V'5!-< The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031271350 Cornell University Librarv arV17323 Text-book of mechanics. ,. 3 1924 031 271 350 olin.anx BY Tan SAWXl AUTBOH, A TREATISE ON THE ADJUSTMENT OF OBSERVATIONS, With Applications to Measures of Precision. New York: D. VAN NOSTRAND COMPANY. TEXT-BOOK MECHANICS With Numerous Examples. THOMAS WALLACE WEIGHT, Pbofessob in Uniom College. The concrete in which lies always the perennial. — Cablyle. NEW YORKr D. VAN NOSTEAND COMPANY, 33 Murray and 27 Wakren Steeets. JS90. ® LI PR Copyright, 1890, By Thomas Wallace Wriqht. Robert Druhhond, Eleotrotyper, m. & IdC Feai'l Street, New York. CONTENTS. PAGE Intkodttction 7 Chap. I. Motion, '. . . . 9 II. I'oKCB AND Motion, 30 III. Dynamics of a Particle, 44 IV. Statics oi' a Rigid Body, 86 V. Friction, 129 VI. Work and Energy, 148 VII. BJNETics OF A Rigid Body, 174 VIII. Elastic Solids, 301 IX. Statics of Ei^uids 312 X. Kinetics of Fluids 343 vii PREFACE. The science of mechanics, as all other departments of physical science, has made great advances in recent years. lu this book I have endeavored to unfold the principles of modern mechanics systematically, and also to call the atten- tion of the student to tlie more useful applications of the su.bject. It has long seemed to me that the more practical parts might with advantage be presented to the beginner at least as fully as the more abstract, and that, too, without any sacrifice of scientific precision. The book is so arranged that the student, whether intending to make a specialty of engineering, physics, or astronomy, can branch out in his special direction without difficulty. It has been the aim to make the examples as practical as possible. Thus while writing the Strong locomotive (No. 444) was brought out by the L. V. R. E. Co., and the West- inghouse air-brake tests (1887) were being made. Many problems founded on these and other mechanisms have been introduced. It has been the aim also to make the examples typical. Instead of making them mere numerical illustrations of formulas, the idea has been to encourage independent thought. In many cases different methods of solution have been indicated to encourage the student to trust rather to an independent investigation than to an answer so called. Considerable attention has been paid to the graphical method of solution, guarding, however, against making it a "complicated weapon with which one can attack all sorts of problems which are more easily solved in other ways. " In several applications of the subject approximate for- mulas are of the utmost importance, and many such for- VI PREFACE. mulas are here developed. In all cases tlie rigorous formula has been given first, and the approximate deduced from it. In this way the degree of approximation can be estimated. As regards the nomenclature of mechanics, I have en- deavored to be modern and at the same time conservative. One or two terms have been intj'oduced, the better to illus- trate the second law of motion and the appropriateness of the expression moment of inertia. The words weight and pound have been used in the double sense employed in ordinary life. No confusion need arise from this, as the context is always sufficient to show the sense intended. (See p. 40.) Other departures from the traditional treatment will be noticed. Thus the usual chapter on the Mechanical Powers has been emitted, though the " Powers" themselves have been discussed in their proper places. All dynamical equations have been expressed in terms of the absolute units. By doing this it has been possible to ignore the ex- pression w = mg, the source of so much confusion in dynamics. The Calculus has been used in all cases where its use is attended with marked advantage. In the earlier chapters two parallel courses are given, one with and one without Calculus. This I think an advantage in that it shows the student the oneness of what are called elementary and analytical mechanics. If thought advisable, a course may easily be selected into which no Calculus enters. A few historical notes have been interspersed as tending to add a living interest. The use of the symbol / as the sign of division has added to the compact form of the book by preventing spacing about formulas. I have to thank several friends for assistance rendered, particularly Prof. Klein of Lehigh, Pres. Staley of Case, and Prof. Ziwet of Michigan. T. W. W. Schenectady, N. Y., June, 1590, MECHANICS. INTEODUCTIOK 1. If a body occupies a certain position at one time and another position at another time we say that the change of position must be due to the operation of some cause which we call force. If the form of the body has undergone a change, we say that the change of form is due to the action of force. The fundamental idea of force is derived from the use of our muscles, and to muscular effort or to any cause producing a like efEect we give the name force. Muscular effort is exerted as a push or a pull or some com- bination of them. We will therefore for the present con- sider force in the sense of a push or a pull no matter by what exerted. The science which treats of the different effects of force on bodies is called Mechanics.* 2. Bodies exist in various states which may roughly be classed under two general divisions, the solid and the fluid. Experiment shows that the result of the action of a force on a solid is very different from the result of the action on a liquid. Hence it will tend to clearness and simplicity if * " Strictly speaking, the derivation of this word should have prevented the use of it as the designation of a pure science. It has been, however, employed for a long time in English speech in the identical sense that the French attach to Meeanique pure or the Germans to reine Meclianik. These terms are employed to denote what we should prefer to call abstract dynamics — the pure science which treats of the action of force upon matter, which is correctly the science ef matter and niotjon,'' (Tait,) 7 8 INTRODUCTION. we study the Mechauics of Solids and the Mechanics of Fluids separately. Experiment shows, too, that the effects of force are very different on different solids and on different fluids. Com- pare, for example, iron and putty, molasses and water. Hence still more minute subdivisions are necessary for pur- poses of study. 3. On account of the infinite variety of solids and fluids it becomes necessary as a first step to seize on some phase of the problem applicable to all, and then to discuss the special cases in detail. Now every substance in nature may be conceived to be composed of an indefinitely great number of minute portions or particles so small that any external force acting on such a particle will act equally on all its parts. By considering the action of forces on a single particle we may develop principles applicable to bodies in all forms considered as composed of particles. Still further. Conceive a particle acted on by forces. It will in general move in a path of some kind, either straight or curved. The particle is so small that the path may be regarded as a geometrical line. If we leave the forces acting out of view and consider only the positions of the particle in this path at different times, the relations between these different positions will be in the nature of a geometrical problem. Hence, as every particle is endowed with at least some of the properties of the body which it goes to make up, we may conceive these properties ab- stracted from it, thus converting it as it were into a geo- metrical point, and consider the motion of this point only as it traces out its path during an assigned time and thus be able to study motion in its simplest form. To this idea of motion in which neither the nature of the particle moved nor of the force acting is considered the name Kinematics (= science of motion) is given. Kinematics is therefore motion in the abstract, and its principles are developed by MOTION". 9 an extension of the principles of geometry by the introduc- tion of the idea of time. 4. "From this bird's-eye view of the subject it is seen that the course to be followed is in general outline something of this kind : First geometrical motion, next the action of force on a single particle and on a system of two or more particles either independent or forming a body, and last the modifications in body motion resulting from the pe- culiar constitution of the various states of matter solid or fluid. CHAPTER 1. > MOTION. 5. The elements of a motion which characterize it are change of position {displacement) with change of time. We know the position of a body only by noting its relations to Pj , p other bodies in its neighborhood. Thus the position of a point P situated in the plane of the paper is defined with refer- ence to a point chosen as the point of reference, either by measuring the distance OP and the angle XOP made by OP with a y Fig. 2 known line OX (polar coordinates) or by measuring the distances PM, MO drawn parallel to two known lines OY, OX in the plane (Cartesian co- o m x ordinates). Any displacement in the plane of the paper * would be measured by the changes in these coordinates. The motions of difEerent points may therefore be com- pared by comparing the changes in their coordinates that * If the displacement does not take place in one plane the point Is similarly referred to three coordinate axes in space, but we shall not at present enter into a discussion of this, 10 MOTION. take place in the same time. To render this possible, it is necessary to fix on units of angle, time, and distance. The unit of angle is the Degree, or the ninetieth part of a right angle. The unit of time is the Second. The unit of distance is the Foot or Meter, according as we follow the British or the metric system. We shall give both systems, as it can be done without confusion, and both are in use in the application of mechanical principles. Thus the ciyil engineer uses the British system of the foot and its sub- divisions, the physicist the metric system, and the electrical engineer both systems. For convenience of statement distances measured in one direction along a line are considered positive, and distances measured in the opposite direction negative. The direc- tions are arbitrary, but it is usual to take distances to the right +, to the left — ; distance upwards -|- , down- wards — . 6. The successive positions occupied by the point in its motion trace its Path. This path may be either a straight line or a curve of some kind. As most simple, we shall first of all consider motion in a straight line (rectilinear motion). MOTION IN A STBAIGFHT LISTS. Motion in a straight path may be uniform, or it may varj from time to time. Thus an engine on a straight track may run for a time with the same speed, or it may vary its speed. 7. Velocity. — The rate at which a moving point changes its position is called its velocity. If the rate of change is constant, that is, if equal distances are passed over in equal times, the velocity is said to be uniform or constant. Thus if the point moves uniformly over a distance s in a time t its velocity v ig s/t. This may be abbreviated into V = s/t. VELOCITY. 11 if we place s = I, t = 1, y/e &nd v = 1; or, the unii of veloc- ity is the distance described in unit time. The measure of avelocity is therefore expressed as so many feet (meters) per second. Clearing of fractions, we may write s = vi, or the number of units of distance is equal to the product of the number of units of velocity by the number of units of time. As the point may move in the direction of positive dis- tance or in the opposite direction, it is necessary to regard the sign of the velocity. It is agreed that the same rule of signs shall apply to velocity as to distance. The velocity is -(- if the direction of motion is in the + direction, and — if in the opposite direction. The term speed is sometimes used to denote the magni- tude only of a velocity. Ex. 1. A point moves uniformly with a velocity of 2 meters per second. In what time will it pass over a kilo- meter? Ans. 8 min. 30 sec. 2. A passenger sitting in a railroad car counts 44 tele- graph poles (distant 100 ft.) passed in one minute: show that the train is running at 50 miles an hour. 3. A man a ft. in height walks along a level street at the rate of c miles an hour in a straight line from an electric light b ft. in height: find the velocity of the end of his shadow. Ans. bc/{b — a) miles per hour. 4. The "limited" trains on the N. Y. 0. & H. E. K. E. running on parallel tracks pass at a certain place in 6^ seconds. If each train has the same velocity, and consists of 8 Pullman coaches of 53 ft. 9 in. length, find the rate per hour. Ans. 46 miles per hour. 5. Show that the velocity of a point on the equator aris- ing from the earth's rotation on its axis is about 463 meters per second. 13 MOTION. 8. Curve of Velocity. — A velocity,' being distance per second, may be appropriately represented by a straight line whose length will show the magnitude of the velocity on any assigned scale, and whose direction as indicated by an arrow will show the direction of the motion. We may hence give a geometrical representation of constant velocity. Along a straight line OX (Fig. 3) lay off equal distances OA, AB, ... to any convenient scale (as 10 sec. = 1 in.) for as many seconds as the Fig. 3 vel.bwrve ° '* " *" '^ " motion has continued. Let the velocities at the points 0, A, B, . . . \yQ represented by OY, Aa, ... to any scale (as 10 ft. = 1 in.). Now since the velocity is constant, the lines OY, Aa, . . . are equal to one another, and the curve of velocity Yd is a straight line parallel to OX, the time line. Also Dist. passed over = OF X OD = area YD; that is, the number of feet (meters) in the distance de- scribed would be represented by the number of square feet (square meters) in the area of the rectangle YD. Ex. Plot a velocity of 60 miles an hour on a scale of 11 inches = 1 ft. per second, and also the distance passed over in one minute. [For plotting it is convenient to use cross-section paper.] 9. If a point in motion does not move over equal dis- tances in equal intervals of time, the motion is said to be variable. Variable motion is thus not uniform in every part of its path. The velocities Oy, Aa, Bb, ... at the times 0, OA, OB, ... if plotted and their extremities y, a,b . . . joined would form a line yab . . , not parallel VELOCITY. 13 to the time line OX, as in the case of uniform motion (Fig. 4). This line would be the velocity curte. The points A,B, . . . may be conceived to be taken so close together that the velocities between may be considered uniform and the figures Oa, Al . . . formed to be rectangles. Each rectangle represents a distance passed over, and therefore the total distance would be represented by the sum of these rectangles, that is, by the area 01. Conceive now a point to start from 0, and moving with uniform velocity Oz, to describe the same distance s in the time t as the point moving with the variable velocity. The distance would be represented by the rectangle OM. But with the variable velocities it is represented by the figure 01. Hence the rectangle OM is equal to the figure 01, which can happen only when Oz is the average (mean) of the values Oy, Aa, Bi, . . . Hence, if we can find the average velocity v, or the velocity of a point which, moving uni- formly, passes over the same distance in the same time t as the point moving with variable velocity, we can find the distance s described, from the equation s — vt. 10. A velocity curve may also be constructed by laying off OA, OB, . . . along the line OX to represent the distances passed over, and Aa, Bb, ... at right angles to OX to represent the corresponding veloci- (jfetonce ties. The line through the points a,i, . . . would represent the curve of velocities. A familiar illustration is afforded by the motion of the 14 MOTION. piston of a steam-engine. At the beginning and end of its stroke the velocity is zero, at the middle of the stroke the velocity is greatest, and it varies from this value to the end values. The curve of velocity, if plotted, is found to be of a form such as Oabc ... i in Fig. 5. 11. In uniform motion the velocity v at any instant is equal to s/t. In variable motion the corresponding relation would be ds/dt, since the velocity v may be assumed to be uniform for an indefinitely small distance ds, and during an indefinitely smaU time dt. Hence ds = vdt. The total distance passed over in the time t is found by summing the distances ds, that is, by the integration (— summation) of the expression vdt ; or =/ t vdt. which includes all possible cases. If, for example, the velocity v is constant, then s=-vt -\- c when c is the constant of integration. If we suppose that the point starts from rest (or s = when ^ ;= 0) we have s = vt, as found" above. Ex. 1. A velocity of 60 miles an hour is 88 ft. per sec. 3. A bicycles rider makes m miles in h hours : find his average velocity. Ans. 22m/15h ft. per sec. 3. A train runs 29 miles for 2 hours, 30 miles for 3 hours, and 32 miles for 1 hour : find its average velocity. J ns. 30 m. an hour. 4. The velocity of a particle in a rectilinear path varies ACCELERATION. 15 as it"! distance from the starting-point (= c/i): find its posi- tion at the end of a time t. Ans. s — e"^, where e is the base of the natural system of logarithms. 12. Acceleration. — In variable motion the rate of change of velocity is called the acceleration. If the velocity changes uniformly with the time, the motion is said to be uniformly accelerated. The measure of an acceleration is the change of velocity in unit time. Thus, if a point moves with a velocity of 1-ft. per sec, 3 ft. per sec, 5 ft. per sec, etc., in consecutive seconds, the acceleration is 2 ft. per sec. in every second of the motion. Hence acceleration is distance- per-second per second, the unit being 1 foot (or 1 meter) per-second per second. This method of expressing acceleration, though strictly correct, is cumbersome. If it is understood that both velocity and rate of change of velocity are referred to the same unit of time, it will be sufficient if we express accel- eration as distance per second. This is the more ordinary method, and with the above understanding can cause no confusion.* If the velocities of the point had been 5 ft. per sec, 3 ft. per sec, . . . the change of velocity is numerically the same as before, but in the opposite sense to that of the original velocity. The nature of an acceleration may therefore be indicated by the signs + and — , as the change of velocity is in the same sense or in the opposite sense to that of the original velocity. Hence an acceleration is + if the velocity increases algebraically, and — if it decreases algebraically. To a negative acceleration the name retardation is some- times given. 13. Just as rate of change of position or velocity is uni- form or variable, so may rate of change of velocity or accel- * A single word for unit-acceleration is needed, and one also for Ifnit-velocity. 16 MOTION. eration be uniform or Tariable. In uniformly accelerated motion, tlie rate of change of velocity being constant, if u denotes the initial velocity, and v the final velocity, the change in the whole interval of time t being v — u, the change per second or the acceleration a is (« — u)/t, and therefore V = u -\- at (1) Also, since the rate of change of velocity is constant from beginning to end, the average velocity occurs half-way ; that is, average vel. = initial vel. -|- | change of vel. = M + ^{v — u) = Uu + v). Hence the distance s passed over is found from s = i{u + v) xt = ut + ^af, from eq. (1). (2) The two equations (1) and (2) contain relations between the quantities involved which are independent of one an- other. Other relations may be deduced from them which are convenient, but which contain no new principle. Thus, eliminating t, we find w" = m' -\- 2as, a useful formula. If the point had started from rest, then u = 0, and the equations become V = at, v' = 2(cs, s = lat\ s — Ivt, ACCELEEATIOK. 17 14. If the acceleration is yariable, the j-elation corre- sponding to a = (w — u)/t would be a = dv/dt, since the acceleration maybe assumed to be uniform during an indefin- itely small change of velocity dv in the indefinitely small time dt. Putting v — ds/dt, we have a = d's/df. The formulas found above follow at once from this equar tion. Thus, let a particle start from a point with a velocity u: it is required to find its velocity v and distance s from at the end of a time t, the acceleration being constant. Here d^s/dt^ = a. Integrating, ds/dt = at -\- c, c being the constant of integration. But when ^ = 0, ds/dt or V = M ; hence V = ds/dt = at -\- u, which gives the velocity at the end of the time t. Integrating a second time, s = -^at' + ^^3 since when t = 0, s = 0, and therefore the constant of inte- gration is 0. 15. Curve of Acceleration. — We have seen how to con- struct the curve of velocity of a point in motion, either by taking the times as abscissas or the distances as abscissas. Both are convenient at times. We proceed now to show 18 Uoiio'S. time how to construct the curve of acceleration when the curv6 of velocity has been plotted. First take the case of a motion in which the times are plotted as abscissas. Let the velocity change uniformly from M to V in the time t. The rate of change of velocity being constant, the velocity curve «& . . . of Pig. 4 be- comes a straight line, as in Fig. 6 Fig. 6. The acceleration a — (w—m)/< is represented in the figure by tan 0, that is, by the tangent of incli- nation of the line «J . . . to JT. Hence if the distances GA, AB, . . . represent one second, the accelera- tions measured on the velocity scale would be represented from second to second by ffrt,, 5S,, ... all of which are equal to one another. If, there- fore, a line be drawn parallel to OX, and at a distance a from it measured on the velocity scale, it will represent the curve of^acceleration. If the rate of velocity is not constant, so that the curve of velocity is curvilinear, then since a — dv/dt, the accel- eration at a point P (whose co- ordinates are v, t) is represented by tan 6 when 6 is the inclina- tion of the tangent at P to OX. Hence, to plot the acceleration curve, draw tangents to the velocity curve from second to second, and lay off as the ordi- nates the rise or fall of the tan- gent measured on the velocity scale, be plotted from point to point. time The curve may thus eUEVE of ACOiELEftAttOlf. 19 Next consider the case in which the distances are plotted as abscissas: for exam- ple, the curve of accel- eration of the piston of a steam-engine. From any point P in the velocity curve (Art. 10), let fall FM perpendicu- lar to OX, and draw PQ at right angles to the tangent at P. Then a = dv/di = ~^X^ = ta.nd xPM= MQ. Prolong PM to A, so that MA — MQ, and ^ is a point on the required curve. Similarly other points may be found. Hence the curve of acceleration is the line AB. Comparing the curves of velocity and acceleration in this case, we see that when the velocity is zero the acceleration is a maximum. As the velocity increases the acceleration decreases; when the velocity is greatest the acceleration is zero, and at that point it changes sign. Ex. 1. If the acceleration is constant, prove that the velocity curve is a parabola. 3. If the acceleration increases uniformly, prove that the velocity curve is a straight line. 16. The relations found for the motion of a particle hold also for the motion of a body, provided its motion is a motion of translation ; that is, if all the particles of the body describe paths that are precisely alike. We therefore determine the translation of a body by determining the translation of a single particle of the body. The kine- matics of body translation is therefore the kinematics of a point. Ex. 1. In 5 seconds the velocity of a point changes from 200 ft. to 100 ft. per sec. Find the acceleration. Ans. a = — 20. 20 MOTION. 2. The velocity of a point changes from 20 ft. to 10 ft. per second in passing over 75 ft. Find the acceleration and time of motion. Ans. a — — 2, t — 5 sec. Draw a figure illustrating the motion. 3. A point starts from rest. Show that the accel. if const, is equal to twice the distance described in the first second. 4. A point describes 160 ft. in the first two seconds of its motion, and 50 ft. in the next second. When will it come to rest ? When has it a velocity of 20 ft. per second ? When of — 30 ft. per second ? Ans. 5 sec; 4 sec. ; 6 sec. 5. A point starts with a velocity u and under a constant acceleration — a. Show that it will come to rest in u/a sec, after describing a distance u'/2a. 6. In the Westinghouse air-brake trials (1887) on the P. E. R., a train of 50 freight cars running at 36 miles an hour was stopped in 593.5 ft. Find the acceleration of the brake. Ans. « = — 2.3 ft. per sec. 7. A point starts from a position A with a velocity u ; to find its velocity and its distance from A at the end of a time t, the acceleration being proportional to the time (= ct) and in the direction of the velocity u. Ans. V — u-\- ^cf, s = ut -\- \ct^. 17. Composition of Motions. — Thus far we have consid- ered the case of a point which has received a displacement in a single direction. But a point may receive several dis- placements at the same tirne either in the same direction or in different directions. As only one single path results, these displacements must combine into a single displacement to which the path is due. To this single displacement the name of Eesultant Displacement is given, and to the separate displacements the name of Components. Thus, conceive a point P in motion along a straight line AB (as a ring sliding along a wire), and that at the same time the line is also a p ' moved. If the displacement of the ' line is in the same direction as that of the point, the point COMPOSITION OF VELOCITIES. 31 receives two simultaneous flisplacements or a single total displacement equal to their sum, if in the opposite direction to their / ^'^- '" / difference. / / If the line is moved in the direc- c/- ?l tion A C while the point is moved / / along A B, the point has two simul- / / taneous motions — one along AB i b and one along A 0. The law of the displacements being known, the position of the point may be found at the end of any assigned time. For the motion along AB alone would carry it a distance AQm this time. But the motion along ^Chas carried the line AB io the parallel position CD, and P will therefore be the position of the point at the end of the motion. Hence the final position P is the opposite angular point to the initial position A of the parallelogram CQ, constructed on the lines AQ, AC, representing the distances due to the single motions. This proposition is called the Parallelogram of Displacements. 18. Now velocities and accelerations being quantities having direction and magnitude, and capable of being represented by finite straight lines, may be treated in a manner analogous to displacements. This we proceed to do. (a) Composition of Constant Veloci- ties. — Suppose a point has two simul- taneous constant velocities m, v in the directions OX, OT not in the same line. At the end of the first second the velocity along OX, if acting alone, would carry it to a, , where Oa^ = u; the velocity along OY, if acting alone, would carry it to c,, where Oc = v; when both act, it arrives at d^, the opposite 22 MOTION. angle of the parallelogram Od^ to 0. Similarly, at the end of the second second it arrives at d.^ , where Oa, = 2u, Oc^ = 2«; and so on. The path is thus some line passing through 0, d^, d^, . . . To find its form we notice that c,^,- = 2c, ^'^ which itself moves with a velocity V, but in the opposite direction to B, and represented by OZ. The V*' point B is now at rest. The ve- locity of A is the resultant of the velocities OZ, OX, that is, is equal to the diagonal W, which therefore represents the velocity of A relative to B. The three velocities are represented by the sides of the triangle OXW, the directions being indicated by arrows. Hence, if in a triangle one side OJT represent the velocity of A, XW a velocity equal to and opposite that of B, and OX, XWa,re in the same sense around the triangle, the third side W taken in the opposite sense around the triangle will represent the velocity of A relative to B. Conversely, if the absolute velocity of A and the relative velocity of ^ to B were given, we should have from the same triangle XW to represent the absolute velocity of B, but in the contrary sense. Any mechanism may be employed to illustrate relative motion by putting a sheet of paper on one of its moving pieces, and a pencil on another moving piece, when the KELATIVE MOTION. 39 curve traced by the pencil on the paper will represent the relative motion of the two pieces. Ex. 1. Two vessels start at the same time from the same harbor, one sailing east at 13 miles an hour, the other south at 9 miles an hour : find the velocity of one relative to the other. Ans. 15 m. an hour. 2. Two railroad tracks intersect at 60°, and two trains start at the same instant from the junction at 30 miles an hour each : find their relative velocity in magnitude and direction. 3. Two bodies A, B move with velocities u, v inclined at an angle 6: show that the velocity of B with respect to A is V u''-\-v'' — %uv cos d, and inclined at an angle tan ~ ' V sin 6 / {v cos ^ — m) to the direction of A. 4. Two railroad tracks intersect at 90°. To a passenger in one train travelling at the rate of 32 miles an hour the other seems to have a velocity of 40 miles an hour: find its absolute velocity. A ns. 24 miles an hour. 5. A boat is propelled at 13 miles an hour across a stream flowing at , 5 miles an hour, in a direction perpen- dicular to the current: find the velocity of the boat with reference to the bottom of the channel. Ans. 13 miles an hour, up stream. 6. A man travelling eastward in a wind apparently from the north, doubles his speed when the wind appears to blow from the northeast. Show that the wind is really southeast, and blowing with a velocity of 4 1^3 miles an hour. CHAPTEE II. FORCE AND MOTION. 26. Common experience shows that to put a body in mo- tion or to stop it if in motion requires a certain muscular efEort (push or pull) or some equivalent. To this effort or its equivalent we give, as already stated, the name /orce. Hitherto we have considered motionapart from the body moved and apart also from the force acting. This was an ideal case, and gave us the geometrical side of the question (Kinematics). In an actual case we must consider the motion with reference to the body moved and to the force acting as well. This gives us the physical side, and is known as Dynamics. In doing this and to lay the foundation for an extension of the kinematical treatment of motion to dynam- ical questions, we are compelled to call in the aid of ex- periment. The science of dynamics rests upon three principles or laws, known as Newton's laws of motion.* Before stating them, certain rude experiments will be indicated whicli are sufficient to suggest the laws but not to establish their truth. 'No direct proof is possible. The proof is indirect, and is made in this way. Assume the laws true, and cer- tain consequences follow which can be tested experiment- ally. This has been done in so many ways and by so many * " . . . Though Newton's laws of motion are a much clearer and more general statement of the grounds of Mechanics than had ap- peared before his time, they do not Involve any doctrines which had not been previously stated or taken for granted by other mathema- ticians. " — Whewell, 30 STRESS. 31 independent observers, particularly in astronomical work, that we are justified in accepting them as true. So com- plete have been the tests and so firm is the conviction of the triith of the laws, that results deduced from them which cannot be verified in any way are accepted without hesitar tion. 27. Stress. — In order to exert force the agent acting must meet a resistance. Thus the hand in motion does not exert force until it meets some object. The object reacts on the hand. Press the table and the table will press the hand. Force is always a mutual action : in other words, forces are never single, but act in pairs — one the action and the other the reaction. This pair of actions is known as a Stress. If it is of the nature of a push, as in cracking a nut, the name compression (or pressure) is given to it ; if of the na- ture of a pull, as in breaking a string, the name tension is given. A stress can never occur except between two bodies or two parts of the same body, and is always exerted over the surfaces coming in contact. If we divide the total stress by the area of this surface we obtain the stress per unit area, as per sq. in., for instance. Thus we are accustomed to speak of the stress of steam in a boiler as pressure per sq. in. Or we may consider stress without reference to area, looking to the total magnitude only, and consider it acting at a point. In some cases the relation between the action of the agent and the reaction of the resistance is sufficiently evident. Thus if one body rests upon another, it will be granted that the pressure exerted by the upper is equal to the counter- pressure exerted by the lower : if a horse hauls a canal-boat to which he is attached by a rope, the pull of the rope on the horse is equal to its pull on the boat, and so on. But when a stone falls from a height it is not evident whether the action of the earth oi; the stone is equal to the action of 32 FOBCB AND MOTION. stone on the earth. Nor is the relation evident between the actions of a magnet and a piece of iron, nor between bodies ■widely separated, as the earth and moon. The results of experiments direct and indirect on the mutual actions of bodies are summed up in what is known as the law of stress : To every action there exists always a reaction equal in magnitude and opposite in direction ; or as it may be expressed : The mutual actions of two todies are always equal, and act in opposite directions. 