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MECHANICS FOR ENGINEERS - > MECHANICS FOR ENGINEERS A TEXT-BOOK OF INTERMEDIATE STANDARD BY ARTHUR MORLEY M.Sc., UNIVERSITY SCHOLAR (VICT.) SENIOR LECTURER IN ENGINEERING IN UNIVERSITY COLLEGE NOTTINGHAM WITH 200 DIAGRAMS AND NUMEROUS EXAMPLES LONGMANS, GREEN, AND CO. 39 PATERNOSTER ROW, LONDON NEW YORK AND BOMBAY All rights reserved PREFACE ENGINEERING students constitute a fairly large proportion of those attending the Mechanics classes in technical colleges and schools, but their needs are not identical with those of the students of general science. It has recently become a common practice to provide separate classes in Mathematics, adapted to the special needs of engineering students, who are in most institutions sufficiently numerous to justify similar provision in Mechanics. The aim of this book is to provide a suitable course in the principles of Mechanics for engineering students. With this object in view, the gravitational system of units has been adopted in the English measures. A serious injustice is often done to this system in books on Mechanics by wrongly denning the pound unit of force as a variable quantity, thereby reducing the system to an irrational one. With proper premises the gravitational system is just as rational as that in which the "poundal" is adopted as the unit of force, whilst it may be pointed out that the use of the latter system is practically confined to certain text-books and exami- nation papers, and does not enter into any engineering work. Teachers of Engineering often find that students who are learning Mechanics by use of the " poundal " system, fail to apply the principles to engineering problems stated in the only units which are used in such cases the gravitational vi Preface units. The use of the dual system is certainly confusing to the student, and in addition necessitates much time being spent on the re-explanation of principles, which might other- wise be devoted to more technical work. Graphical methods of solving problems have in some cases been used, by drawing vectors to scale, and by esti- mating slopes and areas under curves. It is believed that such exercises, although often taking more time to work than the easy arithmetic ones which are specially framed to give exact numerical answers, compel the student to think of the relations between the quantities involved, instead of merely performing operations by fixed rules, and that the principles so illustrated are more deeply impressed. The aim has not been to treat a wide range of academic problems, but rather to select a course through which the student may work in a reasonable time say a year and the principles have been illustrated, so far as the exclusion of technical knowledge and terms would allow, by examples likely to be most useful to the engineer. In view of the applications of Mechanics to Engineering, more prominence than usual has been given to such parts of the subject as energy, work of forces and torques, power, and graphical statics, while some other parts have received less attention or have been omitted. It is usual, in books on Mechanics, to devote a chapter to the equilibrium of simple machines, the frictional forces in them being considered negligible : this assumption is so far from the truth in actual machines as to create a false impres- sion, and as the subject is very simple when treated experi- mentally, it is left for consideration in lectures on Applied Mechanics and in mechanical laboratories. The calculus has not been used in this book, but the Preface vii student is not advised to try to avoid it; if he learns the elements of Mechanics before the calculus, dynamical illus- trations of differentiation and integration are most helpful. It is assumed that the reader is acquainted with algebra to the progressions, the elements of trigonometry and curve plotting ; in many cases he will doubtless, also, though not necessarily, have some little previous knowledge of Mechanics. The ground covered is that required for the Intermediate (Engineering) Examination of the University of London in Mechanics, and this includes a portion of the work necessary for the Mechanics Examination for the Associateship of the Institution of Civil Engineers and for the Board of Education Examination in Applied Mechanics. I wish to thank Professor W. Robinson, M.E., and Pro- fessor J. Goodman for several valuable suggestions made with respect to the preparation and publication of this book; also Mr. G. A. Tomlinson, B.Sc., for much assistance in correcting proofs and checking examples; in spite of his careful corrections some errors may remain, and for any intimation of these I shall be obliged. ARTHUR MORLEY. NOTTINGHAM, June, 1905. CONTENTS CHAPTER I KINEMATICS PAGES Velocity ; acceleration ; curves of displacement, and velocity ; falling bodies ; areas under curves ; vectors ; applications to velocities ; relative velocity ; composition and resolution of acceleration ; angular displacement, velocity, and acceleration 1-26 CHAPtER II THE LAWS OF MOTION First law ; inertia ; weight ; momentum ; second law ; engineers' units ; c.g.s. system ; triangle and polygon of forces ; impulse ; third law ; motion of connected bodies ; Atwood's machine 27-47 CHAPTER III WORK, POWER, AND ENERGY Work ; units ; graphical method ; power ; moment of a force ; work of a torque ; energy potential, kinetic ; principle of work 48-67 CHAPTER IV MOTION IN A CIRCLE: SIMPLE HARMONIC MOTION Uniform circular motion ; centripetal and centrifugal force ; curved track ; conical pendulum ; motion in vertical circle ; simple harmonic motion ; alternating vectors ; energy in S.H. motion ; simple pendulum 68-90 x Contents CHAPTER V STA TICS CONCURRENT FORCES FRICTION PAGES Triangle and polygon of forces ; analytical methods ; friction ; angle of friction ; sliding friction ; action of brakes ; adhesion ; friction of screw 91-113 CHAPTER VI STATICS OF RIGID BODIES Parallel forces ; moments ; moments of resultants ; principle of moments ; levers ; couples ; reduction of a coplanar system ; conditions of equilibrium ; smooth bodies ; method of sections ; equilibrium of three forces 114-139 CHAPTER VII CENTRE OF INERTIA OR MASS CENTRE OF GRAVITY Centre of parallel forces ; centre of mass ; centre of gravity ; two bodies ; straight rod ; triangular plate ; rectilinear figures ; lamina with part removed ; cone ; distance of e.g. from lines and planes ; irregular figures ; circular arc, sector, segment ; spherical shell ; sector of sphere ; hemisphere . . 140-166 CHAPTER VIII CENTRE OF GRAVITY PROPERTIES AND APPLICATIONS Properties of e.g. ; e.g. of distributed load; body resting on a plane ; stable, unstable, and neutral equilibrium ; work done in lifting a body ; theorems of Pappus 167-187 CHAPTER IX MOMENTS OF INERTIA ROTATION Moments of inertia ; particles ; rigid body ; units ; radius of gyration ; various axes ; moment of inertia of an area ; circle ; hoop ; cylinder ; kinetic energy of rotation ; changes in energy and speed ; momentum ; compound pendulum ; laws of rotation ; torsional oscillation ; rolling bodies . . 188-222 Contents xi CHAPTER X ELEMENTS OF GRAPHICAL STATICS PAGES Bows' notation ; funicular polygon ; conditions of equilibrium, choice of pole ; parallel forces ; bending moment and shearing force ; diagrams and scales ; jointed structures ; stress diagrams ; girders : roofs ; loaded strings and chains . 223-252 APPENDIX 2 53~ 2 55 ANSWERS TO EXAMPLES 256-259 EXAMINATION QUESTIONS . 260-273 MATHEMATICAL TABLES ,....; 274-278 INDEX 279-282 MECHANICS FOR ENGINEERS i. KINEMATICS deals with the motion of bodies without reference to the forces causing motion. MOTION IN A STRAIGHT LINE. Velocity. The velocity of a moving point is the rate of change of its position. Uniform Velocity. When a point passes over equal spaces in equal times, it is said to have a constant velocity ; the magnitude is then specified by the number of units of length traversed in unit time, e.g. if a stone moves 15 feet with a constant velocity in five seconds, its velocity is 3 feet per second. If s = units of space described with constant velocity v in t units of time, then, since v units are described in each second, (v x /) units will be described in / seconds, so that s = vt and v = - Fig. i shows graphically the relation between the space described and the time taken, for a constant velocity of 3 feet per second. Note that v=- = - or - or -, a constant Mechanics for Engineers velocity of 3 feet per second whatever interval of time is considered. 1234- Time in- seconds FIG. i. Space curve for a uniform velocity of 3 feet per second. 2. Mean Velocity. The mean or average velocity of a point in motion is the number of units of length described, divided by the number of units of time taken. 3. Varying Velocity, The actual velocity of a moving point at any instant is the mean velocity during an indefinitely small interval of time including that instant. 4. The Curve of Spaces or Displacements. Fig. 2 shows graphically the relation between the space described and T N M in. seconds FIG. 2. Space curve for a varying velocity. the time taken for the case of a body moving with a varying velocity. At a time ON the displacement is represented by Kinematics 3 PN, and after an interval NM it has increased by an amount QR, to QM. Therefore the mean velocity during the interval QR QR A NM is represented by 7 or --^ or by tan QPR, i.e. by the tangent of the angle which the chord PQ makes with a hori- zontal line. If the interval of time NM be reduced indefi- nitely, the chord PQ becomes the tangent line at P, and the mean velocity becomes the velocity at the time ON. Hence the velocity at any instant is represented by the gradient of the tangent line to the displacement curve at that instant. An upward slope will represent a velocity in one direction, and a down- ward slope a velocity in the opposite direction. 5. If the curvature is not great, i.e. if the curve does not bend sharply, the best way to find the direction of the tangent line at any point P on a curve such as Fig. 2, is to take two ordi- nates, QM and ST, at short equal distances from PN, and join QV QS; then the slope of QS, viz. --, is approximately the same as that of the tangent at P. This is equivalent to taking the velocity at P, which corresponds to the middle of the interval TM, as equal to the mean velocity during the interval of time TM. 6. Scale of the Diagram. Measure the slope as the gradient or ratio of the vertical height, say QV, to the hori- zontal SV or TM. Let the ratio QV : TM (both being measured in inches say) be x. Then to determine the velocity represented, note the velocity corresponding to a slope of T inch vertical to i inch horizontal, say y feet per second. Then the slope of QS denotes a velocity of xy feet per second. 7. Acceleration. The acceleration of a moving body is the rate of change of its velocity. When the velocity is in- creasing the acceleration is reckoned as positive, and when decreasing as negative. A negative acceleration is also called a retardation. 8. Uniform Acceleration. When the velocity of a point increases by equal amounts in equal times, the acceleration is said to be uniform or constant : the magnitude is then specified 4 Mechanics for Engineers by the number of units of velocity per unit of time ; e.g. if a point has at a certain instant a velocity of 3 feet per second, and after an interval of eight seconds its velocity is 19 feet per second, and the acceleration has been uniform, its magnitude is increase of velocity TO 3 .- , -. - = b = 2 feet per second in each of time taken to increase 8 the eight seconds, i.e. 2 feet per second per second. At the end of the first, second, and third seconds its velocities would be (3 + 2 ): (3 + 4), and (3 + 6) feet per second respectively (see Fig. 3). K 10 8 8 7 ^ V) 5 M 1234-56 Time- in, seconds FIG. 3. Uniform acceleration. 9. Mean Acceleration. The acceleration from 3 feet per second to 19 feet per second in the last article was sup- posed uniform, 2 feet per second being added to the velocity in each second; but if the acceleration is variable, and the increase of velocity in different seconds is of different amounts, then the acceleration of 2 feet per second per second during the eight seconds is merely the mean acceleration during that increase of velocity time. The mean acceleration is equal to time taken for increase", and is in the direction of the change of velocity. Kinematics 5 The actual acceleration at any instant is the mean acceleration for an indefinitely small time including that instant. 10. Fig. 3 shows the curve of velocity at every instant during the eight seconds, during which a point is uniformly accelerated from a velocity of 3 feet per second to one of 19 feet per second. n. Calculations involving Uniform Acceleration. If u = velocity of a point at a particular instant, and^ = uni- form acceleration, i.e. f units of velocity are added every second then after i second the velocity will be ti +/ and 2 seconds u -f- 2/ jj 3 M ^ 3y / v will be u +ff (i) e.g. in the case of the body uniformly accelerated 2 feet per second per second from a velocity of 3 feet per second to a velocity of 19 feet per second in eight seconds (as in Art. 8), the velocity after four seconds is 3 -f- (* X 4) = 1 1 feet per second. The space described (s) in t seconds may be found as follows : The initial velocity being u, and the final velocity being v, and the change being uniform, the mean or average i . u + v velocity is . u -f v u u -\- ft Mean velocity = SB -U = u 4- ^ft 2 2 ' 2 (which is represented by QM in Fig. 3. See also Art. 2). Hence u + \ft'= - and s=af + \fP ... (2) e.g. in the above numerical case the mean velocity would be =11 feet per second (QM in Fig. 3) and J=nx8 = 88 feet or s = 3 x 8 -f l x 2 x 8 2 = 24 + 64 = 88 feet It is sometimes convenient to find the final velocity in 6 Mechanics for Engineers terms of the initial velocity, the acceleration, and the space described. We have-^ from (i) v = n -\-ft therefore V* = u 1 + zuft +/ 2 / 2 = 2 + /(/ + \ff) and substituting for (ut + |// 2 ) its value s from (2), we have . ****+2/S (3) The formulae (i), (2), and (3) are useful in the solution of numerical problems on uniformly accelerated motion. 12. Acceleration of Falling- Bodies. It is found that bodies falling to the earth (through distances which are small compared to the radius of the earth), and entirely unresisted, increase their velocity by about 32-2 feet per second every second during their fall. The value of this acceleration varies a little at different parts of the earth's surface, being greater at places nearer to the centre of the earth, such as high lati- tudes, and less in equatorial regions. The value of the "acceleration of gravity" is generally denoted by the letter g. In foot and second units its value in London is about 32' 1 9, and in centimetre and second units its value is about 98 r units. 13. Calculations on Vertical Motion. A body pro- jected vertically downwards with an initial velocity u will in / seconds attain a velocity u -)- gt, and describe a space Mf+'fa* In the case of a body projected vertically upward with a velocity //, the velocity after / seconds will be, */ gt, and will be upwards if gt is less than u, but downward if gt is greater than u. When t is of such a value that gt = u, the downward acceleration will have just overcome the upward velocity, and the body will be for an instant at rest : the value of / will then be u -. The space described upward after t seconds will be Sr-i/ft The time taken to rise h feet will be given by the equation h = ut - i/? 2 This quadratic equation will generally have two roots, the Kinematics smaller being the time taken to pass through h feet upward, and the larger being the time taken until it passes the same point on its way downward under the influence of gravitation. The velocity z', after falling through " // " feet from the point of projection downwards with a velocity u, is given by the expression i? = u z + 2^, and if u = o, i.e. if the body be simply dropped from rest, v 2 = 2gh, and v = *J 2gh after falling h feet. 14. Properties of the Curve of Velocities. Fig. 4 shows the velocities at all times in a particular case of a body Time FIG. 4. Varying velocity. starting from rest and moving with a varying velocity, the acceleration not being uniform. (i) Slope of the Curve. At a time ON the velocity is PN, and after an interval NM it has increased by an amount QR to QM; therefore the mean acceleration during the OR OR interval NM is represented by ^rr or p^, i.e. by the tangent of the angle which the chord PQ makes with a horizontal line. If the interval of time NM be reduced indefinitely, the chord PQ becomes the tangent line to the curve at P, and the mean acceleration becomes the acceleration at the time ON. So that tJie acceleration at any instant is represented by the gradient of the tangent line at that instant. The slope will be upward if the velocity is increasing, downward if it is decreasing ; in the latter case the gradient is negative. The scale of accelerations is easily found by the acceleration represented by unit gradient. If the curve does not bend sharply, the direction of the 8 Mechanics for Engineers tangent may be found by the method of Art. 5, which is in this case equivalent to taking the acceleration at P as equal to the mean acceleration during a small interval of which PN is the velocity at the middle instant. (2) The Area under the Curve. If the velocity is constant and represented by PN (Fig. 5), then the distance described in an interval NM is PN.NM, and there- fore the area under PQ, viz. the rectangle PQMN, represents the space de- scribed in the interval NM. If the velocity is not constant, as in Fig. 6, sup- pose the interval NM divided up into a number of small parts such as CD. Then AC represents the velocity at the time represented by OC ; the velocity is increasing, and therefore in the interval CD the space described is greater than that represented by the rectangle AEDC, and less than that represented by the rect- angle FBDC. The total space described during the interval N Tirrve, FIG. 5. M I f f=l ^-**" s A t ' 1 / IN CD M Tinve- FIG. 6. Varying velocity. NM is similarly greater than that represented by a series of rectangles such as AEDC, and less than that represented by a series of rectangles such as FBDC. Now, if we consider the number of rectangles to be increased indefinitely, and Kinematics the width of each to be decreased indefinitely, the area PQMN under the curve PQ is the area which lies always between the sums of the areas of the two series of rectangles, however nearly equal they may be made by subdividing NM, and the area PQMN under the curve therefore represents the space described in the interval NM. The area under the curve is specially simple in the case of uniform acceleration, for which the curve of veloci- ties is a straight line (Fig. 7). F i G . 7 . Here the velocity PN being u, and NM being / units of time, and the final velocity being QM = v, the area under PQ is M PN + QM 2 u 4- v X NM = ST X NM X /(as in Art. u) v u QR And if / is the acceleration / = - (represented by -^ or OR fg, i.e. by tan QPR), .*.// = v u V = U -\-ft .u-\-v . u + u -\-ft and the space described - X / is - X /, which is 2 2 ut + |// 2 (as in Art. TI). 15. Notes on Scales. If the scale of velocity is i inch to x feet per second, and the scale of time is i inch to y seconds, then the area under the curve will represent the distance described on such a scale that i square inch represents xy feet. 1 6. In a similar way we may show that the area PQMN 10 Mechanics for Engineers (Fig. 8) under a curve of accelerations represents the total increase in velocity in the interval of time NM. If the scale of acceleration is i inch to z feet per second per second, and the scale of time is i inch to y seconds, then the scale of velocity is i square inch to yz feet per second. 17. Solution of Problems. Where the motion is of a simple kind, such as a uniform velocity or uniform acceleration, direct calculation is usually the easiest and quickest mode of solution, but where (as is quite usual in practice) the motion is much more complex and does not admit of simple mathematical expression as a function of the time taken or distance covered, a graphical method is recommended. Squared paper saves much time in plotting curves for graphical solutions. Example i. A car starting from rest has velocities v feet per second after / seconds from starting, as given in the following table : V o o 4 iro 9 22'6 17 35-6 24 44'5 30 49-0 32 48-9 40 45 33 7 53 26-8 58 62 24'C Find the accelerations at all times during the first 60 seconds, and draw a curve showing the accelerations during this time. First plot the curve of velocities on squared paper from the given data, choosing suitable scales. This has been done in Kinematics II Fig. 9, curve I., the scales being i inch to 10 seconds and i inch to 20 feet per second. In the first 10 seconds RQ represents 24^2 feet per second f 50 J 40 ^30 S 20 5 *Al N I0 w -KffJI-Utr 20 -S4G6 4-0 50 if Scale, of iTiches FIG. 9. gain of velocity, and OQ represents 10 seconds; therefore the acceleration at N 5 seconds from starting is approximately 24*2 , or 2 '4 2 feet per second per second. Or thus : unit gradient i inch vertical in i inch horizontal represents 20 feet per second " 10 seconds = 2 feet P er second P er Sec nd RQ i '2 1 inch hence -^ = : = 1-21 OQ i inch hence acceleration at N 1 i c T'^T v \ = 2 '4 2 ' eet P er second per second Ib I 2 1 /% 2 (see Art. 14) Similarly in the second 10 seconds, SV which is SM RQ, represents (39'8 24-2), or 15-6 feet per second gain of velocity ; 12 Mechanics for Engineers therefore the velocity at W 15 seconds from starting is approxi- mately -^ ,or 1*56 feet per second per second. 10 Continue in this way, finding the acceleration at say 5, 15, 25, 35, 45, and 55 seconds from starting; and if greater ac- curacy is desired, at 10, 20, 30, 40, 50, and 60 seconds also. The simplest way is to read off from the curve I. velocities in tabular form, and by subtraction find the increase, say, in 10 seconds, thus t O 5 10 IS 20 2? 30 35 40 45 50 55 60 V ... o ^_ 13-5 24'2 32-8 ^/- .39/8:452 49' o 47 '5 40-6 ^- 337 ^^ 29-0 25-1 24/1 Change in v 24-2 iS'6 l 9-2 -8-4 -11-6 -5' 1 for 10 sees. i 9' 3 I2'6 2'I '3*8 -8-6 Acceleration 2-42 i '93 1-56 1-26 0'2I -0-84 -1-38 -1-16 -0-86 -0-51 From the last line in this table curve II., Fig. 9, has been plotted, and the acceleration at any instant can be read off from it. It will be found that the area under curve II. from the start to any vertical ordinate is proportional to the correspond- ing ordinate of curve I. (see Art. 16). The area, when below the time base-line, must be reckoned as negative. Example 2. Find the distance covered from the starting-point by the car in Example i at all times during the first 60 seconds, and the average velocity throughout this time. In the first 10 seconds the distance covered is found approxi- mately by multiplying the velocity after 5 seconds by the time, i.e. 13*5 x 10 = 135 feet. This approximation is equivalent to taking 13-5 feet per second as the mean velocity in the first 10 seconds. In the next 10 seconds the mean velocity being approximately 32'8 feet per second (corresponding to / = 15 seconds), the distance covered is 32*8 x 10 = 328 feet, therefore the total distance covered in the first 20 seconds is 135 + 328 = 463 feet. Proceeding in this way, taking ro-second intervals throughout the 60 seconds, and using the tabulated results in Example i, we get the following results : Kinematics I t o IO 20 30 40 50 60 Space in previous o 135 32 454 475 .337 251 10 sees. Total space 135 463 917 1392 1729 1980 from which the curve of displacements, Fig. 10, has been plotted. 2000 1500 1000 500 10 2O 30 4-0 Time in seconds 50 60 Scale of Inches FIG. 10. Greater accuracy may be obtained by finding the space described every 5 instead of every 10 seconds. The average velocity = space described = 1980 = feet sec time taken 60 Note that this would be represented on Fig. 9 by a height which is equal to the total area under curve I. divided by the length of base to 60 seconds. EXAMPLES I. I. A train attains a speed of 50 miles per hour in 4 minutes after starting from rest. Find the mean acceleration in foot and second units. 14 Mechanics for Engineers 2. A motor car, moving at 30 miles per hour, is subjected to a uniform retardation of 8 feet per second per second by the action of its brakes. How long will it take to come to rest, and how far will it travel during this time ? 3. With what velocity must a stream of water be projected vertically upwards in order to reach a height of 80 feet ? 4. How long will it take for a stone to drop to the bottom of a well 150 feet deep? 5. A stone is projected vertically upward with a velocity of 170 feet per second. How many feet will it pass over in the third second of its upward flight ? At what altitude will it be at the end of the fifth second, and also at the end of the sixth ? 6. A stone is projected vertically upward with a velocity of 140 feet per second, and two seconds later another is projected on the same path with an upward velocity of 135 feet per second. When and where will they meet? 7. A stone is dropped from the top of a tower 100 feet high, and at the same instant another is projected upward from the ground. If they meet halfway up the tower, find the velocity of projection of the second stone. The following Examples are to be worked graphically. 8. A train starting from rest covers the distances s feet in the times seconds as follows : ! r o 5 10 ii 54 18 170 22 260 27 390 31 450 38 504 4 6 55 5 570 Find the mean velocity during the first 10 seconds, during the first 30 seconds, and during the first 50 seconds. Also find approximately the actual velocity after 5, 15, 25, 35, and 45 seconds from starting-point, and plot a curve showing the velocities at all times. 9. Using the curve of velocities from Example 8, find the acceleration every 5 seconds, and draw the curve of accelerations during the first 40 seconds. 10. A train travelling at 30 miles per hour has steam shut off and brakes applied ; its speed after / seconds is shown in the following table : * ... ... v, miles perj hour ...j 30-0 4 26-0 12 2I"S 2O 167 26 i4'o 35 io'4 42 77 5 4-8 Kinematics 1 5 Find the retardation in foot and second units at 5-second intervals through- out the whole period, and show the retardation by means of a curve. Read off from the curve the retardation after 7 seconds and after 32 seconds. What distance does the train cover in the first 30 seconds after the brakes are applied ? II. A body is lifted vertically from rest, and is known to have the following accelerations / in feet per second per second after times t seconds : t ... o 3-0 0-8 2-9 1-9 2-85 3-0 2'6o 3 '9 2'2O 4-8 175 6-0 1-36 6-8 I'2O 8-0 1*04 8-8 0-97 Find its velocity after each second, and plot a curve showing its velocity at all times until it has been in motion 8 seconds. How far has it moved in the 8 seconds, and how long does it take to rise 12 feet ? VECTORS. 1 8. Many physical quantities can be adequately expressed by a number denoting so many units, e.g. the weight of a body, its temperature, and its value. Such quantities are called scalar quantities. Other quantities cannot be fully represented by a number only, and further information is required, e. g. the velocity of a ship or the wind has a definite direction as well as numerical magnitude : quantities of this class are called vector quantities and are very conveniently represented by vectors. A Vector is a straight line having definite length and direction, but not definite position in space. 19. Addition of Vectors. To find the sum of two vectors ab and cd (Fig. n), set out ab of proper length and direction, and from the end b set out be equal in length and parallel to i6 Mechanics for Engineers FIG. 12. cd ; join ae. Then ae is the geometric or vector sum of ab and cd. We may write this ab -f be = ae or, since be is equal to cd ab -j- cd = ae 20. Subtraction of Vectors. If the vector cd (Fig. 12) is to be subtracted from the vector ab, we simply find the sum ae as before, of a vector ab and second vector be, which is equal to cd in magnitude, but is of opposite sign or direc- tion ; then ae = ab + be = ab cd If we had required the difference, cd ab, the result would have been ea instead of ae. 21. Applications: Displacements. A vector has the two characteristics of a displacement, viz. direction and magni- tude, and can, therefore, represent it completely. If a body receives a displacement ab (Fig. n), and then a further dis- placement completely represented by cd, the total displacement is evidently represented by ae in magnitude and direction. 22. Relative Displacements. CASE I. Definition. If 7 a body remains at rest, and a second body receives a dis- placement, the first body is said to re- ceive a displacement of equal amount but opposite direction re- lative to the second. CASE II. Where Two Bodies each receive a Displacement. If a body A receive a displacement represented by a vector ab (Fig- T 3)> an d a body B receive a displacement represented by FIG. 13. Kinematics \ 7 cd, then the displacement of A relative to B is the vector difference, ab cd. For if B remained at rest, A would have a displacement ab relative to it. But on account of B's motion (cd), A has, relative to B, an additional displacement, dc (Case I.) ; therefore the total displacement of A relative to B is ab 4- dc (or, ab cd} = ab + be = ae (by Art. 20) ; where be is of equal ength and parallel to dc. 23. A Velocity which is displacement per unit time can evidently be represented fully by a vector ; in direction by the clinure of the vector, and in magnitude by the number of units of length in the vector. 24. Triangle and Polygon of Velocities. A velocity is said to be the resultant of two others, which are called components, when it is fully represented by a vector which is the geometrical sum of two other vectors representing the two components ; e.g. if a man walks at a rate of 3 miles per hour across the deck of a steamer going at 6 miles per hour, the resultant velocity with which the man is moving over the sea is the vector sum of 3 and 6 miles per hour taken in the proper directions. If the steamer were heading due north, and the man walking due east, his actual velocity is shown by ac in Fig. 14; ab 6 be = 3 ac = J6* + 3 2 = \/45 miles per hour = 671 miles per hour and the angle which ac makes with ab E. of N. is given by tan0=:=i = 26 35' Resultant velocities may be found by drawing vectors to scale or by the ordinary rules of trigonometry./ If the re- sultant velocity of more than two components (in the same plane) is required, two may be compounded, and then a third with their resultant, and so on, until all the components have been added. It will be seen (Fig. 15) that the result is repre- sented by the closing side of an open polygon the sides of c 1 8 Mechanics for Engineers which are the component vectors. The order in which the sides are drawn is immaterial. It is not an essential condition that all the components should be in the same plane, but if not, the methods of solid geometry should be employed to draw the polygon. FIG. 15. Fig. 15 shows the resultant vector af of five co-planor vectors, ab, be, cd, de, and ef. If, geometrically, af = ab + be and ad = ac + cd then ad = ab + be -j- cd and similarly, adding de and ef- af = ab + be + cd + de -f ef In drawing this polygon it is unnecessary to put in the lines ac, ad, and ae. 25. It is sometimes convenient to resolve a velocity into two components, i.e. into two other velocities in particular directions, and such that their vector sum is equal to that velocity. Rectangular Components. The most usual plan is to resolve velocities into components in two standard directions at right angles, and in the same plane as the original veloci- ties : thus, if OX and OY (Fig. 16) are the standard directions, and a vector ab represents a velocity v, then the component in the direction OX is represented by ac, which is equal to ab cos B, and represents v cos 0, and that in the direction OY is represented by cb, i.e. by ab sin 0, and is v sin 0. Kinematics This form of resolution of velocities provides an alternative method of finding the re- y sultant of several velocities. Each velocity may be re- solved in two standard directions, OX and OY, and then all the X com- ponents added algebraically and all the Y components added algebraically. This reduces the components to two at right angles, which may be replaced by a re- ' FIG. 16. sultant R units, such that the squares of the numerical values of the rectangular com- ponents is equal to the square of R, e.g. to find the resultant FIG. 17. of three velocities Vj, V 2 , and V 3 , making angles a, /3, and y respectively with some fixed direction OX in their plane (Fig. 17). Resolving along OX, the total X component, say X, is X = Vj cos a + V 2 cos /3 + V :t cos y Resolving along OY Y = V, sin a + V 9 sin 4- V ;t sin y and R 2 = X 2 + Y 2 or R = /X + Y 2 and it makes with OX an angle 6 such that tan 6 = ^- 2O Mechanics for Engineers Fig. 1 7 merely illustrates the process ; no actual drawing of vectors is required, the method being wholly one of calcu- lation. Exercise i. A steamer is going through the water at 10 knots per hour, and heading due north. The current runs north-west at 3 knots per hour. Find the true velocity of the steamer in magnitude and direction. (i) By drawing vectors (Fig. 18). Set off ab, representing 10 knots per hour, to scale due north. Then draw be inclined 45 to the direction ab, and representing 3 knots per hour to the same scale. Join ac. Then ac, which scales 12*6 knots per hour when drawn to a large scale, is the true velocity, and the angle cab E of N measures 10. ! (2) Method by resolving N. and E. N. component =10 + 3 cos 45 = 10 + -4- knots per hour, V 2 E. = 3 sin 45 = -7- knots per hour, or 2*12 V 2 Resultant velocity R = V( 12 ' 12 ) 2 + ( 2<I2 ) 2 = I2 '6 knots per hour And if is the) 3_ . /, i -/3\ 2 ' 12 angle E. of N.) a ~ ^2 :. = 9 55' RELATIVE VELOCITY. 26. The velocity of a point A relative to a point B is the rate of change of position (or displacement per unit of time) of A with respect to B. Let v be the velocity of A, and u that of B. If A remained stationary, its displacement per unit time relative to B would be u (Art. 22). But as A has itself a velocity v, its total velocity relative to B is v + ( u) or v n, the subtraction to be performed geometrically (Art. 20). The velocity of B relative to A is of course u v, equal in magnitude, but opposite in direction. The subtraction of velocity z> u may be performed by drawing vectors to scale, Kinematics 21 by the trigonometrical rules for the solution of triangles, or by the method of Art. 25. Example. Two straight railway lines cross : on the first a train 10 miles away from the crossing, and due west of it, is ap- proaching at 50 miles per hour ; on the second a train 20 miles away, and 15 E. of N., is approaching at 40 miles per hour. How far from the crossing will each train be when they are nearest together, and how long after they occupied the above positions ? First set out the two lines at the proper angles, as in the left side of Fig. 19, and mark the positions A and B of the first and second FIG. 19. trains respectively. Now, since the second train B is coming from 1 5 E. of N., the first train A has, relative to the second, a component velocity of 40 miles per hour in a direction E. of N., in addition to a component 50 miles per hour due east. The relative velocity is therefore found by adding the vectors pq 50 miles per hour east, and yr 40 miles per hour, giving the vector pr, which scales 72 miles per hour, and has a direction 57^ E. of N. Now draw from A a line AD parallel to pr. This gives the posi- tions of A relative to B (regarded as stationary). The nearest approach is evidently a distance BD, where BD is perpendicular to AD. The distance moved by A relative to B is then AD, which scales 23-2 miles (the trains being then a distance BD, which scales 8' 1 2 miles apart). The time taken to travel relatively 23-2 miles 2 3' 2 at 72 miles per hour is - hours = 0^322 hour. 22 Mechanics for Engineers Hence A will have travelled 50 x o'322 or 16*1 miles and B ,, ,, 40 x 0-322 or 12-9 A will then be 6'i miles past the crossing, and B ,, 7'i short of the crossing. 27. Composition, Resolution, etc., of Accelerations. Acceleration being also a vector quantity, the methods of composition, resolution, etc., of velocities given in Arts. 23 to 26 will also apply to acceleration, which is simply velocity added per unit of time. It should be noted that the acceleration of a moving point is not necessarily in the same direction as its velocity : this is only the case when a body moves in a straight line. If ab (Fig. 20) represents the velocity of a point at a certain instant, and after an interval t seconds its velocity is represented by ac, then the change in velocity in / seconds is be, for ab -\- be = ac (Art. 19), and be = ac ab (Art. 20), representing the change in velocity. Then during the / seconds the mean acceleration is represented by be 4- 1, and is in the direction be. 28. Motion down a Smooth Inclined Plane. Let be the angle of the plane to the horizontal, then the angle ABC (Fig. 21) to the vertical is (90 a). Then, since a body has a downward ver- tical acceleration g, its component along BA will be g cos CBA = g cos (90 a) = g sin a, pro- vided, of course, that there is nothing to cause a re- tardation in this direction, i.e. provided that the plane is perfectly smooth and free from obstruction. If BC = h feet, AB = h cosec a feet. The velocity of a body starting from rest at B and sliding down AB will be at A, \J 2 .^sin 6 X h cosec 6 = */ 2g/i, just as if it had fallen h feet vertically. FIG. FIG. 21. Kinematics Angular Displacement. If P FIG. 22. 29. Angular Motion P (Fig. 22) be the posi- tion of a point, and Q a subsequent position which this point takes up, then the angle QOP is the angular displace- ment of the point about O. The angular displacement about any other point, such as O', will generally be a different amount. 30. Angular Velocity. The angular velocity of a moving point about some fixed point is the rate of angular displacement (or rate of change of angular position) about the fixed point ; it is usually expressed in radians per second, and is commonly denoted by the letter w. As in the case of linear velocity, it may be uniform or varying. A point is said to have a uniform or constant angular velocity about a point O when it describes equal angles about O in equal times. The mean angular velocity of a moving point about a fixed point O is the angle described divided by the time taken. If the angular velocity is varying, the actual angular velocity at any instant is the mean angular velocity during an in- definitely small interval including that instant. 31. Angular Acceleration is the rate of change of angular velocity ; it is usually measured in radians per second per second. 32. The methods of Arts. 4 to n applicable to angular motion as well as to linear motion. 33. To find the angular velocity about O of a point describing a circle of radius r about O as centre with constant speed. Let the path PP' (Fig. 23) be de- scribed by the moving point in / seconds. Let v be the velocity (which, although F|G - 2 3- constant in magnitude, changes direction). Then angular and 1 6 are velocity about O is o> = -. But = Mechanics for Engineers arc PP and arc PP' = vt Vt 6 Tt V :. 6 = - and o> = -= / = - r t r r This will still be true if O is moving in a straight line with velocity v as in the case of a rolling wheel, provided that v is the velocity of P relative to O. If we consider / as an indefinitely small time, PP' will be indefinitely short, but the same will remain true, and we *V should have to = - whether the velocity remains constant in magnitude or varies. In words, the angular velocity is equal to the linear velocity divided by the radius, the units of length being the same in the linear velocity v and the radius /-. Example. The cranks of a bicycle are 61 inches long, and the bicycle is so geared that one complete rotation of the crank carries it through a distance equal to the circumference of a wheel 65 inches diameter. When the bicycle is driven at 15 miles per hour, find the absolute velocity of the centre of a pedal (i) when vertically above the crank axle ; (2) when vertically below it ; (3) when above the axle and 30 forward of a vertical line through it. The pedal centre describes a circle of 13 inches diameter relative to the crank axle, i.e. 13* inches, while the bicycle travels 6$v inches. Hence the velocity of the pedal centre relative to the crank axle is 1 that of the bicycle along the road, or 3 miles per hour. 1 5 miles per hour = 22 feet per second 3 > = 4'4 FIG. 24. Kinematics 25 (1) When vertically above the crank axle, the velocity of pedal is 22 + 4*4 = 26-4 feet per second. (2) When vertically below the crank axle, the velocity of pedal is 22 4'4 = 1 7 '6 feet per second. (3) Horizontal velocity X = 22 + 4-4 cos 30 = 22 + 2*2 ^3 feet per second. Vertical velocity downwards Y = 4-4 x sin 30 = 2 '2 feet per second. Resultant velocity being R R = 22^/(i + ^) + (~ J = 25-8 feet per second and its direction is at an angle below the horizontal, so that _ Y _ 2-2 i i and = 4-85 EXAMPLES II. 1. A point in the connecting rod of a steam engine moves forwards horizontally at 5 feet per second, and at the same time has a velocity of 3 feet per second in the same vertical plane, but in a direction inclined 1 10 to that of the horizontal motion. Find the magnitude and direction of its actual velocity. 2. A stone is projected at an angle of 36 to the horizontal with a velocity of 500 feet per second. Find its horizontal and vertical velocities. 3. In order to cross at right angles a straight river flowing uniformly at 2 miles per hour, in what direction should a swimmer head if he can get through still water at 24 miles per hour, and how long will it take him if the river is loo yards wide ? 4. A weather vane on a ship's mast points south-west when the ship is steaming due west at 16 miles per hour. If the velocity of the wind is 20 miies per hour, what is its true direction ? 5. Two ships leave a port at the same time, the first steams north-west at 15 knots per hour, and the second 30 south of west at 17 knots. What is the speed of the second relative to the first? After what time will they be IOO knots apart, and in what direction will the second lie from the first ? 6. A ship steaming due east at 12 miles per hour crosses the track of another ship 20 miles away due south and going due north at 16 miles per hour. After what time will the two ships be a minimum distance apart, and how far will each have travelled in the interval. 26 Mechanics for Engineers 7. Part of a machine is moving east at 10 feet per second, and after ^ second it is moving south-east at 4 feet per second. What is the amount and direction of the average acceleration during the ^ second ? 8. How long will it take a body to slide down a smooth plane the length of which is 20 feet, the upper end being 37 feet higher than the lower one. 9. The minute-hand of a clock is 4 feet long, and the hour-hand is 3 feet long. Find in inches per minute the velocity of the end of the minute-finger relative to the end of the hour-hand at 3 o'clock and at 12 o'clock. 10. A crank, CB, is I foot long and makes 300 turns clockwise per minute. When CB is inclined 60 to the line CA, A is moving along AC at a velocity of 32 feet per second. Find the velocity of the point B rela- tive to A. 11. If a motor car is travelling at 30 miles per hour, and the wheels are 30 inches diameter, what is their angular velocity about their axes ? If the car comes to rest in 100 yards under a uniform retardation, find the angular retardation of the wheels. 1 2. A flywheel is making 1 80 revolutions per minute, and after 20 seconds it is turning at 140 revolutions per minute. How many revolutions will it make, and what time will elapse before stopping, if the retardation is uniform ? CHAPTER II THE LAWS OF MOTION 34. NEWTON'S Laws of Motion were first put in their present form by Sir Isaac Newton, although known before his time. They form the foundation of the whole subject of dynamics. 35. First Law of Motion. Every body continues in its state of rest or uniform motion except in so far as it may be com- pelled by force to change that state. We know of no case of a body unacted upon by any force whatever, so that we have no direct experimental evidence of this law. In many cases the forces in a particular direction are small, and in such cases the change in that direction is small, e.g. a steel ball rolling on a horizontal steel plate. To such instances the second law is really applicable. From the first law we may define force as that which tends to change the motion of bodies either in magnitude or direction. 36. Inertia. It is a matter of everyday experience that some bodies take up a given motion more quickly than others under the same conditions. For example, a small ball of iron is more easily set in rapid motion by a given push along a horizontal surface than is a large heavy one. In such a case the larger ball is said to have more inertia than the small one. Inertia is, then, the property of resisting the taking up of motion. 37. Mass is the name given to inertia when expressed as a measurable quantity. The more matter there is in a body the greater its mass. The mass of a body depends upon its volume and its density being proportional to both. We may define density of a body as being its mass divided by its volume, or mass per unit volume in suitable units. 28 Mechanics for Engineers If /;/ = the mass of a body, v = its volume, and p = its density, then P = v A common British unit of mass is one pound. This is often used in commerce, and also in one absolute system (British) of mechanical units ; but we shall find it more con- venient to use a unit about 32^2 times as large, f for reasons to be stated shortly. This unit has no particular name in general use. It is sometimes called the gravitational unit of mass, or the " engineer's unit of mass." In the c.g.s. (centimetre-gramme-second) absolute system, the unit mass is the gramme, which is about ? Ib. 453-6 38. The weight of a body is the force with which the earth attracts it. This is directly proportional to its mass, but is slightly different at different parts of the earth's surface. 39. Momentum is sometimes called the quantity of motion of a body. If we consider a body moving with a certain velocity, it has only half as much motion as two exactly similar bodies would have when moving at that velocity, so that the quantity of motion is proportional to the quantity of matter, i.e. to the mass. Again, if we consider the body moving with a certain velocity, it has only half the quantity of motion which it would have if its velocity were doubled, so that the quantity of motion is proportional also to the velocity. The quantity of motion of a body is then proportional to the product (mass) X (velocity), and this quantity is given the name momentum. The unit of momentum is, then, that possessed by a body of unit mass moving with unit velocity. It is evidently a vector quantity, since it is a product of velocity, which is a vector quantity, and mass, which is a scalar quantity, and its direction is that of the velocity factor. It can be resolved and compounded in the same way as can velocity. 40. Second Law of Motion. The rate of change of The Laws of Motion 29 momentum is proportional to the force applied, and takes place in the direction of the straight line in which the force acts. This law states a simple relation between momentum and force, and, as we have seen how momentum is measured, we can proceed to the measurement of force. The second law states that if F represents force F oc rate of change of (m X v) where m = mass, v = velocity ; therefore Fa m X (rate of change v), if m remains constant or F cc m y, f where /= acceleration, w and/ cc m where F is the resultant force acting on the mass m ; hence F = m x f X a constant, and by a suitable choice of units we may make the constant unity, viz. by taking as unit force that which gives unit mass unit acceleration. We may then write force = (mass) x (acceleration) or F = m X / If we take i Ib. as unit mass, then the force which gives i Ib. an acceleration of i foot per second per second is called the poundal. This system of units is sometimes called the absolute system. 1 This unit of force is not in general use with engineers and others concerned in the measurement and calcu- lation of force and power, the general practice being to take the weight of i Ib. at a fixed place as the unit of force. We call this a force of i Ib., meaning a force equal to the weight of i Ib. As mentioned in Art. 38, the weight of i Ib. of matter varies slightly at different parts of the earth's surface, but the variation is not of great amount, and is usually negligible. 1 The gravitational system is also really an absolute system, inasmuch as all derived units are connected to the fundamental ones by fixed physical relations. See Appendix. 30 Mechanics for Engineers 41. Gravitational or Engineer's Units. One pound of force acting on i Ib. mass of matter (viz. its own weight) in London 1 gives it a vertical acceleration of about 32^2 feet per second per second, and since acceleration = - C ce -, i Ib. of mass force will give an acceleration of i foot per second per second (i.e. 32*2 times less), if it acts on a mass of 32*2 Ibs. Hence, if we wish to have force denned by the relation force = rate of change of momentum, or force = (mass) X (acceleration) F = m X f we must adopt g Ibs. as our unit of mass, where g is the acceleration of gravity in feet per second per second in some fixed place; the number 32*2 is correct enough for most practical purposes for any latitude. This unit, as previously stated, is sometimes called the engineers' unit of mass. > Then a body of weight w Ibs. has a mass of W - units, g and the equation of Art. 40 becomes F = X f. Another plan is to merely adopt the relation, force = (mass) X (acceleration) x constant. The mass is then taken in pounds, and if the force is to be in pounds weight (and not in ponndals) the constant used is g (32*2). There is a strong liability to forget to insert the constant g in writing expressions for quantities involving force, so we shall adopt the former plan of using 32*2 Ibs. as the unit of mass. The unit of momentum is, then, that possessed by 32*2 Ibs. moving with a velocity of i foot per second, and the unit force the weight of i Ib. The number 32*2 will need slight adjustment for places other than London, if very great accuracy should be required. Defining unit force as the weight of i Ib. of matter, we may define the gravitational unit of mass as that mass which has unit acceleration under unit force. 42. C.G.S. (centimetre-gramme-second) Units. In this absolute system the unit of mass is the gramme ; the 1 The place chosen is sometimes quoted as sea-level at latitude 45. The Laivs of Motion 31 unit of momentum that in i gramme moving at i centimetre per second; and the unit of force called the dyne is that necessary to accelerate i gramme by i centimetre per second per second. The weight of i gramme is a force of about 981 dynes, since the acceleration of gravity is about 981 centi- metres per second per second (981 centimetres being equal to about 3 2 '2 feet). The weight of one kilogram (1000 grammes) is often used by Continental engineers as a unit of force. Example i. A man pushes a truck weighing 2*5 tons with a force of 40 Ibs., and the resistance of the track is equivalent to a constant force of 10 Ibs. How long will it take to attain a velocity of 10 miles per hour? The constant effective forward force is 40 10 = 30 Ibs., hence the acceleration is force 2*5 x 2240 mass = 3 " ~~^T2~ ~ 0<I 7 2 5 foot per second per second 10 miles per hour = * or - 4 / feet per second The time to generate this velocity at 0*1725 foot per second per second is then * -4- 0-1725 = 85 seconds, or i minute 25 seconds. Example 2. A steam-engine piston, weighing 75 Ibs., is at rest, and after 0*25 second it has attained a velocity of 10 feet per second. What is the average accelerating force acting on it during the 0^25 second ? Average acceleration = 10 * 0-25 = 40 feet per sec. per sec. hence average accelerating force is -~ x 40 = 93-2 Ibs. 3 2 ' 2 43. We have seen that by a suitable choice of units the force acting on a body is numerically equal to its rate of change of momentum ; the second law further states that the force and the change of momentum are in the same direction. Mo- mentum is a vector quantity, and therefore change of momentum must be estimated as a vector change having magnitude and direction. For example, if the momentum of a body is represented by ab (Fig. 25), and after t seconds it is represented by cd, then the change of momentum in / seconds is cd ab = eg (see 32 Mechanics for Engineers Art. 20), where ef = cd and gf = ab. Then the average' rate of change of momentum in t seconds is represented by ^ in magnitude and direction, /.<?. the resultant force acting on the body during the / seconds was in the direction eg. Or Fig. 25 FIG. 25. may be taken as a vector diagram of velocities, and eg as gcr representing change of velocity. Then represents accelera- tion, and multiplied by the mass of the body it represents the average force. Example. A piece of a machine weighing 20 Ibs. is at a certain instant moving due east at 10 feet per second, and after 1*25 seconds it is moving south-east at 5 feet per second. What was the average force acting on it in the interval ? The change of momentum per second may be found directly, or the change of velocity per second may be found, which, when multiplied by the (constant) mass, will give the force acting. Using the method of resolution of velocities, the final component of velocity E. = 5 cos 45 = 4- feet per second initial E. = 10 hence gain of component 1/5 \ / $ \ , .. > = ( --, 10 ) east, or ( 10 -. west velocity / \ /v /2 ) \ Jil Again, the gain of velocity south is 5 sin 45 = / feet per second V 2 The Laws of Motion 33 If R = resultant change of velocity and R = \/54'3 = 7'37 f eet P er second in 1} seconds Hence acceleration = 7*37 -f- i'25 = 5^9 feet per second per second, and average force acting = ~ - x 5*9 = 3' 66 Ibs. in a direction south of west at an angle whose tangent is -4- -5- ( 10 -5- ) v 2 \ v 2 / or 0-546, which is an angle of about 28^ south of west (by table of tangents). 44. Triangle, Polygon, etc., of Forces. It has been seen (Art. 27) that acceleration is a vector quantity having magnitude and direction, and that acceleration can be com- pounded and resolved by means of vectors. Also (Art. 40) that force is the product of acceleration and mass, the latter being a mere magnitude or scalar quantity ; hence force is a vector quantity, and concurrent forces can be compounded by vector triangles or polygons such as were used in Arts. 19 and 24, and resolved into components as in Arts. 25 and 28. We are mainly concerned with uniplanar forces, but the methods of resolution, etc., are equally applicable to forces in different planes ; the graphical treatment would, however, in- volve the application of solid geometry. The particular case of bodies subject to the action of several forces having a resultant zero constitutes the subject of Statics. The second law of motion is true when the resultant force is considered or when the components are considered, i.e. the rate of change of momentum in any particular direction is pro- portional to the component force in that direction. 