point of ascending, and Q of descending. Then, since the motion of each Inxiy is about to ensue, the total resistances, R and ^S', must each make the angle of friction with the corresponding normal (Art 95, Oor.) ; and since the weight, P, is about to move upwards the friction must act downwards, and therefore R must lie below the normal, while, since Q is about to move downwards, the friction must act upwards, and therefore 8 must be above the normal. Fig.4y &ilKiS^&- i to the greatest , if the angle at ) hill is equal to a force acts to it along a hori- led to the plane may also deter- ces of a drawing ;tion. tpeeted, because ;d in lifting the ' be diminished. 1 Plana.— Two a rough double- g wliich passes ver theintersec- ition of equili< F(s.47 i to ensue, the e the angle of 95, Oor.) ; and ds the friction t lie below the ownwards, the must be above DOUBLE-mcUNSD PLAIflS. 187 If T is the tension of the string, we hare for the eqni- libriam of P, (Art. 32), y^j> «»(« + ») cos (d — 1^) And for the equilibrium of Q, ~ ^co8(e' ■\-)' and if P is about to move down the piano, the friction acts in an opposite direction, and therefore the sign of must be changed and we have „ sin (« — 0) _ si n (/? + 0) cos {0 + ) COB (0 + ^) sin (o — 0) cos ((?' — 0) (7) the angles and 6' being connected by (3). There will be equilibrium if ^ be acted on by any force whose magnitude lies between P| and Pg. fc — espectively down case, considering along, and per- (*) (6) i (5) we get (1), ances instead of is usually more ;he latter forces. lat it is about to ho" <*^> Q up, its value, n by any force Kiii iaiiilliMWIM I WipWIWI'l fmm ut'iiimnmHmmimmMmixiKn' FRICTION OF A TRUNNIOie. 159 Pi8.4S 98. FrictioB on Two XncUned Planes.— A beam rt'sts ou two rough inclined planes; find the position of equilibrium. Let a and b be the segments, AG and BG, of the beam ; let be the inclination of the beam to the hori- zon, a and /3 the inclinations of the planes, and R and S the total resist- ances. Suppose that A is on the point of ascending; then the total resistances, R and 8, must each make the angle of friction with the correeponding normal and act to the right of the normal. The three forces, W, R, S, must meet in a point (Art. 62) ; and the angles GOA and GOB are equal to a -f- 0, and (3 — (p, respectively. Hence {a + b) cot BGO = o cot GOA — b cot GOB, or (a + ft) tan e = o cot (a + 0) — i cot (/3 — 0). (1) Cob. — If the planes are smooth, ^ = 0, and (1) becomes {a + b) taxi6 = acota — b cot j3. (See Ex. 7, Art 62.) 99. Friction of a Trunnion.* — Trunnions are the cylindrical projections from the ends of a shaft, which rest on the concave surfaces of cylindrical boxes. A shaft rcEits in a horizontal position, with its trunnions ou rough cylindrical surfaces; find the resistance due to friction which is to be overcome when the shaft begins to turn about a horizontal axis. « BometimM called " JonmaL" 160 mWTrON OP A PIVOT. FiR.4» Let IM and BAED be two right flections of the trunnion and its box; the two circles are tangent to each other internally. If no rotation takes place the trunnion presses upon its lowest point, H', through which the direction of the resulting pressure, R, passes ; if the shaft begins to rotate in tiie direction AH, the trunnion ascends along the inclined surface, EAB, in consequence of the friction on its bearing, until the force, ^S, tending to move it down just balances the friction, /'. Resolving R into a normal force AT and a tangential one, 8, we have, since the t<onent of R in urging the trunnion down the burface = tlie Mction which opposes it. •8=F-(iN; but R* ^ S* + N*; or therefore and the friction R* = ix*2P + I^; R N- fiR Vi +^«' Rtantft or vT-f n* Vl + tanS P =r Ram (p. (Art. 96), Ilonco, fo fiiv^ (he friction upon n (rtinnioti, muUiply the resultant of the forces which act upon it by the sine of the angle of friction. 100. Friction of • Pivot—A heavy circular shaft rests in a vortical position, with its end, which is a circular Fip.49 trunnion ascends isequcnce of tho , tending to move aohing R into a e have, since the >o traimion down t. + N*', rt. 95), )//, multiply the y the sine of the circular shaft til is a circular SSfi^^ FBicnoN Of A prroT. IGI section, on a horisontal plate; And tho resigtimoe due to friction which is to be overcon^e, when tho shaft begins to revolve about a vertical axis. Let a be the radius of the circular section of the shaft; let the plane of ( •. $) be the horizontal one of contact between the end of the shaft and the plate; and let tho centre of the circular area of contact be the pole. Let W = the weight of the siiaft, then the vertical pressure on W each unit of surface is — j; and therefore, if rdrd6 is the area-element, we have the pressure on the element W rra- -, r dr de ; the friction of the element sz n~- rdr dO. no' The friction is opposed to motion, and the diroc- jii oi its action is tangent to the circle descril)ed by the element ; the moment of the friction about the vertical axis through the centre ^Wt^d rde ■^ "~ ita*" ' therefore the moment of friction of the whole circular end «/o t/fl (iWr^drde 2ftWa 8 ' (1) and consequently varies aa the radius. Hence arises the advantage of reducing t« tho smallest possible dimensions llio area of the ba«o of a vertical shaft revolving with its end reiiting on a horizontal be lermine P in the last example if its direction is hoiizontal. Ans. P = 465 lbs. 8. find the force along the piano required to draw a weight of 25 tons np a rough inclined plutte, the coefficient of friction being -j^, and the inclination uf the plnno being snch that 7 tons acting along the plane would siip|)ort the weight if the plane were smooth. Ans. Any force greater than 17 tons. 4. Find I'lo force in the preceding example, supposing it to act at the most advantageous inclination to the plaice. Ans. ]5t^ tons. 5. A ladder inr-Hned at an angle of 60° to the horizon rests between a rouyh pavement and the smooth wall of a house. Show that if the ladder kgin to slide when a man hAB osoopdod so that his centre of gravity in half way up, then the coefficient of friction between the foot of the ladder and the pavement is \ VS. 6. A body whose weight is 20 lbs. ip just sustained on a rough inclined plane by a horizontal force of 2 iba., and a force of 10 lbs. along the piano ; the coefficient of friction is I ; find the inclination uf thu plane. Ans. tan~' (||). lalled the mean iis circular end Hon is called a on a horizontal iircctiou mal es dine P and the I being .62. = 756-9 lbs. its direction is P = 465 lbs. irod to draw a !, the coefficient the plnne l)eing Id support the Lhau 17 tons. i])Ic, supposing to the plaiiC. }5-j^ tons. to the horizon rnootfi wall of ii c when a mun is half way up, foot of the sustained on a 1 2 lt)0., and ti it of frictioa is tan-'(il). i-'mmmmmmm BXAMPLBS. 163 7. A heavy body is placed on a rough pljine whose inclination to the horizon is sin "* (f), and is counected by a string passing over a smooth pulley with a body of equal weight, which hangs freely. Supiwsing that motion is on the point of t-nsu'ng up the plane, find the inclinatiop of the striag to the plane, the coeCaoient of friction being \. Ana. ^ = 2 tan"' (i). 8. A heavy body, acted uiwn by a force equal in magni- tude to its weight, is just about to ascend a rough inclined plane under the influence of this force ; find the inclination, d, of the force to the inclined plane. Am. e = ^ — *■» or 2(;fr + t - 1 where t = inclination 2 * of the plane, and <> = angle of friction, {d is here sup- posed to be measured from the upper side of the inclined plane). If 5 > 2^ -f », * is negative and the applied force will act towards the under side. 9. In the first solution of the last example, what is the magnitude of the pressure on the plane ? Ana. Zero, Explain this. 10. If the shaft, (Art. 100), is a square prism of the weight W, and rotates about an axis in its centre, prove that the moment of the friction of the square end varies as the side of the square. 11. If the shaft is composed of two equal circular cylinders placed side by side, and rotatcH about the line of contact of the two cylinders, show that the moment of the friction of the surface in contact with the horizontal piano _ 3a/mlF -~ 9n ' 12. What is the least coefficient of friction that will allow of a heavy body's being just kept from sliding down EXAMPLES. an inclined piano of given inclination, the body (whoso weight is W) being sustained by a given horizontal force, P ? An,, l^'^P. »K + /' tan i 13. It is obsoi-ved that a body whose weight is known to be fKf'flin bo juet sustained on a rough inclined plane by a liorizontil force P, and tlmt it can also be just sustained on the same plane by a force Q up the plane; express the angle of friction in terms of these known forces. Am. Angle of friction = cos~* • 14. It is o' ved that a force, ^,, acting up a rough inclined pL .11 ju«t sustain on it a body of weight W, and that a for-^e, Q^, acting up the piano will jest drag tho same body up ; find the angle of friction. Ans. Angle of friction — sin"' ^l -T-jiJ 15. A hea>> uniform rod rests with its extremities on the interior of a rough vortical circle; find the limiting position of equilibrium. Ans. If 2« is the angh subtended at tho centre by the rod, and A the angle of friction, the limiting inclination of the rod to the horizon is given by the equation tan 6 sin 2A cos 2A -f- cob 2a 16. A solid triangular prism is placed, with its axia horiisonUil, on a rough inclined i)lane, the inclination of which is gradually increased ; determine tho nature of the kititil motion of the prism. Am. If the ta-ianglc ABO is the section iwrpondionlar to the uxis, aud the gide AR is isi contact with the plane, A B body (whoso zontal force, P ? rtan i — p V + .Tii^' ^ht is known to led plane by a i8t sustained on s; express the ces. PW ig up tt rough 7 of weight W, jest drag tho xtreniities on the limiting centre by the inclination of ith its axis U'linution of atiire of the pudionlar he plane, EXAMPLES. 165 being the lower vertex, the initial motion will be one of tumbling if 5» + 3 4A the sides of the triangle being a, b, c, and its area A. If ft is less than this value, the initial motion will be one of slipping. 17. A frustnaa of a solid right cone is placed with its base on a rough inclined plane, the inclination of which is gradually increased ; determine the nature uf the initial motion of the body. Am. If the radii of the larger and smaller scctionB arc R and r, and h is the height of the frustum, the initial motion will be one of tumbling or slipping according as '*>garfied as positive ; the sign, therefore, of a virtual mo- ment will be the same as that of tho virtual v«'locity. Cob. — If d be the angle between the force and the virtual dibpiacement, we have for the virti?al moment, P . ON = P • OA cos « = P co£ e . OA. OA is called the * Thf- rHw)tpN> 3t Vinnil VeloctUwi ww dlfoovered by Otllleo, mi wm rmj Ailly d'-" ''ped by BnrtiouiUi and La^gnngt. t Soiiu tlniM oatM ' bj UuluuotJ. VirtiuU Work." Tte ovaa " Virtiul Momaat " < i^ren .'JiM**» ^■^ OCITIES* f application of idefinitely small n the direction the force ; and tlocity has been Fig.50 . is called the it on the line P acts, ofi in positive; but tion line pro- 08 are always a virtual mo- looity. nd the virtual OA. lleo, and ww very omont " WM giTen '■ il l Wg ff'WIII I IIIIIII I IWWIU I llll l VIRfUAL VELOCITIES. 107 Now P cos 6 is the projection of the force on the direction of the displacement, and is equal to OM, OP being the force and PM being drawn perpendicular to OA. Hence we may also define the virtual moment of a force as the product of the virtual displacement of its point of applica- tion into the projection of the force on the direction of this displacement; and this definition for some purposes is more convenient than the former. Rbmahk.— A forca is said to do work if It movM the body to wbidi it is applied ; and tlie work doDu by it is measuriMi by tlie product i>{ tlie force into tiie space through wliich it moTce the body. Generally, the work done by any force during an infinitely tuBall dkaplacemeut uf its point of application is tlie product of the rettolved paii of tbc force in the direction of the displacement into tlie displaomnent . hmI thia is the same as the virtutM^ «»>«M«Mt of Um force. 102. Principle of ▼ht—l "W^HocStx^m. — (1) The virtual moment of a force m e'^ttui iv the xum of the virtual moments of its components. Let OR represent a force, ft, act- ing at 0, and let its components be /' and Q, represented by OP and OQ. Let OA be the virtual dis- placement of 0, and let its projec- tions on R, P, and Q, be r, p, and q, respectively. Then the virtual moments of these forces are R • r, P Pm, and Qo, perpendicalar to OA. Fia.M p, Q • q. Draw /?>/, Then On, Om, and Oo (= mn), are the projections of R, P, and Q, on the dirr-c- tion of the displacement ; and henoe (Art. lOX, Cor.) we have i2 . r = OA On ; P.;>=OA.Om; ^ - q ss OA • mn. 168 rinvuAL VELoctTtsa. Hence P • ^ + Q • y = OA (Om + mn) = OA ' On = E-r. (See MinohiD'a Statics, p. 68.) (2) If there are any number of component forces we may compouud them iu order, taking any two of them first, and finding the virtual moment of their resultant as above, then finding the virtual moment of the resultant of these two and a third, likewise the virtual moment of the resi!i>«nt of the first three and a fourth, and bo on to the last ; or we may use the polygon of forcew (Ai-t. 33). The sum of the virtual moments of the forces is equal to the virtual dis- placement mullipliod by the sum of tiic projci (ions on the displacement of the sides of the polygon Avhich rt'pre.^ent the forces (Art 101, Cor.). But the sum of these projec- tions is cyual to thv projection of the remaining side of the polygon,* and this side represents the resultant, (Ar!. 3'6, Cor. J). Therefore, the sum of the virtval moments (if any number of concurring forces is equal to the virtual moment of the resultant. (3) If the forces are in equilibrium, their resultant is equal to zero ; hencA , it follows that ivhen any number of concurring fnrren ur» in equilibrium, the sum of their virtual moments =■ 0. This principle la gon(*rully lin<»wn as the Principle of Virtual Velocities, and Ih of great use in the BoluUon of practical problems in Statics. * Krom Uis lUture of projectiuiia (Anal. Ocom., Art. laB), It |8 Oleir llilt in any terleit of polnU the projection (on a glTen line) of tlio lliio n lili.li Jlijna ilio flret and l8Bt, Ir eq;ukl to the oum of the projection)! of the Ilium which Join the pointH. iwa and two. Thim, If the eldes of a cloxed polfKun. taken In order, be morkcd witli arrows pointing from each vertns to the next one ; and If their projettlon-j be marked with arrows In the same dln^ctlong, then, lined mcasorsd from left to right being oonsldered pocltive, and linos from right to left negative, the swn (ff tKr pro- Jtetionn m left to right the sum of the pro- VntTDAL MOMMJfTS. 169 103. Natnre of the Sisplaeemttnt.— It must be care- fully observed that the displacement of the particle on which the forces act ia virtual and arlilrarv. The word virtual in Statics is used to intimate tha^. tbo displacements are not really made, but only supposed, i. e., they are not actual but imagined displacements ; but in the motion of a particle treated of in Kinetics, the dJsplacemcni is often taken to be that which the particle actually undergoes. In Art. 101, the displacement was limited to an infiniteai- mal. In some cases, however, & finite displacement may be used, and it may be even more convenient to consider a finite displacement. But in very many cases any finite dis- placement is suflScient to alter the amount or direction of the forces, so as to prevent the principle of virtual velocities from \mug applicable. This diflSculty can always bo avoid- ed in practice by assuming the displacement to be infinitesi- mal ; a»d if the virtual disphujement is infinitesimal the virtual volocitieci are all inflnitcsimaL 104 Equation of Virtual Moments. —Let P^, P,, Pf, < !. iiofr the fonos, and ip^. (5jt>,, '5/;,, etc., the vir- taai I lien the priut>ipUs of virtual velocities is expressed (Art 102) by the etiuation ^i • ^Pi + Pt • ^P% + i\ ' ^P, + etc. = 0; or ^P6p = 0, (1) which is called tbo equation of virtu/il moments.* Hon. — If the virtual diHplacemont is at right angles to the direction of any force, it is clear that dp, the virtual velocity, is rijual In m-ro. Hence, when the virtual dia- placement is at ripht angles, to the direction of the force, • Or virtual wort (Set Art. 101, Rem.). Tbl* eqiutlon IiM been in&de hy La. itranKe tho fiiaudation of hia Kreat work on Xeclivilo, " MiJbaahtno /jialytlque,'* (I'rlce'H AiuO. Uocb., ~.'ol. I, p. 14t.) 170 SrSTBM or PAItTICLXa. m. m »M the virtual moment of the force = 0, and the force will not enter into the equation of virtual moments. Such a virtual displacement is always a convenient one to choose when we wish to get rid of some unknown force which acts upon a pari:icle or system. 105. System of Particlea Rigidly Connected.— (1) If a particle in equilibrium, under the action of any forces, be constrained to maintain a fixed distance from a given fixed point, the force due to the constraint (if any) is directed towards the fixed point Let B be the particle, and A the fixed point. Then it is clear that we may substitute for the string or rigid rod which connecta B with A, a smooth circular tube enclosing the particle, with the centre of the tube at A. Now, in order that B may be in equilibrium inside the tube, it is necessary that the resultant of the forces acting upon it should be normal to the tul)e, t. e., direct^ towards A. (2) Let there be any number of particles, mi, m,, 7n,, et^., each acted on by any forces. P,, P,, Pg, etc., and connected with the others by inflexible right lines so that the figure of the system is invariable. Then eacli particle is acted on by all the external forces applied to it, and by all the internal forces proceeding from the internal con- nections of the particle with the other particles of the system. Thus tlie particle, m, is acted on by Pj, Pg, etc., and by the internal forces which proceed from its connec- tion with m^, «!,, «t,, etc., and which act along the lines, mm I, mm^, etc., by (1) of this Article. Denote the forces along the lines »iw,, «i»i„ mm,, etc., by /j, t^, t^, etc., and their virtual velocities by dt^, rf^,, wards A. internal oon- icles of the P,, Pj, etc., its connec- ig the linos, e the forces /,, and the same result is evidently true whatever be the num- ber of particles forming the system. Hence, if any num- ber of forces in a system are in equilibrium, the sum of their virtual moments = 0. The converse is evidently true, that if the sum of the virtual moments of the forces vanishes for every virtual displacement, the system is in equilibrium. The following are examples which are solved by the principle of virtual velocities. ^m i»-'S!?^S4 SMAGE EVALUATION TEST TARGET (MT-S) « ^ / /£ Ua (/. f V 1.0 II u i.25 ^1^ 1^ ■cwr 2.2 1.4 llllim 1.6 li Hiotographic Sciences Corporation 33 WIST MAIN STRUT WGBSTeR.N.V. MHiO (71«) 173-4303 '^V^ ''.^(S -"•f«^"^^fl»WWI ,m iw H.nf. ■ , iii ^,| i ),iwi i ^ n s CIHM/ICMH Microfiche Series. CIHM/ICIVIH Collection de microfiches. Canadian Institute for Historical Microraproductiont / Institut Canadian da microraproductiona historiquaa ,l j » ipiia III I -»■ . iii llJiiw^iT^iS'j *^ I I I lis* 17a MXAMPLXS, EXAMPLES. 1. Determine thv> condition of equilibrium of a heavy body resting on a smooth inoline^l plane under the action of given forces. Let W bo the weight of the body sustained on the plane BC by the force, P, making an angle, 0, with the plane. To avoid bringing the un- known reaction, R, into our equation, we make the displacemeLt of it« point of application perpendicular to its line of action, (Art. 104, Sch.); hence we conceive as receiving a virtual dis- placement, OA, at right angles to R, the magnitude of which in the present case is unlimited. Draw Am and A» perpendicular to W and P respectively, Om and 0» are the virtual velocities of W and P, (Art 101) ; and W • Om and P • On are their virtual moments. Horce (1) of Art. 104, gives W • Of» - - P . On = 0. * ^H. But Om = OA sia a, ^^H On = OA cos & ; ^^BE therefore W sin a — Poos* s= 0; m which agrees with Ex. 3, Art. 41. :h If the Toroo acts parallel to the plane, 9 = 0, aud (1) becomes P = W sin « ; which agrees with Ex. 1, Art. 41. mmemirwai ibrinm of a heavy ander the action action, (Art. 104, Dg a virtual dis- thfl magnitude of Draw Aw and vely, Om and 0» (Art. 101) ; and ents. Horce (1) (1) « = 0, aud (1) MXAMPLga. 178 2. Sappoae the plane in Ex. 1 to be rough, and iha« the body is ou the point of being draggsd up the pUine, find the condition of equilibrium. The normal resistance will now be replaced by the total resistance, C, inclined to the normal at an angle = , tht- argle of friction (Art. 95, Cor. ). Let the virtual d itplacemeut, OA, take pL>i0e perpendicularly to R, then (1) of Art. 104, gives WOm — P.On = 0. But Om = OA sin (o -f ^), i>nd On = OA cos (^ — 6) ; therefore W aic (a + 0) =: P oos (^ — 9) ; which agrees with (3) of Art. 96. ?,. Determine the horisontal force which will keep a particle in a given position inside a circular tube, (1) when the bibe is smooth end (3) when it is rough. (1) Lot the virtual displaoament, OA, be an infinitesimal, = tis, aloiig the tube. Then since dt is infinites* imal the virtual velocity of B = 0. Then the equation of virtual moments ij — W • Om 4- P • On = 0. Wi§,» But Om = da ■ sin 6, and On =z Ha -co* 8] therefore W«in« = PooiS; or P = W An e. EXAMPLSa. {%) Snppoao tho force, P, jnst sustains the particle ; the normal rcsiatanoe must now be replaced by the total resist- ance, making the angle, ^, with the normal at the right of it. Take the virtual displacement, OA', at right angles to the total resistance (Art 105, Sch.), and let it be as before, an infinitosimal d*. Then (1) of Art 104, gives --W-Om + P.0«'=0. But Om = ) — Q\ .'. & siu ^ (^^ = a sin 9 dd. But from the geometry vt the figuro we have d sin ^ = 3a sin 9; .'. 6 cos ^ rfi^ = 2a COS 9 positive, and a resistance is one whose elementary work is negative. The moving force is, for convenience, called the * See Art. 101, Ran. Is, 178 MECHANICAL ADVANTAOE. power ; and becanse the attraction of gravity is the most common form of the force or resistance to bo overcome it is usually called the weight. The weight or radstanoe to be overcome may be the enrth's attrac- tion, aa in raising a weight ; the molecular attractiona between the particles of a body as in stamping or catting a metal, or dividing wood ; or fristlon, aa in drawing a heavy body along a rough road. The power may be that of men, or horses, or the steam engine, etc., and may be jost sufficient to overcome the resistance, or it may be in excess of what is necessary, or it may be too small. If just sufficient, the machine, if in motion, will remain uniformly so, or if it be at rest it will be on the point of moving, and the power, weight, and friction will be in equilibrium. If thr power be in excess, the machine will be set in motion and will continue in accelerated motion. If the power be too small, it will not be able to move the marJiine ; and if it be already in motion it will gradually come to rest. The geueral problem with regard to machines is to find the relation between the power and the weight. Some- times it is most convenient that this relation should be one of equality, t. e., that the power should equal the weight Generally, however, it is most convenient that the power should be very different from the weight Thus, if q man bus to lift a weight of one ton hanging by a rope, it is clear that he cannot do it unless the mechanical contrivance provided enable him to lift the weight by exercising a pull of very much less, say one cwt When the power is much smaller than the weight, as it is in this case, which is a very common one, the machine is said to work at a mechan- ical advantage. When, as in some other cases, it is de8irul)le that the power should be greater than the weight, there is said to be a mechanical disadvantage of the machine. 107. Mechanical Advantage.— (1) Let P and W he the power and weight, and p and w their virtual velocities respectively ; and let friction be omitted. Then fh>m the equation of virtual work (Art 104), we have Pp-Ww=iO, OT ~-y mum OE. gravity is the most ,0 be overcome it is be the eitrth's attrac- itmctions between the : « metal, or dividing y along a rough road, the Bteam engine, etc., lance, or it may be in all. If just sufficient, f BO, or if it be at rest 3r, weight, and friction cesB, the machine will rated motion. If the re the machine ; and if > rest. machines ia to find e weight. Some- Ition should be one equal the weight t that the power Thus, if % man a rope, it is clear nical contrivance exercising a pull he power is much |s case, which is a work at a mechan- laees, it is de8irul>le e weight, there ia e machine. iLet P and »r be virtual velocities Then from the ^ve 10 y pir'j ,11m MSCHANICAL ADVANTAQX. \n which shows that the smaller P is in comparison with W, the smaller w will be in comparison with p. But the smaller P is in comparison with W, the greate; is the mechanical advantage. Hence, the greater the mechanical advantage is the lefc3 will be the virtual velocity of the weight in comparison with that of the power. Now, if motion actually takes place the vi/tual velocities become aciual velocities ; and hence we have the principle what is gained in power is lost in velocity. (3) There are no cases in which the weight and power are the only forceo to be considered. In every movement of a machine there will always be a certain amount of fric- tion ; and this can never be omitted from the equation of virtual work. There are cases, however, as that of a balance on a knife-edge, where the frict'on is very small ; and for these the principle, what is gained in jwwer is lost in velocity, is very approximately true. Where the friction is considerable this is no longer the case. Lot F and / bo the resistance of tiiction and its virtual velocity, then the equation for any machine will take the form Pp — Ww — Ff = 0, which shows us that although P can be made as small sta we wish by taking p large enough, yet the mechanical advantage of diminishing P is restricted by the fact that / increases with/>; and therefore as P diminishes there is' a corresponding increase of the work to be done against fric- tion. Hence if friction be neglected, there is no practical limit to the ratio of P to fF ; but if the friction be con- sidered, the advantage of diminishing P has a limit, since if Pp remains the same, Ww must decrease as /y increases; i. c, the work done against friction increases with the^ complexity of the machine ; and thus puts a practical limit to the mcchanicAl advantage which it is possible to obtain by the use of machines. 180 aiXPLB MACBlNSa. 108. Sln^te MacMnea.— The simple machines, some- times called the Mec/ianical Powers, are generally enumei- ated as six in number ; the Lever, the Wheel and Axle, tho Inclined Plane, the Pulley, the Wedge, and the Screw. The Lever, the Inclined Plane, and the Pulley, may bo considered as distinct in principle, while the others are combinations of them. llie efficiency* of a machiuQ is the ratio of the useful work it yields to the whole amount of work performed by it The use/til work is that which is performed in over- coming useful resistances, while losi work is that wiiich is spent in overcoming wasteful resistances. Useful lesist- anccs are those wliich the machine is specially designed to overcome, while the overcoming of wasteful resistances is foreign to its purpose. Friction and rigidity of cords are wasteful resistances while the weight of the body to be lifted is the useful resistance. Let W be the work done by the moving forces, W^ the useful and Wi the lost work when the machine is moving uniformly. Then If = r. + Wi, and if Jf denote the efficiency of the machine, we have W In a perfect machine, whore there is no lost work, the efficiency is unity ; but in every machine some of tho work is lost in overcoming wasteful resistances, so that the efficiency is always less than unity ; and the object of all improvements in a machine is to bring its efficiency as near unity as possible. The most noticeable of the wasteful resistances are fric- tion and rigidity of cords ; and of these we shall consider * HonmimM ckUad mo^Mn. «* KnviLianitTM OF raw lkvsk. lU bines, eome- illy enumer- ; then we have by (1) of Art. 30 or It* = P* + W» + 2PW COB AFB ; IP = P» + W^ + 2PW cos fc). (8) ic?iich gives the pressure, R, on thefukrum. To find its direction resolve P, W, und R parallel and perpendicular to the lever, and we have for parallel forces, P cos a — W cos fi—R cos = 0; for perpendicular forces, P sm a + W sin (i—R sin = 0; by transposition and division we get P sin « -f- IF sin /3 tan e = F cos a — W cos /3' (8) which gives the direction of the pressure. Cob. — When the lover is bent or curved the condition of equilibrium is the same. Solution by the principle of virtual velocities. Suppose the lever to be turned round in the direction of P through the angle dO, into the position ab; let p and [aires that the iversely as the its dii-oi^tioii ; V, intersect in Id cquilibrinm he reaction of be e<]Tia!l and R88 through Denoto this !i the lever by (1) of Art. 30 ^; (2) ! parallel and ^ cos = 0; i? sin © = 0; (3) condition of he direction let p and BQUILIBRIUM OF TBB LJSVXR. "*1*pi 188 q be the perpendiculars CD and G£ respectively, then the virtual velocity of P will be (Art. 101), Aa sin a — AC-dO-aln a ■=. pd&. Similarly, the virtual velocity of TF is — qdB. Hence, by the equation of virtual work we have P'P-dd-^ W-qdd = 0; .-. P p= W-q. which is the same as (1). (*) (3) With Friction. -In the above wo have supposed fric- tion to be neglected ; and if the lever turns round a sharp edge, like the scale beam of a balance, the friction will be exceedingly small. Levers, however, usually consist of flat bars, turning about rounded pins or studs which form the fulcrums, and between the lever and the pin there will of coni-se be friction. To find the friction let r be the radius of the pin round which the lever turns ; then the friction on the pin, acting tangentially to the surface of the pin and opposing motion, = ^ sin ^ (Art. 99) ; and the virtual velocity of the point of application of the friction = rdO ; and hence the virtual work of the friction = £ sin ^- rdO, Hence the equation of virtual work is p.pd6 — W-qM — ^ sin ^ r' 4- ^j?? COS w 4- g* — »•* sin* (p sin* ta p^ — t^ siu* flk .(6) which gives the relation between the power and the weight when friction is considered, the upper or lowei sign of r sin <) being taken according as P or W ie about to pro- ponderate. Cob. — If the friction is so small that it nay be omitted, r sin ^ = 0, and (6) becomes w~ p m 112.. Th« Common Balano. — In machines generally the object is to produce motion, not rest ; in other words to do woik. The statical investigation shows only tlie limil of force to be a'-plied to put tiio machine on the point of motion, or to give it uniform motion. For any work to bo done, the force applied must exceed this limit, and the grealor the excess, the greater the amount of worL done. Thei-e 's, however, one class of applications of the levt/ where the object is not to do work, but to produce equi- librium, and which !>,re therefore 8))eL'iRlly adapted for treat- ment by statics. This is the class of measuring machines, where the object is not to overcome a particular resistance, but to measure its amount. The t'Osting machine is a gotnt example, measuring the pull which a bar of any material will sustain Ixifore breaking. The common balance and steelyard for weighing, are familiar examples. The common balance is an instrument for weighing ; it is a lever of the hrst kind, with two equal anns, with a Bcale-pr.u suspended from each extremity, the fulcrum being vertically above the centrn of gravity of the borm when the latter is horiscnt«l, and therefore vertically above lin' sin* u» ,„. , (C) EUid the weight lower sign of about to pro- ay be omitted, (7) hines generally n other words 3 only the limit n the point of my work to bo limit, and the jf work done. 3 of the lever produce equi- pted for treat- ing machines, ar ix^sistancc, ine is a goe auapended hj me&na of a Icnifeedge, i. e. , a projecting metallic edge transverse to its len){th, which reatfl open a plate of agate or othei hard sulMtance. The chains which rapport the acaln- pauB should be suspended tiom the extremities of the beam in the same manner. The point of support of the beam (folerum) should be at equal distances from the points of suspension of the scales ; and when the bnlance is not loaded the Iteam s.iould be horisontal. We can ascertain f these conditions are satisfied by olwerving whether tliere is still 'quillbrium when the snbntanoe is transferred to the scale which th ) weight originally occupied and : he weight to thai which the subst inoe originally oocnpind. The chief requisites of a good balance are : (1) When equal weights are placed in the sca1e-3Hui8 the beam should be perfectly horizontal. (2) The balance should poss*;?'^ great misibilitif ; x. «., if two weights which are very nearly equal be placed in the Bcale-pana, the beam should vary t$n«ibly from its horizontal position. ij-i 186 aSQUIBITSa OP a qood balancm. (8) When the bala..^ is disturbed it should readily return to its state of rest, or it should have staMity. L 112. To DetenniiM the Chief RaqniBites of a Good Balanpe. — Let P and W be the weights iu the scale-pans ; O the fliicrum ; h its distance from the straight line, AB, which joins the points of at- tachtuent of the sc»le-parf> to the beam ; O the centre of grayity of the beam ; and let AB be at right angles to 00, the line joining the falcmwt to the centre of gravity of the beam. Let AC = OB = a; OG — k\ w = the weight of the beam ; and 6 = tho angle which the beam makes with the horizon when there is equilibrium. Now the perpendicular from on the direction of P — a cos e-h sin 0', it (( (( W = a cos e + h sin 9\ << « « to — k sin e; therefore tak'ng moments round we have P (a COS 6~h sin 5)— If (a cos e+A sin 6)—wh sin d = ; tan = (P- W)a {P + W)h + wh' a) This equation determines the position of equilibrium. Tho first requisite — tho horizontality of 'the beam when P and W are equtd — is Ba^^isfied by making the arms equal. The second requisite [(2) of Art. Ill], requires that, for a given value of P — W, the inclination of the beam to the horisoii must be «s great as possible, and therefore the sen- sibility is greater the greater tan is for a given value of P — W'y and for a given value of tan 9 the sensibility is - fietwassMsw-v-jMiMrm mam*m f^ff ii H Ki ! m^ii>slsm9S» i' 9 ^wiv* l ^■IIWi'fcl|Wfl extent, at variance with the second requisite. Tlicy may both b .• satisfied, however, by making {P + W)h -\- wk large and a large also ; t. e., by increasing the distances of the fulcrum from the beam and from the centre of gravity of the beam, and by lengthening the arms. (See Todhunter's Statics, p. 180, also Pratt's Mechanics, p. 78.) The comparative importance of these qualities of sensi- bility and ntability in a balance will depend upon the use for which it is intended ; for weighing heavy woightA, stabilitif is of more importance; for use in a chemical laboratory the balance must possess great BMmbiUty ; ar d instruments have been constructed which indicate a varia- tion of weight less than a miUionth part of the whole. In a balance of great delicacy the fulcrum is made as thin as possible ; it is generally a kni/t-tdge of hardened steel or agate, resting on « polished agate plate, which is supported on a strong vertical pillar of brass, t^:m^' -1^ jvn 188 TBS STEBLTASD. 113. The Ste«l]rarcL— This if a kiud of balance in which the arras &ve unequal in length, the longer one beiuti graduated, along which a poist may be moved in order to balance different weignts which 01*6 placed in a scale-pan on the short-arm. While the moment of the substance weighed is changed by increasing or diminishing its quan- tity, its arm remaining constant, that of the poise is changed by altering its arm, the weight of the poise remaming the same. 114. To Qradnate the Common Steelyard.— (1) When the point of suapeiunon is coincident with the centre of gravity. Let AF be the beam of the steel- yard suspended about an axis pass- ing through its centre of gravity, C ; on the arm, CF, place a mov- able weight, P ; then if a weight, W, equal to P, is suspended from A, the beam will balance when P on the long arm is at a distance from C equal to AO. If W equals twice the weight of P, the beam will balance when the distance of P from is twice AO ; and so on in any proportion. Hence if W is successively 1 lb., 2 lbs., 3 lbs., etc., the distances of the notches, 1, 3, 3, 4, etc., where P is placed, are as 1, 2, 3, etc.. I. 0., the arm CF is divided into equal divisions, begin- ning at the fulcrum, 0, as the zero point (2) When the point of suspension is not coincident toith the centre of gravity. Let C b<» the fulcrum, W the substance to be weighed, hanging; at the extremity. A, and P the movable weight. Suppuse that when W is removed, the weight, P, placed at B will balance the long arm, CF, and keep the eleelyard in a horizontal position ; then the moment of the instrument WWi M *w r>n »-<|«ttgw>ya of balance in •nger one beiui-- ved in order to 1 a scale-pan on the substance hing its quon- f the poise is t of the poise teelyard.— (1) vith the centre p-r^ n|.M weight of P, P from is lence if IF is lances of the re as 1, 2, 3, risions, begin- nncident mth be weighed, table weight. P, placed at aleelyard iu instrument MXAMPLXa. 189 itself, aboBt 0, is on the side, CF, and is equal to P' CB. Hence, if W hangs from A, uid P from any point E, then for equilibrium we must have PCE + PBC = r. AC; • or P-BB=fr-AO; BE = -^ . AO. If we make W successively equal to P,2P, SP, etc., then the values of BE will be AG, 2AC, SAC, etc., and these distances must be measured off, commencing at B for the zero point, and the points so determined marked 1, 2, 3, 4, etc. Such a steelyard catinot weigh below a certain limit, corresponding to the first notch, 1. To find the length of the divisions on the beam, divide BE, the distance of the poise from the zero point, by the weight, W, which P balances when at the point E. The steelyard often has tivo fulcrums, one for small and the other for large weights. EXAMPLKS. 1. What force must be applied at one end of a lever 12 ins. long to raise a weight of 30 lbs. hanging 4 ins. from the fulcrum which is at the other end, and what is the pressure on the fulcrum f Ans. 10 lbs. : 20 lbs. 2. A lever weighs 3 lbs., and its weight acts at its middle point ; the ratio of its arras is 1 : 3. If a weight of 48 lbs. , be hung from the end of the shorter arm, what weight must be suspendeu from the other end to prevent motion ? Ans. 15 lbs. f HI IW WHEEL Am) AXLM. 8. The arms of a bent lever are 3 ft. and S ft. and inclined to each other at an angle B = 150°. To the short arm a weight of 7 lbs. is applied and to the long osm a weight of 6 lbs. is applied. Required the inclination of each arm to the horizon when there is equilibrium. Ans. The short arm is inclined at an angle of 18° 32' above the horizon, and the long arm is inclined at an angle of 48° 22' below the horizon. 1.15. The Wheel and Axle.— This machine consists of a wheel, a, rigidly connected with a horizontal cylinder, b, movable round two tmnnions (Art 99), one of which is shown at c. The power, P, is applied at the circumfer- ence of the wheel, sometimes by a cord coiled round the wheel, sometimes by handspikes as in the capstan, or by handles as in the windlass ; the weight, W, hangs at the end of a cord Listened to the axle and coiled round it 116. Conditions of EqcUibrinm of tiie Wheel and Axle. — (1) Let a and b be the radii of the wheel and axle respectively ; P and W the power and weight, supposed to act by strings at the circumference of the wheel and axle perpendicular to the radii a and b. Then either by the principle of virtual velooiti < or by the principle of momenta we have Pa = Wby or P _ radius of axle W ~ radius of wheel' (1) It is evident that, by increasing the radius of the wheel or by diminishing the radius of the axle, any amount of mechanical advantage may be gained. It will also be seen 5 ft. and inclined the short arm a ium a weight of D of each arm to angle of 18° 22' lined at an angle V, hang8 at the round it he Wheal ftud wheel and axlo ;ht, sapposed to wheel and axle m either by the pie of moments (1) ioB of the wheel any amount of rill also be seen DTFTERSIfTIAL WBJBSL AlfD AXLB. 191 that thi" machine is only a modification of the lever ; the peculiar advantage of the wheel and axle being that an end< lc'88 scries of levers are brought into play. In this respect, then, it surpasses the common lever in mechanical advan- tage. In the above we have supposed friction to be neglected, or, what amoui.f« to the same thing, have assumed that the trunnion is indefinitely small In practice, of course, the trunnion has a certain radius, r, and a certain coefficient of friction. Calling R the resultjint of P and If, and taking into account the friction on the trunnion we have for the relation between P and W • Pa = ITJ + r sin ^ V/^lPrMP 1Fcm"w, (2) u being the angle between the directions of P and W exactly as in Art 110. (2) Differential Wheel and Axle. — By diminishing b, the radius of the axle, the strength of the machine is diminished ; to avoid this disadvantage a differential wheel and axle is sometimes employed. In this instrument the axle consists of two cylinders of radii h and b' ; the rope is wound round the former in one direction, and after passing under a movable pulley to which the weight is attached, is wound round the latter in the opposite direc- tion, so that as the power, P, which is applied ae before, tangentially to the wheel of radius, a, moves in its own direction, the rope at h winds up while the rope at b'. unwinds. For the equilibrium of the forces (whether at rest or in uniform motion), the tensions of the rope in hn and b'n Pia.eo ■:>vaiisimmtmg^ 192 TOOTHED WBSBLa. are each equal \Ai\W. Hence, taking moments ronod the centre of the traniiion, e, we have Pa ■{- \Wh' - \m ^ Oi (3) hence by making the difference, b — b', small, the power can be made as small as we please to lift a given weight. Let the wheel turn through the angle dd; the point of application of P will describe a space := add, and the weight will bo liftad through a space = ^ (6 — b') de, which latter will be very small if b — b' is very small. Therefore, since the amount of taork to be done to raise the weight to any given height, is constant, economy of power is accomplished by a loss in the time of performing the work. 117. Toothed Whoels. — Toothed or cogged toheeh arc wheels provided on the circumferences with projections called teeth or cogs which interlock, as shown in the figure, and which are therefore capable of transmitting force, so that if one of the wheels be turned round by any means, the other will be turned round also. When the teeth are on the aides of the wheel instead of the circumference, they are oalldd croton wheels. Wlicn the axes of two wheels are neither perpendicular nor parallel to each other, the wheels take the form of frustums of cones, and are called beveled wheels. When tliero is a pair of toothed wheels on each axle with the teeth of the large one on one axle fitting between the teeth . ifq Fi».M InaiA ^^^^MmMMm j.i;j«a:n«tPn*r.rf-i' .:^^^R~^^l^-^ nomeDts roaod the 0; (3) , small, the power ifb a given weiglit. f performing the cogged wheels arc with projectiona awn in the figure, smitting force, so id by any means, wheel instead of wheels. When f\%.U TOOTHED WHEELS, 193 of the small one on the next axle, the larger wheel of each pair is called the wheel, and e smaller is called the pinion. By means of a combination of toothed wheels of this kind called a train of wheels, motion may be transferred from one point to another and work done, each wheel driving the next one in the aeries. The discussion of this kind oic machinery possesses great geometric elegance ; but it would be out of place in this work. We shall give only a slight sketch of the skimpiest case, that in which the axes of the wheels are all parallel. For the investigation of the proper formfe of teeth in order that the wheels when made shall run truly one upon another the student is referred to other works,* 118. To Find the Relation of the Power find Weight in Toothed Wheels.— Let j* and B be the fixed centres of the toothed wheels on the circumferences of which the teeth are arranged ; QCQ a normal to the sur- faces of two teeth at their point of contact, C. Suppose an axle is fixed on the wheel, B, and the weight, W, suspended from it at E by a cord ; also, suppose the power, P, acts at D with an arm DA; di-aw Aa and BA perpendicular to QCQ. Let Q be the mutual pressure of one tooth upon another at C ; this pressure will be in the direction of the normal QCQ. Now since the wheel, A, is in equilibrium about the fixed axis. A, under the action of the forces, P and Q, we have P.AD = ^.Aa; (1) and since the wheel, B, is in equilibrium abont the fixed axis, B, under the action of the forces, Q and W, we have Fr.BE=: ^B*. (2). • 8m Ouodere'i OimmU tf MtOmUm ; Rankice's AppHtd MetAanie* ; Mom. ley's Oiglnetrbtg; WUlls'* PHmetplm tf UtehamUm; CoUlgnon's SloMfiM,- and • Paf$r qf Mr. Mtf* <« M< Om*. mi. Trmt., Vol. n, p. fn. 9 "F 194 TRAIN OP IT WaSBLa. Dividing (1) by (2) we have P W AD BE Aa B4' or mome nt of P _ Aa moment of Jf ~ B4* If the direction of the normal, QCQ. at the point of con« tact, C, changes ae the action passes from one tooth to the succeeding, the relation of P to TT becomes variable. But, if the teeth are of such form that the normal at their point of contact shall always be tangent to both wheels, the lines Aa and Bd will become radii, and their ratio constant. And since the number of teeth in the two wheels is propor- tional to their radii, we have mome nt of P _ number of tee th on the wheel P moment of fT "" number of teeth on the wheel W' (3) 119. Relation of Power to Weight in a Train of n Wlieela — Let R^, R,, R^, etc., be the radii of the suc- cessive wheels in such a train ; Ti, r„ r„ etc., the radii of the corresponding pinions; and let P, P,, P,, P,, . . . W, be the powers applied to the circumferences of the successive wheels and pinions. Then the first wheel is in equilibrium about its axis under the action of the forces Pand P.^, since the power applied to the circumference of vhe second wheel is equal to the reaction on the first pinion, therefore Similarly p X R^ = Pi X r,. Pt A R, = P, X r, ; p. X /?, = P, X r, ; etc = etc.; Pn-l X Rn =3 w X r» •■ . ■ jpaiMiy^jiiJMLwiMiik^mi!^^ mxisenmmsswf&'i^sirx: m. EXAMPLES. 195 the point of con- Q one tooth to the IBS variable. But, mal at their point ) wheels, the lines 'ir ratio constant. I wheeUi ia propor- khe wheel P . . he wheel W' ^ ' in a Train of n radii of the sac- !., the radii of Pi. Ps, . . . W, of the successive in equilibrium jrcea Pand Pj, ce of the second inion, therefore is Multiplying these equations together and omitting common factors, wo have P W _ '1 X r, X r. X • ■ a • Jii X R, X Jti X (1) It will be observed, in toothed gearing, that the smaller the radius of the pinion as compared with the wheel, the greater will be the mechanical advantage. There is, how- ever, a practical limit to the size that can be given to the pinion, because the teeth must be large enough for strength, and must not be too few in number. Six is generally the least number admissible for the teeth of a pinion. Equa- tion (1) shows that by a train consisting of a very few pairs of wheels and pinions there is an enormous mechanical advantage. Thus, if there are three pairs, and the ratio of each wheel to the pinion is 10 to 1, then P is only one thousandth part of W; but on the other hand, W will only make one turn where P makes one thousand. Such trains of wheels are very useful in machinery such as hand cranes, where it is not essential to obtain a quick motion, and where the power available is very small in comparison to the weight (See Browne's Mechanics, p. 109.) EXAMPLES. 1. What is the diameter of a wheel if a power of 3 Ibe. is just able to move a weight of 12 lbs. that hangs from the axle, the radius of the axle being 2 ins.? Arts. 16 ins. 2. If a weight of 20 lbs. be supported on a wheel and axle by a force of 4 lbs., and the radius of the axle is f in., find the radius of the wheeL Ana. 3^ ins. 3. A capstan is worked by a man pushing at the end of a pole. He exerts a force of 50 lbs., and walks 10 ft. round for every 2 ft. of rope pulled .in. What is the icsistauce overcome ? Ans. 250 lbs. 196 tNCUi^SD PLAlfE. 4. An axle whose diameter is 10 ine., iias on it two wheels the diameters of wliich are 2 ft and 2| ft. respew- tively. Find the weight that would be suppoi ted on the axle by weights of 25 lbs. and 24 lbs. on the smaller and lurgor wheels respectively. Ana. 264 lbs. 120. The Inclined Plane.— This has already been partly considered (Art 96, etc.). Let the power, P, whose direction makes an angle, B, with a rough inclined plane, be employed to drag a weight, W, up the plane. Then if is the angle of friction and i the inclination of the plane, we have from (3) of Art. 96, P _ w wn (t 4- ») • "^ cos (« - e)' If P acts along the plane, = 0, and (1) becomes p - |yB' P(»' + ^) , cos ^ If P acts horizontally, 9 = — t, and (1) becomes P= ff' tan («• + 0). (1) (2) (8) Cor.— If we suppose the friction = 0, (1), (3), and (3) become respectively rsin t P= W COS0* <*) P= TTsini, (5) P = fF tan i. (6) SoH.-— It follows from (4), (5), and (6) that the smaller Mituitttsus»<>iititmmmm ms., lias un it two ft aud 2| ft. respac- *o Buppoited on the on the smaller uud Ann. 264 lbs. } has already boon the power, P, whose agh inclined plane, he plane. Then if oation of the plane, (1) (1) becomes (2) (1) becomes (3) (1), (2), and (3) (4) (6) (6) that the smaller — S! w^mmmmm. TBE PULLEY. mr- 197 the inoKtiation* of the plane to the horison, *he greater will be the mechanical advanta^. If we take in friction there exception to this rule wnen i > ^ — 0. The 18 an 2 gradients on railways are the most common examples of the use of the inclined plane ; these are always made au low iis is convenient in order to enable the engine to lift the lieavieat possible train. 