28. Force being thus always double may be looked at from the point of view of the body acted on or of the body (as agent) exerting the action. The two components being equal and opposite, either will suffice for a numerical meas- ure of the magnitude of the force. Suppose a body at rest on a level floor. If let alone it will remain at rest. If a push is given to it (= force ap- plied) so as to cause it to move along the floor, it will come to rest after going a short distance. If the floor is waxed, it will go a greater distance for the same push. The smoother the surface the farther it goes, and the more nearly in a straight line. If the floor were perfectly smooth, we can conceive of no reason why the body should not con- tinue to move in a straight line forever. Now our minds are so constituted that we cannot con- ceive of a change occurring without a cause. On the rough floor the change of motion is accounted for by the action between the floor and the body — an action outside the body. But on the smooth floor this outside action is removed. StiU, in this case, to make the body change its velocity or change the direction of motion, some outside action is found- to be necessary. "We say then that the body has within itself no power of making any change in its state, either of motion or of rest. To this property of inability of a body to change its state we give the name Inertia.* It * The term inertia or vis inertix was introduced by Kepler (1571- 1630) STEESS. 33 is to be regarded as an inherent or characteristic property of matter. The law of inertia was enunciated by Newton as follows : Every lody continues in its state of rest or of uniform motion in a straight line except in so far as it is compelled hy impressed forces to change that state. It thus appears that force causes not merely change of place in the body moved, but change of velocity as well. We may therefore extend the deiinition from muscular effort to this: Force is whatever causes deviation from uni- formity or rectilinearity of motion in the lody acted on. 29. The force exerted by an agent on a body being one component of the stress, the other component, the reaction of the body or the " kick against change of motion," de- pends on the inertia of the body, and may be called inertia- resistance. Being equal to the action of the agent, if we can measure it we have a measure of the stress, which term includes both components. We have many familiar illustrations of inertia-resistance. Thus, on Jumping from a train in motion, on reaching the ground we are hurled forward. In jumping on a train in motion, a jerk is received. So a sudden change of velocity will make itself known to the passengers by a thrust or jerk. The intensity of the thrust or jerk depends on the difference of velocity of train and passenger, thus showing that the inertia-resistance called into play is proportional to the instantaneous change of velocity or the acceleration a communicated. We may therefore write inertia-resistance = ma when wi is a constant. This constant may be called the coefficient of inertia. 30. The comparison of forces is thus reduced to a com- parison of distances and times, and by assuming a unit of force all forces may be expressed in terms of this unit so long as we keep to the same body. 34 FOKCB AND MOTIOW. -But in general the question is not regarding the comparison of forces acting on the same body, but of forces acting on different bodies. Suppose two passengers to jump on the same train, the larger will receive the greater jerk. We explain this by saying that the two bodies are of difEerent Mass. "We may therefore compare the masses of bodies by communicating the same acceleration to each body when the ratio of the masses will be the ratio of the coefficients of inertia. A rude measurement would be af- forded by having each passenger grasp at the instant of jumping on the train. a spring-balance: the ratio of the pulls would indicate the ratio of the masses. We may observe the effects of the same force on blocks of the same substance of different sizes. It is found that the larger the block the smaller the acceleration imparted by the force, that is, the greater the mass. The mass would in this case depend on the size of the block, or, as is some- times said, the quantity of matter in it. And when we come to blocks of different materials we agree conven- tionally to ascribe the greater mass to the body to which the smaller acceleration is imparted. If the same force gives the same acceleration to two different bodies, we say conven- tionally that they are of the same mass ; and if the accelerar tion given to one is n times that given to the other, we say that the mass of the second is n times that of the first. Hence, the masses of bodies are inversely proportional to the accelerations imparted to them iy the same force. 31. Unit of Mass. — It follows that masses may be com- pared by a measurement of accelerations. The operation may be called massing. It enables us to express all masses in terms of some one standard mass, and so make a quantitative measurement of mass. This one mass we may choose for unit-mass, and we may make it what we please. As in the. case of the other fundamental units already assumed, the units of distance and time, all that is neces- MASS. 35 sary is that having once chosen it we must be consistent in its use. The units employed are the British unit, which is a certain lump of platinum called a Pound (lb.), and the metric unit, which is also a lump of platinum called a Kilogram (kg.). The masses of all bodies may now be ex- pressed in pounds or kilograms by comparing the accelera- tions due to the action of the same force on the unit mass and on the bodies in question. Multiples or submultiples of the standard units, such as the ton, ounce, gram, etc., are often employed as being more convenient in certain cases than the standard units themselves. 32. Unit of Force. — We have seen that forces may be com- pared by comparing the accelerations produced in the same mass. This enables us to express all forces in terms of a unit force. The units of acceleration and mass being already defined, the unit force may be conveniently ex- pressed in terms of them as the force producing unit- acceleration in unit mass. If we take unit acceleration to be one ft. per sec, and unit mass to be one lb., the name Poundal is given to unit force ; if unit acceleration be one cm. per sec. and unit mass be one gram, the name Dyne is given to it. 33. We may find a numerical expression for any force in terms of the unit force, say the poundal. For by our definition, to impart to a mass of one lb. an acceleration of one ft. per sec. requires a force of one poundal, and the preceding makes it reasonable to suppose that to impart to a mass m an acceleration a would require a force of ma poundals. Denoting the force by F, we may write F = ma, which means that the force which produces an acceleration a ft. per sec. in a mass of m lb. is ma poundals. Now the force F and the inertia-resistance of the body 36 FORCE AND MOTION being opposite aspects of the same stress, it follows from Art. 30 that with this system of units the coefacient of inertia is equal to the number of units of mass contained in the body, and the inertia-resistance is equal to the mass- acceleration ma of the body. If now u is the velocity of a body of mass m at the be- ginning of time t and v the velocity at the end of this time under the action of a constant force F, the acceleration a is equal to {v — u)/t (Art. 13). Hence, substituting for a in ^= ma, we have Ft = mv — mu. The product Ft is called the impulse of the force during the time t, and the product mv or mu is called the mo- mentu7n of the mass m in the direction of the velocity v or w. Hence mv — mu is the change of momentum due to the impulse. We may now appreciate more clearly the meaning of New- ton's second law of motion, which states that : The change of momentum of a body is numerically equal to the impulse which produces it, and is in the same direction. 34. From the law we have the relation Ft = mv — mu, as its statement in symbolic form. Writing this in the form F = {mv — 7?iu)/t, we see that force may be defined as rate of change of momentum. Putting the rate of change of velocity {v — u)/t = a, the acceleration, we have F= ma, a convenient form, and which is called the general equation of motion. MASS. 37 Whether the acceleration be uniform or variable, we may write (Art. 14) a = cfs/dt^, and hence F = md's/df, which is also called the general equation of motion. The general equation of motion, which is the algebraic statement of the second law of motion, is the connecting link between motion and force. It enables us to pass from the kinematical properties of motion already laid down to questions involving force and mass. It is the link between the ideal and the actual, the geometrical and the physical. 35. Prom this law we infer, too, that since change of momentum per second is proportional to the magnitude only of the force acting, this change is the same whether the body is at rest or in motion. Hence, too, for the same body force and acceleration are simultaneously constant, being connected by the relation i?'= ma; or, as it maybe stated, a constant force constitutes constant acceleration to the body acted on by it. A variable force maybe considered to consist of a succes- sion of constant forces varying in magnitude and direction, and acting for indefinitely small intervals. The accelera- tions contributed may be considered uniform during these intervals, and the total acceleration in any time found by summation. The law also implies that when two or more forc3S act on a body at the same time, each foi'ce produces an accel- eration in its own direction without reference to the others. It therefore follows thft forces may be combined by the rules already laid down for the combination of accelerations (Art. 18). Ex. 1. Explain why by striking the handle of a hammer against a wall the head may be fixed on firmly. 38 FORCE AND MOTION. 2. A man stumbling can save himself more easily on land than on smooth ice. Explain. 3. In suburban-passenger traffic the trains must stop and start quickly. The boiler and machinery are placed over the driving-wheels. Why ? 4. Show that it necessarily follows from the second law of motion that forces can be represented by straight lines. 5. What are the tests of the equality (1) of two forces, (2) of two masses ? 6. State the parallelogram of momenta. 7. A man with a hod on his shoulder falls off a ladder : find the pressure on His shoulder during the fall. 36. Gravitation Measure of Force. — The unit of mass be- ing assumed, we have seen how all masses maybe expressed in terms of the unit by measuring the accelerations contrib- uted by equal forces. This is a strictly scientific method of measuring mass, and is sufficient. But though easily described in general terms, it is difficult of performance in practice. Accordingly we give another method more easily put in operation. It is a fact of common observation, that a body free to move falls towards the earth. It acts as if the earth at- tracted it. It is assumed as a convenient explanation of the observed phenomenon that there exists an attraction between the earth and the body, and to this attractive force the name /o?'ce of gravity is given. A body free to move if exposed to the action of the force of gravity is uniformly accelerated. Experiment* shows that at the same place it acts on all bodies in the same way; that is, the acceleration g produced by it has no relation to the magnitude of the bodies or to the material of which they are composed. • * Drop simultaneously pieces of lead, iron, paper, etc., etc., from a shelf in a glass vessel from which the air has been exhausted. All will be obsei-ved to strike the bottom of the vessel simultaneously. The only force acting is the force of gravity, and since all strike the bottom at once, they must have the same acceleration, as each hsis passed over fhe same distance in the same tiuie. GEAVITATIOK MEASURE OF FORCE. 39 Experiment shows, too, that the value of g is constant so long as we keep to the same place on the earth's surface. It varies, however, from place to place.* This is explained by the fact that the earth is not a perfect sphere, and is not homogeneous in structure. For latitude 45° at sea-level its value is 33.3 ft., or 9.81 meters nearly. This may be taken as an average value. 37. We may make a rough comparison of any force with the gravity force of a body in this way. Place the body on a spring-balance and note the compression of the spring. If the force produces the same compression of the spring we say it is equal to the gravity force of the body. The unit force is naturally assumed to be equal to the gravity force of the unit mass, one lb. or one gram. This force causes, as explained above, an acceleration g in one lb. But the force causing an acceleration g in one lb. is^ pound- als. Hence the unit force is equivalent to g poundals. As, however, it is convenient to have a distinct name, the unit is called a Pound, so that a force of one pound is equivalent to the attractive force between one pound mass and the earth, and is equal to g poundals. Hence we may convert poundals to pounds by dividing by the value of g at the place in question. 38. The use of the word pound in the double sense of mass and force is objectionable, but it is sanctioned by ordinary custom, and the context must decide in which sense it is used. To aid in making the distinction we shall use for mass the symbol lb., and for force the word jjowrarf.f The pound is called the gravitation unit of force, because it depends on the force of gravity, which is not constant in * "When Halley in 1677 went to the island of St. Helena to observe the stars of the southern hemisphere he found his clock lose so much that the screw at the bottom of the pendulum did not enable him to shorten it sufficiently." t Suggested by Supt. Mendenhall of the United States Coast anij Geodetic Survey. 40 FORCE AND MOTION. value over the earth's surface. The other unit of force, the poundal (or dyne), does not involve g, and is known as the absolute unit. The first being the older and more easily applied, is the unit of daily life ; the second being the more comprehensive is used in astronomical and electrical work, and in precise physical investigations in general. 39. Weight. — The gravity forces of two bodies of masses m, wi, being as ing poundals to m^g poundals or as m to m, are in the ratio of the masses. If the bodies are placed in the scale-pans of a common balance and the balance remains in the same position, the gravity force of each is the same, and the two bodies have equal mass, or, as we say in common language, have equal Weight.* The process is called weigh- ing, and hence, by assuming a body of standard mass, we may express all bodies in terms of this standard. We may there- fore, in comparing masses, substitute for the complicated process of massing (Art. 31) the simple operation of weigh- ing. In ordinary language the word weight is used in the double sense of mass and force. The original signification of the term was what we now call mass, and its extension to force was a later development. ''The word weight must be understood to mean the quantity of the thing as deter- mined by the process of weighing against standard weights." Thus in buying a barrel of flour we buy the mass that weighs 196 pounds. The builder and machinist find it necessary to use it in the other sense, as in computing the force necessary to support a load of given weight (= balancing the gravity force of a load), or in computing the stresses in a structure designed to support a given load. It is unfortunate that the term is ambiguous, but there is no help for it— any more than for the ambiguity of the term * On the effect of "centrifugal force" on the weight of a body see Art. 69, with the examples appended, particularly 9-Jl inclusive. NOTE ON UNITS. 41 pound. In fact the two go together — pound weight (mass) originally, and pound weight (force) secondarily. To prevent confusion, we shall express all dynamical formulas in terms of the absolute units. The passage to the gravitation units and the use of the terms weight, pound, etc., in the examples can give no trouble if the explanations given are kept in mind. The context will always make clear the sense intended. Ex. 1. How many poundals are equal to [the gravity force of] one ton? Ans. 2000 g poundals. 2. Show that one poundal is equivalent to i oz. 3. Would it be advantageous for a merchant to use a spring-balance for buying groceries in New York to sell in Cuba ? How would a pair of scales answer ? 40. Note on TJnits. — The three fundamental units in Mechanics are the units of time, distance, and mass. Being fundamental, they are arbitrary, and are chosen for con- venience, or as the result of circumstances. Their defin- itions have already been given ; but we shall here repeat, collect, and go into a little more detail. The unit of time is the Second. It is derived from ob- servations of the earth's rotation. The assumption of this unit therefore really amounts to making the motion of the earth on its axis the standard motion, and by means of the second all motions are tacitly compared with this standard. The standards of length are the Yard and the Meter. The yard is the distance between two lines on a certain bronze bar kept in London, England, when the bar is at the tem- perature 62° F. The foot is J of this distance, and the incJi ■^ of the same distance. The meter is the distance between the ends of a certain platinum bar kept in Paris, Prance, when at the temperature 0° 0. It was intended to be the ten-millionth part of the meridian distance between the equator and the pole, and is nearly equal to this distance, but not exactly. A meter is equal to 39.37 inches, or 3.28 ft. The centimeter is -j-o o °^ *^^^ meter. 42 SORCE AiTD MOTIOIf. The standards of mass are the Pound (lb.) and the Kilo- gram. The pound is a certain piece of platinum kept at London, England. The ounce is yV o* ^^^ pound. The kilogram is a certain piece of platinum kept at Paris, Prance. The ffram is j-J^^ of the kilogram, and was in- tended to be of the same mass (which it is very nearly) as a' cubic centimeter of water at 3°.9 0., the temperature of maximum density of water. The units used in any investigation are multiples or sub- divisions of the standard units as found most convenient. The system of units involving the centimeter, gram, and second, with the dyne as unit of force, is called the C. G.S. system. It forms a sort of international system, and is being largely adopted by physicists, astronomers, and electricians. The British (absolute) system of the foot, lb., and second, with the poundal as unit of force, is known as the F.F.S. system; and the British (gravitation) system of the foot, lb., and second, with the pound as unit of force, is the sys- tem of every-day life. In engineering work, the inch, ton, and minute are very often the units employed. The following table of relative magnitudes will be found convenient : 1 lb. = 453.59 grams. 1 inch = 3.54 centimeters. 1 foot = 30.48 centimeters. 1 mile = 1609.83 meters. 1 gram = 0.0022 lb. 1 centimeter = 0.3987 inch. 1 meter = 3.2809 feet. 1 gram = 15.482 grains. 41. Dimensions of Units. — All mechanical quantities are expressed in terms of some system of units. The two leading systems in use in this country, the F.P.S. and the O.G.S., have been explained. It is convenient to be able to pass rapidly from the one system to the other, or from one system to any other. Of course the quantity itself is quite inde- pendent of the unit employed to measure it,— just as the matter of this book is in no way affected by the size of the type used by the printer. NOTE ON UNITS. 43 The fundamental units are those of distance, time, and mass. Let L, T, M denote the magnitudes of these units. Then if we say a distance is I units in length, the complete symbol representing this would be IL. Usually it is written I only, the unit being tacitly assumed. All derived units may be expressed in terms of the fun- damental units. Thus unit velocity being the Telocity of a point which describes unit distance in unit time, we have unit vel. = unit dis. / unit time — L/T. Similarly, unit accel. = unit vel. / unit time = L/T'; unit force = unit mass X unit accel. = ML/T''; unit mom. = unit mass X unit vel. = ML/T; and unit impulse being measured by the number of units of momentum generated is of the same dimensions as unit momentum. Ex. ] . How many dynes in a poundal ? [1 poundal = ML/T' = lb. X ft./sec' - 453.59 warns X 30.48 cm./sec' = 13825.3 dynes.] 2. Show that the foot per sec. and mile per hour units of velocity are as 15 : 22. 3. Show that one mile per hour is 44.7 cm. per sec. 4. Reduce a velocity of 100 ft. per min. to cm. per sec. 5. How many dynes in [the force of] an imperial pound? Ans. 445,000 dynes, about = | megadyne. 6. Show that [the force of] 1 gram = 981 dynes. 7. Show that [the force of] 1 grain = 63.58 dynes. 8. Prove that a pressure of 1 pound per sq.- ft. is equiva- lent to 479 dynes per sq. cm. 9. Find the value of g if one minute is taken as the unit of tinie._ ^ Ans. 115,200 ft. 10. Find the unit of time if g is taken as unity, one ft. being the unit of length. Ans. 0.25 V^ sec. 11. How many dynes in the unij; of force if the meter, mmute, and kilogram are taken as the units of distance, time, and mass respectively ? '?* Ans. 100 X 1000 X 60' dynes. CHAPTER III. DYNAMICS OF A PARTICLE. 42. Having considered the geometrical properties of motion, and also the methods of measuring mass and force, we are ready to study the motion produced in a body of given mass by forces of given magnitude. When a body is acted on by forces, experience shows that various forms of motion may arise. If all of the component particles of the body move through equal distances in the same direction, the motion is said to be a motion of Translar tion. The motion of any particle would in this case give the motion of the body. In a motion of rotation the par- ticles do not move through equal distances in the same direction, those nearest the axis of rotation moving the shortest distance. If a body consisted of a single particle, it would, in its rotation about an axis passing through it, remain in the same position. We shall therefore exclude rotation, and be able to study the translation of a body if we consider the motion of a single particle only. We may con- ceive the whole body concentrated as it were into a single particle of equivalent mass. 43. Composition of Forces. — The number of forces acting on a particle may be one or more than one. If we can combine the separate forces into an equivalent single force, we can reduce all cases to that of the action of a single force. The method of combining forces will be our first step, and next we shall consider the motion of a particle under the action of a force. If several forces act on a particle at rest or in motion, the 44 COMPOSITIOlSr OF FORCES. 45 accelerations communicated are the same as if each force acted separately on the particle at rest. In other words, the acceleration produced by a force on a particle is inde- pendent of any motion it may have and independent of motions produced by other forces acting simultaneously. This is involved in the statement of the second law of mo- tion, but is repeated and expanded here for greater clear- ness. It is sometimes known as the principle of the independence of forces. 44. Representation of Force. — When a force acts on a particle its line of action must pass through the particle. The force itself contains a certain number of poundals or dynes. We may say that the elements of the force are three — the geometrical position of the particle acted on, the direction of the force, and the number of units it contains or its magnitude. It may therefore be represented by a straight line AB, the length ot AB representing the mag- nitude of the force, the direction from ^ to -B the direction of the force, and the point A the point of application. Each unit of length of A B will represent unit force. But the length of the unit is arbitrary. Hence we may plot forces to any scale we please, as 1 poundal = 1 inch, 10 poundals = 1 inch, etc. The direction of a force may be indicated by its sign. Thus a force of 2 units acting at A towards the right might be written -\- 2, an equal force in the p, ,9 opposite direction — 2. The choice - ' [ ^^ of signs is arbitrary, and it is only ^ a ^ necessary to remember that if one direction is assumed to be +3 the opposite direction must be — . In a diagram the direction is conveniently indicated by an arrow-head. Ex. On a scale of 100 pounds per in. what force would be represented by a line 20 in. long ? Ans. 2000 pounds. 45. Resultant of Two Forces. — Suppose that F^, F^ are two forces which act on a particle A of mass m. The 46 DYNAMICS OF A PARTICLE. Fifl. 21 accelerations «, , a, produced by these forces are (Art. 34) FJm, FJm respectively, and the resultant acceleration a in direction and magnitude is repre- Fig. 20 sented (Art. 18) by the diagonal of the parallelogram of which a, , a^ are adjacent sides. Hence, if R denotes the force which would produce the acceleration a, that is, if ij! = ma, it must represent in magnitude and direction a force which produces the same acceleration on A as F^ and F^ . To the force R the name Resultant Force is given, and conversely F^ , F, are the Components of R. We may represent this graphically. Let AJB, ^C repre- sent the forces P^ , F, in magnitude and direction (scale, say, 1 poundal = 1 in.); then AD the concurrent diagonal of the parallelogram con- structed on AB, AG as adjacent sides will represent in magnitude and direction the resultant R of the two forces on the same scale. This principle is called the Parallelogram of Forces.* Hence, instead of finding the accelerations due to the forces and combining them into a single acceleration, and thence finding the force which would produce this acceleration, we may combine the forces them- selves directly. Ex. 1. If the two forces act in the same straight line, find their resultant. Ans. F^ ± F,. Why the double sign ? 3. Explain the action of the forces by which an arrow is discharged from a bow. 3. Explain (by a drawing) the action of the forces by which a kite rises in the air. 4. Show by a drawing that the value of R decreases as the angle between the forces increases. * The parallelogram of forces was first formulated by Newton (1642^1737). COMPOSITIO^ir OF FOECES. 47 5. A satchel is carried by a strap slung over the shoulder : show that the longer the strap the less its tension. 46. Instead of finding the diagonal of the parallelogram by a geometrical construction, as just explained, we would obtain the same result if we computed its yalue from the known values of the sides AB, ^Cand the contained angle BAC {— 0). Thus from trigonom- etry we have in the triangle ABD [note AC^BD; BAG + ABD = 180°]. AD' = AB' + BD' -2AB. BD cos ABD = AB' + AC + 2AB .AC cos BAG, or R' = F^' + F," + 2F^F, cos d, which gives the magnitude of the resultant. The direction may be found by solving the triangle ABD to find the angle BAD. The position is known since the force acts at A. Hence R is completely determined. The special case of the forces acting in directions at right angles is important. The parallelogram becomes a rectan- gle, and from the figure R' = F," + F,', cos BAD = FJR, whence both the magnitude and direction of the resultant are determined. Ex. 1. If two equal forces F, i^are at right angles to on& another, then R =^ F V2. 2. If two equal forces F, Fare inclined at an aingle 20, then R = 2Fcos 0. 3. Prove that the value of R increases as the angle be- tween the forces diminishes, and vice versa. 4. When the angle BAC is 180° the forces are in the same straight line, and the formula, if correct, should re- duce to the sum or difference of the forces. Examine, and see if it does. 48 DYKAMICS OF A PAKTICLE. Fig. 23 5. Two equal forces Fact at an angle of 60° : find their resultant. 6. Find the resultant of two equal forces, each of 10 pounds, acting at an angle of 30°. Ans. 19.3 pounds. 47. Consider in Fig. 22 the manner in which the result- ant is formed. The forces F. , F^ drawn to scale are rep- resented by the lines AB, AC. From B the line BD is drawn parallel to A 0, and from G the line CD is drawn parallel to AB. The diagonal ^Z> of the parallelogram represents the resultant in magnitude, direction, and position. This construction is equivalent to that shown in Fig. 23. Plot the forces F^ , F^ as before. From B draw BD parallel and equal to A C. Join AD, which represents the resultant. Still better, by breaking the figure into two parts (Figs. 24,25). Let i^.,i?; be ^.^ ^4 the forces acting at 0. ' From any point A draw AB to scale equal and parallel to F^ . From B draw BD to scale equal and parallel to F^. Join AD, which will represent the resultant in magnitude and direction. To find its position: We know that it must pass through 0, and hence if through we draw a line equal and parallel to AD, we hare M in magnitude, direction, and position. We have therefore a Force Diagram (Fig. 24) and a Construction Diagram purely geometrical (Fig. 25). In Fig. 23 the two overlap, and in simple cases there is no confusion in their being so drawn; but in complicated cases it is better to keep them separate, as we shall see later. 48. Resultant of more than two Forces. — Let F^,F^,F,, F^ represent forces acting on a particle at : it is required to find their resultant. COMPOSITION OF FOKClIS. 49 Fig. 26 Following the method of Art. 47, from a point A We draw AB equal and parallel to F^ , 2? (7 equal and parallel to F^ , CD equal and par- . allel to P, , BE equal and parallel to F^ . Join A E, which will represent the resultant in magnitude and direction. In the force diagram draw R equal and parallel to AE, and we have the resultant in magnitude, direction, and position. . For join AC, AB. Then JC is the resultant of AB, Fig. 28 B BC, that is, of ^ 7^. . .r. :- xi_„ resultant of AC, CB, that is, of i^,, F^, F,; AEis the resultant of AB, BE, that is, of F,,F,,F„F,; which proves the proposition. We might have combined the two diagrams as shown in Fig. 28, or we might have derived the resultant directly from the paral- lelogram of forces as shown in Fig. 39. Notice that in any method we may take the forces in any order, and we shall always find the same Fig- 29 value of AE. Test this statement by making drawings to a large scale. Ex. 1. Three forces of 6, 8, 10 pounds act at angles of 130° with each other : find their resultant. Draw to scale by different methods, and compare results. compare. 3. Forces of 1, 3, 3, 4, 5, 6 act at angles of 60 Test as in Ex. 1. 3. Forces of 30, 30, 31 pounds act at a point. The angle between the first and second is 130° and between the second and third 30°: find R. Ans. 39 pounds. 4. Is it necessary that force and construction diagrams be drawn to the same scale ? Vary order, and find^. 50 DYNAMICS OF A PARTICLE. 5. In the construction diagram the lines are drawn paral- lel to the forces : would it be allowable to draw them per- pendicular to the forces, or inclined at any (the same) angle, for example ? Test by a drawing. 6. If the forces are in the same straight line, what does the force polygon become ? what the construction diagram ? 49. Resolution of Forces. — By means of the parallelogram of forces, a force R can be found equivalent to two forces p. „ F^, F,, acting on a particle A, Conversely, c _' — P the force R acting at A may be resolved J. / R/^/ into two component forces F^ , F^ acting at 7 ^^ I A, by constructing on R as diagonal a par- 1^^ I allelogram, and taking the sides to represent ^ '^' the components. The problem is similar to that already discussed in Art. 20. Ex. 1. If a force is resolved into two components, prove that the greater component always makes the smaller angle with the force. 3. Eesolve a force of 20 pounds into two components each of which makes an angle of 60° with it. Ans. Each = 20 pounds. 3. Eesolve a force of 10 pounds into two equal compo- nents, one of them making an angle of 45° with the force. Ans. 7 pounds, nearly. 4. Find that rectangular component of a force of 10 pounds which makes an angle of 60° with the force. Ans. 5 pounds. 5. Explain the boatmen's saying, that there is greater "power" in hauling a canal-boat with a long rope than with a short one. The values of the rectangular components X, Y of a force R may readily be computed analytically. Thus in the right-angled triangles ABD, c_ ''' d ACD, ■" '^ X=AB = R cos e, Y=AO= Raind = R cos 90° - 6, which give the values of the two compo- nents X and Y. Hence the rectangular component of s, RESOLUTION OF FORCES. 51 force ^ in a given direction is equal to i2 X cos (angle between component and H). As a check, X'+Y'^ R" cos' e-\-R'' sin» e = R% which is also evident from the figure. Ex. 1. Find the components of a force 10 when 6 = 60°, 90°, 120°, 180°, 240°, 270°,, 300°. Ans. 5, 5 V^ ; 0, 10 ; - 5, 5 V3; - 10, ; -5,-5 Vs ; 0, - 10; 5,-5 4/8. [Draw a figure for each case, and explain the sign of the result. ] 2. The rectangular components of a force are each equal top poundals : what is the force? 3. Show that the components of a force F in two direc- tions making angles /?, y with it are F sin ^/sin (/? + y), F&m y/sva. (yS 4- y). 4. In a direct-acting steam-engine the piston pressure P is equivalent to P tan 6 perpendicular to its line of action and P sec 6 along the connecting-rod, d being the angle of inclination of the connecting-rod to the line of action of the piston (see Pig. 119). 