45. Impulse. By the impulse of a constant force in any interval of time, we mean the product of the force and time. Thus, if a constant force of F pounds act for / seconds, the impulse of that force is F X A If this force F has during the interval / acted without resistance on a mass ;;/, causing its velocity to be accelerated from i\ to v, the change of momentum 34 Mechanics for Engineers during that time will have been from mv^ to mv. 2 , i.e. mr 2 mv^ or m(v^ V}). And the change of velocity in the interval / under the constant acceleration f is f x / (Art. 1 i), therefore z/ 2 v l =//, and t(v. 2 vj = m .f.t; but m X/= F, the accelerating force (by Art. 40), hence m(v 2 v^} = F/, or, in words, the change of momentum is equal to the impulse. The force, impulse, and change of momentum are all to be estimated in the same direction. The impulse may be represented graphically as in Fig. 26. If ON represents / seconds, and PN represents F Ibs. to scale, M Time FIG. 26. then the area MPNO under the curve MP of constant force represents F x /, the impulse, and therefore also the change of momentum. Impulse of a Variable Force. In the case of a variable force the interval of time is divided into a number of parts, and the impulse calculated during each as if the force were constant during each of the smaller intervals, and equal to some value which it actually has in the interval. The sum of these impulses is approximately the total impulse during the whole time. We can make the approximation as near as we please by taking a sufficiently large number of very small intervals. The graphical representation will illustrate this point. Fig. 27 shows the varying force F at all times during the interval NM. Suppose the interval NM divided up into a The Laws of Motion 35 A N C D M Tirr^e, FIG. 27. Impulse of a variable force. number of small parts such as CD. Then AC represents the force at the time OC ; the force is increasing, and therefore in the interval CD the impulse will be greater than that repre- sented by the rectangle AEDC, and less than that represented by the rectangle FBDC. The total impulse during the interval NM is simi- larly greater than that represented by a series of rectangles such as AEDC, and less than that represented by a series of rectangles such as FBDC. Now, if we consider the number of rectangles to be indefi- nitely increased, and the width of each rectangle to be decreased indefinitely, the area PQMN under the curve PQ is the area which lies always between the sums of the areas of the two series of rectangles however far the subdivision may be carried, and therefore it represents the total impulse in the time NM, and therefore also the gain of momentum in that time. It may be noticed that the above statement agrees exactly with that made in Art. 16. In Fig. 8 the vertical ordinates are similar to those in Fig. 27 divided by the mass, and the gain of velocity represented by the area under PQ in Fig. 8 is also similar to the gain of momentum divided by the mass. Note that the force represented by -, ~~i ir (^ e - by tne length NM average height of the PQMN) is the mean force or time- average of the force acting during the interval NM. This r u j c j total impulse time-average force may be defined as - =-- . total time The area representing the impulse of a negative or opposing force will lie below the line OM in a diagram such as Fig. 27. In case of a body such as part of a machine starting from rest and coming to rest again, the total change of momentum is zero ; then as much area of the force-time diagram lies below 36 Mechanics for Engineers the time base line (OM) as above it. The reader should sketch out such a case, and the velocity-time or momentum-time curve to be derived from it, by the method of Art. 1 6, and carefully consider the meaning of all parts of the diagrams the slopes, areas, changes of sign, etc. The slope of a momentum-time curve represents accelerat- ing force just as that of a velocity-time represents acceleration (see Art. 14), the only difference in the case of momentum and force being that mass is a factor of each. It is to be noticed that the impulse or change of momentum in a given interval is a vector quantity having definite direction. It must be borne in mind that the change of momentum is in the same direction as the force and impulse. If the force varies in direction it may be split into components (Art. 44), and the change of momentum in two standard directions may be found, and 4he resultant of these would give the change of momentum in magnitude and direction. 46. Impulsive Forces. Forces which act for a very short time and yet produce considerable change of momentum on the bodies on which they act are called impttlsive forces. The forces are large and the time is small. Instances occur in blows and collisions. 47. The second law of motion has been stated, in Art. 40, in terms of the rate of change of momentum. It can now be stated in another form, viz. The change of momentum is equal to the impulse of the applied force, and is in the same direction. Or in symbols, for a mass m m(v 2 7'j) = F . / where # 2 and z> x are the final and initial velocities, the sub- traction being performed geometrically (Art. 20), and F is the mean force acting during the interval of time /. Example i. A body weighing W Ibs. is set in motion by a uniform net force P l Ibs., and in /j seconds it attains a velocity V feet per second. It then comes to rest in a further period of / 2 seconds under the action of a uniform retarding force of P 2 Ibs. Find the relation between P,, P 2 , and V. The Laws of Motion 37 During the acceleration period the gain of momentum in the W direction of motion is .V units, and the impulse in that direction is Pj^, hence W P!/! = .V g During retardation the gain of momentum in the direction W of motion is -- .V units, and the impulse in that direction is P 2 . t z ', hence W and finally . V = P,/, = P,/ t = + the last relation following algebraically from the two preceding ones. Example 2. If a locomotive exerts a constant draw-bar pull of 4 tons on a train weighing 200 tons up an incline of i in 120, and the resistance of the rails, etc., amounts to 10 Ibs. per ton, how long will it take to attain a velocity of 25 miles per hour from rest, and how far will it have moved ? The forces resisting acceleration are Ibs. (a) Gravity T |<j of 200 tons (see Art. 28) = - = 3733 () Resistance at 10 Ibs. per ton, 200 x 10 = 2000 Total ... 5733 The draw-bar pull is 4 x 2240 = 8960 Ibs. ; hence the net accelerating force is 8960 5733 = 3227 Ibs. Let / be the required time in seconds ; then the impulse is 3227 x / units. 25 miles per hour = ^ x 88 feet per second (88 feet per second = 60 miles per hour) W so that the gain of momentum is . V therefore 200 x 2240 5 32'2 * 12 32*2 12 from which / = 1 59 seconds, or 2 minutes 39 seconds 38 Mechanics for Engineers Since the acceleration has been uniform, the average speed is half the maximum (Art. 28), and the distance travelled will be in feet \ x ^Vj x 88 x 159 = 2915 feet Example 3. How long would it take the train in Ex. 2 to go i mile up the incline, starting from rest and coming to rest at the end without the use of brakes ? Let t l = time occupied in acceleration, /.j = time occupied in retardation. During the retardation period the retarding force will be as in Ex. 2, a total of 5733 Ibs. after acceleration ceases. The average velocity during both periods, and therefore during the whole time, will be half the maximum velocity attained. Average velocity = . feet per second *i + *a and maximum velocity = 2 x ~ T feet per second *1 T j 200 V. 2240 . 5280 .*. momentum generated = - - x 2 x / units 32 2 l l -+- 1 2 The impulse = 3227*1 = 5733^ i 3227 and /, + /, = \3227 / 3227 By the second law, change of momentum = impulse. 200 x 2240 5280 ' - - x 2 x , . . = 5733/2 32-2 (A + / 2 ) and substituting for / a the value found x 2240 5280 * x 2 x / 2 agreeing with the last result in Ex. i. hence (^ + / 2 ) = 267 seconds = 4 minutes 27 seconds Example 4. A car weighing 12 tons starts from rest, and has a constant resistance of 500 Ibs. The tractive force, F, on the car after t seconds is as follows : The Laws of Motion 39 / ... F ... 1280 2 1270 5 1 220 8 1 1 10 ii 905 13 800 16 720 19 670 20 660 Find the velocity of the car after 20 seconds from rest, and show how to find the velocity at any time after starting, and to find the distance covered up to any time. Plot the curve of F and /, as in Fig. 28, and read off the force, 1200 1000 v; 600 .g 600 400 200 1 > ~*^ ^ s V ^ ^ . en 24 6 8 10 12 14 16 18 2C . w& secoTtaLs FIG. 28. say every 4 seconds, starting from / = 2, and subtract the 500 Ibs. resistance from each as follows : / 2 6 10 H 18 F Ibs. 1270 1190 980 760 680 F-SCO ... 770 690 480 260 1 80 The mean accelerating force in the first 4 seconds is approxi- mately 770 Ibs., and therefore the impulse is 770 x 4, which is also the gain of momentum, _, , , . 12 x 2240 _ The mass of the car is - = 835 units 32*2 . . .. , momentum The velocity after 4 seconds = 770 x 4 mass 835 = 3*69 feet per second Mechanics for Engineers Similarly, finding the momentum and gain of velocity in each \ seconds, we have t o 4 8 12 16 20 Gain of momentum \ in 4 seconds . . . \ o 3080 2760 1920 1040 720 Momentum 3080 5840 7760 8800 9520 Velocity, feet per ? second \ o 3-69 7 'oo 9'3i IQ'55 11-41 After 20 seconds the velocity is approximately ii'4i feet per second. The velocity after any time may be obtained approxi- mately by plotting a curve of velocities and times from the values obtained, and reading intermediate values. More points on the velocity-time curves may be found if greater accuracy be desired. The space described is represented by the area under the velocity-time curve, and may be found as in Art. 14. EXAMPLES III. 1. The moving parts of a forging hammer weigh 2 tons, and are lifted vertically by steam pressure and then allowed to fall freely. What is the momentum of the hammer after falling 6 feet ? If the force of the blow is expended in 0^015 second, what is the average force of the blow ? 2. A mass of 50 Ibs. acquires a velocity of 25 feet per second in 10 seconds, and another of 20 Ibs. acquires a velocity of 32 feet per second in 6 seconds. Compare the forces acting on the two masses. 3. A constant unresisted force of 7000 dynes acts on a mass of 20 kilo- grams for 8 seconds. Find the velocity attained in this time. 4. A train weighing 200 tons has a resistance of 15 Ibs. per ton, sup- posed constant at any speed. What tractive force will be required to give it a velocity of 30 miles per hour in I '5 minutes ? 5. A jet of water of circular cross-section and 1*5 inches diameter impinges on"a flat plate at a velocity of 20 feet per second, and flows off at right angles to its previous path. How much water reaches the plate per second ? What change of momentum takes place per second, and what force does the jet exert on the plate ? 6. A train travelling at 40 miles per hour is brought to rest by a uniform resisting force in half a mile. How much is the total resisting force in pounds per ton ? 7. A bullet weighing I oz. enters a block of wood with a velocity of 1800 feet pet cecond, and penetrates it to a depth of 8 inches. What is the average resistance of the wood in pounds to the penetration of the bullet ? 8. The horizontal thrust on a steam-engine crank-shaft bearing is 10 tons, The Laws of Motion 41 and the dead weight it supports vertically is 3 tons. Find the magnitude and direction of the resultant force on the bearing. 9. A bullet weighing i oz. leaves the barrel of a gun 3 feet long with a velocity of 1500 feet per second. What was the impulse of the force pro- duced by the discharge ? If the bullet took 0*004 second to traverse the barrel, what was the average force exerted on it ? 10. A car weighing 10 tons starts from rest. During the first 25 seconds the average drawing force on the car is 750 Ibs., and the average resistance is 40 Ibs. per ton. What is the total impulse of the effective force at the end of 25 seconds, and what is the speed of the car in miles per hour ? 11. The reciprocating parts of a steam-engine weigh 483 Ibs., and during one stroke, which occupies 0-3 second, the velocities of these parts are as follows : Time O'O o-o 5 !o-075 O'lOO 0-125 i 0-I500-I75 O'200 0-225 0-250 0-275 0-300 Velocity 1 in feet > O'OO 3'46 6-558-91 IO"22 lO'OX) IO-48 9-32 7'75 6'02 4-14 2-10 O'OO per sec.) ' i Find the force necessary to give the reciprocating parts this motion, and draw a curve showing its values on a time base throughout the stroke. Draw a second curve showing the distances described from rest, for every instant during the stroke. From these two curves a third may be drawn, showing the accelerating force on the reciprocating parts, on the distance traversed as a base. 48. Third Law of Motion. To every action there is an equal and opposite reaction. By the word " action " here is meant the exertion of a force. We may state this in another way. If a body A exerts a certain force on a body B, then B exerts on A a force of exactly equal magnitude, but in the opposite direction. The medium which transmits the equal and opposite forces is said to be in a state of stress. (It will also be in a state of strain, but this term is limited to deformation which matter undergoes under the influence of stress.) Suppose A and B (Fig. 29) are connected by some means (such as a string) suitable to withstand tension, and A exerts a pull T on B. Then B exerts an equal tension T' on A. This will be true whether A moves B or not. Thus A may be a locomotive, and B a train, or A may be a ship moored to 42 Mechanics for Engineers a fixed post, B. Whether A moves B or not depends upon what other forces may be acting on B. Again, if the connection between A and B can transmit a B FIG. 29. Connection in tension. thrust (Fig. 30), A may exert a push P on B. Then B exerts an equal push P' on A. As an example, A may be a gun, and B a projectile ; the gases between them are in compression. - P > < B A FIG. 30. Connection in compression. Or in a case where motion does not take place, A may be a block of stone resting on the ground B ; then A and B are in compression at the place of contact. 49. An important consequence of the third law is that the total momentum of the two bodies is unaltered by any mutual action between them. For since the force exerted by A on B is the same as that exerted by B on A, the impulse during any interval given by A to B is of the same amount as that given by B to A and in the opposite direction. Hence, if B gains any momentum A loses exactly the same amount, and the total change of momentum is zero, and this is true for any and every direction. This is expressed by the statement that for any isolated system of bodies momentum is conservative. Thus when a projectile is fired from a cannon, the impulse or change of momentum of the shot due to the explosion is of equal amount to that of the recoiling cannon in the opposite The Laws of Motion 43 direction. The momentum of the recoil is transmitted to the earth, and so is that of the shot, the net momentum given to the earth being also zero. 50. Motion of Two Connected Weights. Suppose two weights, Wj Ibs. and W. 2 Ibs., to be connected by a light inextensible string passing over a small and perfectly smooth pulley, as in Fig. 31. If W^ is greater than W a , with what acceleration (/) will they move (W 1 downwards and W 2 upwards), and what will be the tension (T) of the string ? ( W \ Consider Wj ( of mass ) : the downward force on it is W l (its weight), and the upward force is T, which is the same throughout the string by the " third law ; " hence the downward accelerating force is Wj T. FIG W Hence (by Art. 40) 1 ./= W, - T (i) o Similarly, on W 2 the upward accelerating force is T W 2 ; W hence 2 ./=T - W 2 . . . . (2) W, adding (i) and (2) and from (i) W 2 W.-W, The acceleration / might have been stated from considering the two weights and string as one complete system. The accelerating force on which is W, W 2 , and the mass of \V -f- W 2 which is : accelerating force _ W t hence / - = W 2 total mass 44 Mechanics for Engineers As a further example, suppose W 2 instead of being suspended slides along a perfectly smooth horizontal table as in Fig. 32, W 2 w FIG. 32. the accelerating force is W 15 and the mass in motion is hence the acceleration / = . accelerating force on W 2 T and since /also = - moeo nf ^ - = Wa -< we have T = ^ mass of W 2 If the motion of W 2 were opposed by a horizontal force, F, W, F the acceleration would be r^tr-fr Wj + W a We have left out of account the weight of W 2 and the reaction of the table. These are equal and opposite, and neutralize each other. The reaction of the pulley on the string is normal to the direction of motion, and has therefore no accelerating effect. Atwood's Machine is an apparatus for illustrating the laws of motion under gravity. It consists essentially of a light, free pulley and two suspended weights (Fig. 31), which can be made to differ by known amounts, a scale of lengths, and clockwork to measure time. Quantitative measurements of acceleration of known masses under the action of known accelerating forces can be made. Various corrections are The Laws of Motion 45 necessary, and this method is not the one adopted for measuring the acceleration g. Example I. A hammer weighing W Ibs. strikes a nail weigh- ing iv Ibs. with a velocity V feet per second and does not rebound. The nail is driven into a fixed block of wood which offers a uniform resistance of P Ibs. to the penetration of the nail. How far will the nail penetrate the fixed block ? Let V = initial velocity of nail after blow. W Momentum of hammer before impact = .V momentum of hammer and nail after impact = . V' g W + w , TI W Vr . , W Tr hence . V = V .'. V = ^r^ . V g g- <w + W Let / = time of penetration. W Impulse P/ = .V (the momentum overcome by P) /- During the penetration, average velocity = V (Arts. 1 1 and 14) hence distance moved by nail = $V x / ' W W WV x + Example 2. A cannon weighing 30 tons fires a icoo-lb. pro- jectile with a velocity of 1000 feet per second. With what initial velocity will the cannon recoil ? If the recoil is overcome by a (time) average force of 60 tons, how far will the cannon travel ? How long will it take ? Let V = initial velocity of cannon in feet per second. Momentum of projectile = - - x 1000 = momentum of cannon looo 30 x 2240 or - x looo = x V = IOOOX roo Let t = time of recoil. and v = IOOOX roo = 14-87 feet per second 30 x 2240 46 MecJianics for Engineers Impulse of retarding force = 60 x 2240 x / = momentum of shot 1000 x 1000 60 x 2240 x t - 32-2 and hence / = 0*231 second 14-87 x 0-231 Distance moved = V x t - - - = 174 feet Example 3. Two weights are connected by a string passing over a light frictionless pulley. One is 12 Ibs. and the other ir Ibs. They are released from rest, and after 2 seconds 2 Ibs. are removed from the heavier weight. How soon will they be at rest again, and how far will they have moved between the instant of release and that of coming to rest again ? First period. . , . accelerating force 1211 g Acceleration = - - = - - x j* = total mass 12+11 23 72*2 velocity after 2 seconds = 2 x * = 2 '8 feet per second Second period. ^ Retardation = ; x g II + IO 21 2 x - velocity 23 time to come to rest = -3 r = = 2 x 4 = 1*825 sec. retardation g_ 21 average velocity throughout = | maximum velocity (Art. n) total time =2+1-825 seconds distance moved = \ x 2'8 x 3*825 = 5-35 feet EXAMPLES IV. 1. A fireman holds a hose from which a jet of water i inch in diameter issues at a velocity of 80 feet per second. What thrust will the fireman have to exert in order to support the jet ? 2. A machine-gun fires 300 bullets per minute, each bullet weighing 1 oz. If the bullets have a horizontal velocity of 1800 feet per second, find the. average force exerted on the gun. 3. A pile-driver weighing W Ibs. falls through h feet and drives a pile weighing w Ibs. a feet into the ground. Show that the average force of W 2 h the blow is ~ : Ibs. W + w a 4. A weight of 5 cwt. falling freely, drives a pile weighing 600 Ibs 2 inches into the earth against an average resistance of 25 tons. How far will the weight have to fall in order to do this ? The Laws of Motion 47 5. A cannon weighing 40 tons projects a shot weighing 1500 Ibs. with a velocity of 1400 feet per second. With what initial velocity will the cannon recoil ? What average force will be required to bring it to rest in 3 feet ? 6. A cannon weighing 40 tons has its velocity of recoil destroyed in 2 feet 9 inches by an average force of 70 tons. If the shot weighed 14 cwt., find its initial velocity. 7. A lift has an upward acceleration of 3'22 feet per second per second. What pressure will a man weighing 140 Ibs. exert on the floor of the lift ? What pressure would he exert if the lift had an acceleration of 3-22 feet per second per second downward? What upward acceleration would cause his weight to exert a pressure of 170 Ibs. on the floor ? 8. A pit cage weighs 10 cwt., and on approaching the bottom of the shaft it is brought to rest, the retardation being at the rate of 4 feet per second per second. Find the tension in the cable by which the cage is lowered . 9. Two weights, one of 16 Ibs. and the other of 14 Ibs., hanging vertically, are connected by a light inextensible string passing over a perfectly smooth fixed pulley. If they are released from rest, find how far they will move in 3 seconds. What is the tension of the string ? 10. A weight of 17 grammes and another of 20 grammes are connected by a fine thread passing over a light frictionless pulley in a vertical plane. Find what weight must be added to the smaller load 2 seconds after they are released from rest in order to bring them to rest again in 4 seconds. How many centimetres will the weights have moved altogether ? 11. A weight of 5 Ibs. hangs vertically, and by means of a cord passing over a pulley it pulls a block of iron weighing 10 Ibs. horizontally along a table-top against a horizontal resistance of 2 Ibs. Find the acceleration of the block and tension of the string. 12. What weight hanging vertically, as in the previous question, would give the lo-lb. block an acceleration of 3 feet per second per second on a perfectly smooth horizontal table ? 13. A block of wood weighing 50 Ibs. is on a plane inclined 40 to the horizontal, and its upward motion along the plane is opposed by a force of 10 Ibs. parallel to the plane. A cord attached to the block, running parallel to the plane and over a pulley, carries a weight hanging vertically. What must this weight be if it is to haul the block IO feet upwards along the plane in 3 seconds from rest ? CHAPTER III WORK, POWER, AND ENERGY 51. Work. When a force acts upon a body and causes motion, it is said to do work. In the case of constant forces, work is measured by the product of the force and the displacement, one being estimated by its component in the direction of the other. One of the commonest examples of a force doing work is that of a body being lifted against the force of gravity, its weight. The work is then measured by the product of the weight of the body, and the vertical height through which it is lifted. If we draw a diagram (Fig. 33) setting off the constant force F by a vertical ordinate, OM, (p N then the work done during Distance ,. , any displacement represented by ON is proportional to the area MPNO, and is represented by that area. If the scale of force is i inch =/ Ibs., and the scale of distance is i inch = q feet, then the scale of work is i square inch = pq foot-lbs. 52. Units of Work. Work being measured by the product of force and length, the unit of work is taken as that done by a unit force acting through unit distance. In the British gravitational or engineer's system of units, this is the work done by a force of i Ib. acting through a distance of i foot. It is called the foot-pound of work. If a weight Work, Power, and Energy 49 W Ibs. be raised vertically through // feet, the work done is WA foot-lbs. Occasionally inch-pound units of work are employed, particularly when the displacements are small. In the C.G.S. system the unit of work is the erg. This is the work done by a force of one dyne during a displacement of i centimetre in its own direction (see Art. 42). 53. Work of a Variable Force. If the force during any displacement varies, we may find the total work done approximately by splitting the displacement into a number of parts and finding the work done during each part, as if the force during the partial displacement were constant and equal to some value it has during that part, and taking the sum of all the work so calculated in the partial displacements. We can make the approximation as near as we please by taking a sufficiently large number of parts. We may define the work actually done by the variable force as the limit to which such a sum tends when the subdivisions of the displacement are made indefinitely small. 54. Graphical Representation of Work of a Variable Force. Fig. 34 is a diagram showing by its vertical ordinates M C D FIG. 34- the force acting on a body, and by its horizontal ones the dis- placements. Thus, when the displacement is represented by E 50 Mechanics for Engineers ON, the force acting on the body is represented by PN. Suppose the interval ON divided up into a number of small parts, such as CD. The force acting on the body is represented by AC when the displacement is that represented by OC. Since the force is increasing with increase of displacement the work done during the displacement CD is greater than that represented by the rectangle AEDC, and less than that represented by the rectangle FBDC. The total work done during the displacement will lie between that represented by the series of smaller rectangles, such as AEDC, and that represented by the series of larger rectangles, such as FBDC. The area MPNO under the curve MP will always lie between these total areas, and if we consider the number of subdivisions of ON to be carried higher indefinitely, the same remains true both of the total work done and the area under the curve MP. Hence the area MPNO under the curve MP represents the work done by the force during the displacement represented by ON. The Indicator Diagram, first introduced by Watt for use on the steam-engine, is a diagram of the same kind as Fig. 34. The vertical ordinates are proportional to the total force exerted by the steam on the. piston, and the horizontal ones are proportional to the dis- placement of the piston. The area of the figure is then pro- portional to the work done by the steam on the piston. In the case of a force vary- Space, N ing uniformly with the displace- FIG. 35. Force varying uniformly ment, the CUrVC MP is a Straight with space. .. /,-,. > , , line (rig- 35), and the area MPNO = - x ON, or if the initial force (OM) is F 3 Ibs., and the final one (PN) is F 2 Ibs., and the displacement p i -p (ON) is d feet, the work done is - . d foot-lbs. In stretching an unstrained elastic body, such as a spring, Work, Power, and Energy 5 i the force starts from zero (or Y l = o). Then the total work done is ^F.y/, where F 2 is the greatest force exerted, and d is the amount of stretch. Average Force. The whole area MPNO (Figs. 34 and 35) divided by the above ON gives the mean height of the area ; this represents the space-average of the force during the displacement ON. This will not necessarily be the same as the time-average (Art. 45). We may define the space-average of a varying force as the work done divided by the displacement. 55. Power. Power is the rate of doing work, or the work done per unit of time. One foot-pound per second might be chosen as the unit of power. In practice a unit 550 times larger is used; it is called the horse-power. It is equal to a rate of 550 foot-lbs. per second, or 33,000 foot-lbs. per minute. In the C.G.S. system the unit of power is not usually taken as one erg per second, but a multiple of this small unit. This larger unit is called a watt, and it is equal to a rate of io 7 ergs per second. Engineers frequently use a larger unit, the kilowatt, which is 1000 watts. One horse-power is equal to 746 watts or 0746 kilowatt. Example i. A train ascends a slope of i in 85 at a speed of 20 miles per hour. The total weight of the train is 200 tons, and resistance of the rails, etc., amounts to 12 Ibs. per ton. Find the horse-power of the engine. The total force required to draw the load is N , 200 x 2240 (200 x 12) -\ ~ ~ = 7670 Ibs. The number of feet moved through per minute is i\ x 88 x 60 = 1760 feet; hence the work done per minute is 1760 x 7670 = 13,500,000 foot-lbs., and since i horse-power = 33,000 foot-lbs. per minute, the H.P. is 13 : ${{$p Q = 409 horse -power. Example 2. A motor-car weighing 1 5 cwt. just runs freely at 12 miles per hour down a slope of i in 30, the resistance at this speed just being sufficient to prevent any acceleration. What horse- power will it have to exert to run up the same slope at the same speed ? In running down the slope the propelling force is that of gravity, which is ;& of the weight of the car (Arts. 28 and 44) ; hence the 52 Mechanics for Engineers resistance of the road is also (at 12 miles per hour) equivalent to 15 x 112 or 56 Ibs. 30 Up the slope the opposing force to be overcome is 56 Ibs. road resistance and 56 Ibs. gravity (parallel to the road), and the total 112 Ibs. The distance travelled per minute at 12 miles per hour is \ mile = 5a 5 HO or 1056 feet ; hence the work done per minute is . 112 x 1056 foot-lbs., and the H.P. is U^ 2L- or y^ H p 33000 Example 3. The spring of a safety-valve is compressed from its natural length of 20 inches to a length of 17 inches. It then exerts a force of 960 Ibs. How much work will have to be done to compress it another inch, z>. to a length of 16 inches ? The force being proportional to the displacement, and being 960 Ibs. for 3 inches, it is fi Q or 320 Ibs. per inch of compression. When 1 6 inches long the compression is 4 inches, hence the force is 4 x 320 or 1280 Ibs. ; hence the work done in compression is '- x i, or 1 120 inch-lbs f . (Art. 54, Fig. 35), or 93-3 foot-lbs. EXAMPLES V. 1. A locomotive draws a train weighing 150 tons along a level track at 40 miles per hour, the resistances amounting to 10 Ibs. per ton. What horse-power is it exerting ? Find also the horse-power necessary to draw the train at the same speed (a) up an incline of I in 250, (t>) down an incline of i in 250. 2. If a locomotive exerts 700 horse-power when drawing a train of 200 tons up an incline of I in 80 at 30 miles per hour, find the road resistances in pounds per ton. 3. A motor-car engine can exert usefully on the wheels 8 horse-power. If the car weighs 16 cwt., and the road and air resistances be taken at 20 Ibs. per ton, at what speed can this car ascend a gradient of I in 15 ? 4. A winding engine draws from a coal-mine a cage which with the coal carried weighs 7 tons ; the cage is drawn up 380 yards in 35 seconds. Find the average horse-power required. If the highest speed attained is 30 miles per hour, what is the horse-power exerted at that time ? 5. A stream delivers 3000 cubic feet of water per minute to the highest point of a water-wheel 40 feet diameter. If 65 per cent, of the available work is usefully employed, what is the horse-power developed by the wheel ? 6. A bicyclist rides up a gradient of I in 15 at 10 miles per hour. The Work, Power, and Energy 53 weight of rider and bicycle together is 180 Ibs, If the road and other resistances are equivalent to T J 5 of this weight, at what fraction of a horse- power is the cyclist working ? 7. Within certain limits, the force required to stretch a spring is proportional to the amount of stretch. A spring requires a force of 800 Ibs. to stretch it 5 inches : find the amount of work done in stretching it 3 inches. 8. A chain 400 feet long and weighing 10 Ibs. per foot, hanging vertically, is wound up. Draw a diagram of the force required to draw it up when various amounts have been wound up from o to 400 feet. From this diagram calculate the work done in winding up (a) the first 100 feet of the chain, (b) the whole chain. 9. A pit cage weighing 1000 Ibs. is suspended by a cable 800 feet long weighing if Ibs. per foot length. How much work will be done in wind- ing the cage up to the surface by means of the cable, which is wound on a drum ? 56. It frequently happens that the different parts of a body acted upon by several forces move through different distances in the same time ; an important instance is the case of the rotating parts of machines generating or transmitting power. It will be convenient to consider here the work done by forces which cause rotary motion of a body about a fixed axis. Moment of a Force. The moment of a force about a point is the measure of its turning effect or tendency, about that point. It is measured by the product of the force and the per- pendicular distance from the point to the line of action of the force. Thus in Fig. 36, if O is a point, and AB the line of action of a force F, both in the plane of the figure, and OP is the perpendicular from O on to AB measuring r units of length, the moment of F about O is F X r. The turning tendency of F about O will be in one direction, or the opposite, according as O lies to the right or left of AB looking in the direction of the force. If O lies to the right, the moment is said to be clockwise ; if to the left, contra-clockwise. In adding moments of forces about O, the clockwise and contra- clockwise moments must be taken as of opposite sign, and the IF) FIG. 36. 54 Mechanics for Engineers algebraic sum found. Which of the two kinds of moments is considered positive and which negative is immaterial. If O lies in the line AB, the moment of F about O is zero. 1 The common units for the measurement of moments are pound-feet. Thus, if a force of i Ib. has its line of action i foot from a fixed point, its moment about that point is one pound-foot. In Fig. 36, if the force is F Ibs., and OP represents r feet, the moment about O is F . r pound-feet. Moment of a Force about an Axis perpendicular to its Line of Action. If we consider a plane perpendicular to the axis and through the force, it will cut the axis in a point O ; then the moment of the force about the axis is that of the force about O, the point of section of the axis by the plane. The moment of the force about the axis may therefore be defined as the product of the force and its perpendicular distance from the axis. In considering the motion of a body about an axis, it is necessary to know the moments,, about that axis of all the forces acting on the body in planes perpendicular to the axis, whtether all the forces are in the same plane or not. The total moment is called the torque^ or twisting moment or turning moment about the axis. In finding the torque on a body about a particular axis, the moments must be added algebrai- cally. 57. Work done by a Constant Torque or Twist- ing Moment. Suppose a force F Ibs. (Fig. 37) acts upon a body which turns about an axis, O, perpendicular to the line of action of F and distant r feet from it, so that the turning 1 Note that the question whether a moment is clockwise or contra- clockwise depends upon the aspect of view. Fig. 36 shows a force (F) having a contra-clockwise moment about O, but this only holds for one aspect of the figure. If the force F in line AB and the point O be viewed from the other side of the plane of the figure, the moment would be called a clockwise one. This will appear clearly if the figure is held up to the light and viewed from the other side of the page. Similarly, the moment of a force about an axis will be clockwise or contra-clockwise according as the force is viewed from one end or the other of the axis. The motion of the hands of a clock appears contra-clockwise if viewed from the back through a transparent face. Work, Power, and Energy 55 moment (M) about O is F.r Ib.-feet. Suppose that the force acts successively on different parts of the body all distant r from the axis O about which it rotates, or that the force acts always on the same point C, and changes its direc- tion as C describes its circular path about the centre O, so as to always remain tangential to this circular path ; in either case the force F is always in the same direction as the displacement it is producing, and therefore the work done is equal to the product of the force and the displacement (along the circumference of the circle CDE). Let the displace- FIG. 37. ment about the axis O be through an angle 6 radians correspond- ing to an arc CD of the circle CDE, so that CD = 0, or CD = r. (The angle is 2w, if a displacement of one complete cir- cuit be considered.) The work done is F X CD = F . rO foot-lbs. But M = F . r Ib.-feet therefore the work done = M X foot-lbs. The work done by each force is, then, the product of the turning moment and the angular displacement in radians. If the units of the turning moment are pound-feet, the work will be in foot-pounds ; if the moment is in pound-inches, the work will be in inch-pounds, and so on. The same method of calcu- lating the work done would apply to all the forces acting, and finally the total work done would be the product of the total torque or turning moment and the angular displacement in radians. Again, if w is the angular velocity in radians per second, the power or work per second is M . o foot-lbs., and the horse- 56 Mechanics for Engineers power is , where M is the torque in Ib.-feet; and if N is the number of rotations per minute about the axis HP = 2 5.N^ 33,000 This method of estimating the work done or the power, is particularly useful when the turning forces act at different distances from the axis of rotation. We may, for purposes of calculation, look upon such a state of things as replaceable by a certain force at a certain radius, but the notion of a torque and an angular displacement seems rather less artificial, and is very useful. The work done by a variable turning moment during a given angular displacement may be found by the method of Arts. 53 and 54. If in Figs. 33, 34, and 35 force be replaced by turning moment and space by angular displacement, the areas under the curves still represent the work done. In twisting an elastic rod from its unstrained position the twisting moment is proportional to the angle of twist, hence the average twisting moment is half the maximum twisting moment ; then, if M = maximum twisting moment, and 6 = angle of twist in radians the work done = |M0 Example i. A high-speed steam-turbine shaft has exerted on it by steam jets a torque of 2100 Ib.-feet. It runs at 750 rotations per minute. Find the horse-power. The work done per minute = (torque in Ib.-feet) x (angle turned through in radians) = 2100 x 750 x 2Tr foot-lbs. 2100 X 750 X 2ir horse-power = = 300 H.P. 33,000 Example 2. An electro motor generates 5 horse-power, and runs at 750 revolutions per minute. Find the torque in pound-feet exerted on the motor spindle. Horse-power x 33,000 = torque in Ib.-feet x radians per minute horse-power x 33,000 hence torque m Ib.-feet = - rad i an y^eFminute^ 5 x 33.000 = ?- - = 35 lb. feet 750 X 27T " Work, Power, and Energy 57 EXAMPLES VI. 1. The average turning moment on a steam-engine crankshaft is 2000 Ib.-feet, and its speed is 150 revolutions per minute. Find the horse-power it transmits. 2. A shaft transmitting 50 H.P. runs at 80 revolutions per minute. Find the average twisting moment in pound-inches exerted on the shaft. 3. A steam turbine develops 250 horse-power at a speed of 200 revolu- tions per minute. Find the torque exerted upon the shaft by the steam. 4. How much work is required to twist a shaft through 10 if the stiffness is such that it requires a torque of 40,000 Ib. -inches per radian of twist? 5. In winding up a large clock (spring) which has completely "run down," 8| complete turns of the key are required, and the torque applied at the finish is 200 Ib. -inches. Assuming the winding effort is always proportional to the amount of winding that has taken place, how much work has to be done in winding the clock ? How much is done in the last two turns ? 6. A water-wheel is turned by a mean tangential force exerted by the water of half a ton at a radius of IO feet, and makes six turns per minute. What horse-power is developed ? 58. Energy. When a body is capable of doing work, it is said to possess energy. It may possess energy for various reasons, such as its motion, position, temperature, chemical composition, etc. ; but we shall only consider two kinds of mechanical energy. 59. A body is said to have potential energy when it is capable of doing work by virtue of its position. For example, when a weight is raised for a given vertical height above datum level (or zero position), it has work done upon it ; this work is said to be stored as potential energy. The weight, in returning to its datum level, is capable of doing work by exerting a force (equal to its own weight) through a distance equal to the vertical height through which it was lifted, the amount of work it is capable of doing being, of course, equal to the amount of work spent in lifting it. This amount is its potential energy in its raised position, e.g. suppose a weight W Ibs. is lifted h feet ; the work is W . // foot-lbs., and the potential energy of the W Ibs. is then said to be W . h foot-lbs. It is capable of doing an amount of work W . h foot-lbs. in falling. 58 Mechanics for Engineers 60. Kinetic Energy is the energy which a body has in virtue of its motion. We have seen (Art. 40) that the exertion of an unresisted force on a body gives it momentum equal to the impulse of the force. The force does work while the body is attaining the momentum, and the work so done is the measure of the kinetic energy of the body. By virtue of the momentum it possesses, the body can, in coming to rest, overcome a resisting force acting in opposition to its direction of motion, thereby doing work. The work so done is equal to the kinetic energy of the body, and therefore also to the work spent in giving the body its motion. Suppose, as in Ex. i, Art. 47, a body of weight W Ibs. is given a velocity V feet per second by the action of a uniform force F] Ibs. acting for ^ seconds, and then comes to rest under a uniform resisting force F 2 Ibs. in 4 seconds. We had, in Art. 47 f W Impulse F^ = V = F 2 / 2 <5 But, the mean velocity being half the maximum under a uniform accelerating force, the distance d^ moved in accelerat- ing, is IVA feet, and that d. 2) moved in coming to rest, is ~Vt z hence the work done in accelerating is W W F 1 xiVA = -V X iV = ^V 2 A A and work done in coming to rest is W W F 2 x !V/ 2 =-VxiV = i^V 2 i> <!> W hence l-V a = F^ = F 2 <r <S These two equalities are exactly the same as those of Ex. / W \ i, Art. 47 f viz. V = F^ = F >2 / 2 1 , with each term multi- V plied by , and problems which were solved from considera- tions of changes of momentum might often have been (alter- natively) solved by considerations of change of kinetic energy. Work, Poiver, and Energy 59 The amount of kinetic energy possessed by a body of W weight W Ibs. moving at V feet per second is therefore \ V' 2 <S foot-lbs. Again, if the initial velocity had been it feet per second instead of zero, the change of momentum would have been W (v it), and we should have had A W Fj/! = (v ?/), v being final velocity o n + "v W. N n + v and the work done = F : X - X /" t = ( ) - W - 1- ( 7 ,2 _ 2\ J <r = change of kinetic energy Similarly, in overcoming resistance at the expense of its kinetic energy, the work done by a body is equal to the change of kinetic energy whether all or only part of it is lost. 61. Principle of Work. If a body of weight W Ibs. be lifted through h feet, it has potential energy W/; foot-lbs. If it falls freely, its gain of kinetic energy at any instant is just equal to the loss of potential energy, so that the sum (potential energy) -f (kinetic energy) is constant ; e.g. suppose the weight has fallen freely x feet, its remaining potential energy is x} foot-lbs. It will have acquired a velocity V 2gx feet W per second (Art. 13), hence its kinetic energy ^ V 2 , will be & W W fh- X2gx=VJx foot-lbs., hence W(/$-*)-r-i V 2 = W^, which > & is independent of the value of x, and no energy has been lost. Note that for a particular system of bodies the sum of potential and kinetic energies is generally not constant. Thus, although momentum is conservative, mechanical energy is not. For example, when a body in motion is brought to rest by a resisting force of a frictional kind, mechanical energy is lost. The energy appears in other forms, chiefly that of heat. Principle of Work. Further, if certain forces act upon 6o Mechanics for Engineers a body, doing work, and other forces, such as frictional ones, simultaneously resist the motion of the body, the excess of the work done by the urging forces over that done against the resistances gives the kinetic energy stored in the body. Or we may deduct the resisting forces from the urging forces at every instant, and say that the work done by the effective or net accelerating forces is equal to the kinetic energy stored. Thus in Fig. 38, representing the forces and work done graphically as in Art. 54, if the ordinates of the curve MP represent the forces urging the body forward, and the ordinates of M'P' re- present the resistances to the same scale, the area MPNO represents the work done ; the work lost against resistances is represented by the area M'P'NO, and the difference between these two areas, viz. the area MPP'M', represents the kinetic energy stored during the time that the distance ON has been traversed. If the body was at rest at position O, MPP'M' represents the total kinetic energy, and if not, its previous kinetic energy must be added toobtain the total stored at the position ON. From a diagram, such as Fig. 38, the velocity M FIG. 38. can be obtained, if the mass of the moving body is known, by the relation, kinetic energy = ^(mass) X (velocity) 2 . Fig. 39 illustrates the case of a body starting from rest and coming to rest again after a distance O^, such, for example, as an electric car between two stopping-places. The driving forces proportional to the ordinates of the curve abec cease Work, Power, and Energy 61 after a distance cc has been traversed, and (by brakes) the resisting forces proportional to the ordinates of the curve def increase. The area abed represents the kinetic energy of the car after a distance 0c, and the area efgc represents the work Distances FIG. 39. done by the excess of resisting force over driving force. When the latter area is equal to the former, the car will have come to rest. The kinetic energy which a body possesses in virtue of its rotation about an axis will be considered in a subsequent chapter. Example I. Find the work done by the charge on a projectile weighing 800 Ibs., which leaves the mouth of a cannon at a velocity of 1800 feet per second. What is the kinetic energy of the gun at the instant it begins to recoil if its weight is 25 tons? The work done is equal to the kinetic energy of the projectile I W I 8<DO , o \9 K.E. = - x xV 2 = -x x (1800)* = 40,200,000 foot-lbs. The momentum of the gun being equal to that of the projectile, the velocity of the gun is 1800 x 800 25 X 2240 and the K.E. = - x = 2571 feet per second x (257 1 ) 2 = 577,ooo foot-lbs. It may be noticed that the kinetic energies of the projectile 62 Mechanics for Engineers and cannon are inversely proportional to their weights. The 1 W i W K.E. is - x x V 2 , or - x - x V x V. which is i x momen- 2 g 2 g turn x velocity. The momentum of the gun and that of the pro- jectile are the same (Art. 52), and therefore their velocities are inversely proportional to their weights ; and therefore the products of velocities and half this momentum are inversely proportional to their respective weights. Example 2. A bullet weighing i oz., and moving at a velocity of 1500 feet per second, overtakes a block of wood moving at 40 feet per second and weighing 5 Ibs. The bullet becomes embedded in the wood without causing any rotation. Find the velocity of the wood after the impact, and how much kinetic energy has been lost. Let V = velocity of bullet and block after impact. Momentum of bullet = ? x -= = 16 g g momentum of block = hence total momentum befo/e ) and after impact f 5 200 ~ x 40 = Total momentum after impact = g 29375 xV = and therefore V = , = 58' i feet per second - 5-0625 x 1500 = 2183 foot- Ibs. = 124 Total K.E. before impact = 2307 Total K.E. after impact = I x 5 25 x 58-1 x 58-1 = 265 foot-lbs. 2 32-2 Kinetic energy of bullet = - x x x 2 16 32-2 Kinetic energy of block = - x. ^- x 40 x 40 Loss of K.E. at impact = 2307 - 265 = 2042 Example 3. A car weighs i2'88 tons, and starts from rest ; the resistance of the rails may be taken as constant and equal to 500 Ibs. After it has moved S feet from rest, the tractive force, F Ibs., exerted by the motors is as follows : s ... F ... 1280 20 1270 50 1220 80 i no no 905 130 800 160 720 190 670 200 660 Work, Power, and Energy 63 Find the velocity of the car after it has gone 200 feet from rest ; nlso find the velocity at various intermediate points, and plot a curve of velocity on a base of space described. Plot the curve of F and S as in Fig. 40, and read off the force every 20 feet, say, starting from S = 10, and subtract 500 Ibs. resistance from each, as follows : S ... 10 30 5 70 90 IIO 13 1^0 170 190 F ... 1275 1260 1 220 1150 1050 905 800 740 6QS 670 F-SOO 775 76o 720 650 55o 405 300 240 195 170 1200 1000 800 1 600 ) S ; *oo 200 -o^ !S k X ^fit- ^*-. " < >--^ ^o-< 20 40 60 80 100 120 140 160 ISO 2C S i-n feet FIG. 40. The mean accelerating force during the first 20 feet of motion is approximately equal to that at S = 10, viz. 775 Ibs. ; hence the work stored as kinetic energy (K.E.), i.e. the gross work done less that spent against resistance, is (1275 x. 20) - (500 x 20), or 775 x 20 foot-lbs. = 15,500 foot-lbs. Then, if V is the velocity after covering S feet, for S = 20 K.E. = and W = 1 2 ' V*= 15,500 x 2240 Ibs. W , . I2'88 x 2240 _ , therefore , the mass of the car is - - - : or 890 units, and 64 Mechanics for Engineers x V 2 = 15,500 1 W W2 1 - x V 2 = i x 2 g V = = 5 '90 feet per second Similarly, finding the gain of kinetic energy in each 20 feet, the square of velocity (V 2 ), and the velocity V, we have from S = 20 to S - 40 gain of K.E. = 760 x 20 = 15,200 foot-lbs. /.total K.E. at S = 40 is 15,500 + 15,200 = 30,700 foot-lbs. and so on, thus S o 20 4O 60 80 IOO 1 20 I4.O 1 60 180 200 Gain of K.E. ) in 20 feet, > o 15500 15200 14400 13000 1 1000 8100 6000 4800 3900 3400 foot-lbs. ) Total K.E., 1 foot-lbs. / c 15500 30700 45100 58100 69100 77200 83200 88000 91900 95300 V 2 or - ' o 34-8 68-5 lOO'O 129-4 154-0 172-1 185-5 196-2 204-8 212-5 V ft. per sec. 8-28 10-03 "'34 12-40 13-12 13-62 14-01 14-30 14-58 These velocities have been plotted on a base of spaces in Fig. 41. 20 60 80 100 120 S. in, feet FIG. 41. 1*0 160 180 200 Work, Power, and Energy 65 Example 4. From the results of Example 3, find in what time the car travels the distance of 20 feet from S = 80 to S = 100, and draw a curve showing the space described up to any instant during the time in which it travels the first 200 feet. At S = 80, V = 11-34 feet per second at S = 100, V = 12*40 feet per second' hence the mean velocity for such a short interval may be taken as approximately 11-34 + 12-40 , or 1 1 -87 feet per second 2 Hence the time taken from S = 80 to S = 100 is approximately 20 11-87 = 1-685 seconds Similarly, we may find the time taken to cover each 20 feet, and so find the total time occupied, by using the results of Ex. 3, as follows. The curve in Fig. 42 has been plotted from these numbers. 200 ISO 100 tt 3 * 50 10 15 Time, in, seconds FIG. 42. 20 25 66 Mechanics, for Engineers s o 20 40 60 80 zoo I2O 140 160 i 80 2 Mean velocity for last 20 ft., o 2'95 7-09 9'IS 10-68 11-87 I2-76 13-37 1381 14-15 '4' feet per sec. Time for last 20 feet, se- o 6.780 2-824 2-188 1-872 1-685 I-S68 1-496 i '4*7 i'4i3 i'3 conds Total time,) / seconds J o 6780 9-604 11-792 13-66415-349 16-917 18-413 19-850 21-263 22'6 EXAMPLES VII. 1. Find in foot-pounds the kinetic energy of a projectile weighing 800 Ibs. moving at 1000 feet per second. If it is brought to rest in 3 feet, find the space average of the resisting force. 2. At what velocity must a body weighing 5 Ibs. be moving in order to have stored in it 60 foot-lbs. of energy ? 3. What is the kinetic energy in inch-pounds of a bullet weighing I oz. travelling at 1800 feet per second? If it is fired directly into a suspended block of wood weighing 1-25 lb., how much kinetic energy is lost in the impact ? 4. A machine-gun fires 300 bullets per minute, each bullet weighing I oz. and having a muzzle velocity of 1700 feet per second. At what average horse-power is the gun working ? 5. A jet of water issues in a parallel stream at 90 feet per second from a round nozzle I inch in diameter. What is the horse-power of the jet ? One cubic foot of water weighs 62*5 Ibs. 6. Steam to drive a steam impact turbine issues in a parallel stream from a jet J inch diameter at a velocity of 2717 feet per second, and the density of the steam is such that it occupies 26'5 cubic feet per pound. Find the horse-power of the jet. 7. A car weighing 10 tons attains a speed of 15 miles per hour from rest in 24 seconds, during which it covers 100 yards. If the space-average of the resistances is 30 Ibs. per ton, find the average horse-power used to drive the car. 8. How long will it take a car weighing 1 1 tons to accelerate from 10 miles per hour to 15 miles per hour against a resistance of 25 Ibs. per ton, if the motors exert a uniform tractive force on the wheels and the horse-power is 25 at the beginning of this period ? 9. A car weighing 12 tons is observed to have the following tractive forces F Ibs. exerted upon it after it has travelled S feet from rest : ? '... o 1440 10 1390 30 1250 50 1060 65 910 80 805 94 760 100 740 Work, Power, and Energy 67 The constant resistance of the road is equivalent to 600 Ibs. Find the velocity of the car after it has covered 100 feet. Plot a curve showing the velocity at all distances for loo feet from the starting-point. What is the space-average of the effective or accelerating force on the car ? 10. From the results of the last question plot a curve showing the space described at any instant during the time taken to cover the first ico feet. How long does the car take to cover 100 feet ? 11. A machine having all its parts in rigid connection has 70,000 foot- pounds of kinetic energy when its main spindle is making 49 rotations per minute. How much extra energy will it store in increasing its speed to 50 rotations per minute ? 12. A machine stores 10,050 foot-lbs. of kinetic energy when the speed of its driving-pulley rises from ico to 101 revolutions per minute. How much kinetic energy would it have stored in it when its driving-pulley is making loo revolutions per minute? CHAPTER IV MOTION IN A CIRCLE: SIMPLE HARMONIC MOTION 62. Uniform Circular Motion. Suppose a particle de- scribes about a centre O (Fig. 43), a circle of radius r feet with uniform angular velocity w radians per second. Then its velocity, ?', at any instant is of magnitude wr (Art. 33), and its direction is along the tangent to the circle from the point in the circumference which it occupies at that instant. Although its velocity is always of magnitude wr, its direc- tion changes. Consider T the change in velocity between two points, P and Q, on its path at an angular distance 6 apart (Fig. 43)- Let the vector cb parallel to the tangent PT represent the linear velocity v at P, and let the vector ab, of equal length to cb and parallel to QT, the tangent at Q, represent the linear velocity v at Q. Then, to find the change of velocity between P and Q, we must subtract the velocity at P from that at Q; in vectors ab cb = ab + be = ac (Art. 27) Then the vector ac represents the change of velocity between the positions P and Q. Now, since abc = POQ = 0, length Motion in a Circle: Simple Harmonic Motion 69 9 ac= 2(ib . sin -, which represents 2V sin.