121. The Pnlley. — The pulley consists of a grooved wheel, capable of revolving freely abo> t^^at of the next 2>P, and so on; and the tension of the »th cord is 2»-»P. Then the sum of all the tensions of the cords attached to the weigl t must equal W. Hence /» + 2P + 2»P + 2»-»P = 2" — 1 In thifl system the weights of the movable pulleys assist P ; in the two former systems they act against it. (CXAMPL.ES. 1. What force ia necessary to raise a weight of 480 lbs. by an arrangement of six pulleys in which the same string passes round each pulley ? Ana, 80 lbs. * 8. Find the power which will support a weight of 800 lbs. with three movable pulleyai, arranged as in the second system. An*. lOOlbs. inOS TWf WMDOU, 3. If thero b« eqoiUbriam between P and W with three pulleys in the third syBtem, what ftdditional weight can be raised if 2 lbs. be added to P? Ana. 14 lbs. 126. The Wedg*. — The wedge w a triangnlar prism, usually isosceles, and is used for separating bodies or parts of the same body by introdacing its edge between them and then thrusting the wedge forward. This is effected by the blow of a hammer or other snoh moaii^ which produces a ▼iolent pressure, for a short time, in a direction perpen- dicular to the back oi the wedge, and the resistance to be overcome consists of friction and a reaction due to the moleoalar attractions of the particles of the body which are being separa>>ed. This reaction will be in a direction perpendicular to the inclined surface of the wedge. 137. The lfeet4aio«l Ad- vantage of the Wedge.-^Let ACB represent a section of the wedge perpendicular to its in- clined fitces, the wedge having been driven into the material a distauce equal to DC by a force, P, acting in the direction DO. Draw DE, DF, perpendicular to AC, BC, and let B denote the reactions along ED and FD ; then liR will he the frictitm acting at E and F in the directions EA and FB. Let the angle of the wedge or ACB = %a. Resolve the foroos which act on the wedge in directions perpendioalar and parallel to the back of the wedge, then we have for peqiendioular forces n9.N P = 2R sin a + 2ftR co» a. (1) 7%M equaiioH may ako i» oHmntd from Uie prinoiple of work m follows : If the wedge has been driv«B inW the W with tbree weight ean be Ana. 14 lbs. ingubir prism, odiee or parts reen them and iflected by the ch produces a action perpen- sistauoe to be n due to the ic body which in a direction redge. I9.M the frictioB i'B. Let the in directions wedge, than (1) fmnoiple tf MBOHANICAL ADVANTAatt OP WJBDGX. 203 material a distance eqnal to DO by a force, P, acting in the direction DO,, then the work done by i* is P x DO (Art 101, Rem.); and since the points £! and F were originally together, the work done against the resistance i? is 22 X DE -f /? X DP = a^ X DE ; and the work done against friction is inR x EC. Hence the equation cf work is P X DO = 2i2 X DE + 2^i2 x EC, (?) which reduces to (1) by substituting sin a and cos a for DE - EC ]T0"*'^D0* Cob. — If friction be neglected, (2) becomes P 2DE DC AB AC that is -i: = P R back o f the we dgg length of one of the equal sides' It follows that the narrower the back of the wedge, the greater will be the mechanical advantage. Knives, chisels, and many other implements are examples of the wedge. In the action of the wedge a great part of the power is employed in cleaving the material into which it is driven. The force required to effec^: this is so great that instead of applying a continuous pushing force perpendicular to the back of the wedge, it is driven by a series of blows. Be- tween the blows there is a powerful reaction, B, acting to puBh the weJge back again out of the cleft, and this is resisted uj the triction which now acts in the directions EC and FO. tience when the wedge is on the point of starting back, between the blows, t.a equation of equi- librium will be from (1) 2S sin a — 2ftR cos a =r ; . • . as tAn~' I*. S04 THM 8CBBW. fi N And tho wedge will fly back or not according as a > or < tan~* /«. (See Browne's Mechanics, p. 117. Also Magnus's Mechanics, p. 157.) 128. The Screw.— The screw consists of a right cir- cular cylinder, on the convex surface of which there is traced a uniform projecting thread, abed .... inclined at L nstant angle to straight lines parallel to the axis of the cylinder. The path of the thread may be traced by the edge AG of an inclined plane, ABC, wrapped ronnd the cylinder; the base of the plane corresponding with the circumference of the cylinder, and the height of the plane with the distance between the threads which is called the pitch of the screw. The threads may be rectangular or triangular in section. The cylinder fits into a blqck, ou the inner sur- face of which is cut a groove which is the exact counterpart of the thread. The block in which the groove is cut is often called the nut. The power is generally applied at the end of a lever fixed to tho centre of tlie cylinder, or fixed to the nut. It is evident that a screw never requires any pressure in the direction of its axis, bnt must ^ made to revolve only ; and this can be done by a force acting at right angles to the extremities of its diameter, or its diameter produced. 129. The Relation between the Fewer end Jie Weight in the Screw. — SuppiiM) the TH)wcr, P, to act in a plane por{)ondicular to the axis of the cylinder and at the end of an arm, DE = a, and suppose the screw to have made one revolution, the power, P, will have moved through the circumference of which a, is the radius, and tlie work done by P will be Py.%ita. During the same Fia.«7 I ^te MM ng as a > or k.180 Magnus's r a right cir> bich there is . iuclined at le axis of the J •7 ooanterpart cat is often »t the eod of i to the nut. ssuro in tUa )volve only ; ht angles to troduced. and Jh.9 P, to act in :* and at the ew to have ave moved radius, and the same I TBS SCREW. 205 time the screw will have moved in the direction of its axis t'.roug 1 the distance, AB = Sfrr tan a, r being the radius of the cylinder, nnd n the angle which the thread of the screw makes with its base. Then as this is the direction in which the resistance is encountered, the work done against the resistance, W, is Winr tan a. Hence if no work is lost the equation of work will be P X 2ffa = ff X 2rrr tan «. (1) That is the power is to the weight as the pitch of the screw is to the circumference described by the power. If there is friction between the thread and the groove, let B bo the normal pressure at any point, p, of the thread, and nR the friction at this poinf^, iihen the work done against the friction in one revolution is ftI,R 2nr sec a, I.R denoting the sum of the normfd reactions at all points of the thread. Hence the equation of work is P %na = W2irr tan « -f ^ 2nr sec «li?. (2) But, for the oqnilibrinm of the screw, resolving parallel to the axis, we have therefore W = I. (R cos a— nJt sin a), W ZR = cc3a — (i sm a which in (2) gives Pa ^ WV tan « H : — ; cos a — /u sm n' or Pa = Wr tan (a -f 0), being the angle of friction. (3) 306 psoyr'a D/jvmsgfmAL sossw. 129a. Fxonj'u DifEurential 8or«w.— If h denote the pitch of a screw (1) becomes aPira = Wh, which expresses the relation between P and W, when fno- tion is neglected. Therefore the mechanical advantage is gained by making the pitch very small. In some cases, however, it is desirable that the screw should work at fair speed, as in ordinary bolts and nuts, and then the pitch must not be too small. In oases where the screw is used specially to obtain pressure, as in screw-presses for cotton, etc., we do not care for speed, but only for pressure. But in practice it is impossible to get the pitch very small from the fact that *' the angle of inclination is very flac, the threads run so near each other as to be too weak, in which case the screw is apt to " strip its thread," that is, to tear bodily out of the hole, leaving the thread behind. Where very great pressure is required a difierential nut- bole is resorted to. Let the screw work in two blocks, A and B, the first of which is fixed and the second movable along a fixed groove, n ; and let A be the pitch of the thread which works in K SOS SSS| ns-M the block, A, and h' the pitch of the thread which works in the block B. Then one revolution of the screw impresses two opposite motions on the block, B, one equal to h in the dii-ection in which the screw advances, and the other equal to /(' in the opposite direction. If then the block, B, is connected with the resistance W, we have by the principle of work 8P»Trted in « horisontal posi- tion by props placed at its extremities : find where a weight of 28 lbs. must be placed so that the pressure on one of the props may be 8 lbs. Ana. Two feet from the e.nd. 6. Two weights of 12 lbs. and 8 Ib& respectively at the ends of a horisontal lever 10 feet long balance : find how far the fulcrum ought to be moved for the weights to bal- ance when each is increased by 2 lbs. Ann. Two inches. 6. A lever is in equilibrium under the action of the forces P and Q, and is also in equilibrium when P is trebled and Q is increased by 6 lbs.: find the magnitude of Q. Ans. 3 lbs. 7( In a lever of the first kind, let the power be 217 lbs., the weight 726 lbs., and the angle between them 196°. Find the pressure on the falcmm. Am. eS2.7 lbs. ao8 BXAMPhEA 8. If the power and weight in a straight lever of the first kind be 17 lbs. and 32 Ib&, and make wi^'a each other an angle of 79° ; find the pressure on the fulcrum. An8. 39 lbs. 9. The length of the beam of a false balance is 3 ft. 9 ins. A body placed in one scale balances a weight of 9 lbs. in the other ; but when placed in the other scale it balanceb 4 lbs.; required the true weight, W, of the body and the lengt,hs, a and b, of the arms. Ans. F = 6 lbs.; « = 1 ft. 6 ins.; i = 2 ft 3 ins. 10. If a balance be false, having its arms in the ratio of 15 to 16, find how much per lb. a customer really pays for tea which is sold to him from the longer arm at 3s. 9d. per lb. ^n«. 4s. per lb. 11. A straight uniform lever whose weight is 50 lbs. and length 6 feet, rests in equilibrium on a fulcrum when a weight of 10 lbs. is suspended from one extremity : find the position of the fulcrum and the pressure on it. Ana. 2^ ft from the end at whioh 10 lbs. is suspended ; 60 lbs. 12. On one arm of a false balance a body weighs 11 lbs.; on the other 17 lbs. 3 02.; what is the true weight ? Ans. 13 lbs. 12 oz. 13. A bent lever is oompoeer of two straight uniform rods of the same length, inclined to each other at 120°, and the fulcrum is at the point of intersection : if the weight of one rod be double that of the other, show that the lever will remain at rest with the lighter arm horizontal. 14. A uniform lever, / feet long, has a weight of W lbs., suspended from its extremity ; find the position of the ful- crum when the long end of the lever balances the short • t lever of the ta each other rum. Ins. 39lbe. lance is 3 ft. s a weight of other scale it , of the body 2 ft 3 ins. in the ratio of ;r really pays arm at 3s. 9d. 4s. per lb. is 50 lbs. and mm when a lity : find the i suspended; ighs 11 lbs.; jht? lbs. 12 oz. fht uniform at 120°, and le weight of .he lever will of W lbs., of the ful- iis the short 9 SXAMPLKS. 209 end with the weight attached to it, supposing each unit of length of the lever to be w lbs. ■^nt. ^T-n-- , J - \ is the short arm. 15. A lever, I ft long, is balanced ,. her it 's placed upon a prop i of its length from the thick end ; when a weight of W lbs. is suspendofl from the small end the prop must be shifted j ft. towards it in order to maintain eqailibiium ; required the weig.t of the lever. Ana. ^W. 16. A lever, I ft. long; is balanced on a prop by a weight of W lbs.; first, when the weight is suspended from the thick end the prop is a ft. from it; secondly, when the weight is suspended from the small end the prop is b ft. from it ; required the weight of the lever. . W{a + b) ,. l — {a -1,-0) 17. The forces, P and W, act at the arms, a and b, respectively, of a straight lever. When P and W make angles of 30° and 90° with the lever, show that when equi- fibW librium takes place P = • a 18. Supposing the beam of a false balance to be nniform, a and b the lengths of the arms, P and Q the apparent weights, and IV the true weight ; when the weight of the beam is taken into account show that a b P -W 19. W-Q If a be the length of the short arm in Ex. 14, what must be the length of the whole lever when equilibrium takes place P /2aW Ana. a + a/ —^^ + aK w 20. A man whose weight is 140 lbs. is just able to snp- port a weight that hangs over an axle of 6 ins. radius, by SIO MXAMTLMS. hanging to the rope that passea over the corresponding wheel, the diameter of which is 4 ft; find the weight sup- ported. Ans. 560 lbs. 21. If the difference between the diameter of a wheel and the diameter of the azte be aiz times the radius of the axie, find the greatest weight that can be sustained by a force of 60 lbs. Ana. 240 lbs. 22. If the radius of the wheel is three times that of the axle, and the string round the wheel can support a weight of 40 lbs. onlj, find the greatest weight that can be lifted. Am. 120 lbs. 23. What force will be required to work the handle of a windlass, the resistaDce to be overcome being 1156 lbs., the radius of the axle being six ins., and of the handle 2 ft. Sins.? Ana. 216.75 lbs. 24. Sixteen sailors, exerting each a force of 29 lbs., push a capstan with a length of lever equal to 8 ft, the radius of the capstan being 1 ft. 2 ins. Find the resistance which this force is capable of sustaining. Ana. 1 ton 8 cwt. 1 qr. 17 lbs. 25. Supposing them to have wound the rope round the capstan, so that it doubles back on itself, the radius of the axle is thus increased by the thickness of the rope. If this be 2 ins. how much will the power of the instrument be diminished. ' Ana. By \, or 12f per cent. 26. The radios of the axle of a capstan is 2 feet, and six men push each with a force of one cwt. on spokes 5 feet long ; €nd the tension they wiU be able to prodnce in the rope which leaves the axle. Ans, m cwt. 27. The difference of the diameters of a wheel and axTe is 2 feet 6 inches ; and the weight is equal to six times the power ; find the radii of the wheel and the axle. Ant. 1%\xul; 3 ina. I sorresponding s weight Bup- 18. 560 lbs. tf a wheel and IS of the axle, by a force of na. 240 lbs. E>8 that of the wrt a weight »n be lifted. w. 120 lbs. e handle of a 1156 lbs., the handle 2 ft. 216.75 lbs. 29 Iba, push the radins of stance which qr. 17 lbs. )e round the radius of the •ope. M this istmment be per cent. eet, and six spokes 5 feet oduce in the IS. 15 cwt. eel and axle six times the e. ns.; dina. I ' itAJtPLSa. an 28. If the rsdins of a wheel is 4 ft., and of the a tie 8 ins., And the power that will balance a weight of 500 lbs., the thickness of the rope coiled round the axle being one inch, the powe; acting without a rope. Ana. 88.54 lbs. 29. Two given weights, P and Q, hang vertically from two points in the rim of a wheel turning on an axis; find the position of the weights when equilibrium takes plaoe, supposing the angle between the radii drawn to the points of suspenrijn to be 90°, and that 6 is the angle which the radius, drawn to /**& point of sus- pension, makes with the Tertical. Ana. tan d = Q 30. What weight can be supported on a plane by a hori- zontal force of 10 lbs., if the ratio of the height to the base isfP Ana. IS^lbs. 31. The inclination of a plane is 30°, and a weight of 10 lbs. is supported on it by a string, bearing a weight at its extremity, which passes over a smooth pulley at its summit ; find the tension in the string. Ana. 5 lbs. 32. The angle of a plane is 45° ;. what weight can be supported on it by a horizontal force ot 3 lbs., and a force of 4 lbs. parallel to the '^Ivxe, both acting together. Ana. 3 -H 4 V2 lbs. 33. A body is supported on a plane by a force parallel to it and equal to | of the weight of the body ; find the ratio of the height to the base of the plane. Ana. 1 : 2 -v/e. 34. One of the longest inclined planes in the world is the road from Lima to Oallao, in S. America ; it is 8 miles long, and the faU is 511 ft. Calculate the inclination. Am. 88' 27", OT 1 yard in 62. 212 SXAMPLBB. 3? 35. If the force required to draw a wagon on a horizontal road be ^th part of the weight of the wagon, what will be the force required tc draw it up a hill, the elope of which is 1 in 43. Ans. -rrVitl' pa^t of the weight. 36. If the force required to draw a train of cars on a IcTcl railroad be yf^th jmrt of the load, find the force required to draw it up a grade uf 1 in 5(3. Ans. xrS-jth part of the load. 37. What force is required (neglecting friction) to roll a ciisk weigliing 964 lbs. into a curt 3 St. high, by means of a plank 14 ft long resting against the cart. Arts. The force must exceed 206 lbs. 38. A body is at rest on a smooth inclined plane when the power, weight and normal pressure are 18, 26, and 12 lbs. respectively ; find the inclination, a, of the plane to the horizon, and the angle, 6, which the direction of the power makes with the plane. Ana. a = 37° 21' 26"; 6 = 28° 46' 54". 39. If the power which will support a weight when act- ing along the plane be half that which will do so acting horizontally, find the inclination of the plane. Ana. 60°. 40. A power P acting along a plane can support W; and acting horizontally can support x ; show that P»= »'»-««. 41. A weight W would be supjwrted by a power P act- ing horizontally, or by a power Q acting parallel tn a horizontal , what will be lope of which the weight. I of cars on a Qnd the force of the load. ;ion) to roll a by means of a eed 206 lbs. id plane when ) 18, 26, and i the plane to •ection of the 58° 46' 54". ht when act- do so acting Ana. 60°. )port W; and (owor P nct- rallel t. weight, W, of each pulley 6 lbs. Ans. W — 15P •^- llw = 1666 lbs.; Strain = 16P + 16u> = 1675 lbs. 49. In a system of pulleys of the third kind, there arc 2 movable pulleys, each weighing 2| lbs. What power is required to support a weight of 6 cwt. ? Aiu. 94.67 Ibc. 60. Find the power th»t mill support a weight of 100 lbs. by means of a system of 4 pulleys, the strings being all attached to the weight, and each pulley weighing 1 lb. Ana. 5\^ lbs. 814 sxjZii'Lsa. i 51. The cinmmfeMnoe of the circle oorrestwiulhig to tibe point of applicatioii of P is 6 feet ; find how many tunui the screw must make on a cylinder 2 feet long, in order that IT may bo equal to 144P.. An». 48. 62. The distance between two consecutive threads of a screw is a quarter of an inch, and the length of the power arm is 5 feet; find what weight will be sustained by a power of 1 lb. Ann. 480Tr Iba. 53. How muiy turns must be giten to h screw formed upon a cylinder whose length is 10 ins., and circumference 6 ins., that a power of % ozs. may overcome a pressnre of IOO0Z8.P Ana. 100. 54. A screw is made to '.evolve by a force of 2 lbs. applied at the end of a le.or 3.5 ft long; if the distance between the threads be \ in., what pressure can be pro* duced r Ant. 9 cwts. 1 qr. 20 lbs. 56. The length of the power-am is 16 inches ; find the distance between two consecutive threads of the screw, that the mechan ol advantage may be 30. Ana. it ins. 66. A weight of IF pounds is suspended f • +^i' btflok of a single movable pulley, and the end f't .: ■ c**vii hi which the power acts, is fastened at the d'!ut<.uii < h ft from the fulcrum of a horizontal lever, a ft long, «.f the second kind ; find the force, P, which must be applied per- p«Qdioakirly at th« extremity of the lever to initain W. Ana. P = -^-. 57. In a steelyard, the weight of the bean is 10 lbs., and the distance of its centre of gravity from the fulcrum it 2 ins., find where a weight of 4 lbs. most be phioed to bal- ance it Ana. At 6 ini. '^nmmm^mmtmmgmm ipoo^hig to tile low many tnnifl i long, in ordef Atu. 48. ve threads of a h of the power eostained by a rm. 4807rlba. a Borew fonned ciroiimferenoe e a pretsare of Ana. 100. force of a lbs. if the distance re can he pro- 1 qr. 20 Ibfl. ches; find the of the Bci-ew, Aru. IT ins. Ir* fh' block Ui ■ c<»rd in :i' • 5 ft long, *A the >e applied per- luttain W. P = I*. ia B 10 lbs., and le fnlcnim it ihioed to bal- At Sim. MXAMPLM8. iUS 68. A body whose weight is V2 lbs., is placed on a rongh plane inclined to the horizon at an angle of 46°. The co- efficient of friction being -p, find in what direction a force of (V3 — 1) lbs. must act on the body in order jnst to support it Am. At an angle of 80° to the plane. 59. A rongh piano is inclined to the horizon at an angle of 60° ; find the magnitni'e and the Jireotion of the least force which will prevent e body weighing 100 lbs. fiom slid- ing down the plane, the coefficient of friction being — • Am. 60 lbs. inclined at 30° to the plane. U^WIV' ■«'««■ IlLJIBUIIIl IMI^U. ■W^ i CHAPTER VIII. THE FUNICULAR* POLYGON— THE CATENARY ATTRACTION. 130. fiqiiilibfilun of the Fnniciilar Poljrgon.— If a cord whose weight is neglected, is saspended From two fixed points, A and B, and if a series of weights, Pj, P„ P„ etc., be saspended from the given points Q^, Q„ Q^, etc., the cord will, when in equilibrinn:, form a polygon in a vertical' plane, which is called the Funicular Polygon. Let the tensions along the successive portions of the cord, AQ I, Qi Q^, QtQ»> ®^ ^ respec- tively r,, T„ r,, etc., and let 9,, 9„ 9„ eta, be the inclinations of these portions to the horizon. Then Qi is in equilibrium under the action of three forces viz., P,, acting vertically, 7",, die tension of the cord AQ^, and 7,, the tension of Qi Q,. Resolving these forces we have, for horizontal foroes, jT, cos 0, — T, cos 9,-0, (1) for vertical forces, P, + T, sin 0, — T, sin 9, = 0, (2) In the same way for the point Q, we hare, for horizontal forces, T, cos 0, — T, cos 9, = 0, (8) for vertical forces, P, + T, sin 9, — T, sin 9, = 0, (4) • The term, ^mtou l w, bM rafcrwM* Alone to Um oord, ud kaa bo ■«A«iilMd ■ignttOHMe. Fia-M Mvomwii .wTRf^f^w" I. CATENARY Poljrgon.— Ifa )d from two fixed ^tfl, P„ P„ />„ ^xj Qt> Qi, etc., a polygon in a %T Polygon. i.n foHMS via., P,, AQ„ and T^, 8 we have. .0, =. :0, (1) «» = 0, (2) ». = 0, (8) (?, = 0, (4) hM no BcchMilMl MQUIUBRIUM or TBtt WUmCULAS POLTOON. 217 Henoe from (1) and (8) we have T^ cog 0, = JT, COB ©, = J*, cos 0, = etc. , that is, the horitontal components of the tensions iv the dif- ferent portions of the cord are constant. Lot this constant be denoted by T; then we have T,^ T» ^.= COB 0, cos tfj * ■*• — cos e, ' which in (2) an ^ (4) give P, + rtan (J, — rtan fi, = 0, P, + r tan e, — rtan 8, = 0, and from (5) and (6) we have P. ; etc.. (6) tan e, = tan e, 4- -^ and Similarly and P, -f" P, T' p tan 0^ = tan fl, 4- -^i etc., etc. tan 6, = tan 0, + tan 0, = tan e^ + (n If we suppose the weights P|, P,, etc., each equal to TT, (7) becomes tan ^i — tan 0g = tan 0, ~ tan 9, = tan 0, — tan 04 - _ FT (8) Hence, /A0 tangents of the successive inclinations form a series in Arithmetic Progression. In the figure 0, = 0, 10 M^IWI IWIKIH II— I IHWI>W 818 coNSTancnotr or rat rmncvLA» polygon. tand, =-^; tan 0, = ^; T *»«»*• = -ST ; tan e^ --fgr\ etc. (9) 131. To Oonstrnct the Fanicnlar Polygon Trhen the Horisontal Progectioiui of the euocesajve Por- tions of the Cord are all eqnaL— Let ^,^4, Q^q^, q^q^, q,qi, etc., be all of constant length = a, and let Q^q^ ■=. c. Then since by (9) of Art. 130, the tangents of 9^, d„ 0,, di, etc., are as 1, 2, 8, 4, etc., we have Q^n — 3Q,qt = 3c; etc. Hence, taking the middle point, 0, of the honzontal portion, QiQ^, as origin, and the horizontal and vertical lines through it as axes of x and y, the co-ordiiateu of Q, are (fa, c) ; those of Q, are ({a, 30) ; those of Q^ .%-<^ (fa, 6c), and those of the nth vertex from Q^ are e ^ idently o 2v 9i Q' X = 2n + 1 » (« + 1] — s a; if g= A 'ft S Eliminating n from these equations we get ■'"4 (1) which, being independent of ft, is ntisfled by all the ver- tices indifi'erently, and is therefore the equation of a oarre passing through all the vertices of the polygon, and denotes a parabola whose axis is the vertical line, OY, and whose vertex is vertically below at a distance = x* The shorter the distances ^4^1, QtQw ®tc., the more nearly does the funiouiar polygon ooinoide with the para- bolic curve. I'OLrodfr. tc. (9) olygon T7hen tceMdive Por- I let Q^q^ = c. the honzontal il and vertical rdiiatei) of Q, of Qt *"» (K e ^idently (1) tj all the rer- ion of a onrre polygon, and ine, OY, and c 8 to., the more th the para- |e = O0M9 aVPFOMVmO LOAA 319 :i32. Cord Bnppoiting a Load TJnUemalj Sis- tribnted owr th* HoriscwtaL— If the number cf vertices of the polygon be very great, and the suBpended weights all equal so that the load is distribated uniformly along the straight line, FE, the parabola which passes through all the vertices, virtually coincides with the cord or chain forming the polygon, and gives the figure of the Suspension Bridge. In this bridge the weights suspended from the successive portions of the chain are the weights of equal portions of the flooring. The weight of the chain itself and the weights of the sustaining bars are neglected in comparison with the weight of flooring and the load which it carries. Fit.7l Let the span, AB, = 2a, and the height, OD, = h. Then the equation of the parabola referred to the vertical and horizontal axes of x and y, respectively, through 0, is y* = 4mx, (1) 4m being the parameter. Because the load between O and A is uniformly dis- tributed over the horizontal, 0£, its resultant bisects OB at 0; therefore the tangents at A and intersect at (Art 63). From (1) we have ^ _ 8» _ y. m I 220 CORD aUPPORTINO LOAD^ which is the tangent of the inolination of the ottxve at any point (a;, y) to the axis of x. Hence the tangent at the p6int of support, A, makes with the horizon an angle, a, whose tangent is — , which also is evident from the tri- angle ACE. Let If be the v^eight on the cord ; then ^ IF is the weight on OA, and therefore is the yertical tension, V, at A. Then the three forces at A are the verticcJ tension V ■= \W, the total tension at the end of the cord, acting alonj^ who tangent AG, and the horizontal tension, T, which is every- where the same (Art. 130). Hence, by the triangle of forces (Art 31) these forces will be represented by tiie three lines, AE, AC, CE, to which their directions are respectively parallel ; therefore we have for the horizontal tension r = AE cot o = W^, and the total tension at A is 4A EX AMPLB, The entire load on the cord in (Fig. 71) is 320000 lbs.; the span is 150 ft. and the height is 15 ft.; find the tension at the points of support and at the lowest point and also the inclination of the curve to the horizon at the points of support. tan « = -— = .04 ; a a ~ 2V 48'. The yertical tension at each point of support is r = i weight = 160000 lbs. ; c t it t tl: cc B tb ca or lei th ax lii P, hi th W( !:he ottxve at any tangeut at the ZOQ an angle, a, t from the tri- r W is the weight F, at A. Then nV=^W, the ;ting aloni^ k.uo , which is every- the triangle of resented by the • directions are r the horizontal is 320000 lbs.; tnd the tension Ut and also the the points of 48'. Irtif raw COMMON CATSXASr. 221 the homontal tension is Tis »r^ = 400000 lbs.; and the total tension at one end is .. VV*+ T* = 430813 lb& 133. The Common Catenary.— Its Equation.— A catenary is the curve assumed by a perfectly flexible cord when its ends are fastened at two points, A and B, nearer together than the length of the cord. When the cord is of constant thickness and density, t. e., when equal portions of it are equally heavy, the carve is called the Common Catenary, which is the only one we shall consider. Let A and B be the fixed points to which the ends of the cord are attached ; the cord will rest in a vertical plane passing through A and B, which may be taken to be the plane of the paper Let be the lowest point of the catenary; take this as the origin of co-ordinates, and let the horizontal line through be taken for the axis of X, and the vertical line through for the axis of y. Let {x, y) be any point, P, in the curve ; denote the length of the arc, CP, by « ; let c* be the length of the cord whose weight is equal to the tension at ; and T the length of the cord whose weight is equal to the tension at P. V H' X . \ " J * 4 o A X T' n » * Tb* weif^t of a nuU of ImgUi of tbe conl balog here Uken m the nuit o( TMa COMMON OATUNART. Then the arc, CP, after it has assumed its permanent fonn of equilibrium, may be considered as a rigid body kept at rest by three forces vis.: (1) T, the tension, acting at P along the tangent, (2) c, the horizontal tension at the lowest point C, and (3) the weight of the cord, CP, acting vertically downward, and denoted by a. Draw PT' the tangent at P, meeting the axis of y at T'. Then by the triangle of forces (Art 31), these forces may be represented by the three lines PT, NP, T'N, to whici* their directions are respectively parallel. Therefore or T'N _ weight of CP ~NP ~ tension at C * ^ — i. tke («) Differentiating, substituting the value of ds, and reducing, we have d (I) y/Mtf e ' Integrating, and remembering that when x = 0, ^ = 0, we obtain where « is the Naperian base. Solving thif equation for 2 = »('--0' (•> ^, we obtain / ed ita pennanent 1 as a rigid body he tension, acting tal tennion at the cord, CP, acting Draw PT' the T'. Then by the ay be represented :u their directions is. and reducing, . = 0,g = 0. X _ * c' ii« eqaation for THM COMMON CATMNAST. 283 and by integration, observing that y = when » = 0, we have 9 = 1(^ + 0"')-^' (8) which is the equation required. We may simplify this equation by moving the origin to the point, 0, at a dis- tance equal to c below 0, by putting y — c for y, so that (2) becomes, =i(.%4 (8) which is the equation of the catenary, in the usual form. The horizontal line through O is called the directrix* of the catenary, and is called the origin. e r Cob. 1.— To find the length of the arc, AP, we have = Y I + i(e^ - e'*} dx, from (1), = iK+e"7<^; (4) .-. « = |( - 2ac (m e ; .*. rdr sz ac Bin 6 d9; (1) Shell— By the law of matter attracts lies directly an the rseltf aa the square I spherical shell of tckness, on a par- t. Let OM = o, ; the angle which I through OP. M (Art. 88) is whole shell sots ary mass at M on ed along OP, y moaoa of (1) ac substituting these values in (1), the attraction ot JM on P along PO = g.-(' + -;-)** w To obtain the resultant attraction of the whole shell, we take the ^integral between the limits and 2n, and the r-integral between c — a and c + a. Hence the resultant attraction of the shell on P along PO npka 1 + «»- ^dr. ^npkn* mass of the shell (3) Since e is the distance of the point P from the centre this shows that the attraction of the shell on tlie paVticIo at P is the same as if the mass of the shell were condensed into its centre. It follows iTom this that a sphere which is either homo- geueous or consists of concentric spherical shells cf uniform density, attracts the particle at P in the same manner as if the whole mass were collootcd at its centre. (2) lict the particle, P, be inside the sphere. Then wo proceed exactly as beioro, and obtain equation (2), which is true whether the particle be outside or inside the sphere ; ■ h mmm MXAKPLMB. but the r-limits in thia eaae gre « — c and a + c. Hence from (2) we have, by performing the ^integration, attraction of sheU = ^ J^ (l - ^-^^r) ^^' = !:^(2c-ac) = 0. therefore a particle within the ehell is equally attracted in every diiection, t. «., is not attracted at all. CoE.— 11 a particle be inside o homogenons sphere at Uhe distance r from its centre, all that portion of the sphere which is at a greater distance ftrom the centre than the particle produces no effect on the particle, while the re- mainder of the sphere attracts the particle in the same manner as if the mass of the remainder were all collected at the centre of the sphere. Thiw th. attraction of the sphere on the particle |7rpr» _ i-T— or 4Trpr Honoe, toiihin a homogeneous sphere the attraction varies as the dittoes from the centre. The propositious respecting the attraction o. » uniform Bphericd lihell on an external or internal particle were given by Newton (Principia, Lib. I, Prop. 70, 71). (See Todhunter's Statics, p. 276, also Pratt's Mechs., p. 187, Price's Anal. Mechs., Vol. I, p. «66, Minohin's Statics, p. 408). BXAM PLBS. 1. The span Ali =mO foet, and CO = 1600 feet, And Uu' length of the curve, CA, the height, CH, and the nda + c. Hence tegration, ««-e» )dr. = 0, [aally attracted in JL noQB sphere at Hbe tion of the sphere 9 centre than the icle, while the re- ticle in the same were all collected attraction of the attraction varies ioD Oi a nniform lal particle were ). 70, 71). (See Mechs., p. 187, inohin's Statics, = 1600 feet, find t, CH, and the MXAMPLMtH 229 iaoliaation, B, id the ewfe te the horisoai ii eithsr point of siupension. (1) Here - = f, and < ==: 3'7183d, c therefore ^ =: (a-71828>* = 1-2840, and • = (2. 71828p = 0- 7788. Substitnting these values in (5) we get i$ = 800 X 0.5052 = 404-16. CA = 404-16 feet. Hence (2) = 800 X 2-0628 — 1600 = 60-24 r^ot («) therefore tan « = ^? = t(^*---*). cte flrom (1), = 0-2526, e = 14° 11'. 8 404-16 Otherwise tan » = -, from (a), = -YaST = 0-2526, as before. 1600 2. The entire load on the cord in Fig. 71 is 160000 lbs., the span is 192 (t, and the height is 15 ft; find the tension at the points of support, and also the tension at the lowest point Ana. Tension at one end = 268208 lbs. Horizontal tension = 256000 "* Ii : ^lr* no MXAMPLXa. 3. A chain, AOB, 10 feet long, and weighing 30 lbs., is aaspended so that the height, CH, = 4 feet ; find the horiaontal tension, and the inclination, 9, of the chain to the horizon at the points of support Am. Horizontal tension = 3| lbs., 9 = 77° 19'. 4. A chain 110 ft long is suspended from two points in the same horizontal plane, 108 ft. apart; show that the tension at the lowest point is 1.477 times the weight of the chain nearly. weighing 30 lbs., ig 4 feet; find the 0, of the chain to M., 9 = 77° 19'. rom two points in t; show that the the weight of the PART II. KINEMATICS (MOTION). CHAPTER I. RECTILINEAR MOTION. 134. Dsflnitioiia.— Velocity. — Kinematics is that branch of Djrnamics which treats of motion without refer- ence to the bodies moved or the foree* producing the mo- tion (Art. 1). Although we do not know motion as free ttom force or from the maiter that is moved, yet there are cases in which it is advantageous to separate the ideas of force, matter, and motion, and to study motion in the abstract, t. «., without any reference to what is moving, or the cause of motion. To the study of pure motion, then, we devote this and the following chapter. The velocity of a particle has been defined to be its rate of motion (Art. 7). The formulae for uniform and variable velocities are those which were deduced in Art. 8. From (1) and (2) of that Art. we have ,1 ds di' (1) (») in which v is the velocity, s the space, and t the time. ■MtJMo. 888 EXAMPLB8. :X AM PI.E8. 1. A body moyes at the rate of 754 yar«l8 per hour. Find ihe velocity in feet per second. Since the velocity is uniform we use (1), hence *' = 7 = H7^ STi = 0.628 ft. per sec, Ans. t bU X oU 2. Find the position of a particle at & given time, /, when the velocity varies as the distance from a given point on the rdctiUnear path. Here the velocity being variable we have from (2) (to , *' = rf-< = *'' where /b is a constant ; d$ therefore = kdt; logs s= kt + e, (1> where c is an arbitrary coustant. Now if we suppose that s^ is the distance of the particle from the given pomt when < = we have c = log »,, which in (1) gives log — = ife/ ; ' or s = s,e^. 3. A railway train tra' Is at the rate of 40 mile«per hour ; find its velocity in feet per second. Ans. 58.66 ft per ieoond. 4. A train takes 7 h. 31 m. to travel 300 miles ; find its velocity. Ans. 39.02 ft per eec. 5. If « = 4^, find the velocity at the end of five seconds. Ans. 300 f t^ per sea 6. Find the position of the particle in Ex. 2, when the Telocity varies as th« time. Ans. s sx s^ + ^kfi. fl per hour. Find , bence per sec, Ans. I giren time, t, •m a given point I from (2) he, {1} of the particle m c = log «„ >f 40 milMper peraeoond. miloB ; find its ft per eec. of five seconds. it. per 860. X. 2, when the AOCMLMSATnur MMMO. 233 7. Find tb« diertiance the particle will more in one minute, when the Telocitj is 10 ft. at the end of one second and Ti « = «,< + «•» .1 r 'f* .rt'2] I' i tt4 AOOELMnATIOIf CONSTAltT. 1< 3 i i in which 9, is the space which the body has passed over when ^ = 0. If < is computed from the time the body stairtj from rest, then a = v^t The student will observe that this is a case of uniform velocity. 137. Th« Relation (1) between tiie Space and Time, and (2) between the Space and Veloeity, when the Acceleration ia Oonatant (1) Let A be the initial position of cT the particle supposed to be moving p Fig.73 toward the right, P its position at any time, t, from A, v its velooity at that time, and / the constant acceleration c* its velocity. Take any fixed point, 0, in the line of motion as origin, and let OA = «, ; OP = s. Then the equation of motion ia ••• rf/ =/" + «• Suppose the velocity of the particle, at the point A to be V,, then when ^ = 0, v = Vs;* hence e ■=■ v„ and .«. « = !//« + v^t + c'.. But when / = 0, « = «,; hence c' = «,, and » = !//» + v,< + «„ (2) (3) Hence if a particle moves from rest from the origin O, with a constant acceleration, we have * OOM MIM ▼•loelty and ipaoe rwpMtlTclr, or the velocity the pwtide lia% •nd epMe It hM moved over at Uie Inatant ( begini to be nekosed. T. iy has passed over te time the body dent will observe tlM Space and and Velosity, Flo.73 ime, /, from A, v it acceleration c* ;he line of motion hen the equation (1) point A to be v„ and (8) and (8) origin 0, with tdty tbe pwtlele haM, MMd. ACCMJJISATION VABtABLM. s:Aj,,-jr--';J;55.i; 235 (4) and thus the space described varies as the square of the time. (2) From (1) we have ... g = 2/, + 0. But when « = #„ t> = t», ; hence t) = t>,« — 2/5jo, and therefore v» i:^ 2/« + Vo" - 2/8,. (6) Equations (2) and (3) give the velocity and position of the particle in terras of t ; and (5) gives the velocity in terms of «. 138. When the Acceleration Vaxiea dirjcily aa tiie Time from a State of Rest, find the Velocit7 and Space at ttie end of the Time t. Here ' ' dt where «, is the initial velocity ; .-. » r= !«<• + V, the initial space being since / is estimated from rest. 139 When the Acceleration ▼ariea direotiiy aa the IMatance ftam a giwen Point bt the line of Mo- tion, and ia negattve, find the Relation between the Space and Time. 1 :|;:vj -'. ?■ m Jl itL.y.iy«iWiiiJiuii,v 286 Here J BXAMPLBS. ==-ks; by calling «, the valae of $ when the particle is at rest da V^o* - «• ifcirf/. the negative sign being taken since the particle is moving towards the origin ; . • . cos-' — = kU, if « = «, when / = ; i^t. EXAMPLES. 1. A body commences to move with a velocity of 30 ft. per sec, and its velocity is increased in each second by 10 ft. Find the spoce de«cribed in 5 seconds. Here / = 10, v, = 30, s^ = 0, and / = 6, thei^fore from (2) we have « = |.10.a6 + 30.5 = 275, Ans. 2. A body starting with a velcoity of 10 ft per sec, and moving with a constant acceleration, describes 90 ft. in 4 sees.; find the acceleration. Ana. 6^ ft. per sec. 3. Find the velocity of a body which starting fipom reit with an acceleration of 10 ft per sec., has described a space of 20 a Ana. Wtt, ;le ifl at rest rticle is moying locity of 30 ft lach second by : 5, therefore Ins. per sea, and jbea 90 ft. in 1^ ft per sec. ing from rest bribed a space Ana. 30 ft FALLtJtB BODina. 4. Throngh what space mast a body pan nnder an accel- eration of 6 ft per sec, so that its velocity may increase from 10 ft to 20 ft per sec. ? Ans. 30 ft 5. In what time will a body moving* with an cxxselera' tion of 25 ft per sec., acquire a velocity of 1000 ft per second? ' Ana. 40 sees. 6. A body starting flrort rest has been moving for 5 min- utes, and has acqaired a velocity of 30 miles an hour; what is the acceleration in feet per second ? Ana. W ft. per sec. 7. If a body moves from rest with an acceleration of { ft per sec, how long mast it move to acquire a velocity of 40 miles an hour? Ans. 88 sees. 140. Equations of Motion for Falling Bodies.— The most important case of the motion of a particle with a constant uccoleration in its line of motion is that of a body moving under the action of gravity, which for smtdl dis- tances above the earth's surface may be considered constant. When a body is allowed to fall freely, it is found to acquire a velocity of about 32.2 feet per second during every second of its motion, so that it moves with an acceleration of 32.2 feet per second (Art. 21). This acceleration is less at the summit of a high mountain than near the surface of the earth ; and less at the equator than in the neighborhood of the poles ; i. e., the velocity which a body acquires in falling freely for one nocond varies with the latitude of the place, and vrith its aliitueh above the sea level ; but is independ- ent of the site of the body and of its masa. Practically, however, bodies do not fall J>eely, as the resistance of the air opposes their motion, and therefore in practical cases at high speed (0. g., in artillery) the resistance of the air must be taken into account But at present we shall neglect * In each cate tbe body ia «nppoMd to etart from rect nnleH otherwise stated. iij ! II 238 FALLTNO BODIES. this resistance, and consider the bodies ad moving in vacuo under the action of gravity, t. e., with a constant accelera- tion of about 32.2 feet per second. As neither the substance of the body nor the cause of the motion needs to be taken into consideration, all prob- lems relating to falling bodies may be reganled as cases of accelerated motion, and treated from purely geometric considerations. Therefore if we denote the acceleration by g, as in Art. 23, and consider the particle in Art. 137 to be moving vertically downwards, then (2), (3), (5) of Art 137 become, by substituting jr for/, s = yp + vj + So, v* = 2flf« 4- Vo' — 2jrso» (A) s being measured as before from a fixed point, 0, in the line of motion. Suppose the particle to be projected downward from O, then A commences with and «, = 0. Hence (A) be- comes V =zgt + v^, (1) 8 =yi* + v^t, i^ = 2g& + «,». (2) (3) As a particular case suppose the particle to bo dropped from rest at (Fig. 71). Thro A coincides with 0, and s, = 0, Vj = 0. Hence eqiov-ions (A) become v = gt. (4) » = igi', (6) tf> = 2ga. (6) WsmKiasmr- 'jn-rn ovmg m vacuo istant accelera- or the cause of ation, all prob- rckd as cases of rely geometric acceleration by I Art. 137 to be (5) of Art. 137 (A) oint, 0, in the iward from O, Hence (A) be- (1) («) (8) to be dropped with 0, and me (») («) PARTIOLX PROJECTED UPWARDS. 239 141. When the Particla im Projected Vextioally TTpwards. — Here if we measure s upwards from the point of projection, 0, the acceleration tends to diminish the space and therefore the acceleration is negative, and the equation of motion is (Art 135) = -9- In other respects the solution is the same. Taking therefore «, = in (A) and changing the sign of g,* we obtain t> = Vo — gt, (1) t)» = Vo» - ^gif. (2) (3) Gob. 1. — The time during which a particle rises when projected vertically upwards. When the particle reaches its highest point, its velocity is zero. If therefore we put v = in (1), the corresiwnd- ing value of t will be the time of the particle ascending to a state of rest. 9 Cor. 2. — The time qf flight hefwe returning to the dart- ing point. From (3) we have the distance of the particle from the starting point after ^ seconds, when projected vertically upwards with the velocity v,. Now when the particle has risen to its maximum height and returned to the point of projection, « = 0. If, therefore, we put « = in (2), and solve for /, we shall get the time of flight. Therefore, • g in poaltlve or mgaUye Moordin^ aa the particle is deaoendlng or aa- cendingr. ■^ r i 240 FABTICLM PROJSCTKD UPWAMl^H. which givM / = 0, or tv. I The first yaluo of t shows the time before the part'cle starts, the latter shows tine time when it has returned. 2r Hence, the whole time of flight is -^, which is just double the time of rising (Cor. 1) ; that is, the time , = 200, « = 100 ; therefore from (3) we have »8 = 40000 - 6400 = 33600 ; .-. v = 40'v/2T. 3. A man is ascending in a balloon with a uniform velocity of 20 ft. per sec., when he drops a stone which reaches the ground in 4 sees.; find the height of the balloon. Here v, = 20, and f := 4 ; therefoi-e from (2) we have, after changing the sign of the second mcrabor to make the result positive. fl = - (80 - 256) = 176, which was the height of the balloon. 4. A body is projected upwards with a velocity of 80 ft. ; after what time will it return to the band ? Ana. 5 second*. 8. With what velocity must a body be projected ver- tically upwards that it may rise 40 ft. P Atts. 10 VlO ft. per SHJ. a ,? . !' -.^>J^.,-Jtf.,........:.|^.«^ 'mmmamm 242 COMPOSITION or VBLOCITISa. 6. A body projected vertioallj upwards paBses a oertain point with a velocity of 80 fl. per sec. ; how much higher will it ascend P Ans. ICJ ft. 7. Two balls are dropped from the top of a tower, one of them 3 sees, before the other ; how far will they be apart 5 sees, after the first was let £eiI1 ? Am. 336 ft. 8. If a body after having fallen for '3 sees. breakF a pane of glass and thereby loses one-third of its velocity, find the entire space through which it will have fallen in 4 sees. Ans. 224 ft. 142. Composition of Velocities.— (1) From the Par- allelogram of Velocities, (Art. 29, Fig. 2), we see that if AB represents in magnitude and direction the space which would be described in one second by a particle moving with a given velocity, and AC represents in magnitude and direction the space which would be described in one second by another particle moving with its velocity, then AD, the diagonai of the parallelogram, reprenenta the resultant velocity in magnitude and direction. (2) Hence the resultant of any two velocities, as AB, BD, (Fig. 2), is a velocity represented by the third side, DA, of the triangle ABD; and if a point have simultaneously, velocities represented by AB, BC, CA, the sides of a trian- gle, taken in the name o-der, it is at rest. The linos which are taken to represent any given forces may clearly be taken to represent the velocities which measure these forces (Art, 19), therefore from the Polygon and Parallelopiped of Forces the Polygon and ParaUel- ojtiped of Velocities follow. (3) Hence, if any number of velocities be represented in magnitude and direction by the sides of a closed polygon, taken all in the same order, the resultant is tsro. (4) Also, if three velocities be represented in magnitude passes a certain w much higher Ana. ICt) ft. ' a tower, one of I they be apart Atu. 336 ft. . breake a pane ilocity, find the m iu 4 sees. Ana. 224 ft. Prom the Par- ) see that if AB le space which 3le moving with magnitude and 1 in one second then AD, the the reaullant es, aa AB, BD, side, DA, of nmullaneoualy, ka of a triau- ly given forces tlocitiea which the Polygon and Parallel- repreaented in ihaed polygon, ro. in magnitude ■W'^^>:iS?^^J!jf»'A*^*fi9ft!j¥JV»>:|^ BEaOLUnON OF VBLOCPtlKS. 243 and direction by the thre* edges of a parallelepiped, the re- aultant velocity will be repreaented by the diagonal. (5) When there are two velocities or three velocities in two or in three rectangular directions, the resultant is the square root of the sum of their squares. Thus, if -£, ^-, ^j, -^, are the velocities of the moving point and its components parallel to the axes, we have from (2) of Art. 30, and from (1) of Art. 34, (1) da dt =V6?)V(I)^(I)* <') 143. Resolution of Velooitie>.— Af the diagonal of the parallelogram (Fig. 2), whose sides re])re8ent the com- ponent velocities was found to represent the resultant velocity, so any velocity, represented by a given straight line, may be resolvrd into component velocities represented by the sides of the parallelogram of which the given lino \ the di'\gonaL It will b< rtasily seen that (2) of Art. 134 is equally applicable whether iiie point bo considered as moving in a straiglit line or in a curved line ; but since in the lalter ase the direction of motion continually changes, the mere liount of the velocity is not sufficient to describe the ii 'tion completely, so it will be necessary to know at every instant the direction, &b well as the magnitude, of the point's velocity. In such oasec as thie Ihe method commonly cm- ployed, whether we deal with velocities or accelerations, consist.) mainly in studying, not the velocity or acceleration, directly, but its components parallel to any three assumed rectangular axes. If the {mrticle be at the point (x, y, «), ' , i n 1 ■ a: ) " f 1 i 1- i I I'- Jj!l if} . i -J I tfiiMMn ^jMW imitn iili Hiiii'ftyiJiiMiMHiiiri ■ n iii_rnm-i Ti-Tn-iiriT-TT'iir""'V iiirir-rfT"" i- 2U BXAMPLS9. at the time /, and if we denote its relocitiet parallel respectively to the three axes by «, v, w, we hare dx dy dz dt dt dt Denoting by v the velocity of the moving particle along the curve at the time t, we have as above and if o, /3, y be the angles which the direction of motion along the curve makes with the axes, we have, as in (2) of (Art 34), • dx ds ^ =s ^ oca a = V 008 « = u; dt da ^ = J-. cos y = » cos y = «>. m dt . , . d!* rfv rf' • 1. u Hence c«*ch of the components ^, ^, j^ is to be found from the whole velocity by retolving the velocity, I. e., by multiplying the velocity by coeim of the angle between the direction of motion and that of the compo- nent. EXAMPLES. 1. A body moves under the influence of two velocities, at right angles to each other, equal respectively to 17.14 ft. and 13.11 ft. per second. Find the magnitude of tb" resultant motion, and the angles into which it divides right angle. Ans. 81.679 ft. per sec. ; 37° «ft' and 68° ities parallel ITC •article along I (1) )n of motion > as in (2) of h if is to be the velocity, of the angle the compo- ▼elocities, to 17.14 ft. ode of tb" divide* dSJ" MOnON ON AH tlfCUNSD PLANB, %. A ship Bails due north at the rate of 4 knots per hour, and a ball is rolled towards the east, across her deck, at right angles to her motion at the rate of 10 ft. per second. Find the magnitude and directicoi of the real motion of tlie ball. Am. 12.07 ft per soc.; and N. 66° E. 3. A boat moves N. 30° E., at the rate of 6 miles per hour. Find its rate of motion northerly and easterly. Ans. 5.2 miles per hour north, and 3 miles per hour east. 144. Motion on an ZncUnod Plane.