50. The analytical solution leads us to a method of com- iiiiing forces which is often more conven- / lent than the graphic method given in Art. 48. The two methods may be used to check one another. Take three forces F^, F^, F^, acting on a particle 0. Through draw any two lines OX, 01^ at right angles to each other, and let 6*,, (9,, 6^ denote the angles which ° ^-^"'^ * the directions of F^,F^, F, make with OX. The compo- nents of F, are F^ cos 6^ along OX; F^ sin 6^ along OY; F, are F, cos 0^ along OX; F^ sin 6', along OF; F, are F^ cos 0^ along OX; F, sin 0^ along OY. 53 DYNAMICS OF A PABTICLE. The components along OX being in the same straight line, may be combined by addition into a single force X (Art. 48) ; that is, F, cos B^ + F, cos (9, -\-F, cos d,= X. . . . (1) Similarly, the components along OY, being in the same straight line, may be combined into a single force Y, or F, sin e^ + F, sin d, + F, sin 6,-= Y. . . . (3) Hence the original forces are equivalent to two forces X, Y acting in directions OX, OT" at right angles to each other. The resultant of X, 1' must therefore be the resultant of the original forces. Call it B, and let B be the angle it makes with the axis of X ; then i? cos 6* = X, i2 sin = y, (3) Square and add (remembering that cos' B + sin' B = 1), and R = l/X" + Y', which gives the magnitude of the resultant. Divide the second of equations (3) by the first, and tan B = Y/X, which gives the direction of the resultant. Hence, since the resultant acts at 0, it is known in posi- tion, magnitude, and direction, and is completely determ- ined. If we equate the values of X, Yin equations 1, 2, 3, we find RcosB= F, cos (9, + F, cos (9, + F, cos B, ; It sin B=: F^ sin (9, + F, sin (9, + F, sin B, . Now ox, 01^ are any two rectangular axes. Hence the component in any direction of the resultant of a number of forces is equal to the sum of their components in the same direction. Ex. 1. Three forces of 6, 8, 10 pounds act on a particle at angles of 120° to each other : find the resultant in magni- tude and direction. RESOLUTION OF FOECBS. 53 /O '» ^5*- [Since the direction of OX is arbitrary, we may take it along any of the forces. Ftg. 33 (1) Take OX along the force 6. Then X=6-8cos60° -10 cos 60° = - 3; Y= 8 cos 30 ° - 10_cos 30° = - Vd ; , and B = VW+J = 3 i/3 ; tan (9 = — 4/3/ - 3 = 1/ 1/3 or 6/ = 310°. (3) Take OA" to fall along the force 8. Then X=8-6 cos60° -10cos60°=_0; / F = 10 sin 60° - 6 sin 60° = 2 f 3; E = 2V^, as before; tan 6/ = 3 Vs/O = 00 and 6 = 90°; or the resultant is perpendicular to the force 8, showing it to be in the same relative position with reference to the other forces as before.] 2. Solve with OX along the force 10. 51. Having now the means of combining the forces that act on a particle into a single force giving the same motion. Fig. 34 / we proceed to study this motion. R^ Conceive a particle acted on by a number of forces whose result- ant is R. If a force equal and. opposite to R be added, the whole system of forces acting on the particle will balance. The resultant of the forces is 7iil, and the system is said to be in Equilibrium. From the relation F= ma, it follows when F=0 that a — 0. Hence a system of forces in equilibrium implies that there is no acceleration. Thus the velocity of the body, if it had any before the forces commenced to act, would be unchanged, and the motion would continue uniform and in a straight line; if at rest, it would remain at rest. Equilibrium, therefore, does not imply rest, but rest implies equilibrium. The branch of dynamics which considers the circumstances for which equilibrium is possible is called Statics. When the forces do not balance, an acceleration arises from the resultant force, and the particle moves with a 54 DYNAMICS OB A PARTICLE. motion compounded of the motion in its original path, and that due to the resultant. The branch of dynamics which considers the circumstances under which change of motion takes place is called Kinetics. STATICS OF A PARTICLE. 52. When a particle is in equilibrium under forces acting in the same straight line, the total acceleration produced by the forces is nil, and therefore the sum of the acceler- ations in one direction is equal to the sum in the opposite direction. Hence the sum of the forces acting in one di- rection must be equal to the sum in the opposite direction. In other words, the forces must reduce to two forces equal in magnitude and opposite in direction. Next, let three forces not in the same- straight line act on 2 the particle. Find the Fig. 35 F^^\ resultant R of any two P g^^^^ R ^^^>^ -^1 J -^s • ^or equilibrium ' '^ *l\ '''^° *° e,iX^\., R and i? must ^s^ ^^^' be equal and opposite. ' ^' Hence if three forces J, acting on a particle are ^/"iv p in equilibrium, any one ^.^ I, X^ of them is equal and op- A F posite to the resultant of the other two. The sides of the construction triangle A CD are parallel to the three forces F^, F^, F,, and proportional to them in magnitude (Art. 47). Notice that their directions are the same way round the triangle. Hence three forces act- ing on a particle loill ie in equilibrium if they can be rep- resented by the three sides of a triangle drawn parallel to the forces, and taken the same way round. This proposi- tion is known as the Triangle of Forces. It may be expressed analytically. If a, /?, y be the POLYGON OI" FOBCES. 55 angles between the directions of the forces, then in the construction triangle the angles are evidently 180° — a, 180° — /3, 180° — y; and since the sides of a triangle are as the sides of the opposite angles, DC Ism (180° - a) = C^/sin (180° - /?) = AD/wa. (180° - y), or FJsm a = FJwa. /? = FJ&va. y; Fig. 36 that is, when three forces acting on a particle are in equi- librium, each is proportional to the sine of the angle between the directions of the other two forces. Illustration. — Take a piece of board, and drive in three smooth pegs A, B, C, or place three pulleys ?i,\> A,B, C. Eun strings over the pegs, and knot together as at 0. Suspend weights from the strings. Draw lines along the strings on the board, and plot the triangle abc with sides parallel to these lines. The sides of this tri- angle will be found to be pro- portional to the weights. Ex. 1. Make the weights 3, 4, 5 oz., and it will be found that one angle of the triangle abc will be 90°. 2. Could the three weights be equal to one another? Plot abc in this case, if possible. 53. Polygon of Forces. — If in the force diagram of Art. 48 the direction of B be reversed, the particle will be in equilibrium under the action ot F,, F^, F^, F^, — R. In- dicate the directions of these forces on the construction diagram, and notice that they are the same way round. Hence any number of forces in the same plane acting on a particle ivill be in equilibrium if they can be represented by 56 DYNAMICS OF A PAKTICLE. the sides of a polygon drawn parallel to the forces, and taken the same way round. This is known as the Polygon of Forces. Ex. In the polygon of forces any side represents in mag- nitude and direction the resultant of the remaining forces, but with sign reversed. 54. The analytical equivalent of the polygon of forces may be deduced from Art. 50. For if the forces acting at are in equilibrium, the resultant R must be equal to zero. Hence JP + Z' = 0, which, since X^ and F' are both positive, can only be satis- fied by X = 0, F = 0, that is, by i?;cos i9, +^, cos ^, + . . . = 0, F, sin ^, + i^^ sin ^, + . . .^ 0, Hence if any number of forces in the same plane acting on a ijarticle are in equilibrium, the sums of the compo- nents of the forces along any two straight lines at right angles to each other through the particle are equal to zero. Ex. 1. State the analytical conditions of equilibrium, when 2, Z, . . . n forces act on a par- ticle. 3. A rod AB whose weight may be neglected is hinged at A, and supports a weight W at B. It is held up by a string BG fastened to a fixed point -"^ vertically above ^. If ^5 is horizontal and angle ABO = 30°, find the tension jT, of the string, and the thrust T, along [The point B is in equilibrium under T^, T^, W. Re- solve vertically and horizontally, then T, cos 60" - Vf^=0, r, cos 30° - 2\ = 0; .-. T^ = 2W, T,= W VZ.] 3. In a canal with parallel banks, a boat is moored by two ropes attached to posts on the banks. If the ropes are in- Fig. 37 KINETICS OF A PARTICLE. 57 cliaed at angles of 30°, 60° to the banks, compare the pulls on them, both ropes being in the same horizontal plane. Ans. liVS. KINETICS OF A PABTICLB. 55. If a number of forces act on a particle and the re- sultant be found, a certain motion is due to this resultant. If the particle has this motion, it is said to he free; if it has some other motion,- the deviation must be owing to the entrance of some cause not accounted for, and the motion is said to be constrained. In free motion the particle is isolated from all causes tending to affect its motion except the acting forces, while in constrained motion this is not the case. "We have seen (Art. 5) that the position of a particle is defined by its coordinates with reference to certain axes assumed to be fixed. A change in position is represented by changes in these coordinates. Hence the coordinates being either a distance and two angles or three distances, a point is said to have three degrees of freedom to move. If the point is compelled to remain in an assigned plane (as the plane of the paper), its position is defined by two coordinates, and it is said to have two degrees of freedom and one degree of constraint. , Similarly, if compelled to remain at the same distance from a fixed point it would move on the surface of a sphere, and have two degrees of freedom and one of constraint. Again, if the point were compelled to remain in two planes, that is, in their line of intersection, it would have one degree of freedom ^nd two of constraint: so also if compelled to remam in one plane and keep at the same distance from a fixed point, that is, to move in a circular path. If compelled to remain in three planes, it can have only 58 DYNAMICS OF A PARTICLE. one position, their point of intersection, and is therefore wholly constrained. Ex. How many degrees of freedom has a curling-stone on smooth ice; a stone in a sling; a compass joined to a tripod by a ball-and-socket joint ? 56. Free Motion. — By means of the relation F = ma connecting force acting, mass acted on, and acceleration produced, we are able to extend the geometrical properties of motion to particles acted on by given forces. Various paths may result, depending on the motion of the particle at the time the force begins to act. We shall first of all consider the particle to be unconstrained, and have all de- grees of freedom. Suppose the particle to have an initial velocity u, and that the force F acts in the direction of this velocity, caus- ing an acceleration a. Then the resultant velocity at the end of a time t is composed of that due to u and that due to the acceleration «. Hence if v is the final velocity and s the distance passed over, we have (Art. 13) V = u -\- at, s = ut -\- ^af, = u + Ft/m, = ut + ^Ft'/m, in terms of the absolute units. Ex. 1. A mass of 10 lbs. is moved along a smooth table by a weight of 6 pounds attached to a string which passes over a smooth peg on the edge of the table: find the dis- tance passed over in 2 sec, and the velocity acquired. [Effective force = 6 pounds = 6^ poundals ; mass moved = 10 + 6= 16 lbs.; .-. a = 6f//16 = 12ft.;t) = 3 X 13 = 24 ft. per sec; s = J X 13 X 2" = 24 ft.]. 2. An ice-boat weighing 1000 lbs. is driven for 30 sec by a force of 100 pounds : find the velocity acquired and the distance passed over, supposing it starts (1) with a ve- locity of 10 ft. per sec, (2) from rest. 3. Find the tension P of the string in Ex. 1. [EfEective force on mass 10 = 10^ — P; .: accel, = (10^ - i')/10i FBBB MOTION. 59 Effective force on mass 6 = P — 6ff', .: accel. = {P -6g)/6. Hence P = 7^ pounds.] 4. An elevator weighing m lbs. is lifted by a force of n pounds: find the acceleration and tension of the lifting . m — n ^ 2mn chain. Ans. a = ; — a; F = j — g. 5. A bucket weighing 25 lbs. is let down into a well with a uniform velocity: find tension of rope. Ans. 25 pounds. 6. Two bodies of weights Wjand w^ pounds are Fig. 38 fastened to a string which passes over a smooth peg: find the acceleration and tension of the strmg. Ans. a = ^—; — -g ; rn = — + ~- [If the value of a is observed, we have the value of a from — —, all of the quanti- ■^ • w^ — w^ ^ ties in this expression being now known. By making the differences between w^ and w^ small, the acceleration a may be made as small as we please. The smaller it is the easier it is to ob- serve. In this consists the advantage of using two weights instead of a single weight falling freely to determine g. To find a we observe the distance s passed over in t sec. by means of a graduated scale placed vertically. Then, since s = ^af, we have a at once. For example, let iv^ — 21 oz., w, = 20 oz., and suppose that -in 5 sec. the weight w has fallen 9.8 ft. Then 1 Q fi 90-1-91 « = i|^ = 0.784, and bgl the particle will perform com- plete revolutions ; if u' < ^gl it will oscillate in an arc less than a semi-circumference; and if li' > 3^Z and< bgl it will cease to describe a circle and the motion become parabolic. 77. Centrifugal Pendulum. — Suppose a particle of mass m sus- pended by a string from a point and caused to swing about the vert- ical axis OA with a uniform velo- city V in a circular path. Such an arrangement is called a centrif- ugal pendulum. Let B be the position of the par- ticle at any time. Denote the angle between OB and the vertical OA by d. 84 DYNAMICS 01? A PAKTICLJE. The particle is acted on by two forces, the gravity force mg and the pull P in the string directed to the point 0. Since the resultant motion is the same as iu the case of a particle acted on by a centripetal force (directed to A), it follows that the resultant of the acting forces must be di- rected along the radius BA (= r) and form a centripetal force. Hence, there being no resultant vertical force, we have F cos d — mg — 0; centrip. force = P sin ^ = mg tan 6; and the acceleration a due to the centripetal force is thus g tan d. But since v is the velocity in the circular path, a = v'/r. Eliminating a, v'' = rg tan d, or ]iv^ = gf', if we put the height OA = h ; which gives the relation between li, v, r. The time T in which the particle makes a revolution, or the period, is given by T = circum. of path/velocity — 27[r/v = 2/7- VJi/g seconds. Also, we may determine A so that the number of revolu- tions per second may be any desired number, n for example. Then V = 27rr X n. But ■ V Vh = r Vg. Whence eliminating v, and putting g = 32.2, hi' = 0.815 feet, the relation required. CONSTBAINED MOTION. 85 Fig. 52 Ex. In going round a ring 100 ft. in diameter on a 50- in. bicycle a velocity of 10 ft. per sec. is attained. Find the distance of the highest point of the wheel from the vert- ical through the lowest point. Ans. 3 in. nearly. 78. The centrifugal pendulum may be used as a regu- lator of mechanical motion. An apparatus of this kind, known as the Governor, was applied by James Watt to the steam-engine.* In Fig. 52, which represents a Corliss-engine governor, as the speed of the engine increases the spindle, A revolves more quickly, and the balls separate ; as it diminishes, the balls come together. The slide rises and falls accordingly, and by means of a set of levers C the steam- valves of the engine are acted on, and the supply of steam ad- mitted to the cylinder regu- lated. Ex. 1. Explain clearly the pendulum motion in the gover- nor. 2. Does the Watt governor prevent increase of speed ? 3. Find the length of a Watt governor that will run 60 revolutions per miimte. Ans. 9.78 in. * "If a pan- of common fire-tongs suspended by a cord from the top be made to turn by the twisting or untwisting of the cord the legs will separate from each other with force proportioned to the speed of rotation. Mr. Watt adapted this fact most ingeniously to the regulation of the speed of his steam-engine." CHAPTER IV. STATICS OF A RIGID BODY. 79. Having studied the behavior of a particle under the action of forces, we proceed to study the behavior of a body of finite size, a body being regarded as a collection of par- ticles. The directions of the forces applied to a particle must necessarily all pass through the particle. In a body the directions need not all pass through one point. Besides, forces applied to a body may cause it to change its shape as well as to change its position. To exclude the former, we shall for the present assume that the body cannot be made to change its form or be distorted by the action of the forces applied. To such a body the name of Rigid Body is given. Though bodies differ more or less as regards •rigidity, we are not acquainted with one perfectly rigid; so that a result deduced on the hypothesis of a body being perfectly rigid can only be regarded as an approximation to the actual state of the case in practice. The hypothesis is made only for convenience of study, as it is simpler to dis- cuss the properties of bodies one at a time than to attempt to grasp all at one time. As in particle motion, the first step will be to combine the acting forces, all of which are supposed to lie in the same plane. 80. Composition of Forces. — Suppose forces F,, F,, F^, F, in one plane to act on a body at ''■'s- "/f, different points A, B, C, B. Each y/ cA ^/ particle acts on the particle next it, / /,^tff^' and is acted on by it in return. These — *-r5 '\ J internal forces forming actions and l a/ reactions occur always in pairs, and ^— ""^ V being equal in magnitude and opposite in sense, are themselves in equilibrium. COMPOSITION- OF FOKCES. 87 Hence we need only consider the external forces F^, F^, F^, F^ so far as the motion of the body is concerned. Prolong the directions of all the forces, and suppose these directions to meet in a common point 0. All of the forces may be conceived to act at this point. The motion will consequently be the same as if the whole body were concentrated into a single particle at and the resultant force R would be found graphically as in Art. 47 by plotting the polygon dbcde, whose closing side ae would be this resultant in magnitude and direction; or by analysis, as in Art. 49, by resolving the forces along two axes through In directions at right angles to each other, and making the sums of the components in each direction equal to zero. 81. If the directions of the forces do not all meet in a point, we can still find the resultant by Fig. 55 repeated applications of the parallelogram of forces. For the resultant of F^ at A and F^ at B is the resultant R^ of these forces acting at D; the resultant of R^ at D and F^ at C is the resultant R^ of R^ and .^3 at E. Hence R^ is the resultant of F^, F^, F^ acting at A, B, G respec- tively. This construction is often inconvenient, modification or rather combination of it and the preceding is more practical : Plot to scale the forces F^, F^, F^ in order, the line ab representing F^ , be representing F, , and cd representing F^ . The line ad closing the polygon will on the same scale represent the resultant R of the forces in magnitude and direction. ' To prove this, join ac. Then ac is the resultant of al, Ic, and ad is the resultant of ac, cd. The following 88 STATICS 01' A RIGID BODY. that is, of al, Ic, cd. Hence the resultant is determined in magnitude and direction. To find its position, that is, some point in its line of action. At any point p in the line of action of F^ apply any two equal forces i2, , i2, in opposite directions.' Let R, along qp be the resultant of F^ and R, ; R, along rq of F^ and R, ; and i?, of F, and R^ . Hence the two forces J?, , R^ are equiva- Fig. 56 lent to the three forces F^,F^,F^. The resultant RotR,, R^ passes through the point t, in which their directions inter- sect. Hence if through t a line equal and parallel to ad be drawn, it will represent the resultant B oi F^, F,, F, in magnitude, direction, and position. Again, since db represents the force F^ , if we draw rtO, 10 parallel to R^,R^, the sides of the triangle Oab will represent the forces R^,F^, R, acting at p. Similarly, the sides of the triangles Obc, Ocd will represent the forces at q, r. But /?, is any force. Hence Oa is any line, and the position of the point is arbitrary. The point is called the Pole. This gives us the key to the following rule for finding graphically the resultant of any number of forces acting in one plane on a rigid body. Construct a polygon ahcd to scale, whose sides are parallel to and in the same sense as the forces ; the closing side will represent the resultant in magnitude and direction. COMPOSITION 01' FORCES. 89 From any point 0, draw lines Oa, Ob, Oc, Od to the angu- lar points of the force polygon. From any point p in F^ draw pq parallel to Oh to meet F^ in q, and draw qr paral- lel to Oc. A line through t, the intersection of pt parallel to Oa, and rt parallel to Od, will give the resultant in posi- tion. Hence it is completely determined. 82. It is evident that a force equal and opposite to the resultant R would hold the forces F^, F^, F^ in equilib- rium. Hence forces in equilibrium in a plane may be rep- resented by the sides of a closed polygon ahcd, whose sides are parallel and in the same sense as the forces. The converse of this, that if forces acting in a plane can be represented by the sides of a closed polygon which are parallel to and in the same sense as the forces, they are in equilibrium, is not true. For the polygon would be the same, no matter what the position of the forces may be. This condition, in fact," provides against translation only. An additional condition to provide against rotation is neces- sary. (See Art. 88.) Ex. 1. Three forces P, Q, R are represented in direc- tion by the sides of an equilatei-al triangle taken the same way round: show that their resultant is V{P'+ Q'+ R'-PQ-QR- RP). 2. If R is the resultant of two forces P arid Q, a«d S the resultant of P and R, show that the resultant of Q and Sis 2R. 3. In a jib-crane a weight of 30 tons hangs at rest: iind the pull P of the chain AB it AC=2AB. Ans. P = 10 tons. 4. A square frame has a force 4 acting from Ato B, 5 from B to C, 6 from A to B, and 7 from Z? to 0: find the resultant in magni- tude and position. 5. An interestingapplication of the triangle of forces is afEorded by the suspension-bridge Fig. 57 90 STATICS O'e A RIGID BODY. with the roadway uniformly loaded, is represented in the figure. Fig. 58 One half of the bridge [Let AB represent the pier, AC the suspension cable, BB the roadway, and AB the anchorage cable. The roadway is suspended by rods from the cable, and the weight on the cable may therefore be assumed to be uniform per foot length, and its direction GH to- pass through H, the mid- dle point of BB. If I is the span and w the weight per unit of length, th e load acting at G is wl/2. The forces acting on AC are the weight wZ/ 2 and the tensions P^ at its ends, which act along the tangents at those points. The tangent CO at C is horizontal, and the direction G OH of wl/2 is vertical. Hence^ is the direction of i'. If. d is the deflection CL of the span, draw the triangle of forces, and show that Q = tvl'/Sd. "What is the value of P ? Also prove that the curve of the chain ^C is a parabola. (Take as origin, CL, CO axes; then equation to curve will be found to be 4:dx'' = Py, a parabola.)] 83. Parallel Forces. — The case of the forces being paral- lel is of special importance. Draw ab, be to scale, to rep- \ \, U A ^^ '"' c f) R' Fig. S9 X \ ->0 resent the parallel forces F^, F^: the line en closing the polygon (in this case a straight line) will represent a force equal and op- posite to the re- sultant. Hence the magnitude of the resultant is equal to the sum of tho forces. COMPOSITIOK 01? FORCES. 91 To find its position. Take any point 0, and join Oa, Ob, Oc. Draw CA parallel to Oa, AB parallel Oh, ^C parallel Oc. Then C, the intersection of A C and 5 C, is a point on the resultant. A line through C equal and parallel to ac, that is, parallel to F^ or F^ , will therefore give the result- ant in magnitude, direction, and position. ^ We may readily find an expression for the position of the point D, in which the resultant cuts AB. From similar triangles Oa J, ^ CZ); Och, BCD; CD /AD = ab/Ol, CD/BD = cl/Oh. Eliminate CD and 01, and BD/AD = al/ch = F, /F„ or F,xAD = F,x BD; which, since the whole distance AB is known, gives the position of D. Illustration. — Make an ap- paratus as in Fig. 60, and find the point A of balanc- ing of known weights by trial. Compute its position and compare results. Compare W with the sum of the weights strung along ~ the rod. 1. Draw the figure corre- sponding to Fig. 59 when F^ , "^ F^ act in opposite directions. Show that .S — J', — F^. 2. A pulley is a wheel or sheaf with a groove round its outer edge, and capable of re- volving freely about an axis through its centre 0. This axis is fixed in a frame or block to which a hook is attached. 92 STATICS OF A RIGID BOBY. Fig. 62 . ,P "w e In & fixed pulley let a cord passing over the sheaf, sup- port a weight W. The pull F on the string being the same throughout its length, i^= W; and if P is the pi-essure on the support, we have the pulley acted on by 4 forces, F, W, P, and its weight w. Hence for equilibrium P = F^ W+ w-2 W+ to, In a movable pulley the block is supported by a cord passing under the sheaf. 3. InFig.63wehave '^'S' *' a single fixed and a single movable pulley. If the ropes 3 and 3 are parallel, and W is the weight of the movable pulley, |C prove W-\- 10 = 2F. Find the pull on the hook at 0, 4. In a pulley tackle the upper and lower blocks each contain two sheaves, and the same rope passes round all: prove Fig. 64 W-\- wt. lower block - iF, supposing all of the cords to be parallel. 5. A weight of 400 lbs is being raised by 1 a pair of double pulley-blocks. The rope is fastened to the upper block, and the parts of the rope (whose weight may be disregarded) are con- sidered vertical. Bach block weighs 10 lbs. Find the pressure on the axle of the upper block. Ans. 532.5 pounds. 84. Moments. — If, besides acting in opposite directions, F, , F, are equal, the points a and c in the construction dia- .>F MOMEJTTS. 93 gram (Fig. 59) coincide, and the resultant is zero. The lines AC, BG become parallel, and therefore do not inter- sect. A consideration of Fig. 65 will show that the Fig. k tendency of the forces is to turn the body round an axis. We are thus led to discuss the case of forces that cause turning. F" Suppose a force F in the plane of the paper acting on a body, and causing it to turn about an axis through a point perpendicular to the plane of the paper. The turning effect '^' depends on the magnitude of the force and on its distance from the axis, and the prod- uct of the two may be regarded as a ~*o measure of the importance of the force in producing turning. This product is called the Moment of the Force about the Point, so that we may define the moment of a force about a point to be numerically equal to the ijroduct of the force and the perpendicular let fall from the point on its line of action. Thus the moment of i'' about a fixed point is JFp, p being the perpendicular let fall from on F. The unit of moment depends on the units of force and distance, and is named a foot-poundal, a foot-pound, an inch- ton, etc., according to the units of force and distance em- ployed. (See Art. 130.) It is cTident that the direction of turning about is as indicated by the arrow in the figure. Reverse the direc- tion of the force, and the direction of turning is reversed. To indicate the sense of the turning it is usual to call the moment of a force about a point negative when the ten- dency to produce turning is in the direction in which the hands of a clock move, and positive when the tendency is 04 STATICS OP A EIGID BODY. in the opposite direction. Thus in the figure the moment is —Pp. 85. The moment of a force F with refer- ence to a fixed point may be represented graphically. For if ab represent F plotted to scale and Oc the distance jO, then the moment Fp is represented by ab X Oc, that is, by twice the area of the triangle Oab, which has ab for base and Oc for altitude. .Fig. 67 Fig. 68 . We may now find the relation between the moment of a force AD about a given point 0, and the moments of its components AB, AC about the same point. For from geometry GAD = 046'+ OOD + ADC = OAG+ OAB; that is, the moment ot AD about is equal to the sum of the moments of AB, AC about 0. This very important proposition may be proved more generally as follows: Let F^, F^,... be the forces acting at A, R their result- ant, and any point in the plane of the forces. From let fall the perpendic- ulars Oa, Ob, ... 01 on F,, F,,... R. Join A 0. Then, since the component of R in any direction is equal to the sum of the components ot F^ , F^, . . . in the same direction (Art. 49), take the direction A Fat right angles to A 0, and we have 7? sin OAl = F^ sin OAa + F^sin OAb + . . ., (1) or R X 01 ^ F, X Oa -\- F^ X Ob -{■ . . .; (2) Fig. 69 MOMBNTS. 95 which shows that the moment of the force R about is equal to the sum of the moments of its component forces about the same point. Two important consequences follow : (a) If the direction of R is reversed, the forces F^,F^, . . . — R are in equilibrium, and we have = iJ X 01-^ F,y, Oa-^ P^X Ob + .. .; or, when forces acting at a point are in equilibrium, the algebraic sum of their moments about any point in their plane is zero. {b) When forces acting at a point A are in equilibrium, their moment about any point gives the same relation as the resolution of the forces about an axis through A perpendicular to A 0. One equation is a multiple of the other, equation (2) being deduced from (1) by multiplying by A 0. See Art. 100 for an illustration of this. Ex. 1. Find the moment of a force about any point in its line of action. 3. Compare in magnitude and direction the moments of two forces about any point on their resultant. 2a. Hence find the algebraic sum of the moments of two or more forces about any point situated on their re- sultant. 3. Is there any reason why a man should put his shoulder to the spoke rather than to the body of a wagon in helping it up hill ? 4. If R is the resultant of two parallel forces F^, F^, and able is any line per- pendicular to their lines of action, prove independently of Art. 85 that RX al = F, Xab + F, X ac. [Follows from F^ X U = F, X cl; , p^j^p^^ R. Art. 83.] 5. Show that the dimensions of the moment of a force about a point are MU/T'. Fig. 70 Fa \ 96 STATICS OF A RIGID BODY. 86. Couples. - Fig. 71 Let us return to the two equal and op- posite parallel forces, Art. 84. The moments of the forces about any point aa.re — Fxab and + F X ac, respect- ft_ [6 \q ively. Hence the measure of the turning effect of the two forces would lie - F X ab + F X ac or F X be, that is, the product of one of the forces by the distance between the directions of the forces. To the system of two equal parallel forces acting in op- posite directions bub not in the same straight line the name Couple is given. The line be being the distance be- tween the lines of action of the forces is called the Arm of the couple, and the product Fx be, or force X arm, is called the Moment of the couple, or the Torque. An example of a couple is seen in the copying-press. The handle is pushed at A and pulled at B, the push and pull forming a couple. In conse- quence the screw rotates. It requires a couple to wind a watch. Fig. 72 87. The moment of a couple depending only on the magnitude of the forces and the distance between them, the effect of a couple is not altered by turning the arm through any angle about one end, nor by moving the arm parallel to itself in the plane of the couple, nor by changing the couple into another couple having the same moment. It hence follows that the resultant of a number of couples in a plane is a couple whose moment is equal to the sum of their moments;. COUPLES. 97 It also follows that a single force F and a couple P, P acting on a body cannot be in equi- librium. For let the moment of '^' '^ / the couple be Pa, a being its arm. Eeplace the couple P, P \>j &, couple F, F ot arm l, so that Fb = Pa, and placa it in the plane so that one of its forces F is opposite P" to the single force F. The two forces F, F at C are in equilibrium, leaving the single force F at £> unbalanced. Hence there cannot be equilibrium. Ex. 1. Three forces are represented in magnitude, direc- tion, and position by the sides of a triangle taken the same way round: show that they form a couple whose moment is numerically equal to twice the area of the triangle. 3. Four forces are represented in magnitude, direction, and position by the sides of a square taken the same way round: prove that they form a couple whose moment is numerically equal to twice the area of the square. 88. Eememberiug that the -turning efEect of a force is measured by the moment of the force, we can now find the conditions of equilibrium when any number of forces F^ , F^ , ... act on a rigid body at different points A, B, . . ., all of the forces being in the same plane. For the reasons stated in Art. 80, it is necessary to consider the external forces only. At any point introduce two forces F^ , F/, each equal to F, , and of opposite directions. This will F, '^' J f: not disturb the equilibrium. Now F^ and F^' form a couple of moment F^p^ Up, is the distance of from the line of action of F, . Thus F, at A is equivalent to F, at 0, and a couple F^p,. Treating the other forces in the same way, we have the forces , J. F,&t A, F^&tB, . . . equivalent to F^, F^, '^ "* ... at 0, and the couples Fj}, , F^p, , . . . . The forces at may be combined into a single resultant 98 STATICS OF A RIGID BODY. R and the couples into a single couple Q whose moment is equal to the sum of their moments. But a single force and a single couple cannot be in equilibrium. Hence for equilibrium we must have 72 = 0, G = 0, the conditions sought. These conditions may be put in a form -more convenient for computation. Through draw two lines OX, OY, forming axes of coordinates, and let each force be resolved into components parallel to these axes. Denote the com- ponents by X, , F, ; X,, y, ; . . . re- Fig- 75 spectively. Now if "SX, '2 Y denote the sums of theforcesJr,,X,,...; F,, Y„ . . . . along the axes of X and Y, then (,'2XY + {'SYf = m and i2 = can only be satisfied by JSX^O, ^F=0. Also, since the moment of the resultant about any point is equal to the sum of the moments of its components about the same point, the moment Fj}^ must be equal to FjS;, — Xj]^ where x^, «/, are the coordinates of A. Hence Q ='2 (Yx — Xy), and if G = 0, we must have :S{Yx - Xy) = 0. Hence the conditions of equilibrium may be stated — (a) T%e sums of the components of the forces along lines parallel to each of two rectangular axes drawn through any point in their plane is zero. (b) The sum of the moments of the forces about any point in the plane is zero. Prom these conditions three unknowns may be determ- ined, and no more. Hence, in order that a problem of COUPLES. 