[0B^t g^tttjj ®lmjr^t0tt 1903 Cornell University Library arW3888 The elements of analytical mechanics. 3 1924 031 364 296 olln,anx Cornell University Library The original of this book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031364296 THE ELEMENTS OP Analytical Mechanics. DeVOLSON wood, A.M., C.E., PROFESSOR OF MATHEMATICS AND MECHANICS IN STEVENS', INSTITUTE OF TECHNOLOfay ; AUTHOR OF RESISTANCE OF MATERIAIiS; ROOFS, AND BRIDGES; REVISED EDITION OP MAHAN'S CIVIL ENGINEERING; AMERICAN EDITION OF MAGNUS' LESSONS ON ELEMENTARY MECHANICS. NEW" YOKK: JOHl^ WILEY & SONS. 1876. COPIBIGHTED, 1876, BT devolson wood. PREFACE. This work was designed especially for students who are beginning the study of Analytical Mechanics. It involves tlie use of Analytical Geometry and Calculus. To those who desire a more elementary work, Magnus' Lessons on Element- ary Mechanics^ which avoids the use of the Calculus, and is full of practical examples, is especially recommended. This work, though analytical, is in a certain sense the reverse of Legendre's celebrated Mecanique Analytiqiie. Legendre, at the outset, deduced a general equation from which all others were derived ; but this work at first establishes the equation for motion due to a single force, and by adding principle after principle the most general equations become established. The latter method is the one by which every science is at first devel- oped, and presents great advantages over the former in a course of instruction ; but the former, as a method of pure analysis, cannot be excelled. I have avoided the use of the terms " Force of Inertia," " Impulsive Force," and " Instantaneous Force," except to refer to them and define the senses in which they are generally used ; for they are not only useless, but harmful. The student has a vague and confused idea as to their real meaning. lie considers them as forces, when they are not, and thinks that he is to consider a variety of forces, when really he is to consider but one, that which is equivalent to a pressure. _ I do not believe that inertia is a force, and in this am supported by many em- inent writers ; but there are those, also eminent, who state iv PEEFACE, distinctly that it is a force. This antagonism existing among those who have made the subject a study, and who ought to agree, if agreement were possible, indicates clearly that after a student has spent valuable time in seeking to determine the essential nature of the expression, he can do no more than take one side or the other, and leave the question still open for debate. The term Instantaneous Force is not open to the same objection, but the conditions which it indicates are physically impossible ; for forces do not produce their effects instantly, but require time for action. Tlie quantity which it is used to represent is not a. force, but the effect of a force after it has acted during a time, although that time may be exxseedingly short. It is, therefore, ambiguous, and hence is dropped. Regardless, however, of the merits of the case, I consider it j)olitio to drop these terms, since they are liable to produce unprofitable discussion by students. I have also avoided the expression " no force," now some- what common with certain English writers in speaking of the motion of a body on which all accelerating forces have ceased to act. Forces may act upon such a body and be in equili- brium among themselves. The terms moment of inertia and centrifugal force may have arisen from erroneous conceptions as to what they really mean, but they cannot be dropped without devising new names to take their places, and as they answer our purpose, they are retained. I have sought to present the subject in such a manner as to familiarize the reader with analytical processes. For this reason the solution of examples has been treated as applications of general formulas. The solution of many of the more elemen- tary examples might be much shorter and more simple by special methods, but if the reader is already familiar with these, his attention can be confined more closely to the analytic cal processes. These general methods make the solution of PEEPACE. V difficult examples more easy, while the solution of elementary examples may be more tedious by them. I have sought to make prominent a few of the more impor- tant equations by following them with a large number of ex- amples. These examples are graded, beginning with the more simple and gradually increasing in difficulty, until, in some cases, those are reached which cannot be solved. The instruc- tor can stop with any one desired and pass to the next article. The data of some of the examples are intentionally left in- complete, such, for example, as No. 6, page 91. In such cases the teacher may supply such data as he desires, or require the student to do it. They may -serve as the basis for several examples. Thus, in the example referred to, the base of the solid may be square or rectangular, the edges may be vertical or inclined, the weights may be arranged in different orders, etc. Instead of requiring deiinite solutions, it may be better to permit the student to state how he would proceed to solve them. Occasionally a remark or problem is given which is intended as a hint to the student that the subject may be extended in that direction. I trust that this work, though elementary, will make a gcJod foundation for those who prosecute the subject further. De Volson Wood. HoBOKEN, Sept., 1876. TABLE OF CONTENTS. CHAPTER I. DEFINITIONS AND THE LAWS OP FOKCBS WHICH ACT ALONG A STBAIGHT LINE. ARTICLES PAGB 1-14 — Definitions 1 15-20— Velocity ; Gravity ; Mass 6 21-33— Force ; Mass ; Density 16 34 — Examples of Accelerated Motion 31 25-37 — Work ; Energy ; Momentum 44 38-33— Impact 53 34-36 — Statics ; Power ; Inertia 58 37-38 — Kewton's Laws of Motion. Eccentric Impact 63 CHAPTER II. COMPOSITION AND EESOLUTION OF FORCES. 39-43 — Conspiring Forces 65 43-46 — Composition of Forces 65 47 — ^Resolution on Three Axes 70 48-49 — Constrained Equilibrium of a Particle 71 50-60— Statical Moments 79 Examples 84' CHAPTER III. PARALLEL FORCES. 61-63— Resultant 86 64 — Moments of Parallel Forces 87 65-68— Statical Couples 88 69-74— Centre of Gravity of Bodies 93 75 — Centrobaric Method , 103 77-81 — General Properties of the Centre of Gravity 108 82— Centre of Mass 110 viii TABLE OF CONTENTS. CHAPTER IV. SONCONCUBBBNT POItCES. ARTICLES PAGB 83- 85 — General Equations HI 86— Problems 115 CHAPTER V. STRESSES. 87- 90— Stresses Resolved 143 91- 92— Sheaxing Stresses— Notation 144 93- 94— Resultant Stress 140 95- 96— Discussion 158 97- 98 — Conjugate Stresses. General Problem 156 CHAPTER VI. VIRTUAL VELOCITIES. 100 — Concurring Forces 159 101 — Nonconcurring Forces 160 Examples 161 CHAPTER VII. MOMENT OF INBRTIA. 102-104 — Examples. Formula of Reduction 165 105 — Relation between Rectangular Moments 169 106— Moments of Inertia of Solids 172 107— Radius of Gyration 173 CHAPTER VIII. MOTION OF A FREE PARTICLE. 108 — General Equations ; 175 109 — ^Velocity and Living Force 176 110-117— Central Forces 180 CHAPTER IX. CONSTRAINED MOTION OF A PARTICLE. 118-121— General Equations 191 122-124— Centrifugal Force on the Earth 198 TABLE OF CONTENTS. ix CHAPTEE X. ROTATION OP A BODY WHEN THE FORCES ARE IN A PLANE. ARTIOLFS TAQE 120-131— General Equations 200 133— Reduced Mass 204 133-135 — Spontaneous Axis ; Centre of Percussion ; Axis of Instantaneous Rotation 209 Examples 310 136-138— Compound Pendulum. Captain Kater's Experiments 218 CHAPTER XI. FORCES APPLIED AT DIFFERENT POINTS AND ACTING IN DIFFERENT DIRECTIONS. 139— General Equations 331 140 — Conservation of the Centre of Gravity 335 141 — Conservation of Areas 325 143 — Conservation of Energy 336 143 — Composition of Angular Yelocities 330 144— Euler's Equations 382 145 — Rotary Motion from au Impulse 233 140 — No Straru on Principal Axes 233 147 — Axes not Permanent Constantly Changing 334 148 — ^At every Point Three Principal Axes. 334 GREEK ALPHABET. Letters. A a B ^ Fry A 8 E e z r H r, & S0 I I K K A \ M /i Names. Alpha Beta Gamma Delta Epsilon Zeta Eta Theta Iota Kappa Lambda Mil LetteiB. Names, N V Nu s? Xi Omicron n IT Pi p p Rho 5 0- s Sigma T T Tau r V Upsilon qp Phi Xx Chi W ■yjr Psi 12 CO Omega ANALYTICAL MECHANICS. CHAPTER I. DEFINITIONS, AND PRINCIPLES OF ACTION OF A SINGLE EOECE, AND OF FORCES ACTING ALONG THE SAME LINE. 1. Mechanics treats of the laws of forces, and the eqni- librinra and motion of bodies under the action of forces. It has two grand divisions. Dynamics and Statics. 2. Dynamics treats of the motion of material bodies, and the laws of the forces which govern their motion. 3. Statics treats of the conditions of the equilibrium of bodies under the action of forces. There are many subdivisions of the subject, such as Hydrodynamics, Hy- drostatics, Pneumatics, Thermodynamics, Molecular Mechanics, etc. That part of mechanics which treats of the relative motion of bodies which are so connected that one drives the other, such as wheels, pulleys, links, etc., in machinery, Ls called Cinematics. The motion in this case is independent of the intensity of the force which produces the motion. • Theoretic Mechanics treats of the effect of forces applied to material points or particles regarded as without weight or magnitude. Somatology is the application of theoretic mechanics to bodies of definite form and magnitude. 4. Matter is that which receives and transmits force. In a physical sense it possesses extension, divisibility, and impene- trability. Matter is not confined to the gross materials which we see and handle, but includes those ethereal substances by which sound, heat, light, and electricity are transmitted. It is unnecessary in this connection to consider those refined speculations by which it is sought to determine the essential nature of matter. According to some of these speculations, matter does not exist, but is only a conception. 1 2 DEFINITIONS. [5-6.] According to this view, bodies are forces, within the limit of which the attrao- ti-ve exceed the repulsive ones, and at the Umlts of which they are equal to each other. But observation, long continued, teaches practically that matter is inert, that it has no power within itself to change its condition in regard to rest or motion ; that when in motion it cannot change its rate of m.otion, nor be brought to rest without an external cause, and this cause we call FORCE. One also learns from observation that matter will transmit a force, as for instance a pull applied at one end of a bar or rope is transmitted to the other end ; also a moving body carries the effect of a force from one place to another. 5. A Body is a definite poi'tion of matter. A particle is an infinitesimal portion of a body, and is treated geometrically as a point. A molecnle is composed of several particles. An atom is an indivisible particle. 6. FoEOE is that which tends to change the state of a body in regard to rest or motion. It moves or tends to move a body, or change its rate of motion. "We know nothing of the essential nature of force. We deal only with the laws of its action. These laws are deduced by observations upon the effects of forces, and on the hypothesis that action and reaction are equal and opposite / or, in other words, that the effect equals the cause. In this way we find that forces liave different intensities, and that a relation mav be established between them. It is necessary, therefore, to establish a unit. This may be assumed as the effect of any known force, or a multiple part thereof. The effect of all known forces is to produce a pull, or push, or their equivalents, and may be measured by pounds, or by something equivalent. The force of gravity causes the weight of bodies, and this is measured by pounds. We therefore assume that a standakd POUND is the UNIT of force. The standard pound is established by a legal enactment, and has been so fixed that a cubic foot of distilled water at the level of the sea, at latitude 45 degrees, at a temperature of 62 degrees Fahrenheit, with the barometer at 30 inches, will weigh about 62.4 pounds avoirdupois. The English sUndard pound was originally 5,760 troy grains. The grain was the weight of a certain piece of bias.s which was deposited with the clerk of the House of Commons. This was destroyed at the time of the burning of the House of Commons in 1834, after which it was decided that the legal [7.] STANDARD UNITS. 3 pound should be the weight of a certain piece of platinum, weighing 7,000 grains. This is known as the avoirdupois pound, and the troy pound ceased to be the legal standard, although both have remained in common use. The legal standard pound in the United States is a copy of the English troy pound, and was deposited in the United States Mint in Philadelphia, in 1827, where it has remained. The avoirdupois pound, or 7,000 grains, is used in nearly all commercial transactions. The troy pound is a standard at 63 de- grees Fahrenheit and 30 inches of the barometer. The weight of a cubic inch of water at its maximum density, as accepted by the Bureau of Weights and Measures of the United States, is stated by Mr. Hasler, in a report to the Secretary of the Treasury, 1843, to be 253.7453 grains. Mr. Hasler determined the temperature at which water has a maxi- mum density, at 39.83 degrees Fahrenheit, but Playfair and Joule determined it to be 39.101° F. The exact determination of the equivalent values of the units is very diffi- cult, and has been the subject of much scientific investigation. — (See T/ie Me- tric System, by F. A. P. Barnard, LL.D., New York, 1872.) When a quantity can be measured directly, the unit is generally of the same quality as the thing to be measured : thus, the unit of time is time, as a day or second ; the unit of length is length, as one inch, foot, year, or metre ; the unit of volume is volume, as one cubic foot ; the unit of money is money ; of weight is weight ; of momentum is momentum ; of work is work, etc. When dissimilar quantities are used to measure each other a proportion must be established between them. It is commonly said that "the arc mea- sures the angle at the centre," but it does not do it directly, since there is no ratio between them. The arc is a linear quantity, as feet or yards, or a num- ber of times the radius, while the angle is the divergence of two lines, and is usually expressed in degrees. But angles are proporti 2r or. dy and, _ds _dy ds _ I 2y , — 1. — 1, ' — V 7. ' dt dt dy 14 GEAVITT. [19] If j, = 0, '^=0; y — T, V = V^v' ; y = 2r,v =' 2v' ; y = ^r,v = v'. Hence, at the instant that any point of the wheel is in con- tact with the straight line, it has no velocity, and the velocity at the highest point is twice that of the centre. The velocity at any point of the cycloid is the same as if the wheel revolved about the point of contact, and with the same angular velocity as that of the generating circle. For, the length of the chord which corresponds to the ordi- nate y is y/^ry, and hence, if v:2v':: VWy :2r; V2y we have v = -^ v', as before found. 19. Geavitation is that natural force which mutually draws two bodies towards each other. It is supposed to extend to every particle throughout the universe according to fixed laws. The force of gravity above the surface of the earth diminishes as the square of the distance from the centre increases, but within the surface it varies directly as the distance from the centre. If a body were elevated one mile above the surface of the earth it would lose nearly ^ttVtt of its weight, which is so small a quantity that we may consider the force of gravity for small elevations above the surface of the earth as practically constant. But it is variable for different points on the surface, being least at the equator, and gradually increasing as the latitude increases, according to a law which is approxi- mately expressed by the formula g = 32.1726 - 0.08238 cos 2Z, in which Z = the latitude of the place, ff = the. acceleration due to gravity at the latitude Z, or simply the force of gravity, and 32.1726 ft. = the force of gravity at latitude 45 degrees. [19.] EXAMPLES. 15 From this we find that at the equator g = g^ =z 32.09022 feet, and at the poles g = g^— 32.25498 feet. The varying force of gravity is determined by means of a penduhim, as will be shown hereafter. It is impossible to de- termine the exact law of relation between the foi'ce of gravity at different points on the surface of the earth, for it is not homogeneous nor an exact ellipsoid of revolution. The delicate observations made with the pendulum show that any assumed formula is subject to a small error. (See Mecanigue Celeste, and Puissanfs Geodesie.) Substituting g for/' in equations (11), (12), and (13), we have the following equations, which are applicable to bodies falling fi'celj' iu vacuo : — , — 2s vz=gt=V2gs = -; 8 = h^=^^ = h>f; > (16) Examples. 1. A body falls through a height of 200 feet; required the time of descent and the acquired velocity. Let g = 33^ feet. Am. < =: 3.53 seconds. V = 113.13 feet. 2. A body is projected upward with a velocity of 1000 feet per second; required the height of ascent when it is brought to rest by the force of gravity. Ans. s = 15,544 feet, nearly. 3. A body is dropped into a well and four seconds afterwards it is heard to strike the bottom. Required the depth, the velocity of sound being 1130 feet per second. Am. 231 feet. 4. A body is projected upward with a velocity of 100 feet per second, and at the same instant another body is let fall from a height 400 feet above the other body ; at what point wiU they meet f 16 DYNAMIC MEASURE OP A FORCE. [30, 21.] 5. With what velocity must a body be projected downward that it may in t seconds overtake another body which has already fallen through a feet ? a Ana. V = . + VZag. 6. Required the space passed over by a falling body during the ra* second. 20. Mass is quantity of matter. If we conceive that a quantity of matter, say a cubic foot of water, earth, stone, or other substance, is transported from place to place, without expansion or contraction, the quantity will remain the same, while its weight may constantly vary. If placed at the centre of the earth it will weigh nothing ; if on the moon it will weigh less than on the earth, if on the sun it will weigh more; and if at any place in the universe its weight will be directly as the attractive force of gravity, and since the acceleration is also directly as the force of gravity, we have — = constant, 9 for the same mass at all places. This ratio for any contem- poraneous values of TFand g may be taken as the measure of the mass, as will be shown in the two following articles. The weight in these cases must be determined by a spring balance or its equivalent. 21. Dynamic measure of a force. Conceive that a body is perfectly free to move in the direction of the applied force, and that a constant uniform force, which acts either as a pull or jjush, is applied to the body. It will at the end of one second produce a certain velocity, which call v^iy If now forces of different intensities be applied to the same body they will produce velocities in the same time which are proportional to the forces ; or fx t)(i„ in which f is the applied force. Again, if the same forces are applied to bodies having differ- ent masses, producing the same velocities in one second, then will the forces vary directly as the masses, or, f QC Jf. [31.] DYNAMIC MEASURE OF A FORCE. 17 Hence, generally, if uniform, constant forces are applied to different masses producing velocities ^d) in one second, then f 00 Mv^l) ; or, in the form of an equation, f^cif^d,; (17) whei'e c is a constant to be determined. If the forces are constantly varying, the velocities generated at the end of one second will not measure the intensities at any instant, but according to the above reasoning, the rate of varia- tion of the velocity will be one of the elements of the measure of the force. Hence if F = a, variable force ; J^ = the mass moved ; -=- —f=^ the rate of variation of the velocity; or velocity-increment ; and, -^ be substituted for W(ij in equation (17), reducing by equation (6), we liave F=^cMf=cM^^=.cM§, (18) From this we have F cM= ^; hence, the value of cM is expressed in terras of the constant ratio of the acceleration f produced by a force i^upon a mass M. To determine this ratio experimentally I suspended a weight, W, by a very long line wire. The wire should be long, so that the body will move practically in a straight line for any arc through which it will be made to move, and it should be very small, BO that it will contain but little mass. By means of suitable mechanism I caused a na. s. constant force, F, to be applied horizontally to the body, thus causing it to move sidewise, and determined 2 18 UNIT OF MASS. [33. J the space over which it passed daring the first second. This equalled one-half the acceleration (see the iii'st of equations (12) when t = l). I found when F= ^W, that /= 1.6 feet, nearly ; and for i^= -^^W, f= 3.2 feet, nearly ; and similarly for other forces ; hence eM = -jV W, nearly. But the ratio of F to _/ is determined most accurately and conveniently by means of falling bodies; for ,/"= p' = the acceleration due to the force of gravity, and W the weight of the body (which is a measure of the statical effect of the force of gravity upon the body), hence W cM=j; (19) in which the values of W and g must be determined at the same place ; but that place may be anywhere in the universe. The value of c is assumed, or the relation between e and M fixed arbitrarily. If c = 1, we have W Jf=y; (20) ■and this is the expression for the 7nass, which is nearly if not quite universally adopted. This in (18) gives and hence the dtwamic mbastjeb of the peesstjee which moves A BODY is the product of the mass multiplied ly the aoGelera- tion. If there are retarding forces, such as friction, resistance of the air or water, or forces pulling in the opposite direction; then the first member, F, is the measure of the unbalanced forces, and the second member determines its value. 22. Unit of Mass. If it is assumed that c = 1, as in the the preceding article, the unit of mass is virtually fixed. In (20) if W— 1 and ^ = 1, then M= 1; that is, a unit of mass ds the quantity 'of matter which will weigh one pound at that [32.] UNIT OP MASS. 19 place in the nniverse where the acceleration dne to gravity is one. If a quantity of matter weighs 32^ lbs. at a place where g = 32^ feet, we have hence on the surface of the earth a body which weighs 32^ ]iounds (nearlj') is a tmit of mass. It would be an exact unit if the force of gravity were exactly 32| feet. In order to illustrate this subject fnrther, suppose tliat we make the unit of mass that of a standard pound. Then equa- tion (19) becomes c.l=l, ffo in which ffg is the value of g at the latitude of 45 degrees. This value resubstituted in the same equation gives M=^ W, g and these values in equation (18) give „ 1 ^_ Wd?s the final vahie of which is tlie same as (21). Again, if the unit of mass were the weight of one cubic foot of distilled water at the place where ^o = 32.1801 feet, at which place we would have TF"= 62.3791, and (19) would give 62.3791 '^ " 32.1801' and this in the same equation gives 32.1801 W ^ ~ 62.3791 ' g ' and these values in (18) give ,, WdJ's . . i' = 77, , as before. g dv' 20 DENSITY. [33.] 23- Density is the mass of a unit of volume. If i(f = the mass of a body ; Y = the volume ; and D — the density ; then if the density is uniform, we have jy — y-. If the density is variable, let S = the density of any element, then . dM :. M-- fSdV (22) from which the mass may be determined when S is a known function of V. Examples. 1. In a prismatic bar, if the density increases uniformly from one end to the other, being zero at one end and 5 at the other, required the total mass. Let I = the length of the bar ; A = the area of the transverse section ; and X = the distance from the zero end ; then will 5 J = the density at a unit's distance from the zero end ; 5 jx = the density at a distance a? ; and dV— Adx ; .:M=A —j- = -Al. 2. In a circular disc of uniform thickness, if the density at a unit's distance from the centre is 2, and increases directly as the distance from the centre, required the mass when the radius is 10. 124.] APPLICATIONS OF EQUATION (31). 21 3. Ill the preceding problem suppose that the density in- creases as the square of the radius, required the mass. 4. Ill the preceding problem if the density is two pounds per cubic foot, required the weight of the disc. 5. If in a cone, the density diminishes as the cube of the distance from the apex, and is one at a distance one from the apex, requii-ed the mass of the cone. Having established a unit of density, we might properly say that mass is a certain number of densities. 24- Applications of Equation (21). [Obs. — If, for any cause, it is considered desirable to omit any of the matter of the following article, the author urges the student to at least establish the equations for the acceleration for each of the 31 examples here given. This part belongs purely to mechanics. The reduction of the equations belongs to mathematics. It would be a good exercise to establish the fundamental equa- tions for all these examples, before making any reductions. Such a course serves to impress the student with the distinction between mechanical and mathematical principles.] 1st. If a body whose weight is 50 pounds is moved horizon- tally hy a constant force of 10 vounds. required the velocity ao- ^^^- qmred at the end of 10 seconds ^ ^ ^g — , and the »pace passed over during -^^ 9 that time, there being no friction nor other maternal resistance, and the body sta/rting from rest. Here M-- W g ' 50 lbs., and F = 10 lbs. ; d^s F 193 df~ 'M" 30" hence (21) gives 193 ' <■' ■/ Multiply by dt and integrate, and ds _ di~ The second integral is ds 193 J# = ^= 30 ^+(^^ = ^^- 22 ACCELERATING FOKCES. [34.] 1 Q^ and hence for t = 10 seconds, we have V = 64.33 + feet. s = 321.66 + feet. 2c?. Suppose the data to he the same as in the preceding example, and also that the friction hetween the body and the plane is 5 pounds. Mequired the space passed over in 10 seconds. Here F= (10 — 5) pounds. ^«_^_193 ^ jC^ ^^ '''df~ M~ 60* " ' Bd. Suppose that a iody whose weight is ^0 pounds is moved horizontally iy a weight of 10 lbs., which is attached to an inex- tensible, hut perfectly flexible string which passes over a witeel and is attached at the other end to the hody. Required the distance passed over in 10 seconds, if the string is without weight, and no resistance is offered hy the wheel, plane, or string. so Us. ^ Xly II W lis. Fig. 10. In this case gravity exerts a force of 10 pounds to move the jriass, or F= 10 lbs., and the mass moved is that of both bodies, or M= (50 + 10) ~ 32f dh_F_im ^l df ~M ~ "36"- The integration is performed as before. Ans. s = 268.05 feet. Uh. Find the tension of the string in the preceding example. [24] MOVING MASSES. 23 The tension will equal tliat force which, if applied directly to the body, as in Ex. 1, will produce the same acceleration as in the preceding example. Let P = 10 pounds ; TT = 50 pounds ; T = tension ; P + TF = the mass in the former example ; and W — = the mass moved by the tension. Hence, from Equation (21), LJ:Jlf^_p. and W g-' Eliminate y^ and we find WP T ■ .: r= 8.33 lbs. "What must be the value of P so that the tensid^li will be a maximum or a minimum ? &th. In exanvple 3, what must J>e tlie weight of P so thai the tension shall he (^) part of P f ^ - >rS ^ -'^^^ ^-- ^=(—1) ^- Qth. If a body whose weight is W falls freely in a vacuum by the force of gr amity, determine tlie formulMS for the motion. Here Mg = W and the moving force F^= W\ W d^s g dt^' = W d^s = g- The integrals of this equation will give Equations (16), when the initial space and velocity are zero. Let the student deduce them. 24 PROBLEMS OF [24-] 'Jth. Suppose that the moving pressure {pull or push) equals tlie weight of the body, required the velocity and space. Here Mg — W and i^= W, hence the cireninstances of motion will be the same as in the preceding example. The forces of nature produce motion without apparent pressure, but this example shows that their effect is the same as that produced by a push or pull whose intensity equals the weight of the body, and hence both are measured by pounds, or their equivalent. Sth. If the force F Is constant, show that Ft = Mv ; also Fs = iMv^, and \Ff = Ms. If F is variable we have Mv =fFdt. 9th. Suppose that a piston, devoid of friction, is driven Iry a constant steam- pressure through a portion of the length of a cylinder, at what point in the stroke tnust the prressure be instantly reversed Fig. 11. ^'"^ ^0 that the full stroke shall equal the length of the cylinder, the cylinder being horizontal ? At the middle of the stroke. Whatever velocity is gen- erated through one-half the stroke will be destroyed by the counter pressure during the other half. XOth. If the pressure upon the piston is 500 pounds, weight of the piston 60 pounds, and the friction of the piston in the cylinder 100 pounds, required the point in the stroJce at which the pressure must be reversed. The uniform effective pressure for driving the piston is 500 — 100 = 400 lbs., and the uniform effective force for stopping the motion is 500 + 100 = 600 pounds. The velocity generated equals the velocity destroyed, and the velocity destroyed equals that which would be generated in the same space by a f oi-ce equal to the resisting force ; hence if F^^ the effective moving force ; s — the space through which it acts ; V = the resultant velocity ; F' = the resisting force ; and «' = the space through which it acts ; [34.] AOOELEKATING FORCES. 23 then, from the expression in Example 8, we have Fs = iMv^, and F's' = iMv\ .: Fs = F's', or, F: F'::s' : s. In tlie example, F= 400 lbs., and F'— 600 lbs. Let x = the distance from the starting point to the point where the pressure must be revei-sed. Then 600 : 400 ::« : 12 — a;, .-. ce = 7| inches. 11/A. j^in the preceding example the j)iston moves vertically up and down, required the jpoint at which the pressure must he instantly reversed so that the full strohe shall he 12 inches. The efiEective driving pressure upward will be 500 — 100 — 50 = 350 pounds, and the retarding force will be 500 + 100 + 50 = 650 pounds, and during the down-stroiie the driving force is 500 + 50 — 100 = 450 pounds, and the retarding force is 500 — 50 + 100 = 550 pounds. 12^^. A string passes over a wheel and has a weight P attached at one end, and W at the other. If there are no resistances fro-m the string or viheel, and the string is devoid of weight, required the resulting motion. Suppose W > P; then F= W- P, and M= ; g . dh_F_ ■'■ df " M By integrating, we find W-P W+ P V = W + P^' fft, and. s — W-P ^ 26 PEOBLEMS OF [34.J 13th. Required the tendon of the string in the precedvng example. The teDsion equals the weight P, plus the force which will produce the acceleration W-P W+ P^ when applied to raise P vertically. The mass multiplied by the acceleration is this moving force, or P W-P hence the tension is " ' ''K P + W-P p _ 2WP WTp ~w+p- Similarly, it equals TT minus the accelerating force, or W- W-P W^ 2WP W+P " W+P' A complete solution of this class of problems involves the mass of the wheel and frictions, and will be considered here- after. 14:#A. A string passes over a wheel and has a weight P attached to one end and on the other side of the wheel is a weight W, which slides along the string. Requi/red the friction hetween the weight W and the Mrvng, so that the weight P will remain at rest. Also re- qui/red the acceleration of the weight W. Fia. 13. The friction = P ; Mg= W; and, i^= W-P; d^s F W-P df hence, and, M W-P W W 4^: p w' ge. 134.] ACCELERATING FORCES. 27 15th. In the preceding exam/pie, if P were an animal whose weight is less than W, required the acceleration with which it must ascend, so that W loill remain at rest. IQth. If the weight W descend along a rough rope with a given acceleration, required the acceleration with which the body P must ascend or descend on the ojpposite rope, so that the rope m,ay remain at rest, no allowance being made for friction on the wheel. (Tlie ascent mnst be due to climbing up on the cord, or be produced by an equivalent result.) Vlth. A particle moves in a straight line under the action of a uniform acceleration, and describes spaces s and s' in t and t' seconds respectively, determine the accelerating force and the velocity of projection. Let u = the velocity of projection, and f — the acceleration ; then . s' -s and If IQth. If a perfectly flexible and perfectly smooth rope is placed upon a pin, find in what time it will run itself off . If it is perfectly balanced on the pin it will not move, unless it receive an initial velocity. If it be unbalanced, the weight of the unbalanced part will set it in motion. Suppose that it is balanced and let Vo = the Initial velocity, 2Z = the length of the rope, w = the weight of a unit of length, and t — the time. Take the origin of coordinates at the end of the rope at the instant that motion begins. When one end has descended s feet, the other has ascended tlic same amount, and hence the J - ~ t' -. f 0/, - _ s\2t -1)- - s{2t' - -1) Vv - 2(# -t') 8 2#'- " 2t - — , then u = :0. 28 PROBLEMS OF [24.] unbalanced weight will be 2ws. 2wl -T- g ; hence we have (Ps _F_ _ 2ws Multiply by ds and integrate, and we have The mass moved will be Is dr I Integrating again, gives •=\/i-% =^i% if s = Z. 19th. If a particle moves towards a centre of force which ATTRACTS directly as the distance from the force, determine the motion. Let ft = the absolute force ; that is, the acceleration due to the force at a unit's distance from the centre ; and s = the distance, then dr'=-'^- The second member is negative, because s is an inverse func- tion of t, that is, as t increases s diminishes Generally attrac- tive forces are negative and repulsive ones positive, in reference to the centre of the force. This is the same equation as in Example 5, Article 17; hence, if a is the initial value of «, and V = V n {a^ — «') ; t = n-\ (sin-^ - —\n); Cb [24.] ACCELERATING FORCES. 29 and the velocity at the centre of the force is found by making s — 0, for which we have, V — a V/j; 113 and t = - -fi-in, -^-in, 2^"*'^, etc., hence, the time is independent of the initial distance. It may be proved that within a homogeneous sphere the attractive force varies directly as the distance from the centre. Hence, if the earth were such a sphere, and a body were per- mitted to pass freely through it, it would move with an accele- rated velocity from the surface to the centre, at Mdiich point the velocity would bo a maximum, and it would move on with a retarded velocity and be brought to rest at the surface on the opposite side. It would then return to its original position, and thus move to and fro, like the oscillations of a pendulum. The acceleration due to gravity at the surface of the earth being g, and r being the radius, the absolute force is r * and the time of passing from surface to surface on tlie equator would be P" oi^in /20,923-161 .„ ,„ t = n^- = 3.1416 \/'i^^ = 42m. 1.6 sec. The exact dimensions of the earth are unknown. The semi-polar axis of the earth is, as determined by Besael 30,853,662 ft. Airy S0,853,810ft. Clarke 30,853,429 ft. The equatorial radius is not constant, on account of the elevations and depressions of the surface. There are some indications that the general form of the equator is an ellipse. Among the more recent determinations are those by Mr. Clarke, of England (1873), and his result given below is considered by him as the most probable mean. The equatorial radius, is according to Bessel 20,923,596 ft. Airy 30,923,713 ft. Clarke 20,923,101ft. 30 PROBLEMS OP P4i] The determiuation of the force of gravity at any place is subject to small errors, and when it is computed for different places the result may differ from the actual value by a perceptible amount. The force of gravity at any particular place is assumed to be constant, but all we can assert is that if it is variable the most delicate observations have failed to detect it. But it is well known that the surface of the earth is constantly undergoing changes, being elevated in some places and depressed in others, and hence, assuming the law of gravitation to be exact and universal, we cannot escape the conclusion that the force of gravity at every place on its surface changes, and although the change is exceedingly slight, and the total change may extend over long periods of time, it may yet be possible, with apparatus vastly more delicate than that now used, to measure this change. It seems no more improbable than the solution of many problems already attained — such for instance, as determining the relative velocities of the earth and stars by means of the spectroscope. 20tk. Suppose that a coiled spring whose natural length is A B,is com/pressed to B O. If one end rests against an im- movable iody B, and the other against a hody at C, vjhich is perfectly free to 7nove horizontally, what will Fig. 14. be the time of movem.ent from C to A, and what will be the velocity at A ? It is found by experiment that the resistance of a spring to compression varies directly as the amount of compression, hence the action of the spring in pushing the body, will, in reference to the point A, be the same as an attractive force which varies directly as the distance, and hence it is similar to the preceding example. But if the spring, is not attached to the particle the motion will not be periodic, but when the particle has reached the point A it will leave the spring and proceed with a uniform velocity. If the spring were destitute of mass, it would extend to A, and become instantly at rest, but because of the mass in it, the end will pass A and afterwards recoil and have a periodic motion. If the body be attached to the spring, it will have a periodic motion, and the solution will be similar to the one in the Author's Resistance of Materials, Article 19. Take the origin at A, s being counted to the left ; suppose that 5 pounds will compress the spring one inch, and let the total compression be a = 4 inches. Let W = the weight of the body = 10 pounds. [24.] ACCELERATING FORCES. 