F ROM -TH E - L! BRARY- OF W1LL1AM-A HILLEBRAND ',;;... o NOTES AND EXAMPLES IN MECHANICS; WITH AN APPENDIX ON THE GRAPHICAL STATICS OF MECHANISM ; BY IRVING P. CHURCH, C. E., Professor of Applied Mechanics and Hydraulics , College of Civil Engineering, Cornell University. SECOND EDITION, REVISED AND ENLARGED. NEW YORK : JOHN WILEY & SONS. LONDON : CHAPMAN & HALL, LIMITED. 1903. * 3 COPYRIGHT, 1891, BY P. CHURCH. ERRATA. P. 63. In Fig. 76 for " 12.2'" read " 12.5'." P. 69. I/ast line but one, for " 4937 Ibs." read " 4973 Ibs." P. 70. Second line, for " 7063 Ibs." read " 7027 Ibs." Plate V of Appendix ; in Fig. 17 [A] for " S " read " 82 ' PEEFACE. THE following pages form a companion volume to the writer's Mechanics of Engineering, and contain various notes and many practical examples, both algebraic and numerical, serving to illustrate more fully the application of fundamental principles in Mechanics of Solids ; together with a few paragraphs relating to the Mechanics of Materials, and an Appendix on the "Graphical Statics of Mechanism " Advantage has been taken of the use of preliminary impressions in the classroom to make corrections in the electrotype plates; and it is therefore thought that the present complete edition is comparatively free from typo- graphical errors. In the Appendix are presented many of the problems of Prof. Herrmann's ' ' Zur Qraphischen Statik der Maschinengetriebe " in what seems to the writer a clearer form than in the original (for reasons stated on the first page of Appendix). In this part of the work, the text and diagrams not being adjacent, alternate pages have been left blank in such a way that any diagram and its appropriate text can be kept in view simultaneously. Besides his indebtedness to Prof. Herrmann's work, the writer would grate- fully acknowledge the kindness of the Messrs. Wiley in securing a higher order of excellence in the execution of the diagrams than had at first been contemplated. In references to the writer's Mechanics of Engineering the abbreviation M. of E. is used. CORNELL UNIVERSITY, ITHACA, N. Y., March, 1892. PEEFACE TO SECOND EDITION. FOB, this second edition the plates of the first have been carefully revised and corrected (except as indicated in the errata on the opposite page), and an entirely new chapter added (Chap. VIII, pp. 119-133), containing various notes and explanations, as also many examples for practice. ITHACA, January, 1897. 995884 CONTENTS. CHAP. I. DEFINITIONS. PRINCIPLES. CENTER or GRAVITY. PAGES 1-15. Illustrations of Forces. Mass, Weight, Equilibrium. Fundamental Theorem of the Integral Calculus. Centers of Gravity. Simpson's Eule. 1-J4 CHAP. II. PRINCIPLES AND PROBLEMS INVOLVING NON- CONCURRENT FORCES IN A PLANE. 16-40a. Conditions of Equilibrium. Classification of Rigid Bodies. Two- force Pieces ; Three-force Pieces, etc. Levers ; Bell-crank ; Cranes, Simple and Compound. Eedundant Support. Two Links ; Rod and Tumbler ; Door ; Wedge and Block. Roof Truss, and Cantilever Frame. 15-42 CHAP. III. MOTION OF A MATERIAL POINT. 41-51. Velocity. Acceleration. Momentum. Cord and Weights. Lifting . a Weight. Harmonic Motion. Ballistic Pendulum. Balls and Spring. Cannon as Pendulum. Simple Circular Pendulum 43-53 CHAP. IV. NUMERICAL EXAMPLES IN STATICS OF RIGID BODIES AND DYNAMICS OF A MATERIAL POINT. 52-72a. Center of Gravity. Toggle Joint. Crane. Door. Roof Truss. Train Resistance. Motion on Inclined Plane. Block Sliding on Circular Guide, etc. Harmonic Motion of Piston. Conical Pendulum. Weighted Governor. Motion in Curve. Motion of Weighted Piston with Steam used Expansively. Ball Falling on Spring 54-76 CHAP. V. MOMENT OF INERTIA OF PLANE FIGURES. 72b-76. Section of I-beam ; of Box-beam. Irregular Figures, by Simp- son's Rule. Graphical Method 77-82 -.' t '- CHAP. VI. DYNAMICS OF A RIGID BODY. 77-97. Rotary Motion. Pendulum. Speed of Fly-wheel. Centrifugal Action on Bearings. " Centrifugal Couple." Piles. Kinetic Energy of Rotary Motion. Work and Energy. Numerical Examples. Action of Forces in Locomotive. The Appold and Carpentier Dynamometers. Boat Rowing. Solutions of Numerical Examples. Work of Rolling Resistance. Strap Friction ; Examples , 83-106 CHAP. VII. MECHANICS OF MATERIALS AND GRAPHICAL STATICS. 98-107. Stresses in a Rod in Tension. Rivet-spacing in a Built Beam. I-beams; without using Moment of Inertia. "Incipient Flexure "in a Column. Tests of Wooden Posts. The Pencoyd Experiments in Columns. Graphical Constructions 107-118 CHAP. VIII. MISCELLANEOUS NOTES. 108-116. Center of Gravity. The Time- velocity Curve. Reduction of Moment of Inertia of Plane Figure. Miscellaneous Examples. The " Imaginary System " in Motion of Rigid Body. Angular Motion 119-133 APPENDIX ON THE GRAPHICAL STATICS OF MECHANISM. 1-34 (See p. 28 of the Appendix for its Table of Contents.) NOTES AND EXAMPLES IN MECHANICS. CHAPTEE I. \ t /;, DEFINITIONS. PRINCIPLES. CENTRE OF GRAVITY. 1. Applied Mechanics is perhaps a more common term for the same thing than u Mechanics of Engineering." " Pure Mechan- ics" is another name for Analytical Mechanics, which deals with the subject entirely from a mathematical point of view. 2. Abstract Numbers. In experimental investigations in which formulae are to be deduced, it is best to throw experimental coefficients into the form of abstract numbers, if possible, for these are immediately comparable with those of another experi- menter in the same field, if the latter follows the same plan., whether he uses the same units for space, force, and time, or not. Thus : if the coefficient of friction be defined as the ratio of the friction [force] to the normal pressure [force] producing it, we obtain the same number for it in a definite experiment, whether we express our forces in pounds or in kilograms. 3. Forces. One of the most important things to be acquired in dealing with the practical problems of this study is a proper conception of forces. We do not use the word force in any abstract general sense, nor in any popular sense, such as is in- stanced in the Note of 15c, M. of E. It should always mean the pull, pressure, rub, attraction (or repulsion), of one body upon another, and always implies the existence of a simultaneous, equal, and opposite force exerted by that other body on the first body, i.e., the reaction / but this reaction will not come up for consid- eration in any problem unless this "first" body is under treat- 2 NOTES AND EXAMPLES IN MECHANICS. ment as regards the forces acting on it. In most problems in Mechanics we have one or more definite rigid bodies under con- sideration, one at a time, in whose treatment we must form clear conceptions of the forces acting on it ; and these always emanate from other bodies. Hence in no case should we call anything a force unless we, can conceive of it as capable of measurement by a spring-balance, &n&.are able do say from what other ~body it comes. \ *>tFor vx&mpl&,*&'\)ody said to weigh 30 Ibs. lies at rest on a ;6&i)<>th;iVel' table, which is the only body with which it is in 'c'orit'acft." "'Wlten* 6ortsrdered by itself this body is acted on by only two other bodies in a manner which justifies the use of the word force ; viz., the action of the earth upon it is a vertical downward attraction (force) of 30 Ibs. ; while the action of the table upon it is an upward pressure (force) of 30 Ibs. (We here ignore the atmosphere whose pressures on the body are balanced in every direction.) But suppose the same body and the table with which it is in contact to be allowed to fall, from rest, in a vacuum. The two bodies, during the fall, remain apparently in as close contact as before ; but now the upper body is under the action of only one force, viz., the downward attraction of the earth, 30 Ibs. ; and there is no pressure of the upper body against the table, and con- sequently no pressure of the table against the upper body. As another instance, an iron rod rests horizontally on two level-faced supports, at its extremities, and bears a load of 60 Ibs. in the middle. "When this rod is considered "free" i.e., when those other bodies which act on it in a u /bm?-able" way are supposed removed (their places being for present purposes taken by the respective forces with which they act on the first body), we find it to be under the action of four forces, viz. : a pressure on its middle, vertical and downward, of 60 Ibs. from its load ; the downward attraction of the earth on it, i.e., its own weight, say 10 Ibs. (which is really distributed among all of its particles, but which, so far as the equilibrium, or state of rest or motion, of the body is concerned, is the same as if applied at the centre of gravity, viz., the middle of the rod) ; and the two upward press- ures of the two supports against the ends of the rod, these being FORCES. 35 Ibs. each. If the nature of the investigation requires it, we may go on and consider one of the supports by itself, or " free" ; in which case, whatever the actions of other bodies on it may be, that of the rod will be a downward force of 35 Ibs., the equal and opposite of the 35 Ibs. upward pressure of the support against the rod. These pressures of the two supports against the rod are usually called the ' Reactions of the Supports." As another instance : a ball of 10 Ibs. weight hangs at rest by a cord attached to a support above. The cord is of course ver- tical. This ball is under the action of two forces, viz., a down- ward attraction of 10 Ibs. emanating from the earth, and an upward pull of 10 Ibs. emanating from the cord. A portion of the above cord, taken in the part under ten- sion, is under the action of two forces, thus : the part just above it exerts an upward pull of 10 Ibs. upon it, and that below it exerts a downward pull of 10 Ibs. upon it. (We here neglect the weight of the portion of cord considered as presumably very small.) In such a case the tension of the cord is said to be 10 Ibs. (not 20 Ibs.). Further illustration. Fig. 1 shows a prismatic rod CB lean- ing against the smooth vertical side of a block. Both rest on a rough horizontal plane. The rod is under three forces, viz. : its weight G acting vertically downwards through its middle; the pressure of the wall against it, P (which, since the wall-surface is perfectly smooth, must be horizontal and points toward the right) ; and a third force, Q, the pressure of the floor against the rod. Since the rod and block are at rest, P and G intersecting at J., it must be that the floor is sufficiently rough to enable the pressure Q to deviate from the vertical (that is, from the normal to plane of floor) by as much as the angle AB F, at least; for, as will be proved later, if three forces act on a body and it remains at rest, the three lines of action must intersect in a common point. We next consider the block, or wall, by itself, and find it to FIG. l. 4 NOTES AND EXAMPLES IN MECHANICS. be under the action of P' , the equal and opposite of jP, and therefore pointing horizontally toward the left ; of 6r', its weight; and a pressure, R, from the floor, whose line of action is deter- mined by the fact that it must be the same line of action as that of the (ideal) resultant of the forces G-' and P\ since if three forces balance, i.e., are in equilibrium, any one of them must be the- equal and opposite of the resultant of the other two and have the same action-line. Forming a parallelogram, therefore, on the forces P' and 6?', first conceiving each of the two forces to be transferred in its line of action to their point of intersection, A', the diagonal of this parallelogram represents the equal and oppo- site of ^, and has the same line of action. If this diagonal intersects the floor on the left of the lower left-hand corner of the block, the supposed stability is impossible, unless the block is cemented to the floor. 