-, and the time 2 2 Q taken between the positions P and Q is - seconds (Art. 33). Therefore the average change of velocity per second is . 6 Sln 2 2 v sin - -f- or <av . a 2 0) U 2 which is the average acceleration. Now, suppose that Q is taken indefinitely close to P that is, that the angle is in- sin - 2 definitely reduced ; then the ratio & has a limiting value 2 unity, and the average change of velocity per second, or average acceleration during an indefinitely short interval is v* w, or w'V or -, since v = <ar. This average acceleration during an indefinitely reduced interval is what we have defined (Art. 9) as actual acceleration, so that the acceleration at P i? is wV or feet per second per second. And as the angle is diminished indefinitely and Q thereby approaches P, the vector al>, remaining of the same length, approaches cb (a and c being always equidistant from />), and the angle bai increases and approaches a right angle as 6 approaches zero. Ultimately the acceleration (o>V) is perpendicular to PT, the tangent at P, i.e. it is towards O. 63. Centripetal and Centrifugal Force. In the previous article we have seen that if a small body is describing a circle of radius r feet about a centre O with angular velocity w radians per second, it must have an acceleration oV towards O; hence the force acting upon it must be directed towards the centre O and of magnitude equal to its (mass) X w'V or W - orr Ibs., where W is its weight in pounds This force causing 7O Mechanics for Engineers the circular motion of the body is sometimes called the centri- petal force. There is (Art. 51), by the third law of motion, a reaction of equal magnitude upon the medium which exerts this centripetal force, and this reaction is called the centrifugal force. It is directed away from the centre O, and is exerted W upon the matter which impresses the equal force o>V upon 0> the revolving body ; it is not to be reckoned as a force acting upon the body describing a circular path. A concrete example will make this clear. If a stone of weight W Ibs. attached to one end of a string r feet long describes a horizontal circle with constant angular velocity u> radians per second, and is supported in a vertical direction by a smooth table, so that the string remains horizontal, the force W which the string exerts upon the stone is o>V towards the d centre of the circle. The stone, on the other hand, exerts on W the string an outward pull urr away from the centre. In A other cases of circular motion the inward centripetal force may be supplied by a thrust instead of a tension ; e.g. in the case of a railway carriage going round a curved line, the centri- petal thrust is supplied by the rail, and the centrifugal force is exerted outward on the rail by the train. 64. Motion on a Curved " Banked " Track. Suppose a body, P (Fig. 44), is moving with uniform velocity, v, round a CL FIG. 44 . smooth circular track of radius OP equal to r feet. At what angle to the horizontal plane shall the track be inclined or Motion in a Circle : Simple Harmonic Motion 7 [ "banked" in order that the body shall keep in its circular path? There are two forces acting on the body (r) its own weight, W ; (2) the reaction R of the track which is perpendicular to W v 2 the smooth track. These two have a horizontal resultant g r towards the centre O of the horizontal circle in which the body moves. If we draw a vector, ab (Fig. 44), vertically, to represent W, then R is inclined at an angle a to it, where a is the angle of banking of the track. If a vector, be, be drawn from b inclined at an angle a to ab, to meet ac, the perpendicular to ab from a, then be represents R, and ac or (ab + be) represents W z; 2 the resultant of W and R, viz. , and g r ' ac W V* V* tan a = -7 = -f- W = ab g r gr which gives the angle a required. 65. Railway Curves. If the lines of a railway curve be laid at the same level, the centripetal thrust of the rails on the wheels of trains would act on the flanges of the wheels, and the centrifugal thrust of the wheel on the track would tend to push it sideways out of its place. In order to have the action and reaction normal to the track the outer rail is raised, and the track thereby inclined to the horizontal. The amount of this " superelevation " suitable to a given speed is easily calculated. Let G be the gauge in inches, say, v the velocity in feet per second, and r the radius of the curve in feet. Let AB (Fig. 45) represent G; then AC represents the height in inches (exaggerated) which B stands above A, and ABC is the angle of banking, as in Art. 64. Then AC = AB sin a = AB tan a nearly, since a. is always very small ; hence, by Art. 64, AC v 2 - represents G tan a, or G ff . inches. Mechanics for Engineers W FIG. 46. 66. Conical Pendulum. This name is applied to a combination consisting of a small weight fastened to one end of a string, the other end of which is attached to a fixed point, when the weight keeping the string taut, describes a horizontal circle about a centre vertically under the fixed point. Fig. 46 represents a conical pendulum, where a particle, P, attached by a thread to a fixed point, O, describes the hori- zontal circle PQR with con- stant angular velocity about the centre N vertically under O. Let T = tension of the string OP in Ibs. ; w = angular velocity of P about N in radians per second ; W = weight of particle P in Ibs. ; r = radius NP of circle PQR in feet ; / = length of string OP in feet ; . a = angle which OP makes with ON, viz. PON ; // = height ON in feet ; g- = acceleration of gravity in feet per second per second. At the position shown in Fig. 46 P is acted upon by two forces (i) its own weight, W; (2) the tension T of the string OP. These have a resultant in the line PN (towards N), the vector diagram being set off as in Art. 64, ab vertical, representing the weight W, of P, and be the tension T. Then W the vector ac = ab -f bc> and represents the resultant force - X wV along PN ; hence we- ., , TT tan a. = -, = w V W = ab g g Also ON or h = NP 4- tan a. = r 4- = ^ feet hence the height h of the conical pendulum is dependent only on the angular velocity about N, being inversely proportional to the square of that quantity. Motion in a Circle: Simple Harmonic Motion 73 Since h or / cos a = -, a> 2 = , and w = t.\* ft FIG. 47. Also the time of one complete revolution of the pendulum is angle in a circle _ 27r _ /^ angular velocity ~~ w '\/ ^ the period of revolution being proportional to the square root of the height of the pendulum, and the number of revolutions per minute being therefore inversely proportional to the square root of the height. This principle is made use of in steam- engine governors, where a change in speed, altering the height of a modified conical pendulum, is made to regulate the steam supply. 67. Motion in a Vertical Circle. Suppose a particle or small body to move, say, contra- clockwise in a vertical circle with centre O (Fig. 47). It may be kept in the circular path by a string attached to O, or by an inward pressure of a circular track. Taking the latter instance Let R = the normal inward pressure of the track ; W= the weight of the rotating body in pounds; v = its velocity in feet per second in any position P such that OP makes an angle to the vertical OA, A being the lowest point on the circum- ference ; ?; A = the velocity at A ; r = the radius of the circle in feet. W Then the kinetic energy at A is \ z ; A 2 d At P the potential energy is W X AN, and the kinetic energy W is r, - v*, and since there is no work done or lost between A *g and P, the total mechanical energy at P is equal to that at A (Art. 61). Therefore 74 Mechanics for Engineers W W I .zr + W.AN = 1-V 2 <sr V hence z> 2 4- 2^. AN = v^ . . . . (i) Neglecting gravity, the motion in a circle would be uniform, and W z' 2 would cause a reaction from the track (Art. 63). And in addition the weight has a component W cos 6. in the direction OP, which increases the inward reaction of the track by that amount ; hence the total normal pressure W ?; 2 R-.- 4-WCOS0 . . . . (2) g r The value of R at any given point can be found by sub- stituting for v from equation (i) provided V A is known. The least value of R will be at B, the highest point of the circle, where gravity diminishes it most. If V A is not sufficient to make R greater than zero for position B, the particle will not describe a complete circle. Examining such a case, the condition, in order that a complete revolution may be made without change in the sign of R, is R B > o W 7, 2 i.e. B + W cos 180 > o . g r or since cos 180 = i g r or z/ B 2 > gr and since v s 2 = z> A 2 2g . AB = v? 4gr, substituting for t' B 2 , the condition is w . 2 A (ri- ^> (fr i.e. the velocity at A must be greater than that due to falling through a height fr, for which the velocity would be J $gr (Art. 28). For example, in a centrifugal railway (" looping the loop") the necessary velocity on entering the track at the Motion in a Circle: Simple Harmonic Motion 75 lowest point, making no allowance for frictional resistances, may be obtained by running down an incline of height greater than two and a half times the radius of the circular track. If the centripetal force is capable of changing sign, as in the case of the pressure of a tubular track, or the force in a light stiff radius rod supporting the revolving weight, the condition that the body shall make complete revolutions is that Z' B shall be greater than zero, and since ' B 2 = Vj? 4gr, the condition is i.e. the velocity at A shall be greater than that due to falling through a height equal to the diameter of the circle. Similarly, the position at which the body will cease to describe a circular track (in a forward direction) if # A is too small for a complete circuit, when the force can change sign and when it can not, may be investigated by applying equations (i) and (2), which will also give the value of R for any position of the body. The pendulum bob, suspended by a thread, is of course limited to oscillation of less than a semicircle or to complete circles. Example i. At what speed will a locomotive, going round a curve of looo-feet radius, exert a horizontal thrust on the outside rail equal to T of its own weight ? Let W = the weight of loco, v = its velocity in feet per second. W Z/ 2 Centrifugal thrust = . - - - = T J W jp" I OOO IOOO X .*. #2 _ - - 6. = 322 100 v = 1 7 '95 feet per second, equivalent to 12*22 miles per hour Example 2. A uniform disc rotates 250 times per minute about an axis through its centre and perpendicular to its plane. It has attached to it two weights, one of 5 Ibs. and the other of 7 Ibs., at an angular distance of 90 apart, the first being I foot and the second 2 feet from the axis. Find the magnitude and direction of the resultant centrifugal force on the axis. Find, also, 76 Mechanics for Engineers where a weight of 12 Ibs. must be placed on the disc to make the resultant centrifugal force zero. ~, i 250 X 2ir 2?ir The angular velocity is , = -- radians per second The centrifugal pull Fj (Fig. 48) is 32-2 \ 3 and the centrifugal pull F 2 is 7 -5 Ibs. s x IT ' hence the resultant R of Fj and F 2 at right axis' i' " ou>3 - angles is FIG. 4 8. R = \fio6 z + 297 2 = 315 Ibs. at an angle tan" 1 = tan" 1 0*357 = ig'6 to the direction of F 2 (Arts. 24 and 44) To neutralize this, a force of 31 5 Ibs. will be required in the opposite direction. Let x radius in feet of the i2-lbs. weight placed at 180 ig'6 or i6o'4 contra-clockwise from F 2 . . - hence x 1-23 feet Example 3. Find in inches the change in height of a conical pendulum making 80 revolutions per minute when the speed increases two per cent. The increase in speed is T gg x 80 = r6 revolutions per minute to 8 1 '6 revolutions per minute. The height is 4r (Art. 66), where is the angular velocity in CO radians per second. At 80 revolutions per minute the angular velocity is 2ir X 80 Sir ,. = radians per second 60 3 or ^2*2 X Q hence the height h^ = ^ = zr~a u) OA7T = 0-4585 foot Motion in a Circle: Simple Harmonic Motion 77 At 8 1 -6 revolutions per minute the angular velocity is 2T x 8r6 ir r radians per second = 0-441 1 foot i = 0-6174 foot or 0*207 inch W and the height is ^ 81 . 6 = hence the decrease in height is 0-4585 - 0-4411 Example 4. A piece of lead is fastened to the end of a string 2 feet long, the other end of which is attached to a fixed point. With what velocity must the lead be pro- jected in order to describe a hori- zontal circle of 2 feet diameter ? Let OP, Fig. 49, represent the string; then the horizontal line PN is to be i foot radius. In the vector triangle abc, ab N 1 represents W, the weight of lead, FlG - 49- be the tension T of the string OP, and ac their resultant ; then NP _ ac _ W it _ v n ~ ON""^-^"^" 4 " J1TF where v = velocity in feet per second ; hence ^= = = 18-57 and v = 4-309 feet per second Exercise 5. A stone weighing } Ib. is whirling in a vertical circle at the extremity of a string 3 feet long. Find the velocity of the stone and tension of the string (i) at the highest position, (2) at lowest, (3) midway between, if the velocity is the least possible for a complete circle to be described. If the velocity is the least possible, the string will just be slack when the stone is at the highest point of the circle. Let v be the velocity at the highest point, where the weight just supplies the centripetal force ; V* = 3 x 32*2 = 96-6 and z'o = 9*83 feet per second. (2) At the lowest point let the velocity be v^ feet per second. 78 Mechanics for Engineers Since there is no loss of mechanical energy, the gain of kinetic energy is \ x 6 foot-lbs., hence ill ill i - x - x v,* = - . - . -v? + - . 6 24^ 2 4 <? - r 4 and z>, 2 = r/ 2 + 2 .g . 6 = 96-6 + 386-4 = 483 (or 5 x g x 3) v l = V483 = 22 feet per second (nearly) and the tension is \ , i i i 483 )>=- + - - = i'5lbs., or six times the weight ~ -4- . 4 I2o'o 4 4 32-2 3 J of the stone (3) When the string is horizontal, if v' = velocity in feet per second similarly i. - V 2 = - . - . -v 4-^-3 y '2 4 g 2-4'^ ^4 J t-' 2 = r/ 2 + ?.g x 3 = 96-6 + I93'2 v' = .v/289'8 = 17 feet per second and the tension is i i 289/8 >= 075 lb., or three times the weight of the 4 '32-2 X 3 J stone EXAMPLES VIII. 1. How many circuits per minute must a stone weighing 4 ozs. make when whirled about in a horizontal circle at the extremity of a string 5 feet long, in order to cause a tension of 2 Ibs. in the string ? 2. At what speed will a locomotive produce a side thrust equal to g'g of its own weight on the outer rail of a level curved railway line, the radius of the curve being 750 feet ? 3. What is the least radius of curve round which a truck may run on level lines at 20 miles per hour without producing a side thrust of more than T ig of its own weight ? 4. How much must the outer rail of a line of 4 feet 8.J inches gauge be elevated on a curve of 800 feet radius in order that a train may exert a thrust normal to the track when travelling at 30 miles per hour ? 5. The outer rail of a pair, of 4 feet 85 inches gauge, is elevated 25 inches, and a train running at 45 miles per hour has no thrust on the flanges of either set of wheels. What is the radius of the curve ? 6. At what speed can a train run round a curve of 1000 feet radius without having any thrust on the wheel flanges when the outer rail is laid i '5 inches above the inner one, and the gauge is 4 feet 8^ inches ? 7. To what angle should a circular cycle-track of 15 laps to the mile be Motion in a Circle: Simple Harmonic Motion 79 banked for riding upon at a speed of 30 miles per hour, making no allow- ance for support from friction ? 8. A string 3 feet long, fixed at one end, has attached to its other end a stone which describes a horizontal circle, making 40 circuits per minute. What is the inclination of the string to the vertical ? What is its tension ? 9. What percentage change of angular speed in a conical pendulum will correspond to the decrease in height of 3 per cent. ? 10. The revolving ball of a conical pendulum weighs 5 Ibs., and the height of the pendulum is 8 inches. What is its speed ? If the ball is acted upon by a vertical downward force of i lb., what is then its speed when its height is 8 inches ? Also what would be its speed in the case of a vertical upward force of I lb. acting on the ball ? 1 1 . \Vhat will be the inclination to the vertical of a string carrying a weight suspended from the roof of a railway carriage of a train going round a curve of 1000 feet radius at 40 miles per hour ? 12. A body weighing \ lb., attached to a string, is moving in a vertical circle of 6 feet diameter. If its velocity, when passing through the lowest point, is 40 feet per second, find its velocity and the tension of the string when it is 2 feet and when it is 5 feet above the lowest point. 68. Simple Harmonic Motion. This is the simplest type of reciprocating motion. If a point Q (Fig. 50) describes a circle AQB with constant angular velocity, and P be the rectangular projection of Q on a fixed diameter AB of the circle, then the oscillation to and fro of P along AB is defined as Simple Harmonic Motion. Let the length OA of the radius be a feet, called the amplitude of oscillation. Let w be the angular velocity of Q in radians per second. Let 6 be the angle AOQ in radians, denoting any position ofQ. Suppose the motion of Q to be, say, contra-clockwise. A complete vibration or oscillation of P is reckoned in this country as the path described by P whilst Q describes a complete circle. Let T = the period in seconds of one complete vibration ; then, since this is the same as that for one complete circuit made by Q radians in one circle 2ir *~r* / - \ ~~ radians described per second <> 80 Mechanics for Engineers Let x - distance OP of towards A, the / v 7" o , *~ 1 r -: ' *L Trs 1 s, _ r "j ! S "" T^ ^^ *~ \ 7 S i ^ "^ ^ flL S / ^^ ,' ^ ~' ^ J jf s. jt "^ . JJ ~ ^ ' \ . ~. 5 j 5 "I ' ^k d s *-* : ^i " . / $ y 'S-K c ' < * ^ j^* / r ^ i \v / > 7 s \ ^ ^ ^ ~? cr "" \ s _ / . s ^ . s j ' 5s. -i lt- y % s e" ' ~ >^ o- 8 - 7 S u. ' - ^ ^ L ^ ^ . ^ ^ s ^_ "~ ^2T^iCiJK2L2Lj c/ ^ - ^ j "~ ^ ! . / > "s ~7 \ ^ ~7 ' \ . 7 i L A J \ f\ ^ . ' r *** ^ r jj ^ , - (/C ^ _$ ^ ,, \ j/ ^ V. ~*~ - - -3r* ^-~ . ! _ "^ of P from O in feet, reckoned positive then x = a cos ; and let v velocity of P in feet per second in position 6. Draw OS perpendicular to OQ to meet the circum- ference of the circle AQSB in S, and draw SM perpen- dicular to AB to meet it in M. Then for the position or phase shown in the figure, the velocity of Q is wa (Art. 33) in the direction perpen- dicular to OQ, i.e. parallel to OS. Resolving this velocity along the diameter AB, OSM being a vector triangle, the component velocity of Q OM parallel to AB is -~c X wa, vJo or <aa sin 6, or w . OM. This is then the velocity of P towards O, the mid-path. Since sin 6 = OM OS v = <i>a sin 6 which gives the velocity of P in terms of the amplitude and position. Or, if OS represents geo- metrically the velocity of Q, then OM represents that ot P to the same scale. Motion in a Circle: Simple Harmonic Motion 81 Acceleration of P. The acceleration of Q is u?a along QO towards O (Art. 62). Resolving this acceleration, the PO component in direction AB is w 2 x pr or o> 2 0.cos % 0, or w 2 . x, towards O ; and it should be noted that at unit distance from O, when x = i foot, the acceleration of P is w 2 feet per second per second. The law of acceleration of a body having simple harmonic motion, then, is, that the acceleration is towards the mid-path and proportional to its distance from that point. When the body is at its mid-path, its acceleration is zero ; hence there is no force acting upon it, and this position is one of equilibrium if the body has not any store of kinetic energy. Conversely, if a body has an acceleration proportional to its distance from a fixed point, O, it will have a simple harmonic motion. If the acceleration at unit distance from O is /x, feet per second per second (corresponding to w 2 in the case just considered), by describing a circle with centre about its path as diameter, we can easily show that the body has simple harmonic motion, and by taking w = vV> !" corresponding to w 2 in the above case, we can state its velocity and acceleration at a distance x from its centre of motion O, and its period of vibration, viz, velocity v at x feet from O is vV V a * ~ x *> or V P-( a * ~ * 2 )- Acceleration at x feet from centre O is p . x, and the time 27T of a complete vibration is == VJK- Alternating: Vectors. We have seen that, the displace- ment of P being OP, the acceleration is proportional also to OP, and the velocity to OM ; so that OP and OM are vectors representing in magnitude and direction the displacement and velocity of P. Such vectors, having a fixed end, O, and of length varying according to the position of a rotating vector, OQ or OS, are called " alternating vectors." It may be noted that the rate of change of an alternating vector, OP, of ampli- ade a is represented by another alternating vector, OM, of the same period, which is the projection of a uniformly rotating vector of length OS = w . OQ or <a (to a different scale), and one right angle in advance of the rotating vector OQ, of which G 82 Mechanics for Engineers OP is the projection. A little consideration will show that the rate of change of the alternating vector OM follows the same law (rate of change of velocity being acceleration), viz. it is represented by a third alternating vector, ON, of the same period, which is the projection of a uniformly rotating vector of length OQ' = o> . OS or u?a (to a different scale), and one right angle in advance of the rotating vector OS, of which OM is the projection. The curves of displacement, velocity, and acceleration of P on a base of angles are shown to the right hand of Fig. 50. The base representing angles must also represent time, since the rotating vectors have uniform angular velocity <o. The fi ft time / = - seconds, since w = -. The properties of the curves 0) / of spaces, velocities, and accelerations (Arts. 4, 14, and 16) are well illustrated by the curves in Fig. 50, which have been drawn to three scales of space, velocity, and acceleration by projecting points 90 ahead of Q, S, and Q' on the circle on the left. The acceleration of P, which is proportional to the displacement, may properly be considered to be of opposite sign to the displacement, since the acceleration is to the left from P to O when the displacement OP is to the right of O. The curves of displacement and acceleration are called " cosine curves," the ordinates being proportional to the cosines of angle POQ, or 6, or w/. Similarly, the curve of velocity is called a " sine curve." The relations between the three quantities may be expressed thus Displacement (x) : velocity (v) : acceleration = a cos wt : aw sin w/ : w 2 cos o>/ Curved Path. If the point P follows a curved path instead of the straight one AB, the curved path having the same length as the straight one, and if the acceleration of the point when distant x feet from its mid-path is tangential to the path and of the same magnitude as that of the point following the straight path AB when distant x feet from mid- path, then the velocity is of the same magnitude in each case. This is evident, for the points attain the same speeds in the Motion in a Circle: Simple Harmonic Motion 83 same intervals of time, being, under the same acceleration, always directed in the line of motion in each case. Hence the periodic times will be the same in each case, viz. vV where ju, is the acceleration in feet per second per second along the curve or the straight line, as the case may be. 69. There are numerous instances in which bodies have simple harmonic motion or an approximation to it, for in perfectly elastic bodies the straining force is proportional to the amount of displacement produced, and most substances are very nearly perfectly elastic over a limited range. A common case is that of a body hanging on a relatively light helical spring and vibrating vertically. The body is acted upon by an effective accelerating force proportional to its distance from its equilibrium position, and, since its mass f force \ does not change, it will have an acceleration I - - 1 also \ mass / proportional to its displacement from that point (Art. 40), and therefore it will vibrate with simple harmonic vibration. Let W = weight of vibrating body in pounds. e = force in pounds acting upon it at i foot from its equilibrium position, or per foot of displace- ment, the total displacement being perhaps less than i foot. This is sometimes called the stiffness of the spring. Then e . x = force in Ibs. x feet from the equilibrium position and if p = acceleration in feet per second per second i foot from the equilibrium position or per foot of displacement accelerating force _ . W _ eg ^ mass g ~ W hence the period of vibration is - ^ or 2ir A / ; (Art. 68) Vj"- * The maximum force, which occurs when the extremities of the path are reached, is e.a, where a is the amplitude of 84 Mechanics for Engineers the vibration or distance from equilibrium position to either extremity of path, in feet. The crank-pin of a steam engine describes a circle ABC (Fig. 51), of which the length of crank OC is the radius, with FIG. 51. fairly constant angular velocity. The piston P and other reciprocating parts are attached to the crank-pin by a con- necting-rod, DC, and usually move to and fro in a straight line, AP, with a diameter, AB, of the crank-pin circle. If the connecting-rod is very long compared to the crank-length, the motion is nearly the same as that of the projection N of the crank-pin on the diameter AB of the crank-pin circle, which is simple harmonic. If the connecting-rod is short, however, its greater obliquity modifies the piston-motion to a greater extent. 70. Energy stored in Simple Harmonic Motion. If c = force in pounds at unit distance, acting on a body of weight W Ibs. having simple harmonic motion, the force at a distance x is ex, since it is proportional to the displacement. Therefore the work done in displacing the body from its equili- brium through x feet is \eo? (Art. 54 and Fig. 35). This energy, which is stored in some form other than kinetic energy when the body is displaced from its equilibrium position, reaches a maximum \ecf- when the extreme displacement a (the amplitude) has taken place, and the effective accelerating force acting on the body is ea. In the mid-position of the body (x = o), when its velocity is greatest and the force acting on it is nil, the energy is wholly kinetic, and in other inter- mediate positions the energy is partly kinetic and partly otherwise, the total being constant if there are no resistances. B Distance. A FIG. 52. Motion in a Circle: Simple Harmonic Motion 85 Fig. 52 shows a diagram of work stored for various dis- placements of a body having simple harmonic motion. The amplitude OA = a, and therefore the force at A is ae, which is represented by AD, and the work done in moving from O to A is represented by the area AOD(Art.54andFig. 3 5). At P, distant x feet from O, the work done in motion from O is ^ex 1 , represented by the area OHP, and the kinetic energy at P is therefore represented by the area DAPH. 71. Simple Pendulum. This name refers strictly to a particle of indefinitely small dimensions and yet having weight, suspended by a perfectly flexible weightless thread from a fixed point, about which, as a centre, it swings freely in a circular arc. In practice, a small piece of heavy metal, usually called a pendulum bob, suspended by a moderately long thin fibre, behaves very nearly indeed like the ideal pendulum defined above, the resistances, such as that of the atmosphere, being small. Let 0, Fig. 53, be the point of suspension of the particle P of a simple pendulum. Let OP, the length of thread, be / feet. Let 6 = angle AOP in radians which OP makes with the vertical (OA) through O in any position P of the particle. Draw PT perpendicular to OP, i.e. tangent to the arc of motion to meet the vertical through O in T. The tension of the thread has no component along the direction of motion (PT) at P. The acceleration along PT is FIG. 53- 86 Mechanics for Engineers then g sin 0, since FT is inclined 6 to the horizontal (Art. 28). If 6 is very small, sin 6 may be taken equal to 6 in radians. (If does not exceed 5, the greatest error in this approxi- mation is less than i part in 800.) Hence the acceleration along PT is g6 approximately. And 6 = T. ^^ ; therefore acceleration along PT =* j - , and the acceleration is proportional to the distance AP, along the arc, of P from A, g being -, per foot of arc. Hence the time of a complete oscilla- tion in seconds is ^ Vf =2 x/^ (Art 68) and the velocity at any point may be found, as in Art. 68, for any position of the swinging particle. In an actual pendulum the pendulum bob has finite dimen- sions, and the length / will generally be somewhat greater than that of the fibre by which it is suspended. The ideal simple pendulum having the same period of swing as an actual pen- dulum of any form is called its simple equivalent pendiilum. For this ideal pendulum the relation /= 2tr\ /-holds, and V g Pg therefore / = -~g, from which its length in feet may be calculated for a given time, /, of vibration. The value of the acceleration of gravity, g, varies at different parts of the earth's surface, and the pendulum offers a direct means of measuring the value of this quantity g, viz. by accurate timing of the period of swing of a pendulum of known length. The length of an actual pendulum, i.e. of its simple equivalent pendulum, can be calculated from its dimensions. Example i. A weight rests freely on a scale-pan of a spring balance, which is given a vertical simple harmonic vibration of period o'5 second. What is the greatest amplitude the vibration may have in order that the weight may not leave the pan ? What is then the pressure of the weight on the pan in its lowest position ? Let a = greatest amplitude in feet. Motion in a Circle: Simple Harmonic Motion 87 The greatest downward force on the body is its own weight, and therefore its greatest downward acceleration is g, occurring when the weight is in its highest position and the spring is about to return. Hence, if the scale-pan and weight do not separate, the downward acceleration of the pan must not exceed g, and therefore the acceleration must not exceed per foot of dis- a f placement. /2ir\ 2 The acceleration per foot of displacement is f ) ; therefore a or a > 3p feet l6ir 2 i.e. a ;j> o'2O4 feet or 2*448 inches If the balance has this amplitude of vibration, the pressure between the pan and weight at the lowest position will be equal to twice the weight, since there is an acceleration upwards which must be caused by an effective force equal to the weight acting upwards, or a gross pressure of twice the weight from which the downward gravitational force has to be subtracted. Example 2. Part of a machine has a reciprocating motion, which is simple harmonic in character, making 200 complete oscilla- tions in a minute ; it weighs 10 Ibs. Find (i) the accelerating force upon it in pounds and its velocity in feet per second, when it is 3 inches from mid-stroke; (2) the maximum accelerating force; and (3) the maximum velocity if its total stroke is 9 inches, i.e. if its amplitude of vibration is 4^ inches. Time of i oscillation = = o'3 second 200 therefore the acceleration per foot) /2 w\ 2 4001^ > = I - ) = - feet per second distance from mid-stroke J \o'3/ 9 per second and the accelerating force 0-25 foot from mid-stroke on 10 Ibs. is 10 4OOT 2 x 0-25 x * - = 34'oS Ibs. 3* y and the maximum accelerating force 4^ inches from mid-stroke is i'5 times as much as at 3 inches, or 34 - o8 x 1*5 = 5i'i2 Ibs. 88 Mechanics for Engineers The maximum velocity in feet per second occurring at mid-stroke = amplitude in feet x >J Acceleration per foot of displacement (Art. 68) = amplitude in feet x - period 1 21T IjTT or = =j x = = 7*85 feet per second 8 o'3 2 Velocity at 3 inches) = V^LEI*' (Art . 68 ) from mid-stroke \ 4' 5 = 7-85 x 'V " 25 = 5-85 feet per second Example 3. The crank of an engine makes 150 revolutions per minute, and is 1-3 feet long. It is driven by a piston and a very long connecting rod (Fig. 51), so that the motion of the piston may be taken as simple harmonic. Find the piston velocity and the force necessary to accelerate the piston and recipro- cating parts, weighing altogether 300 Ibs., (i) when the crank has turned through 45 from its position (OB) in line with and nearest to the piston path ; (2) when the piston has moved forward o'65 foot from the end of its FIG. 54 . stroke. Let ABC (Fig. 54) be the circular path 1*3 feet radius of the crank-pin, CN the perpendicular from a point C on the diameter AB. The angular velocity of crank OC is 7 = 5*- radians per second (i) The motion of the piston being taken as that of N, the acceleration of piston when the crank-pin is at C is (57r)' 2 x i'3 x cos 45 (wVcos 6, Art. 68) and the accelerating force is 300 / y, i X (Sir) 2 X I '3 X ->- = 2IIolbs. 32-2 ,/a The velocity is 5ir x i '3 x sin 45 = H'43 feet per second Motion in a Circle: Simple Harmonic Motion 89 (2) When BN = 0*65 foot, ON = OB - BN = 1-3 - 0-65 = 0-65 ON foot, and CON = cos" 1 _- = cos" 1 \ = 60. The accelerating force is then \ x (SO 2 x 1-3 x \ = 1493 Ibs. jz z and the velocity is 5 IT x i '3 x sin 60 = 17*67 feet per second Example 4. A light helical spring is found to deflect 0*4 inch when an axial load of 4 Ibs. is hung on it. How many vibrations per minute will this spring make when carrying a weight of 15 Ibs.? The force per foot of deflection is 4 * = 120 Ibs. 1 2 is 2w / _]> = o'3Q2 second V 32-2 x 120 hence the time of vibration 32'2 X 1 2O and the number of vibrations per minute is Tr;^ = I53' 2 Example 5. Find the length of a clock pendulum which will make three beats per second. If the clock loses i second per hour, what change is required in the length of pendulum ? Let / = length of pendulum in feet. Time of vibration = ^ second ' f Qxf* 32:2 fcet = inches 4iH 36*-* The clock loses i second in 3600 seconds, i.e. it makes 3599 x 3 beats instead of 3600 x 3. Since / oc / 2 oc y where n = number of beats per hour, therefore correct length _ 3599** _ ( . 1-09 inches ~ 3600" ~ V 1 ** } i YgW approximately therefore shortening required = --^ inches = 0*000606 inch EXAMPLES IX. I. A point has a simple harmonic motion of amplitude 6 inches and period 1-5 seconds. Find its velocities and accelerations OT second, O'2 second, and 0*5 second after it has left one extremity of its path. go Mechanics for Engineers 2. A weight of 10 Ibs. hangs on a spring, which stretches O'lS inch per pound of load. It is set in vibration, and its greatest acceleration whilst in motion is i6 - i feet per second per second. What is the ampli- tude of vibration ? 3. A point, A, in a machine describes a vertical circle of 3 feet diameter, making 90 rotations per minute. A portion of the machine weighing 400 Ibs. moves in a horizontal straight line, and is always a fixed distance horizontally from A, so that it has a stroke of 3 feet. Find the accele- rating forces on this portion, ( I ) at the end of its stroke ; (2) 9 inches from the end ; and (3) 0^05 second after it has left the end of its stroke. 4. A helical spring deflects \ of an inch per pound of load. How many vibrations per minute will it make if set in oscillation when carrying a load of 12 Ibs. ? 5. A weight of 20 Ibs. has a simple harmonic vibration, the period of which is 2 seconds and the amplitude 1*5 feet. Draw diagrams to stated scales showing (i) the net force acting on the weight at all points in its path ; (2) the displacement at all times during the period ; (3) the velocity at all times during the period ; (4) the force acting at all times during the period. 6. A light stiff beam deflects I'I4S inches under a load of I ton at the middle of the span. Find the period of vibration of the beam when so loaded. 7. A point moves with simple harmonic motion ; when 0*75 foot from mid -path, its velocity is 1 1 feet per second ; and when 2 feet from the centre of its path, its velocity is 3 feet per second. Find its period and its greatest acceleration. 8. How many complete oscillations per minute will be made by a pendulum 3 feet long? g= 32 '2. 9. A pendulum makes 3000 beats per hour at the equator, and 3011 per hour near the pole. Compare the value of g at the two places. CHAPTER V STA TICS CONCURRENT FORCES FRICTION 72. THE particular case of a body under the action of several forces having a resultant zero, so that the body remains at rest, is of very common occurrence, and is of sufficient importance to merit special consideration. The branch of mechanics which deals with bodies at rest is called Statics. We shall first consider the statics of a particle, i.e. a body having weight, yet of indefinitely small dimensions. Many of the conclusions reached will be applicable to small bodies in which all the forces acting may be taken without serious error as acting at the same point, or, in other words, being con- current forces. 73. Resolution and Composition of Forces in One Plane. It will be necessary to recall some of the conclusions of Art. 44, viz. that any number of concurrent forces can be replaced by their geometric sum acting at the intersection of the lines of action of the forces, or by components in two standard directions, which are for convenience almost always taken at right angles to one another. Triangle and Polygon of Forces. If several forces, say four, as in Fig. 55, act on a particle, and ab, be, cd, de be drawn in succession to represent the forces of 7, 8, 6, and 10 Ibs. respec- tively, then ae, their geometric sum (Art. 44), represents a force which will produce exactly the same effect as the four forces, i*. ae represents the resultant of the four forces. If the final point e of the polygon abcde coincides with the point 0, then the resultant ae is nil, and the four forces are in equilibrium. This proposition is called the Polygon of Forces, and may be 92 Mechanics for Engineers stated as follows : If several forces acting on a particle be represented in magnitude and direction by the sides of a closed polygon taken in order, they are in equilibrium. By a closed polygon is meant one the last side of which ends at the point FIG. 55. from which the first side started. The intersection of one side of the polygon with other sides is immaterial. The polygon of forces may be proved experimentally by means of a few pieces of string and weights suspended over almost frictionless pulleys, or by a number of spring balances and cords. This proposition enables us to find one force out of several keeping a body in equilibrium if the remainder are known, viz. by drawing to scale an open polygon of vectors corresponding to the known forces, and then a line joining its extremities is the vector representing in one direction the resultant of the other forces or in the other direction the remaining force neces- sary to maintain equilibrium, sometimes called the cqvilibrant. For example, if forces Q, R, S, and T (Fig. 56) of given magnitudes, and one other force keep a particle P in equili- brium, we can find the remaining one as follows. Set out vectors ab, be, cd, and de in succession to represent Q, R, S, and T respectively ; then ae represents their resultant in magnitude and direction, and ea represents in magnitude and direction the remaining force which would keep the particle P in equilibrium, or the equilibrant. Statics Concurrent Forces Friction 93 Similarly, if all the forces keeping a body in equilibrium except two are known, and the directions of these two are known, their magnitudes may be found by completing the ----'e FIG. 56. open vector polygon by two intersecting sides in the given directions. In the particular case of three forces keeping a body in equilibrium, the polygon is a triangle, which is called the Triangle of Forces. Any triangle having its sides respectively parallel to three forces which keep a particle in equilibrium represents by its sides the respective forces, for a three-sided closed vector polygon (i.e. a triangle) with its sides parallel and proportional to the forces can always be drawn as directed for the polygon of forces, and any other triangle with its sides parallel to those of this vector triangle has its sides also pro- portional to them, since all triangles with sides respectively parallel are similar. The corresponding proposition as to any polygon with sides parallel to the respective forces is not true for any number of forces but three. 74. Lami's Theorem. If three forces keep a particle in equilibrium, each is proportional to the sine of the angle between the other two. Let P, Q, and R (Fig. 57) be the three forces in equilibrium acting at O in the lines OP, OQ, and OR respectively. Draw any three non-concurrent lines parallel respectively to OP, OQ, and OR, forming a triangle abc such that ab is parallel to OP, be to OQ, and ca to OR. Then angle abc = 180 - POQ, angle 94 Mechanics for Engineers bca = 1 80 QOR, and angle cab = 180 ROP, and there- fore A A. sin abc = sin POQ sin b'ca = sin QOR sin cab = sin ROP In the last article, it was shown that any triangle, such as P Q R ab be ca , ab be ca * . .^ . A _ sin oca sin cab ab be sin abc ca FIG. 57- abc, having sides respectively parallel to OP, OQ, and OR, has its sides proportional respectively to P, Q, and R, or (I) " sin QOR sin ROP sin POQ and multiplying equation (i) by equation (2) P Q R sin QOR "~ sin ROP ~ sin POQ that is, each of the forces P, Q, and R is proportional to the sine of the angle between the other two. This result is sometimes of use in solving problems in which three forces are in equilibrium. 75. Analytical Methods. Resultant or equilibrant forces of a system, being representable by vectors, may be found by the rules used for resultant velocities, i.e. (i) by drawing Statics Concurrent Forces Friction 95 vectors to scale ; (2) by the rules of trigonometry for the solu- tions of triangles ; (3) by resolution into components in two standard directions and subsequent compounding as in Art. 25. We now proceed to the second and third methods. To compound two forces P and Q inclined at an angle 6 to each other. Referring to the vector diagram abc of Fig. 58 (which need I (Q) not be drawn, and is used here for the purpose of illustration and explanation) by the rules of trigonometry for the solution of triangles (ac)* = (ab)* + (be)* - 2 ab.bccos abc = (ab)* + (be)* + 2 ab.be cos Q hence if ab and be represent P and Q respectively, and R is the value of their resultant R 2 = P 2 + Q 2 + 2PQ cos Q from which R may be found by extracting the square root, and its inclination to, say, the direction of Q may be found by considering the length of the perpendicular ce from c on ad produced Since ec = dc sin Q and de = dc cos & ec dc sin 6 P sin 6 ~ ae ~ ad -+ dc cos 6 ~ Q -f P cos which is the tangent of the angle between the line of action of ie resultant R and that of the force Q. When the resultant or equilibrant of more than two concurrent forces is to be found, the method of Art. 25 is g6 Mechanics for Engineers sometimes convenient. Suppose, say, three forces F 15 F 2) and F 3 make angles a, /3, and y respectively with some chosen fixed direction OX, say that of the line of action of Fj, so that a = o (Fig. 59). Resolve F 1} F 2 , and F 3 along OX and along OY perpen- dicular to OX. Let F x be the totalof the components along OX, and let F Y OY. Let R be the resultant force, and 6 its inclination to OX ; then F x = Fj + F 2 cos |8 + F 3 cos y F Y = o + F 2 sin /3 + F 3 sin y and compounding F x and F Y , two forces at right angles, R is proportional to the hypotenuse of a right-angled triangle, the other sides of which are' proportional to F x and F Y ; hence R2 = F x 2 + F Y 2 and R = The direction of the resultant R is given by the relation If the forces of the system are in equilibrium, that is, if the resultant is nil R 2 = o or F x 2 + F Y 2 = o This is only possible if both F x = o and F Y = o. Statics Concurrent Forces Friction 97 The condition of equilibrium, then, is, that the components in each of two directions at right angles shall be zero. This corresponds to the former statement, that if the forces are in equilibrium, the vector polygon of forces shall be closed, as will be seen by projecting on any two fixed directions at right angles, the sides of the closed polygon, taking account of the signs of the projections. The converse statement is true, for if F x = o and F Y = o, then R = o ; therefore, if the com- ponents in each of two standard directions are zero, then the forces form a system in equilibrium, corresponding to the statement that if the vector polygon is a closed figure, the forces represented by its sides are in equilibrium. Example i. A pole rests vertically with its base on the ground, and is held in position by five ropes, all in the same horizontal plane and drawn tight. From the pole the first rope runs due north, the second 75 west of north, the third 15 south of west, and the fourth 30 east of south. The tensions of these four are 25 Ibs., FIG. 60. 15 Ibs., 20 Ibs., and 30 Ibs. respectively. Find the direction of the fifth rope and its tension. The'directions of the rope have been set out in Fig. 60, which H 98 Mechanics for Engineers represents a plan of the arrangement, the pole being at P. The vector polygon abate, representing the forces in the order given, has been set out from a and terminates at e. ae has been drawn, and measures to scale 18-9 Ibs., and the equilibrant ea is the pull in the fifth rope, and its direction is 7 north of east from the pole. Example 2. Two forces of 3 Ibs. and 5 Ibs. respectively act on a particle, and their lines of action are inclined to each other at an angle of 70. Find what third force will keep the particle in equilibrium. The resultant force R will be of magnitude given by the relation R2 = 32 + 52 + 2 . 3 . 5 C os 70 = 9 + 25 + (30 x 03420) = 34 + 10*26 = 44*26 R = -v/44'4 2 = 6*65 Ibs. And R is inclined to the force of 5 Ibs. at an angle the tangent of which is (,. 3 sin 70 3 x 0-9397 5 + 3 cos 70 5 + (3 x 0-3420) 2-8171 = -; -r = 0*467 6-026 which is an angle 25. The equilibrant or third force required to maintain equilibrium is, therefore, one of 6*65 Ibs., and its line of action makes an angle of 180 25 or 155 FIG. 6r. with the line of action of the force of 5 Ibs., as shown in Fig. 61. Example 3. Solve Example i Ly resolving the forces into components. Taking an axis PX due east (Fig. 60) and PY due north, component force along PX F x = - 15 cos 15 20 cos 15 + 30 cos 60 = ( 35 x '9659) + (30 x 0-5) = -i8'8o6 Ibs. Component force along PY F Y = 25 + 15 cos 75 - 20 cos 75 - 30 cos 30 = 25 (5 x 0-2588) 30 x o'866o = -2-274 Iks. hence R 2 = (i8-8i) + (2-27)2 = 359*3 R = \/359'3 = 18-96 Ibs. Statics Concurrent Forces Friction 99 R acts outwards from P in a direction south of west, being inclined to XP at an acute angle, the tangent of which is Fy 2'274 F x = l8-8o6 = ' 121 which is the tangent of 6 54' ; i.e. R acts in a line lying 6 54' south of west. The equilibrant is exactly opposite to this, hence the fifth rope runs outwards from the pole P in a direction 6 54' north of east, and has a tension of i8'g6 Ibs. EXAMPLES X. 1. A weight of 20 Ibs. is supported by two strings inclined 30 and 45 respectively to the horizontal. Find by graphical construction the tension in each cord. 2. A small ring is situated at the centre of a hexagon, and is supported by six strings drawn tight, all in the same plane and radiating from the centre of the ring, and each fastened to a different angular point of the hexagon. The tensions in four consecutive strings are 2, 7> 9> and 6 Ibs. respectively. Find the tension in the two remaining strings. 3. Five bars of a steel roof-frame, all in one plane, meet at a point ; one is a horizontal tie-bar carrying a tension of 40 tons ; the next is also a tie-bar inclined 60 to the horizontal and sustaining a pull of 30 tons ; the next (in continuous order) is vertical, and runs upward from the joint, and carries a thrust of 5 tons ; and the remaining two in the same order radiat-2 at angles of 135 and 210 to the first bar. Find the stresses in the last two bars, and state whether they are in tension or compression, i.e. whether they pull or push at the common joint. 4. A telegraph pole assumed to have no force bending it out of the vertical has four sets of horizontal wires radiating from it, viz. one due east, one north-east, one 30 north of west, and one other. The tensions of the first three sets amount to 400 Ibs., 500 Ibs., and 250 Ibs. respectively. Find, by resolving the forces north and east, the direction of the fourth set and the total tension in it. 5. A wheel has five equally spaced radial spokes, all in tension. If the tensions of three consecutive spokes are 2000 Ibs., 2800 Ibs., and 2400 Ibs. respectively, find the tensions in the other two. 6. Three ropes, all in the same vertical plane, meet at a point, and there support a block of stone. They are inclined at angles of 40, 120, and 1 60 to a horizontal line in their common plane. The pulls in the first two ropes are 150 Ibs. and 120 Ibs. respectively. Find the weight of the block of stone and the tension in the third rope. 76. Friction. Friction is the name given to that pro- perty of two bodies in contact, by virtue of which a resistance ioo Mechanics for Engineers is offered to any sliding motion between them. The resistance consists of a force tangential to the surface of each body at the place of contact, and it acts on each body in such a direction as to oppose relative motion. As many bodies in equilibrium are held in their positions partly by frictional forces, it will be convenient to consider here some of the laws of friction. 77. The laws governing the friction of bodies at rest are found by experiment to be as follows : (1) The force of friction always acts in the direction opposite to that in which motion would take place if it were absent, and adjrists itself to the amount necessary to maintain equilibrium. There is, however, a limit to this adjustment and to the value which the frictional force can reach in any given case. This maximum value of the force of friction is called the limiting friction. It follows the second law, viz. (2) The limiting friction for a given pair of siirfaces defends upon the nature of the surfaces, is proportional to the normal pressure between them, and independent of the area of the sur- faces in contact. For a pair of surfaces of a given kind (i.e. particular sub- stances in a particular condition), the limiting friction F = p. . R, where R is the normal pressure between the surfaces, and p, is a constant called the coefficient of friction for the given surfaces. This second law, which is deduced from experiment, must be taken as only holding approximately. 78. Friction during Sliding Motion. If the limiting friction between the bodies is too small to prevent motion, and sliding motion begins, the subsequent value of the frictional force is somewhat less than that of the statical friction. The laws of friction of motion, so far as they have been exactly investigated, are not simple. The friction is affected by other matter (such as air), which inevitably gets between the two surfaces. However, for very low velocities of sliding and moderate normal pressure, the same relations hold approxi- mately as have been stated for the limiting friction of rest, viz. F = ^R where F is the frictional force between the two bodies, and R Statics Concurrent Forces Friction 101 is the normal pressure between them, and ju, is a constant coefficient for a given pair of surfaces, and which is less than that for statical friction between the two bodies. The friction is also independent of the velocity of rubbing. 79. Angle of Friction. Suppose a body A (Fig. 62) is in contact with a body B, and is being pulled, say, to the right, the pull increasing until the limiting amount of frictional re- sistance is reached, that is, until the force of friction reaches a limiting value F = ju,R, where R is the normal pressure between \S T & FIG 62. the two bodies, and p, is the coefficient of friction. If R and F, which are at right angles, are compounded, we get the resultant pressure, S, which B exerts on A. As the friction F increases with the pull, the inclination of the resultant S of F and R to the normal of the surface of contact, i.e. to the line of action of R, will become greater, since its tangent is always cb F equal to -7 or (Art. 75). Let the extreme inclination to the normal be X when the friction F has reached its limit, j,R. tanA = R = R = V- This extreme inclination, A, of the resultant force between IO2 Mechanics for Engineers two bodies to the normal of the common surface in contact is called the angle of friction, and we have seen that it is the angle the tangent of which is equal to the coefficient of friction tan A = //,, or A = tan" 1 p, 80. Equilibrium of a Body on an Inclined Plane. As a simple example of a frictional force, it will be instructive here to consider the equilibrium of a body resting on an inclined plane, supported wholly or in part by the friction between it and the inclined plane. Let |, be the coefficient of friction between the body of weight W and the inclined plane, and let a be the inclination of the plane to the horizontal plane. We shall in all cases draw the vector polygon of forces maintaining equilibrium, not necessarily correctly to scale, and deduce relations between the forces by the trigonometrical relations between the parts of the polygon, thus combining the advantages of vector illustra- tion with algebraic calculation, as in Art. 75. The normal to the plane is shown dotted in each diagram (Figs. 63-68 inclusive). i. Body at rest on an inclined plane (Fig. 63). a W I w FIG. 63. If the body remains at rest unaided, there are only two forces acting on it, viz. its weight, W, and the reaction S of the plane; these must then be in a straight line, and therefore S must be vertical, i.e. inclined at an angle a 1o the normal to the plane. The greatest angle which S can make to the normal Statics Concurrent Forces Friction 103 is A, the angle of friction (Art. 79) ; therefore a cannot exceed A, the angle of friction, or the body would slide down the plane. Thus we might also define the angle of friction between a pair of bodies as the greatest incline on which one body would remain on the other without sliding. Proceeding to supported bodies, let an external force, P, which we will call the effort, act upon the body in stated directions. 2. Horizontal effort necessary to start the body up the plane. Fig. 64 shows the forces acting, and a triangle of forces, abc. a, FIG. 64. When the limit of equilibrium is reached, and the body is about to slide up the plane, the angle dbc will be equal to X, the maximum angle which S can make with the normal to the plane; then or P = W tan (a. + A) which is the horizontal effort necessary to start the body up the plane. 3. Horizontal effort necessary to start the body sliding down the plane (Fig. 65). When the body is about to move down the plane, the angle cbd will be equal to the angle of friction, A ; then P ca .-,. = , = tan (A a ) W ab or P = W tan (A - a) 104 Mechanics for Engineers If a is greater than A, this can only be negative, i.e. c falls to the left of a, and the horizontal force P is that necessary to just support the body on the steep incline on which it cannot rest unsupported. 