31 Then the absolute force is found from the equation J/ = i^ or H./^ 5 ; .-. / = 16^ feet. Hence, for any point we have ---16 1s- from which we find that ^n nearly, and, ■« = 64 inches nearly. 21st. Suppose that in the preceding problem a body whose weight is M' is at B, and another M" at C, loth heing perfectly free to move horizontally, required the time of movement that the distance between them shall he equal to A£ ; and the resultant velocities of each. It is convenient in this case to take the origin at the centre of one of tlie bodies — say that of B — and remain at the centre during motion. The origin will be movable. Let n = the absolute force of the spring ; that is, the force which will compress the spring a unit of length — say one inch — ; a = the total compression ; and b =^ the length after compression. Then jxa = the total reaction of the spring when motion begins ; fi{a—{s—b)) = the reaction (or moving force) when the spring has expanded an amount equal to s — 5 ; and M+ M' = the total mass moved ; hence ^__ F_ fija + b-s) , df~M~ M' + M" ' ds' _ fi{2as + 2fe - s>) ^ dl~ M^T~S" '■ 32 REPULSIVE FOECES. [34.J {2a + b)b But « = for s = 5 ; .-. Ci = — ft v/; dt- M'+ M" ' ds -M" V'2(s - b)a - (s - hf and integrating gives ^/ Jf ' + M" a and, making s = a + 5, we have which, as in the preceding example, is independent of the amount of compression of the spring. To find the relation between the absolute velocities, Let s' = the space passed over Ijy M', and s" = the space passed over by M"; then since the moving force is the same for both, we have df df Integrating, gives M'v' = M"v". 22d. Suppose that the form varies directly as the distance from the centre of force and is eepulsive. Then d^'s df='''' in which u is the initial velocity. 2Sd. Swpjpose tJiat the force varies inversely as the square of the distance from the centre of the force and is attraotivk. [This is the law of universal gravitation, and is known as the law of the inverse squares. While it is rigidly true, so far as we know, for every [34.] ATTRACTIVE FORCES. 33 particle of matter acting upon any other particle, it is not rigidly true for finite bodies acting upon other bodies at a finite distance, except for homoge- neous spher/s, or spheres composed of homogeneous shells. The earth being neither homogeneous nor a sphere, it will not be exactly true that it attracts external bodies with a force which varies inversely as the square of the distance from the centre, but the deviations from the law for bodies at great distances from the earth will not be perceptible. We assume that the law applies to all bodies above the surface of the earth, the centre of the force being at the centre of the earth.] Let the problem be applied to the attraction of the earth, and H = the radius of the earth ; g = the force of gravity at the surface ; /i := the absolute force ; and s = the distance from the centre ; then 9 . /" = -73 and Multiply by ds and integrate ; observing that for s = a, v = 0, and we have ='4s-l) (^) df „ as — s^ 2i" \i 7, — sds ■■•(f) {as — 6-^)* ■ using the negative sign, because t and s are inverse functions of each other. The second member may be put in a convenient form for integration by adding and subtracting -J a to the numerator and arranging the terms. This gives ^J^J^ds (as — s'p a — 2.9 , , ads ds + 2{as-if'ji 2{as-s')i ' 34 ATTRACTIVE FOECES. [24.] the integral of which is _i 2s (5) {as — ^)i — id, versin ' — +C. But when s = a,t = .-. 0= ia^ ; From the circle we have tt —TerBiD-' _ = tt — cob~W1 — — J = "•-( -('-?))=="-■('"-•)• Prom trigonometry, we have 3 cos^ j/ — 1 = cos 3 y. Let 3 y= cos ~' I 1 J, then COS 22/ = _ — 1 ; . ■. cos'' y = --, and 2^ = cos-' , / £ ; and 'i' fls fa Sj/ = 3 cos~'4/ — ; orjT — Tersin-'— . From (a) it appears that for s = 0, w = 30 ; hence the velocity :at the centre will be infinite when the body falls from a finite •distance. If s = df = Qc , -y = 0. If a body falls freely from an infinite •distance to the earth, we have in equation («) a = 00 ; and s = r = the radius of the earth ; for the velocity' at the surface. But ^ = ^ ; .-. V = {2gr)h Jig— 321 ^eet and r = 3962 miles, we have _ (6^ X 3962\i . ''-I- 5280- J --=6.90 miles. Hence the maximum velocity with which a body can reach the earth is less than seven miles per second. [24.] ATTRACTIVE FORCES. 3o 2iM. Suppose that the force is atteaotive and varies in- versely as the n''^ power of the distance. Then d^s _ fi de~ ~ ?' ■ ■ d^" n-\ Is"-! ^^1 / ' and integrating, gives \ 2 f« / •'a ' According to the tests of integrability this may be integrated when 5 3 1 ,3 5 ''=■■■ • 7> 5. 3>-l,n>or-....etc., 3 2 1 . „ 3 orn= ....-,-, -,0,2, or- etc. 2Hh. Let the force vary inversely as the square root of the distance and he atteactivb. (This is one of the special cases of the pi'eceding example.) We have d?s _ __fi d£''"~ S' The negative sign is taken because i and s are inverse functions of each other. Add and subtract — -pz — 7 and we have 3 y« Va*— »4 r__Vj__ gy/a _ ay/a ~| cfo 36 ATTRACTIYE FORCES. 134.] t - 26i?A. Swppose that the force is aiteactive and varies in- versely as the distance. Hence 'de~ ~ 7' dg^ n 1 « ••■^=2, log-; in which s = a for v = 0. Hence the time from s =a to s = 0, is ]_ /"O ^s V'iA'./rYi^^^U ^2^/ -s.y»(i„,j) =«&)*■ Let I log —1=2^; then f or « = a, jr = and f or « = 0, jr — oo . Squaring and passing to exponentials, we have , log — = If .•.— — e" , 01 s = a e " ; .'. ds = ae ^ .2y dy; This is called a gamma-function, and a method of integrating it is as follows : — Since functions of the same form, integrated between the same limits are independent of the yariahles and have 'ihe same value, therefore [24.] ATTRACTIVE FORCES. 37 Jo Jo and / e-*^<;y / r^ at = / e'^'^rfy . Also the left hand member -will be of the same value if the sign of integra- tion be placed over the whole of it, since the actual integration Vfill be performed in the same order ; hence ■^dy at in which y = tu\ .'. dy = t du. Integrating in reference to t, we have _ -i^ (1 + m 2 (1 + m2) du. du which for i = oo becomes zero, and for i = becomes r- , and the iu- tegral of this is ^ tan~' m, which is zero for m = 0, and \ it for m =: oo ; -•00 (See also Mee. Celeste, p. 151 [1534 0].) Or we may proceed as follows : — J! „ dx Let «-" =2! .■.(/* = —(— log a!)2 g^ ; .-. f e~^dt= f -ii-logxf^ dm. Jo J\ Let !B = a» and consider a less than unity ; then log a will be negative, and —log X = jf log a ; 2 . •. dx — — oiH 2y dy log a ; 38 ATTRACTIVE FORCES. [34.] ■whicli substituted above gives «-* = / - i (- log xY^dx = (-log as)"* / fls^/ dy. •fl •'0 Dividing by (— log a)^ and multiplying both sides by —da, we have J' -i (- log ar*M(-j ; (h) and, s= — log (l + ^ u^j. (0 Substitute this value of « in equation (c) of the preceding example, and we have v=JUl 1 \ ^n iog(i + ««2)j *^ M i + i^A which gives the velocity in descending to the point from which it started ; and as it is less than «, the velocity of return will be less than that with which it was thrown upward. This is because the resistance of the air is against the velocity during the entire movement, both upwards and down- wards. The same value of s {Eq. (i) ) substituted in (d) of the preceding example gives the time of descent, 1 Vg + ku^ + w Vk * ~ o — 7= ""S ■ — 7=- ; * V KQ 'V g X KV^ —UVK which differs from the time of the ascent, as given by (/») above. 31 St. Supjjose that the force is attractive and varies inversely as the cube of the distance, and that the medium resists as the square of the velocity, and as the square of the density, the density varying inversely as the distance from the origin. Let K = the coefficient of resistance, being the resistance for a unit of density of the medium and a unit of velocity ; 44 WORK. [35.J K Ids \^ then "2 l-^l ds = the resistance at any point. ^' ^ K (dsV ■'■ df ~ ^^ ^\dt)' Multiply by Ms, and we have (ds\i 2ic (dsV , 2^ This is a linear differential equation of which the integrating factor is e *. The initial values are i = 0, and « = as for » = M ; I 3k - « — 3k — (I — ; « « « » y, ^0-"-^H which gives the velocity in terms of the space. The final integral cannot be found. 25. "WoEK AND Vis Yiva {or living force.) — ^Eesuming equation (21), and multiplying both members by ds, we have Fds = M-~ds. Integrating between the limits, v — Vo for s = ; and v = v for s = 5, we have fFds = Wiv" - V). (23) If V(, = 0,we have fFds = iMiP. (24) The expression J/?/ is called the vis viva (or living force) of a body whose mass is M and velocity v. Its physical im- portance is determined from the first member of the equation, which is called the work done by a force F in the space s. Hence the vis viva equals twice the worJc done hy the moving force. WoEK, mechanically, is overcoming resistance. It requires a certain amount of work to raise one pound one foot, and twice t35.] WORK. 45 that amount to raise two poniuls one foot, or one pound two feet. Siiuilarlj, if it requires 100 pounds to move a load on a horizontal plane a certain amount of work will be accomplished in moving it one focit, twice that amount in moving it two feet, and so on. Hence, generally, if i^= a constant force which overcomes a constant resist- ance, and s = the space over which i'^acts, then ^Vor7c = Is; (25) and similarly, if F= a variable force, then Work = 'S Fds ; (26) and if Fl?> a function of s we have Work =fFds. The TJXiT of work is one pound raised vertically one foot. This nnit being small for most practical purposes, another unit called the horse-poioer is used for determining tlie efficiency of motors and machines. The horse-power is 33,000 pounds raised one foot per minute. The total work, according to equation (25), is independent of the time, since the space may be accomplished in a longer or shorter time. But implicitly it is a function of the time and velocity. If the work be done at a uniform rate, we have 8 — vt, and F.s = F.v.t. If # = 1, we have, for the work done in a unit of time, F.v. (28) which is called the Dynamic Effect, or Mechanical Power. Every moTing; body on the surface of the earth does work, for it overcomes a resistance, whether it be friction or resistance of the air, or some other resistance. The same is true of every body in the universe, unless it moves 46 WORK. [35.1 IE, a non-resisting medium. * Animals work not only as beasts of burden, but in their sports and efforts to maintain life ; water as it Rourses the stream wears its banks or the bed, or turns machinery ; wind fills the sail and drives the vessel, or turns the windmill, or in the fury of the tornado levels the forest, and often destroys the works of man. The raising of water into the air by means of evaporation ; the wearing down of hiDs and mountains by the operations of nature; the destruction which follows the lightning-stroke, etc. , are examples of work. "Work may be useful or prejudicial. That work is useful which is directly instrumental in producing useful effects, and prejudicial when it wears the machinery which produces it. Thus in drawing a train of cars, the useful work is performed in moving the train, hut the prejudicial work is overcoming the friction of the axles, the friction on the track, the resistance of the air, the resistance of gravity on up grades, etc. It is not always possible to draw a practical line between the useful and prejudicial works, but the sum of the two always equals the total work done, and hence for economy the latter should be reduced as much as possible. In order to determine practically the work done, the inten- sity of the force and the space over which it acts must be measured simultaneously. Some form of spring balance is commonly used to measure the force, and when thus employed is called a Dynamometer. It is placed between the moving force and the resistance, and the reading may be observed, or autographically registered by means of suitable mechanism. The corresponding space may also be measured directly, or secured automatically. There are many devices for securing these ends, and not a few make both records automatically and simultaneously. If the force is not a continuous function of the space, equa- tion (26) must be used. The result may be shown graphically by laying off oil the abscissa, AB, the distances ac, ce, etc., proportional to the spaces, and erecting oi-dinates ab, cd, ef, etc., proportional to the corresponding forces, and joining their upper ends by a broken line^ or, what is better, by a line which * All space is filled with something, since light is transmitted from all directions. But is it not possible that there may be a sometliing through which bodies may move without resistance ? [25.] WORK. 47 Fio. 16. is slightly curved, the amount and direction of curvature being indicated by the broken line previously constructed ; and the area thus inclosed will represent the Nvork. The area will be giv- A. en by the formula Simpson's rule for determining the area is: — Divide the abscissa AB into an even number of equal parts, erect ordinates at the points of division, and number them in the order of the natural numbers. Add together four times the- even ordinates, tivice the odd ordinates and the extreme ordinates, and multiply the sum by one third of the distance between any two consecutive ordinates. If 2/o) Vi, Vi: etc., are the successive ordinates, and I the distance between any two consecutive ones, the rule is expressed algebraically as follows : — Area = U {y, + 2?/i + iy^ + 2^3 + 4^^ + 2j/g + - - - y„) (29) If the applied pressure, F, is exerted against a body whicli is perfectly free to move, generating a velocity v, then the work which has been expended is, equation (24), ^Mv^. This is called stored work, and the amount of work whicli will be done by the moving body in being brouglit to rest will be the same amount. If the body is not perfectly free the quantity ^Mv^ is the quantity of work which has been expended by so much of the applied force as exceeds that which is necessary 'in overcoming the frictional resistance. Thus a locomotive starts a train from rest, and when the velocity is small the power exerted by the locomotive may exceed considerably the resist- ances of fiiction, air, etc., and produce an increasing velocity, until the resistances equal constantly the tractive force of the locomotive, after which the velocity will be uniform. The work done by the locomotive in producing the \elocity v in excess of that done in overcoming the resistances will he^Mv^, in which Mi& the mass of the train, including the locomotive. 4:8 EXAMPLES. [35.] We see that double the velocity produces four times the work. This is because twice the force produces twice the velocity, and hence the body will pass over twice the space in the same tiine, so that in producing double the velocity we have 2^2s = 4:F.s, and similarly for other velocities. [We have no single word to express the unit of living force. If a unit of mass moving with a velocity of one foot per second be the unit of living force, and be called a Dynam, then would the living force for any velocity and mass be a certain number of Dynams.'] Since wo7'7c is not force, but the effect of a force exerted through a Certain space, independently of the time, we call it, for the sake of brevity, space-effect. Vis viva, or living force, is not force, but it is twice the work stored in a moving mass. It equals twice the space- effect. Some writers have called \Mi?' the vis viva. The advantage is that it enables one to say that the work equals the vis viva instead of twice the vis viva; but the great majority of writers have employed the former definition, and it is not desirable to change it. Examples. 1. A body whose weight is 10 pounds is moving with a velocity of 25 feet per second ; required the amount of work which will be done in bringing it to rest. An^. 97.4 foot-pounds. 2. A body falls by the force of gravity through a height of A feet ; required the work stored in it. Let W = the weight of the body, M = the mass of the body, g = acceleration due to gravity, and v = the final velocity, then t? - 2gh, and Mg — W; W .: IMv-' = g- • ^gh ^ Wh. [26.] ENERGY. 49 3. A body ■whose weight is 100 pounds is moving on a hori- zontal plane with a velocity of 15 feet per second ; how far will it go before it is brought to rest, if the friction is con- stantly 10 lbs 1 Ans. = T0.4- ft. 4. A hammer whose weight is 2000 pounds has a velocity of 20 feet per second ; how far will it drive a pile if the constant resistance is 10,000 pounds, supposing that the whole vis viva is expended in driving the pile 1 5. If a train of cars whose weight is 100,000 pounds is moving with a velocity of 40 miles per hour, liow far will it move before it is brought to rest by the force of friction, the friction being 8 pounds per ton, or ^-^^-^ of the total weight ? 6. If a train of cars weighs 300 tons, and the frictional resistance to its movement is S pounds per ton ; required the horse-power which is necessary to overcome- this resistance at the rate of 40 miles per hour. Ans. 256. 7. If the area of a steam piston is 75 square inches, and the steam pressure is 60 pounds per square inch, and the Nelocity of the piston is 200 feet per minute, required the horse-power developed by the steam. 8. If a stream of water passes over a dam and falls through a vertical height of 16 feet, and the transverse section of the stream at the foot of the fall is one square foot, required the horse-power that is constantly developed. Let ff =32| feet, and the weight of a cubic foot of water, 62i lbs. Ans. 21.89. 9. A steam hammer falls vertically through a height of 3 feet under the action of its own weight and a steam pressure of 1000 pounds. If the weight of the hammer is 500 pounds, required the amount of work which it can do at the end of the fall. 26. Energy is the capacity of an agent for doing work. The energy of a moving body is called actual or Kinetic energy, and is exj)ressed by ^Mv^. But bodies not in motion may have 4 50 MOMENTUM. [27.] a capacity for work when the restraining forces are removed. Thus a spring under strain, water stored in a mill-dam, steam in a boiler, bodies supported at an elevation, etc., are examples of stored work which is latent. This is called Potential energy. A moving body may possess potential energy entirely distinct from the actual. Thus, a locomotive boiler containing steam, may be moved on a track, and the kinetic energy would be expressed by ^Mv^, in which M is the mass of the boiler, but the potential energy would be the amount of work which the steam is capable of doing when used to run machinery, or is otherwise emj^loyed. These principles have been general- ized into a law called the Conservation of energy, which implies that the total energy, including both Kinetic and Poten- tial, in the universe remains constant. It is made the funda- mental theorem of modern jjliysical science. (See page 226.) The energy stored in a moving body is not changed by changing the direction of its path, provided the velocity is not changed ; for its energy will be constantly expressed by ^Mv^. Such a change may be secured by a force acting continually normal to the path of the moving body; and hence we say that a force which acts continually perpendicular to the path of a moving body does no work upon the hody. Thus, if a body is secured to a point by a cord so that it is compelled to move in the circumference of a circle ; the tension of the string does no work, and the vis viva is not affected by the body being constantly deflected from a rectilinear path. MOMENTUM. 27- Resuming again equation (21), multiplying by dt, and integrating gives, Pu,^Mj:§ = M^ = M,. (30) The expression Mv is called momentum., and by comparing it with the first member of the equation we see that it is the effect of the force F acting during the time t, and is indepen- dent of the space. For the sake of brevity we may call the momentum a time-effect. [27.] MOMKNTUM. 51 / If the body has an initial velocity we have Fdt=^M{v-Vo); (31) 'h which is the momentum gained or lost in passing from a velocity v^ to v. Momentum is sometimes called quantity of motion, on account of its analogy to some other quantities. Thus the intensity of heat depends upon temperature, and is measured in degrees; but the quantity of heat depends upon the volume of the body containing the heat and its intensity. The inten- sity of light may be uniform over a given surface, and vs'ill be measured by the light on a unit of surface ; but the quantity is the product of the area multiplied by the intensity. The intensity of gravity is measured by the acceleration which is produced in a falling body, and is independent of the mass of the body; but the quantity of gravity (or total force) is the product of the mass by the intensity (or Mtj). Similarly with momentum. The velocity represents the intensity of the motion, and is independent of the mass of the body ; but the quantity of motion is the product of the mass multiplied by the velocity. Differentiating (30) and reducing, gives which is the same as (18), and in which -r^ is a velocity-incre- ment ; hence the momentum, impressed each instant is a measure of the moving force. If the force F is constant we have from (30), Ft = Mv ; and for another force F' acting during the same time F't-= M'v' ; .-. F:F'\\Mv: M'v' ; Iience, the forces are directly as the momenta produced by them respectively. 53 IMPACT. [38. j If the forces are variable, let I Fdt = P ^ Mv, and / F'dt = F' = M 'v' ; then F:P'::Mv: M'v' ; hence the time-efeets are dh-ectly as the momenta impressed. We thus have several distinct quantities growing out of equation (21) of which the English units are as follows : — The unit of force, F, \s 1 ft- The unit of work or space effect is 1 Tb x 1 ft. The unit of mechanical effect is 1 n> x 1 ft- The unit of vis-viva is 32^ lbs x 1' ft. The unit of momentum 33 ^bs X 1 ft. Work and mechanical effect are of the same quality, and properly have the same unit. DIEEOT CENTKAL IMPACT. 28, If two perfectly smooth bodies impinge directly against each other, whether moving in the same or opposite directions, they mi^tnally displace the particles J*' p in the vicinity of the point of con- — tact, producing compression whieh goes on increasing until it becomes a maximnm, at which instant they Fig. 17. have a common velocity. A com- plete analysis of the motion during contact involves a knowledge of the motion of all the particles of the mass, and would make an exceedingly complicated problem, but the motion at the instant of maximnm compression may be easily found if we assume that the compression is instantly distributed through- out the mass. Let M^ and J/^ be the respective masses of the bodies ; v-i and ^2 the respective velocities before impact ; Vi and Vi the respective velocities at the instant of maximum compression, and Pi and P^ the momenta gained or lost respectively by the bodies dnring compression. [29.1 IMPA6T. 53 Then from (31) Pi = M^ K-t,/) ; which is the momentum lost bj^ ifj on account of the action of J/j. Similarly which, as indicated, will be negative if the bodies move in the same directitni, and will be the momentum gained by M^ on account of the action of M^. But at the instant of greatest compression «/ = v^ ; and because they are in mutual contact during the same time, their time-effects are equal, or A = P,. Combining these four equations, we find by elimination ^' =-:^rr^ =^'^' (^^) which veh^city remains constant for perfectly non-elastic bodies after impact, since such bodies have no power of resti- tution, and will move on with a common velocity. DIEECT CENTEAL EMPACT OF ELASTIC BODIES. 29. Elastic bodies are such as regain a part or all of their distortion when the distorting force is removed. If they regain their original form they are caWedi perfectly elastic, but if only a part, they are called imperfectly elastic. At every point of the restitution there is assumed to be a constant ratio between the force due to compression and that to restitution. But it is unnecessary for present purposes to trace tliese effects, for by equation (31) we may determine the result when the bodies finally separate from each other. 54 IMPACT. [30.1 Let iSj = the ratio of the force of compression to that of restitution of one body, which is called the modulus of elasticity, e^ — the corresponding value for the other ; Fi ~ the velocity of M^ at the instant when they separ- ate from each other ; and "Fg = the corresponding velocity for M^. Then from equation (31) e,P,= M,{V,-v^); (34) -e,.F,==^M,{V,-v,'). (35) As before Pi = Jf^ and we will also assume that 6^ = 6.^ — e. These combined with (32) and (33) give Fi: J/i-yj + Jl^Wa eM^ M, + M, - M, + M, (^1-^^); (36) 30, Discussion of EQtrA'noNS (36) and (37). 1°. If the bodies are perfectly non-elastic, « = 0. which is the same as (33). 2°. If they are perfectly elastic e— 1. •■• ^^ = ^1 - m:^-M, (^1 - ^^) 5 (39) TT , 2jri , Fa = t;,-f-^-j-^(^,,-^,,). (40) [30.] IMPACT. ■ 55 From (38) we have M aT,d, r,-v,= ^~-^^ K-«,). Similarly from (39) and (40) hence, the velocity lost by one body and gained by the other is twice as much when the bodies are perfectly elastic as when they are perfectly uon-elastic. 3°. If Ml = M2 , then for perfectly elastic bodies we have ^2 = ^2 + -^:^-^ (.vt-v^) = vr, that is, they will interchange velocities. 4°. If Ml impinges against a fixed body, we have M2 = (*i) EXAMTLES. (1.) A mass M\ with a Telocity of 10, impinges on M^ moving' in an opposite direction, moying with a velocity 4Nand has its velocity reduced to 5 ; required the relative magnitudes of M\ and Jf^. (2.) Two inelastic bodies, weighing 8 and 5 pounds respectively, move in the same direction with velocities 7 and 3 ; required the common velocity after impact, and the velocity lost and gained by each. (3.) If J/i weighs 12 pounds and moves with a velocity of 15, and is im- pinged upon by a body M^ weighing 16 pounds, producing a common velocity of 30, required the velocity of M^ before impact if it moves in the same or opposite direction. (4.) If 5J/i = ^Mi , frill = — Sbj , ®2 = 7, and e = § ; required the velocity of each after impact. (5.) If J/i = %M^ , V^ - f»„ and »2 = ; required e. (6.) If Bi is 26, Ml is moving in an opposite direction with a velocity of 16 ; Ml = 2M2 , e—^; required the distance between them 5^ seconds after impact. (7.) Two bodies are perfectly elastic and move iu opposite directions ; the weight of Ml is twice M2, but ®2 = 2», ; required the velocities after impact (8.) There is a row of perfectly elastic bodies in geometrical progression whose common ratio is 3, the first impinges on the second, the second on the third and so on ; the last moves off with ^f the velocity of the first. What is the number of bodies ? Ana. 7. LOSS OF VIS VIVA IN THE IMPACT OF BODIES. 31. Before impact the vis viva of both bodies was and after impact [33.] KELATIONS OF FOECE, MOMENTUM, AND WORK 57 which by means of (36) and (37) becomes M, V,' + M, Vi = J^W + M,vi - ^l^l^(^^-^^y, (42) For perfectly elastic bodies e = 1 and the last term disap- pears ; hence in the iinpact of perfectly elastic iodies no vis viva is lost. If the bodies are imperfectly elastic e is less than 1, and since («i — ■'■3)^ is always positive, it follows that in the impact of imperfectly elastic hodies vis viva is always lost, and the greatest loss is suffered when the hodies are perfectly non-elastic. If e = 0, (42) becomes M^^ - V,^) + Mlvi - F/) =^f-~\ {V, - v,f • (43) in which each member is the total loss by both bodies. It is also the loss up to the instant of greatest compression when the bodies are elastic. If M2 is very large compared with M^ we have from (38) Fi = v^ nearly, = V^ , and (43) becomes ifit>;^ - i/i Fi^ = M,{vi -- V,f, the second member of which is frequently used in hydraulics for finding the vis viva lost by a sudden change of velocity. These investigations show the great utility of springs in vehicles and machines which are subjected to impact. EELATION8 OF rOEOE, MOMENTUM, WOEK, AND VIS VIVA. 32. We may now deteeminb the exact office in the same problem of the quantities ;— force, momentum, wor/c, and vis viva. Suppose that a force, whether variable or constant, impelsa body, it will in a time t generate in the mass M a certain velocity v. This force Tnay at any instant of its action be measured by a co-tain number of pounds or its equivalent. 58 AN IMPULSE. [33.] Suppose that this mass impinges upon another body which inaj be at rest or in motion. In order to determine the effect upon their velocities we use the principle of momentum, as has been shown. But the bodies are compressed during impact and hence work is done. The amount of worl?: which they are capable of doing is one half the sum of their vis vivm ; and if they are brought to rest all this work is expended in compressing them. If the velocity of a body after impact is less than that before, it has done an amount of work repre- sented by iM{v^ — V^), and similarly if the other body has its velocity increased it has lining force imparted to it. The distor- tions of bodies represent a certain, artiount of worJe expended. And this explains why in the impact of imperfectly elastic bodies vis viva is always lost, for a portion of the distortion remains. Is any_£c«:ee lost? One of the grandest generali- zations of physical science is, that no fam^ee'in nature is lost. In the case of impact, compression develops heat, and this passes into the air or surrounding objects, and the amount of energy which is stored in the heat, electricity or other element or elements, which is developed by the compression, exactly equals that lost to the masses. We thus see that in the case of moving bodies, force impels, momentum determines velocity after impact, and work or vis viva p^p^f^esemts- the resistance iphich the particles offer to being displaced. 33. An impulse is the effect which a force produces in an imperceptibly short time. In the impact of bodies it is of the nature of a blow. It is sometimes considered as an instantane- ous force; that is a force which produces its effect instantly, requiring no time for its action / but no such forces exist in nature, and hence are purely ideal. Every force requires time in which to produce its effect. Resuming equation (31), we have J, Mt = M{v — Vo); /: in which if we consider F to be indefinitely large, and t indefinitely small, the value of the definite integral will still be expressed by the second member of the equation. But in this [33.] AN IMPULSE. 59 case the first member is tlie time-effeot of an impulse, and the second member measures its effect in producing a cliange of velocity. Calling this value Q, -we have Q =: Mi^v - V,) = MV. (U) Hence, the measure of an impulse in producing a change of veloc'itij of a hody is the increased {or decreased) rnomentum produced in the body. This is the same as when the foi-ce and time are finite. If the force were strictly instantaneous tlie velocity would be changed from ■«„ to v witliout moving the body, since it would have no time in which to move it. Similarly from equation (29) we have / Fds = iM{v^ - v^") ; in which for an impulse ^ will be indefinitely large; and hence the work done by an impidse is measured in the same way as for finite forces. All the effects therefore of an impulse are measured in the same way as the total effects produced by a finite force. In regard to forces, we investigate their laws of action ; or having those laws and the initial condition of the body we may determine the velocity, energy, or position of the body at any instant of time or at Any point in space, and hence we may determine final results ; but in regard to impulses we deter- mine only certain final results without assuming to know axiy- thinjr of the laws of action of the forces, or of the time or space occupied in producing the effect. The terms "■Impulsive force, " and ^'' lnstard,aneous force^'' are frequently used to denote the effect of an "Impact; " but since the effect is not a force, they are ambiguous, and the term "Impact " appears to be more appropriate. An incessant force may be considered as the acition of an infinite number of infinitesimal impulses in a finite time. The question is s(;metiraes asked, " What is the force of a 60 EXAMPLES. , L33-] blow of a hammer ? " If by the force is meant the pressure ill pounds between the face of the hammer and the object struck, it cannot be determined unless tlie law of resistance to compression between the bodies is known during the con- tact of the bodies. But this law is generally unknown. Tlie pressure begins with nothing at the instant of contact and increases very rapidly up to the instant of greatest compression, after which the pressure diminishes. The pressure involves the elasticity of both bodies ; the rapidity with which the force is transmitted from one particle to another ; the amount of the distortion ; the pliability of the bodies ; the duration of the impact ; and some of these depend upon the degree of fixed- ness of the body struck ; and several other minor conditions ; and hence we consider it impossible to tell exactly what the force is. Examples. 1. Two bodies whose weights are W and W^ are placed very near each other, and an explosive is discharged between them ; required the relative velocities after the discharge. 2. A man stands upon a rough board which is on a perfectly smooth plane, and jumps off from the board ; required the relative velocities of the man and board. [Obs. The common centre of gravity of the man and hoard will remain the same after they separate that it was before. After separating they would move on forever if they did not meet with any obstacle to prevent their motion. ] 3. A man whose weight is 150 pounds walks from one end of a rough board to the other, which is twelve feet long, and free to slide on a perfectly smooth plane ; if the board weighs 50 pounds, required the distance travelled by the man in space. 4. In example 3 of article 24, suppose that the weight 10 pounds is permitted to fall freely through a height A, when it produces an impulse on the body (50 pounds) through the intermediate inextensible string ; required the initial velocity of the body. [33.] EXAMPLES. 61 Let Vo = V^jcA = the velocity of the weight just before tlie impulse ; and V = the velocity immediately afterward, which will be the common velocity of the body and weight ; then ^ = 50 v = 10 (wo - -y) ; .-. V = l^'o • The subsequent motion may be found by equation (21), observing that the initial velocity is v. The tension on the string will be infinite if it is inextensible, but practically it will be finite, for it will be more or less elastic. [Some writers have used the expression impulswe tension of the string instead of momentum.'\ 5. If a shell is moving in a straight line, in vacno, with a velocity v, and bursts, dividing into two parts, one part moving directly in advance with double tlie velocity of the body ; what must be the ratio of the weights of the two parts so that the other part will be at i-est after the body bursts? 6. Explain how a person sitting in a chair may move across a room by a series of jerks without touching the floor. (Can he advance if the floor is perfectly smooth 1) 7. A person is placed on a perfectly smooth plane, show how he can get off if he cannot reach the edge of the plane. The same impulse applied to a small body will impart a greater amount of energy than if applied to a large one. Thus, in the discharge of a gun, the impulse imparted to the gun equals that imparted to the ball, but the work, or destruc- tive effect, of the gun is small compared with that of the ball. The t{m.e of the action of the explosive is the same upon both bodies, but tlie space moved over by tlic gun will be small compared with that of the ball during that time. The product Mv, being the same for both, as if decreases v increases, but the work varies as tlie square of the velocity. 62 STATICS. POWEE. INERTIA. [34-37.] 34. Statics is that case in which the force or forces which would produce motion are instantly arrested, resulting in pressure only. The expression for the elementary work which a force can do is Fds, but if the space vanishes, we have, Fds — 0. This, as we shall see hereafter, is a special case of " virtual velocities." The forces which act upon a body may be in equilibrium and yet motion may exist, but in such cases the velocity is uniform. 35. PowEE implies an ability to do. It refers to the acting agent. Thus we speak of a water power, the power of the wind, the power of a lens, animal power, etc., without implying any particular measure. It is a more general term than_/o/"ce, though too often used synonymously with it. 36. Inertia implies passiveness or want of power. It means that matter has no power within itself to put itself in motion, or when in motion to change its rate of motion. Unless an external force be applied to it, it would, if at rest, remain for- ever in that condition ; or if in motion, continue forever in motion. Gravity, which is a force apparently inherent in matter, can produce motion only by its action upon other matter. Ineetia is not a force, but because of the property above explained, those impressed f(jrces which produce motion are measured by the product of the mass into the acceleration as explained in preceding articles ; and many writers call this measuee the force of inertia. 