4. Mass and Weight. The question of mass will be further discussed in a subsequent chapter, p. 53, M. of E. By weight we are always to understand the force of the attraction which the earth exerts upon the body, and not the amount of matter (mass) in it. This weight will therefore be different in different latitudes and at different distances from the centre of the earth, and re- quires a spring -balance for its determination. Physicists also use the word weight in this sense (force). 5. The Heaviness of a body, in the sense used in this study, is something quite different from its total weight. For instance, if the substance of the body is not uniform in composition and den- sity, we cannot speak of the heaviness of the body as a whole, since its various portions have not a common heaviness ; however, we may speak of its average heaviness, which might be of some use in certain problems, and would be the quotient of its total weight divided by its volume. Since heaviness is not an abstract number, it would not be sufficient to say that a certain substance has a heaviness of 40, for instance, nor even 40 Ibs. ; the full statement must be 40 Ibs. per cubic foot ; which is equivalent to the statement that the heaviness is 0.540 ton per cubic yard. 6. Rigid Body. As an illustration of the definition in 10, EQUILIBRIUM TRANSMISSIBILITY OF FORCE. 5 M. of E., it may be said that if a horizontal bar, supported at its extremities, is so moderately loaded that the deflection or sinking of the central point is only about one 3-hundredth part, for in- stance, of the span or distance between the supports, it is suffi- ciently accurate, for most purposes, to consider that there has been no change in the length of the horizontal projection of any dis- tance measured along the bar. (Where such a consideration is inadmissible, attention will be called to it.) 7. Equilibrium. Besides speaking of a system of forces being in equilibrium, the phraseology is also sometimes used that the rigid body is in equilibrium under the forces acting on it ( 40a). The reservation made in 11, M. of E., as to state of motion refers to the fact that any alteration in the distribution of forces acting on a rigid body will usually cause a difference in the inter- nal strains and stresses produced, though the state of motion may or may not be affected, according as any second system of forces applied to the body, on removal of the first, has a different resultant from that of the first, or the same resultant. 8. Division of the Subject. As to the division given in 12, M. of E., Sir William Thomson, the noted English physicist, has adopted a different nomenclature, which is getting into wider and wider use. He makes the term Dynamics include both statics and dynamics (i.e., what is here and by Rankine and Continental writers called Dynamics), and replaces their word Dynamics by Kinetics. 9. Transmissibility of Force. Resultant. The principle of the transmissibility of force refers only to the state of motion of the rigid body. For instance (see Fig. 2), as far as the rest or motion of the sickle-shaped body is concerned, it is immaterial whether the force P balance P' (being equal and opposite to it and in the same line) by being applied at 0, or by being applied at A ; but in the former case the part ABO would be under a bending strain, and in the latter would be under no strain. It must be remembered that the resultant of a given system of forces is always a purely imaginary force ; that is, all we mean 6 NOTES AND EXAMPLES IN MECHANICS. is this : that if the given forces were removed and their resultant acted in their stead, in proper position as well as magnitude, the state of motion of the rigid body would not be any different from what it would be without; the replacement. 10. Parallelogram of Forces; or Triangle of Forces. By some it is contended that this geometrical construction, or relation, the Parallelogram of Forces, should give place to the triangle of forces ; i.e., that the resultant (a line laid off to scale representing it) is equal to the third side of a triangle whose other two sides are the two given forces ; but a construction of that nature does not show the resultant as acting through the same point as the two com- ponents, which should be the case if the construction is to give the position, as well as the magnitude of the resultant. It is true that the systematic application of the triangle of forces gives rise to the methods of Graphical Statics, as will be seen, but in that case the magnitudes of all the forces (or rather lines of definite length representing them and parallel to them) are drawn on a separate part of the paper from that containing the figure of the body acted on. 11. General Remarks on Forces. It is not such a simple matter as it might at first appear, to bring to bear upon a given body a force of prescribed magnitude and direction at a specified point. For example, if we have the idea that a given tension can be produced in a vertical cord by hanging upon it a body whose weight is the given amount, we must remember that the tension in the cord will be equal to the body's weight only in case the body is at rest or moving in a right line with unchanging velocity (i.e., describing equal spaces in equal times). While we can always be sure that the weight of the body itself (or action of the earth on it) is '(x}\Ax. . (3) Project all the lengths like ab on a convenient vertical line, as ED, and note that their sum is, of course, not quite equal to ED, or NC, which is y n y<>> This sum we may express as and hence state that is almost equal to y n y . . . (4) THEOREM OF THE INTEGRAL CALCULUS. 9 Or, since y n y<> ma J De written \4>(x)] x = Xn [4>(x]\ x=X( ^ = 0(#), we may re-state the fact thus : LO rx n ^ N \_'(x) x*) ; while as to the second, although the limit of 2 N (x)Ax is / n (x)dx, and hence is not zero, yet the outside factor, Ax, is t/X zero and hence the second term vanishes. Therefore the total moment of the paraboloid is M = 2y?rp f Xn \x*~]dx', and in this CENTRE OF GRAVITY. 11 we note that the \x) of the general form of eq. (6) is a? a , and that the x anti-derivative of a? 2 is Ja? 3 + const. ; and hence, finally, M = This quantity divided by the total weight (= ynpx^) of the solid will give the distance of its centre of gravity from the vertex, or origin, ; i.e., x = %x n . As to notation, it is customary to anticipate the fact that the desired result justifies the use of the notation / n \(f) f (x)\dx, and t/X to employ at once dx for Ax in making out the form of one term of the series, dx is then called an infinitesimal, which simply means that Ax is finally to become zero; in other words, that the result sought is the limit which the sum of the series approaches as Ax diminishes. 14. Position of Centre of Gravity of Various Geometrical Forms. (Homogeneous, etc.; the plane figures representing thin- plates of uniform thickness.) Obelisk. Fig. 6 shows a homogeneous obelisk, or solid bounded by six plane faces, of which two are rectangular (hori- zontal in this figure) with cor- responding edges parallel (and hence these rectangular faces are parallel, and may be con- sidered as bases). Required the distance, z, of the centre of gravity, C' , of the obelisk, from the base EEGF. See Fig. 6 for notation, h is the perpendicular distance be- tween the bases. By passing a plane through the edge AD \\ to face BCGF; a plane through AB \\ to DCGH\ and noting their intersections (dotted lines) with the faces of the obelisk, we subdivide it into the following geometric forms : A parallelepiped ABCD-JMGL, of volume F, = blh and 12 NOTES AND EXAMPLES IN MECHANICS. having its centre of gravity at a distance z l = ^h above base 1-1F\ A triangular prism AD-IJLII, of volume = Y t = i&(^ l)h and for whose centre of gravity z^ -JA ; Another triangular prism AB-KFMJ, of volume = F 3 = \l(bi b)/i and for whose centre of gravity z s = -J/< ; and finally, K pyramid A-EKJI, whose volume is F 4 J-A(5 1 5)(^ I) and whose u mean &" is ^ 4 = \h. Hence by eq. (3) of p. 19, M. of E., with z^ etc., instead of ? x , etc., w^e have, after reduction, 77 I O77l 77l 77 7 o.L -+- 6t)L -4- O.I H- Off, fl 11 I II II z =. vbl, + m + 1,1 + Triangular Plate. Fig. 7. Bisect AB in M. Join 6>J/. Bisect 0^ in JV and join JL^V. The intersection, O. is the centre of gravity of the triangular plate or plane figure. [The centre of gravity of the mere perimeter of the triangle (slender wires, homo- geneous and of same sectional area) is the centre of the circle inscribed in a triangle formed by joining the middles of the sides.] Parabolic Plate. Fig. 8. AN being -j to axis ON. ~ Upper Half of preceding Parabolic Plate. Fig. 9. x and y = %AN. FIG. 8. FIG. 9. FIG. 10. $ /Semi-ellipse. Fig. 10. Semi-axes are a and I- x 5. O/T Sector of a Sphere. AEBO, Fig. 11. Let h = the altitude of the zone, or cap, of the sector ; i.e., let h r OK\ then OC := $r -fA ; C being the centre of gravity. SIMPSON'S EULE. 13 Segment of a Sphere. AK, Fig. 12. Let the altitude, of the segment be A, and C the centre of gravity; then (Zr - Solid Spherical Segment. FIG. 12. FIG. 13. Join the vertex with MN. From D lay off FIG. 14. Any Pyramid (or Cone). Fig. 13. Z>, the centre of gravity of the base CD = fDO. C is the centre of gravity of the solid. Zone on Surface of Sphere. Fig. ]4. (Thin shell, homogeneous and of uniform thickness.) The centre of gravity lies at the middle of the altitude, A, in the axis of symmetry. The small circles of the sphere, CD and AB. lie in parallel planes. 