— By an exten- sion of the equations of Art. 140, we may treat the case of a pai-tiole sliding from rest down a smooth inclined plane. As tb>is is a very simple case in which an acceleration is resolved, it is convenient to treat of it in this part of our work ; yet as it properly belongs to the theory of con- strained motion, we are unable to give a complete solution of it, until the principles of such motion have been ex- plained iu a future chapter. Let P bo the position of the particle at any time, t, ou the inclined plane OA, OP sr «, its distance from a fixed point, O, iu the line of motion, and let « be the inclina- tion of OA to the horizontal line AB. Let Pi represent g, the vertical acceleration with rfj-M which the body would move if free to fall. Resolve this into two components, Po = ^ sin a along, and Vc — g cos a perpendicular to OA. The component g cos a pro- duces pressure on the plane, but does not affect the motion. The only acceleration down the plane is that component of the whole acceleration which is parallel to the plane, vie., g sin «c. The equation of motion, therefore, if ^ = ^ sin «, (1) — f- j».t* ^^..^^■liiHrWBIIilllttfi MIMIMM 240 DBCOENT DOWN CHORDS OF A OIBCLX. the Bolation cf which, aag nintt is constant, is incladed in that of Art. 140; and all the resalts for particles moving vertically aa given in Arts. 140 and 141 will be made to apply to (1) by writing g sin et for g. Thns, if the particle be projected down or up the plane, we get from (1), (2), (3) of Arts. 140 and 141, by this means V = Vo ± jf sin u't, s = Vft ± \g sin a- 1, V* = v,» ± 2^ sin «•«, (3) (4) in which the -f or — sign is to be taken according as the body is projected down or up the plane. If the particle starts from rest from 0, we get from (4), (5), (6) of Art. 140 V = ^ sin a* t, (6) a = 4jr sin a- ^, »» = 2jf sin «•«. (8) (7) CoE. 1. — Tk« velocity acquired by a particle in falling down a given inclined plane. Draw PO parallel to AB (Fig. 74), then if » be the velocity at P, wo have from (7) 1^ = 2g sin a*« Hence, from (6) of Art. 140 the velocity is the same at P as if the particle had fullen throngh the vertical space OC ; that is, tlte velocity acquired in falling down a smooth inclined plane is the same as would be acquired in falling freely through the perpendicular height of the plane. tCLg. is incladed in rticles moving ill be made to if tbe {Hirticle >m(l),(a),(3) (2) (3) (4) ;ording as the } get from (4), (ft) (8) (7) cfo in falling if V be the le same at P kl space OC ; V« a stnoolh in falling \lan0. DBaCSNT DOWN CBOBDS OF A CtBCLX. CJOB. 2. — When the particle is projected up the plane teith a given veloef^y, to find how high it will ascend, and the time ofascetU. From (4) we have 1^ = w,' — 2^ sin a'S. When V = the particle will stop ; henoe, the distance it will ascend will be given by thb equation = r,» — Zgaina'S, 8 = 2g sin « To find the time we have from (2) V = v^ —g aina-t; and the particle stops when v = 0, in which case we have i = g ana From (6) we derive the following canons and nseful result 145. The Times of Descent down all Chords drawn from the Highest Point of a Vertical Circle are eqnaL— Let AB be the vertical diameter of the circle, AG any cord through A, a its inclination to the horizon ; join BO ; then if ^ be the time of descent down AG we have from (6) of Art. 144 Fi»7S Bat AC = yP sin d. AC 'js AB sin u ; aMiWNtaMliUHWUM' ^mmmmmimiammttm iiUfM OF qmoKMar onacavt. .'. AB»5fcf#», /2AB which is constant, and shows that the time of falling down any chord is the same as the time of falling down the diameter. Cor. — Similarly it may be shown that the times of descent down all chords drawn to B, the lowest point, are equal ; that is, the time down OB is eqaal to that down AB. 146. The Straight Idna of QoickoBt Descent from (1) a Qiven Point to a Qiven Straight lone (2) from a Given Point to a Given Onrve. (1) Let A be the given point and BO c the given Uqq. Through A draw the horizontal line AO, meeting OB in 0; bisect the angle ACB by 00 which inter- sects in the vortical line drawn through A ; from draw OP perpendicular to BO; Join AP ; AP is the required line of quick- est descent For OP is evidently equal to OA, and therefore the circle described with as centre and with OP (= OA) for radius, will touch the line BO at P, and since the time of falling down all chords of this circle from A is the same, AP must be the line of quickest descent. (2) To find the straight line of quickest descent to a given ewrvt, all that is required is to draw a circle having the given point as the upper extremity of its vertical diameter, and tangent to the curve. Hence if DE (Fig. 76) be the curve, A the point, draw AH vertical ; and, with centre in AH, describe a circle p^wsing through A, and IHI m 1 falling down ing down the the times of I lowest point, equal to that Descent from U»9 (2) from n«.7« ^ therefore the }P {- OA) for I the time of ia tiie same, descent to a circle having jf its Tertical if DE (Fig. Ell ; and, with >agh A, «ad XXAMPLSa. 249 touching DE at P, then AP is the required line. For, if we take any other point, Q, in DE, and draw AQ cutting the circle in q, then the time down AP = time down A.qtl(K!ity is 19 ft. per sec. Aeeuming its acceleration to be uniform, what was its velocity at the end of 4 sees., and what will be its velocity at the end of 10 sees. ? Ana. 16- 8 ; 30. 8. A body is moving at a given instant with a velocity of 30 miles an hoar, and comes to rest in II sees.; if the retardation is uniform what was its velocity 5 sees, before it stopped ? Ans. 20 ft. per sec. 9. A body moves at the rate of 12 fi a sec with a nniform acceleration of 4; (1) state exactly what is meant the number 4 ; (2) suppose the aci deration to go on for i 6 .sees., and then to cease, what distance will the body 1; describe between the ends of the 6th and 12th sees.? Ans. 224 ft. 10. A body, whose velocity undergoes a uniform retarda- tion of 8, describes in 2 sees, a distance of 30 ft.; (1) what was its initial velocity ? (2) How much longer than the 2 sees, would it move before coming to rest ? Am. (1)23; (2) | sec. 11. A body whose motion is uniformly retarded, changes its velocity from 24 to 6 while describing a distance of 12 ft.; in what time does it describe the 12 ft.? Ans. 0-8 sec. 12. The velocity of a body, which is at first 6 ft. a sec, nndergoos a uniform acceleration of 3 ; at the end of 4 sec& the acceleration ceases ; how far does the body move in 10 sees, from the beginning of the motion ? Ans. 156 ft 13. A body moves for a quarter of an hour with a uni- form acceleration ; in the first 5 minutes it describes 350 yards; in the second 5 minutes 420 yards; what is the whole distance described in a quarter of an hour ? Ans. 1260 yds. » at the rate of 3ity is 19 ft. per m, what was its I be its velocity w. 16-8; 30. ith a velocity of I I sees.; if the 5 sees, before it 20 ft. per sec. ^ a sea with a what is meaut ion to go on for will tho body ith sees.? Ans. 284 ft. oiform retarda- 50 ft.; (1) what onger than the S3; (2) I sec. ardcd, changes distuuce of 12 ms. 0-8 sec. frst 6 ft. a sec, end of 4 sees; move in 10 ins. 156 ft |ur with a nni- describes 3.50 what is the jur i 1260 Yds. MXAMPLSS. Ml 14. Two sees, after a body is let fall another body is projected yertioally downwards with a velocity of 100 ft per sec. ; when will it overtake the former i* Ana. 1| sees. 15. A body is projected upwards with a velocity of 100 ft per sec.; find the whole time of flight Ans. li^ sees. IG. A balloon is rising uniformly with a velocity of 10 ft. per sec, when a man drops from it a stone which reaches the ground in 3 sees.; find tho height of the balloon, (1) when the stone was dropped ; and (2) when it reached the ground. Ans. (1) 114 ft; (2) 144 ft 17. A man is standing on a platform which descends with a uniform acceleration of 6 ft per sec. ; after having descended for 2 sees, he drops a bull ; what will be the velocity of the ball after 2 more seconds ? Ans. 74 ft. 18. A balloon has been ascending vertically at a uniform rate for 4.6 sees., and a stone let fall from it reaches the ground in 7 sees.; find the velocity, v, of the balloon and the height, «, fh)m which the stone is let fall. Ans. V = 174| ft per sec.; s = 784 ft If the balloon is still ascending when the stone is let fall v = 68-17 ft. per sec; s = 306-76 ft? 19. With what velocity must a particle bo projected downwards, that it may in t sees, overtake another particle which has already fallen through a f t ? Ans. V -. T + VZc^. 20. A person while ascending in a balloon with a vertical velocity of V ft per sec, lets fall a stone when he is k ft. above the ground ; required the time in which the ntony will reach the ground. ^ y + ^yt — 2gh Ans. 20S XXAMPLS& 21. A body, A, is projected vertically downwards from the top of a tower with the velocity V, and one sec. after- wards another body, B, is let fall from a window a ft. from the top of the tower ; in what time, t, will A overtake B ? 2a + g Ana. t = ^(y + g) 32. A stone let fall into a well, is heard to strike the bottom in t seconds ; required the depth of the well, sup- posing the velocity of sound to be a ft. per sec. ^ a_~|» 'iff V¥gJ ' Ans. \/ at + 23. A stone is dropped into a well, and after 3 sees, the sound of the splash is heard. Find the depth to the surface of the water, the velocity of sound being 1127 ft. per sec. 24. A body is simultaneously impressed with three uniform velocities, one of which would cause it to move 10 ft. North in 2 sees. ; another 12 ft. in one sec. in the same direction ; and a third 21 ft. South in 3 sees. Where will the body be in 5 sees. ? Am. 50 ft. North. 25. A boat is rowed across a river 1^ miles wide, in a directioi^ making an angle of 87° with the bank. The bviat travels at the rate of 5 miles an hour, and the river nu\s at the rate of 2.3 miles an hour. Find at what point of tlie opposite bank the boat will land, if the angle of 87° be made against the stream. Ans. 898 yards down the stream from the opposite point. 26. A body moves with a velocity of 10 ft. yter sec. in a given direction ; find the velocity in a direction inclined at an angle of 30° to the original direction. Am. 6 V3 ft. per sec. i(^^!^p»i^S!W««!^ta^'''i EXAMPLES. 963 wrnwards from one sec. af ter- dow a ft. from overtake B ? 1 to strike the the well, aup- c. ter 3 sees, the depth to the being 1127 ft. I with three se it to move le sec. in the sees. Where ft. North. C8 wide, in a )ank. The ind the river what point angle of 87° the opposite t)er sec. m a u inclined at rt. per sec. 27. A smooth piano is inclined at an angle of 30° to the horizon ; a body is started up the plane with the velocity Hg; find when it is distant 9^ from the starting point Arts. 2, or 18 sees. 28. The angle of a plane is 30° ; find the velocity with which a body mnst be projected up it to reach the top, the length of the plane being 20 ft. Ana. 8 VlO ft. per sec. 29. A body is projected down a plane, the inclination of which is 45°, with a velocity of 10 ft.; find the space described in 2^ sees. Ant. 95.7 ft. nearly. 30. A steam-engine atarts on a downward incline of 1 in 200* with a velocity of 7|^ miles an hour neglecting friction ; find the space traversed in two minutes. Ana. 824 yards. 31. A body projected up an incline of 1 in 100 with a velocity of 15 miles an hour just reaches the summit ; find the time occupied. Ana. 68.75 sees. 32. From a point in an inclined plane a body is made to slide up the plane with a velocity of 16.1 ft. per sec. (1) How far will it go before it comes to rest, the inclination of the plane to the horizon being 30° ? (2) Also how far will the body be from the starting point after 5 sees, from the beginning of motion ? Ana. (1) 8.05 ft. ; (2) 120.75 ft. lower down. 33. The inclination of a plane is 3 vertical to 4 hori- zontal ; a body is made to slide up the incline with an initial velocity of 36 ft. a sec. ; (1) how far will it go before beginning to return, and (2) after how many seconds will it return to its starting point? Ana. (l)33f ft.; (2) 3| sees. , * An Incline 6r 1 In SOO mMns here 1 fnot vcrttcally to • length of iOO ft., Uumgh U to nMd b; Bnglneen to mean 1 foot vertically to nO ft. AoH«mta^. MMHMiMi Ui MXAMPLM$, 34. There is so inclined plane of 6 vertical to 12 hori- zontal, a body «lidea down 52 ft of its length, and tLo passes without b« of velocity on to the horizoittl pluuT after how long from the beginning of the motion will it hj at a distauce of 100 f L from the foot of the incline ? Ans. 6.7 sees. i/iu^ I^^ '' projected up an inclined plane, whose length IS 10 times its height, with a velocity of si tiZ «ec. ; m what time will its velocity be destroyed ? ' Am. 9| sees., if ^r = 32. 30. A body falls from rest down a given inclined plane; compare the tiwes of describing the first and last halves -4ns, As 1 : ^2 -f 1. 37. Two bodies, projected down two pianos in-^lmed to he honzon at anglee of 45» and 60°. describe in the same time sp-oces re8{.ectively as ^/i : ^3; find the ratio of the initial vokKjities of the projected bodies. Ana. V2 : 1/3. 38. Thmugh what chord of a circle must a body fall to Zeter^ ''' "'"'^-^ '''"'' '' '*"*"^ "^-V th^ Am. The chord which is inclined at 60' to the vertical il\ ^^^ ^^^ ""'^."^''^^ ""'^^ ^^'"'^ * ^y «^««5<1 be pro. jocted down an inclined plane, /. «o that the time of running down the plane shall be equal to the time of falling down the height, h. . A * «in «\ nf *r;. ^'f!''^^ inclination of this plane, when a velocity of fths that duo to the height is sufficient to render the timoH of running down the piano, and of falling down the height, equal to eacli othor. ^„,, 30°. tical to 12 hori- eugth, and tlieu lorizoQtal piano ; aotion will it be incline? Ans. 6.7 sees. d plaue, whose ity of 30 ft. per iyed? 38., if ^ ~ 33. inclined plane; and last halves li Vi -h 1. nes in'ilined to be in the eanio :he ratio of the f. V2 : Vs. a body fall to through the the rerticaL ihould be pro- the tiiuo of the time of o h Bin n\ leu a velooitj to render the ng down the Ans. 80°. SXAMPLJB8. 41. Through what chord of a circle, drawn from tlie bottom of the yertical diameter mu^t a body descend; so sm to acquire a velocity equal to -th part of the velocity acquired in falling down the vertical diameter ? A»s. If denoi' aie angle between tho required chord and the vertical diameter cos = - • 42. Find the incUnation, d, of the radius of a circle to the vertical, such that a body running down will describe the radius in the same time that another bodji n>quir&a to full down the vertical diameter. Ans. 6 = 60°. 43. Find the inclination, 0, to the vertical of tho diam- eter down which a body falling will describti the last half in the same time as the vertical diamekvr. 8a/|j-4 /8 "" Ah$. co« 6 = 44. Show that the times of descent down all the radii of curvature of tho cycloid (Fig. 40, Calculus) ore equal; that 8r is, the time down PQ is equal to the time down O'A = y 46. Find the Inclination, ^, to the horizon of an inclined plane, so that the time of descent of a particle down the length may be » tiroes that down tho height of tho plane. Ans. = sin .1 46. Find the line of quickest descent from tho focus to a parabola whose axis is verticul and vertex upwards, and show that its length is equal to that of the latus rectum. 47. Find the lino of quickest descent from the foous of a parabola to the curve when the axis is horisMirital. "IMII 256 SXAMPLSa, 48. Find geometrically the line of quickest descent (1) tfom a point within a circle to the circle : (2) from t\ circle to a point without it. 49. Find gpometrically the straight line of longest descent from a circle to a point without it, and which lies below the circle. 60. A man six feet high walks in a straight line at the rate of four miles an hour away from a street lump, the height of which lis 10 foot : eupposing the man to starh from the L*mp-i)ost, finu the rate at which the end of his shadow travels, and also the rate at which the end of bis shadow Be])arate8 from himself. Ana. ;j! ,dow travels 10 miles an hour, and gains on himseii (3 miles an hour. '•i. Two bodies fall in the same time from two given pointB in space in the same vertical down two straight J-n. « drawn to any point of a surface ; show that the sur- laco is an equilateral hyperboloid of revolution, having the given jwintfl as vertices. 62. Find the form of a curve in a vertical plane, such that if ht'iivy pa/tioles be simultaneously let full from ca^h point Mt it 80 as to nlide freoly along the normal at that point, they may all reach a given horizontal straight Hue at the same instant. 53. Show that the time of quickest descent down a focal chord of a parabola whoso axis is vertical is m where I is the latus rectum. 54. Particles slide from rest at the highest point of a vertical circle down chords, and ore then allowed to moye «<'■«' t deacent (1) from n circle i of longest t, and which it line at the •eet lump, the man to Btarti le end of his le end of bis md gains on om two given two straight lit the sur- huving the ! I;ine, such 1 from each rmul at that 'uighk Hue at lown a focal fioint of a to move EXAMPLES. freely ; show that the locns of the foci of their paths is a circle of half the radius, and that all the paths bisect the vertical radius. 55. If the particles slide down chords to the lowest point, and be then suffered to move freely, the locus of the foci is a cardioid. 66. Particles fall down diameters of a vertical circle ; the locus of the foci of their subsequent paths is the circle. ii: CHAPTER II. CURVILINEAR MOTION. 147 Remarlw on CnrTillnear Motion.— The mo- tion, vrliioh waa coneidered in the laat chapter, wa8 that of a partiola describing a rectilinear path. In this chapter V.\e circnmstances of motion in which the path is ctirvilinear will he considered. The conception and the definition of Telocity and of acceleration which were given in Artfl. 134, 135, are evidently ad applicable to a particle describing a curvilinear v" ha to one moving along a straight line; and coDsequ i. Jie fo>-mulaB for velocity in Arts. 143, 143, are ap{>licablc eiiher to rectilinear or to curvilinear motion. In the last chapter the effects of the corapoeition and the resolutiou of velocities were considered, when the path taken by the particle in consequence of them was stmight ; we have now to investigate the effects of velocities and of accelerations in a more general way. 148. Composition cf Unifonn Velocit7 and Ac- o«ler«tlon.—SupiwBO a body tends to move in one direc- tion with a uniform velocity which would carry it from A to B in one second, and also suV>ject to an acceleration thai would carry it from A to in one second ; then at the end of the second the body will bo at D, the opposite end of the diagonal of the par- allelogram ABDC, just as if it had moved from A to B and then from B to D in the second, bnt the body will move in th^ nivve and not along the diagonal. For, the body in its motion is making progress uniformly iu the direction AB, at the same rate iw if it had no other motion; and at the Raniw time it is being accelenUed in the OOMPOBtTION OF AOCMLMMATIOKa, N. ion. — The mo- ■jN, waH that of this chapter the h is curvilinear he definition of m in Artfl. 134, ilo describing a I straight line ; J Arts. 142, 143, rilinear motion. »itiou and the rhen the path was sti-aight ; ocitit'B and of ity and Ac- in one direc- rry it from A cond, bnt the the diagonal. ■em uniformly had no othor united in the 9M direction AC, aa fast as if it had no other motion. Hence the body will reach D as far from the tine AO as if it bad moved over AB, and as far fi'om AB as if it had moved over AC ; but since the velocity along AC is not uniform, the spaces described in equal intervals of times will not be equal along AC while they are equal along AB, and there- fore the points a^, a,, o,, will not be in a straight line. lu this case, therefore, the path is a curve. 149. Composition and Resolution of Accelera- tions. — If a body i« Bnbjeot to two different accelerations in different directions the sides of a parallelogram may bo taken to represent the ComponeiU Accelerations, and the diagonal will represent the Resultant Acceleration, although the path of the body may be along some other line. Rem. — These results with those of Arts. 142, 148, may be summed up in one general law: When a body tmds to move with several different velocities ^'n different directions, the body will be, at the end of any given time, at the same point, as if it had moved with each velocity separately. This is the fundamental law of the composition of veloci- ties, and it shows that all problems which involve tenden- cies to motion in different direotionH simultaneously, may be treated as if those tendcDcios were successive.* (jPg If -js be the acceleration along the curve, and («, y, t) be the place of the moving particle at the time, ij it is evident that th(> component accelerations parallel to the llf*' ^' 3^' ^^^°**'''°» ^^^^ ^y "» **" ««i ares are W8 have df a*'* dP «»; di* — «•; and V<«(^ -I- o/ + «i^ i« the rumUani acceleration. • flwi H^irairkii oo Newlon'pi M Ur, Art. I6E>. IKHBHI 360 coMPoamoif or AccELSRATwna. Also if a, P, y, be the angles which the direction of motioQ makes with the axes, we have ^ = ^-,0080 = 0,; d»2 The acceleration is not generally the complete resultant of the threr component acceleratious, bat is eo only when the path ig a 9trai}i;'>^ line or the velocity ia lero. It is, however, the only part (P* of thew iHwultant which has any effect on the velocity, -r-^ is the sum of the njaolved partB of the component accelerations in the direc- tion of motion, as the following identical equation showa: which follows immediately from (1) of Art. 187 by diil^rentlation. Accelerations are therefore Bubject to the same laws of composition and resoluti'm as velocities ; and ronseqaently the acceleration of the particle along any lino is the sum uf the resolved |)art8 of the axial accelerations along that line. Thus to find .^, the acceleration along •, ^ (' dj? \,^ has to be multiplied by v-, which is the direction cosine of the dC '^ ' d$ small arc ds. 'Hie other part of the resultant is at right biihUk to this, and Its only effect iii to change the dircetion of the motion of the l>oint. (S<><> Tait and Steele's Dynamics of a Partlcl.', also Thomson and Tait's Nat Phil.) The following are examples in wluch the preceding ex- priHsio! > ar>' applii-d to onaei in which the laws of velocity and of ac<^ler»tion ore given. ONB. the direction of 9 resultant of the ten the path ia a fer, the only part locSty. ^ ia the tiona in the direc tiowa: y diflerenttation. of composition cel«ratiun of the •arts of the axial eleration along », on coalne of the right ati(^li« to le motion of the t), also Thomaon )roceding cx- vrs of velocity EXAMPLES. 1. A particle moves so that the axial compononts of its velocity vary as the corres wnding co-ordinates ; it is required to find the equation of its path ; and the accel- erations along the axes. Here dx di .— kx dy dt = h> .'. ^ = ^ = kdt; ' y if (a, b) is the initial »faice of tb« particle, is the equation of the path. And the axial accelerations are H-m di» dfl 2. A wheel rolls along a straight line with a nniform velocity ; compare the velocity of a given point in the ilr- cumference with that of the centre of the wheel. Let the line along which the wheel rolls be the axis of x, and let v be the velocity of its centre; then a point in its circumference describes a cycloid, of which, the origin being taken at its starting {mint, the equation is a — a vera a (2ay - y)*; iiiiii It !i- i ,1 EXAMPLMa. ^ _ d^ __ _ ^ — *•»> show that the path is an equi- lateral hyperbola and that the axial components arc 4. A particle describes an eDiiwe so that the a^component of its velocity is a constant, a ; find the y-com|)onent of \U vt'l(Kjity and acceleration, and the time of describing the ellipse. Ijet the fif/iia«i/(n of the ellipse be ^i \\ and lot {x, y) bo the position of the parHfiht at the time / j then ^--. "-' ^V ^ dx . dy hh; W^"' and J=-;^; ^ ^dy (h _ dt dx' dt ~ «» y' which is the y-component of the velocity. dy jircumference of lest point of the entre, while the ?ht line has no [,p.416.) path is ao oqui- lenta are bo ^-Component )m])onent of itti doBcribiug the at the timo t wmmm mmmm^. MXAMPLB. 363 Also 9y _ dP ~ dx fly hence the acceleration parallel to the axis of y varies inversel; as the cube of the ordinate of the ellipse, and acts towards the axis of x, as is shown by the negative sign. The time of passing from the extremity of the minor axis to that of the major axis i& fonnd by dividing a by a, the constant velocity parallel to the axis of x, giving , and the time of describing the whole ellipse is — • If the orbit is a circle b = a, and the acceleration par- allol to the axis of w is 3— If the v'olooity parallel to the ^-axis is constant and equal to P, then di *» • V ' dh: and the periodica time s= 4ft a? 1 ; find 5. A particle descrihes the hyperbola -, — ^ (1) the acceleration parallel to the aiis of a; if the velocity parallel to the axis of ^ is a constant, 0, and (3) find the acceleration parallel to the y-axis if the velocity parallel to the X-axis is a constuut a. (1) Here we have di -^' and t>m'>mi-iriysr,«isim>fMi'' •■ iUM u 264 EXAMFhBS. dx dt dx dy dy ^ /3rt» y which is the velocity parallel to the a;-axi& Also % dx _ ^ hence the acceleration parallel to the ar-axis varies inversely as the cube of the abscissa, and *he »-component of the velocity is increasing. (2) Here we have , dx dt = «; dy_ dt ~ Z dh, _ dt» ~ : — ay and hence the acceleration parallel to the y-axis is negative and the y-component of the velocity is decreasing. 6. A particle describes the parabola, «» + y* = a*, with a constant velocity, c ; find the accelerations parallel to the axes of X and y. Here we have da and d'X dt ■A- = c; ds y * {x + y)^* m varies inversely mponent of the is negative and 'g- - y^ = a*, with B parallel to the '^MM m 0mmw EXAMPLES. .♦. X X + y X i*X y' and d^ ~ dfi ' y _ X -\- y ~^ X + y' differentiating we get d^ 2{x + yr ^ _ d^ ~ (?{ax)^ ■ 2 (a; + vY 865 7. A particle describes a parabola with such a varying velocity that its projection on a line peipendicular to the axis is a constant, v. Find the velocity and the accelera- tion parallel to the axis. Let the ec^uation of the parabola be then and y» = %px; dy ~dt = v, dx _ dx dy dt ^ dy dt vy. which is the velocity parallel to x Also dfi P which shows that the particle is moving away from the tangent to the curve at the vertex with a constant accelera- tion. ,YS' >-«'**«*«>t*tti«iM«iife^a >.'^..\ IMAGE EVALUATION TEST TARGET (MT-3) ^ ^ #/. o A [/ ^^ ^ fA^ 1.0 I.I 1.25 •>5 2.2 IM iia. U. 11.6 Hiotographic Sciences Corporation 33 WIST fciAIN STRUT WnSTiR.N.Y. 143*0 (7*6) &>a-4S09 ■• iwuij m w— CiHM/ICMH Microfiche Series. CIHIVI/ICIVIH Collection de microfiches. Canadian Institute for Historical Microreproductions / Institut Canadian da microraproductions hiatoriquaa i *■•*' ,.?a#t#"*^' ■••■■ !^a?f**** PROJKCTIhM IN VACUO. Hence ae the earth acts on partiolee near its surface with a constant acceleration in vertical lines, if a particle is projected with a velocity, r, in a hoiiiontal line it wU move in a parabolic path. 160. Motiun of Protjeetiles in Vacua— If a particle be projected in a direction obliqne to the bozixon it i« called a Projectile, and the path which it describes is called it« Trajectory. The ca»e which we ghali here considt-r is that of a particle moving in vacuo under the action of gravity; so that the problem is that of the motion of a firojeciile in vacuo; and hence, as gravity does not »ii\ct its horizontal velocity, it resoivea itself into the purely kinematic problem of t particle moving so that its hori- zontal acceleration i? and its vertical acceleration is the constant, g, (Art 140). 151. TlM Patji of • Projectile in Vacuo is a Parabola.— Let the plane in which the particle is pro- jected be the piano of zy ; let the axis of x bo horizon- tal And the axis of y vertical and positive upwards, tb3 origin being at the point of projection; let the velocity of projection =: v, and let the line of projection be inclined at an angle a to the axis of x, so that t; co« «, and v bin « are the rcbolved parts of the velocity of projection along the axes of r and y. It is evident that the particle will oon- tijue to move in the plane of xy, as it is projected in i*, and is subject to no toroo which would tend to withdraw it from that plane. Lot {x, y) be the plaoe, P, of the partide at the time / ; then the equations of motioa arj ' its snrface with if a particle is line it vnll move O. — If a particle le bomon it ic lescribes is called here consider is it the action of the motion of a ' does not aij:>>ct into the purely 30 that its hon- ioeleration is the ^ c 3tion be inclined M a, and V bin a lection along the )artiole will oon- 1 projected in i*, tnd to withdraw e at tb0 time / ; PMOJBOTILg IN VACUO. 867 the aeceleration being negative since the y-component of the velocity is decreasing. The first and second integrals of these equations will then be, taking the limits corresponding to < = / uid < = 0, _^„cos«;f = g 5= vQMoti y vmaa — gt\ (1) vmuut-- yfl. (it) Eqoationa (1) and ^2) give the coordinates of the particle and its velocity parallel to either axis at any time, t. Eliminating t between equations (3) we obtain 9) y = *to»«-2jfe- which is the equation of the traject<>ry, and shows that the particle will move in a parabola. 153. TheParaawfeer; th« Range JB; tbaGhrMtaak Height H; Height of the Directrix— Equation (3) of Art. 1^1 may be written . 2v* sin a cos a 2t/* coi^ «e a^ _ x=s r !/, or ti* sin « COS o\' 2t)* cos* a i 9 o\' 2t)*co8'a/ »"8ln»«\ .,. 9 ^ By comparing this with the equation of a parabola referred to its vertex as origin, we find tor ., . , ... . V* sin a cos a ,a\ the absouM of the ^ «)rtex = —— } (2) ft' ^^ I -Jill ! \ 268 PROJSCTTLS IN VACUO. the ordioate of the vertex = "° " ; (3i the parameter (latus rectum) = ^^ — ^- (4) And by transferring the origin to the vertex (1) becomes . 2w» co8» a «•= :; — y (5) i which is the equation of a parabola with its axis vertical and the vertex the highest point of the curve. The distance, OB, between the point of projection and the point where the projectile strilies the horizontal plane is called the Range on the horizontal • plane, and is the value of X when y = 0. Putting y = in (3) of Art. 161 and solving for x, we get tlui horizontal range /2 =: OB = ^ "° ^" ; (C) which is evident, also, geometrically, as OB = 200 ; that is, the range is eqiml to twice the abscissa of the vertex. It follows from (6) that the range is the greatest, for a given velocity of projection, when a = 45°, in which case the range = — • Also it appears from (6) that the range is the same when o is replaced by its complement ; that is, for the same velocity of projection the range is the same for two differ- ent angles that are complements of each other. If « = 46° the two angles become identical, and the range is a maximum. OA is calltsd the greatett height, H, of the projectile, and ia given by (3) which, when a = 45° becomee 7-. (7) f X I* sin' g ''iff ' 2 »» cos» g (3) (*) tex (1) becomoB (5) h its axis vertical irve. of projection and e horizontal plane plane, and ia the in (3) of Art. 151 w» sin 2a (C) OB = 200; that of the vertex, the greatest, for a 5°, in which case is the same when is, for the same ne for two differ- (ther. If « = 46° the range ia a ;he projectile, and 4^' (y Daes vsLOcirr or tbs psojjbottlk. I%e height of the directrix = CD ''iff' — —^*Jt A. X ^'^ C OS* « - 2g -^i ~g 269 (8) Hence when « = 45° the focns of the parabola lies in tlie horizontal line through the point of projection. 153. Tlie Velocity of the Partiole at any Point of its Path. — Let V be the velocity at any point of its path, then V^ = (|)V (I)*, or by (1) of Ait 151 = w* cos' « + (v* sin* a — 2v sin agt + g*P) s=v> — 2gi/. To acquire this veWity in falling from rest, the particle y» must have fallen through a height -^, (6) of Art. 140, or itsoqual " ~-ff = M8-MPhy(8) = PS. Hence, the velocity at any point, P, on the curve is that which the particle would acquire in falling freely in vacuo down the vertical height 8P; that is, in falling from (ho directrix to the curve; and the velocity of projection at O is that which the particle would acquire in falling freely through the height CD. The directrix of the parabola is therefore determined by the velocity of projection, and is ai u vortical distance above the point of projection equal to that down which a particle flailing would have the velocity of projection. 154. The Time of Flight, T, along a Horizontal Plane.— Put y = in (a) of Art. 151, and solve for a-, the i: ':. •mta or ruosr op pnojsoi'tMs. %ifi sin K COB CE 870 values of which are and 9 ' But the horizon- %v sin a tal velocity ia v cos «. Hence the time of flight = which varies as the sine of the inclination tx) the axis of z. 155. To FinA the Point at n^ch a PrOjoctlla will Striko a Oiveii IseliiMd FbuM paacing tbrongh the Point of Projeotion, and the Time of Flight— Let the inclined plane make an angle /3 with the horizon ; it is evident that we have only to eliminate y between y =: a: tan /3 and (3) of Art. 151, which gives for the abscissa of the '' point where the projectile meets the plane _ 2t>' cos a sin (« — jB) , ^coe/3 * and the ordinate i« x^ z= (1) _ a»» COB CT t a n /3 sin (« -- <3) '* ~" g cos /3 Hence the time of flight T = Xf _ 2v sin (g — /3) V cos « g cos /3 (2) 156. The Direotlon of Protjeotloii whiek giTea the Oreateat Range on a GUven Plane. — The range on the horizontal plane is t^ sin 2a which for a given value of v is greatest when « = 7 {^^ 152). The range on the inclined plane = a^j sec 0' — 8*^ COB « «'P ( « "" <^) . (1) • But the horizon- f flight = —^ D \a the axis of x. a Prdjdctlle wlU icing tisirongh the ifriight— Letthe the horizon ; it is between y — x tan ' the abscissa of the le (1) "&) -J) 8 (2) k ^hiek giTtts th« —The range on the when « = T (Art. , Bee fi (1) ANOLX or MLBVATtOH Of PBOjaOTtLg. 271 To find the value of a which makes this a maximum, we must equate to zero ita derivative with respect to «, which gives ooe (2a — /3) = 0; and henoe (9) (8) which is the angle which the direction of projection makes wi*h the inclined plane when the range is a maximum ; that is, the projection bisects the angle between the inclined plane and the vertical. In this case by subttitating in (1) the values of « and (« — (3) as given in (2) and (3) and reducing, we get the g«ate.t range =:^-j^j. (*) 1S7. The «agl« fti Bltvatloii uo th«t tiui Partlcio nwy pUM through a Given Point— From Art. 153, there are two directions in which a particle may be pro^ jected so as to reach a given point; and they are equally inclined to the direction of projection (« = -). Let the given point lie in the plane which makes an angle ft with the horizon, and Btijipoge its abscissa to be A ; then we must have from (1) of Art. 155 2w» ^-^^cos«gin(a-/J)=A. -f «' and «" be the two values of a which satisfy this equation, we must have ooi «' sin (•' - /I) = 008 «" sin («" ~ 0) ; ■"■•P|*|. IIL iT 27a EQUATION OF TBAJBCTOST, SECOND METHOD. and therefore a" — ^ = 5 — «', I i or »"-ig + P)=i(i + /')-«' (1) But each member of (1) is the angle between one of the directions of projection and the direction for the greatest range [Art. 156, (2)]. Hence, as in Art 152, the two directions of projection which enable the particle to pass through a point in a given plane through the point of pro- jection, are equally inclined to the direction of projection for the greatest range along that plane. (See Tait and Steele's Dynamics of a Particle, p. 89.) 158. Second Method of Finding the Equation of the Trujectory.— By a somewhat simpler method than that of Art. 151, we may find the equation of the path of the projectile as the resultant of a uniform velocity and an acceleration (Art 148). Take the direction of projection (Fig. 78) as the axis of X, and the vertical downwards from the point of projection as the axis of y. Then (Art. 149, Rem.) the velocity, v, due to the projection, will carry the particle, with uniform motion, parallel to the axis of x, while at the same time, it is carried with constant acceleration, g, parallel to the axis of y. Hence at any time, t, the equations of motion along the axes of x and y respectively are X — vt, That is, if the particle were moving with the velocity v, alone, it would in the time t, arrive at Q , and if Jt were then to move with the vertical acceleration g alone it would in the same time arrive at P; therefore if the velocity v f' W MBTHOP. — a. (1) etween one of the u for the greatest Lrt 152, the two 18 particle to pass the point of pro- ition of projection I. (See Tait and the Equation of pier method than tion of the path of m velocity and an 78) as the axis of point of projection n.) the velocity, v, ;icle, with uniform it the same time, it parallel to the axis 18 of motion along rith the velocity v, Q , and if It were on g alone it would re if the velocity v EXAMPLES. S73 and the acceleration g are simuUaneotis, the particle will in the time t arrive at P (Art. 149, Rem). Eliminating t we have g " which is the equation of a parabola referred to a diameter and the tangent at its vertex. The distance of the origin from the directrix, being J^th of the coefficient of y, is V* H-, asm Art. 163, (8). BXAMFLES. 1. From the top of a tower two particles are projected at angles a and /3 to the horizon with the same velocity, v, and both strike the horizontal plane passing tlirough the bot- tom of the tower at the same point; find the height of the tower. Let A = the height of the tower; v = the velocity of projection ; then if the particles are projected from the edge of the top of the tower, and x is the distance from the bottom of the tower to the point where they strike the horizontal plane we have from (3) of Art. 151 - A = a; tan b - 1^ (1 + tan» a), A = X tau /3 - g (1 + tan» /3), (1) (2) by subtraction X = 2t»» Sw* cos « cos i9 ^ g (tan a + tanT^ "" ^sin (a + /3) ' which in (1) or (2) gives k — ^^ ^'^^ " "*^" ^ ^ ^ ^^ "*" ^^ ■■i AlWi.SWl.i'Sirl ^■?su^i«< eiis^^T^ss^s^Sie^E^aSj^^Si^^ wmm H i rf^ 874 vuLooirr of mscuARoE of skslls. 2. Piurticles are projected with a giTen veiocitj in all lines in a Tertioal plane £rom the point 0; it ia required to find the loona of their highest points. Let {x, y) be the highest point ; then from (2) and (3) of Art 162, we have «* sin o cos rt g- ; »» Bin» a therefore sin* a = -^, and coa* « = —--• 1^ 2v^y g ' ff Adding 4^ + a:* which is the equation of an ellipse, wliose major axis = and t,he minor axis = „-; and the origin is at the extremity of the minor axis. 8. Find the angle of projection, a, so that the area con- tained between the path of the projectile and the hori- zontal line may be a muximnm, and find the value of the maximum area. Ans. tt = 60° and Max. Area (3)*. 4. Find the ratio of the areas A, and A, of the two parabolas described by projectiles whose horizontal ranges are the same, and the angles of projection are therefore complements of each other. ^ A, , . Ana. -jT^ = tan' a. At 159. Velocity of Dischaxge of Balls and Shells from the Mouth of a GNul— As the result of numerous SHSLLS. n velocity in all it ia required to ■om (2) and (3) of major axis = - ; 8 at the extremity hat the area con- ic and the hori- i the value of the rea »ff . (3)*- 1 Aj of the two horizontal ranges ion are therefore -r^ = tan' a. A, ills and Shells lult of numerous m ANGULAR VELOCITY. 2T5 experiments made at Wcolwich, the following formula was regarded as a correct expression fov the velocity of balls and shells, on quitting the gun, and fired with moderate charges of powder, from the pieces of ordnance commonly used for military purposes : t; = 1600>^^, where v is the velocity in feet per second, P the weight of the charge of powder, and W the weight of the ball. For the investigation of the path of a projectile in the atmosphere, see Chap. I of Kinetics. 160. Angular Velocity, and Angnlar Aocelera< tion. — Hitherto the method of resolving velocities and wcelerations along two rectangular axes has been employed. It remains for us to investigate the kinematics of a particle describing a curvilinear path, from another point of view and in relation to another system of reference. Before we consider velocities and accelerations in reference to a system of polar co-ordinates, it is necessary to enquire into a mode of measuring the angular velocity of a particle. Angular Velocity way be defined as the rate of angular motion. Thus let (r, 0) be the position of the point P, and suppose that the radius vector has revolved uniformly through the angle 9 in the time t, then denoting the angular velocity by w, we shall have, ae in linear velocity (Art. 8) 6 a = t If however the radius vector doe-a not revolve uniformly through the angle 6 we may always regard it as revolving uniformly through the angle dd in the infinitesimal of time dt ; hence we shall have as the proper value of w, •"aWBWaffihffMl 276 SXAMPLSa. i !i dB dt (1) Hence, whether the angular velocity bo uniform or variable, it is the ratio of the anglp described by the mdius vector in a given time to the time in which it is described ; thus the increase of the angle, in angular velocity, takes the place of the increase of the distance frcm a fixed point, in linear velocity, (Art 8). Angular Acceleration is the rat- of tncretse of angular velocity : it is a velocity increment, and is measnred in the same way as linear accelerafion (Art. 10). Thus, whether the angular acceleration is uniform or variable, it may always be regarded as uniform during the infinitesimal of time dt in which time the increment of the velocity will be du. Hence denoting the angular acceleration at any time, t, by (p, we have . dot d /de\, .,. d£*' (2) and thus, whether the increase of angular velocity is uniform or variable, the angular acceleration is the increase of angular velocity in a unit of time. The following examples are illustrations of the preceding mode of estimating velocities and accelerations. EXAMPLES. 1. If a particle is placed on the icvolving line at the distance r from the origin, and the line revolves with a uniform angular velocity, w, the relation between the linear velocity of the particle a&d the angular velocity may thus be found. •^mmm"- BgWW i WgllJliJ EXAMPLES. m (1) bo uniform or ed by the rudius h it is described ; IT velocity, takes ;m a fixed point, rease of angular measured in the Thus, whether variable, it may I infinitesimal of ! velocity will be ion at any time, (2) :nlar velocity is •n is the increase of the preceding ons. ring line at the revolves with a itween the linear locity may thus ( Let dd be the angle through which the radius revolves in the time dt, and let ds be the path described by the particle, so that ds = rdO ; then iug (3) and (4) by sir. 6 and cos 6 respectively, and subtracting the former from the latter, we get OOS0 d*x .. ^dr dd m -dfi''''^ = ^di'dt-^*^Jfi _1 rf/.d dr dr dd h dr dt ~ d)' dt •— r> de* and '•* dt* ~ r*' de» VW' which in (6) of Art. 161 gives the acceleration along the radius vector '-~t*dm ^r»\dsi} f«' an expression which is independent of t. (l) This may be put into a more convenient form as follows: I«t f = - ; then dr de du de' «mm. 28!^ CONSTANT ANeULAB VBLOCITT. tPr — 1 ^ J. ^ l^^ which in {1) and reducing, gives the acceleration along the radias vector (2) From these two forranlse the 'iaw of acceleration along the radias vector may be deduced when the curve is given, and the curve may be deduced when the law of accelera- tion along the radius vector is given. Examples of these processes will be given in Chap. (2), Part III. 164. When th« Angular Velocity is Constant- Let the angular velocity be constant = « suppose. Then = w; therefore from (5) of Art 161 the acceleration along the radius vector d*r , The acceleration perpendicular to the radius vector dr = 2u dr (1) (2) and both of these are independent of ft The following example is an illustration of these formulte : A particle describes a path r7ith a constant angular velocity, and without acceleration along the radius vector ; find (1) the equation of the path, and (2) the acceleration perpendicular to the radius vector. "^*»-, mmt acirr. a (2) acceleration along ;he curve is given, law of accelera- ilxamples of these III. is Constant. — ► suppose. Then adins vector (1) (2) itration of these constant angular the radius vector ; I the acceleration j( Ba W tg ! iKWi'tn i |»JIWI | i|IW i W IL IWIIMI I II»^ COMITAA-T ANOULAR VELOCITY, (1) From (1) we have, from the conditions of the question. dp — w»r = 0. Integrating ve have dr' if r = a when jj = 0. at Therefore dr — udt; if r = o when t = 0, r = ^ie^ + -^^ga'<*^)«*^^''8''>'y- Ans. JBXAMPLBA 386 ) ^; (5) *r to the radins M is Bnfficient for lema which might [iserted here; but >int of view, .ties of a particle see Price's Anal, ele's Dynamics of y = /fc»; find (1) X if the velocity and (2) find the } velocity parallel '*•; (2) ^V. f = iax; find f if the velocity Ans. -- carve, y :=: of; 3n if the y-com- »nd (2) find the imponent of the !)a» (log a)«y. 4. A particle describes the cycloid, the starting point being the origin; find (1) the a^component of the accel- eration if the y-component of the velocity is (i, and (2) find the y-component of the acceleration if the a;-componont of the velocity is o. ^^ . ^^^ ffl«y . ^jj) _ ^. Ata, (1) (2«y-y)*' ^ 5. A particle describes a catenary, y = s(«*' + e "1; find (1) the a;-component of the acceleration if the y-com- ponent of the velocity is /3, and (2) find the y-component of the acceleration if the aMsomponent of the velocity is «. Am. (1) 6. Determine how long a particle takes in moving from the point of projection to the farther end of the latus rectum. ^ ** / • , \ Ans. - (sm a + cos a). 9 7. A gun was fired at an elevation of 60°; the ball struck the ground at the distance of 2449 ft. ; find (1) the velocity with which it left the gun and (2) the time of flight. (y = 32i). Ana. (1) 200 ft. per sec; (2) 19.05 sees. 8. A ball fired with velocity u at an inclination a to the horizon, just clears a vertical wall which subtends an angle, (i, at the point of projection; determine the instant at which the ball just clears the wall. « sin a — \gt _ Ana. u cos a im(3. 9. In the preceding example determine the horizontal distance between the foot of the wall and the point where the ball strikes the ground. Ana. — co8» « tan /3. 8 vT 386 BXAMPLSa, 10. At the diatanco of a quarter of a mile from the bot- tom of a oUff, which ia 120 ft. high, a shot ia to be fired which ahall just clear the clifiF, and paaa over it horizon- tally ; find the angle, a, and velocity of projection, v. Ant. a = 10° 18'; t> = 490 ft. per aec. 11. When the angle of elevation ia 40° the range is 2449 ft. ; find the range when the elevation is 29|°. Ana. 2131.6 ft. 12. A body is projected horizontally with a velocity of 4 ft. per see. ; find the latua rectum of the parabola de- Bcribed, (g = 32). Ana. 1 foot. 13. A body projected from the top of a tower at an angle of 45° above the horizontal direction, fell in 5 seca. at a distance from the bottom of the tower equal to its altitude ; find the altitude in feet, (g = 32). Ana. 200 feet 14. A ball ia fired up a hill whoae inclination ia 15°; the inclination of the piece is 45°, and the velocity of pro- jection is 500 ft. per sec. ; find the time of flight before it strikes the hill, and the distance of the place where it falls from the point of projection.* Ana. T = 16.17 aecs.; R = 1.121 miles. 15. On a deacending plane whose inclination is 12°, a ball fired from the top hits the plane at a distance of two miles and a half, the elevation of the piece ia 42° ; find the velocity of projection. Ana. v = 579.74 ft. per aec. 16. A body ia projected at an inclination a to the hori- zon; determine when the motion is perpendicular to a plane which is inclined at an angle /3 to the horizon. u sin a Atu. u COB a ^ = ±oot|9. * The nnge on the Inclined plane. mile from the bot- Bbot is to be fired )8 over it horiaon- >rojectiun, v. 490 ft. per sec. 40" the range is m is 29|°. Ans. 2131.5 ft, ffith a velocity of '. the parabola de- Ans. 1 foot. tower at an angle ill in 5 sees, at a lal to its altitude ; Ans. 800 feet. iclination is 15°; 3 velocity of pro- 5 of flight before e place where it = 1.121 miles. ination is 12°, a . distance of two is 42° ; find the L74 ft. per sec. 1 a to the hori- rpendicutar to a e horizon. ^' = ± oot /3. nHMmii MXAMPLS& 287 17. Calculate the maximum range, and time of flight, on a descending plane, the angle of depression of which w 15°, the velocity of \ ejection being 1000 ft, per sec. Ana. Max. range = 7.98 miles ; T = 51.34 sec. 18. With what velocity doe^ the ball strike the plane in the last example ? Ans. V = 1303 feet. 19. If a ship is moving hori«on tally with a velocity = 3 T = '^?'* 3 ' - 9 2'4. Given the velocity of sound, V; find the horizontal range, when a ball, at a given angle of elevation, «, is so projected towards a person that the ball and sound of the discharge reach him at the same instant Ans. — tan «. ff 23. A body is projected horizontally with a velocity of 4g from a point whose height above the ground is 16jr ; find the direction of moHon, d, (1) when it has fallen half-way to the ground, and (2) when half the whole time of falling has elapsed. ^^ (l)e = 46°; (2) e = tan-» -^-=. 288 EXAMPLBS. I,-. I; :• 24. Particles are projected with a given velocity, v, in all lines in a vertical plane from the point ; find the locus of them at a given time, /. Ans. a:* + (y + io^Y — ^^> which is the equation of a circle whose radius is vt and whoso centre is on the axis of y at a distance ^P below the origin. 