37, Newton's Theee Laws of Motion. Sir Isaac Newton expressed the fundamental principles of motion in the form of three laws or mechanical axioms ; as follows : — 1st. Every body continues in its state of rest or of uniform motion in a straight line unless acted upon by some external force. 2d, Change of motion is proportional to the force impressed, and is in the direction of the line in which the force acts. [38.] ECCENTRIC IMPACT. 63 3d. To every action there is opposed an eqnal reaction. [As simple as these laws appear to the student at the present day, the science of Mechanics made no essential progress until they were recognized. See Whewell's Inductive Sciences, 8d Ed., Vol. 1, p. 311.] DIRECT ECCENTEIC IMPACT. 38. In all the problems thus far considered, it has been assumed that the action-line of the force or forces passed through the centre of the mass, producing a motion of transla- tion only. But if the action line does not pass through the centre, it will pro- duce both transla- tion and rotation. 'y If an impulse, P, wJiose action-line is nor7nal to the surface at the point of impact, hut which does not pass through the centre of the body, be imparted to the body, it loill produce the same translation as if its action-line passed throxigh the centre and also impart a rotary motion to the body. In Figure 18, let P = Mv be the impulse imparted to the body ; in which M k the mass of the body and v the velocity of the centre. Let this impulse be imparted at a. At b, a distance from the centre =^ cb = ac, let two equal and oppo- site impulses be imparted, each equal to ^P The impulse, P, equals ^P -f ^P. The four impulses evidently produce the same effect upon the body as the single impulse /-*. If now one of the impulses, ^/', above the centre is combined with the equal and parallel one acting in the same direction below the centre, their effect will be equivalent to a single one, equal to /-* applied at the centre c. This produces translation only. The other i^P above the centre combined with the equal and opposite ^P below the centre, produces rotation only ; and it 64 ECCENTRIC IMPACT. [38.] is evident that the greater the distance a, the point of im{)act, is from the centre, the greater will be the amount of rotation. It should be observed that the impulse imjaarted is not the same as that of the body producing the impact. To impart tlie same impulse at a as at e by the impact of a moving body, the latter must act through a greater space, or move with a greater velocity iu the former than in the latter case. An impulse at a will produce more vis viva than if imparted at c ; for there will be the same vis viva in both cases due to translation, and in tlie former case there will be an additional amount due to rotatiou. CHAPTEE II. COMPOSITION AND RESOLUTION OF FORCES. CONCUERENT FOECES. 39. If two or more forces act upon a material particle, they are said to be concurrent. They may all act towards the particle, or from it, or some towards and others from. 40. If several forces act along a material line, they are called conspiring forces, and their effect will be the same as if all were applied at the same point. 41. The Eesultant of two or more concurrent forces is that force which if substituted for the system will produce the same effect upon a particle as the system. Therefore, if a force equal in magnitude to the resultant and acting along the same action-line, but in the opposite direction, be applied to the same particle, the system will be in equilib- rium. If the resultant is negative, the equilibriating force will be positive, and vice versa. Hence, if several concurrent forces are in equilibrium, any one may be considered as equal and opposite to the resultant of all the others. 42. The resultant of several conspiring forces, equals the algebraic sum of the forces. That is, if i^ , ^ , i^ , etc., are the forces acting along the same action-line, some of which may be positive and the others negative, and li is the resultant ; then R = F^ + Fi + F^ + etc. = :SF. (45) 43. If two concurring forces be represented in nn,agnitude and direction by ths adjacent sides of a parallelogram,, the resultant will be represented in magnitude and direction by; the diagonal of the parallelogram. This is called the paral- lelogram of forces. If each force act upon a particle for an element of time it will generate a certain velocity. See equation (44). Let 5 66 PARALLELOGRAM OF FORCES. [43.] Pro. 19. the velocity which ^would produce be represented by AB ; and that of F hj AD = BO. These represent the spaces over which the forces respectively would move the particle in a unit of time if each acted sepa- rately. If we conceive that the force F moves it from J. to ^ and that the motion is there ar- rested, and that P is then applied at B, but acting parallel to AD, then will the particle, at the end of two seconds, be at O. If, next, we con- ceive that each force acts alternately during one-half of a second beginning again at A, the particle will be found at a in one-half of a second ; at 5 at the end of one second ; at c at the end of one and one-half seconds ; and finally at C at the end of two seconds. If the times be again subdivided the path will be Ad, de, ef,fi, ^9, gK ^S ^°ll Fig. 20. If exceeds 90 degrees, we have B^ = F^ + F'- 2PFcos 0. If ^ = 90 degrees, we have Also, if ^ = 90°, and a be the angle between R and P ; and ^ between R and F\ then (47) P = R cos a ; I F = R cos ^ =z R sin a. j Squaring and adding, we have P^ + F^ = R", as before. The forces P and T'^are called component forces, or simply cow/ponents. 44, Telajstgle of Foeces. If two forces are rep>res6nted in Tnagnittids and direction hy two sides of a triangle the restilt- ant will he represented in magnitude and direction hy the third side. Thus, in Fig. 19, if AB and BG lepresent two forces in magnitude and direction, A G will represent the resultant. parallelogram of pressures ; but we have seen that a pressure when acting -upon a free body will produce a certain amount of motion, and that this motion is a measure of the pressure, and hence its use in the proof appears to be admissible. But the strongest proof of the correctness of the proposition is the fact that in all the problems to which it has been applied, the results agree with those of experience and observation. 68 POLYGON OF FOKCBS. [45-46.] Since the sides of the angles of a triangle are proportional to the sides opposite, we have F _ P R (48) A A sinPji? sini^i? sin^i' POLYGON OF FOECKS. 45. If several con- cwrrent forces are re- presented in magni- tude and direction hy the s-ides of a closed polygon, taken in their order they vnU be in equilibrium. This may be proved by finding the resultant of tv?o forces by means of the triangle of forces ; then the resultant of that resultant and another force, and so on. Fio 21o. PAEALLELOPIPED OF FORCES. 46. If three concurrent forces not in the same plane are represented in magnitude and direction hy the adjacent edges of a paralleopiped, the resultant will be represented in magnitude and direction by the diago- nal / and conversely if the diagonal of a parallelopiped represents a force, it may be considered as the resultant of three forces represented' by the adjacent edges of the parallelopiped. In Fig 22, \i AD represents the force F^ in magnitude and direction, and sirpilarly DB represents i^^, and BG, F^; then according to the triangle of forces AB will represent the resultant of Fx and F^; and AO the resultant of AB and /'g, and hence, it represents the resultant of i'l , i^ , and F^ . Fio. 33. [4G.J EXAMPLES. 69 If F[, F^, and i^ are at right angles with each other, we have A A A and if « is the angle Ii,Fl , ^ of HyFi , and y of Ii,I'l ; then Fx — R cos « ; ] F2 = B(iOS§; V (49) i^ = ^ cos y. ) So. 23. direction-lines of the respec- tive forces make with the axis of x ; ^1 5 ft ) etc., the angles which they make with y ; fi, 78 , etc., the angles which they make with s ; and X, JT, and Z, the algebraic sum of the components of the forces when resolved parallel to the axes x, y, and s, respectively. Then, according to equations (45) and (49), we have for equilibrium ; X= i^cosai-|-i^cosa2 + -^cos «3-j-etc. = 2' Z^cosa = 0; ) J"=^cos|Ji-|-i^cos|i(2-|-i^3COS^3-l-etc.=2'i?'cos^= 0; I (50) Z = i^cos)'i+i^cos3'2-hi's«os3'3+etc.=^i?'cos7 = 0; ) If they are not in equilibrium, let JS be the resultant, and by introducing a force equal and opposite to the resultant, the system will be in equilibrium. [48.] CONSTRAINED EQUILIBRIUM. 71 Let a, h and o be the angles which the resultant makes with the axes x, y and z i-espectively ; then X = li cos a ; T-Bcosh; (51) Z — M cos . Squaring and adding, we have X^ + Y^ + Z" ^ IS? (52) If jff = equations (51) reduce to (50). When the forces are in equilibrium any one of the F-forces may be considered as a resultant (reversed) of all the others. Equations (50) are therefore general for concurring forces. The values of the angles a, §, y, etc., may be determined by drawing a line from the origin parallel to and in the direction of the action of the force, and measuring the angles from the axes to the line as in Analytical Geometry. The forces may always be considered as positive, and hence the signs of the terms in (50) will be the same as those of the trigonometrical func- tions. In Fig. 23 the line Oa is parallel to F^ , and the corre- sponding angles which it makes with the axes are indicated. If all the forces are in the plane x y then y^, y^, etc. = 90°, and (50) becomes X= 2'i^ cos « = 0; \ Y^SFcoa^ = 0. ) (53) CONSTEAINED EQUILIBKIUM. 48. A body is constrained when it is prevented from moving freely under the action of applied forces. If a particle is constrained to remain at rest on a surface under the action of any number of concurring forces, the resultant of all the applied forces must be in the direction of the normal to the surface at that point. For, if the resultant were inclined to the normal, it could be resolved into two components, one of which would be Fio. S4. 72 CONSTRAINED EQUILIBRIUM. £48.] tangential, and would produce motion ; and the other normal, which would be resisted by the surface. Let ir= the normal reaction of the surface, which will be equal and opposite to the resultant of all the impressed forces ; 6^ = the angle {N'^) ; e„ = the angle {J}/',y) ; ^2 = the angle {JV,3) ; Z= (a?, y, z) — 0, be the functional equation of the sur- face; and II, Fi,J^s, etc., be the impressed forces. Then from (51) and (52), we have X= iTcos 6^; Y= iVcos^j,; Z = N cos 0^; From Calculus we have COS^a, n/^ - m- (if (54) ■and similar'ly for eos 0y and cos 6^. These values in (64) readily give X _ T _ Z \dx) \^) \d^) (55) (56) After substituting the values of cos 0^ , cos d^ , and cos 5, in [48.] CONSTEAINED EQUILIBRIUM. 73 (54), multiply the first equation by dx, the second by dy, the third by ds, add the i*esults, and reduce by the equation which is the total differential of the equation Z = ; and we have X<& + Fdy + Zdz = 0. (57) Equations (56) give two independent simultaneous equations which, combined with the equation of the surface, will deter- mine the point of equilibrium if there be one. Equation (57) is one of condition which will be satisfied if there be equilibrium. To deduce (55) let / (x^y) = 0, and /' (a', e) = 0, be the equations of the normal to the surface at the point where the forces are applied. In Fig. 25 let Oa be drawn through the origin of co-ordinates parallel to the required normal, then will dx\ dy' and dz' be directly propor- tional to the co-ordinates of a ; cos aOx = cos Bx = Oa dx' Fio. 25. \/dx"'+ dy'^+dz'^ 1 But the normal is perpendicular to the tangent plane, and hence the pro- jections of the normal are perpendicular to the traces of the tangent plane. The Equation of Condition of Perpendicularity is of the form 1 -1- iza' =; (See Analytical Geometry) ; in which a' = -iL , and a= -^ \ the latter of dx dx which is deduced from the equation of the surface ; .: 1 + ~,~ =0: and similarly dx dx 1 + dz' dz dx' dte 0; 74 hence CONSTRAINED EQUILIBEITTM. 149,3 dx (&;' (fe' da; and -=-7 ^ — 3— da> dz wr the last terms of which contain the partial differential co-efficients deduced from the equation of the surface. These, substituted in the value of cos g^ above, and reduced, give equation (55). CONSTBAINED EQUILIBRIUM IN A PLANE. 49- If all the forces are in the plane of a curve, let the plane yx coincide with that plane ; then Z = and (56) becomes /dL\ ldL\ \dx I \dy I (58) or, Xdx = — Ydy ; and, Xdx + Tdy = ; (59) in which the first of (58) may be used when the equation of the curve is given as an implicit function ; and the second of (58), or (69), when the equation is an explicit function. When the particle is not constrained it has three degrees of freedom (equations (50) ) ; when confined to a surface, two degrees (equations (56) ) ; and when confined to a plane curve, only one degree (equation (58) ). Examples. 1. A body is suspended vertically by a cord which passes over a pulley and is attached to another weight which rests upon a plane; required the position of equilibrium. In Fig. 26, let the pulley be at the upper end of the plane and the cord and plane perfectly smooth. The Pia. 26. weight P is equivalent to a force [49.] EXAMPLES. 75 which acts parallel to the plane, tending to move the weight W up it. Let W = the weight on the plane, which acts vertically downwards ; I* = the weight suspended by the cord ; i — the inclination of the plane to the hcmzontal ; and I, = — y + ax + b = (J, he the equation of the plane. Then X=Pcosi; Y= -W + P&mi; a = tan i = sin ^ (^) = -^'"°'^(&r) = ^5 \dy and these in (58) give P = T^sin^■; which only establishes a relation between the constants, and thus determines the relation which must exist in order that there may be equilibrium ; and since the variable co-ordinates do not appear, there will be equilibrium at all points along the plane when P = TTsin *. The equation of the line, given explicitly, is y = ax + 1); .: dy — a dx; which in the 2°* of (58), or in (59), gives, P = TF sin * as before. 2. Two weights P and TFare fastened to the ends of a cord, which passes over a pulley ; the weight W rests upon a vertical plane curve, and P hangs freely ; required the position of equilibrium. The applied forces at W are the weight W, acting vertically downward ; the ten- sion P on the string ; and the normal reaction of the curve. Let d= wo A I r= OW; x = AW; y=OA; ,1 Y A \ /■= ft-V ^N\ yr t \^ ^r PW '1 , \ Y X N Pio. at. 76 EXAMPLES. [49.] Then, g, „ sin 5 = — : cos ^ = -; r' = a?+ v'; r r r = TF + P cos aOB - W— F cos ; X= P cos cba= — P sin 6 ; and (59) becomes P sin edx + {W- P cos e)dy=0; or, Wdy = P^^^^=Pdr; which integrated gives Wy=Pr+ C; {a) and this, combined with the equation of the curve, gives the required co-ordinates. 3. Let the given curve be a parabola, in which the origin is at the focus. The equation of the curve will be as^ = %> (y + \p) ; but it is unnecessary to use it, since, by a well-known property of the curve, r equals the distance from the focus to the directrix = y + p. Substituting this value of r in equation (a), we have Wy^P{y+j>)+ C. To iind C we observe that when the weight TF, Fig. 27, is in the horizontal line through the focus, y = .-. C = — Pp ; and this value substituted above gives Wy = Py; or W=P; from which it appears that if W = P, the weights will be in equilibrium at all points on the curve. The result holds true when the parabola reduces to the particular case of two vertical straight lines passing through the focus. This problem may also be proved by observing that the normal bisects the angle formed by the radius vector and diameter passing through any point, and, since the forces along the diameter and radius vector are equal, their resultant will be perpendicular to the tangent at that point. 149] EXAMPLES. 77 4. Let the curve be a .circle in whicli tlie distance of from the centre is a ; and the equation of the circle is {a-yf + ^= I^. Ans. r = ^f=. a. W 5. Let the curve be an hyperbola, having the origin of coordinates at the centi-e of the hyperbola. The equation of the curve will be aV — jy = — «'S', and if e is the eccentricity, we have _ aW 6. Eequired the curve such that the weight W may be in equilibrium with any weight P at all points of the curve. This requires that the relation between y and r (or y and a?) in equation («) shall be true for all assumed values of W and P. In Fig. 27, let OB = a, be the distance of the origin of coordinates from the vertex of the curve, then when W is at B, we have y =■ r = a, which, in equation {a), gives C={W-P)a; and from the figure we have y = r cos 6 ; which values in (a) finally give F r =^ ^— — ^— a, 1 ^ a 1 — -^ cos which is the equation of a conic of which the focus is at the pole O. (Discuss the equation and determine all the species of the conic.) 7. A particle is placed on the concave surface of a smooth sphere and acted upon by gravity, and also by a repulsive 78 EXAMPLES. [49.J force, which varies inversely as the square of the distance from the lovirest point of the sphere ; find the position of equilibrium of the particle. Take the lowest point of the sphere for the origin of coordi- nates, y positive upwards, and the equation of the surface will be Z = a!»-|-2^ + s'- 2^y = 0. Let T be the distance of the particle from the lowest point ; then r' = fJtf + ^ = 2Ry. {by Let ft be the measure of the repulsive force at a unit's dis- tance ; then the forces will be -^ = 5"^-! ^^^ '"^9 = w = the weight of the particle. . x = -J^ — Y^J^ 1- ^--t —• 2Ry T ' lliy' r ' ^iRy' r ' which in (56) give, after reduction y M» which in (5) gives, r* = -^ B,. w To see if these values satisfy equation (57), substitute in it the values of X, Y, Z, and the final values of y and r, and we find, xdx + ydy — Rdy + sds = ; which is the differential of equation (5), and hence is true. [This is the theory of the Electroscope.'] 8. A particle on the surface of an elipsoid is attracted by forces which vary directly as its distance from the principal planes of section ; determine the position of equilibrium. [50, 51.] Let MOMENTS OP FOECES. Z = {x, y, 3,) =^ + ^ + ~ -1= 0, 79 i'-* '^ d" be the equation of the surface ; ■'• \dx/ a" ' \dyl b^ ' \dBJ~ , and it is evident tliat this effect varies directly as F. If the action- line of F passed through it would have no tendency to move the particle about that point, and the greater its dis- tance from that point the greater will be its effect, and it will vary directly as that distance ; hence, the measure of the effect of a moment varies as the product of the force and. perpendi- cular / or as cFf; where c is a constant. But as c may be chosen arbitrarily, we make it equal to unity, and have simply Ff as given above. 52. Def. The point from which the perpendiculars are drawn is chosen arbitrarily, and is called the origin of mo- ments. When the sj-stem is referred to rectangular coordinates, the origin of moments may, or may not, coincide with the origin of coordinates. The solution of many problems is simplified by taking the origin of moments at a particular point. 53. The levek aem, or, simply, the arm,, of a force is the perpendicular from the origin of moments to tlie action-line of the force. Thus, in Fig. 29, Oa is the arm of the force F^ ; Oc that of the force F^ , etc. Generally, the arm is the per- pendicular distance of the action-line from the axis about which the system is supposed to turn. 54. The sign of a moment is considered positive if it tends to turn the system in a direction opposite to that of the hands of a watch ; and negative, if in the opposite direc- tion. This is arbitrary, and the opposite directions may be chosen with equal propriety ; but this agrees with the direction in which the angle is computed in pilane trigonometry. Gen- erally we shall consider those moments as positive which tend to turn the system in the direction indicated by the nat- ural order of the letters, that is, positive from + a; to -{- y / from + y to + z ; and from -f- 3 to + x; and negative in the reverse direction. The value of a moment may be represented by a straiglit line drawn from the origin and along the line about which [55-59.] MOMENT OP A FORCE- 81 rotation tends to take place, in one direction fir a ^positive value, and in the opposite direction for a negative one. 55. The composition and resolution of moments may be effected in substantially the same manner as for forces. They may be added, or subtracted", or compounded, so that a resultant moment shall produce the same effect as any num- ber of single moments. The general proof of tliis- proposi- tion is given in the next Chapter. 56. ^ moment axis ie a line passing through the origin of moments and pei*pendicular to the plane of the force and arm. 57. The moment of a fokoe in eefeeencb to' an axis is the product of the firce into the perpendicular distance of the force from the axis. If, in Fig. 29, a line is drawn through O perpendicular tO' the plane of the force and arm, it w^ill be a moment axis, an^ the turning effect oE j?'^ upon that axis will be the sarnie wher- ever applied, providing that its arm Oa remains constant.. If the force is not perpendicular tO' the' arbitrarily chosen a:xis, it may be resolved into two forces, one of which will be perpendicular (but need not intersect it)' and the other parallel to the axis. The' moment of the former component will be the same as that given above, but the latter will have no mo- ment in reference to that axis although it may have a moment in reference to another axis perpendicular to the former. 58. The moment of a foeoe. in befeeence to a plane to which it is paeali.el. is the product of the force into the distance of its action-line from the plane. 59. If any nuiniber of concurring forces are in equilibrium, the algebraic sum of their moments will be 0. Let i^ , i^, i^ , etc.. Fig. 29, be the forces acting upon a particle at A ; and the assumed origin of moments. Join and A, and let fall the per- pendiculars Oa, Oh, Oc, etc., upon the action-lines of the respective forces, and let Oa =/t ; Ob -f^ ; Oc =f ; etc. 82 RESULTANT MOMENT. [60.] Eesolve the forces perpendicularly to the line OA ; and since they are in equilibrium, the algebraic sum of these com- ponents will be zero ; hence, Fx sin OAF^ -^F^ sin OAF^ + F^ sin OAF^ + etc. = ; -,0a ^ Ob ^ Oe , „ or,i^^ + i^^+i^s^ + etc. = 0. Multiply by OA, and we have F^Oa + F^Ol + F^dc + etc. = ; or, i^/i + F,f, + iij/s + etc. = 2F.f^ 0. (60) It is evident that any one of these moments may be taken as the resultant of all the others. MOMENTS OF CONCUEEmG FOEOES WHEN THE SYSTEM IS EEFEKEED TO EEOTANGULAE AXES. y V 60. Let A, Fig. 30, be the point of application of the forces Fi,, Fi, Fi, etc., and the ori- gin of coordinates, and also the origin of moments. Let x, y, and 3 be the coordinates of the point A. Resolving the forces parallel to the coordinate axes, we have, from equation (50), > C y *. ,/ \ ^ \yv, B '^ \ 1 / / z Pio.i X = 2 Fcos a ; T=2F(io&^; Z=2FGoay. The X-forces prolonged will meet the j?Z(zne of yz'\nB\ and will tend to turn the system about the axis of y, in reference to which it has the arm^C'^s; and also about a, in refer- ence to which it has the arm BD = y. Hence, employing the notation already established, we have for the moment of the sum of the components parallel to a?. — Xy, and -f- Xz. [60.] MOMENTS OF CONCXTREING FORCKS. 83 Similarly for the y-oomponents we find the moments, + Yx, and — Ys ; and for the s-convponents, — Zx, and + Zy. The moment Xy tends to turn the system one way about the axis of s, and Yx tends to turn it about the same axis, but in the opposite direction ; and hence, the combined eifect of the two will be their algebraic sum ; or Yx - Xy. But since there is equilibrium the sum will be zero. Com- bining the others in the same manner, we have, for the Tnoments of concurring forces, the following equations : In reference to the axis oix . . . . Zy — Yz = Q; " " " " " " y . . . . Xa - Za; = ; l (61) " " " " " " s . . . . F* - Xy= 0. The third equation may be found by eliminating s from the other two ; hence, when X, Y, and Z are known, they are the equations of a straight line ; and are the equations of the resultant. If the origin of moments be at some other point, whose coordinates are x', y\ and s' ; and the coordinates of the point A in reference to the origin of moments be «i, 3/1, and %; then will the lever arras be Xi = x — x' ; y\ = y — y' ; and % == s — s'. When the system is referred to rectangular coordinates the arm of the force is, y cos a — X cos /8, in which y and x are tlie coordinates of any point of the action-line of the force ; and a is the angle which the action-line makes with the axis of x, and /3 the angle which it makes with y. Flo. 31. 84 EXAMPLES. [60.] In Fig. 31, let AF be the action-line of the force F, O the origin of coordinates, A any point in the line AF, ai which the coordinates x = Oh, and y — Ah. Draw Oa and he perpendicular to AF, and Od from parallel to AF. The origin of moments being at 0, Oa will be the arm of the force. We have dOh — a = chA, cAh = ^- Obd, ch=-y cos a, dh=^x cos /8 ; .-. aO 1= ch — db= y cos a — x cos p. (^1«) If there are three coordinate axes, this will be the arm in reference to the axis of s ; and if there be many forces, the sum of their moments in reference to that axis, will be 2F{y cos a — a; cos /3). Examples. 1. A weight W is attached to a string, which is secured at A, Fig. 32, and is pushed from a vertical by a strut CB ; required the pressure FonBG wlien the angle CAB is d. The forces which concur at B are the weight W, the pressure F, and the tension of the string AB. Take the origin of moments at A, and we have W.BG + F.AC+ tension x = ; Fio. 32. .F= . BO AC'' W tan e. 2. A brace, AB, rests against a vertical wall and upon a horizontal plane, and supports a weight W at its upper end ; required the compression upon the brace and the thrust at A when the angle CAB w0. Fio. sa [60.] EXAMPLES. 85 Tlic concurring forces at A, are W, acting vertically down- ward, the reaction of the wall JV acting horizontally, and the reaction of the brace IK Take the origin of moments at £, we have - Iir.DB + W.OB +F.0 = 0; .-.JV^Wtand. Taking the origin of moments at J), we have W. AD-F. DB sin <9 + iV; = ; .•.F= TFsec6'. 3. A rod whose length \& BG—l is secured at a point B, in a horizontal plane, and the end C is held up by a cord A so that the angle ^^C is d, and the distance AB = a ; required the tension on AC and compression on BG, due to a weight W applied at C. Ans. Tension = — TF cot 0. a 4. A cord whose length AC = lis secured at two points in a horizontal line, and a weight W is suspended from it at B y required the tension ou each part of the cord. CHAPTEK III. PAEAIXEL FOKOES. Fig. 34. 61. Bodies are extended masses, and forces may be applied at any or all of their points, and act in all conceivable direc- tions, as in Fig. 34. 62. SUiPPOSE THAT THE ACTION- LINES OF ALL THE FOECES ARE PARAL- LEL TO EACH OTHEB. This is a spe- cial case of concurrent forces, in which the point of meeting of the action-lines is at an infinite distance. In Fig. 35, let the points a, b, G, etc., which are on the action- lines of the forces and within the body, be the points of applica- tion of the forces, and the point where they would meet if pro- longed. If the point O recedes from the body, while the points of application a, h, c, etc. remain fixed, the action-lines of the forces will approach parallelism with each other, and at the limit will be parallel. 63. Resultant of parallel forces. The forces being par- allel, the angles which they make with the respective axes, including those of the resultant, will be equal to each other. Hence, [64.] MOMENTS OF PAEALLEL FORCES. 87 a= Oi — a^ =03, etc. = a ; J = /3i = A-=i8s,etc. = /8; c = 7i = 72 = 7s , etc. = 7 ; and these, in equations (.50) and (51), give X= B(iOsa = {Fi + E,+ Fi + etc.) cos a = ; Y^RcoaP = {F. + F^ + F^ + etc.) cos ^S = ; - (62) Z = E cos, r/ = {F^ Jf F2 + F^ + etc.) cos 7 = 0, From either of these, we have B^F^ + F^ + F3+ etc. = XF. (63) Hence, the resultant of parallel forces equals the algebra/io sum of the forces. From (62), we have B =VX^+ Y^+Z% which is the same as (52). MOMENTS OF PABAXLEL FOECES. 64. Let Fi, F2, F3, etc., be the forces, and a^, y^, s^ ; os^, y^, 02, etc., be the coordinates of the points of application of the forces respectively (which, as before stated, may be at any point on their action-lines). Then the moments of F^ will be, according to (61), in reference to the axis of x, Fx, cos 7 . y^— F^ cos fi .Si= ;. " " " " « " y,i^,cosa.%-i^cos7.a!i=0; " " " " " " 2, i^,cos^.(Bi-i^cosa.yi=:0; and similarly for all the other forces. Hence, the sum of the moments in reference to the respective axes will be ; (i^2/i + ^2 + i^3 + etc.) cos 7 j -(^1% + F^ + F^ + etc.) cos ^ i (7^2i + i^2 + i'^ + etc.) cos a ) _ —{F^^ + F^ + i^«3 + etc.) cos 7 ) ' ;h»^ {F^x^ + i^ + ife + etc.) cos /Si _ -(i^^i + i^yj + i^sj/s + etc.) cos ^1^ 88 THREE PAEAIiLEL FORCES. [65.] Since meither a, y3, nor 7 is necessarily equal to zero, and ai's independent of their coefficients, the latter will equal zero. Hence, weiiave Fv^ + F^ + F^ + etc. = SFx = ; F,y^ + F^i + F^s + etc. = XFy=0; F^si + F^ + F~^ + etc. = "ZFz — 0. (64) Any one of the forces which hold the system in equilibrinm may be considered as directly opposed to the resultant of all the others. Let ai, y, and "i be the coordinates of the point of application of the resultant. Then (64) gives Rx=%Fx; Ey = XFy; m = XFz (65) in which, if we substitute the value of S, equation (63), we find y = XFx SF' SFy SF' SFs SF (66) which, being independent of tlie angles a, )3, and 7, it follows that the point of ajyplication of the resultant is not changed by changing the directions of the action-line^ of the forces. This point is also called the centre of parallel forces. SYSTEM OF THEEE FOECES. 65. If the SYSTEM CONSISTS OF THKEE FOECES ONLY, and are in the plane xy, we have B=F^±.F^; _ FyX,±F^^ F^yt±F^^ (67) f' f ^ i. c X [66.] STATICAL COUPLES. 89 1st. Consider the positive signs. The resultant will equal the arith- metical sum of the forces. Take the origin at a Fig. 36, where the resultant cuts the axis of x; then X = 0, and the second of (66) gives and hence, if i^ > i^, a^ will exceed tCi ; that is, the resultant is nearer the greater force. 2d. Consider the negative signs. In this case the resultant equals the difference of the forces. Take the origin at a, Fig. 37, and we have and hence both forces are. either at tho right or left of the resultant. 3d. Let Fi= Ei — F, and one of the forces be negative, then B = F-F^Q; x= p^^^ = x ; and y = oo ; ■ (68) that is, the resultant is zero, while the forces may have a finite moment equal to Fix^ ± x^. Such systems are called R Pio. 37. STA.TIOAL COUPLES. 66. -4 couple consists of two equal parallel forces acting »n opposite directions at a finite distance from each other. A statical couple cannot be equilibriated by a single force. It does not produce translation, but simply rotation. A c»uple can he equilibriated only by. an equivalent couple. Equivalent couples are such as have equal moments. The resultant of several couples is a single couple which vill produce the same effect as the component couples. 90 AXIS or A COUPLE. [67-68.] 67- The arm of a cowple is ' the j>erjpendicula/r distance between the action-Unes of the forces. Thus, in Fig. 38, let O be the origin of coordinates, and the axis of x per- pendicular to the action-line of J^; — X then will the moment of one force be X, FoRy, and of the other Fx^, and hence the resultant moment will he Fis. 38. F{xy-x^ = F.ab; (69) hence, ah is the arm. If the origin of coordinates were between the forces the moments would \>e F{xx + x^ = Fab as before. If the origin be at a we have FO + Fab — Fab as before. 68. The axis of a statical couple is any line perpen- dicular to the plane of the couple. The length of the axis may be made proportional to the moment of the couple, and placed on one side of the plane when the moment is positive, and on the opposite side when it is negative ; and thus com- pletely represent the couple in magnitude and direction. If couples are in parallel planes, their axes may be so taken that they will conspire, and hence the resultant couple equals the algebraic sum of all the couples. If the planes of the couples intersect, their axes may intersect. Let = F.db = the moment of one couple ; 6*1= F-^hhi = the moment of another couple; = the angle between their axes ; and Os — the resultant of the two couples ; then and this resultant may be combined with another and so on until the final resultant is obtained. Examples. 1. Three forces represented in magnitude, direction and position, by the sides of a triangle, produce a couple. I.] EXAMPIiES. 91 Fig. 39. 2. In the eleventh example, page 25, show that there is a couple whose moment equals the tension of the string multi- plied by the radius of the wheel. 3. On a straight rod are suspended several weights ; J*l — 5 lbs., i^ = 15 lbs., jFI = 7 lbs., i^ = 6 lbs., 11 = 9 lbs., at dis tances AjS = 3 ft., £D = 6 ft., DE= 5 ft., and EF = ^ ft. ; required the distance ^C at which a fulcrum must be placed so that the weights will balance on it ; also required the pressure upon it. 4. The whole length of the beam of a false balance is 2 feet 6 inthes. A body placed in one scale balances 6 lbs. in the othei-, but when placed in the other scale it balances 8 lbs. ; required the true weight of the body, and the lengths of the arras of the balance. 5. A triangle in the horizontal plane aj, y has weights at the sjveial angles which are proportional respectively to the opposite sides of the triangle ; required the coordinates of the centre of the forces. Let a?! , yi be the coordinates of A, X, y of the point of application of the resultant ; then we have (a + 5 + c) S = axi + Sa^ + gx^ ; and [a+h + c)y =ayi + hy^ + cy^. 6. If weights in the proportion of 1, 2, 3, 4, 5, 6, Y and 8 are suspended from the respective Fa angles of a parallelopiped ; re- quired the point of application F, of the resultant. 7. Several couples in a plane, whose forces are parallel, are applied to a rigid right line, as in Fig. 40 ; required the re- sultant couple. Pjg Jg^ F, Fy 92 CENTRE OP GRAVITY [69.] 8. Several couples in a plane, whose respective arms are not parallel, as in Fig. 41, act upon a rigid right line ; required the resultant couple. CENTRE OF GEAVITT OF BODIES. 69. The action-lines of the force of gravity are normal to the surface of the earth, but, for those bodies which we shall here consider, their convergence will be so small, that we may consider them as parallel. We may also consider the force as the same at all- points of the body. The centre of gravity of a hody is the point of application of the resultant of the force of gravity as it acts upon every particle of the body. It is the centre of parallel forces. K this point be supported the body will be supported, and if the body be turned about this point it will remain constantly in the centre of the parallel forces. Let M = the mass of a body ; m = the mass of an infinitesimal element ; V = the volume of the body ; Z> — the density at the point whose coordinates are X, y, and s ; Ji — W = the resultant of gravity, which is the weight ; and X, y, and i be the coordinates of the centre of gravity. We have, according to equations (63) and (20), and (65) becomes ' ^ X Sgjn = Sgmx ; or JUx = Smx ; y Xgm — "Zgmy ; or My = Smy : z Sgm, = Sgmz ; or Ms = Sms. If the density is a continuous function of the coordinates of the body we may integrate the preceding expressions. The '! (70) OF BODIES. 93 complete solution will sometimes require two or three integra- tions, depending upon the character of the problem ; but, using only one integral sign, (21) and (70) become '^/DdV=fBxdV; y/DdV=fDxdV; '^/DdV^fDxdV. (71) If the origin of coordinates, be at the centre of gravity, then 85 = 0; y =.0 •, z — 0; and hence, Xmx =fDxdV^(i; and similarly for the other values. If D be constant, this becomes f,','xdV- 0; (ria) (715) the limits of integration including the whole body. If the mass is homogeneous, the density is uniform, and D being cancelled in the preceding equations, we have - fxdV ■^ Y~ ' y — y 1 fsdV 3 = - V (72) Many solutions may be simplified by observing the following principles : 1. If the hody has an axis of symmetry the centre of gravity will he on that axis. 94 CENTRE OF GRAVITY [70.] 2. If the hodyhas aj>lane of symmetry the centre of gravity will he in that plane. 3. If the hody has two or more axes of symmetry the centre of gravity will be at their intersection. Hence, the centre of gravity of a physical straight line of uniform density will be at the middle of its lengtJi ; that of the circumference of a circle at the centre of the circle ; that of the circumference of an ellipse at the centre of the ellipse ; of the area of a circle, of the area of an ellipse, of a regular polygon, at the geometrical centre of the figures. Similarly the centre of gi-avity of a triangle will be in the line joining the vertex with the centre of gravity of the base ; of a pyramid or cone in the line joining the apex with the centre of gravity of the base. There is a certain inconsistency in speaking of the centre of gravity of geometrical lines, surfaces, and volumes ; and when they are used, it should be understood that a line is a. physical or material line whose section may be infinitesimal ; a surface is a materiydxdy. \ ^^^^ The integrals are definite, including the whole area. The order of integration is immaterial, but after the first integra- tion the limits must be determined from the conditions of the problem. If Z) is constant and the integral is made in respect to y, we have ■ _ fyxdx fydx'' _ \ffdx y = \ (^6) fydx ' and if x be an axis of symmetry, the first of these equations will be sufficient. If the surface is referred to polar coordinates, then = Og, then, gd = dp, gf^pdd, gh — p COS 6d, ^ also, Fig. 45. a? = /3 COS 6 COS (f); y = p sin 6 ; and z = p cos 6 sin ;^ Vy =,Jf/' p'' cose sin ddpded<}} ;[ (80) ys = f/f p"" cos" e si/nj>d pde d i. J If the volume be one of revolution about the axis of x, wo have [73.1 EXAMPLES. 