15. Simpson's Rule (Fig. 15). If ABCDEFG is a smooth curve and ordinates be drawn from its extremities A and G to the axis X, an approximation to the value of the area so enclosed, A..D..G..N..O..A, between the curve and the axis X, is obtained by Simpsons Rule, now to be demonstrated. Divide the base ON into an even number, n, of equal parts, each = Ax (so that ON=n.4x), and draw an ordiriate from each point of division to the curve, the lengths of these ordinates being u l9 u^ , etc. ; see figure. 14 NOTES AND EXAMPLES IN MECHANICS. Consider the strips of area so formed in consecutive pairs ; for example, CDEE" C" is the second pair in this figure (count- ing from left to right). Conceive a parabola, with its axis vertical, to be passed through the points C\ D, and E. It will coincide with the real curve between C and E much more closely than would the straight chords CD and DE\ and the segment CDE^ considered as the segment of this parabola, has an area equal to two thirds of that of the circumscribing parallelogram CO'EE. Hence, since the area of this pair of strips = trapezoid CEE" C" -{-parabolic segment CDE, we may put tfltips P CE> \ which reduces to ... \Ax\u^ + 4w s -f- ^J- Treating all the -^ pairs of strips in a similar manner, we have finally, after writing Aw (x n o? ) -=- n, Whole area } x n AG" (appro*.} \ = The approximation is closer the more numerous the strips and the more accurate the measurement of the ordinates u^ u l9 u 9 , etc. If the subdivision on the axis JT were " infinitely small," an exact value for the area would be expressed by the calculus form /a: = X I ud *J x X Q udx. Hence for any integral of this form, udx, if we are only able to determine the particular values (u , u t , etc.) of the variable u corresponding respectively to the abscissae a? , x -\- Ax , a? -j- 2^/aj, etc. (where Ax = (x n a? ) -f- n, n being an even number), we can obtain an approximate value of the integral or summation by writing As to the meaning of n> note that the first ordinate on the left is not u^ but u ; also that while there are n strips, the num- ber of points of division is n -j- 1, counting the extremities O and^T. FORM FOR ANALYTICAL CONDITIONS OF EQUILIBRIUM. 15 CHAPTER II. PRINCIPLES AND PROBLEMS INVOLVING NON-CONCURRENT FORCES IN A PLANE. 16, Most Convenient Form for Analytical Conditions of Equi- librium (Fig. 16). Let P, , P 2 , P 8 , etc., constitute a system of non-concurrent forces in a plane acting on a rigid body and in equilibrium. Of the actual system, P, and P 2 are the only forces shown in the figure. As- suming a convenient origin, 0, introduce into the system two opposite and equal forces, P/ and P/', both acting at and equal and parallel to P t . Evidently the presence of these two forces does not destroy the equilibrium of the original system. Similarly, introduce at the mutually annulling forces jy and P 2 " bearing the same relation to P 2 (parallelism and equality) that P/ and P/' do to P, ; and so on for each of the remaining forces of the system. Drop a perpendicular from on each of the forces of the original system, the lengths of these perpendiculars being a } , a a , a 3 , etc. (a l and & 2 are shown in the figure). We now note that for each force P of the original sys- tem we have in the new system a single force at 0, equal and parallel to P and similarly directed, and also a couple, of moment Pa. For example, the force P 2 of the original system is now replaced by the force P 3 " parallel and equal to P, and similarly directed, but acting at the point ; and by the couple formed of the two forces P 2 and P 2 ', the arm of this couple being a 2 . It follows, therefore, that the new system consists of a set of forces (P/', P/', P,", etc.), all meeting at O (and hence forming a concurrent system in a plane *), and a set of couples, of moments P l a l , P a # a , etc. Since no single force can balance a couple ( 29, M. of E.) or set of couples, the forces of the concurrent system at must be in equilibrium among themselves ; i.e. , Y being any two directions at right angles, we must have * And therefore, (unless balanced,) equivalent to a single force or resultant; see M. of E., p. 8. 16 NOTES AND EXAMPLES IN MECHANICS. and 2Y separately equal to zero for the concurrent system at ; and the set of couples must be in equilibrium among themselves, whence it follows that the moment of their resultant couple must equal zero. Since the couples are in the same plane, the moment of their resultant couple is the algebraic sum (see 34, M. of E.), i.e., P fl , + P,a, + . . . , or 2(Pa), = 0. For the equilibrium^ therefore, of a system of non-concurrent forces in a plane, we must have not only 2X and 2 Y = 0, but also 2(Pa) = 0. That is, in practical language (the body being originally at rest), the forces of the system so neutralize each other that they not only do not tend to move the body sideways or vertically, but also do not produce rotation. In the practical application of these conditions of equilibrium in solving problems it is not necessary to introduce the pairs of equal and opposite forces at (the conception of which is needed only for purposes of proof), since the sum of the X-components (or ^-components) of the actual forces of the system is equal to that of the Jf-components of the auxiliary forces introduced at O\ while to form the moment-sum of the auxiliary couples, we have only to multiply each force of the actual system by the per- pendicular distance of its line of action from the origin, whose position is taken at convenience. The student should now read the latter half of p. 33, M. of E. 17. The Rigid Bodies Dealt with at Present. Each rigid body now to be considered is one whose dimensions perpendicular to the paper are supposed to be very small, and therefore may be considered to lie in the plane of the paper. An actual structure is made up of such pieces, or members, which are provided with forked joints, or duplicated in such a way that the above supposition (each piece lying in the plane of the paper) is practically justified. All surfaces of contact between any two contiguous pieces are supposed perpendicular to the paper, and friction between two such parts of a structure is disregarded ; i.e., the pressure between two contiguous pieces (or " members") is in the plane of the paper and normal to the surfaces of contact / for it is a matter of com- mon experience that pressure can be exerted at the smooth sur- faces of contact of two bodies only in a direction normal to those surfaces. CLASSIFICATION OF EIGID BODIES. 17 In problems where the weights of one or more bodies con- nected with the structure are considered, the plane of the struc- ture will be vertical, and then (considering what has already been postulated) the system of forces acting on each member, or piece of the structure, is a system of forces in a plane (a " uniplanar" system of forces). 18. Contact Forces or Pressures. If one of the two bodies in contact is rounded at the point of contact, while the other is quite flat at that point, the action-line of their mutual pressure neces- sarily lies in a perpendicular, or normal, to the latter flat surface, and passes through the point of contact. Hence, the shapes of the bodies being known, this action-line becomes known on inspection. Let this be called " flat-contact pressure" (N.B. As a better definition of flat-contact pressure, we might describe it as a pressure occurring in such a way that its action-line can- not be materially changed by any slight motion of one piece rela- tively to the other during the small alteration of form and posi- tion which actually takes place when a load is gradually placed on the structure.} But if the mode of connection of the two bodies is & pin- connection, that is, if one body carries a round pin or bolt fitting (somewhat loosely) in a corresponding ring forming part of the other body, we are unable to say in advance just where, on the inner circumference of the ring, the contact (and accompanying pressure) is going to be. Wherever the point is, all that we can immediately say as to the action-line of the force is that it passes through the centre of the circle, its direction (if determinate at all) being found from a consideration of the other forces acting on the piece in question. This will be illustrated later. Such a pressure may be called a hinge-pressure. 19. Classification of Rigid Bodies under Uniplanar Systems of Forces. As conducive to clearness in subsequent matter, the rigid bodies composing a structure will be named according to the number of forces acting on each ; and these forces consist of gravity actions (i.e., weights) and of the pressures exerted by neighboring pieces on the piece in question. The resultant gravity action on a single piece, or member of the structure, is a single force, called its weight, acting vertically downward through its 18 NOTES AND EXAMPLES IN MECHANICS. FIG. 17. centre of gravity. (Of course, the action of gravity is distributed over all the particles of a body, but the above-mentioned single force is the full equivalent of these distributed forces as far as the equilibrium of the piece is concerned; though not such as re- gards the straining action on the piece. With these straining actions we cannot deal here ; they will be treated later in the proper place. In many cases the whole weight of a piece is so small in comparison with any of the other forces of the system acting on that piece that no appreciable error is made in regard- ing it as without weight. JSTotice is always given in such cases.) 