4. Effort required parallel to the plane to start the body up the plane (Fig. 66). angle acb = 90 X ab = W ca = P a, When the body is about to slide up the plane, the reaction S will make its maximum angle A (die) to the normal. P _ ac _ sin (A + a) rrii J- '* ' I Then = - = - W ab or P = W W ab sin (90 - A) sin (A + a) cos A which is the effort parallel to the plane necessary to start the body moving up the plane. 5. Effort required parallel to the plane to start the body down the plane (Fig. 67). Statics Concurrent Forces Friction 105 When the body is just about to slide down the plane, ebd=- A. i _ sin (A - 1) sin (90 sin (A a) Then P -**- sin (A -a) J.11CI1,,, , / o _ * W ab sin (90 A) or P = W' cos A which is the least force parallel to the plane necessary to start the body moving down the plane. If a is greater than A, this W angle acb = 90 \ ab= W be- S ra = P FIG. 67. force, P, can only be negative, i.e. c falls between a and d, and the force is then that parallel to the plane necessary to just support the body from sliding down the steep incline. 6. Least force necessary to start the body up the incline. Draw ab (Fig. 68) to represent W, and a vector, be, of indefinite length to represent S inclined A to the normal. Then the vector joining a to the line be is least when it is perpendicular to be. Then P is least when its line of action is perpendicular to that of S ; that is, when it is inclined 90 A to the normal, or A to the plane ; and then P ca , w%7 =sm( a + A) Note that when a = o, P = W sin A io6 Mechanics for Engineers which is the least force required to draw a body along the level. = S 7. Similarly, the least force necessary to start the body down a plane inclined a to the horizontal is P = W sin (A - ) if A is greater than a. If a is greater than A, P is negative, and P is the least force which will support the body on the steep incline. In either case, P is inclined 90 A to the normal or A to the plane. 8. Effort required in any assigned direction to start the body up the plane. Let be the assigned angle which the effort P makes with the horizontal (Fig. 69). angle bac = 90 6 cub = 90 - a - \ + FIG. 69. Statics Concurrent Forces Friction 107 TKr, P - ca s ' n ( A + <*) _ s i n (* + ) i nen . ; ^- . - w <* sin acb cos w (a + A)} or P - W cos [0 - (A + a)} which is the effort necessary to start the body up the plane in the given direction. 9. The effort in any assigned direction necessary to pull the body down the plane may be similarly found, the resultant force S between the body and plane acting in this case at an angle A to the normal, but on the opposite side from that on which it acts in case 8. 81. Action of Brake-blocks : Adhesion. A machine or vehicle is often brought to rest by opposing its motion by a frictional force at or near the circumference of a wheel or a drum attached to the wheel. A block is pressed against the rotating surface, and the frictional force tangential to the direction of rotation does work in opposing the motion. The amount of work done at the brake is equal to the diminution of kinetic energy, and this fact gives a convenient method of making calculations on the retarding force. The force is not generally confined to what would usually be called friction, as frequently considerable abrasion of the surface takes place, and the blocks wear away. It is usual to make the block of a material which will wear more rapidly than the wheel or drum on which it rubs, as it is much more easily renewed. If the brake is pressed with sufficient force, or the coefficient of "brake friction" between the block and the wheel is sufficiently high, the wheel of a vehicle may cease to rotate, and begin to slide or skid along the track. This limits the useful retarding force of a brake to that of the sliding friction between the wheels to which the brake is applied and the track, a quantity which may be increased by increasing the proportion of weight on the wheels to which brakes are applied. The coefficient of sliding friction between the wheels and the track is sometimes called the adhesion, or coefficient of adhesion. 82. Work spent in Friction. If the motion of a body is opposed by a frictional force, the amount of work done io8 Mechanics for Engineers against friction in foot-pounds is equal to the force in pounds tangential to the direction of motion, multiplied by the distance in feet through which the body moves at the point of applica- tion of the force. If the frictional force is applied at the circumference of a cylinder, as in the case of a brake band or that of a shaft or journal revolving in a bearing, the force is not all in the same line of action, but is everywhere tangential to the rotating cylinder, and it is convenient to add the forces together arith- metically and consider them as one force acting tangentially to the cylinder in any position, opposing its motion. If the cylinder makes N rotations per minute, and is R feet radius, and the tangential frictional force at the circumference of the cylinder is F Ibs., then the work done in one rotation is 2?rR . F foot-lbs., and the work done per minute is 27rRF . N foot-lbs., 2?rR . F . N , and the power absorbed is - horse-power (Art. 155). 33,000 In the case of a cylindrical journal bearing carrying a resultant load W Ibs., F = ^W, where p is the coefficient of friction between the cylinder and its bearing. 83. Friction and Efficiency of a Screw. The screw is a simple application of the inclined plane, the thread on FIG. 70. either the screw or its socket (or nut) fulfilling the same functions as a plane of the same slope. For simplicity a square-threaded screw (Fig. 70) in a vertical position is considered, the diameter Statics Concurrent Forces Friction 109 d inches being reckoned as twice the mean distance of the thread from the axis. Let/ = the pitch or axial distance, say in inches, from any point on the thread to the next corresponding point, so that when the screw is turned through one complete rotation in its fixed socket it rises / inches. Then the tangent of the angle of slope of the screw thread at its mean distance is ., which corresponds to tan a in Art. 80. Hence, if a tangential hori- zontal effort P Ibs. be applied to the screw at its mean diameter in order to raise a weight W Ibs. resting on the top of the screw where tan A = ju, (Art. 80 (2)) ; or, expanding tan ( + A) P _ tan a + tan A _ ird p + [tsird W i tan a tan A p.p ird p,p ird which has the value , or tan for a frictionless screw. Again, the work spent per turn of the screw is P x vd = W(tan a + A) . ird inch-lbs. The useful work done is W . p inch-lbs. ; therefore the work lost in friction is W tan (a + X)ird W/ foot-lbs., an expression which may be put in various forms by expansion and substitu- tion. The " efficiency " or proportion of useful work done to the total expenditure of work is Vffl tan a W tan (a + X)-rrd tan (a -f A) which may also be expressed in terms of /, d, and /x. The W quantity -p is called the mechanical advantage ; it is the ratio of the load to the effort exerted, and is a function of the IIO Mechanics for Engineers dimensions and the friction which usually differs with different loads. 84. Friction of Machines. Friction is exerted at all parts of a machine at which there is relative tangential motion of the parts. It is found by experiment that its total effects are such that the relation between the load and the effort, between the load and the friction, and between the load and the efficiency generally follow remarkably simple laws between reasonable limits. The subject is too complex for wholly theoretical treatment, and is best treated experimentally. It is an important branch of practical mechanics. Example i, A block of wood weighing 12 Ibs. is just pulled along over a horizontal iron track by a horizontal force of 3^ Ibs. Find the coefficient of friction between the wood and the iron. How much force would be required to drag the block horizontally if the force be inclined upwards at an angle of 30 to the horizontal ? If /* = the coefficient of friction ju x 12 = 3^ Ibs. /x = = O'2QI 12 Let P = force required at 30 inclination ; S = resultant force between the block and the iron track. ttlbs 12U>s FIG. 71. abc (Fig. 71) shows the triangle of forces when the block just reaches limiting equilibrium. In this triangle, cab = 60, since P is inclined 30 to the horizontal ; and Status Concurrent Forces Friction in hence sin tan abc = /* = 0-291 or -fa i cot 2 7 = = sin A and cos x = f * P__ <** . 12 rt sin sin sin A. sin \ sin (A. + 60) i sin x + -- cos A 2 7x2 -= = 0-289 7 + 24^/3 P = 1 2 x 0*289 = 3-46 Ibs. Or thus- Normal pressure between block > _ . and track / horizontal pull P cos 30 = /*(i2 P sin 30) p ( + &="*** \ 2 4o/ hence P = 3-46 Ibs. Example 2. A train, the weight of which, including locomotive, is 1 20 tons, is required to accelerate to 40 miles per hour from rest in 50 seconds. If the coefficient of adhesion is i, find the necessary weight on the driving wheels. In what time could the train be brought to rest from this speed, (i) with continuous brakes (i.e. on every wheel on the train) ; (2) with brakes on the driving-wheels only ? The acceleration is f X 88 X ^ = 1*173 ^ eet P er sec - P er sec - i **o The accelerating force is 1-173 x ~~r~ = 4*37 tons The greatest accelerating force obtainable without causing the driving-wheels to slip is \ of the weight on the wheels, therefore the minimum weight required on the driving-wheels is 7 x 4-37 = 30-6 tons. (i) The greatest retarding force with continuous brakes is 120 x 1 tons. Hence, if / = number of seconds necessary to bring the train to rest, the impulse 120 x 1 x / = x ton and second units. Hence x , the momentum in _ 7 x 3 x 32-2 = 12-75 seconds 1 1 2 Mechanics for Engineei's (2) If the brakes are on the driving-wheels only, the retarding force will be restricted to \ of 30-6 tons, i,e. to 4*37 tons, which was the accelerating force, and consequently the time required to come to rest will be the same as that required to accelerate, i.e. 50 seconds. Example 3. A square-threaded screw 2 inches mean diameter has two threads per inch of length, the coefficient of friction between the screw and nut being o - o2. Find the horizontal force applied at the circumference of the screw necessary to lift a weight of 3 tons. The pitch of the screw is \ inch. If a = angle of the screw, tan a = = 0*0794 27T and if A. = angle of friction, tan A. = o'o2 Let P = force necessary in tons. P _ , . _ tan o + tan A. _ 0*0794 + 0*02 3 "~ I tan o tan A. i 0x3794 x 0*02 = 0- hence P = 0-2987 ton EXAMPLES XI. 1. A block of iron weighing n Ibs. can be pulled along a horizontal wooden plank by a horizontal force of 1-7 Ibs. What is the coefficient of friction between the iron and the plank ? What is the greatest angle to the horizontal through which the plank can be tilted without the block of iron sliding off ? 2. What is the least force required to drag a block of stone weighing 20 Ibs. along a horizontal path, and what is its direction, the coefficient of friction between the stone and the path being 0*15 ? 3. What horizontal force is required to start a body weighing 15 Ibs. moving up a plane inclined 30 to the horizontal, the coefficient of friction between the body and the plane being 0*25 ? 4. Find the least force in magnitude and direction required to drag a log up a road inclined 1 5 to the horizontal if the coefficient of friction between the log and the road is 0*4. 5. With a coefficient friction 0*2, what must be the inclination of a plane to the horizontal if the work done by the minimum force in dragging 10 Ibs. a vertical distance of 3 feet up the plane is 60 foot Ibs. ? 6. A shaft bearing 6 inches diameter carries a dead load of 3 tons, and the shaft makes 80 rotations per minute. The coefficient of friction between the shaft and bearing is o - oi2. Find the horse-power absorbed in friction in the bearing. Statics Concurrent Forces Friction 1 1 3 7. If a brake shoe is pressed against the outside of a wheel with a force of 5 ton?, and the coefficient -of friction between the wheel and the brake is 0-3, find the horse-power absorbed by the brake if the wheel is travelling at a uniform speed of 20 miles per hour. 8. A stationary rope passes over part of the circumference of a rotating pulley, and acts as a brake upon it. The tension of the tight end of the rope is 120 Ibs., and that of the slack end 25 Ibs., the difference being due to the frictional force exerted tangentially to the pulley rim. If the pulley makes 170 rotations per minute, and is 2 feet 6 inches diameter, find the horse-power absorbed. 9. A block of iron weighing 14 Ibs. is drawn along a horizontal wooden table by a weight of 4 Ibs. hanging vertically, and connected to the block of iron by a string passing over a light pulley. If the coefficient of friction between the iron and the table is O'i5, find the acceleration of the block and the tension of the string. 10. A locomotive has a total weight of 30 tons on the driving wheels, and the coefficient of friction between the wheels and rails is O'I5. What is the greatest pull it can exert on a train 7 Assuming the engine to be sufficiently powerful to exert this pull, how long will it take the train to attain a speed of 20 miles per hour if the gross weight is 120 tons, and the resistances amount to 20 Ibs. per ton ") if. A square-threaded screw, 1*25 inches mean diameter, has five threads per inch of length. Find the force in the direction of the axis exerted by the screw when turned against a resistance, by a handle which exerts a force equivalent to 500 Ibs. at the circumference of the screw, the co- efficient of friction being o'o8. CHAPTER VI STATICS OF RIGID BODIES 85. THE previous chapter dealt with bodies of very small dimensions, or with others under such conditions that all the forces acting upon them were concurrent. In general, however, the forces keeping a rigid body in equilibrium will not have lines of action all passing through one point. Before stating the conditions of equilibrium of a rigid body, it will be necessary to consider various systems of non-concurrent forces. We shall assume that two intersecting forces may be replaced by their geometric sum acting through the point of intersection of their lines of action ; also that a force may be considered to act at any point in its line of action. Its point of application makes no difference to the equilibrium of the body, although upon it will generally depend the dis- tribution of internal forces in the body. With the internal forces or stresses in the body we are not at present concerned. 86. Composition of Parallel Forces. The following constructions are somewhat artificial, but we shall immediately from them find a simpler method of calculating the same results. To find the resultant and equilibrant of any two given like parallel forces, i.e. two acting in the same direction. Let P and Q (Fig. 7 2) be the forces of given magnitudes. Draw any line, AB, to meet the lines of action of P and Q in A and B respectively. At A and B introduce two equal and opposite forces, S, acting in the line AB, and applied one at A and the other at B. Compound S and P at A by adding the vectors Kd and de, which give a vector A^, representing Rj, the resultant Statics of Rigid Bodies 1 1 5 of S and P. Similarly, compound S and Q at B by adding the vectors B/and/i, r , which give a vector sum Eg, representing R.,, the resultant of Q and S. Produce the lines of action of RI and R 2 to meet in O, and transfer both forces to O. Now resolve Rj and R 2 at O into their components again, and we Vector de represents P. Vector^' represents Q. Vectors Kd and B/" represent equal and opposite forces S. FIG. 72. have left two equal and opposite forces, S, which have a resultant nil, and a force P + Q acting in the same direction as P and Q along OC, a line parallel to the lines of action of P and Q. If a force P + Q acts in the line CO in the opposite direction to P and Q, it balances their resultant, and therefore it will balance P and Q, i.e. it is their equilibrant. Let the line of action of the resultant P + Q cut AB in C. Since AOC and Ked are similar triangles CA OC ae ii6 Mechanics for Engineers and since BOC and B?/are similar triangles CB B/ S and dividing equation (2) by equation (i) CB_ P CA~Q or the point C divides the line AB in the inverse ratio of the magnitude of the two forces ; and similarly the line of action OC of the resultant P + Q divides any line meeting the lines of action of P and Q in the inverse ratio of the forces. To find the resultant of any two given unlike parallel forces, i.e. two acting in opposite directions. Let one of the forces, P, be greater than the other, Q (Fig. 73). By introducing equal and opposite forces, S, at A Vector de represents P. Vector fg represents Q. Vectors A</ and B/~ represent equal and opposite forces S. FIG. 73. and B, and proceeding exactly as before, we get a force P Q acting at O, its line of action cutting AB produced in C. Since AOC and hfd are similar triangles _CA_A^_S M OC ~ de ~ P (3; Statics of Rigid Bodies 117 and since BOC and Eg/ are similar triangles CB_B/_^ CQ-y^-Q ' Dividing equation (4) by equation (3) CB_ P CA~Q or the line of action of the resultant P Q divides the line AB (and any other line cutting the lines of action of P and Q) externally, in the inverse ratio of the two forces, cutting it beyond the line of the greater force. If a force of magnitude P Q acts in the line CO in the opposite direction to that of P (i.e. in the same direction as Q), it balances the resultant of P and Q, and therefore it will balance P and Q ; i.e. it is their equilibrant. This process fails if the two unlike forces are equal. The resultants Rj and R 2 are then also parallel, and the point of intersection O is non-existent. The two equal unlike parallel forces are not equivalent to, or replaceable by, any single force, but form what is called a " couple." More than two parallel forces might be compounded by successive applications of this method, first to one pair, then to the resultant and a third force, and so on. We shall, however, investigate later a simpler method of compounding several parallel forces. 87. Resolution into Parallel Components. In the last article we replaced two parallel forces, P and Q, acting at points A and B, by a single force parallel to P and Q, acting A _/_ at a point C in AB, the posi- tion of C being such that it divides AB inversely as the mag- nitudes of the forces P and Q. Similarly, a single force may be *..* FIG. 74. Resolution into two like replaced by two parallel forces parallel components, acting through any two given points. Let F (Fig. 74) be the single force, and A and B be the two given points. Join AB 1 1 8 Mechanics for Engineers and let C be the point in which AB cuts the line of action of F. If, as in Fig. 74, A and B are on opposite sides of F, then F may be replaced by parallel forces in the same direction as F, at A and B, the magnitudes of which have a sum F, and which are in the inverse ratio of their distances from C, viz. a CB AC force F X -^5 at A, and a force F X -TJ, at B. The parallel equilibrants or balancing forces of F acting at A and B are CB AC then forces F X -ps and F X -^ respectively, acting in the opposite direction to that of the force F. If A and B are on the same side of the line of action of the force F (Fig. 75), then F may be replaced by forces at A and B, 'F FIG. 75. Resolution into two unlike parallel components. the magnitudes of which have a difference F, the larger force acting through the nearer point A, and in the same direction as the force F, the smaller force acting through the further point B, and in the opposite direction to the force F, and the magnitudes being in the inverse ratio of the distances of the /'p forces from C, viz. a force F X -y-n at A, in the direction of F, AC and an opposite force F X vg at B. CB The equilibrants of F at A and B will be F X -^g in the AC opposite direction to that of F, and F X -T-JT in the direction of F, respectively. Statics of Rigid Bodies 119 B As an example of the parallel equilibrants through two points, A and B, on either side of the line of action of a force, we may take the vertical up- ward reactions at the supports of a beam due to a load con- centrated at some place on the beam. Let W Ibs. (Fig. 76) be the load at a point C on a -^ beam of span /feet, C being x feet from A, the left-hand support, and therefore / x feet from the right-hand support, B. Let R A be the supporting force or reaction at A ; R B be the supporting force or reaction at B. k JC--*i i j : FIG. 76. Then R A = W x = Ibs. More complicated examples of the same kind where there is more than one load will generally be solved by a slightly different method. 88. Moments. The moment of a force F Ibs. about a fixed point, O, was measured (Art. 56) by the product F X d Ib.-feet, where d was the perpen- dicular distance in feet from O to the line of action of F. Let ON (Fig. 77) be the perpen- dicular from O on to the line of action of a force F. Set off a vector ab on the line of action of F to represent F. Then the product ab . ON, which is twice the area of the triangle Oat>, is proportional to the moment of F about O. Some convention as to signs of clockwise and contra-clockwise moments (Art. 56) must be adopted. If the moment of F about O is contra-clockwise, i.e. if O lies to the left of the line /a, FIG. 77. 1 20 Mechanics for Engineers of action of F viewed in the direction of the force, it is usual to reckon the moment and the area Oafr representing it as positive, and if clockwise to reckon them as negative. 89. Moment of a Resultant Force. This, about any point in the plane of the resultant and its components, is equal to the algebraic sum of the moments of the components. Let O (Fig. 78) be any point in the plane of two forces, P and Q, the lines of action of which intersect at A. Draw Qd parallel to the force P, cutting the line of action of Q in c. Let FIG g the vector Ac represent the force Q, and set off Kb in the line of action of P to represent P on the same scale, P i.e. such that Kb = Kc X Q. Complete the parallelogram Kbdc. Then the vector Kd = Ac + cd Ac + Kb, and represents the resultant R, of P and Q. Now, the moment of P about O is represented by twice the area of triangle AO^ (Art. 88), and the moment of Q about O is represented by twice the area of triangle AO, and tlje moment of R about O is represented by twice the area of triangle AOrf. But the area KOd = area AcJ + area AO<r = area Kbd -f- area AOc Kbd and Acd being each half of the parallelogram Abdc \ hence area AO./ = area AOZ> + AO, since AO<5 and Abd are between the same parallels ; or twice area AQd = twice area AO + twice area AO^. and these three quantities represent respectively the moments of R, P, and Q about O. Hence the moment of R about O is equal to the sum of the moments of P and Q about that point. If O is to the right of one of the forces instead of tc the left Statics of Rigid Bodies 121 of both, as it is in Fig. 78, there will be a slight modification in sign ; e.g. if O is to the right of the line of action of Q and to the left of R and P, the area AO and the moment of Q about O will be negative, but the theorem will remain true for the algebraic sum of the moments. Next let the forces P and Q be parallel (Fig. 79). Draw a line AB from O perpendicular to the lines of action of P and Q, cutting them in A and B respectively. Then the resultant R, which is equal to P + Q, cuts AB in C such that BC _ P AC~Q' Then P . AC = Q . BC The sum of moments of P and Q about O is P . O A + Q . OB, and this is equal to P(OC - AC) + Q(OC + CB), which is squal to (P + Q)OC - P . AC + Q . CB = (P + QJOC, since P . AC = Q . CB. And (P + QJOC is the moment of the resultant R about O. Hence the moment of the resultant is equal to the sum of moments of the two component forces. The figure will need modification if the point O lies between the lines of action of P and Q, and their moments about O will be of opposite sign, but the moment of R will remain equal to the algebraic sum of those of P and Q. The same remark applies to the figure for two unlike parallel forces. The force equal and opposite to the resultant, i.e. the equilibrant, of the two forces (whether parallel or intersecting) has a moment of equal magnitude and opposite sign to that of the resultant (Art. 88), and therefore the equilibrant has a moment 122 Mechanics for Engineers about any point in the plane of the forces, of equal magnitude and of opposite sign to the moments of the forces which it balances. In other words, the algebraic sum of the moments of any two forces and their equilibrant about any point in their plane is zero. 90. Moment of Forces in Equilibrium. If several forces, all in the same plane, act upon a body, the resultant of any two has about any point O in the plane a moment equal to that of the two forces (Art. 89). Applying the same theorem to a third force and the resultant of the first two, the moment of their resultant (i.e. the resultant of the first three original forces) is equal to that of the three forces, and so on. By successive applications of the same theorem, it is obvious that the moment of the final resultant of all the forces about any point in their plane is equal to the sum of the moments of all the separate forces about that point, whether the forces be all parallel or inclined one to another. If the body is in equilibrium, the resultant force upon it in any plane is zero, and therefore the algebraic sum of the moments of all the separate forces about any point in the plane is zero. This fact gives a method of finding one or two unknown forces acting on a body in equilibrium, particularly when their lines of action are known. When more than one force is unknown, the clockwise and contra-clockwise moments about any point in the line of action of one of the unknown forces may most conveniently be dealt with, for the moment of a force about any point in its line of action is zero. The Principle of Moments, i.e. the principle of equation of the algebraic sum of moments of all forces in a plane acting on a body in equilibrium to zero, or equation of the clockwise to the contra-clockwise moments, will be most clearly under- stood from the three examples at the end of this article. Levers. A lever is a bar free to turn about one fixed point and capable of exerting some force due to the exertion of an effort on some other part of the bar. The bar may be of any shape, and the fixed point, which is called the fulcrum, may be in any position. When an effort applied to the lever is just sufficient to overcome some given opposing force, the lever has just passed a condition of equilibrium, and the relation Statics of Rigid Bodies 123 \4-tons between the effort, the force exerted by the lever, and the reaction at the fulcrum may be found by the principle of moments. Example I. A roof- frame is supported by two vertical walls 20 feet apart at points A and B on the same level. The line of the resultant load of 4 tons on the frame cuts the line AB 8 feet from A, at an angle of 75 to the horizontal, as shown in Fig. 80. The supporting force at the point B is a vertical one. Find its amount. The supporting force through the point A is unknown, but its moment about A is" zero. Hence the clockwise moment of the 4-ton resultant must balance the contra-clockwise moment of the vertical supporting force RB at B. Equating the magnitudes of the moments 4 x 8 sin 75 = 20 x R B (tons-feet) therefore R B = HJlILZ-L _ r g x 0^9659 = i'545 tons Example 2. A light horizontal beam of 12-feet span carries loads of 7 cwt., 6 cwt., and 9 cwt. at distances of i foot, 5 feet, and 10 feet respectively from the left-hand end. Find the reactions of the supports of the beam. If we take moments about the left-hand end A (Fig. 81), the FIG. 80. AC = AD - 17 'Curt/. \6curu \ C ID \f)curt-. TE B AE = 10 feet AB = 12 feet, i \ RA FIG. Si. vertical loads have a clockwise tendency, and the moment of the reaction R B at B is contra-clockwise ; hence R B x 12 = (7 x i) + (6 x 5) + (9 x 10) I2R B = 7 + 30 + 90 = 127 RB = V/- = 10*583 cwt. / /30 / B 1 24 Mechanics for Engineers R A , the supporting force at A, may be found by an equation of moments about B. Or since RB + RA = 7 + 6 + 9 = 22 cwt. RA = 22 10-583 = 11-416 CWt. Example 3. An L-shaped lever, of which the long arm is 1 8 inches long and the short one 10 inches, has its fulcrum at the right angle. The effort exerted on the end of the long arm is 20 Ibs., inclined 30 to the arm. The short arm is kept from moving by a cord attached to its end and perpendicular to its length. Find the tension of the chord. Let T be the tension of the string in pounds. Then, taking moments about B (Fig. 82), since the unknown reaction of the hinge or fulcrum has no moment FIG. 82. about that point AB sin 30 x 20 = BC x T 18 x | x 20 = 10 x T T = 18 Ibs. EXAMPLES XII. 1. A post 12 feet high stands vertically on the ground. Attached to the top is a rope, inclined downwards and making an angle of 25 with the horizontal. Find what horizontal force, applied to the post 5 feet above the ground, will be necessary to keep it upright when the rope is pulled with a force of 120 Ibs. 2. Four forces of 5, 7, 3, and 4 Ibs. act along the respective directions AB, BC, DC, and AD of a square, ABCD. Two other forces act, one in CA, and the other through D. Find their amounts if the six forces keep a body in equilibrium. 3. A beam of 15-feet span carries loads of 3 tons, \ ton, 5 tons, and I ton, at distances of 4, 6, 9 and 13 feet respectively from the left-hand end. Find the pressure on the supports at each end of the beam, which weighs | ton. 4. A beam 20 feet long rests on two supports 1 6 feet apart, and over- hangs the left-hand support 3 feet, and the right-hand support by I foot. It carries a load of 5 tons at the left-hand end of the beam, and one of 7 tons midway between the supports. The weight of the beam, which may be looked upon as a load at its centre, is I ton. Find the reactions at the Statics of Rigid Bodies 125 FIG. 83. supports, i.e. the supporting forces. What upward vertical force at the right-hand end of the beam would be necessary to tilt the beam ? 5. A straight crowbar, AB, 40 inches long, rests on a fulcrum, C, near to A, and a force of 80 Ibs. applied at B lifts a weight of 3000 Ibs. at A. Find the distance AC. 6. A beam 10 feet long rests upon supports at its ends, and carries a load of 7 cwt. 3 feet from one end. Where must a second load of 19 cwt. be placed in order that the pressures on the two supports may be equal ? 91. Couples. In Art. 86 it was stated that two equal unlike parallel forces are not replace- able by a single resultant force ; they cannot then be balanced by a single force. Such a system is called a cotiple, and the perpendicular distance between the lines of action of the two forces is called the arm of the couple. Thus, in Fig. 83, if two equal and opposite forces F Ibs. act at A and B perpen- dicular to the line AB, they form a couple, and the length AB is called the arm of the couple. 92. Moment of a Couple. This is the tendency to pro- duce rotation, and is measured by the product of one of the forces forming the couple and the arm of the couple ; e.g. if the two equal and opposite forces forming the couple are each forces of 5 Ibs., and the distance apart of their lines of action is 3 feet, the moment of the couple is 5x3, or 15 Ib.-feet ; or in Fig. 83, the moment of the couple is F x AB in suitable units. The sum of the moments of the forces of a couple is the same about any point O in their plane. Let O (Fig. 84) be any point. Draw a line OAB per- pendicular to the lines of action of the forces and meeting them in A and B. Then the total (contra-clockwise) moment of the two forces about O is F . OB - F . OA = F(OB - OA) = F . AB FIG. 84. 126 Mechanics for Engineers This is the value, already stated, of the moment of the couple, and is independent of the position of O. A couple is either of clockwise or contra-clockwise ten- dency, and its moment about any point in its plane is of the same tendency (viewed from the same aspect) and of the same magnitude. 93. Equivalent Couples. Any two couples in a plane having the same moment are equivalent if they are cf the same sign or turning tendency, i.e. either both clockwise or both contra-clockwise; or, if the couples are equal in magnitude and of opposite sign, they balance or neutralise one another. The latter form of the statement is very simply proved. Let the forces F, F (Fig. 85) constitute a contra- clockwise couple, and the forces F', F' constitute a clock-wise couple having a moment of the same magnitude. Let the lines of action of F, F and those of F', F' intersect in A, B, C, and D, and let AE be the perpen- dicular from A on BC, and CG the perpendicular from C on AB. Then, the moments of the two couples being equal F X AE = F' x GC F X AB sin ABC = F X CB sin ABC F x AB = F X CB F = C_B F' AB Hence CB and AB may, as vectors, fully represent F and F respectively, acting at B. And since ABCD is a parallelogram, CD = AB, and the resultant or vector sum of F and F is in the line DB, acting through B in the direction DB. Similarly, the forces F and F' acting at D have an equal and opposite resultant acting through D in the direction BD. These two equal and opposite forces in the line of B and D balance, hence the two couples balance. Statics of Rigid Bodies 127 F, B It has been assumed here that the lines of action of F and F' intersect ; if they do not, equal and opposite forces in the same straight line may, for the purpose of demonstration, be introduced and compounded with the forces of one couple without affecting the moment of that couple or the equilibrium of any system of which it forms a part. 94. Addition of Couples. The resultant of several couples in the same plane and of given moments is a couple the moment of which is equal to the sum of the moments of the several couples. Any couple may be replaced by its equivalent couple having an arm of length AB (Fig. 93) and forces F^ F 1} pro- vided Fj X AB = moment of the couple. Similarly, a second couple may be replaced by a couple of arm AB and forces F 2 , F 2 , provided F 2 X AB is equal to the moment of this second couple. In this way clockwise couples must be replaced by clock- wise couples of arm AB, and contra- clockwise couples by contra-clock- wise couples of arm AB, until finally we have a couple of moment (F! + F 2 + F 3 + . . . etc.)AB = Fj x AB + F 2 X AB -f F 3 X AB + . . . etc. = algebraic sum of moments of the given couples the proper sign being given to the various forces. 95. Reduction of a System of Co-planar Forces. A system of forces all in the same plane is equivalent to (i) a single resultant force, or (2) a couple, or (3) a system in equi- librium, which may be looked upon as a special case of (i), viz. a single resultant of magnitude zero. Any two forces of the system which intersect may be jplaced by a single force equal to their geometric sum acting irough the point of intersection. Continuing the same process FIG. 86. 128 Mechanics for Engineers of compounding successive forces with the resultants of others as far as possible, the system reduces to either a single re- sultant, including the case of a zero resultant, or to a number of parallel forces. In the latter case the parallel forces may be compounded by applying the rules of Art. 86, and reduced to either a single resultant (including a zero resultant) or to a couple. Finally, then, the system must reduce to (i) a single resultant, or (2) a couple, or (3) the system is in equilibrium. 96. Conditions of Equilibrium of a System of Forces in One Plane. If such a system of forces is in equi- librium, the geometric or vector sum of all the forces must be zero, or, in other words, the force polygon must be a closed one, for otherwise the resultant would be (Art. 95) a single force represented by the vector sum of the separate forces. Also, if the system is in equilibrium (i.e. has a zero re- sultant), the algebraic sum of all the moments of the forces about any point in their plane is zero (Art. 90). These are all the conditions which are necessary, as is evident from Art. 95, but they may be conveniently stated as three con- ditions, which are sufficient (i) and (2) The sum of the components in each of two directions must be zero (a single resultant has a zero component in one direction, viz. that perpendicular to its line of action). (3) The sum of the moments of all the forces about one point in the plane is zero. If conditions (i) and (2) are fulfilled the system cannot have a single resultant (Art. 75), and if condition (3) is ful- filled it cannot reduce to a couple (Art. 92), and therefore it must reduce to a zero resultant (Art. 95), i.e. the system must be in equilibrium. These three conditions are obviously necessary, and they have just been shown to be sufficient, but it should be remem- bered that the algebraic sum of the moments of all the forces about every point in the plane is zero. The above three con- ditions provide for three equations between the magnitudes of the forces of a system in equilibrium and their relative posi- tions, and from these equations three unknown quantities may be found if all other details of the system be known. Statics of Rigid Bodies 129 97. Solution of Statical Problems. In finding the forces acting upon a system of rigid bodies in equilibrium, it should be remembered that each body is in itself in equi- librium, and therefore we can obtain three relations (Art. 96) between the forces acting upon it, viz. we can write three equations by stating in algebraic form the three conditions of equilibrium ; that is, we may resolve all the forces in two directions, preferably at right angles, and equate the com- ponents in opposite directions, or equate the algebraic sums to zero, and we may equate the clockwise and contra-clockwise moments about any point, or equate the algebraic sum of moments to zero. The moment about every point in the plane of a system of co-planar forces in equilibrium is zero, and sometimes it is more convenient to consider the moments about two points and only resolve the forces in one direction, or to take moments about three points and not resolve the forces. If more than three equations are formed by taking moments about other points, they will be found to be not independent and really a repetition of the relations expressed in the three equations formed. Some directions of resolution are more convenient than others, e.g. by resolving perpendicular to some unknown force, no component of that force enters into the equation so formed. Again, an unknown force may be elimi- nated in an equation of moments by taking the moments about some point in its line of action, about which it will have a zero moment. "Smooth " Bodies. An absolutely smooth body would be one the reaction of which, on any body pressing against it, would have no frictional component, i.e. would be normal to the surface of contact, the angle of friction (Art. 79) being zero. No actual body would fulfil such a condition, but it often happens that a body is so smooth that any frictional force it may exert upon a second body is so small in comparison with other forces acting upon that body as to be quite negli- gible, e.g. if a ladder with one end on a rough floor rest against a horizontal round steel shaft, such as is used to transmit power in workshops, the reaction of the shaft on the ladder might K 130 Mechanics for Engineers without serious error be considered perpendicular to the length of the ladder, i.e. normal to the cylindrical surface of the shaft. Example i. A horizontal rod 3 feet long has a hole in one end, A, through which a horizontal pin passes forming a hinge. The other end, B, rests on a smooth roller at the same level. Forces of 7, 9, and 5 Ibs. act upon the rod, their lines of action, which are in the same vertical plane, intersecting it at distances of 1 1, 16, and 27 inches respectively from A, and making acute angles of 30, 75, and 45 respectively with AB, the first two sloping downwards towards A, and the third sloping downwards towards B, as shown in Fig. 87. Find the magnitude of the supporting forces on the rod at A and B. 9 Ibs. FIG. 87. Since the end B rests on a smooth roller, the reaction RB at B is perpendicular to the rod (Art. 97). We can conveniently find this reaction at B by taking moments about A, to which the unknown supporting force at A contributes nothing. The total clockwise moment about A in Ib.-inches is 7 * = 270^2 Ib.-inches The total contra-clockwise moment about A is R B x 36. Equating the moments of opposite sign RB x 36 = 270*2 Ib.-inches Statics of Rigid Bodies 131 The remaining force R A through A may be found by drawing to scale an open vector polygon with sides representing the forces 7, 9, 5, and 7-5 Ibs. (Rs); the closing side then represents R A . Or we may find R A by resolving all the forces, say, horizontally and vertically. Let H A be the horizontal component of R A estimated positively to the right, and V A its vertical component upwards. Then, by Art. 96, the total horizontal component of all the forces is zero ; hence H A 7 cos 30 9 cos 75 + 5 cos 45 = o HA = 7 x o'866 + 9 x 0*259 5 x 0707 = 4-85 Ibs. Also the total vertical component is zero, hence V A 7 sin 30 - 9 sin 75 - 5 sin 45 + 7-5 = VA = 7 x + 9 x 0-966 + 5 x 0-707 - 7*5 = 8-23 Ibs. Compounding these two rectangular components of RA RA = A/KfSs) 2 + (8'2 3 ) 2 } (Art. 75) RA = V9 1 ' 2 = 9'54 Ibs. Example 2. ABCD is a square, each side being 17-8 inches, and E is the middle point of AB. Forces of 7, 8, 12, 5, 9, and 6 Ibs. act on a body in the lines and directions AB, EC, BC, BD, 8 FIG. CA, and DE respectively. Find the magnitude, and position with respect to ABCD, of the single force required to keep the body in equilibrium. 132 Mechanics for Engineers Let F be the required force ; H A be the component of F in the direction AD ; V A be the component of F in the direction AB ; P be the perpendicular distance in inches of the force from A. Then, resolving in direction AD, the algebraic total component being zero H A + 8 cos OEC +12 + 5 cos 45 - 9 cos 45 \ - 6 cos EDA / ~ H A + 8 x -4j= + 12 - 4 x -7= - 6 x -= = o Vs V2 Vs H A + (2 x 0-895) + 12 - 4 x 0707 = o H A = 10-96 Ibs. Resolving in direction AB V A + 7 + 8 cos EEC - 5 cos 45 9 cos 45 \ _ + 6 cos AED / = V A + 7 + 14 x -*= - 14 x -^ . = o V5 V 2 V A = -7 - 6-26 + 9-90 = -3-36 then F = J{( 10-96)2 + (3-36)2 } = 11-46 Ibs. and is inclined to AD at an angle the tangent of which is -3-36 - 10-96 = 0-3065 i.e. at an angle 180 + 17 or 197. Its position remains to be found. We may take moments about any point, say A. Let p be reckoned positive if F has a contra- clockwise moment about A. 1 1*46 x p + 6 x AD sin ADE - 5 x OA - 12 ) _ x AB - 8 x AE sin EEC / " 89 , , - + 213-6 v/5 292-3 p = i = 25-51 inches 11-46 This completes the specification of the force F, which makes an angle 197 with AD and passes 25-51 inches from A, so as to have a contra-clockwise moment about A. The position of F is shown in Fig. 89. Statics of Rigid Bodies 133 The force might be specified as making 197 with AD and cutting it at a distance 25'5i -4- sin 197 or 86'5 inches from A ; t\e. 86-5 inches to the left of A. 197' FIG. 89. 98. Method of Sections. The principles of the pre- ceding article may be applied to find the forces acting in the members of a structure consisting of separate pieces jointed together. If the structure be divided by an imaginary plane of section into two parts, either part may be looked upon as a body in equi- librium under certain forces, some of which are the forces exerted by members cut by the plane of section. For example, if a hinged frame such as ABCDE (Fig. 90) is in Fic - equilibrium under given forces at A, B, C, D, and E, and an imaginary plane of section XX' perpendicular to the plane of the structure be taken, then the portion kBzyw is in equilibrium 134 Mechanics for Engineers H 12 FIG. 91. under the forces at A and B, and the forces exerted upon it by the remaining part of -the structure, viz. the forces in the bars BD, BC, and AC. This method of sections is often the simplest way of finding the forces in the members of a jointed structure. Example. One end of a girder made up of bars jointed together is shown in Fig. 91. Vertical loads of 3 tons and 5 tons are carried at B and C respectively, and the vertical supporting force at H is 12 tons. ;' The sloping bars are inclined at 60 to the horizontal. Find the forces in the bars CD, CE, and FE. The portion of the girder ACFH cut off by the vertical plane klm is in equili- brium under the action of the loads at B and C, the supporting force at H, and the forces exerted by the bars CD, CE, and FE on the joints at C and F. Resolving these forces vertically, the forces in CD and FE have no vertical component, hence the downward vertical component force exerted by CE on the left- hand end of the girder is equal to the excess upward force of the remaining three, i.e. 12 - 3 - 5 = 4 tons; hence Force in CE x cos 30 = 4 tons 2 or force in CE = 4 x -= = 4"6l tons V3 This, being positive, acts downwards on the left-hand end, i.e. it acts towards E, or the bar C E pulls at the joint C, hence the bar CE is in tension to the amount of 4'6i tons. To find the force in bar FE, take a vertical section plane through C or indefinitely near to C, and just on the right hand of it. Then, taking moments about C and reckoning clockwise moments positive 12 x AC - 3 x BC + A /3 x FE x (force in FE) = o 12x2 3x1 + ^3 x (force in FE) = o 2 1 and force in FE = = = 12*12 tons V3 The negative sign indicates that the force in FE acts on F in the opposite direction to that in which it would have a clockwise moment about C, i.e. the force pulls at the joint F ; hence the member is in tension to the extent of 12' 12 tons. Statics of Rigid Bodies 135 Similarly, taking say clockwise moments about E, the force in CD is found to be a push of 14*42 tons towards C, i.e. CD has a compressive force of 14*42 tons in it, as follows : 12x3-3x2-5x1 + ^3 (force in CD) = o force in CD = 14*42 99. Rigid Body kept in Equilibrium by Three Forces. If three forces keep a body in equilibrium, they either all pass through one point (i.e. are concurrent) or are all parallel. For unless all three forces are parallel two must intersect, and these are replaceable by a single resultant acting through their point of intersection. This resultant cannot balance the third force unless they are equal and opposite and in the same straight line, in which case the third force passes through the intersection of the other two, and the three forces are concurrent. The fact of either parallelism or concurrence of the three forces simplifies problems on equilibrium under three forces by fixing the position of an unknown force, since its line of action intersects those of the other two forces at their intersection. The magnitude of the forces can be found by a triangle of forces, or by the method of resolution into rectangular com- ponents. Statical problems can generally be solved in various ways, some being best solved by one method, and others by different methods. In the following example four methods of solution are indicated, three of which depend directly upon the fact that the three forces are concurrent, which gives a simple method of determining the direction of the reaction of the rough ground. Example i. A ladder 18 feet long rests with its upper end against a smooth vertical wall, and its lower end on rough ground 7 feet from the foot of the wall. The weight of the ladder is 40 Ibs., which may be looked upon as a vertical force halfway along the length of the ladder. Find the magnitude and direction of the forces exerted by the wall and the ground on the ladder. The weight of 40 Ibs. acts vertically through C (Fig. 92), and the reaction of the wall Fj is perpendicular to the wall (Art. 97). These two forces intersect at D. The only remaining force, F 2 , on 136 Mechanics for Engineers the ladder is the pressure which the ground exerts on it at B. This must act through D also (Art. 99), and therefore its line of action must be BD. F l may be found by an equation of the moments about B. Ft x AE = 40 x |BE. - 7 2 ) = 40 x | V(275) = 8-44 Ibs. And since F 2 balances the horizontal force of 8'44 Ibs. and a vertical force of 40 Ibs. F 2 = V{(8'44) 2 + 4Q 2 } = 40-8 Ibs. and is inclined to EB at an angle ^ EBD, the tangent of which is AE _ 2 x ^(275) _ .._. ~ - which is the tangent of 78-1. A second method of solving the problem consists in drawing a vector triangle, abc (Fig. 92), representing by its vector sides Fj, F 2 , and 40 Ibs. The 4o-lb. force ab being set off to scale, and be and ca being drawn parallel to F 2 and F l respectively, and the magnitudes then measured to the same scale. A third method consists (without drawing to scale) of solving the triangle abc trigonometrically, thus Fj : F 2 : 40 = ca : cb '. ab =HB:BD:HD from which Fj and F 2 may be easily calculated, viz. 7 -^ 8-44 ibs. 2 X F 2 = 40 x = 40-8 Ibs. \/275 Fourthly, the problem might be solved very simply by resolving the forces F t and F 2 and 40 Ibs. horizontally and vertically, as in Statics of Rigid Bodies 137 this particular case the 4o-lb. weight has no component in the direction of F,, and must exactly equal in magnitude the vertical component of F 2 ; the horizontal component of F 2 must also be just equal to the magnitude of Fj. Example 2. A light bar, AB, 20 inches long, is hinged at A 50 as to be free to move in a vertical plane. The end B is sup- ported by a cord, BC, so placed that the angle ABC is 145 and AB is horizontal. A weight of 7 Ibs. is hung on the bar at a point D in AB 13 inches from A. Find the tension in the cord ind the pressure of the rod on the hinge. FIG. 93. Let T be the tension in the cord, and P be the pressure on the linge. Taking moments about A, through which P passes (Fig. 93) T x AF = 7 x AD T x 20 sin 35 = 7 x 13 U-47T = 91 T = 7-94 Ibs. The remaining force on the bar is the reaction of the hinge, vhich is equal and opposite to the pressure P of the bar on the linge. The vertical upward component of this is 7 T sin 35 = 2*45 Ibs., and the horizontal component is T cos 35 = 6-5 Ibs. Hence P = V(6'5) 2 + (2 45) 2 = 6-93 Ibs. The tangent of the angle DAE is -~ = 0^377, corresponding o an angle of 20 40'. The pressure of the bar on the hinge is then 6 93 Ibs. in a 138 Mechanics for Engineers direction, AE, inclined downwards to the bar and making an angle 20 40' with its length. EXAMPLES XIII. 1. A trap door 3 feet square is held at an inclination of 30 to (and above) the horizontal plane through its hinges by a cord attached to the middle of the side opposite the hinges. The other end of the cord, which is 5 feet long, is attached to a hook vertically above the middle point of the hinged side of the door. Find the tension in the cord, and the direction and magnitude of the pressure between the door and its hinges, the weight of the door being 50 Ibs., which may be taken as acting at the centre of the door. 2. A ladder 20 feet long rests on rough ground, leaning against a rough vertical wall, and makes an angle of 60 to the horizontal. The weight of the ladder is 60 Ibs. , and this may be taken as acting at a point 9 feet from the lower end. The coefficient of friction between the ladder and ground is o'25. If the ladder is just about to slip downwards, find the coefficient of friction between it and the wall. 3. A ladder, the weight of which may be taken as acting at its centre, rests against a vertical wall with its lower end on the ground. The coefficient of friction between the ladder and the ground is , and that between the ladder and the wall \. What is the greatest angle to the vertical at which the ladder will rest ? 4. A rod 3 feet long is hinged by a horizontal pin at one end, and supported on a horizontal roller at the other. A force of 20 Ibs. inclined 45 to the rod acts upon it at a point 21 inches from the hinged end. Find the amount of the reactions on the rod at the hinge and at the free end. 5. A triangular roof-frame ABC has a horizontal span AC of 40 feet, and the angle at the apex B is 120, AB and BC being of equal length. The roof is hinged at A, and simply supported on rollers at C. The loads it bears are as follow : (l) A force of 4000 Ibs. midway along and perpen- dicular to AB ; (2) a vertical load of 1500 Ibs. at B ; and (3) a vertical load of 1400 Ibs. midway between B and C. Find the reactions or supporting forces on the roof at A and C. 6. Draw a 2-inch square ABCD, and find the middle point E of AB. Forces of 17, 10, 8, 7, and 20 Ibs. act in the directions CB, AB, EC, ED, and BD respectively. Find the magnitude, direction, and position of the force required to balance these. Where does it cut the line AD, and what angle does it make with the direction AD ? 7. A triangular roof-frame ABC has a span AC of 30 feet. AB is 15 feet, and BC is 24 feet. A force of 2 tons acts normally to AB at its middle point, and another force of I ton, perpendicular to AB, acts at B. There is also a vertical load of 5 tons acting downward at B. If the sup- porting force at A is a vertical one, find its magnitude and the magnitude and direction of the supporting force at C. Statics of Rigid Bodies 139 8. A jointed roof-frame, ABCDE, is shown in Fig. 94. AB and BC are inclined to the horizontal at 30, EB and DB are inclined at 45 to the FIG. 94. horizontal. The span AC is 40 feet, and B is 10 feet vertically above ED. Vertical downward loads of 2 tons each are carried at B, at E, and at D. Find by the method of sections the forces in the members AB, EB, and ED. 9. A jointed structure, ACD . . . LMB (Fig. 95) is built up of bars all FIG. 95. of equal length, and carries loads of 7, 10, and 15 tons at D, F and L re- spectively. Find by the method of sections the forces on the bars EF, EG, and DF. CHAPTER VII CENTRE OF INERTIA OR MASS CENTRE OF GRA VITY 100. Centre of a System of Parallel Forces. Let A, B, C, D, E, etc. (Fig. 96), be points at which parallel forces FI> F 2 , F 3 , F 4 , F 5 , etc., respectively act. The position of the resultant force may be found by applying successively the rule FIG. 96. of Art. 86. Thus Fj and F 2 may be replaced by a force AX F FJ + F 2 , at a point X in AB such that ^^ = =? /Art. 86). Ar> I 1 ! This force acting at X, and the force F 3 acting at C, may be replaced by a force F! + F 2 + F 3 at a point Y in CX such XY F Centre of Inertia or Mass Centre of Gravity 141 Proceeding in this way to combine the resultant of several forces with one more force, the whole system may be replaced by a force equal to the algebraic sum of the several forces acting at some point G. It may be noticed that the positions of the points X, Y, Z, and G depend only upon the positions of the points of application A, B, C, D, and E of the several forces and the magnitude of the forces, and are independent of the directions of the forces provided they are parallel. The point of application G of the resultant is called the centre of the parallel forces F 1} F 2 , F 3 , F 4 , and F 5 acting through A, B, C, D, and E respectively, whatever direction those parallel forces may have. 101. Centre of Mass. If every particle of matter in a body be acted upon by a force proportional to its mass, and all the forces be parallel, the centre of such a system of forces (Art. 100) is called the centre of mass or centre of inertia of the body. It is quite independent of the direction of the parallel forces, as we have seen in Art. 100. Centre of Gravity. The attraction which the earth exerts upon every particle of a body is directed towards the centre of the earth, and in bodies of sizes which are small compared to that of the earth, these forces may be looked upon as parallel forces. Hence these gravitational forces have a centre, and this is called the centre of gravity of the body ; it is, of course, the same point as the centre of mass. The resultant of the gravitational forces on all the particles of a body is called its weight, and in the case of rigid bodies it acts through the point G, the centre of gravity, whatever the position of the body. A change of position of the body is equivalent to a change in direction of the parallel gravitational forces on its parts, and we have seen (Art. 100) that the centre of such a system of forces is independent of their direction. We now proceed to find the centres of gravity in a number of special cases. 