101 dV=f',rfdx; I Vx = Trfy' xdx. ] ^ ^ Example. Find the centre of gravity of one-eighth of the volume of a homogeneous ellipsoid, contained within the three jprincipal planes. Let the equation of the ellipsoid be , ' , a?^ y^ s^ ' -2 + 32 + ;y-l-0; then, equation (79) gives, X I I / dxdydz— I I xdxdydz. Jo Jo Jo Jq Jo Jo Performing the integration, we have \-irabc ^ = tV ira^bCy .". !« = •§• a. Similarly, y — -fj, and z = %o. Performing the above integration in the order of the letters x, y and e, and using the limits in the reverse order as indicated, we have for the Xrli/mUa. a; = 0, and as = a|/l _ ^_!^ ~x, and the limits for y will be those which correspond to these values of x, or toix = 0,y = by 1-^T= ^; and for a = flsw l-ff-^. y = 0. For the Jir«{ member of the equation, we have I ¥ ■' ■'■ Consider t / 1 — -j- = S, a constant in reference to the y-integration, and we'have 102 EXAMPLES. [74.J b , For the seamd member, we have -tX Consider z as constant in performing the y-integration, and we have - -^va''be .: x= ^ i— = |a: ^irabe ' as given above. 74. When the centre of gravity of a body is known and the centre of gravity of a part is also known, the centre of the remaining part may be found as follows : — Let W= the weight of the whole body ; X = the distance from the origin to the centre of gravity of the body ; Wi = the weight of one part ; a?! = the distance of the centre of w^ fi-om the same origin ; Wi = the weight of the other part ; and a^ = the distance of the centre of w^ from the same origin ; then WiODi + w^ = (wi + w~^x= Wx; and hence, Wx — WiX, (75.] THEOEBMS OF PAPPUS. 103 If the body is homogeneous, the volumes may be substituted for the weight. Example. Let ABC be a cone in which the line BE joins the vertex and the centre of gravity of the base; and the cone ADC, having its apex Z>, on the line GE, and the same base AB be taken from the for- mer cone, required the centre of gravity of the remaining part, ADCB. Let F = the volume of A CB, fm. ^ a = BE, oi - DE, X = Ea — ^a = the distance of the centre of ABC from E, Xi= Eb = ^ai— the distance of the centre of ABC from E) then, a," -3- F = the volume of A CD, and (82) becomes r.ia- F^.K Xt = V- V ^ a? a* — a^ c^ — af Centeobaeio Method, oe 75. Theoeems of Pappus oe of GiiLDnirus. Multiply both members of the second of (74) by 2w, and it may be reduced to ^irys - ^fyds, (83) the second member of which is the area generated by the revo- lution of a line whose length is s about the axis of x, and the 104 EXAMPLES. ^ [75.], first member is, the circumference described by the centre of gravity of the lines, multiplied by the length of the line;. hence, the area generated hy the revolution of a line about a fixed axis equals the length of the Une multiplied ly the d/rcumfer- ence described by the centre of gravity of the line. This is one of the theorems, and the following is the other. From the second of (76), we find %Tfy A=firfdAe. (81) The right-hand member, integrated between the proper limits, is the volume generated by the revolution of a plane area about the axis of x. The plane area must lie wholly on one side of the axis. In the first member of the equation, A is the area of the plane curve, and 27r y is the circumference described by its centre of gravity. Hence, the volume gener- ated by the revolution of a plane curve which lies wholly on one side of the axis, equals the area of the curve multiplied by the circumference described by its centre ofgramty. i Examples. 1. Eind the surface of a ring generated by the revolution of a circle, whose radius is r, about an axis whose distance from the centre is c. Ans. 4fn^rc. 2. The surface of a sphere is 47r/^, and the length of a semicircumf erence is tt r- ; required the ordinate to tlie centre of gravity of the arc of a semicircle. 3. Required the volume generated by an ellipse, whose semi- axes are a and b, about an axis of revolution whose distance from the centre is c ; c being greater than a or b. (Observe that the volume will be the same for all positions of the axes a and b in reference to the axis of revolution.) 4. The volume of a sphere is ^ i^, and the area of a semi- circle is -Jir 7^ ; show that the ordinate to the centre of gravity — 47" of the semicircle \%y — -^. [76.] EXAMPLES. 105 76. Additional Examples. 1. Find the centre of gravity of the quadrant of the ciroumferenoe of a circle contained between the axes a and y, the origin being at the centre. 2r Ans. X = — =y. v 2. Find the distance of the centre of gravity of the arc of a cycloid from the vertex, r being the radims of the generating circle. Ans. y = ^. 3. Find the centre of gravity of one-half of the loop of a lemiaiscate, of which the equation is r' = 2a cos 39, I being the length of tte half loop. «2 _ a» Vg _ 1 AtIS. X = rr; V — -i ^— . 4. Find the centre of gravity of the helix whose equations are a = a cos (f>,y = a sin f, z = na cji, the helix starting on the axis of x. A ~ y - a — X ^-, Ana. X = na - ; y =ina ; and z = -Jz. 5. Find the centre of gravity of the perimeter of a triangle in space. 6. If Xi, and y^ are initial points of a curve, find the curve such that tnx = X — Xo, and ny = y — y^. 7. A curve of given length joins two fixed points ; required its form so that its centre of gravity shall be the lowest possible. (This may be solved by the Calculus of Variations). Ans, A Catenary. 8. Find the centre of gravity of a trapezoid. Let ADEB be the trapezoid, in which BE and /, AB are the parallel sides. Produce AB and BE / I '\ until they meet in U, and join O with E, the mid- dle point of the base ; then the centre of gravity will be at some point ff on this line. The centre of the triangle A CB will be on C'E, and at a distance of iCF from E; and similarly that of BGE will be _^^ on the same line, and at a distance of J-CC? from O ; then, by (79) we may find Eg. Ans.Eg = iFG-^^^^j^. (If BE is zero, we have Eg — iEO for the centre of gravity of the triangle ABC.) 9. Find the centre of gravity of the quadrant of an ellipse, whose equation is aY + b'!!^ - a'V. , - a - . b Am. x = i — \ y = i—' B / \ iwn > Fig. 47. .AB + WE 106 EXAMPLES. [76.] 10. Find the centre of gravity of the oircular sector ABC. Let the angle AOB = B; then _ „ r sine x=Cg = i — -—■ 9 11. Find the centre of gravity of a part of a circular annnlus ABED. Let AC = r, DG = r^, xdA ACB = e; then Bin 9 r^ + jTi + ri' Ans. Cgi = f r+Ti 13. Find the centre of gravity of the circular spandril FOB. 13. Find the centre of gravity of a circular segment. Am. Bist. frm C = ^^ area of segmmf 14. Find the distance of the centre of gravity of a complete cycloid from its vertex, r being the radius of the generating circle. Ang. y = \t. 15. Find the centre of gravity of the parabolic spandril OCB, Fig. 49, in which OC = y, and CB = x. _ Ans. x = -?aX\ ~y = ly- 16. Find the centre of gravity of a loop of the leminisoate, whose equation is r' = a' cos 39. FlQ. 49. 17. Find the centre of gravity of a hemispherical surface. 3t Ans. a = ir. 18. Find the centre of gravity of the surface generated by the revolution of a semi-cycloid about its base, a being the radius of the generating circle and x the distance from the centre. Ans. X = \ia. 19. The centre of gravity of the volume of a paraboloid of revolution is x = lx. 30. The centre of gravity of one half of an eUipsoid of revolution, of which the equation is ffiV + ^^^^ = «'*°i is x=ia. 31. The centre of gravity of a rectangular wedge is a — }o. [76.] EXAMPLES. 107 33. The centre of gravity of a semioiroular cylindrical wedge, whose radius is r, is FiQ. 50. PiQ. 61. 33. The vertex of a right circular cone is in the surface of a sphere, the axis of the cone passing through the centre of the sphere, the base of the cone being a portion of the surface of the sphere. If 38 be the vertical angle of the cone, required the distance of the centre of gravity from the apex. . 1 — cos'e Ana. r. 1 — cos* e 34 Find the distance from G, Fig. 48, to the centre of gravity of a spheri- cal sector generated by the revolution of a circular sector GGA, about the axis 6C. Am. i{GG+ GH). 35. A circular hole with a radius r is cut from a circular disc whose radius is B ; required the centre of gravity of the remaining part, when the hole is tangent to the circumference of the disc. 36. Find the centre of gravity of the frustum of a pyramid or cone. It win be in the line which joins the centre of gravity of the upper and lower bases. Let h be the length of this line, and a and b be corresponding lines in the lower and upper bases respectively, required the distance, measured on the line A, of the centre from the lower end. a« + 3aJ + 34« Ans. x=ift- a« + aJ + b^ If J = 0, we have the distance of the centre of a pyramid or cone from the base equal to ih. 37. Find the centre of gravity of the octant of a sphere in which the density varies directly as the rath power of the distance from the centre, r being the radius of the sphere. - n + 3r Ana. X = pr =y = ^- n + % " 38. Find the centre of gravity of a paraboloid of revolution of uniform density whose axis is a. _ Arm. a — Ja. 108 GENERAL PROPERTIES [77-79.J SOME GENEEAL PROPERTIES OF THE OENTEB OF GRAVITY. 77. When a hody is at rest on a surface, a vertical through the centre of gravity wlUfaU within the oujpport. For, if it passes without the support, the reaction of the snr- faco upward and of the weight downward form a statical couple, and rotation will result. 78. When, a hody is suspended at a point, and is at rest, the centre of gravity will be vertically under the point of sus- pension. The proof is similar to the preceding. When the preceding conditions are fulfilled the body is in equilibrium. 79. A body is in a condition of stable equilibrium when, if . its. position be slightly disturbed, it tends to return to its former position when the disturbing force is removed ; of unstable equilibrium if it tends to depart further from its position of rest when the disturbing force is removed ; and of indifferent equilibrium if it remains at rest when the disturbing force is removed. Example. A paraboloid of revolution rests on a hori- zontal plane ; required the inclination of its axis. Let P be the point of contact o'f the para- boloid and plane, then will the vertical through P pass through the centre of gravity G, and r T" PG will be a normal to the paraboloid. ^^°' ^*" The equation of a vertical section through the centre is y^ =■ '2ipx, in which x is the axis, the origin being at the vertex. Let a — AX = the altitude of the paraboloid ; 6 = GPP = the inclination of the axis ; then, AG = fa, (see example 28 on the preceding page) ; AJ^=%a-p; hence, •dx y y 2ic V 2{^a-p) which will be positive and real as long as %a exceeds^. In [80-81.] OF THE CENTEE OF GRAVITY. 109 this case the equilibrium is stable. When ^a exceeds j) it will also rest on the apex, but the equilibrium for this position is unstable. "When \a — p, d — 90°, and the segment will rest only on tlie apex. Wlien \a is less than p, tan d becomes imaginary, and hence, this analysis fails to give the position of rest; but by independent reasoning we find, as before, that it will rest on the apex, and that the equilibrium will be stable. 80. In a plane material section the sum of the prodiMCts found by multi- plying each elementary m,a,ss by the square of its distance from an axis, equals the sum of the sim,ilar products in reference to a parallel axis passing through the centre, plus the mai,s mul- tiplied by the square of the distance between the axes. ^"*- ^• Let AB be an axis through the centre, CD a parallel axis, D = the distance between AB and CD, dm — an elementary mass, yi — the ordinate from AB to m, y = the ordinate from CD to m,, and M = the mass of the section. Then f=(^, + Dy~ y^ + 2y,D + V. Multiply by dm and integrate, and we have ffdm =fy?dm + 2Dfy-^dm + D^dm. But since AB passes through the centre, the integral of yxdm, when the whole section is inclnded, is zero (see Eq. 715), «a.^fdm = M; hence, ffdm =fyi^dm + MD\ (82) Similarly, if dA be an elementary area, and A the total area, we have ffdA =fyNA + AD'. (83) 81. In am/ plane^a/rea, the sum of the products of each ele- mentary a/rea multiplied by the square of its distance from an axis, is least when the axis passes through the centre. 110 CENTRE OF MASS. [83.] This follows directly from the preceding equation, in which the first member is a minimum for J) = 0. OKNTEB OF THE MASS. 82, The centre of the mass is such anoint that, if the whole mass he multiplied by its distance from an axis, it will equal the sum of the products found by multiplying each elementary area by its distance from the same axis. Let m, — a,n elementary mass; Jif — the total mass ; Xi, yi, and % be the respective coordinates, of the centre of the mass, and x,y, and s the general coordinates, then, according to the definition, we have Mxi = Smx; \ My, = Smy;V (84) Msi = Sms ; ) which being the same as (70) shows that when we consider the force of gravity as constant for all the particles of a body, the centre of the mass coincides with the centre of gravity. This is practically true for finite bodies on the surface of the earth, although the centre of gravity is actuaUy nearer the earth than the centre of the mass is. If the origin of coordinates be at the centre of the mass, we have 2mx — ; Zmy = ; 2mz = ; (84a) which are the same as (71a). CHAPTEE lY. NON-CONCUEEENT F0ECE8. 83. EqTJILIBEIUM of a SOLED BODY ACTED UPON BT AOTT NUMBEE OF FOEOES APPLIED AT DIFFEEENT POINTS AND ACTING IN DIFFEEENT DIEBCTIONS. Let A be any point of a body, at which a force F is applied, and the origin of coordinates, wliich, being chosen arbitrarily, may be within or without the body. On the coordinate axes construct a parallelopipedon having one of its angles at 0, and the diagonally opposite one at A, Let the typical force Fhe. in the first angle and acting away from the origin, so that all of its direction-cosines will be posi- tive ; then will the sign of the axial component of any force be the same as that of the trigonometrical cosine of the angle which the direction of the force makes with the axis. 112 GENERAL EQUATIONS \8S.] Let a = the angle between ^and the axis of x, /3 = iC u a •( a a ii "y. 7 = (I (( (( a it u (( "^, then will the -Z", Y, and Z-components of the force ^be X =^ F(iO& a, Y=Ftios^, Z = ^C08 7. The point of application of the X-com/ponent, being at any point in its line of action, may be considered as at Z?, where its action-line meets the plane yz. At E introduce two equal and opposite forces, each equal and parallel to X, and since they will equilibrate each other, the mechanical effect of the system will be the same as before they were introduced. Combining the force + JT at D with — ^ at E, we have a couple whose arm is DE = y — the y-ordinate of the point A. This couple, according to Article 54, will be negative, hence, its moment is -Xy. Hence, a force + ^ at ^ produces the same effect upon a body as the couple — Xy, and a force + ^ at ^. At the origin O introduce two equal and opposite forces, each equal to X, acting along the axis of x. This will not change the mechanical effect of the system. Combining + X at O with — ^ at ^, we have the couple + Xz, and the force -H X remaining at 0. Hence, a single force +X at A is equivalent to an equal pq/rallel force at the origin of coordin- ates, and the two couples, — Xy and + Xs. Treating the Y-cpmponent in a similar manner, we have the force + Y at the origin, and the moments, + Yx and — Yz ; and similarly for the Z-convponent, the force + Z at the origin, and the moments, — Zx and + Zy. But the couples + Zy and — Ys, have the common axis x, [84.] OF STATICS. 113 and hence are equivalent to a single couple which is equal to the algebraic sum of the two ; and similarly for the others ; hence, the six couples may be reduced to the three following : Zy — Yz, having x for an axis : Xs — Zx, " y " " « Yx—Xy, « z " " " hence, for the single force F acting at A there may he substi- tuted the three axial components of the forces acting at the origin of coordinates, and three pairs of couples having for their axes the respective coordinate axes. If there be a system of forces, in which Fi, F2, 1*1, etc., are the forces, ^i Vi} ■%) tlie coordinates of the point of application of Ft, ^^ y-ii ^1 -fi, etc., etc., etc., ai, Oj, 03, etc., the angles made by F-^, F^, etc., respectively with the axis of x, A> A) A; Gtc, the angles made by the forces with y, and 7i' 72) 73) etc., the corresponding angles with 3 ; then resolving each of the forces in the same manner as above, we have the axial components X = Fi cos at + F2 cos 02 + ^ cos 03 + etc. = SFcoa a ; \ Y= Ft cos I3t + F2 cos A + i^ cos A + etc. = SFcos 13 ; I (85) Z = i^ cos 7i + i^ cos 72 + i^ cos 73 + etc. — HFcos y ; ] and the component moments Zy — Ys- %{Fy cos 7 — ^ cos y8) = Z n Xz— Zx- %{Fz cosa — Fx cos 7) = Jf ; I (86) Yx — Xy= S{Fx COS ^— Fy cos a)-JV;) in which Z, M, and N are used for brevity. EESIXLTANT MOMENT AND EE8IJLTANT COtTPLE. 84. Let R = the resultant of all the forces whose point of application is at the origin, and whose mag- nitudes and directions are the same as those of the given forces ; 8 114 DISCUSSION OF EQUATIONS. [83.] a, h, and c = the angles which it makes with the axes x, y, and z respectively ; Q = the moment of the resultant couple ; d, e, and f — the angles which the axis of the resultant couple makes with the axes x, y, and s re- spectively ; then X= B co5a;\ r=J?cos5;V (87) Z = i? cos c ; ; L = G (ioad;^ M^Gcose-A (88) J!f =G cos/ ) If a force and a couple, equal and opposite respectively to the resultant force and resultant couple, be introduced into the system, there will be equilibrium, and B and G will both be ^ero. Hence, for equilibrium, we have X=0; F=0; Z=0; (89) Z = 0; M=Q; W==0. (90) 85. Discussion of equations (87) and (88). 1. Supjpose that the iody is perfectly free to move in any manner. a. If the forces concur and are in equilibrium, equations • (87) only are necessary, and are the same as equations (60) ; hence, we will have x=o, r = o, Z=0. b. Ji B = and G is finite, equations (88) only are necessary. c. If B and G are both finite, then all of equations (87) and (88) may be necessary. 2 If one point of the body isficed, there can be no trans- lation, and equations (88) will be sufficient. 3. If an axis parallel to x is fixed in the body, there may be translation along that axis, and rotation about it ; hence, the 1st of (87) and the 1st of (88) are sufficient. !.] PROBLEMS. 115 i. If two points arefii'ed, it cannot translate, bnt m&y rotate ; and by taking x so as to pass througli the two points, the equa- tion Z = is sufficient. 5. If oi^e point only is confined to the plane xy, the body will have every degree of freedom except moving parallel to z, and hence, all of equations (S7) and (88) are necessary except the 3d of (87). 6. If tJiree jwints, not in the same straight line, are confined to theplane ,v(/, it may rotate about s, but cannot move parallel to 3 ; hence, the 1st and 2d of (87) and the 3d of (88) are necessary and sufficient. 7. If two axes parallel to x are fixed, the body can move only parallel to x, and the 1st of (87) is sutlicient. 8. If the forces are parallel to the axis of y, there can be translation parallel to y only, and rotation about x and b. 9. If the forces are in the plane xy^ the equations for equi- librium become X — SP cos a = ^ cos a = ; j I^= 2'PcoSi8= i?cos5 = 0; I (91) Yx- Xy = ^{Fxao&^-Fyco&a)^- 0. J [Oes. In a mechanical sense, whatever holds a body is a force. Hence, when we say " a point is fixed," or, '' an axis is fixed," it is equivalent to in- troducing an indefinitely large resisting force. Instead of finding the value of the resistance, it has, in the preceding discussion, been eliminated. When we say "the body cannot translate," it is equivalent to saying that finite, active forces cannot overcome an infinite resistance.] 86. Applications of equations (91). a. peoblems in which the tension of a steing is involved. 1. A hody AB, whose weight is W, rests at its lower end upon a perfectly smooth horizontal plane, and at its upper end against a perfectly smooth vertical plane : the lower end is pre- vented from sliding by a string CB. Determine the tension on the strhig, and the pressure upon the horizontal and vertical planes. 116 EXAMPLES. [86.] Take the origin of coordinates at C, the axis of x coinciding with CIS, andy with AG, x being positive to the right, and y positive upwards. Let W = the weight of the body whose centre of gravity is at 6^; Ji = the reaction against the vertical wall, and, since there is no friction, its direction will be perpen- dicular to ^C; ]V= the reaction of tlie horizontal plane, which will be perpendicular to OS ; I = the horizontal distance from O to the vertical through the centre of gravity ; t = the tension of the string ; then equations (91) become X= B + t = 0; Y^ JSr+ W^ 0; Xy - Yx = R.AG + #.0 + N.GB + W.l = Q; in wliich all the quantities are treated as positive. Solving these equations, gives N = - W, and If the centre of gravity is at the middle of AB, then GB R = 2Z^ 1^ = i ^tan BAG. We thus see that the horizontal plane sustains the whole weight, and that the tension of the string is equal and opposite to the reaction of the vertical plane. The reaction iVand the weight W, being equal and parallel, and acting in opposite directions, constitute a couple whose arjn is equal to GB — I. Similarly the tension i and the reac- tion R constitute another couple M-hose arm is -4.6'; and since there is equilibrium in reference to rotation, we have W{GR) - 1 = R.AG; ■ 7? ^^-^ W ■■^=-AC~ ^' as before. [86.] EXAMPLES. 117 The direction of the forces will generally he hnownfrom the c inditions of the problem^ and it is generally hest to enter them in the equations with their proper signs ; but, as we have seen above, this is not always necessary. When only three forces are involved in a system their relative ^\g\i9, maybe determined tVom the analysis. [Obs. It will be a profitable exercise for the student to solve the same ex- ample by taking the origin of coordinates at different points.] 2. -A ladder rests on a smooth horizontal plane and against a vertical wall, the lower end being held by a horizontal string; a pei-son ascends the ladder, required the pressure against the wall for any position on the ladder. 3. A uniform beam, whose length is AB and weight TF", is held in a liorizontal position by the inclined string CD, and carries a weight P at the extremity ; required the tension of the string. AB DC_ AI)' AC Ans. t — 4. A prismatic piece AB is per- mitted to turn freely about the lower end A, and is held by a string CE\ given the position of the centre of gravity, the weight TFof the piece, the inclination of the piece and string, and the point of attachment E\ required the tension of the string, and the fi^j. st. pressure against the lower end of the beam at A. 5. A heavy piece AB is supported hy tvjo cords which pass over pulleys C emd D, and have weights P]^ and P attached to them; required the inclination to the horizontal of the line AB joining the points of attachment of the cord. (Consider the pulleys as reduced to the points and D.) {P + iW). lis EXAMPLES. 8.] Let O, the centre of gravity of AB, be on the line join- ing the points of attachment A and B ; a^ AG; l = BG; i = the angle DCM; S = the inclination of BA to DC; a = BOA ; and j8 = CBB. Eesolving horizontally and vertically, we have X = Pi cos MOa"- P cos JVBB + TTcos 90° = ; = Pi cos (a—i) — P cos (/3 + «■) = ; {a) Y-P^ sin {a-i) + P sin (/8 + *) - W ^ 0. \h) Taking the origin of moments at G, making Gp perpendi- cular to A C, and Gp^ perpendicular to DB, we have PiX Gp- P X Op^- TFxO = 0; or Pi . a sin (a + S) - P . 5 siu (/3 - S) = 0. (c) The angle i is given by the conditions of the problem ; hence the three equations {a), ib), and (c) are sufficient to determine the angles a, /3, and S, when the numei-ical values of the given quantities are known. The inclination will be S + i. 6. Suppose, in Fig. 58, that the lengths of the strings XC and BB are given, required the inclination of AB. [The solution of this problpm involves an equation of the 8th degree] . c „A Y. A heavy piece AB, Fig. 59, is free to swing about one end A, and is supported by a string BG which passes over a pulley at U, and is attached to a weight P ; find the angle A GB when they are in equilibrium. 8. A weight W rests on a' plane whose inclination to the horizontal is i, and is held by a string whose in- clination to the plane is 9 ; I'equired the relation between the tension P and the weight, and the value of the normal pressure upon the plane. Fig. 60. P=^TF; cos^ Normal pressure = ,, ' W. cos 6 6.] EXAMPLES. 119 h. equilibhutm of perfectly smooth bodies in contact with EACH OTHEE. 9. A heavy ieam rests on two smooth inclined planes, as in Mg. 61 ; required the inclination of the ieam to the horizontal, and the reactions of the respective planes. Let AC aud CB be the in- clined planes; AB the beam whose centre of gravity is at O. When it rests, the reactions of the planes must be normal to the planes, for otlierwise they would have a component parallel to the planes which would produce mo- tion. Let a^ = AG; a^= GB ; R = the reaction at A R'= « " "^; Tr= the weight of the beam ; a = the inclination oi AO to the horizon ; Q —. li « « ^6' " " " a —■ a u « AB " " " r \#^ ^^ / ^^ J\ C Take the origin of coordinates at the centre of gravity G of the body, x horizontal and y vertical. The forces resolved horizontally give X = 2? sin a — ^' sin /3 + TTcos 90° = ; and vertically, Y=B(iOsa + B' (io&P- Fsin 90° = 0. {a) % The moment of ^ sin a is, . Hence Xy + B sin a X J- , and if the beam be suspended at D by means of the two cords DA and DB it will retain its position when the planes AC and CB are re- moved. If /3 = 90°, the plane GB will be vertical, and we find Ox sin a o^cos a 7-> <^2 -nr Ttr (h sin a -_. a, ,„ ^ = ^.Trseca; B' -~ 1^=^-17 tan a; Oi 0-2 cos a a, ' If ffli = flta, then _ sin^ ^ ^ sin a tan^=,^!?i^^. z sm a sm yS [86.] EXAMPLES. 121 If jS^ 90° and a =0°, then ^' = 0, 6=0, and B=W. A special case is that in which the beam coincides with one of the planes. The formulas do not apply to this case. 10. Two equal, smooth cy- lindei-s rest on two smooth 2ilanes whose inclinations are a and /8 respectively ; required the inclination, 6, of the line joining their centres. Fia. 63. Ans. Sin 6 = -Kcot a — cot /3). 11. A heavy, uniform, smooth beam rests on one edge of a box at 0, and against the vertical side opposite ; required its inclination to the vertical. Let g be the centre of gravity. Ans. Si„« = y^, 12. Three equal, smooth cylinders are placed in a box, the two lower ones being tangent to the sides of the box and to each other, and the other placed above them and tangent to both ; required the pressure against the bottom and sides of the box. Fig. 64. Ans. Pressure on the 'bottom = total weight of the cylin- ders. Pressure on one side = \ weight of one cylinder y. tan 30°. 13, Two homogeneous, smooth, prisma- tic bars rest on a horizontal plane, and are prevented from sliding upon it; required their position of equilibrium when leaning against each other. Let AB and CP be the two bars, resting against each other AJi Fio. 6B. 122 Examples. [86.] at B ; then will they be in equilibrium when the resultant of their pressures at B is perpendicular to the face of CD. Let I = AB; o^ CD; x = BD ; a = AD ~ the distance between the lower ends of the bars; W = the weight of AB ; PTi = the weight of CD; Ea.uA tr the respective centres of gravity of the bars, which will be at the middle of the pieces ; then we have 2 {a? + 5^ - a^) a^ W= c {a? -P + a?) {-a' + P + a?) W^; which is an equation of the fifth degree, and hence always ad- mits of one real root. 14. The upper end of a heavy piece rests against a smooth, vertical plane, and the lower end in a smooth, spheri- caXiowl ; required the ^iosition of equi- librium. Let AB be the piece, BF the verti- cal surface, EA the spherical surface, and g the centre of gravity of the piece. When it is in equilibrium, the reaction at the lower end will be in the direction of a normal to the surface, and hence will pass through C, the centre of the sphere, and the reaction of the ver- tical plane will be horizontal. Let W = the weight of the piece ; r = the radius of the sphere ; a = J.^ ; I = Bg; I — AB ; d = CF; H = the reaction of the vertical plane ; N = the reaction of the spherical surface ; i = the inclination of the beam to the horizontal ; 6 — the inclination of the radius to the horizontal. Take the origin of coordinates at g, x horizontal and y ver- tical ; and we have X= iV^cos^ - ^ = 0; T = iTsin 9 - TF = ; Moments = ~ R.h sin i + R.a sin {9 — i) ; Pig. 66. [86.] INDETERMINATE PROBLEMS. 123 and the geometj'ical relations give, Icos i = KB = RF= d -\- r co'& 6. From these equations, we have iV^=TFcoseo 6; E =Trcot^; a sin {6 — i) —b cos ^ sin * = 0, which, by developing and reducing, becomes {a + h) tan i — a tan d ; this, combined with the fourth equation above, will determine i and Q. The position is independent of the weight of the piece, but depends upon the position of its centre of gravity. 15. A heavy prismatic bar of infinitesimal cross-section rests against the concave arc of a vertical parabola, and a pin placed at the focus ; required the position of equilibrium. Let I = AB = length of tlie bar ; -p = CD = one-half the parameter of the parabola, C being the focus, and d = A CD. Am. 6 = 2 16. Required the form of the curve such that the bar will rest in all positions. Ans. The polar equation is r = ^l + c sec 0, in which I is the length of the bar, and o an arbitrary constant. It is the equation of the conchoid of Nichomedes. C. INDBTEEMINATE PEOBLEMS. 17. To determine the pressures exerted hy a door ujpon its hinges. Let W = the weight of the door ; a = the distance between the hinges ; b = the horizontal distance from the centre of gravity of the door to the Vertical line which passes through the hinges ; ^= the vertical reaction of the upper hinge ; Fi = the vertical reaction of the lower hinge; S = the horizontal reaction of the upper hinge ; Hi = the horizontal reaction of the lower hinge ; 124 INDETERMINATE PROBLEMS. then X = ^ + ^1 = ; T= F+ F^-Wz= 0; Xy - Tx= Ha-Wi = 0; •which give B:= - H^^ —W; m.d. a F^ F,=W. The result, therefore, is indeterminate, but we can draw two general inferences : 1st, The horizontal pressures upon the hinges are equal to each other hut in opposite directions / and, 2d, The vertical reaction upon both hinges equals ths weight of the door. It is necessary to have additional data in order to determine the actual pressure on each hinge. The ordinary imperfec- tions of workmanship will cause one to sustain more weight than the other, but as they wear they may approach an equality. The horizontal and vertical pressures being known, the actual pressures may be found by the triangle- of forces. If the upper end sustains the whole weight, the total pressure W npon it will be — yo? + J^. If each sustains one-half the weight, the pressure on each will be one-half this amount. 18. A rectangular stool rests on four legs, one being at each corner of the stool ; required the pressure on each. (^The data are insufficient.) P ] 9. A weight P is supported by three un- equally inclined struts; required the amount which each will sustain. ^_ Fio. 68. [Obs. If more conditions are given than there aie qaantities to be deter- mined, they wm either be redundant or conflicting.] [86.1 EXAMPLES. 325 d. STRESS ON TBAMKR 20. Suppose that a trumgulo.r truss, Fig. 69, is loaded with equal weights at the upper apices ; it is required to find the stress upon any of the pieces of the truss. [The stress is the pull or push on a piece.] Let the truss be supported at its ends, and let I = Aa ^ ah ^= etc., = the equal divisions of the span AB; ]V= the number of bays in the chord AB ; L= Nl = AB, the span ; Pi,Pi,Pi, etc., be the weights on the successive apices ; which we will suppose are equal to each other ; hence p=p^-p^^ei(i.; Np — the total load ; V^ the reaction at ^ ; and Vi= " " " B. 1st. There will he equilibrium among the external forces. All the forces being vertical, their horizontal components will be zero, hence X=0; F=F+Fi-.Si? = F+Fi-7Vp = 0; {a) and taking the origin of moments at B, observing that the moment of the load is the total load multiplied by the horizontal distance of its centre of gravity from B, we have V.AB - Np.^AB^^O; or, V.L-Np.^L^O; .:Y^\Np; 1-26 STRESS ON [86.J which in (a) gives "Fi also equal to ^JVp ; hence the supports sustain equal amounts, as they should, since the load is syme- trical in reference to them, and is independent of the form of trussing, 2d. To determine the internal forces. — Conceive that the truss is cut hy a vertical plane and either part removed while we consider the remaining part. To the pieces in the plane section, aj^ly forces actin.g in such a manner as to produce the same strains as existed iefore they were severed. Consider the forces thus introduced as- exterruil, and the problem is reduced to that of determining their value so that there sJiall he equilibrium among the new system of external forces. Let CB, Fig. 69, be a vertical section, and suppose that the right-hand part is removed. Introduce the ex- ternal forces in place of the strains, as shown in Fig. 70. Fig. 70. Let H = the compressive strain in the upper chord ; Hi =: the tensile strain in the lower chord ; i'' = the pull in the inclined piece ; 6 =: the inclination of i^to the vertical ; n = the number of the bay, iJ), counting from A (which in the figure is the 3d bay) ; and D = CD = the depth of the frame. The origin of coordinates may ie taken at any point. Take it at A, X being horizontal and y vertical. Resolving the forces, we have X= B,- B + Fsin 6=0; (a) Y = V — Pi, — Pi — etc., to^n — i^cos = 0; = V - np - Fcos e = 0; (5) the moments = V.O —pi-il —pi-^l — etc.+ S.B — F.AE=0, or, Wp + SD -Fnl cos 6 = (c) 86.] FRAMES. _ 127 Eliminating i^^ between equations (i) and (c), substituting the value of F = inp, and reducing, give that is, the strains on the hays of the upper chord vary as the product of the segments into which the lower chord is dvoided hy the joint directly under the hay considered. From (5) we have i^cos e = V- np = f (i\^ - 2w) J? ; (e) and since 6 is constant, the stress on the inclined pieces decreases uniformly from the end to the middle. At the middle n = ^Iff, and F =^ Q ; hence, /or a uniform load, there is no stress 07i the central braces. Those pieces which incline from the ends towards the middle will be subjected to compression, and the others to tension. Eliminating iZ^and i'^from {a), we have H,= \^N{N-^)-n{n-l)\^ (/) For forces in a plane the conditions of statical equilibrium give only three independent equations, {a), Q>) and {c) ; (or Eqs. (91) ); hence, if a plane section cuts more than three indepen- dent pieces in a frame, the stresses in that section are indeter- minate, unless a relation can be establislied among the stresses, or a portion of them be determined by other considerations. 21. If iV^= 7,_p =i?i =i>2 = etc. = 1,000 lbs., AB = 56 feet and Z> = 4 feet ; required the stress on each piece of the frame. 22. In Fig. 69, iipi andj?2are removed, and^g = Pt = Ps ~ 1,000 lbs., find the stress on the bay 2 — 3, and the tie 2 — 5. 23. If all the joints of the lower chord are equally loaded, and no load is on the upper chord, required the stress on the n'* pair of braces, counting from A, Fig. 69. Ans. i(ir— 2n -\- l)p sec 0. 128 STRESS ON A [86.] 24. A roof truss ADB is loaded with equal weights at the equidistant joints 1, 2, 3, eto. ; required the stress on any of its members. [Obs. The weight at J? is the same as the others. This load will produce the same stress as that of a load uniformly distributed, except that the latter would produce cross strains upon the rafters, which it is not our purpose to discuss in this place.] Let the tie AB be divided into equal parts, Aa, db, etc., and the joints connected as shown in the figure. The joints are 1 > > i n \< h > . \ \ X A a i m. c B Fid. 71. assumed to be perfectly flexible. The right half of Fig. 71 may be trussed in any manner by means of ties or braces, or both, and yet not affect the analy- sis applied to the left half. Conceive a vertical section nm, and the right-hand part removed. In- troduce the forces H., H^ and i^as previously explained, and the condi- tions of the problem will be repre- sented by Fig. 72. The letters of reference given below involve both figures. Let N = the number of equal divisions (bays) in AB • n — the number of the bay Ic counting from A • I — Aa = ab, etc. ; jp = the weight on any one of the joints of the rafter; V = the vertical reaction at ^ or ^ ; /> = DC, the depth at the vertex; " 52c; and* = DAG. e Then (iT- l)j> = the total load ; F=i(ir-l)_p. I.] ROOF TRUSS. 129 Take the origin of coordinates at A, and the origin of mo- ments at the joint marked 2. Resolving the forces shown in Fig. 72 horizontally and vertically, we have X = - ^cos DAG + El- Fsini^G = 0; T = V— np - ^sin DAG + Fcos b2o — ; or. H^ 308 i + Hx~ Fsin = 0; V — np — H%va.i + Faoi e := also the moments 1 HM - V.ab +(n- -l)p.