25. How much powder will throw a 13-inch shell* 40G0 ft. on an inclined plane whose angle of elevation is 10° 40' ; the elevation of the mortar being 35". Ans. Charge = 4.67 lbs. 26. A projectile is discharged in a horizontal direction, with a velocity of 450 ft. per sec, from the summit of a conical hill, the vertical angle of which is 120° ; at what distance down the hillside will the projectile fall, and what will be the time of flight? Ans. Distance = 2812.5 yards; Time = 16.23 sees. 27. A gun is placed at a distance of 500 ft. from the base of a cliff which is 290 ft. high ; on the edge of the cliff there is built the wall of a castle 60 ft. high ; And the elevation, «, of the gun, and the velocity of discharge, v, in order that the ball may graze the top of the castle wall, and fall 120 ft. inside of it Ans. « = 63° 19' ; v = 165 ft. per sec. 28. A piece of ordnance burst when 50 yards from ^-^ wall 14 ft. high, and a fragment of it, originally in con- tact with the groand, after grazing the wall, fell 6 ft. beyond it on the opposite side ; find how high it rose in the air. Ans. 94 ft. * The weight of • la-inch ihell U IM Ibe. Ire : m mi iven velocity, v, in ifcO; find the locus ! the equation of a e is on the tuis of a 13-inch shell* gle of elevation is g35". irge = 4.67 lbs. rizontal direction, the summit of a is 120° ; at what tile fall, and what le = 16.23 sees. [) ft from the base edge of the cliff t high; find the 7 of discharge, v, )f the castle wall, 165 ft. per sec. 50 yards from ^ originally in con- B wall, fell 6 ft. »w high it rose in An8. 94 ft. Iba. PART III. KINETICS (MOTION AND FORCE). CHAPTER I. LAWS OF MOTION— MOTION UNDER THE ACTION OF A VARIABLE FORCE— MOTION IN A RESISTING MEDIUM. 165. Deflnitioiui. — Kinetics is that branch of Dynamics which treats of the motion of bodies under tfie action of forces. In Part I, forces were considered with reference to the pressures which they produced upon bodies at rest (Art. 16), i. e., bodies under the action of two or more forces in equilibrium (Art 26). In Part II we considered the purely geometric properties of the motion of a point or particle without any reference to the causes producing it, or the properties of the thing moved. We are now to consider motion with reference to the causes which produce it, and the things in which it is produced. The student must here review (Chapter I, Part I, and obtain clear conceptions of MomeiUutn, Aeeelsrattini tf Mimuintum, and the KinHie meamre of Fane (Arts. 18. 14. 19, and 90), as this is necessary to a fall understanding of the fundamental laws of motion, on the trutli of which all our succeeding investigations are founded. 166. Newton's Laws of Motioa— The fundamental 13 1' \ ■1. t. mm 290 A^rrojvs LAWS or motion. :• I : principles in aooordancfi with wliich motion takes pliicc are embodied in three statements, generally known as New/on^ Laws of Motion. These laws must be considered as resting on convictions drawn from observation and experiment, and not on intuitive perception.* The laws are the fol- lowing: Law \.— Every body continues in its state of rest or of uniform motion in a straight line, except in so far as it is compelled hy force to change that state. Law II.— Change of motion is proportional to the force applied, and takes place in the direction of the straight line in which the force acts. Law IIL— 2b every action there is always an equal and contrary reaction; or, the mutual ac- tions of any two bodies are always equal and oppo- sitely directed. 167. Remarks on Law l— Uw i snppUeB m with » definition of force. It indicates that force is that which tends to change a body's state of rest or of uniform motion in a straight line ; for if a body does not continue in its state of rest or of uniform mo- tion in a straight line it mupt bo under the action of force. A body has no power to change its own state as to rest or motion ; when it is at rest, it has no power of putting itself in motion ; when in motion it hr« no power of increasing or diminishing its velocity. Matter is inert (Art. 8). If it is at rert, it will mnain at rest ; if it is moving with a given velocity along a rectilinear path, it will continue to move with that velocity along that path. It Is alike natural to mat(*r to be at rest or in motion. Whenever, therefore^ a body's state is changed either from rest to motion, or fr far as the direction or velocity of i» motion is chonged by forces bctwet-n the psivicles and some other matter not Moiiging to the gyHem : also the centre of gn-vity of any system of particles moves Just as all the matter of the system, if .».»centrated in a point, would move under tho lntiuenc(> of forces (>*]Ual ami parallel to ihe foreve ivsHy miiag on its difll>rent parts. (For furthitr -(._^, wm TWO LA Wa OP MOTTOIf. 295 tideratood as a fac- e X is the a'-com- i force exerted on as with A definition >w to oompouud, aod r to investigate the particle subjected to n^ to Law III, if one drawn hy tliis other (Art. 10). A horse a force equal to that If one body strikes uther body, its own and in the opposite i the bodle8 exert un momenttim that one tain foix^e, while the The result is that W( oiint of its attrac- der the influoncii o* e earth being enor- le forces on the two he motion produced than that produced tinn parallel to any ucncing one another leir mutual action. yRt«>m of mntually Hiiir untform'v in a alooity of i's motion Tie other matter not y of any systom of a, it ..tincontrated in (Mjual ami parallel rts. (For fortliMr remarkr on tbese laws see Tait and Steele's Dynunics of a Particle, Thomson and Tait's Nat. Phil., Pratt's Mechanics, etc.) 170. Two Laws of Motion in the Fronch Troa- tises. — Newton's Laws of motion are not adopted in the principal French treatises ; bet we fiud in them two prin- ciples only as borrowed from experience, viz.: First. — ^The Law of Inertta, that a body, not acted upon by any force, would go on for ever with a uniform velocity. This coincides With Newton's First Iaw. Second. — That the velocity communicated is proportional to the force. The second and third Laws of Motion are thus reduced to this second priticiple by the French writers, especially Poisson aud Laplace.* 171. Motion of a Particle under the Action of an Attractive Force. — A particle moves under a force of attraction which is in its line of motion, and varies directly as the distance of the particle from the centre of force; it is required, to determine the motion. The point whence the it fluenco of a force emanates is called the centre of force ; and the force is called an attrac- tive or a repulsive force according as it attracts or repels. jjet be the centre of force, P the position of the particle at any time, t, v ~ its velocity at that time, and let OP = x, and OA = a, where A is the position of the particle when / = ; let ^ = the absolute force ; that is, the force of attraction on a unit of mass at a unit's distance from 0, which is supposed to be known, and is sometimes called the strength of the attraction. At preaeo* we shall suppose * ParkiDaou'i MMbaaiei, p. 187. See paper by Dr. Whewell on ttip principle* of DTiuunks. iMitUcnlarty as alalsd by FrMieb wiiten, Is tbo Idinboiili jonrsal gf Hci*w». Vol. VaL .-^ Fi|.M M» ^m 39i aSMAMKS ON LAW Ut. the former, the unit of mass must be imilerstood aa a fac- tor on the left-hand side, in which case X ia the x-com- ponent, for the unit of 'Tiass, of the whole force exerted on the moving body. The flnt two Uwb, h»ve, therefore, furniahed aa with a definition and a meature of force ; aud they alau ahow how to oompouud, and lherefoi« how to reeolve. forcea; and also how to inveetigata the conditiona of equilibrium or motion of a aingle particle aubjectod to given forces. 169. Ztomarka on Law ni.— According to Law III, if one bodjr presaea or drawa another, it ia prnsaed or drawn by thia other with an equal force in tlie oppoeite directior (Art. 10). A horse towing a boat on a canal, is pulled backwards by a force equal to that whic>i he impreasee ou the towing-rope forwarda. If one l)ody strikes another Ixhdy and changes the motion of the other body, ita own motion will l>o changed in an equal quantity and in the opposite dirucUun; fur at each instant during the impact the Ixidiea exert <>n each othi>r oqnal and opposite pressures, and the momentum that one liody luBcs is equal to that which the other gains. The eorth attracts a falling pebble with a certain force, while the pebble nttracta the eurth with an equal force. The reault is that while the pebble moves towards the earth on account of its attrac- tion, the <>arth also moves towards the pebble under the infauenoe of the nttractitm of the latter ; but the mass of the earth being enor- WDUiily greater than that of the pebble while the forces on the two arising from their mutual attn ^tions are equal, the motion produced thereby in the earth is almost incomparably leas than thai produced iu the pebble, and is consequently insensible. It follows thai the sura of the quantities of mai.ion parallel to any fixed direction of the partlclro of any system influondng one another in any poHsible Avsy, remains unchanged by their mntual action. Therefhro if the c-entro of gravity of any system of mntnally influencing particles is in motion, it oontinuea movir^|r uniformly in a straigRt lino, unless in bd fur us the direction or velocity of i's motion is changed by forcea between the particles and some other matter not hflonging to the tyntem : also the centre of gravity of any system of ]iartirles moves Just as all the matter of the system, if concentrated in 11 |M>lnt, would move under the influenc«< of .orres •■<)ual and parallel to ihe forcva leaHy acttHg on its ditlcrr>at parts. (For furtliMT ' ^ '' fl B Sfc^ i^ ' flt P i lESB fflll TWO LA wa OP Morrox; 395 derstood as a fac- X ia the ar-com- force exerted on 38 with a tkfinition V to eompouud, »ad to investigats the article subjected to Of to Law III, if one Irawn by tliia other Art. 16). A horse I force equal to that If one body atrikea ther body, its own id in the opposite the IxHiiea exert on lomentum that one lin force, while the The result is that count of its attrac- er the infauencti of earth being enor- I forces on the two e motion produced than thai, produced nn parallel to any pncing on<» another >ir mutual action. Rtem of routnally rfr uniformly in a ority of i's motion o other matter not of any systom of . if concentrated in Hfual and parallel ts. (Por furtliMf remarks on these laws see Tait and Steele's Dynamics of a Particle, Tikomaon and Tait's Nat Phil., Pratt's Mechanics, etc.) 170. Two Laws of Motion in the French Troa- tiaoa — Newton's Laws of motion are not adopted in the principal French treatises ; bnt we find in them two prin- ciples only as borrowed from experience, viz.: First. — ^The Law of Inertia, that a body, not acted upon by any force, would go on for ever with a uniform velocity. ThiS coincides with Newton's First T^iw. Second. — That the velocity communicated ia proportional to the force. The second and third Laws of Motion ai'e, thup reduced to this second principle by the French writers, especially Poisson and Laplace.* 171. Motion of a Pazticto nnder tho Action of an Attr a ct i ve Force. — A particle moves under a force of attraction whtck is in its line of motion, and varies directltf as the distance of the particle from the centre nf force; it is required, to determine the motum. Tho point whence the influence of a force emanates is called the centre of force ; and the force is called an attrac- tive or a repulsive force according as it at/racla or repels. Ijet be the centre of forco, P the ^, position of the particle at any time, /. v its velocity at that time, and let OP = x, and OA = a, where A is tho position of the particle when t — 0; let /i = the absolute force ; that is, the force of attraction on a unit of mass at a unit's distance from O, which is supposed t,o be known, and is sometimes called the strength of the attracti .i. At present we shall suppose 4-^ Ft|.N M» Ail* 1 n 9.1 ! * ParWnion'* Xachanles, p. Wl. See paper bj Dr. Wkcwell on ibe principle* of Djmamiea, parttcalarly •■ ■tated by Viench wrlten, In the Bdlnborgh joomai of •, Vol. VUL S96 A VARIABLE ATTRACnVB tORCU. the maas of the particle to be unity, aa it aimplifles the eqaatiouB. Then fix is the magnitude of the force at the distance x on the particle of unit mass, or it is the accelerr.- tion at P ; and the equation of motion is (1) the negative sign being taken because the tendency of the force 18 to diminish x\ idzcPx = — %\ix dx. Integrating, we get d^ dp = fi (a» - a^), (2) if the particle be at rest when x = a and t — 0, .'. -^^===V^dt, Va* - a? the negative sign being taken, because x decreases as t increases. Integrating again between the limits correspond- ing to ^ = / and t = 0, COS" i? = H*/. 1 -1* < = -j- COS ' -• (3) Prom (2) it appears that the velocity of the particle is zero when a; = a and - o ; and is a maximum, viz.: «^*, when a; = 0. Hence the particle moves ftom rest at A : its velocity increases until it reaches where it becomes a r rOHCll. 88 it simplifies the of the force at the or it is the accelei^- is the tendency of the (2) i< = 0, e X decreases as t e limits correspond- (3) ty of the particle is laximaro, viz.: mju*, from rest at A ; its here it becomes a -. -«• 'jMiaiim'""* •^I'liiin 'MmxxmsiattKnx^itri/msmmsimieKiiiienis^j A VAMlABm ATTRACTIVM Wn^BGJl, S97 masimnm, and where the force is zero ; the putiole passes through that point, and its velocity decreases, and at A', at a distance = — a, becomes zero. From this point it will return, under the action of the force, to its original posi- tion, and continually oscillate over the space 2a, of which O is the middle point. From (3) we find when a; == a, < = and when x = 0, t = — T ; 80 that the time of passing irom A to = — r . and the time from to A' is the same, so that the time of oscillation from A to A' is -r* Tbis result is remarkable, as it shows that the time of oscillation is independent of the velocity and distance of projection, and depends solely on the strength of the attraction, and is greater as that is less. This problem includes the motion of a particle within a homogeneous sphere of ordinary matter in a straight shaft through the centre. For the attraction of such a sphere on a particle within its bounding surface varies directly as the distance from the centre of the sphere (Art. 1338). If the earth were such a homogeueoua sphere, and if AOA' (Pig. 80) represented a shaft running straight through its centre from surfitce to surface, then, if a particle were free at one end. A, it would move to the centre of the earth, 0, where its velocity would be a maximum, and thence on to the opposite side of the earth. A', where it would come to rest ; then it would return through the centre, O, to the j- ie. A, from where it suirted ; and its motion would continue to bo oscillatory, and thus it would move backwards and forwards from one side of the earth's surface to the other, and the time of the oscillation would bo independent of the earth's radius; that is, at whatever point within the earth's sur&ce the particle be placed it would reach the centre in the same time. Sit ■ t ' 298 A VARtABhM BSPITLBIVK t^jXOX. in jjifr i OoB. — To find this time. Since /i is the attraction at a unit of distance and g the attraction at the distance B^ we hare fi = -^, which in t =. — j gives for the time it wonld take a body to moTe from any point within the earth's surface to the centre. If we put g = Z^ feet and R = 3963 miles we get ^ = 21 m. 6 s. about, which would be the time occupied in passing to the earth's centre, however near to it the body might be placed, or however far, so long as it is within th<) surface. 172. Motton of a Particle under the Action of a Variable RspnlaiTe Force. — Let the force be one of repulsion and vary as the distance, then the equation of motion is dhi ~ dt» = fix. Let us suppose the particle to be projected from the cen- tre of force with the velocity v^ ; then we have dt* F«^ + »,» ; (1) As t increases x also increases, and the particle recedes further and further from the centre of force; and the velocity also increases, and ultimately equals ao when x =r t = cc. Thus in this case the motion is not oscillatory. ii ffjRCB. 9 the attraction at a b the distance R^ we OTe from any point 3 miles we get wiing to the earth'g aight be placed, or surface. r the Aetioii of a he force be one of en the equation of ected from the oen- ire have (1) *'> the particle recedes of force; and the equals 00 when x =r is not oscillatory. '■imm ivmtim mmmm^j i i '-'Mi i i i ii iii i i i i|| •mm A VASTABIX ATTHACnva POSOX. 390 X73. Motton of « Particle under the Aotioa of an Attractive Force which ia in the line of motion, and which variea Invervely aa the Square of the Uatance from the Centre of Force. Let O (Fig. 80) be the centre of force, P the position of the particle at the time t; and A the position at rest when < = 0, 80 that the particle starts from A and moves to- wards 0. Let OP = X, Ok = a, and /« = the absolute force as before or the acceleration at unit distance firom 0. Then the equation of motion is ft dx Multiplying by 2 tj and integrating, we get dt* =^e-^. (1) which gives the velocity of the particle at any distance, x, from the origin. From (1) we have 2/u\/< ax dt the negative sign being taken because in the motion to- wards 0, X diminishes as / increases. This gives ^J^dt^ — xd x = u [' a — 2x Vox — afl 2 y/ax _ ^^ J \dx. ■'* # 800 VMLOCtrr IN FALLING. Integrating and taking the lunits corresponding to ^ s ^ and / = 0, we have (2) which gives the valne of t When the particle arrives at 0, a; =: 0, therefore the time of felling to the centre from A is From (1) we see that the velocity = when a; = a ; and = « when a; = ; hence the velocity increases as the particle approaches the <»ntre of forco, and ultimately, when it arrives at the centre., becomes infinite. And although at any point very near to there is a very great attraction tending towards 0, at the point itself there is no attraction at all; therefore the particle, approaching the centre with an indefinitely great velocit;y, must pass through it. Also, everything being the same at equal distances on either side of the centre, we see that the motion must be retarded as rapidly as it was accelerated, and therefore the particle will proceed to a point A' at a distance on the other side of equal to that from which it started ; and the motion will continue oscillatory. 174. Vdloeiiy acquired in Falling throng a Oraat Height above the Earth.— The preceding case of motion includes that of a body falUng iVom a great height above the earth's surface towards its centre, the distance through which it falls being so great that the variations of the earth's attraction due to the distance must be taken into account For a sphere attracts an external particle with a force which varies inversely as the square of the distance of the particle a. ■espondrng tot x t 1+™] (') = 0, therefore the ) when a; = a ; and Y increases as the 0, and ultimately, kes infinite. And lere is a very great nt O itself there is rticle, approaching relocit;y, must pass ;he same at equal , we see that the 1 it was accelerated, to a point A' at a that from which it scillatory. ; tliroii^ a Or««t ling case of motion great height above B distance through ations of the earth's taken into account ) with a force which iuce of the particle fit il.lWIHH ivapii vtLocirr m fallino. 301 from the centre of the sphere (Ait 133a); therefore if i? is the earth's radius^ g the kinetic measure of gravity on a unit of mass at the earth's surface (Arts. 20, 23), and z the distance of a body from the centre of the earth at the time t, then the equation of motion is • if ^' which is the same as the equation in Art 173 by writing n for gR" , therefore the results of the last Art. will apply to this case. Substitu'ang gIP for /* in (1) of Art 173 we have /« -- x\ ^y «" = ^*(~> When the body reaches the earth's surfoce, x = R and (1) becomes ^^%gR(^^). P) H o is infinite (3) becomes so that the velocity can never be so great as this, however fiir the body may fall; and hence if it were possible to project a body vertically upwards with this velocity it would go on to infinity and never stop, supposing, of course, that there is no resisting medium nor other disturbing force. If in (2) we put g = 32^ feet and R = 3963 miles we get , V = [2- 321-3963. 6280]* feet = 6-95 miles; BO that the greatest possible velocity which a body (»n acquire in falling to the earth is less than 7 miles per second, and if a body were projected upwards with that "■^s^^as^agyAjEFgCi^'S mm^mmmm 303 MOTION IN A RXaiSTlNQ MSDIUJl. I*; veloci V, and were to meet with no resistance except gravity, it would never return to the earth. Cob.— To find the velocity which a body would acquire in falling to the earth's surface from a height h above the surface, we have from (1) by putting z = Rm^a = h + R, If A be small com^)ared with Ji, this may be written which agrees with (6) of Art 140. The laws of force, enumerated in Arts. 171, 173, are the only laws that are known to exist in the universe (Pratt's Mecbs., p. 212). 175. Motion in a Resiatisg Modiun.— In the pre- ceding discussion no account is taken of the atmospheric resistance. We shall now consider the motion of a body near the surfoce of the earth, taking into account the resistance of the air, which we may assume varies as the square of the velocity. A particle under the action of gravity, as a constant force, moves in the air supposed to be a resisting medium of uniform density, of which the resistant^ varies as the square of the velocity required to determine the motion. Suppose the particle to descend towards the earth from rest Take the origin at the starting point, let the line of its motion be the axis of x ; and let x be the distance of the particle from the origiii at the time t,' and for con- venience let gifi be the resistance of the air on the particle for a unit of velocity; gJi^ is called the eoeffioient of resist- ance. Then the resistance of the air at the distance x fh>m rtte MBDIUM. resistance except rth. body would acquire height ft above the = iff and a = A + i?, fiffRh R + h' Y be written ». 171, 173, are the le UDiTerse (Pratt's iiun. — ^In the pre- )f the atmospheric motion of a body into account the lume Taries as the M a constant force, fisting medium of 'aries as the square otion. is the earth from nt, let the line of be the distance of e t,' and for con- ^ron the particle ^ffii^ient of resist- te distance x fh>m >>«;W«V« = 0, we get 1 + h- t = ^log ^, (Calculus, p. 259, Ex. 5). 2k ^ "dt Passing to exponentials we have dt ke^g* + e-^*' («) which gives the velocity in terms of the time. To fina it in terms of the space, we hav« from (1) ««(§)' -^=:2gi^dx; observing the proper limits; m if L fit ■■-'WfiiiifflWifflgitl^^ 3C4 MOTION or ASCENT IN TETM AlB, which gives the velocity in terms of the distance. (4) Also, integrating (2) taking the same limits as before, we get gli^x = log (e*»* + C"***) — log 2 ; 2«ff»'« = «*»« + e-*Ot, (5) which gives the relation between the distance and the time of falling through it. As the time increases the term e-*v diminishes and from (5) the space increases, becoming infinite when the time is infinite; bat from (2), as the time increases the velocity becomes more nearly uniform, and when < = oo, the velocity = r ; and although this state is never reached, yet it is that to which the motiuu approaches. 176. Motion of a Particle Aaeending in the Air agsdnst tbe Action of Oravity.— Let ns suppose the particle to be projected upwards, that is, in a direction contrary to that of the action of gravity, with a given velocity, v, it is required to detennine the motion. Let UB suppose the particle to be of the same form and siso as before, and the same coefficient of resistance. Then, taking x positive upwards, both gravity and the resistance of the air tend to diminish the velocity as t increases; so tliat the equation of motion is tPx -,<)•; (1) (4) istance. limits as before, log 2; (5) mce and the time linishes and from when the time is aases the velocity len ^ = 00, the never reached, yet ling in the Air as suppose the 8, in a direction ty, with a given motion. le same form and it of resistance, gravity and the the velocity as t is (1) MOTION or ASCMXT IX TBM .AIM, dx dh dl i + KI) .'. tan-> 1e^ — *»«"* (H - O^d » (Calculus, p. 244, Ex. 3), since the initial velocity is v. Taking the tangent of both members and solving for dx . ^ _ 1 v l — tan kgt . di ~ k' 1 + vk toD kgt* (8) which gives the velocity in terms of the time. To find it in terms of the difltanoe, w» 1mv(> from (1) -(f)' -, s= -«jfifc»' which gives the velocity in terms of the distance. Also, integrating («) irfter sabstituting sine and cosine for tangent, and taking the same limits w before, we get ^/L*» =3 log (»* sin /is'' +• oo* M) ; (*) which gives the space desori'jed by the particle in terms of tiie tim«. i UBI mozaon or AsoMifT isr tsjb air. OoB. 1.— To find the greatest height to which the par- tide will ascend put the velocity, -^ = 0, in (3) and get which is the distance of the highest point da Putting ^ = v> in .^s Te ge*; («) kg m which is the time required for the particle to reach the highest point. TIaving reached the greatest height, the particle will begin to foil, and the circnmfitances of the fall will be giyen by the equations of Art 175. OoB. 2.— Since h is the same in this and Art 176, we may compare the velocity of projection, t', with that which the parcicle would acquire in descending to the point wh< nee it was projected. Denote by v^ the velocity of the particle when it reaches the '^nint of starting. From (3) of Art 176 we have and placing this value of x equal to that given in (6), we get, r-w:*^^-^^' .'. », = (l + *»l^)*' which is less than t>; hence the velocity »^aired in the •Ai f.W"WtW»K5?2?l m S AIR, t to which the par- 0, in (3) and get (6) t (7) irtiole to reach the latest height, the cnm^tances of the ;. 175. and Art 176, we t>, with that which ling to the point t'o the velocity of )f starting. From that given in (6), ty »^aired in the wmmmm UOTIOH OF A PSOJBCTtLB, 307 descent is less than that lost in the ascent, as might have been inferred. Cob. 3.— Substituting (6) in (5) of Art 176, we get for the time of the descent. / ^log(Vl + iV + *»), which is difFennt from the time of the ascent as given in (7). (See Pri je's Anal. Mech's, Vol. I, p. 406 ; Venturoli's Mech's, p. 81) ; Tait and Steele's Dynamics of a Particle, ■AHi. ) 177. Motion of a Protjectile in • Resisting M^- dimn. — The theory of the motion of projectiles in vacuo, which waf; examined under the head of Kinematics, affords results which differ greatly from those obtained by direct experiment in the atmosphere. When projectiles move with but small velocity, the discrepancy between the para- bolic theory, and what is found to occur in practice, is small ; but with increasing velocities, as those with which bftils and shells traverse their paths, the air's resistance increases in a higher ratio than the velocity, so that the discrepancy becomes very great. The most important application of the theory of projec- tiles> is that of Gunnery, in which the motion takes place in the air. If it were allowable to negWct the resistance of the air the investigations in Part II would explain the theory of gunnery ; but when the velocity is considerable, the atmospheric resistance changes the nature of the tra- jectory so much as to render the conclusions drawn from the theory of projectiles in vacuo almost entirely inap- plicable in practice. The problem of gunnery may be stated as follow^: Given a projectile of known weight and dimensions, starting with a known velocity a^ a known angle of eleva- 308 XOTtON or A PKOJBCTTLE. tion in a calm atmosphere of approximately known density ; to find its range, time of flight, velocity, direction, and position, at any moment ; or, in otb'jr words, to construct its trajectory. This problem is not yet, however, suscepti- ble of rigorous treatment ; mathematics has hitherto proved unable to furnish complete formulae satisfying the condi- tions. The resistance of the air to slow movements, say of 10 feet per second, seems to vary with the first power of the velocity. Above this the ratio increases, and as in the case of the wind, is usually reckoned to vary as the square of the velocity ; beyOnd this it increases still further, till at 1200 feet per second the resistance is found to vary as the cube of the velocity. The ratio of increase after this point is passed is supposed to diminish again ; but thoroughly satisfactory data for its determination do not exist. From experiments* made to determine the motion of cannon-balls, it appears that when the initial velocity is considerable, the resistance of the air is more than 20 times as great as the weight of the ball, and the horizontal range is often a small fraction of that which the theory of pro- jectiles in vacuo gives, so that the form of the tngectory is very different firom that of a parabolic path. Such experi- ments have been made with great care, and show how little the parabolic theory is to be depended upon in determining the motion of military projectiles. 178. Motion of « Prc^otile in the Atmosphere Snppoeing its Resistance to wary ss the Square of the Telocity. — A particle under the action of gravity ia projected from a given point in a given direction loith a given velocity, and moves in the atmosphere whose resistance is assumed to vary as the square of the velocity ; to deter- mine the motion. * B«e BDcrolo)Midto BrllMiiiiea, Art ChuuMrr ; Hnttoa'f Tncu. •iw Bobln'i Qnaasrj, and IJP. 3ly known density; ity, direction, and ords, to construct however, suscepti- las hitherto proved isfying the condi- movements, say of the first power of ses, and as in the rary as the square till farther, till at nd to vary as the se after this point ; but thoroughly not exist, le the motion of initial velocity ia lore than 20 times ; horizontal range the theory of pro- I the trajectory is th. Such experi- d show how little m in determining M Atmoaphera ■ the Square of fion of gravity ia direction with a a whoae reaiatanca 'eiocitjf ; to deter- Bobln'i OnniierT, aaA mwaOH OP A PROJEOfiLA Take the given point as origin, the axis of x horizontal, the axis of y vertical and positive upwards, so that the direction of projection may be in the plane of xy. Let v be the velocity of projection, g the acceleration of gravity, a the angle between the axis of x and the ^'le of projection, and let the resistance of the air on the particle be i for a nnit of velocity ; then the resistance, at any time, t, in the line of motion, is h ijA ; and the x- and y-components of thie resistance art, respectively, , da dx J * jj • TV, and dt dt da dy dt dt Then the equationfl of motion are, resolving horizontally and vertically, (Pa; _ ,da dx .^v dfl- ~^dt rr ^ ' dP ~ 9 Prom (1) we have tdx\ dx dt = -h-la; dx log ^dt dt dm dt voosa (») — ha\ since when < = 0, -^ =:^ v c(x a', dx di = v cos « «-■ (8) Maltildying (1) and (2) by dy and dx, respectively, and aabtraoting the former from the latter we have d^dx — iPxdy _ ^ dfi gdx. (4) ii 310 MonoN or a projbctilm. Substituting in (4) for dP ita value from (3) we get Substitu ting in t he second member of (6) for dx its value dxV ^ da«/ i;» co8» - '^'"* (6) Pat ^ = P' and (6) becomes "+'■>** = pi?-. •**• Integrating, and remembering that when « = 0,^ = tan <^ we get P (1 + J»«)* + log [p + (l+ JB»)*] -^,-e« (7) Ar* Cf /8» a where e is the constant of integration whose value = tan o sec « + log (tan o + sec «) + , -^ . . fg^ ' «t)»COS»o ^ ' From (5) we have t>»oos»a dxXdxf' which in (7) gives i'a +i^)* + log [i» + (1 +i,»)*] - c = ^|, , dp " /'(l+jJ^+l0g[iB + (l+^)*]_c^**'' ^*^ ttHH tM. (3) we get 9 % (6) i) for dx its value «»»»(fo. (6) »rf». « = 0,jp = tan «> /^)*] Mseralne ifct)»C08»tt (7) (8) — = kdx, (9) .uu l.i'.JCJiW gWWWff^^^^Wff!— W^gl'^ff'wW and MOTtON or A PROJMCTILa. pdp 311 =:Hy. (10) /» (1 +i^* + log O + (1 + pO*] - c From (4) ..e have dX'dp = — gdfi. Sabgtituting this value of dx in (9) and solving for dt we get |c - p (1 + p»)* ~ log [/» + (1+P»)*] f* the negative sign of dp being taken because j» is a decreas- ing function of ^. Bephicing the value of i> = ^, (9), (10), and (U) become (fo = T 4^ dx *i(i^^Wwl^(^*]-«' , 1 dx dx 'l('+g)*-'-[|+(' + g)*]-«' -d dt=: dx ♦^{-K'+gr-Hi^^OT" (A) (B) , (0)- from which equations, were it possible to integrate them, X, y, and t might be found in terms o* ^ > *»^ *' ^ ""^^ eliminated from the two intends, of (A) and (B), the re- sulting equation in terms of x and y would be that of the I 1 1 1 312 Monoir or a pxwjwmjk required trajectory. But these eqnatiors cannot be inte- grated in 6nit6 terms; only approximate .eolations of them can be made ; and by means of these the path of the pro- jectile may be constructed approximately. (See VentoroM's Mechs., p. 92.) Squaring (A) and (B), and dividing their suit by the square of (0) we get di* 9 k 1+^ -n'^%r-^<^^\^ + V ^m (D) which gives the velocity in terms of ^. 179. Motion of a Projectile in the Atmosphere under a small Angle of Elevation.— The case fre- quently occurs in phwtice where the angle of projection }* very small, and where the projectile rises but a very little above the horizontal line. In this case the equation of the part of the trajectory that lies above the horizontal line may easily bo found; for, the angle of projection being dy very small, ^ will be very small, and fterefore, throughout the path on the upper side of the axis of x, powers of ^ higher than the first may be neglected. In this case then diszix', .•. » = «} which in (6) of Art. 178, becomes rfy d^ = «»co«*« «** which in (4) and (5) respectively of (Art. 10), gives tn ; and « = 4 -^ /*. * m 2. A body weighing n lbs. is moved by a constant foroe which generates in the body in one second a velocity of a feet per second ; find the force in pounds. Ana. 5? lbs. 9 3. Find in what time a force of 4 lbs. wonld move a weight of 9 lbs. through 49 ft along a smooth horizontal plane ; and find the velocity acquired. 21 4. Find the number of inches through which a force of one ounce, constantly exerted, wiU move a mass weighing one lb. in half a second. ^„,, 3^ /ijs, 6. Two weights, P and Q, are connected by a string which passes over a smooth peg or pulley ; required to determine the motion. Since the peg or pulley is perfectly smooth the tension of the string is the same throughout; hence the foroe which causes the motion is the difference between the weights, P and Q, the weight of the string being neglected. The moving force therefore is P — 0; but the weight of the mass moved is P + ^. Hence substituting in (1) of Art 25, we get P+ Q 9 X Q* ■p-e 'f. Fig. SOo. •re from (1) of Art m i (Art. 10), gives by a constant force Jond a velocity of a is. Ana. — lbs. lbs. would move a emooth horizontal 21 V^' V = ^t. gh which a force of e a mass weighing Am. 3ffH)\ leoted by a string alley ; required to Btly the lich een the >rce the ing X Q* Fig. SOo. •mtt 'm i't w mv ta gmm mw' i m i mi P i f smr miUKa XXAMPLSa. .-. /= P-Q. which is the acceleration. Substituting this in (4) and (5) of Art 10, we have 815 (1) V = -i^.— jf 20 feet, when the 10, we have (calling feet before it begins horizontal plane at team is turned off ; e jfy of the weight, le will run before it BXAMPhK3. »17 Let ff be the weight of the engine; then the rosistance W of friction is -r^t and tl»8 is directly opposed to motion, 400 w m 9^' 400 30x1760x8 =:44 The velocity, v, is 30 miles an hour = — ^ ^ g^ feet per second. Substituting these Vfiues of / and v in the equation v = //, we get 44 = ^«; ,'. t = 550 sees., which is the time it will take to bring the engine to rest if the velocity be retarded ^ feet per second. Also r» = Ufa, therefore g — « «yy ^x*oo — 12100 feet. 11. A man whose weight is W, stands on the platform of an elevator, as it descends a vertical shaft with a uniform acceleration of |jf ; find the pressure of the man upon the platform. Let P be the pressure of the man on the platform when it is moving with an acceleration of \g ; then the moving force is W — P; and the weight moved is W', therefore W W^P = ^\9; tr. J2. A plane supporting a weight of 12 ozs. is descending with a uniform acceleration of 10 ft. per second ; find ,the pressure that the weight exerts on the plane. Am. S\ ozs. 318 SXAMPLSa. 13. A weight of 24 lbs. hanging over the edge of a smooth table drags a weight of 12 lbs. along the table; find (1) the acceleration, and (2) the tension of the string. Aw. (1) 5^ ft. per sec. ; (2) 20 lbs. ||; !*• A weight of 8 lbs. rests on a platform; find its pressure on the platform (1) jf the latter is de- scending with an acceleration of \g, and (2) if it is ascending with the same acceleration. Ans. (1) 7 lbs.; (2) 9 lbs. 15. Two weights of 80 and 70 lbs. hang over a smooth pulley as in Ex. 5 ; find the space through which they will move from rest in 3 sees. A7i8. 9| ft. 16. Two weights of 16 and 17 ounces respectively hang over a smooth pulley as in Ex. 5 ; find the space de- scribed and the velocity acquired in five seconds from rest Ana. » = 26, V = 10. 17. Two weights of 5 lbs. and 4 lbs. togethei pull one of 7 lbs. over a smooth fixed pulley, by means of a con- necting string; and after descending through a given space the 4 lbs. weight is detached and taken away without interrupting the motion ; find through what space the remaining 6 lbs. weight will descend. Ans. Through J of the given space. 18. Two weighta are attached to the extremities of a string which is hung over a smooth pulley, and the weights are observed to move through 6.4 feet in one second ; the motion is then stopped, and a weight of 5 lbs. is added t.) the smaller weight, which then descends through the same space i\» it ascended before in the same time ; deter- mine the original weights. Ans. | 11)8. ; V lbs. 19. Find what weight must Iw added to the Bnmller wcigltt in Ex. 5, so that tlie acceleration of the systen. may over the edge of a t)e. along the table; jnaion of the string, jr sec. ; (2) 20 lbs. a platform ; find the latter is de- , and (2) if it ia I 7 lbs.; (2) 9 lbs. hang over a smooth igh which they will Ans. 9| ft. es respectively hang find the space de- ! scoouds from rest » = 26, V = 10. . togethei pull one by means of a cou- ; through a given taken away without B[h wi)at space the f the given space. 10 extremities of a ey, and the weights in one second ; the of 5 lbs. is added icends through the same time ; deter- s. I 11)8. ; V lbs. I'd to the Bnjaller of the syateiL may XX' nBS. 810 have the same nnmerioal value aa before, but may l)e in the opposite directisn. Ana. 20. A body \s. projected np a rough inclined plane with the velocity which would be acquired in faUing freely through 12 foet, and just reaches the top of the plane ; the inclination of the plane to the horizon is 60°, and the coeflScient of friction is equal to tan 30°; find the height of the plane. Ana. 9 feet. 21. A body is projected up a rough inclined plane with the velocity ^g ; the inclination of the plane to the horizon is 30°, and the coeflHcient of friction is equal to tan 15° ; find the distance along the plane which the body will describe. ^na. g (v^S + 1). 22. A body is projected up a rough inclined plane ; the inclination of the plane to the horizon is «, and the coef- ficient of friction ie tan e ; if m be the time of ascending, and n the time of descending, show that \« / ~ sin (a + e)' 23. A weight P is drawn up a smooth plane inclined at an angle of 30° to the horizon, by means of a weight Q which descends vortictjUy, the weights being connected by a string passing over a small pulley at the top of the plane ; if the acceleration be one^fourth of that of a body fa]lh.j freely, find the ratio of ^ to P. Ant. Q = P. 24. Two weights P and Q are connected by a string, and Q hanging over the top of a smooth plane inclined at 30° to the horizon, can draw P up the length of the plane in just half the time that P would take to draw up Q ; show that Q U, half m heavy again as P. 830 MXAMPhaa. 25. A particle moyes in a straigfat line undo* the action of an attraction varying inversely as the (|)th power of the distance; sho\» that the velocity acquired hy falling from an infinite distance to a distance a from the centre is equal to the velocity which would be acquired in moving from rest at a distance a to a distance ^• m te nndear the action the (|)th power of acquired by falling [ from the centre is acquired in nioTing CHAPTER II. CENTRAL FORCES.* 180. Definitlona. — A central force is one which acts directly towards or from a fixed point, and is called an attractive or a repulsive force according as its action on any particle is attraction or repulsion. The flxed point is called the Centre. The intensity of the force on any par- ticle is some function of its distance from the centre. Since the case of attraction is the most important af>plica- tion of the subject, we shall take that as our standard case ; but it will be seen that a simple change of sign will adapt our general formnlsa to repulsion. If the centre be itself in motion, we may treat it as flxed, in which case the term "actual motion "of any particle means its motion "rela- tive" to the ct itre, taken as fixed. The line firom the centre to the particle, is called a Radius Vector. The path of the particle under the action of an attraction or repulsion directed to the centre is called its Orbit.^ All the forces of nature with which we are acquainted, are central forces ; for this reason, and be- cause the motion of bodies under the action of central forces is a branch of the general theory of Astronomy, we shall devote this chapter to the consideration of their action. 181. A PuHele under the Aotion of a Central Attnotion; Reqnired the Polar Equation of the Path. — The motion will clearly take place in the plane passing through the centre, and the line along which the • TbiM chapter conUln* the Ant prinolplei of lUtbWDaticai Attronomy. It nuy, howaver, be omlltad bj the •tudent of KngtMering. t OkU«d Omtnl Urbiu. :il M CKlfTSAL ATTRACnO'cf. particle is initially projected, as there is nothing to with- draw the particle from it. Let the centre of attraction, 0, be the origin, and OX, OY, any two lines iirough ftt right angles to each other, be the axes of co- ordinates. Let (x, y) be the position of the particle M at the time t, and (r, 9) its position referred to polar co-ordinatos, OX being the initial line. Then, calling P the central attractive force, we have for the components parallel to the axes of x and y,* respectively, -- P-, — P^, the forces being nega- tive, since they tend to diminish the co-ordinates. There- fore the equations of motion are (1) Multiplying the former by y, and the latter by x, and iubtracting, wo have ; — 2 _ « — a;^-y,^ =0. (3) Integrating we have .% dt "dt where A is .a undetermined constant Since x =5 r oos 0, and y = r sin 0, we hav« dx = COS B dr — r Bin do, dy =z nine dr + rcoaO d9, which in (3) gives (8) (*) o:y. is nothing to with- itie of attraction, 0, Fli.M llel to tlie axes of x e forooB being nega< D-ordinates. There- ■ Pf. (1) he latter by x, and (3) (8) ire iuiye ide, )d$, <*) CMlfTBAL ATTSACnOJf, d( ^ '*• 898 (5) Again, multiplying the first and second of (1) by 2tbe and 2dy respectively, and adding, we get 2dxdh: + %dy dhf _ 2P{xdx + ydy) d0^ ~ r * d da* ,5? "•■<«« -2/Wn m Substituting in (6) the values of dafi and Jy* from (4), we have (7) Put r ss - ; and . •. dr as — — | ; and (7) becomes 2P ./dvF .A 2P , performing the differentiation of the fint member> and dividing by idu, and transposing, we get dht + «♦- ]?u* m which i$ th* differential equation of the orbit described ; and as, in any particular instance, the force P will be given in terms of r, and therefore in terms of u, the integral of this equation will be the polar equation of the requijr^ path. 6 the required centre as pole ; we must then differentiate . twice with respect to 0, uid substitute the result in the expression for P, eliminating 9, if it occurs, by means of the relation between « and 9. In this way we shall obtain P in terms of u alone, and therefore of r alone. OoB. 8.— When we know the relation between r and fl from (9), we may by (6) determine the time of describing a given portion of the orbit ; or, conversely, find the posi- tion of the particle in its orbit at any time.* * IM lUt ud StMle'i DyuuniM of > I^wtkla, p. IM; alw Pm(t'« Ifack't, V. ; (9) by a different pro- along the radius I will contain four introduced in (5), integration of (8). from the integral equation be then itrodnced, and the ; any time will be I determined from viz., the initial two independent I direction of pro- certain the law of ;icle to canse it to re must determine liar equation of the ole ; we must then and aubstitnte the ng 6, if it occurs, >. In this way we ind therefore of r I between r and 6 time of describing sly, find the posi- ic.* lU; atao Pmtt'* Ifach't, THB BBCTIONAL AREA. OoB. 4, — If p is the perpendicular from the origin to the tangent we have from Oalcnlus, p. 176, xdy — ydx =: pd9 i which in (3) gives and this in (6) gives ds dt P' (10) ,h* Differentiating, and solving for P, we have A»rfr f dp' (11) which is the equation of the orUt between the radius vector and the perpendicular on the tangent at any point. 182. The Etoctorial Area Swept over by tiie Raditis Vector of the Particle In any time is Pro- portional to the Time. — Let A denote this area ; then we have £rom Calculus, p. 364, if A and t be both measured from the commencement of the motion. Therefore the areas swept over by the radius vector in different times are proportional to the times, and equal areas will be described in equal times. Cob. — If t = 1, we have A = |A. Hence h = twice the sectorial arja described in one unit of time. 1B3. The Velocity of the Particle at any Point of ita OrMt— We have for the velooity, A = i/r»rfd = ^fhdt,hj (5) of Art. 181, i- = iAf, 1 886 vsLocmr at ant poi^rr or rmi obbit. = j^ by (10) of Art 181. (1) Hence, the velocity of the patticle at each point of its path is inversely proportional to the perpendicular from the centre ott the tangent at that point. OoB. 1.— We have, by Calculus, p. 180, 2- _ 1 1 rff* = «» + ^', since f = - (Art. 181), which in (1) gives •'=1 = »•(«•+ a. another important expression for the velocity. Cob. 2.— From (6) of Art. 181, we have (2) (8) I*t V be the velocity at the point of projection, at which let r =: B, and since P is some function of r, let P = /(r), then integrating (3) we get which is another expression for the velocity ; and since this is a ftinotion only of the corresponding distance*, R and r, it follows that th$ velocity at any point qf tho orbit is rjfV OBBJT. of Art 181. (1) t each point of its pendicular from the 30, 36 f = - (Art 181), ). (2) locity. IT© Fdr. (8) t of projection, at e fanotion of r, let (r)], {*) sity ; and since this diatanoM, E and r, nnt qf the orbU %$ vBLOcrtr AT Atrr potsr of wr« oMBtt. 88? independent of the path described, and dtpmdt iokljf on the magnitude of the atttaetion, the dietanoe of th$ point from the centre, and the velocitif and distance of projection. From (4) it appears that the velocity is the same at all points of the same orbit which are equally distant from the centre ; if r =: i?, the velocity = V; and thus if the orbit is a reentering cunrc, the particle always, in its successive revolutions, passes through the same point with the same velocity. If the velocity vanishes at a distance a from the centre (4) becomes t;» = 3[A(a)-A(r)] (6) and a is called the radius of the circle of zero Velocity. OoB. 3.- -From (3) we have rf(r») = - 2Pdr; .*. vdv = — Pdr. Taking the logarithm of (1) we have log w = log A -log/. Differentiating we get dv P («) (*J Dividing (6) by (7), we get = 2P X J chord of curvature* through the centre ; (8) • To prove that f ~ U one-f onrth the cbord of OWmttM. 8 dp Let MD (Fig. 81), ba tbe Ungent to the orbit, end C the MDtre of cnrratara ; let OD =|), CM = p, tbe radliM of curvature ; end the angle HSS - ^. Then MS, the 338 VSLOCJTT AT Afrr POINT OP THE OBBIT. and, comparing this with (6) of Art. 140, it appears that the particle at finy point has the same velocity which it would have if it mov^ from rest at that point towards the centre of force, under the action of the force continuing constant, through one-fourth of the chord of the circle of curvature. Hence, the velocity of a particle at any point of a central orbit is the same as that which would be acquired by a particle moving freely from rest through one-fourth of the ciiord of curvature at that point, through the centre, under the action of a constant force whose magnitude is equal to ihat of the central attraction at the point. Cob. 4. — If the orbit is a circle having the centre of force part of the radlns vector CM, which la intercepted by the circle of cnrratON is Called the chord tf curvature. Ita valae is determined as follows ; We have (Fig. 81) * = fl + OMD = » + sln-i rVr'—tf Viom Calcoluf, p. 180, (10), we have pdr a$ = — r Vr" — p" and Subatitiitlng CI) in (1) we get <$, = '■ rOr m HE OBBIT. 10, it appears that velocity which it point towards the I force coutiDuing rd of the circle of point of a central ' be acquired by a i one-fourth of the h the centre, under titude is equal to the centre of force tbe circle of cnmtara ia FoUowb; 0) W tare. TBB ORBIT mrDER VARtABLB ATTRACTION. 320 in the centre, and R, V, P, are reapeotiyely the radius, velocity and central force, we have r» = PR. OOR. 5.— From (5) of Art 181, we have de dt (9) The first number, being the actual velocity of a point on the radius vector at the unit's distance from the centre, is the angular velocity of the particle (Art. 160). Hence the angular velocity of a particle varies inversely as the square of the radius vector. ScH.— A point in a central orbit at which the radius vector is a maximum or minimum is called an Apse ; the radius vr^tor at an apse is called an Apsidal Distance ; and the angle between two consecutive apsidal distances is called an Apsidal Angle of the orbit The analytical conditions for an apse are, of course, that ^ = 0, ai.d that the first derivatiye which does not vanish should be of ati even order. The first condition ensures that the radius vector at an apse is perpendicular to the tangent. 184. The Orbit when fhe Attraction Vaxiee In- v-ersely as the Square of the Diatanca— J particle is projected from a given point in a given direction with a given velocity, and moves under the action of a central attraction varying inversely as the square of the distance ; to determit.e the orbit. Let the centre of force be the origin; V = the velocity of projection ; i? =r the distance of the point of projection from the origin ; /3 = the angle between R and the line of Hi 880 TBlt OBBIT Umxn VABIABhW ATTBACTIOB. projeotioii ; and let ^ = the absolate force and < s= when the particle is projected. Then since the relooity s - (Art 183), and at the point of projection /> = ^ sin /3, wc have K Sin p As the force yaries inversely as the square of the distance, wehaye p SB ^ SB fiv^, ^since »" = z)- (2) which in (9) of Aft 181 gives Multiplying by ddu and integrating, we get S + «* = 2?.