1 02. Centre of gravity of two particles of given weights at a given distance apart, or of two bodies the centres of gravity and weights of which are given. Let A and B (Fig. 97) be the positions of the two particles 142 Mechanics for Engineers (or centres of gravity of two bodies) of weights w^ and w 2 A G B FIG. 97. respectively. The centre of gravity G is (Art. 86) in AB at such a point that GA w or GA = and GB = . AB .AB In the case of two equal weights, AG = GB = AB. A convenient method of finding the point G graphically may be noticed. Set off from A (Fig. 98) a line AC, making any angle with AB (preferably at right angles), and proportional to w z to any scale; from B set off a line BD parallel to AC on the opposite side of AB, and proportional to w l to the same scale that AC represents w z . Join CD. Then the intersection of CD with AB determines the point G. The proof follows simply from the similarity of the triangles ACG and BDG. 103. Uniform Straight Thin Rod. Let AB (Fig. 99) be the uniform straight rod of length AB : it may be supposed to be divided into pairs of particles of equal weight situated at equal distances from the middle G b D point G of the rod, since there will be as many such particles between A and G as between G <B FIG. 99. and B. The e.g. (centre of gravity) of each pair, such as the particles at a and l>, is midway between them (Art. 102), viz. at the middle point of the rod, G, hence the e.g. of the whole rod is at its middle point, G. Centre of Inertia or Mass Centre of Gravity 143 B 1 FIG, 100. 104. Uniform Triangular Plate or Lamina. The term centre of gravity of an area is often used to denote the e.g. of a thin lamina of uniform material cut in the shape of the particular area concerned. We may suppose the lamina ABC (Fig. 100) divided into an indefinitely large number of strips parallel to the base AC. The e.g. of each strip, such as PQ, is at its middle point (Art. 103), and every e.g. is therefore in the median BB,' i.e. the line joining B to the mid-point B' of the base AC. Hence the e.g. of the whole triangular lamina is in the median BB'. Similarly, the e.g. of the lamina is in the medians AA' and CC. Hence the e.g. of the triangle is at G, the intersection of the three medians, which are concurrent, meeting at a point distant from any vertex of the triangle by f of the median through it. The perpendicular distance of G from any side of the triangle is \ of the perpendicular distance of the oppo- site vertex from that side. Note that the e.g. of the triangular area ABC coincides with that of three equal particles placed at A, B, and C. For those at A and C are statically equivalent to two at B', and the e.g. of two at B' and one at B is at G, which divides BB' in the ratio 2 : i, or such at B'G = BB' (Art. 102). Uniform Parallelogram. If a lamina be cut in the shape of a parallelogram, ABCD (Fig. 101), the e.g. of the triangle ABC is in OB, and that of the triangle ADC is in OD, therefore the e.g. of the whole is in BD. Similarly it is in AC, and therefore it is at the intersection O. 105. Rectilinear Figures in General. The e.g. of any lamina with straight sides may be found by dividing its area up into triangles, and finding the e.g. and area of each triangle. FIG. roi. 144 Mechanics for Engineers Thus, in Fig. 102, if G 1? G 2 , and G 3 are the centres of gravity of the triangles ABE, EBD, and DEC respectively, the e.g. of the area ABDE is at G 4 , which divides the length GiG 2 inversely as the weights of the triangles AEB and EDB, and therefore inversely as their areas. Simi- larly, the e.g. G of the whole figure ABCDE divides G 3 G 4 in- versely as the areas of the figures ABDE and BCD. The inverse division of the lines G : G 2 and of G 3 G 4 may in practice be performed by the graphical method of Art. 102. 106. Symmetrical Figures. If a plane figure has an axis of symmetry, i.e. if a straight line can be drawn dividing it FIG. 102. FIG. 103. into two exactly similar halves, the e.g. of the area of the figure lies in the axis of symmetry. For the area can be divided into indefinitely narrow strips, the e.g. of each of which is in the axis of symmetry (see Fig. 103). If a figure has two or more axes of FIG. 104. symmetry, the e.g. must lie in each, hence it is at their intersection, e.g. the e.g. of a circular area is at its centre. Other examples, which sufficiently explain themselves, are shown in Fig. 104. Centre of Inertia or Mass Centre of Gravity 145 107. Lamina or Solid from which a Part has been removed. Fig. 105 represents a lamina from which a piece, B, has been cut. The centre of gravity of the whole lamina, including the piece B, is at G, and the e.g. of the removed portion B is at g, The area of the remaining piece A is a units, and that of the piece B is b units. It is required to find the e.g. of the remain- ing piece A. Let G' be the required FIG. 105. e.g. ; then G is the e.g. of two bodies the centres of gravity of which are at G' and g, and which are proportional to a and b respectively. Hence G is in the line G'g, and is such that GG' :Gg:: b : a (Art. 102) That is, the e.g. G' of the piece A is in the same straight line gG as the two centres of gravity of the whole and the part b B, at times their distance apart beyond the e.g. of the whole lamina. The point G' divides the line G^ externally in the ratio ^ or G'G ; G'g:: b \ a + b. The same method is ap- plicable if A is part of a solid from which a part B has been removed, provided a repre- sents the weight of the part A, and b that of the part B. Graphical Construc- tion. The e.g. of the part A may be found as follows : from g draw a line gP (Fig. 1 06) at any angle (preferably at right angles) to Gg and proportional to a -f b. From G L 146 Mechanics for Engineers draw GQ parallel to p and proportional to b. Join PQ, and produce to meet gG produced in G'. Then G' is the e.g. of the part A. 1 08. Symmetrical Solids of Uniform Material. If a solid is symmetrical about one plane, i.e. if it can be divided by a plane into two exactly similar halves, the e.g. evidently lies in the plane, for the solid can be divided into lamina? the FIG. 107. e.g. of each of which is in the plane of symmetry. Similarly, if the solid has two planes of symmetry, the e.g. must lie in the intersection of the two planes, which is an axis of the solid, as in Fig. 107. If a solid has three planes of symmetry, the line of inter- section of any two of them meets the third in the e.g., which is FIG. 108. a point common to all three planes, e.g. the sphere, cylinder, etc. (see Fig. 108). 109. Four Equal Particles not in the Same Plane. Let ABCD (Fig. 109) be the positions of the four equal particles. Join ABCD, forming a triangular pyramid or tetrahe- dron. The e.g. of the three particles at A, B, and C is at D', the e.g. of the triangle ABC (Art. 104). Hence the e.g. of the four particles is at G in DD', and is such that D'G : GD = i : 3 (Art. 102) or D'G = i DD' Centre of Inertia or Mass Centre of Gravity 147 FIG. 109. Similarly, the e.g. of the four particles is in AA', BB', and CC', the lines (which are concurrent) joining A, B, and C to the centres of gravity of the triangles BCD, ACD, and ABD respectively. The distance of the e.g. from any face of the tetrahedron is \ of the per- pendicular distance of the opposite vertex from that face. no. Triangular Pyramid or Tetrahedron of Uniform Material. Let ABCD (Fig. no) be the triangular pyramid. Suppose the solid divided into indefinitely thin plates, such as abc, by planes parallel to the face ABC. Let D' be the e.g. of the area ABC. Then DD' will intersect the plate abc at its e.g., viz. at d, and the e.g. of every plate, and there- fore of the whole solid, will be in DD'. Simi- larly, it will be in AA', BB', and CC', where A', B', and C' are the centres of gravity of the triangles BCD, CDA, and DAB respectively. Hence the centre of gravity of the whole solid coincides with that of four equal particles placed at its vertices (Art. 109), and it is in DD', and distant \ DD' from D', in CC' and 5 CC' from C', and so on. It is, therefore, also distant from any face, \ of the perpendicular distance of the opposite vertex from that face. in. Uniform Pyramid or Cone on a Plane Base. If V (Fig. in) is the vertex of the cone, and V the e.g. of the base of the cone, the e.g. of any parallel section or lamina into which the solid may be divided by plates parallel to the base, will be in VV. Also if the base be divided into an indefinitely large number of indefinitely small triangles, the solid is made up of FIG. no. 148 Mechanics for Engineers an indefinitely large number of triangular pyramids having the triangles as bases and a common vertex, V. The e.g. of each small pyramid is distant from V f of the distance from its base to V. Hence the centres of gravity of all the pyramids lie in a plane parallel to the base, and distant from the vertex, f of the altitude of the cone. The e.g. of a right circular cone is therefore in its axis, which is the intersection of two planes of symmetry (Art. 108), and its distance from "the base is \ the height of the cone, or its distance from the vertex is f of the height of the cone. Example i. A solid consists of a right circular cylinder 3 feet long, and a right cone of altitude 2 feet, the base coinciding with one end of the cylinder. The cylinder and cone are made of the same uniform material. Find the e.g. of the solid. If r = radius of the cylinder in feet the volume of cylinder _ irr 2 x 3 _ 9 volume of cone irr 2 x ^ x 2 2 hence the weight of the cylinder is 4*5 times that of the cone. The e.g. of the cylinder is at A (Fig. 112), the mid-point of its axis (Art. 108), i.e. i'5 feet from the plane of the base of the cone. FIG. in. AC FIG. 112. The e.g. of the cone is at B, \ of the altitude from the base (Art. m), i.e. 0^5 foot from the common base of the cylinder and cone. Hence AB = AD + DB = 1-5 + 0-5=2 feet 2 And G is therefore in AB, at a distance 2+9 .AB from A (Art. 102), i.e. AG = ^ of 2 feet = j^ foot, or 4-36 inches. Centre of Inertia or Mass Centre of Gravity 149 Example 2. A quadrilateral consists of two isosceles triangles on opposite sides of a base 8 inches long. The larger triangle has two equal sides each 7 inches long, and the smaller has its vertex 3 inches from the 8-inch base. Find the dis- tance of the e.g. of the quadrilateral from its 8-inch diagonal. Let ABCD (Fig. 113) be the quadrilateral, AC being the 8-inch diagonal, of which E is the mid- point ; then ED = 3 inches EB= 7^- : 4 2 inches FIG. 113. The e.g. of the triangle ABC is in EB and \ EB from E ; or, if Gj is the e.g. EGj = = 1-915 inches Similarly, if G 2 is the e.g. of the triangle ADC EG 2 = i of 3 inches = I inch therefore GjG 2 = 1-915 + i = 2*915 inches This length is divided by G, the e.g. of the quadrilateral, so that G 2 G _ area of triangle ABC _ BE _ I'9i5 GjG area of triangle ACD ED I G 2 G _ 1*915 _ 1*915 G!^ I + 1-915 2-915 G 2 G = 1-915 inches and EG = G 2 G G 2 E = 1*915 I = 0-915 inch which is the dista'nce of the e.g. from the 8-inch diagonal. Example 3. A pulley weighs 25 Ibs., and it is found that the e.g. is 0*024 inch from the centre of the pulley. The pulley is required to have its e.g. at the geometrical centre of the rim, and to correct the error in its position a hole is drilled in the pulley with its centre 6 inches from the pulley centre and in the same diameter as the wrongly placed e.g. How much metal should be removed by drilling ? Let x be the weight of metal to be removed, in pounds. 150 Mechanics for Engineers Then, in Fig. 114, OA being 6 inches and OG 0*024 inch, the removed weight x Ibs. having its e.g. at A, and the remaining FIG. 114. 25-.tr Ibs. having its e.g. at O, the e.g. G of the two together divides OA, so that r\r. OG _ x GA 25 -"x OG _ _^ )r OA 25 25 x OG 0-024 hence x - ^QT = 25 x ~ - o'i Ib. EXAMPLES XIV. 1. A uniform beam weighing 180 Ibs. is 12 feet long. It carries a load of 1000 Ibs. uniformly spread over 7 feet of its length, beginning I foot from one end and extending to a point 4 feet from the other. Find at what part of the beam a single prop would be sufficient to support it. 2. A lever 4 feet long, weighing 15 Ibs., but of varying cross-section, is kept in equilibrium on a knife-edge midway between its ends by the application of a downward force of I '3 Ibs. at its lighter end. How far is the e.g. of the lever from the knife-edge? 3. The heavy lever of a testing machine weighs 2500 Ibs., and is poised horizontally on a knife-edge. It sustains a downward pull of 4 tons 3 inches from the knife-edge, and carries a load of I ton on the same side of the knife-edge and 36 inches from it. How far is the e.g. of the lever from the knife-edge ? 4. A table in the shape of an equilateral triangle, ABC, of 5 feet sides, has various articles placed upon its top, and the legs at A, B, and C then exert pressures of 30, 36, and 40 Ibs. respectively on the floor. Determine the position of the e.g. of the table loaded, and state its horizontal distances from the sides AB and BC . 5. Weights of 7, 9, and 12 Ibs. are placed in the vertices A, B, and C respectively of a triangular plate of metal weighing 10 Ibs., the dimensions of which are, AB 16 inches, AC 16 inches, and BC n inches. Find the e.g. of the plate and weights, and state its distances from AB and BC. 6. One-eighth of a board 2 feet square is removed by a straight saw-cut through the middle points of two adjacent sides. Determine the distance of the e.g. of the remaining portion from the saw-cut. If the whole board before part was removed weighed 16 Ibs., what vertical upward force Centre of Inertia or Mass Centre of Gravity 151 applied at the corner diagonally opposite the saw cut would be sufficient to tilt the remaining \ of the board out of a horizontal position, if it turned about the line of the saw-cut as a hinge ? 7. An isosceles triangle, ABC, having AB 10 inches, AC 10 inches, and base BC 4 inches long, has a triangular portion cut off by a line DE, parallel to the base BC, and 7 '5 inches from it, meeting AB and AC in D and E respectively. Find the e.g. of the trapezium BDEC, and state its distance from the base BC. 8. The lever of a testing-machine is 15 feet long, and is poised on a knife-edge 5 feet from one end and 10 feet from the other, and in a horizontal line, above and below which the beam is symmetrical. The beam is 16 inches deep at the knife-edge, and tapers uniformly to depths of 9 inches at each end ; the width of the beam is the same throughout its length. Find the distance of the e.g. of the beam from the knife-edge. 9. A retaining wall 5 feet high is vertical in front and 9 inches thick at the top. The back of the wall slopes uniformly, so that the thickness of the wall at the base is 2 feet 3 inches. Find the e.g. of the cross-section of the wall, and state its horizontal distance from the vertical face of the wall. 10. What is the moment of the weight of the wall in Question 9 per foot length, about the back edge of the base, the weight of the material being 120 Ibs. per cubic foot ? What uniform horizontal pressure per square foot acting on the vertical face of the wall would be sufficient to turn it over bodily about the back edge of the base ? 1 1 . The casting for a gas-engine piston may be taken approximately as a hollow cylinder of uniform thickness of shell and one flat end of uniform thickness. Find the e.g. of such a casting if the external diameter is 8 inches, the thickness of shell \ inch, that of the end 3 inches, and the length over all 20 inches. State its distance from the open end. 12. A solid circular cone stands on a base 14 inches diameter, and its altitude is 20 inches. From the top of this a cone is cut having a base 3-5 inches diameter, by a plane parallel to the base. Find the distance of the e.g. of the remaining frustum of the cone from its base. 13. Suppose that in the rough, the metal for making a gun consists of a frustum of a cone, 10 feet long, 8 inches diameter at one end, and 6 inches at the other, through which there is a cylindrical hole 3 inches diameter, the axes of the barrel and cone being coincident. How far from the larger end must this piece of metal be slung on a crane in order to remain horizontal when lifted ? 14. A pulley weighing 40 Ibs. has its e.g. 0-04 inch from its centre. This defect is to be rectified by drilling a hole on the heavy side of the pulley, with its centre 9 inches from the centre of the pulley and in the radial direction of the centre of gravity. What weight of metal should be drilled out ? 15. A cast-iron pulley weighs 45 Ibs., and has its e.g. 0*035 mcn from its centre. In order to make the e.g. coincide with the centre of the 152 Mechanics for Engineers pulley, metal is added to the light side at a distance of 8 inches from the centre of the pulley and in line with the e.g. What additional weight is required in this position ? If the weight is added by drilling a hole in the pulley and then filling it up to the original surface with lead, how much iron should be removed, the specific gravity of lead being ii'35, and that of iron being 7-5 ? 112. Distance from a Fixed Line of the Centre of Gravity of Two Particles, or Two Bodies, the Centres of Gravity of which are given. Let A (Fig. 1 15) be the position of a particle of weight w lt and let B be that of a particle of weight w. 2 , or, if the two bodies are of finite size, let A and B be the positions of their centres of gravity. Then the centre of gravity of the M N Q FIG. 115. two weights w l and w. 2 is at G in AB such that AG or AG = .AB and GB = AB Let the distances of A, B, and G from the line NM be x lt x 2 , and x respectively, the line NM being in a plane through the line AB. Then AN = x lt BM = x 2 , and GQ = x. GR AG w z Now ' ES = AB = ^+^ or GR = - . BS and GQ or x = RQ + GR = AN + *!* *&i ~r hence x = x l iu n , BS Distance of the e.g. from a Plane. If ^ the respective distances of A and B from any plane, then NM Centre of Inertia or Mass Centre of Gravity 153 may be looked upon as the line joining the feet of perpen- diculars from A and B upon that plane. Then the distance x of G from that plane is _ This length x is also called the mean distance of the two bodies or particles from the plane. 113. Distance of the e.g. of Several Bodies or of One Complex Body from a Plane. Let A, B, C, D, and E (Fig. 116) be the positions of 5 par- ticles weighing w lt w 2 , u> 3 , ze/ 4 , and w & respectively, or the FIG. 116. centres of gravity of five bodies (or parts of one body) of those weights. Let the distances of A, B, C, D, and E from some fixed plane be x lt x 2 , x 3 , x 4 , and x 5 respectively, and let the weights in those positions be w lt w. 2 , w 3 , w 4 , and w & respectively. It is required to find the distance x of the e.g. of these five weights from the plane. We may conveniently consider the plane to 154 Mechanics for Engineers be a horizontal one, but this is not essential ; then x lt x%, x 3 , x 4 , and x 5 are the vertical heights of A, B, C, D, and E respec- tively above the plane. Let a, b, c, d, and e be the projections or feet of perpendiculars from A, B, C, D, and E respectively on the plane, so that Aa, B, O, ~Dd, and ~Ee are equal to x lt x^ x 3 , 0.4, and x s respectively. Let G! be the e.g. of w^ and > 2 , and let i be its projection by a vertical line on the plane ; then , . f .. (Art. 112. (i)) l 2 Let G 2 be the e.g. of (u\ 4- w 2 ) and w 3 , and let g% be its projection by a vertical line on the plane ; then G 2 divides G X C so that 1V 3 GiG.) = 7 ; ^ ,~ GiC l Z/. and G^ = ~ ^~+T -* (Art. 112. (i) ) and substituting the above value of Gii Similarly, if G 3 is the e.g. of w lt w. lt w 3 , and ze/ 4 , and g 3 is its projection on the plane, then 4- ss > and so on and finally Gg or x = - -* . . (2) Wi 4~ w.2 4~ ^'3 ~r w/4 4~ if which may be written where 2 stands for " the sum of all such terms as." If any of the points A, B, C, etc., are below the plane, their distances from the plane must be reckoned as negative. Centre of Inertia or Mass Centre of Gravity 155 Plane -moments. The products u\x^ w^x.^ #ve 3 , etc., are sometimes called plane-moments of the weights of the bodies about the plane considered. The plane-moment of a body about any given plane is then the weight of the body multiplied by the distance of its e.g. from that plane. Then in words the relation (3) may be stated as follows : " The distance of the e.g. of several bodies (or of a body divided into parts) from any plane is equal to the algebraic sum of their several plane-moments about that plane, divided by the sum of their weights." And since by (3), x X 2(w) = 2(w), we may state that the plane-moment of a number of weights (or forces) is equal to the sum of their several plane-moments. This statement extends to plane-moments the statement in Art. 90, that the moment of the sum of several forces about any point is equal to the sum of the moments of the forces about that point. It should be remembered that a horizontal plane was chosen for convenience only, and that the formulae (2) and (3) hold good for distances from any plane. 114. Distance of the e.g. of an Area or Lamina from a Line in its Plane. This is a particular case of the problem of the last article. Suppose the points A, B, C, D, and E in the last article and Fig. 116 all lie in one plane perpendicular to the horizontal plane, from which their distances are x lt x. 2 , x 3 , x 4 , and x s respectively. Then their projections a, l>, c, d, and e on the horizontal plane all lie in a straight line, which is the inter- section of the plane containing A, B, C, D, and E with the horizontal plane, viz. the line OM in Fig. 117. Thus, if x u x. u x 3) etc., be the distances of the centres of gravity of several bodies all in the same plane (or parts of a lamina) from a fixed line OM in this plane, then the distance of the e.g. of the bodies (or lamina?) from the line being x . . . , etc. _ ^(wx) >, 4- . . , etc. " So/ 156 Mechanics for Engineers This formula may be used to find the position of the e.g. of G, a lamina or area by finding its distance from two non-parallel fixed lines in its plane. If the lamina is of irregular shape, as in Fig. 118, the dis- tance of its e.g. from a line OM in its plane may be found approximately by dividing it into a number of narrow strips of equal width by lines parallel to OM, and taking the e.g. of each strip as being midway between the parallel boundary-lines. The weight of any strip being denoted by w FIG. us. w = volume of strip X D M where D = weight of unit volume of the material of the lamina, or- w = area of strip X thickness of lamina X D If the weight of the first, second, third, and fourth strips be /!, w 2 , w 3 , and w 4 respectively, and so on, and their areas be a lt a 2) a 3 , and a respectively, the lamina consisting of a material of uniform thickness /, then u\ = aj. D, w z = a./. D, and Centre of Inertia or Mass Centre of Gravity 157 so on. And if x is the distance of the e.g. of the area from OM, then by equation (4) . . . , etc. w 3 4- etc. h . . . i etc. (5) 4- a 3 tD 4- j etc. or, dividing numerator and denominator by the factor /D x 4 4- ... , etc. 83X3 #1 + 2 + 3 4- <* 4 + . o . , etc. = SM or SM (6) where A = total area of the lamina, and 2 has the same meaning as in (3), Art. 113. Similarly, the distance of the e.g. of the area A from another straight line may be found, and then the position of the e.g. is completely determined. Thus in Fig. 119, if x is the distance of the e.g. of the lamina from OM, and y is its distance from ON, by drawing two lines, PR and QS, parallel to OM and ON and dis- tant "x and y from them respectively, the inter- section G of the two lines gives the e.g. of the lamina or area. Moment of an Area. The products a^ lt etc., may be called moments of the areas a lt etc. Regular Areas. If a lamina consists of several parts, the centres of gravity of which are known, the division into thin strips adopted as an approximate method for irregular figures FIG. 119. 158 Mechanics for Engineers is unnecessary. The distance x of the e.g. from any line OM is ^(product of each area and distance of its e.g. from OM) whole area or- x = S(plane mo. of each area about a plane perpend, to its own) whole area The product of an area and the distance of its e.g. from a line OM may be called the " line moment " of the area about OM, and we may write 2(line moments of each part of an area) whole area For example, in Fig. 120 the area ABECD consists of a triangle, EEC, and a rectangle, ABCD, having a common side, BC. Let the height EF = h ; let AD = / and AB = d. Then the area ABCD = d x I, and the c area EEC = -j X / X h, and if G! is the e.g. of the triangle EEC, and G 2 that of the rectangle ABCD, the distance x of the e.g. of the area ABECD from AD is FIG. 120. found thus _ zd+h 115. Lamina with Part removed. Suppose a lamina (Fig. 121) of area A has a portion of area a, removed. Let x = distance of e.g. G of A from a line OM in its plane ; let x l be the distance of the e.g. of the part a from OM; and let x z be the distance of the e.g. of the remainder (A a) from OM. Centre of Inertia or Mass Centre of Gravity 159 - x^a -f- x. 2 (A. a} . Then* = - ^ -(Art. 114) x . A = and x 9 = a) A a In this way we can find the distance of the e.g. of the part A a from OM, and similarly we can find the distance from M FIG. 121. any other line in its plane, and so completely determine its position as in Art. 114. This method is applicable particularly to regular areas. 1 1 6. Solid with Part removed. The method used in the last article to find the e.g. of part of a lamina is applicable to a solid of which part has been removed. If in Fig. 122 A is a solid of weight W, and a portion B weighing w is re- moved, the distance of the e.g. of the remainder (W w) from any plane is x. 2 where rW -x ?e> *A- V V ^^ i-vitc' M W w FIG. 122. by (i) Art. 112 and the method of Art. 115, where x = distance of e.g. of A from the plane, and x 1 = distance of e.g. of B from the plane. i6o Mechanics for Engineers 117. Centre of Gravity of a Circular Arc. Let ABC (Fig. 123) be the arc, OA being the radius, equal to a units of length, and the length of arc ABC being / units. If B is the middle point of the arc, OB is an axis of symmetry, and the e.g. of the arc is in OB. Draw OM parallel to AC. Let the arc be divided into a B number of small portions, such as PQ, each of such small length as to be sensibly straight. Let the weight of the arc be w per unit length. The e.g. of a small portion PQ is at V, its mid-point. Draw VW parallel to OM, and join OV. Draw PR and QR parallel to OM and OB respectively. Then, if x = distance of e.g. of arc from the line OM, as in Art. 114 _ 2(PQ X w X OW) _ S(PQ X OW) _ S(PQ . OW) 'S(PQ) "~t~ Now, since OV, VW, and OW are respectively perpen- dicular to PQ, RQ, and PR, the triangles PQR and OVW are similar, and PQ_ RP "OV"" OW~ or PQ . OW = OV . RP = a . RP hence S(PQ . OW) = S( . RP) = S(RP) = a X AC and therefore S(PQ.OW) AC x = - ~ ~ 7 AC, or 7- X a LI & The e.g. of the arc then lies in OB at a point G such that OG = OB X - - or radius X - / arc or, if angle AOC = 2, i.e. if angle AOB = a (radians) AC 2AD 2 . a sin a sin a OG = a X -j- = a X r = ax - - = a . I I a X 2tf a Centre of Inertia or Mass Centre of Gravity 161 When the arc is very short, OG is very nearly equal to OB. 118. Centre of Gravity of Circular Sector and Segment. Let the sector ABCO (Fig. 124) of a circle centred at O and of radius a, subtend an angle 20. at O. The sector may be divided into small parts, such as OPQ, by radial lines from O. Each such part is virtually triangular when PQ is so short as to be regarded as a straight line. The e.g. of the triangle OPQ is on the median OR, and # from O. Similarly, the centres of gravity of all the constituent triangles, such as PQO, lie on a concentric arc abc of radius and subtending an angle 20. at O. The e.g. of the sector coincides with the e.g. of the arc abc, and is therefore in OB and at a FIG. 124. distance - from O (Art. 117) ; e.g. the e.g. of a semi- circular area of radius " a " is at a distance \a - or from 2 37T its straight boundary. The e.g. of the segment cut off by any chord AC (Fig. 124) may be found by the principles of Art. 115, regarding the segment as the remainder of the sector ABCO when the triangle AOC is removed. 119. Centre of Gravity of a Zone of a Spherical Shell. Let ABCD (Fig. 125) be a zone of a spherical shell of radius a and thickness /, and of uniform material which weighs w per unit volume. Let the length of axis HF be /. Divide the zone into a number of equal smaller zones, such as abed, by planes perpendicular to the axis OE, so that each has an axial length //. Then the area of each small zone is the same, ftf 1 62 Mechanics for Engineers viz. ZTraft, and the volume of each is then zirah . t, and each has its e.g. on the axis of symmetry OE, and midway between the bounding planes, such as a*/ and be, if h is indefinitely short. Hence the e.g. of the zone coincides with that of a large number of small bodies each of weight w . z-rrah . /, having their centres of gravity uniformly spread along the line FH. Hence the e.g. is at G, the mid-point of the axis FH of the zone, or e.g. the distance of the e.g. of a hemispherical shell from the plane of its rim is half the radius of the shell. 120. Centre of Gravity of a Sector of a Sphere. Let OACB (Fig. 126) be a spherical sector of radius a. If the sector be divided into an indefinitely great number of equal small pyramids or cones having a common vertex O such that their bases together make up the base ACB of the sector, the c.g.'s of the equal pyramids will each be |<7 from O, and will therefore be evenly spread over a portion acb (similar to the surface ACB) of a spherical surface centred at O and of radius \a. The e.g. of the sector then coincides with that of a zone, acb, of a thin spherical shell of radius f a, and is midway between c and the plane of the boundary circle ab, i.e. midway between d and c. Solid Hemisphere. The hemisphere is a particular case of a spherical sector, and its e.g. will coincide with that of a hemispherical shell of f, where a is the radius of the solid hemisphere. This is a point on the axis of the solid hemisphere, and half of #, or \a from its base. Example i. The base of a frustum of a cone is 10 inches diameter, and the smaller end is 6 inches diameter, the height being 8 inches. A co-axial cylindrical hole, 4 inches diameter, is bored through the frustum. Find the distance of the e.g. of the remaining solid from the plane of its base. The solid of which the c g. is required is the remaining portion Centre of Inertia or Mass Centre of Gravity 163 of a cone, ABC (Fig. 127), when the upper cone, DBE, and a cylinder, FGKH, have been removed. Since the cone diameter decreases 4 inches in a height of 8 inches The height BM = 8 + 8xf = 2O inches and the e.g. of the conei . , .. AT ,~ . . , >= 5 inches from AC ABC is J x 20 inches j volume of cone ABC = n- . (5)2 . 2 .^- = TT . 5-Q& cubic inches distance from AC of e.g. 1 . , c i- j -c-rvu I = '* 4 inches of cylinder FGKH ) volume of cylinder) , . FGKH }=T.2.8 = 3 2T cublc inches volume of cone DBE = TT . 3 2 . * = 36^- cubic inches distance from AC of e.g. \ , 12 . , f 1^-0 T? ( = + 4 = JI WChCS of cone DBE then volume of remaining frustum is TT( A | Q 32 - 36) = TT . 2 fi cubic inches Let h = height of e.g. of this remainder from the base. Then equating the plane-moments about the base of the three solids, BDE, FGKH, and the remainder of frustum, to the plane- moment of the whole cone (Art. 1 13) (and leaving out of both sides of equation the common factor weight per unit volume) TT . S X 5 = rr{(32 X 4) + (36 X II) + (^ X //)} 833'3 = 524 + -l~h h = -2 He x 39"3 = 3 -I 35 inches Example 2. An I-section of a girder is made up of three rectangles, viz. two flanges having their long sides horizontal, and one web connecting them having its long side vertical. The top flange section is 6 inches by i inch, and that of the bottom flange is 12 inches by 2 inches. The web section is 8 inches deep and i inch broad. Find the height of the e.g. of the area of cross-section from the bottom of the lower flange. Fig. 128 represents the section of the girder. Let x = height of the e.g. of the whole section. The height of the e.g. of BCDE is i inch above BE ; FGHKis 2 + f = 6 inches above BE ; LMNP is 2 + 8 + = 10-5 inches above BE. 1 64 Mechanics for Engineers Equating the sum of the moments of these three areas about A to the moment of the whole figure about A, we have (12 x 2)1 + (8 x i)6 + (6 x 1)10-5 = ~r{(i2X2) + (8xi) + (6xi)} 24 + 48 + 63 = I<24 + 8 + 6) ~* = W = 3' 5 5 inches ;*- B A * / 2' FIG. 128. which is the distance of the c g. from the bottom of the lower flange. Example 3. Find the e.g. of a cast-iron eccentric consisting of a short cylinder 8 inches in diameter, having through it a cylin- drical hole 2-5 inches diameter, the axis of the hole being parallel to that of the eccentric and 2 inches from it. State the distance of the e.g. of the eccentric from its centre. This is equivalent to finding the e.g. of the area of a circular lamina with a circular hole through it. In Fig. 129 AB = 8 inches CD = 2 inches EF = 2'5 inches Let the distance of the e.g. from A be x. If the hole were filled with the same material as the remainder of the solid, the e.g. of the \\ hole would be at C, its centre. S- 2 FIG. 129. Centre of Inertia or Mass Centre of Gravity 165 Equating moments of parts and the whole about A AC x (area of circle AB) = (AD x area of circle EF) + (x x area of eccentric) 4X 64 = 6x6 25+ -r(64 - 6-25) 256 37 5 x = :2i - 2 = 3783 5775 hence the distance of the e.g. from C is 4 3783 or 0*217 inch. Example 4. A hemispherical shell of uni- form material is 6 inches external radius and i '5 inches thick. Find its e.g. Let ABC (Fig. 130) be a solid hemisphere 12 inches diameter, from which a concentric solid hemisphere abc, 9 inches diameter, has been cut, leaving a hemispherical shell ACB&rtf i '5 inches thick. Let x = distance of its e.g. (which is on the axis of symmetry OC) from O. Equating moments of volumes about O (i.e. omitting the factor of weight per unit volume) Volume of solid \ ., . r i-j L -m \ , / i r i. i ARr 3OT / = ( vomrne f solid acox.%Oc) + (volume of shell x.r) |7r6 3 X f X 6 = |TT X () 3 X X f + |7r{6 3 (|) 3 }^ from which x = 2*66 inches The e.g. of the shell is on the axis and 2'66 inches from the centre of the surfaces. H-C FIG. 130. EXAMPLES XV. 1 . The front wheel of a bicycle is 30 inches diameter and weighs 4 Ibs. ; the back wheel is 28 inches diameter and weighs 7 Ibs. The remaining parts of the bicycle weigh 16 Ibs., and their e.g. is 18 inches forward of the back axle and 23 inches above the ground when the steering-wheel is locked in the plane of the back wheel. Find the e.g. of the whole bicycle ; state its height above the ground and its distance in front of the back axle when the machine stands upright on level ground. The wheel centres are 42 inches horizontally apart. 2. A projectile consists of a hollow cylinder 6 inches external and 3 inches internal diameter, and a solid cone on a circular base 6 inches diameter, coinciding with one end of the cylinder. The axes of the cone and cylinder are in line ; the length of the cylinder is 12 inches, and the 1 66 Mechanics for Engineers Ct -40 height of the cone is 8 inches. Find the distance of the e.g. of the projectile from its point. 3. A solid of uniform material consists of a cylinder 4 inches diameter and 10 inches long, with a hemispherical end, the circular face of which coincides with one end of the cylinder. The other end of the cylinder is pierced by a cylindrical hole, 2 inches diameter, extending to a depth of 7 inches along the cylinder and co-axial with it. Find the e.g. of the solid. How far is it from the flat end ? 4. The profile of a crank (Fig. 131) consists of two semicircular ends, CED and AFB, of 8 inches and 12 inches radii respectively, centred at points P and O 3 feet apart, and joined by straight lines AC and BD. The crank is of uniform thick- ness, perpendicular to the figure, and is pierced by a hole 10 inches diameter, centred at O. Find the distance of the c.g of the crank from the axis O. 5. Find the c.g. of a T girder section, the height over all being 8 inches, and the greatest width 6 inches, the metal being \ inch thick in the vertical web, and I inch thick in the horizontal flange. 6. An I-section girder consists of a top flange 6 inches by I inch, a bottom flange 10 inches by 1*75 inches, connected by a web 10 inches by 1*15 inches. Find the height of the c.g. of the section from the lowest edge. 7. A circular lamina 4 inches diameter has two circular holes cut out of it, one I '5 inches and the other I inch diameter with their centres I inch and I '25 inches respectively from the centre of the lamina, and situated on diameters mutually perpendicular. Find the c.g. of the remainder of the lamina. 8. A balance weight in the form of a segment of a circle fits inside the rim of a wheel, the internal diameter of which is 3 feet. If the segment subtends an angle of 60 at the centre of the wheel, find the distance of its c.g. from the axis. 9. If two intersecting tangents are drawn from the extremities of a quadrant of a circle 4 feet diameter, find the distance of the c.g. of the area enclosed between the tangents and the arc, from either tangent. 10. A balance weight of a crescent shape fits inside the rim of a wheel of 6 feet internal diameter, and subtends an angle of 60 at its centre. The inner surface of the weight is curved to twice the radius of the outer surface, i.e. the centre from which its profile is struck is on the circumference of the inside of the wheel. The weight being of uniform thickness perpendicular to the plane of the wheel, find the distance of its c.g. from the axis of the wheel. N.B. The profile is equivalent to the sector of a circle plus two triangles minus a sector of a larger circle. CHAPTER VIII CENTRE OF GRAVITY: PROPERTIES AND APPLICA TIONS I2i. Properties of the Centre of Gravity. Since the resultant force of gravity always acts through the centre of gravity, the weight of the various parts of a rigid body may be looked upon as statically equivalent to a single force equal to their arithmetic sum acting vertically through the centre of gravity of the body. Such a single force will produce the same reactions on the body from its supports ; will have the same moment about any point (Art. 90) ; may be replaced by the same statically equivalent forces or components ; and requires the same equilibrants, as the several forces which are the weights of the parts. Hence, if a body be supported by being suspended by a single thread or string, the e.g. of the body is in the same vertical line as that thread or string. If the same body is suspended again from a different point in itself, the e.g. is also in the second vertical line of suspension. If the two lines can be drawn on or in the body, the e.g., which must lie at their intersection, can thus be found experimentally. For example, the e.g. of a lamina may be found by suspending it from two different points in its perimeter, first from one and then from the other, so that its plane is in both cases vertical, and marking upon it two straight lindl which are continuations of the suspension thread in the two positions. Fig. 132 shows G, the e.g. of a lamina PQRS, lying in both the lines of suspension PR and QS from P and Q respectively. The tension of the cord acts vertically upwards on the lamina, and is equal in magnitude to the vertical downward force of 1 68 Mechanics for Engineers the weight of the lamina acting through G. The tension can only balance the weight if it acts through G, for in order that two forces may keep a body in equilibrium, they must be con- FIG. 132. current, equal, and opposite, and therefore in the same straight line. A " plumb line," consisting of a heavy weight hanging from a thin flexible string, serves as a convenient method of obtaining a vertical line. 122. Centre of Gravity of a Distributed Load. If a load is uniformly distributed over the whole span of a beam, the centre of gravity of the load is at mid-span, and the reactions of the supports of the beam are the same as would I be produced by the whole load concentrated at the middle of the beam. Thus, if in Fig. 133 a beam of 20-feet span carries a load of 3 tons per foot of span (including the weight of the beam) uniformly spread over its length, the reactions at the supports. A and B are each the same as would be produced by a load of 60 tons acting at C, the middle section of the beam, viz. 30 tons at each support. Next suppose the load on a beam is distributed, not evenly, but in some known manner. Suppose the load per foot of span at various points to be B FIG. 133. Centre of Gravity : Properties and Applications 1 69 shown by the height of a curve ACDEB (Fig. 134). The load may be supposed to be piled on the beam, so that the curve ACDEB is its profile, and so that the space occupied is of constant thickness in a direction perpendicular to the plane of the figure. Then the e.g. of the load is at the e.g. G of FIG. 134. the area of a section such as ACDEB in Fig. 134, taken halfway through the constant thickness. The reactions of the supports are the same as if the whole load were concen- trated at the point G. The whole load is equal to the length of the beam multiplied by the mean load per unit length, which is represented by the mean ordinate of the curve ACDEB, i.e. a length equal to the area ACDEB divided by AB. Example. As a par- ticular case of a beam carrying a distributed load not evenly spread, take a beam of 2o-feet span carry- ing a load the intensity of which is 5 tons per foot run at one end, and varying uniformly to 3 tons per foot at the other. Fig. 135 represents the distribu- tion of load. Find the reactions at A and B. FIG. 135. The total load = 20 x mean load per foot = 20 x - = 80 tons 2 Let I- be the distance of the e.g. of area ABCD from BD. jr(area ACFB + area CDF) = (loxarea ACFB) + (-Sfxarea CDF) X3 x 20 + 5 20 X 2 ) = ( 10 X 20 X 3) + * X 3p X 2 600 + 133*3 ,. * = 8 -^ = 9-i6 feet and distance of e.g. from AC = 20 9' 1 6 = io'8.3 feet Mechanics for Engineers If R A and R B be the reactions at A and B respectively, equating opposite moments about B of all the forces on the beam R A x 20 = 80 x 9-16 R A = 80 X * 20 R B = 80 36-6 = 43'3 tons R A = 80 x 2 = 36-6 tons 20 123. Body resting upon a Plane Surface. As in the case of a suspended body, the resultant of all the supporting forces must pass vertically through the e.g. of the body in order to balance the resultant gravitational forces in that straight line. The vertical line through the e.g. must then cut the surface, within the area of the extreme outer polygon or curved figure which can be formed by joining all the points of contact with the plane by straight lines. If the vertical line through the e.g. fall on the perimeter of this polygon the solid is on the point of overturning, and if it falls outside that area the solid will topple over unless supported in some other way. This is sometimes expressed by saying T G I I i FIG. 136. that a body can only remain at rest on a plane surface if the vertical line through the e.g. falls within the base. From what is stated above, the term " base " has a particular mean- ing, and does not signify only areas of actual contact; e.g. in Fig. 136 are two solids in equilibrium, with GN, the vertical line through G, the e.g., falling within the area of contact; Centre of Gravity : Properties and Applications 1 7 1 but in Fig. 137 a solid is shown in which the vertical through the e.g. falls outside the area of contact when the solid rests upright with one end on a horizontal plane. If, however, it falls within the extreme area ABC, the solid can rest in equilibrium on a plane. Plan. FIG. 137. Two cases in which equilibrium is impossible are shown in Fig. 138, the condition stated above being violated. The first is that of a high cylinder on an inclined plane, and the second FIG. 138. that of a waggon-load of produce on the side of a high crowned road. It will be noticed that a body subjected to tilting will topple over with less inclination or more, according as its e.g. is high or low. Example. What is the greatest length which a right cylinder of 8 inches diameter may have in order that it may rest with one end on a plane inclined 20 to the horizontal ? 172 Mechanics for Engineers The limiting height will be reached when the e.g. falls vertically over the circumference of the base, i.e. when G (Fig. 139) is FIG. 139. vertically above A. Then, G being the mid-point of the axis EF, the half-length of cylinder GE = AE cot AGE = AE cot ACD or GE = AE cot 20 = 4x2 7475 = 10-99 inches The length of cylinder is therefore 2 x 10^99 = 21 '98 inches. 124. Stable, Unstable, and Neutral Equilibrium. A body is said to be in stable equilibrium when, if slightly disturbed from its position, the forces acting upon it tend to cause it to return to that position. If, on the other hand, the forces acting upon it after a slight displacement tend to make it go further from its former position, the equilibrium is said to be unstable. If, after a slight displacement, the forces acting upon the body form a system in equilibrium, the body tends neither to return to its former position nor to recede further from it, and the equilibrium is said to be neutral. ' A few cases of equilibrium of various kinds will now be con- sidered, and the conditions making for stability or otherwise. 125. Solid Hemisphere resting on a Horizontal Plane. If a solid hemisphere, ABN (Fig. 140), rests on a Centre of Gravity: Properties and Applications 173 horizontal plane, and receives a small tilt, say through an angle 0, the e.g., situated at G, of ON from O and in the radius ON, takes up the position shown on the right hand of FIG. 140. the figure. The forces acting instantaneously on the solid are then (i) the weight vertically through G, and (2) the reaction R in the line MO vertically through M (the new point of contact between hemisphere and plane) and normal to the curved surface. These two forces form a " righting couple," and evidently tend to rotate the solid into its original posi- tion. Hence the position shown on the left is one of stable equilibrium. Note that G lies below O. 126. Solid with a Hemispherical End resting on a Horizontal Plane. Suppose a solid consisting of, say, FIG. 141. a cylinder with a hemispherical base, the whole being of homogeneous material, rests on a plane, and the e.g. G (Fig. 741) falls within the cylinder, i.e. beyond the centre O of the hemispherical end reckoned from N, where the axis cuts the 1/4 Mechanics for Engineers curved surface. On the left of Fig. 141 the solid is shown in a vertical position of equilibrium. Now suppose it to receive a slight angular displacement, as on the right side of the figure. The weight W, acting vertically downwards through G, along with the vertical reaction R of the plane, forms a system, the tendency of which is to move the body so that G moves, not towards its former position, but away from it. The weight acting vertically through G and the reaction of the plane acting vertically through O form an " upsetting couple " instead of a "righting couple." Hence the position on the left of Fig. 141 is one of unstable equilibrium. Note that in this case G falls above O. If the upper part of the body were so small that G is below O, the equilibrium would be stable, as in the case of the hemisphere above (Art. 125). The lower G is, the greater is the righting couple (or the greater the stability) for a given angular disturbance of the body. While in the case of in- stability, the higher G is, the greater is the upsetting couple or the greater the instability, and we have seen that such a solid is stable or unstable according as G falls below or above O. 127. Critical Case of Equilibrium neutral. If G coincides with the centre of the hemisphere (Art. 126), the equilibrium is neither stable nor unstable, but neutral. Suppose the cylinder is shortened so that G, the e.g. of the whole solid, falls on O, the centre of A^ /--^| R ^e hemisphere. Then if the solid receives a slight angular displacement, as in the right side of Fig. 142, the reaction R of | w the plane acts vertically F, G . I42 . upwards through O, the centre of the hemisphere (being normal to the surface at the point of contact), and the resultant force of gravity acts vertically downward through the same point. In this case the two vertical forces balance, and there is no couple formed, and no tendency to rotate the body towards or away from its former position. Hence the equilibrium is neutral. FlG - Centre of Gravity : Properties and Applications 175 In each of the above instances the equilibrium as regards angular displacements is the same whatever the direction of the displacement. As further examples of neu- tral equilibrium, a sphere or cylinder of uniform material resting on a horizontal plane may be taken. The sphere is in neutral equilibrium with regard to angular displacements in any direction, but the horizontal cylinder (Fig. 143) is only in neutral equilibrium as regards its rolling displacements ; in other directions its equilibrium is stable. Example. A cone and a hemisphere of the same homogeneous material have a circular face of i foot radius in common. Find for what height of the cone the equilibrium of the compound solid will be neutral when resting with the hemi- spherical surface on a horizontal plane. The equilibrium will be neutral when the e.g. of the solid is at the centre of the hemi- sphere, i.e. at the centre O (Fig. 144) of their common face. Let h be the height of the cone in feet. Then its e.g. G l is \h from O, and its volume is \h x - x 2 2 = \irh cubic feet. 4 The e.g. G 2 , of the hemisphere is at volume is TT cubic feet. Then FIG. 144. foot from O, and its G t O _ \h _ weight of hemisphere _ TT G 2 O weight of cone \-rrh and \h = 7 h h = 2 feet If h is greater than ^3 feet the equilibrium is unstable, and if it is less than ^3 f eet the equilibrium is stable. Mechanics for Engineers 128. In the case of bodies resting on plane surfaces and having more than one point of contact, the equilibrium will be stable if the e.g. falls within the area of the base, giving the word the meaning attached to it in Art. 123 for small angular displacements in any direction. If the e.g. falls on the perimeter of the base, the equilibrium will be unstable for displacements which carry the e.g. outside the space vertically above the " base." The attraction of the earth tends to pull the e.g. of a body into the lowest possible position ; hence, speaking generally, the lower the e.g. of a body the greater is its stability, and the higher the e.g. the less stable is it. In the case of a body capable of turning freely about a horizontal axis, the only position of stable equilibrium will be that in which the e.g. is vertically below the axis. When it Unstable Stable, FIG. 145. is vertically above, the equilibrium is unstable, and unless the e.g. is in the axis there are only two positions of equilibrium. If the e.g. is in the axis, the body can rest in neutral equilibrium in any position. Fig. 145 represents a triangular plate mounted on a hori- zontal axis, C ; it is in unstable, stable, or neutral equilibrium according as the axis C is below, above, or through G, the e.g. of the plate. 129. Work done in lifting a Body. When a body is lifted, it frequently happens that different parts of it are lifted through different distances, e.g. when a hanging chain is wound up, when a rigid body is tilted, or when water is raised from one vessel to a higher one. The total work done in lifting the Centre of Gravity: Properties and Applications 177 body can be reckoned as follows : Let w^ w 2 , 7f 3 , w 4 , etc., be the weights of the various parts of the body, which is supposed divided into any number of parts, either large or small, but such that the whole of one part has exactly the same displace- ment (this condition will in many cases involve division into indefinitely small parts). Let the parts w^ w^ w 3 , etc., be at heights #1, x. 2 , x 3 , etc., respectively above some fixed horizontal plane ; if the parts are not indefinitely small, the distances x ly x. 2 , x 3 , etc., refer to the heights of their centres of gravity. 2/K'jv) Then the distance x of the e.g. from the plane is - S(w) (Art. 113). After the body has been lifted, let .r/, x/, x 3 ', etc., be the respective heights above the fixed plane of the parts weighing w l} w z , w s , etc. Then the distance x' of the e.g. V/7/v'\ above the plane is -*/ r (Art. 113). Z(w) The work done in moving the part weighing w^ is equal to the weight w l multiplied by the distance (xj x^} through which it is lifted ; i.e. the work is w^Xi x^ units. Similarly, the work done in lifting the part weighing ?