20. " Two-force Pieces " and their Treatment. A " two-force" piece being a piece on which only two forces act, if the weight of the piece is considered there is but one other force. For example, Fig. 17, a body of weight G hangs by a stem and ring which form a rigid part of it, on a pin projecting from a fixed sup- port. Since, evidently, the equilibrium of a two- force piece requires that the two forces shall be equal and opposite and act in the same line, the piece A . . B will not be in equilibrium unless the centre of gravity c lies in a vertical line drawn through the centre of the circular section of the pin at A. Here we have an instance of the final full deter- mination (by the necessity of its being vertical) of the action- line of a hinge-pressure, concerning which we know in advance, only that it passes through the centre of the circle at A. We find, therefore, that the pressure of the pin at A against the ring of the rigid body A . .B, hanging at rest, must have a direction verti- cally upward, and an amount, V, numerically equal to G, while its action-line, c . . d, passes vertically through the centre of the hinge-circle. In Fig. 18 is another two-force piece, in which, for equilibrium, the same result is reached, but, the stem being curved, the straining action in it is of a bending nature ; whereas, in Fig. 17 it is a simple tension, or stretching action. The case of a two-force piece whose weight is neglected is FIG. 18. THREE-FORCE PIECES. 19 Sett Crank -G FIG. 19. instanced in Fig. 19, where the two-force piece A . . B is sub- jected to tension through the action of a suspended weight, G, and a bell-crank lever, BCD. The hinge-pressure at A, from the support 8 against the piece A . . B, must pass through the centre of the circle at A ; while also the hinge-pressure at B, of the bell-crank against the piece A . . B, must pass through the centre of the hinge-circle at B. But these two hinge-pressures are the only forces acting on the piece A . . B, and for the equilibrium of a two-force piece must be equal, opposite, and coincident as to action-line ; that is, these pressures must both act in the line joining the centres of the two circles. A B' shows this piece represented as a free body, with the equal and directly opposite forces P r and P' act- ing at the ends. As to the value, or amount, of this force P r , it cannot be found until the case of the bell-crank has been treated, depending, as it does, not only upon the design of the bell-crank and the amount of the load #, but also upon the position of the piece AB itself. 21. Three-force Pieces. If a rigid body is at rest (i.e., in equilibrium) under the actions of three forces, it is evident that these forces must have action-lines intersecting in a common point, and that each force must be the equal and opposite of the diagonal of the parallelogram formed on the other two as sides (laid off to scale) ; for any one of them 'must ~be the anti-resultant of the other two. (See 15, M. of E.) [In the particular case where the forces are parallel the intersection-point is at infinity and the value of any one of the forces is numerically equal to the sum of the other two (algebraic sum).] 22. Example of a Three-force Piece. The bell-crank BCD of Fig. 19 furnishes an instance. This body is subjected to the three hinge-pressures at B, C, and D, respectively. That at D is a vertical downward pressure 6r', equal to the weight G of the 20 NOTES AND EXAMPLES IN MECHANICS. two-force piece DE. The action-line of the hinge-pressure at B has been found by the previous consideration of the two-force piece AB, and is a . . b. The hinge-pressure P' ', at C\ passes through the centre of the corresponding circle, but its action-line is as yet unknown. The problem, then, stands thus : Of the three forces, G', P\ and P" ^ under whose action the bell-crank is in equilibrium, G' is known in amount and line of action, P' is known as to action-line but not in amount, while as to P" we know neither its amount nor its direction but simply one point, 6", of its action-line. However, since the three action-lines must meet in a common point, we need only note the intersection, 0, of the known action-lines of P 1 and G' 9 and join o with C (centre), in order to determine C . . o, the action-line of P" . Next, as to finding the amounts of both P' and P" , consider that P" is the anti-resultant of P' and G', and that therefore the (ideal) resultant of P' and G' must act along o . . C; hence lay off o . . m by scale to represent G' and through m draw a line || to o . . A intersecting o . . C in some point r, and draw r . . k || to G', to determine ~k on the line o . . B. Then o . . n, laid off along O . . o, and = o . . r, but in the opposite direction from 0, gives the amount and direction of P". For o . . r is the resultant of P' and G' , and o . . n is its equal and opposite. Of course P' and P" must be measured by the same force- scale that was used in laying off o . . m = G' . We can see by inspection of the figure that if the position of the link, or two-force piece, AB, were changed in such a manner that, while the line A . . B continues to pass through 0, the pin- joint, or hinge, B is caused to approach nearer and nearer to C 9 the forces P" (always equal to the ideal o . . r) and P' both increase without limit ; for the point r moves out from 0, m being fixed (i.e., the load G remains invariable) until, when B is infinitesimally near to C, m . . r is || to o . . C and r is at infinity. The method just pursued is a graphic one ; analytically, we would proceed thus : since the system of forces G', P r , and P" is balanced, i.e., in equilibrium, the algebraic sum of their moments about any point in the plane must vanish, i.e., = 0. THREE-FORCE PIECES ANALYTIC TREATMENT. 21 (See 16.) Take an origin, or centre of moments, at C and denote by d and c the lengths of the perpendiculars let fall from C upon the action-lines of G' and P ', respectively. (These lengths may be obtained trigonometrically from given distances and angles, but are most easily, and with sufficient precision, scaled off from an accurate drawing.) With C as origin the lever-arm of the force P" is zero and hence the moment of this force is zero ; consequently this force does not enter the moment- equation, which therefore will contain but one unknown quantity, viz., the amount of the force P 1 '. The resulting moment-equation (following the routine recom- mended at the foot of page 33, M. of E.) is P' .G- G f .d + P"x = 0. .(1); whence P' = -G' . . . (1) G becomes known ; and since P" is the anti-resultant of P' and G-', we have also, a being the angle between the action-lines of the latter forces, P" = ^P j ^^ 7 ^^P j W~^~a ..... (2) (See p. 7, M. of E.) 23. Other Examples of Three-force Pieces. The ordinary straight lever, with flat-contact supports, is shown in Fig. 20. Since the pressures (or reactions] of the supports against the lever must be ~| to the axis of the latter, and hence parallel, in this case the action-line of the third force P' must be made "| to the lever. Otherwise equilibrium could FlG - 2 - not he maintained, for the point of intersection of the three force- lines is fixed by the intersection of the fiat-contact pressures P" and Q ; at infinity in this instance. Given P', we determine Q and P" by considering the lever as a free body under a system of three forces in equilibrium (in a plane), taking a moment-centre at B so as to exclude the unknown force P" from the equation ; and obtain first, as a moment equation, Qb-P'a = 0, . . or, Q = P t ' 22 NOTES AND EXAMPLES IN MECHANICS. and then, by summing all the components of the three forces in a direction at right angles to the lever, + P"-0-P' = 0; whence P" ' = Q + P 1 ; and thus the two unknown forces have been found in amount (and are already known in position). Obviously, if Q is given, being, e.g., the weight of a body to be sustained, we compute in a similar manner the necessary force P' to be applied at C and the resulting pressure, P" = P' + ft at the support or fulcrum B. It is also evident that the smaller the distance 1} is made in comparison with a, the greater the pressures P" and ft for a given P 1 '; in fact, as 5 approaches zero, P" and Q increase with- out limit. For a given Q and diminishing value of the ratio 5 : a, the necessary force P 1 decreases toward zero. (N.B. In these cases the weight of the lever itself is neglected.) As showing how the possibility of equilibrium may be de- pendent in some cases on the design and position of the support- ing surfaces, let us consider the curved lever in Fig. 21, where the supporting surfaces A and B are capable of furnishing only flat- contact pressures or reactions, whose direc- tions, A . . O and B . . O, are fixed, being normal to the respective smooth and flat surfaces of contact. (N.B. Smooth sur- faces are postulated in all the present prob- lems ; rough surfaces will be considered later.) The intersection, O, of their action- lines is therefore fixed, and if a force P' is to be applied at a given point C to induce pressures at A and B, its line of action must be taken along C . . O ; otherwise the lever will begin to move out of its pres- ent position (weight of lever neglected). Given, then, the force P' along C . . O, we determine P" and Q for equilibrium, in the same manner as before shown, by filling out the parallelogram O . m . r . Ic, in Fig. 21, precisely as was done in Fig. 19 (except that the P' of this problem corresponds to the G r of Fig. 19). CURVED LEVER. REDUNDANT SUPPORT. FIG. 22. A we have an un- action-line A . . O. However, if we assume any direction at pleasure for the action-line of P' through the point (7, and wish to secure equilib- rium, we have only to change the mode of support at A or at B (say B\ into a pin- joint or hinge-support ; for the direction of a hinge-pressure is not determined solely by the nature of this mode of support ; the only restriction upon it known in advance is that it must pass through the centre of the hinge-circle, its direction being determined by other relations. This change being made, we have Fig. 22, in which P' is given, or assumed, both in amount and position. At known flat-contact pressure Q in a known The intersection O of the two action-lines A . . and O . . must be a point in the action-line of P" , the unknown hinge-pressure ; i.e., B . . is the action-line of the latter, while its amount, as well as that of Q, is found by a construction like that in Fig. 19, which need not be explained again. The results are that P" = . . ft, and Q = . . n (the equal and opposite of . . r\ these lines, or rather lengths, representing forces on the same scale as that on which . . m represents the first given force P f . 