99 this kind be determinate, the number of unknown forces that enter cannot exceed three. An important applica- tion of this principle occurs in finding stresses in roof and bridge trusses. Ex., 1. What are the conditions of equilibrium of two forces ? Also of three forces, two of which are parallel ? 2. Show that parallel forces are in equilibrium when the sum of the forcee = 0, and the sum of their moments about every point in their plane = 0. 3. In the single movable pulley (Fig. 63) if the ropes are inclined at an angle 6, prove W =2F cos 8/2. 4. If the sum of the moments of a number of forces acting at a point in a plane about each of three points not in the same straight line is zero, the forces are in equilib- rium. Prove this. 5. Let AB represent a rigid rod (as a crowbar) turning on a fixed support 0. Let a force i^be applied at A, and let W be the '"'s- ''^, resistance to be balanced at C. Given i" the lengths of A G, GB, it is required a c! n to find the relation between P, W I S when in equilibrium. j [Neglect for the present the weight I of the rod. Let i^and TFbe vertical, If w-f and let R denote the vertical pressure ' at G; then R must be the resultant of i^and W, and there- fore R = F+ W. Take moments about G, and Fx AG-WxBG=0. Hence the ratios of F, W, R are found. The rod AB is known as a lever, the support G the ful- crtwi, and the distances A G, GB the arms of the lever. The equation F X AG — W X BG = is sometimes called the principle of the lever.*] 6. Show by sketch the positions of force, resistance, and * The properties of the lever were first given by Archimedes (b.c. 387-312). 100 STATICS OF A RIGID BODY. Fig. 77 fulcrum in the following levers : wheelbarrow, spade, claw- hammer, rowboat, pair of scissors, pair of nut-crackers, the forearm. 7. A lever is 3 ft. long: where must the fulcrum be placed that 10 pounds at one end may balance 30 pounds at the other end ? 8. From a pole resting on the shoulders of two men a weight W is suspended. It is m times as far from one man as ixom the obher; what does each support ? Ans. W/{n + 1), nW/{n+l). 9. Find the relation between F and PTin a bell-crank lever. A and B the bell wires, the pivot about which the lever turns. [The directions of F, W, and the reaction R of the pivot meet in a point 0. Hence take moments about C] 10. A pair of nut-crackers is a inches in length and a pressure oip pounds will crack a nut placed h inches from the hinge : what weight placed on the nut would crack it ? A ns. pa/b pounds. 11. Show that a single fixed pulley is equivalent to a lever with equal arms (Fig. 63). 13. Two cylinders fastened together move freely on a common axis which is horizontal. A force F acts by a cord coiled round the larger cylinder (or wheel), and balances a weight W hang- Fig- 78 ing from a cord coiled round the smaller cylinder (or axle). [The apparatus is equivalent to a lever with unequal arms, the axis correspond- ing to the fulcrum of the lever, and the radii to the arms. It is called the wheel arid axle. We have p -\- W= pressure on axis, F X radius wheel — W X radius axle.] OOUPLRS. 101 13. The axle of a capstan is 2 ft. in diameter. If four sailors push with a force of 40 pounds each at the ends of spikes 4 ft. long, find the weight of the anchor that is lifted. 14. If the force is transmitted by toothed wheels, the teeth work in each other, so that the motion takes place as Fig. 80 if between two circles (called pitch circles) in rolling con- tact. We therefore replace the wheels by the pitch 103 STATICS OF A RIGID BODY. Fig. 81 circles, and consider the force to be tangential to these circles at the successive points of contact. [In a winch we have a lever AB combined with toothed wheels, and a drum round which the rope attached to the weight W is moved. Eeplace the toothed wheels by pitch circles, and the mechanism in outline with the forces acting is as in Fig. 81. We have FXAC=EX CH, RXHD= WxDE; F W'' DE CH rad. drum no. teeth in pinion no. teeth in wheel AG^ HD X •] Fijg, 82 length arm 15. Given that the cranks have 18 in. leverage, the gears are 4 to 1, the drum 6 in. in diameter, and the capacity with two-man power is 3 tons : find the force exerted by each man. 89. Centre of Parallel Forces. — We have seen (Art. 83) that two parallel forces F^, F^ acting at A, B are equivalent to a single force F^ -{• F, acting at a point i> on the line AB, such that F,XAI) = F,X BD. Similarly, if F^ is a third force acting at C, the three forces are equivalent to F^ -\- F^atB and F^ at G, or to a single force F,+F^ + F, at G, such that {F, + F,) xDG = F,x GG; and so on. This result is entirely independent of the directions of the forces, so that the point G will be in the same position if the forces are turned in the same sense through the same CENTRE OF PARALLEL FORCES. 103 angle about J, B,so as to remain parallel to one another. The fixed point G being the centre of the points of applica- tion of the parallel forces, is called the Centre of Parallel Forces. It is convenient to express the coordinates x, y of G, the centre of parallel forces, in terms of the coordinates x^, y^; x^, y^; . . . oi A, B, . . . referred to axes OX, Y drawn through any point in the plane of the forces. Since G is in the same position no matter what the directions of the_ forces may be, let them be parallel to OY. Take moments about 0, and {F^ -\-F, + F,)^= F,x, + F,x, + F,x, , and X is found. Next take them parallel to OX, and take moments about 0; then {F, + i^, + F,) y = F,y, + F,y, + F,y, , and y is found. The values of x, y may be written for any number of forces, X = 2Fx/2F, y = 2Fy/2F, when 2 is the symbol of summation. It is evident that the same reasoning would apply if the points A, B, . . . were not in the same plane, the forces still being parallel. If z^ , z^ denote their distances from a fixed plane, the distance z of G from this plane would be given by J^ 2 Fz/2F. 90. Centre of Gravity. — As an illustration of parallel forces, consider the force of gravity 3,cting on the particles 104 STATICS OF A RIGID BODY. of a body. A body may be regarded as built up of parti- cles, the weights of the particles forming a system of forces whose lines of action, pass- ^'^' '^ ing through the earth's centre, are so nearly parallel in a body of ordinary size that we may con- sider them to be so. If m^ , m^ , \^g ... lbs. are the masses of the X particles, the parallel forces act- ing downwards on them are m.g, mj, . . ^ poundals, and the resultant force would be found by adding the forces. To G, the centre of these parallel forces, the name of Centre of Gravity* is given. Its distance z from the plane of X, Y may be written 'z — '2mgz/'2mg. This may also be written z =: ^mifSm, and hence the centre of gravity is also called the centre of mass, or by' some the centroid. 91. That the line of action of the resultant force of gravity passes through the centre of gravity in all positions of a body, suggests an experimental method of finding the centre of gravity. Thus conceive the body suspended by a string from a point p. The forces acting are the resultant force of gravity at the centre of gravity, and the tension (pull) of the string. The lines of action of these forces must lie in the vertical through P. Hence, to find G, suspend from P, and strike the vertical PH; next suspend from any other point Q, and strike the vertical QK: the point * The idea of the centre of gravity of bodies is due to Archimedes. CENTRE OP PARALLEL FORCES. 105 of intersection of PH and QK will be the centre of gravity G required. 92. When the particles of a body are so distributed that there are always the same number in the same volume, the body is said to be of uniform density. The unit of density is taken to be the mass of unit volume. Thus, if m is the mass and V the volume, then the density p = m/ V. If the density is not uniform, we must estimate it at every point. Thus for an indefinitely small mass dm of vol- ume dv about a point, the density at the point is dni/dv. We shall consider bodies of uniform density only. Ex. 1. The 0. of G. of a uniform straight rod is at its middle point. [The rod being uniform, is such that the number of particles on one side of the centre C is equal to that on the other side, and the 0. of G. of every pair being at C, the 0. of G. of the whole is at C] 2. Find the 0. of G. of a triangular lamina of uniform thickness and density. [Conceive it divided into strips by lines parallel to AB. The 0. of G. of each strip will be at its middle point. Hence the 0. of G. of the whole will lie on the line GD joining C to D, the middle point of A B. Simi- larly, it will lie on the line joining A to E, the middle point of BC Hence it is at G, the intersection of CD and AE. Join DE. The triangles DGE, AGO are similar, and DE=^AO. .-. DG = iCG = iCI).] 3. Prove that the C. of G. of a triangle (triangular lamina, strictly) coincides with that of three equal weights placed at its angular points. 4. Three men support a heavy triangular board at its corners: compare the weights supported by each man. 5. The sides of a triangle are 3, 4, 5 : find the distance of the 0. of G. from the angles. Ans. Vu/d, 3 VIsTs", 5/3. 6. Explain how an instrument which stands on three 106 STATICS OF A RIGID BODY. screws can be "levelled" by means of these screws, a galvanometer, for example. Take Fig. 86 m,gV Ymo 93. We pass at once to the centre of gravity of a system of bodies rigidly connected, by con- sidering that each body may be conceived concentrated into a par. tide of equal mass acting at the centre of gravity of the body. Thus, reasoning as in Art. 89, if m^, m^, . . . represent the masses of two bodies, the centre of gravity Fig,. 87 I IC G of the system will be found from m, X G^G = m, X G^G. Similarly, if G, G, are given, G, may be found. Ex. 1. Weights of 1, 2, 3, 4, 5 pounds are strung on a uniform rod AB, whose weight is 3 pounds, at distances of 4 in. from each other : find the point at which the rod will balance. Ans. AG = 101 ill- 2. Find the C. of G. of a T-iron, depth = d; depth of web = d^; breadth of flange = b; breadth of web = b^ . Ans. CG{bd - dXb - b,)]= bd X |- <(5 - 5>) (^- f ) " 3. A common form of cross-sec- tion of a reservoir wall or embank- ment wall is a trapezoid whose top and bottom sides are parallel. If top side = a, bottom = b, and height = h, show that ^=4 + ^-^); Fig. 88 a _hf 2a-\- b \ a + bh ^ = Z\ CENTRE OF PARALLEL FORCES. 107 Fig. 90 4. Hence show, and also show independently, that if one-fourth part of a Pig. gg triangle is cut off by a p _c k d line parallel to the base, """"""""—--, ' the 0. of G. of the re- mainder is at 2/9 of the line joining the vertex a h to the middle point of the base. 5. From a circular disc another circular disc described on its radius as diameter is cut: show that the C. of G. is distant 1/6 radius from the centre. 6. Prove the following construction for finding the C. of G. of a trapezoid ABCD. (Fig. 89.) Prolong ^^, making BE = OD; prolong DC, making CF = AB. The point of intersection of EF and HK, the line joining the middle points of AB, CD, is the 0. of G. 7. Prove the following rule for finding the 0. of G. of a quadrilateral: (Fig. 90.) Draw the diagonals. Make^i^ = DE. The C. of G. of the triangle CFB is also that of the quadrilateral. 94. The application of the general formula of Art. 90 to finding the C. of G. of bodies may be further illustrated by the following examples: Ex. 1. To find the C. of G. of a circular arc AB. [The C. of G. must lie on a line OCX joining the centre to the middle point Cof the arc. Let ds denote the length of the indefinitely small part PQot the arc, h its cross-section, p its density. The mass of PQ = kpds. The coordi- nates of the C. of G.of PQavQ the same as those of P, or x, y. Hence x{=OG) =fkpds X x/fkpds = fxds/fds. If COP = e and OP = r, then z — r cos 6, ds = rdd, and x=fr cos 6'dd/fdd. Fig. 91 108 STATICS OF A EIGID BODY. If angle A OB = 2/f, then, integrating between the limits + A - A X = r sin J3//3.] 2. To find the C. of G. of a semicircle. . Ans. x — Ir/n. 3. To find 0. of G. of a circular sector A OB if angle AOB-2 §. (See last figure^^) [Divide the sector into triangular pieces as OPQ. Area OPQ = \rd6. xr^ ir'dO. If G, is 0. of G. of OPQ, then x= 0M= OG, cos 6- ir cos ft . • . a; = / f r cos (9 X ir'dd/ / ^rVZi9 = f r sin ^/p.] 4. Find the 0. of G. of a circular ring, radius OA = r , 00 ^i\, angle A0B = 2 ^. [A practical application would be the front surface of a circular arch of which r, is the radius of intrados and r J radius of extrados. } Ans.x-^{r,'-r^')sin/3/ir,'-r,')/3. 5. For a semicircle, x = 4r/3;r; for a quadrant of a circle, x = 4 V2r/37C. 95. Stability. — If a body suspended from a point be slightly displaced from its position of equi- librium and let go, it will turn about the point of support. If the centre of gravity G is beloio the point of support, the ten- dency is for the body to return to its original position. This may perhaps be made more evident by resolving the weight W into F^ along GO and F^ at right angles to F^. The force F^ gives the pressure on the sup- port, and iJ^jthe-tiirning force which tends to swing the body to its former position. In this case the body in its original position is said to be in stable equilibrium. Fig. 93 •r STABILITY. 309 In Fig. 94 the centre of gravity is alove the point of sup- port, and the tendency for the body when pjg, 94 disturbed is to move farther from its origin- al position. In this case the equilibrium is unstable. If, however, the point of sup- port is at the centre of gravity G, the body will remain at rest in any position, and the equilibrium is neutral. Similarly, if the body, instead of being suspended from a point, rests at a point on a fixed support the equilibrium may be stable, unstable, or neutral, according as the forces acting on the body in its displaced position tend to restore it to its original position or make it move farther from that position, or are in equilibrium. When one point of a body is fixed the resultant of the external forces tending to cause translation is balanced by the reaction of the constraint at 0. Also, if there is equi- librium the tendency to turn about in one direction must be balanced by the tendency to turn in the opposite direction, or the sum of the moments about must be zero. This, therefore, is the condition of equilibrium of a system of forces acting on a body with one point fixed. Ex. 1. In suburban passenger traffic the train must stop and start quickly. The engine is built with a large wheel base. Why ? 2. A circular table weighing to lbs. has three equal legs at equidistant points on its circumference. The table is placed on a level floor. Neglecting the weight of the legs, find the smallest weight which, when placed on the table, will upset it. Ans. w lbs. 3. If the table has four legs at equidistant points, find the least weight that will upset it. • 4. Suppose A BCD (Fig. 95) to be the cross-section of a wall built to withstand the pressure of earth or water on one side: for example, the wall of a reservoir or of a railroad embankment. Such a wall is ciilled a rctaining-ivall. 110 STATICS OF A EIGID BODY. Fig. 93 c D E -; / /r / G /A H W B [We assume that the wall is built so that it cannot give way except in one mass, and by being overturned about the edge A, Let P be the resultant of the ' forces pressing on the side BD, and let be its point of application. The weight W acts vertically down- ward through the centre of gravity G. The stability depends on the difference of Px^^andPFx^^, the first tending to over- turn, the second to restore. Hence, when the wall is just on the point of turning, we have the relation PTX AH—P X AE.'\ 5. The centre of gravity of a ladder weighing 50 lbs. is 13 ft. from one end, which is fixed. What force must a man apply at a distance of 6 ft. from this end to raise the ladder to a vertical position ? Ans. 100 pounds. 6. Find the proper elevation BE of the outer rail on a railroad track for a given velocity v of engine weighing m lbs., and on a curve of radius r, in order that there may be no flange or lateral pressure on the rails. [The forces acting are the weight tng of the engine, the centrifugal force C(= mv'/r) and the reac- tions iVj, iVj of the rails. Since there is no flange pressure the reac- tions are perpendicular to the track AB. Let be the inclination of AB io the horizon. Eesolving the forces along AB, we have (7cos ^ = mg sin d^ or tan 6 = C/mg = v'/gr, BEAM SUPPORTED AT ONE POINT. Ill But if is smallj tan 6 = BE/AB = elevation/gauge; . • . elevation of rail = gauge X v'/gr. For standard gauge of 1 ft. 8^ in., tills gives elevation of rail = lv''/\:v inches, nearly, V being expressed in feet per second, and r in feet. J 7. If in (6) the velocity V is expressed in miles per hour, show that elevation of rail =15 Fy4r inches, nearly, the radius r being, as before, expressed in feet. 8. Find the greatest velocity v a locofliotive can have to be just on the point of overturning on a curved level track of radius r ft., the centre of gravity of the locomotive being 6 ft. above the rails, and the gauge of the track 4 ft. 8|- in. Ans. 3.55 V^' ft. per sec. nearly. 96. Beam Supported at One Point. — Suppose a beam of wood or metal of length 2a suspended at a point 0. The pressure on the support being equal to the weight W of the beam, the centre of gravity G of the beam will lie in the vertical OG. A weight P placed on the beam or suspended Fig. 98 B 97 o >'W from it anywhere except in the line Offwill cause the beam to take an inclined position, as in Fig. 98. Suppose P suspended at A, where AO = a. The forces P and W are parallel, and therefore their resultant must be equal to their sum. Being balanced by the rea,ction of the support 113 STATICS OF A EIGID BODY. Fig. 99 0, this resultant must pass through 0. Hence the moment of W about is equal to that of F about 0, or PX 00=Wx GH, or P X acQsd — W Xli sin 6, or F= tan P. a By attaching a pointer to the beam free to move over a graduated arc we have a means of compar- ing weights. An example is afforded by thft common letter-scale (Fig. 99). 97. Balance. — If in the beam repre- sented in Fig. 97 we place two equal weights P, P at the same distance A from 0, equilibrium will not be disturbed. For the moment about is the same, being = Pa in both cases. Hence the beam may be used for comparing equal weights. At- taching pans to A, B, the weights for com- parison may be placed in these pans and the operation facilitated. Such an arrange- ment is the common Balance. The parts of a balance should be arranged so as to secure the greatest accuracy in making the com- Fig. loi parisons, and with the least loss of time. How can this be done? Suppose the weights P, Q to be unequal, and that the points of support A, B are equidistant from and in the BEAM SUPPORTED AT ONE POINT. 113 same straight line with it. Let AB — 2a and OG = h. Take moments about 0, and P X aoos6 — Wh sin 6 — Qa cos 6 = 0, or tan d = {P - Q)a/WJi. E"ow the balance will indicate small differences P — Q the more clearly the greater the angle d through which it swings for these differences. But tan ^ or 6' is greatest when a/ Wh is greatest, that is, when a is large or the beam has long arms, when W is small or the beam light, when h is small or the centre of gravity is just below the point of suspension. Such a balance has great sensihility, and is suitable for delicate inyestigations in Chemistry, Physics, Assaying, etc. In scales for weighing large masses stability rather than sensibility is wanted; that is, for small differences of P and Q the angle of deviation of the beam from the horizontal as shown by tan 6 should be small. This requires Wli to be large or the beam to be heavy, with a long distance be- tween the centre of gravity G and the point of suspension 0. By making the arms long, a balance may be constructed which shall possess in a measure both sensibility and stabil- ity. As the two conditions are at variance, the amount of compromise must be decided by the use to which the bal- ance is to be put. Ex. 1. If the centre of gravity coincides with the point of suspension, the balance is in equilibrium in all positions. 3. The arms of a balance are equal, but the scale pans are not of the same weight. If a body weighs P lbs. in one and Q, lbs. in the other, find the true weight. Ans. \{P ^r Q) lbs. 3. A body placed in one scale pan appears to weigh P lbs. and in the other Q lbs. If the pans are of equal weight but the arms are not of equal length, show that its true weight = ^PQ lbs. 114 STATICS OS A EIGID BODY. Fig. 102 98. Steelyard. — Suppose (Pig. 103) a beam suspeiided from a point di- rectly above its centre of gravity 0. The tipper edge ABC be- ing straight and at right angles to OG wiU be horizontal. If ©/__P_J from the beam we suspend two bodies of unequal weights P, Q, it will still remain horizontal if the moments of P and Q about are equal, or PxAO:=^Qx£0. Let the weight P suspended from ^ be a scale pan. If to P we add an unknown weight W, we shall still have equi- librium, provided Q is suspended from a point C such that {P+ W)AO^ Qx GO. Subtract these equations, and WxAO=QxBC, which gives the unknown IF as soon as BC is measured. To save measurements of BG at every weighing of a body, it is convenient to graduate the beam in the first place. Thus suppose P = 1 lb., Q = Z lbs., and y4 = 4 in. Then OB = 2 m., and a notch can be made at B, which, as the weight Q then balances the pan P only, would be marked 0. W Let now W= 1 lb.; then BG^jr X A0 = 2 in., and G would be the position of the 1 lb. mark. Make W=2 lbs. and BD = 4 in., giving D the 2 lb. mark, and so on. Hence in weighing a body it is only necessary to place it in the pan and move the weight Q until the notch is found where the beam will remain horizontal. The number at the notch indicates the weight. This instrument is called a Steelyard. Ex. 1. Graduate a steelyard to weigh half-pounds. BEAM SUPPORTED AT TWO POINTS. 115 2. If the point of suspension be not over the centre of gravity and the movable weight Q placed at a point H holds the steelyard in a horizontal position, show that HB — W X AO/Q, and hence show how to graduate the steelyard. 3. A steelyard beam weighs 3 lbs., the wt. § is 4 lbs., and the distance of the centre of gravity from is 3 in., and of the point of suspension of the scale A from 5 in. : show that the 1 lb. gi-aduation marks are at intervals of | in. 4. A steelyard weighs PFlbs. and is correctly graduated for a movable weight Q: prove that a weight 2Q may be used provided a fixed weight W \s suspended at the centre of gravity of the steelyard. 5. A piece is broken off the longer arm of a steel- yard: show that the customer is defrauded. Fig. 104 99. Beam Supported at Two Points. — We naturally pass from a body supported at a single point to one in which two points A and 5 are supported, as for example a beam supported by two smooth horizontal pins A and B. The forces acting are the weight W vertically downwards through G and the 116 STATICS OF A RIGID BODY. ..N, i Fig. 105 . .N, Fig. 106 reactions iV, , W^ of the supports A, B in directions which are unknown. If we assume that they are in the same plane, the forces being three in number must meet in some point in the vertical through G, Until we are able to fix the directions of iV, or N^, the problem is indeterminate, as there is an indefinitely great number of points in this line. 100. If, for example, the beam rests on two props A, B in the same horizontal plane, the reactions of the props are vertical. Hence, resolving vertically, and taking mo- ments about G, N^-\-N,- W=Q, N.xAO - W,XBG^0, from which iV", , JV, are found. Or take moments about A and B in succession, and ^r,xAB-W X BC= 0, N,x AB-W X AO^O, from which the same values result as before. The pressures iVj, JV, on the supports may also be determined graphically by the method of Art. 81. Suppose the beam ^i? to carry besides its own weight W a load W, at C. Draw ab, be to scale to represent Jf,, W, and parallel to their directions. The force R SCALE OF DIMENSIONS jb: ;LE OP PORCEB. BEAM SUPPOKTBD AT TWO POINTS. 117 represented by the line cba which closes the polygon will hold W^, Win equilibrium. Take any pole and Join Oa, Ob, Oc. From any point p in the direction of iV, draw pq parallel to aO; draw qr parallel to 50 and rs parallel to cO. The intersection t of pq and sr gives the position of the resultant R. Its direction is parallel to cba, and is therefore vertical. Join sp and draw 01 parallel to sp. Now cba (or R) may be resolved into Oa along pq and c along sr. Also Oa may be resolved into 01 along sp and la^dlong pA. And cO may be resolved into cl along s^ and 10 along ^s. But the forces 01, 10 being equal and in opposite directions, must be in equilibrium. Hence cba (or R) is equivalent to la at A and cl &\,B. But c5a holds TFj, PT in equilibrium. Hence ?a, cZ hold PT,, IF in equilibrium,and are the reactions A'",, iV^ at^ and B. Hence the rule, (a) Form the force polygon by laying off the forces to scale. {b) Select a convenient pole and form the polygon pqrs. (c) Draw 01 parallel to the closing line sp dividing ca into parts la, cl, which will represent the reactions N^ , N^&t A and B. Ex. 1. A beam of 20 ft. span carries a weight of 10 tons 8 ft. from one end : find the pressures on the end supports. 2. A highway bridge 25 ft. long weighs 6 tons: find the , pressures on the abutments when a 2^-ton wagon is \ of the distance across. Ans. 5 tons; 3.5 tons. 3. A beam of 40 ft. span weighs 1 ton per running foot. One half of it carries a uniform load (as a train of coal cars) of 2 tons and the other of 3 tons per running foot : find the pressures on the end supports. Ans. 65 tons; 75 tons. 4. A truss of 60 ft. span and weigh- Fig- 107 ing 100 tons car- 1 § 1 § § ries an Brie con- S 1 S s S solidation engine A lo' ^i'B^ 4V s? s'l^ A as m the figure: find the pressures on the supports. Ans. 66.6 tons; 84.9 tons. 118 STAllCS 0]? A KIGID BODY. 101. Two Beams Hinged. — Again, suppose to the beam AB axi equal beam '''s- '08 £0 attached at B by a hinge or pin, and the two beams to be supported by two pins at J, C in the same hori- zontal plane. The pin B being in equilibrium, the re- action Not AB on it is equal and op- posite to that of BC on it. Since the beams are equal the direction of N must be horizontal. The weight PF of the beam AB acts vertically through the middle point G. Hence the reaction JV", of the pin A must pass through ff, the intersection of the directions of ^ and W. Similarly for the beam 5 C The forces acting on the beam AB are parallel to the sides of the triangle AHK, and N, JV, may be scaled off directly if HK be taken to represent the weight W. Put the length AB—l, the height BO — h, the span AC =2a, and we have N=W>^ AE/EH = Wa/2h, N=WxA ff/EH = W Va' + U'/2h, which give the algebraic values of the reactions. These values may also be found by taking moments about two BEAMS ttlNGED. 119 A and £ in succession. Thus taking moments about A. NxHK= W X AK, giving N &s before. Similarly for the value of iV, . The reaction N^ may also be found by resolving it into two components, X horizontal and Y vertical. Then Y=W, X= N=Wa/U, and N^ = ^/X" ~\- Y' ^W Va' -\- W/U, as before. 102. The reaction N^ at A may be resolved into two components — a longitudinal force AL along the beam AB and a transverse force AE perpendicular to it; also iV at 5 into a longitudinal force BL and a transverse force BF. The longitudinal force diminishes from A L at A to BL at B. The transverse forces AE, BF being each equal to HL, are equal to one another. Also HL = HG cos 6 = W -^- cos ^, so that the transverse forces are equal to half the /v component of W at right angles to the beam. The values of the transverse forces are the greater the smaller the angle d is, or the more nearly the beam is hori- zontal. When the beam is horizontal and the external forces are vertical, the longitudinal forces disappear and the transverse forces alone enter. In the study of struc- tures it is necessary to consider the efEect of both of these classes of forces. 103. The computation of the longitudinal and trans- verse forces, even in the simple case given, is tedious. It can readily be understood that in a complicated framework it would become intolerably so. Accordingly among archi- tects and engineers a method of greater simplicity is fol- lowed, leading to results practically close enough. This is 120 STATICS Ot A RIGID BODY. Fig. 109 done by considering the loads carried by the frame, includ- ing its weight, to be concentrated at the joints of the frame, and applying the conditions of equilibrium to each joint in succession. Thence the stresses along each piece, meeting at-a joint, are found by the triangle of forces. The method belongs to a special branch of mechanics known as Graphical Statics. We add a short sketch of its application to the determination of the longitudinal stresses in simple trusses, forming as it does a good illustration of the triangle of forces. 104. Jointed Frames. — As the simplest possible example of a jointed frame, let us consider three beams hinged by pins at A, B, C, and resting on supports at A, C in the same horizontal plane. This is known as a Triangular Truss. Sup- pose the beams all alike, and of weight W each. The reactions of the supports balance these weights and act vertically upwards, the sup- ports being horizontal. Hence the external forces acting and keeping the truss in equilibrium are as in Pig. 109. Next transfer the weights to the pins. Thus Wat G, =iWatA + iWatB, Wat G, = lWatB+iWat 0, Wat G, = lWatA + lWa\ C; . • . sum = Tfat^ + WatB+ Wat C; and each reaction being equal to half the total weight 3 W, we have the forces as in Fig. 110. Combining forces and reactions at A and O, we have finally the forces as in Fig. 111. JOINTED FEAMES. 121 Fig. Ill > W We have thus transferred the weights of the beams to the joints, and can now consider the beams as without weight, and indi- cating direction only. The result- ing stresses in the pieces we next find. Since the weights on each pin are in equilibrium with the stresses pro- duced in the pieces meeting at the pin, we consider the pins one at a time. (cf) Pin A. The forces acting are W/2 vertically up- wards, and the unknown stresses in AB, A C. Draw (Fig. 112) Oa to scale to represent T'r/2. Prom a draw al parallel to BA and b parallel to A C. Then al, b represent on the Fig. 112 same scale the stresses in AB, AC. Their directions are indicated by the arrow-heads. {b) Pin B. The forces acting are the stress m. AB and W which are known, and the stress in BO which is un- known. Draw la to represent the stress in AB, ac to repre- sent W; then cb will represent the stress in BG. Notice that the arrows on ab point in opposite directions in Fig?. A and B. This is because the stress on the pin A is opposite to that on the pin B, 133 STATICS OF A EIGID BODY. (c) Pin C. The forces are the stress In BG, W/2, and stress in AC, all of whicli are known. For check the dia- gram may be drawn as in Fig. 0. 105. It is erident that we should have saved labor by adding the second figure to the first and the third to the sum as in the fourth figure, which is the complete Stress Diagram. In practice it is convenient to consider the stress dia- gram as in two parts. Thus the line ac is the polygon of external forces, ac being the downward force at B bal- anced by the upward forces cO at C and Oa at J, and is complete in itself. The closing of this polygon shows that the reaction forces have been properly estimated. On this polygon as base the stress diagram is added step by step until complete. In case the truss is symmetrical, as in our example, it is only necessary to consider the first half of the pins. But it is safer to consider all of them, as the symmetry of the drawing will furnish a test of its accuracy. 106. In Fig. A we see that in the beam AB the stress acts towards the pin A and also towards the pin B. The beam is thus compressed between the pins, and is called a Strut. Similarly for BC. Itl AO the stress is from the pin A and also from C, and the beam is in a state of tension. It is called a Tie, and in practice a rod would be used. Hence in designing a structure to carry an assigned load a study of the stress diagram will show not only the amount of stress but the kind of stress, and therefore whether a strut or tie should be employed. 107. For tracing the connection between the pieces them- selves and the stresses in them as shown by the stress dia- gram, an exceedingly convenient system of notation has been devised.* * Due to Prof. Henrici, London, but usually known as Bpw'g notation. JOINTED FRAMES. 123 A beam or a stress is named by letters placed on either side of it. Thus (Fig. Ill) Oh is the tie A C, Oa the reac- tion W/2 at the left support, ab the strut AB, ac the load W at B, and so on. These letters carried into the stress diagram (.Fig. 112) give us ab the stress in the piece ai, cb the stress in cb, Ob the stress in the rod Ob. The letters ■ A, B, C at the pins do not enter the stress diagram. 108. In the following examples first draw the truss to scale (inches to the foot) from the dimensions given, next compute the pin loads, next the reactions of the supports, next draw the force polygon (pounds to the inch), and finally the stress diagram. Scale off the stresses, and tabu- late under the heads Compression and Tension. Ex. 1. In a triangular roof truss the rafters are 2^ ft. apart and the roofing material weighs 20 lbs. per sq. ft. The span is 24 ft. and height 5 ft. : find the stresses on the rafters. Ans. 590 pounds. 2a. Show that the stress diagram for the truss repre- Fig, II J aCALE OF DIMENSIONS SCALE OF STRESSES sented in the figure, loaded at the centre over the vertical piece ab, known as the king-post, is as in the margin, , 2b. A foot-bridge (Fig. 113) 18 ft. span and 6 ft. breadth has a crowd of people on it equal to 100 lbs. per sq. ft. of fioor surface. The king-posts of the two trusses are 3 ft, in depth : find the stresses. Ans. Stress on post al = 2700 pounds, 3, In (2a) the span is %l, depth (?; show that the compres- 124 STATICS OF A RIGID BODY. sion in Aa is Wl/2d, and find the tension in Oa and the stress in the Tertical ab. 4. In a roof of 30 ft. span and height 10 ft. the trusses Fig. 114 are 10 ft, apart, and the pieces ah, cd come to the middle points of the rafters. If the weight of the roof covering is 25 lbs. per sq. ft. surface, draw the stress diagram and scale ofE the stresses. 