\{n - 2)1 = But from Fig. 71 we have b2 Ah {n-\) I CD ~ AG ~ iJVl "• Substituting in the equation of moments the value of 52 found above, of F= i{N' — T)jp, of Ah — (n — 1) I, and reducing, give By means of the other two equations, we find E = i(iV" — n)p cosec i ; F= i(n — l)j?sec 6. e. STRESS IN A LOADED BEAM. 25. Swppose that a heam, is firm,ly fixed in a wall at one end, and that the pt'ojecting end is loaded with a weight P ; required the forces in a vertical section mn, Fig. 73. Take the origin of coorcli- nates at A, x horizontal and y vertical. Take the plane sec- tion perpendicular to the axis of 3?. Without assuming to know the directions in which the Fio. 73. 130 LOADED COED. 6.] forces in the section act, we may conceive them to be resolved into horizontal and vertical components. Let F be the typical horizontal force, then will X = %F^ 0; hence, some of the F-forces will be positive, and the others negative. Neglecting the weight of the beam, and letting Y^ be the sum of the vertical components in nm, we have r= ri-P = .-. 7i = P; as shown in the figure. The forces, + P and — P, constitute a couple whose arm is Aa ; and since the F-forces are the only remaining ones, the resultant of the + F''s and the — F^s must constitute a couple whose moment equals P-Aa with a contrary sign. [Obs. Investigations in regard to the distribution of the forces over the plane section belong to the Resistance of Material. \ f. LOADED COED. 26. Suppose that a 'perfectly flexible, inextensible cord is ■fixed at two points and loaded continuously, according to any law ; it is required to find the equation of the curve and the tension of the cord. Assuming that equilibrium has become established, we may treat the problem as if the cord were rigid, by consider- ing the curve which it assumes as the locus of the point of ap- plication of the resultant. The resultant at any point will be in the direction of a tangent to the curve at that point ; for otherwise it would have a normal component which would tend to change the form of the curve. Take the origin of coordinates at the lowest point of the curve. Let a be any point whose coordinates are x and y; X = the sum of the x-components of all the external forces between the origin and a ; Y = the sum of the y-components ; [86.] LOADED CORD. 131 t = the tension of the cord at a ; 4 = the tension at the origin. Eesolving the tension {t) by multiplying it by the direction- cosine, we ha\e < T- = the aj-component of t, and t-~ = the j'-eomponent. For the part Ca, equations (91) become {a) \Obs. In the problems which we shall consider, the third of these equations win be unnecesEary, since the other two furnish all the conditions necessary for solving them: ] Let OH' the applied forces he vertical. Then X = 0, and the first two of equations {a) become ds T+t 0. (5) From the first of these we have {7CG t-^ = to = a constant ; hence, the horizontal component of the tension will he constant throughout the length for any law of vertical loading. From the second of (5), we have 132 PARABOLIC CORD. S.] hence, the vertical cnrrvponent of the tension at any point equals the total load between the lowest point and the point considered. 27. Let the load he uniformly dis- tributed over the horizontal. (This is approximately the condition of tlie ordinary suspension bridge. ) Let w = the load per unit of length, then X = wx; and (b) becomes '■-''£="■' - wx+t$ = 0. ds Eliminating t gives t^dy = wxdx; and integrating gives t^ = iwa^ + (C = 0) ; .-. gff' = —- y; w ^ ' («) {d) hence, the curve is a parabola whose axis is vertical, and whose 2# parameter is —5-. The parameter will be constant when t^ — w is constant ; hence the tension at the lowest point will he the same for all parabolas having the same parameter and the same load per unit along the horizontal, and is independent of the length of the curve. To find the tension at the hwest point, substitute in equa- tion {d)t\\e value of the coordinates of some known point. Let the coordinates of the point J. be aJi and y\ ; then (d) gives 4 = WXi ie) 186.] THE CATENARY. 133 To find the tiiiswn at any 2>oint we have from the first of equations (c) and the Theory of Curves ds _ . Vda? + df _ / dil' To find the tension at the highest point A, from (d) find dx x^ ■= yr substitute in {^f), and we obtain ' ' ^' ' t,--^-~yx? + ^,\ (To find So by the Theory" of Momentg, take the origin at A. The load on Zi will be iiXi, and its arm the horizontal distance to the centre of gravity of the load, or i^i ; hence, its moment wUl be iwx^'. The moment of the tension will be toyi ; hence, toWi = iOTse'i or to = s — , as before.) The slope (or inclination of the curve to the horizontal) may be found from equation {g) ; which gives tan % — -^ . Xy 28. The Catenary. A catenary is the curve assumed by a perfectly flexible string of uniform section and density, when suspended at two points not in the same vertical. Mechanically speaking the load is uniformly distributed over the arc, and hence varies directly as the arc. To find the equation, let w — the weight of the cord per unit of length ; .•. Y =■ W8{s being the length of the arc) ; and equations (5) become ws+t^=^0. ds > W 134 THE CATENARY. [86.] Transposing and dividing the second by the first, gives dy w -Y- = -SI ax ta and differentiating, substituting the value of ds and reducing, gives d .-. - dx = to \dx} Integrating gives i^-%.[i.v/:;i]; or, passing to exponentials, gives dx^y ^d^ dx^ dx' ^*' ^ da?- [/' dx] ' from virhich we find which integrated gives = 4^ [^•- -«-'•"]; which is the equation of the Catenary. Eliminating -^ between equations (%) and {j), we find {k) (Z) ds ,r f. -;?.-j ^ = *L' ""' J' [88.] THE OATENABY. 135 tixe integral of which is which gives the length of the curve. The following equations may also be found v/ 2w If 5 — the inclination of the curve to the vertical, then X = s tan 6 log^ cot id. The tensions, t and tg, are so involved that they can be de- termined only by a series of approximations. The full devel- opment of these equations for practical purposes belongs to Applied Mechanics. The catenary possesses many interesting geometrical and mechanical prop- erties, among which we mention the following : — The centre of gravity of the catenary is lower than for any other curve of the same length joining two fixed points. If a common parabola be rolled along a straight line, the Iocub of the focus will be a catenary. According to Eq. (k) it appears that if the origin of coordinates be taken directly below the vertex at a distance equal to tg -i- w, the constant of integra- tion will be zero. (This distance equals such a length of the cord forming the catenary as that its weight will equal the tension at the lowest point of the curve). A horizontal line through this point is the directrix of the catenary. The radius of curvature at any point of the catenary equals the normal at that point, limited by the directrix. The tension at any point equals the weight of the cord forming the cate- nary whose length equals the ordinate of the point from the directrix. If an indefinite number of strings (without weight) be suspended from a catenary and terminated by a horizontal line, and the catenary be then drawn out to a straight line, the lower ends of the vertical lines will be in the arc of a parabola. If the weight of the cord varies continuously according to any known law the curve is called Catenarian. 136 LAW OF LOADING. [86.] 29. To determine the equation of the Catenarian curve of uniform density in which the section varies directly as the tension. Let Jc = the variable section ; S = the weight of a unit of volume of the cord ; = the ratio of the section to the tension ; then Y=fBkds, Tc = ct, .-.T =^ hoftds; which substituted in (5) and reduced, gives hey ■= loge sec chx, for the required equation. g. LAW OF LOADING. 30. It is required to find the law of loading so that the action-line of the resultant of the forces at any point shall he tangent to a given curve. Assume the loading to be of uniform density, and the variations in the load- ing to be due to a variable depth. In Fig. 76, let O be the origin of coordi- nates ; Z =ab — the depth of loading' over a point whose abscissa is a; ; 6? = the depth of the loading over the ori- gin, and S = the weight per unit of volume of the loading, then T=fhZdx; which in Eq. (b) gives dy Transposing, and dividing the latter by the former, gives [86.] LAW OF LOADING. 137 which, differentiated, gives But, from the Theory of Curves, we have (l + ^f cPy _ \ d3?l _ sec' * dx^ ~ p "" p in which p is the radius of curvature, and 4 is the angle between a tangent to the curve and the axis of x. From these we read- ily find „ to sec'*' At the origin p = po, i = 0, and Z = d; which values sub- stituted in the preceding equation give :. Z — dpa . (n) p Discussion. For all curves which have a vertical tangent, we have at those points i = 90° ; .•. sec ^■ =:; oo , and, if p is finite Z ^ ao -^ hence, it is practically impossible to load such a curve through- out its entire length in such a manner that the resultant shall be in the direction of the tangent to the curve. A portion of the curve, however, may be made, to fulfil the required con- dition. Let the given cv/rve he the are of a cvrcle / then p = p^, and equation (n) becomes Z = d sec' i, from which the upper limit of the loading may be found. For 138 EXAMPLES OP A [86.] small angles sec'i will not greatly exceed unity, and hence, the upper limit of the load will be nearly parallel to the arc of the circle for a short distance each side of the highest point. At the extremities of the semicircle, * — 90°, and Z = dx — the sc-component, and / j)dy = the y-component of the pressures, hence, equations {a), p. 131, become , dx ta + I pdx + t ^— = ; J:pdy + 4=»^ differentiating which, gives dx pdy^td{^) = 0. [86.] NORMALLY PRESSED ARC. 139 Transposing, squaring, adding and extracting the square root, give that is, the normal pressure varies inversely as the radius of curvature. 1. If a string be stretched upon a perfectly smooth curved surface by pulling upon its two ends the normal pressure upon the surface will vary inversely as the radius of curvature of the surface, the curvature being taken in the plane of the string at that point. 2. If p be constant p will be constant ; hence, if a circular cylinder be immersed in a fluid, its axis being vertical, the nor- mal pressure on a horizontal arc will be uniform throughout its circumference. h. THE LAW OF LOADING ON A NOEMAXLT PEESSED ARC BEING GIVEN, EEQDIKED THE EQUATION OF THE AEG 32. The ties of a suspension ^ 5 hridgeheing normal to the curve /'^Vs^ n ^^f^^^K of the cable, and the load uni- / / /Tl \\ \_\ form along the span, requi/red d e F the equation of the curve of fio. n. the cable. ($W^)-4=li^\/(i) Ans. I 1 + the origin being at O, x horizontal and y vertical d^ dx If tan * = ^, and />o = the radius of curvature at the vertex, then X = i/3o (1 + cos' i) sin i, y = JyOj sin' i cos i. (See solution by Prof. S. "W. Robinson, Journal of the Franklin Institute, 1863, vol. 46, p. 145 ; and its application to bridges anJ arches, vol. 47, p. 152 and p. 361.) 140 NOEMALLT PEESSED ARC. [86.] 33. A jyerfectly flexible, inextensible trough of indefinite length is filled with a fluid, the edges of the trough being par- allel and supported in a horizontal plane ; required the equa- tion of a cross section. The length is assumed to be indefinitely long, so as to elimin- ate the effect of the erul pieces. The pressure of a fluid against a surface is always normal to the surface, and varies directly with the depth of the fluid. The actual pressure equals the weight of a prism of water whose base equals the surface pressed, aud whose height equals the depth of the centre of gravity of the said surface below the surface of the fluid. The problem may therefore be stated as follows : — Jiequired the equation of the curve assumed by a cord fixed at two joints in the same hori- zontal, and pressed normally by forces which vary as the verti- cal distance of the point of application below the said hori- zontal. Let A and B be the fixed points. Take the origin of the coordinates at D, midway be- tween A and £, and y positive downwards. Let 8 be the weight of a unit of volume ; then p = Sy, which in equation (o) gives t = Syp, and for the lowest point t — SJ)pg ; in which I) is the depth of the lowest point and po the radius of ciirvatui-e at that point ; .-. Syp = SDpo, or ^ = -. But from the Theory of Curves we have P~\^dx^) dj?' which substituted above, and both sides multiplied by dy, may be put under the form [86.] HYDROSTATIC THOUGH. 141 the integral of which is (-fr=4-« But ^ = 0, for y = Z> ; .-. C = -^^^ ; which substituted and tlie equation reduced gives (dx" + dv']*- ds - ^Po^^y {OX + ay) -as- ^^^jj^p.^^^Dp^-jy^+fy] Squaring and reducing, gives These may be integrated by means of Ellvptio Functions. Making y ^ D cos <^, and c— j—, they may be reduced to known forms. Using Legendre^s notation, we have (See Article by the Author in the Journal of the Franklin Institute, 1864, vol. 47, p. 289. CHAPTEE V. EELATION BETWEEN THE INTENSITIES OP F0ECE8 ON DIFFEEENT PLANES WHICH OUT AN ELEMENT- 87. DisTErBUTED FoEGES are those whose points of applica- tion are distributed over a surface or throughout a mass. The attraction of one mass for another is an example of the latter, some of the properties of which have been discussed in the Chapter on Parallel Forces ; similarly, when one part of a body is subjected to a pull or push, the forces are transmitted through the body to some other part, and are there resisted by other forces. If the body be intersected by a plane, the forces which pass through it will be distributed over its surface. Planes having different indinations ieirhg passed throuyh an element, it is proposed to find the relation between the intensi- ties of the forces on the different planes. 88. Definitions. Stresses are forces distributed over a sur- face. In the previous chapters we have assumed that forces are applied at points, but in practice they are always distrib- uted. A strain is the distortion of a body caused by a stress. Stresses tend to change the form or the dimensions of a body. Thus, apuU elongates, a push compresses, a twist produces tor- sion, etc. (See Hesistanoe of Materials.) A simple stress is a pull or thrust. Stresses may be com- pound, as a combination of a twist and a pull. A dieect simple stress is a pull or thrust which is normal to the plane on which it acts. A. pull is considered ^os^^we, and a thrust, negative. The intensity of a stress is the force on a unit of area, if it be constant ; but, if it be variable, it is the ratio of the stress on an elementary area to the area. To form a clear conception of the forces to which an element is subjected, conceive it to be removed from the body and then [S9, 90.] EESOLVED STRESSES. 143 subjected to such forces as will produce the same strain that it had while in the body. 89. Formulas fok tub intensitt of a stress. Let i^ be a direct simple stress acting on a surface whose area is A, and j? the measure of the intensity, then j? = A' (92) when the stress is uniform, and dF J? dA , when it is variable. If the stress be variable we will assume that the section is so small that the stress may be considered uniform over its sur- face. 90. Direct stress resoltep. Let the prismatic element AB, Fig. 79, be cut by an oblique plane DF. Let the stress F be simple and direct on the surface CJi, and JV = the normal component of i'^on DF; T = the component of F along the plane DF, which is called the tangential component ; 6 = FON = the angle between the action-line of the force and a normal to the plane DF, and is called the obliquity of the plane ; A = the area of VB, and A' that of DF. Then, according to equations (62), we have N = i^cos 6 ; T = Fsm e. From the figure we have A' — A sec 6, hence, on the plane DE, we have Fig. 79. 144 SHEARING STRESS. [01.] 717- 7 • . •. J^ -^cos e , „ Jyormal'mtens%ty,pn'= —ti = —a a ~ P ''^^ ' * .o. -a. sec V Tangentialintensity, 2>t = "T7 = 'a -n=psiii.dcosd. (93) Pass another plane perpendicular to DE, having an obli- quity of 90° — 6; then, accenting the letters, we have y„=i?sin»^; ) , p't = p cos6 sin 6. ) This result is the same as if a direct stress acting upon a plane perpendicular to GB, having an obliquity of 90° — in reference to DE, be resolved normally and tangentially to the latter. Combining equations (93) and (94) we readily lind '«+y«=i^;] (95) that is, when an element {or hody) under a direct simple stress is intersected iy two planes the sum of whose obliquities is 90 degrees, the sum of the intensities of the normal components of the stress equals the intensity of the direct simple stress, and the intensities of the tangential stresses are equal to each other. 91. Sheaking stress. The tangential stress is corainonly called a shearing stress. It tends to draw a bodj' sidewise along its plane of action, or along another plane parallel to its plane of action. Its action may be illustrated as follows : — Suppose that a pile composed of thin sheets or horizontal layers of paper, boards, iron, slate, or other substance, having friction between the several layers, be acted upon by a horizontal force applied at the top of the pile, tending to move it sidewise. It will tend to draw each layer upon the one immediately beneatli it, and the total force exerted between each layer will equal the applied force, and the resistance at the bottom of the pile will be equal and opposite to that of the applied force. If other horizontal forces are applied at different points along the ver- tical face of the pile, the total tangential force at the base of the pile will equal the algebraic sum of all the applied forces. A shearing stress and the resisting force constitute a couple, [91.] SHEAEING STRESS. 145 »jr and as a single couple cannot exist alone, so a pair of shearing stresses necessitate another pair for equilibrium. When the direct simple stresses on the faces of a rectangular paraUelopipedon are of equal intensity, the shearing stresses will he of equal intensity. Let Fig. 80 represent a paralellopipedon with direct and shearing stresses applied to its several faces. At pre- sent suppose that all the forces ai-e parallel to the plane of one of the faces, as ahfe, and call it a plane of the forces ; then will the planes of action, which, in this case, will be four of the faces of the parallelopipe- don, be perpendicular to a plane of the forces. If the direct stress + F — — F, and + i'" = — F', they will equilibrate each other. The moment of the tangential force T, will be Pt X areafc x ah ; and of T' p't X area ac x hf. The couple T.ah tends to turn the element to the right and T' to the left, hence, for equilibrium, we have Pt x a/reafc x ah = p't x area ao X hf; but area fcxah — a/rea ao x hf = the volume of the ele- ment, hence Pt^p'f (96) The effect of a pair of shean^ing , stresses is to distort the element, changing a rectangular one into a rhomboid, as shown in Fig. 81. Direct stresses are directly opposed to each other in the same plane or on opposite surfaces ; shearing stresses sm. 8i. act on parallel planes not coincident. 10 146 KOTATION. [93, 93.1 92. Notation. A very good notation was devised by Oo- riolis, wliicli has been used since 1837, and is now commonly employed for the general investigations on this subject. It is as follows : — Let J? be a typical letter to denote the intensity of a stress of some kind ; jp^ the intensity of a stress on a plane normal to X ; pxx the intensity of a stress on a plane normal to x and in a directiou parallel to x, and hence indicates the intensity of a direct simple stress ; and p^ the intensity of a stress on a plane normal to x but in the direction of y, and hence indi- cates the intensity of a shearing stress. Or, generally, the first sub-letter indicates a narmal to the plane of action and the second one the direction of action. Hence we have rNTENSITtBS OF THE FOECES parallel to X JPx yx y Pm Pm ■ on a plane normal to x; y\ z. i'ee P^ P^ If direct stresses only are considered, one sub-letter is suffi- cient; as^a,,j?j„ or j»,. 93. Tangential stress eesolved. Let T be the tangential stress on the right sec- tion AB = A, the section being normal to y, then Let GD be .an oblique section, normal to the axis y' ; x' and x being in the plane of the axes y and y' ; then will the angle be- tween y and y' be the obliquity of the plane CD. This we will denote by iyy'\ Let the tangential force be parallel to the axis of x. Kesolving this force, we have Normal component on OD = T sin iyy') ; Tangential component on GD = T cos iyy'). [94.1 RESOLVED STRESSES. Dividing each of these by arm CD have 147 AB -^ cos (yy'), we Normal intensity Tangential -Pv'v' T sin {yy') OPS (yy') area AB = Pyxsm (yy') COS d/y'); Tcoa' ^^''''"^i^aAB =py-<'°^'^yy'y^ (97) and for a tangential stress on a plane normal to x, resolved upon the same oblique plane OJD, we have P'v'y' ^P^ cos iyy') sin {yy'); (98) If the tangential stresses on both planes (one normal to y, and the other normal to x) are alike, and the obliquity of the plane CD less than 90°, the resultant of their tangential components will be the difference of the two components, as given by equations (97) and (98) ; that is, it will he j>y,^ — P'y'x>; but the normal intensity will be the sum of the components as given by the same equations. The reverse will be true in re- gard to the direct stresses. 9i. Let a hody he subjected to a direct simple stress ; it is required to find the stresses on any two planes perpendicula/r to one another and to the plane of the forces ; also the intensity of the stress on a thvrd plane perpendicular to the plane of the forces ; and the normal and tangential components on that pla/ne. Let the forces be parallel to the plane of the paper ; A and OB, planes perpendicular to one another and to the plane of the paper, having any obliquity with the forces. Let the axis of x coincide with OB, and y with A 0. Let AB be a third plane, also perpendicular to the plane of the paper, cutting the other fio. sa. planes at anj' angle. Take y' perpendicular to AB and x' parallel to it and to the plane of tlae paper. 148 RESOLVED STRESSES. \S5.] — > The oblique forces may be resolved normally and tangen- tially to the planes AO and 0£, by means of equations (93) and (94). The problem will then be changed to that shown in Fig. 84, in which one set of stresses is simple and direct, and the other set tangential; and, according to Article 91, the intensity of the shearing stress on the two planes will be the t =-{pyv —Pxx)- sin. (^yy') cos {yy') + pxv{ cos' [yy') - Bm= (yy') }• . J The resultant stress oxv AB will be, according to equation (46), e being 90°, Pr = '^fv'y' + fy'^ \ (100) and the inclination of the resultant stress to the normal, y', will be tan (to') -^^ (101) 95. DiSOTTSSION OF EQUATIONS (99). A. Find the indication of the plane on which, there is no tangential stress. In the 2d of equations (99) make jjj^y = 0, and representing this particular angle by (tt'), we find tan 2fTT'^ - si" ^V') ^'os W) _ ^p..„ ,., f.„. [95.] PRINCIPAL STRESSES. 149 which gives two angles differing from each other by 90°, or, the planes will be perpendicular to one another. Hence, hi every case of a direct simple stress upon a pair of planes perpendicular to one another and to a plane of the stresses, there are tioo planes, also perpendicular to one another and to the plane of the stresses, on which there is no tangential itress ; and this is true whether there is a shearing stress on the original planes or not. These tvn) directions are czWeA principal axes of stress. Principal axes of stress are the normals to two planes per- pendicular to one another on which there is no tangential stress. Principal stresses are such as are parallel to the principal axes of stress. (In some cases there is a third principal stress perpendicular to the plane of the other two.) The formulas for the stresses become most simple by refer- ring them directly to the principal axes. a. Let one of the direct stresses he zero, or pyy = 0. Equation 102 gives tan 2(ty') = ^ (103) h. Let one of the direct stresses le a pull, and the other a push. Then j?j^ becomes negative, and we have tan 2(TT') = -^ (104) c. Let them act in opposite senses amd equal to each other. Then tan2(TT')=-^. (105) JPxx d. Let the/re be no tangential stress on the original plames, orp^ = 0. Then, tan 2(yy') = ; .-. (yy') = or 90° ; and the original planes otq principal pla^nes. 150 RESOLVED STRESSES. [95.] e. Let there he no direet stresses. Then, tan 2(tt') = qo ; or (tt') = 45° or 135° ; (106) that is, if on two planes, perpendicular to one another and to the plane of the stresses, there are no direct stresses, then will the stress on two plants, perpendicular to one another and to the plane of the stresses, whose inclination with the original planes is 45°, ie simple and direct. f. Let the direct stresses he equal to one another and act in the same sense, and let there he no shearing on the original planes. Then tan2(YT')=^; and (yy') is indeterminate; hence, in this case every plane perpendicular to a plane of the stress will be a principal plane. Examples. 1. A rough cube, whose weight is 550 pounds, rests on a horizontal plane. A stress of 150 pounds applied at the upper face pulls vertically upward, and another direct stress of 125 pounds, applied at one of the lateral faces, tends to draw it to the right, while another direct stress of 50 pounds tends to draw it to the left ; required the position of the planes on which there are no tangential stresses. If the cube is of finite size it will be necessary to modify the problem, in order to make it agree with the hypothesis under which the formulas have been established. The force of gravity being distributed throughout the mass, would cause a variable stress, and the surface of no shear would be curved instead of plane. We will therefore assume that the cube is without weight, and the 550 pounds is applied directly to the lower swrface. Then the vertical stress will be 150 pouiids, the remaining 400 pounds being resisted dh-ectly by the plane oti which it rests, and 80 far as the present problem is concerned, only produces friction for resisting the shearing stress. The direct horizontal [95.] EXAMPLES. 151 Stress will be 50 pounds, the remaining 75 pounds producing a shearing on the horizontal plane. The former force tends to turn the cube right-handed by rotating it about the lower right- hand corner, thus producing a reaction or vertical tangential stress of 75 pounds. Let the area of each face of the cube be unity, then we have jp^ = 75 pounds ; i?^ = 50 pounds ; ;pyy = 150 pounds ; and these in (102) give .-. (tt') = - 28° 9' 18", or -f 61° 50' 42". If the body be divided along either o^^ these planes, the forces will tend to lift one part directly from the other without producing sliding Upon the plane of division. 2. A rough body, whose weight is 100 pounds, rests on an inclined plane ; required the normal and tangential components on the plane. (Use Eq. (93).) 3 A block without weight is secured to a horizontal plane and thrust downward by a stress whose intensity is 150 pounds, and pulled towards the right by a stress whose intensity is 150 pounds, and to the left with an intensity of 100 pounds ; re- quired the plane of no shear. 4. A cube rests on a horizontal plane, and one of its vertical faces is forced against a vertical plane by a stress of 200 pounds applied at the opposite face, and on one of the other vertical faces is a direct pulling stress of 75 pounds, which is directly opposed by a stress of 50 pounds on the opposite vertical face ; required the position of the plane of no shear. In this case the weight of the cube would be a third princi- pal stress, but it is eliminated by the conditions of the problem. The shearing stress is 25 pounds; and because the direct stresses are unlike, we use Eq. (104). 5. A rectangular parallelopipedon stands on a horizontal plane, and on the opposite pairs of vertical faces tangential 152 PLANES OP [05.] stresses of equal intensities are applied ; required the position of the plane of no shear. (See Eq. (106).) 6. In the preceding problem find the intensity of the direct stress on the plane of no shear. (Substitute the proper quanti- ties in the 1st of (99).) B. To find the planes of action for maximum andm,vnimum, normal stresses, and the values of the stresses. Equate to zero the first differential coefficient of the 1st of Equations (99), and we have 2p^ cos iyy') sin {yy') + ^p^ sin {jjy') cos ij^y') - '^JP^ sin^ (S^') + %>»!!, cos^ W) = ; .-. tan 2(2/y') = I (107) ,. _ _^^ Psac Pyy which, being the same as (102), shows that on those planes which have no shearing stress, the direct stress will be either a maximum or a minimum. Testing this value by the second differential coefficient, we find that one of the values of (yy') gives a maximum and the other a minimum. Comparing (107) with the second of (99), shows that thefi/rst differential coefficient of the value of the direct stress on any plane equals the shearing stress on that plans. From (107), observing that cos (yy') =; 4/I — &v[^{yy'), we find and these values in the 1st of (99), and the maximum and minimum values designated by p^t, give p^, - \{p^ + j?j,^) ± ^{lip^ -Prnf +1^^]; (109) in which the upper sign gives the maximum, and the lower the minimum stress. These are principal stresses, and we denote them by one sub-letter. [95.] MAXIMUM STEESS. 153 a. Ifpxy = 0, we have iyy') = 0° or 90°, as we should. i. Ifjpyy = 0, we have mam.mum,p^, = \p^ + Vlp'^'rf^; \ minim urn, p^ = \jp^ - /jy^+y^; j ^^^^^ hence, the niaximnm normal stress will be of the same kind as the principal direct stress, j),^; that is, if the latter is a. pull, the former will also be a pull, and the minimum principal stress will be of the opposite kind. c. ff there are no direct stresses p^x will also he zero, and we have (tt') = 45° or 135° ; and maximum p^i =Pxy = —J>y> for minimum / that is, the principal stresses will have the same intensity as the shearing stresses, and act on planes perpendicular to one another, and inclined 45° to the original planes. Examples. 1. Suppose that a rectangular box rests on one end, and that one pair of opposite vertical sides press upon the contents of the box with an intensity of 20 pounds, and the other pair of vertical faces press with an intensity of 40 pounds, and that horizontal tangential stresses, whose intensities are 10 pounds,. are applied to the vertical faces, one pair tending to turn it to the right, and the other to the left ; required the position of the vertical planes of no shearing, and the maximum and minimum values of the direct stresses. 2. For an application of Equations (103) and (110) to the stresses in a beam, see the Author's Resistance of Materials, 2d edition, pp. 236-240. C. To find the position of the planes of maximum and minimum shearing. (Ill) 154 PRINCIPAL STRESSES. L95.J Equate to zero the first differential coefficient of the second of (99) and reduce, denoting the angles sought by {YY'), and we find, - cot 2{YY') = tan 2(yt'); .-. 2ZJ"-2(tt') + 90°; or, YY' = ty' + 45° ; that is, the plames of maximum and minimum shear make angles of 4:5 degrees with the PEmciPAi planes. D. Let the planes he peinoipal sections. Then the stresses will be principal stresses, and p^ = 0. Using a single subscript for the direct stresses, equations (99) become Pv' =T^ cos^ W) +i'j^ -I>y) sin iyy') «os {yy'). a. Let^^ =iV5 t^^n Py'=p^; and^j/^- = 0; that is, when two principal stresses are alike and equal on a pair of planes perpendicular to the plane of the stresses, the normal intensity on every plane perpendicular to the plane of the stresses will he equal to that on the principal planes, and there will he no shearing on any plane. This condition is realized in a perfect fluid, and hence very nearly so in gases and liquids, since they offer only a very slight resistance to a tangential stress. If a vessel of any liquid be intersected by two vertical planes perpendicular to one another, the pressure per square inch will be the same on both, and will be normal to the planes ; hence, according to the above, it will be the sarae upon all planes traversing the same point. This is only another way of stating the fact that fluids press equally in all directions. h. To find the planes on which there will be no normal pres- sure. For thU py/ in (111) will be zero ; [96.] PROBLEM. 155 ■Wy') = \/^ ••• tan [yy) = s^^-f ^TZTY; which, being imaginary, shows that it is impossible when the stresses ai-e alike ; but if they are unlike, we have 11 p^ = -py, then (jjy') - 45°, and the 2d of (111) gives which shows that when the direct stresses are unlike and of eqvMl intensity on planes perpendicular to one another, the shearing stress on a plane cutting both the others at an angle of 45 degrees, will be of tlie same intensity. Let {yy') = 45°, or 135°, then (111) become Pi/x' = ^k{p=c-Py);) in the latter of which the upper sign gives a maximum, and the lower a minimum value. Using the upper sign, we find Px=Py'+Pyv;\ Pv-Pv'-JPv'^- I 96. PROBLEM. Fimd the plane on which the obliquity of the stress is greatest, the intensity of that stress, and the angle of its ohliquity. Let the stresses be principal stresses and of the same kind, and the angle of obliquity of the required plane to the stress ; then sin<^=-t^^ — —; the intensity = i/(pxPy)', and the angle be- px '^ Py tween the principal plane x and the required plane = 45°— \j>. 156 CONJUGATE STRESSES. [97, 98.] If the principal stresses are unlike, then sin ^' =:^"L^; the intensity = V—p^^y, and the angle be- tween the principal plane x, and the oblique plane = 45° — i<^'. Example. If a body of sand is retained by a vertical wall and the intensity of the horizontal push is 25 pounds, and of the ver- tical pressure is 75 pounds ; required the plane on which the resultant has the greatest obliquity, and the intensity of the stress on that plane. CONJUGATE STEKSSES. 97. A pair of stresses, each acting parallel to the plane of action of the other, and whose action-lines are parallel to a plane which is perpendicular to the line of intersection of the planes of action, are called conjugate stresses. Thus, in Fig. 85, one set of stresses acts on the plane YY, parallel to the plane XX, and the other set on XX, parallel to YY. In a rigid body the intensities of these sets of stresses are independent of each other; for each set equilibrates itself. Principal stresses are also conjugate. There may be three conjugate stresses in a body, and only three. For, in Fig. 85, there may be a third stress on the plane of the paper, which may be parallel to the line of intersection of the planes XX and YY, and each stress will be parallel to the plane of the other two. A fourth stress cannot be introduced which will be conjugate to the other three. Conjugate stresses may be resolved into normal and tangen- tial components on tlieir planes of action, and treated according to the preceding articles. The fact that the stresses have the same obliquity, being the complement of the angle made by the planes, simplifies some of the more general problems of stresses. GENERAL PROBLEM. 157 GENERAL PROBLEM. 98. Oimn the stresses on the three rectangular coordinate planes ; required the stresses on any oUigioe plane in any re- quired direction. As before, the element is supposed to be indefinitely small. Let abc be the oblique plaue, the normal to which designate by n. The projection of a unit of area of this plane on each of the coordinate plains, gives respectively cos (nic), cos {ny), cos [m). The direct stress parallel to a? l^ acting on the area cos {nx) will y/ give a stress of p^x cos {nx), and fig. 86. the tangential stress normal to y and parallel to x will produce a stress jjj^ cos (ny), and siiiiilai-ly the tangential stress normal to z and parallel to x gnGBp^c cos {ns) ; hence the total stress on the unit normal to n and parallel to x will be Pnx = Pxx COS {nx) + py^ cos {ny) + p^^ cos {m) ; - (114) similarly. Pnv = Pxv cos (nx) +Piiy COS (ny) + p^y cos (nz) ; j Pni = Pxi COS {m) + pyi iiO%{ny) + p>zz cos {m). J Let these be resolved in any arbitrary direction parallel to s. To do this multiply the first of the preceding equations by cos (sx), the second by cos {sy), and the third by cos {sz), and add the results. For the purpose of abridging the formulas, let cos (nx) be written Gnx, and similarly for the others. Then we have p^ =1 p^QrvxQsx + pyyGnyGsy + p^^GnzGsx \ +Pi/z(CnyCs3 + GnsGsy) +p2x{Gji2Gsx v (115) + GmcGsz) + pxy (GnxGsy + GnyGsx). ) This expression being typical, we substitute x' for n and s, and thus obtain an expression for the intensity on a surface normal to x' and parallel to x'. Or generally, substitute suc- cessively x', y', z' for n and s, and we obtain the following formulas : 158 GENERAL FORMULAS. [98.] DIRECT STEESSBS. + ^:,J^zx'Cxx'+ '2^^Caix'Oyx' ; + 2j?«C32/' Qixy'+ ^p^Cxy'Cyy' ; p^^ =pJJxz' +Pm(yyz' +pjyez' + '^ySyyz'Gzz' + 2^„C33'0a3s' + 'ip^Gx^Oyz' ; TAKGENTIAL STRESSES. IPy'z' = ^K«Ca;y'Caj3' + pyy Oyy'Qyz' + i^^ Csy'Gzz' + Czz'Qxy') + p^{Qiiiyy'Cyz' + Qxz'Cyy') ; p^^ = ^aa-Ctca'Ciea?' + pyyCyz'Gyx' + j>JGzz'Gzxf ■^PyJ{Qx2'Qzx,' + Cyx'Czy') + p^JGzz'Qxx' + Csas'Cte^') ' +P==u {Gxz'Cyz' + Ca^'Cya') ; p^y, = p^Cxx'Cxy' + pyyCyx'Cyy' + pj:izx'Qzy' +PylGyx'Cizy' + Oyy'Gzx') +p,J,CzaiGxy' + Czy'Gxx') + Jpxv^xx'Gyy' + Cxy'Cyx'). It may be shown that for every state of stress vn a hody there are three planes perpendicular to each other, on vjhich the stress is entirely normal. [These equations are useful in discussing the general Theory of ths Elasticity of Bodies.'] These formulas apply to oMique axes as well as right, only it should be observed when they are oblique that pyig/ is not a stress on a plane normal to y' , parallel to z', but on a plane nor- mal to x' resolved in the proper direction. CHAPTEE VI. VIET0AL VELOCITIES. 