24. Redundant Support. If the two supports, A and B, of the lever are l>oth hinge-joints, as shown in Fig. 23, the body or lever ABC is redundantly supported ; for now the hinge-pressures at A and B are indeterminate, from simple Statics alone, but depend on the form and elasticity of the lever itself and upon the degree of loose- ness of fitting of the hinge-rings around the pins of the supports A and E, and upon any slight elastic yielding of the latter ; as well as upon the amount and position of FIG ' 23 ' theforceP'. In fact, if the body is elastic, and we have to " spring" it to NOTES AND EXAMPLES IN MECHANICS. cause the rings to fit over the pins, pressures are produced at the supports before the application of any force at C. From simple Statics, then, all that we can claim is that the action-lines of the hinge-pressures P" and Q must intersect in some point situated on the given action-line of the given force P' ; but have no means of fixing the position of this point 0., Problems of this nature, therefore, cannot be treated until the theory of Elasticity is presented, and then, as will be seen, only in comparatively simple cases. In attempting an analytical treat- ment in this case we should find that it presents four unknown quantities ; whereas from simple Statics only three equations (independent equations) can be obtained for the equilibrium of a system of forces in a plane ; hence the indetermi nation. 25. Four-force Pieces. If the rigid body is in equilibrium under a system of four forces one of which is given both in amount and position, while the action-lines of all the other three are known, the amounts of those three can be determined. A simple graphic method for solving this case is based on the obvious principle that if four forces are in equilibrium the (ideal) resultant of any two must be equal and opposite to the resultant of the other two and have the same action-line. 26. The Simple Crane. A convenient example of a four-force piece is presented by the simple kind of crane, ABC, in Fig. 24, consisting of a single rigid body, of curved form. Its lower extremity rests in a shallow socket, while at B the edge of the (wharf) floor furnishes lateral support. We neglect the weight of the crane, and assume that no pressures are induced -at A and B unless the crane bears a load ; i.e., that the parts are FIG. 24. loosely-fitting at A and B. Placing now a known load at (7, viz., P l , we note that in preventing the overturning of the crane the right-hand edge of the floor at B reacts against the crane with some horizontal pressure P^ (horizontal, since the surface of contact is vertical), while at A there are two surfaces under pressure, one of which is horizontal, Simple Crane SIMPLE CRANE GEAPHICALLY AND ANALYTICALLY. 25 while the other is the left-hand vertical side of the socket ; there- fore at A we have the vertical and horizontal reactions, P t and P 3 , both unknown in amount. Now pair off the four forces at convenience ; for example, as in our figure, pair off P l with P % , noting the intersection, 0, of their action-lines ; P s and jP 4 constitute the other pair and inter- sect at o' . Since o is a point in the action-line of the resultant of JP, and PI, while o r is a point in that of the resultant of P t and P 4 ; and since these two resultants must have a common action-line for equilibrium, that common action-line must be o . . o'. Pro- ceed therefore as follows : Prolong G . . o and make o . . m equal to PI by any convenient scale. We know that the diagonal of the parallelogram formed on JP, and jP a must lie on the line o . . o f (prolonged, here). This parallelogram is found by making m . . r || to P z to determine r on o . . o' ; then drawing r . . k \\ to m . . o to intersect A . . o in some point k. Now prolong o . . o' beyond ; while by summing vertical components V + V n - G- , = 0; hence V =G+G 1 -V n . 29b. Pressures at the Joints ; D, B, and A (Fig. 25). By in- spection we see that the tie-rod DB is a two-force piece (its own weight neglected) ; that is, the hinge-pressures at D and B must have a common action-line, viz., D . . B. DB is a straight two- force piece or "straight lin~k" as we shall hereafter call it ; and is subjected to a tensile action along its axis (tension, here ; as we note by inspection ; a straight link under compressive action is called a strut, or compression-member). The Jib as a free Body. Fig. 26. It is a three-force piece, being acted on by the known vertical down- a v.^ ward force G at 6 7 , by an unknown horizontal force T (directed toward the left) at B (T ._ being the pull of the tie-rod), and an unknown hinge-pressure P at A, making an unknown angle a with the horiozntal. NOTE. Since the mast which acts on the jib at A is not a two-force piece, we have no means of knowing the position of the action- line of P from a mere inspection of the mast, as we did with the tie DB and the hinge-pressure at B. Of course, graphically, P passes through the point m where the action-lines of the other two forces intersect ; but as we are now TENSION IN THE TIE-KOD. 29 using analytical methods, we shall replace P, which is unknown in amount and in position, by its (ideal) horizontal and vertical components, P h and P v (i.e., by two unknown forces in known action -lines). We thus have a four-force piece to deal with. If we take A as a centre of moments, the force T will be the only unknown quantity in the corresponding moment-equation, which is Tl} Got ; whence we have, for the tension T in the tie-rod, T= G(a ~- 1). (This T is the value of the hinge- pressure at D and also that at B, in Fig. 25.) Assuming an axis X horizontal and _F vertical, we now have from 2X= 0, P h - T= ; or, P h = T\ while from 2Y=0, P V -G=0', or, P v = G ; and we now p easily find P itself, since P = VPJ + P; ; while tan a = ^~. ** 29c. Tension in the Tie-rod FE. As with DB, so with FE we note by inspection that it is a two-force piece, so that the hinge-pressure, T ', at E, Fig. 25, though unknown in amount as yet, must have E . . jFas action-line. The force T f and the pressures (or supporting forces) II and V exerted by the left-hand side and bottom, respectively, of the shallow socket U against the foot of the mast, are the three unknown forces in known action-lines acting on the free body shown in Fig. 27, consist- ing of the jib, mast, and tie-rod DB. The mutual actions between these three bodies are internal to the free body taken and are hence omitted (see note in 29a), the external system consisting of T r , II, V, and G\ all in known action-lines, G being the only force known in amount. By moments about the point 7, we have T'n - Ga = ; whence T' = - . G. . . . (1) From ^X= 0, IT- T' cos a' = 0, and . . JI= - G cos a'. (2) n = 0, F-^-r'sinar'^O; .-. V= G+ T sm a'. (3) 30 NOTES AND EXAMPLES IN MECHANICS. FIG. 28. It will be noticed that we have not made use of the mast as a separate free body. This might have been done as a means of finding the three forces just determined, T ', H, and Y', since the hinge-pressures at D and A had already been de- duced ; but the process would have been more roundabout. However, as a reminder of the principle of T action and reaction and of the definition of force, Fig. 28 is presented, showing the system of forces we should have to deal with in treating the mast as a free body ; and also Fig. 29, representing the car-platform and the two pairs of wheels as a single free body, with the external forces acting. The student will note that the H and V in Fig. 28 are the equals and opposites, respectively, of those in Fig. 29. A similar statement may be made for the T' of those figures. Again, the P and T of Fig. s 28 are the equals and oppo- sites of the P and T of Fig. 26 (action and reaction). The plane of the crane being sup- posed to be midway between the two wheels of each pair, the pressure V is equally divided between the two wheels forming the pair on the left. Similarly at JV, with V n . 30. Simple Roof and Bridge Trusses ; Hitter's Method of Sections. A truss is an assemblage of straight pieces jointed together in one plane. If the joints consist of pins (one in each joint) inserted through holes or rings in the ends of the pieces (or u members") the truss is said to be "pin-connected"; while if the ends of the pieces meeting at each joint are rigidly riveted together (a favorite method in Europe) the truss is said to have riveted connection. In the first case, pin-connection, each piece is free to turn about the pin, independently of all other pieces, during the gradual, though slight, change of form which the truss undergoes in the gradual settling of a load upon it, and the stresses induced in the pieces are called " primary stresses" (whereas, with riveted FIG. 29. RITTER'S METHOD OF SECTIONS EXAMPLE OF TRUSS. 31 joints, other, and additional, stresses, called " secondary stresses," are caused in the pieces, from the constraint exerted on each other by the members meeting at each joint). Confining ourselves to the consideration of pin-connected trusses, constructed so that each piece connects no more than two joints, and loaded only at the joints (its own weight being con- sidered as concentrated at the various joints), we note that in such a case each member must be a straight two-force piece ^ or straight link (neglecting its own weight) ; i.e., it is subjected to a simple tension or com pression (according as it is acting as a tie or a strut) along its axis. NOTE. If such a straight link be conceived divided into two parts (for separate treatment) by any imaginary transverse plane or surface passing between the joints connected by that piece, the action or force exerted on one of these parts by the other is a pull (if tension) or thrust (if compression), applied at the section and directed along the axis or central line of the piece. For example, in the truss of Fig. 30, pin-connected, and composed of t " straight links," if we wish to consider free the portion on the left of the imaginary cutting surface A . . B, the system of forces acting on the ideal body so obtained (see Fig. 31) consists of the abutment reaction F" , the loads G, G' , and G" (at the joints #, &, and YI + ^ ~ ^ ""^ = 5 ( and JLJi -\- 1 ,0 Cr 1 a 1 6r 2 2 = 0. FIG. 34. For Fig. 34 | T ^ ^%7 f ' ^ ^ l G * ~ ; ( atid 7})' X,ti + G 3 a 9 = 0. The elimination, by which to find separately the six unknowns, JT , y o , JT/, r,, 2T a , and I 7 ,, is left to the student. (Treated graphically, this case would come under Class B of p. M. ofE.) 458, 34 NOTES AND EXAMPLES IN" MECHANICS. 