5. Draw the stress diagram for a Grerman truss loaded as in Pig. 115. 6a. Draw a stress diagram for a queen-post truss (Fig. 116). The queen-posts ab, he divide the span into three equal parts, and the truss is loaded at the joints with weights W. 6b. A foot-bridge (queen-post) of span 25 ft., breadth 7 ft., length of queen-posts 3 ft., car- ries a load of 150 lbs. per sq. ft. of floor: find the stresses developed, the queen-posts Fig. ii6 dividing the span into three equal parts. 7. Draw stress diagrams for the roof trusses repre- sented; the first being that of the Eock Island Arsenal and the second o the Masonic Temple, Philadelphia (Figs. 117, 118). JOINTED FRAMES. Fig. 117 125 Fig- 119 109. The graphical method may be conveniently applied to finding the stresses in a mechanism. Take, for example, the steam-engine. Let P be the pressure ex- erted by the piston on the pin A (Fig. 120) of the cross-head. It is transmitted by the connecting-rod to the crank-pin B, and thence to the crank axis C. If now the machinery is driven by a wheel CG on the axis C working in a,nother wheel at G, the resistance R would be tangent to the pitch circles of these wheels. [This includes the case of a locomotive when the traction is exerted by adhesion to the rail XY. The amount of this adhesion corresponds to i?.] Consider the pin A. It is in equilibrium under the pressure P along the axis of the piston-rod, the thrust Q 126 STATICS OF A RIGID BODY. along the connecting-rod, and the reaction JV" of the guide- bar of the cross-head. Hence plot P to scale, and complete the triangle of forces, from which scale off Q and N. Again, the wheel GO is in equilibrium under the action of Q, R and the reaction S of the crank axis C. All three must meet in the point D, where Q and R meet. Hence plot the triangle of forces, and scale off S and R. The Fig. 120 relation between P the piston pressure and R the force transmitted to the mechanism is therefore determined. We have neglected the weights of the pieces. The only one important to consider is the weight resting on the crank axis 0. Call, it W. It acts vertically. Combine Q and W into one resultant /?, . The reaction S will now pass through the intersection of E^ and R. Hence complete tlie triangle of forces for R, S, R^, and .scale off S and R. Thus the relatioa between P and R is found. Similarly, the weights of all of the pieces may be taken into account if desired. ATTE ACTION. 137 110. Attraction. — In astronomy and mathematical elec- tricity an important application of the composition of forces is to find the resultant attractive force between a finite body and a particle, the law of force being that known as the " law of inverse squares. " This law may be stated as follows: Every particle of matter in the universe attracts every other particle with a force whose direction is in the line joining them, and whose magnitude varies di- rectly as the mass of each particle and inversely as the square of the distance letween them. Thus if the masses are m, m^, and r the distance between them, the mutual attraction F is given by F = cmmjr"^, when c is a constant. The attraction of one particle on another being an inde- pendent attraction, the total attraction of a number of par- ticles (or body) on a particle is a problem of summation which is most readily solved by means of the integral cal- culus. We shall develop one problem to illustrate the method. Ex. 1. To find the attraction of a uniform thin circular disc, radius a, thickness /;, and density S, on a particle of unit mass situated on a line OG through the centre of the disc and perpendicular to its plane. Let the distance OC =1. Conceive the disc divided into an indefinitely great number of concen- tric rings, centre C, and let x be the radius of any one of the rings and dx its width. Then mass of ring = %nxdxhS, and dis- tance of all points of the ring from = W + x\ Hence attr. of ring along OC=cX ^nShxdx cos COD/{W-\-x^, and result, attr. of disc = 'indcbhl^ xdx/{W + x^Y = 27i6ch\l-{l/F + r'')^, an important result in mathematical electricity. 128 STATICS OP A EIGID BODY. 2. If the disc is of indefinitely great extent, show that the attraction = 2i7tdcli. 3. If the particle is indefinitely near to the disc, show that the attraction = ^nSch. (See Thompson's Electricity and Magnetism.) 4. Hence show that all parallel discs of the same thick- ness, forming the bases of cones having the same vertical angle, exert the same force on a particle at the vertex. 5. Hence show from (4) that the resultant attraction of a thin spherical shell of uniform density on a particle situ- ated within it is nil. CHAPTEE V. FRICTION. 111. As already stated, greater simplicity and clearness are secured by considering the properties of bodies one at a time, and thus leading up to the actual state of the case— which is quite complicated, since bodies in nature possess many properties. Thus far the surface of a body has been assumed to be perfectly smooth, that is (Art. 63), to exert a pressure in a normal direction only, or what is the same thing, to ofEer no resistance to the motion of a body pressing against it. But in reality we know that if one body be moved along another (as a book along a table) a certain resistance will be offered to the motion. The resistance arises from irregularities in the surfaces in contact, from elevations and depressions which fit more or less closely into one another. To it the name Friction is given. Suppose a body of weight W to rest on a horizontal table, and to be pressed by a vertical force Q (as of a load resting upon it). The total pressure F {= Q-\-W) is balanced by the vertical reaction JV of the table. If now a force is ap- plied parallel to the table, and gradually increased, a magnitude i^will be reached when the body is Just on the point of moving. The body is held in equilibrium by the force P, the force F, and the reaction B, which can be no longer vertical. Let R be resolved into vertical and horizontal components. The vertical component must 12fl Fig. 122 130 FEICTION. be equal to P, and is therefore equal to the reaction N. The horizontal component / must be equal to the hori- zontal force F. This horizontal component arises from the frictional resistance, so that we may treat friction as if it were a force precisely similar to the acting force i^, but of the contrary sense. The body being just on the point of moving, the friction resulting is named Static Friction. From experiments carried out with this form' of apparatus it is found that between surfaces with little or no lubrication, under moderate loads, and just beginning to slide on one another, the amount of static friction is — (1) Proportional to the normal pressure between the sur- faces in contact. (2) Independent of the areas in contact. (3) Dependent on the material of toMch the bodies are composed. These are known as the laws of static friction. They are roughly true, not only when the motion is on the point of occurring, but also when the velocity of motion is small. Hence the friction / corresponding to the normal pressure iVmay be found from where /< is a constant. It is called the Coefficient of Friction, and its value must be determined by experiment. 112. The above " laws" of friction were deduced from experiments made with surfaces having little or no lubrica- tion, and moving with low velocities ; and for such condi- tions only are they to be depended on. In machines, how- ever, surfaces without lubrication and moving with low velocities are the exception, and we there have an entirely different set of conditions. The friction is now Kinetic* * The distinction between static and kinetic friction was first pointed out by Coulomb (1736-1806). The laws of static friction were enunciated by Coulomb, and confirmed by the later experi- ments of Moriu at Metz'ia 1837-1838. ROLLING FRICTION. 131 Recent experiments show that even with surfaces of the same material, the character of the lubrication, the load, the velocity, the form of the surfaces, whether fiat or curved, the areas as in contact, and the temperature of the surfaces have each great influence on the friction produced. No general relation between the fj'iction (/) and the pressure (iV) producing it has yet been deduced depending on these conditions. Special experiments are necessary in all cases where the conditions differ in any of the points mentioned from those entering into experiments already made. We may, however, in general, write when the value of ;m is determined by the conditions of the problem. These conditions will show whether we may as- sume the so-called laws of statical friction or have recourse to special experiment. Eoughly, the coefficient of friction /t for ivell-luhricated surfaces is from ^ to -^ that for dry surfaces; and if the pressure and velocity are not very large, it varies inversely as the pressure, directly as the area in contact, and inversely as the temperature. Ex. In a locomotive the engineer applies the brakes suffi- ciently to prevent slipping of the drivers in order to obtain the maximum brake retardation. Explain. 113. When one surface rolls on another, the resistance encountered is termed Rolling Friction. In mechanisms it is in general small, and need only be considered in very special cases. In a locomotive the rolling friction between the wheels and the rails, like the joui'nal friction of the wheels, tends to diminish the adhesion of the engine on the rails. 114. The following coefficients of friction may be re- garded as average values, to be used when no special experi- 133 FBICTIOK. ments covering the cases under consideration are possible. The circumstances may be such that the tabular values are very far from the truth in these cases. Indeed, at present we may be said to be acquainted with no quantitative laws of friction of much value. Metal on metal, 0.10 to 0.30 Metal on wood, 0.10 to 0.60 Wood on wood, 0.10 to 0.70 Wood on stone, 0.30 to 0.60 Stone on stone, 0.40 to 0.75 115. In the experiment of Art. Ill, suppose the friction / and the reaction JV combined into a single resultant R. Let ^ denote the angle which R makes with the normal to the table. Draw to scale and com- plete the triangle of forces F, P, R. Then tan d> =F/P =f/N=ii. Fig. 123 Hence has always the same value so long as the substances in contact are the same and under the same conditions. It is called the Angle of Friction. The fact that the reaction force R, which holds the applied forces i?'and P in equilibrium, makes with the normal to the surfaces in contact an angle equal to the angle of friction, gives a key to the application of the graphical method of the polygon of forces to problems involving fric- tion. The method is especially valuable in cases where the forces are interlaced as in mechanisms. Ex. 1. The weight on the driving-wheels of a locomotive is twenty tons, and the coefficient of friction is 0.2: find the greatest pull the engine is capable of. Ans. 4 tons. 3. A train moving at 40 miles an hour is brought to rest ANGLE OF REPOSE. 133 by friction in half a minute. Prove that the coefficient of friction is 11/180. 116. If the reaction iV" is constant the direction of E will depend on the value of /, that is, of F. Hence, if F is such that the inclination of R to the normal exceeds the angle of friction, motion will take place. This gives us the clue to an experimental method of determining Fsin (6* -)- 0). Get the same answer by resolution of forces.] 7. If the force is horizontal, prove F^ r tan (6^ + 0), 8. If the force is parallel to the plane, prove F= Tf (sin(9 ± jxcos 6) as the body is on the point of moving up or down the plane. 9. In Ex. 6 find the force necessary to push the weight down the plane. Ans. P cos (yS + 0) = (f sin {& — 0). 10. The force necessary to haul a train at uniform speed on a l^ grade is 3.5 times that on the level. Show that the coefiBcient of friction is 1/250. 11. The angle of a wooden incline is 68°. Show that it is impossible to drag a wooden block up the plane by a hori- zontal force, the coefficient of friction being 0.4. 12. The foot of a ladder of length I rests on the ground at A and the top at B against a rough vertical wall : find its inclination when on the point of sliding, the coefiB- cient of frictjou iri each case being 0. 5. 4ns, t,m = \. MOTION ON A ROUGH SUEFAGE. 135 117. Motion on a Rough Surface. — Suppose a particle of mass m to slide on a rough horizontal plane of which the coefficient of friction is )x ; then the normal pressure being mg and the friction pimg, we have for the accelera- tion of motion fl ;= — nmg/m — — fxg. If u is the initial Telocity, we have, by substituting in the equations of Art. 13, V == u — fxgt, v^ = u^ — 2 figs, s = ut — \fjigf, and the motion is completely determined. If the plane is inclined at an angle B, then resolving mg into its components mg sin 6 along the plane, and mg cos 6 at right angles to it, the latter repre- sents the pressure on the plane. Hence the friction is fAmg cos 6, and the force causing motion down the plane is mg sin — jjLmg cos 0. Therefore the acceleration down the plane is found from a = {mg sin — fxmg cos 0)/m — g (sin d — .fA. cos 0) = g sin {8 — 0)/ cos 0, if is the limiting angle of friction. Ex. 1. A body with initial velocity u slides along a rough horizontal plane on which the coefficient of friction is /<: show that it will come to rest in u/yg seconds after pass- ing over a distance ii'/2iug. 2. If a body is projected up a plane whose inclination is 0, show that a = — g sm (ff -\- 0)/ cos 0. 3. A train of w lbs. is hauled along a horizontal track by a const;int pull of p pounds. If the resistance of friction 136 rEicTiosr. is/ pounds, find the velocity of the train in t seconds after starting from rest, and the distance passed over in that time. Ans. s = \(p —f)gt''/w. 4. A train of 100 tons (excluding engine) runs up a 1^ grade with an acceleration of 1 ft. per sec. If the friction is 10 pounds per ton, find the pull on the drawbar between engine and train. Ans. 4| tons. AVe now pass to various practical applications of the principles laid down. Fig. 127 118. Axle, Journal, or Pin Friction. — A beam AB is pierced by a shaft 0, about which as an axis it is acted on by two forces P, Q, of which P is the driving force and Q the resistance. AXLE, JOURNAL, OR PIN FRICTION. 137 Suppose P to be on the point of overcoming Q. The pressure N between the two surfaces being normal, passes through the centre 0. The resultant R of iVand the fric- tion / must pass through C, the intersection of P and Q, and at the same time be inclined to the direction of i\^at an angle . Hence plot P to scale, construct the triangle of forces, and scale off Q and R. The point A, through which the resultant R passes, is the point where the pressures on the axles are concentrated. The reaction N and friction / at A may be found by completing the tri- angle of forces to the left. Notice that if we drop a perpendicular OB on R and describe a circle with radius OB, the direction of 7? is a tangent to this circle. But OB = r sin 8 = 0.1, tan = 0.1, "' " =0.5; tan /3 = 0.1, tan = 0.3, " " =0.3. Hence show the importance of lubrication. CHAPTEK VI. WORK AND ENERGY. 127. When au agent exerts force on a body the effects produced are change of position and change of form. The first of these lias been considered in the preceding chapters, the effect of force being measured by the acceleration pro- duced. All of the results found depend on the experiment- al principles assumed, the laws of inertia, mass^accelera- tion, and stress. We now proceed to study the effects of force from another point of yiew, and to lay down a method of treating these effects more general than that stated hitherto, in that it applies to change of form as well as to change of position. The comparison of results reached by the two methods affords a test of the truth of the new method just as already pointed out, that the truth of the laws of motion is not capable of direct demonstration, but must be tested by con- sequences arising from assuming their truth. In order that change of position or change of form may occur by the action of a force, the body must yield to the force, or, in other words, the point of application of the force must be displaced. When this displacement occurs. Work is said to be done by the force and on the resistance offered. Thus when a body is falling freely, work is done by the force of gravity on the body; while a spring is being bent, work is done by the acting force on the spring; and so on. If a body is lifted vertically upwards, work is being done by the lifting force and against the force of gravity. 148 MEASUKB OF WORK. 149 If, then, we take work in the dh-ection of a force as +, that done against a force must be taken as — . In lifting a weight W (that is, in overcoming the gravity force of W) tlai'ough a height 7i, the work done depends on the values of W and A, and is measured by the product Wh. lu general, the work IT of a force F is measured by the product of the force and the displacement s of its point of application i7i the direction of the force, or K=Fs. This is the case whether the displacement is actually made or is only conceived to be made. In the latter case the work is said to be virtual {— hypothetical). 128. In the deiinition nothing is said about the path of the point of application of the force. If, then, the point of application A of a force ^acting along the line OA be dis- placed to B in any path, and i?C be let fall perpendicular to OA, the distance AB is the total displacement, and the distance ^C is the displacement s ot A in the direction of the force. Hence by the definition K= Fx AC = F X AB cos d = Fcos e X AB; or, the work done hy a force acting obliquely to the path of a body is measured by the product of the force and the projection on its direction of the total displacement, or by the product of the component of the force along the total dis- placement by that displacement. When e = 90°, then cos (9 = and K=0. Hence when the displacement is at right angles to the direction of the force, the work of the force is nil. 150 "VVOKK AND ENEKGY. Ex. In a pendulum find the work done by tlie pull of the rod on the bob as it swings to and fro. A}is. nil. 129. In the general formula for work K=Fs; taking F—1, s = 1, we have K—1; and therefore the tiiiit of work is taken to be the work done by unit force acting through unit distance. Two forms are in common use — the scientific or laboratory unit^ and the engineering unit or unit of every-day life. The scientific unit of work in the F. P. S. system is the work done by a force of one poundal acting through a dis- tance of one foot, and is called the Foot-poundal ; in the 0. Gr. S. system it is the work done by a force of one dyne acting through one centimeter, and is called the Erg. Thus a force of 10 poundals acting through 3 ft. will do a work of 30 f t.-poundals, and 10 dynes through 1 meter a work of 1000 ergs. To avoid inconveniently large numbers, an enlarged unit, the Joule, equal to 10' ergs, is often used in the 0. G. S. system. The engineering unit of work is based on the gravitation measure of force, and is the work done by a force of one pound acting through a distance of one foot. This is usually called a Foot-pound. Thus the work done in raising 100 lbs. vertically through 6 feet is the work done in over- coming a weight (= gravity force) of 100 pounds through 6 feet, and is 600 foot-pounds. In the metric system the engineering unit is the kilogrammeter (kgm.), which is the work done by a force of one kilogram acting through a dis- tance of one meter. 130. The dimensions of the unit of work will be the dimensions of F X s. The dimensions of F are ML/T' (Art.41) and of s, L, Hence the dimensions of Kure ML'/T'. CONDITION OF EQUILIBRIUM. 151 This is the same result as that obtained for the dimensions of the moment of a force* (p. 95). Ex. 1. Prove 1 kilogrammeter = 7.23 foot-pounds. 2. ProYe 1 foot-pound = 13.56 X 10° ergs, approx. 3. Prove 1 foot-poundal = 0.42 X 10° ergs, nearly. 4. The tractive force of a consolidation engine is 10 tons. Pind the work done in hauling a train one mile. Alls. 105600000 ft. -pounds. 131. When the direction of displacement of the point of application is not in the line of action of the force considered, the body must be acted on by other forces, and the work done estimated for each separately. Thus consider a particle of weight Wto rest against a smooth vertical wall, and to be raised vertically by a force F acting at an angle to the vertical. Let AB he the displacement. The work done against gravity is W X AB. The work done by the reaction i\^is nil. The work done by FisFx BG. If the motion is uniform, that is, if the forces acting are in equilibrium, we have, by resolving vertically, i?'cos e ~ W-Q. Multiply each member by AB, and F cos 6 X AB - Wx AB = 0, or Fx BC - Wx AB = Q; or, when equilibrium exists, the sum of the works of the forces acting at the point is zero. * There is really a difference between the dimensions of the two, as unit work is the product of a force into a displacement in the direc- tion of the force, and unit moment is the product of a force into a distance perpendicular to the direction of the force, so that the latter is strictly \^'^MU>/T\ 152 WORK AND ENERGY. 132. More generally, let several forces i^^ , -F,, ... act p. ijg at a point causing a displacement OA. Let i2 be the resultant of the forces. From A let fall perpendiculars on the directions of the forces, and let 6^, 0^, . . . be the inclinations of these directions to OA, and 6 the inclination of the resultant E. Then (Art. 50) 72 cos ^ = F, cos ^, + i?; cos ^, + . . . , or i2 X Oa/OA = F, X Ob/OA +F,X Oc/OA + . . . , or Ex Oa = F,X Ob + F^X Og+ . . . ; or the work done by the forces is equal to that done by their resultant. The equation may be written 0=- Ex Oa + F,X Ob + F^X Oc+ . . . , which shows that the algebraic sum of the works done by a system of forces acting at a point and in equilibrium is equal to zero. Conversely, if any number of forces act at a point, the condition of equilibrium is that the sum of the works done for every displacement shall be equal to zero. For the sum of the works is equal to the work of the resultant, and for equilibrium the work done by the resultant must be equal to zero for each and every displacement. This is the Princi- ple of Work as applied to forces acting at a point, or to forces acting on a particle. PRINCIPLE OF WORK. 153 133. The principle of work may be regarded as included in the law of stress (Art. 27), the work done by the forces F^, F^, ... corresponding to the action, and the work done by —R, or rather the work done against R, to the reaction. In this sense we have merely the law in another form, the action of the agent being measured by the work done by it, and the reaction of the resistance by the work done against it. Looked at in this light, the principle of work falls within our old lines, and is therefore consistent with them. 134. It follows by summing np from particle to particle of a rigid body, that the condition of equilibrium is that the sum of the works done by the forces acting is for every dis- placement, equal to zero. For, the body being rigid, no work is done as the forces are transferred from particle to particle, since there is no yielding. Hence it is necessary to take into account the external forces only. The same principle will evidently apply to a system of bodies rigidly connected,or so connected that the geometrical relations existing among the parts are not disturbed by the displacement, and hence to what we call a Machine. A machine is so constructed that the configuration of the parts is not disturbed when it is in operation. The external forces, that is, the driving force and the resisting force, form a system in equilibrium; and hence the Fig. I40 relation between them follows . .i,b at once by equating to zero ' '~ ' the sum of the works done. The work done by the driving and resisting forces being con- trary in character, will be de- noted by opposite signs. \^ Ex. 1. In the straight lever to find the relation between F and W. -""' W-f. 154 WORK AND ENEEGY. [Let the point of application A of the force F descend a distance x, and £ consequently ascend a distance y. Then Fx-Wy = 0. But from similar triangles x:y = AO:Ba .-. FxAC=WX BG, which is the "principle of the lever" (Art. 88).] 3. In a pulley tackle the driving force descends 1 ft., while the' weight to be raised ascends 1 in. What force will raise 1 ton? Ans. 166f pounds. 3. In a bell-bottom jack-screw (Fig. 136) with force F applied at the end of a lever arm I a body of weight W is being raised with uniform velocity. Prove F X2nl — W X pitch of screw. [Notice that while the lever arm makes one turn, the weight is raised a distance equal to the pitch of the screw.] 4. Find the relation between F and W in the copying- press (Fig. 73), 21 being the length of the handle. Ans. F=W X pitch of screw/4;rZ. 5. In a telescopic jack-screw a smaller screw G works in a companion nut cut in the larger screw D, which latter works in a nut in the fixed block B. The block A being fixed, the upper screw does not rotate. If I is the length of the lever arm, find the re- lation between F and W. Ans. Fx 2;r; = lf Xdifl. of pitch of screws. 6. In a hoisting machine (Fig. 142) the gears are 36 to 36 teeth, the drum 21 in. dia- meter, and the load for one horse 1^ ton. Find the pull exerted by the horse at the end of a 7 ft. horizontal lever. Ans, 375 pounds. Fig. 141 WOKK AGAINST FRICTION. 155 7. In a derrick winch (Fig. 80) the crank is I in. leverage, Fig. 142 Find the Fig. 143 the gears n to 1, and the drum d in. diameter, two-man-power capacity, each man exerting a force of p pounds. Ans. ^pln/d pounds. 8. In a combination of single- threaded worm and wheel used in hoists, the worm wheel has n teeth, the lever handle is I in. long, and the radius of the drum around which the lifting rope winds is r in. Find the relation between F and W. Ans. Fin — Wr. 135. To overcome friction, as to overcome any resistance, a certaiin driving force or effort is necessary. In a machine, therefore, work must be done to balance the friction that arises. Part of the effort commu- nicated is taken up in doing this work, which is, as it were, absorbed. This work may serve no useful purpose, and is therefore said to be wasted. This does not mean that the work is lost, as we shall see presently. The work absorbed by friction is measured as the work 156 WOBK AND ENERGY. Fig. 144 ,, done by any other resistance. We have Work against fric. = fric. X dist. desc. = Pfxs, where P is the normal pressure on the journal, fi the coef- ficient of friction, and s the distance described. Thus the work done in overcoming the frictional resist- ance in one revolution of a journal of diameter d and carry- ing a load of W lbs. is TT/^ X Tcd ft.-pounds. In case the journal revolves n times per minute the work absorbed is PF)^ X Ttdx n ft.-pounds per minute. Ex. 1. Find the work done in hauling a sled weighing 500 lbs. half a mile, the coefl&cient of friction being 0.3. Ans. 264,000 ft.-pounds. 2. Show that the work done in hauling a body of weight PF up a rough incline AG is equal to that done in hauling it along the level AB, the coefficient of friction being the same, and in raising it through the height BG. [For Wis equivalent to IF sin 6 .along the plane, and W cos 6 at right angles to it. Hence the force of friction along the plane is fxW cos 6. The total efEective force required to move the body up the plane is W siu e -[- )xW cos e. Work done in moving from ^ to C = (Fsin 6* + /ipFcos e)AC = WX BG+fiWxAB, which proves the proposition. For moderate inclines the work done may be taken equal to WxBG+mWxAG.] 3. Find the work done in hauling a train of 100 tons one mile up a 1^ grade, the resistance being 8 pounds per ton.- Ans. 2800 X 5380 ft.-pounds, nearly. "WORK Olf A VAllIABLE FOllCE. 157 136. We have considered the force acting and the resist- ance to be overcome as uniform in their action. It more frequently happens that they are variable. For example^ the pressure of a compressed spring, the pressure on the ■ piston of a steam-engine before and after the steam is cut off, etc. In cases of varying resistance, the work done is the same as would be done by a force acting uniformly, and which is equal to the average of the varying forces. This average as in the case of velocities is most conveniently estimated graphically (Art. 8). Thus let AL plotted to scale represent the distance passed over, by the point of application of the force, and let Aa plotted to scale represent the force at ^. If the force is uniform throughout, the work done [^= AL X Aa) is repre- sented by the area of the rectangle A I. If the force is variable, let the distance AL he divided into a large number, n, of Fig. 145 F!g. 146 BC to \>el{Bb-{-Cc), etc. equal parts AB, BO, . . . and let Aa, Bl, . . . represent the corresponding forces. The average force acting through the distance AB may be taken to be \{Aa + Bb), through Hence, by addition. Work done = ^^(^a + 2^5 + 2Cc + . . . + LI). If AB, BC, . . . ure indefinitely small, the curve abc . . .1 becomes continuous, and represents the varying action of the force. It \& called the Curve of Resistance. The total work done would be represented by the area AalL. By means of certain contrivances the curve of resistance may be plotted mechanically by the resistance itself, as, for 158 WOEK AND ENERGY. example, in the steam-engine by the Indicator, Having the curve, the mean resistance A may be found by stretching a string so as to have equal areas above and below it. Or the area may be read off at once by an Amsler polar planim- eter, and the work done found directly. Or the indicator drawing or "card" may be divided up by drawing equi- distant ordinates, the lengths of these ordinates scaled off, and the formula above applied. All of these methods are at times useful. 137. Power. — It is important to notice that the term work, as defined above, is not in all respects the same as what is called work in ordinary language. Ordinarily the idea of time enters, and the idea of motion is not essential. A man merely supporting a load does not come under the mechanical definition, no matter how long he may support it, though he is doing work in the ordinary sense of the term. The definition is somewhat arbitrary, which is quite allow- able so long as we use the term consistently with the defini- tion. It is plain from the definition that any small force can do work of any magnitude provided sufiBcient time is given. Hence, in order to compare agents which do work, it is necessary to take tl^e time employed into consideration. To indicate the amount of work performed in a given time, the term Power is used. By the unit of power we mean the power of an agent which can do unit work in one second. This is expressed as foot-poundals per second, or ergs per second, or foot-pounds per second, according to the system of units employed. Thus to raise 1 ton of coal through 100 feet in 10 min. would require an expenditure of 2000 X 100 ft. -pounds of work in 10 min., or of 20,000 ft.-pounds per minutfe, which would be the expression of the power of the agent. When the power of an agent doing work is great, no very definite idea is conveyed by the large numbers of the above POWER. 159 units that indicate this, and accordingly multiples of the units are used. In engineering work the multiple unit employed is called a horse-power. The term horse- power was introduced by James Watt. As horses formerly did the work done by steam-engines, it was natural to institute a comparison. Watt found that a Clydesdale horse could walk 2J miles an hour, and at the same time raise a weight of 150 lbs. This is equivalent to 2| X 5280 X 150/60 = 33000 foot-pounds per minute. The agent, therefore, which could do a work of 33000 ft.- pounds per minute, or 550 ft. -pounds per second, was named a horse-poioer. The enlarged unit used in electrical work is the watt. It is the power of an agent which can do a work of one joule (= 10' ergs) per second. The relations between these units will be seen in the ex- amples. Ex. 1. Show that one H. P. = 550 ft. -pounds per sec. = 396,000 inch-pounds per min. = 198 inch-tons per min. 2. Show that one H. P. = 550^ ft.-poundals. per sec. 3. Show that one H. P. = 746 watts. 4. Show that one joule = f ft. -pound, approx. 5. The French H. P. is 75 kilogrammeters per ^c. ; show that it is about t^ less than the British H. P., and that it is equal to 735f watts. 6. Show that one watt = | ft.-pound per sec, approx. 7. 23 tons of coal are to be hoisted through 50 yards in 10 min.: find the H. P. of engine necessary. Ans. 20 H. P. 8. How many gallons of water would be raised per minute from a mine 600 ft. deep by an engine of 175 H. P. ? Ans. 1152 gals. 9. A belt passing round two pulleys moves with a velocity of 10 ft. per sec. : find the H. P. transmitted if the differ- ence of tension of the belt above and below the pulleys is 1100 pounds, Ans. 20 H, P. 160 WOEK AND ENEKGT. 10. Show that the dimensions of power or rate of doing work, that is, of Fs/t, are MU/T\ 11. A shaft 14 ft. in diameter is to be sunk in gravel in 10 days of 10 hours each. Taking the weight of the gravel at 100 lbs. per cubic ft., find the H. P. required. 138. For determining the power developed by a steam- engine or other machine, the Prony Brake is used. The idea is to balance the work done by the machine by a fric- tional resistance, compute this resistance, and thence find the power of the machine. The brake ahsorhs the work to be measured. Let be a shaft of radius r, to which the brake AB is fastened. By means of the screws a, i, the friction of the Fig. 147 brake on the shaft may be regulated. Suppose it adjusted so that the engine develops a friction /, just sufficient to balance a body of weight W placed at the end A of the beam. Then the moments of /and IF about must be equal, or fr= Wl. Suppose the shaft to revolve uniformly n times per min- ute. Then, assuming that the friction for uniform motion of the shaft is the same as at the point of just beginning to move, we have Work done in one min. = friction developed in n revs. = /X 2n-r X n = 27cnWl. POWER. 