<" 2 is w. 2 (x. 2 x. 2 ). Hence the total work done is WT(XI - x^ + w z (x. 2 - x. 2 ) + w 3 (xs - x 3 ) +, etc. which is equal to (w^Xi + Wix. 2 + WaXs' -r, etc.) (w^x-i + W 2 x. 2 + Wi +, etc.) or 1(wx) ^(wx) But 2,(wx') =~x'2(w) and 2(w,r) =~x'$(w) therefore tne work done = x"^(w} x$(w) = (x 1 - x)1(w) The first factor, x ' x, is the distance through which the e.g. of the several weights has been raised, and the second factor, 2(w), is the total weight of all the parts. Hence the total work done in lifting a body is equal to the weight of the body multiplied by the vertical distance through which its e.g. has been raised. N 178 Mechanics for Engineers Example i. A rectangular tank, 3 feet long, 2 feet wide, and i '5 feet deep, is filled from a cylindrical tank of 24 square feet horizontal cross-sectional area. The level of water, before filling begins, stands 20 feet below the bottom of the rectangular tank. How much work is re- quired to fill the tank, the weight of i cubic foot of water being 62-5 Ibs. ? The water to be lifted is 3 x 2 x 1-5 or 9 cubic feet, hence the level in the lower tank will be lowered by ^ or i of a foot, i.e. by a length BC on Fig. 146. The 9 cubic feet of water lifted occupies first the position ABCD, and then fills the tank EFGH. In the former position its e.g. is -|BC or j 3 ^ foot below the level AB, and in the latter position its e.g. is |GH or | foot above the level EH. Hence the e.g. is lifted (^ 6 + 20 + f ) feet, i.e. 2o}f feet, or 20*9375 feet, feet of water lifted is 9 x 62*5 FIG. 146. The weight of the 9 cubic = 562-5 Ibs. Hence the work done is 562-5 x 20-9375 = 11,770 foot-lbs. Example 2. Find the work in foot-pounds necessary to upset a solid right circular cylinder 3 feet diameter and 7 feet high, weighing half a ton, which is resting on one end on a hori- zontal plane. Suppose the cylinder (Fig. 147) to turn about a point A on the circumference of the base. Then G, the e.g. of the cylinder, which was formerly 3-5 feet above the level of the hori- FIG. 147. zontal plane, is raised to a position G', i.e. to a height A'G' above the horizontal plane before the cylinder is overthrown. Centre of Gravity : Properties and Applications \ 79 The distance the e.g. is lifted is then A'G' EG A'G' = + EG2 ) = V0'5 2 + 3'5 2 ) = 3 - 8o7 feet The e.g. is lifted 3-807 3-5 = 0-307 foot and the work done is 1120 x 0*307 = 344 foot-lbs. Example 3. A chain 600 feet long hangs vertically ; its weight at the top end is 12 Ibs. per foot, and at the bottom end 9 Ibs. per foot, the weight per foot varying uniformly from top to bottom. Find the work necessary to wind up the chain. It is first necessary to find the total weight of the chain and the position of its e.g. The material of the chain may be considered to be spread laterally into a sheet of uniform thick- ness, the length remaining unchanged. The width of the sheet will then be proportional to the weight per foot of length ; the total weight, and the height of the e.g. of the chain, will not be altered in such a case. The depth of the e.g. below the highest point (A) of the chain (Fig. 148) will be the same as that of a figure made up of a rect- angle, ACDB, 600 feet long and 9 (feet or other units) broad, and a right-angled triangle, CED, having sides about the right angle at C of (CD) 600 feet and (CE) 3 units. The depth will be (600 x 9 x 300) + (\ x 600 x 3 x a 8 a ) , A , -. S-2 (Art. 114) (600 x 9) + (f x 600 x 3) which is equal to 2857 feet. The total weight of the chain will be the same as if it were 600 feet long and of uniform weight - - or 10-5 Ibs. per foot, viz. 600 x 10-5 = 6300 Ibs. Hence the work done in raising the chain all to the level A is 6300 x 2857 = 1,800,000 foot-lbs. 130. Force acting on a Rigid Body rotating uni- formly about a Fixed Axis. Let Fig. 149 represent a cross-section of a rigid body of weight W rotating about a fixed axis, O, perpendicular to the figure. For simplicity the body will be supposed symmetrical i8o Mechanics for Engineers about the plane of the figure, which therefore contains G, the e.g. of the body. In the position shown, let w^ be the weight of a very small portion of the body (cut parallel to the axis) situated at a distance r from O. Let W be the uniform angular velocity of the body about the axis O. Then the force acting upon the small portion of weight w^ in order to make it rotate about O is -wV, directed towards O (Art. 63), and it evidently acts at the middle of the length of the portion, i.e. in the plane of the figure. Resolving this force in any two perpendicular directions, XO and YO, the components in these two directions are m*r cos and wV sin 6 respectively, where 6 is the angle which AO & makes with OX. These may be written - . w 2 . x and 'w 2 . y respectively, <$ O where x represents r cos and y represents r sin 0, the projections of r on OX and OY respectively. Adding the components in the direction XO of the centri- petal forces acting in the plane of the figure upon all such portions making up the entire solid, the total component F x = S =te - -l(u,x) = =-*SM = ~ . to' . * *' / g S g and the total component force in the direction YO is W 2 = . or where x and y are the distances of G, the e.g. of the solid (which is in the plane of the figure), from OY and OX re- spectively. Centre of Gravity : Properties and Applications 1 8 1 Hence the resultant force P acting on the solid towards O is = . w . (+ = w2 R vvhere R = V x' z + / 2 , the distance of the e.g. from the axis O. Hence the resultant force acting on the body is of the same /W \ magnitude as the centripetal force ( -^w 2 R ) which must act on a weight W concentrated at a radius R from O in order that it may rotate uniformly at an angular velocity w. Further, ~if the tangent of the angle which P makes with XO is ^r- (Art. 75), which is equal to - , or where GN is perpen- y GN . nr X dicular to OX. Hence the force P acts in the line GO, and therefore the resultant force P acting on the rotating body is in all respects identical with that which would be required to make an equal weight, W, rotate with the same angular velocity about O if that weight were concentrated (as a particle) at G, the e.g. of the body. It immediately follows, from the third law of motionj that the centrifugal force exerted by the rotating body on its con- straints is also of this same magnitude and of opposite direction in the same straight line. Example. Find the force exerted on the axis by a thin uniform rod 5 feet long and weighing 9 Ibs., making 30 revolutions per minute about an axis perpen- < dicular to its length. The distance from the axis O to G, the eg. of the rod (Fig. 150), is 2*5 feet, the e.g. being midway between the ends. The angular velocity of the , . 3O X 27T ,. rod is "* > = TT radians per second. The cen- 60 trifugal pull on O is the same as that of a weight of 9 Ibs. concentrated at 2-5 feet from the axis and describing about O, TT radians per second, which is - x Ti- 2 x 2-5 = 6-89 Ibs. 32-2 1 82 Mechanics for Engineers 131. Theorems of (iu Id in us or Pappus. (a) The area of the surface of revolution swept out by any plane curve revolving about a given axis in its plane is equal to the length of the curve multiplied by the length of the path of its e.g. in describing a circle about the axis. Suppose the curve ABC (Fig. 151) revolves about the axis OO', thereby generating a surface of revolution of which OO' is the axis. Let S be the length of the curve, and suppose it to be divided into a large number of small parts, s v s 2 , s 3) etc., each of such short length that if drawn straight the shape of the curve is not appreciably altered. Let the distances of the parts s 1} s 2 , s 3 , etc., from the axis be x lt x. 2) x 3 , etc. ; and let G, the e.g. of the curve which is in the plane of the figure, i.e. the plane of the curve, be distant x from the axis OO'. The portion s l generates a surface the length of which is 2irx l and the breadth ^i ; hence the area is 2Trx- i s l . Similarly, the portion s 2 gene- rates an area 2irx. 2 . s 2 , and the whole area is the sum , etc., or 2ir2,xs If the portions s lt s 2 , s 3 , etc., are of finite length, this result is only an approximation ; but if we understand 2(xs) to represent the limiting value of such a sum, when the length of each part is reduced indefinitely, the result is not a mere approximation. Now, since 2<(xs) = x X S(j') = x X S, the whole area of the surface of revolution is 2irx . S, of which zirx is the length of the path of the e.g. of the curve in describing a circle about OO', and S is the length of the curve. (b) The volume of a solid of revolution generated by the revolution of a plane area about an axis in its plane is equal to the enclosed revolving area multiplied by the length of the path of the e.g. of that area in describing a complete circle about the axis. Suppose that the area ABC (Fig. 152) revolves about the axis OO', thereby generating a solid of revolution of which Centre of Gravity : Properties and Applications 183 OO' is an axis (and which is enclosed by the surface generated by the perimeter ABC). Let the area of the plane figure ABC be denoted by A, and let it be divided into a large , number of indefinitely small parts #1, a 2 , a 3 , etc., situated at distances jc 1} x 2 , x 3 , etc., from the axis OO'. The area a- 1} in revolving about OO', generates a solid ring which has a cross- section a^ and a length 27TJCJ, and therefore its volume is 2irx 1 a l . Similarly, the volume swept out by the area a z is 2irx. 2 a. 2 , and so on. The whole volume swept out by the area A is the limiting value of the sum of the small quantities , etc., , etc.,) or 2ir'2 l (ax) And since 2(ax) = x^(a) = x . A (Art. 114 (6)), the whole volume is 2irx . A, of which 2irx is the length of the path of the e.g. of the area in describing a circle about the axis OO', and A is the area. FIG. 152. or 2ir(a l x l + a%x z + a 3 x B Example. A groove of semicircular section 1*25 inches radius is cut in a cylinder 8 inches diameter. Find (a) the area of the curved surface of the groove, and (b) the volume of material removed. (a) The distance of the e.g. of the semicircular arc ABC (Fig. 153) from AB is ( 1*25 x - ) or \ It / 7T inches. Therefore the distance of the e.g. of the arc from the axis OO' is (4 --5 ) inches. The 0' FIG. 153. length of path of this point in making one complete circuit about 1 84 Mechanics for Engineers OO' is 271/4 - ^ = (877 - 5) inches. The length of arc ABC is i'257T inches, hence the area of the surface of the semicircular groove is r257r(87r 5) square inches = icw 2 6 - 257r = 987 i9'6 = 79' I square inches (6) The distance of the e.g. of the area ABC from AB is x I '25 = 0*530 inch, and therefore the distance of the e.g. from OO' is 4 0-53 = 3-47 inches. The length of path of this point in making one complete circuit about OO' is 2rr x 3*47 = 21*8 inches. The area of the semicircle isi(i*25)' 2 7r = 2'453 square inches, hence the volume of the material removed from the groove is 2r8 x 2-453 = 53-5 cubic inches 132. Height of the e.g. of a Symmetrical Body, such as a Carriage, Bicycle, or Locomotive. It was stated in Art. 121 that the e.g. of some bodies might conveniently be found experimentally by suspending the bodies from two different points in them alternately. This is not always con- venient, and a method suitable for some other bodies will now be explained by reference to a particular instance. The c.g. of a bicycle (which is generally nearly symmetrical about a FIG. 154. vertical plane through both wheels) may be determined by first finding the vertical downward pressure exerted by each wheel on the level ground, and then by finding the vertical pressures when one wheel stands at a measured height above the other one. Suppose that the wheels are the same diameter, and that the centre of each wheel-axle, A and B (Fig. 154), stands Centre of Gravity : Properties and Applications 185 at the same height above a level floor, the wheels being locked in the same vertical plane. When standing level, let W A = weight exerted by the front wheel on a weighing machine table ; let W B = weight exerted by the back wheel on a weighing machine table ; then W A + W B = weight of bicycle Let AB, the horizontal distance apart of the axle centres, be d inches. If the vertical line through the e.g. G cuts AB in C, then W A BC = m rfw- </(Art. 87) Next, let the weight exerted by the front wheel, when A stands a distance " h " inches (vertically) above B, be W ; and let CG, the distance of the e.g. of the bicycle above AB, be H. FIG. 155. Then, since ABE and DGC (Fig. 155) are similar triangles GC _ BJ _ V"(<* a - yfe 2 ) CD ~ AE ~ h W W andCD = BC-BD = TT // ~ hence GC or H = - WA-W, W A +~W B In an experiment on a certain bicycle the quantities were d = 44 inches, // = 6 inches, weight of bicycle = 32'QO Ibs., pressure (W A ) exerted by the front wheel when the back wheel 1 86 Mechanics for Engineers was on the same level = 14-50 Ibs., pressure (W a ) exerted by the front wheel when the back wheel was 6 inches lower = 1 3 '84 Ibs. TT 44 ^ 14x0 ir84 v Hence H = L X J ^ * X 44 6 32-90 = 6-54 inches or the height of the e.g. above the ground is 6-54 inches plus the radius of the wheels. The distance BC of the e.g. horizon- tally in front of the back axle is ^ v 44, or 10-4 inches. 32-90 A similar method may be applied to motor cars or locomotives. In the latter case, all the wheels on one side rest on a raised rail on a weighing machine, thus tilting the locomotive sideways. EXAMPLES XVI. 1. A beam rests on two supports at the same level and 12 feet apart. It carries a distributed load which has an intensity of 4 tons per foot-run at the right-hand support, and decreases uniformly to zero at the left-hand support. Find the pressures on the supports at the ends. 2. The span of a simply supported horizontal beam is 24 feet, and along three-quarters of this distance there is a uniformly spread load of 2 tons per foot run, which extends to one end of the beam : the weight of the beam is 5 tons. Find the vertical supporting forces at the ends. 3. A beam is supported at the two ends 15 feet apart. Reckoning from the left-hand end, the first 4 feet carry a uniformly spread load of I ton per foot run ; the first 3 feet starting from the right-hand end carry a load of 6 tons per foot run evenly distributed, and in the intermediate portion the intensity of loading varies uniformly from that at the right- hand end to that at the left-hand end. Find the reaction of the supports. 4. The altitude of a cone of homogeneous material is 18 inches, and the diameter of its base is 12 inches. What is the greatest inclination on which it may stand in equilibrium on its base ? 5. A cylinder is to be made to contain 250 cubic inches of material. What is the greatest height it may have in order to rest with one end on a plane inclined at 15 to the horizontal, and what is then the diameter of the base? 6. A solid consists of a hemisphere and a cylinder, each 10 inches diameter, the centre of the base of the hemisphere being at one end of the axis of the cylinder. What is the greatest length of cylinder consistent with stability of equilibrium when the solid is resting with its curved end on a horizontal plane ? Centre of Gravity : Properties and Applications \ 87 7. A solid is made up of a hemisphere of iron of 3 inches radius, and a cylinder of aluminium 6 inches diameter, one end of which coincides with the plane circular face of the hemisphere. The density of iron being three times that of aluminium, what must be the length of the cylinder if the solid is to rest on a horizontal plane with any point of the hemispherical surface, in contact? 8. A uniform chain, 40 feet long and weighing 10 Ibs. per foot, hangs vertically. How much work is necessary to wind it up ? 9. A chain weighing 12 Ibs. per foot and 70 feet long hangs over a (frictionless) pulley with one end 20 feet above the other. How much work is necessary to bring the lower end to within 2 feet of the level of the higher one ? 10. A chain hanging vertically consists of two parts : the upper portion is 100 feet long and weighs 16 Ibs. per foot, the lower portion is 80 feet long and weighs 12 Ibs. per foot. Find the work done in winding up (a) the first 70 feet of the chain, (b) the remainder. 11. A hollow cylindrical boiler shell, 7 feet internal diameter and 25 feet long, is fixed with its axis horizontal. It has to be half filled with water from a reservoir, the level of which remains constantly 4 feet below the axis of the boiler. Find how much work is required to lift the water, its weight being 62*5 Ibs. per cubic foot. 12. A cubical block of stone of 3-feet edge rests with one face on the ground : the material weighs 150 Ibs. per cubic foot. How much work is required to tilt the block into a position of unstable equilibrium resting on one edge? 13. A cone of altitude 2 feet rotates about a diameter of its base at a uniform speed of 180 revolutions per minute. If the weight of the cone is 20 Ibs., what centrifugal pull does it exert on the axis about which it rotates ? 14. A shaft making 150 rotations per minute has attached to it a pulley weighing 80 Ibs., the e.g. of which is o'l inch from the axis of the shaft. Find the outward pull which the pulley exerts on the shaft. 15. The arc of a circle of 8 inches radius subtends an angle of 60 at the centre. Find the area of the surface generated when this arc revolves about its chord ; find also the volume of the solid generated by the revolu- tion of the segment about the chord. 16. A groove of V-shaped section, i'5 inches wide and i inch deep, is cut in a cylinder 4 inches in diameter. Find the volume of the material removed. 17. A symmetrical rectangular table, the top of which measures 8 feet by 3 feet, weighs 150 Ibs., and is supported by castors at the foot of each leg, each castor resting in contact with a level floor exactly under a corner of the table top. Two of the legs 3 feet apart are raised 10 inches on to the plate of a weighing machine, and the pressure exerted by them is 66" 5 Ibs. Find the height of the e.g. of the table above the floor when the table stands level. CHAPTER IX MOMENTS OF INERTIA ROTATION 133. Moments of Inertia. (i) Of a Particle. If a particle P (Fig. 156), of weight w w and mass , is situated at a distance r from an axis OO', then o p its moment of inertia about that axis is denned as the quantity ant . r 2 , or (mass of P) X (distance from OO') 2 . 0' (2) Of Several Particles. If several particles, P, Q, R, and S, etc., of weights a' lf a/ 2 > w z> '4> etc., be situated at distances r \, f'zt r& and r^ etc., respectively from an axis OO' (Fig. 157), FIG. 156. r S Q ^X o 1 / _x^*7 R 4C 3 FIG. 157. End view of axis OO'. then the total moment of inertia of the several particles about that axis is denned as Moments of Inertia Rotation 189 7f'j o , W. 2 o . W., o , WA _Jr* -f ~ V 2 2 + V 3 - + V 4 2 -f , etc. " or 2{(mass of each particle) x (its distance from OO') 2 } (3) Rigid Bodies. If we regard a rigid body as divisible into a very large number of parts, each so small as to be regarded as a particle, then the moment of inertia of the rigid body about any axis is equal to the moment of inertia of such a system of particles about that axis. Otherwise, suppose a body is divided into a large but finite number of parts, and the mass of each is multiplied by the square of the distance of some point in it from a line OO' ; the sum of these products will be an approximation to the moment of inertia of the whole body. The approximation will be closer the larger the number of parts into which the body is divided ; as the number of parts is indefinitely increased, and the mass of each correspondingly decreased, the sum of the products tends towards a fixed limiting value, which it does not exceed however far the subdivision be carried. This limiting sum is the moment of / ' IV \ inertia of the body, which may be written ^(wr 2 ) or 2( ?" 2 ) Units. The units in which a moment of inertia is stated depend upon the units of mass and length adopted. No special names are given to such units. The " engineer's unit " or gravitational unit is the moment of inertia about an axis of unit mass (yz Ibs.) at a distance of i foot from the axis. 134. Radius of Gyration. The radius of gyration of a body about a given axis is that radius at which, if an equal mass were concentrated, it would have the same moment of inertia. (7/' \ -f 1 ) of a body about some <S ' axis be denoted by I, and let its total weight 2(/) be W, and (w\ W - } = g) S Mechanics for Engineers Let k be its radius of gyration about the same axis. Then, from the above definition I W 7 o ^fiv 2 \ = ft* = 2 -r 2 ) \ ' w 135' Moments of Inertia Y W of a Lamina about an Axis perpendicular to its Plane. Let the distances of any particle, P (Fig. 158), of a lamina from two perpen- dicular axes, OY and OX, in its plane be x l and ji re- spectively, and let w t be its weight, and i\ its distance from O, so that r? = x^ + y^. Then, if I x and I Y denote the moments of inertia of the lamina made up of such particles, about OX and OY re- spectively FIG. 158. and adding I g , etc. or g g g , which may be denoted by I . Then I = I x 4- IY (0 This quantity I is by definition the moment of inertia about an axis OO' perpendicular to the plane of the lamina, and through O the point of intersection of OX and OY. Moments of Inertia Rotation 191 Hence the sum of the moments of inertia of a lamina about any two mutually perpendicular axes in its plane, is equal to the moment of inertia about an axis through the intersection of the other two axes and perpendicular to the plane of the lamina. Also, if < x > ^Y> an d k Q be the radii of gyration about OX, OY, and OO' respectively, OO' being perpendicular to the fui\ W plane of Fig. 158, and if 21 ) = , the mass of the whole ^ o lamina . W W W and I Y = > Y 2 . and therefore, since I x + Iy = W by (i) (2) Or, in words, the sum of the squares of the radii of gyration of a lamina about two mutually perpendicular axes in its plane, is equal to the square of its radius of gyration about an axis through the intersection of the other two axes and perpendicular to the plane of the lamina. 136. Moments of Inertia of a Lamina about Parallel Axes in its Plane. Let P, Fig. 159, be a 0' constituent particle of weight ii\ of a lamina, distant x-^ from an axis ZZ' in the plane of the lamina and through G, the e.g. of the lamina, the distances being reckoned positive to the right and negative to the left of ZZ'. Let OO' be an axis in the plane of the lamina parallel to ZZ' and distant d from it. Then the distance of P from OO' is d x^ whether P is to the right or left of ZZ'. FIG. 159. I9 2 Mechanics for Engineers Let I be the moment of inertia of the lamina about OO' ; and let I z ZZ'. Then I = |V - Xl )* + V - xtf + W (d - xtf +,etcj 0/2*2 + a'3#3 +, etc.) The sum 7# 1 ;c 1 4- ^2-^2 + ^3X3 +, etc., is, by Art 114, equal to (z^! + 7> 2 -r- 7/ 3 +, etc.) X (distance of eg. from ZZ') which is zero, since the second factor is zero. Hence <S ^ <b 6 w orI =-rf a + I z ............. (i) o where W is the total weight of the lamina. And dividing each W term of this equation by (2) where and k z are the radii of gyration about OO' and ZZ' respectively. I37. 1 Extension of the Two Previous Articles to Solid Bodies. (a) Let ZX and ZY (Fig. 160) represent (by their traces) two planes perpendicular to the plane of the paper and to each other, both passing through the e.g. of a solid body. Let P be a typical particle of the body, its weight being w^ 1 This article may be omitted on first reading. The student acquainted with the integral calculus will readily apply the second theorem to simple solids. Moments of Inertia Rotation 193 and its distances from the planes ZY and ZX being x and y l respectively. Then, if r^ is the distance of P from an axis ZZ', which is the intersection of the planes XZ and YZ, and passes through the c.g. r? - xf + y?. Let I z be the moment of inertia of the body about ZZ', and I that about a parallel axis OO'. Let OO' be distant d from ZZ', and distant p and q from planes ZY and ZX respectively. Let other constituent particles of the body of weights 7i'. 2 , ' 3 , iv I, etc., be at distances x. ta x s , x 4 , etc., from the plane ZY, and distances y^ y*, y-> etc., from the plane ZX respectively, the x distances being reckoned positive to the right and negative to the left of ZY, and the y distances being reckoned positive above and negative below ZX. Let ;:,, r. A , r 4 , etc., be the distances of the particles from ZZj. Let w l + w 2 -f- / :i + , etc. = 2(w) or W, the total weight of the body. By definition FIG. 160. therefore I = ~{ + it>i(f - Ji) 2 + , etc.} etc. - <z 7v. 2 y 2 -f- , etc.) - 2 +, etc.)} I = / <*> ' 3 ;- 3 2 +, etc.) *94 Mechanics for Engineers since the planes XZ and YZ pass through the e.g. of the body (Art. 113). Hence \ = -d* 4- I z ..... (,) i W and dividing both sides of (i) by <b where / = radius of gyration about OO', and k z = radius of gyration about ZZ'. (b) Also 1t\ o i '< > i MS o Z== I l 7" r " / ; ' 3 +>etc ' = T(*i a + J'l 2 ) + ?<*-' +^ + ^ + ^> +' etc " o o o - -(W + w^ 2 4- ^s^s 2 +, etc.) + - (w-j? + w 2 y 2 2 + b O v>s}'3* +, etc.) W i i -kj=-*w*)+j(wf) ......... (3) d ^> ^> , 2 _ 5(o'*5 , a ^ = which may be written ...... (4) where j; 2 and _y 2 are the /<?^ squares of the distances of the body from the planes YZ and XZ respectively. The two quantities x 1 and _j^ are in many solids easily calculated. 138. Moment of Inertia of an Area. The moment of inertia I of a lamina about a given axis OO' in its plane / *iet \ is 2( - *' 2 ) (Art. 133), where w is the weight of a constituent Moments of Inertia Rotation 195 particle, and r its distance from the axis OO'. This quantity W is equal to . & (Art. 134), where k is the radius of gyration ^> about this axis OO', and W is the total weight of the lamina, so that g In a thin lamina of uniform thickness /, the area a (Fig. 161) occupied by a particle of weight w is proportional to w, for w = a . t . D, where D is the weight per unit volume of the material ; hence ^wr 1 and similarly, W = A . t . D, where A is the total area of the lamina ; FIG. 161. A/D Thus the thickness and density of a lamina need not be known in order to find its radius of gyration, and an area may properly be said to have a radius of gyration about a given axis. The quantity 2(ar*) is also spoken of as the moment of ineiiia of the area of the lamina about the axis OO' from which a portion a is distant r. The double use of this term " moment of inertia " is un- fortunate. The "moment of inertia of an area" ^(ar*) or 2 . A is not a true moment of inertia in the sense commonly used in mechanics, viz. that of Art. 133 ; it must be multiplied by the factor " mass per unit area " to make it a true moment of inertia. As before mentioned, the area has, however, a radius of gyration about an axis OO' in its plane defined by the equation ig6 Mechanics for Engineers Units. The units of the geometrical quantity 2(r 2 ), called moment of inertia of an area, depend only upon the units of length employed. If the units of length are inches, a moment of inertia of an area is written (inches) 4 . 139. Moment of Inertia of Rectangular Area about Various Axes. Let ABCD (Fig. 162) be a rectangle, AB = */, ~ BC =.b. The moment of inertia of the B C area ABCD about the axis OO' in the side U* AD may be found as follows. Suppose AB ^ divided into a large number , of equal parts, and the area ABCD divided into n equal narrow strips, each of width -. The Fi whole of any one strip EFGH is practically at a distance, say, FA from AD, and if EFGH is the/th strip from AD, FA => X -. n Multiplying the area EFGH, viz. b x -, by the square of its distance from AD, we have (area EFGH) x FA 2 = b X - X n There are n such strips, and therefore the sum of the products of the areas multiplied by the squares of their distances from OO', which may be denoted by ^(ar 2 ), is or J* = - X n 3 6 6 2 I When n is indefinitely great, = o, and - 2 = o, and the sum // ft- ^(ar 2 ) becomes ^ X 2 or This is the " moment of inertia 6 3 of the area " about OO' ; or, the radius of gyration of the area about OO' being k Moments of Inertia Rotation 197 If ABCD is a lamina of uniform thickness of weight w, its w w true moment of inertia about OO' is k z - \ . d 2 . S ' g The radius of gyration of the same area ABCD about an axis PQ (Fig. 163) in the plane of the figure and parallel to OO' and distant - from it, dividing the rectangle into halves, can be found from the formula (2), Art. 136, viz. 2 -Q where k f = radius of gyration about PQ ; whence / P 2 = (i The sum ^(ar 1 O C iR D 0' FIG. 163. about PQ is then 2(<*) X k\ = I'd X = ^ 12 Similarly, if s is the radius of gyration of the rectangle about RS B and therefore, if k G = radius of gyration about an axis through G (the e.g.) and perpendicular to the figure W = ^ + k* or T V( 2 + d*) (Art. 135 (2)) which is also equal to yVB(7 or JGB 2 . Example. A plane figure consists of a rectangle 8 inches by 4 inches, with a rectangular hole 6 inches by 3 inches, cut so that the diagonals of the two rectangles are in the same straight lines. Find the geometrical moment of inertia of this figure, and its radius of gyration, about one of the short outer sides. Let IA be the moment of inertia of the figure about AD (Fig. 164), and k be its A radius of gyration about AD. Moment of inertia of abed '\ ahnnt AD / = T2( area abcd ^ X ( Slde *) + ( area abcd ) FIG. i 4 . x (JAB)* (Arts. 139 and 136) Moment of inertia of "I _ t 3 ABCD about AD / ~ :J X 4 X 198 Mechanics for Engineers Hence I.\ = moment of inertia of ABCD moment of inertia of abed = i . 4 8 3 - (^ x 6 3 x 3 + 6 x 3 x 4*) = * a - (54 + 288) = 340-6 (inches) 4 The area of the figure is 8x4-6x3 = 14 square inches therefore k z = _ = 24*33 (inches) 2 14 and k = 4'93 inches 140. Moment of Inertia of a Circular Area about Various Axes. (i) About an axis OO' through O, its centre, and perpendicular to its plane. Let the radius OS of the circle (Fig. 165) be equal to R. Suppose the area divided into a large number , of circular or ring-shaped "D strips such as PQ, each of width Then the distance of the /th strip from TJ O is approximately / X -, and its n area is approximately FIG. 165. R R R 2 27T X radius X width = 2?r x /-- = 2irp^ n n n The moment of inertia of this strip of area about OO' is then R 4 . R 2 ^ //R\ 2 -2 X (' ) = & \ n / 27T- and adding the sum of all such quantities for all the n strips R4 . , ov ** / o i 8 I *? I 3 I J\& I I T \\ 71 + 2n 3 + n- R4 (( + i)) 2 R 4 ;/ 4 = 27 vr~Y- r =27r -/7-- 2 I When is indefinitely great, - = o and -^ = o, and the Moments of Inertia- -Rotation 199 rR 4 becomes , which is the " moment of inertia of sum the circular area " about OO'. And since 2(ar 2 ) about OO' = , if we divide each side of the equation by the area of the circle R 2 where k is the radius of gyration of the circular area about an axis OO' through its centre and perpendicular to its plane. (2) About a diameter. Again, if A and k c are the radii of gyration of the same area about the axes AB and CD respectively (Fig. 166) hence k* = k<? = R 2 R 2 from which the relations between the moments of inertia about AB, DC, and OO' may be found by multiplying each term by TrR 2 . That is, the moment of inertia of the circular area about a diameter is half that about an axis through O and perpendicular to its plane. Example. Find the radius of gyration of a ring-shaped area, bounded outside by a circle of radius a, and inside by a concentric circle of radius b, about a diameter of the outer circle. The moment of inertia of the area bounded by the outer circle, about AB (Fig. 167) is ; that of the inner circular area about 4 200 Mechanics for Engineers the same line is - - ; hence that of the ring-shaped area is ^(rt 4 -^ 4 ). The area is ^(d 1 $} ; hence, if k is the radius of gyration of the ring-shaped area about AB 7- IT (a 2 l>-) + Note that & = * "+* = ("-+.} \ 2 / Y /' so that when a and b FIG. i6 7 . are nearly equal, i.e. when rt b is a small quantity, the radius of of the inner and outer radii. gyration / , about the axis O, approaches the arithmetic mean a_+ b 2 141. Moment of Inertia of a Thin Uniform Rod. The radius of gyration of a thin rod d units long and of uniform material, about an axis through one end and perpendicular to the length of the rod, will evidently be the same as that of a narrow rectangle d units long, which, by Art. 139, is given by the relation h* = % d 2 , where k is the required radius of gyration. Hence, if the weight of the rod is W Ibs., its moment of inertia , . W,_ W J* about one end is k* or g S 3 Similarly, its moment of inertia about an axis through the W d" 2 - middle point and perpendicular to the length is 142. Moment of Inertia of a Thin Circular Hoop. (i) The radius of the hoop being R, all the matter in it is at a distance R from the centre of the hoop. Hence the radius of gyration about an axis through O, the centre of the hoop, and perpendicular to its plane, is R, and the moment w of inertia about this axis is . R 2 , where W is the weight of A the hoop. Moments of Inertia Rotation 201 (2) The radius of gyration about diameters OX and OY (Fig. 1 68) being k x and k u respectively hence k% = k^ = R 2 FIG. and the moment of inertia about any diameter of the hoop is W R 2 143. Moment of Inertia of Uniform Solid Cylinder. (i) About the axis OO' of the cylinder. The cylinder may be looked upon as divided into a large number of circular discs (Fig. 169) by planes perpendicular to the axis of the cylinder. The radius of gyration of each disc about the axis of the cylinder R 2 is given by the relation k z = > where k is radius of gyration of the disc, and R the outside radius of the cylinder and discs. If the weight of any one disc is /, and that of the whole cylinder is W, the moment of inertia of one disc is w R 2 and that of the whole cylinder is R 2 \ R 2 va/ R a \ R* l = _ \g 2 / 2g W R 2 R 2 and the square of the radius of gyration of the cylinder is . 2 (2) About an Axis perpendicular to that of the Cylinder and through the Centre of One End. Let OX (Fig. 170) be the axis about which the moment of inertia 202 Mechanics for Engineers of the cylinder is required. Let R be the radius of the cylinder, and / its length. Let o? = the mean square of the distance of the constituent particles from the plane YQO'Y' ; f = the mean square of the distance of the constituent particles from the plane OXX'O' ; = the radius of gyration of the cylinder about OO'. Then k* = j? + j? by Art. 137 (4) and from the symmetry of the solid, o? = y' R 2 hence k* or = 2oc = z* and x 1 = = y* The cylinder being supposed divided into thin parallel rods all parallel to the axis and / units long, the mean square of the FIG. 170. distance of the particles forming the rod from the plane VOX of one end, is the same as the square of the radius of gyration of a rod of length / about an axis perpendicular to its length /2 and through one end, viz. - (Art. 141). The axis OX is the <J intersection of the planes XOO'X' and VOX, the end plane ; hence, if x is the radius of gyration about OX TJ2 . 137(2)) Moments of Inertia Rotation 203 (3) Also, if k G is the radius of gyration about a parallel axis through G, the e.g. of the cylinder P R 2 / 2 or k<? = k K -- = -- 4 4 12 The moments of inertia of the cylinder about these various axes are to be found by multiplying the square of the radius of w gyration about that axis by the mass , where u> is the weight cS of the cylinder, in accordance with the general relation I t. 134). Example. A solid disc flywheel of cast iron is 10 inches in diameter and 2 inches thick. If the weight of cast iron is o'26 Ib. per cubic inch, find the moment of inertia of the wheel about its axis in engineers' units. The volume of the flywheel is IT x 5 2 x 2 = SOTT cubic inches the weight is then 0*26 x 50^- = 40^9 Ibs. and the mass is ~ i'27 units 32-2 The square of the radius of gyration is i^o) 2 (feet)-. Therefore the moment of inertia is 1*27 x oVs 0*1104 ur >it EXAMPLES XVII, 1. A girder of I-shaped cross-section has two horizontal flanges 5 inches broad and I inch thick, connected by a vertical web 9 inches high and I inch thick. Find the " moment of inertia of the area " of the section about a horizontal axis through its e.g. 2. Fig. 171 represents the cross-section of a cast-iron girder. AB is 4 inches, BC I inch, EF I inch ; EH is 6 inches, KL is 8 inches, and KN is I '5 inches. Find the moment of inertia and radius of gyration of the area of the section about the line NM. 3. Find, from the results of Ex. 2, the moment of inertia and radius of gyration of the area of section about an axis through the e.g. of the section and parallel to XM. 204 Mechanics for Engineers B i c ( 1 F C G L 1 E H M 4. Find the moment of inertia of the area enclosed between two con- centric circles of 10 inches and 8 inches diameter respectively, about a diameter of the circles. 5. Find the radius of gyration of the area bounded on the outside by a circle 12 inches diameter, and on the inside by a concentric circle of 10 inches diameter, about an axis through the centre of the figure and perpen- dicular to its plane. 6. The pendulum of a clock con- sists of a straight uniform rod, 3 feet long and weighing 2 Ibs., attached to which is a disc o - 5 foot in diameter and weighing 4 Ibs. , so that the centre of the disc is at the end of the rod. Find the moment of inertia of the pendulum about an axis perpendicular to the rod and to the central plane of the disc, passing through the rod 2*5 feet from the centre of the disc. 7. Find the radius of gyration of a hollow cylinder of outer radius a and inner radius b about the axis of the cylinder. 8. Find the radius of gyration of a flywheel rim 3 feet in external diameter and 4 inches thick, about its axis. If the rim is 6 inches broad, and of cast-iron, what is its moment of inertia about its axis? Cast iron weighs O'26 Ib. per cubic inch. 144. Kinetic Energy of Rotation. If a particle of a body weighs w^ Ibs., and is rotating with angular velocity w about a fixed axis 1\ feet from it, its speed is o>i\ feet per second (Art. 33), and its kinetic energy is therefore 1 1U l . (<ai\Y foot-lbs. (Art. 60). Similarly, another particle of 2 ,s" the same rigid body situated r z feet from the fixed axis of rotation, and weighing w 2 Ibs., will have kinetic energy equal to -.-. (o>r 2 ) 2 ; and if the whole body is made up of particles weighing w^ w 2 , w 3 , ze> 4 , etc., Ibs., situated at r 1} r 2 , r 3 , r^ etc., feet respectively from the axis of rotation, the total kinetic energy of the body will be etc or W( -V, 8 + -V + -V 3 2 -K etc.) foot- - \^ g g J lbs. Moments of Inertia Rotation 205 The quantity (*V + ^V + % +, etc.) or O <b u ' has been defined (Art. 133) as the moment of inertia I, of the body about the axis. Hence the kinetic energy of the body is W W Iw 2 , or \ . K 2 or, or \ V 2 foot-lbs., where K = radius of <^ S gyration of the body in feet about the axis of rotation, and V = velocity of the body in feet per second at that radius of gyration. This is the same as the kinetic energy ^MV 2 or W W - -V 2 of a mass M or , all moving with a linear velocity V. The kinetic energy of a body moving at a given linear velocity is proportional to its mass ; that of a body moving about a fixed axis with given angular velocity is proportional to its moment of inertia. We look upon the moment of inertia of a body as its rotational inertia, i.e. the measure of its inertia with respect to angular motion (see Art. 36). 145. Changes in Energy and Speed. If a body of moment of inertia I, is rotating about its axis with an angular velocity w n and has a net amount of work E done upon it, thereby raising its velocity to w.>; then, by the Principle of Work (Art. 61) il(o> 2 2 - Wl 2 ) = E W or i-K 2 ( W3 2 - Wl 2 ) = E & W or i-(V 2 2 - V, 2 ) = E where K = radius of gyration about the axis of rotation, and V 2 and Vj are the final and initial velocities respectively at a radius K from the axis. Hence the change of energy is equal to that of an equal weight moving with the same final and initial velocities as a point distant from the axis by the radius of gyration of the body. If the body rotating with angular velocity o> 2 about the axis is opposed by a tangential force, and does work of amount E in overcoming this force, its velocity will be reduced 2o6 Mechanics for Engineers to MU the loss of kinetic energy being equal to the amount of work done (Art. 61). 146. Constant resisting Force. Suppose a body, such as a wheel, has a moment of inertia I, and is rotating at an angular velocity w 2 about an axis, and this rotation is opposed by a constant tangential force F at a radius r from the axis of rotation, which passes through the centre of gravity of the body. Then the resultant centripetal force on the body is zero (Art. 130). The particles of the body situated at a distance r from the centre are acted on by a resultant or effective force always in the same straight line with, and in opposite direction to, their own velocity, and therefore have a constant retardation in their instantaneous directions of motion (Art. 40). Hence the particles at a radius r have their linear velocity, and therefore also their angular velocity, decreased at a constant rate ; and since, in a rigid body, the angular velocity of rotation about a fixed axis of every point is the same, the whole body suffers uniform angular retardation. Suppose the velocity changes from o> 2 to wj in / seconds, during which the body turns about the axis through an angle radians. The uniform angular retardation a is - 2 - \ Also the work done on the wheel is Fr x 6 (Art. 57), hence F . r. 6 = I(w 2 2 w a 2 ) = loss of kinetic energy . (i) The angle turned through during the retardation period is Note that F . r is the moment of the resisting force or the resisting torque. Again, o> 2 2 c^ 2 = (o> 2 + wj)^ wj and w 2 o>i = a/ and o)j + u> 2 = twice the average angular velocity during the retardation Moments of Inertia Rotation 207 Hence the relation F.r.0 = ll(<o 2 2 - V) may be written YrO = 11. a./. ~ or F . r = I . a . . , . . . . (2) i.e. the moment of the resisting force about the axis of rotation is equal to the moment of inertia of the body multiplied by its angular retardation. Similarly, if F is a driving instead of a resisting force, the same relations would hold with regard to the rate of increase of angular velocity, viz. the moment of the accelerating force is equal to the moment of inertia of the body multiplied by the angular acceleration produced. Compare these results with those of Art. 40 for linear motion. We next examine rather more generally the relation between the angular velocity, acceleration, and inertia of a rigid body. 147. Laws of Rotation of a Rigid Body about an Axis through its Centre of Gravity. Let w be the weight of a constituent particle of the body situated at P (Fig. 172), distant r from the axis of rota- tion O ; let a> be the angular velocity of the body about O. Then the velocity v of P is tar. Adding the vectors repre- senting the momenta of all F ' G - 172- such particles, we have the total momentum estimated in any particular direction, such as OX (Fig. 172), viz. ^./w \ w.. 2( - v cos 6), or -2(wr cos 6) But 2 (wr cos 0) is zero when estimated in any direction if r cos 6 is measured from a plane through the e.g. Hence the total linear momentum resolved in any given direction is zero. Moment of Momentum, or Angular Momentum of 2o8 Mechanics for Engineers a Rigid Body rotating about a Fixed Axis. This is defined as the sum of the products of the momenta of all the particles multiplied by their respective distances from the axis, ^ or 2 .v r / Yw \ ^(w \ But 3(- .v.r)= ^- . cor . r) = or the angular momentum is equal to the moment of inertia (or angular inertia) multiplied by the angular velocity. Suppose the velocity of P increases from i^ to z> a , the angular velocity increasing from ^ to w. 2 , the change of angular momentum is If the change occupies a time / seconds, the mean rate of change of angular momentum of the whole body is where/" is the average acceleration of P during the time /, and w f or F is the average effective accelerating force on the o> particle at P, acting always in its direction of motion, i.e. acting always tangentially to the circular path of P (see Art. 40). Also 3 (F . r) is the average total moment of the effective or net forces acting on the various particles of the body or the average effective torque on the body. If these average accelerations and forces be estimated over indefinitely small intervals of time, the same relations are true, and ultimately the rate of change of angular momentum is equal to the moment of the forces producing the change, so that rate of change of Iw = 2(Fr) = M = total algebraic moment of effective forces, or effective torque Moments of Inertia Rotation 209 Also rate of change of Iw = I X rate of change of o> or I . a, where a is the angular acceleration or rate of change of angular velocity. Hence = M = la a result otherwise obtained for the special case of uniform acceleration in (2), Art. 146. Problems can often be solved alternately from equation (i) or equation (2) (Art. 146), just as in the case of linear motion the equation of energy (Art. 60) or that of force (Art. 51) can be used (Art. 60). Example i. A flywheel weighing 200 Ibs. is carried on a spindle 2*5 inches diameter. A string is wrapped round the spindle, to which one end is loosely attached. The other end of the string carries a weight of 40 Ibs., 4 Ibs. of which is necessary to overcome the friction (assumed constant) between the spindle and its bearings. Starting from rest, the weight, pulling the flywheel round, falls vertically through 3 feet in 7 seconds. Find the moment of inertia and radius of gyration of the flywheel. The average velocity of the falling weight is f foot per second, and since under a uniform force the acceleration is uniform, the maximum velocity is 2 x 2 or f foot per second. The net work done by the falling weight, i.e. the whole work done minus that spent in overcoming friction, is (40 - 4)3 foot-lbs. = 1 08 foot-lbs. The kinetic energy of the falling weight is *.^ 2 .(?) 2 = 0-456 foot-lb. If I = moment of inertia of the flywheel, and a = its angular velocity in radians per second. By the principle of work (Art. 61) I 2 4- 0-456 = 108 foot-lbs. ilw 2 = 108 - 0*456 = 107-544 foot-lbs. The maximum angular velocity <a is equal to the maximum linear velocity of the string in feet per second divided by the radius of the spindle in feet, or P 2io Mechanics for Engineers 12 / i-25 875 = 8*22 radians per second therefore |I x (8 - 22) 2 = 107-544 T 107-544 x 2 215*1 1 = ,o \o -- = -TT-T = 3*io units (o 22) 15 67-0 And if k = radius of gyration in feet, since the wheel weighs 200 Ibs. 32-2' 2 = 0-5l8 (foot) 2 k 0716 foot or 8'6 inches Example a. An engine in starting exerts on the crank-shaft for one minute a constant turning moment of 1000 Ib.-feet, and there is a uniform moment resisting motion, of 800 Ib.-feet. The flywheel has a radius of gyration of 5 feet and weighs 2000 Ibs. Neglecting the inertia of all parts except the flywheel, what speed will .the engine attain during one minute ? (1) Considering the rate of change of angular momentum The effective turning moment is loco 800 = 200 Ib.-feet The moment of inertia of the flywheel is x 5 2 = 1552 units 32-3 Hence if a = angular acceleration in radians per second per second 200 = 15520 (Art. 146 (2)) o = - - = 0-1287 radian per second per second And the angular velocity attained in one minute is 60 x 0*1287 = 774 radians per second 774 x 60 or - - = 74 revolutions per minute 27T (2) Alternatively from considerations of.energy. If u = angular velocity acquired CO - = mean angular velocity Total angle turned through ) , u> ,. . t = DO x - = 3Ow radians in one minute 2 Net work done in one minute = 200 x 300- foot-lbs. 200 x 3oco = JIw 2 6ooow = J. 1552. to 2 w = I200 = 7*74 radians per second X 55 2 as before Moments of Inertia Rotation 211 Example 3. A thin straight rod of uniform material, 4-5 feet long, is hinged at one end so that it can turn in a vertical plane. It is placed in a horizontal position, and Q ~ then released. Find the velocity of the free end (i) when it has described an angle of 30, (2) when it is vertical. (i) After describing 30 the centre of gravity G (Fig. 173), which is then at G ON. has fallen a vertical distance ON = OGj cos 60 = AOG, = I x 2-25 = 1-125 feet FIG. 173. If W is the weight of the rod in pounds, the work done by gravitation is W x i 'i 25 foot-lbs. /W \ The moment of inertia of the rod ( ^*J is W (4-5)2 _ 27 W If ta l is the angular velocity of the rod, since the kinetic energy of the rod must be ri25\V foot-lbs. Q ^ - . , ' g w i 2 = I x 2 8 T x 3 2 '2 = 1073 = 3'28 radians per second the velocity of A in position A l is then 3'28 x 4'5 = 1474 feet per second (2) In describing 90 G falls 2 25 feet, and the kinetic energy is then 2'25\V foot-lbs. And if <a. 2 is the angular velocity of the rod J . 23-. _ a, 2 _ ->">r\y s ' wf = % x ./- x 32'2 = 2 1 '47 o) 2 = 4-63 radians per second and the velocity of A in the position A 2 is 4^63 x 4'5 = 20*82 feet per second 212 Meclianics for Engineers 148. Compound Pendulum. In Art. 71 the motion of a " simple pendulum " was investigated, and it was stated that such a pendulum was only approximated to by any actual pendulum. We now proceed to find the simple pendulum equivalent (in period) to an actual pendulum. Let a body be suspended by means of a horizontal axis O (Fig. 174) perpen- dicular to the figure and passing through the body. Let G be the e.g. of the body in any position, and let OG make any angle 6 with the vertical plane (OA) through O. FlG - *- Suppose that the body has been raised to such a position that G was at B, and then released. Let the angle AOB be <, and OG = OB = OA = //. The body oscillating about the horizontal axis O constitutes a pendulum. Let / = length of the simple equivalent pendulum (Art. 71) ; I = the moment of inertia of the pendulum about the axis O ; = radius of gyration about O ; k G radius of gyration about a parallel axis through G. Let W be the weight of the pendulum, and let M and N be the points in which horizontal lines through B and G respectively cut OA. When G has fallen from B to G, the work done is W x MN = W(ON - OM) = W(A cos 0-A cos <) = W/*(cos - cos <) Let the angular velocity of the pendulum in this position be at, then its kinetic energy is --Ito 2 (Art. 144), and by the principle of work (Art. 61), if there are no resistances to motion the kinetic energy is equal to the work done, or s - cos and therefore (cos 6 cos 0) . . . . (i) Moments of Inertia Rotation 213 Similarly, if a particle (Fig. 175) be attached to a point O' by a flexible thread of length /, and be released from a position B' such that B'6'A = <, O'A being vertical, its velocity v when passing G' such that G'O'A = 6 is given by V* = 2 g , M'N' = 2,/(cOS 6 COS <) and its angular velocity o> about O f being FIG. 175. = -(cos cos The angular velocity of a particle (or of a simple pendulum) given by equation (2) is the same as that of G (Fig. 174) given by equation (i), provided i.e. provided / = This length r is then the length of the simple pendulum equivalent to that in Fig. 174, for since the velocity is the same at any angular position for the simple pendulum of length / and the actual pendulum, their times of oscillation must be the same. Also, since k? = k<? -f h 2 (Art. 137 (2)) r The point C (Fig. 174), distant -f- + h from O, and in the line OG is called the " centre of oscillation" The expression k ' l k(? -j- + h shows that it is at a distance -r- beyond G from O. A particle placed at C would oscillate in the same period about O as does the compound pendulum of Fig. 174. 214 Mechanics for Engineers Example. A flywheel having a radius of gyration of 3*25 feet is balanced upon a knife-edge parallel to the axis of the wheel and inside the rim at a distance of 3 feet from the axis of the wheel. If the wheel is slightly displaced in its own plane, find its period of oscillation about the knife-edge. The length of the simple equivalent pendulum is ffacV 3 + ^ *' = 3 + 3-5208 = 6-5208 feet Hence the period is 2H-* / = 276 seconds V 32*2 149. The laws of rotation of a body about an axis may be stated in the same way as Newton's laws of motion as follows : Law T. A rigid body constrained to rotate about an axis continues to rotate about that axis with constant angular velocity- except in so far as it may be compelled to change that motion by forces having a moment about that axis. Law 2. The rate of change of angular momentum is pro- portional to the moment of the applied forces, or torque about the axis. With a suitable choice of units, the rate of change of angular momentum is eqtial to the moment of the applied forces, or torque about the axis. Law 3. If a body A exerts a twisting moment or torque about a given axis on a body B, then B exerts an equal and opposite moment or torque about that axis on the body A. 150. Torsional Simple Harmonic Motion. If a rigid body receives an angular displacement about an axis, and the moment of the forces acting on it tending to restore equilibrium is proportional to the angular displacement, then the body executes a rotary vibration of a simple harmonic kind. Such a restoring moment is exerted when a body which is suspended by an elastic wire or rod receives an angular displacement about the axis of suspension not exceeding a certain limit. Let M = restoring moment or torque in Ib.-feet per radian of twist ; I = moment of inertia of the body about the axis of suspension in engineer's units ; p. = angular acceleration of the body in radians per second per second per radian of twist. Moments of Inertia Rotation 215 Then M = I . //. (Art. 147) M or //. = Then, following exactly the same method as in Art. 68, if Q (Fig. 176) rotates uniformly with angular velocity \f p. in a circle centred at O and of radius OA, which represents to scale the greatest angular displacement of the body, and P is the projection of Q on OA, then P moves in the same way as a point distant from O by a length representing the angular displacement 0, at any instant to the same scale that OA represents the extreme displacement. The whole argument of Art. 68 need not be repeated here, but the results are FlG '7& Angular velocity for an angular displacement 0, represented by OM, is Angular acceleration for an angular displacement 0, repre- sented by PO, is vV 0. T = time of complete vibration = = seconds or, since M Example. A metal disc is 10 inches diameter and weighs 6 Ibs. It is suspended from its centre by a vertical wire so that its plane is horizontal, and then twisted. When released, how many oscillations will it make per minute if the rigidity of the suspension wire is such that a twisting moment of i Ib.-foot causes an angular deflection of 10 ? The twisting moment per radian twist is i 80 573 lb.-feet The square of radius of gyration is ICi 5 ;-) 2 = 0*0862 (foot) 2 2l6 Mechanics for Engineers The moment of inertia is ----- x o'o862 = o'oi6i5 unit Hence the time of vibration ion is 277. / = 2n- / O ' OI " I 5 V M ~ -'V 573 573 second The number of vibrations per minute is 60 \ = 178 then Q'337 151. It is evident, from Articles 144 to 150, that the rotation of a rigid body about an axis bears a close analogy to the linear motion of a body considered in Chapters I. to IV. Some comparisons are tabulated below. Angular or Rotational. Moment of inertia, I. Angular displacement, 0. Angular velocity, w. Angular acceleration, a. Moment of force, or torque, M. Angular momentum, I . w. Average angular velocity, - Average angular acceleration, Linear. W Mass or inertia, - or m. g Length, /. Velocity, v. Acceleration,/. Force, F. W Momentum, . v or mv. Average velocity, - . Average acceleration, - t w (z\ z> 2 ) Average force, - - - - o Work of constant force, F . /. f UO Kinetic energy, \ v* or \ mi z . * Period of simple vibration, Im / * / or 27TA / \f e \/ m w , or 27TA -.where e e = force per unit displace- ment. Average moment or torque, !((! fa) 2 ) / Work of constant torque, M . 0. Kinetic energy, |Iw 2 . Period of simple vibration, 27r \/ ;r?, where M = torque per radian displacement. Moments of Inertia Rotation 217 The quantities stated as average values have similar mean- ings when the averages are reckoned over indefinitely small intervals of time, or, in other words, they have corresponding limiting values. 