99. Def. If the point of application of a force be moved in the most arbitrary manner an indefinitely small amount, the projection of the path thus described on the original action-line of the force is called a h virtual velocity. The product of the force /H ^ into the virtual velocity is called the tivrtual '' ^ m moment. Thus, in Fig. 87, if a be the point of application of the force F, and ah the arbitrary displacement, ac will be the virtual velocity, and F.ac the virtual moment. The path of the displacement must be so short that it may be considered a straight line ; but in some cases its length may be finite. If the projection falls upon the action-line, as in Fig. 87, the virtual velocity will be considered positive, but if on the line prolonged, it will be negative. 100. Peop. If seDeral concurring forces are in equilibrium, the algebraic sum of their virtual rnoments will he zero. Using the notation of Article (47), and in addition thereto let I be the length of the displacement ; and jp, q, and r the angles which it makes with the respective coordinate axes; then will the projections of I on the axes be I cos^, I cos q, I cos r, respectively. Multiplying equations (50), by these respect- ively, we have i^ cos ai I co%p + F2 cos tta I cos^ -I- etc. = ; F. cos ^, I cos q + Fi cos /Sj I cos q f etc. = ; Fi cos 7i I cosr + Fi cos 73 Icxtsr + etc. = 0. 160 VIRTUAL VELOCITIES. flOl.] Adding these together term by term, observing that cos a cos J) + cos ^ coa q + cos 7 cos r- = cos {Fl) ; which is the cosine of the angle between the action-line of F and the line I; and that I cos (Fl) = Sf (read, variation/) is the virtual velocity of F, we have F,df, + FSf, + FSf, + etc. = SFBf = ; (116) which was to be proved. 101. If any numher of forces in a system are in equilib- riitm, the mm of their virtual moments will he zero. Conceive that the point of application of each force is con- nected with all the others by rigid right lines, so that the action of all the forces will be the same as in the actual problem. If any of the lines thus introduced are not subjected to strain, they do not form an essential part of the system and may be cancelled at iirst, or considered as not having been introduced. Let the system receive a displacement of the most arbitary kind. At each ^"'" ^' point of application of a foi'ce or forces, the strains in the rigid lines which meet at that point, combined with the applied force or forces at the same point, are. necessarily in equilibrium, and by separating it from the rest of the system, we have a system of concurrent forces. Hence, for the point B, for in- stance, we have, according to (116), Fhf + F^hf + etc. + BCSBG+BA8BA + BDIBD = ; in -which BCy etc., are used for the tension or compression which may exist in the line. But when the point C is consid- ered, we will have .5 6*85 C with a contrary sign from that in the preceding expression, and hence their sum will be zero. Proceeding in this way, as many equations may be estab- lished as there are points of application of the forces ; and adding the equations together, observing that all the expressions which represent strains on the lines disappear, we finally have 5i^'S/=0. (117) [101.] EXAMPLES. 161 The converse is evidently true, that when the sum of the mr- tual moments is zero the system is in equilihrium. Equations (116) and (117) are no more than the vanishing equations for work. If a system, of forces is in equilibrium it does no work. This principle is easily extended to Dynamics. For, the work which is stored in a moving body equals that done by the impelling force above that which it constantly does in overcoming resistances. Thus, when friction is overcome, the impelling forces accomplish work in overcoming this resist- ance, and all above that is stored in tlie moving mass. Letting a be the resultant of all the impressed forces producing motion, and s the path described by the body, we have ESr- ^m^Ss^O. (118) This is the most general principle of Mechanics, and M. i] Ogon(k- e made it the fundamental principle of his celebrated work on Mihwni^ue Analytiqu^, which consisted chiefly of a discussion of equation (118). Examples. 1. Determine the conditions of equilibrium of the straight lever. Let AB be the lever, having a weight P at one end and W at the other, in equilibrium on the fulcrum as a centre. From F' draw F'd parallel to AA!, and from F drop a per- pendicular Fe upon dF' . Then, from similar triangles, we have F6 = ^- Ec, a FF' will be perpendicular to BE, and Fe perpendicular to dF', hence eFF' = ADE^ ^•,dFe=. 90° a; .-. dFF' = 90°-a + ;3; and FF' = FeBecB = - Ecaec ^8. The triangle EdE' gives FF' sin a AA' = dF' sin (90'' — a + /B)' hence, Eg _ a sin a cos AA! ~ 1 cos {a-^y which, substituted in equation {a), above, gives p _ Tn- « sin a cos /3 I cos (a — y8)' 8. Deduce the formula for the triangle of forces from the principle of Virtual Yelocities. CHAPTER YII. MOMENT OF INE&TIA. (This chapter may be omitted until its principles are needed hereafter (see Ch. X) Although the expression given below, called the Moment of Inertia, comes directly from the solution of certain mechanical problems, yet its prin- ciples may be discussed without involving the idea of force, the same as any other mathematical expression. The term probably originated from the idea that inertia wa£ considered a force, and in most mechanical problems which give rise to the expression the moment of a force is involved. But the expres- sion is not in the form of a simple moment. If we consider a moment as the product of a quantity by an arm, it is of the form of a moment of a moment. Thus, dA being the quantity, ydA would be a moment, then considering this as a new quantity, multiplying it by y gives y^dA, which would be a moment of the moment. Since we do not consider inertia as a force, and since all these problems may be reduced to the consideration of geometrical magnitudes, it appears that some other term might be more appropriate. It being, how- ever, universally used, a change is undesirable unless a new and better one be univenally adopted.) DEFINITIONS. 102. The expression,/y'(^J., in which dA represents an ele- ment of a body, and y its ordinate from an axis, occurs fre- quently in the analysis of a certain class of problems, and hence it has been found convenient to give it a special name. It is called the moment of inertia. THE MOMENT OF INERTIA OF A BODY is the sum of the products obtained ly multiplying each element of the hody by the squa/re of its distance from an axis. The axis is any straight line in space from which the ordinate is measured. The quantity dA may represent an element of a line (straight or curved), a surface (plane or curved), a volume, weight, or mass ; and hence the above definition answers for all these quantities. 166 MOMENT OF INERTIA. [103.] The moment of inertia of a plane surface, when the axis lies in it, is called a rectangular moment ; but when the axis is perpendicular to the surface it is called ^jpolar moment. 103. Examples. 1. Find the moment of inertia of a rect- angle in refei-euce to one end as an axis. Let h = the breadth, and d = the depth of the rectangle. Take the origin of coor- dinates at 0. We have dA = dydx ; I y^dydx — ^ I y^dy = JSc?. Fio 95. and /= X- 2. What is the moment of inertia of a rectangle in reference to an axis through the centre and parallel to one end ? "■^ Ans. ^fid^. 3. What is the moment of inertia of a straight line in reference to an axis through one end and perpendicular to it, the section of the line being considered unity \ Ans. \1?. 4. Find the moment of inertia of a circle in reference to an axis through its centre and perpendicular to its surface. We represent the polar Tuioment of inertia by I^. Let r = the radius of the circle ; p = the radius vector ; = the variable angle ; then dp = one side of an elementary rectangle ; pd6 = the other side ; and dA = pdpdd ; , and, according to the definition, we have pHpdd = ^r*. '0 -^0 is>^ [104.] EXAMPLES. 167 5. Wliat is the moment of inertia of a circle in reference to a diameter as an axis? (SeeArticle 105.) Ans. ^r*. 6. What is the moment of inertia of an ellipse in reference to its major axis ; a being its semi- major axis and 5, its semi-minor? ^'o- ^■ Ans. ^ab^. 7. Find the moment of inertia of a triangle in reference to an axis through its vertex and parallel to its base. Let i be the base of the triangle, d its altitude, and x any width parallel to the base at a point whose ordinate is y ; then dA — xdy, and we have 1= I I ^xdy = -^ \ fdy = \hd?. 8. What is the moment of inertia of a triangle in reference to an axis passing through its centre and parallel to the base ? Ans. -^^hd^. 9. What is the moment of inertia of an isosceles triangle in reference to its axis of symmetry ? Ans. -^-gb^d. 10. Find the moment of inertia of a sphere in reference to a diameter as an axis. The equation of the sphere will he a? + y^ + ^ == li^. The moment of inertia of any section perpendicular to the axis of x will be ^Try* ; hence for the sphere we have /•4-r rr r= / ^y'dx = TT / {E"- a?f dx = ^E^. U —r «-'^ ' FOEMULA OF EEDUOTION. 104. The moment of inertia of a hody,in reference to any axis, equals the moment of inertia in reference to a parallel axis passing through the centre of the body plus the product of the area {or volume or mass) hy the square of the distance between the axes. 168 FOEMTJLA OF REDtTCTION. [104.J This proposition for plane areas was proved in Article 80. To prove it generally, let Fig. 98 re- present the projection of a body upon the plane of the paper, B the projection of an axis passing through the centre of the body, A any axis parallel to it, O the projection of any element; AC — r, BG = n, the angle CBB — 6, and V = the Yolume of the body. Then ii = fr^d V will be the moment of inertia of the volume in reference to the axis through the centre; and I =fr^dV, the moment in reference to the axis through A. Let AB ^= D, then AE = I) + ricoee, and r^ = n^ sin^ + (Z> + /"i cos 6f ; .-. f'^dV^fr^dV + 2i)//-i cos BdV ^ B^fdV. But fr^ cos OdV = 0, since it is the statical moment of the body in reference to a plane AD passing through the centre perpendicular to the plane of the paper, and the preceding equation becomes /= /i -f- VI?; (119) which is called iheformida. of reduction. From this, we have k = I- FZ)l (120) Examples. 1. The moment of inertia of a rectangle in reference to one end as an axis being i^i^^ required the moment in reference to a parallel axis through .the centre. Equation (120) gives /i = \ld? - Id {^df = }^hd\ 2. Given the moment of inertia of a triangle in reference to ,an axis through its vertex and parallel to the base, to find the moment relative to a parallel axis through its centre. [105.J EXAMPLES. 169 Example 7 of the preceding Article gives /= JW ; hence equation (120) gives I, = Ihd^ - \ld {\df = ^M» 3. Find the moment of the same triangle in reference to the base as an axis. Equation (119) gives / = ^\M' + \M {Uf = ^d\ 105. To FmD THE EELATION BETWEEN THE MOMENTS OF IN- EETIA IN KEFEKENCE TO DIFFERENT PAIRS OF EEOTANGULAE AXES HAVING THE SAME ORIGIN. Let X and y be rectangular axes, a3i and y^, also rectangular, having the same origin ; a = the angle between x and Xi ; Jx = the moment of inertia relatively to the axis X, similarly for I^,Ix, and Jyj _B — pxydA ; and A =fXiyidA. For the transformation of coordinates we have »! = a? cos a — y sin a ; y^ z=^ X sin a + y cos a ; Also Hence, x^ + y^ = a? + ^. dA = dxdy = dx^dy^. Ix^ ^fy^ dA — TxKOS^ o- + ly siii^ a —2B cos a sin o ; Jy^ = I^ sin^ a + ly cos^ a + 'hli cos a sin a ; By = {Ix — ly) cos a sin a + ^ (cos^ a — sin' a) ; • '• J-Xi T J-yi '^^ J-x "V iy ^ -'iJ ) (121) the last value of which is found from the expression J'y^dA + fx^dA — f{y^ + xl) dA —fpHA — Ij,; which shows that the 170 MOMENT OF INERTIA. [105.] polar moment equals the sum of two rectangular moments, the origin being the same. If the rectangular moments equal one another, we have Ip = 2Ix ■ Thus, in the circle, Jp — iirr*. (See Ex. 4, Article 103), hence I^ = ^r*: The last of equations (121) is an isotropic function; since the sum of the moments relatively to a pair of rectangular axes,equals the sum of the moments relatively to any other pair of rectangular axes having the same origin ; or, in other words, the sum of the moments of inertia relatively to a pair of rect- angular axes, is constant. To find the maximum or minimum moments we have, from the preceding equations, da and = — {Ix — ly) cos a sva. a — B (cos^ a — svo? a) = 0; -j^ — + {Ix— ly) C0& a &m a + B (cos^ a — sin^ a) = ; .-. ^1=0. ' From the first or second of these we have 2B 2 cos a sin a Ix — ly cos^ a — siu^ a = tan 2a. It may be shown by the ordinary tests that when I^^ is a maximum, ly^ will be a minimum, and the reverse ; hence there is always a pair of rectangular axes in reference to one of which the moment of inertia is greater than for any other axis, and for the other it is less. These are called principal axes. Thus, in the case of a rectangle, if tlie axes are parallel to the sides and pass through the centre, we find r r\id B= xydA = 0; i " ., ■^ -^ -id hence x and y are the axes for maximum and minimum moments ; and if ^ > 5, -^bd^ is the maximum, and ^^b^d a minimum moment of inertia for all axes passing through the origin. In a similar way we find that if the origin be at any [105.] EXAMPLES. in other point the axes innst he parallel to the sides for maximum and ininimmn moments. The preceding analysis gives the position of the axes for maximum and minimum moments, when the moments are known in reference to any pair of rectangular axes. But if the axes for maximum and minimum moments are known as I^ and ly, then ^ = 0; and calling these I^' and ly,, Eqs. (121) become I^^ = I^ cos^ a + ly' sin^ a ; ) Z,, = X/ sin^ a + lyi cos^ a ; '^Vi (121) B^ = {Ix' — lyi) COS a sin a. ) In the case of a square when the axes pass through the centre Ix' — ly' ; .•. I^^ = I^, (cos^ a + sin^ a) — I^/; Ty^ = ly,, and A = 0; hence the moment of inertia of a square is the same in refer- ence to all axes passing through its centre. The same is true for all regular polygons, and hence for the circle. Examples. 1. To find the moment of inertia of a rect- angle in reference to an axis through its cen- ^ ^ tre and inclined at an angle a to one side, we have Ix = -i^hd^ and ly = -^^h^d :. Ix^ = -^hd {d^ cos^ a + ¥■ sin^ a) ; ly^ = ^^hd [d^ %iv? a + ¥ cos^ a). fig. lOO. 2. To find the moment of inertia of an isosceles triangle in reference to an axis through its centre and inclined at an angle a to its axis of symmetry. We have I^ = ^\M^ and ly = ^jb^d, in which h is the base and d the altitude ; .-, I - J^ld {d"" C0B« a + W sill" a) I, = ^\l)d {d-' sin^ a + W cos* a). The moment of inertia of a regular polygon about an axis 172 MOMENT OF INERTIA. [106.] through its 'centre may be found by dividing it into triangles having their vertices at the centre of the polygon, and for bases the sides of the polygon ; then finding the moments of the triangles about an axis through their centre and parallel to the gi\-en axis and reducing them to the given axis by the for- inula of reduction. If R be the radius of the circumscribed circle, r that of the inscribed circle, and A the area of the polygon ; then, for a A B I'egular polygon, we would find that For the circle R ^= r, as before found. For the square, r ^=: ^h, R ^= h V2, and A as before found. I — -i¥ Fis. 102. 106. Examples of the moment of iNEE-riA of solids. (The following results are taken from Moslej's Mechanics and Engineering.') 1. The moment of inertia of a solid cylinder about its axis of symmetry, r being its radius and h its height, is ^TrAr*. 2. If the cylinder is hollow, c the thickness of the solid part and R the mean radius (equal to one-half the sum of the external and internal radii) then 7= 27rAc^ (E" + \(?). ' 3. The moment of inertia of a cylinder in reference to an axis passing through its centre and perpendicular to its axis of symmetry is \irhr^ {j^ + \W). 4. The moment of inertia of a rectangular paral- lelopipedon about an axis passing through its cen- tre and parallel to one of its edges. Let a be the length of the edge parallel to the axis, and b and c the lengths of the other edges, then 1= ^abc (P + 6'*) = yV "/ the volume multiplied hy the Fio. 103. square of the diagonal of the base. [107.] EADIUS OF GYRATION. 173 HA 5. The moment of inertia of an upright triangular prism liaving an isosceles triangle for its base, in reference to a vertical axis passing througli its centre of gravity. Let the base of the triangle be a, its altitude b, and the altitude of the prism be h, then Via. 104. 6. The moment of inertia of a cone in reference to an axis of symmetry is -^^irrh^. {r being the radius of the base and h the altitude.) Fie. 106. Fig. 107. 7. The moment of inertia of a cone in reference to an axis through its centre and perpendicular to its axis of symmetry is -^^■m^h (/^ + Ifi). 8. The moment of inertia of a sphere about one of its diam- eters is -^-girli^. 9. The moment of inertia of a segment of a sphere about a diameter parallel to the plane of section. Let i? be the radius of the sphere, and h the distance of the plane section from the centre, then / = ^tjTT (1 QJ^ + 157^*5 + IQE'ly' - %% Pia. lus. EADIUS OF GTEATION. 107, We may conceive the mass to be concentrated at such a point that the moment of inertia in reference to any axis will be the same as for the distributed mass in reference to the same axis. The radius of gyration is the distance from the moment axis to a point in wliich, if the entire mass be concentrated, the moment of inertia will be the same as for the distributed mass. 174 EXAMPLES. [107.] The princvpal radius of gyration is the radius of gyration in reference to a moment axis through the centre of the mass. Let k — the radius of gyration ; ki = the principal radius of gyration ; M = the mass of the body ; and D = the distance between parallel axes ; then, according to the definitions and equation (119), we have Mk? = Smt^ (122) = ^mr,^ + MD^ = Mh^ + Miy; .-. ¥=k^ + IP; (123) from which it appears that ^ is a minimum, for 2? = 0, in which case k = k^; that is, the priruyipal radius of gyration is the minimum radius for parallel axes. We have I^= Mk^; . Z.2 _ A . hence, the square of the principal radius of gyration equals the moment of inertia .in reference to a moment axis through the centre of the body divided by the mass. Examples. 1. Find the principal radius of gyration of a circle in refer- ence to a rectangular axis. Example 5 of Article 103 gives, I^ = ^i"^, which is the moment of an area, hence, we use wt^ for M, and have h^=^-^. 2. For a circle in reference to a polar axis, k^ = ^. 3. For a straight line in reference to a moment axis perpen- dicular to it, k^ — ylj?. 4. For a sphere, k^ — f r*- 5. For a rectangle whose sides are respectively a and h, in reference to an axis perpendicular to its plane, k^=^^ {a^+V). 6. Find the principal radius of gyration of a cone when the moment axis is the axis of symmetry. CHAPTEE VIII. MOTION OF A PARTICLE FEEE TO MOVE IN ANT DIRECTION. 108. A free, material particle, acted upon by a system of! forces which are not in equilibrium among themselves, will describe a path which will be a straight or a curved line. The direction of motion at any point of the path will coincide with that of the acticn-iine of the resultant of all the forces which have been impressed upon the particle prior to reaching the point, which will also coincide with the tangent to the path at that point. Let ds be an element of the path described by the particle in an element of time dt ; R the resultant of the impressed forces, and m the mass of the particle; then, according to Article 21, we have Let a be the angle between the action-line of the resultant B (or of the arc ds) and the axis of x ; multiplying by cos a, we have i?co8a »^ ^ cos a = ; in which R cos a is the x-cooiiponent of the resultant, and according to equation (51) equals X; or, in other words, it is the projection of the line representing the resultant on the axis of a; ; dJ^s cos a is the projection of dh on the axis of x, and is d'x. Hence, the equation becomes. m d^x = 0; and similarly, ^ df~ ' (124) 176 MOTION OF A [109.] which are the equations for the motion of a particle along the coordinate axes ; and are also the equations for the motion of a body of finite size when the action-line of the resultant passes through the centre of the mass. They are also the equations of translation of the centre of any free mass when the forces produce both rotation and translation ; in which case m should be changed to M to represent the total mass. See Article 38. VELOOITT AND LmNG FOECE. 109. Multiplying the first of equations (124) by dx, the second by dy, and the third by ds, adding and reducing, give Xdx + Ydy + Zdz — ^mdl— ^ ^ j = ^d-rr^; and integrating gives / {Xdx + Ydy + Zdz) = Jto ^^ = i"^'"^ + ^'• The first member is the work done by the impressed forces ; for if JR be the resultant, and s the path, then, according to Article 25, equation (26), the work will be fUds, and by pro- jecting this on the coordinate axes and taking their sum, we have the above expression. The second member is the stored energy plus a constant. Let X, Y, Z be known functions of x, y, z, and that the terms are integrable. (It may be shown that they are always integrable when the forces act towards or from fixed cenires.) Performing the integration between the limits «„, y^, s„, and Xi, y-i, Si, we have "^ {^0^ 2/o> ^o) - K, 2/1, si) = im {v„ - V,) ; (125) hence, the work done by the impressed forces upon a body in passing from one point to another equals one-half the difference of the living forces at those points. It also appears that the velocity at two points will be independent of the path desci-ibed ; also, that, when the body arrives at the initial point, -it will have the same velocity and the same energy that it previously had at that point. [109.] FREE PARTICLE. 177 Examples. l.Ifa lody is projected into space, and acted upon only ly gravity and the impulse / required the curve described by the projectile. Take the coordinate plane xy in the plane of the forces, x horizontal and y vertical, the origin being at the point from which the body is projected. Let W = the weight of the body ; F= the velocity of projection ; and a = BAX = the angle of elevation at which the projection is made, We have, X=0; Y^-mg; Z = ; and equations (124) become £ EX Fia. lO'J. 3=0; cpy ~di^ = 0. Integrating, observing that v cos a will be the initial velocity along the axis of x, and v sin a that along y, we have, dx dt = V cos a ; -J —v^vo.a — gt', Cut and integrating again, observing that the initial spaces are zero, we have, x = vte,\\i a; ] y = vtco& a — igt\ J Eliminating t from these equations, gives y = xta.na — r—f- — 5— ; (J) ^ 2^^cos^.-' ^ ' which is the equation of the common parabola, whose axis is parallel to the axis of y. 12 178 PROJECTILES. ri09.] Let h be the height through which a body must fall to acquire a velocity «, then i? — 2gh, and the equation (5) becomes, a? y = X tan a — ti ■ (c) ^ 4A cos a ^ ^ To find the range AE, make y = in equation (5), and we find x = 0, and X ^ AE=^ 4A cos a sin a = 2h sin 2a ; (cl) which is a maximum for a — 45°. The range will be the same for two angles of eleyation, one of which is the complement of the other. T/ie greatest height, will correspond to a; = A sin 2a, which, substituted in (b), gives, h sin^ a. (e) The velocity at the end of the time t IS, ¥= ■or, by eliminating t by means of the first of equations (a), we V=v^-2gxt^na+^^^. ig) The direction, of motion at any point is found by differentiating equation (c), and making dy X tan = -ji^ = tan a — j-j . (h) dx 2Acosa ^ ' At the highest point Q — 0, :.x — h sin^ a, as before found. For x — 2h sin 2a, we have, tan 5 =: — tan a, or the angle at the end of the range is the supplement of the angle of projection. 2. A body is projected at an angle of elevation of 45°, and has a range of 1,000 feet ; required the velocity of projection, the time of flight, and the parameter of the parabola. [109.] EXAMPLES. 179 3. Wlaat must be the angle of elevation in order that the liorizontal range may equal the greatest altitude ? What, that it may equal n times the greatest altitude ? 4. Find the velocity and the angle of elevation of a projectile, so that it may pass throngh the points whose coordinates are x^ = 400 feet, y^ — 50 feet, aij = 600 feet, and y^ = 40 feet. 5. If the velocity is 500 feet per second, and the angle of elevation 45 degrees ; required the range, the greatest elevation, the velocity at the liighest point, the direction of motion 6,000 feet from the point of projection, and the velocity at that point. 6. If a plane, whose angle of elevation is *, passes through the origin, find the coordinates of the point C, Fig. 109, where the projectile passes it. 7. In the preceding problem, if * is an angle of depression, find the coordinates. 8. Find the equation of tlie path when the body is projected horizontally. 9. If a body is projected in a due south- erly direction at an angle of elevation a, and is subjected to a constant, uniform, horizontal pressure in a due easterly direc- tion ; required the equations of the path, neglecting the resist- ance of the air. "We have X = ; Y= — rrig; Z = F{a. constant). The projection of the path on the plane xy will be a para- bola, on X2 also a parabola. 10. If a body is projected into the air, and the resistance of the air vaiies as the square of the velocity ; required the equa- tion of the curve. (The final integrals for this problem cannot be found. Approximate solu- tions have been made for the purpose of determining certain laws in regard to gunnery. It is desirable for the student to establish the equations and make the first steps in the reduction. ) (The remainder of this chapter may be omitted without detriment to what follows it. It, however, contains an interesting topic in Mechanics, and is of vital importance in Mathematical Astronomy and Physios.) 180 CENTRAL [110,111.1 CENTRAL FOECES. 110. Central forces are such as act directly towards or from a point called a centre. Those which act towards the centre are called attractive, and are considered negative, while those which act from the centre are repulsive, and are considered positive. The centre may be fixed or movable. The line from the centre to the particle is called a radius vector. The path of a body under the action of central forces is called an orbit. The forces considered in Astronomy and many of those in Physics, are central forces. GENERAL EQUATIONS. 111. Consider the force as attractive, and let it be represented by - F. Take the coordinate plane an/ in the plane of the orbit, the origin being at the centre of the force, and OP = r, the radius vector, then X=-i^cosa = -F-\ r and the first two of equations (124) become dr r (126) To change these to polar coordinates, first modify them by multiplying the first by y and the second by as, and subtracting, and we have my [113.] rORCES. 181 and multiplying the first by x and the second by y, and adding, we have Let Q — POM— the variable angle, then a; = ?• cos ^, y = r sin 6, and differentiating each twice, we find d^x = (««/• - rd^) cos e — (2c?/-<^0 + rd''&) sin 6* ; dry = (<^=/- - r<^^2) sin 6 + (2c?/-(3^0 + rd'O) cos <9 ; which substituted in the preceding eiinations, give r Ide-^ F dj'r ldd\^ di - drdQ d?6 ^ 2 \- r = • dt ^ di' ' which may be put under the form 1 d/ ,de\ ^ rdlf -dt) = '■ (128) Equation (127) shows that the acceleration along r is the force on a unit of mass ; and (128) sliows that there is no acceleration perpendicular to the radius vector. 112. jPrvncipIe of equal areas. — Integrate equation (128), and we have ^f=C; (129) and integrating a second time, we have fi»dd = Ct ; (130) the constant of integration being zero, since the initial values of t and Q are both zero. But from Calculusyr't?^ is twice the sectoral area POX\ hencie the sectoral area swept over by the radius vector increases directly as the time ; and equal ar*as will he passed over in equal tiw.es. 182 CENTEAIi [113.] Making t = 1,-we find that C will be twice the sectoral area passed over in a unit of time. The convei-se is also true, that if the areas are proportional to the tiTTies the force will he central. For, multiplying the first of (126) by x and the second by y and taking their difference, we have or, Xx- Yy=.0; which is the equation of a sti-aight line, and is the equation of the action-line of the resultant, and since it has no absolute term it passes through the origin. 113. To find the equation of the orbit, eliminate dt from equations (127) and (128). For the sake of simplifying the final equation, make /• = — , and (129) becomes Differentiating and reducing, gives , du y^dudt ■ur ad o^' d^_ _ pdu dt~ dd' the first member of which is the velocity in the direction of the radius. Differentiating again, gives Substituting these in equation (127) and at the same time making m equal to unity, since the unit is arbitrary, we have [114.] FOECES. 183 or, — the perpendicular from the centre on the tangent, then from Calculus we have ^ ~ ds" ~ di^ + r'd^ = -2/f* which integrated gives the time. To find it in terms of the angle, V7e have from (129) \-S^de; from which r must be eliminated by means of the equation of the orbit, and the integration performed in reference to Q. 117. To find the com/ponents of the force along the tangent and radius vector. Let T = the tangential component ; iV = the normal component ; and resolving them parallel to x and y, we have ^ ,dx _ j^dy ' df "^ '^ ds ds' m^ = X=r^-i\^^, m^=Y:^Tf-^Jvf-. dor ds ds Eliminating iV gives d^x , d^v , m or d? '^^^ ^"' df ^^ ^ ^*^*' ^^0! dy _]go_^^ 186 CENTRAL FORCES. [117.] Eliminating T gives „ dx d?y dy d^x _ mdi /daxPy — dyd^x\ ~ dMs \ H / _ P m— , (141) hence the component of the force in the direction of the nor- mal is dependent entirely upon the velocity and radius of cur- vature. This is called the centrifugal force. It is the measure of the force which deflects the body from the tangent. The force directed towards the centre is called centripetal. (It is attractive). When the normal coincides with the radins vector the centrifugal force is directly opposed to the centripetal. If &> = the angnlar velocity of the radius of curvature, then, V = pa, and equation (141) becomes N = rnM^p. (142) Examples. 1. If a body is attached to a string and compelled to revolve in a horizontal circle ; required the number of revolutions per minute, that the tension of the string may be twice the weight of the body. (Use equation (142).) 2. A body whose weight is ten pounds, revolves in a hori- zontal cii'cle whose radius is five feet, with a velocity of forty feet per second ; required the tension of the string which holds it. (Use equation (141).) 3. Required the velocity and periodic time of a body re- volving in a circle at a distance of n radii from the earth's centre. The weight of the body on the surface being mg, at the dis- (r \^ 7no — I = —J-, and this is a [117-] EXAMPLES. 187 measure of the force at that distance. (Use equation (141) or (139).) An.., = ief, .= (*) (This is substantially the problem -wMoh Sir Isaac Newton used to prove the law of Universal Gravitation. See Whewell's Inductive Saienees.) 4. A particle in projected from a given point in a given direction with a given velocity, and moves under the action of a force which varies inversely as the square of the distance from the centre ; required the orbit. Let ft = the force at a unit's distance, then and equation (132) becomes d^u (I or, the fii-st integral of which becomes by reduction - ^ (^ - 772-) de = —rJ ^=^,; in which A is an arbitrary constant, and the negative vahie of the radical is used. Integrating again, making d^ the arbitrary constant, we have , J « - 77^ t . 0- ^„ = cos-'|— j^ j ' which by reduction gives 1 a / AO^ \ = ^ = "^ (i + IT "^' (^ - ^4 (^) u 188 ' ORBITS OP [117.] which is the general polar equation of a conic section, the origin being at the focns. As this is the law of Universal Gravitation, it follows that the orbits of the planets and comets are conic sections having the centre of the sun for the focus. In equation {a), 6^ is the angle between the major axis and a line drawn through the centre of the force, and is the eccentricity = e ; hence the equation may be written .= -^,(i + ecos{e-e,)y (5) The magnitude and position of the orbit will be determined from the constants which enter the equation, and these are determined hy knowing the position, velocity, and direction of motion at some point in the orbit. Draw a figure to represent the orbit, and make a tangent to the curve at a point which we will consider the initial point. Let /8 be the direction of the projection with the radius vector at the initial point, r^ the initial radius vector, and V^ the initial velocity ; then at the initial point u — —, d = 0, cot 8 = —j^ = -j^, (c) To ' ' Todd vdd ^ ' and from equation (5) e cos ^0 = Ij {ci\ du fie . . which, combined with equation (c), gives H-ro - = — e sm 0^. ie) From equation (137) = ForosinyS; (/) [117.] THE PLANETS. 189 which, substituted in equations {d) and {e), and the latter divided bj- the former, gives FoVo sin /S cos /8 tan dn = juati( tiou if), give /* — Vo\ sin^ /3' Squaring equations (of) and (e), adding and reducing by equa- ^ J _ JWsin»y8/2 fe - T')- ^^^ Hence, when >. Fq' > — , e > 1, and the orbit is a hyperbola, '0 2/i e = 1, and the orbit is a parabola, 2/t Fo^ < — , e < 1, and the orbit is an ellipse. ^0 or (see example 25, page 30) the orbit will be a hyperbola, a parabola, or an ellipse, according as the velocity of projection is greater than, equal to, or less than the velocity from infinity. As the result of a large number of observations upon the planets, especially upon Mars, Kepler deduced the following laws: 1. The planets desci-ibe ellipses of which the Sun occupies a focns. 2. The radius vector of each planet passes over equal areas in equal times. 3. The squares of the periodic times of any two planets are as the cnl)e3 of the major axes of their orbits. The first of these is proved hf the preceding problem, since the orbits are reenterent curves. The second is proved by equation (130). The third we will now prove. 5. Required the relation l)etween the time of a complete circuit of a pa/rticle in cm ellipse, and the major axis of the orhit. Let the initial point be at the extremity of the major axis near the pole, then ^j in equation {d) will be zero, and we have C^ = fir, (1 + «) ; 190 CENTRAL FOECES. [117.] but from the ellipse, r^ = a — ae := a {1 — e); .-. 0= VfM (1 - e»). (») Equation (130) gives _, _ 2 area ofellvpse 1 - jj Stt a? ^/ (1 - ^ ~ V fj,a (1 - e^ .'. T'oc aK 6. The orbit being an ellipse, required the law of the force. The polar equation of the ellipse, the pole being at the focua, is _1 _ 1 + g cos (g - dp) '^~r~ a{l- e") ' which, differentiated twice, gives d^u _ e sin (0 — 0^ W " a (i- ^) ' and these, in equation (132), give, Tr_ G' 1 ■ a (1 - e^) r» ' hence the force varies inversely as the square of the distance. 7. Find the law of force by which the particle may describe a circle, the centre of the force being in the circumference of the circle. (Tait and Steele, Dynamics of a Particle.) Ans. Fx —. 8. If the force varies directly as the distance, and is attrac- tive, determine the orbit. (This is the law of molecular action, and analysis show that the orbit is an ellipse. The problem is of great importance in Physics, especially in Optica and Acoustics.) CHAPTER IX. CONSTRAINED MOTION OF A PAETIOLE. 118. If a body is compelled to move along a given fixed curve or surface, it is said to be constrained. The given curve or surface will be subjected to a certain pressure which will be normal to it. If instead of the curve or surface, a force be substituted for the pressure which will be continually normal to the surface, and whose intensity will be exactly equal and opposite to the pressure on the curve, the particle will describe the same path as that of the curve, and the problem may be treated as if the particle were free to move under the action of this system of forces. Let iV= the normal pressure on the surface, and L ~f{x, y, z) — 0, be the equation of the surface ; ^ic, ^v, ffz the angles between JV&nd the respective coor- dinate axes. Then X+ J^cosdx—'m,^^ = 0; d^ Y+ Ncosdy -^^ = df Z + iVcos ^z — w^ :j3 = ; d?z 'dt' (143) in which the third terms are the measures of the resultants of the axial components of the applied forces. We will confine the further discussion to forces in a plane. Take xy in the plane of the forces, then we have m '^ = X-N^ de d^y "^dJ^ ds ' 7 + J^p. ds {lU) 192 CONSTEAINED MOTION OF [119,] EliminatiDg JV, we find m. ,.„ . ^^, or 1^(S)'-rdy; and integrating, making % the initial velocity along the path, and V the velocity at any other point, we have ^ (^ _ ^,^2) = 2f{Xdx + Ydy) ; (145) hence, the living force gained or lost in passing from one point to another equals twice the worh done by the impressed forces ; also, the velocity is independent of the path described, and is dependent upon the coordinates of the points ; also, the velocity is independent of the normal pressure. This is an example of the conservation of energy. If friction is one of the forces, a portion of the work will be used in overcoming it. 119. To find the normal pressure multiply the first of equations (IM) by dy, the second by dx, subtract, observing that di? + dy'' = d^, and we find dx d?y dy d^x ) _ 'fh~df~ds'df\~ ,.dx^ _dyd^x /_ ^dx _ -^dy „, or in which p is the radius of curvature at the point. The first and second terms of the second member are the normal com- ponents of the impressed forces. The total normal pressure [120, 121.] A PARTICLE. 