161 If W is expressed in pounds and I in feet, then the H. P. = 37rwM/33000 = 0.00019W Wl. Similarly, in the mechanism shown in Fig. 131, Art. 132, if the H. P. transmitted by the belt and the velocity v of the belt per second were given, then the tensions P, Q of the belt would be found from H. P.= (P - Q)v/550; P = Qei^^ or, if the H. P. and the number of revolutions per minute n of the driving pulley were given, the tensions would be found from H. P. = (P - Q)7rdn/33000, P = Qe''^ when d is the diameter of the pulley. A common case is when /t = 0.3, and from 0.4 to 0.5 of the smaller pulley is embraced by the belt. Then d varies from 0.8;r to tt, and P = 2Q, nearly. Also, the H. P. = 0.00005Pdn, P being the tension of the belt on the taut side. 139. In a machine, owing to friction between the pieces, part of the work done by the driving force is wasted, so that the resulting useful work done is less than the total work done by the effort in the first place. We have, in fact. Total work = useful work + useless work. The ratio of the useful work to the total work is known as the EflEiciency of the machine. Or, since the total work is given by the indicated horse-power (Art. 136) and the useful work by the braked horse-power (Art. 138) we may define efificiency to be the ratio of the B. H. P. to the I. H. P., or EfBciency = B. H. P./I. H. P. 163 WORK AND ENERGY. Ex. 1. In testing a Corliss engine running at 100 revolu- tions per minute, the lever arm was 10^ ft., and the weight at A 2000 Ibsi : find the H. P. developed. Ans. 400 H. P. 3. "AC. and 0. electric motor shows on a Prony brake a pull of 5 ounces on a one-ft. lever, that is, 2 ft. -pounds per revolution, or about ^ H. P. at 1500 revolutions per min- ute." Check the conclusions in this statement. 3. In a Corliss engine running at 100 pounds pressure and 100 revolutions per minute, the diameter of the cylin- der is 18 in., and length of stroke 43 in. If the brake was used on a pulley 6 ft. in diameter, and keyed to the engine shaft, find the friction on the face of the pulley. Ans. f — 9450 pounds. 4. A 6-ton fly-wheel on a 14-in. axle makes 90 revolutions per minute. Find the H. P. absorbed in friction, the co- efficient of friction being 0. 1. Ans. 13 H. P. 5. A steam hoist of 3 H. P. is found to raise a weight of 10 tons to a height of 50 ft. in 20 min. How many ft.- pounds of work are wasted by f i-iction in a day of 10 hours ? Anx. 39,400,000 foot-pounds. 6. The tractive force of a consolidation engine is 10 tons : find the H. P. exerted in hauling a train one mile in 3 min. Ans. 1600 H. P. 7. A pumping engine of piston area 100 sq. in., steam- pressure 60 pounds per sq. in., length of stroke 3 ft., and number of revolutions per min. 35, raises 500 gallons of water per minute a height of 50 ft. : find the efficiency. Ahs. 0.25, nearly. 8. A traction engine weighing 5 tons hauls a load of 10 tons at 8 miles an hour, the resistance being 20 pounds per ton : find the H. P. exerted. Ans. 6.4 H. P. 9. A train weighing 100 tons runs at 43 miles an hour on ■ a level track, the resistance being 8 pounds per ton: find its speed up a 1^ grade (1 ft. rise in 100 ft.) if the engine power is unchanged. [Total resist. = 8 + 2000/100 = 28 pounds per ton. /. 8 X 43 = 28 X a;, and a; = 12 miles an hour.] 10. A traction engine weighing 5 tons can haul 15 tons on a level, the coefficient of friction being 0.02 : find the net load it can haul up a 1^ grade. Ans. 8^ tons. POWER. 1G3 11. A train of 100 tons is hauled by an engine of 150 H. P. The resistance is 14 pounds per ton: find the great- est velocity that the engine can attain. Ans. 60 miles an hour, nearly. 12. Check this statement : " 55 pounds meau effective pressure at 600 ft. piston speed gives 1 H. P. for each sq. in. of piston area." 13. Prove H. P. of an engine = SNAP / :i200Q, where S= stroke in ft., iV^ = number of strokes per mm., A = area of piston in sq.in.,/' = mean steam pressure in pounds per sq. in. of piston area. 14. Find the work done per hour at the crank-pin of an engine revolving 40 times a minute and acting against a re- sistance of 7000 lbs., the radius of the crank being 18 inches. (See Fig. 119.) Ans. 27r X li X 7000 X 40 X (10 ft.-pounds. 14a. In the Strong locomotive 444, built for the L. V. R. R., the cylinders are 20 in. in diameter, the stroke is 24 in., and the diameter of the driving wheels 62 in. At 160 lbs. steam-pressure per sq. in. find the work done at each stroke. Ans. ^ X 20' X 2 X 160 foot-pounds. 145. Find the tractive force P of the engine. [In each stroke of the piston ihe drivers make a half revolution. Hence, there being two cylinders, 1^ X 20' X 24 X 160 = I X 62 X P,and P is found.] 15. Show that the cylinder diameter of an engine that will produce n horse-power at a piston velocity of s ft. per minute under a mean effective pressure of p pounds per sq. in. is 210 y — inches, nearly. 16. The driving pulley (Fig. 131) runs at 100 revolutions per minute, and is 2 ft. in diameter. The engine is 3 H. P. Find the tensions of the belt if ^ of the circumference of the driving pulley is covered, and the coefiBcieutof friction is 0.3. 164 WORK AND ENERGY. 17. In the California Street cable road, San Francisco, the total H. P. required to haul the cable alone is 84, and the speed is six miles an hour : iind the total pull transmitted by the driving drum. 140. Energy. — We have seen that when the forces acting on a body are in equilibrium, or the body moves with a uniform motion, the sum of the works done is zero, that is the work done by the resultant force or effort is equal to that done on the resistance. Now the action of a force is to cause acceleration. If, then, the motion is uniform, the acceleration caused by the effort is balanced by the equal and opposite acceleration caused by the resistance. But if the acceleration caused by the effort exceeds that caused by the resistance, velocity is gained, and the motion is not uniform. We proceed to inquire as to the work necessary to be done in order to change the velocity of a body of mass m from say u ft. per sec. to v ft. per sec. Let P denote the effort or acting force, a the acceleration produced, and s the distance passed over in the line of action of the force; then (Art. 13) as = Jw" — ^m'. But (Art. 34) F = ma. Hence, eliminating a, Fs = Imv' — ^mu". Now Fs is the work done by F in passing over a distance s in its line of action, and therefore a mass m in having its velocity changed from ?« to v feet per second must have Imv' — ^mti" units of work done on it. If the force F does not act uniformly, we have, from Art. 34, F=md's/dt', and ENERGY. 1(55 or, tlie loorTc done depends on the initial and final velocities, atid is independent of the intermediate velocities. If M = 0, or the body starts from rest, Fs — ^mv', and therefore the work done in giving a body of mass via, velocity v is ^»u'^ units of work. The force F which will generate a velocity v in acting through a distance s will destroy the same velocity if acting through the same dis- tance in the opposite direction ; in other words, the body by virtue of its velocity v can do a work Fs units in giving up that velocity and coming to rest. This capacity which the body possesses of doing work in consequence of its velocity known as its vis-viva or Kinetic Energy. Hence the meas- ure of the kinetic energy possessed by a body of mass m and velocity v is ^mv^ units of work. In acquiring the velocity . V by the work done on the body energy may be said to be stored in it, to be restored in doing work as it parts with this velocity and returns to its original condition. We may therefore state the general relation Fs = \mv'' — ^mu" in the form : If a body or system of bodies with configura- tion remaining the same is in motion under the action of force, the work done in passing from one position to another is equal to the corresponding change of the kinetic energy. This is called the Principle of Kinetic Energy. Ex. 1. Find the work done in stopping a 100-lb. shot moving with a velocity of 1000 ft. per sec. Ans. 1,562,500 ft. -pounds. 2. Find the force exerted in stopping a train of 250 tons in 1000 ft. from a velocity of 30 miles an hour. Ans. 15,125 pounds. 3. A shot pierces a target of a certain thickness h : show that to pierce one of 4 times the thickness twice the veloc- ity is necessary, [irnv'h = l-mv^' xih. . • . v, = 2w,.] 166 WORK AND ENERGf. 4. A blacksmith's helper using a 16-lb. sledge strikes 20 times a minute, and with a velocity of 30 ft. per sec: find his rate of woi;k. Aois. 3/22 H. P. 5. A stone is thrown with a horizontal velocity of 50 ft. per sec. : find the velocity with which it strikes the ground which is horizontal and 6 ft. below the point of projection. Alls. 53.7 ft. per sec. 6. Show that to give a train a velocity of 20 miles au hour i-equires the same energy as to lift it vertically through a height of 13. 3 feet. 7. A hoisting engine lifts an elevator weighing 1 ton through 50 ft. when it attains a velocity of 4 ft. per sec. If the steam is shut off, how much higher will it rise ? Ans. ^"^'^-^ 1 = 2000^ X dist. 8. In (7) find the time of rising 50 ft., supposing the mo- tion uniformly accelerated, and also find the H. P. of the engine. Ans. 25 sec; 7.3 H. P. 9. Show that the energy stored in a train of weight W lbs. and moving with a velocity of V miles per hour is WVyso foot-pounds. 10. A train of 100 tons is running at 30 miles an hour up a 2^ grade : find the H. P. required, the resistance on a level being 10 lbs. per ton, due to axle friction chiefly. Ans. 400 H. P. 11a. In the Strong locomotive (L. V. R. R. 444) running on the level at 30 miles an hour the tractive force is 8 tons. Taking the resistance of friction as 10 lbs. per ton, find the number of 20-ton cars that can be hauled if engine and tender weigh 100 tons. Atis. 75 cars. lib. Find the number that would be hauled up a 2^ grade. Ans. 11 cars. lie. Find the H. P. exerted in the former case. Ans. 1280 H. P. 12a. An engine exerts on a car weighing 20,000 lbs. a net pull of 2 lbs. per ton : find the energy stored in the car after going 2^ miles. Ans. 264,000 ft. -lbs. 12b. If shunted to a level side-track when the frictional ENERGY. 167 resistance is 10 lbs. per ton, find how far it will run before coming to rest. Ans. 264,000/10 X 10 ft. = | mile. 12c. If shunted on a side track with a 1^ grade, how far will it run before coming to rest ? A71S. 264,000/(20 + 10) 10 ft. = \ mile. 12^. If there are brakes on half the wheels, and these are applied with a pressure of half a wheel load, how far will the car run up a Ifo grade, the coefficient of friction be- tween wheel and brake-shoe being 0.2. [Total resist. = brake + grade + fric. = 130 lbs. per ton. Ans. 203 ft., nearly.] 13. In the Westinghouse brake tests (Jan. 1887) at Weehawken a passenger train moving 22 miles an hour on a down grade of 1^ was stopped in 91 ft. There was 94^ of the train braked. Taking the frictional resistance as 8 lbs. per ton, find the net brake resistance per ton and the grade to which this is equiyalent. [The brake has to overcome the energy due to the ve- locity and the i esistance due to the grade, but is aided by the resistance due to friction. Hence brake resist, per ton = 2000 X 227(30 X 91) -f 20 - 8 = 367 pounds. As only 94^ was braked, we have the net brake resist, per ton = 367/0.94 = 390 pounds, which is equivalent to a grade of 390/20 {- 19.5) per cent.] 14. The tractive foi'ce of an engine is P tons. If the weight of engine and train is W tons and the frictional resistance n lbs. per ton, show that in going up ana^ grade the velocity acquired in t seconds from rest will be Qgt ft. per sec. and the energy Q.bWQ^gL'' ft. -tons, where Q-P/W- a/100 - m/3000. 141. The term energy arises from the consideration of muscular exertion in the first place. In doing what is called work in ordinary language we recognize that efPort is needed, and that exhaustion follows after a time. It is necessary to store up work-capacity or energy by consum- ing food in order to be able to continue doing work. The same idea is extended to machines where the force exerted by the expansion of steam, by water in motion, by air in motion, etc., produces effects which can also be ob- 168 WORK AND ENERGY. tained by muscular exertion directly applied. The machine is then said to be doing work, and from analogy the term agent is applied to it as well as to man. The agent thus forms, as it were, the converter of energy, whether of food, or of coal, etc., into work.* The energy existing in a stored- up condition is ready to be called on, and is hence known as Potential Energy. Suppose we call on the potential energy of steam to give motion to operate a mechanism, as for example a locomotive. The energy of the steam has enabled the locomotive to overcome the resistance offered by the friction of the wheels, the resistance of the air, etc., and to attain besides a certain velocity. If the steam be shut off the locomotive will not at once come to rest, but will continue for a time to overcome resistance as before. The energy communi- cated in excess of that required to overcome the resistances is not lost, but is stored in the form of motion. The energy existing in virtue of the motion, which thus continues to do work until exhausted, is, as stated above, called Kinetic Energy. It would thus seem that the energy of motion is produced at the expense of energy stored, and conversely the energy of motion may be made to do work of some kind to be stored in some form or other. The subject is a very large one, and we cannot go into it in detail. It is sufficient to * "Now. Buckland," said Stephenson, "I have a poser for you. Can you tell me what is the power that is driving that train?" "Well," said the other, " I suppose it is one of your big engines." " But what drives the engine?" " Oh, very likely acanny Newcas- tle driver." "What do you say to the light of the sun?" "How can that be?" asked the doctor. "It is nothing else," said the en- gineer; " it is light bottled up in the earth for tens of thousands of years — light absorbed by plants and vegetables being necessary_ for the condensation of carbon during the process of their growth, if it be not carbon in another form; and now, after being buried in the earth for long ages in fields of coal, that Intent light is again brought forth and liberated, made to work as in that locomotive, for great human purposes." — Smiles. POTENTIAL. 169 state, that, as a result of observation and experiment, the conclusion is arrived at, that energy may be transferred from one form to another, but can neither be created nor de- stroyed. In a word, energy is indestructible; or, as it may be expressed : The total energy of any lody or system of bodies is a quantity wliich can neither be increased nor diminished by any mutual action of these bodies, though it may be transformed into any of the forms of which energy is susceptible. This principle, known as the Conservation of Energy, is the general principle premised at the beginning of this chapter. It involves the laws of motion and the principle of work as special cases, and consequently on it may be made to rest the whole subject of mechanics. 142. Potential.* — An important application of the prin- * A graphical illustration of the fact that iu the case of two elec- trified spheres the potential function is a measure of work, is given by Prof. A. M. Mayer as follows: Suppose an electrically charged sphere fixed in space with its centre at 0, and that another sphere charg- ed with a unit of similar electricity I pj- 143 is pushed towards from an infinite * distance along the line OX, and that the electric strain on the mov- ing sphere causes (without work) a vertical rod to slide out of its top in proportion as the stress between the spheres increases. As the sphere progresses along OX it will thus mark at each point of its progress the repulsive force existing between it and the fixed sphere. The end of the sliding-rod during the motion of the sphere from X towards will have traced out the curve DFGO, whose ordinates are as the inverse squares of their distances from 0. The potentiaV at any point reached in the progress of the charged body towards — work done = resistance overcome in pushing body from infinite distance to that point; and this work done is measured by the sum of the resistances at each point of the path X length of path. But this product is equal to the area included between the 170 WORK AND ENERGY. ciple of work is the determination of the work done dur- ing the passage of a body of given mass from one position to another under the action of a force of attraction or of repulsion on the body. The region in which the force acts is called the Field of Force. We shall confine ourselves to fields of force, in which the law of force is that of the inverse square of the distance (Art. 110), and the forces themselves are gravitational, electrical, or magnetic. Thus suppose a particle of mass m placed at to exert an attractive force on a unit mass ^3" ' in its motion from A to B, the law of force being that of the inverse square of the distance. The force F of attraction between the particles is cm./r'', when they are at a distance r from one an- other, c being a constant. The work done by the force as the particle moves the indefinitely small distance CD in its path is CYj-jy X DJS, when OU is perpendicular to OD. Or putting OD = x, DE — dx, and noting that the force ordinate (say B) of path, the axis of X, and the curve, both indefin- itely extended; or say CBAD. The equation of the curve is y = a/ai'. Area ABGD, indefinitely extended; = ydx = / ata/a? = a/m. ox Or area indefinitely extended, which represents the work, is in- versely as the distance of y, the bounding ordinate of area, from or K = Q/d, when Q = quantity and d = distance of the gentres of the two charged spheres. POTENTIAL. 171 being attractive dx is negative, this may be written til/ — —J dx. Hence the work done in bringing the particle from ^ to ^ is C'- dx (I IN — cm I —cm[ , tya X \r al when OA — a, OB = r. If the starting-point A is at an indefinitely great distance from the source of attraction 0, then 1/a = zero, and the work done is cm/r, a result independent of the form of the path AB. The work necessary to bring a particle of unit mass from a position of zero attraction, that is, from an indefinitely great distance, to a position B, where the force of attrac- tion is finite and the distance from the attracting particle r, is called the Potential at the point B. It is usually denoted by the letter F, so that V — cm/r, when r is the distance of B from 0. If the field at consists of several particles of masses m^, m^, . . ., with distances r, , r^, . . . from the unit mass B, the potential of B will be c{inJr^-\-mJr^+. . .). It follows that if F^ denote the potential at any other point C the work done in bringing the particle from C to B by any path is F — F, . 143. Conceive a surface at every point of which the potential has the same value. Such a surface is called an Equipotential Surface. If a particle is moved from any one point to any other point on this surface no work ig done. Thus, if a railroad track is located on a " level " no work is done against the forc^ of gravity in hauling a train op thjs track. 172 WOEK AND ENEKGY. An equipoteiitial surface for tlie force of gravity could be determined by finding the places at which the same mass would compress a spring the same amount. Eoughly, it would be a sphere concentric with the earth. The work done on any body by the force of gravity in falling to the earth from any point of this equi potential surface would be the same in amount. So also the work done in falling from one such surface to another would be constant in value. This suggests a method of finding the relation between the resultant force R and the potential V at any point. For suppose the potential at points A, B situated on two equi- potential surfaces to be V^, F^, respectively. The result- ant force i? at ^ (as force of gravity on a falling body) will be in a line AC, normal to the surface at A. The work done in moving a particle of unit mass from AtoB will be the same as from A to C, when Cis the point in which the normal at A meets the second surface, and is equal to R xAC. But this work is by definition equal to the difference of potential at A and B, or V — V Hence EXAG=V^-V^. If the distance between the surfaces is indefinitely small (= dr), this may be written R:==dV/dr, or the resultant force is the rate of change of potential at the point in question, the direction of the force being normal to the equipotential surface through the point. Potential being expressed in terms of work done, the wiii of potential is the same as the unit of work, the foot- poundal or erg. . The theory of. the potential is of great use in magnetic and electrical investigations. Ex. 1. Show that the dimensions of potential are ML/T. POTENTIAL. 173 3. Show that (theoretically) an equipotential surface eonld be determined by finding the points at which a pend- ulum beating seconds had the same length. 3. To find the potential of a particle of unit mass on the line through the centre Cand perpendicular to the plane of a uniform circular disc of radius a, thickness h, and density 6, the distance OC being = i. [As in Art. 110, mass of ring = 27tr X dr X h X S. Hence F= fc X %Tth8rdr/0D . = r c X 27rhdrdr/ Vlf + r" = 27cchd{ V¥T^ ~ i)-] 4. In (3) find the potential at the centre of the plate. Ans. ZrtchSa. 5. Find the potential at the centre of a circular wire of density d and indefinitely small thickness h. Ans. ZncliS. 6. Find the potential at any point within a spherical shell of mass w( and radius a. Ans. cm/ a. 7. Find the potential at a point without a spherical shell of mass m, radius a, and distant h from the centre. Ans. cm/i. 8. Assuming the earth to be a sphere of 8000 miles diameter, prove that the potential of the unit mass 1 lb. situated on the earth's surface is — 21,120,000 foot-pounds. 9(7. In a series of concentric spherical equipotential sur- faces the distance between any two is proportional to the square of the geometric mean of the distances from the centre. 95. Hence show that at great distances from the earth's centre the pound mass must be moved over a long path in order to do a ft.-pound of work on it. 9c. For example, at the moon, 240,000 miles distant from the earth's centre, find the shortest path.. Ans. 3600 feet. CHAPTER VII. KINETICS OF A RIGID BODY. 144. The term rigid body is used in the sense already defined of a body regarded as composed of particles so con- nected that no part of the body can be moved relatively to any other part. As in the case of a single particle, we shall consider the nature of the motions of the particles of the body, without reference to the forces causing the motions. This forms a problem in kinematics. A rigid body is fixed in position by fixing three points J, B, C, not in the same straight line. These points determine a plane ABC with reference to which the positions of all points in the body may be defined. Hence the displace- ment of any point in the body may be determined by not- ing the change in the positions of these three points, since the point in question must keep in a fixed position relative to the three points A, B, G during the motion. If the two points A, B remain unchanged in position during the motion, the third point G must describe a circle about an axis through A, B, and the motion is one of rotation. If only one point A is fiLxed, the points B, G may be brought from their initial to their final positions B^ , C, by two rotations. For by one rotation AB may be brought into the position AB^, and by revolving about AB^ as an axis the plane A B^G may be brought into the position AB,G^. If no point is fixed, the new position A^B^O^ may be reached by a translation of J to ^,, and by two rota- tions bringing .1,Z?(7 successively into the positions A^B^C 174 PLANE MOTION. 175 and A^B^G^ . Hence every motion of a rigid body is either a motion of translation or of rotation, or some combination of the two. The most important case is that in which the particles of the body move in parallel planes. Such a motion is called plane mo- / ^' tion. Fig. ISO / Let us consider the motion of a /' g line joining the points A, B in the /' /"x ^ plane of the paper. A translation ^^ / / / /"'> AA^ and a rotation through an P"---, / 1/ ,-^'^^^ angle B'A^B^ will displace AB into J, -/---^^o the position A^B^ . So far as the j /y-^ displacement itself is concerned, it v-^ makes no difference whether the translation and rotation occurred simultaneously, or not. But this displacement might have been produced by a rotation only. For bisect AA^, BB^ in a, b, and let the perpendiculars aO, bO intersect in 0. Then evidently 0A = OA,,OB = OB, , and angle AOB =A,OB,. Hence the displacement may be produced by rotation about the point 0. It follows therefore that any plane motion may be regarded as a motion of rotation about a centre in the plane of the motion, or, in other words, that a translation and a rotation about an axis perpendicular to the direction of rotation may be combined into a single rotation giving an equivalent motion. 145, In many cases the final displacement is difficult to arrive at. This is particularly the case in mechanisms where the connections of the parts are often very complicated. Besides, the final displacement may not define clearly the intermediate displacements. It is therefore necessary to study the displacement from instant to instant of the mo- tion. 176 KINETICS OF A EIGID BODY. Suppose two points A, B ot a body to have any motion in the plane of the paper. The points A, B will each trace out a path. Consider A. The line joining two consecutive positions of A will give the direction of motion in the path. This line is the direction of the tangent to the path at A . Since an indefin- itely great number of curves may have a common tangent at a point, it follows that this tangent is quite independent of the form of the path. Hence for the instant we may consider the path to be a circular arc. The perpendicular AO to the tangent will pass through the centre of the circle, and conversely, the direction of motion at A for the instant will be perpendicular to the radius of the circle. Hence the instantaneous motion of A is the same as if it took place in a circle with centre somewhere on A 0. Similarly, the motion of B is the same as if in a circle with centre somewhere on BO. But is common to AO and BO. Hence the instantaneous motion of A and B, and therefore of the line ^-8, is a motion of rotation about a point 0, which is called the Instantaneous Centre. An axis through perpendicular to the plane of the paper is the Instant- aneous Axis. The points A, B are any two points in the body. Hence, whatever the 'form of the body, and whatever its plane mo- tion, it is always possible to find a point such that for the instant the motion about it shall represent the actual motion, in other words, a± any instant one point is at rest, and the other points are moving in directions perpen- dicular to the lines joining them to this point. If in Fig. 151 the radii ^0,^0 do not intersect, the tangents to the paths at A, B are parallel, and the motion is a motion of translation. The radii being parallel may be said to intersect at infinity, and hence a motion of transla-' ISrSTANTANEOUS AXIS. 177 tion may be regarded as a rotation about a centre at an in- finite distance. 146. In general, the instantaneous centre will vary in position from instant to instant. The locus or path de- scribed by it is called a Centrode.* But in case the radii A0-, £0 continue to intersect in the same point 0, as the motion progresses the instantaneous centre becomes a permanent or fixed centre. For example, a wagon wheel revolves about the axle as a permanent centre, but with reference to the ground it revolves about the point of con- tact as an instantaneous centre. The path traced by the wheel on the ground is the centrode. Ex. 1. A ladder BG slides between a vertical wall and the ground, which is horizontal: find the instantaneous centre and the centrode. [The paths are along AB, A V. Hence the instantaneous centre is at the intersec- tion of the perpendiculars 50, CO. It is evident that AO = BO, the length of the ladder, and .■. is at a constant dis- tance from A. Hence the centrode is a circle, with A as centre.] 2. A lever moves about a fulcrum : find the nature of the centre of motion. 147. In the case of moving bodies rigidly connected to- gether, the determina- tion of the velocity of one with respect to an- other may be based on the preceding. For illustration take the ordinary steam-engine. The mechanism itself has been already shown in Fig. 119. Suppose "■ '' we wish to find the ■ Term introduced by Clifford (1845-1879J. Fig. 153 178 KINETICS OF A RIGID BODY. velocity of the piston P, relative to the crank-pin B. The velocity of the piston is the same as that of- the extremity A of the connecting-rod AB. The velocity of the crank- pin B is the same as that of the extremity B of the con- necting-rod. Hence the relation sought is the same as that between the velocities of the extremities A. and B of the connecting-rod. The bed-plate ^ C is fixed. The extremity A moves in a straight line PC, and the direction of motion being along PC, the instantaneous centre is in a line 4 at right angles to PC. The extremity B moves in a circle of centre C, and thei-efore the instantaneous centre is in the line CB. Hence the connecting-rod is fur the instant in the condition of a wheel turning about an axis through 0, the intersection of OA and CB. Consequently, velocity A : velocity B = OA : OB, or the velocities are as the distances from the instantaneous centre. If therefore the velocity of one of the two, piston or crank- pin, is given, that of the other follows at once. Thus, suppose the crank-pin to have a velocity 10 ft. per sec. Lay off to scale a distance BH — 10 ft., and draw HO parallel to BA. Then, since HB : GA= OB : OA, we scale off GA as the velocity of the piston. By drawing the crank in different positions, and finding the corresponding positions of G, a curve will result, the ordiftates of which will give the velocity of the piston through its stroke. Ex. 1. In the above example draw the complete curve of piston velocity oh a scale of velocity 5- ft. per sec. = 1 in. and of dimensions 1 ft. = 1 in. AKGULAR VELOCITY. 179 2. Find the height above the track of a point on a 6-ft. locomotive wheel running on a straight track, that has half the velocity of the highest point of the wheel. Ans. 1.5 ft. 148. Angular Velocity. — If the motion of the body is a motion of rotation about a fixed axis, each particle describes a circumference, whose centre is in the axis. Since each circumference is described in the same time, the velocities of the particles must be proportional to the distances of the particles from the axis, the greater the distance the greater the velocity. It is therefore clear that we can attach no meaning to the phrase " velocity of a body" as in the case of the motion of translation. In a word, we have a new kind of motion, and we must introduce new modes of measurement. Thus suppose the body to revolve about an axis through 0, and that it moves through an angle AOa in t sec. Then the angle de- scribed in one second, or the Angular Velocity oo, would be measured by AOa/t. This angle is described by every point in the body, so that one characteristic of rotation is that the angular velocity of every point is the same. The unii of angular velocity is naturally taken to be unit angle described in one second. The unit angle employed is the radian or the angle AOi, whose arc Ab \s equal in length to the radius A 0. The angular velocity oa will therefore be denoted by the number of radians described per second. Thus if a body revolves 60 times per minute, or once per second, the number of radians described per second is 27t. The relation between the angular velocity oo of the body and the linear velocity v of any particle A situated at a 180 KINETICS OF A RIGID BODY. distance r from the axis of rotation follows at once. For the time of motion being f, we have 00 = A Oa/t, arc Aa=-vt. But AOa : AOb = arc Aa : arc' Ab, or cot :1 radian = vt : r, and GO — v/r radians, the relation sought. Ex. 1. Show that the dimensions of angular velocity are l/T. 2. If oa is expressed in degrees, show that i = 27rcar/360°. 3. A body makes n revolutions per second: show that the linear velocity of a particle 1 ft. from the centre is 27rn ft. per sec. and the angular velocity 27rn radians per sec. 4. A belt passes over a pulley d ft. in diameter and mak- ing n revolutions per min. Find its velocity. Ans. Ttdn ft. per min. 5. A wheel 4 ft. in diameter revolves 420 times per min. Find the angular velocity and the linear velocity of a point 1.5 ft. from the centre. Ans. Utt radians per sec. : 2l7r ft. per sec. 6. The crank of an engine makes ?i revolutions per min. Its radius is r ft. Find the linear velocity of the crank- pin. Ans. 7rrn/30 ft. per sec. 7. In the driving wheel of a locomotive show that for an instant one point is moving twice as fast as the locomotive and in the same direction. 8. Find the ratio of the angular velocities of the hour and second hands of a watch. Ans. 1/720. 9. A locomotive is running at 45 miles an hour. The driving wheels are G ft. in diameter, and the stroke is 2 ft. Find the piston velocity. Ann. H/tt ft. per sec. EQUATIONS OF MOTION. 