« + C; (3) d0» ¥ 1 1 du* F* when « =: 0, tt =± - i= -g, and ^ + «» sa -p-, (Art 183, Cor.l); therefore ' ~ A» A»i2 A»i? Substituting thl« valtte for c We get Therefore (Art. 183, Cor. 1) we have (velocity)' = F» + 2^ (^ - ^ (4) (5) ' ATTBACTIOir. te force aod < r= since the Telocity =z jection /> = i2 sin /3, f2 sin /3. (1) qnare of the distance, (8) re get .tt»s=-^, (Art.188, -2p »i2 8p« -i) w 0) TJ» OSBtT tniJ>MB VABtABLM ATfttAOTtOlf. 381 which showa that the velocity it th« grmtest whm ritth* least, and the leaet when r is the greatest. Changing the fonn of (4) we Uave d0* - h»R ■^h*-\h*-'V' (6) To express this in b bimpler form, let g = *, and ^^^~ + ^ = <^.; a^^ (6) becomes _=c!-.(«-J)a; d9» — du [c» -. (» - J)«]* = dO, the negatiTe sign of- the radical being taken. Integrating we have. COS" where c' is an arbitrary constant; .'. tt =a 4 + C DOS (« — C*). (7) Replacing in (7) the values of h and c, and the value of h, from (1), and dividing both terms of the second number by Hf we have for the equation of the path. 1 + ri ( F»iJ - 8^) /? r» Bin» /3 + 1 J 008 (»-«') ti = 1 _____^___- , ^r»8in»/J (8) which is the equation of a conio section, the pole being at the focus, and the angle (9 — c') being measared from the THE ORBIT UNDER VARIABLE AVtRACTION. shorter length of the axis major. For if is the eccentricity of a conic section, r the focal radias vector, and ^ the angle between r and that point of a conic section which is nearest the focus, we hare, ' 1 1 + c cos A — H* ta^m ' ■ ■ ■ '■ - • Comparing (8) aud (9), we see that fl» = -(FJ72-2^)i?r«3in»/3 + 1; 4» = d e\ (») (10) (11) Now the couis section is an ellipse, parabola, or hyper- bola, according as e is less than, equal to, or greater than unity ; and from (10) e is leas than, equal to, or greater than, unity according as V*B — 2fi is negative, zero, or positive ; therefore we see that if 2n F* < -^, e < 1, and the orbit is an ellipse, 2u F» = -p, e = 1, and the orbit is a parabola. (12) (13) 2u F» > ~, « > 1, and the orbit is a hyperbola. (14) CoR. 1.— By (1) of ;^rt. 173, we see that the square of the velocity of a particle falling from infinity to a distance Ji from Ihe centre of force, for the la^ of atti-action we , , . . 2« arc considering, is -^. Hence the above conditions may be expressed more concisely by saying that the orbit, d^mihed about this centre 0/ force, wiil be an eUipse, a Mfe tirRACfioir. 6 is the eccentricity vector, and ^ the aic section which is (») */3 + l; (10) (11) parabola, or hyper- to, or greater than iqnal to, or greater 1 uogative, zero, or an ellipse, (12) a parabola, (13) a hyperbola. (14) that the square of finity to a distance h^ of atti'action we ve conditions may ng that the orbit, ill be an ellipse, a TBS ORBIT AN BLLIPSJB. 883 parabola, or a hyperbola, according as the velocity is ksa than, equal to, or greater than, the velocity from infinity. The species of conic section, therefore, does not depend on the position of the line in which the particle is pro- jected, but on the velocity of projection in reference to the distance of the point of projection fh)m the centre of force. Oor. 2.— From (11), we see that 6 — c' is the angle between the focal radius vector, r, and that part of the principal axis which is between the focus and the point of the orbit which is nearest to the focus ; i. e., it is the augle PFA (Fig. 82) ; and therefore if the principal a^Js is the initial line c' = 0. 185. Suppose the Orbit to be an EUipse.— Here p-s ^ ^ . eo that from (10) we have c» = 1 - i (2?* - F»/Z) i? r» sin» /3. (1) Now the equation of an ellipse, where r is the focal radius vector, 9 the angle Iwtween r and the shorter seg- ment of the miyor axis, 2o the major axis, e the eccon- '■"" 1 -f- eome* 1 fl Cos •■* »* = o (1 _ «») + ^JT^ e») ' comparing (2) with (8) of Art, 184, we have 1 ± . (») 4 834 TBB ORBIT Air SLUPSK. substituting f or 1 — «» its yaliw from (1), and aolTingfor a, we have " - 2^ _ yJR' (8) which shows that ths major axis is independent of the direc- tion of projection. We may explain the several quantities which we have used, by Fig. 82. B is the point of projection; FB = 5; DB is the line along which the particle is projected with the velocity V; FBD = p, the angle of projection; FP = r; PFA = d; FD = /2 «n /3; if (9 = 90°, the particle is projected from an apse, i. «., from A or A'. Cob. 1.~To determine the apsidal diitances, FA and FA', we must put ^ = 0, (Art 183, 8oh.), and (4) of Art 184 give us the quadratic equation (*> the two roots of which are the reciprocals qf the two apsidal diitances, a{l —e) and a (1 -f e). Cor. a.— Since the coefficient of the second term of ( ) is the sum of the roots with their signs changed, we have a(l-e) ^ a^l-e) «(l-^=^'l which gives the latus rtetum of the «rUi. (») rj& (1), and lolTing for (3) mdeni of the direc- es whioh we have 2; DB is the line ith the velocity V; = r; PFA = 6; is projected from ittanccB, FA and Soh.), and (4) of : 0. (4) jff the two apaidal econd term of \ > langed, we have in. (») MBPLgR'S LAWS. CJOB. 3.— From Art 183 we have, calling Tthe time, T=: aA where A is the area swept over by the radius vector in the time T. Therefore for the time of describing an ellipse, we have _ a area of ellipse 2na* Vl — e« , from (5), Sff Vf' which is the time oceupiad bif the particle in passing from any point of the ellipse around to the same point cgain.* l^'i. KflplftX** Xiaws. — By laborious calculation from an immense series of observations of the planets, and of Mars in ptsrticular, Kepler enanoiattd the following as the Ifiws of the planetary motions aboat the Sun. /, 'A%e orbits of ihe planed are ellipses, of whioh the Sun occupies a focus. II. The radius vector of each planet describes equal areas in equal times. III. The squares of the periodic times of the planets are a (1) • • ""dd = iji - aJ sine cos », (») But [Art. 181, (9)] we have MXAMPLWa, lerefore ju tiiiist be a of the Snn must not only is the law t the abaotule force )rce8 is all that we iries farther would 3 especially entitled bo wishes to pursue Steele's Dynamics , VoL I, or to any We shall conclude nder an attraction ired to find the law at of the orbit, and • e, the pole at the (1) J 9, (2) - Bin' 6). (3) by (3), (COS* e — sin* d)], by (2), by factoring, A* 1 1 u /iv ** - m and therefore the attraction varies directly as the distance. If ^ = the absolute force we have, by (4), A» = ^a»ft». (6) (2) If t; = the velocity, we have, by Art. 188, t»» = ^ = ^J (Anal. Geom., p. 133) = tib\ by (6), where V is the semi^diameter conjugate to t. .'. V tsh' V/ii. (3) If 2* = the periodic time, we have, by Art. 182, _ 2iTfl& %n V^ , by (6), and hence the periodic time is independent of the magni- tude of the ellipse, and depends only on the absolute central attraction. (;See Tait and Steele's Dynamics of a 840 BXAMPLSa. Particle, p. 144, also Prioe'fl AnaL Mech's, Vol. I, p. 516.) %. A particle describes an ellipse ander an attraction always directed to one of the foci ; it is required to find the law of attraction, the velocity, and the periodic time. (1) Here we have 1 + 6 cos 0(1 -c)» ' u = and tPu dp du _ dd - ' ecos 9 e sin (1) a(l-e»)' which in (9) of Art. 181 gives a(l -fl») ~ o(l-c») r»' (2) hence the attraction varies inversely as the square of the distance. Ufi = the absolute force, we have by (2) (2) By Art 183, Cor. 1, we have 1 - , dM» 20W-1 . ... j^ = *^' + ^ = 5Mrr^'*'y<^>' . h* ^(2o« — 1) . ,_. , ... • *• *^ = p = ~~S ' y ^^^ ^^^ ^*^' (3) If r = the periodic time we have (Art. 182) y^ 2rroMl-<^)* - g^g* (1 -- <**)* _ ^ i "[^(l-e»)]*~ v^"' (3) (4) (6) (6) . Mech's, Vol. I, ander an attraction required to find the periodic time. e Bin (1) _ i (8) ,8 the square of the e have by (2) (3) ry^JWi (4) 3) and (4). (6) e (Art. 182) 3ff 1 = — a*, (8) SXAMPLSA 841 and hence the periodic time varies aa the square root of the cube of the major axis. 3. Find the attraction by which a particle may describe a circle, and also the velocity, and the periodic time, (1) when the centre of attraction is in the centre of the circle, and (2) when the centre of attraction is in the circum- ference. (1) Let a = the radiiu; then the polar equation, the pole at the centre, is Also 1 rfu = and «• = If T the periodic time, we have fir* 2« na' (See Priced Anal. Mech., Vol. Ill, p. 518.) 4. Find the attraction by which a particle may deseribe the lemnisoate of Bernouilli and also the velocity, and the time of describing one looj^i, the centre of attraction being in the centre of the lemn ideate, and the equation being r» = c» cos 3ff. Ans. P = «2 —' T 5. Find tlie attraction by which a particle may describe the cardioid and also the velocity, and the periodic time, the equation being r = a (1 + cos 0). (>. Find the attraction by which a particle may describe a i)arabola, and also the velocity, the centre of attraction being at the focus, and the equation being r = = i« Compare (13) of Art. 184. 7. Find the attraction by which a particle may describe a hyperbola, and the velocity, the centre of attraction being at the focus, and the equation being r = z — ^• ' ** 1 + e cos © Ans. P = _ JL_ \. ^- M(2a« + 1) a ( 1 - e») fS ' *^ ~ a solute force, we III, p. 518.) le may describe jlocity, aud the .ttraction being equation being e may describe periodic time, = (Sfia')K. le may describe 9 of attraction *" ~" 1 + cos e' ) of Art. 184. e may describe ittraction being (g*-!) ■f fl cos 6 o MAAMPl^MS. 343 8. K the centre of attraction is at the centre of the hyperbola, find the attraction, and velocity, the equation . . eos^ 9 sin* d Deing — ^ ^ = «». AtU. P =z 55*' = ■fir', v» =r ^ (f» -> a» + ¥). 9. Find the attraction to the pole under which a particle will describe (1) the curve whose equation is r = 2a cos nd, and (2) the curve whose equation is r = -"- - -. 1 — e cos hS (1 - n») A» n»)A». That is, the attraction in the first curve varies partly as the inverse fifth power, and partly as the inverse cube, of the distance ; and in the second it varies partly as the inverse square, aud partly as the inverse cube, of the distance. 10. A planet revolved round the sun in an orbit with a major axis four times that of the earth's orbit ; determine the periodic time of the planet. Ans. 8 years. 11. If a satellite revolved round the earth close to its surface, determioe the periodic time of the satellite. Ana. ——J of the moon's period. 13. A body deecrihefl an 'ellipse under the action of a force in a focus : compare the velocity when it is nearest the focus with its velocity when it is furthest from the focus. Am. As 1 + c : 1 — c, where e is the eccentricity. 13. A body describes an ellipse under the action of a force to the focus 8', if if be the other focus show that the 844 MXAMPLES. velocity at any point P may be resolved into two velocities, respectively at right angles to 8P and HP, and each vary- ing as HP. 14. A body describes an ellipse under the action of a force in the centre : if the greatest velocity is three times the least, find the eccentricity of the ellipse. Ans. | \/2. 15. A body describes an ellipse under the action of a force in the centre : if the major axis is 20 feet and the greatest velocity 20 feet per second, find the periodic time. Ans. It seconds. 16. Find the attraction to the polo under which a par- ticle may describe an equiangular spiral. ^^^^^ 17. M P = ^ (5r» + 8c»), and a particle be projected from an apse at a distance c with the velocity firom infinity ; prove that the equation of the orbit ia r = |(c«'-e-«'). 18. M P = 2fi iX — ^, and the particle be projected from an apse at a distance a with velocity — , prove that it will be at a distance r after a time + y/i» — a* -f r Vh"^^)' !s«)SW!?wse?»«s»wiSMP«w»«*«^ i into two velocities, HP, and each vary- tder the action of a locity is tliree times llipse. Ana. | ■>/2. nder the action of a i is 20 feet and the id the periodic time. Ans. It seconds. under which a par- al. . » 1 Ana. P<^^' particle be projected relooity firom infinity ; particle be projected ocity —, prove that r Vr» — en' CHAPTER III. CONSTRAINED MOTION, 188. Definitioiis. — A particle is constrained in its mo- tion when it is compelled to move along a given fixed curve or surface. Thus far the subjects of motion have been particles not constrained by any geometric conditions, but free to move in such paths as are due to the action of tho impressed forces. We come now to the ctvse of tho motion of a particle which is constrained ; that is, in which the motion is subject, not only to given forces, but to unrieter- mined reactions. Such cases occur when the particle is in ft small tube, either smooth or rough, the bore of which is supposed to be of the same size as the particle ; or when a small ring slides on a curved wire, with or without friction ; or when a particle is fastened to a string, or moves on a given surface. If we substitute for tho curve or surface a force whoso intensitj'^ and direction are exactly equal to those of the reaction of the curve, the particle will describe the same path as before, and we may treat the problem as if the particle were free to move under the action of this system of forces, and therefore apply to it the general equa- tions of motion of a free particle. 189. Kinetic Energy or Vis Viva (Living Force), and Work. — A particle is constrained to move on a given smooth plane curve, under given forces in the plane of the curve, to determine the motion. Let APC be the curve along which the particle is com- l)elled to move when acted upon by any given forces. Let Ox and Oy be the rectangular axes in the plane of the ^i III! 346 KTNSTIC ENERGY. curve, ilie axis y positive up- wards, and {x, y) the place of the particle, P, at the time t ; let X, Y, parallel respectively to the axes of x and y, be the axial components of the forces, the mass of the particle being m ; let R bo the pressure between the cuvve and particle, which acta in the norma.1 to the curve, since it is smooth. Then the equations of motion are FI«.83 (IP as (1) (3) Multiplying (1) and (2) respectively by dx and dy, and atlAiag, we have „d^^y^fl = Xdx + Ydy. lutograting between the limits t and i^, and calling v^ the initial velocity, we have !J ,^ - 'I r,« = CiXlx + Ydy m The t«rm ^ ^ »« called the vis vivo/*, or Kinetic Energy of the mass w; that is, vis viva or kinetic energy is a quantity which varies as the product of the mass of the particle and the square of its velocity. There is particular advantage in def.ning vis viva, or kinotic ener gy, as half t IB smooth. Then ; (1) (3) y by dx and dy, and J, and calling v^ the + rrfy (3) , or Kinetic EniTgy kinetic energy is a of the mass of Ihe There is purlicnlar notic energy, a« half til, p. MB. KINETIC SNSR&r> 847 the product of the mass and the »quaro of its velocity.* The lirst member, therefore, of (3) is the via viva or kinetic energy of m acquired in its motion from (x,, y^) to {x, y) under the action of the given foi-ces. The terms Xdx and Ydy are the products of the axial compoL-nts of the forces by the axial displacements of the mass in the time dt, and are tlierefore, tlie elements of work done by the accelerating forces X and Y in the time dt, according to the definition of work given in Art 101, Eeni.; 80 that the second member of (3) expresses the work done by these forces through thfe spaces over which they moved the mass in the timo between t^ and /. This equation is called ihe equation of kinetic energy and of work; it shows that the work done by a force exerting action through a given distance, is equal to tl»o increase of kinetic energy which has accrued to the mas« in its motion through that distance. If in the motion, kinetic energy is lost, negative work is done by the force ; i. e., the work i;- -.orcd up as potential work in the mriss on which the foicp has acted. Thus, if work is spent on winding up a wulch, that work is stored, in the coiletl spring, and is thus potential and ready to be '•estored under adajjted circumstances. Also, if a weight is raised through a vertical distance, work is Bi)ent ui raising it, and that work may l)e recovered by lowering the weight through tlie same voitieal distance. Tliis theorem, in its most general form, is the modern principle of nmservation of eneryy ; and is made the funda- mental theorem of abstract dyiiamics as applied to natural philosophy. In this case we have an instance of spaee-intecirah, which, as wo have seen, gives uw kinetic energy and work ; tlio Holution (»f problems of kinetic energy and work will 1)6 oxplaiuod in Chap. V. • Si)ni(> wnUTH iltifliiK viR Ttvk a* llii> whiil. KM. 'jTtyiiffli'wwrontU Bfffl w SB ACTION OF TBS COMiTRATNTNO CUB VS. Now if X and Y are functions of the co-ordinates x and y the second member of (3) can be integrated ; let it be the differential of some function of x and y, as ^ (ar, y). Inte- grating (8) on this hypothesis, and supposing w and v^ to bb the velocities of the particle at the points (x, y) and (*«> Vit) corresponding to t and t^, "ve have "c v«») = <^ (^, y) - ^ («o. yo) (4) »^'i i-i'i which shows that the kinetic energy gained by the particle constrained to move, under the forces X, F, along any path whatever, from the point (a-^, y,) to the point {x, y), is entirely independent of the path pursued, and depends only upon the co-ordiniites of the points left and unived at; the reaction R does not appear, which is clearly as it should be, since it does no work, because it acts in a line perpendicular to the direction of motion. 190. To Find the Reaction of the Constraining Curve. — For conveniens, the muss oi the particle iniiy bo taken as unity. Multiplying (1) and (2) of Art. 189 by ^ and -V-; subtracting the former from the latter, and solving for R, we have, „ _ (Pydx — rf»x d y -^dy ydx ""- " ilPda "^ ds ' ds «^-. + ^% - Y%^'^^n^)<^^ ^rLm (1) in which p is the radius of curvature at the point P. The lu«t two terms of (1) are the nornml conipoiionts of the impressed fonrs; and (liorefore, if the purticlf were at rej^t, tlioy would donule I be whole prcsaure on the curve ; but Mto fO CUB vs. , co-ordiuates x and •ated ; let it be tho as ^ (x, y). lutc- posing V and v, to poiuts (x, y) and ve («o. yo) (4) led by the particle jes X, F, along j»,) to tho point )ath pursued, and the points left and ir, which is clearly icause it acts in a bion. ;he Constraining he particle nuiy bo 2) of Art 189 by in the latter, and at :;i) of Art. 102 (1) the point P. The •onipononts of the iirtich' Were at rest, on the curve ; but REACTION OF THE CONSTRAWirfO CURVE. 349 the particle being in motion, there ;.> an additional pressure on the curve expressed by -• In the above reasoning we have considered the particle to be on the concave side of the curve, and the resultant of X and Y to act towards the convex side along some line as PF so as to produce pressure against the curve. If ou the contrary, this resultant acts towards tho concave side, along PF' for example, then, whether the jiartiole be on tho concave or convex side, the pressure against the curve will .3 bo the diflEerence between - and tho normal resultant of X 9 and Y. 191. To Find the Point where the Particle WiU Leave the Constraining Curve. — It is evident that at that point. R = 0, as there will be no pressure against the curve. Therefore (1) of Art. 190 becomes - = Y~- X^ p ds ds = F' cos F'PR if F' be tho resultant of X and Y. .-. «)»= F'p con F'PR = 2F'-i chord of curvature in the direction PF'. Comparing this with (6) of Art. 140, we see that the particle will have the curve at the point where its velocity is such as timtild be produced by the resultant force then acting on it. if continued constant during its fall from rest through a space equal to J of the cnord of curvature parallel to that resultant. (See Tait and Steele's Dynamics of a Particle, p. 170.) m ■/miii i>u a»!aiM .i »t v i « 350 COySTRAlNSD UOTtOK. 192. Constrained Motion Undor the Action Gravity. — When gravity is the only force acting on thi particle, the formnlsB are simplified. Taking the axis of y vertical and positive downwards, the forces become X=0, and r= -{-^; and for the velocity we have, by (3) of Art 189, where y, is the initial space corresponding to the time t^. "^r the pressure on the curve we have, by (1) of Art. 190, p ^ da m If the origin be where the motion of the particle begins, the initial velocity and space are zero, and (1) becomes i«^==i?y. (8) This shows that the velocity of the particle at any time is entirely independent of the form of the cnrve on which it moves ; and depends fjolely on the perpendicular distance through which it falls. 193. Motion on a Circnlar Aro in a Vertical Plane. — Take the vertical dianiete' a? axis "of y, and its lower extremity m origin ; then the equation of the circle is 3f = 2ay~f; dT a-y X dy __ ds j4 ^■' "(PP ■ -^f^. a) ' the AotLon irce acting on thi ;ing the axis of y B8 become r* rt 189, I (1) g to the time t^. by (1) of Art. 100, (3) le particle begins, I (1) becomes (3) tide at any time le cnrve on which endicalar distance In a Vertical nxis'of f/y and its ion of the oirclo is (1) 1 MOTTON ON A CIXCULAR ABC Let (ife, A) be the point K where the particle starts from rest, and (x, y) the point P where it is at the time /. Then the particle will have fallen through the height HM — h — y, and hence from (3) of Art. 492 we have 351 (8) Hence the velocity is a minimum when y = h, and a maximum when y = 0; and this maximum velocity will carry the particle through to iT' at the distance h above the horizontal line through 0. To find the time occupied by the particle in its descent from K to the lowest point, 0, we have from (2) dt da Vajr (* - y) — ady Vfi9{h-y)(iay-f) by(i) (3) the negative sign being taken rinoe / is a decreasing func- tion of 8. This expression does not admit of integration ; it may be reduced to an elliptic integral of the first kind, and tables are given of the approximate values of the integral for given values of y.* If, however, the radius of the circle is large, and the greatest distance KO, over which the particle moves, is Biurtll, wo may develoiw (3) into a series of teruis in ascend- ing powers of | , and thus find the integral approximately. • See Leuondre i Twit* des, PoncU: the j article to move on a curve, wo may imagino it r,aa- pended by a string of invariable length, or a thin rod considered of no weight, and moving iu a vertical plane ftlmnt Mil' point 0: for, whether the force acting on the jmrticle be the reaction af the curve or the tension of the string, its intensity is the Hame, while its direction, in either caae is along the uormai Lu the curve. MM if. icle from K to K', : h again, then (3) •] dy ^hy — y» ' •4/ W ] (4) le time of moving of the verticiil to this is called an "ol. in., p. 588). nail in comparison will be very small, (5) toad of supposing ly imagine it ^ias- th, or a thin rod 1 a vertical piano rce acting on the the tension of the e its dtrection, in •ve. S^^^m^^^^^E^B^Sn RSLATION OF TIME, LBNOTH, STC. 353 When the particle is supposed to be snspended by a thread without weight, it bocomes what is termed a simple pendulum ; and although such an instrument can never be jjerfectly attained, but exists only in theory, yet approxima- tions may be made to it sufficiently near for practical pur- poses, and by means of Dynamics we may reduce the calculation of the motion of such a pendulum to that of the simple pendulum. If I is the length of the rod, the time of an oscillation is approximately given by the formula = ''v/l 0) when site angle er of vibratioacs performed dur- ing ^seconds, and iTthe time of one vibration, then = ^byetermiiiecl with s carried to the top hi of the mountain illation. dt red spherical , h rface ; I the length of gravity ou the intiiiu respectively. SB^HssMSssw^rfj&a^ss^caB mmm BEI6BT or MOUXTAIlf DMTKRMINED. 9 g'~\ r /' 9 a' - ^ • y - (r -f hy* 355 (1) which is the force of gravity at the top of the mountain. Let n — the number of oscillations which the seconds pendulum at the top of the mountain makes in 24 houi-s ; 24 X 60 X 60 then the time of osciUatioo (1) of Ai-t. 195, we have 24 X 60 X 60 Hence from h r 24 X 60 X 60 n -1, (since Tr'Y/- = l), (2) which gives the height of the mountain in terms of the radius of the earth. For the sake of an example, suppose the pendulum to lose 5 seconds 1n a day ; that is, to make '< oscillations less than it would make on the surface of the earth. Then » = 24 X 60 X 60 — 5 ; ii in (2) gives A = ('- 24 X 00 X _U X 60 X 60 24 X 60 X 00 - I -1 h = 4000 24 X 60 X 12 = J mile, nearly, nearly ; »4 X 6») X 12 r being 4000 miles (approximately). 197. The Depth of a Mine Oetenniued by Ob- serving the Change of Oscillation in a Seconds Pendulum.— [jf I r ho the radius of the earth ae in the 356 CENTRIPETAL FORCE. last case ; h the depth of the mine ; g and g' the values of ' gravity on the earth's surface and at the bottom of tlio mine. Then (Art. 171) we have g' r~h (1) Let n = the number of oscillations which the S'econds pendulum at the bottom of the mine makes in 24 hours. Then 24 X 60 X 60 n _ / Ir r — h' r ~ V24 X 60 X 60/ ' from which h can be found. If, as before, the pendulum loses 5 seconds a day, we have * = '-(■ 24 __j y X 60 X 12/ nearly. 12 X 60 X 12 . • . h = ^ mile nearly. (See Price's Anal. Mech's, Vol. I, p. 590, also Pratt's Mech's, p. 376.) 198. Centripetal and Centrifugal Forces. ;iuce the pressure — , at any point, depends entirely upon tlie velocity at that point and the radius of curvature, it would remain the same if the forces X and Y wer^ *K)th zero, in which case it would be the whole norma pressure, B, I g' the values of lie bottom of tlu; (1) hich the seconds es in 34 hours. -h) re, the penduhim 3 _, rJ 590, also Pratt's Forces. — Face jntirely upon tlie rvfjture, it would yer« fwth zero, in mil; pressure, /?. ,,«,, r.'»>mmimmmKgmmtmmmKt&iKKmm^ CBN'^ rrVQAL FORCS. 867 against the curve. It is easily seen, therefore, that this pressure arises entirely from the inertia of the moving particle, i. e., from it- tendency at any point, to move in the direction of a tangent ; and this tendency to motion along the tangent uecessurily causo-- it to exert a pre iire against the deflecting curve, and wluJi requires the curve 4 to oppose the resistance -• Hence, since tl- particle if left to itself, or if left to the action of a force uong the tan- gent, would, by the law of inertia, continue to niovi' along that tangent, - is the effect of the force which deflects the particle from its otherwise rectilinear path, and draws it towards the centre of curvature. This force is called the Centripetal Force, which, therefore, may be defined to be the force which deflects a particle from its otherwise recti- linear path. The efjual and opposite reaction exerted away from the centre is called thi' Centrifugal Force, which may be defined to be the resistance which the inertia of a particle in motion opposes to whatever deflects it from its rectilinear path. Centri[)etal and centrifugal are tlei-efore the same quantity under different aspects. The action of the former is towards the centre of curvature, while that of the latter wfron the centre of curvature. The two are called central forces. They determine the direction of motion of the par- ticle but do not affect the velocity, since they aet continu- ally at right angles to its puth. If a particle, attached to a string, be whirled about a centre, the intensity of these central forces is measured by the tension of the string. If the string be cut, the particle will move along a tangent to the curvo with unchanged velocity. Cob. 1.— If m be the mass moving with velocity v, its centrifugal force is »i -• If w bo the angular velocity ^f^i^ifv^" ,'WrTr^ 'iWS'^J--- *#«**i«»(i»rtt«|Sft^gw^i. MiHm IMAGE EVALUATION TEST TARGET (MT-3) A^^. /. y. «(S z= AOP = the latitude of P; J? the radius, AC, of the earth ; and I^ the radius PD of the parallel of lati- tude passing through P. Th«m we have i2' :^ i2 cos . Let the centrifagal force at tha point P, whidi is exerted in tho direction of the radius DP, be repwier.ted by the lino PB. Resolve this into the two components Pf, act- ing along the tangent, and PE, acting along the normal. Then by (2) ol Art 198 we have s n|.« (1) , by (1). w 360 CSITTRtmjftAL PORCJS. Hence, the centrifugal force at any point on (he earth's surface varies directly as the cosine of the latitude of the place. For the normal component we have PB = PB cos ^ i-n*R co8» . ... = jiT— ^7 (2) = / co8» ^, by (1) of Art 197. (3) Honce^ the component of the centrifugal force which directly opposes the force of gravity, at any point on the earth's sur- face, is equal to the centrifugal force at the equator, mul- tiplied by the square of the cosine of the latitude of the place. Also PF = PB Bin — ^^^ **'" cos ^ . by (2) .kl . = J sin 20. by (1) of Art. 197 ; (4) that is, the component of the centrifugal force which tends to draw particles from any parallel of laltfnde, P, towards the equator, and to cause the earth to assume the figure of an oblate spheroid, varies as the sine of twice the latitude. The preceding calculation is made on the hypothesis that the earth is a perfect sphere, whereas it is an oblate spheroid ; and the attraction of the eaith on particles at its surface decreases as we pass from the poles to the equator. The t>endalum furnishes the most accurate m m'nt on (he earth's ' the latitude of the by (2) (1) of Alt 197. (3) force which directly ! on the earth's sur- it the equator, mul- the latitude of the (l)of Art. 197;(4) I force which tends lattinde, P, towards } assume the figure sine of twice ike the hypothesis that as it is an oblate arth on particles at 1 the poles to the the most accnrate TBJl CONICAL PENDULUX. 861 na-N method of determining the force of gravity at different places on the earth's surface. 201. The Conical Pendn- lum.— The Gtovemor. — Suppose a particle, P, of mass m, to be at- tached to one end of a string of length /, the other end of which is fixed at A. The particle is made to describe a horizontal circle of radius PO, with uniform velocity rouml the vertical axis A0,&o that it makes n revolutions per second. It is required to find the inclina- tion, 0, of the string to the vertical, and the tension of *he string. The velocity of P in feet per second = 2rrn • OP = 2ir« I sin e. The forces acting upon it are the tension, 7, of the string, the weight m, of the particle, and the centrifugal force, m — ^j— ^ — (Art 198). Hence resolving, we have for horizontal forces, T* sin (9 = »j. 4n»»»» / sin e ; (1) for vertical forces, iTcos© = mg. (%) From (1) 7'=m.4r.V?, (3) which in (2) givea whore 7* and 9 ere completely determined. If the string be replaced by u rigid rod, which can turn about ^ in a ball and socket joint, the instrument is called a conical pendulum, and occurs in the governor of the steam-engine. 16 d6d MXAMrXtMB. EXAMPLES. 1. If the length of the seconds pendalam be 39.1393 inches in London, find the valne of jr to three places of decimals. Ana. 82.191 feet. 2. In what time wiU a pendnlum nbratc whose length is 15 inches ? Ans. 0.63 sec. nearly. 3. In T.hat time will a pendnlnm vibrate, whose length is double that of a seconds pendulum P Ang. 1.41 sees. 4. How many vibrations will a pendalam 3 feet long make in a minute ? Ant. 62.65. 5. A pendulum which beats seconds, is taken to the top of a mountain one mile high ; it is required to find the number of seconds which it wiP lose in 12 hours, allowing the radius of the earth to be 4000 miles. Ans. 10.8 sees. 6. What is the length of a pendulum to beat seconds at the place where a body falls 16^ ft in The first second 7 Ann. 39.11 ins. r early. 7. If 39.11 ins. be taken as the length of the seconds pendulum, how long must a pendulum be to beat 10 times in a minute ? Ans. 117^ ft 8. A particle slides down the arc of a circle to the lowest point ; find the velocity at the lowest point, if the angle described round the centre is 60°. Ana. Vgr. 9. A pendulum which oscillates in a second at one place, is carried to another place where it makes 120 more oscil- lations in a day; compare the force of gravity at the latter place with that at the former. Ana, (|fH)** 10. Find the number of vibrations, n, which a pendulum will gain in Jf seconds by shortening the length of the peudulum. m ilam 1)6 39.1393 ) three places of 1. 32.191 feeL whose length is 63 sec. nearly. , whoso length is [ng. 1.41 sees. urn 3 feet long Ans. 62.66. aken to the top ired to find the honre, allowing Ins. 10.8 sees. beat seconds at first Hecond ? II ins, r«arly. I of the seconds bo beat 10 times Ans. m^it a circle to the est point, if the Ana. Vffr. •nd at one place, 120 more oscil- ity at the latter ^»*. (IIH)'. _ ioh a pendulnm le length of the Let the length, i, be decreased by a small qnantity, 1 1, and let n bo increased by »( ; then from (2) of Art 196 we get which, divided by (2) of Art 195, gives Ilenoe «, = — 1. 11. If a pendulnm be 45 inches long, ho«r many vibra- tions will it gain in one day if the bob* be screwed up one turn, the screw having 32 threads to the inch P Ans. 30. 13. If a clock loses two minutes a day, how many turns to the right hand must we give the nut in order to correct its error, supposing the screw to have 50 threads to the inch? • Ans, 5*4 tarns. 13. A mean solar day contains 24 hours, 3 minates, 56 '5 seconds, sidereal time; calculat«d the length of the pendulum of a clock beating sidereal seconds in London. See Ex. 1. Ans. 38-925 inches. 14. A heavy ball, scspended by a fine wire, vibrates in a small arc ; 48 vibrations are counted in 3 minutes. Cal- culate the length of the wire. Ans. 45 -87 feet 15. The height of the cupola of St Paul's, above the 6oor, is 340 ft; calculate the number of vibmtions a heavy body would make in half an hour, if suspended from the dome by a fine wire which reaches to within 6 inches of the floor, Ans. 176-4. • The kwer extremity ot the pendnhua. 'I :!, .■I :, wm 16. A seconds pendulnm is carried to the top of a mountain m milea high ; aasuming that the force of gravity varies inversely as the square of the distance from the centre of the earth, find the time of an oscillation. /4000 + m\ 17. Prove that the lengths of pendulums vibrating dar- ing the same time at the same place are inversely as the squares of the number of oscillations. 18. In a series of experiments made at Harton coal-pit, a pendulum which beat seconds at the surface, gained 2| beats in a day at a depth of 12C0 ft. ; if g and g' be tho force of gravity at the surface and at the depth mentioned, show that 9'-9 _ 1 • "- itto'o- 19. A pendulum is found to make 640 vibrations at the equator in the same time that it makes 641 at Greenwich; if a string hanging vertically can just sustain 80 lbs. at Greenwich, how many lbs. can the same string sustain at tho equator ? Am. 80J lbs. about. 20. Find the time of descent of a particle down the arc of a cycloid, the axis of the cycloid being vertical and vertex downward ; and show that the time of descent to the lowest point is the same whatever point of the curve the particle starts from. /^ \ 9 21. If in Ex. 20 the particle begins to move from the extremity of the baae of the cycloid find the prtssure at the lowest point of the curve. Ans. 2^; i. e., the pressure ia twice the weight of tho particle. to the top of a bat the force of the distance from m oscillation. tOOQ + m\ 4000 ) Bees. ms vibrating dnr- ft inversely as the Harton coal-pit, a iirface, gained 2| if g and g' be the depth mentioned, Tibrations at the 541 at Greenwich ; sustain 80 lbs. at string snetain at 80^ lbs. about. tide down the arc irertical and vertex Micnt to the lowest curve the particle Am. Wy to move from the he pressure at the he weight mm m m smmmmmmm HXAMPLSa. d65 %%. Find the prepare on the lowest point of the carve in Art 193, (1) when the particle starts from rest at the highest point. A, (Fig. 84), (ii) when it starts from rest at the point B. Ana. (1) 5^; (2) 3^; i.e., (1) the pressure is five times the weight of the particle and (2) it is three times the weight of the particle. 23. In the simple pendulum find the point at which the tension on the string is the same as when the particle bangs at rest Am. y = f.'., where h is the height from which the pendulum has fallen. 24. If a particle be compelled to move in a circle with a velocity of 300 yards per minute, the radius of the circle being 16 ft, find the centrifugal forco. Ana. 14' 06 ft. per sec. 25. If a body, weighing 17 tons, move on the circum- ference of a circlo, whose radius is 1110 ft., with a velocity of 16 ft per sec., find the centrifugal force in tons (take g = 32-1948). Am. 0-1217 ton. 26. If a l)ody, weighing 1000 lbs., be constrained to move in a circle, whose radius is 100 ft, by means of a string capable of sustaining a strain not exceeding 450 lbs., find the velocity at the moment the string breaks. Ana. 38-06 ft per sec. 27. If a railway carriage, weighing 7-21 tons, moving at the rate of 30 miles per hour, describe a portion of a circle whose radius is 460 yards, find its centrifugal force in tons. Am. 0-314 ton. 28. If the centrifugal force, in a circle of 100 ft radius, be 146 ft per sec, find the periodic time. Ana. 6-2 sees. 4^1 ; 366 XXAMPnSA 29. If the oentrifa^l force be 131 oss., and the radias of the circle 100 ft., the periodic time being one hour, find the weight of the body. Ans. 386- 309 tons. 30. Find the force towards the centre required to make a body move uniformly in a circle whose mdius is 5 ft., with Huch a velocity as to complete a revolution in 6 sees. 5 ■ Ans. 31. A stone of one lb. weight is whirled round horizon- tally by a string two yards long having one end fixed ; find the time of revolution when the tension of the string is 3 lbs. Ana. 2rT VI sees. 38. A weight, w, is placed on a horieontal bar, OA, which is made to revolve round a vertical axis at 0, with the angular velocity «; it is required to determine the position, A, of the weight, when it is upon the point of sliding, the coefficient of friction being /. Am. OA 33, Find the diminution of gravity at the Sun's equator caused by the centrifugal force, the radius of the Sun being 441000 miles, and the time of levolation on his axis being 607 h. 48 m. Ans. 0- 0192 ft. per 8e<;. 84. Find the centrifugal force at the equator of Mercury, the radius being 1670 miles, and the time of revolution 24 h. 6 m. Ans. 0- 0435 ft. per sec 35. Find the centrifugaJ force at the equator, (1) of Venus, radius being 3900 miles and time of revolution 23 h. 21 m., (2) of Mars, radius being 2050 miles and iwriodic time 34 h. 37 m., (3) of Jupiter, radius being 43500 miles and periodic time 9 h. 56 m., and (4) of Saturn, radius being 39580 miles and periodic time 10 h. 29 m. (8., and tbe radias ;ing one hour, find s. 386-309 tons. required to make 086 rndiuB is 5 ft., olution in 6 sees. Ana. -=-' ed round horizon- ne end fixed ; find ! the string is 3 lbs. s. 2n VI sees. rieontal bar, OA, cal axis at 0, with to determine the ipou tbe point of Ans. OA the Sun's equator 8 of the Sun being 1 on his axis being Qin ft. per 8e<;. jnator of Mercury, ime of revolution 0435 ft. per sec le equator, (1) of mo of revolution r 2050 miles and iter, radius being and (4) of Saturn, le 10 h. 29 m. mmmimmmmmmsmmmim MXAMPLXa. 867 Ana. (1) 0> 11504 ft. per sec.; (2) 0>0544 ft per sec.; (3) 7-0907 ft per 8oa; (4) 6- 7924 ft per sea 36. Find the effect of centrifngal force in diminishing gravity in the latitude of 60°. [See (3) of Art. 200). Ana. 0-028 ft. per sec. 37. Find (1) the diminution of gravity caused by cen- trifugal force, and (2) the component which urges particles towards the equator, at the latitude uf 23°. Ans. (1) 0-09 ft per sec; (2) 0-04 ft per sec. 38. A railway carriage, weighing 12 tons, is moving along a circle of radius 720 yards, at the rate of 32 miles an hour; find the horizontal pressure on the nils. Ana. 0-39 ton, nearly. 39. A railway train is going ouoothly along a curve of 500 yards radius at the rate of 30 miles an hour ; find at what angle a plumb-line hanging in one of the carriages will be inclined to the vertical Am. 2° 14' nearly. 40. The attractive force of a mountain horizontally is/, and the force of gravity is ^; show that the time of vibra- tion of a pendulum will be ww j - y^ ; of the pendulum. 41 1 In motion of a particle down a cycloid prove that the vertical Telocity is greatest when it has completed half its vertical descent 42. When a partide falls from the highest to the lowest point of a cycloid show that it describes half the path in two-thirds of the time. 43. A railway train is moving smoothly along a curve at the rate of 60 miles an hoar, and in one of the carriages a pendulum, which would ordinarily oscillate seconds, is observed to oscillate 121 times in two minutes. Show that the radiOB of tiia ourve is very nearly a qiiarter of a mile. a being the length 368 EXAMPLES. 44. One end of a string is fixed ; to the other end a particle is attached which describes a horissontal circle with uniform velocity so that the string is always inclined at an angle of 60" to the vertical ; show that the velocity of the particle is that which would be acquired in falling freely from rest through a space equal to three-fourths of the length of the string. ■15. The horizontal attraction of a mountain on a particle at a cciiain place is such as would produce in it an accelera- tion denotfld by "• Show that a seconds pendulum at that , .,, . 21600. , . , , place will gam — j— beats ma da>, very nearly. 46. In Art 201, suppose I equal to 2 ft. and m to be 20 lbs., and that the system makes 10 revolutions per sec., and ^ = 32; find e and T. Ans. e — co8-» — ^; T = SOOtt* pounds. 47. A tube, bent into the form of a plane curve, revolves with a given angular velocity, about its vertical axis ; it is required to determine the form of the tube, when a heavy particle placed in it remains at rest in all parts of the tube. (Take the vertical axis for the axis of y, and the axis of x horizontal, and let e whole process is probably finished within a thousandth of 11 socond.* The impulsive forces are bo much more intense than the • TbomMD tud TtliV Nkt. Pbtl., p. fti. SdiP 9n DIBBCT AND CJSNTRAL IMPACT. ordinary forces, that daring the brief time in which the former act^ an ordiuar}' force does not produce an effect comparable in amount with that produced by an impulsive force. For example, an impulsive force might generate a velocity of 1000 in less time than one-U^nth of a second, while gravity in one-tenth of a second would generate a - elocity of about three. Hence, in dealing with the effects of impulses. Unite forces need not be conaidei'ed. 204. Direct and Central Impact— When two bodies impinge on oach other, so that their centres before impact are movic^? in the same straight line, and the common tan- gent at the point of contact is perpendicular to the line of motion, the impact is said to be direct and central. When these conditions are not fulfilled, the impact is said to be obligue. When two bodies impinge directly, one upon the other, the mutual action between them, at any instant, must be in the line joining tbeir centres ; and by the third law (A?*t. 166), it must be equal in amount on the two bodies. Hence, by Law II, they must experience equal changes of motion iu contrary directions. We may consider the impact as consisting of two parts ; during the first part the bodies are coming into closer con- tact with each other, mntvally displacing the particles in the vicinity of the point of contact, producing a compres- sion and distortion about that noint, which increases till it reaches a maximum, when the molecular reactions, thns called into play, are sntfioient to resist farther compression and distortion. At this iusiant it is evident that the points in contact are moving with the same velocity. No body in nature is perfectly itietantic ; and henoe, at the instant of greatest compression, the elastic force* of resti- tution are brought into action ; and during -the second part of the impact the mutual pressure, produced by the elastic forcoB, which were brought into action by the compression «■« ACT. nc in which the produce an cfFect by an impnlsive might generate a nth of a second, vonid generate a % with the effects dei-ed. -When two bodies es before impact the common tan- lar to the Hne of I central. When uct is said to be e upon the other, instant, must be by the third law I tlie two bodies, equal changes of ng of two i»rt« ; f into closer con- the particles in icing a compres- b increases till it T reactions, thus ■ther compression evident th&t the me velocity. No nd henoe, at the c fortm of resti- gthe second part led by the elastic the compression SLASTICJTy OF BODIMS. mmmsmamn mmi j. 373 during the first part of the impact, tend to separate the two bodies, and to restore them to their original form. 205. Elasticity of Bodiea.-^o«Aci«at of Resti- tution. — It appears from experiment that bodies may be compressed in various degrees, and recover more or less their original forms after the compreMing force has ceased ; this property is termed ekiHticity. The force urging the approach of bodies is called the force of compression ; the force causing the bodies to separate again is called the force of reditution. Elastic bodies are such as regain a part or all of their original form when the compressing force is removed. The ratio of the force of restitution to that of compression is called the Coefficimt of JiestituHon.* It has been found that this ratio, in the, same bodies, is constant whatever may be their velocities. When this ratio is unity the two forces ore equal, and the body is said to lie perfectly elaeticj when the ratio is aero, or the force of restitution is nothing, the body is said to bo non-elastic; when the ratio is greater than zero and less than unity, the body is said to be imperfectly elastic. There are no bodies either perfectly elastic or perfectly non-elas- tic, all being more or less elastic. In the cases discussed the bodies will be supposed ppher- ical, and in the case of direct impact of smooth spheres it is evident that they may be considered as particles, since they are symmetrical with respect to the line joining their centres. The theory of the impact of bodiea ia chiefly due to Newton, who found, in his experiments, that, provided the impact is not so violent as to make any seuHiblo iidenttition in eitiier body, the relative velocity of separation after the impact bears a ratio to the relative velocity of approach before the impact, which is constant for the same two • BoneUmea called CmIIIcIsd( of IhMtlcitr. To <" -"')• p Similarly from (8) wo have M *'•'=•»'' ^ft:f'(i + ')(*'-»^)5 (3) (*) which are the velocities of the balls when finally eeparated These results may be more easily obtained by the oon- sidorution that the whole impulse is (1 + e) R; for this gives at once the whole momentum lost by IT or gained by M' dnring oomprMnon and restitution as follows : M(v - V,) = (1 -f- «) J?, (6) ■\?ia»»-'ijMWW«Bi*w«wi»»»ww» 'w iiii i i w'» i M n w ^ tt tendency which il fonn, begins to lomentum that it reflsure is propor- ; then daring the eR, acta on each 18 B acted daring i velocities of the irated. Then we ■ v.) = en, (1) - F) = eB. (2) -V) d (5) of Art 306, "'). (3) t^h (*) uiUy eeparakd. ined by the oon- +• «) -S ; for this Mot gained by Pollows : and DISKCT IMPJLOT Of IttBhASTIO BODTKS. 377 M'{v,'^v')=z{l + e)Jt. (6) Substituting in (5) and (6) the value of It from (5) of Art. 206, we have the values of v and v,' immediately. Cob. 1. — If the balls are moving in opposite directions* «' becomes negative. If the balls are non-elastic, e = 0, and (3) rad (4) reduce to (4) of Art 206, as they should. Cob. 2.— If the balls are perfectly elaatie, « = 1, and (3) and (4) become ^^' = V+-J^,(v-v'). (8) M+ M Cob. 3,— Subtracting (4) from (3) and reducing, we get », — »,' = r ~ t>— (1 + e) {v r- v'), Hence, tks relative velocity after impact «« — e times the relative velocity before impact. Cob. 4.— Multiplying (8) and (4) by M and M', respeet- ively, and adding, we get Mv, + M'v,' =r ift? ^. M'v'. (10) Hence, as in Art. 206, the algebraic turn of the mometUa after impact ie the same at before; i. e., there i» no mo- mentum lost, which of conne is a direet oonseqnence of the third law of motion (Art 169). Cob. 6. — Suppose «* = 0, so ^t the body ot mass M, moving witii velocity v, impinges on a.tody of mass M' at' rest, th$n (3) and (4) become 878 LOSB OP KINSTW BNXROT. M-eM' . , i/ (1 + «) V, and t», = jy^jyrt'. (11) Hence the body wJiich is struck goes onwards ; and the striking body goes onwai-ds, or stops, or goes backwards, according as Jif is gfreater than, equal to, or less than eM'. If M' = eM, then (11) becomes V, = (1 — e) V, and »,' = v. (12) OOB. 6.— If M = M and e = 1 ; that is, if the balls are of equal mass, and perfectly elastic,* then (7) and (8) become, respectively, Vx = »', and < = t>; (13) that is, the balls interchange their velocities, and the motion is the same as if they had passed through one another without exerting any mutual action whatever. Cob. 7.— If M' be inflnite, and v' = 0, we have the case of a ball impinging directly upon a fixed surface ; substi- tnting these values in (3) it becomes Vt= —w, (14) that is, the baU rebounds from the fixed surface unth a veloc- ity e times that with which it impinged. 208. Lou of Kinetic Bnexgyf in the Zmpaet of Bodies.— Squaring (9) of Art 207, and multiplying it by MM', we have MM' (r, - »,')» = MM' ^(v-i/y = MM'{v- v'y -(!-«») MM' (V - »')». (1) • ThU it the nioal phnMoiogy, bat iiiialaadiiiK, Incy. Brit, Vid. XV, Art. Xceh'f. t 8m Art. 189. kmf oth Spheres.— •ections and with impulse and the i their centres at mpact): GA and iging sphere, M, I'B' those of the CD and A'C'D, ike with the line D'D, which their are smooth, the ilace in the line t. Let R be the ion. Resolve all at right angles impact, and the ne way as if the Telocities in the I j3, «,' cos P", we 207, i—i/ COB a'), (1) SXAMPLSa. M 383 v,'co8/}' = v'oosa'+^-^y, (l-i-e) («oo8«-r'co8 «'), (a) which are the final velocities c/the two spheres along the line of impact ED. Also, from (6) of Art. 306, we obtain by substitution, ~ W+l^' '" °°'' ** — *' ''OS a'), (8) (See Tait and Steele's Dynamics of a Particle, p. 323.) Cob. 1.— Multiplyin/j (1) by M, and (8) by M', and add- ing we get Mv, cos ^ + M'vt' cos i3' = Mv cos a + M'v^' cos «', (4) which shows that the momentum of the systtm resolved along the Urn of impact is the same after impact as before. Cor. 2.