152. Kinetic Energy of a Rolling Body. We shall limit ourselves to the case of a solid of revolution rolling along a plane. The e.g. of the solid will then be in the axis of revolution about which the solid will rotate as it rolls. Let R be the extreme radius of the body at which rolling contact with the plane takes place (Fig. 177); let the centre O be moving parallel to the plane with a velocity V. Then any point P on the outside circumference of the body is moving with a velocity V relative to O, the angular velocity of P and of the whole body about O being TJ, or say w radians per second. Consider the kinetic energy of a particle weighing w Ibs. at Q, distant OQ or r from the axis of the body. Let OQ make an angle QUA = with OA, the direction of motion of O. Then the velocity v of Q is the resultant of a velocity V parallel to OA, and a velocity wr perpendicular to OQ, and is such that v 1 = (w;-) 2 + V 2 + 2 cor . V . cos (90 + 6) Hence the kinetic energy of the particle is 1 W |-(V a + V 2 - o sin 0) The total kinetic energy of the body is then V 2 ) = S| (o>V 2 + V 2 - 2torV sin 0)} 2 J la ') V 2 , *s' *s . r sin 0) 21 8 Mechanics for Engineers Now, 2(*w. sin (9) = o (Art. 113 (3)) (W 2\ and 2{~ r ) = I, the moment of inertia of the solid g w I TV hence 21 - \2j about the axis O W = kinetic energy of rotation about O + kinetic energy of an equal weight moving with the linear velocity of the axis. This may also be written iW/ 79 . ^,o i-( 2 + R or + ^ 2 \ J where k is the radius of gyration about the axis 0. The W / / 2 \ kinetic energy V 2 ( i +^ 2 ) ' s tnen tne same as that of a weight w( i + ITS) moving with a velocity V of pure trans- lation, i.e. without rotation. In the case of a body rolling down a plane inclined to the horizontal (Fig. 178), using the same notation as in the previous case, the component force of gravity through O and parallel to the direction of motion down W the plane is . sin 0. In g rolling a distance s down the plane, the work done is W sin . s. Hence the Fie. 178. kinetic energy stored after the distance s is ,W T + jp) = W sin 6 . s (Art. 61) a R2 or V 2 = zsg sin "nirT~p This is the velocity which a body would attain in moving Moments of Inertia Rotation 219 without rotation a distance s from rest under an acceleration R 2 g sin ^irrnrw 1 Hence the effect of rolling instead of sliding down the-plane is to decrease the linear acceleration and linear velocity attained by the axis in a given time in the ratio R2-+Ta < See Art ' 28 >' We may alternatively obtain this result as follows ; Resolving the reaction of the (rough) plane on the body at T into components N and F, normal to the plane and along it respectively, the net force acting down the plane on the body is W sin 6 F ; and if a = angular acceleration of the body about O, and/= linear acceleration down the plane . But la = FR (Art. 146 (2)) F being the only force which has any moment about O ; la I/ hence F = R = ^2 and the force acting down the plane is W sin 6 ^2- force acting down the plane /,.. I/\ . W Hence /=- f , ,- - = I W sin 0--^ 2 } - mass of body \ R 2 / g 3 R 2 or/= gsm X R 2 + * Example. A solid disc rolls down a plane inclined 30 to the horizontal. How far will it move down the plane in 20 seconds from rest ? What is then the velocity of its centre, and if it weighs 10 Ibs., how much kinetic energy has it ? The acceleration of the disc will be R 2 32*2 x sin 30 x 3 = 32*2 x i x if R 2 + - 2 = 1073 feet per second per second 22O Mechanics for Engineers In 20 seconds it will acquire a velocity of 20 x 1073 = 214-6 feet per second Its average velocity throughout this time will be 214-6 ^ = 107-3 feet per second It will then move 107-3 x 20 = 2146 feet corresponding to a vertical fall of 2146 sin 30 or 1073 feet. The kinetic energy will be equal to the work done on it in falling 1073 f eet j *> e ' I0 73 x 10 = 10,730 foot-lbs. EXAMPLES XVIII. 1. What is the moment of inertia in engineer's units of a flywheel which stores 200,000 foot-lbs. of kinetic energy when rotating 100 times per minute ? 2. A flywheel requires 20,000 foot-lbs. of work to be done upon it to increase its velocity from 68 to 70 rotations per minute. What is its moment of inertia in engineer's units ? 3. A flywheel, the weight of which is 2000 Ibs., has a radius of gyration of 3-22 feet. It is carried on a shaft 3 inches diameter, at the circumference of which a constant tangential force of 50 Ibs. opposes the rotation of the wheel. If the wheel is rotating 60 times per minute, how long will it take to come to rest, and how many rotations will it make in doing so? 4. A wheel 6 feet diameter has a moment of inertia of 600 units, and is turning at a rate of 50 rotations per minute. What opposing force applied tangentially at the rim of the wheel will bring it to rest in one minute ? 5. A flywheel weighing I -5 tons has a radius of gyration of 4 feet. If it attains a speed of 80 rotations per minute in 40 seconds, find the mean effective torque exerted upon it in pound-feet ? 6. A weight of 40 Ibs. attached to a cord which is wrapped round the 2-inch spindle of a flywheel descends, and thereby causes the wheel to rotate. If the weight descends 6 feet in 10 seconds, and the friction of the bearing is equivalent to a force of 3 Ibs. at the circumference of the spindle, find the moment of inertia of the flywheel. If it weighs 212 Ibs., what is its radius of gyration ? 7. If the weight in Question 6, after descending 6 feet, is suddenly released, how many rotations will the wheel make before coming to rest ? 8. A flywheel weighing 250 Ibs. is mounted on a spindle 2'5 inches Moments of Inertia Rotation 221 diameter, and is caused to rotate by a falling weight of 50 Ibs. attached to a string wrapped round the spindle. After falling 5 feet in 8 seconds, the weight is detached, and the wheel subsequently makes 100 rotations before coming to rest. Assuming the tangential frictional resisting force at the circumference of the axle to be constant throughout the accelerating and stopping periods, find the radius of gyration of the wheel. 9. A rod is hinged at one end so that it can turn in a vertical plane about the hinge. The rod is turned into a position of unstable equilibrium vertically above the hinge and then released. Find the velocity of the end of the rod (i) when it is horizontal; (2) when passing through its lowest position, if the rod is 5 feet long and of uniform small section throughout. 10. A circular cylinder, 3 feet long and 9 inches diameter, is hinged about an axis which coincides with the diameter of one of the circular' ends. The axis of the cylinder is turned into a horizontal position, and then the cylinder is released. Find the velocity of the free end of the axis (i) after it has described an angle of 50! (2) when the axis is passing through its vertical position. n. A flywheel weighs 5 tons, and the internal diameter of its rim is 6 feet. When the inside of the rim is supported upon a knife-edge passing through the spokes and parallel to its axis, the whole makes, if disturbed, 21 complete oscillations per minute. Find the radius of gyration of the wheel about its axis, and the moment of inertia about that axis. 12. A cylindrical bar, 18 inches long and 3 inches diameter, is suspended from an axis through a diameter of one end. If slightly disturbed from its position of stable equilibrium, how many oscillations per minute will it make ? 13. A piece of metal is suspended by a vertical wire which passes through the centre of gravity of the metal. A twist of 8'5 is produced per pound-foot of twisting moment applied to the wire, and when the metal is released after giving it a small twist, it makes 150 complete oscillations a minute. Find the moment of inertia of the piece of metal in engineer's or gravitational units. 14. A flywheel weighing 3 tons is fastened to one end of a shaft, the other end of which is fixed, and the torsional rigidity of which is such that it twists o'4 per ton-foot of twisting moment applied to the flywheel. If the radius of gyration of the flywheel and shaft combined is 3 feet, find the number of torsional vibrations per minute which the wheel would make if slightly twisted and then released. 15. The weight of a waggon is 2 tons, of which the wheels weigh \ ton. The diameter of the wheels is 2 feet, and the radius of gyration 0*9 foot. Find the total kinetic energy of the waggon when travelling at 40 miles per hour, in foot-tons. 16. A cylinder is placed on a plane inclined 15 to the horizontal, and is allowed to roll down with its axis horizontal. Find its velocity after it has traversed 25 feet. 222 Mechanics for Engineers 17. A solid sphere rolls down a plane inclined a to the horizontal. Find its acceleration. (NOTE. The square of the radius of gyration of a sphere of radius R is R 2 .) 1 8. A motor car weighs W Ibs., including four wheels, each of which weigh w Ibs. The radius of each wheel is a feet, and the radius of gyration about the axis is k feet. Find the total kinetic energy of the car when moving at v feet per second. CHAPTER X ELEMENTS OF GRAPHICAL STATICS 1 53- I N Chapter VI. we considered and stated the condi- tions of equilibrium of rigid bodies, limiting ourselves to those subject to forces in one plane only. In the case of systems of concurrent forces in equilibrium (Chapter V.), we solved problems alternatively by analytical methods of resolu- tion along two rectangular axes, or by means of drawing vector polygons of forces to scale. We now proceed to apply the vector methods to a few simple systems of non-concurrent forces, such as were considered from the analytical point of view in Chapter VI., and to deduce the vector conditions of equilibrium. When statical problems are solved by graphical methods, it is usually necessary to first draw out a diagram showing correctly the inclinations of the lines of action of the various known forces to one another, and, to some scale, their relative posi- tions. Such a diagram is called a diagram of positions, or space diagram ; this is not to be confused with the vector diagram of forces, which gives magnitudes and directions, but not positions of forces. 154. Bows' Notation. In this notation the lines of action of each force in the space diagram are denoted by two letters placed one on each side of its line of action. Thus the spaces rather than the lines or intersections have letters assigned to them, but the limits of a space having a particular letter to denote it may be different for different forces. The corresponding force in the vector diagram has the same two letters at its ends as are given to the spaces separated by 224 Mechanics for Engineers its line of action in the space diagram. We shall use capital letters in the space diagram, and the corresponding small letters to indicate a force in the vector diagram. The notation will be best understood by reference to an example. It is shown in Fig. 179, applied to a space diagram and vector polygon for Space Diagram FIG. 179. five concurrent forces in equilibrium (see Chapter V.). The four forces, AB, BC, CD, DE, of 5 Ibs., 6 Ibs., s| Ibs., and 6-g- Ibs. respectively, being given, the vectors ab, be, cd, de are drawn in succession, of lengths representing to scale these magnitudes and parallel to the lines AB, BC, CD, and DE respectively, the vector ea, which scales 57 Ibs., represents the equilibrant of the four forces, and its position in the space diagram is shown by drawing a line EA parallel to ea from the common intersection of AB, BC, CD, and DE. (This is ex- plained in Chapter V., and is given here as an example of the system of lettering only.) 155. The Funicular or Link Polygon. To find graphically the single resultant or equilibrant of any system of non-concurrent coplanar forces. Let the four forces AB, BC, CD, and DE (Fig. 180) be given completely, i.e. their lines of action (directions and positions) and also their magnitudes. First draw a vector ab parallel to AB, and representing by its length the given magnitude of the force AB ; from b draw be parallel to the line BC, and representing the force BC com- pletely. Continuing in this way, as in Art. 73, draw the open Elements of Graphical Statics 225 vector or force polygon abcde; then, as in the case of con- current forces, Art. 73, the vector ae represents the resultant (or ea, the equilibrant) in magnitude and direction. The problem is not yet complete, for the position of the resultant is unknown. In Chapter VI. its position was determined by finding what moment it must have about some fixed point. The graphical method is as follows (the reader is advised to FIG. 180. draw the figure on a sheet of paper as he reads) : Choose any convenient point o (called a pole) in or about the vector polygon, and join each vertex a, b, c, d, and e of the polygon to o; then in the space diagram, selecting a point P on the line AB, draw a line PT (which may be called AO) parallel to ao across the space A. From P across the space B draw a line BO parallel to bo to meet the line BC in Q. From Q draw a line CO parallel to co to meet the line CD in R. From R draw a line DO parallel to do to meet the line DE in S, and, finally, from S draw a line EO parallel to eo to meet the line AO (or PT) in T. Then T, the intersection of AO and EO, is a point in the line of action of EA, the equilibrant, the magni- tude and inclination of which were found from the vector ea. Q 226 Mechanics for Engineers Hence the equilibrant EA or the resultant AE is completely determined. The closed polygon PQRST, having its vertices on the lines of action of the forces, is called a funicular or link polygon. That T must be a point on the line of action of the resultant is evident from the following considerations. Any force may be resolved into two components along any two lines which intersect on its line of action, for it is only neces- sary for the force to be the geometric sum of the components. (Art. 75). Let each force, AB, EC, CD, and DE, be resolved along the two sides of the funicular polygon which meet on its line of action, viz. AB along TP and QP, BC along PQ and RQ, and so on. The magnitude of the two components is given by the corresponding sides of the triangle of forces in the vector diagram, e.g. AB may be replaced by components in the lines AO and BO (or TP and QP), represented in magni- tude by the lengths of the vectors ao and ob respectively, for in vector addition ao + ob = ab (Art. 19) Similarly, CD is replaced by components in the lines CO and OD represented by co and od respectively. When this process is complete, all the forces AB, BC, CD, and DE are replaced by components, the lines of action of which are the sides TP, PQ, QR, etc., of the funicular polygon. Of these component forces, those in the line PQ or BO are represented by the vectors ob and bo, and therefore have a resultant nil. Similarly, all the other components balance in pairs, being equal and opposite in the same straight line, except those in the lines TP and TS, represented by ao and oe respectively. These two have a resultant represented by ae (since in vector addition ao + oe = ae), which acts through the point of intersection T of their lines of action. Hence finally the resultant of the whole system acts through T, and is represented in magnitude and direction by the line ae; the equilibrant is equal and opposite in the same straight line. 156. Conditions of Equilibrium. If we include the equilibrant EA (Fig. 180, Art. 155) with the other four forces, we have five forces in equilibrium, and (i) the force or vector Elements of Graphical Statics 227 polygon abcdc is closed ; and (2) the funicular polygon PQRST is a closed figure. Further, if the force polygon is not closed, the system reduces to a single resultant, which may be found by the method just described (Art. 155). It may happen that the force polygon is a closed figure, and that the funicular polygon is not. Take, for example, a diagram (Fig. 181) similar to the previous one, and let the FIG. 181. forces of the system be AB, BC, CD, DE, and EA, the force EA not passing through the point T found in Fig. 180, but through a point V (Fig. 181), in the line TS. If we draw a line, VW, parallel to oa through V, it will not intersect the line TP parallel to ao, for TP and VW are then parallel. Re- placing the original forces by components, the lines of action of which are in the sides of the funicular polygon, we are left with two parallel unbalanced components represented by ao and oa in the lines TP and VW respectively. These form a couple (Art. 91), and such a system is not in equilibrium nor reducible to a single resultant. The magnitude of the couple is equal to the component represented by oa multiplied by the length represented by the perpendicular distance between the lines TP and VW. A little consideration will show that it is also equal to the force EA represented by ea, multiplied by the distance represented by the perpendicular from T on the 228 Mechanics for Engineers line VX. Or the resultant of the forces in the lines AB, BC, CD, and DE is a force represented by ae acting through the point T; this with the force through V, and represented by ra, forms a couple. Hence, for equilibrium it is essential that (i) the polygon of forces is a closed figure ; (2) that the funicular polygon is a closed figure. Compare these with the equivalent statements of the analytical conditions in Art. 96. Choice of Pole. In drawing the funicular polygon, the pole o (Figs. 1 80 and 181) was chosen in any arbitrary posi- tion, and the first side of the funicular polygon was drawn from any point P in the line AB. If the side AO had been drawn from any point in AB other than P, the funicular polygon would have been a similar and similarly situated figure to PQRST. The choice of a different pole would give a different shaped funicular polygon, but the points in the line of action of the unknown equilibrant obtained from the use of different poles would all lie in a straight line. This may be best appre- ciated by trial. Note that in any polygon the sides are each parallel to a line radiating from the corresponding pole. 157. Funicular Polygon for Parallel Forces. To find the resultant of several parallel forces, we proceed exactly as in the previous case, but the force polygon has its sides all in the same straight line ; it is " closed " if, after drawing the various vectors, the last terminates at the starting-point of the first. The vector polygon does not enclose a space, but may be looked upon as a polygon with overlapping sides. Let the parallel forces (Fig. 182) be AB, BC, CD, and DE of given magnitudes. Set off the vector ab in the vector polygon parallel to the line AB, and representing by its length the magnitude of the force in the line AB. And from b set off be parallel to the line BC, and representing by its length the magnitude of the force in the line BC. Then be is evi- dently in the same straight line as ab, since AB and BC are parallel. Similarly the vectors cd t de, and the resultant ae of Elements of Graphical Statics 229 the polygon are all in the same straight line. Choose any pole o, and join a, b, t, d, and e to o. Then proceed to put in the funicular polygon in the space diagram as explained in a FIG. 182. Art. 155. The two extreme sides AO and EO intersect in T, and the resultant AE, given in magnitude by the vector ae, acts through this point, and is therefore completely deter- mined. 158. To find Two Equilibrants in Assigned Lines of Action to a System of Parallel Forces. As a simple example, we may tatae the vertical reactions a FIG. 183. at the ends of a horizontal beam carrying a number of vertical loads. Let AB, BC, CD, and DE (Fig. 183) be the lines of action 230 Mechanics for Engineers of the forces of given magnitudes, being concentrated loads on a beam, xy, supported by vertical forces, EF and FA, at y and x respectively. Choose a pole, o, as before (Arts. 156 and 157), and draw in the funicular polygon with sides AO, BO, CO, DO, and EO respectively parallel to ao, bo, co, do, and eo in the vector diagram. Let AO meet the line FA (i.e. the vertical through x) in /, and let q be the point in which EO meets the line EF (i.e. the vertical through y). Join pq, and from o draw a parallel line of to meet the line abcde in/. The magnitude of the upward reaction or supporting force in the line EF is represented by ef, and the other reaction in the line FA is represented by the vector fa. This may be proved in the same way as the proposition in Art. 155. of and fe represent the downward pressure of the beam at x and y respectively, while fa and ef represent the upward forces exerted by the supports at these points. 159. In the case of non-parallel forces two equilibrants can be found one to have a given line of action, and the other to pass through a given point, i.e. to fulfil altogether three conditions (Art. 96). FIG. 184. Let AB, BC, and CD (Fig. 184) be the lines of action of given forces represented in magnitude by ab, be, and cd respec- tively in the vector polygon. Let ED be the line of action of Elements of Graphical Statics 231 one equilibrant, and p a point in the line of action of the second. Draw a line, dx, of indefinite length parallel to DE. Choose a pole, o, and draw in the funicular polygon corre- sponding to it, but drawing the side AO through the given point p. Let the last side DO cut ED in q. Then, since the complete funicular polygon is to be a closed figure, ]<ympq. Then the vector oe is found by drawing a line, oe, through o parallel to pq to meet dx in e. The magnitude of the equili- brating force in the line DE is represented by the length de, and the magnitude and direction of the equilibrant EA through / is given by the length and direction of ea. 1 60. Bending Moment and Shearing Force. In con- sidering the equilibrium of a rigid body (Chapter VI.), we have hitherto generally only considered the body as a whole. The same conditions of equilibrium must evidently apply to any part of the body we may consider (see Method of Sections, Art. 98). For example, if a beam (Fig. 185) carrying loads Wj, W 2 , W 3 , W 4 , and W 5 , as shown, be ideally divided into two vy, Y 2 vy, % vy : I t> I f I :A : Y///W//////// FIG. 185. parts, A and B, by a plane of section at X, perpendicular to the length of the beam, each part, A and B, may be looked upon as a rigid body in equilibrium under the action of forces. The forces acting on the portion A, say, fulfil the conditions of equilibrium (Art. 96), provided we include in them the forces which the portion B exerts on the portion A. Note that the reaction of A on B is equal and opposite to the action of B on A, so that these internal forces in the beam make no contribution to the net forces or moment acting on the beam as a whole. For convenience of expression, we shall speak of the beam 232 Mechanics for Engineers as horizontal and the loads and reactions as vertical forces. Let R A and R B be the reactions of the supports on the por- tions A and B respectively. Considering the equilibrium of the portion A, since the algebraic sum of the vertical forces on A is zero, B must exert on A an upward vertical force Wj + W. 2 R A . This force is called the shearing force at the section X, and may be denoted by F x . Then F x = W> + W 2 - R A , or W, + W 2 - R A - F x = o If the sum Wj + W 2 is numerically less than R A , F x is negative, i.e. acts downwards on A. The shearing force at any section of this horizontal beam is then numerically equal to the algebraic sum of all the vertical forces acting on either side of the section. Secondly, since the algebraic sum of all the horizontal forces on A is zero, the resultant horizontal force exerted by B on A must be zero, there being no other horizontal force on A. Again, if x lt d v and </ 2 are the horizontal distances of R A , W 15 and W 2 respectively from the section X, since W ls W. 2) and R A exert on A a clockwise moment in the plane of the figure about any point in the section X, of magnitude R*.* 1 --W a .4-W v .4 B must exert on A forces which have a contra- clockwise moment M x , say, numerically equal to R A . x l W^ W.^, for the algebraic sum of the moments of all the forces on A is zero, i.e. X = o Or M x = R A . #, - W,*/, - This moment cannot be exerted by the force F x , which has zero moment in the plane of the figure about any point in the plane X. Hence, since the horizontal forces exerted by B on A have a resultant zero, they must form a couple of contra-clockwise moment, M x , i.e. any pull exerted by B must be accompanied by a push of equal magnitude. This couple M x is called the moment of resistance of the beam at the section X, and it is numerically equal to the algebraic sum of Elements of Graphical Statics 233 moments about that section, of all the forces acting to either side of the section. This algebraic sum of the moments about the section, of all the forces acting to either side of the section X, is called the bending moment at the section X. 161. Determination of Bending Moments and Shear- ing Forces from a Funicular Polygon. Confining our- selves again to the horizontal beam supported by vertical forces at each end and carrying vertical loads, it is easy to show that the vertical height of the funicular polygon at any distance along the beam is proportional to the bending moment W, W 2 FIG. i at the corresponding section of the beam, and therefore repre- sents it to scale, e.g. that xl (Fig. 186) represents the bending moment at the section X. Let the funicular polygon for any pole o, starting say from z, be drawn as directed in Arts. 155 and 157, eg being drawn parallel to zp or GO, the closing line of the funicular, so that R 15 the left-hand reaction, is represented by the vector ga and R 2 by fg, while the loads W lf W 2 , W 3 , \V 4 , and W 5 are repre- sented by the vectors at>, be, cd, de, and ^respectively. Con- sider any vertical section, X, of the beam at which the height of bending-moment diagram is xl. Produce xl and the side zw to meet in y. Also produce the side win of the funicular 234 Mechanics for Engineers polygon to meet xy in n, and let the next side mq of the funicular meet xy in /. The sides zw, wm, and mq (or AO, BO, and CO) are parallel to ao, bo, and co respectively. Draw a horizontal line, zk, through z to meet xy in k, a horizontal line through w to meet xy in r, and a horizontal 0H through o in the vector polygon to meet the line abcdef in H. Then in the two triangles xyz and gao there are three sides in either parallel respectively to three sides in the other, hence the triangles are similar, and ag ao Also the triangles zky and oHa are similar, and therefore zy _ zk ao 0H Hence from (i) and (2) xy zk ag . zk = -^, or xy.oH = ag X zk, or xy = g- Therefore, since ag is proportional to Rj, and zk is equal or proportional to the distance of the line of action of R! from X, ag . zk is proportional to the moment of Rj about X, and 0H being an arbitrarily fixed constant, xy is proportional to the moment of Rj about X. Similarly ab . wr 1777 ?~~ _ ~ and therefore yn represents the moment of Wj about X to the same scale that xy represents the moment of Rj about X. Similarly, again, nl represents the moment of W 2 about X to the same scale. Finally, the length xl or (xy ny In) represents the algebraic sum of the moments of all the forces to the left of the section X, and therefore represents the bending moment at the section X (Art. 160). Elements of Graphical Statics 235 Scales. If the scale of forces in the vector diagram is i inch to/ Ibs. and the scale of distance in the space diagram is i inch to q feet ; and if 0H is made h inches long, the scale on which xl repre- sents the bending moment at X is i inch top.q.h. foot-lbs. A diagram (Fig. 187) showing the shearing force along the AJB FIG. 187. length of the beam may be drawn by using a base line, sf, of the same length as the beam in the space diagram, and in the horizontal line through g in the force diagram. The shearing force between the end of the beam s and the line AB is con- stant and equal to Rj, i.e. proportional to ga. The height ga may be projected from a by a horizontal line across the space A. A horizontal line drawn through b gives by its height above g the shearing force at all sections of the beam in the space B. Similarly projecting horizontal lines through c, d, e, and f we get a stepped diagram, the height of which from the base line st gives, to the same scale as the vector diagram, the shearing force at every section of the beam. 236 Mechanics for Engineers EXAMPLES XIX. 1. Draw a square lettered continuously FQRS, each side 2 inches long. Forces of 9, 7, and 5 Ibs. act in the directions RP, SQ, and QR respectively. Find by means of a funicular polygon the resultant of these three forces. State its magnitude in pounds, its perpendicular distance from P, and its inclination to the direction PQ. 2. Add to the three forces in question I a force of 6 Ibs. in the direction PQ, and find the resultant as before. Specify it by its magnitude, its distance from P, and its inclination to PQ. 3. A horizontal beam, 15 feet long, resting on supports at its ends, carries concentrated vertical loads of 7, 9, 5, and 8 tons at distances of 3, 8, 12, and 14 feet respectively from the left-hand support. Find graphically the reactions at the two supports. 4. A horizontal rod AB, 13 feet long, is supported by a horizontal hinge perpendicular to AB at A, and by a vertical upward force at B. Four forces of 8, 5, 12, and 17 Ibs. act upon the rod, their lines of action cutting AB at I, 4, 8, and 12 feet respectively from A, their lines of action making angles of 70, 90, 120, and 135 respectively with the direction AB, each estimated in a clockwise direction. Find the pressure exerted on the hinge, state its magnitude, and its inclination to AB. 5. A simply supported beam rests on supports 17 feet apart, and carries loads of 7, 4, 2, and 5 tons at distances of 3, 8, 12, and 14 feet respectively from the left-hand end. Calculate the bending moment at 4, 9, and 1 1 feet from the left-hand end. 6. Draw a diagram to show the bending moments at all parts of the beam in question 5. State the scales of the diagram, and measure from it the bending moment at 9, II, 13, and 14 feet from the left-hand support. 7. Calculate the shearing force on a section of the beam in Question 5 at a point 10 feet from the left-hand support ; draw a diagram showing the shearing force at every transverse section of the beam, and measure from it the shearing force at 4 and at 13 feet from the left-hand support. 8. A beam of 2O-feet span carries a load of 10 tons evenly spread over the length of the beam. Find the bending moment and shearing force at the mid-section and at a section midway between the middle and one end. 162. Equilibrium of Jointed Structures. Frames. The name frame is given to a structure consist- ing of a number of bars fastened together by hinged joints; the separate bars are called members of the frame. Such structures are designed to carry loads which are applied mainly at the joints. We shall only consider frames which have just a sufficient number of members to prevent deformation or collapse under the applied loads. Frames having more Elements of Graphical Statics 237 members than this requirement are treated in books on Graphical Statics and Theory of Structures. We shall further limit ourselves mainly to frames all the members of which are approximately in the same plane and acted upon by forces all in this same plane and applied at the hinges. Such a frame is a rigid body, and the forces exerted upon it when in equilibrium must fulfil the conditions stated in Art. 96 and in Art. 156. These "external" forces acting on the frame consist of applied loads and reactions of supports ; they can be represented in magnitude and direction by the sides of a closed vector polygon ; also their positions are such that an indefinite number of closed funicular polygons can be drawn having their vertices on the lines of action of the external forces. From these two considerations the complete system of external forces can be determined from sufficient data, as in Arts. 155 and 159. The "internal" forces, i.e. the forces exerted by the members on the joints, may be determined from the following principle. The pin of each hinged joint is in equilibrium under the action of several forces which are practically coplanar and concurrent. These forces are : the stresses in the members (or the " internal " forces) meeting at the particular joint, and the " external " forces, i.e. loads and reactions, if any, which are applied there. If all the forces, except two internal ones, acting at a given joint are known, then the two which have their lines of action in the two bars can be found by completing an open polygon of forces by lines parallel to those two bars. If a closed polygon of forces be drawn for each joint in the structure, the stress in every bar will be determined. In order to draw such a polygon for any particular joint, all the concurrent forces acting upon it, except two, must be known, and therefore a start must be made by drawing a polygon for a joint at which some external force, previously determined, acts. Remembering that the forces which any bar exerts on the joints at its two ends are equal and in opposite directions, the drawing of a complete polygon for one joint supplies a means of starting the force polygon for a neighbouring joint for which at least one side is then known. An example of the determination of 238 Mechanics Jor Engineers the stresses in the members of a simple frame will make this more easily understood. Fig. 1 88 shows the principles of the graphical method of finding the stresses or internal forces in the members of a simple frame consisting of five bars, the joints of which have been denoted at (a) by i, 2, 3, and 4. The frame stands in (b) For joint I For join I Z the vertical plane, and carries a known vertical load, W, at the joint 3 ; it rests on supports on the same level at i and 4. The force W is denoted in Bow's notation by the letters PQ. The reactions at i and 4, named RP and QR respectively, have been found by a funicular polygon corresponding to the vector diagram at (b), as described in Art. 158. Elements of Graphical Statics 239 Letters S and T have been used for the two remaining spaces. When the upward vertical force RP at the joint i is known, the triangle of forces rps at (c) can be drawn by making rp proportional to RP as in (b), and completing the triangle by sides parallel to PS and SR (i.e. to the bars 12 and 14) respectively. After this triangle has been drawn, one of the three forces acting at the joint 2 is known, viz. SP acting in the bar 12, being equal and opposite to PS in (c). Hence the triangle of forces spt at (</), for the joint 2 can be drawn. Next the triangle tpq at (e) for joint 3 can be drawn, tp and pq being known; the line joining qt will be found parallel to the bar QT if the previous drawing has been correct ; this is a check on the accuracy of the results. Finally, the polygon qrst at (/) for joint 4 may be drawn, for all four sides are known in magnitude and direction, from the previous polygons. The fact that when drawn to their previously found lengths and directions they form a closed polygon, constitutes a check to the correct setting out of the force polygons. The arrow-heads on the sides of the polygons denote the directions of the forces on the particular joint to which the polygon refers. 163. Stress Diagrams. It is to be noticed in Fig. 188 that in the polygons (), (c), (d), (e), and (/), drawn for the external forces on the frame and the forces at the various joints, each side, whether representing an external or internal force, has a line of equal length and the same inclination in some other polygon. For example, sr in (c} corresponding to rs in (f), and // in (d) with tp in (<?). The drawing of entirely separate polygons for the forces at each joint is unnecessary ; they may all be included in a single figure, such as (g), which may be regarded as the previous five polygons superposed, with corresponding sides coinciding. Such a figure is called a stress diagram for the given frame under the given system of external loading. It contains- (i) a closed vector polygon for the system of external forces in the frame, (2) closed vector polygons for the (con- current) forces at each joint of the structure. As each vector representing the internal force in a member of the frame represents two equal and opposite forces, 240 Mechanics for Engineers arrow-heads on the vectors are useless or misleading, and are omitted. Distinction between Tension and Compression Members of a Frame. A member which is in tension is called a " tie," and is subjected by the joints at its ends to a pull tending to lengthen it. The forces which the member exerts on the joints at its ends are equal and opposite pulls tending to bring the joints closer together. A member which is in compression is called a "strut;" it has exerted upon it by the joints at its ends two equal and opposite pushes or thrusts tending to shorten it. The member exerts on the joints at its ends equal and opposite " outward " thrusts tending to force the joints apart. The question whether a particular member is a " tie " or a " strut " may be decided by finding whether it pulls or thrusts at a joint at either end. This is easily discovered if the direction of any of the forces at that joint is known, since the vector polygon is a closed figure with the last side terminating at the point from which the first was started. E.g. to find the kind of stress in the bar 24, or ST (Fig. 188). At joint 4 QR is an upward force ; hence the forces in the polygon qrst must act in the ->--> -> directions qr, rs, sf, and tq ; hence the force ST in bar 24 acts at joint 4 in the direction s to /, i.e. the bar pulls at joint 4, or the force in ST is a tension. Similarly, the force in bar 23, or PT, acts at joint 3 in a direction ty, i.e. it pushes at joint 3, or the force in bar 23 is a compressive one. Another method. Knowing the direction of the force rp at joint i (Fig. 1 88), we know that the forces at joint i act in the directions rp, ps, and sr, or the vertices of the vector polygon rps lie in the order r p s. The corresponding lines RP, PS, and SR in the space diagram are in clockwise order round the point i . This order, clockwise or contra-clockwise (but in this instance clockwise) is the same for every joint in the frame. If it is clockwise for joint i, it is also clockwise for joint 2. Then the vertices of the vector polygon for joint 2 are to be taken in the cyclic order s p /, since the lines SP, PT, and TS lie in clockwise Elements of Graphical Statics 241 order round the joint 2, e.g. the force in bar 23, or PT, is in -> the direction//, i.e. it thrusts at joint 2. This characteristic order of space letters round the joints is a very convenient method of picking out the kind of stress in one member of a complicated frame. Note that it is the character- istic order of space letters round a joint that is constant not the direction of vectors round the various polygons constituting the stress diagram. 164. Warren Girder. A second example of a simple stress diagram is shown in Fig. 189, viz. that of a common type FIG. 189. of frame called the Warren girder, consisting of a number of bars jointed together as shown, all members generally being of the same lengths, some horizontal, and others inclined 60 to the horizontal. Two equal loads, AB and BC, have been supposed to act at the joints i and 2, and the frame is supported by vertical reactions at 3 and 4, which are found by a funicular polygon. The remaining forces in the bars are found by completing the stress diagram abc . . . Mm. Note that the force AB at joint i is downward, />. in the direction ab in the vector diagram corresponding to a contra- clockwise order, A to B, round joint i. This is, then, the characteristic order (contra-clockwise) for all the joints, e.g. to find the nature of the stress in KL, the order of letters for joint 5 is K to L (contra-clockwise), and referring to the vector 242 Mechanics for Engineers diagram, the direction k to / represents a thrust of the bar KL on joint 5 ; the bar KL is therefore in compression. 165. Simple Roof=frame. Fig. 190 shows a simple roof-frame and its stress diagram when carrying three equal vertical loads on three joints and supported at the extremities of the span. FIG. 190. The reactions DE and EA at the supports are each obvi- ously equal to half the total load, i.e. e falls midway between a and d in the stress diagram. The correct characteristic order of the letters round the joints (Art. 163) is, with the lettering here adopted, clockwise. 1 66. Loaded Strings and Chains. Although not coming within the general meaning of the word " frame," stress Elements of Graphical Statics 243 diagrams can be drawn for a structure consisting partly of perfectly flexible chains or ropes, provided the loads are such as will cause only tension in flexible members. Consider a flexible cord or chain, Xi23Y (Fig. 191), sus- pended from points X and Y, and having vertical loads of W, Wj, W 2 , and W 3 suspended from points i, z, and 3 respectively. Denoting the spaces according to Bow's notation by the letters A, B, C, D, and O, as shown above, the tensions in the strings Xi or AO and i 2 or BO must have a resultant at i equal to Wj vertically upward, to balance the load at i. If triangles of forces, abo, bco, and cdo, be drawn for the points 1,2, and 3 respectively, the sides bo and co appear in two of them, and, as in Art. 163, the three vector triangles maybe included in a single vector diagram, as shown at the right-hand by the figure abcdo. The lines ao, bo, co, and do represent the tensions in the string crossing the spaces A, B, C, and D respectively. If a horizontal line, 0H, be drawn from o to meet the line abed in H, the length of this line represents the horizontal component of the tensions in the strings, which is evidently constant through- out the whole. (The tension changes only from one space to the neighbouring one by the vector addition of the intermediate vertical load.) The pull on the support X is represented by 244 Mechanics for Engineers ao, the vertical component of which is aU. ; the pull on Y is represented by od, the vertical component of which is Hd. A comparison with Art. 157 will show that the various sections of the string Xi23Y are in the same lines as the sides of a funicular polygon for the vertical forces W ls W 2 , and W 3 , corresponding to the pole o . If different lengths of string are attached to X and Y and carry the same loads, W 1( W 2 , and W 3 , in the lines AB, BC, and CD respectively, they will have different configurations ; the longer the string the steeper will be its various slopes corresponding to shorter pole distances, H0, i.e. to smaller horizontal tensions throughout. A short string will involve a great distance of the pole o from the line abed, i.e. a great horizontal tension, with smaller inclinations of the various sections of the string. The reader should sketch for himself the shape of a string connecting X to Y, with various values of the horizontal tension H0, the vertical loads remain- ing unaltered, in order to appreciate fully how great are the tensions in a very short string. A chain with hinged links, carrying vertical loads at the joints, will occupy the same shape as a string of the same length carrying the same loads. Such chains are used in sus- pension bridges. The shape of the string or chain to carry given loads in assigned vertical lines of action can readily be found for any given horizontal tension, H<?, by drawing the various sections parallel to the corresponding lines radiating from o, e.g. AO or Xi parallel to do (Fig. 191). Example i. A string hangs from two points, X and Y, 5 feet apart, X being 3 feet above Y. Loads of 5, 3, and 4 Ibs. are attached to the string so that their lines of action are i, 2, and 3 feet respectively from X. If the horizontal tension of the string is 6 Ibs., draw its shape. The horizontal distance ZY (Fig. 192) of X from Y is \/5 2 - 3" = 4 feet so that the three loads divide the horizontal span into four equal parts. Let Vx and V Y be the vertical components of the tension of the string at X and Y respectively. Elements of Graphical Statics 245 The horizontal tension is constant, and equal to 6 Ibs. Taking moments about Y (Fig. 192) Clockwise. Contra-clockwise. V X x 4 = (4 x i) 4- (3 x 2) + (5 x 3) + (6 x 3) Ib.-feet 4V X = 4 + 6+ 15 + 18 = 43 V x = -^ =1075 Ibs. Since the vertical and horizontal components of the tension of the string at X are known, its direction is known. The direction of each section of string might similarly be found. Set out the vector polygon abed, and draw the horizontal line Ho to represent 6 Ibs. horizontal tension from H, aH being measured along abed of such a length as to represent the vertical component 1075 Ibs. of the string at X. Join o to , l>, c, and d. Starting from X or Y, draw in the lines across spaces A, B, C, and D parallel respectively to ao, bo, co, and do (as in Art. 157). The funicular polygon so drawn is the shape of the string. Example 2. A chain is attached to two points, X and Y, X being I foot above Y and 7 feet horizontally from it. Weights of 20, 27, and 22 Ibs. are to be hung on the chain at horizontal distances of 2, 4, and 6 feet from X. The chain is to pass through a point P in the vertical plane of X and Y, 4 feet below, and 3 feet 246 Mechanics for Engineers horizontally from X. Find the shape of the chain and the tensions at its ends. Let V x and V Y be the vertical components of the tension at X and Y respectively, and let H be the constant horizontal tension throughout. Taking moments about Y (Fig. 193) Clockwise. Contra-clockwise. V x x 7 = (H x i) + (20 x 5) + (27 x 3) + (22 x i) 7V X = H + 203 Ibs.-feet (i) Taking moments about P of the forces on the chain between X and P Clockwise. Contra-clockwise. V x x 3 = H x 4 + (20 x i) 3V X = 4H + 20 (2) and 28V X = 4H + 812 from (i) hence 2$V X = 792 V x = 3i'681bs. H = 7V X - 203 = 22176 - 203 = 1876 Ibs. Draw the open polygon of forces, abed (a straight line), and set Elements of Graphical Statics 247 off am from a to the same scale, 31*68 Ibs. downwards. From in set off mo to represent 1876 Ibs. horizontally to the right of m. Then the vector ao am + mo = tension in the string XZ, which pulls at X in the direction XZ. By drawing XZ parallel to ao the direction of the first section of the chain is obtained, and by drawing from Z a line parallel to bo to meet the line of action BC, the second section is outlined. Similarly, by continuing the polygon by lines parallel to co and do the complete shape of the chain between X and Y is obtained. The tension ao at X scales 37 Ibs., and the tension od at Y scales 44 Ibs. 167. Distributed Load. If the number of points at which the same total load is attached to the string (Fig. 191) be increased, the funicular polygon corresponding to its shape will have a larger number of shorter sides, approximating, if the number of loads be increased indefinitely, to a smooth curve. This case corresponds to that of a heavy chain or string hanging between two points with no vertical load but its own weight. If the dip of the chain from the straight line join- ing the points of the attachment is small, the load per unit of horizontal span is nearly uniform provided the weight of chain per unit length is uniform. In this case an approximation to the shape of the chain may be found by dividing the span into a number of sections of equal length and taking the load on each portion as concentrated at the mid-point of that section. The funicular polygon for such a system of loads will have one side more than the number into which the span has been divided ; the approximation may be made closer by taking more parts. The true curve has all the sides of all such poly- gons as tangents, or is the curve inscribed in such a polygon. The polygons obtained by dividing a span into one, two, and four equal parts, and the approximate true curve for a uniform string stretched with a moderate tension, are shown in Fig. 194. Note that the dip QP would be less if the tensions OH, OA ? etc., were increased. 1 68. The relations between the dip, weight, and tension of a stretched string or chain, assuming perfect flexibility, can 248 Mechanics for Engineers more conveniently be found by ordinary calculation than by graphical methods. Assuming that the dip is small and the load per horizontal Q FIG. 194. foot of span is uniform throughout, the equilibrium of a portion AP (Fig. 195) of horizontal lengths, measured from the lowest point A, may be considered. l FIG. 195. Let w = weight per unit horizontal length of cord or chain ; y = vertical height of P above A, viz. PQ (Fig. 195) ; T = the tension (which is horizontal) at A ; T = the tension at P acting in a line tangential to the curve at P. Elements of Graphical Statics 249 The weight of portion AP is then wx, and the line of action of the resultant weight is midway between AB and PQ, i.e. at OC a distance - from either. 2 Taking moments about the point P T x PQ = wx X - 2 wy? orTX7 = WX' S TT This relation shows that the curve of the string is a parabola. If d = the total dip AB, and / = the span of the string or chain, taking moments about N of the forces on the portion AN B which gives the relation between the dip, the span, and the horizontal tension. Returning to the portion AP, if the vector triangle rst be drawn for the forces acting upon it, the angle 6 which the tangent to the curve at P makes with the horizontal is given by the relation xw st - Y = - = tan6 Also the tension T' at P is T sec 6, or T' = Ti+tan 2 = and at the ends where x = - 2 T' = 250 Mechanics for Engineers And since T = -7, the tension at N or M is wl I / 2 , wl' 1 ~2\i i6d 2 f 1 wl* d which does not greatly exceed -x-j (or T), if -= is small. Example. A copper trolley-wire weighs \ Ib. per foot length ; it is stretched between two poles 50 feet apart, and has a horizontal tension of 2000 Ibs. Find the dip in the middle of the span. Let d the dip in feet. The weight of the wire in the half-span BC (Fig. 196) is 25 x \ = 12-5 Ibs. The distance of the e.g. of the wire BC from B is practically I2'5 feet horizontally. Taking moments about B of the forces on the portion BC 2OOO X d = I2'5 X I2'5 d = 0^078 1 2 foot = 0-938 inch EXAMPLES XX. i. A roof principal, shown in Fig. 197, carries loads of 4, 7, and 5 tons in the positions shown. It is simply supported at the extremities of a span FIG. 197. of 40 feet. The total rise of the roof is 14 feet, and the distances PQ and Elements of Graphical Statics 251 RS are each 5'4 feet. Draw the stress diagram and find the stress in each member of the frame. 2. A Warren girder (Fig. 198), made up of bars of equal lengths, carries a single load of 5 tons as shown. Draw the stress diagram and scale off FIG. 198. the forces in each member ; check the results by the method of sections (Art. 98). 3. Draw the stress diagram for the roof-frame in Fig. 199 under the 4lons JOfeet given loads. The main rafters are inclined at 30 to the horizontal, and are each divided by the joints into three equal lengths. 4. A chain connects two points on the same level and 10 feet apart ; it has suspended from it four loads, each of 50 Ibs., at equal horizontal intervals along the span. If the tension in the middle section is 90 Ibs., draw the shape of the chain, measure the inclination to the horizontal, and the tension of the end section. 5. Find the shape of a string connecting two points 8 feet horizontally apart, one being I foot above the other, when it has suspended from it weights of 5, 7, and 4 Ibs. at horizontal distances of 2, 5, and 6 feet respectively from the higher end, the horizontal tension of the string being 6 Ibs. 6. A light chain connects two points, X and Y, 12 feet horizontally apart, X being 2 feet above Y. Loads of 15, 20, and 25 Ibs. are suspended from the chain at horizontal distances of 3, 5, and 8 feet respectively from X. The chain passes through a point 7 feet horizontally from X and 4 feet 252 Mechanics for Engineers below it. Draw the shape of the chain. How far is the point of suspension of the is-lb. load from X ? 7. A wire is stretched horizontally, with a tension of 50 Ibs., between two posts 60 feet apart. If the wire weighs 0^03 Ib. per foot, find the sag of the wire in inches. 8. A wire weighing o'oi Ib. per foot is stretched between posts 40 feet apart. What must be the tension in the wire in order to reduce the sag to 2 inches ? 9. A wire which must not be stretched with a tension exceeding 70 Ibs. is to be carried on supporting poles, and the sag between two poles is not to exceed i'5 inches. If the weight of the wire is 0^025 Ib. per foot, find the greatest distance the poles may be placed apart. APPENDIX UNITS AND THEIR DIMENSIONS Units. To express the magnitude of any physical quantity it has to be stated in terms of a unit of its own kind. Thus by stating that a stick is 275 feet long, we are using the foot as the unit of length. Fundamental and Derived Units. We have seen that the different quantities in common use in the science of mechanics have certain relations to one another. If the units of certain selected quantities are arbitrarily fixed, it is possible to determine the units of other quantities by means of their relations to the selected ones. The units arbitrarily fixed are spoken of as fundamental units, and those depending upon them as derived units. Fundamental Units. There are two systems of units in general use in this country. In the C.G.S. system (Art. 42), which is commonly used in physical science, the units chosen as funda- mental and arbitrarily fixed are those of length, mass, and time, viz. the centimetre, gramme, and second. In the British gravitational system the fundamental units chosen are those of length, force, and time, viz. the foot, the pound, i.e. the weight of i Ib. of matter at some standard place, and the mean solar second. The latter system of units has every claim to the name " absolute," for three units are fixed, and the other mechanical units are derived from them by fixed relations. The weight of a body of given mass varies at different parts of the earth's surface in whatever units its mass is measured. The value of i Ib. force, however, does not vary, since it has been defined as the weight of a fixed mass at a fixed place. Dimensions of Derived Units. (a) Length Mass Time Systems. In any such system other than, say, the C.G.S. system, let the. unit of length be L centimetres, the unit of mass M grammes, and the unit of time be T seconds. 254 Mechanics for Engineers Then the unit of area will be L x L or L 2 square centimetres, i.e. it varies as the square of the magnitude of the unit of length. Similarly, we may derive the other important mechanical units as follows : Unit volume L x L x L or L 3 cubic centimetres, or unit volume varies as L 3 . Unit velocity is L centimetres in T seconds = ~ centimetres per second, or LT -1 centimetres per second. Unit acceleration is = centimetres per second in T seconds = ^2 centimetres per second, or LT~ 2 centimetres per second. Unit momentum is that of M grammes moving ~ centimetres per ML second, i.e. '-^- C.G.S. units of momentum, or MLT~ : C.G.S. units. MLT -1 Unit force is unit change of momentum in T seconds, or ^ units in one second, or MLT~ 2 dynes (C.G.S. units offeree). Unit impulse is given by unit force (MLT~ 2 C.G.S. units) acting for unit time, T seconds generating a change of momentum (or impulse) MLT" 1 C.G.S. units. Unit work is that done by unit force (MLT~ 2 dynes) acting through L centimetres, i.e. ML 2 T~ 2 centimetre-dynes or ergs. Unit kinetic energy is that possessed by unit mass, M grammes moving with unit velocity (LT- 1 ),*.*. |M(LT- 1 ) 2 =1ML 2 T- 2 C.G.S. units. ML 2 T~ 2 Unit power is unit work in unit time T seconds, or , or ML 2 T~ 3 ergs per second. x , . , L units of arc ,. , . . , Note that unit angle T ; j-. = i radian, and is inde- L units radius pendent of the units of length, mass, or time. Unit angular velocity is unit velocity ^ divided by unit radius L centimetres, or LT -1 -i- L = T Appendix 255 Unit moment of momentum or angular momentum is unit momentum MLT" 1 at unit perpendicular distance L, or ML 2 *!