181 149. Angular Acceleration. — Angular velocity, like linear Telocity, may be uniform or variable. If variable, the rate of change is called the Angular Acceleration. The unit of angular velocity being one radian jDor sec- Fjg. 155 ond, the unit of angular acceleration is one radid,n-per-second per second. If the body rotates with a uniform an- gular acceleration a, the gain of circum- ferential velocity per second (or the linear acceleration) of a particle A at a, distance r from the axis is measured by an arc Aa equal to ar. Hence we may write linear accel. = ang. accel. X rad. The direction of the acceleration at A for an indefinitely small arc is normal to A 0, that is, along the tangent at A, so that we have tang, accel. = ang. accel. x rad. Hence, if m denotes the mass of the particle A, tang, force = mass X tang, accel. (Art. 34) = mra. Ex. When steam is shut off, the fly-wheel of an en- gine is making 90 rev. per min. If the coefficient of fric- tion is 0. 1, find the time in which the wheel will come to rest. Ans. 30;r sec. Thus far we have considered the kinematics of a rigid body. We now proceed to discuss its kinetics. 150. Equations of Motion. — The result of the action of a force on a particle of a rigid body is different from that on the particle if free. For besides this force other forces act on the particle resulting from the action of the adjacent particles of the body on one another. The 183 KINETICS OF A RIGID BODY. real acting force is therefore the resultant of the external and internal forces, and the motion is the same as if we considered the particle free, but subject to the action of a force equal to the resultant of all external and internal forces. To this resultant the name of Effective Force is given. Thus, suppose P to be a particle of a body of mass wi, , F^ the component in a certain direction of the external force impressed on it, and F^' the component of the re- sultant internal forces in the same direction. If a, is the component of the acceleration in this direction due to the resultant of the forces F^ , F^', the effective force acting on the particle must be equivalent to m^a^ . Hence p^ 4- F^' = mfi^ . Similarly, F^ + F^' — wi.a, . By addition, 2F + :SF' = Sma. But from the law of stress the internal forces occur as pairs of forces equal in magnitude and opposite in sense, or 2F' = 0, and therefore 2F=2ma, or the system of impressed forces is in equilibrium with the system of effective forces reversed. Now if a is the acceleration of the centre of gravity of the body, then (Art. 90) a = 'Sma/'Sm, and . • . "SF = '2ma ; or the linear acceleration of the centre of gravity is the same as that of a particle of piass ^m ^Qt^d on by a force :sjF., ' EQUATIONS OF MOTION'. 183 Consider now the rotation of a rigid body about a fixed axis under the action of external forces. Let the axis through be perpendicular to the plane of the paper, and let at an assigned instant F^, F^, . . . be the components of the external forces on the particles A,B, . . . acting at distances i\, r^, . . . from the axis, and parallel to the plane of the paper. Also, let F^', F^, ... be the components in their respective planes of the internal forces on the particles A, B, . . . , and denote by J?, , i?^ , . . the resultant forces a.t A, B, . . . These resultant forces may be resolved into components along the tangent and radius to the paths ot A, B, . . . The tangential accelera- tion is due to the tangential component. If, therefore, a denotes the angular acceleration of the body, the linear tan- gential acceleration at A is i\a, and the tangential force (or iiiertia-resistance) mj-^a. Denote the normal compo- nent at ^ by iV, . Similarly, we have in^i\a and N, at B, and so on for the other points. These are the effective forces. Taking moments about 0, we have, \t p^,p^, . . .; p/ ,p,', • . . ; ... denote the perpendiculars let fall from Oon F,,F^, ...; F,', F^' ;...;.. . respectively, = F,p, + i?;^^ + . . . + f^'p: + f,'p: + . . . or 2mr^a - 2Fp + 2F'p'. But the internal forces consisting of pairs of equal and opposite forces, we must have 2F'p' = 0. Also, since a is the same for all the particles, 2mr'a = a2mr^ ; and we have, .finally, a2mr' = 2Fp. The right-hand member of this equation is the ordinary expression for statical moment. The left-hand member is the sum of the moments of the inertia-resistances m^r^a. 184 KINETICS OF A RIGID BODY. ni^r^a, . . . about the axis^ and may therefore be appropri- ately called the Moment of Inertia* of the body about tlie axis. But the term moment of inertia is more usually ap- plied to the factor 2mr', which is constant for the body considered and independent of the varying acceleration a. This factor is denoted by the letter /, so that / = 2mr'. We may therefore write the equation ang. accel. = mom. of ext. forces/mom. inertia. The relations 2F = a2mr, 2Fp = a'Smr" , the first giving the acceleration of the centre of gravity in terms of the external forces and the mass, the second the angular acceleration about a fixed axis, are called the Equa- tions of Motion of a rigid body. Having found the accelerations from these equations, and being given the initial velocities, the velocities at the end of any given time and the distances passed over may be determ- ined. 151. The conditions of equilibrium of a rigid body already found in Art. 88 follow at once. For for equilibrium the acceleration of the centre Of gravity must be zero and the angular acceleration must be. zero. Hence, since a — Q, « = 0, we have 2F= 0, 2Fp = 0, or, the sum of the components of the external forces in any direction is zero, and the sum of the moments of the external forces about any point is zero. In fact all static problems may be regarded as limiting cases of kinetic problems, and may be treated accordingly. * The term was introduced by Euler (1707-1783), MOMEKT OP IKEllTIA. 185 152. Moment of Inertia. — It will be convenient in this place to give some examples of moments of inertia that are of frequent occurrence. We shall for simplicity and con- venience make use of the integral calculus. Indeed, the finding of moments of inertia is a problem of summation or integration, and is to be regarded as such, and not a mechan- ical question at all. This summation may, however, some- times be effected without the calculus, and as an illustration Ex. 1 is solved in both ways. First take the axis through the centre of gravity of the body. Ex. 1. To find the moment of inertia / of a thin uniform rod, mass M, length I, about an axis Y pig. 155 through its centre 0, and at right an- y gles to the rod. [Conceive the rod cut into elements of indefinitely small length dx, and let X be the distance of any one of these x ^^ elements from 0. Then moment of o •"* x inertia of element = Sdx x »' when d is the linear density, and . • . 6dx the mass of length dx. Hence =/: Sx'dx = 6r/12 = Mr/12. Or thus : Suppose the rod divided into a large number 2n of equal parts l/2n. The distances of these parts from may be taken to be the distances of their centres of gravity from 0, that is, l/in, 3?/4w, . . . Hence, taking half the rod, (V + 3'+ ... ton terms) 33?i 4m' = — (■ ~ 24\ 186 KINETICS OF A RIGID BODY. Fig f57 Y « h » h X X 61' = -—, -when 11 is indefinitely great, and , / = Miyi2, as before.] 2. A thin rectangular lamina or plate of breadth b and depth h, about an axis through its centre of gravity and parallel to li. [Conceive the lamina cut into strips parallel to the axis, and of breadth dx. Let x denote the dis- tance of one of these strips from the axis, S its density ; its mass is Sh X dx. Hence dh xdxxx' = ^ShV = ^Ml\'\ 3. If in (3) the axis is parallel to the side b, show that / = i\lhyi2. 4. A thin circular plate, radius r, about a diameter YOT' as axis. [Conceive the plate cut into strips parallel to the axis, and of breadth dx. Then the equation to the circle being x" + y' = '>"', the length of a strip at a distance X from is 2 Vr" — x^, and mass of strip = (J X 2 Vr" - x'dx. Hence Ux'\(? r I = I 2(^3;' V.r" - x'dx = dn-rV4 = Mr'/i. 5. A square plate of side a about a diagonal. Ans. 1 = May VI. 6. A hexagon of side a about a diagonal. Ans.I-hMa^L MOMENT Off INERTIA. 187 7. A triangle of base h, height //, about an axis through its centre of gravity and parallel to the base. Ans. I = MV36. Fig. 159 Fig. 160 :i rrg,,.. 8. A channel iron or I-iron, base I, depth h, thickness of web 1)^ , depth of web h^ , about axis h through 0. of G. Ans. l={bh' -{b- b,)\'}/l2. 9. A T-iron with dimensions as in Pig. 160 about axis through C. of G-. Ans. I = {bh' + bji,')/n. 153. The form of the expression for the moment of inertia about a fixed axis or a fixed point, 27nr', shows that we may define it as the sum of the pl-oducts of the masses of the particles of a rigid body into the squares of tlieir distances r from the fixed axis or from the fixed point.* If the body consist of an indefinitely thin plane lamina, and be referred to rectangular axes OX, 01^ in its plane, we may write }■■■ — x'-i-y' and I :^ 2m{x' -{- y") for the moment relative to or to an axis through per- pendicular to the plane of the lamina. But 2mx', 2my^ are the moments of inertia relative to the axes OY, OX, respectively. Hence, for a body in the form of an indefin- itely thin lamina, if /j , /, denote the moments of inertia relative to two rectangular axes OX, OY, in the plane, and I the moment relatire to an axis through perpen- dicular to the plane, we have 1=1.^ h- * This is the usual definition. Il falls to show the slgniflcance of the term moment of inertia. (See Art. 150.) 188 KINETICS OF A RIGID BODY. Ex. 1. Axectangularlamina, breadths, depth A, about an axis through its centre of gravity and perpendicular to its plane. Ans. I = M{b' + h')/12. 2. A circulai- plate, radius r, about an axis through its centre and perpendicular to the plate. Ans. I = Jfr'/S. 3. A ring of radii r, , ?•, about an axis through its centre and perpendicular to its plane. A7is. /= M{r^^ + O/^- 4. A sphere, radius r, about a diameter as axis. [Conceive the sphere divided into plates of width dx by planes perpendicular to the axis. Let the distance of any plate from the centre be a;. Then radius of plate = Vr' — a", and mass of plate = 6n{r^— x')dx. Hence, from Ex. 2, 1= J^^\Sn{r' - x')dx X {r' - x'), 5. Show, by differentiating the result of Ex. 2, that the moment of inertia of an indefinitely thin ring of radius r about an axis through its centre and perpendicular to its plane is Mr^. 6. Show by differentiating the result of Ex. 4 that the moment of inertia of an indefinitely thin spherical shell about a diameter is 2 J/irys. 154. The moment of inertia of a body about an axis not passing through the centre of gravity may be readily re- ferred to a parallel axis through that point. For suppose the two parallel axes through a point and the centre of gravity G to lie in a plane perpen- dicular to the plane of the paper. Take as origin, the plane of the paper the plane of X, J', and OOXilas axis of X. Let x, y denote the coor- dinates of any particle P of mass m. Call the distance Fig. rei 'f Y // / / 1 > / 1 C > X MOMENT OF INERTIA. 189 00= d. Then if /denote the moment of inertia about an axis through 0, we have I = ^m{y' + {x -{- dy] ' = 2m{y' + x')+ 2d2mx + d':Sm, since the distance d is constant. Of the three expressions in the right-hand member, the first is equal to /, , the moment of inertia about 0; the sec- ond is equal to zero, since is the centre of gravity (Art. 90) ; and the third is equal to Md^, where M is the total mass. Hence I=I, + Md\ or the moment of inertia about any axis is equal to the mo- ment of inertia about a parallel axis through the centre of gravity, together ivith the product of the mass into the square of the distance bet'ween the two axes. Ex. 1. Given that the / of a rod of lengbh I about an ' axis through its centre is MP/12 (Ex. 1, p. 185), show that about one end it is MTfi. 2. A rectangular lamina about A GB has I=Mhyi2 (Ex. 3, p. 186): show that about CD it is = Mh'/i. 3. A circular lamina, radius r, about its centre has / = Mr''/2 (Ex. 2, p. 188) : show that about any point in the circumference / = ^Mr^/2. 4. A triangular lamina of base b, height //, about an axis (1) coincident with the base, (2) through the apex and parallel to the base. Ans. bV/\2 ; bJf/4:. 5. Prove that / is the same for all parallel axes situated at equal distances from the centre of gravity. 6. Of all parallel axes the / of that which passes through the centre of gravity is the least. 155. Radius of Gyration. — The general expression for the moment of inertia of a body relative to an axis is Fig. 162 C D A G B h b 190 EiNEtlCS OF A KiGlD BODY. 2mr', the sum of the products of the mass of each particle into the square of its distance from the axis. Now instead of the particles being distributed in this way, we may con- ceive them concentra'ted into a single particle of mass M equal to the whole mass, and at such a distance h from the axis that the product MV — 2mr'. To this distance k tlie name Eadius of Gyration * is given. If the axis passes through the centre of gravity, we write Mk^' = 2mr^ where k^ is called the principal radius of gyration. The relation between k and k^ follows at once from the preceding article. For 2j?m'" = '2im\^ -f Ma', or M¥ = Mk;- + Ma\ or ¥= k,^-\- a% the relation sought. Also, since I^ — Mk', the value of k^ is at once found. Ex. 1. Show that the principal radius of gyration is a minimum radius for parallel axes. 3. For a straight line of length I with reference to its centre, show that k^ — r/12. 3. For a rectangle of breadth b, depth ?i, with reference to its centre of gravity, show that k^' = {b' + h')/12. 4. For a circle of diameter d with reference to its centre show that k,' = dyi6. 5. For a triangle of base b, height h, about an axis through its centre of gravity parallel to the base, show that k," = 7*718. 6. Find k^' for a right cylinder about its axis, r being the radius of cross-section. [Since the cylinder may be conceived to consist of an in- finite number of plates, each of which has the same radius of gyration with respect to an axis through the centre and perpendicular to their planes, the radius of gyration of the cylinder is the same as that of any plate, and .". k^' = r'/2.] * Called also radius of inertia. THE PHYSICAL PENhULUM. 191 7. For a right cylinder about an edge k^ — 3»-y3. 8. For a hollow cylinder, inner radius 9\ , outer radius r^ , relative to its axis, k^^ = {r^" -\- r/)/2. 9. For a vertical cylinder, radius r, length I, about a horizontal axis through its centre of gravity show that k; = r'/i + r/l2. 10. For a rectangular prism of dimensions a, b, c about the edge a, show that k; = (5' + t'')/13. 156. We shall now give some applications of the general equations of motion to special cases in which the axes of rotation are fixed, and in which they are instantaneous. (a) The Physical Pendulum. — In Art. 73 was considered the problem of the time of oscillation of a heavy particle P suspended from a fixed point by a weightless rod, and acted on by the force of gravity. This problem is purely hypo- thetical, as no such apparatus can be constructed. But just as a rigid body is regarded as built up of parti- cles joined together, so an actual or physical pendulum may be regarded as composed of simple pendulums whose oscillations so act on one another as to result in a common oscillation. The duration of this oscillation will give the length of a simple pendulum which fulfils exactly the same conditions of mo- tion. Suppose the physical pendulum to be a body of any form, and let the two be placed side by side, and both swung through an angle 8. The external forces on the physical pend- ulum are the weight ilf^ acting at the centre of gravity G, and the reaction R at the point of support C. Hence, since the moment of 192 KINETICS OP A RIGID BODY. R about G is nil, we have ang. accel. = Mg X QH/I = Mgh sin 6/1, where / is the moment of inertia with respect to the axis through C, and li is the distance CG. For the simple pendulum, Mg being the weight of the particle at P, ang. accel. = Mg X P§/mom. inertia = Mgl sin e./M? = g sin 6/1. But the angular acceleration being the same in the two cases, 'U sin 6/1 = g sin 6/1, or I = I/Mh, which gives the lengtli,of a simple pendulum with the same motion as the compound. Hence the time of oscillation t of the compound pendu- lum is (Art. 73) t = n VJfg - Tt Vl/Mgh. A point jD at a distance I from G, the point of suspension, is called the Centre of Oscillation,* for the reason that the time of oscillation of the whole pendulum is the same as that of a simple pendulum of length I and swinging about C. Denote the distance DGhj k, so that h -\- k = I. Suppose now the pendulum inverted, and suspended at B instead of at G. The time of an oscillation is n^/r/Mgh, * First determined by Huygens. THE PHYSICAL PENDULUM. 193 where I' is the moment of inertia about D. To find the centre of oscillation if the time of oscillation is the same as if suspended at C, we must find the length l^ of the sim- ple equivalent pendulum. We have . I=I,-\-M]i\ where /j is the moment of inertia about G (Art. 154). But I^Mhl, l'=-Mk\, and .•. by au easy reduction l = \, or the centre of oscillation in the latter case is the centre of suspension in the former. Hence the points of suspension and of oscillation can be interchanged without changing the time of oscillation, and appropriately therefore a pendulum with points of suspen- sion situated as C, D is known as a Reversion Pendulum. The principle of reversion is employed to determine the length I of the simple equivalent pendulum experimentally. Theoretically, I can be found from / = MM, but practi- cally it is difficult to make the measurements required by that equation. Hence the experimental method is gen- erally employed. A pendulum of given form is suspended from a point C (on a knife edge) and caused to oscillate. By trial another point D is found, from which if it is sus- pended it will oscillate in the same time. The distance between C and D is carefully measured, thus giving the length I, 194 KINETICS OF A KIGID BODY. Again, the value of I being knowu, and the position of the centre of gravity found as by balancing on a knife edge, we have at once the moment of inertia of the pendulum from / = MM. This method of finding moments of inertia is particularly useful with solids of irregular figure, or if not perfectly homogeneous throughout. The solid is mounted as a pend- ulum. Still further : The length I being known and the time of os- cillation being observed,we have from the relation ^ = tt VJJg the value of g, the acceleration due to gravity. The length of the seconds pendulum was used in England for a time as the standard of length, but was afterward abandoned for a certain brass rod called the standard bar, for the reason that several of the elements of reduction of pendulum experiments are doubtful. Ex. 1. A rod of length I is suspended at one end, and caused to oscillate : find the length of the equivalent simple pendulum. Ans. 2Z/3. 2. If the rod in Ex. 1 is suspended at ^ of its length from one end, find the time of an oscillation. 3. If a pendulum is suspended at the principal centre of gyration, prove that the time of oscillation is a minimum. [For t — 7t V{h'' -\- h^)/glh, and t is a min. when /* = A^.] [^ Fig. 164 {b) The Steam Engine. —To illustrate motion about an instantaneous axis, we shall confine our- selves to the steam en- gine, and the still more simple case of the rolling disc or sphere. The relation between the piston pressure P and the crank-pin resistance ^, wheij STEAM ENGINE. 195 the connecting-rod is inclined at any angle, has already been solved in Art. 109, bnt may be solved more simply by aid of the instantaneous centre. For the velocity v, of the cross-head is to the velocity v^ of the crank-pin as the instantaneous radii OA, OB, directly (Art. 147). But by the principle of work, P xv^= Qy.v^. Hence P •>iOA= Qx OB; or, P is to Q as the instantaneous radii inversely. The value of Q for a given piston pressure will thus vary according to the position of the connecting-rod. It may be represented graphically, as in the case of the indicator diagram (Art. 136). The average of the values of Q for a complete revolution of the crank corresponding to a given piston pressure P will be found by equating the work done by each of the two forces. We have, if r is the radius of the crank arm and 8 the length of the stroke, - QX^Ttr^Py. 28. But 8 = 2r; .'. 7tQ = 2P, the relation required. Ex. 1. In a ISTorris engine the diameter of the cylinder is 14 in., and the steam-pressure 75 pounds per sq. in.; find the average value of the force acting on the crank-pin. Ans. 2 X 7" X 75 pounds. 2. In (1) find the force acting when the crank stands at 60°, and the ratio of the connecting-rod to the crank is 5^, 3. In a steam riveting machine the piston pressure P is applied at the joint a, and the rivet squeezed between the jaws c, d. Find the relation between P and the force Q ex- erted on the rivet. [The instantaneous centre is at Of 196 KINETICS OF A RIGID BODY. where ia and the perpendic- ular through c to the sliding surface 8 intersect. Hence F XaO= QxcO. As a approaches g, cO dimin- ishes; and when a reaches g, Q becomes indefinitely great. Hence the advantage of the apparatus in that an enormous pressure may be produced by a moderate foi'ce acting through a small distance. This is an example of the Tog- gle Joints a mechanism of very considerable importance. It is applied, for example, in the Westinghouse air brake on locomotive drivers, in cider, oil, and other presses, etc., etc. In Fig. 166 is shown part " of a power screw oil press. Fig. 166 vd," Fiq. 165 ~t. \ 1^ vV f . '// fx \ \ \ \ \ -P i e^- ■■"\ a \ \ Q ^Jiffi.1 BNEEGY or KOTATIOif. 197 4. A cylinder rolls down an inclined plane of height h from rest : find its velocity v at the bottom. [0 is the instan. centre ; r the radius of the cylinder. The forces acting are the wt. Mg and the reaction N. ^mg . •. ang. accel. a = Mgr sin B/\Mr^ = %g sin 9/r ; . \ linear accel. of centre — ra — ^g sin 6, and v' = 2as = 3 X Iff sin ^ X -= — 7, = 4ffA.] ''^ sin & ^^ 5. A spherical shot rolls down a plane 70 ft. long and inclined at 30° to the horizon : find its velocity at the bottom. Ans. 40 ft. per sec. 6. A sphere will roll and not slide down an inclined plane if the coefficient of friction is greater than -f tan a where a is the inclination of the plane. 7. The kinetic energy acquired by a sphere in moving from rest down a smooth plane is to that acquired by an equal sphere rolling down a rough plane of the same incli- nation and length as 7 to 5. 157. Energy of Rotation.— The energy stored in a body rotating with a given angular velocity go will be found by add- ing together the energy of each particle. If v is the linear velocity of a particle and r its distance from the axis of rotation, we have v — cor, and the energy stored = imv' = ^noa'r'- Hence the energy stored in the body is ^■JmosV where 2 is the symbol of summation. If n is the number of revolutions made per minute, ao = 27rn/60, and we have finally energy of rotation = S^moo'r' = ^loo' = 7^7180, nearly. We shall confine ourselves to the cases of a circular disc and of a ring rotating about axes through their centres, per- 198 KINETICS OS A RIGID BODY. pendicular to their planes. Common examples are a car- wheel and a fly-wheel. For a uniform disc of radius r, / = ^Mr', and for a ring with r, , r, the inner and outer radii / = iM{r^' + r,'), or if r„ r, are nearly equal, / = MrJ', where r^ is the mean radius of the ring. It is seldom that a wheel is in the form of a simple uni- form disc. In general the greater part of the weight is contained in a ring next to the rim of the wheel. This is more the case in a locomotive-wheel than in a car-wheel, and more in the fly-wheel of a stationary engine than in a locomotive-wheel. The two extremes are the uniform disc and the ring. In order to find readily the mass if of a fly-wheel, we notice that since 1 cub. in. cast-iron weighs 0.261 lb., we have, if ^ is the area of the cross-section of the rim in square inches, and d the diameter in inches, M—7tdX A X 0.261 = 0.82dA lb., nearly; and .-. 7= 0.205 Ad\ nearly. 158. In a stationary engine the fly-wheel is introduced to give steadiness. It does this by giving up when called on part of the energy stored in it.* To compute the dimen- sions of a wheel, we must decide on the greatest amount of energy that should be demanded of it, and also on the maximum range, of velocity that can be allowed. Suppose * " The propvietor was showing to a friend the method of punch- ing holes in iron plates. He held in his hand a piece of iron | in. thick, which he placed under the punch. Observing after several holes had been made that the punch made its perforations more and more slowly, he called to the engine-man to know what made the engine work so sluggishly, when it was found that the fly-wheel and punching apparatus had been detached from the steam-engine just at the comuiuucemeut of bis experiment,"— J5w!iJ'ag'«. ENBKGY OF ROTATION. 199 we wish to call on it for 36000 ft.-popndsj and that the ve- locity may change from 48 to 53 revolutions per minute. Then 7(52= - 48=)/180 = 36000^ and / = 16300^. If we decide to make the wheel 14 ft. in diameter, Mr' = 16200^ and M= 5 -{- tons, whence the cross-section of the rim may he computed from the formula given above. 159. In many cases a body possesses both an energy of translation and an energy of rotation, and their compara- tive amounts or their sum may be required. Take, for ex- ample, a railroad car in motion. Each wheel of the car acts as a fly-wheel in which energy is stored to be given out before the car comes to rest. If v is the velocity of the train, the tangential velocity cor of the wheel is equal to the velocity of translation v. Hence the energy stored in the wheel considered as a disc = iloo' = ^ X ^Mr'' X v'/r' = ^Mv', being one half that due to the forward motion of the wheels. Suppose the total (loaded) weight of the car to be 40000 lbs., and that the eight wheels weigh 4500 lbs., and the velocity is 30 miles an hour. Then the total energy stored in the car is i X 40000 X 44" + i X 4500 X 44" foot-poundals, which may be reduced to foot-pounds by dividing by 33.2. Ex. 1. In a disc revolving about its axis the^radius of gy- ration is nearly 0.7 of the actual radius. 2. The rim of the fly-wheel of a Norris engine is 14 ft. in diameter^ and weighs 11400 lbs, : find its I about the centre, 300 KINETICS OF A RIGID BODY. 3. The rim of a fly-wheel weighs 15 tons, and its diameter is 20 ft. ; the wheel makes 60 revolutions per minute : find the energy stored. Ans. 1,875,000 ft. -pounds. 4. A fly-wheel of a tons wt. and b ft. diameter makes c revolutions per minute : find the energy accumulated. Ans. O.OSTaS'c" ft. -pounds, nearly. 5. The weight of a fly-wheel is W lbs., and its diameter d^ inches. If it is making n revolutions per minute, find in how many revolutions it will be stopped by the friction of the axle if its diameter is d^ inches and the coefficient of friction /i. Ans. 7rn^d^y7200 /jgd, . 6. Examine the following statement: " Every engineer knows that a thing so balanced as to stand in any position is not necessarily balanced for running : that a 4-lb. weight at 3 in. from the axis of rotation though balanced statically by a 1 lb. weight at 12 inches from the axis is not balanced by it dynamically. On the contrary, a 4-lb. weight at 5 in. is balanced by a 1-lb. weight at 10 in. from the axis." CHAPTER VIII, ELASTIC SOLIDS. 160. Ik laying down the foundations of the subject it was stated as the result of observation and experiment that forces acting on a body might change its motion or its form, or that both changes might take place. The efEect of the force depends on the nature of the body acted on. As most simple, we have considered first of all changes of mo- tion only. The action of the external forces was conceived to be resisted by the Internal reactions of the particles on one another in such a way that the particles retained their original distances from one another so that change of form did not take place, or the body acted on was rigid. The conditions of equilibrium and of claange of motion on this hypothesis have been developed in the preceding chapters. Experience shows that no perfectly rigid body exists in nature. The internal reactions do not prevent changes of form, the body yields to the external forces and changes its form or its size. If it returns towards its original configura- tion on the removal of the forces, it is said to be Elastic — a body possessing elasticity of form being called a solid, and one altogether devoid of elasticity of form a fluid. Conceive an elastic body (as a bent spring) in equilibrium under the action of given forces. When in this condition, nothing will be changed by supposing it to become rigid. The conditions of equilibrium of a rigid body may there- fore be applied to it in its distorted form, and the problem solved as for a rigid body. So that just as we use the par- ticle as a stepping-stone to the rigid body, we use the rigid body as a stepping-stone to the elastic body. SOI 202 ELASTIC SOLIDS. 161. The form of the body when in the position of equilibrium being distorted, differs from the original form. Hence, before we apply the conditions of equilibrium we must first of all inquire into the changes of form capable of being produced by the external forces. To do this it is neces- sary to appeal to ex- periment. Conceive a uniform beam with horizontal axis and resting on two supports under the action of external forces. These forces are transferred from particle to par- ticle of the beam, and give rise to internal forces or reac- tions. Let Fig. 168 represent a vertical section through the axis of the beam. For simplicity consider first the external forces to be parallel to this plane and let them be resolved into components along and at right angles to the axis. Conceive the beam divided, by a plane AB perpendicular to the axis, into two parts Xand Y. The equilibrium will remain as before, provided forces equal and opposite to the internal forces at the section are applied. Let these forces be resolved into vertical and horizontal components, and let these be combined' into single resultants, Q, Q; R, R. The forces of which Q, R are the resultants being distributed over the surfaces of the section and forming pairs in equilibrium, are known as Internal Stresses. The part X is in equilibrium under the action of the external forces and the stresses -{- R, — Q representing the action of Fupon X, and the part Y under the external forces and the stresses — R, -\- Q representing the action STEESS AND STRAIN. 203 of X upon T. In finding the relation between the exter- nal forces and the internal stresses the equilibrium of either X or F may be considered. We shall take X, the section to the left of the cutting plane. 162. To the stresses various names are given. Thus the stresses along the beam may tend to pull the pieces apart or to push them together. To the pull the name of tensile stress or Tension is given, and to the push the name of compressive stress or Compression. The transverse stress (consisting of equal and opposite forces -{-It, — R, considered acting indefinitely near the plane of section, but on opposite sides) causes the pieces to slide along the plane, and forms the shearing stress or Shear. The transverse stress may cause X, Y to turn about an axis in the plane AB, and forms the Bending Stress. If the external forces are not in the same plane, the bending stress may take place about an axis perpendicular to the plane AB, and forms the twisting or Torsion Stress. This generally occurs in machine shafting. These are the simple forms of stress. Usually stresses are compound, but may be resolved into two or more of the simple forms. 163. When a beam is subjected to the action of a stress it yields. The change of dimension is known as a Strain. Eemove the stress, and the beam returns to its original dimensions. This is observed to be true for stresses up to a certain amount. When that amount is exceeded, the beam will not return to its original form on removing the stress, but will assume another form between the two, or take a Permanent Set, as it is called. Increase the stress still further, and the beam will be finally ruptured. The discussion of stresses and strains forms the Mechan- ics of Materials, and will be found in special treatises. 204 ELASTIC SOLIDS. Fig. 169 164. Impact. — Suppose a sphere of mass m to come in contaet with another of mass m^ , a collision or Impact takes place. This impact is said to be direct if the hodies are moving in the same straight line, and the com- mon tangent plane at the point of meeting is perper dicular to the direction *of motion: if not, the impact is said to be obhque. 165. Direct Impact. — In conse- quence of the impact there is a mutual pressure produced, which increases the velocity of one of the bodies and diminishes that of the other. The velocity u of m is changed to v say, and the velocity u^ of »m, to w, . The impact action requires a certain time, which may be conceived divided into small intervals. Let P be the press- ure developed in any one of these intervals, and a, a, the accelerations produced in the masses m, m, , respectively. Then (Art. 34) P = ma, P ■■ «(,a,, and . • . a : «, = wi, : m. Similarly for all of the intervals. Hence, by addition, total accel. of m. : total accel. of »i, = m, : m, or w — M : M, — w, = wij : m, or mti + m,M, = mv + ^n^v, ', ... (1) that is, the sum of the momenta before impact is equal to the sum of the momenta after impact, — a statement indeed implied in the law of stress. This equation contains. two unknowns, v and v^. We must therefore have another relation between them, as two equa- tions are necessary to determine two unknowns. Now it is IMPACT. 305 formd by experiment "Ihat -vvhen two bodies impinge direct- ly their relative Telocity after impact bears a constant ratio to their relative velocity before impact so long as the material of the bodies is the same, but is in the opposite direction. This constant ratio is called the Coefficient of destitution of the two bodies, and is denoted by the letter e. We have therefore w — v, = — e{u — u^ .... (2) as the second relation between v and v, . Solving (1) and (3), we find the values of v and v^ . The value of e depends on the material composing the bodies. From its definition it follows that the extreme values of e are and 1. If e = 0, or the bodies are In- elastic, then V = {inu + m^u^)/{m + mj = w^, . . (3) or the bodies move together with a common velocity after impact. It e — 1, then v-v^= -{u-u,), or the velocity of one body relative to the other after im- pact is the same as it was before impact, but in the oppo- site direction. In this case the bodies are Elastic. No examples of either perfectly inelastic or of perfectly elastic bodies occur in nature. But some bodies with very little elasticity, as clay for example, may be regarded as be- longing to the first class, and others, as glass, to the second class. Ex. 1. Two inelastic balls are brought to rest by the im- pact: prove that they must have been moving in opposite directions, with velocities inversely proportional to their masses. 