— Subtracting (2) from (1) we obtain, t>, COS ^ — r/ cos /3' = — e (» cos a — v' cos «'). (6) That is, the relative velocity, resolved along the line of impact, after impact is — e times its value before. EXAMPLES. 1. A body* weighting 3- lbs. moving with a velocity of 10 ft. per second, impinges on a body weighing 2 lbs., and moving with a velocity of 8 ft per second ; find the com- mon velocity after impact Ans. 7\ ft per second. 2. A body weighing 7 lbs. moving ll ft per second, impinges on another at rest weighing 15 lbs.; find the com- mon velocity after impact Ans. 3^ ft per second. • The liodiM are inelaaUc aalen otherwiM itatcd. The flnt W ezamplee are in Mnet Impact. 384 EXAMPLES, hiu} \ 3. A body weighing 4 lbs. moving 9 ft per second, impinges on another body weighing 2 lbs- and moving in the opposite direction with a velocity of 5 .ft. per second ; find the common velocity after impact. Ana. 4^ ft. per second. 4. A body, M', weighing 6 lbs. moving 7 ft per second, is impinged upon by a body, M, weighing 6 lbs. and mov- ing in the same direction; after impact the velocity of M' is doubled : fi ad the velocity of M before impact An$. 19f ft. per second. 6. Two bodiea, weighing 2 ll»., and 4 lbs., and moving in the same direction with the velocities of 6 and 9 ft. respec- tively, impinge upon each other; find their common velocity after impact Ans. S ft per second. 6. A weight of 2 lbs., moving with a velocity of 20 ft per second, overtakes one of 6 lbs., moving with a velocity of 5 ft per second ; find the common velocity after impact. Ana. 94 ft per second. 7. If the same bodies met with the same velocities find the common velocity after impact Ans. 24 ft per second iu the direction of the first 8. Two bodies of different mtiSRcs, are moving towards each other, itnth velocities of 10 ft and 12 ft. per second respectively, and continue to move after impact with a velocity of !• 2 ft per second in the direction of the greater; compare their masses. Arts. As 3 to 2. 9. A body impinges on another of twice its mass at rest: ■how that the impinging body lo, -s two-thirds of its velocity by the impact 10. Two bodies of unequal masses 1 viag in opposite directions with momenta numerically equal meet; show that the momenta are numerically equal after impact. Mai ii J ft per second, 3. aud moving in 5 .ft. per second; ft per second. 7 ft. per second, g 6 lbs. and mov- lie velocity of M' mpact ft. per second. s., and moving in and 9 ft. respec- their common ft per second. velocity of 20 ft : with a velocity nty after impact ft per second. te velocities find on of the first. moving towards 2 ft. per second ' impact with a n of the greater; \n3. As 3 to 2. its mass at rest: ro-tbirds of its ring in opposite lal meet; show her impaot EXAMPLS& 11. A bodr, M, weighing 10 lbs. moving 8 £1. per second, impivjges on M', weighing 6 lbs. and moving in the same direction 5 ft. per se^wnd ; find tbeir velocities after impact, supposing 6 = 1. Am. Velocity of if = 6| ; velocity otM' = 8f. 12. A body, M, weighing 4 lbs. moving 6 ft per neoond, meets M' weighing 8 lbs. and moving 4 ft per second; find their velocities after impact, « = 1. Ans. Each body is reflected back, M with a velocity of 7| andif ' with a velocity of 2|. 13. Two balls, of 4 and 6 lbs. weight, impinge on each other when moving in the same direction with velocities of 9 and 10 ft respectively ; find their velocities after impact, supposing e = ^. Am. 10-08 and 9-28. 14. Find the kinetic energy lost by impact in example 5. Ans. f^. IL Two bcJies weighing 40 and 60 lbs. and moving in the same direction with velocities of 16 and 26 ft respec- tively, impinge on each other: find the loss of kinetic energy by impact. Ana. 37 -3. 10. An arrow shot from a bow starts off with a velocity of 120 ft per second; with what velocity will an arrow twice as heavy leave the bow, if sent off with three times the force ? Ana. 180 ft per second. 17. Two balls, weighing 8 ozs. and 6 ozs. respectively, arc feimultanoously projected upwards, the former rises to a height of 324 ft. and the latter to 266 ft ; compare the forces of projection. Ana. As 3 to 2. 18. A freight train, weighing 200 tons, and traveling 20 miles per hr. runs into a passenger train of 60 tons, stand- ing on the same track; find the velocity at which the remains of the passenger train will be prcpelleO along the tnwk, snpposing « = f Aiu. 19- 3 miles per hr. 386 ^^J[AMFL£8. 19. Thejie is a row of ten perfectly elastic bodies whose masses increase geometrically by the constant ratio 3, and the first impinges on t second with the velocity of 6 ft per second ; find the velocity of the last body. Ana. -yf , ft per second. 20. A body weighing 5 lbs. moving with a velocity of 14 ft. per second, impinges on a body weighing 3 lbs., and moving with a velocity of 8 ft. per second; find the veloci- ties after impact supposing e = \. Ana. 11 and 13. 21. Two bodies are moving in the same direction with the velocities 7 and 6 : and after impact their velocities are 6 and 6; find c, and the ratio of their masses. Ans. e = 4; J/' = 2M. 22. A body weighing two lbs. impinges on a body weigh- ing one lb.; e is ^, show that v, = r ^- v', and that v,' = v. 23. Two bodies moving with numerically equal velocities in opposite dirv'KJtious, impinge on each oth r; the«result is that one of them turns hack with its original velciity, and the othgr follows it with half that velocity; show that one body is four times as heavy as the other, and tluit e = J. 24. A strikes B, which is at reist, and after impact the velocities are Dumerically equal ; if r be the ratio of B'g 2 ma9s to A'b mass, show that e is 7, and that B's mass r — 1 is at least three times A's mass. 25. A body impinges on an equal body at rest ; show that the kinetic energy before impact caunot be greater than twice the kinetic energy after im^Mct 26. A beries of perfectly elastic balls are arranged in the same straight line ; one of them impinges on the next, then this on the noxt and so on ; show that if their masses form a geometric progression of which the oommou ratio atic bodies whose tant ratio 3, and the velocity of wt body, ft per second. 1 a velocity of 14 ;hiiig 3 lbs., and ; find the velocl- [ns. 11 and 13. le direction with !t their velocities masses. 4; J/' = 2M. on a body weigh- and that v/ = v. y equal velocities n:v; theaoe through which it moves the body (Art 101, BciD.). Thus, the work done in lifting a weight throagh a ver- tical distance is proportional to the weight lifted and t)ie vertiukl distance through which is is lifted. The unit of work used in England imd in this country «'« thtU which is required to overcome the weight of a pound through the vertical height of a foot, and is called a foot-pound. For instance, if a weight of 10 lbs. is raised to a height of 5 ft., or 5 lbs. raised to a height of 10 ft, 50 foot-ponnds of work must have been expended in overcoming the resist- ance of gravity. Similarly, if it requires a force of 50 lbs. to move a load on a horizontal plane over a distance of 100 ft., 5000 foot-pounds of work must have been done. If a carpenter urges forward a plane through 3 ft. with a force of 13 Ibe., he does 36 foot-pounds of work ; or, if a weight of 1 lbs. descends through 10 ft., gravity does 70 foot-pounds of work on it, Her^e, the number of units of work, or foot-pounds, ne gethcr make up the finite (lispliicemeut, we obtain the whole work done by the force during such finite displacement. TliuB let a force, P, tot a( a p<^t, O, In the direction OP (Fig. ffO), and let us Buppose tlie point, O, to move into any other poaitiun, A, very near 0. If Iw tiie angl'^ betwwn the direction, OP, of tlie I'orce and tlie direction, OA, of the diiplaoement of tlie point of appli- cation, then the product, POA ocmH,\B called the work done by the forco. If we drop a porp«>ndlcalar, AN, on OP, the work done l>y the force i8 alio equal to the pruduet P ■ ON, where ON ii to lie eati- :jiia"^:iPMMiiJiifly4&Myi: iCS. MEABURE OF WORK. 391 oves on any smooth which the Burfaoe icient to constitute orts a load without I which that term is jolumn does which t. opposite to that in it does work u[)on attraction is nega- a is negative, /. c, tiie direction oppo- , this is frequently ) agaim,. the force, he force Jifting the M by a Force.— of a force varies, or he force during any 8 in Art. ail. In lefinitely small dis- the magnitude and B displacfment, and len taking the sum ag the consecutive [lake up the finite done by the force direction OP (Fig. SO). vay other poritiun, A, direction, OP, of tlie 1 "f tJie point of appli- the work done by the the work done liy the lere ON is to l>e wti- mated as positive when in the direction of the lorce. If several forces act, the work done by each can be found in the same wiiy ; and the sam of all tlieoe is the work done by the whole gyatem of force». It appears from this that tlie work done by any force during an infinitesimal displacement of the point of application, is the product of the resolved part of the force in the direction of the displacement into the displacement ; and this is the same as the virtual momtnt of tlie force, which has been described in Art 101. In Statics we are concerned only with the small hjfputJutkai displacement which we give the point of application of the force in appIyiMg the principle of virtual velocities. But in Kinetics tlie liodies are in motion ; the force aetitaUff disphues its point of application in such a manner that the displacement has a projection along the direction of the force. If ds denote the projection of any elementary arc of a curve along the direction of P, the work done by P in this displacement is Pd». The sum of all these elements of work done by P in its motion over a finite space is the whole work found by taking the integral of Pd$ between proper limits. Hence generally, if « be an arc of the path of a particle, P the tangential component of the for«»c •-aich act on it, the work done on the particle between any two points of its path is /Pd», (1) the integral being taken between limits corresponding to the initial and final positions of the particle. 213. Work on an Inclined Plane.— Let a be the inclination of the plane to the horizon, W the weight moved, « the distance along the plane through which the weight is moved. Resolve W into two components, one along the plane and the other perpendicular to it; the former, W sin a, is the component which resists motion along the plane. Hence the amount of work required 'o draw the weight up the plane = W sin « • s = Wxtho vertical height of the plane ; t. e., the amount of work required is unchanged by the substitution of the oblique path for the vertical. Hence the work in moving a body up an incUntnl plane, without friction, is equal to the product of the weight of the body by the vertical height through which' it ia raised. 4 \'i 393 WORK ON AN INCLINHD PLANW. I Cob. 1. — If the plane be rough, let ft = the coefficient of friction ; then since the normal component of the weight is FF COB a, the resistance of friction is /i fF cos a (Art 92). The work required consists of two parts, (1) raising the weight along the plane, and (2) overcoming the resistance of friction along the plane, the former = (F sin « • «, and the latter is ^ H^ cos « • «. Hence the whoh work necessary to move the weight up the plane is FT (sin a 4- |t* cos cc) «. (1) Since » sin a represents the vertical height through which the weight is raised, and » cos a the korwmtal space through which it is drawn, tiiis result may be stated thus : The work expended it the same as that which would be required to raise the weight through the vertioal height of the plane., together with that which would be required to draw the body aUmg the base qf the plam horitojUaUif against friction. GoR. 3. — If a body be dragged through a space, s, down an inclined plane, which is too rough for the body to slide down by itself, the work done is fT (^ cos a — sin «c) s. (») Cor. 3. — If h — the height of the inclined plane, and d = its horizontal base, then the work done against gravity to move the body up the plane = Wh ; and the work done against friction to move the body along the plane, suppos* ing it to be horixontal, = nb W. Hence (Cor. 1) the total work done is Wh + yhW. (8) If the body be drawn down the phne, the total work expended (Cor. 2) is - Wh + ftbW. (4) ~ : the coefficient at of the weight C08 a (Art. 92). (1) raising the [ the resiBtance tF sin « . 9, and teork necessary (1) leight through \or%»ontai spuoe e stated thas : ohi<^ would be HmI height of be required to M horitoiUaUif ^Mce, 8, down ' body to slide (2) led plane, and igainst gravity the work done plane, euppoB- r. 1) the total (3) le total work JSJCAMl'LBS. If in (4) the former term la greater than the latter, gravity does more work than what is expended on friction, and the body elides down the plane with accelerated velocity. ScH. 1.— If the inclination of the plane is small, as it ia in most cases which occur iu pmctice, as in common roads and railroads, cos a may without any important error be taken as equal to unity, and the expression for the work becomes (Coi-s. 1 and 2) W^ ()">'' ± « Bin a), («) the upper or lower sign being taken according as the body is dragged up or down the plane. ScH. 2.— If the inclination of the piano is small, as in the case of railway gradients, the pressure upon the plane will be very nearly equal to the weight of the body ; and the total work in moving a body atong an inclined plane wiU be from (3) and (4), 1^1 W± Wh, («) where nlW k the work due to friction along the plane of length I, and Wh is the work due to gravity, the proper sign being taken as in (5). EXAMPLES. 1. How much Work is done in lifting 150 and 200 lbs. through the heights of 80 and 120 fl. respectively. The work done = 150 x 80 -f 200 x 120 = 3G000 foot-pounds, Ans. 2. A body weighing 500 lbs. elides on a rough horizontal plane, the coefficient of friction being 0.1 ; how much work must be doue against friction to move the body over 100 ft. ? P ■i 894 sxAitPLsa. Here the friction is a force of 60 lbs. acting directly o|)iK>site to the motion ; hence the work done againgt firio- tion to move thd body ovef 100 ft it 60 X 100 = 6000 foot-pounds. Am, t. A tniin Weigtll lOO totisi the total resistance is 8 lbs. tier ton ; how mnoh mot\ mart be expended in l«ising it to the top of aa inclined plane a mile long, the inclination of the plane behig 1 rertical to 10 horizontal. Here the work done against friction (Sch. 2) = 800 X 6380 =r 4224000 foot-pounds, and the work done against gravity = 8a4000* X 6280 x V» = 16896000 foot-ponnds, 10 that the whole work = 21120000 foot-pouncis. 4. A train weighing 100 tons moves 30 miles an hout along a horizontal rood ; the resistances are 8 lbs. per ton ; find the quantity of work expended each hour. Aha. 126720000 foot-pounds. 6. If 26 cubic feet of water are pumped every 6 minutes from a mine l40 fathoms deep, required the amount of work expended per minute, a cubic foot of water weighing 62}^ lbs. Am. 262500 foot-iK>undB. 6. How much work is done when an engine weighing 10 tons moves half b mile on a horizontal road, if the total resistance is 8 lbs. per ton. Atu. 211200 foot-pounds. 7. If a weight of 1120 lbs. be lifted up by 20 men, 20 ft. high, twice in a minute, huw much work does each man do per hour P Am. 1344000 foot-ponnds. * Uue lou being 3MB Wtti. uoIudh otht-rwlM: stated. 1^ BOXSJt POWXM. 890 aotihg directly oe againtt firio- poonds, Am. Istanoe is 8 11^ 3d in inising it the inclination il. 2) rands, bot-poanda, uncis. miles an houf 8 lbs. per ton ; ur. I foot-pounds. ivery 5 minutes the amount of water weighing foot-|H>undB» ngine weighing ai road, if the foot-pounds. 30 men, 20 ft. does each man foot-pounds. •ted. 8. A body falls down the whole length of an inclined plane on which tiie coefficient of friction is 0.3. The height of the plane is 10 ft and the base 30 ft. On reach- ing the bottom it rolls horisonti^y on a pluie, having the same coefficieDt of friction. Find how far it will roll. Ans. 20 ft. 9. How much work will be requiisd to pump 8000 cubic feet of water from a mine whose depth is 600 fathoms. Atu. 1600000000 horse-power. 10. A hv.se draws 150 lbs. out of a well, by means of a rope going over a fixed pulley, moving at the rate of 2^ miles an hour; how many units of work does this horse perform a minute, neglecting friction. Ana. 83000 units of work. 214. Hone Power. — It would be inconvenient to express the power of an engine in foot-pounds, since Lhis unit is so small ; the term Horse Potoer is therefore used in measuring the performance of steam v<)ngines. From experiments made by Boulton and Watt it was estimated that a horse could raise 33000 lbs. vertically through one foot in one minute. This estimate is probably too high on the average, but it is still retained. Whether it is greater or lets than the power of s horse it matters little, while it is a power so well defined. A Morse Power there/ore means a power which can perform SSOOO foot-pounds of work in a minute. Thus, when we say that the actual horse power of an engine is ten, we mean that the engine u able to per- form 330000 foct-poantl^ of work per minute. It has been eatimated that | of the 88000 foot-pounds would be about the work of a hone of avenge Btreng^h. A mule will perform f the work of a horse. An aaa will perform about \ the work of a horse. A man will do about ^ the work of a horse, or about 8800 units of work per mluiute. See Even' Applied Mech's; also Byrne's Practical Meoh's, 390 WORK OF RA1SISO A SVSTSM OF W BIO UTS. 21S. Work of Rntoing a System of Weights.— Let P, Q, R, bfc any three weights at the distances, p, q, r, rospeotivoly above a fixed horizoutal plane. Then [Art. 50 (3)] or (Art 73, Oor. 3), the distances of the ceutie of gravity of P, Q, B, above this fixed horizoutal plane is Pp + Qq + Jir P+Q-{ Ji ' (1) Now suppose that the weights are raised vertically tlirongh the heights o, b, c, respectively. Then the dis- tance of the centre of gravity of the three weights, in the new position, above the same fixed hcrizontal plane is P(p + a) + Q{q + b) + R{r + c) P+Q+Ji Subtract! ug (1) from (2), we have Pa + Q b + lie P+ Q + Ji ' (2) (3) for the vertical distance between the two positions of the centre of gravity of the three bodies. Now the work of raising vertically a weight equal to the sum of P, Q, Ji, through the space deuoted by (3) is the product of the sum of the weights into the space, which is Pa+ Qb + Re, (4) but (4) is the work of raising the three weights P, Q, R, through the heights a, b, c, respectively. In the same way this may be shown for any number of weights. Herux when several toeighta are raised vertically through different heights, the whole work done is the same as that of raising a weight equal to the sum of the weights vertically from the first position of their centre of gravity to the last position. (See Todhuutcr's Mech'a, p. 338.) Weights.— istunces, p, q, . Then [Art. ' the ceiiti'o of \\ plaue is (1) led vertically rhcn the dis- eigbts, in the plane is £). (2) (3) jitioDs of the equal to the by (3) is the yfx, which is (4) hts P, Q, R, the same way cally through me as that of hts vertically ■y to the last ' ■iMmmmk^-mmmm^Mo^KmMm^ SXAMPLSa. S97 EXAM PLES. 1. How many horse-power would it take to raise 3 c«fc of coal a minute from a pit whose deptii is 110 fathoms? Dopt!i = 110 X 6 = C60 feet. 3 cwt. = 112 X 3 = 336 lbs. Hence the work to be done in a minute = 600 X 336 = 221760 foot-pounds. Therefore the horse-power = 221760 -^ 33000 = 6.72, Ans. 2. Find how many cable feet of water an engine of 40 horse-power' will raise in an hour from a mine 80 fathoms deep, supposing a cubic foot of water to weigh 1000 ozs. Work of the engine per hour = 40 x 33000 x 60 foot- pounds. Work expended in raising one cubic foot of water through 80 fathoms = i^f^ x 80 x 6 = 30000 foot- pounds. Hence the number of cubic feet raised in an hour = 40 X 33000 X 60 -!- 30000 = 2640, Ana, 3. Find the liorse-power of an engine which is to move at the rate of 20 miles an hour up an incline which rises 1 foot in 100, the weight of the engine and load being 60 tons, and the resistance from friction 12 lbs. per ton. The horizontal space passed over in a minute = 1760 ft. ; the vertical space is one-hundredth of this = 17.60 ft. Hence from (6) of Art. 213, we have 12 X 1760 X 60-1-60 x 2240 x 17.6=1760 x 2064 foot-pounds. Therefore the horse-power = 1760 X 2064 + 83000 = 110.08, Am. 4 A well ig to be dng 20 ft. deep, and 4 ft. in dmmetor ; find the work in raising the material, snppw'.ng that a cubic foot of it weighs 140 lbs. » «^ e> Here the weight of the material to bo raised = 47r X 20 X 140 = 140 X 80rr lbs. The work done is eqaiyalcKt to raising this throngh the height of 10 ft (Art 316). Hanoe tU whole work = 140 X 80rr X 10 = lUJOOOrr foot-pounds, Ane. 5. Find the horse-power of an. engine QMt would nuso 7 tons of coal per hour from a pit whose depth is a fathoms. Work per minute = ZLiH^J^liL? .== ^ZiaT; .-. the horse-power = ^^, Atu. 6 Required the work in raising water ftom thi«o different levels whose depths are a, *. c fathoms respoctirely ; from the first A, from the second £, 'rom the third C, cubic feet of water are to be raised per minute. Work in raiting water from the first kfvel = 62.6 J X c X 6 = 376 A-a; and 80 on for the work in the other lev»li ; . •• work per min. =r ^76 (^ .a-|. 5.*+ C-c) foot-pounds. 7. Find the horso-powor of an engine which draws a load of r tons along a level road at the rate of m miles >, Ant. ft in dicmetor ; ppocing that a Bd lis throQgh the b work mds, Ane. \9t would nieo om depth is a 8. three different 5ctiTely; from bird C, cubic foot-pounds. lioh draws a e of M miltHi fa[A.WPM,lbi, an how, lOie Wcdon being p pooadi per to», all other resistances being ne^ciotdL Work of tbe wgine per minute . . U.-F. ~ gj^QQQ - 3000 » ^"« 8. Bequired the nnmber of horse-power to nuse 2260 onWc ft. of water an hour, feenii a mine whose deptii a 63 fathoms. J«*. 26i, 9. What weight of coal will an engine of 4 horse-power raise in one hour from a pit wnose depth is 200 ft. P Am. 89600 lbs. 10. la what time will an engine of 10 ho?»^ppwer raise 5 tons of material from the depth of 182 ft.? Ans. 4>48 minutes. 11. How many cubic feet of water will an engine of 86 horse-power raise in an hour from a mine whose depth is 40 fathoms P ^»»*- 4762 cubic feet 12. The piston of a steam engine Ut U ins. in (Hametw 5 its stroke is 2^ ft. longi it makes 40 strokes per minute ; the mean pressure of the ateam on it is 16 lbs. pw sqnara inch; -^vhat number of foot-pouada is done by the steam per minute, and what is the hone-power of the engine ? Ans. 866072 foot-po«nds ; 8-08 H.-P. 18. A w>rmod in m«»ving the train, but the lost work is that which is xlone in overcoming the friction of the train, the resistance of gravity on up grades, the resist- ance of the nir. etc. The work appliinl to a machine is equal to the whole work done by tl.e machine, both useful ond lost, therefore the useful work is always less than the work Hp|>lied to the machine. * Domotinet oailad Hffidojcj. - second ; what space, «, will the ball move over before it comes to a state of rest, the coefBcient of friction being/? Here the resistance of friction is fw, which acts directly opposite to the motion, therefoK the work done by friction while the body mov% over 8 feet = fws ; the work t ored up in the ball = ^t^ = ^ ; therefore from (1) we have /W8 = 8 = 3. A railway train, weighing 7* tons, has a velocity of v ft. per second when the steam is turned off ; what distance, s, will the train have moved on a level rail, whose friction iap lbs. per ton, when the velocity is v^ ft. per second ? Here the work done by friction = pT8', hence from (2) we have _ . 2240 7*.. „ 1120»-V) ~ 9P 8 = 4. A train of T tons descends an incline of « ft in , length, having a total rise of A ft.; what will be the velocity, V, ocquued by the train, the friction being/* lbs. per ton P 408 KINETIC BNBROY OF A RIGID BODY. Here we have (Art. 213, Sell. 2), the work done on the train = the work of gravity — the work of friction = Z%iQ Th — pTs', which is equal to the work stored up in the train. Hence 2240 7V» ^ = %UQTh—pT8\ • •• V = y/'igh — rh^P'- 5. If the velocity of the tnun in the last example be t\ ft. per second when the steam is turned off, what will be its velocity, v, when it reaches the bottom of the incline ? Ans. V = Vn'o^ + ^h — rhvgP'' 6. A body weighing 40 lbs. is projected along a rough horizontal plane with a velocity of 150 ft. per sec. ; the cneflicient of friction is |; find the work done against friction in five seconds. Ans. 3500 foot-pounds. 7. Find the work accumulated in a body which weighs 300 lbs. and has a velocity of 64 ft. per second. Ans. 19200 foot-pounds. 2ia Kinetie Bnergy of a Rigid Body revolving round an Axis. — L«t m be the mass of any iMtrticle of the body at the distance r firom the axis, and lev u bo the angular velocity, which wil; be the same for every particle, since the body is rigid; then the kinetic energy of m = \m (rw)«. The kinetic energy of the whole body will bo found by taking the sum of these expressions for every particle of the body. Hence it may be written £ Jf?ir«w» = •a'S, mi*. « (1) •'-^"^w^iffimiwwiii^^ M ) BODY. vork done on the of friction be train. Hence ; last example be i off, what will be » of the incline ? ed along a rough I ft. per sec. ; the ork done against >00 foot-pounds. >ody which weighs jcond. too foot-pounds. Body revolving of any i)article of and leu u bo the for every particle, c energy of f?» = hole body will bo ressious for every ritteu (1) EX A MP LBS. 409 S wr» is called the moment of inertia of the body about the axis, and will be explained in the next chapter. Hence the kinetic energy of any rotating body = f In*nM 150^(12IFA + na/w) 4. The weight of a fly-wheel is 8000 lbs., the diameter 20 feet, diameter of axle 14 inches, coefBcient of fHction 0.2 ; if the wheel is separated from the engine when mak- ing 27 revolutions per minute, find how many revolutions it will make before it sti^s {g taken = 381.2). Ana. 16.9 revocations. 'riction on the axle heel make before it f". [8 irhich the fl^-wheel d = P X ffir, and the energy stored up in the hammer =: ^„^ = »' = 6.2. o4 Since the work done in forcing the nail into the wood mnst be equal to all the work stored in the hammer, (Art. ai7), wehave — = 6.2; .'. F=7441b* 412 EXAMPLES, Hence the force which the hammer exerts on the head of the nail is at least 744 lbs. 2. If the hammer in the last example forces the nail into the wood only 0.01 of an inch, find the force exerted on the nail. Ans. 7440 lbs. Heace, we see that, nccoiding as the wood m hanlor, i. e., accord- ing e.s the noil enters I088 at each stroke, the force of the blow beconios greater. So that when we speak of the " force of a blow,"' vnc. must B|)ccify how aeon the body givinf; the blow will come t) r^^t, otherwise the term is meaningless. Thus, suppose n ball of 100 11/6. weight have a velocity that will cause it to ascend 1000 ft. ; ir' t^io UII is to be stopped at the end of 1000 ft., a force of 100 ll«. will do it ; but If it is to be stopped at the end of one foot, it will u((h1 a force of 100000 lbs. to do it ; and to 9top it at the cud ot one inch will require t/ force of 1200000 Iba, and h) on. 220. Work of a Water-FalL—When water or any body falls from a given height, it is plain that the work which is stored up in it, and which it is capable of doing, is otjnal to that winch would Ikj required to raise it to the height from which it has fallen ; i. e., if 1 lb. o1^ water descend tl^|pfgh 1 foot it must accumulate as much work m would be required to raise it through 1 foot. Hence when a fall of wz^m is employed to drive a wator-whc"l, or any other hydraulic niichiue, whose modulus is given, the work done upon the maciiino is equal to the weight of the water in pounds x its fall in feet x the modulus of the machine. BX AMPLBS. 1. The breadth of a stpoam in h feet, depth n feet, mean Telocity r feci per minute, and the height of the fall A feet ; find (1) the horse-power, N, of the water-wooel whose modulus is m, and (2) find the number of cubic feet. A, which the wheel will pump per minute from the Iwttom of the fall to the height of At feoU erts on the head of 9 forces the nail into the force exerted on Ans. 7440 lbs. iB harder, i. e., acconi- , the force of the blow f the " force of a Wow, ' the blow win come t^^ hu8, suppose n ball of le it to ascend 1000 ft.: ft., a force of 100 Ite. ! end of one fo»>t, it will top it at the end ol one con. VTien water or any plain that the work 8 capable of doing, is id to raise it to the <•., if 1 lb. o1^ water lulate as much work ugh 1 foot. Hence ive a water-whc"l, or odulus is given, tlie to the weight of the the modulus of the depth a feet, mean fht of tlie fall A feet ; watcr-Wiioel whose tcr of cubic feet, A, from the Iwttom of SXAMPLSS. 413 62.6 Aht .-. A - (8) Weight of water going over the i*t;l per min. = 62.6 abv. .'. Work of wheel per min. = 62.5 abvhm. (1) •*' ~33000 ' ' ' Work in pumping water per min. = 62.5 J A, ; which must = the work of the wheel per min.; hence from (1) we have 62.5 abvhm ; abvhm 2. The mean section of a stream is 6 ft. by 2 ft. ; its mean velocity is 35 ft jjer muiate ; there is a fall of 13 ft. on this stream, at which is erected n water-wheel whoso modulus is 0.65 ; find the horse-power of the wheel. Ans. 6.6 H.-P. 3. In how miiny hours would the wheel in Ex. 2 grind 8000 bushels of wheat, supposing each horse-power to grind 1 bushel per hour ? Ans. 1428i^ hours. 4. How many cubic feet of water must be made to descend the full per minute in Ex. 2, that the wheel may grind at the rate of 28 bushels jwr hour ? Am. 1749.5 cu. ft, 5. Given the stream in Ex. 2, what must be the height of the fall to grind 10 bushels per hour, if the modi'lus of the wheel is 0.4 ? Ana. 87.7 feet. C. Find the useful horsc-jwwer of a wat«r-wheel, lujp- l)08ing the stream to bo 5 ft. broad and 2 ft. deep, and tx> ilow with a velocity of 30 ft. per minute ; the 'uight of tho full being 14 ft.., and tliu modulus of tho machine 0.05, Ans. 5.2 nearly. I 4U MXAMPLES. 221. The IDntjr of an 'Ba^aam.'-Tkt duty of an engine is the number of units of toork which it is capable cf doing by burning a given quantity of fud. — It has been found by experiment that, wha^<)Ter may be the pressure at v, hich the steam is formed, the quantity of fuel necessary to evaporate a given volume of water is always nearly the same ; hence it is most advantageous to employ steam of a high presavre.* In good ordinary engines the dnty variea between SOOOGO and 600000 units of work for a lb. of eoal. The extent to which tbe eoonumy of fool may be carried is well illuBtrated by the engines eni> ployed to drain the mines in Cornwall, England. In 1816, the average duty of then engines was 20 million units of work for a boshelf of coal : in 1848, by reason of snocnsive improvements, tlie averagv duty had beoome 60 millions, eflSecting a saving of £80000 per aunun. It Is stated ^at in ths ease of oa» engine, the duty waa raised to 196 millions. The duty of the engine depends largely on tlie oonatraotloa of the boiler ; 1 lb. of eoal in the Coniish boHer evaporates 11} lbs. of water, while la a diflferently-shaped boiler 8.7 IB the maximum.) EXAMPLES. 1. An engine burns 2 lbs. of coal for each horse-^iower per hour ; find the duty of the engine for a lb. of coal. Here the work done in one hour = 60 X 830C0 foot-po ands ; therefore the duty of the engine = 30 x 33000 foot-pounds, = 990000 foot-iwunda. % How many bushels of coal must be expended in a day of 24 hours in raising 160 cubic ft. of wttt«r \vir minute -^^ tm TaM In EeshanlM' Macaslna, in the .tmut IMl. t On« bstbol of coal s 64 or M tbi., depcndliif upon wbura It la. a (0 which id the ordinary formula of mensuration. Now it is evident that when the curve is always concave to the line AG (1) will give too small a result, and if con- vex it will give too large a result Let Fig. 90 represent the space between any two odd consecutive ordinates, as Cc and Ee(Fig. 89); divide CE into three equal parts, CK = KL = LE, , JlAJ and erect the ordinates Bufc and U, dividing the two trapezoids CcdD and Drf«E into the three trapezoids CckK, KklL, and LfcE. The sum of the areas of these three trapezoids K OL ng.M = ICK (Cc + 2Kk + 2U + E«) = ^l (Oc + 2Kt 4- 2LZ 4- Ee), (since ^CK = iCD = \l) = \l (Cc + 4Do + Ec), (since 2K* + 2U = 4Do), (2) which is a closer approximation for the area of CceE than (1). Now when the curve is concave towards CE, (2) is smaller than the area between CE and the curve ckdle ; if wo substitute foi Do, the ordinate Drf, which is a little greater than Do and which is given, (2) becomes \l (Cc + 4D(/ + E«), whi'-' is u still closer api>roximatiuu than (2). (8) distanco between iking tlie sum of B for the area of + y*) + ete. f-2y, +y,); (i) tion. 18 always concave 38ult, and if oon- ?cn any two odd ;. 89) ; divide CE [iE, ing [the eE. >ids K OL E d = 4Do), (2) le area of CwE rards CE. (2) is c curve ckdle ; if which is a little icomca (8) (2). Simil -y we have for the areas of AccC and "EetfO, V (Aa + 4B6 + Cc), and \l (Ee + 4F/ + G^). Adalag (3) and (4) together, we have for an approximate value of the whole area, V [yi + ^7 + 2 (f/s + y,) + Hy,+y, + ,/,)], (5) which is SiinpaoH's Formula. Hence Sirapsou's rule for finding the area approximately is the following : Divide the abidssn, AG, into an even number of equal parts, and erect ordinates at the points of division j then add toijethcr the first and last ordinates, twice the sum of all the other odd ordinates, and four times the sum of all the even ordinates ; multiply the sum by otie-tfiird of the comuon distance between any two adjacent ordinates. (See Todbuntcr's Mensuration, also Tate's Geometry and Meusurution, also Morin's Mech's, by Bouuett.) EXAMPLES. 1. A variable force has acted through 3 ft. : the value of the force taken at seven successive oqnidistaut points, including the first and the last, is in lbs. 189, 151.2, 126, 108, 94.6, 84, 75.6 ; find tiie whole work done. Ans. 346,4 foot-jwunds. 2. A variable force has acted through 6 ft. ; the Ytduo of the force taken at seven successive equidistant pointij, including the first and the lust, is in lbs. 3, 8, 15, 24, 35, 48, 63 ; find the whole work done. Ans. 162 foot-j)onnds. 8. A variable force has a<'tod through 9 ft.; the value of the force taken at se^en successive equidistant points, including the first and the lost, is in lbs. 6.082, 6.164, 6.245, 6.403, 6.481, 6.557; find the whole work done. Ans. 56.907 foot-puund«. ':m. MMMnppMKMMT^cwu^ .- 416 aXAMPLMS. Should any of the ordinates become zero, it will not pre- vent the use of Simpeon's rale. Simpson's rule is a^Uoable t:) other investigations as well as to that of work done by a variable fOTce. For example, if we want the velocity generated in a given time in a particle by a variable force, let the straight line AG represent the whole time during which the fcnrce actii, and let the stTMght lines at right angles to AG reprei>6ui the force at the corresponding instants; then the area will represent the whole space described in the given time. BXAMPL.BS. 1. The ram of a pile-driving engine weighs half a ton,* jmd has a fall of 17 ft. ; how many units of work are per- foi-med in raising this ram P Ana. 19040 foot-pounds. a. How many units of woA are required to raise 7 cwt of coal from a mine whose depth is 13 fathoms ? Ans. 61162 foot-pounde. 3. A horse is used to lift the earth fi-om a trench, which he does by moans of a cord going over a pulley. He pulls up, twice v^ery 5 minutes, a uan weighing 130 lbs., and a barrowf ul of earth weighing 860 lbs. Each time the horse gooa forward 60 ft. ; find the units of woik done by the horse per hour. Ana. 374000. 4. A railway i ain of T tons asoends an inclined plane which has a rise of e ft. in 100 ft., with a uniform speed of m miles per hour ; find the horse-power of the engine, the friction being/) lbs. per ton. Ans. ^y(P+/^-^ ) H..P. 6. A railway train of 80 tons ascrnds an incline which rises one foot in 50 ft., with the uniform rate of 16 miles * Ont lull -- tW ewt. - 9S40 lbs. wmmmmtmtm mi rot it will not pre- ' investigations as riable force. For d in a given time straight line AG he force acts, and A.G repreucui the len the area will e given time. reighs half a ton,* of work are per- '40 foot-pounds. d to raise 7 cwt loms ? 52 foot-ponnde. 1 a trench, which pulley. He pulls g 130 lbs., and a h time the horse oi'k done by the Am. 374000. m inclined plane miform speed of ' the engine, the + 22^) „ p 76 "'^^ an incline which rate of 16 miles SXAMPCES. 419 per hour ; find the horse-power of the engine, the friction being 8 lbs. per ton. Ans. 169.96 H.-P. 6. If a horse exert a traction of t lbs., what weight, v>, will he pull up or down a hill of small inclination which has a rise of in 100, the coefficient being/? . 100/ Ana. w = -— r-; • 100/±« 7. From what depth will an engine of 22 horse-power raise 13 t«ins of coal in an hour ? Am. 24.9 ft. 8. An engine is observed to raise 7 tons of material an hour from a mine whose depth is 85 fothoms ; find the horse-power of the engine, supposing | of its work to be lost in transmission. Ans. 4.4829 H.-P. 9. Required the horse-power of an engine that would supply a city with water, working 12 hours a day, the water to bo raised to a height of 50 ft. ; the number of inhabitants being 130000, and each person to use 5 gallons of water a day, the gallon weighing 8| lbs. nearly. Aug. 11.4 H.-P. 10. Prom what depth will an engine of 20 horse-power raise 600 cubic feet of water per hour P Ans. 1056 feet 11. At what rate per hour will an engipe of 30 horse- power draw a train weighing 90 tons gross, the resistance being 8 lbs. per ton ? Ant. 16.628 miles. 12. What is the gross weight of a train iHliioh an engine of 25 horse-power will draw at the rate of 26 miles an hour, resistances being 8 lbs. per ton ? Ans. 46.875 tons. 13. A train whose gross weight is 80 tons travels at the rate of 20 miles an hour; if the resistance is 8 Uw. per ton, what is the horse-power of the engine ? ' Am. Ui\ H.-P. r i'. 420 X^ AMPLSa. 14. What must be the length of the stroke of a piston of an engine, the surfuco of wliich is 1500 square iiicliee;, which makes 20 strokes per minute, so that with u weuii pressure of 12 lbs. on each square aua of the piston, the engine may be of 80 horse-power ? Ans. 7| ft. 15. The diameter of the piston of an engine is 80 ins., the length of the stroke is 10 ft, it makes 11 strokes per minute, and the mean pressure of the steam on the piston is 12 lbs. per square inch ; what is the horse-power ? Ans. 201.0CH..P. 16. The cylinder of a steam engine has an internal diameter of 3 ft, the length of the stroke is 6 ft., it miikcs 6 strokes per minute; under what efFective pressure ^ar square inch would it have to work in order that T5 hoi'sc- power may be done on the piston? Ans. 67' 54 lbs. 17. It is said that a horse, walking at the rate of 2^ miles an hour, can do 1G50000 units of work in an hour ; what force in pounds does ho continually exert ? Ans. 126 lbs. 18. Find the horse-power of an engine which is to move at the rate of 30 miles an hour, the weight of the engine and load being 50 tons, and the resistance from friction 10 lbs. jier ton. Ans. 64 H.-P. 19. There were 0000 cubic ft. of water in a mine whoso depth is 60 fathom^, when an engine of 50 horse-power began to work the pump ; the engine continued to work 5 hours before tho mine was cleared of the water ; required the number of cubic ft of water wliich had run into the mine during the 5 hours, supposing \ of the work of the engine to be lost by transmission. Ans. 10500 cubic ft 20. Find the horse-jiower of a steam engine which will raise 30 cubic it. of water ^Ksr minute from a mine 440 ft doop. Ann. 25 ll.-P. "oke of a piston square iiiclictj, it with u lucuii ' the piston, tbo Ans. 7|ft. igiue is 80 ins., 11 strokes per 1 on the piston -power ? J01.00H.-P. as an internal 6 ft., ft miikcs e pressure ^ ar r that 75 hoi-se- 8. 67. 54 lbs. ^to of 2^ miles an hour ; what ns. 125 lbs. ich is to move of the engine from friction |w. 64H.-P. a mine whoso horse-powor uod to work 5 iter ; required run into the ifi work of the 00 cubic ft. 36 which will a mine 440 ft. 6. aoU.-P. ,jjji|i,|gi!a > ;iii | ^p i! ij^^ i iyffyy?tjMW SXAMPLSS. 21. If a pit 10 ft. deep with an area of 4 square ft. be excavated and the earth thrown up, how much work will have been done, supposing a cubic foot of earth to weigh 90 lbs, Ans. 18000 ft-lbs. 22. A well-shaft 300 ft. deep and 5 ft. in diameter is full of water ; how many units of work must be expended in getting this water out the well ; (t. «., irrespectively of any other water flowing in)? Ans. 165223850 ft.-lbs. 23. A shaft o ft. deep is full of water; find the depth of the surface of the water when one-quarter of the work required to empty the shaft has been done. . a „ i 24. The diameter of the cylinder of an engine is 80 ins., the piston makes per minute 8 strokes of lOJ ft. under a mean pressure of 15 lbs. per square inch ; the modulus of the engine is 0'65; how many cubic ft of water will it raise from a depth of 112 ft. in one minute? Ans. 485. 78 cub. ft. 25. If in the last example the engine raised a weight of 66433 lbs. through 90 ft. in one miuute, what must bo the mean pressure jier square inch on the piston ? Ans. 26.37 lbs. 26. If the diameter of the piston of the engine in Ex. 24 had been 85 ins., what addition in horse-power would that make to the useful power of the engine ? ^»«. 13-28 H.-P. 27. If an engine of 60 horse-power raise 2860 cub. ft. of water per hour from a mine 60 fathoms deep, find the modulus of the engine. Atu. '65, 28. Find at what rate an engine of 30 horse-power could draw a train weighing 50 tons up an incline of 1 in 280, the resistance from friction being 7 lbs. per ton. , Ans, 1320 ft, per minute. 422 SXAUPLSS. 29. A forge hammer weighing 300 lbs. makes 100 lilts a minate, the perpendicniar height of each lift being 2 fL; what is the horse-power of the engine that gives motion to 20 such hammers? Ans. 36*30 H.-P. 30. An engine of 10 horse-power raises 4000 lbs. of coal from a pit 1200 ft. deep in an hour, and also gives motion to a hammer which makes 50 lifts in a minute, each lift having a perpendicular height of 4 ft.; what is the weight of the hammer? Ans. 1250 lbs. 31. Find the horse-powor of the engine to raise T tons of coal per hour from a pit 'vhose depth is a fathoms, and \jA the same time to give motion to a forge hammer weighing to lbs., which makes n lifts per minute of h ft each. 224flr + nhw „ _ Ana, 7^-^t ■ xl.-r. 32. Find the useful work done by a fire engine per second which difioharges every second 13 lbs. of water with a velocity of 60 ft. per second. Am. 508 nearly. 33. A mil way truck weighs m tons ; a horse draws it along horizontally, the resistance being « lbs. per ton ; in passing over a sjiace a the velocity changes from « to f ; find the work done by the horse in this space. 2240W , . J, , Ana. —X — («» — «') + mns. 34. The weight of a ram is 600 lbs., and ai the end of the blow has a velocity of 32J ft.; what work has Ijcen done in raising it P Am. 9G50. 35. Retiuired the work stored in a cannon ball whose weight is 32^ lbs., and velocity 1500 ft. Am. 1126000. 86. A ball, weighing 20 lbs., is projected with a velocity of 60 ft. a second, on a bowling-green ; what space will the ball move over before it comes to rest, allowing the friction to be T«5 the weight of the ball? Am. 1007.3 ft. ':'mimimimmmmmmf>mm akes 1001iit« a lift being 2 ft.; jives motion to 36.36 H.-P. OOO lbs. of coal 10 gives motion linuto, each lift t is the weight ns. 1250 IbH. raise T tons of ithoms, and "ut nmer weighing t. each. ■h nhw H.-P. ire engine per . of water with 508 nearly. Iiorse draws it per ton; in from u to V ; «') + mns, at the end of 'ork has Ixjen Ans. 9650. ball ^.vhose 1125000. ith a Telocity space will the the friction 1007-3 ft. SXAMPLKS. m 87. A train, weighing 198 tons, haa ft velocity of 80 miles an hour when the steam is turned off; how far will the train move on a level nul before coming to rest, the friction being 6^ Iba per ton !* Am. 12266 ft. 38. A train, weighing 60 tons, has a velocity of 40 miles an hour, when the steam is turned off, how Ux will it ascend an incline of 1 in 100, taking friction at 8 lbs. a ton ? Jim. 3942i ft 39. A carriage of 1 ton moves on a level rail with the speed of 8 ft a second; through what space must the carriage move to have a velocity of 2 ft., supposing friction to be 8 lbs. a ton? J >w. 348 ft 40. If the carriage in the last example moved over 400 feet before it comes to a state of rest, what is the resistance of friction per ton ? Ans. 6.57 lbs. 41. A force, P, acts upon a body parallel to the plane; find the space, % moved over when the body has attained a given velocity, v, the coefficient of friction being/, and the body weighing «? lbs. a*,, , ?£??__. Ana. " - 2g{P -fw) 42. Suppose the body in the last example to bo moved for t seconds ; reqair«)d (1) the velocity, v, acquired, and (2) the work stored. A.ta. (1) ■1 to w t9\ (2) 2w 43. A body, weighing 40 lbs., is projected along a rough horizouta! plane with a velocity of 150 ft per second ; the coefficient of friction is \ ; find the work done against fric- tion in 5 seconds. Ans. 3500 foot-pounds. 44. A body weighing 60 lbs., is projected along a rough horizontal plane with the velocity of 40 yards per second ; find the work expended when the body comes to rest. Ah8. 11260 ft-lbs. ■ ■y f ^ « »'» , ' ' «■*■ < 424 -.. - EXAJtPLKS. ' 45. If a train of cars weighing 100000 lbs. is moving on ft horizontal truck with a velocity of 40 milos an hour when the Bteam is turned off ; through what space will it move before it ig brought to rest by friction, the friction being 8 lbs. i)er ton ? Am. 13374.8 ft. 40. What amount of energy is acquired by a body weigh- ing 30 lbs. that falls through the whole length of a rough inclined plane, the height of which is 30 ft., and the base 100 ft., the coefficient of friction being \ ? Ans. 300ft.-lb8. 47. If a train of cars, weighing 7' tons, ascend an in-line having a raise of e ft. in 100 ft., with the velocity v^ ft. per second when the steam is turned off; through what space, X, will it move before it comes to a state of rest, the friction being j» lbs. per ton ? . liaOi--^ Ann. X = — TT-r-.— ^ g {•l-i.\e, + ;)) 48. Suppose the train, in Ex. 4, Art. 217, to be attached to a rope, passing round a wheel nt the top of the incline, which has an empty train of T, tons attached to the other extremity of the rope: what velocity, r, will the train acquii-e iu descending s ft. of the incline ? Ans, r Ti gps 1126" 49. An engine of 35 horse-power makes 20 revolutions per minute, the weight of the Hy-wheel is 20 tons and the diameter is 20 ft.; what is the accumulated energy in foot- pounds? Ans. 307000. 50. If the fly-wheel ui the last example lifted a weight of 4000 lbs. through 3 ft., what part of its angular velocity woitlu it lose ? Ans. ^. 51. If the axis of tlie above fly-wheel be 6 ins. in diameter, the coefficient of friction 0-075, what fraction, . IS moving on 3 an hour when aco will it move 3 friction being r. 13374.8 ft. T a body weigh- >gth of a rough , and the base ?. 300ft.-lbs. fend an in':!line )eity j'p ft. per jh what space, )st, the friction _l]20j;o2 {•i'ZAe+p)' to be attached nf the incline, to the other tvill the train 9P« 1126' !0 revolutions tons and the lergy in foot- n». 307000. d a weight of pilar velocity Ans. ■^. be 6 ins. in 'hat fraction, Hi EXAMPLES. approximately, of the 35 horse-power is expended in turn- ing the lly-wheel ? Ans. ^. 52. In pile driving, 38 men raised a ram 12 times in an hour ; the weight of the ram was 12 cwt., and the height through which it wan raised 140 ft.; find the work done by one man in a minute. Ans. 990 ft-lbs. 53. A battering-ram, weighing 2000 lbs., strikes the head of a pile with a velocity of 20 ft per second ; how far will it drive the pile if the constant resistance is 10000 lbs.? Ans. 1.25 ft. 54. A nail 2 ins. long was driven into a block by suc- cessive blows from a monkey weighing 5.01 lbs.; after one blow it was found that the head of the nail projected 0.8 of an inch above the surface of the block ; the monkey was then raised to a height of 1.5 ft, and allowed to fall uiwn the head of the nail ; after this blow the head of the nail was 0.46 of an inch above the surface ; find the force which the monkey exerted upon the head of the nail at thia blow. Ans. 265.66 lbs. 55. The monkey of a pile-driver, weighing 500 lbs. is raised to a height of 20 ft, and then allowed to foil upon the head of a pile, which is driven into the ground 1 inch by the blow; find the force which the monkey exerted upon the head of the pile. Ans. 120000 lbs. 56. A steam hammer, weighing 500 lbs., falls through a height of 4 ft. under the action of its own weight and a steam pressure of 1000 lbs.; find the amount of work which it can do at the end of the fall. Ans. 6000 ft-lbs. 57. The mean section of a stream is 8 square ft.; its mean velocity is 40 ft per minute ; it has a fall of 17^ ft.; it is required to raise water to a height of 300 ft by means of a water-wheel whose modulus is 0.7 ; how many cubic ft will it raise per minute ? Ans. 13.07 cub. ft. -t' 426 XXAMPLBA 68. To what height would the wheel in the last example raise 2^ cub. ft of water per minute ? Ans. 1742} ft 59. The meaa eectioo of a stretam is 1| ft by 11 ft; ita ntean velocity is 2^ miles an hour ; there is on it a fall of 6 fk. on which is erected a wheel whose modulus is 0.7 ; this wheel is employed to raise the hammers of a forge, each of which weighs 2 tons, and hw a lift of l^ft.j how many lifts of a hammer will the wheel ,^eld per minute P Am. 142 nearly. 