193 will, therefore, he that due to the impressed forces jplus that due to the force necessary to deflect the hody from the tangent. The last term is called the centrifugal force, as stated in Article 117. If the body moves on the convex side of the curve, the last term should be subtracted from the others ; hence it might be written ±m—\ in vrhich + belongs to movement on the P ° concave arc, and — on the convex. 120. To find the time of movement, from equation (145), we have /* mds V2f{Xdx + Ydy) + vf (1*'^) 121. To find where the particle will leave the constrain- ing curve. At that point i\7"= 0, which gives v dy -^dx Mio\ ^j = ^is-^d^-^ ('*'> which, combined with the equation of the curve, makes known the point. If a body is subjected only to the force of gravity, we have X = in all the preceding equations. Examples'. 1. A lody moves down a smooth inclined plane under the force of gravity ; required the formulas for the motion. Take the origin at the upper end and let the equation of the plane be y = ax; 13 194 ' EXAMPLES. flSl.] y being positive downward. Then we have X= 0, Y = mg, dy = adx, v^ = 0, and equation (145) becomes ■y* = 2/gadx = ^^cub = 2gy ; (a) hence, the velocity is the same as if it fell vertically through the same height. To -find the time, equation (147) gives that is, if the altitude of th£ plane iy) is constant the timie varies directly as the length, s. We may also find s — t Vigy — igf sin a. (a) 2. Prove that the times of descent down all chords of a ver- tical circle which pass through either extremity of a vertical (diameter are the same. '3. Find the straight line from a given point to a given in- (clined plane, down which a body will descend in the least time. 4, The time of descent down an inclined plane is twice that down its height ; required the inclination of the plane to the horizon. 5.. At the instant a body begins to descend an inclined plane, another body is projected up it with a velocity equal to the velocity which the first body will have when it reaches the foot of the plane ; required the point where they will meet. 6. Two bodies move down two inclined planes from the same point in a curve to two points in the same horizontal in the same time, the lines all being in a vertical plane ; required the curve. T. A given weight, P, draws another weight, W, up an in- clined plane, by means of a cord parallel to the plane ; through what distance must P act so that the weight, W, will move a feet after P is separated from it. [121.] SIMPLE PENDTTLTIM. 195 8. Hequired a curve such that if it revolve with a uniform angular velocity about a vertical dlarneter, and a smooth ring of infiniteshnal diameter ie jcilaced upon it at any point, it will not slide on the curve. Let 6) be the angular velocity, then we have Y ^ -mg, mafx. V = 0, and equation (145) becomes wV - 2gy + 0=0, which is the equation of the common parabola. If the origin be taken at B, O will be zero. (AsoTHER Solution. — Let NR be a normal to the curve, MB = the oen- trifngal force, NM = the force of gravity ; but the latter is constant, hence NM, the subtangent, is constant, which is a property of the common parabola.) 9. Eind the normal pressure against the curve in the pre- ceding problem. 10. The Pendulum. — Find the time of oscillation of the simple pendulumj. This is equivalent to finding the time of descent of a particle down a smooth arc of a vertical circle. Take the origin of coordinates at A, the lowest point. Let the particle start at D, at a height AC = h\ when it has arrived at P, it will have fallen through a height GB = h — y, and, according to equation {a) on the preceding page, will have a velocity V = V2g{h-y) = £ The equation of the arc is a^ = 2ry — j/*; (a) hence 2ry- f dy". 196 EXLAaiPLES OF [131.] But d^ = dos^ + d^, rdy ds = V2ry — y^ Considering this as negative, since for the descent the arc is a decreasing function of the time, we have from (a) dy V¥gJ^V{h-y){^ry-f) r_ r ^^9 J. This may be put in a form for integration by Elliptic Func- tions ; but by developing it into a series, each term may be easily integrated. In this way we find by means of which the time may be approximated to, with any degree of accuracy. When the arc is very small, all the terms containing ^ will be small, and by neglecting them, we have for a complete oscillation (letting I be the length of the pendulum), T^^t^irJl- (J) ^ 9 that is, for very small arcs the osoillations may he regarded as isochronal, or performed in the same time. For the same place the times of vibration are directly as the square roots of the lengths of the pendulums. For any pendulum the times of vibration vary inversely as the square roots of the force of gravity at different places. If t is constant I'x. g. 11. What is the length of a pendulum which will vibrate three times in a second ? 12. Prove that the lengths of pendulums vibrating during the nSl.] CONSTRAINED MOTION 197 same time at the same place, are inversely as the square of the iinmber of vibrations. 13. Find the time of descent of a particle down the arc of a cj'cloid. The diiterential equation of the curve referred to the vertex as an orighi, x being horizontal and y vertical {r being the radius of the generating circle), is dx = , ^ - dy. V'2,ry - f Ans. IT \ -. ^ 9 The time will be the same from whatever point of the curve the motion begins, and hence, it is called tautochronal. 14:. In the simple pendulum, find the point where the tension of the string equals the weight of the particle. 15. A partid^e is placed in a smooth tube which revolves horizontally about an axis through one end of it ; required the equation of the ouroe traced by the jparticle. The only force to impel the particle along the tube is the centrifugal force due to rotation. Letting r = the radius vector of the curve ; r^ = the initial radius vector ; (a = the uniform angular velocity ; we have d'^r „ which, integrated, gives r — \ra{e + e ), hence, the relation between the radius vector and the arc de- scribed by the extremity of the initial radius vector, is the same as between the coordinates of a catenary. (See example 28, p. 134.) 15. To find a curve joining two jpoints down which a par- ticle will slide by the force of gravity in the shortest time. The curve is a cycloid. This problem is celebrated in the 198 CENTRIFUGAL FORCE [123,123.] history of Dynamics. The solution properly belongs to the Calculus of Variations, although solutions may be obtained by more elementary mathematics. Such curves are called Bret- oMstochrones, and, in many cases, their equations may be deter- mined for forces which act under different laws. PEOBLEMS PKETAINING TO THE EAETH. 122. To find the value of g. We have, from example 10 of the preceding Article, ^=5- (149) Making T =\ second and I = 39.1390 inches, the length of the pendulum vibrating seconds at the Tower of London, we have for that place, g - 32.1908 feet. The relation between the force of gravity at different places on the surface of the earth is given in Article 19. The determination of I depends upon the compound pendu- lum, and is explained in the next chapter. 123. To find the centrifugal force at the equator. We have, from equation (142), for a unit of mass, f^<^B = ^E; (a) in which R, the equatorial radius, is 20,923,161 feet; T, the time of the revolution of the earth on its axis, is 86,164 seconds, and TT = 3.1415926. These values give /= 0.11126 feet. The force of gravity at the equator has been found to be 32.09022 feet (Article 19) ; hence, if it were not diminished by the centrifugal force, it would be G = 32.09022 + 0.11126 ^ 32.20148 feet, and /_ 0.11126 _ J^ G ~ 32,20148 ~ 289 °®^^"^y' U34.] ON THE EARTH. I99 hence the centrifugal force at the equator is j^ of the un- diminished force of gravity. Example. In what time must the earth revolve that the centrifugal force at the equator may equal the force of gravity ? Ans. yV 0/ its present time. 124. To find the effect of the centrifugal force at different latitv^es on the earth. Let Z = P6>^ = the latitude of the point P ; R — OQ— OP — the radius of the earth; then will the radius of the parallel of lati- tude pp' be -,./::rzxp^. B^ = R cos L. L ^i';^ -^^ The centrifugal force will be in the V J plane of motion and may be represented ^"^kTik by the line Pr, or Pr —f = ci^E^ = ay'E cos L ; therefore, the centrifugal force varies directly as the cosine of the latitude. But the force of gravity is in the direction PO. Resolving Pr parallel and perpendicular to PO, we have Pp = ay'E cos' L = -^^-^ G cos^ L ; Pq =z ca^P cos L sin L = -^\^ O sin 2Z ; the former of which diminishes directly the force of gravity, and the latter tends to move the matter in the parallel of lati- tude PP', toward the equator. Such a movement has taken place, and as a result the earth is an oblate spheroid. In the present form of the earth the action-line of the force of gravity is normal to the surface (or it would be if the earth were homo- geneous), and hence, does not pass through the centre 0, except on the equator and at the poles. The preceding formulas would be true for a rigid homogeneous sphere, but are only approximations in the case of the earth. CHAPTEK X. FORCES IN A PLANE PEODUCING EOTATION. 125, Angular motion of a paetiole aboitt a fixed axis. Let the body C, on the horizontal arm AB, revolve about the IT. vertical axis ED. Consider the body reduced to the centre of the mass, and the force F-^ applied at the centre and acting continually tangent to the path described by the particle. This may be done as shown in Fig. 121. In this case the force will be measured in tlie same way as if the path were rectilinear, for the force is applied Hence, according to equation (21), w ^« along the path. in which s is the arc of the circle. If 6 be the angle swept over by the radius, and /"i the radius ; then ds = Ti d0, dh^r^cPO; which, substituted in the equation above, gives F^^Mn df dj'e Fi ■■■W = Mr- (ISO) If a force, F, be applied to the ai-m AB, at a distance r from [126,127.] ANGULAR ACCELERATION. 201 the axis ED, prodiuiing tlie same movement of the mass (7, we have, from tlie equality of moments, Article 65, and the value of F^ deduced from this equation substituted" in the preceding one, gives d^e _ Fr that is^ the angular acceleration produced by a force, F, on a particle, m, equals the moment of the force divided hy the mo- ment of inertia of the mass. (For moments of inertia, see Chapter VTI.) "We observe that, when the force is applied directly to the particle, it produces no strain upon tlie axis, but that it does when applied to other points of the arm. In both cases there will be a strain of due to centrifugal force, a being the angular velocity. 128. Angular motion of a _,.-' ""-•-,, FINITE MASS. Let a body, AB, turn about a vertical axis at A, under the action of constant forces, F] acting horizontally ; mi, m^, etc., masses of the ele- ments of the body at the re- \.,_ ...-■'' spective distances r^, r.^, etc., fig. lie. from the axis A ; and consid- e)-ing equation (151) as typical, we have ^ % Fr ^ %Fr df ~ m^Ti + m^r^ + m^r^ + etc., ^mr^ _ moment of the forces ,.. -„, moment of ine7'tia 127. Energy of a rotating mass. Multiply both membera 202 FOEMULAS FOE [138, 139.] of the preceding equation by dd, integrate and reduce, and we find \^m7»{^^ =fF.rdd; (153) in which rdd is an element of the space passed over by F, and F.rdd is an element of work done by F; hence the second member represents the total work done by ^upon the body, all of which is stored in it. Therefore, the energy of a iody rotat- ing about o,n axis equals the moment of inertia of the mass multi^plied hy one-half the square of the angular velocity. If tlie body has a motion of translation and of rotation at the same time, the total energy will be the sum due to both motions ; for it is evident that while a body is rotating a force may be applied to move it forward in space, Article 38, and that the work done by this force will be independent of the rotation. liv =z the velocity of translation of the axis about which the body rotates ; Q> = the angular velocity ; and Ijn = the moment of inertia of the m^ass ; then the total work stored in the body will be W>^ + 4/m«' (154) 128. Ail iMPOLSE. Multiply both members of equation (134:) by dt, integrate, and we find ^ - _ '^f^^* dt ~'^ ~ Xm7^- But according tc Article 33, fFdt is the measure of an im- pulse and is represented by Q, hence Or _ moment of the impulse moment of inertia 129. The time required to pass over n circumferences will be, in the case of an. impulse, [130, 131. ANGULAR ROTATION. 203 2mr __ 2nTrZ ~ CO ~ Qr ' In the case of accelerated forces the time will be found by integrating equation (153). 130- MovixG AXIS. The body may have a motion of translation and of rotation at the same time. If the axis is rigid, the body may be made to rotate about any point in it, but if the body is free the axis of rotation will pass through the centre ; see Article 38. In either case, the force which pro- duces the translation may be considered as applied directly to the axis, and hence, the two motions may be considered as existing independently of each other. As we have seen in Article 38, the velocity of translation depends upon the magni- tude of the force or impulse, but the velocity of rotation depends upon the arm of the force. For a general demonstra- tion of this principle, see page 221, 131. GE.NEKA.L FOEMULAS. The preceding formulas have been deduced from the consideration of special problems, but they may be deduced in a more general manner, as follows : — • Let the origin of coordinates be taken on the axis of rotation, the axis of s coinciding with that axis, and x coinciding with the direction of translation. Resolving the forces into couples, and forces applied at the origin, as in Chapter III., and the first members of the first of equation (85) will be the impressed forces whicth produce translation, and the third of (86) those which produce rotation ; hence, according to D'Alerabert's principle we have % (Xy -rx)-S {my ^, - mx ^ ■ (156) The expression S {Xy — Yx) is the sum of the moments of the impressed forces = XFr (Article 60). Transforming the second terra into polar coordinates, we have L ' -^ ^y ^/ d^x d^ii\ ^ .d^e ^[my^-rax-^) = %m,^~^; 204 REDUCED MASS. [132.1 heuce, the equations become df dW df ' ZFr (157) 132. Reduced mass. A given mass may be concentrated at such a point, or in a thin annulus, that the force or impulse will have the same effect upon it as if it were distributed. To accoraplish this it is only necessary that Xmr^ in the second of (157) should have an equivalent value. Let M be the mass of the body, k the distance from the axis to the required point, then in which k is the radius of gyration, as defined in Article 107. Bat any other point may be assumed, and a mass deterrained such that the effect shall be the same. Let a be the distance to the point (or radius of the annulus), and M^ the required mass, then we have a*' (158) which is called the reduced mass. Examples. 1. A prismatic bar, AB, falls through a height, h, retaining its horizontal position until one end strikes a fixed obstacle,G ^ required the angular velocity of the piece and the linear velo- city of tlie centre immediately after the impulse. Let M. — the mass of the bar ; I = its length ; V = ^'^gh, the velocity at the instant of impact ; Vi = the velocity of the centre immediately after. The bar will rotate about a hori- zontal axis through the centre, as shown by Article 38 ; and, as shown by Articles 33 and 38, the Y m c Fio. 117. P33.] EXA.MPLES. 205 impulse will he Q =z M {v - Vi) ; that is, it is the change of velocity at the centre multiplied hj the mass. The impact will entiiely an-est the motion of the end, A, at the instant of the impact, and hence at that instant the angular velocity of A in reference to G will be the same as G in reference to A. Equation (155) gives moment of impulse m,oinent of inertia _ M {v- V,) jl = ^ I ■ But at the instant of the impact solving these give Vi = ilm, We now readily find Q = iMv. To find the velocity of any point in a vertical direction at the instant of the impact, we observe that it may be considered as composed of two parts ; a linear velocity Vi downward, and a right-Iiunded rotation. The actual velocity at A due to rota- tion will be ilco = |?i, which will be upward, and the linear velocity downward will be ""i = 1^) hence the result will be no velocity. Similarly, the velocity at B will he ^v + fv — Iv. Also, for any point dis- tant X from G, we have at the left of G ^ - (OX = |wu - ^ J-); and to the right of G we have When the bar cornea into a vertical position, we easily find 206 ASTGULAE MOVEMENT. [133.] that A has passed below a horizontal through C. Every point, therefore, has a progressive velocity, except the point A, at the instant of impact. After the impact the centre will move in the same vertical and witli an accelerated velocity, while the angular velocity will remain constant. 2. Suppose that impact takes place at one-quarter the length from A, required the angular velocity. 3. At what point must the impulse be made so that the velocity of the extremity Ji will be doubled at the instant of impact ? 4. An inextensible string is wound around a cylinder, and has its free end attached to a ■fixed point. The cylinder falls through a cer- tain height (not exceeding the length of the free pai-t of the string), and at the instant of the im- pact the cord is vertical and tangent to the cylinder ; all the forces being in a plane ; re- Fio. 118. quired the angular velocity produced by the impulse, and the momentum. Ans. I - ; Q = iMv. Ap-m v.- 5. In the preceding problem, let the body be a homogeneous sphere, the string being wound around the arc of a great circle. 6. A homogeneous prismatio har, AB, in a horizontal posi- tion constrained to revolve ahout a vertical fixed axis, A, receives a direct impulse from a sphere whose momentum is Mv ; re- quired the angular velocity of the har. The momentum imparted to the bar will depend upon the elasticities of the two bodies. Consider them perfectly elastic. The effect of the impulse will be the same as if the mass of the bar were concentrated at the extremity of the radius of gyration ; hence an equivalent- mass at the point (7 may be determined. Fia. 119. P33.] EXAMPLES. 207 Let Jfi = the mass of AB ; Mi = the reduced mass ; V2 = the velocity of the reduced mass after impact • a = AC. ' Then, by equation (15S), the mass of the bar reduced to the point C, will be a' By equation (40) the velocity of Jfg after impact will be „ _ 2Jf 2ifa» hence, the momentum imparted will be and the moment will be ,^ 2MMJi?a According to equation (155), we have moment of the impulse moment of inertia 2MMJ^a _M£+M£^^ __ 2Ma ~ Ma' + Mlk^^- This result is the same as that found by dividing equation {a) by a, as it should be. 7. Swppose, in the preceding problem, that there is no fixed axis, but that the body is free to rotate,- find where the impact must he made that the initial velocity at the end A shall be zero. 208 AIJGULAE MOVEMENT. [133.] Let I£v be the impulse vmjpaHecL to the body ; Mh^ = the principal moment of inertia ; A = the distance from the centre of the bar to the required point ; then moment of the vmmilse o) = — ir-. — =-; moment of inert%a {a) _ Mvh _ vh and the movement at A in the circular arc will be „ vlh and the initial linear movement will be which, by the conditions of the problem, will be zero ; hence vlh __ A = ^. ® or, A — L I The distance from A will be The bar being prismatic, h^ = -^P ; .: h + il = f Z. The result is independent of the magnitude of the impulse. From (i) we have hence, x and fZ are convertible, and we infer that if the im- pulse be applied at A the point of no initial motion will be at [183.] AXIS OF SPONTANEOUS ROTATION. 209 the point given by equation (5), where the impact was previ- ously applied. 8. In the 'preceding problem, find where the impulse must he applied so that the point of no initial velocity shall he at a distance h' from, the centre. The initial linear velocity due to the rotary movement found from (a) of the preceding example, will be V h'co = j-^ hh', and the initial movement of the required point being zero, we have If tlie point of impact be at h, the point a, where the initial movement is zero, will be on the other ABC side of the centre of the body. Let B ' i % ' be the centre, then h = hB, h' = aB, and from (159), we have hh' = hB.aB = k^ ; (IGO-) and as ^ is a constant, the points a and h are convertible. AXIS OF SrONTAlfEOUS BOTATIOX. 133. In the preceding problem the initial motion would have been precisely the same if there had been a fixed axis through a perpendicular to the plane of motion, and hence the initial motion may be considered as a rotation about that axis. If a fixed axis were there it evidently would not receive any shock from the impulse. The axis about which a quiescent body tends to turn at the instant that it receives an impulse is called the axis of spon- taneous rotation. 14 210 INSTANTANEOUS AXIS. [134, 135.] OENTEE OF PEECUSSIOJST. 134. When there is a fixed axis and the body is so struck that there is no impulse on the axis, any point in the action^- line of the force is called the centre of percussion. Thus in Fig. 120, if a is the fixed axis, h will be the centre of percus- sion. It is also evident that, if 5 be a fixed object, and it be struck by the body AG, rotating about a, the axis will not receive an impulse. AXIS OF INSTANTANEOUS EOTATION. 135. An axis through the centre of the mass, parallel to the axis of spontaneous rotation, is called the axis of instantaneous rotation. A free body rotates about this axis. In regard to the spontaneous axis, we consider tJuit as fixed in space for the instant ; but at the same time the body really rotates about the instantaneous axis which moves in space with the body. EXAMPLES UNDEE THE PEECEDING EQUATIONS CONTINUED. "8. A horizontal uniform disc is free to revolve about a ver- tical axis through its centre. A man walks around on the outer edge ; required the angular distance passed over hy the rnan and disc when he has vjalked once around the circumfer- en,ce. Let W = the weight of the man ; w = the weight of the disc ; r = the radius of the disc ; wi = the angular velocity of the man in reference to a fixed line ; o) = the angular velocity of the disc in reference to the same fixed line ; ^ = the force exerted by the man against the disc ; The result will be the same whether the effort be exerted suddenly, or with a uniform acceleration, or irregularly. We will, therefore, treat it as if it were an impulse. Tlie weights are hero used instead of the masses, for they are directly pro- [135.] EXAMPLES. 211 portional to each other, and it is more natural to speak of the weight of a man than the mass of a man. We have r (U : Tnoment of the impulse inoment of inertia _ Fr ~ wk^ _ TF-y?* ~ wk^ 2W = Q)i. 10 But when the man arrives at the initial point of the disc, we have w -f- Q)i ^ 27r ; which, combined with the preceding equation, gives 2w7r JiW =: w, we have "i = hr, for the angular space passed over by the man, and for the distance passed over by the disc. 9. In Fig. 115 let the force Fie constant; required the nuniber of complete turns which the body O will maJce about the axis DE in the time t. Let r = the radius of the circle passed over by F\ /"i = the distance of the centre of the body from the axis of revolution ; hy ■=■ the principal radius of gyration of the body in refer- ence to a moment axis parallel to DE\ Tc = the radius of gyration of the body in reference to the axis DE; then, according to equation (123), 212 ROTATION OF [135.] and, according to equation (151), ^ cP6 _ moment of forces df ~ moment of inertia _ Fr Multiply by dt and integrate, and we have de _ Fr 'dt~'^^"* the constant being zero, for the initial quantities are zero. Multiplying again by dt, we find Fr ., e = '■2.MW' which is the angular space passed over in time t ; and the num- ber of complete rotations will be 1. - ^'^^ 10. If the body were a sphere 2 feet in diameter, weighing 100 pounds, the centre of which was 5 feet from the axis ; F, a force of 25 pounds, acting at the end of a lever 8 feet long ; required the number of turns which it will make about the axis in 5 minutes. 11. If the data be the same as in the preceding example ; required the time necessary to make one complete turn about the axis. 12. Suppose that an ideflnitely thin body, whose weight is TF, rests upon the rim of a horizon- tal pulley which is perfectly free to move. A string is wound around the pulley, and passes over another pulley and has a weight, P, attached to its lower end. Supposing that there is no resistance by the pulleys or the string, required the distance passed over by P in time t. 213 [135.1 SOLID BODIES. According to equation (152), we bave — = /'/■ _ Pg df W -\- P ^~ (^W+ F)r' 9 from which it may be solved. (This is equivalent to applying the weight P directly to the weight W, as in Pig. 10, and hence we have, according to equation (21) W+ P d^s = P; g de latea, we ] reduces the equation directly to that in the text.) but referring it to polar coordinates, we have r~:=~, which substituted 13. A disc whose weight is W is free to i-evolve about a horizontal axis pass- ing tlirongh its centre and perpendicular to its plane. A cord is wound around its circumference and lias a weight, P attached to its lower end ; required the distance through which P will descend in t seconds. We have Pm. 123 ^^ _ Prg df ~ Wk^ + Pr" ' from which Q may be found, and the space will be rd. (This may be solved by equation (SI). The mass of the disc reduced to an equivalent at the circumference will be — -p , and that equation wiU become g r^ 1 / h '^\ cPs — IP-I- ^—^ I -315 = P; which, by changing to polar coordinates, may be reduced to the equation in the text.) 14. If in the preceding e.xaraple the body were a sphere revolving about a horizontal axis, the diameter of the sphere being 16 inches, weight 500 pounds, moved by a weight of 100 pounds descending vertically, the cord passing around a groove in the sphere the diameter of which is one foot ; required the number of revolutions of the sphere in five seconds. 2U ROTATION OF [135.] Fig. 123. 15. Two weights, P and TF, are stispended on two pulleys by means of cords, as shown in Fig. 123 ; the pulleys being attached to the same axis C. No resistance being allowed for the pulleys, axle, or cords ; required the circumstances of motion. We have d'^d moment of the forces df moments of inertia P.AC-W.BC ~ P{AOf + W{BC'f+disoAC.k^ 4 disc B CJc^^ ' in which disc A G, etc., are used for the weights of the discs. Let the right-hand member be represented by M, then we have e = iMf. 16. In the preceding example let the discs be of uniform density, AC= 8 inches, ^C= 3 inches; the weight of ^ C = 6 pounds, of OB = 2 pounds, of P = 25 pounds, and of TF = 60 pounds ; if they start from rest, required the space passed over by P in 10 seconds, and the tension of the cords. 17. A homogeneous, hollow cylinder rolls down an inclined plane hy the force of gravity / required the time. The weight of the cylinder may be resolved into two c*mponent8, one parallel to the plane, which impels the body down it, the other nor- mal, which induces friction. The friction acts parallel to the plane and tends to prevent the movement down it, and is assumed to be suificient to prevent sliding. Let W = the weight of the cylinder; i = the inclination of the plane to the horizontal ; i\r = TF cos i = the normal component ; ^-^-^ 1 N ^^ r ^\^ Fie. 124. nSS.] SOLID BODIES. 215 <^ = the coefficient of friction ; T = ^N = the tangential component ; T' — W sin i = the component of the weight parallel to the plane ; r^ = the internal radius of the cylinder ; r = the external radius ; 6 = the angular space passed over by the radius ; s =^ AO, the space. This is a case of translation and rotation combined, and equations (157) give -irg =.T' - T= WBini- T; g dt" d^e _ T.r _ ^gTr and from the problem s = r6. Eliminating .« and T from these equations, we get dW 2gr&m i df "~ Zr^ + r?" Integrating and maMag the initial spaces zero, we have _ gr sin i ^ _ '6r^ + r^ ' ...<=s/: 37^ + T-i^ S. gr^ sin i If Ti = 0, the cylinder will be solid, and V ^ sin *' and hence, the time is independent of the diameter of the cylinder. If Ty — r, the cylinder will be a thin annulus, and t =\J —■ — •; V ^ sm * hence, the time of descent will be V^^ times as long as 216 THE COMPOUND [135.] when the cylinder is solid ; the weight being the same in both cases. If it slide down a smooth plane of the same slope, we have / 2« t— \l : . , V g sm ^ which is less than either of the two preceding times. THE PBNDTJLTIM. 18. Zet a body he suspended on a Jwrizon- tal axis by a string without weight, it will oscillate in the same time as the body Od. Or, it is the point at which, if the entire mass be concentrated, it will oscillate about the axis in the same time as when it is distributed. The point 0, when the axis pierces the plane xy, is called the centre of suspension. 137. Results. The centres of oscillation and of percussion coincide. (See Article 134.) According to equation (163), the centres of oscillation and of suspension are convertible. According to the same equation the principal radius of gyra- tion is a mean proportional between the distances of oscillation and of suspension from the centre of gravity. Equation (161) indicates a practical mode of determining the principal radius of gyration. To find it, let the body oscillate, and thus find T, then attach a pair of spring balances to the lower end and bring the body to a horizontal position, and find how much the scales indicate ; knowing which, the weight of the body and the distance between the point of attachment and the centre of suspension 0, the value of h may easily be computed. The ^alue of g being known, all the quantities in equation (161) become known except Jc^, which is readily found by a solution of the equation. 218 CAPTAIN KATER'S [138.1 Example s . 1. A prismatic bar oscillates about an axis passing through one end, and perpendicular to its length ; required the length of an equivalent simple pendulum. 2. A homogeneous sphere is suspended from a point by means of a fine thread, find the length of a simple pendulum which will oscillate iu the same time. 138. Captain Kater used the principle of the convertibility of the centres of suspension and ' w oscillation for determining the length of a simple Fig. 126. seconds pendnlum, and hence the acceleration due to gravity. — Phil. Trans., 1818. Let a body, furnished with a movable weight, be provided with a point of suspension U (figure not shown), and another point on which it may vibrate, fixed as nearlj- as can be esti- mated in the centre of oscillation O, and in a line with the point of suspension and the centre of gravity. The oscillations of the body must now be observed when suspended from C and also when suspended from 0. If the vibrations in each posi- tion should not be equal in equal times, they may readily be made so by shifting the movable weight. When this is done, the distance between the two points 6' and is the length of the simple equivalent pendulum. Thus the length L and the corresponding time T of vibration will be found uninfiuenced by any irregularity of density or figure. In these experiments, after numerous trials of spheres, etc., knife edges were pre- ferred as a means of support. At the centres of suspension and oscillation there were two triangular apertures to admit the knife edges on which the body rested while making its oscil- lations. Having thus the means of measuring the length Z, with accuracy, it remains to determine the time T. This is efi^ected by comparing the vibrations of the body with those of a clock. The time of a single vibration or of any small arbitrary number of vibrations cannot be observed directly, because this would require the fraction uf a second of time, as shown by the clock, to be estimated either by the eye or ear. The vibrations of the tl3&] EXPERIMENTS. 219 body may be counted, and here there is no fraction to be esti- mated, but these vibrations will not probably jSt in with the oscillations of the clock pendulum, and the differences must be estimated. This defect is overcome by "the method of coinci- dences." Supposing the time of vibration of the clock to be a little less than that of the body, the pendulum of the clock will gain on the body, and at length at a certain vibration the two will for an instant coincide. The two pendulums will now be seen to separate, and after a time will again approach each other, when the same phenomenon will take place. If the two pendulums continue to vibrate with perfect uniformity, the number of oscillations of the pendulum of the clock in this in- terval will be an integer, and the number of oscillations of the body in the same interval will be less by one complete oscilla- tion than that of the pendulum of the clock. Hence by a simple proportion the time of a complete oscillation may be found. The coincidences were determined in the following manner : Certain marks made on the two pendulums were observed by a telescope at the lowest point of their arcs of vibration. The field of view was limited by a diaphragm to a narrow aperture across which the marks were seen to pass. At each succeeding vibration the clock pendulum follows the other more closely, and at last the clock-mark completely covers the other during their passage across the field of view of the telescope. After a few vibrations it appears again preceding the other. The time of disappearance was generally considered as the time of coincidence of the vibrations, though in strictness the mean of the times of disappearance and reappearance ought to have been taken, but the error thus produced is very small. (Encyo. Met. Figure of the Eai'th.) In the experiments made in Hartan coal-pit in 1854, the Astronomer Koyal used Eater's method of observing the pendulum. {Fliil. Trans. 1856.) The value of T thus found will require several corrections. These are called " Reductions." If the centre of oscillation does not describe a cycloid, allowance must be made for the altera- tion of time depending on the arc described. This is called " the reduction to infinitely small arcs." If the point of sup- port be not absolutely fixed, another correction is required 220 A STANDARD OF LENGTH. [138.] {Phil Trans. 1831). The effect of the bncivancy and the resistance of the air must also be allowed for. This is the "redaction to a vacuum." The length Z must also be cor- rected for changes of temperature. The time of an oscillation thus corrected enables us to find the value of gi'avity at the place of observation. A correction is now requii-ed to reduce this result to what it would have been at the level of the sea. The atti-action of the intervening land must be allowed for by Dr. Young's rule {Phil. Trans. 1819). We thus obtain the force of gra\ity at the level of the sea, supposing all the land above this level were cut off and the sea constrained to keep its present level. As the sea would tend in such a case to change its level, further corrections are still necessary if we wish to reduce the result to the surface of that spheroid which most nearly represents the earth. (See Camb. Phil. Trans., vol. x.) There is another use to which the experimental determina- tion of the length of a simple equivalent pendulum may be applied. It has been adopted as a standard of length on account of being invariable and capable at any time of recov- ery. An Act of Parliament, 5 Geo. IV., defines the yard to contain thirty -six such parts, of which parts there are 39.1393 in the length of the pendulum vibrating seconds of mean time in the latitude of London, in vacuo, at the level of the sea, at temperature 62° F. The Commissioners, howevei-, appointed to consider the mode of restoring the standards of weight and measure which were lost by fire in 1834, report that several elements of reduction of pendulum experiments are yet doubt- ful or erroneous, so that the results of a convertible pendulum are not so trustworthy as to serve for supplying a standard for length ; and they recommend a material standard, the distance, namely, between two marks on a certain bar of metal under given circumstaiices, in preference to any standard derived from measuring phenomena in nature. {Report, ISil.) All nations, practically, use this simple mode of determining the length of the standard of measure, that of placing two marks on a bar, and by a legal enactment declaring it to be a certain length. CHAPTEE XI. A BODY ACTED UPON BT FORCES OF DIEFEKENT mTENSITIES HAVING DIFFEEENT POINTS OF APPLICATION AND ACTING IN DIFFERENT DIEECTIONS. (We have reserved certain general demonstrations for this Chapter.) 139- Let X, y, s be the coordinates of any particle of a body in reference to a Jixed origin in space, and X, Y, Z the re- solved accelerating forces acting on the same particle, then by D'Alenibert's principle ^-"^w ^-"^w ^-"^w will be zero if the full effect of the impressed forces is ex- pended upon the particle, as they would be if the particle were free ; but as it forms a part of the body, the differences (if any) in the preceding expressions will, by the same principle, be the amount which the impressed forces exert upon the remainder of the body. Considering every particle in the system, and resolving the forces applied to each in the same manner, we ha.