2. Two balls of equal mass are perfectly elastic : prove that after impact they will exchange velocities. 3. A row of equal elastic balls are placed in contact in a 206 ELASTIC SOLIDS. straight line. An equal ball impinges directly on them. Show that all will remain at rest but the last, which will fly off. Verify experimentally. 4. Find the elasticity of two balls of masses in and M in order that if M impinges on m at rest it will itself be brought to rest. ^ns. M/in. 166. Oblique Impact of a Sphere against a Fixed Smooth Plane AB. — Let lo be the ve- •"js- '^' locity before impact, and v the velocity after impact ; d the in- clination of u to the normal, and ^.^ fi the inclination of v. — ^ Resolve the velocities along and * ^ normal to the plane. The plane being smooth, it exerts a normal pressure only. Hence the impact may be considered direct, with velocity u cos a be- fore and V cos yff after impact ; and .• . V cos y3 = — ew cos d. Also, since the pressure is normal, the action along the plane is unchanged by the impact. Hence t; sin yS = M sin d, and V and /J are found. Ex. 1. What are the values of v and /S above ? 2. If the elasticity be perfect, show that the angle of in- cidence 6 is equal to the angle of reflection /?. 3. To hit a ball § by a ball P after reflection from the edge CA of a billiard table. " Aim at a point B as far be- hind the edge GA as Q is in front of it." Prove this. 4. Deduce a rule for reflection at two edges of the table. 5. A ball impinges on an equal ball at rest at an angle of 45° to the line of impact: prove that if both are per- fectly elastic their velocities will be equal after impact. 6. Two balls of given masses and moving in given direc- tions with given velocities impinge on one another : find the resultant velocity of each ball and it§ direction, IMPACT. 207 167. When two bodies impinge, the Energy of Impact is broken into two parts, one being taken up in producing changes of form, heat, light, sound, etc., and the other the resultant motion of the bodies. With our usual notation we have for direct impact, energy before impacb = {inu' + m^u^^)/2, energy after itnpact = (mw" -\- m^v')/2, and the change during impact is therefore the difference between these two expressions. Prom Equations 1, 3, Art. 165, we fiud by a simple reduction i{mu' + m^u;)-i{mv'' +m,v,') = i{l-e') "'^' {u-uV, which gives the change of kinetic energy produced by the impact. The change is greatest when e = 0, or the bodies are inelastic. The expression being a positive quantity (since e" <1) would seem to indicate a loss of energy dur- ing impact. Whether the change is to be so regarded or not, depends on the end to be attained. If that is the propulsion of a missile or the driving of a pile, then change of form, heat, etc., are prejudicial, and the energy used in producing then; is lost. If, on the other hand, change of form is the main thing, as in moulding metal under a hammer or in riveting, this so-called loss becomes the use- ful energy, and the energy of motion useful in the former case becomes prejudicial in this. Ex. Two trains of equal weight, moving with velocities of 30 miles an hour each and in opposite directions, collide: show that the loss of energy produced by the impact is the same as in the case of a train moving at 60 miles an hour striking another at rest. In the latter case find the velocity with which the debris will be moved along the track. Also, show that before impact the total energy in the one gase is double that in the other, 208 ELASTIC SOLIDS. 168. The case of impact that occurs most frequently in practice is when the bodies are inelastic, and one is at rest before impact. Placing e — 0, m, = 0, and substituting for V, «;, their values from (3) Art. 165, we have for the col- lective energy of the two bodies before impact mu^/2, and after impact m'u'/2{'m -\- mj, making the change during impact ■mtn^u'/2{m +wi,). Take for illustration the Pile Driver.* The principle is the same as in driving a nail with a hammer, except that the motion is always vertical, gravity being the force acting. When the ram of mass m in falling through a height h im- pinges on a pile of mass m, with .velocity u, the pile is driven down- wai'd a certain distance s. Let F denote the resisting force of- fered by the ground. The work done on the resistance is Fs foot- poundals. The energy of impact is m''u''/2{m + m,) foot-poundals. The work done after impact by the force of gravity on ram and pile is {ni -\- tnj)gs foot-poundals. Hence m'u^/2{m + m,) + {m + m^)ffs = Fs, and F is found. In gravitation units, it W = wt. of ram in pounds, w = weight of pile in pounds, and P = the resistance in * The figure represents a piledriver made by the Vulcan Iron Works, Chicago. IMPACT. 209 pounds (or the ultimate load the pile will carry), then (re- membering that tt' = 2g7i) the above equation reduces to W'h/{W+w) + {W-j-w)s = Ps, the standard form. At the last blow, the value of s being small, the second term may be disregarded in comparison with the first, and we have WVi/{W+w)=Ps. Still more approximately, by neglecting the weight of the pile in comparison with that of the ram, Wh = Ps. For example, to find the ultimate load a pile weighing 500 lbs. could carry if the last blow from a height of 25 ft. of a one-ton ram sinks the pile one inch. The three form- ulas give 482,500; 480,000; 500,000 pounds, respectively. The Safe Load to be carried by the pile is some frac- tional part of this, which experience has determined. Thus with the second of the above formulas Col. Mason, U.S.A., in a series of experiments at Fort Montgomery, N. Y., found the fraction to be J, so that he proposed safe load = WVi/4:{W+w)s-; and Major Sanders, U. S. A., in an " extensive series of ex- periments made in establishing the foundations of Fort Delaware," concluded that the third formula was to be depended on in the form safe load = W7i/Ss. The fact is, so much depends on local circumstances, particularly on the condition of the head of the pile at the last blow, that pile-driving formulas must remain essentially empirical. 210 ELASTIC SOLIDS. It is useful to notice that the energy of the ram before impact being Wh, the loss of energy will be less tjie more nearly W/{ W + iv) is equal to unity, that is, the greater W is in comparison with iv. Hence the ram should be large in weight compared with the pile. With the riveting hammer, steam hammer, etc., in which change of form is the end to be attained, the useful work done by the hammer depends on Wwh/{ W + w), which is the more nearly equal to Wh the greater w is in comparison with W, that is, the heavier the anvil is in comparison with the hammer. Ex. 1. Find the safe load for a pile weighing 500 lbs. to carry if the pile sinks 0.1 ft. at the last blow under the 5-ft. fall of a 500-lb. ram. Ans. 312fribs. 3. A. steam hammer weighing 500 lbs. has a stroke of 3 ft. If the piston pressure is 1000 pounds and the blow is vertical, find the work delivered in 6 blows. Ans. 270,000 foot-pounds. 3. A pile is driven s ft. vertically into the ground by n blows of a steam liammer fastened to the head of the pile. Given ^ the mean pressure of the steam in pounds per sq. in., d the diameter of the piston in inches, I the length of the .stroke in ft., w the weight in lbs. of the moving parts- of the hammer, and iv^ the weight in lbs. of the pile and fixed parts of the hammer attached to it, and 11 the mean resistance of the ground in pounds, prove mv{w + 7t pd''/^)l = Rs{w -\- to^). 4. In firing from a rifle of weight w lbs. a bullet of weight Wj lbs. with velocity v ft. per sec, show that the energy of recoil is gw^d^ /1w. 169. In the preceding sections the impact considered has been measured by the momentum developed. In most cases the time of impact is so small that it is im- possible to measure it. If it can be measured or esti- IMPACT. 2ll mated, the acting force F is at once found from the rela- tion (Art. 34) Ft = mv. Thus suppose a hammer weighing 4 lbs. to strike a blow with a velocity of 16 ft. per sec, and that the time elapsing from the first contact to the destruction of the motion is 1/1000 second, the average force F acting for this time would be given by 2?^= 4 X 16 X 1000 ^ 32 = 3000 pounds. The effect produced results from the enormous force developed in the short time. If the time be increased the force is diminished in the same proportion. Prom this circumstance many familiar phenomena may be easily ac- counted for. Ex. 1. Why do we receive a severe jar from a step down- ward when one expects to step on the level ? 2. Account for the different effects of a cannon-ball in striking a granite wall and an earth wall. 3. The ram of a pile-driver of 250 lbs. falls through 10 ft. and is stopped in jL second : find the average force ex- erted. 4. The head of a steam hammer weighs 3 tons. If steam is admitted on the under side only for lifting, and the head has a drop of 4 ft., find the average compressive force exerted during a blow if its duration is yV second. Ans. 15 tons. CHAPTER IX. STATICS OF FLUIDS (HYDROSTATICS). 170. According to the general scheme outlined in Art. 4, we now pass on to discuss the action of force on bodies in the fluid state. Our ideas of a fluid are derived from com- mon experience. Such substances as water, oil, air, steam, we call fluids. The one characteristic property by which they are distinguished from solids is the more or less ease with which the particles move among themselves; or in other words, fluids have less elasticity of form than solids. Some fluids offer more resistance to separation of the parti- cles from one another, or to flow, than do others — thus mo- lasses more than water, and water more than ether. This resistance to flow is known as Viscosity. We may, however, conceive of a fluid of such a kind as would offer no resist- ance whatever in any direction, the elasticity of form being altogether wanting. A fluid possessing this property has no existence in nature. It is an abstraction, just as a per- fectly rigid solid or a perfectly elastic solid is, and our only reason for introducing the idea is that it leads to simplicity of treatment. The more any fluid in nature differs from the hypothetical fluid, the greater the modifications neces- sary in the results derived from the hypothesis. As the basis of our reasoning, then, we lay down this definition : A fluid is a substance the particles of which can he moved by any force however small, and which act on one another or on any surface without friction. 171. Two Kinds of Fluids.— It follows from the definition that a fluid offers no resistance to change of form. Hence 212 TWO KINDS OF FLUIDS. 313 the fluid must be confined in a solid Tessel or envelope either rigid or elastic. Suppose a fluid completely confined, and a force P applied to a piston entering the vessel. With some fluids (as water) the change of volume resulting from the force P will be so small as to be practically nil; with others (as air) the change will be dependent on the force, and such that when the force is removed it returns to its former volume. A rough illustration is afforded by the ac- tion of a closed syringe and of a pop-gun. We therefore subdivide fluids into non-compressible or inelastic and compressible or elastic. The former are known as Liquids, the latter as Gases. We shall assume the absolute incompressibility of liquids and the perfect elasticity of gases, which assumptions though not strictly true are very nearly so, as repeated experiments have shown. 172. The results of the a;ction of forces on fluids not equally applied over the whole surface of the fluid is change of form or flow. If the forces acting over the whole surface balance, the fluid is in equilibrium. We are thus led to distinguish the statics of fluids and the kinetics of fluids. The term Hydromechanics is used to designate the me- chanics of fluids. It is divided into Hydrostatics, which treats of fluids at rest ; and Hydrokinetics, which treats of fluids in motion. The term Pneumatics includes the prop- erties of gases as distinct from liquids. By some writers the term Hydraulics, originally applied to the motion of water through pipes, is taken to mean the mechanics of fluids. 173. From the conception of perfect mobility among the particles of a fluid we infer that if a fluid is in equilibrium under the action of external forces — («) Each particle must he equally pressed in every direc- tion. For if not, the particles would move in the direction 214 STATICS OF FLUIDS. of least pressure, which contradicts the hypothesis of equi- librium : — {b) The force exerted ly the Auid on any surface with which it is in contact {or the pressure on the surface) is nor- mal to the surface. For if not, friction must enter, which contradicts the hypothesis of perfect mobility. These are the two fundamental principles of the statics of fluids. 174. Measurement of Pressure. — Suppose that a fluid is in equilibrium under the action of a number of external forces, and let P denote the pressure exerted by the fluid on an area A of the-surface of the vessel containing it or of a surface immersed in it. Then P/A would be the press- ure exerted on a unit of surface if the pressure were uni- form over the surface, and would be the average pressure per unit if it were not uniform. Pressures are expressed in poundals per sq. in., pounds per sq. in., dynes per sq. cm., etc. By making the unit of surface indefinitely small, we have the pressure on an indefinitely small surface, and can there- fore speak in this sense of the pressure at a point. The expression is a conventional one. The pressure on a point is of course nil. 175. Transmission of Pressure. — Since every particle of a fluid in equilibrium is pressed equally in all directions, any assigned particle mu^t exert an equal pressui-e in all direc- tions on its adjoining particles, each of these an equal pressure on those adjoining, and so on throughout the fluid. Hence a pressure applied at any point is transmitted un- changed to every other point of the fluid, and a pressure applied to a surface is transmitted unchanged to every other unit of surface in contact with the fluid. This is sometimes called the. princi2}le of Pascal. TKAS'SMlSSiON OF PEESStEE. 215 Suppose a vessel full of water and furnished with two pistons fitting in two openings A and B. Let A contain A square inches of surface, and B contain B square inches. Suppose a force of P poundals applied at A bal- anced by a force of Q poundals applied at B. The pressure per sq. in. at A is P/A, and at B is Q/B. Hence P/A = Q/B, from which relation, when any three of the quantities P, Q, A, B are given, the fourth can be found. Experiment confirms this result. Again, suppose a plane CD to divide the fluid in to two parts. The pressures on the two sides of this plane are equal, and are normal to the plane. There is equilibrium between the pressure on the surface CBD and the pressure on one side of CD, also between the pressure on CAD and the pressure on the other side of CD. But CD is the projection of CBD or of CAD, no matter what the form of these surfaces. Hence the pressure on a surface in a given direction is equal to the pressure on the projection of this surface on a plane normal to the given direction. Ex. Show that the pressure on a pump plunger is the same whether the end of the plunger is rounded or flat. Application: (1) The safety-valve of a water or steam engine. The pressure on the valve will show the pressure in the boiler, and by suitably placing the weight on the lever, this pressure may be adjusted to any amount desired. When it exceeds this amount the valve will be lifted and the steam escape. 316 STATICS OF FLUIDS. Fig 174 Let W = wt. of valve in lbs., A its area in sq. in., 1^-= dis- tance of centre of valve from fulcrum F in inches ; 'W = wt. of lever ; l^ = dist. of 0. of Gr. of lever from fulcrum ; W — wt. on lever; l^ = dist. of centre of weight from fulcrum ; p — pressure in pounds per sq. in. of blowing-ofE. Take moments about the .fulcrum, and wX + w^l^ + loX = Ph^, the relation required. Ex. 1. The valve weight is 3 lbs., diameter of valve 4 in., distance from fulcrum to centre of weight 36 in., distance from fulcrum to centre of valve 4 in., weight of lever 7 lbs., distance from fulcrum to centre of gravity of lever 15.5 in. : find the weight at the end of the lever to make the blowing-ofE pressure 80 pounds per sq. in. Ans. 108.3 lbs. 2. With the same data find how far the weight must be TRANSMISSION- OF PBESSUEB. 217 Fig. 175 placed from the fulcrum to make the blowing-ofl pressure 75 pounds per sq. in. ^«*'- 33.7 in. (2) The Lifting-jack.— The figure represents a section of a Tangye jack. Water is forced from the cis- tern by the force-pump CG under the ram which ■works in a water-tight collar H. The weight to be raised is placed on the ram at W. By unscrewing the " lowering screw" the water is returned to the cistern. By- means of the "air-screw" air is admitted when the jack is in use. Ex. 1. In a jack the plunger is 1 in. diameter CHARGING SCREW and the ram 10 in. diam- eter. A weight of 10 tons is to be raised : what pressure must be applied at the end of the lever, the lever- age being as 10 to 1. Ans. 20 pounds. 2. In a jack the leverage is a : i, the pressure applied at the end of the lever P pounds, and the weight on the rdJm W lbs. Compare the diameters D, d of ram and plunger. Ans. D/d = VWbJPa. 176. It is evident from the above apparatus that water may be used to transmit force. Not only so, but it may be used for the storing of energy. For suppose the ram fixed, then the pressure by the action of the ram will remain stored until the ram is freed, when it will rise to the same height that it would have done if not interfered 218 STATICS OF FLUIDS. with. A vessel for the storing of energy in water is known as a Hydraulic Accumulator. The mode of action of an accumulator may be more evi- dent from Fig. 176. J is a solid ram working water-tight in a vertical cylinder, and car- rying a heavy weight W. This weight, which in an accumu- lator usually consists of cast- iron blocks, is carried by a platform B, which latter is supported by rods from a cross-piece C, fastened to the top of the ram ^. Z* and Z> are vertical guides. Water is driven into the chamber B through the pipe F by means of a force-pump, thus raising the ram. The mechanism to be operated is connected with the pipe G, through which the energy stored is communicated. In this way hydraulic cranes, elevators, etc., may be worked from a central source by means of water-pressure. Ex. 1. The ram is 21 in. in diameter and the load 120 tons : show that the water-pressure per sq. in. in the accumu- lator is 700 pounds, nearly. 2. In the hydraulic machinery for opening and closing the lock-gates of the St. Mary's Palls Canal, at the entrance to Lake Superior, the ram of the accumulator has a diame- ter of 31 in., and carries a load of 20.76 tons. The water- engine driven by this accumulator makes one revolution per minute, the diameter of the piston is 15 in., and the length of stroke 96 in. Find the H. P. developed. Ans. 5.1, nearly. 177, In the last two sections we have considered the trans- mission of pressure in vessels completely filled with fluid PKESSURBS AT DIFFERENT DEPTHS. 219 and acted on by external forces. We now proceed to con- sider the influence of the weight of the fluid on the surface pressed. 178. Surface of a Liquid at Rest. — Consider a liquid in a Tessel with the upper surface free and acted on by gravity only. This free surface is a horizontal plane. For the force of gravity on each particle is vertical, and as friction is wanting, the surface must arrange itself at right angles to the pressures, — that is, horizontally, — otherwise the parti- cles would glide over one another. This is the characteristic property of liquids as distin- guished from gases. The surface is an example of an equi- potential surface (Art. 143). 179. Pressures at Different Depths. — Let be a point in the liquid at a depth h below the surface. F'g- '77 Suppose a small horizontal circle al of area A described about as centre, and conceive the liquid contained in the verti- cal cylinder described on ab as base and extending to the surface cd to become solid. The equilibrium existing will not be disturbed thereby. The pressures on the sides of the cylinder being normal to the curved surface, are horizontal. Resolving vertically, we have press, on ai = wt. (grav. force) of cylr. ad. Let 2^ — press, per unit area of ab, S — mass of cubic unit of the liquid or its density, and this becomes p X A = SAlig poundals, or p = gdh poundals = 6h pounds. Thus the pressure varies directly as the depth below the surface of the liquid. 230 STATICS OF FLUIDS. If the base ah is not horizontal the pressure will vary from point to point of the base, and p becomes the average pressure per unit surface. If we conceive the base to become smaller and smaller until it becomes indefinitely small, then p will represent the pressure at the point 0. Hence the pressure at any point wWiin a liquid is proportional to the depth of the point ielow the free surface of the liquid. The weight (mass) of a cubic foot of water at the tem- perature 4° C. is 62.424 lbs., or 1000 oz., nearly. A cubic inch of mercury weighs 3429.5 grains, a cubic ft. 13,600 oz. Ex. 1. Find the pressure at a depth of 100 ft. in Lake Superior. Ans. 43.3 pounds per sq. in. 2. A vessel is filled with mercury. At what depth is the pressure 20 pounds per sq. in. ? Ans. 20 = hX 3429.5 X 7000. 180. Total Pressure on a Surface immersed. — This fol- Fig. 178 lows from the preceding by adding the pressures on all the unit surfaces contained in the given surface. Thus, it h^,7i, . . . 7i„ are the depths of the unit surfaces, the total pressure = ff^^h + g^K + • • ■ + gSh^ Then, as in Art. 90, if A is the number of units of area in the surface and h the depth of its centre of gravity, A = (1 X A, + 1 X /t, + . . . + 1 X h„)/A, or Ah = h^-^\-\- . . . -\-h„, and the total pressure on the surface =gdAh. That is, the total pressure on a surface immersed is equal to the weight {— gravity force) of a column of liquid whose lase is the area pressed, and height the depth of the centre of gravity below the liquid level. TOTAL PKESSUKB ON A SUKFACB IMMERSED. 221 Notice (a) that in this discussion nothing is said about the shape of the vessel. The result does not therefore de- pend on the shape. Also the surface pressed on may be placed anywhere in the vessel. It may therefore form its base. Hence tlie pressure on the base of a vessel contain- ing a liquid is independent of the shape of the vessel. For example, the pressure against a dock would be the same if the dock were exposed to the Atlantic Ocean or were situ- ated in a land-locked harbor, provided the depth of water at the dock were the same in both cases and the water at rest. Ex. Explain the paradox, " Any quantity of liqui-d, how- ever small, may be made to support any quantity, however large." Notice (5) that a pressure P on a surface A may be con- sidered as arising from a column of liquid whose base is the surface and whose height h is found from P — gS Ah. This height is called the Head of Water. For a base of one sq. in. and head h ft. the pressure = 62.4347«/144 = 0.434 h pounds ; and conversely, a pres- sure of p pounds per sq. in. corresponds to a head of 2.304 p feet. Ex. 1. Prove that the water pressure on a surface 1 ft. wide, h ft. deep, is 31^ ¥ pounds. 2. Find the resultant pressure on a sluice gate, the water standing 10 ft. on one side and 6 ft. on the other. Ans. 1 ton. 3. Compare the pressures on the upper and lower halves of a sluice gate. Ans. 1 to 3. 4. The slope of a reservoir wall is 1 in 8 and the height 35 ft. If the water reaches to the top of the sloping face find the horizontal pressure. Ans. 19,500 pounds. 5. If two liquids which do not mix are placed in a bent tube open at the ends, prove that their respective heights are inversely as their weights. Hence, by attaching a graduated scale, show how to find 232 STATICS OF FLUIDS. the relative weights of two liquids. Try water and mercury and see if the relation is 1 : 13.6. 6. A sphere is filled with liquid. Account for the total pressure of the liquid on the surface being greater than the weight of the liquid. What is the relation between the pressure and the weight ? 181. Centre of Pressure on a Plane Surface Immersed. — Suppose the plane surface of area A divided into areas of one unit each, and let A, ,li^,...h^ denote their respect- ive depths. The pressures on these areas are gSh^ , gSh^ , . . . gSh^, and being all normal to the surface, and there- fore parallel, may be combined into a single resultant press- ure equal to their sum. The point in which this resultant meets the surface is called the Centre of Pressure. To find its depth A, below the surface of the liquid proceed as in Art. 90, and K = (5'«^^^ X \ + gSli^ X A, + . . . ■^gdK X li,:)/{gSh, + gdU,-\- . . . +gSK). Let 6 denote the inclination of ,the plane surface to the surface of the liquid, r, , r,, . . . r^ the distances of the unit^reas from the line of intersection of the two surfaces, and r the distance of the centre of grayity from this line ; then, since A, = r. sin 6*, . . . , we have K = {r: + r; + . . . + r„^)/(rH- ?-,+ ...+ r„) = ('-i' + i\' + . . . + r„')/Ar (from Art. 90), = mom. inertia/area surf. X dist. of 0. of G., a convenient formula. A case of common occurrence is a canal lock-gate stand- ing vertically. Suppose it to form a rectangle of breadth b, depth A, and with the upper edge in the liquid surface. Then h^ = IbUybh X Ih = f/j, or the point of application of the resultant pressure is at f of the depth of tb© rectangle, CENTRE OP PRESSURE OUT A PLANE SURFACE. 333 182. We can now discuss the Stability of a Wall sub- jected to water pressure on one of its faces, — as a reservoir wall, for example. Suppose the cross-section of the wall rectangular or trape- zoidal, AB=-a,DC =1}, the face exposed to the water vertical, the height h, and that the water reaches to the top of the wall. The forces acting on the wall are the water pressure, the weight of the wall and the reac- tion of the ground sup- port. The resultant water pressure P on one foot length of wall =gdh X A/3, and acts horizontally through iTwhen HG = /t/3. The weight W of one foot length of wall = ^(a + iy^ffS, , and acts vertically through G, the centre of gravity of the cross-section, tf , being the density of the wall. Let the directions of P and W intersect in 0. Complete the parallelogram of forces to scale. The resultant E is equal and opposite to the reaction of the ground. Now, assuming that the wall is a single block, if the resultant R cuts the base between D and Cthe wall will stand; if not, it will be overturned. (Art. 96.) The relation between the forces P and W for a certain assumed position of the resultant may be found by taking moments about the point in which its direction cuts GD. Thus if we assume E to pass through a point U we must have PX 0K= WxEK as a Qondition to be gatisfled, 234 STATICS OF FLUIDS. The wall may also yield by sliding along its base. The frictional force / between the wall and foundation is the product of the weight of the wall by the coefficient of fric- tion fx between wall and foundation (Art. Ill), or f^l{a + l)hgS,ix. Hence, assuming the wall to be incapable of sliding along any joint other than the base, we have, if the base is hori- zontal, and therefore P and / parallel, the condition to be satisfied when the wall is just on the point of sliding. Sometimes a wall is to be designed which shall have a certain Factor of Safety, the meaning being that the thrust necessary to overturn the wall or to cause it to slide shall be an arbitrary multiple of the theoretical thrust. This mul- tiple is the factor of safety. Ex. 1. Which is at the greater depth in a rectangular area immersed in water, the centre of pressure or the centre of gravity ? Ans. The c. of p. 2. The cross-section of an embankment which weighs 125 lbs. per c. ft. is trapezoidal, with one face vertical. It is 5 ft. wide at the top, 11 ft. wide at the bottom, and 15 ft. high. The water is to press against the vertical side, reach- ing to its top. Will the embankment stand ? ^ Ms. Yes. 3. The depth of the c. of p. of a vertical right-angled tri- angular lamina whose base lies in the surface is ^ the alti- tude: if the vertex is in the surface and base horizontal, it is f of the altitude. 4. A reservoir wall, cross-section rectangular, height h, weight per cubic ft. w, , water reaches to top of wall: find its thickness t that it may be on the point of being over- turned by the water pressure about the outer edge C. Ans. ^wW X-^ — wjit X X. UPWARD PRESSURE. 225 5. In (4) find the thickness of the wall that it may be just on the point of sliding on the base CD, the material weighing 135 lbs. per cubic ft., and fi — 3/3. Ans. t = 0.S8h. 6. Show that whether a wall of rectangular cross-section is more likely to yield by rotation or by sliding depends on the coefficient of friction. 7. The height of water on one side of a canal lock is a ft., and on the other side b ft. Show that the resultant pressure acts at a height {a' + ab+b')/Ha+b)it. 8. The figure represents the cross-section of a dam 120 ft. high. The masonry is divided into three principal sections. It is required to find where the resultant pressures cut CD{ = 16 ft.), I!F{=60 ft.), GH(=100 ft.). The dimensions are marked in the figure. The stone weighs 144 lbs. per cubic ft., and the water is on the left face. A71S. Da = 3.9 ft., Fb = 23.1 ft.. He = 43.8 ft. 183. Upward Pressure. — Conceive a portion ABC of a Fig. 181 ^^^A at rest to become solid. The equilib- rium will remain undisturbed. The forces acting on the solid portion are its weiglit ^^s vertically downward through its centre of I gravity, and the fluid pressures normal to the ^j surface at every point. The resultant of S^^E these pressures balances the weight, and must v/ therefore act vertically upward through the 0. of G. otABO. Now if we place in the fluid a solid which displaces the same volume ABC, the upward pressure is the same as 226 STATICS OF FLUIDS. before, because the conditions of equilibrium are the same; that is, the upward pressure on a solid which displaces a portion of fluid is equal in magnitude to the weight of the fluid displaced, and its direction passes through the centre of gravity of the fluid displaced. This is known as the principle of Archimedes* Ex. Infer from the principle of Archimedes that the loss of weight of a body immersed in a fluid is equal to the weight of the fluid displaced. 184. A solid placed in a fluid will either float partially immersed or float.wholly immersed, or sink. Two applica- tions of the principle of Archimedes to these cases are important. (1) Specific gravity ( = relative weight) . Take a solid and p. 132 let its weight in air be W. Suspend it by a fine wire from a hook attached to one scale-pan of a pair of scales, and find its weight W, when immersed in a liquid. The loss of weight W' — Tf, is the upward pressure, and is equal to the weight of an equal ^^ra volume of the liquid. The ratio of 1^^ the weight PFof the solid to the weight ^~ "^ W — IF, of the equal volume of liquid is called the specific gravity a of the solid with reference to the liquid taken. Thus 0-= W/{W-W,). * " Hiero, King of Syracuse, had a quantity of gold made into a crown, and suspecting that the workmen had abstracted some of the gold and used a portion of alloy of the same weight in its place, applied to Archimedes to solve the difficulty. Archimedes, while re- flecting over this problem in his bath, observed the water running over the sides of the bath, and it occurred to him that he was dis- placing a quantity of water equal to his own bulk, and therefore that a quantity of pure gold equal in weight to the crown would displace less water than the crown, the volume of any weight of alloy being greater than that of an equal weight of gold. It is jelated that he immediately ran out into the streets, crying out, ' evpriKoc, fvpr/Ka 1 ' "—Besani, HYDROMETER. 227 As in what is called "weighing" we really compare masses (Art. 39), so in specific gravity the idea is a relative weighing or a comparison of masses. The term may, therefore, be defined — The specific gravity of a body is the ratio of the mass of the iody to the mass of an equal volume of some standard body ; or since the density (J of a body is the mass of unit volume, it may be put in the equivalent form : the specific gravity of a body is the ratio of its density to the density of sotne standard body. The standard taken is usually distilled water at its tem- perature of greatest density, 4° 0. This standard is arbi- trary, but .is chosen on account of its convenience. Note that specific gravity does not depend on volume, but on equality of volume. As we may assume any volume of the standard, take a cubic unit of water (cu. ft. or cu. cm.). Denote its density by S. Then if tr be the sp. gr. of a body of density p, , we have from the definition a = SJS, and the mass M ota, volume U of the body is given by M = r/d, = UaS. Since 1 cubic ft. of water at 4° 0. weighs 62.4 lbs., then Weight or mass M in pounds = 62.4 Ua, Weight or pressure W in pounds = 62.4 Ucr, Weight or pressure W in poundals = 62.4 Ucrg, when U is expressed in cubic feet. Ex. 1. The true weight of a body is 25 grams. It weighs 15 grams when immersed in water : find its specific gravity. Ans. 2.5. 2. If Wj , W^, W, be the apparent weights of a body in three liquids the specific gravities of which are cr^ , cr^, cr^, prove that WA<^, - a,) + W,{cr, - a^) + W,{