60. In the last example determine the mean depth of the stream if the wheel yields 136 lifts per minute. Ana. 1.43 ft 61. In Ex. 59, how many cubic ft of water must descend the fall per minute to yield 97 lifts '>f the hammer per minute ? Ans. 2483 cub. ft 62. A stream is a ft broad and b ft deep, and flows at the rate of e ft per hour ; there is a fall of A ft ; t lie water turns a machine of which the modulus is e ; And the num- ber of bushels of corn which the machine can grind in an hour, supposing that it requires m units of work per minute for one hour to grind a bushel. ^ _ lOOOabche Ans. 16 X 60m 63. Down a l4-ft fall 200 cnK ft. of water descend every minute, and turn a wheel whose modulus is 0.6. The wheel lifts water from the bottom of the fall to a height of 64 ft; (1) how many cubic ft will be thus raised jkt minute? (2) " the water were raised from the top of the fall to the same [loint, what would the numl)cr of cubic ft then be? Ans. (1) 31.1 cub. ft.; (2) 34.7 cub. ft. In tlut ispcond caim the numbnr of cub. ft. of water taken from the top of tho fnll bi-lDg n it a tall of las is 0.7 ; this forge, each of !t.j how many lute? 142 nearly. leao depth of nute. fu. 1.43 ft. ' mast descend I hammer per 1483 cub. ft. and fiowH at ft; die water Ind the num- grind in an of work per lOOOabcJte 16 X 60m' descend every is 0.6. The to a height of us raised jwr ic top of the r of cubic fU 4.: cub. ft. takeu from the the wheel will Drcd up iu a sm I HiiiJiMIIUL, KXAMPhSa. 427 mill-pond which is 100 ft. long, 60 ft. br^, «nd 3 ft. deep, and has a fall of 8 ft. Ans. 7600000. 66. There are three distinct levels to be pamped in a mine, the flrst 100 fathoms deep, the second 120, the thiid 150 ; 30 cub. ft. of w&ter are to come from the first, 40 from the second, and 60 from the third per minute ; the duty of the engine is 70000000 for a bushel of coal Determine (1) its working horse-power and (2) the consumption of coal per hour. Ana. (1) 191 H.-P. ; (2) 5.4 bushels. 66. In the last example suppose there is another level of 160 fathoms to be pumped, that the engine docs as much work as before for the other levels, and that the utmost r»ower of the engine is 276 H.-P. ; find the greatest number of cub. ft. of water that can be raised from the fourth level. An$. 46^ cub ft. 67. A variable force has acted throngh 8 ft.; the value of iho force taken at nine successive equidistant points, including the first and the last, is in lbs. 10.204, 9.804, 9,434, 9.090, 8.771, 8.475, 8.197, 7.937, 7.692; find the whole work done. Ana. 70.641 foot-pounds. 68. The value of a variable force, taken at nine succes- sive equidistant points, including the first and the last points, is in lbs. 2.4849, 2.6649, 2.fS391, 2.7081, 2.7726, 2.8332, 2.8904, 2.9444, 2.9967, the common distance between the points is 1 ft ; find the whole work done. An». 22.0967 foot-ponnda 69. A train whose weight is 100 tons (including the engine) is drawn by an engine of 160 horse-power, the fric- tion being 14 \h% per ton, and all other resistances neglect- ed ; find the maximum speed whicli the engine is capable of sustaining on a level rail. An». 40^ miles per hour. 7d. If the train described in the hist example bo movipg, tit a particular instant with a velocity of 16 ^ailes per hour. EXAMPLES. and the engin? workirig at full power, what ia the accelera- tion at that instant ? (Call g = 3*^.) Ans. -^^V 71. Find the horse-power of an engine required to drjig a train of 100 tone up an incline of 1 in 60 with a velocity of 30 miles an hour, the friction being 1400 lbs. . Ans. The engine must be of not lees than 470f horse- IK)wer. This is somewhat above the power of most locomo- tive engines. 72. A train, of 200 tons weight, is ascending an incline of 1 in 100 at the rate of 30 miles per hour, the friction being 8 lbs. per ton. The steam boiug shut off and the break applied, the train is stopped in a quarter of a mile. Find the weight of the break-van, the coefficient of fric- tion of iron on iron being |. Atis. 11-^ tons. is the acceltra- Ans. ^^. [uired to drag u th a velocitj' of 1. . lan 470| horse- >f most locomo- ling an inclino lur, the friction mt oflf and the rter of a mile, flicient of fric- s. 11^ tona. CHAPTER VI. MOMENT OF INERTIA* 224. Moments of In6rtia.'-The quantity Lmr' in which m is the mem of. an element of a body, and r its ilistance from an axis, occurs frequently in problems of rotiition, so that it becomes necessary to consider it in detail ; it is called the moment of inertia of the body about the axis (Art, 218). Hence, "moment of inertia" may be defined as follows : If the mass of every particle of a body be multipiied by the square of its distance from a straight line, the sum of the products so formed is called the Moment of Inertia of the body about that line. If the mass of every particle of a bjdy be multiplied by the square of its distance from a ^iven plane or from a given pointy tiie sum of the products so formed is called the moment of inertia of the body with reference to that plane or that point. If the body bo referred to the axes of x and y, and if the mam of each parttcio be multiplied by its two co-ordinates, .r, y, i\> sum of the products so formed is called the product of inerd-a of Ihe body about those two axes. If dm denote 'he mass of an element, p its distance from the axis, and / tl>e moment of inertia, we have • / = l.jMm. (I) If the body Ix? leforred to rectangular axes, and x, y, «, l)e the co-ordinatob of any ''lomont, then, according to the dcfiuitioDs, the moments of inertia alM)ut the axes of x, y, z, respectively, will be * Thit term wm Introduced b; Bulor, knd bav now gect to an axis through its centre and perpendicular to il8 surface. Ijet the radius = a, and fi the mass of a unit of area as before, then wo have t/o ''0 • In all MM« mv shall anamo the thlcknens of tho laminin nr plateit lo he liiaultcalmal. ..■ mamm is* 433 PARALLSC AXSS, 4. Find tho moment of inertia of a oircnlar plate (1) about u diameter as an axis, and (2) about a tangent. Ans. (1) imtfi; (2) Jma*. 5. Find the moment of inertia of a square plate, (1) about an axis through its centre and perpendicular to itH plane, (2) about an axis which joins the middle points of two opposite sides, und (3) about an axis passing tlirough an angular point of the plate, and perpendicular to its plane. Let a = the side of the plate and ^ the miiss of a unit of area. (2) T»,»irt3; (3) |wfl«. 6. Find the moment of inertia of an isosceles triangular plate, (1) about an axis through its vertex and perpen- dicular to its plane, and (2) about an sxis which passes through its vertex and bisects the base. Let 2b = the base and a = the altitude, then ^0 ''0 m fi (a^ + y») rfy dx = ^ (3«» + b>) ; (2) ^»»i». 225. Moments of Inertia relative to Parallel Axes, or Planes. — The moment ofimrtia qfa body about any axis is equal foils moment of inertia abopt a parallel axis through the centre of gravity of the body, plus the product of the mass of the body into the square of the dis- tance between the axes. Let the piano of the paper pass tlirough tho centre of gravity of tho body, and be perpendicular to tho two parallel axes, meeting them in and O, and let P be the projection of any element on the plane of the pa])er. a oircalar plate (1) oat a tangs^nt. L) im^; (2) |»ia«. a square plate, (1) perpendicular to itn ihe middle points of sis passing tlirough )orpeudicuIar to its and n the mass of a 6 ~ c" ■ I isosceles triangular vertex and pcrpen- sxis which passes ido, then k» + «») ; (2) ^mb*. tive to Parallel tia qfa body about rtia nbo^t a parallel the body, plus the he square of the die- EH BXAMPLES. 433 Take the centre of gravity, O, as origin, the fixed axis through it perpendicular to the plane of thepapw a» tlie axis of z, and the plane through this and the pai-allcl axis for that of 2*; and let /, be the moment of inertia abont tlio axis through G, /that for the parallel axis (hrougn 0, a the distance, OG, between the axes, and (.r, //) any point, /'. Then we shall have /, = 2 (.i;» + yi) dm ; / = X [(a; + a)^ + f] dm. Hence I — I^ = 2« Zxdm + a'lrfm = a'hti, aineo I.xdin ~ 0, as the centre of gravity is at the origin. .-. /--= 7, +rt»m, (1) wliich is called the formula of reduction. llonco the moment of inertia of a body relative to anv axis can be found when tiiat for the parallel axis through its centre of gravity is known. Cor. 1. — The moments of inertia of a body are iho same for all pai-allel axes situated at the same distance from its centre of gravity. Also, ot all jjarallel axes, that which passes through the centre of gravity of a body has the least moment of inertia. Cor. 2.— It is evident that the same theorem holds if the moments of inertia bo taken with respect to i)arallel planen, instead of parallel axes. A similar proi)erty also connects the moment of iiupiiii relative to any point with that relative to the centre of gravity of the body. EXAMPLES. 1. The moment of inertia of a rectangle* in reference to an axis through its centi-o and parallel to one end is * Sec Note to R«. t, Art. *M ; strictly fpt^WiiR, an area bat s moment of inertta no more tUau it bw weight. ■ mmxiini i Jnm 434 SADIU8 or' aVBATtOff. ■^^mtP ; find tho motoent of iaertia in rdercnee to a parallel axis throngh one end. From (1) we have /, = ^m

int of oonoentra- ineipal centre of BADWa OF aniATIO/f. Let A:, = the {Nrinoipal radius of gyration and r, the distance of an element fivm the axis through the centre of f^ravity; then from (1) we have = £ Ti^dm + mlane, by (Ex. 3, Art. 334), is S^Sa^^S 4E3SaaswTOiasar 438 XXAMPLXS. therefore the moment of inertia of the whole solid is ' f/[/W^; (1) the integration being takeji between proper limits. EXAMPLES. 1. Find the moment of inertia of a right circular cone about its axis. let h =r. the height and * = the radius of the base ; then the equation of the generating curve is y = * a-, which in (1) gives for the moment of inertia, 10 m* Jo ^ ~ = A'"**, (since m = ^f^hlA' I since m Therefore *,« = ^. 2. Find the moment of inertia (1) of a solid cylinder about its axis, b being its radius and h its height, and (2) of a hollow cylinder, b and b' being the external and internal nsdU. Am. (1) f«*» ; (2) ^m {W + b^ 3. Find the moment of inertia of a paraboloid about its axis, h being its altitude and b the radius of the base. Am, -g-. 229. Moment of Inertia of a Solid of Revolntion, with respect to an Axis Perpendicnlar to its Geo- metric Aada.— Take the origin at the iiitcrscctiou of the rholo solid is por limita. right oircalar coiic dias of the base; b curve 18 tf = T X, irtia, >^y t a solid cylinder ts height, and (2) the external and 2) *»« (i» + b'^. raboloid about its I of the base. TTfihb* 6 Ant. 1 of ReTOlntion, Dlar to its Ooo- iitcrsc'Ctiou of the EXAMPLSS. 489 axis of revolution with the axis about which the moment of inertia is required ; and denoting by x the distance of the centre of any droular plate from the origin, y its radius and dx its thickness, we have for the moment of inertia of this circular plate, about a diameter, by Ex. 4, Art 224, therefore (Art 225) the moment of inertia of this plate about the pandlel axis at the dutanoe x from it is therefore the moment of inertia of the whole solid is ''t'f(l+!f^)^^> (1) the integration being taken between proper limita. EXAMPLES. 1. Find the moment of inertia of a right circular cone about an axis through its vertex and perpendicular to its own axis. Let A = the height and b = the radius of the base, then the moment of inertia from (1) 2. Find the moment of inertia of a cone, whose altitude = h, and the radius of whose base = b, about an axis through its ceuire of gravity and perpendicular to its own axis. Ans. ^m{h* + 46»). *fci« i> 440 EXAMPLES. 3. Find the moment of inertia of a pamboloid of revolu- tion about an axis through its vertex and periHjndicular to its own axis, the altitude being A and the radius of tho b 230. Moment of Inertia of Varioua Solid Bodies. EXAMPLES. 1. Find tho moment of inertia of a rectangular parallel- epiped about an axis through its centre of gravity and par- allel to an eilge. Let the edges be a, b, c; !»inee a parallelopiped may bo conceived as consisting of an infinite number of rectangular laminje, Oaeii of which has tiie same mdius of gyration relative to an uxi« ixfrpendicilar to its plane, it foUowa that the radius of gynition of the parallelopiped is tho same as that of tho lamina!. Hence, the moments of inertia roltUivo to three a.\e8 through the centre and par- allel to the edges a, h, c. n^spectively, are by Ex. 1, Art. 237, ^m (ft» + c^), ^m («» -f r^), ^,tn (a* + 6»). 2. Find the moment of inertia of a rectangular parallel- opiped about an edge. This may be obtained immediately from the last oxam* pie by using Art. 225, or otherwise iudefx-ndently afi follows : Take the thi-ee edges a, b, c for the axes of x, y, «, ro8i)ectively ; let [t be the mass of a unit of volume, then tho moment of inertia relative to the edge a is = r I A (»»+«») rf« rfy ''« •/Q •■ '0 'm- iboloiil of revolu- peri^ndicular to the radius of tho Sclid Bodies. tanpular parallel- gravity aud par- lelopijwd may bo xsr of retitangulur idius of gyration plauo, it follows illelopiped is tlio the moments of contro and par- e by Ex. 1, Art. lingular parallel- m the last oxam* indofK-ndcntly as axes of X, y, z, of volume, then a is ,'/ 'f^ c'U MOTIOy OF IS'ERTIA OF A LAMIXA. 441 and similarly for tho moments of inertia about the edLn^s /' and c. * ' The moment of inertia of a cube whose ed^jo is a with respect to ono of its edges m \iia^ = ^i„u\ a. Find the u.jraent of inertia of a segment of a -phero relative to a diameter parallel to tho pkue of flection ^ho nidms of tho sphere being « and the distance of tho lilane section from the ct-ntre b. Am. i^TT (IGflS + \^b + 0)0^1^ ._ y^), 231. Moment of Inertia of a Lamina with respect to any Aaris.— When the moment of inertia of a plane fijruro about any axis is known, we easily find tho moment; oL inertia about any {)arullel axis (Art. 225) ; ,il,wo. whoa the moments of inertia about two rectangular axes in tin- j.lano of the fignrc are known, the moment of inertia about i li(} straight line at right angles to the plane tf these axes it their intersection is known immotliately. (A^ Ui) ; wo now proceed to find the moment of inertia about 'any .straight lino- in the plane inclined to these axes at anv JillglO. Through any point, 0, as ■rigin, draw two rectangular )>;es, OX, OY, in the ])lane of Pig.w !he lamina; and draw any iraight line, OX, in the plane. ii is reiH^i*^^.^-,-— -Hr^--..---..^:^^-^^^-.^^-. ^:;^A..:;^;i^Vft..-..-.t/ntYi"--->--'i-'-'i r-T\i.i "1 442 PRINCIPAL AXJSS OT' A BODY. fil! 7 = X ;)«Jm = £ r» 8in« (0 - a) rfm = 1 (y cos « — a; sin «)* rfm = cos* a £ y*i ^ycotp\■^ ooa y 'Ih. 0P» = «• 4- »• + i •, and 1 =co»>- a + «»•/? + ooa* y. The rnomert of inertU I atjont Ote = SmPN" - ttn (OP» - 0K») = SW It' + V' * »' (* cos « f V com /f + • eo* y)«] =• l>»»[(*'+|f' + »')(COB«afC0«»^KC0B« ^)-(irCOB« + jrCO«/J+fC eosa - ycodocoii/?. (1) To rapreaont lhi» Ruometrleally, take a polut Q on ON ; and let Ita dJatue* from O be r, and Ifcs co-ordinate* be*,, y,, I,. Then «, -recMa. jr, - rcuK^, «. ^rco*). Fig. 814 ■■T--^'it''-^iii'iii'wi».i i'-'^ h'f: 444 TURSS PRtSCIPAL AXXS. ScH. — In many cases the position of the principal can bo seen at once. Suppose, for example, we wish principal axis for a rectangle when the given point is the centre. Dravr through the centre straight lines parallel to the sides of the rectangle ; then these will be the principal [Mil axes I mh the I f ia flw> Tbaufor* (1) becomef I - m, -ytj.y. But the equation dpnoleg a^vflT^iXfJit' srivjgit tavbf- Is sU O; becanw a, ti, e are nccegsBrily posUlvo, eiuee » mr- I at lovrtift U MWfitlally positivu, being the sain u( • bobiIwi' of Miuarr 1 ({ U tt point on tbU uUlpcoid, (3) Ut'oomee I c SmPn* M>id, which coloeidos wllh the liuo. '{Ilia li ckUed Uiit niommtat cS^/mld, and ww first aaed by Oauchy, Mrtrcimt d» Hath., Vol, n. It tuti no phyxical oxlcteDoe, hat In an artiRoi to bring under Ute rauthodH of (jeoinctry tho pro|)ert5en of inomcTiln of tiioKla. The mouienlal elllp. gdid han a drtlulte form tor cvory polBt of a rigid body. Now every elli|if>old hoa thvoe axoc^, to which If il \* ivlbrred, the coefficients of y», cr, xy vanlidi, aud therefore (!)), when tnuiBfbriued to those asus takes the form . _ A/,' ( By," \ 0,« = 1; («) Biul hence (1) or (») wh'in raAnred to these axcB, becoroee 1 « A eoe* • + B to** ^ + COB" »-, m whoro A, B, C, are Uie niomentti of inertia of the body about then! a: m. When three recxangalar axe;!, meeting In a t^veu point, arc chuRon w) that ihe prodncM of Inertia all v«Hl»h, tlie> are c*lle eallw' itio prindija/ }>!cme^ at lltt' ^Ivon point. The monionts of inertia ahont lh« piincjial nxwi at o«iy point are railed the prin- fi;in/ tncnumtJi tf i»*rHa at thai p«ilnt. If the (hreo prlnii|)al moinenti> of Inertia ofa body are Mjniil to one another, the oUl|iHoi'l '4) I«comeii a Hp'jere, ulni^c A =. B - V ; 3nJ therofotc the inumeut of iMrlla about every other atlt it oqiuU to L-nae, tor (R) becoiuua 1 ■-'< A rii] Dynamic*, p. I«, Price's AuftV S.cliv Vui, a. 1. IW, I-irWrJUglilDyuaiulcc, p TO, iie - # principal oxoa 1, we wish tlio a point is tliu 11C8 parallel to I tbo priuci(Ml *. = I. m (8) 9i'0a««riiy postllro, un uf t DDiDlter of td by the aqnsro of coloeldM with the ^auchy, Mtereitet d» I to bring under tlie lie mouienlal I'lliji- Ihc cocfllcleiit« of boDe iLYuii takuD tlio (*) (») iP.ni a: :m. churan no that the aiM at the givvu ilio prlneiiMtl planti tarct-«nei1 IhopHn- lo on« another, tbo 'ore (ho inomeut of laniic*, It. 1«. IMco's TBRSE PBTNCIPAI, AXES. ««8 axes; becaase for orery element, dm, on one side of the axis of X at the point («, y), there is another elenient of equal mass on the other side at the point {x, —y). Hence, yLxydm consists of terms which may bo arranged in pairs, so that the two terms in a pair arc numerically equal but of opposite signs ; and therefore £ xy din — 0. Agiiin, if in any uniform body a straight lino can Imj drawn with respect to which the body is exactly symmetri- cal, this must be a principal axis at every point in its length. Any diameter of a uniform circle or sphere or the axis of a parabola or ellipse or hyperbola is a principal axis at any point in its line; but the diagonal of a rectangular ])lat« is not for this reason a principal axis at its middio point, for every straight line drawn perpendicular to it is not equally divided by it. Let the body be symmetrical about the plane of xy, then for every element dm^ on one side of the plane a*: the point {x, y, «), tliero is another elem nt of equal mjusa on the other side at the point {x, y, — *). Hence, for such a body 2 xz dm = and £ yz dm ■=. 0. If the body bo a lamina in the plane of xy, then z of every element is zero, and wo have again £ xz dm = 0, !• yt dm = 0. Thufi, in tlic case of the ollipsoid, the three principal sections are all planet of symmetry, and therefore the three axes of (he ellipsoid are principal axes. Also, at every jKiint in a lamina one principal axis ia the pcrpendioular to tlie pltme of the lamina, EXAMPLES. 1. Find the moment of inertia of a rectangular lamina :il)uut a diagonal. From Ex. 2, Art. 5S24, the moments of inertia about two linos through the centre parallel to the sides (principal moments of inertia) ore fm 446 BXAMPLSa. • » I ^m<2* and VfWii*; ■whoro h ami rf aro the breadth and dopth respectively. Also, if « be the angle which the diagonal raakee with the tiide i, we have «n» a = I>> + (P' 00^ rt = d2 + <^ SnVwtituting these values for A, B, sin* «, cos» «, in (2) of Art. 231, wo have = ^m 2. Find the moment of inertia of an isosceles trfangnlar plat« about an axis through its centre and inclined at an angle « to its axis of syrametry, a being its altitude and 2ft its baae. Ans. ^ (^a* cos* « H- J» sin* «). 3, Find the moment of inertia of a square plate about a diagonal, « being a sidi! of the square. Jns. ^maK 233. ProductB of Inertia.— The value of the product of inertia at any iwint may be made to depend on the value of the product of inertia for parallel axes through the cen- tre of gravity. Let {x, y) lie the position of any element^ dm, referred to axes through any a88ign<'d point ; {x', y') the jwsition of the element referred to parallel ixes through the centre of gravity, and (A, k) the centre of gravity referr(Hl to the firat pair of axea Then ar = «' + A, y =. y' + h) tlM»fore "£ xydm^I. {x' + h) {y' + h) dm = Y,r!y' dm + hk^dm, (1) since S wwf' = 0, and S my' 0. J,#Sv SXAMPLSA U1 KJCfcively. U makes with a« «, in (2) of elea triangular inclined at an Ititude and 2b H- J» 8in» «). plate about a Ans. '^ftnaK of the prodnct id on the value rough the cen- f any element, iut ; {x', tj') the il ixes through itre of gravity Im (1) ScH.— By (1) we may often find the product of inartia for an assigned origin and axes. Thus, suppose we roqnire the product of inertia in the case of a rectangle, when the origin is at the comer, and the axes are the edges which meet at that ooruer. By Art 332, Sch. we have I.z'y'din = 0; therefore from (1) we have and as h and k are known, being half the longthn of the edges of the rectangle to which they are rcsiiectivcly parallel, the product of inertia is known. BXAM PL.BS. Find the expressions for the moments of in«tia in the following, tiie hwiies beiag wipposed homogeoeous in all cases. 1. The moment of inwtia of a rod of length a, with respect to an axis perpendicular to tite rod and at a distance d from ita middle point , /a" \ * Ans. m{^~ + dn- 2. Tlie moment of inertia of an arc of a circle whose radius is a and which subtends an angle 2« at the centre, (1) about an axis through its centro perpendicular to its plane, (2) about an avis through its middle point perpendicular to its plane, (3) about the diameter which bisects the arc. Ans. (l)«,a- (2) 2„. (!_?!£_«)«.; (8) «, (l -.!l|^ «*. 3. The moment of inertia of the arc of a complete cycloid whose length is a with respect to its base. Ans. ^ffna', 4. The moment o( inertia of an equilateral triai.gle, of aide a, relative to a line in its plane, parallel to a side, at the disUutoo d twiu its oeuke of gravity. . ' Am. m (I + ..W.t.■l.,.L.-^:.ljiA^t>fc•^.^^|J.^,■^.,,, .y:...r..^^-....^-^- . 418 SXAMPLSS. h 6. Oivcn a triangle whoBc- sides are a, b, e, and whoso jwrpendiculars on these sides, from the opposite vortices are p, q, r, respectively ; find the moment uf inertia of the triangle about a line drawn through each vertex and parallel respectively, (1) to the side a, (2) to the side b, {?,) to the side e. Arts. (1) Impf^; (3) |tw7« ; (3) ^m,^, C. Find the moiTient of inertia of the h'iiinglo in the last example relative to the three lines drawn through the centre of gravity of the triangle and parallil respectively to the sides a, b, c. Ann. ^mf\ if,m(j*\ i-xiniK T. Find the moment of inertia of the triangle iu Ex. 5, relative to the three sides a, h. c, respectively. Am. {nlf^'f ifUKf; \mr'K 8. The monioiit of inertia of a right angled triangle^ oi hvputlienose c, relative to a perpendicular to its plane J)a8>itig through the right augle. Av.^. \m^, !t. Tiic moment of inertia of n ring vvhcisc outer atttf inner radii aie a and b respectn ely, (1) with respect lo a IM)lar axis tiiroiigh its centre, and (2) with respect to a diameter. Am. (1) ^m («» -|- h^) ; {%) \m («» + S^). 10. The mof'ienf of inertia of an ellipse, (1) with respect to ')\f> miijor axis, (i) with respect to its mhior axis, and (3) with reaiiet to an tt>«iH through its centre and pc-riiendieular U) it« plane. 4n». (\) ^110 \ |l) Imn'i; (.1) \m («» + A«). 11. The mmmui uf inertia of Hie surface of a rfphero of radius a ahout it.s diameter. Aiis. ImaK li. The moment of inertia of a ri(i)i( pri«m whose haae is !i rigiit anglal triangle, witli l'OB[wot to au axis passing {limutrh thr eentres of gravity of the eud% tUo tldoa cou- laining the right angle of the triangular base being a and b and the height of the prism c. Ans. ^m{a^+¥). ifeite MMMMMM ', c, and whoso jposite vortices L inertia of tho 3h vertex and tlie side h, (3) r« ; (3) \m,^. iglo in the last 1 through the It'l respectively angle in Ex. 5, r. \ /«'/' ; f/Hr*. led ihimgh, of ' to its phe u.-ir outer s»j«t h respect to a I respect to a [m (rt» + HI). I) witli rosjwct »r axis, and (3) ptriiendicuUir ni {(ii + *«). I if (I riphore of Alts. |i//«^, (III whoge baao lu avis pasHtng tUo tidofl oou- being a and b XXAMPLJB& M» 13. The m(»D(«t of ioertia of a right pmnt vhoae height is c, about an axis passing throngh the centres of gravity of the ends, tho base of the piistQ being an isosceles triangle whose base is a and height b. /a> Nt\ ■^-id + r)- 14. The moment of inertia of » sphere < f. radins a, (1) relative to a diameter, and (3) relative to a tangent. Ans. (l)fwo»; (3) |»ja». 15. Tho moment of inertia, about its axis of rotation, (1) of a prolate spheroid, and (3) of an oblate spheroid. Ans. (l)|«ift»; (3)|7»fl2. 16. Tho moment of inertia of a cylinder, relative to an axis perpendicular to its own axis and intersecting it, (1) at a distance c from its end, (3) at the end of the axis, and (3) ■.hi f ho middle point of the axis, the altitude of the cylinder boujg fi and radius of its base a. Ans. (1) ima» + ^m (A» + 8Ac + c») ; (3) ^m (3«» + 4/t«) ; (3) ^m (7*3 + 3«») moment of inortirt of an ollipso about a central radius vector r, making an angle « with the major-axlB. Am. im-^. 18. The moment of inertia of the area of a parabola cut off by any ordinate at a distance a, from the vertex, (1) aliout the tangent at tho vertex, and (3) about the axis of tho iMuiibola. Ans. fwa? ; (3) ^my^ where y is the ordinate correspond- ing to «. l». The moment of inertia of the area of the lomniscatey »■» =z «' COS m, about a lini* through the origin in its plane uiid periRiudicular to it^ a^is. Am. ^ftn (3rr f 8) «». wg,?i i i wwgig flFgy *■ ■■immii fin fiifrfrrtiii 460 MXAMP/.ES. 20, The moment of inertia of the ellipfsoid, ^ + ^ + ^-^- about the axis a, ft, c, respectively. Am. (1) im(*» + c») ; (2) i«i (c» + a») ; (3) iw (a» + J»). j^ s 1, J«i (c» + a») ; ''s^rtliBSUBSSH CHAPTER VII. ROTATORY MOTION. 234. Impressed and Infective Forces.— All wes acting on a body other than tho mutual actionft , tho particles, are called the Impressed Forces that act on the body. ThuB, when a ball is thrown in vacuo, the impresst*! force is gravity; if a ball is rotating about a vertical axis, the impressed forces are gravity and the reaction of tho axis. The impnssod or external forces are the cause of the motion and of all tlio other forces. Which are the impraosed forces depends npon the particular system which is under consideration. Tho same force may be external to one system and inn. rnal to another. Thus, tho pressure between the foot of a man and tlie deck of a ship on which he ifl, is external to tho ship and also to the man and is thfl cause of his own forward motion and of a slight backward motion o, Jio ship ; but if the man and ship are considered as parts of one system the pressure is internal. When a particle is moving as part of a rigid body, it is acted on by the external impressed forces and also by the molecular reactions of the other particles. Now if ".his particle were considered as separated from the rest of the body, and all the forces removed, there is some one force which, singly, would move it in the same way as before. This force is called the Effective Force on the particle; it is evidently the rtsultant of the impressed and molecular forces on the particle. Thus, the effective force is that part of the impressed force which is effective in causing actual motion. It is tho force which is required' for prodncing tUo deviation from the straight line and the change of I <«M^M«wM«^Mi^ii 4ii«S«*WWW. '■ima IMAGE EVALUATION TEST TARGET (MT-S) //I . C/j 1.0 I.I ijiiiia i2.5 .JI ilk 12.2 M 1.8 1 1.25 1.4 1.6 ^ 6" ► ^/ / m " ^^ ^ ^/» ••';■ '/ Photographic Sciences Corporation 91 WESY MAIN STRieT WEBSTE.'.N./. MS80 (716) &73-4503 # :\^^ iV \ "9) V IT- o l\? Vr^^ ^f^ ■^■"i^"^ <» L.„ C'. t^ 4 *^^ ^ . CIKM/ICMH Microfiche Series. CIHM/ICMH Collection de microfiches. Canadian Institute for Historical Microreproductlons / Institut Canadian de mlcroreproductlons historiques ft I ^. n il i«r i ij t ;iM i i i| »i ii rn;Tiini i i r wM i r( i 452 D 'A LEMnKR T 'S PRINCIPLE. vplocitv. If a particle is revolving with constant velocity round a fixed axis, the effective force is the ceatripetal forc« (Art. 198). If a heavy Ixxly fulls wiihoiit rotation, the whole force of gravity is effective ; but if it \» rotating about a liorizoutal axis the weight goeti partly to kiJauce the prcHsure on the axis. If we suppose tlie particle of mass m to bo at the point (r, y. z) at niiy time, /, flnd resolve the forces acting on it into the tiireo axial com])onent8, X, Y, Z, the motion may be found [Art. 108 (2)] by solving the Bimultj.neou8 equa- tions (Pi/ = r m ^, := Z. a) ■^« ite- If we regard a rigid body as one in which the particles retain inviiriable po-sitions with respect lO one another, so that no external force can alter thon (Art 43), we might write down the equations of the several particles in accord- ance with (1), ii" all the forces wore known. Such, how- ever, is not the ease. We know nothing of ihe mutual actions of the particles, and ccn80y y'Alenibcrt lii n«. D 'ALEMBERT'S PRI^CtPLB. 458 velocity round a (Art. 198). If a CO of gravity is the weigbt goes I at the point J acting on it e motion may ituneous cqvia- \ the particlcd )ne another, so 43), we might cles in accord- I. Such, hew- )f the mutual inot determine ! motion of its 111 rigid bodies or tlie problem D'Alombert's fttions may be 18 of motion of ption a8 to the •llowing. which of the laws of system of rigid ■temselws. \ The axial accelerations of the particle of mass m, whicli cPx dt^v (Pz is moving as part of a rigid body, ai-e ^, -^i^ -^^' Let/ be their resultant, then the eflfeotive force is measured by mf Let ^and R be the resultants of the impressed and molecular forces, i-espectively, on the particle. Then mf is the resultant of F and R. Hence if mf bo reversed, tlio three foi-ces, F, R, and mf, are in equihbrium. The same reasoning may be applied to every particle of eiich body of the system, thus furnishing three groups of forces, similar, respectively, to F, E, and mf: and these three groups will form a systeir of forces in equilibrium. Now by D*Alembert's principle the group 7^ will itself form a system of forces in equilibrium. Whence it follows that the group F will be in equilibrium with the group mf. Hence, If forces equal and exactly opposite to the effective ^forces were applied at each parfick of the system, they would l>e in equilibrium with the impressed forces. That is, H'A'emhcrCs principle asserts that the lohoU effective forces of a system are together eqnivaknc to the impressed forces. SoH. — By this principle the solution of a problem in Kinetics is reduced to a problem in Statics ha follows : We first choose ihe co-ordinates by means of which the position of the system in space may be fixwi. Wo then express the effecii^o forces on each element in terms of its cc-oi'dinates. Those effective forces, reversed, will be in equilibrium with the given impressed forces. Lastly, the equations of motion for each body may b^ formed, as is usually done in Statics, by resolving in three directions and taking mo- ments abont- throe straight linos. (See llouth's Rigid Dynamics, Pirio's Rigid DynamicB, Pratt's Mech's, Price's Anal. Mech's, Vol. IL) ^.^•i^ ^HMMWKMMMWMM 464 ROTATION OF A liWID BODY. 236. Rotation of a Rigid Body about a Fixed ii^B under the Action of any Forces.— Let any plaue pagRing through the axis of rotation und fixed in space be taken as a piano of reference. Let m be the mass of any element of the body, r its distance from the axis, and the angle which a pkuo through the axis and the element makes with the plane of rekronce. Then the velocity of m in a direction yrpendicular to the plane containing the element and the axis is r-^-. The moment of the momenluni* of this particle about the axis ia mr' --,-• Hence the moment of the momenta of at all the particles is d6 at 0) Since the particles of the body are rigidly connected, it is clear that -=j is the same for every particle, and is the angular velocity of the body. Hence the moment of the momenta of all the particles of the body about the axis is the moment of in^tia of the body about the axis multiplied by the angular velocity. The acceleration of m perpendicular to the direction in whicii r is measured is r -^ , and therefore the moment of the moving forces of m about the axis is 'wr*-,^- Hence, the moment of fhe effective forces of all the particles of the body about the axis is Iwr» (m (2) which is the moment of inertia of the body about tJie axis mkltiplied by the angular acceleration. * Called alHi) Angular Momentum. (Soo Itrlo'ii Rlcld Dy niunlce, p. 44.) . >ut a Fixed ses.— Let any 1 and fixed in m be the mass from the axis, axis and the ^rpendicular to . . do LC axis \a ''j-.' particle about ;he momenta of (1) idly connected, icle, and is the moment of the t the axis is the is multiplied by the direction in the moment of fir^-jp- Hence, 1 particles of the dy about the axis DyniunlcB, p. 44.) ROTATION OF A RIGID BODY. 455 (1) Let the forces be impulsive (Art. 203) ; let 6), w', be the angular velocities just before and just after the acaon of the forces, and N the moment of the impressed forces about the axis of rotation, by which the motion is pro- duced. Then, since by D'Alembert's principle the effective forces when reversed are m equilibrium with the impressed forces, we have from (1) w' S mi* — w X rnr* = JV"; « (J := N "Lmt* ' moment of imp ulse abo ut axis , ~ momo^IofTncrtia about axis ' (8) that is, the change in the ai^^^^r velocity of a body, pro- duced byan impulse, is equal to the m^^ient of the impulse divided by the moment of inertia of the body. (2) Let the forces be finite. Then taking tiicr.'^^nts about the axis as before, we have from (2) (Pd dt^ JL ^mi^ moment of fo r ces about axis . ~ moment of inertia about axis ' (4) that is, tU angular acceleration of a body, produced by a force, is equal to the mommt of the force divided by the moment of inertia of the body. By integrating (4) we shall know the angle through which the body has revolved in a given time. Two arbi- trary constants will appear in the integrations, whose values are to be determined from the given initial value? of and ~- Thus the whole motion can bo L\mA, and 456 MXAMPLE. WO shall conseqaontly be able to determine the position of the body at any instant ScH. — It appears from (3) and (4) that the motion of a rigid body round a fixed axis, under the action of any forces, depends on (1) the moment of the forces about that axis, and (2) the moment of inertia of the body about the axis. If the whole mass of the body were concentrated into its centre of gyration (Art. 226), and attached to the fixed axis of rotation by a rod without mafls, whose length is the radius of gyration, and if this system were acted on by forcoe having the same moment as before, and were set in motion with the same initial values of 6 and the angular velocity, then the whole subsequent angular motion of the rod would be the same as that of the body. Hence, we may say br'ofly, that a body turning about a fixed axis is Irinetically given when its maaa ana radius of gyration are known. EX ADA? L F. A rough circular horizontal boi. i is capable o£ vixoh']ng freely round a vertical axis through its centre. A man walks on and round at the edge of the board; when he has completed the circuit what will be his position in space ? Let a be the radius of the board, M and M' the maeses of the board and man respectively ; B and 0' the angles described by the board and man, and i^^the action between the feet of the man and the board. The equation of motion of the board by (4) is Fa - ^*.»^' ( Since the action between the man and the board is con- tinually tangent to the path described by the man, the equation of motion of the man is, by (5) of Art. 20, mmmm* the position of tat the motion le action of any be forces about the body about ere concentrated attached to the !8, whose length oa weie acted on re, and were set and the angular ar motion of the Hence, we may a fixed axis is I of gyration are able o£ Vixohing centre. A man board; when he I his position in id M' the masses and 0' the angles e action between (4)i8 the board is con- by the man, the ,f Art. «0, "'^wnmmP'- m!m»!»s^mmmmm>efmBii>i^>m^f>f»»!" TBI! COMPOUND VENDULUM. IP V ^ m Eliminating i^and integrating twice, the constant being zero in both cases, because the man and board start from rest, wo get Mk?0 - M'aW. (1) When the man has completed the circuit we have + 0' = %iT\ also ^'i* = -. Substituting these in (1) we get V * — 2nM -iM' + M' which giTOs the angle in space described by the man. If M — M', this becomes and d' = \ir; = f7r, which is the angle in space described by the board. (See Rye^b's Rigid Dynamics, p. G7.) 237. The Compound Pcc<3nlum.— ^1 body moves aiont n fixed horizontal axis acted on by gravity (?vli/, to determine the motion. Lot ABO be a section of the body made by the iilane of the paper passing through O, tlie centre of gravity, and cutting the uxis of rotation iierpondioularly at 0. Lot — the angle which 00 makes with the vertical OY; and let h = 00, t, = the principiil radius of gyration, and M = the mass of the body. Then hy (4) of Art. 336, we have 20 •iMfl-TT-- TlWiai ''''"•""' I'i ii "r"'n'r i - rn' ' ■imrr 468 TH£ COirPOUND PENDULVM. "~f Mgh ain d _ Mgh Bin 6 = ~ ^;nnra sin e [by (2) of Art. 226], (1) the negative sign being taken because 9 is a decreasing function of the time. This etjuation cannot be integrated in finite terms, but if the oscillations be small, we may develop sin f> and reject all powers above the first, and (1) will become (Id gh *,» + A« e. (2) Multiplying by 2 -^ and integrating, and supposing that the body began to move when 6 was equal to «, (2) becomes rffl* _ gh dP M(«»-n *,^ + A« Hence denoting the time of a complete oscillation by T, we have /h" + k{' (8) which gives the time in seconds, when h and Jfe, are meas- ured in feet and g = 32.18. When a heavy body vibrates about a horizontal axis, by the force of gravity, it is called a compound pendulum. CoE. 1. — If we suppose the whole mass of the compound pendulum to be concentrated into a single point, and this point connected with the axis by a medium without weight, it becomes a simple pendulum (Art. 194). Denoting the distance of the point of concentration from the axis by I, wc luvvc for the time of an oscillation, by (1) of Art. 194, afl of Art. 326], (1) is a decreasing inite terms, but sin and reject me («) i supposing that equal to «, (2) oscillation by T, (3) and yfci are meas- orizontal axis, by d pendulum. 1 of the compound ;le point, and this m without weight, t). Denoting the •om the axis by I, y (1) of Art. 194, ■ ' '^fSISWI^SWiSSW' CBXTSES OF OSCILLATION AND SUSPENSION. 459 "V^- If the point be so chosen that the simple pendu- lum will perform an oscillation in the same time as the comijound pendulum, these two expressions for the time of an oscillation must be equal to each other, and wo shuU liave 1 = h A + ^ = 00', (4) (0' being the point of concentration). Cob. 2. — This length is called the length of the simple equivalent pendulum ; the point is called the centre of suspension ; the point 0', into which the mass of the com- pound pendulum must be concentrated so that it will oscillate in the same time as before, is called the centre of oscillation; and a line through the centre of oscillation and ps-i-allel to the axis of suspension is called an axis of oscillation. From (4) we have {l-7i)7i = *,»; or GO'. GO = *,». (5) Now (5) would not be altered if the place of and 0' were interchanged;* hence if O' be made the centre of suspension, then will be the centre of oscillation. Thus tlie centres of oscillation and of suspension are convertible, and tfie time of oscillation about each is the same. CoR. 3. — Putting the derivative of I with respect to h in (4) equal to zero, and solving for h, wo got h — k^, 460 EXAMPLES. wliich makes I a minimum, and therefore makes / a mini- mum. Hence, when the axis of suspension jjasses thmuijh the principal centre of gyration the time of oseillaiion is a minimum. Rem. — The problem of dctnrmining the law under which a heavy body swings about a liorizontftl nxia is one of tlie most iniiwrtnnt in tlic lilgtory of science. A simple pendulum is a thing of theory ; our accurate Imowlodge of the acceleration of gravity depends tberoforo on our understanding the rigid or compound pendulum. This was th<> first problem to which IVAlembert applied his principle. The problem was called in the days of D'Alembert, the " centre of oscillation." It was required to find if there were a point at which the whole mass of the body might be concentrated, so as to form u simple pendulum whose law of ot^cillation was the same. The jxiRitlon of the centre of oscillation of a body was first correctly determined by Huyghens and published at Wans in 1678. As D'Alembert's principle was not known at that time, Huyghens had to discover some principle for himself.* EXAMPLES. 1. A material straight line oscillates about an axis jier- pendicnlar to its length ; find the length of the equivalent simple pendulum. Let 'ia = the length of the line, and h the distance of its centre of gravity from the point of suspension. Then since ij« = ^-, we have from (4) l = h + U (1) Cor. 1. — If the point of suspension be at the extremity of the line (1) becomes * Rontb's Rigid DTDiimlcB, ]>, 69, EXAMPLES. 461 makes t a mini- fy 2Jasses through f oscillation is a ndor which a heavy ! most iniiKirtnnt in liing of tlieory ; our y depends therefore oduluiu. This was I principle, ibert, the " centre of e a point at which ted, BO as to form a 9 same. dy was first correctly tans in 1678. As le, Huyghens had to xbout an axis -pev- of the equivalent the distance of ite ision. Then sine© (1) le at the extremity that is, the length of the equivalent simple pendulum is two-thirds of the length of the rod. Cob. 2.— Let /t — |a ; then (1) becomes I = !«.. Hence, the time of an oscillation is the same, whether tho line bo suspended from one extremity, or from a point one- third of its length from the extremity. This also iilustnites the convertibility of the centres of oEoillation and of sus- pension (See Cor. 2). Cob. 3.— If h = 10a, then (1) becomes 1 = 2. A circular arc oscillates about an axis through its middle point i^erpendicular to the plane of the arc. Prove that the length of the simple equivalent pendulum is independent of the length of the arc, and is equal to twice the radius. From Ex. 2, Art. 333, we have P ^ /^ sin <*\ o From Ex. 1, Art. 79, we have h — a — a sin a Therefore (4) becomes « .. /^ sin «\ /i sin «\ „ I = 20" ^1 ^— J -7-a\i ~j = 2rt. ,finn Iff LENOTII OF TBTS SECONffS PEXDVLCM. 0jl m' Sll 3. A right cono oscillates about an axis poasiug through its vertex and perpendicular to its own axis ; it is required to find the length of the simple equivalent pendulum, (1) when h is the altitude of the cone and b tlie radius of the baflo, and (;j) when the altitude = the radius of the base = h. Ans. (1) —gy- ; (3) h. That is, in t!ie second cono, the centre of oscillation is in the centre of the bawc ; so that the times of oscillation are equal for axes through the vertex and the centre of the base perpendicular to the axis of the cone. 4. A sphere, radius a. oscillates about an axis ; find the length of the simple eciuivalent pendulum, (1) when the axis is tangent to the sphere, (3) when it is distant 10« from the centre of the sphere, and (3) when it is distant - fi'om the centre of the sphere. , Ans. {I) la; (2) W«; O^) V«. 238. The Length of the Second's Penduluir. Determined Experimentally. — The time of oscillulion k * of a compound pendulum depends on h + -,- by (4) of Art. 337. But there are difficulties in the way of determin- ing /» and ky The centre, G, can not be got at, and, as every body is more or less irregular and v&riablo in density, h^ cannot bo calculated with sufficient accuracy. These (piantities must therefore be determined from experiments. Bessel observed the times of oscillation about different axes, the distances between which wore very accurately known. Captain Kator employed the property of the convertibility of the centres of suspension and oscillation (Art. 237, Cor. 2), as follows : ' Let the pendnlum consiHt of an ordinary stralglit bar, CO, and a small weight, m, wliich may be clamped to it l>y means of a screw, uud shifted fruiu ouu positiuu to auuiher cm the peuduluiu. At tho (2) h. DVLOM. paasiug through 8 ; it is required it pendulum, (1) Llie radius ai the 8 of the base = h. f oscillation ia in of oscillation are he centre of the m axis ; find the m, (1) when the it is distant lOrt vhen it is distant W«; (:i) id's Penduluir. nie of oscillation I + ^^ by (4) of way of detcrmin- be got at, and, as iriablo in density, iiccnraoy. These Prom experiments. I about different 9 very accurately property of the m and oscillation aigbt bar, CO, and a ly means of a screw, B penduluiu. At llio "ipwn' ."r^TO tKV^^r^pKV LENGTH OF THE SECOND'S PENDULtm. 403 [Tin* o I >ifl l1]»» Fls.94 points C and O in tvjo triangular aper- tures, at the distance I apart, let two knife edges of hard steel bo placed parallel to each other, and at right angles to the pendulum, so that it mny vibrate on either of them, OS in Fig. 04. Let m be shifted till it is found that the times of oscillatiuu about and O are exactly the same. It remains only to measure CO, and observe the time of oscHlation. The distance be- tween the two points C and O is the length of the simple equivalent pendulum. This distance between the knife edges was measured by Captain Kater with the greatest care. The mean of three measurements differed by less than a ten-thoustindth of an inch from each of the separate measurements. The time of a single vibration cannot be observed directly, because this would require the fraction of a second of time as shown by tho clock, to be estimated either by the eye or ear. The difflouity may be overcome by observitg the time, say of a thousand vibrations, and thus the error of the time of a single vibration is divided by a thousand. The labor of so much counting may however bo avoided by the use of " the method of coincidences." The pendulum is placed iu f.-ont of a clock pcudulum whob- time of vibration ia slightly different. Certain marks made on the two pendulums are observud by a telescope at the lowest point of their arcs of vibratior The field of view is limited by a diaphragm to a narrow aperture "cross which the marks are seen to pass. At each succeeding vibration one pendulum follows the other more closely, and at Inst its mark is completely covered by the other during their passage acrosr the field of view of the telescope. After a few vibrations it appears again preceding the other. In the interval from one disappearance to the next, ono pendulum has made, as nearly as possible, one complete oscillation more than the other. In this manner 530 half -vibrations of a clock pendulum, each equal to a second, were found to correspond to 533 of Captain Eater's pendulum. The ratio of the times of vibra- tion of the pendulum and the clock pendulum may thus be calculated with extreme accuracy. The rate of going of the clock must then bo found by astronomical means. The time of vibration thus found will require several corrections which are called "reductions." For instance, if the oscillation be not so small that we can put sind — in Art. 237, wo must make a reduction to infinitely small arcs. Another reduction is necessary if fill m r<,l M' j%t*«3»irioi«maa4K»S;»iK<»»a^ JaSJ)^ 4G4 MOTION OF A BODY WHEN UNCO NSTBAI NED. IIP we wish to reduce the rpsult to what It would liAve been at the level of the sea. Tha altfaction uf the interveDing land Toay be allowed for by Dr. Young's rule, (Phil. Trans., 1819). We laay thuu obtain the force of gravity at the level of the sea, annpoeing all the laud above this level were cut ott' and the sea cuaBtraincd to keep its present level. As the level of the sea is altered by the attraction of the land, further correcticas au) still necessary if we wish to reduce the result to the snrfaco of that spheroid which most nearly repre- sents the earth. See Routh's lligid Uynamios, p. 77. For the details of this experiment the Htudent \a referred to the Plill. Trans, for 1818, and to Vol. X. < 239. Modon of a Body when Unconstrained.— If an impulf'o be commuuicated to any point of a freo body ill a direction not pfwsing through the centre of gravity, it will ])roditce both translation and rotation. Let P be the impulse imparted to the body at A. At B, ou the opposite eide of the centre G, a distance GB ^ r= AG, let two opposite impulses be > j applied, each equal to \P ; they will not alter the effect. Now if \P applied at A is combined with the ^7' at B which acts in the same diraotion, their resultant is P, acting at G and in the «.ame direction, and this produces translation only. The remaining ^P at A combined with the remaining \P at B, which acts in the opposite direc- tion, form a couple wuich produces rotation about the centre G. Hence, when a body receives an impulse in a 'lirection which doer not pnifS through the centre of gravity, thxt centre ml' assume n motton of trnndalion as ihongh the ImpuUe were applied immediately to it ; and the body will have a motion of rotation about the centre of gravity, as though Uiat jwint were fixed. 240. Centre of Fercusaion. -Axis of Spontaneous Rotation. — Lot Jli> rcprcsuut the impulse imprensod upon IP FI0.9S BAILED. 3een at the level I tnay be allowed !uay thiu obtain sing all the laud aiued to keep Uh - the attraction of we wisb to reduce noBt nearly repre- , For the dutoila I. Trana. for 1818, aBtrained.— II of a free body re of gravity, it rihjmmmm'mm^ ■ 'fm If Ftg.9S rc&ultant is P, id this produces L combined with I opposite direfi- xion about the J tu a 'lirecHon ivity, th tt centre ugh the mptihe 'jody will have a vily, as though f BpontaneouB imprt'ttsiod ui>ou CENTRE OP PERCUSSION. 4G5 the body (Fig. 95) whose mass is M, and h the perpendicular distance, 00, from the centre of gravity, G, to the line of action, OP, of the impnlse. The centre of gravity will assume a motion of trans- lation with the velocity v, in a direction parallel to that of the impulsive force. Then from (3) of Art 236, we havo for the angular velocity Mvh _ vh Fl0:M w = The absolute velocity of each point of the body will bo compounded of the two velocitieft of translfition and rota- tion. TL " point 0, for example, to which the impulse is applied, haa a velocity of translation, Oa, equaUto that of the cen*^^re of gravity, and a velocity of rotation, ab, about the centre of gravity ; so that the velocity of any point at tf distance a from the centre, 0, will bo expressed by I? i: aw ; the upper x - lower sign being taken according as the point is, or is uot, on the same side of the centra of gravity as the point 0. Thus, if we consider the motion of the body for a vevy short interval of time, the line 000 will assume the position bOC, the point C remaining at rest during this interval ; that is, while the point would ho carried forward over the line Cc by the motion of trans- lation, it would bo carried backward through the same distance by the motion of rotation. Ueuce, since the abso- lute velocity of O is zero, wo have ow .