33. Problem of Rod and Cord (Prob. 4 and Fig. 41 of p. 37 of M. of E.). Consider the rod free by cutting the cord and removing the pin of the hinge ; that is, besides the forces 6rj and r 2 , we must insert the unknown tension P along the axis of the cord, at an angle a with the rod, and the horizontal and vertical components, X and Y , of the hinge-pressure, whose action-line is unknown, and thus have a complete svstem of forces in equilibrium. There are only three unknown quantities, P\ X, , and Y . From 2 hor. comps.=0, we have X P'coso^O ; . . . (1) 2 vert. " =0, u " Y +P' sin a- 6^- #,=0 ; (2) 2 (moms, about hinge) ... P'c 6>, 6X=0. ... (3) Since eq. (3) contains only one unknown, P', we have at once P f =(G l a l + G,fi3 + c; and knowing P ', we obtain X from (1), and Y from (2) ; and finally the hinge-pressure, R = V X* + Y * 9 making an angle whose tan = Y -r- X with the horizontal. 34. Problem of Simple Roof-truss (Frob. 5 and Fig. 40 of p. 37, M. of E.). The right-hand support is supposed to furnish all the horizontal resistance. Hence the system of forces acting on the whole truss, considered free, will be as shown in Fig. 36, in which there are three unknown reac- tions (or pressures, of the support- ing surfaces), V , V n , and H. II becomes known from 2X= 0, viz., 2 TFsin a - H = 0. . By moments about point (9, we have VJ-PJ-P^tf- Wb + + + = 0, which can be solved for V n , while from 2 moms, about B _ yj + Pl + W . I cos a + W(l cos a - 1) + P, . 1 = 0, (3) from which F can be obtained. FIG. 36. (i) (2) SIMPLE KOOF-TRUSS. 35 [NOTE. Having now made use of three independent equations, based on the laws of equilibrium of forces, and by their aid de- termined three quantities originally unknown, the student should Dot imagine that by putting 2 Y = 0, or by writing another moment summation about the point A (for example), he thereby secures another independent equation, from this same free body, capable of determining a fourth unknown quantity. He would find that such, an equation could be established by mere algebra, from the first three above, without further reference to the figure, and hence would be useless as regards determining any other unknown quantities. See top of p. 33, M. of E.] All loads or forces being considered to act at the joints, and no piece extending beyond a joint, we note that this roof- truss is composed entirely of straight two-force pieces (each in simple tension or compression along its - axis), so that portions of the truss may be considered free, isolated by the passing ot' one or more cutting surfaces. For ex- ample, to find the stress in piece AO and that in CO consider the free body in Fig. 37, where S and T are the stresses required, figure form a concurrent system, for which FIG. 8? The forces in this = gives 8 cos or+Tcos /?+ TFsin a=0, . . . (4) and 2 Y = " S sin a+T sin /?+ F P TFcosr=0. . (5) Solve these for S and T. In a numerical case one or both of these will come out negative, indicating compres- sion, not tension as assumed in the figure. To find the stresses in A C and AB, the free body shown in Fig. 38 may be taken. Here the forces form a non-concurrent sys- tem. Taking moments about B, we have Ta- U. iZ + + = 0; . (6) from which, T being already known, U can be obtained. For R, put 2Y = 0, and it will be the only un- known quantity ; or, put moments about C = 0. -_______>, FIG. 38. 36 NOTES AND EXAMPLES IN MECHANICS. 35. Problem. Rod and Tumbler (Fig. 39). The tumbler is smooth-edged with vertical sides. The rod has smooth sides and weighs G Ibs. C is its centre of gravity, at a distance a from the end of rod. Given the distances a and d, at what inclination a with the horizontal should the rod be placed, in contact with tumbler at two points as shown, that the P 8 iti n of the rod may be stable, i.e., that FlG - 39 - the rod may remain in equilibrium ? G, a, and d are known ; H, P, and a unknown ; R and P being the pressures of the tumbler against the rod at the two points of contact. II must be horizontal (why ?), and P 1 to side of rod and hence at angle a with the vertical. The rod being the free body, from 2Z = we have H P sin a = 0; (1) u 2 T = " " G + P cos a = ; (2) " 2 (morns, about 0) we have Pd sec a Ga cos' a = 0. . (3) Now sec a 1 -~ cos a, and, from (2), G = P cos a. Hence (3) becomes /7 3 [d P Pa cos 3 a = ; .-. cos a = \ / -. cos a V a a being now known, P and H are easily found from (2) and (1). (A brief mode of finding a alone is based on the fact that the three action-lines concerned must meet in a common point m. If, therefore, a figure be drawn in which the action-line of P in- tersects that of II in a point m' not coincident with m, that of II and G, we have only to form a trigonometrical expression for the distance ~mm' , involving , d, and a, write it = 0, and solve for cos cf or sec .) 36. Problem. Pole and Tie (Fig. 40). Given the load P, the weight G of the pole, and all the distances and angles marked in the figure, it is required to find the tension P' induced in the chain, whose weight is neglected, and which thus serves as a straight tie. The pole is hinged at 0. The pole may be considered free, as already shown in the POLE AND TIE. 37 Pole and figure, by inserting the pull P' exerted on it at ~k in a known action-line, Jc . . Z, by the chain, and the horizontal and vertical components, X Q and Y , of the hinge-pressure at 0. The action-line of this hinge-pressure makes an unknown angle with the horizontal, but must pass through the centre of the hinge at 0. There are three un- knowns in the system, viz., X , Y P f . From 2X = we have and FIG. 40. P f cos a - X = ; . . (1) 2Y=0 " " Y.-P-G P'sin a = 0-,. . (2) 2 (moms, about 0) = 0,Pa + Gb P'a' = 0. . . (3) The value of P' is easily found from (3), then that of X from (1), and that of Y 9 from (2). Hence the amount of the hinge-pressure, P = VX* + Y*, becomes known, and the tan- gent, = Y -f- X Q , of the angle between its action-line (On) and the horizontal. Graphic Solution. The action-line of the resultant, It, = P + 6r, of the known parallel forces P and G may easily be determined by a construction like that of Fig. 10, M. of E. ; or by applying the principle of the foot-note, p. 14, M. of E. Since this action-line intersects that of the force P' in some point n, the hinge-pressure at must act along the line O . . n and must be the equal and opposite of the resultant of P' and R. 37. Problem, Three Cylinders in Box (Fig. 41). Three solid homogeneous circular cylinders, of equal weight, = G, and of the same dimen- sions, rest in a box, as shown, of hori- zontal bottom and vertical sides. Th^ two lower cylinders barely touch each other ; i.e., there is no pressure between them, as the box is an easy fit. The centres of the three cylinders form the vertices of an equilateral triangle. It is required to find the pressures at all points of contact (points, in Three Cylinders. NOTES AND EXAMPLES IN MECHANICS. FIG. 42. FIG. 43. this end mew ; really, lines of contact) between the cylinders and the box ; also between the cylinders themselves. All surfaces smooth. Fig. 42 shows the upper cylinder as a free body, there being three forces acting on it, viz., (?, the action of the earth, or gravity, and the two press- ures, P' and P") from the cylinders be- neath. From symmetry, P f and P" must be equal. From 2 (vert, comps.) = 0, 2P' cos 30 G = ; i.e., P' (or P") = G + V3. Taking now the lower right-hand cylinder free in Fig. 43, we find it under the action of its weight G ; of the pressure P' (the equal and opposite of the P' in Fig. 42) now known ; of the un- known horizontal pressure or reaction P'" from the side of the box ; and of the vertical pressure JP from the bottom, also unknown. (Concurrent system, with two unknown quantities.) There is no pressure at A. (See above.) From 2 (hor. comps.) = 0, P' s i n 30 - P'" = ; .*. P'" = G( V$ -r- 6). From 2 (vert, comps.) = 0,P -G -P r cos 30= ; .-. P = f #. It thus appears that the sum of the two pressures on the bottom of the box is 3r, i.e., is equal to the combined weight of the three cylinders ; if the sides of the box were not vertical, however, this would not be true, necessarily. 38. Problem of the Door. Fig. 44 shows an ordinary door as a free body. Support is provided by two verti- cal hinge-pins, of smooth surfaces, whose dis- tance apart is such that the lower one alone receives vertical pressure from the door (i.e., furnishes a vertical supporting force, V , at its horizontal upper face). The upper hinge-pin pro- vides only lateral support, as seen in the horizon- tal reaction H which prevents the door from fall- ing away to the right, from that hinge. Simi- larly, the right-hand vertical edge of the lower FIG. 44. DOOR WEDGE AND BLOCK. 39 hinge-pin, by its reaction H , offers lateral support at that point. Given the weight G of the door (considered as concentrated in its centre of gravity) and the distances a and A, it is required to determine the three pressures, H, H , and V . From 2 (vert, comps.) = we have V G = ; i.e., V 6r. From 2 (moms, about 0), #a = ; i.e., H=-j-G. From 2 (hor. comps.) = 0, J7 H = ; i.e., #~ = #. A Since V 9 is parallel and numerically equal to 6r, and IT to 7/ , the system of forces acting on the door is seen to consist of two couples of equal moments of opposite sign, thus balancing each other. The smaller the distance h the greater the value of H (and of its equal, If \ if G and a remain unchanged. (Unless the fitting of the parts is very accurate, only one hinge of a door receives vertical pressure i.e., carries the weight, in practical language.) If more than one receives vertical press- ure, the share carried by each depends on the accuracy of fitting and on the slight straining or change of form of the parts under the forces acting.) 