\e a series of expressions of the same form as those above. Adding these and the sura for each axis will be zero, when all the forces which act upon the body are included. Similarly, taking the moments of the forces in reference to the axes, we have expressions of the form Zy-Yz-[my^-mz^), and similarly for the others. Taking the sum of these expressions for all the forces, and they will be zero in reference to each axis. Writing these results, we have the following six equations ; 222 GENERAL EQUATIONS OF [139.] df XY ^-S=« ^Z-5..f = (164) 5(Zy - Yz) - ^•(^yj- m^g) = 0; ^{Xz - Zx) -X [mz ^ - ™^ ^) = ; (165) Let aJi, ^1, Si be the coordinates of any particle of the body referred to a movable system whose origin remains at the centre of the mass, and whose axes are parallel to the fixed axes, and X, y, z the coordinates of the centre of the mass referred to the fixed origin ; then we have a? = ic + cci; Xmx = Xinx + XTnxy ; Xm^,=Xm-^ + Xm-^. Bnt the origin of the movable system being at the centre of the mass, we have, from equations (71a) or (84a), XmXi = ; V™ dx^ .■.Xm^ = Q; and the last of the preceding equations becomes „ d^x ^ d^x df Xm, [139.] MOTION. 223 since df is a common factor to all the particles m; but 5m = Jf; and similarly for the others. Hence, equations (164) become Similarly, in equations (165), we have ^"^y de = ^, ,- ^ /d^3 dJ'sA (166) y-Sn,+ y^ru^,+ ^,^my,+ ^my,^^'-; de ' '•'■"^^df ' ""'"^^ df But y and i are common factors in their respective terms, therefore the expression becomes y. im -I- fit >. -m df but, Smyi = ; ^ d s^ . ,._ ^^i „ d\ hence, we finally have ^ d^s Zmy^,-^.,^^ and similarly for other terms. In this way the first of equa- tions (165) becomes 224 GEKERAL EQUATIONS OF [130.] Multiply the third of equations (166) by y, the second by i, subtract the latter from the former, and we have ^3^ ^Zy-'^Tz; which, substituted in the preceding equation, gives Dropping "Z before X, Y, Z, and letting those letters repre- sent the total axial components upon the entire body, and Zy-i, — Ysi, etc., the resultant moments of the applied forces, we have the six following equations for the motion of a solid body: (167) -s= ^; -s= r; ^s= Z\ '{ d^3i Ai my,^^-mB,^^ 'p)^Zy,-YB,i ( d?x^ Zlmsi mxi dPz{\ l™^"^ - myi-:^] = Ya^- (168) Equations (167) do not contain the coordinates of the point of application of the forces, heuce, the motion of translation of the centre of a mass is independent of the point of applica- tion of the force or forces ,' or, in other words, it is indepen- dent of the rotation of the mass. Equations (168) do not contain the coordinates of the centre of the mass, and being the equations for rotation, show that the rotation of a mass is independent of the translation of the centre. [140, 141.] MOTION. 225 These equations are sufficient for determining all the circum- stances of motion of a solid. In their further use the dashes and subscripts will be omitted. 140. If X, Y, Z are zero, we hare /d'x dx ^ ^ 5? = ^+^^ = «5 and similarly for the others. Transposing, squaring, adding and extracting the square root, give Ida^ + df + ds" , " = V ^ = ^a^ + Ci + C, ; (169) which, being constant, shows that the motion of the centre of the mass is rectilinear and uniform. This is the general principle of the conservation of the CENTRE OF GRAVITY. CONSERVATION OF AREAS. 141. The expression, ydx — xdy, is, according to Article 112, twice the sectoral area passed over by the radius vector of the body in an instant of time. Hence, if X{mydx — m,xdy) = dA^ ; differentiating, we find dJ'x d''y\ _ d?A^ ^1 dPx d'y\ X[my^-mx^j = If there are no accelerating forces m-^^ = 0; and similarly for the others ; hence dt^ ^x d¥ -W-^'' ~d?' ' 'df -^' .: A^ = c,t; A^ = c^t ; A, = c^t ; (170) 1.5 226 CONSERVATION OF [142.] the initial values being zero. These are the projections on the coordinate planes of the areas swept over by the radius vector of the body. They establish the principle of the ooNSEEVA-noN OF AEEAS. That is, In any system of hodies, moving without accelerating forces and having only mutual actions upon each other, the projections on any plane of the areas swept over Iry the radius vector are proportional to the times. CONSEEVATION OF ENERGY. 142. Multiply each of equations (167) by dt, add and inte- grate, and we have W - Mv^ = 2/(X^£c + Ydy + Zdz) : and for a system of bodies, we have %{Mi?) - %{Mv^) = ^SfiXdx + Ydy + Zdz). (171) The second member is iiitegrable when the forces are directed towards fixed centres and is a function of the distances between them. Let a, 5, c be the coordinates of one centre, «^i! ^1) Ci; of another, etc. ; X, y, z, the coordinates of the particle m ; r, ri, etc., be the distances of the particle from the res- pective centres; F, Fx, etc., be the forces directed towards the respective centres ; then, resolving the forces parallel to the axes, we have -?= i^eos a'r F-t^ cos a + etc. ^a — X —,a,—x = F + Fr -+ etc. ; r r T=F^^ + F^-^^+etc.; r r Z - F + Fx ~ + etc. [143.] ENERGY. 227 Multiplying by dx, dy, ds, respectively, and adding, we have Xdx + Ydy + Zdz=F\^^^^d^+'-^^dy + "-^ K r r T ^pA^-^^dx^^-^^dy^ -dz + etc. — ds But r» = {a - «/ + (J -yf + {o- sf ; (172) and by differentiating, we find similarly, dr= dx ^y - ■ ds ; r r r dr^ = - dx - - — ^dy - dz : n n n etc., etc., etc. These substituted above, give Xdx + Ydy + Zdz=^ —Fdr — F^dr^ - F^dr^ — etc. Therefore, if F, etc., is a function of r, etc., and /*, /ij, etc., the intensities of the respective forces at a distance unity from the respective centres, or F=i.4>{r); etc., etc. ; the second member, and hence the first, will be integrable. In nature, if a particle m attracts a particle m^^, the particle TOj will attract m, each being a. centre of force in reference to the other, and ioth centres will be movable in reference to a fixed origin. But one centre may be considered fixed in ref- erence to the other, and consequently the proposition remains true for this case. The second member of equation (171) being integrated be- tween the limits x, y, z and a?,, y^, Sj, we have Z {Mv") - 2: (Mvo') = 2Z,ji {x, y, z) - 2Z/i

y, 0>Z, be the angular velocities about the respective axes. Eesolve ac par- allel to X and y, ab being the former component, and be the latter. Then, at the instant will dc (or its equal ab) equal to^y, and (X^ = ; (175) which are the equations of a straight line through the orighi, and are the equations of the instantaneous axis. Let a, /3, 7 be the angles which it makes with the axes x, y, z resjiectively, then {Anal. Geom.), (Ox cos a = S/ai + w/ + Tr ^=^^^=4/; Wx + «/ + 0>i cos 7 = 4/( x + Wy + «/ To determine the angular velocity of the body, take any point in a plane perpendicular to the instantaneous axis. Let the point be on the axis of x, and from it erect a perpendicular to the instantaneous axis, and we have (Oy + Wz p = xaina=x Vl — cos^ a = , 3 ^ -— — j^- 232 ROTARY MOTION. ^ [144.] For this point y = and s = in equations (174), and we find for the actual velocity, ¥ — and hence dt m = - = Vmi + V + ml; (176) "which represents the diagonal of a rectangular parallelopipe- don, of which the sides are <»a;, coy, ca^. 144. General equations of rotation about a fixed point. Take the origin of coordinates at the fixed point. For this case equations (167) vanish. Substitute in (168), the values of d?x -=-^, etc., obtained from (174), and reduce. We have d^x dcDy d(Oz , , , , and similarly for the others. Let Z, Jif, JV^he substituted for the last terms respectively of equations (168), and substituting the above values in the last of these equations, we find —j^ Sm{a?+ y')+ tOxCOy Xm{o^ — y^ dt 'da,, 4- lft>j,«a ^\ Xmxz. - = m (177) The other two equations may be treated in the same manner. Bat they are too complicated to be of use. Since the position of the axes is arbitrary, let them be so chosen that Sinssy — 0, Xmxz = 0, Xmyz = ; (178) in which case the axes are called principal axes / and we will show that for every point of a body there are at least three [155, 146.] EULER'S EQUATIONS. 233 principal axes, each of which is perpendiciilar to tlie plane of the otlier two. Also, let A = Xraiy^ + s^), the moment of inertia of the body about x ; B = Sm{^ + a?), moment about y ; C = Sm{a? + y*), moment about 3 ; and substituting these several quantities in (177), we have B^ -{A- C)m^^ = M; <^w-(^ B ) »a;6)y These are called Enler's Equations. N. (179) NO ACCELEEATING FORCES. 145. If there are no accelerating forces, Z = 0, J!f = 0, iV^= 0. A, B, C are constant for the same body, and if {B — C) -T- A = D, etc., we have dwx — Basyos^t = ; "j dcoy - Em^co^t = ; V (180) dta^ — FmxtOydt = Q. ) 146. IfO' ipdy revoVoe about one of the prinovpal axes pass- ing through the centre of gravity of the hody, that axis loill suffer no strain from the centrifugal force. Let 2! be a principal axis, about which the body rotates. The centrifugal force of any particle will be m<^p = mo? \/^ + y^ ; which, resolved parallel to x and y, gives m<^x, ma?y ; and the moments of these forces to turn about s are, for the whole body, ^mai'^xy, Xma?yx ; iJ34 PRINCIPAL [147, 148.] but these, according to equations (178), are zero. If the body revolves about this axis it will continue to revolve about it if the axis be removed. For this reason it is called an axis of perma/nent rotation. 147' If the body he free, and the initial rotation be not about a principal axis, the centrifugal force will cause the instan- taneous axis to change constantly, and it will never rotate about the permanent axis if free from extraneous forces. If, therefore, we observe that a free body revolves about an axis for a short time we infer that it revolved about it from the beginning of the motion. Example." If the earth were a homogeneous sphere, at what point must it be struck, and with what momentum, that it shall have a velocity of translation of F"and of rotation of w ? PEINCIPAl AXES. 148. At every potnt of a iody there are at least three prin- cipal axes perpendicular to each other. When three axes meeting at a point in a body are perpen- dicular to each other, and so taken that Smxy = 0, Xmyz = 0, Smax = ; ihey are called Principal Axes. The planes containing the principal axes are called Princi- pal Planes. The moments of inertia in reference to the principal axes at any point are called the Principal Moments of Inertia at that point. Let OJVhe any line drawn through the origin, making angles a, yS, 7 with the respective coordinate axes. Fii-st find the moment of inertia about tlie line Olf. From any point of the line OJV, erect a perpendicular JVP. The coordi- nates of P will be X, y, s. Hence we have OI^ = x' +f + 0>; FiQ. 128. OJF = X cosa + yeos ^ +' s cos 7 ; 1 = cos'a + cos^/8 + cos'7. [^48.J AXES. 235 The moment of inertia of the body in reference to OJ\f will be J= •LmPN' = 2ot (OF' - OJV') = 2m ^ (.x^ + y'' + e'' — (xcoscL + y coa0 + z cos yf }■ = 2ot ■{ (x'+y^ + z') (oosV+ooa'3+cosV)-(tB cos a+y co3 /3+s cos 7)' }- = Sot (y^ + x') cosV + Sm (a;^ + s") oos'S + 2ra (3;= + y') cos«7 — 22w(/^ cos /3 cos 7 — 3Sm«a! cos 7 cos a — ZSmxy cos a cos (3 =A cos'o+Bcos'^+CcosV— S-DoosScosY— 3£'cos7coso— S-foosiicos^S ; in which A^B, C have the values given on page 233, and Z>, E, F are written for the corresponding factors of the preceding eqnation. This maj be illustrated geometrically. Conceive a radius vector, ?', to move about in space in such a manner that for all angles a,;8, 7 corresponding to those of the line ON, the square of the length shall be inversely proportional to the moment of inertia of the body. Then in which c is a constant. Hence the polar equation of the locus is -j=JLco8''a+Scos'3+ Goosey— 2I> coa^ 0037— 2Scos 7COS0— SJi^cosocos/S. Multiplying by r', we have c= Aa? + Bf + C^ - 2Dy3 - 2E3x - 2Fxy, which is the equation referred to rectangular coordinates, and is a quadric. Since A,B, and O are essentially positive, it is the equation of an ellipsoid, and is called the momental ellip- soid. Therefore, the moment of inertia about every line which passes through any point of a body may be represented by the radius vector of a certain ellipsoid. But every ellipsoid has at least three principal diameters, hence every material system has, at every point of it, at least three ^jrincipal axes. If the ellipsoid be referred to its principal diameters the coefficients of yz, sx, xy vanish, which proves the proposition ; and the equation of the ellipsoid becomes G^A^^Bf+C^. In many cases the principal diameters may be determined by inspection. A CATALOGUE OF THE PUBLICATIONS OF JOHIsr ^^ILEY & SON 8, CONSISTrNG OF SCIEITIFIC AID OTHEE TEIT-BOOKS, AlTD PEAOTIOAL WOEKS, INCLUDING THE WORKS OF JOHN RUSKIN. WITH A COMPLETE INDEX AND CLASSIFICATION OF SUBJECTS. 15 Astor Place, Ifew York. 1876. *»* JOHN WILEY & SONS' Complete Classified Catalogues of the most valuable and latest scientific publications, Parts I. and II., 8ro, mailed to order on the receipt of 10 cents. INDEX AND CLASSIFICATION SUBJECTS. Advanced Guard and Out-Pest Service. 2Iahan. Agricultural Chemistry. Liebig. Agriculture, Jlodem. Liebig. American House Carpenter. Hatfield. Analytical Hebrew and Chaldee Lexicon. Davidson. Anilin e Colors, Manufacture and Use of. Urace-C'altert. Reimann. Aratra Penteleci. Muskin. Ariadne Ploreutina. Buskin. Architectural Iron Work. Fryer. Architecture. Eatfidd. Buskin. Wightwick. " Mechanicjd Principles of. " and Painting, Lectures on. BiLskin. Art Culture, Hand-Book of. Buskin. " in Decoration and Manufacture. (Two Paths.) Buskin. " " Natural Science. (Eagle's Nest. )5i«(^m. Art of Memory. Omi/roAid. " " " Dictionary of. Oouraud. " Political Economy of. Bi.iskin. Assaying. Bodemann. MUeJull. Astronomy, Treatise on. Norton. Bible. For Manuscript Notes. Blank-Paged. " References printed under each verse. Commenta/ry wlwUy Biblical. Birds, Greek and English (Love's Meine). Buskin. Bleaching. O'Neil. Book-keeping and Accountantship. Jones, Books, Women, &o. (Sesame and Lilies.) Buskin. Boston Machinist. Fitzgerald. Breath of Life, The. Catlin. Bridge Engineers, Hand-Book for. Herschel. BoUer. Wood. Du Bois. Her- Bronchitis, Treatise on. Oreen. Calculus, The Differential. Bice. Calico Printing. O'Neil. Grace- Oahert. Macfarlane. Capital Punishment. Oheever. Carpenter, American House. Batfield. ■ " and Joiners' Hand-Book. Holly. Cast and Wrought Iron. Fairbaim. Cements, Hydraulic and Calcareous. Austin. Charcoal Burners, Hand-Book of. Svedelius. - Chemical Physics. MiUer. Chemistry, Agricultural. Liebig. ' ' Applied to Arts and Manufactures Muspratt. ' ' Inorganic. Miller. " Organic. Miller. " Qualitative, Analysis of. Craft. Johnson. Fresenius. Perkins. " Quantitative, Analysis of. Fres- enius. Johnson. Thovpe. " Theoretical and Practical. Miller. Chemist's Reports on Lead Pipe. Kirkwood. Child's Book of Favorite Stories. Chinese Empire and its Inhabitants. (Middle Kingdom.) Williams. Choice Selections. Buskin. Clock and Watch Making. Booth. Coal-Tar Dyes. Beimann. Coloring Matters, Manufacture of. O'Neil. Complete Angler, The. Walton <6 Cotton. Concordance, English. Manual. " Greek. Englishman's. " Hebrew. Englishman's. Construction of Sheepfolds, Notes on. Bus- kin. Consumption or Pulmonary Tuberculosis. Oreen. Cottage Eesidences. Downing. Cotton Manufacture, Hand-Book of. Gddard. Crooke's Metallurgy. Kerl. Crown of WHd Olives. Buskin. Croup, Pathology and Treatment of. Oreen. Descriptive Geometry. Malian. Wa/rren. " Mineralogy. Dana. Deucalion. Buskin. Diseases of the Air-Passages. Green. Drafting Instruments, &c. Warren. Drawing. Ooe. Malian. Bwskin. Smith. Warren. " Free-Hand. Warren. " Industrial. Mahan. " Machine. Wa/rren. " Topographical. Smith. Mahan. Dyeing and Calico Printing. Grace- Calvert. ONeU. MoAfwrlane. e:. Eagle's Nest. Buskin. Electricitj and Magnetism. Miller. Engineering, Bridge. Du Bois. Boiler. Ber- scJiel. " CivU. Mahan. Mosdy. Wheeler. Wood. " Mechanical Principles of . Mosdy. " Military. Mahan. Engraving on Wood and Metal. (Ariadne Florentina.) Buskin. Ethics of the Dust. Bus/an. P. Favorite Prescriptions of Living Practitioners. Green. Flower-Garden, Companion to. Loudon. Fors Clavigera. BusMn. Fortifications, Field. Malian. " Permanent. Mahan. Free-Hand Drawing. Warren. Free Town Libraries. Edwards. Frondes Agrestes. Buskin. Fruits and Fruit Trees. Downing. " Culture and Propagation. Downing. " Encyclopedia of. Downing. " Guide to Eipening of. Downing. " Insects Injurious to. " Selected. Downing. Foundry, A Model Iron. Fryer. Gardening for Ladies. Loudon. Gardens, How to Lay Out. Kemp. " New Styles of. Sargent in Down- ing's Cottages. " Trees, Shrubs, and Plants for. Sar- gent, in Downing's Cottages. Geometry, Descriptive. Mahan. Warren. " Plain Problems in. Warren. Gold, Silver, and other Metals, Assaying of. Bodemann. MiteheS. " " Metallurgy of . Kerl. Graphical Statics of Bridges, &c. Ihi Bois. Grsecum Novum Testamentum. Greenfield. Greek Concordance. Englishman's. " Myths of Cloud and Storm (Queen of the Air). Buskin. Testament. Greenfield. " vrith Lexicon. " Hand-Book of Grammar to. Green. and English Testament. " " " and Lexicon. " " " and Concordance. New Testament, Grammar of. Oreen. Lexicon. Analytical. Oreen. Green- field. H. Heat as a Source of Power. Trowbridge. Hebrew Bible. Letteris. Hebrew-English Lexicon. Hebrew, Chaldee, and English Lexicon. Ana- lytical. Gesenius. Tregdles. ' ' Chrestomathy. Green. ' ' Concordance. Englishman's. " Grammar. Oreen. " " Elementary. Oreen. Heroes and Hero-Worship. Gu/rlyle. Highway Bridges. BoUer. Hill Difficulty. Oheever. Hints to Mothers during Pregnancy. BvU. Hints to Persons about Building in the Country. Downing. Hints to Young Architects. Wight icick. Horticulture. Liiidley. How to Lay Out Garden and Lands, from one quarter to 100 Acres. Kemp. Hydraulic Limes and Cements. Austin. Hydraulic Motors. Bresse. DiiBoia. Wets- bach. Hydropathy . Francke. I. Industrial Drawing. Mahan. Lifidelity, Causes of. Cmiyheare. Iron for Building, Cast and Wrought. Fair- bairn. " Trade Manual. Wdey. " " History of. French. " Work, Architectural. Fi-yer. Iron Work. Specifications for Iron. Fryer. J. Journal of the Pilgrims at Plymouth Rock. Cheecer. TTing of the Golden River. Ruskin-. King's Treasuries and Queen's Gardens, &o. (Sesame and Lilies. ) Ruskin. Landscape Gardening. Kemp. Lapse of Waves and Life of Stones. (Deuca- lion. ) Ruskin. Larynx and Glottis, Treatment of. Green. Laws of Work. (Time and Tide. ) Ruskin. Lead Pipe for Service Pipe. Kirkwood. Lectures to Little Housewives. (Ethics of the Dust.) Rtcskin. Leila Ada. Heighway. " " Relatives of. Heighway. Letters to the Workmen, &c. (Pors Clavi- gera. ) Ruskin. Libraries, Free Town, The Formation of. Edwards. Limes, Hydraulic. AvMin. Locomotive Engineering. Oolburn. Love's Meine. Ruskin. M. Machine Drawing. Warre7i. Machinist, Boston. Fitzgerald. Magnetism and Electricity. Miller. Mai-Respiration. CaUin. Materials, Resistance of. Wood. Mechanics, Analytical. Wood. Elementary. Magnus. Mechanism of Railways. Colburn. " Principles of. Willis. Medical. BuU. Franake. Green. Von Buhen. Metallurgy. Orookes. Eerl. Microscopical Diagnosis. Von Duben. Middle Kingdom, The. WUlimns. Mineralogy, Descriptive. Bana. " Determinative. Brush. Bana. " Text-Book of. Bana. Modem Painters. Ruskin. Mornings in Florence. Ruskin. Munera Pulveris. Ruskin. N". Naval Ordnance and Gunnery. Cooke. New Tale of a Tub. Bailey. New Theory of Disease. Francke. o. Oil Painting for Young Artists. Bouvier. Ordnance and Gunnery, Naval. Oooke. Out-Post Service. Mahan. Painting. Lectures on. Ruskin. " in OU. Bouvier. " Landscape and Portrait. Bomier. Pentateuch, The, Vindicated from Colehso. Green. Perspective. Ruskin. Smith. Warren. Perversion. Gonybeare. Pocket Bible, Story of. Poetry of Architecture. Ruskin. Political Economy. (Unto This Last. ) Ruskin. Precious Thoughts. Ruskin. Pre-Raphaelitism. Ruskin. Prescriptions, Medical. Green. Projection Drawing. Warren. Proserpina. Ruskin. Proverbial Philosophy. Tupper. Quadrature of the Circle. Parker. Queen of the Air. Ruskin. R. Railway Carriages and Plant. Colburn. Railways, Mechanism of. Colburn. 6 Readings on Modern Painters. (Fronde? Agrestes.) Buskin. Relation of Natural Science to Ait. (Eagle's Nest. ) Buskin. Resistance of Materials. Wood. s. Saw-Filing, Art of. Holly. Saws, Directions for Putting in Order. HoUy. Screw Propeller, Treatise on. Bourne. Sculpture, Elements of. (Aratra Penteleoi.) Buskin. Sesame and Lilies. Buskin. Seven Lamps of Architecture. Buskin. Shades and Shadows. Warren.. Ship-Building. Watts. Bankine, &e. Wilson. Soap, Mamifacture of. Morfit. Steam Engine. Trowbridge. Stereotomy, or Stone-Cutting. Malum. Wa/r- ren. Stone- Cutting. Mahan. Wa/rren. Stones of Venice. Buskin. Studies on Christian Art. (Mornings in Florence.) Buskin. T. Textile Colorist, The. A Monthly Journal. O'Neil. " Fabrics. O'Neil. Timber, Preservation of. (Wood's Materials. ) Time and Tide. Buskin. Topographical Drawing. Mahan. Smith. Warren. Treasury Bible, with 500,000 extra Refer- ences to, Bible., The. Trees, Shrubs, and Plants for Gardens. Downing's Cottages. TregeUes' Hebrew and Chaldee Lexicon. True and Beautiful. Buskin. Two Paths in Art. Buskin. Turning, Wood and Metal. (Lathe and its Uses.) TJ. Unto This Last. Buskin. V. Ventilation, Treatise on. Leeds. Beii. w. Wanderings of a Pilgrim in the Alps. Wanderings of the River of the Water of Life. Oheever. Watch Making. Booth. Water Wheels. Bresse. Da Bois. Wayside Flowers. (Proserpina.) Buskin. Work, Traffic, and War. (Crown of WHd Olives.) Buskin. TEXT BOOKS AND PRACTICAL WORKS POK SCHOOLS, COLLEGES, POLYTECHNIC INSTI- TUTES, ENGINEERS, ARCHITECTS, &c., Will be found in this CATALOGUE on the following SUBJECTS, under the names of the AUTHORS. AGRICULTURE. Downing — Kemp — Liebig — Lindley — Loudon. ARCHITECTURE. Downing — Hatfield — Holly — Euskin — Wightwick — ^Wood. ASSAYING. Bodemann & Kerl— Mitchell. ASTRONOMY. Norton. BOOK-KEEPING. Jones. CHEMISTRY. Crafts — Fbesenius — Johnson — Kibkwood — Miller — Muspeatt — Per- kins — Thorpe. DRAWING AND PAINTING. BouviER — CoE — Mahan — RusKiN— Smith — Warren. DYEING. Crace-Calvert — Macfaklane — Eeimann — Beesse. ENGINEERING. Austin — Boller — Colburn — ^DuBois — Herschel — Mahan — Mosely — Warken — Wood . IRON AND METALLURGY. BoDEMAjfN — Crookes — Fairbairn — Feench — ^Fryee — KiBirwooD. MACHINISTS— MECHANICS. Fitzgerald — Holly — ^Magnus — Willis^ Wood. MINERALOGY. Brush — Dana. SHIP-BUILDrNG AND NAVAL ORDNANCE. Bourne — Watts — Wilson — Goose. STEAM ENGINE. Trowbridge. VENTILATION. Leeds — ^Reed. Also for THEOLOGICAL SEMINARIES. Hebrew and Greek Bibles — Concordances — Graumars — Lexicons, Etc., Etc., Etc. New York, July, 1876. CATALOGUE OF THE PUBLICATIONS OF JOHI"^" WILEY & SOI^fS, 15 ASTOR PLACH. Books marked vith an * an sold at net prices to the Trade. ANAI,YTICAI, GREEK I^GXTCON, (The.) Consistiog of an alphabetical arrangement of every occurring inflexion of every word contained in the Greek New Testament Scriptures. With a grammatical analysis of each word and lexicographical illustration of the meanings. A complete series of paradigms, with grammatical remarks and explanaTions. 1 vol. small 4to, ^ mor $6 50 AUSTIN JAS. G. A PRACTICAI, TRE.'ITISE: OST THE PREPARA- TION, CO.TIBINATION, AND APPl,ICATION OF CAIiCAREOVS- AND HirDRAI7I.IC lilJHES AND CEMENTS. Compiled and aJrranged from the best authorities and from the practical experience of the compUer during a long professional career. To which is added many useful recipes for various scientific, mercantile, and domestic purposes. 1 vol. 12mo $3 00 BAIIiET^ F. \r. N. THE NEW TAIiE OF A TUB. An adventure in verse. By F. W. N. Bailey. With illustrations. 1 vol. 8vo $0 75 •BIBLE, THE TREASURE, dec. Containing the authorized English version of the Holy Scriptures, interleaved with a Treasury of more than 500,000 Parallel Passages from Canne, Brown, Blavney, Scott, and others. With numerous illustrative notes. 1 vol. half bound $7 50 1 vol. morocco 10 00 BliANK-PAGED BIBLE— THE HOLT SCRIPTURES OF THE OLD AND NEW TESTAMENTS, ~ With copious references to parallel and illustrative passages, and the alternate pages ruled for MS. notes. This edition of the Scriptures contains the Authorized Version, illustrated by the references of "Bagater's Polyglot Bible," and enriched with accurate maps, useful tables, and an Index of Subjects. 1 vol. Sw), morocco extra $9 50 1 voL 8vo, full morocco 11 00 BODEMANN T. A TREATISE ON THE ASSAYING OF liEAD, SIL- VER, COPPER, GOLD AND MERCURY. From the German of Th. Bodemann and Bruno Kerl. Translated by W. A. Goodyear, Ph.B. 1 vol. lamo, cloth, plates $3 50 BOI.LER A. P. PRACTICAL TREATISE ON THE CONSTRUCTION OF IRON HIGH^VAY BUIDGEl. For the use of Town Committees, together with a short Essay upon the ap- plication of the principles of the Lever to a ready analysis of the strains upon the more customary forms of Beams and Trusses. With many fine wood engravings. By A. P. Boiler, A.M., C.E. 1 vol. 8vo, cloth $3 50 10 PUBLICATIONS OF JOHN -VVILET & SONS. BOOTH M. I.. NEW AND COMPIiEXE CLOCK ANO WATCH MAKERS' MANUAI.. Comprising descriptions of the various gearings, escapements, and compen- sations now in use in French, Swiss, and English Clocks and Watches, Patents, Tools, etc. , with directions for cleaning and repairing. Compiled from the French, with an Appendix containing a History of Clock and Watch Making in America. Numerous plates. 1 vol. 12mo, cloth, $3 00 BOITRNE J. A TREATISE ON THE SCREW PROPELBULL THOS. HINTS TO MOTHERS FOR THE MANAGEMENT OF HEALTH DURING THE PERIOD OF PREGNANCY, AND IN THE LYING IN ROOM. With an exposure of popular errors in connection with those subjects. By Thomas Bull, M.D. Ivol 12mo, cloth $1 00 'CARLYLE THOS. ON HEROES, HERO-W^ORSHIP, AND THE HE- ROIC IN HISTORY. Six Lectures. Reported, with emendations and additions. 1 vol. 12mo, cloth $0 75 CATLIN GEO. THE BREATH OF LIFE, OR M.AL-RESPIRATION AND ITS EFFECTS UPON THE ENJOYMENTS AND LIFE OF MAN. With numerous wood engravings. 1 vol. 8vo ; $0 75 CHEEVER REV. GEO. B., D.D. PUNISHMENT BY DEATH; ITS AUTHORITY AND EXPEDIENCY. 13mo. cloth $0 50 HILL DIFFICULTY, AND OTHER MISCELLANIES. 1 vol. 13mo, cloth $1 00 JOURNAL OF THE PILGRIMS AT PLYMOUTH ROCK. 1vol. 12mo, cloth $1 00 WANDERINGS OF A PILGRIM IN THE SHADOW OF MONT BLANC. 1 vol. 12mo, cloth |1 00 WANDERINGS OF THE RIVER OF THE W^ATER OF LIFE. 1 vol. ISmo, cloth |1 00 ^CHILD'S BOOK OF FAVORITE STORIES. Large colored plates. 4to, cloth $1 50 PUBLICATIONS OF JOHN WILEY & SONS. 11 COE BENJ. H. PROGRESSIVE DRAWING-BOOK. 1 vol. cloth |3 50 DRAWING FOB lilTTIiE FOLKS, OB FIRST LESSONS FOB THE NlTRSERir. 30 drawings. Neat cover $0 20 FIRST STUDIES IN DRAW^ING. Containing Elementary Exercises, Drawings from Objects, Animals, and Rustic Figures. Complete in three numbers of 18 studies each, in neat covers. Bach . . |0 20 COTTAGES. An Introduction to Landscape Drawing. Containing 73 Studies. Complete in four numbers of 18 studies each, in neat covers. Each. .$0 20 EASY LESSONS IN LANDSCAPE. Complete in four numbers of 10 studies each. In neat 8vo. cover. Each $0 20 BEADS, ANIMALS, AND FIGURES. Adapted to Pencil Drawing. Complete in three numbers of 10 studies each. In neat 8vo covers. Bach |0 20 COPY BOOK, WITH INSTBUCTIONS $0 37i COLBirBN X. LOCOMOTIVE ENGINEERING AND THE MECHANISM OF RAILWAYS. A Treatise on the Principles and Construction of the Locomotive Engine, Kailway Carriages, and Railway Plant, with examples. Illustrated by Sixty- four large engravings and two hundred and forty wood-cuts. By Zerah Colburn. Complete, 20 parts, $15.00; or 2 vols, cloth $16 00 Or, halt morocco, gilt top |30 00 COMMENTARY WHOLLY BIBLICAL, (The.) Contents : — The Commentary : an Exposition of the Old and New Testa- ments in the very words of Scripture. 3264 pp. II. An outline of the Geography and History of the Nations mentioned in Scripture. III. Tables of Meafcures, Weights, and Coins. IV. An Itinerary of the Children of Israel from Egypt to the Promised Land. V. A Chronological comparative Table of the Bangs and Prophets of Israel and Judah. VI. A Chart of the World's History from Adam to the Third Century, A.D. VII. A complete Series of Illustrative Maps. IX. A Chronological Arrangement of the Old and New Testament* X. An Index to Doctrines and Subjects, with numer- ous Selected Passages, quoted in full. XI. An Index to the Names of Per- sons mentioned in Scripture. XII. An Index to the Names of Places found in Scripture. XIII. The Names, Titles, and Characters of Jesus Christ our Lord, as revealed in the Scriptures, methodically arranged. 3 vols. 4to, cloth $30 00 3 vols. 4to, half morocco, gilt edges 30 00 3 vols. 4to, morocco, gilt edges 40 00 CONCOBDANCE. A MANUAL TO THE ENGLISH NEW TESTAMENT. 18mo, i bd $1 25 CONYBEARE W. I. PERVERSION, OR THE CAUSES AND CON- SEaUENCES OF INFIDELITY. A Tale of the Times. 12mo |1 00 COOKE A. P. A TEXT-BOOK OF NAVAL ORDNANCE AND GUN- NERY. Prepared for the Use of the Cadet Midshipmen at the United States Naval Academy. By A. P. 'Cooke, Com. TJ. S. N. One thick volume, illustrated by about 400 fine cuts. Cloth $12 50 CRACE-CALVEBT DR. F. DYEING AND CALICO PRINTING. Including an account of the most Recent Improvements in the Manufacture and Use of Aniline Colors. Edited by Dr. Stenhouse and C. E. Groves. Illustrated with wood engravings and numerous specimens of printed and dyed fabrics, l.vol. 8vo $8 00 CROOKES WM. METALLURGY. See Kerl. V2 PUBLICATIONS OF JOHN WILET & SONS. CRAFTS J. JIK. A SHORT COURSE! IX dVAIilTATIVX; CHEMICAIi ANAIiYSIS. With the new notation. By Prof. J. M. Crafts. Second edition. 1 vol. 12mo, cloth $1 50 DANA J. D. DESCRIFTIVX; mitVERAE.OG'X'. Comprising the most recent Discoveries. Fifth edition. Almost entirely re-written and greatly enlarged. Containing nearly 900 pages 8vo, and up- wards of 600 wood engravings. By Prof. J. D. Dana. Aided by Prof. Geo. J. Brush. Including Appendices by Profs. Brush and E. S. Dana, complet- ing the work up to 1875. Cloth |10 00 " Wo have used a good many works or Mineralogy, but have met with none that begin to compare with this in fulness of plan, detail, and execution." — American Jowrnal uf Mining. ArPEivmcES TO ditto. Completing the work to 1875, by Prof. Geo. J. Brush and E. S. Dana. Svo f 1 00 DETERtllNATIVE PIIINERAI.OGY. (See Brush's Blow-Pipe, etc). 1 vol. 8vo, cloth $3 00 A TEXT-BOOK OF MINERAKiOOY. 1 vol. (In preparation.) DOWNING A. J. & C. COTTAGE RESIDENCES. NEW EDITION. A Series of Designs for Rural Cottages and Cottage Villas, and their Gardens Grounds. By A. J. Downing. Containing a revised List of Trees, Shrubs, and Plants, and the most recent and best selected Fruit, with some account of the newer style of Gardens. By Henry Wiuthrop Sargent and Charles Downing. With many new designs in Rural Architecture. By George Harney, Architect. 1 vol. 4to $6 00 ENCYCliOPEDIA OF FRUITS. Or, Fruits and Fruit Trees of America. Part I. — Apples. With Supple- ments, containing many new varieties, and brought down to 1876. By Chas. Downing. With numerous outline engravings. 8vo, full cloth $3 50 ENCYCIiOPEDIA OF FRUITS. Or, Fruits and Fruit Trees of America. Part 3. — Chekries, Grapes, Peaches, Pears, &c. With Supplements, containing many new varieties, and brought down to 1876. By Charles Downing. With numerous out- line engravings. Svo, full cloth |3 50 FRUITS AND FRUIT TREES OF AMERICA. 1 , the Culture, Propagation and Management in the Garden and Orchard Fruit Trees generally, with descriptions of aU the finest varieties of Fruit, Native and Foreign, cultivated in this country. By A. J. Downing. Second revision and correction, with large additions. By Charles Downing. 1 vol. Svo, over 1100 pages, with several hundred outline engravings, includ- ing. Suppiements for 1873 and 1876 $5 00 "As a work of reference it has no equal in this country, and deserves a place in the library of every pomologist in America." — Marshall P. Wilder. HORTICUliTURE. See Lindley. HINTS TO PERSONS ABOUT BUIUDING IN THE COUNTRY.- AND HINTS TO VOUNG ARCHITECTS. Calculated to facilitate their practical operations. By George Wightwick, Architect. Wood engravings. Svo, cloth $3 00 liADIES) FliOWER GARDEN. See Loudon. SELECTED FRUITS, From Downing's Fruits and Fruit Trees of America. With some new va- rieties, including tlieir Culture, Propagation, and Management in the Gar- den and Orchard, with a Guide to the selection of Fruits, with reference to the Time of Ripening. By Charles Downing. Illustrated with upwards of four hundred outlines of Apples, Cherries, Grapes, Plums, Pears, &c. lTol.l3mo $2 50 PUBLICATIONS OF JOHN WILEY & SONS. 13 ♦BAVSDSON B. ANALYTICAL IIEBREUr AND CIIALDX:!!; LEXICON, Consisting of an alphabetical arrangement of every word and inflection con- tained in the Old Testament Scriptures, precisely as they occur in the sacred text With a grammatical analysis of each word and lexicographical illus- tration of the meanings. A complete series of Hebrew and Chaldee para- digms, with grammatical remarks and explanations, by B. Davidson. 1 vol. small 4to. half-bound |11 00 DIJ BOISA. J. ELEMENTS OF CRAPHICAL STATICS. And their Application to Framed Structures. With numerous practical ex- amples. Cranes ; Bridge, Roof, and Suspension Trusses ; Braced and Stone Arches; Pivot and Draw Spans ; Continuous Girders, etc., together with the best methods of calculation, and containing also new and practical formulae for the Pivot or Draw Span, Braced Arch, Continuous Girder, &c. 2 vols. 8vo, 1 vol. text and 1 vol. plates $5 00 HYDRAULIC MOTORS. With numerous practical examples for the calculation and construction of Water Wheels, including Breast, Undershot, Back-jitch, Overshot Wheels, &c , as well as a special discussion of the various forms of Turbines, translated from the fourth edition of Weisbach Mechanics, profusely illus- trated. 1 vol. 8vo. EDTT.VRDS E. FREE TOWN LIBRARIES. The Formation, Management and History in Britain, France, Germany, and America Together with brief notices of book-collectors, and of the respective places of deposit of their surviving collections. By Edward Ed- wards. 1 vol. thick 8vo $4 00 'ENGLISHMAN'S, THE, GREEK. CONCORDANCE OF THE NETT TES TAMENT. Being an attempt at a verbal connection between the Greek and the English Texts ; including a concordance of the Proper Names, with Indices, Greek- English and English-Greek. Sixth edition. Svo $10 00 'ENGLISHMAN'S, THE, HEBREW AND CHALDEE CONCORDANCE OF THE OLD TESTAMENT. Being an attempt at a verbal connection between the Original and the Eng- lish Translation, with Indices, a List of Proper Names, and their Occur- rences, &c. Third edition. Two vols. Svo, cloth $35 00 FAIRBAIRN "WM. ON THE APPLICATION OF CAST AND WROt'GHT IRON FOR BUILDING PURPOSES. 8vo, cloth $3 00 FITZGERALD WALTER, THE BOSTON MACHINIST. Being a complete School for the Apprentice as well as the Advanced Machin- ist, showing how to make and use every tool in every branch of the business ; with a treatise on Screws and Gear-cutting. 1 vol. 18mo, cloth $0 75 FRANCKE. OUTLINES OF A NEW THEORY OF DISEASE, Applied to Hydropathy, showing that water is the only true remedy. With observations on the errors committed in the practice of Hydropathy, notes on the cure of cholera by cold water, and a critique on Priessnitz's mode of treatment. Intended for popular use. By the late H. Francke. Trans- lated from the German by Robert Blakie, M.D. 1 vol. 13mo, cloth f 1 50 FRESENIUS C. REMIGIUS, A MANUAL OF QUALITATIVE CHEMI- CAL ANALYSIS. By C R. Fresenius. Translated into the New System, and newly edited by Samuel W. Johnson, M.A., Prof, of Theoretical and Agricultural Chemistry, in the Sheffield Scientific School of Yale College, New Haven. 1 vol. Svo, cloth. 1875 f * 50 Ditto. Edition of 1S74. Svo, cloth $3 00 A SYSTEM OF INSTRUCTION IN CtUANTITATIVE CHEMICAL ANALYSIS. From latest editions, edited, with additions, by Prof. S. W. Johnson. With Chemical Notation and Nomenclature, old and new, $6 00 14 PUBLICATIONS OF JOHN WILEY & SONS. FBENCH B. F. HISTORf OF THE RISF AND PROGRESS OF THE IRON TRADE IN THE UNITED STATES FROM 1621 TO 1S57. With numerous statistical tables relating to the manufacture, importation, exportation, and prices of iron for more than a centnry. 8vo, cloth $3 00 FRTER TT. I. ARCHITECTITRAI. IRON TFORK.-i A practical work for Iron Workers, Architects and Engineers, and all whose trade, profession, or business connects them with architectural iron work, showing the organization and mechanical and financial management of a foundry and shops for the manufacture of iron work for buildings, with specifications of iron work, useful tables and valuable suggestions for the successful conduct of the business, with many fine wood engravings. 8vo, doth $3 50 GEIiDARD JAS. HANDBOOK ON COTTON MANUFACTURE; OR, A GUIDE TO aiACHINE-BUIUDING, SPINNING, AND TTEAVING. With practical examples, all needful calculations, and many useful and im- portant tables. The whole intended to be a complete yet compact authority for the manufacture of cotton. With steel engravings. 1 vol. 12mo, cloth |2 50 GESENIUS Dr. "Wm. HEBRE'tf AND CHAI.DEE UEXICON TO THE OI/D TESTAMENT SCRIPTURES. Translated and edited, with additions and corrections from the author's The- saurus and other works, by S. P. Tregelles, LL.D. In thiR edition great care ha=4 been talcen to gaard the student from NeoLogian tendencies by snita- ble remarks whenever needed. '^ The careful revisal to which the licxicon has been subjected hy a faithful and Orthodox transla- tor exceedingly enhances the practical value of this edition." — Ediiityurgh Ecclestastical Journal, Small 4to, in large clear and perfect type, ^ bd. $7 00 GOURAUD F. F. PHKENO-MNEMOTECHNT ; OR, THE ART OF MEMORY. The series of Lectures explanatory of the principles of the system. 1 vol 8vo, cloth |2 00 PHRENO-MNEMOTECHNIC DICTIONARY. Being a Philosophical Classification of all the Homophonic Words of the Snglish Language. To be used in the application of the Phreno-Mnemo- technic Principles. 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