-= ; « = - = ^ ; w A ' (1) lud hence denoting 00 Ity I we havo ^uM# 466 AXIS OF SPOyTAffXOUS fiOTATIOff. 1 = (3) Now if there had been a fiied axis throngh C perpen- dicular to the plane of motion^ the initial motion would liave been precisely the same, and this fixed axis evidently would p.ot have received any pressure from the impulse. "When a rigid body rotates about a fixed axis, and the bod^ can be so struck that there is no pressure on the axis, any point in the line of action of the force is called a centre of percussion. When the line of action of the blow is given a^ d the body is free from all constraint, so that it is capable of translation as well as of rotation, the axis about which the body begins to turn is called the axis of spontamous rota- tion. It obviously coinoidcs with the position of the fixed axis iu tbtv first case. Cou. 1. — From (1) we have * ah = GC'OQ = *,»; hence the points and C are convertible, that is, if the (ms of rotation be supposed to pass throvgh the point 0, the centre of spontaneous rotation will coincide with the sen- tre of percussion. Cob. 3, — From (2) it follows, by comparison with (4) of Art. 237, that if the axis of spontaneous rotation coincides wi*h the aids of suspension, the centre of percussion coin- cides with the centre of oscillation. Son. — It is evident that if there be a fixed obstacle at 0, and '<- be struck by the body OC rotating about a fixed axis through C, the obstacle will recei"e the whole force of the moving body, and the axis will not receive any. Hence the centre of percuss Icn also dct?rrainos the position ' in which a fixed obBtuclo niunt bo ])laccd, on which if the rotating body impinges nud iu brought to rest, the axis of rotation will suflior no pretwure. TIOH. (3) rough C perpcn- ial motion would jd axis evidently 1 the impulse, ced axis, and the jsure on the axis, I is called a centre is given a' d the it it is capable of about which the spontamous rota- litiou of the lixed EXAMPLES. 467 lie, that is, if the ovgh the point O, icide with the cen- arison with (4) of rotation coincides \f percussion coin- ixed obstacle at 0, ,ing about a fixed the whole force 1 not receive any. mines the position 1, on which if the ,0 rest, the axis of An axis through the centre of gravity, parallel to the axis of spontaneous rotation, is called the axis of instantane- ous rotation. A free body rotateo about this axis (Art, 239). * EXAMPLF. S. 1. Find the centre of percussion of a circular plate of radius a capable of rotating about an axis which touches it. Here /;," = j, and h = a. Hence from (3) we have I = a + ■T = {a. 2. A cylinder is capable of rotating about the diameter of one of its circular ends ; find the centre of percussion. Let a = its length, a.nd b = the raf'ius of ltd base. 3i» + 4a^ Ans. I =z Ga Henca if 2&' = 2a\ the centre of iwrcussion wiE be at the end of the cylinder. If b is very small compared with a, I = ^i; thus if a straight rod of small transverse section is held by one end in the hand, I gives the point at which it may be struck so that the hand will receive no jar. 2^1. The Principfil Radius of Gyratior Deter- mined Practioally. — Mount the body upon an axis not passing through the centre of gravity, and cause it to oscillate ; from the number of oscillations performed in a given time, say an hour, the time of one oscillation ia known. Then to find h, which is the distance from the axis to the centre of gravity, attach a spring balance to the lower onil, and bring the centre of gravity to a horizontal ])lane through the axis, which position will be indicated by the maximum reading of (he balance. Knowing the maxi mum reading, R, of the balance, the weight, H", of the body, and the distance, a, from the axis of snKiHjusion to IP" i 468 TUB BALLISTIC PENDTTLVM. tho point of attachment, we have from the principle of moments, Ra = Wh, from which h is found. Substitut- ing in (3) of Art 237, this value of h, and for T the time of an oscillation, kt becomes known. 242. The Ballistic Fendnlnm. — An interesting aj)- plication of the principles of the compound pendulum is the old way of determining the velocity of a bullet or can- non-ball. It is a matter of considerable importance in tho Theory of Gunnery to determine the velocity of a bulk ' as it issues from the mouth of a gun. It was to determine this initial velocity that Mr. Robins about 1743 invented the Ballistic Pendulum. This consists of a large thick heavy mass of wood, suspended from a horizontal axis in tiie shape of a knife-edge, after the manner of a compound pendulum. The gun is so placed that a ball projected from it horizontally strikes this pendulum at rest at a cer- tain point, and gives it a certain angular velocity about its axis. The velocity of the ball is itself too great to be measured directly, but the angular velocity communicated to tho pendulum may be made as small as we please by increasing its bulk. The arc of oscillation being meas- ured, the velocity of the bullet can be found by calcu- lation. Tho time, which the bullet takes to penetrate, is so short that we may suppose it completed before the pendulum has sensibly moved from its initial position. Let M be the riass of the })eudulum and ball ; m that of tho ball ; v tho velocity of tlie ball at the instant of impact ; h the distance of the centre of gravity of tiio pen- dulum and ball from the axis of suspension ; a the distance of tho point of impact from the axis of susfwusion ; w tho angular vi'octity due to tho blow of the ball, and k the radius of gyration of the pendulum and ball. Then since t h(> initial velocity of tiio bullet is v, its impulse is nieaaurod l.v iiiv, au(i therefore from (3) of Art. ^30 wu have for the TUB UAhLlSTIC PEXDVhUM. 469 be principle of iid. Substitut- for T the time interesting ajv id pendulum is I bullet or can- jortance in the y of a bulk ' as s to deterniino 1743 invented I a large thick rizontal axis in of a compound ball projected t rest at a cer- locity about its too great to be communicated as we please by on being meas- bund by calcii- rate, is so short 3 pendulum has L and ball; m it the instant of niy of the pon- ; a the distance pension ; w the l)all, and k the ,11, Then since iilse is meuaurcd ve have for the initial angular velocity generated in the pendulum by this impulse, mva 6) = M-a' (1) and from (I) of Art. 237 we have for the subsequent motion 0^ qh . ^ (8) Integrating, and observing that, if « bo the angle through which the iiendulum moves, we have ^ = w when © = 0, and ^ = when = «, (2) becomes cos a). Eliminating w between (1) and (3) we have %Mh r-r . a V = Vffh sm - , ma " 2 (3) (4) from which v becomes known, since all the quantities in the second member may be observed, or are known. Wo may determine a as follows : At a point in the jwn- dulum at a distance h from the axis of suspension, attach the end of a tape, and let the rest of the tape be wound tightly round a reel ; as the pendulnm ascends, let a length c be unwound from the reel ; then c is the chord of tlio angle « to the radius h, so that c ~ 2h sin |, which in (4) givce Mkc fa The values of k and // may bo dot«rmmed as in Art. 241. If the mouth of the gun is placed near to the pendulum. 470 ROTATION OF A HEAVY BODY. m the value of v, given by (5), must be nearly the velocity of projection. The velocity may also be determined in the following manner : Let the gun be attached to a heavy pendulum ; when the gun is discharged the recoil causes the pendulum to turn round its axis and to oscillate through an arc which can be measured ; and ihe velocity of the bullet can be deduced from the magnitude of this arc. (See Price's Anal. Mech's, Vol. II, p. 231.) Before the invention of the balliatic pendulnm by Mr. Robins in 174.'}, but little progress bad been made in the true theory of mLitary projectiles. Robins' New Principles of Gunnery was soon translated into several languages, and Euler added to his translation of it into German an extensive commentary ; the work of Baler's being again tranalatSd into English in 1784. The experiments of Robins were all conducted with musket balls of about an ounce weight, but they wore afterwards continued during several years by Dr. Button, who used cannon-balls of from one to nearly three pounds in weight. Ilutton used to suspend his cannon as a pendulum, and measure the angle through which it was raised by the discliarge. His experi- monts ar<) still regarded as some of the most trustworthy on smooth- bore guns. See Routh's Rigid Dynamics, p. 04, also Encyclop»dia Britannica, Art. Gunnery. 243. Motion of a Heavy Body about a Horizon- tal Axle through its Centre. — Let the body be a spliere whose radius is R, and weight W, and \z'c a weight P be attaclied to a cord wound round the circumference of a wheel on ^ho same axle, the radius of the wheel being r ; rofiuircd the dif;taiice ptissed over by P in t seconds. From (4) of Art. 230 wo have cPd _ Prg d^ ~ Wkf+ Pr^' I Multiplying by dt and integrating twice, we have Prgt^ $ = i{Wk* + Pr^y (1) m >r. y the velocity of n the following !avy pendulum ; cs the pendulum through an arc ii the bullet can re. (See Price's 1 by Mr. Robi-s in e theory of mLitary eras soon translated ranslation of it into Euler's being again its of Robins were ce weight, but they by Dr. Hutton, who I pounds in weight, m, and measure the charge. His experi- stworthy on smooth- ;, also Encyclopsedia )OQt a Horizon- e body be a sphere 1; a -weight P be ircumfercnoe of a le wheel being r ; t seconds. ii|gB;i^ j MI|> ! «j ! MMW !e, we have (1) FXAMPLBS. 471 the constants being zero in both integrations, since the body starts from rest when t = 0. The space will be rd. EXAMPLES. 1. Let the body be a sphere whose radius is 3 ft. and weight 500 lbs.; let P be 50 Ibe., and the radius of the wheel 6 ins.; required the time in which the weight P will descend through 50 ft. (Take g = 32.) Ans. 21 seconds. 2. Let the body be a sphere Whose radius is 14 ins. and weight 800 lbs.; let it be moved by a weight of 200 lbs. attached to a cord wound round a wheel the radius of which is one foot ; find the number of revolutions of the sphere in eight seconds. (Take g = 32.) Ans. 51.3. 244. Motion of a Wheel and Azle when a Given Weight P RaiseH a Given Weight W.— Let the weights P and W be atfached to cords wound round the wheel and axle, respectively, (Fig. 97) ; let R and r be be the radii of the wheel and axle, and w and w' their weights; required the angulai' distance passed over in t seconds. From (4) of Art. 236, we have I w Fig.S7 PR-Wr PR' + Wt^ + itoR^ + \w'r^ and let fall so that at the moment when the string becomes tight it is vertical, and tangent to tJie reel. The whole motion being supposed to take place in one plane, determine the effect of th^ impulse. The reel at first will fall vertically without rotation. Let V be the velocity of the centre at the momer^t when the string becomes tight ; f ', « the velocity of the centre and the angular velocity just after the impulse ; T the impul- sive tension ; m the mass of the reel, and a its radius. Just after the impact the part of the reel in contact with the string has no velocity, and at this instant the reel rotates about this part; but it may be considered as rotating about its axis as fixed, and the angular velocity of its axis, at this instant, in reference to the part in contact is the same as that of the latter in reference to the former. The impulsive tension is T — m {v — v'). (1) Hence from (4) of Art 236, we have for the motion of rotation . myfcj'w =z m{v — v'). (8) »Mmi lUtfian ■ i/i-'>-:l.<,if^^-^-/r\^-,',.:' 478 IMPULSIVE FORCE. u lit' Sinco the part of the rod in contact with the string has no velocity at the instant of impact, we have V = aw. Solving (3) and (3) we have 6) =: av a> + *,» (3) (4) «.a If the reel bo a homogeneous cylinder, ft^' = -, and we « have from (3) and (4) w = f-, v' = Iv, (5) I* and from (1) we have for the impulsive tension, T = ^mv. Cob. — To find the subsequent motion. The centre of the reel begins to descend vertically ; and as there is no hori- zontal force on it, it will continue to descend in a vertical straight line, and throughout all its subsequent motion ?;ho string will be vertical. The motion may therefore be easily investigated, as in Art 246, since it is similar to the case of a body rolling down an inclined plane which is vertical, the tension of the string taking the place of the friction along the plane. Hence putting « = S' '^^^ letting the friction F = the finite tension of the string, we have, from (1) and (7) of Art. 346, ■ ' F= \mg, and ^ = fer ; that is, the finite tension of the string is one-third of the jjggg^^jIMi li the Btring has (8) (4) ,» = -, and we (6) lion. he centre of the bere is no hori- ind in a vertical aent motion ^ho ay therefore be is similar to the [ plane which is the place of the ig « = I' ^^^ of the sti-ing, we one-third of the EXAMPLES. 477 weight, and the reel descends with a nniform acceleration Since the initial velocity of the reel from (5) is \v, we have, for the space descended in the time t after the impact, from (8) of Art ''46, \v + \gfl. (See Kouth's Rigid Dynamics, p. 131.) X EXA.MPL.BS. 1. A thin rod of steel 10 ft long, oscillates about an axis passing through one end of it ; find (1) the time of an oscillation, and (3) the number of oscillations it makes in a day. Ans. (1) 1.434 sec. ; (3) 60254. 2. A pc-ndulum oscillates about an axis passing through its end ; it consists of a steel rod 60 ins. long, with a rect- angular section i by J of an inch ; on this rod is a steel cylinder 2 in. in diameter and 4 in. long; when the ends of the rod and cylinder are set square, find the time of an oscillation. Am. 1.174 sees. 3. Determine the radius of gyration with reference to the axis of suspension of a body that makes 73 oscillations in 2 minutes, the distance of the centre of gravity from the axis being 3 ft 2 in. Ans. 5.267 ft. 4. Determine the distance between the centres of suspen- sion and oscillation of a body that oscillates in 2J^ sec. Ans. 20.264 ft 5. A thin circular plate oscillates about an axis jfassing through the circumference ; find the length of the simple equivalent pendulum, (1) when the axis touches the circl? and is in its plane, and (2) when it is at right angles to the plane of the circle. Ans. (1) Jw ; (2) |«. 6. A cube oscillates about one of its edges; find the length of the simple equivalent pendulum, the edge being = ^«- Ans. |« ^2. m :t3i!i"' liil ■■m m ■■W'[ 478 EXAMPLES. 7. A prism, whose cross section is a square, each sido being = 2n, aud whose length is I, oscillates about one of its upper edges; find the length of the simple equivalent pendulum. Ans. f V'to* + P. 8. An elliptic lamina is such that when it swings about one latus rectum as a horizontal axis, the other Intus rectum passes through the centre of oscillation ; prove, that the eccentricity is ^. 9. The density of a rod varies as the distaiice from one end j find the axis perpendicular to it about which the time of oscillation is a minimum, I being the length of the rod. Ans. The distauce of the axis from the centre of gravity is \ Va. 10. Find the axia about which an elliptic lamina must oscillate thai the time of oscillation may bo a minimum. Alls. The axis must be parallel to the major axis, and bisect the semi-minor axis. 11. Find the centre of percussion of ■\ cube w'lich rotate? about an axis parallel to the four parallel edges of the cube, and equidistant from the two nearer, as well as from the two farther edges. Let 'ia bo a side of the cube, and let ■:. be the distauce of the rotation-axis fn m its centre of gravity. I = c ■\- -5-, where / is the distance from the rota- /1ns. 3c' tion-iixis to the centre of percussion. 12. Find .the centre of ix^rcussion of a sphere which rotates about an axis tangent to its surface. A,h9. I ss \a. ' 13. Tift the body in Art. 243, be u sphere whoso radius is 17 ins. and weight 1200 lbs. ; let it be moved by a weight of 260 Iba atf'"^hcd to a cord, wound round a wheel whose KXAMPLES. 479 quare, each sido ,C3 about one of imple equivalent . f Vi^T" P. it swings about the other Ifitus BciUation; prove. istapce from one ibout which the the length of the centre of gravity ptic lamina must :)e a minimum. 3 major axis, and lube w'lich rotates edges of the cube, well as from the 10 cube, and let :7 « m its centre of ice from *;he rotsi- F a sphere which ce. A, IS. I = \a. ere whose radius is noved by a weight nd a wheel whose revolutions of the Ans. 5b V. radius Is 15 ins.; And the number of sphere in 10 seconds, (g = 33.) • 14. Let the body in Art. 243 be a sphere of radius 8 ins. and weight 500 lbs. ; let it be moved by a weight of 10(^ lbs. attached to a cord wound round a wheel whose radius is 6 in.; find the number of revolutions of the sphere in 5 seconds, {ff = 32^.) 4»s. 28.09. 15. In Art 244, let the weight i^ = 40 lbs., W = 100 lbs., w = 12 lbs., and w' = lbs.; and let E and r be 12 ins. and 7 ins. ; required (1) the space T>assed over by P in 16 sees, if it starts from rest, and (2) the tensions T and T' of the cords supporting P and W. {(/ = 32). Ans. (1) 926.5; (2) T = 49.04 lbs., T' = 86.81 lbs. 16. In Art. 344, let the weight P - 25 lbs., W - 60 lbs., w = 6 lbs., and -lo' = 2 lbs. ; and let R and r be 8 in. and 3 in.; required (1) tlie space passed over by P in 10 sees, if it starts from rest, and (2) the tensions T and T' of the cords supporting P and W. {g = 32|.) Ans. (1) 109.92 ft; (2) T = 23.29 lbs.; T'= G1.54 lbs. 17. In Art 245, let the body be a sphere whose radius is 3 ft, whose weight is 800 lbs., and the distance of whose centre from the axis is 9 ft. ; and let F he a force of 60 lbs. acting at the end of an arm 12 ft long; find (1) the num- ber of revolutions which the body will make about the axis in 12 min., and (2) the time of one revolution. {g = 32.) Ans. (1) 14043.6 ; (2) 6.07 sees. 18. In Ex. 17, let the radius = one foot, the weight = 100 lbs., the distance of centre from axis = 5 ft., and F ■= 25 lbs. acting at end of arm 8 ft long; find (1) the number of revolutions which the body will make about the axis in 5 min., and (2) the time of one revolution. (<, = 32|.) Ans. (1) 18139.09 ; (2) 2.23 sees. 19. If the body in Art 247 be a homogeneous sphei-e, the s*.riiig being round the circumference of a great circle, tmme&am^sss 3S£££S 480 EXAMPLB8. find (1) the angular velocity just after the impulse, and (2) the impulsive tension. ^ Sy 2" A bar, I feet long, falls vertically, retaining its hori- zon^l position till it strikes a fixed obstacle at one-quarter the length of the bar from the centre ; find (1) the angu- lar velocity of the bar, (2) the linear velocity of its centre just after the impulse, and (3) the impulsive force, the velocity at the iustan . of the impulse being v. 12« 21. A bar, 40 ft. long, falls through a vortical height of 60 ft., retaining its horizontal position till one end strikes a fixed obstacle 60 ft. above the ground ; find (1) its anjju- lar velocity, (2) the linear velocity of its centre just after the impulse ; (3) the number of revolutions it will make before reaching the g;ound, (4) the whole time of falling to the ground, and (5) i\& linear velocity on reaching the ground. ^ Ann. (1) 2.12; (2) 42.43; (3) 0.345; (4) 2.79; (5) I ' 1 npulso, and (2) aining its hori- I at one-quartor (1) the angu- ty of its centre Isive force, the |v ; (3) Htnv. rtical height of one end strikes id (1) its angu- entre just after 18 it vill make I time of falling )U reaching the (4) 2.79; (5) CHAPTER VHI. MOTION OF A SYSTEM OF RIGID BODIES IN SPACE. 248. The EqnationB of Motion of a System of Rigid Bodies obtained by D'Alemberf s Principle.— Let {x, y, z) be the position of the particle m at the time / referred to any set cf rectangular axes fixed in space, and X, r, Z, the axial components of the impressed accelera- ting forces acting on the same particle. Then -ttj , j^f > ;^. are the axial components of the accelerations of the parti- cle ; and by D'Alembert's Principle (Art. 235) the forces, «(x-'^). ,„(K-g). ,«(z-f;). acting on m together with pimilur forces acting on every f article of the system, are in equilibrium. Hence by the principles of Statics (Art G5) we have the following six equations of motion : (1) (2) 21 482 TRAXSLATION AND ROTATIOIT. w\ \M m By means of these six equations the motion of a rigid body acted on by any finite forces, may be determined. They lead immediately to two important propositions, one of which enables us to calculate the motion of translation of the body in space ; and the other the motion of rotation. L49. Independe&oe of the Motion cf Translation of ths Centre of Gravity, and of Rotation about an Axis Passing through it.- -Let (i, y, i) be the position of the centre of granty of the body at the time t, referred to fixed axes, {x, y, z) the position of the particle m referred to the same axes, («', y, z') the position of m referred t^ a Bystcui of axes passing through the centre of gravity and parallel to the fixed axes, and M the whole mass. Then 1. x=zi + x', y = y + y', z — l + z'. (1) Since the origin of the movable system is at the centre of gravity, we have (Art. 59) Ima;' = 7.my' = Smz' = ; (2) ^ z „« Substituting these values in (1) of Art. 248, wo have dr (4) ON. notion of a rigid be determined, propositions, one n of translation otion of rotation. cf Translation ;aiion about an ) be the position e time /, referred (article m, referred I m referred t.:. » re of gravity and 3 maas. Then 5 + «'. (1) is at the centre of 0; (2) =:a (3) . 248, we have (4) MOTION OF A BO or. 488 These three eqnations do not contain the co-ordinatos of the point of apphcation of the forces, and are the same as those which would be obtained for the motion of the contre of gravity supposing the forces all applied at that point. Hence The motion of the centre of gravity of a system acted on by any forces is the same as if all the mass were collcded at the centre of gravity and all the forces were applied at that point parallel to t/ieir former directions. 2. Differentiating {] ) twice we have W ~ dt^ ■*" rf^«' di^ ~ dt» "^ dt^' dt* ~ dt^^ dP' Substituting these valras in the first of equations (2) of Art. 248, we have 2m[(y+>')Z-(5 + On Performing the operations indicated we get Uii>- f II 11 484 TRAySLAriON AND ROTATION. I Omitting the Ist, 2d, 4th, 5th, Gth, and 8th terms which vanish by reason of (2), (3), and (4), .we have similarly from the other two equations o' (2) \ e have/ (•^) j = I.mix'r-y'X). These three equations do not contain the co-ordinates of the centre of gravity, and are exactly the equations we would have obtained if we had regarded the centre of gravity as a fixed point, and taken it as the origin of moments. Hence ne motion of a body, acted on by any forces, about its centre of gravity is the samp, as if the centre of gravity were fixed and the same forces acted on the body. That is, from (4) the motion of translation of the centre of gravity of the body is independent of its rotation ,• and from (5) the rota- tion of the body is independent of the translation of its centre. These two important propositions are called respectively, the principles of the conservation of the motions of transla- tion and rotation. Sen.— -By the first principle the problem of finding the motion of the centre of gravity of a system, however com- plex the system may be, is reduced to the problem of finding the motion of a single particle. By the second principle the problem of finding the angular motion of a fiw body in space is reduced to that of determining the motion of that body about a fixed jmint ON. 3th terms which ive •2) y e have^ x'Z), CONSBltVATION OF CENTRE OF ORAVITT. 485 (&) y'X). Rem.— In using the first principle it should be noticed that the impressed forces are to be applied at the centre of gravity parallel to their former directions. Thus, if a rigid body be moving under the influence of a central force, the motion of the centre of gravity is not generally the same us if the whole mass were collected at the centre of gravity and it were then acted on by the same central force. What the principle asserts is, that, if the attraction of the central force on each element of the body be found, the motion of the centre of gravity is the same as if these forces were applied at the centre of gravity parallel to their original directions. le co-ordinates of le equations we d the centre of as the origin of forces, about its e of gravity were t. That is, from I of gravity of the from (5) the rota- ranslation of its lied respectively, Hions of transltt- n of finding the m, however com- the problem of By the second ular motion of a determining the 250. The Principle of the Conservation of the Centre of Gravity.— Suppose that a material system is acted on by no other forces than the mutual attractions of its parts ; then the impressed accelerating forces are zero, which give 2X=SF=SZ=0; thereiore from (4) of Art. 249, we get 0. ^ = 0; ^^ = r„cos«, ^ = »o COS ft ^ = «o COS y. (1) where i;„ is the velocity of the centre of gravity when f = 0, and «, ft y, are the angles which its direction makes with the axes. Therefore, calling v the valocity of the centre of gravity at the time /, we liave V=:^J'■ iV? + dy^ + d? dC^ = v, 0* (3) lohich is evidently coiistunt. 486 CONSERVATION 'OF AREAS. If Vj = 0, the centre of gravity remainx at rest. Integrating (1) we get i =z v^t COB a ■+■ a, y = v^t 008/3 + 4, i = v„f COS y + c ; a — a _ y — b _ z — c cos a "~ cos /3 ~ cos y (3) (ff, b, c) being the platje of the centre of gravity of the system when / = 0. As (3) ar: the equations of a straight line it follows that the motion of the centre of gravity is rectilinear. Hence token a material system is in motion under the action of forces, none of which are external to the system, then the centre of gravity moves uniformly «i a straight line or remains at rest. REM.^Thns the motion of the centre of gravity of a system of particles is not altered by their mutual collision, Avhatever degree of elasticity they may have, because a reaction always exists equal and opposite to the action. If an explosion occurs in a moving body, whereby it is broken into pieces, the line of motion and the velocity of tlie centre of gravity of the body are not changed by the explosion ; thus the motion of the centre of gravity of the earth is unaltered by earthquakes ; volcanic explosions on the moon will not change its motion in space. The motion of the centre of gravity of the solar system is not affected by the mutual and reciprocal action of its several members ; it is changed only by the action of forces external to the system. ( 251. The Principle of the Conservation of Areas.— If X, y be the rectangular, and r, the polar co-ordiuatcs of u particle, we have 1 I. at rest. 113 + *, m [)f gravity of the tions of a straight ntre of gravity is motion under the 'tial to the system, in a straight line e of gravity of a mutual collision, have, because a to the action. If loreby it is broken le velocity of the ; changed by the e of gravity of the ;anic explosions on pace. The motion em is not affected 6 several members ; cea external to the ation of Areas. — 3 polar co-ordinates coNsesvATioy of axsas. r» co8» e ^ (tan e) = /^^. dt^ ' dt 487 (1) Now \i/^dO is the elementary ai-ea described round the origin in the time dt by the projection of the radius vector of the pai-ticle on the plane of xy, (Art. 182.) If twice this polar area be multiplied by the mass of the particle, it is called the area conserved by the particle in the time dt round the axis of z. Hence is called the area conserved by the system. Let dAx, dAy, dAg be twice the areas described by the projections of the radius vector of the particle m on the planes olyz, zx, xy, respectively ; then from (1) we have ^ I dy dx\ ^ dAg and differentiating we get „ / d^/ €Px\ „ d^A, (2) If the impressed accelerating forces are zero the first member of (2) is zero, from (5) of Art. 24i/; therefore the second member is zero. Hence v^.^-^' A. dt^ aM 488 CONSEBVATTON OF VIS VIVA. Similarly . 1?» -~ — 0, Sw -^ = ; and therefore by integration at dt at «", h, h', h" being constants. . • . ^mAx = ht, ^mAy = h't, ImAg — h"t ; the limits nf integration being snch that the areas and the time begin simultaneously. Thus, the sum of the products of the mass of every particle, and the Drojection of the area described by its radius vector on each co-ordinate plane, varies as the time. This theorem is called the principle of the conservation of areas. That is, When a material system is in motion under the action offerees, none of which are external to the systetn, tJten the sum of the products of the mass of each partich by the pro- jection, on any plane, of the area described by the radius vector of this particle measured from any fixed point, varies as the time of motion. 252. Conservation of Vis Viva or Energy.'*'— Let {x, y, z) be the place of the particle m at the time t, and let X, Y, Z be the axial components of the impressed accelerating forces acting on the particle, as in Art. 348. Tlie axial components of the efEective forces acting on the same particle at any time / are dhi d^ ^ "^d?' "'Sp' "^di^- ' If the efEective forces on all the particles be reversed, * See Art. 189. "hU 1 h"t', e areas and the of the products stion of the area i-ordinate plane, the principle of under the action system, tJien the iiclo by the pro- id by the radius Ixed point, varies Energy.*— Let b the time t, and f the impressed I, as in Art. 248. 3s acting on the cles be reversed, ?«r?Ri I WW Iiy i « ^ l| JM il MPM;. l i W^i '.t pM^J t t l HMIIf m J- CONSSRVATION OF VIS VIVA. 489 they will be in equilibrinm with the whole gronp of im- pressed forces (Art. 236). Hence, by the principle of virtual velocities (Art. 104), we have .».[(x-*)^^(r-g)., + (z-g).«]=„.,:, where dx, 6y, dt are any smaU arbitrary displacements of the particle m parallel to the axes, consistent with the con- nection of the parts of the system with one another at the time t. Now the spaces actually described by the particle m dur- ing the instant after the time t ijarallel to the axes are consistent with the connection of the parts of the system witli each other, and hence we may take the arbitrary dis- placements, Sx, 6y, 6z, to be respectively equal to the actual displacements, ^ 6t, ^ 6t, j^ 6t, of the particle.* Making this substitution, (1) becomes ^"^ w dt ^ d(^ dt'^ m^ It) -(^f+4f+4)- Integrating, we get £»««;» - 27«v„» = 2Sm / {Xdx + Ydy + Zdz), (2) vhere v and v^ are the velocities of the particle m at the times t and t^. The first member of (2, is twice the vis viva or kinetic euL '^ of the system acquired in its motion from the time ^0 to the time t, under the action of the given forces. , * '^i^' /"• t'"»>"'8'» «« ^ •«»» eqoal to ~y«!»^g g i'i;« ^^^ ^ CONSERVATION OP VIS VIVA, 491 m on m' is Pm, and the whole attractive force of »? ' on to is Pm' ; and we have X — *M ^ iL Z-lif X=.m~LF, Y=m^-=^P, Z=m^-P; r=-m^^P, Z'=z. m J r r ■ r Also r* = (« - «')» + (y - y')* + (« - «')'• Therefore for these two particles, we have • m (Xdx + Ydy + Zdz) + to' (X'da;' + F'rfy' + Z'dz') = ~^ [(« - *') ('^a; - rfo;') + (y _ y') (rfy _ rfy') + {z- z') {dm — < ground, it huu fallen aa far as the ciruuui- )f no external 36 the vis viva acting on the positive dowii- g. Hence (2) )• me of xy to the ! and t^, and if ,). (4) m depends only ntre of gravity m whenever the izontal plane. d bj Huyghens in a body (Art. 337, at onco a relation id tbe coordinates »t when, from the 38 may be made to I sufficient to deter- it h above tl:n mr- the force of gravity gbt has acquirwl a ich is wjual to »/»,, 6), about three axea at right angles, they are to /ether equivalent to a single angalar velocity a», where w = Vwi'-j-Wi'+Wj^ about an axis inclined to the given axes at ungles whose cosined are respectively --,-*, -1. 254. Motion of a Rigid Body referred to Fixed Ax- '■- — Let U8 suppose that one point in the body is flxed. I.,: ..... point be taken as the origin of co-ordinates, and let tlie axes OX, OY, OZ be any directions fixed in space and at right angles to one another. The body at the time t is turning about some axis of instantaneous rotation (Art. 240). Let its angular velocity about this axis be <•>, and let this be resolved into the angular velocities <>>,, 6>„ fa>, about the co-oixlinat^ axes. It is required to find the roaolved linear velocities, jft ^■>-jf> parallel t) the axes of co-ordinates, of a particle wi at the point P, {x, y, z), in terms of the angular velocities about the axes. These angular velocities are sup- posed positive when they tend the same way round the axes that positive CO' .; - *'jnd in Statics (Art. 06). 'i' i the positive direction^ Oi ■• - ,, w, are re- spectively froi. ;^ to z about x, from « to a; about y, and from x to y about t ; and those negative which act in the opposite direc- tions. Tjot us determine the velocity of P parallel to the axis of t. Let PN bo the ordinate z, na.ioo with the given , w,, w, about • equivalent to A angles whose red to Fixed e body is fixed. >-ordinateB, and i fixed in space ody at tht time ineons rotation this axis be <•>, jloeities <•>,, w», red to find the el i > the axes of P, («, y, «)» »» Les. ria.iQO the ordinate z, AXIS OF INaTAJfTANSOUS BOTATION. 495 and draw PM perpendicular to the axis of x. The velocity of P doe to rotation about OX is u^PM, Resolving this parallel to the axes of y and z, and reckoning those linear velocities positive which tend from the origin, and vice verm, we have the velocity along MN = — t^^PM cos NPM =—«,«; and along NP = UiPM sin JVPJf = w,y. Similarly the velocity dne to the rotation about OF par- allel to OX is iifZ, and parallel to OZ is — u^x. And that due t3 the rotation about OZ parallel to CX is — «jy, and parallel to OF is ,y-u>,x. 255. Azi« of Xnatantaneoiu Rotatioa-^Every par- ticle in the axis of instantaneous rotation is at rest relative to the origin ; hence, for these particles each of the first members of (1) in Art. 264, will reduce to zero, and wc have «i« — w«y = 0, ' w,« " «,« = 0, (1) 496 ANOULAK VELOCITY. which are the equations of the axis of instantaneous rota- tion, the third equntion being a tiecessary consequence of the first tvio; hence. (li. u. « = --«, y = -a « ; u. w. (2) that is, tho instantaneous axis is a straight line passing through the origin which is at rest at the instant con- sidered ; and the whole body must, for the instant, rotate about this line. Cob. — Denote by a, /3, y the angles which this axis makes with tho co-ordinate axes », y, «, respectively, then (Anal- Geom., Art. 175) we have cos « = «4 Vw,» + Wg* + w,»' COS |3 s= (t). V, we have V = — = V«i* + w,* + w,», wAtcA »« /Ae angular telodty required. 257. Euler'B Equations. — To determine the general equations of molio^n of a body about a fixed point. Let the fixed point be taken as origin ; let {x, y, f) be ^he place of any jiarticle m, at the time /, referred to any rectangular axes fixed in spaoe, a.. ^ let Ox^, Oyi, Ozi be the rrincipal axes of the body (Art. 231). Piffereutiating (1) of Art. 254 with respect to t, we have d?~'^~d~^~dt'^^* <"»y ~ ^^'^^ "■ "' (<^8«~'^i«)j „ fa),, since they are equal to Uo;, (^, (•>,, the three , for the chanRM In the two angolar Tclocltloe, u, and <■<«, dnrinir a W ~' of tfiven BDitll time after the axle of a), colneide* with the axis of x, will dUhr only by a unantlly which depends npon the angle passed throngh by the axis of x, dnring timt given khisII time ; the dUferenoo between u, and oi* will therefore be an Inflnitesimal of the second order and thc.cfore their derlratWes will bo eqnal. (Hee Prntt's Mech., p. 438. For flirthor demoi'stratlon of this equality, tbesindcnt la roftorred to Ruuth's BiKid Dynamics, pp. 181' and 188.) r n = N. (1) le same way. moments and axes fixed in e as the body velocities about 1 space are per- it the principal ent under con- ^rt 232), = 0; meuts of inertia >me8 ; and similarly oing the letters iy, tOt, the three OS, w, and <<>«, darins a of z, wiU dUftr only by )y the axil of x, during I* will therefore be an veg will bo eqnal. (See jquallty, the (indent ta SULES'S SqUATTONS. 499 equations of motion of the body referred to the principal axes at the fixed point are dt /?!!■;;»_ (C - 4) UjO), =2 M, (2) These are called Eulor's Equations. SoH. — ^If the body is moving so there is no point in it which is fixed in space, the motion of the body about its centre of gravity is the same as if that point were fixed. It is clear that, instead of referring the motion of the body to the principal axes at the fixed point, as Enler has done, we may use any axes fixed in the body. But these are in general so complicated as to be nearly useless. 258. Motion of a Body about a Principal Asia through its Centre of Gra^ty. — If a body rotate about one of its principal axes pamng through the centre of gravity, this axis mil suffer no pressure from the centrifu- gal force. Let the body rotate about the axis of « ; then if u be its angular velocity, the centrifugal force of any particle m will be (Art. 198, Cor. 1) nw^p — WW* V'*' + y', which gives for the x- and y-components muh; and tmchf ; and the moments of these forces with respect to the axes of y and x are for the whole body S,tnuh;x, and £m<>>>y«. 600 AXIS OF PERMANENT ROTATION. But these are each equal to zero when the axis of rotation is a principal axis (Art. a32) ; hence, the centrifugal force will have no tendency to incline the axis of z towai-ds the plane of xy. In this case the only effect of the forces mtJ^x and mul^y on the axis U to move it parallel to itself, or to translate the body in the directions of x and y. But the sum of all these forces is ^Lmurhi and I.miJ'y, each of which is equal to zero when the axis of rotation passes through the centre of gravity ; hence we conclude that, wfien a body rotates about one of its principal axes passing through its centre of gravity, the rotation causes no pressure upon the axis. If the body rotates about this axis it will continue to rotate about it if the axis be removed. On, this account a principal axis through the centre of gravity is called an axis of permanent rotation.* SoH. — If the body be free, and it begins to rotate about an axis very near to a principal axis, the centrifugal force will cause the axis of rotation to change continuidly, inas- much as the foregoing conditions cannot obtain, and this axis of rotation will either continually oscillate about the principal axis, always remaining very near to it, or else it will remove itself indefinitely from the principal axis. Hence, whenever we observe a fVee body rotating about an axis during any time, however short, we may infer that it has continued to rotate about that axis from the beginning of the motion, and that it will continue to rotate about it for ever, unless chocked by some extrajieous obstacle. (See Young's Mechs., p. 230, also Venturoli, pp. 135 and 160.) ♦ Prstt's Mochs., p. 489. Called also a natural axU qf rotatUm, seo Tonng's Hectw., p. sao ; aim (?n intartailt aicU, we Price'« Mechs., Vol. n, p. 997. m. ixis of rotation ntrifugal force if z towards the the forces m(>)^x I to itself, or to id y. But the ixis of rotation ce we conclude ( principal axes lation causes no will continue to this account a ity is called an to rotate about entrifugal force ontinuaUy, inas- obtain, and this cillate about the to it, or else it ) principal axis. >tating about an lay infer that it m the beginning » rotate about it 18 obstacle. (See .. 136 and 160.) r rotation, «eo Tonng'B Vol. n, p. m- VBLOCmr ABOUT A PRINCIPAL AXIS. 501 259. Velocity about a Principal Axis when there are no Accelerating Forces. — In this case L — M = JV = in (2) of Art. 257 ; also A, B, axQ constant for the same body ; and if we put B-o^^^ £^ = 0, ^ G = H, (1) A -" B (2) of Art. 257 becomes dui^ = Htii^u^dt. Put u^fti^u^dt = dip, and we have a^dt'ix = Fdtp., <>),fifu>g = Od^, (^idu^ = ffdip; and integrating,^ we get w,» = 2i?V> + o', w,» == 2(70 + **, w,« = 2Zr0 + c«. (2) where a, &, c are the initiil yalues of w„ w,, w,; hence from (1) and (2) dt = d^ V'(2/l^ + a») (2(70 + IP) {2H

2 'S/^OS V^Fp + a» Replacing 2i^ + a' by its value W|«, and rf0 by its value F i-, we have <« = dt^^^ Voir Wi' i8' 602 THB INTEORAL OF EULER'a XQUATIONS, and integrating, we get ti. —a 0+ tVGff=~ log^^~. w. e*i(7 . giat ^oa — rii w, +o (4) The constant (7 mnst be determined so that when ^ = 0, Wj is the initial velocity a ; hence ^^ = or (7 = — oo , which makes the first member of (4) zero for every value of L Hence, at any time t, we mnst have u^ = a; and therefore from (2) ^ = 0, and w, = u, = 0. Conse- qumily tits impressed velocity about one of the principal axes of rotation continues perpetttal and uniform, as before shown (Art 258). 260. The Integral of Euler's Equations.—^ body revolves about its centre of gravity acted on by no forces but such as pass through that point; to integrate the equations of motion. As the only forces acting on the body ai'e those which pass through its centre of gravity, they create no moment of rotation about an axis passing through that centre; and therefore (2) of Art. 257 become A^-(B-C)u>,o>,=0/ B^-iC-A)o>,o>,=0,] (1) the principal axes being drawn through the centre of gravity. Tiosa. (4) at when ^ = 0, or C = — oo , for every value i u^ = a', and = 0. Conse- f the principal iform, as before dona— ^ body >y no forces but le the equations !u'e those which ate no moment bat centre ; and (1) the centre of SraW INTEQBAL OF EULSR'S EQUATIONS. 503 Multiply these equations severally (1) by Wj, w,, Wj ; and (2) by Au)^, -Bw,, Cwg, and add ; then we have di»>t (?6), du>t dt = 0; ^ + (^.^ = 0,1 (2) integrating, we have where A« and ** are the constants of integration. Eliminating Wj' from (3), we have A{A-C)i^* + B{B- C) 0),' = *» - a»; (8) I "•' ~ £ (5 - C) [ifc9_oa»_^(^_C)V]; (*) anda,,» = ^^^^[i»-5A»-^M-P)«,«]. (5) Substituting these values of u, and Wj in the first of equa- tions (1), we have dt ^L dt (A- O)iA-B) EG r-"- A{A-C) ) G^ISj-')]*- <«) which is generally an elliptic transcendent, and so does not admit of integration in finite terms. In certain particular cases it may be integrated, which will give the value of <«>, in terms of /, and if this value bo substituted in (4) and (5), 604 APPLICATION OF TBS OSNERAL EqUATIONS, the values of w, and w, in tonus of I will be known, and thus, in these caaes, the problem admits of compiute solu- tion. Cob. — I^t Wa, jwy, 0)8 bo the axial components of the initial angular velocity about the principal axes when < = 0; then integrating the first of (2), and taking tiie limits corresponding to / and 0, we have ^w*» -I- 5a»,a + Cw3» = Jw,2 + 5a),» + Cio?. (7) Ijet a, /J, y be the direction-angles of the instantaneous axis at the time t relative to the principal axes ; so that, if 6) is the instantaneous angular velocity, and I.mr^ is the moment of inertia relative to that axis, we have (Art. 253), b), = u cos a, 6), = (i) cos /3, 0), = (i> cos y, which sub- -tuted in (7), gives J + B(^y* + Cw," = 6)« ( J cos^ « + 5 cos" /3 + C COS* y) ' ' ? = l'^ ■'.:■ = the vis viva of the body ; from which it appears that the vis viva of the body is con- stant throughout the whole motion. , Rem.— An application of the general equations of rotatory motion (Art. 267), which is of great interest and impor- tance, is that of the rotatory phenomena of the earth under the action of the attracting forces of the sun and the moon, tlie rotation being considered relative to the centre of gravity and an axis passing through it, just as if the centre ef gravity was a fixed point (Art. 249, Sch.) ; and the problem treated as purely a mathematical one. Also, in addition to the sun and the moon, the problem may be qVATIONS. be known, and )f complete solu- nponenta of the jipal axes when , and taking tl)c •>? + Ci^.\ (7) the instantaneous axes ; so that, if and 'Lmr^ is the have (Art. 253), los y, which eub- coB'/S + Cco^y) , Cor.) e bodj ; )/ the body is con-r ations of rotatory crest and impor- »f the earth nnder in and the moon, to the centre of 9t as if the centre Sch.) ; and the al one. Also, in problem may be '•■^rssiz iiiimi t ri.iKiv.i.mi/fnj^'f « ir",) ^ «^v • I ' ^^/i- i ii , . " i i ^yfJi SXAMPLES. 605 extended so aa to include the action of all the other bodies whose influence affects the motion of the eaith's rotation. In fact the investigation of the motion of a system of bodies in space might be continued at great length ; but such investigations would be clearly beyond the limits proposed in this treatise. The student who desires to continue this interesting subject, is referred to more extended works.* EXAMPLES. 1. A hollow spherical shell is filled with fluid, and rolls down a rough inclined plane ; determine its motion. Let M and M' l/C the masses of the shell and fluid respectively, h and h' their radii of gyration respectively alHJut a diameter, and a and a' the radii of the exterior and interior surfaces of the shell ; then using the same nota- tion aa in Art. 246, we have (Jf + if ') S = (-^ + M') gAna- F. df> (1) As the spherical shell rotates in its descent down the plane, the fluid has only motion of translation ; so that the equa- tion of rotation is JfP^ = F«. (3) Multiplying (1) by a' and (2) by a, and adding, we have [{M + M') d> + JfF] H = (J/ + M') a\j sin «. (3) If the interior were solid, and rigidly joined to the shell, the equation of motion would be • See Price's Mech'g, VoL II, Pmtt'a Xeob's, Bonth's Rigid Dynamics, La Place's Hgcanique C61este, ete. asB 506 KXAKPLMS. [{M+M') fl«+^*»+if' /fc'»] g ^ {M+M') ct-g sin «. (4) Integrating (3) and (4) twice, and denoting by s and «' the BIM1C08 thr tugli which the centre moves during the time / in these two cases respectively, we have (6) 80 that a greater space is described by the sphere which has the fluid than by that which has the solid in its interior. If the densities of the solid and the fluid are the same, we have fiom (5), by Art. 233, Ex. 14, ~, = y . _ q ,, • (Price's Anal Mechs., Vol II, p. 368). 2. A homoguneons spliere rolls down within a rongh spherical bowl ; it is ro<{uired to determine the motion. liOt a be the radius of the sphere, and b the radius of the bowl ; and let us suppose the sphere to be placed in the bowl at rest. Lot OCQ = ^, QPA = e, BCO = a, u> = the angular velocity of the ball about an axis throagh its centre P, k = the correspond- ing radius of gyration ; 0M=: X, MP = y ; «i = the mass of the ball. Then Fig. mm TO jTj =r — 7? sin + /* cos ; ♦» T^ = /< 008 ^ -f Fnn ^ — mg\ (1) ')(^ga.na. (4) by s and s' the ing the time t k^ (6) , ihcre which has its interior. ) are the same, 3l II, p. 368). itliin a rough he motion. le radins of the I placed in the i»; mg (1) BXAMPhES. at 507 (8) Also X — (J — a) sin ^ ; if — b — (b — a) cos ^ ... g=(J-a)co,*g-(S-.)«.*(f); (4) g = (»- a) eiu ** + (»-.) 00. *(*)•. (6) — CO80 + ^8in^ = (i-fl)^. m (d — o) ^ = JT— fw^- sin ^ (7) Now to determine the angnlar velocity of the ball, we mnst estimate the angle described by a fixed line in It, as PA, from a lino fixed in direction, as PM, and the ratio of the infinitesimal increase of this angle to that of the time will be the angular velocity of the ball. ti dMPA d4> . de dt ~ dt "^ dt Since the sphere does not slide, aO = ft (a — ^) ; a — b d^ a dt dit> a — b d^ '** dt ~~ir dp' from (3), (7), and (8) we get (*-o)^~ -^8in^; (•) (») S08 BXAMPTjBB. (10) (11) . • . {b-a) (^p = — ^ (cos - COB «). Substituting (9) in (7) we have F = ^mg Bin 0. Substituting (4), (9), (10), (11) in (1) we have 5 = ^ (17 COB ^ — 10 COB «) J therefore the pressure at the lowest point = ^(17-10coH«); and the pressure of the ball on the bowl vanishes when cos ^ = ^ cos a. Cob. — If the ball rolls over u small arc at the lowest part of the be ivl, 80 that « and are always small, cos «, and cos oth sides of 0, the lowest point of the bowl ; and the periodic time is iS a). (10) (11) ave •)i '< mishcs wben A tho lowest part small, COB «, and - ^ respectively ; dt; t\ ) angular distance of the bowl ; and SXAMPLSS. 509 therefore the oscillations are performed in the same time aa those of a dimple pendulum whose length is \ {b — a), (Art. 194). (Price's Anal. Mech's, Vol. II, p. 369.) 3. A homogeneous sphere has an angular velocity w about its diameter, and gradually contrs^cts, remaining constantly homogeneous, till it has half the original diameter ; required the final angular velocity. Ans. 4w. 4. If the earth were a homogeneous sphere, at what point must it be struck, that it may receive its prtsent viilocity of translation and of rotation, the former being 68000 miles per hour nearly ? Ana. 24 miles nearly from the centre. &. A homogeneous sphere rolls down a rough inclined plane; the inclined plane rests on a smooth horizontal plane, along which it slides by reason of the pressure of the sphere ; required the motions of the inclined plane and of the centre of the sphere. Let m = the mass of the sphere, M = tho mass of the inclined plane, a = the radius of the sphere, rt =. the angle of tho inclined plane, Q its apex ; the place of Q when t z=z 0; 0' the point on the plaiio which was in contact with the point A of the sphere when / = 0, at which time we may sup- pose all to be at rest ; A CF = 6, the angle through which tho sphere has revolved in the time i. Let be the origin, and let the horizontal and vertical lines through it be tiio axes of x and y ; OQ = x' ; and -let {x, y) {h, k) be the places of the centre of tho sphere at the times ( — t and / = respectively. Then the equations of motion of the sphere are w j^ = F cos « — li sin a, FI8.I02 610 BXAMPLES. «» ^ = ^ein a + /J COS o — mgt and the equation of motion of the plane is . M -^ = — F co^ a ■{■ R sax a. From the geometry we have a; = A + «' — aO cos o, y = ifc — oO sin a. From these equations we obtain , wi cos « - __ 6ct sin a cos a gfi _ ~ 7 (m + My— bm cos*^ ' T ' « = A — y = *- Slfsinacosa gfi 6{m + M) sin* a gfi 7 (m + if ) — 5»« co8» a ' aT which give the values of a; and y in' terms of /. Also wo obtain (m + M) {x — h) sin o — if (y — /fc) cos a = ; which is the equation of the path describeti by the centre of tlie sphere ; and therefore tliis path is a straiglit line. 6. A heavy solid wheel in tlie form of a right circular cylinder, it) composed uf two substunces, whuau volumes are -mg. u