39. Problem of the Wedge and Block (Fig. 45). The shaded parts represent a smooth horizontal table or bed-plate eh, and a flat and smooth ver- tical guide md, both immovable. W is the wedge whose angle of sharpness is a, and B the block, on which rests a weight (not shown) whose pressure on B is verti- cal and = Q. The weights of wedge and block are neglected. There is supposed to be no friction, otherwise t!,c results would be quite different (see 9 of Graph. Stat. of Mechanism, in this book, and p. 171, M. of E.). Given Q and a, what force P must be applied horizontally at the head of the wedge to prevent the block B from sinking (or to raise the block with constant velocity if the latter has an upward motion)? Supposing the required force P to be in action, and that there FIG. 45. 40 NOTES AND EXAMPLES IN MECHANICS. FIG. 46. FIG. 47. is no friction, the mutual pressure, N, between block and wedge is normal to the surface of contact and hence makes an angle a with the vertical, while the pressures on surfaces cd and eh, viz., 8 and J?, are horizontal and vertical, respectively. Hence, when the block B is considered free (Fig. 46) we hrve equilibrium between the three forces, Q, N, and /&; the two last are unknown, but are determined thus : From 2 (hor. comps.) = 0, /S = JV" sin a ; From 2 (vert, comps.) = 0, N cos a = Q. Similarly, we have the wedge free in Fig. 47 under the action of jV, now known, and the unknown and the required P. Hence R = N cos a, (= Q\ from 2 Y = ; and P = N sin a, = Q tan oc, from 2X 0. Evidently, the sharper the wedge the smaller the force P necessary for a given Q. 40. Cantilever Frame (Fig. 48). This frame consists of eleven straight pieces in a vertical plane, pin-connected, and supposed without weight. Pieces (7, Z>, E, and / are horizontal, G and H vertical, the others oblique. The vertical rod at A is anchored, as shown, but no stress is induced in it (nor in any other piece) unless a load P is placed at w (no other joint to be loaded). Since, furthermore, no piece connects more than two joints, each piece is a straight two-force piece, subjected to a tensile or compressive stress along its axis, and any one may be conceived to have a portion removed, when desired, in the isolation of the various "free bodies" to be considered. The necessity of the rod and anchorage is evident from the fact that the rectangle m'nom is left unbraced. The load P being given, it is required to find the stress in each piece of the frame and in the rod at A. FIG. 48. WEDGE CANTILEVER FRAME. 41 FIG. 49. The free body in Fig. 49 enables us to find the stresses E and F. Since only three forces are concerned, meeting at a point, a simple procedure is to resolve the known P into two components, E' *-* ^ "/ffi""*^ and F') along the action-lines of E and F, as /&\ / shown by dotted lines. E r and F f are equal and opposite to the required stresses E and f\ respectively ; i.e., E = P cot a = P j , and /^=jPcosec or= P -=- sin a. (The summation of horizontal and vertical components put equal to zero would give the same result.) ^is tension; F, compression. Next take the free body in Fig. 50, involving the known force P and the unknown stresses 7>, 7, and Z, assuming for them the characters implied by the pointing of the arrows. Taking moments about centre of pin at 0, we have FIG. 50. -Pa = Q; whence _Z> = + P, and is tension. From 2 (vert, comps.), P = L cos ft, or L = + ~ COS fJ is therefore compression as assumed ; while, from ^(hor. comps.) = 0, I D L sin ft = ; i.e., /"= + P (tan /? + cot a\ and is compression. To find the stresses in A, B, and G we use the free body in Fig. 51. By moments about O, m' e . . A = -f- P -j , and is tension. X f /M ^-^-.^..VH From 2 (hor. comps.) = 0, D ^cos y 0; whence B = 4- -7- , and is compression, as assumed. ' fi cos y ' FIG. 51. 42 NOTES AND EXAMPLES IN MECHANICS. In Fig. 4$ we note that since pieces C and D are in the same straight line and G is the only other piece connecting with the joint m' (there being also no load on m'\ the stress in the piece G must be zero and the stress in C must equal that in D ; similarly, the stress in piece II is zero and E= D (as already found above). Hence G is marked = in Fig. 52, showing a free body FIG. 52. in which the stresses J and K are the only unknown forces, C being = D, = P j , and G = zero. There- fore, by elimination between the equations J cos -f- C I R cos d (from sum of hor. comps.), and J sin -\-l sin # A = (from sum of vert, comps.), we obtain the stresses J and K, both of which are assumed to be compressions in this figure. If the joints m' and n were not both in the horizontal line joining w and v, Fig. 48, the stresses in G and H would not be zero, as they are in this instance. 40a. Real and Ideal Forces. " Balanced Forces." The student should be careful to distinguish between real and ideal forces. If a body A receives pressures P and Q from bodies B and C re- spectively, the resultant of P and Q is purely id-eal, being merely conceived to take the place of P and Q, if any useful and legiti- mate purpose can thereby be served. Again, the X and Y components of an actual, or real, pressure P are ideal ^ serving a mathematical purpose only, when we suppose P to be removed and its components inserted in its stead. Such customary phrases as "balanced forces," "forces in equilibrium," etc., are unfortunately worded, as they seem to imply that forces act on forcos, which is an absurdity. In reality, bodies act on bodies, force being the mere name given to the action (if it is a push, pull, etc.) ; so that, instead of stating that " the forces are balanced," we should more logically say : "The rigid body is in equilibrium, or balanced, under the actions of certain other bodies." (See correspondence in the London Engineer of June, July, and Aug., 1891.) CHAPTEK III. MOTION OF A MATERIAL POINT. 41. Velocity and Acceleration. Any confusion between the ideas of velocity and acceleration is fatal to a clear understanding of the subject of motion. Just as velocity may be defined as the rate at which distance is gained, so acceleration may be defined as the rate at which velocity is gained, thus : If at a certain instant a material point has a velocity of 10 feet per second, we mean that it has such a speed of motion that it would pass over a distance of 10 feet ^that rate of speed remained constant dur- ing the whole of the next second of time. Similarly if, at the same instant, the acceleration of the material point is said to be 20 feet per " square second " (or " 20 feet per second, per second "), we mean that if the velocity were to continue to change, during the whole of the next full second, at the same rate at which it is changing at the instant mentioned, the velocity at the end of that second would be greater by 20 units of velocity than at the in- stant mentioned, i.e., the velocity would be 30 feet per second. According to this definition, then, if the velocity of the material point remains unchanged, the acceleration is zero at every point of the motion. At each point, then, in the path of a material point moving along a right line, we have to do with Sj its distance from some convenient origin in that line ; v 9 its velocity, or rate at which that distance is changing, = -.- ; dt p, its acceleration, or rate at which the velocity is changing, dv There is no need of defining still another quantity as the rate at which the acceleration is changing, for the reason that the 43 44 NOTES AND EXAMPLES IN MECHANICS. force which at any instant occasions the peculiarity of the motion of the material point is determined from the acceleration, viz., from the relation force =. mass X accel. In the uniformly accelerated motion of a free fall in vacuo, the following are the values of the above quantities at beginning of the motion, and at the end of each full second thereafter, thus : s, = Distance, v, Yeloc. p, Acceleration = g. At beginning, ... 0.0 ft. 0.0 ft. p. sec. 32.2 ft. p. (sec.) 2 ; At end of 1st full sec., 16.1ft. 32.2 " " 32.2 " At " 2d " " 64.4ft. 64.4 " " 32.2 " At " 3d ' " " 144.9 ft. 96.6 " 32.2 " " A recent English writer is so desirous that acceleration shall not be confused with velocity that he calls the unit of velocity a " Speed " and the unit of acceleration a " Hurry" For example, using the foot and second as units, he would say that at the end of the second full second of a free fall in vac.uo the velocity is 64.4 "speeds" while the acceleration at the same instant is 32.2 "hurries." These words are quite suggestive and should be borne in mind. To say that at a certain instant the acceleration is zero does not imply that the body is not moving, but simply that its veloc- ity, however small or great, is not changing; and again, the statement that the velocity is zero at a certain instant does not imply that the acceleration is zero also, but only that the velocity, as its value changes, is then passing through the value zero, while the rate at which it is changing (i.e., the acceleration) is not de- termined until some further statement is made. 42. Momentum. This word is used quite largely in some works on Mechanics, but may be considered a superfluity, liable to give rise to confusion of ideas ; though sometimes useful, it need rarely be used. By definition, it is a name given to the product of the mass of a material point by its velocity at any instant, i.e., Mv. Of course, the mass of the body is a constant quantity, while its velocity may be continually changing; hence the momentum is always proportional to the velocity. In accordance with this definition, the value of a force which is accelerating the velocity of a material point in its path is some- MOMENTUM COKD AND WEIGHT. 