- 1 C.G.S. units. Unit moment of force is unit force MLT~ 2 at unit distance L centimetres, or ML 2 T~ 2 C.G.S. units. Unit rate of change of angular momentum is ML 2 T -1 C.G.S. units in unit time T seconds = ML 2 T~ 2 C.G.S. units. Unit moment of inertia is that of unit mass M grammes at unit distance L centimetres, which is ML 2 C.G.S. units. Thus each derived unit depends on certain powers of the magnitudes of the fundamental units, or has certain dimensions of those units. (U) The dimensions of the same quantities in terms of the three fundamental units of length, force, and time may be similarly written as follows : Quantity. Length. Force. Time. Velocity. Acceleration. Mass. Momentum. Impulse. Work. Kinetic energy. Power. Angular momentum. Moment of force. Rate of change of angular momentum. Dimensions. L. F. T. or LT- 1 . -:- T or LT- 2 . Force T- Acceleration FT. FT. FL. or FL-'T 2 . FLT- 1 . FLT. FL. FL. ' Symbolical formulae and equations may be checked by testing if the dimensions of the terms are correct. Each term on either side of an algebraic equation having a physical meaning must necessarily be of the same dimensions. ANSWERS TO EXAMPLES EXAMPLES I. (i) 0-305 foot per second per second. (2) 5*5 seconds; 121 feet. (3) 7i'77 feet per second. (4) 3-053 seconds. (5) 89-5 feet ; 447-5 feet ; 440 '4 feet. (6) 5 '63 seconds after the first projection ; 278 feet. (7) 567 feet per second. (8) 4-5, 14-6, and 11*4 feet per second. (10) o'57 and 0-393 foot per second per second ; 880 feet. (n) 77 '3 feet ; 2-9 seconds. EXAMPLES If. (1) 4-88 feet per second ; 35 23' to the horizontal velocity. (2) 405 feet per second ; 294 feet per second. (3) 53 up-stream ; 2 minutes 16-4 seconds. (4) 10 6 south of west. (5) 19*54 knots per hour ; 5 hours 7'2 minutes ; 12 8' west of south. (6) 48 minutes ; 9*6 miles ; I2'8 miles. (7) 154-2 feet per second per second ; 2i"5 south of west. (8) 2-59 seconds. (9) 5-04; 4716. (ic) 16-83 feet per second, (n) 35*2 radians per second ; 2*581 radians per second per second. (12) 135 revolutions and 1-5 minutes from full speed. EXAMPLES III. (l) 2735 units; 182,333 Ibs. or 8i"4 tons. (2) $ or 1-172 to I. (3) 2*8 centimetres per second. (4) 9802 Ibs. (5) I 5'33 Iks. 5 9'53 units per second in direction of jet ; 9*53 Ibs. (6) 45-3. (7) 4720 Ibs. (8) 10-43 tons inclined downwards at 16 40' to horizontal. (9) 2-91 units ; 727-5 Ibs. (10) 8750 units ; 8-57 miles per hour. EXAMPLES IV. (1)67-8 Ibs. (2) 17-48 Ibs. (4) 34-54 feet. (5) 23-44 feet per second ; 255,000 Ibs. (6) 1005 feet per second. (7) 154 Ibs. ; 126 Ibs. ; 6*9 feet per second per second. (8) ii'243cwt. (9) 9-66 feet; 14-93 Ibs. (10) 4*69 grammes ; 477 centimetres. (n) 6-44 feet per second per second ; 4 Ibs. (12) i -027 Ibs. (13) 48-9 Ibs. Answers to Examples 257 EXAMPLES V. (1) 1 60 horse-power ; 303 '36 horse-power ; 1 6 '64 horse-power. (2) 15*75 Ibs. per ton. (3) 22' 15 miles per hour. (4) 929; 1253. (5) 147-5 horse-power. (6) 0*347 horse-power. (7) 60 foot-lbs. (8) 350,000 foot-lbs. 800,000 foot-lbs. (9) 1,360,000 foot-lbs. EXAMPLES VI. (i) 57' i horse-power. (3) 6570 Ib.-feet. (5) 534O inch-lbs. ; 2220 inch-lbs. (2) 39,390 Ib.-feet. (4) 609 inch-lbs. (6) i2'8 horse-power. EXAMPLES VII. (i) 12,420,000 foot-lbs. ; 4,140,000 Ibs. (3) 37>74 inch-lbs. ; 35,940 inch-lbs. (5) 7'O2 horse-power. (7) 19-6 horse-power. (9) 10*5 feet per second ; 467 Ibs. (ii) 2886 foot-lbs. EXAMPLES VIII. (2) 27-8 feet per second. (4) 25-5 horse-power. (6) 7-25 horse-power. (8) 8'65 seconds. (10) 15-3 seconds. (12) 500,000 foot-lbs. (i) 68-5 (2) 1 1-85 miles per hour. (4) 4-25 inches. (5) 3052 feet. (7) 47 to horizontal. (8) 52'5 ; i '64 times the weight of the stone. (9) !5 P er cent, increase. (10) 66'4 ; 72-7 ; 59-3 revolutions per minute. (12) 38'33 ; 35'68 feet per second, 7*79 ; 6'28 Ibs. EXAMPLES IX. (3) 2672 feet. (6) 20 miles per hour. (ii) (1) o 855, 1*56, i'8i feet per second ; 8-05, 5-96, 4-4 feet per second per second. (2) f inch. (3) 1654, 827, 1474 Ibs. (4) X 53'3- (6) 0*342 second. (7) 1*103 second ; 67*3 feet per second per second. (8) 31-23. (9) i to 1-0073. EXAMPLES X. (i) 14*65 Ibs. ; 17-9 Ibs. (3) 9-6 tons tension ; 55*6 tons tension. (5) 2250 Ibs. ; 2890 Ibs. (2) 3 Ibs. ; 13 Ibs. (4) 4i c "7 south of west ; 720 Ibs. (6) 220 Ibs. ; 58*5 Ibs. 258 Mechanics for Engineers EXAMPLES XI. (i) 0-154; 8 '8 (2) 2-97 Ibs. ; 8 -5 to horizontal. (3) 14-51 Ibs. (4) O'6 times the weight of log ; 36'8 to horizontal. (5) io 0> 4 (6) 0-3066 horse-power. (7) 179 horse-power. (8) 3*84 horse-power. (9) 3 '4 feet per second per second ; 3-57 Ibs. (10) 4-5 tons; 31-9 seconds. (n) 3820 Ibs. EXAMPLES XI [. (I) 261 Ibs. (2) 16-97 Ibs. ; 4-12 Ibs. (3) Left, 5*242 tons ; right, 5*008 tons. (4) Left, 10 tons ; right, 3 tons ; end, 2*824 ions. (5) I '039 inches. (6) 5-737 feet from end. EXAMPLES XIII. (1) Tension, 2i'68 Ibs. ; pressure, 33-4 Ibs. ; ig"j to vertical (2) 0-1264. (3) 36. (4) 15*3 Ibs. at hinge ; 8*25 Ibs. at free end. (5) 395 Ibs. at A ; 2954 Ibs. at C. (6) n-2 Ibs. cutting AD 2'i inches from A, inclined I9'3 to DA. (7) 4*3 tons ; 3-46 tons ; 46-7 to horizontal. (8) 8*2 tons compression ; 4-39 tons tension ; 4 tons tension. (9) 8*78 tons tension ; 25-6 tons compression ; 2P22 tons tension. , EXAMPLES XIV. (i) 1-27 feet from middle. (2) 2 '08 inches. (3) 43 inches. (4)' 1*633 f eet > I<22 5 feet. (5) 4-18 inches ; 4*08 inches. (6) lo'i inches ; 5-5 Ibs. (7) 2-98 inches. (8) 27-2 inches. (9) 975 inches. (10) 1293 Ib.-feet ; 103-5 Ibs. per square foot, (n) 11-91 inches. (12) 4-82 inches. (13) 4 feet 5-1 inches. (14) 0-17" Ib. (15) 0-197 Ib. ; 0-384 Ib. EXAMPLES XV. (i) 19*48 inches ; 16-98 inches. (2) I2'i6 inches. (3) 6"o8 inches. (4) 15-4 inches. (5) 2-52 inches from outside of flange. (6) 4-76 inches. (7) O'2O2 inch from centre. (8) 16*6 inches. (9) 5'36 inches. (10) 33-99 inches. Answers to Examples 259 EXAMPLES XVI (i) 16 and 8 tons. (2) 25 and 16 tons. (3) Left, 16-5 tons ; right, 33-4 tons. (4) 53 10'. (5) 16-43 inches; 4-41 inches. (6) 3-53 inches. (7) 3-67 inches. (8) 8000 foot-lbs. (9) 1 188 foot-lbs. (10) 140,000 ; 74,400 foot-lbs. (1.1) 75,600 foot-lbs. (12) 2514 foot-lbs. (13) 1 10-3 Ibs. (14) 5-1 1 Ibs. (15) 37'6 square inches. (16) 7-85 cubic inches. (17) 4 feet 3-9 inches. EXAMPLES XVII. (i) 312 (inches) 4 . (2) 405 (inches) 4 ; 4-29 inches. (3) 1 9S (inches) 4 ; 2^98 inches. (4) 290 (inches) 4 . (5) 5'5 2 3 inches. (6) 0-887 gravitational units. .(7) (8) 16*1 inches ; 35*15 gravitational units. EXAMPLES XVIII. (0 3 6 47 gravitational units. (2) 13,215 gravitational units. (3) 10 minutes 46 seconds ; 323. (4) 17-48 Ibs. (5) 35olb.-feet. (6) 2-134 gravitational units ; 6-83 inches. (7) 141 '3- (8) 7-71 inches. (9) 22 feet per second ; 31*06 feet per second. (10) 14*85 feet per second ; 16-94 f eet per second. (11) 3-314 feet ; 3819 gravitational units. (12) 53-7. (13) 0-0274 units. (14) 125-5. (15) 117-5 foot-tons. (16) 167 feet per second. (17) 23 sin a feet per second per second. ( EXAMPLES XIX. (1) 6-47 Ibs. ; 0-016 inch ; IO2'6. (2) 7-8 Ibs. ; 0-013 inch ; 36. (3) 17-7 right ; II'3 left. (4) 21 -6 Ibs. ; 134 measured clockwise. (5) 3'4 38*i5 34'85 tons-feet. (6) 38-15. 34'85, *9'6, 25-95 tons-feet. (7) 1-65 tons; 2-35 tons; 3-65 tons. (8) 25 tons-feet ; nil ; 18-75 tons-feet ; 2-5 tons. EXAMPLES XX. (4) 48 ; 134-5 Ibs. (6) 4-06 feet. (7) 3*24 inches. (8) 12 Ibs. (9) 53 feet. EXAMINATION QUESTIONS Questions selected from the Mechanics Examinations Intermediate (Engineering) Science of London University. r. What is implied in the rule : the product of the diameter of a wheel in feet, and of the revolutions per minute, divided by 28, is the speed in miles an hour ? Also in the rule : three times the number of telegraph posts per minute is the speed in miles an hour ? (i93-) 2. Continuous breaks are now capable of reducing the speed of a train 3^ miles an hour every second, and take 2 seconds to be applied. Show in a tabular form the length of an emergency stop at speeds of 3f, 7^, 15, 30, 45, and 60 miles an hour. Compare the retardation with gravity ; express the resisting force in pounds per ton ; calculate the coefficient of adhesion of the break shoe and rail with the wheel ; and sketch the arrangement. 093-) 3. Prove that the horse-power required to overcome a resistance of R Ibs. at a speed of S miles an hour is RS -i- 375. Calculate the horse-power of a locomotive drawing a train of 200 tons up an incline of i in 200 at 50 miles an hour, taking the road and air resistance at this speed at 28 Ibs. a ton. ( I 93-) 4. If W tons is transported from rest to rest a distance s feet in / seconds, being accelerated for a distance jj and time t l by a force Pj tons up to velocity v feet per second, and then brought to rest by a force P 2 tons acting for / 2 seconds through s 2 feet _ p g Wz/ 2 S 9 S = 2 = 2 7 = 2 7 * Examination Questions 261 A train of 100 tons gross, fitted with continuous breaks, is to be run on a level line between stations one-third of a mile apart, at an average speed of 12 miles an hour, including two-thirds of a minute stop at each station. Prove that the weight on the driving wheels must exceed 22| tons, with an adhesion of one-sixth, neglecting road-resistance and delay in application of the breaks. (1903-) 5. Give a graphical representation of the relative motion of a piston and crank, when the connecting rod is long enough for its obliquity to be neglected ; and prove that at R revolutions a o minute, the piston velocity is times the geometric mean of the distance from the two ends of the stroke. Prove that if the piston weighs W Ibs., the force in pounds which gives its acceleration is W ir 2 R 2 (distance in feet from mid-point of stroke). g 900 (1903-) 6. Write down the formula for the time of swing of a simple pendulum, and calculate the percentage of its change due to i per cent, change in length or gravity, or both. Prove that the line in Question 4 could be worked principally by gravity if the road is curved downward between the stations to a radius of about 11,740 feet, implying a dip of 33 feet between the stations, a gradient at the stations of i in 13, and a maximum running velocity of 31 miles an hour. ( I 93-) 7. Prove that if a hammer weighing W tons falling h feet drives a pile weighing w tons a feet into the ground, the average resistance of the ground in tons is W 2 h W + w a Prove that the energy dissipated at the impact is diminished by increasing W (1903.) w 8. Prove that the total kinetic energy stored up in a train of railway carriages, weighing W tons gross, when moving at v feet per second is where W t denotes the weight of the wheels in tons, a their radius, and / their radius of gyration. 262 Mechanics for Engineers Prove that W in the equations of Question 4 must be increased W ? by V- to allow for the rotary inertia of the wheels. (1903.) 9. Determine graphically, by the funicular polygon, the reaction of the supports of a horizontal beam, loaded with given weights at two given points. Prove that the bending moment at any point of the beam is represented by the vertical depth of the funicular polygon. (1903-) 10. A wheel is making 200 revolutions per minute, and after 10 seconds its speed has fallen to 150 revolutions per minute. If the angular retardation be constant, how many more revolutions will it make before coming to rest? (!9O4-) 11. A piston is connected to a flywheel by a crank and con- necting rod in the usual manner. If the angular velocity of the flywheel be constant, show that if the connecting rod be sufficiently long the motion of the piston will be approximately simple-harmonic ; and find the velocity of the piston in any position. If A, B, C, D, E be five equidistant positions of the piston, A and E being the ends of its stroke, prove that the piston takes twice as long to move from A to B as it does from B to C. (1904.) 12. A train whose weight is 250 tons runs at a uniform speed down an incline of i in 200, the steam being shut off and the brakes not applied, and on reaching the foot of the incline it runs 800 yards on the level before coming to rest. What was its original speed in miles per hour ? [The fractional resistance is supposed to be the same in each case.] (1904.) 13. A weight A hangs by a string and makes small lateral oscillations like a pendulum ; another weight, B, is suspended by a spiral spring, and makes vertical oscillations. Explain why an addition to B alters the period of its oscillations, whilst an addition to A does not. Also find exactly how the period of B varies with the weight. (1904.) 14. A steel disc of thickness t and outer radius a is keyed on to a cylindrical steel shaft of radius b and length /, and the centre of the disc is at a distance c from one end of the shaft. Find the distance from this end of the shaft of the mass-centre of the whole. (1904.) 15. A uniform bar 6 feet long can turn freely in a vertical plane about a horizontal axis through one end. If it be just started from Examination Questions 263 the position of unstable equilibrium, find (in feet per second) the velocity of the free end at the instant of passing through its lowest position. (1904-) 1 6. Explain why, as a man ascends a ladder, the tendency of its foot to slip increases. A man weighing 13 stone stands on the top of a ladder 20 feet long, its foot being 6 feet from the wall. How much is the horizontal pressure of the foot on the ground increased by his presence, the pressure on the wall being assumed to be horizontal ? (1904.) 17. A horizontal beam 20 feet long, supported at the ends, carries loads of 3, 2, 5, 4 cwts. at distances of 3, 7, 12, 15 feet respectively from one end. Find by means of a funicular polygon (drawn to scale) the pressures on the two ends, and test the accuracy of your drawing by numerical computation. (1904.) Questions selected from the Associate Members' Examinations of the Institution of Civil Engineers. 1. A beam 20 feet long is supported on two supports 3 feet from each end of the beam ; weights of lolbs. and 20 Ibs. are suspended from the two ends of the beam. Draw, to scale, the bending- moment and shearing-force diagrams ; and, in particular, estimate their values at the central section of the beam. (I.C.E., February, 1905.) 2. The speed of a motor car is determined by observing the times of passing a number of marks placed 500 feet apart. The time of traversing the distance between the first and second posts was 20 seconds, and between the second and third 19 seconds. If the acceleration of the car is constant, find its magnitude in feet per second per second, and also the velocity in miles per hour at the instant it passes the first post. (I.C.E., February, 1905.) 3. In a bicycle, the length of the cranks is 7 inches, the diameter of the back wheel is 28 inches, and the gearing is such that the wheel rotates -2\ times as fast as the pedals. If the weight of the cyclist and machine together is 160 Ibs., estimate the force which will have to be applied to the pedal to increase the speed uniformly from 4 to 12 miles an hour in 20 seconds, frictional losses being neglected. (I.C.E., February, 1905.) 4. A thin circular disc, 12 inches radius, has a projecting axle \ inch diameter on either side. The ends of this axle rest on two parallel inclined straight edges inclined at a slope of i in 40, the 264 Mechanics for Engineers lower part of the disc hanging between the two. The disc rolls from rest through I foot in 53^ seconds. Neglecting the weight of the axle and frictional resistances, find the value of g. (I.C.E., February, 1905.) 5. A gate 6 feet high and 4 feet wide, weighing 100 Ibs., hangs from a rail by 2 wheels at its upper corners. The left-hand wheel having seized, skids along the rail with a coefficient of friction of \. The other wheel is frictionless. Find the horizontal force that will push the gate steadily along from left to right if applied 2 feet below the rail. You may solve either analytically or by the force- and-link polygon. (I.C.E., October, '1904.) 6. A ladder, whose centre of gravity is at the middle of its length, rests on the ground and against a vertical wall ; the co- efficients of friction of the ladder against both being \. Find the ladder's inclination to the ground when just on the point of slipping. (I.C.E., October, 1904.) 7. The faceplate of a lathe has a rectangular slab of cast iron bolted to it, and rotates at 480 revolutions per minute. The slab is 8 inches by 12 inches by 30 inches (the length being radial). Its outside is flush with the edge of the faceplate, which is 48 inches diameter. Find the centrifugal force. (Cast iron weighs \ Ib. per cubic inch.) Where must a circular weight of 300 Ibs. be placed to balance the slab? (I.C.E., October, 1904.) 8. A boom 30 feet long, weighing 2 tons, is hinged at one end, and is being lowered by a rope at the other. When just horizontal the rope snaps. Find the reaction on the hinge. (I.C.E., October, 1904.) 9. A beam ABCD, whose length, AD, is 50 feet, is supported at each end, and carries a weight of 2 tons at B, 10 feet from A, and a weight of 2 -5 tons at C, 20 feet from D. Calculate the shearing force at the centre of the span, and sketch the diagram of shearing forces. (I.C.E., October, 1904.) 10. Referring to the loaded beam described in the last question, how much additional load would have to be put on at the point B in order to reduce the shearing force at the centre of the span to zero? (I.C.E., October, 1904.) 11. Estimate the super-elevation which ought to be given to the outer rail when a train moves round a curve of 2000 feet radius at a speed of 60 miles an hour, the gauge being 4 feet 8| inches. (I.C.E., February, 1904.) 12. Show that in simple-harmonic motion the acceleration is proportional to the displacement from the mid-point of the path, and that the time of a small oscillation of a simple pendulum of Examination Questions 265 length / is 2ir. /-. Deduce the length of the simple pendulum \i *> which has the same time of oscillation as a uniform rod of length L suspended at one end. (I.C.E., February, 1904.) 13. To a passenger in a train moving at the rate of 40 miles an hour, the rain appears to be rushing downwards and towards him at an angle of 20 with the horizontal. If the rain is actually falling in a vertical direction, find the velocity of the rain-drops in feet per second. (I.C.E., February, 1904.) 14. If it take 600 useful horse-power to draw a train of 335 tons up a gradient of i in 264 at a uniform speed of 40 miles an hour, estimate the resistance per ton other than that due to ascending against gravity, and deduce the uniform speed on the level when developing the above power. (I.C.E., February, 1904.) 15. In a steam-hammer the diameter of the piston is 36 inches, the total weight of the hammer and piston is 20 tons, and the effective steam pressure is 40 Ibs. per square inch. Find the acceleration with which the hammer descends, and its velocity after descending through a distance of 4 feet. If the hammer then come in contact with the iron, and compress it through a distance of i inch, find the mean force of compression. (I.C.E., February, 1904.) 16. A uniform circular plate, i foot in diameter and weighing 4 Ibs., is hung in a horizontal plane by three fine parallel cords from the ceiling, and when set in small torsional oscillations about a vertical axis is found to have a period of 3 seconds. A body whose moment of inertia is required is laid diametrically across it, and the period is found to be 5 seconds, the weight being 6 Ibs. Find the moment of inertia of the body about the axis of oscillation. (I.C.E., February, 1904.) 17. The acceleration of a train running on the level is found by hanging a short pendulum from the roof of a carriage and noticing the angle which the pendulum makes with the vertical. In one experiment the angle of inclination was 5 : estimate the acceleration of the train in feet per second per second and in miles per hour per hour. (I.C.E., February, 1904.) 1 8. Two weights, one of 2 Ibs. and the other of i lb., are con- nected by a massless string which passes over a smooth peg. Find the tension in the string and the distance moved through by either weight, from rest, in 2 seconds. (I.C.E., February, 1904.) 19. A solid circular cast-iron disc, 20 inches in diameter and 2 inches thick (weighing 0*25 lb. per cubic inch), is mounted on ball bearings. A weight of 10 Ibs. is suspended by means of a 266 Mechanics for Engineers string wound round the axle, which is 3 inches in diameter, and the weight is released and disconnected after falling 10 feet. Neglecting friction, find the kinetic energy stored in the wheel, and the revolutions per minute the wheel is making when the weight is disconnected, and also the time it would continue to run against a tangential resistance of \ Ib. applied at the circumference of the axle. (I.C.E., February, 1904.) 20. A girder 20 feet long carries a distributed load of i ton per lineal foot over 6 feet of its length, the load commencing at 3 feet from the left-hand abutment. Sketch the shearing-force and bending-moment diagrams, and find, independently, the magni- tude of the maximum bending moment and the section at which it occurs. (I.C.E., February, 1904.) 21. Explain what is meant by centripetal acceleration, and find its value when a particle describes a circle of radius r feet with a velocity of v feet per second. (I.C.E., October, 1903.) 22. In an electric railway the average distance between the stations is \ mile, the running time from start to stop ITT minutes, and the constant speed between the end of acceleration and beginning of retardation 25 miles an hour. If the acceleration and retardation be taken as uniform and numerically equal, find their values ; and if the weight of the train be 1 50 tons and the frictional resistance 1 1 Ibs. per ton, find the tractive force necessary to start on the level. (I.C.E., October, 1903.) 23. The mass of a flywheel may be assumed concentrated in the rim. If the diameter is 7 feet and the weight 2\ tons, estimate its kinetic energy when running at 250 revolutions per minute. Moreover, if the shaft be 6 inches in diameter and the coefficient of friction of the shaft in the bearings be 0-09, estimate the number of revolutions the flywheel will make before coming to rest. (I.C.E., October, 1903.) 24. A plane inclined at 20 to the horizontal carries a load of 1000 Ibs., and the angle of friction betweeen the load and the plane is 10. Obtain the least force in magnitude and direction which is necessary to pull the load up the plane. (I.C.E., October, 1903.) 25. State the second law of motion. A cage weighing 1000 Ibs. is being lowered down a mine by a cable. Find the tension in the cable, (i) when the speed is increasing at the rate of 5 feet per second per second ; (2) when the speed is uniform ; (3) when the speed is diminishing at the rate of 5 feet per second per second. The weight of the cable itself may be neglected. (I.C.E., October, 1903.) Examination Questions 267 26. Show that when a helical spring vibrates freely under the action of a weight, its periodic time is the same as that of a simple pendulum having a length equal to the static extension of the spring when carrying the weight, the mass of the spring itself being neglected. (I.C.E., October, 1903.) 27. A flywheel, supported on an axle 2 inches in diameter, is pulled round by a cord wound round the axle and carrying a weight. It is found that a weight of 4 Ibs. is just sufficient to overcome friction. A further weight of 16 Ibs., making 20 Ibs. in all, is applied, and two seconds after starting from rest it is found that the weight has descended a distance of 4 feet. Estimate the moment of inertia of the wheel about the axis of rotation in gravitational units. (I.C.E., October, 1903.) 28. With an automatic vacuum brake a train weighing 170 tons and going at 60 miles an hour on a down gradient of I in 100 was pulled up in a distance of 596 yards. Estimate the total resistance in pounds per ton ; and if the retardation is uniform, find the time taken to bring the train to rest. (I.C.E., October, 1903.) 29. A string, ABCD, hangs in a vertical plane, the ends A and D being fixed. A weight of 10 Ibs. is hung from the point B, and an unknown weight from the point C. The middle-portion BC is horizontal, and the portions AB and CD are inclined at 30 and 45 to the horizontal respectively. Determine the unknown weight and the tensions in the three portions of the string. (I.C.E , October, 1903.) 30. Two masses, of 10 Ibs. and 20 Ibs. respectively, are attached to a balanced disc at an angular distance apart of 90 and at radii 2 feet and 3 feet respectively. Find the resultant force on the axis when the disc is making 200 turns a minute ; and determine the angular position and magnitude of a mass placed at 2'5 feet radius which will make the force on the axis zero at all speeds. (I.C.E., October, 1903.) 31. A crane has a vertical crane-post, AB, 8 feet long, and a horizontal tie, BC, 6 feet long, AC being the jib. It turns in bearings at A and B, and the chain supporting the load passes over pulleys at C and A, and is then led away at 30 to AB. Find the stresses in the bars and thrusts in the bearings when lifting i ton at a uniform rate. (I.C.E., February, 1903.) 32. A man ascends a ladder resting on a rough horizontal floor against a smooth vertical wall. Determine, graphically or other- wise, the direction of the action between the foot of the ladder and the floor. (I.C.E., February, 1903.) 33. A train on a horizontal line of rails is accelerating the speed 268 Mechanics for Engineers uniformly so that a velocity of 60 miles an hour is being acquired in 176 seconds. A heavy weight is suspended freely from the roof of a carriage by a string. Calculate, or determine graphically, the inclination of the string to the vertical. (I.C.E., February, 1903.) 34. Explain the meaning of the term " centrifugal force." With what speed must a locomotive be running on level railway lines, forming a curve of 968 feet radius, if it produce a horizontal thrust on the outer rail equal to fa of its weight ? (I.C.E., February, 1903.) 35. Explain how to determine the relative velocity of two bodies. A is travelling due north at constant speed. When B is due west of A, and at a distance of 21 miles from it, B starts travelling north-east with the same constant speed as A. Determine graphically, or otherwise, the least distance which B attains from A. (I.C.E., February, 1903.) 36. Two men put a rail way- waggon weighing 5 tons into motion by exerting on it a force of 80 Ibs. The resistance of the waggon is 10 Ibs. per ton, or altogether 50 Ibs. How far will the waggon have moved in one minute ? Calculate at what fraction of a horse- power the men are working at 60 seconds after starting. (I.C.E., February, 1903.) 37. State and explain fully Newton's third law of motion. A loo-lb. shot leaves a gun horizontally with a muzzle velocity of 2000 feet per second. The gun and attachments, which recoil, weigh 4 tons. Find what the resistance must be that the recoil may be taken up in 4 feet, and compare the energy of recoil with the energy of translation of the shot. (I.C.E., February, 1903.) 38. An elastic string is used to lift a weight of 20 Ibs. How much energy must be exerted in raising it 3 feet, supposing the string to stretch i inch under a tension of i Ib. ? Represent it graphically. If the work of stretching the string is lost, what is the efficiency of this method of lifting ? (I.C.E., February, 1903.) 39. Explain how to determine graphically the relative velocity of two points the magnitudes and directions of whose velocities are known. Find the true course and velocity of a steamer steering due north by compass at 1 2 knots, through a 4-knot current setting south-west, and determine the alteration of direction by compass in order that the steamer should make a true northerly course. (I.C.E., October, 1902.) 40. Find, by graphic construction, the centre of gravity of a section of an I beam, top flange 4 inches by i inch ; web, between flanges, 14 inches by i| inches ; bottom flange 9 inches by 2 inches. (I.C.E., October, 1902.) Examination Questions 269 41. A crankshaft, diameter \i\ inches, weighs 12 tons, and it is pressed against the bearings by a force of 36 tons horizontally. Find the horse-power lost in friction at 90 revolutions per minute (coefficient of friction = o'o6.) (I.C.E., October, 1902.) Questions selected from the Board of Education Examinations in Applied Mechanics. (Reprinted by permission of the Controller of His Majesty's Stationery Office.) 1. A truck, weighing 5 tons without its wheels, rests on 4 wheels, which are circular discs. 40 inches in diameter, each weighing | ton, and moves down an incline of i in 60. Find the velocity of the truck in feet per second after moving 100 feet from rest, if the resistance due to friction is I per cent, of the weight. What percentage of the original potential energy has been wasted in friction ? (Stage 3, 1905.) 2. A flywheel is supported on an axle aj inches in diameter, and is rotated by a cord, which is wound round the axle and carries a weight. It is found by experiment that a weight of 5 Ibs. on the cord is just sufficient to overcome the friction and maintain steady motion. A load of 25 Ibs. is attached to the cord, and 3 seconds after starting from rest it is found that the weight has descended 5 feet. Find the moment of inertia of this wheel in engineers' units. If the wheel is a circular disc 3 feet in diameter, what is its weight ? (The thickness of the cord may be neglected.) (Stage 3, 1905.) 3. The angular position D of a rocking shaft at any time / is measured from a fixed position. Successive positions at intervals at g\j second have been determined as follows : Time t, se- ) conds j Position D, | radians ) O'O o'lo6 O'O2 0'208 0-04 Q'337 o'o6 0-487 0-08 0-651 O'lO O'SlQ O'I2 0-978 O-I4 mi 0*16 I '201 0-18 I '222 Find the change of angular position during the first interval from / = o to / = 0*02. Calculate the mean angular velocity during this interval in radians per second, and, on a time base, set this up as an ordinate at the middle of the interval. Repeat this for the 2/0 Mechanics for Engineers other intervals, tabulating the results and drawing the curve show- ing approximately angular velocity and time. In the same way, find a curve showing angular acceleration and time. Read off the angular acceleration in radians per second per second, when / = 0x175 second. (Stage 2, 1905.) 4. A motor car moves in a horizontal circle of 300 feet radius. The track makes sideways an angle of 10 with the horizontal plane. A plumb-line on the car makes an angle of 12 with what would be a vertical line on the car if it were at rest on a horizontal plane. What is the speed of the car? If the car is just not side- slipping, what is the coefficient of friction ? (Stage 3, 1904.) 5. A body whose weight is 350 Ibs. is being acted upon by a variable lifting force F Ibs. when it is at the height x feet from its position of rest. The mechanism is such that F depends upon x in the following way ; but the body will stop rising before the greatest x of the table is reached. Where will it stop ? i X ... o 15 *j 50 70 100 125 150 180 2IO F 530 525 56 490 425 300 210 160 no 90 Where does its velocity cease to increase and begin to diminish ? (Stage 3, 1904.) 6. Part of a machine weighing i ton is moving northwards at 60 feet per second. At the end of 0-05 second it is found to be moving to the east at 20 feet per second. What is the average force (find magnitude and direction) acting upon it during the interval ox>5 second ? What is meant by " average " in such a case ? What is meant by force by people who have to make exact calculations? (Stage 3, 1904.) 7. A flywheel and its shaft weigh 24,000 Ibs. ; its bearings, which are slack, are 9 inches diameter. If the coefficient of friction is ox>7, how many foot-pounds of work are wasted in overcoming friction in one revolution ? If the mean radius (or rather the radius of gyration) is 10 feet, what is the kinetic energy when the speed is 75 revolutions per minute ? If it is suddenly disconnected from its engine at this speed, in how many revolutions will it come to rest ? What is its average speed in coming to rest? In how many minutes will it come to rest ? (Stage 2, 1904.) 8. A train, weighing 250 tons, is moving at 40 miles per hour, and it is stopped in ten seconds. What is the average force during Examination Questions 271 these ten seconds causing this stoppage ? Define what is meant by force by people who have to make exact calculations. (Stage 2, 1904.) 9. A tram-car, weighing 15 tons, suddenly had the electric current cut off. At that instant its velocity was 16 miles per hour. Reckoning time from that instant, the following velocities, V, and times, /, were noted : V, miles per hour ... ... 16 14 12 10 /, seconds o 9^3 21 35 Calculate the average value of the retarding force, and find the average value of the velocity from / = o to / = 35. Also find the distance travelled between these times. (Advanced, 1903.) 10. A projectile has kinetic energy = 1,670,000 foot-lbs. at a velocity of 3000 feet per second. Later on its velocity is only 2000 feet per second. How much kinetic energy has it lost? What is the cause of this loss of energy? Calculate the kinetic energy of rotation of the projectile if its weight is 12 Ibs., and its radius of gyration is 075 inch, and its speed of rotation is 500 revolutions per second. (Advanced, 1903.) 1 1. A weight of 10 Ibs. is hung from a spring, and thereby causes the spring to elongate to the extent of 0^42 foot. If the weight is made to oscillate vertically, find the time of a complete vibration (neglect the mass of the spring itself). (Advanced, 1903.) 12. A flywheel weighs 5 tons and has a radius of gyration of 6 feet. What is its moment of inertia? It is at the end of a shaft 10 feet long, the other end of which is fixed. It is found that a torque of 200,000 Ib.-feet is sufficient to turn the wheel through i. The wheel is twisted slightly and then released : find the time of a complete vibration. How many vibrations per minute would it make ? (Honours, Part I., 1903.) 13. A flywheel of a shearing machine has 150,000 foot-lbs. of kinetic energy stored in it when its speed is 250 revolutions per minute. What energy does it part with during a reduction of speed to 200 revolutions per minute ? If 82 per cent, of this energy given out is imparted to the shears during a stroke of 2 inches, what is the average force due to this on the blade of the shears ? (Advanced, 1902.) 14. A weight of 5 Ibs. is supported by a spring. The stiffness of the spring is such that putting on or taking off a weight of i Ib. produces a downward or upward motion of 0^04 foot. What is the time of a complete oscillation, neglecting the mass of the spring ? (Advanced, 1902.) 272 Mechanics for Engineers 15. A car weighs 10 tons : what is its mass in engineers' units? It is drawn by the pull P Ibs., varying in the following way,/ being seconds from the time of starting : P ... t IO20 980 2 882 5 720 8 702 10 650 13 713- 16 722 19 805 22 The retarding force of friction is constant and equal to 410 Ibs. Plot P 410 and the time /, and find the time average of this excess force. What does this represent when it is multiplied by 22 seconds ? What is the speed of the car at the time 22 seconds from rest? (Advanced, 1902.) 16. A body weighing 1610 Ibs. is lifted vertically by a rope, there being a damped spring balance to indicate the pulling force F Ibs. of the rope. When the body had been lifted x feet from its position of rest, the pulling force was automatically recorded as follows : X F o 4010 ii 3915 20 3763 34 3532 45 3366 55 3208 66 3100 76 3007 Using squared paper, find the velocity v feet per second for values of x of 10, 30, 50, 70, and draw a curve showing the probable values of v for all values of x up to 80. In what time does the body get from x = 45 to x = 55 ? (Honours, Part I., 1901.) 17. A machine is found to have 300,000 foot-lbs. stored in it as kinetic energy when its main shaft makes 100 revolutions per minute. If the speed changes to 98 revolutions per minute, how much kinetic energy has it lost ? A similar machine (that is, made to the same drawings, but on a different scale) is made of the same material, but with all its dimensions 20 per cent, greater. What will be its store of energy at 70 revolutions per minute ? What energy will it store in changing from 70 to 71 revolutions per minute ? (Honours, Part I., 1901.) 1 8. A body of 60 Ibs. has a simple vibration, the total length of a swing being 3 feet ; there are 200 complete vibrations (or double swings) per minute. Calculate the forces which act on the body at the ends of a swing, and show on a diagram to scale what force acts upon the body in every position. (Advanced, 1901.) 19. An electric tramcar, loaded with 52 passengers, weighs altogether 10 tons. On a level road it is travelling at a certain Examination Questions 273 speed. For the purpose of finding the tractive force, the electricity is suddenly turned off, and an instrument shows that there is a retardation in speed. How much will this be if the tractive force is 315 Ibs. ? If the tractive force is found on several trials to be, on the average 342 Ibs. when the speed is 12 miles per hour 3'5 10 j) 2 94 8 what is the probable tractive force at 9 miles per hour ? (Advanced, 1901.) 274 LOGARITHMS. 1 2 3 4 5 6 7 8 9 123 4 5 21 20 6789 10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 4 9 13 17 4 8 12 16 25 30 34 38 24 28 32 37 11 12 0414 0792 0153 0828 0492 0864 0531 C899 0569 0934 0607 0969 0645 1004 0682 1038 0719 1072 0755 1106 4 8 12 15 4 7 11 IS 3 7 11 14 3 7 10 14 19 19 18 17 2327 31 35 2J 26 30 33 21 25 28 32 20 24 27 31 13 14 1139 1461 1173 1492 1206 1523 1239 1553 1271 1581 1303 1614 1335 1641 1367 1673 1399 1703 1430 1732 3 7 10 13 3 7 10 12 3 6 9 12 3 6 9 12 16 16 15 15 20 23 26 30 19 22 2i 29 18 21 24 28 17 20 23 26 15 1761 1790 IfclS 1847 1875 1903 1931 1959 1987 2014 3 6 9 11 3 5 8 11 14 14 17 20 23 26 16 19 22 25 16 17 2041 2304 2068 2330 2095 2355 2122 2380 2148 2405 2175 2430 2201 2455 2227 2480 2253 2504 2279 2529 3 5 8 11 3 5 8 10 3 5 8 10 2 5 7 10 14 13 13 12 16 19 22 24 15 18 21 23 15 18 20 23 15 17 19 22 18 19 2553 2788 2577 2810 2601 2833 2625 2856 2648 2878 2672 2900 2695 2923 2718 2945 2742 2967 2765 2989 2579 2 5 7 9 2479 2468 12 11 11 11 11 14 16 19 21 14 16 18 21 13 16 18 20 13 15 17 la 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 2468 13 15 17 19 21 22 23 24 3222 3424 3617 3802 3243 3444 3636 3820 3263 3461 3655 3838 3284 3483 3674 3356 3304 3502 3692 3874 3324 3522 3711 3892 3345 3541 3729 3909 3365 3560 3747 3927 3385 3579 3766 3945 3104 3598 3784 3962 2468 2468 2467 2457 10 Id 9 9 12 14 16 18 12 14 15 17 11 13 15 17 11 12 14 16 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 2367 9 10 12 14 15 26 27 28 29 4150 4314 4472 4624 4166 4330 4487 4639 4183 4316 4502 4654 4200 4362 4518 4669 4216 4378 4533 4683 4232 435)3 4548 4698 4249 4409 4564 4713 4265 4125 4579 4728 4281 4440 4594 4742 4298 4456 4609 4757 2357 2356 2356 1346 8 8 8 7 10 11 13 15 9 11 13 14 911 12 14 9 10 12 13 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 1346 7 9 10 11 13 31 32 33 34 4914 5051 5185 5315 4928 5065 5198 6328 4942 5079 5211 5340 4955 5092 5224 5353 4969 5105 5237 5366 4983 5119 5250 5378 4997 5132 5263 5391 5011 6145 5276 5103 5024 5159 5289 5416 5038 5172 5302 5428 1346 1345 1345 1345 7 7 6 6 8 10 11 12 X 9 11 12 8 9 10 12 8 9 10 11 35 5441 5453 5465 5478 5490 5502 5514 5527 6539 5551 1245 6 7 9 10 11 36 37 38 39 5563 5682 579b 5911 5575 5694 6809 5922 5587 6705 5821 5933 5599 5717 6832 5944 5611 5729 5843 5955 5623 5740 5855 5906 5635 5752 5866 5977 5647 5763 5877 5988 5658 5775 5888 5999 5670 5786 5S99 6010 124 5 1235 1235 1234 6 6 6 5 7 8 10 11 78 1) 10 7 8 9 JO 7 8 9 10 40 6021 6031 6042 6053 6064 6U75 6085 6096 6107 6117 1234 6 8 9 10 41 42 43 44 6128 6232 6335 6435 6138 6243 6345 6444 6149 6253 6355 6454 6160 6263 6365 6464 6170 6274 6375 6474 6180 6284 6385 6484 6191 6294 6395 6493 6201 6301 6405 503 6212 6314 6415 6513 6222 6325 6425 6522 1234 1234 1234 1234 5 5 5 6789 6789 6789 6789 45 6532 6542 6551 6561 6571 05-0 6590 6599 6609 6618 1234 5 6789 46 47 48 49 50 6628 6721 6812 6902 6637 6730 6821 (-'.II 1 6646 6739 6830 6920 6656 6749 6839 6928 6665 6758 6848 6937 6675 6767 6857 6946 6684 6776 6566 6955 6693 6785 6S75 6964 6702 6794 6884 6972 6712 6803 6893 6*81 1234 1234 1234 1234 5 5 4 4 4 6778 5 6 7 H 5678 5678 5678 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 1233 LOGARITHMS. 275 1 2 3 4 5 6 7 8 9 123 4 5 6789 51 52 53 54 7076 7160 7243 7324 7084 7168 7251 7332 7093 7177 7*59 7340 7101 7185 7267 7348 7110 7193 7275 7356 7118 7202 7284 7364 7126 7210 7292 7372 7135 7218 7300 7380 7143 7226 7308 7388 7152 7235 1233 1223 4 4 5678 5677 73M6 1223 4 5667 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 1223 4 5567 56 57 58 59 7482 7559 7631 7709 7490 7566 7642 7716 7497 7574 7649 7723 7505 7582 7657 7731 7513 7589 7664 7738 7520 7597 7672 7745 7528 7604 7679 7762 7536 7612 7686 7760 7543 7619 7694 7767 7551 7b27 7701 7774 122 3 1123 1123 4 4 4 556V 4567 4 5 G 7 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 1123 4 4566 61 62 63 64 7853 7924 7993 8062 7860 7931 8000 8U69 7868 7938 8007 8075 7875 7945 8014 8082 7882 7952 8021 8089 7889 7959 80^8 8096 7896 7966 8035 8102 7903 7973 8041 8109 7910 7980 8048 8116 7917 7987 8055 8122 1123 1123 1123 1123 4 3 3 3 4566 4566 4556 4556 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 1123 3 4556 66 67 68 69 70 8195 8261 8325 8383 8202 8267 8331 8395 8209 8274 8338 8401 8215 8280 344 84U7 8222 8287 8351 8414 8228 8293 8357 8420 8235 8299 8363 8426 8241 8306 837u 8432 8248 8312 8376 813!) 8254 8319 8382 8445 1123 1123 1123 112 2 3 3 3 3 4556 4556 4456 4456 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 1122 3 4456 71 72 73 74 8513 8573 8633 8692 8519 8579 8639 869s 8525 8585 8645 8704 8531 8591 8651 8710 8537 8597 8657 8716 8543 8603 8663 8722 8549 8609 8669 8727 8555 8615 8675 8733 8561 8621 3681 3739 8567 8627 8686 8745 11 1 2 2 1122 1 1 li -.2 112 2 3 3 3 3 4465 4465 4455 4455 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 1122 3 3455 76 77 78 79 8808 8865 8921 8976 8814 8871 8927 8982 8820 8876 8932 8987 8825 8882 8938 8993 8831 8887 8943 8998 8837 8893 8949 9004 8842 8899 8954 9009 8848 8904 8960 9015 8854 8910 8965 9020 8859 8915 8971 9025 1122 1122 1 1 2, 2 112 2 3 3 3 3 3455 3 4 1 f) 3446 3445 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 1122 3 3445 81 82 83 84 85 9085 9138 9191 9243 9090 9143 9196 9248 9096 9149 9201 9253 9101 9154 9206 9258 9106 9159 9212 9263 9112 9165 9217 9269 9117 9170 9222 9274 9122 9175 9227 9279 9128 9180 9232 9284 9133 9186 9238 9289 1122 112 2 1122 112 2 3 3 3 3 3445 3445 3445 3445 9294 9299 9304 9309 9315 9320 9325 9330 9336 9340 1122 3 3445 86 87 88 89 9345 9395 9445 9494 9350 9400 9450 9499 9355 9405 9455 9504 9360 9410 9460 9509 9365 9415 9465 9513 9370 9420 9469 9518 9375 9425 9474 9523 9380 9430 9479 9528 9385 9435 9484 9533 9390 9440 9489 9538 1122 0112 0112 0112 3 2 2 2 3445 3344 3344 3344 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 0112 2 3344 91 92 93 94 9590 9638 9685 9731 9595 9643 9689 9736 9600 9617 9694 9741 9605 9652 9699 9745 9609 9657 9703 9750 9614 9661 9708 9754 9619 9666 8713 9759 9624 9671 9717 9763 9628 9675 9722 9768 9633 9680 9727 9773 0112 0112 II 1 1 2 0112 2 2 2 2 3344 3344 3344 3344 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 0112 2 3344 96 97 98 99 9823 9868 9912 9956 9827 9872 9917 9961 9832 9877 9921 9965 9836 9881 9926 9969 9841 9886 9930 9974 9845 9890 9934 9978 9850 9894 9939 9983 9864 9899 9943 'J9*7 9859 9903 9948 9991 9863 9908 9952 9996 0112 0112 0112 0112 2 2 2 I 3344 3344 3344 3334 2/6 ANTILOGARITHMS. 1 2 3 4 5 6 7 8 9 1234 5 6789 00 1000 1002 1005 1007 1009 1012 1014 1016 1019 1021 0011 i 1222 01 1023 1026 1028 1030 1033 1035 1038 1040 1042 1045 0011 i 1222 03 04 107^ 1096 1074 1099 1076 no; 1079 1104 1081 1107 1084 1109 1086 1112 1089 1114 1091 1117 1094 1119 0011 0111 i i 1222 2222 05 1122 1125 1127 1130 1132 1135 1138 1140 1143 1146 0111 i 2222 06 07 08 09 1148 1175 1202 1230 1151 1178 1205 1233 1153 1180 1208 1236 1156 1183 1211 1239 1159 1186 1213 1242 1161 1189 1216 1245 1164 1191 1219 1247 1167 1194 1222 1250 1169 1197 1225 1253 1172 1199 1227 1256 0111 0111 0111 0111 i i i i 2222 2222 2223 2223 10 125!) 1262 1265 1268 1271 1274 1276 1279 1282 1285 0111 i 2223 11 12 13 14 1288 1318 1349 1380 1291 1321 1352 1384 1294 1324 1355 1387 1297 1327 1358 1390 1300 1330 1361 1393 1303 1334 1365 1396 1306 1337 1368 1400 1309 1340 1371 1403 1312 1343 1374 1406 1315 1346 1377 1403 0111 0111 0111 0111 2 2 2 2 2223 2223 2233 2233 15 1413 1416 1419 1422 1426 1423 1432 1435 1439 1442 0111 2 2233 16 17 18 19 1445 1479 1514 1549 1449 1483 1517 1552 1452 1486 1521 1556 1455 1489 1524 1560 1459 1493 1528 1563 1462 1496 1531 1567 1466 1500 1535 1570 1469 1503 1538 1574 1472 1507 1542 1578 1476 1510 1545 1581 0111 0111 0111 0111 2 2 2 2 2233 2233 2233 2333 20 1585 1589 1592 1596 1600 1603 1607 1611 1614 1618 0111 2 2333 21 22 23 24 1622 1660 1698 1738 1626 1663 1704 1742 1629 1667 1706 1746 1633 1671 171U 1750 1637 1675 1714 1754 1641 1679 1718 1758 1644 1683 1722 1762 1648 1687 1726 1766 1652 1690 173(1 1770 1656 1694 1734 1774 0112 0112 0112 0112 2 2 2 2 2333 2333 2334 2334 25 1778 1782 1786 1791 1795 1799 1803 1807 1811 1816 0112 2 2334 26 27 28 29 1820 1862 1905 1950 1824 1866 1910 1954 18^8 1871 1914 1959 1832 1875 1919 1963 1837 1879 1923 1968 1841 1884 1928 1972 1845 1888 1932 1977 1849 1892 1936 1982 1854 1897 1941 1986 1858 1901 1945 1991 0112 0112 0112 0112 2 2 2 2 3334 3334 3 3 4 4 3344 30 1995 2000 2004 2009 2014 2018 2023 2028 2032 2037 0112 2 3344 31 32 33 34 2012 2089 2138 2188 20461 2094 2143 2193 2051 2099 2148 2198 2056 2104 2153 2203 2061 2109 2158 2208 2065 2113 2163 2213 2070 2118 2168 2218 2075 2123 2173 2223 2080 2128 2178 2228 2084 2133 2183 2234 0112 0112 0112 1122 2 2 2 3 3344 3344 3344 3445 35 2239 2244 2249 2254 2259 2265 2270 2275 2280 2286 1122 3 3445 36 37 38 39 2291 2344 2399 2155 2296 2350 2404 2460 2301 2355 2410 2466 2307 2360 2415 2472 2312 2366 2421 2477 2317 2371 2427 2483 2323 2377 2432 2489 2328 2382 2438 2495 2333 2388 2443 2500 2339 2393 2449 2506 1122 1122 1122 1122 3 3 3 3 3445 3445 3445 3455 40 2512 2518 2523 2529 2535 2541 2547 2553 2559 2564 1122 3 4455 41 42 43 44 2570 2630 2692 2754 2576 2636 2698 2761 2582 2642 2704 2767 2588 2649 2710 2773 2594 2655 2716 2780 2600 2661 2723 2786 2606 2667 2729 2793 2612 2673 2735 2799 2618 2679 2742 2805 2624 2685 2748 2812 1122 1122 1123 1123 3 3 3 3 4455 4456 4450 4456 45 2818 2825 2831 2838 2844 2851 2858 2864 2871 2877 1123 3 4556 46 47 48 49 2881 2951 3020 3090 2891 2958 3027 3097 2897 2965 3034 31U5 2904 2972 3041 3112 2911 2979 3048 3119 2917 2985 3055 3126 2924 2992 3062 3133 2931 2999 3069 3141 2938 3006 3076 3148 2944 3013 3083 3155 1123 1123 1123 1123 3 3 4 4 4556 4556 4566 4566 ANTILOGARITHMS. 277 1 2 3 4 5 6 7 8 9 1 234 5 6789 50 3162 3170 3177 3184 3192 3199 3206 3214 3221 3228 1123 4 4567 51 52 53 54 3236 3311 3388 3467 3243 3319 3396 3475 3251 3327 3404 3483 3258 3334 3412 3491 3266 3342 3420 3499 3273 3350 3428 3508 3281 3357 3436 3516 3289 3365 3443 3524 3296 3373 3451 3532 3304 3381 3459 3540 1223 1223 1223 1223 4 4 4 4 5567 5567 5667 b 6 6 7 55 3548 3556 3565 3573 3581 3589 3597 3606 3614 3622 1223 4 5677 56 57 58 59 3631 3715 3802 3890 3639 3724 3811 3899 3648 3733 3819 3908 3656 3741 3828 3917 3664 3750 3837 3926 3673 3758 3846 3936 3681 3767 3855 3945 3690 3776 3864 3954 3698 3784 3873 3963 3707 3793 3882 3972 1233 1233 1234 1234 4 4 4 5 5678 5678 5678 5678 60 3981 3990 3999 4009 4018 4027 4036 4046 4055 4064 1234 5 6678 61 62 63 64 4074 4169 4266 4365 4083 4178 4276 4375 4093 4188 4285 4385 4102 4198 4295 4395 4111 4207 4305 4406 4121 4217 4315 4416 4130 4227 4325 4426 4140 4236 4335 4436 4150 4246 4345 4446 4159 4256 4355 4457 1234 1234 1234 1234 5 5 5 5 6789 6 7 8 !i 6789 6789 65 66 67 68 69 4467 4477 4487 4498 4508 4519 4529 4539 4550 4560 1234 5 6789 4571 4677 4786 4898 4581 4688 4797 4909 4592 4699 4808 4920 4603 4710 4819 4932 4613 4721 4831 4943 4624 4732 4842 4955 4634 4742 4853 4966 4645 4753 4864 4977 4656 4764 4875 4989 4667 4775 4887 5000 1234 1234 1234 1235 5 5 6 6 6 7 9 10 7 8 9 10 7 8 9 10 7 8 9 10 70 5012 5023 5035 5047 5058 5070 5082 593 5105 5117 1245 6 6 6 6 6 7 8 9 11 71 72 73 74 5129 5248 5370 5495 5140 5260 5383 5508 5152 5272 5395 5521 5164 5284 5408 6534 5176 5297 5420 5546 5188 5309 5433 5559 5200 5321 5445 5572 5212 5333 5458 5585 5224 5346 5470 5598 5236 5358 5483 5610 1245 1245 1345 1345 7 8 10 11 7 9 10 11 8 9 10 11 8 9 10 12 75 6623 5636 5649 5662 5675 5689 5702 5715 5728 5741 1345 7 8 9 10 12 76 77 78 79 5754 5888 6026 6166 5768 5902 6039 6180 5781 5916 6053 6194 5794 5929 6067 6209 5808 5943 6081 6223 5821 5957 6095 6237 5834 5970 6109 3252 5848 5984 6124 6266 5861 5998 6138 6281 5875 6012 6152 6295 1345 1345 1346 1346 7 7 7 7 8 9 11 12 8 10 11 12 8 10 11 13 9 10 11 13 80 6310 6324 6339 6353 6368 6383 6397 6412 6427 6442 1346 7 9 10 12 13 81 82 83 84 6457 6607 6761 6918 6471 6622 6776 6934 6486 6637 6792 6950 6501 6653 6808 6966 6516 6668 6823 6982 6531 6683 6839 6998 6546 6699 6855 7015 6561 6714 6871 7031 6577 6730 6887 7047 6592 6745 6902 7063 2356 2356 2356 2356 8 8 8 8 9 11 12 14 9 11 12 14 9 11 13 14 10 11 13 15 85 7079 7096 7112 7129 7145 7161 7178 7194 7211 7228 2357 8 10 12 13 15 86 87 88 89 7244 7413 7586 7762 7261 7430 7603 7780 7278 7447 7621 7798 7295 7464 7638 7816 7311 7482 7656 7834 7328 7499 7674 7852 7345 7516 7691 7870 7362 7534 7709 7889 7379 7551 7727 7907 7396 7568 7745 7925 2357 2357 2457 2457 8 9 9 9 9 10 12 13 15 10 12 14 16 11 12 14 16 11 13 14 16 90 7943 7962 7980 7998 8017 8035 8054 8072 8091 8110 2467 11 13 15 17 91 92 93 94 8128 8318 8511 8710 8147 8337 8631 8730 8166 8356 8551 8750 8185 8375 8570 8770 8204 8395 8590 8790 8222 8414 8610 8810 8241 8433 8630 8831 8260 8453 8650 8851 8279 8472 8670 8872 8299 8492 86SIO 8892 2468 2468 2468 2468 9 10 10 10 11 13 15 17 12 14 15 17 12 14 16 18 12 14 16 18 95 8913 8933 8954 8974 8995 9016 9036 9057 9078 9099 2468 10 12 15 17 19 96 97 98 99 9120 9333 9550 8772 9141 9354 9572 9795 9162 9376 9594 9817 9183 9397 9616 9840 9204 9419 9638 9863 9226 9441 9661 9886 9247 9462 9683 9908 9268 9484 9705 9931 9290 9506 9727 9954 9311 9528 9750 9977 2468 2479 2479 2579 11 11 11 11 13 15 17 19 13 15 17 20 13 16 18 20 14 16 18 20 2/8 Angle. Chord. Bine. Tangent. Co- tangent. Cosine De- grees. Radians. 000 8 1 1-414 1-5708 90 89 88 87 86 1 2 3 4 0175 0349 0524 0698 017 035 052 070 0175 0349 0523 0698 0175 0349 0524 0699 57-2900 28-6363 19-0811 14-3007 9998 9994 9986 9976 1-402 1-389 1-377 1-364 1-5533 1-5359 1-5184 1-5010 5 6 7 8 9 0873 1047 1222 1396 1571 087 0872 0875 11-4301 9962 1-351 1-4835 85 105 122 140 157 1045 1219 1392 1564 1051 1228 1405 1584 9-5144 8-1443 7-1154 6-3138 9945 9925 9903 9877 1-338 1-325 1-312 1-299 1-4661 1-4486 1-4312 1-4137 84 83 82 81 10 1745 174 1736 1763 5-6713 9848 1-286 1 -3963 80 79 78 77 76 11 12 13 14 1920 2094 2269 2443 192 209 226 244 1908 2079 2250 2419 1944 2126 2309 2493 5-1446 4-7046 4-3315 4-0108 9816 9781 9744 9703 1-272 1-259 1-245 1-231 l-a788 1 3614 1-3439 1-3265 15 2618 261 2588 2679 3-7321 9669 1-218 1 '3090 75 16 17 18 19 2793 2967 3142 3316 278 296 313 330 2756 2924 3090 3256 2867 3057 3249 3443 3640 3-4874 3-2709 3-0777 2-9012 9613 9563 9511 9455 1-204 1-190 1-176 1-161 1-2915 1-2741 1-2566 1-2392 74 73 72 71 20 3491 347 364 382 399 416 3420 2-7475 9397 1-147 1-2217 70 21 22 23 24 3665 3840 4014 4189 3584 3746 3907 4067 3839 4040 4245 4452 2-6051 2-4751 2-3559 2-2460 9336 9272 9205 9135 1-133 1-118 1-104 1-089 1-2043 1-1868 1-1694 1-1519 69 68 67 66 25 4363 433 4226 4663 2-1445 9063 1-075 1-1345 65 64 63 62 61 26 27 28 29 4538 4712 4887 5061 450 467 484 501 4384 4540 4695 4848 4877 5095 5317 5543 2-0503 1-9626 1 8807 1-8040 8988 8910 8829 8746 1-060 1-045 1-030 1-015 1-1170 1-0996 1-0821 1-0647 30 5236 518 5000 5774 1-7321 8660 1-000 1-0472 60 31 32 33 34 5411 5585 5760 5934 534 551 568 585 5150 5299 5446 5592 6009 6249 6494 6745 1-6643 1-6003 1-5399 1-4826 8572 8480 8387 8290 986 970 954 939 1-0297 1-0123 9948 9774 59 58 57 56 35 6109 601 618 635 651 668 5736 7002 1-4281 8192 923 9599 55 54 53 52 51 36 37 38 39 6283 6458 6632 6807 6878 6018 6157 6293 7265 7536 7813 8098 1-3764 1-3270 1-2799 1-2349 8090 7986 7880 7771 908 892 877 861 9425 9250 9076 8901 40 41 42 43 44 6981 684 6428 8391 1-1918 7660 845 829 813 797 781 8727 50 7156 7330 7505 7679 700 717 733 749 6561 6691 6820 6947 8693 9004 9325 9667 1-1504 1-1106 1-6724 1-0355 7547 7431 7314 7193 8552 8378 8203 8029 49 48 47 46 45 7854 765 7071 1-0000 1-0000 7071 765 7864 45 Cosine. Co- tangent. Tangent. Sine. Chord. Radians. De- greesw Angle. INDEX ( The numbers refer to pages) Acceleration, 3 Adhesion, 107 Alternating vectors, 81 Amplitude, 79 Angular acceleration, 23 momentum, 207 motion, 23 velocity, 23 Atwood's machine, 44 Average force (space), 51 force (time), 35 B Bending moment, 231 diagram, 233 Bicycle, centre of gravity, 184 Bows' notation, 223 Brakes, 107 Centre of gravity, 141-165 of mass, 141 of parallel forces, 140 Centrifugal force, 69, 181 Centripetal force, 69 C.g.s. units, 30 Chains, loaded, 243 Circular arc, 160 motion, 68 sector and segment, 161 Coefficient of adhesion, 107 of friction, 100 Compound pendulum, 212 Conditions of equilibrium, 97, 128, 226 Conical pendulum, 72 Couple, 125 Curve, motion on, 70, 71 D Density, 27 Derived units, 253 Displacement curve, 2 , relative, 16 Distributed load, 168, 247 E Efficiency of machines, 1 10 of screw, 108 Energy, 57 in harmonic motion, 84 , kinetic, 58 Equilibrant, 92 Equilibrium, conditions of, 97, 128, 226 , stability of, 172 280 Mechanics for Engineers (The numbers refer to pages) First law of motion, 27 Force, 27, 29 Forces, coplanar, 127 , parallel, 114 , resolution and composition of, 91 -, triangle and polygon of, 33, 91 Frames, 236 Friction, 99 , angle of, 101 , coefficient of, 100 , laws of, 100 of machines, I IO of screw, 1 08 , sliding, 100 -, work spent in, 107 Fundamental units, 253 Funicular polygon, 224, 228, 233, 243 Gravitational units, 30, 253 Gravity, acceleration of, 6 Guldinus, 182 H Harmonic motion, 79 Hemisphere, 162, 172 Horse-power, 51 Impulse, 33 Impulsive force, 36 Inclined plane, 102 , smooth, 22 Indicator diagram, 50 Inertia, 27 Inertia, moment of, 188 , (areas), 194 K Kinematics, Chapter I. Kinetic energy, 58 of rotation, 204 of rolling body, 2 Lami's theorem, 93 Laws of motion, Chapter II. Levers, 122 Lifting, work in, 176 Limiting friction, 100 Load, distributed, 168 Locomotive, centre of gravity, 186 M Machines, HO Mass, 27 Mechanical advantages of screw, 109 Method of sections, 133 Moment, 53, 119-122 of an area, 157 of inertia, 1 88 of areas, 194 of momentum, 207 Momentum, 28 Motion, first law of, 27 of connected weights, 43 , second law of, 28 , simple harmonic, 79 , third law of, 41 Motor-car, centre of gravity, 1 86 N Neutral equilibrium, 172, 174 Newton's laws of motion, 27 Index 281 ( 77/<? numbers refer to pages) Pappus, 182 Parallel axes, moment of inertia about, 191, 192 forces, 114 Pendulum, compound, 212 , conical, 7 2 , simple, 85 , simple equivalent, 86, 213 Plane-moments, 155 Polygon offerees, 33, 91 , funicular or link, 224 of velocities, 17 Pound, unit of force, 29, 253 Poundal, 29 Power, 51 Principle of moments, 122 of work, 59 R Radius of gyration, 189 Railway curve, 71 Reduction of forces, 127 Relative displacement, 1 6 velocity, 20 Resolution of accelerations, 22 of forces, 9 1 of velocity, 18 Rolling body, 217 Roof, 242 Rotation about axis, 179, 204, 207 Screw friction, 108 Second law of motion, 28 Sections, method of, 133 Sector of circle, 161 of sphere, 162 Segment of circle, 161 Shearing force, 231 Shearing-force diagram, 235 Simple equivalent pendulum, 213 harmonic motion, 79 , torsional, 214 86, pendulum, 85 Smooth body, 129 Space-average force, 51 curve, 2 diagram, 223 Spherical shell, 161 Spring, vibrating, 83 Stable equilibrium, 172 Statics, 91 Stress diagram, 239 , tensile and compressive, 240 String, loaded, 243 polygon, 242 Strut, 240 T Theorem of Guldinas or Pappus, 182 Third law of motion, 41 Tie, 240 Time-average force, 35, 36 Torque, 54 Torsional oscillation, 214 Triangle of forces, 33, 91 of velocities, 17 Twisting moment, 54 U Uniform circular motion, 68 Units, 253 Unstable equilibrium, 172 V Vector diagram, 223 Vectors, 15 Velocity, I 282 Mechanics for Engineers ( The numbers refer to pages) Velocity, angular, 23 , component, 18 curves, 7 , polygon of, 17 , relative, 20 Vertical circle, motion in, 73 motion, 6 Vibration of spring, 83 W Warren girder, 133, 241 Weight, 28 Work, 48 in lifting, 176 of a torque, 54 , principle of, 59 THE END PRINTED BY WILLIAM CLOWES AND SONS, LIMITED, LONDON AND BECCLES. 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