45 times stated to be equal to the rate at which the momentum is changing ; by which is simply meant the following: From eq. (IV), p. 53, M. of E., we have P = Mp, = (mass X dv accel.), butj? = -j- ; hence we may write cut ^dv Mdv d\_Mv\ _ ("change of momen- dt ~ ' dt dt ~ L turn in time dt i.e., P = the rate at which the momentum is changing. There- fore the above statement as to the value of the accelerating force is nothing more than what we already have in the form of P = Mp. 43. Cord and Weights. There are few mistakes more common than rushing to the conclusion that the tension in a vertical cord to which a weight is attached is equal to that weight. This may be true (and is true if the weight is at rest and has no other sup- port), but should not be assumed without thought. For example, Fig. 53, having two weights attached to the same cord, if the point A of the cord were fastened, A the tension in B would be = G ; and that in (7, = G', if G and G' are the respective weights of the two bodies. But if, the cord bein^ continuous and not fastened, with no friction at the pulley-axles, an accelerated motion begins (assume G' > G), the tension on each side depends on that acceleration, FIG. 53. as well as on the weights of the bodies. To find this accel- eration, which is common to both sides, neglecting the masses and weights of the pulleys and cord (by which we mean that the tension S in the cord at A may be taken equal to that at B and C} let us consider the body G' free. We note that this body has a downward accelerated motion, and that the forces acting on it are 6r', directed vertically downward, and S, the tension in the cord, pointing vertically upward, i.e., acting as a resistance; G' hence, calling the acceleration p, we have G' /S = p. As t7 for the other weight, it is rising with an upward acceleration = p. 46 NOTES AND EXAMPLES IN MECHANICS. under an upward force 8 and a downward resistance G, whence S G = p. From these two relations we obtain by elimina- y tion G' - G and *= The value of p might have been obtained directly by consid- ering that the acceleration of the motion is just the same as if the G+G' whole mass -- : - were moving in the same right line under i/ the action of a single accelerating force G r G. 44. Example. Lifting a Weight. A rigid mass weighing 100 Ibs. is to be lifted vertically through a distance of 80 ft. in 4 sec- onds of time, and with uniformly increasing velocity, from a condition of rest (i.e., veloc. 0). "What tension must be main- tained in a vertical cord attached to it, to bring about this result? The average velocity is 20 ft. per second, and since the initial velocity is zero and the rate of increase of velocity (i.e., the acceleration, p) is to be constant (uniformly accelerated motion), the final velocity will be double the average, viz., 40 ft. per sec. (see also eq. (4), p. 54, M. of E.). If, then, 40 velocity-units are gained in 4 seconds, the acceleration is 40 -=- 4 = 10 ft. per second per second (10 " hurries"), and an upward accelerating force of Mp = 2^2 X 10 = 31 Ibs. must be provided. But the only forces actually present, acting on this mass, are the action of the earth, viz., 100 Ibs. pointing vertically downward, and the ten- sion, S, in the cord, directed vertically upward ; and their (ideal) resultant, which is S 100 Ibs. (since they have a common action- line), is required to be 31 Ibs. and to act upward ; whence we have the required cord-tension, = S, = 131 Ibs. To lift the 100-lb. weight, then, under the conditions imposed, requires a cord-tension of 131 Ibs. [If the cord be allowed to become slack at the end of the 80-ft. distance, the body does not immediately come to rest, as it then has an upward velocity of 40 ft. per second. Its further progress is an " upward throw" HAKMONIC MOTION. 47 (p. 52, M. of E.) with 40 ft. per second as an initial velocity, the only force acting at this stage being the downward attraction of 100 Ibs., which will gradually reduce the velocity to zero.] If a smaller height of lift had been assigned with the given time, or the same height with a longer time, the difference be- tween the requisite tension S and the gravity force (or weight) of 100 Ibs. would have been smaller ; and vice versa. It is convenient to note that in using the foot-pound-second system of units for force, space, and time, the mass of a body is obtained by multiplying its weight in pounds by 0.0310 (which is the reciprocal of 32.2) ; thus, in the preceding example the mass of the 100-lb. weight is = 100 X .0310 3.10 mass-units. 45. Harmonic Motion. From eq. (3), p. 59, M. of E., remem- bering that G and a are constants, we note that s, the " displace- ment," or distance of the body from the origin (middle of the oscillation), is proportional to the sine of an arc or angle, and that this arc is proportional to the time, t, elapsed since leaving the origin ; whence arise the following graphical relations : In Fig. 54, let O be the origin or middle of the oscillation, c , and the horizontal line CD the path of the body ; OD, = /, being the extreme displace- ment, i.e., the " semi-amplitude," of the mo- tion. At any instant of time the body is at some point m between C and D, i.e., Om = the variable s (dis- placement). With as centre, and OD, = r, as radius, describe a circle and erect a vertical ordinate at m at whose intersection n with the curve draw a radial line On, making some angle 6 with the vertical, and the abscissa nk. Now 0m, or s, = r sin 8 ; and the length of curve En is proportional to the angle 0. Hence the linear arc En is proportional to the time occupied ~by the body in describing the distance s = Om ; that is, if n be re- garded as a moving point, confined to the circumference of the circle, and always in the vertical through m, its motion being thus controlled by the harmonic motion of w, the velocity of n in its circular path is constant (equal linear arcs in equal times). The constant velocity of n must be equal to the initial velocity 48 NOTES AND EXAMPLES IN MECHANICS. of ra, i.e., c, as the latter leaves the origin, 0\ for at the two points are both moving horizontally. Conversely, therefore, if a point n move in the circumference of a circle with a constant velocity c (continuously in one direc- tion), the foot of its ordinate, i.e., the point m, moves with har- monic motion along the horizontal diameter ; this is the proof called for in the line below Fig. 64 of M. of E. The of Fig. 54 is t Va of the formulae on p. 59, M. of E. Since the acceleration (as also accelerating force) in harmonic motion is proportional to the displacement, the length of the line 0m, or kn, represents the acceleration at any instant, while the length, mn, of the ordinate is proportional to the velocity of the body or material point, m (since by eq. (4) of p. 59, M. of E., that Velocity varies as the cosine of 0). Of course, in obtaining values, from the drawing, of these variables -y, p, and , for different positions of the body m, due regard must be paid to the scales on which the lengths marked in the figure represent these variables, none of which is a linear quantity. 46. Numerical Example of Harmonic Motion (using the foot, pound, and second). With the apparatus and notation of p. 58, M. of E., suppose that by previous experiment with the elastic cords we find that a tension T l of 4 Ibs. is required in either of them to maintain an elongation of 3 in., i.e., of a foot ; then for any elongation s (or " displacement" of the body during its motion) the tension (and retarding force) is T = T,s -=-*, = 16,9. Let the small block weigh 128.8 Ibs., then its mass is 128.8 X .0310 = 4 = M ; and hence the constant quantity called a is a = (T v ~ Ms^) = 16 M ; i.e., a = 4. Let an initial velocity of c = 4 ft. per sec., from left to right, be given to the block at ; required the extreme distance attained by the block from (i.e., the semi-amplitude," = /), and the time occupied in describing the semi-amplitude (i.e., required the time of a quarter-period). From p. 59, M. of E., r = c ~ \f~cu; .-. r = 4 ~- V~ = 2 ft. ; while for a quarter-period, or half-oscillation, we have the time \n -r- Va = %7t = 0.7854 sec. BALLISTIC PENDULUM. 49 Hence in the diagram of Fig. 54 we make OD == r = 2 f t. Then, since the linear arc ED represents 0.7854 seconds, the time represented by En for any position m of the block will be on a scale of sec. to each foot of En (since 0.7854 -f- (\7t x 2 ft.) = J). Since OE, or 2 ft., represents the velocity of m on leaving O, i.e., 4 ft. per second, mn represents the velocity at m on a scale of 2 ft. per second to each foot of mn. Similarly, since the acceleration OD of m at D (from jp as) is (4 X 2) = 8 ft. per (sec.) 2 , that at m will be kn on a scale of 4 to each foot of kn. It must be noted that after m reaches D and begins to return toward the left, the point n in the curve will have passed below the horizontal through D ; that is, the time is now greater than ED, and the velocity has changed sign. When n passes under- neath the acceleration and the displacement change sign ; and so on indefinitely. (Tension in cord at D ?) 47. The Ballistic Pendulum. This old-fashioned apparatus for determining the velocity of a cannon-ball consisted of a heavy body B, of mass M z , suspended, by a rod or rods, from a horizontal hinge on a fixed support. A cavity in its side is partly filled with clay, so as to make the impact inelastic. This body being initially at rest (see Fig. 55), the ball is shot horizontally into the cavity, in which it adheres. As a result of the impact the centre of gravity of the combined masses receives a velocity C, with which it begins its ascent along the circular arc J3m<, finally reaching a vertical height II above its initial position before coming to rest (for an instant). H being measured, or computed from the observed angle ft, we have O= + VZyH (see foot of p. 80, M. of E.). The known mass of the ball being M 1 and its unknown velocity just before impact -|- c l , eq. (4) of p. 65, M. of E., enables us to write (In this theory the suspended mass has been treated as a material point, and the result is only approximate. The longer 50 NOTES AND EXAMPLES IN MECHANICS. and lighter the suspension-rod and the smaller the dimensions of the suspended body the more accurate the results. The strictly correct theory is rather complicated.) 48. Apparatus for the Determination of the u Coefficient of Restitution." It will be noted, by an examination of eqs. (2), (3), (5), and (6), on pp. 64 and 65, M. of E., that the coefficient qf restitution, w ^h cen * res at tne same level. Being allowed to swing simultane- ously from rest at a and J respec- tively, their impact takes place at O FIG. 56. (or very nearty so), their velocities just before impact being c l = + VZg^ and ?- equal to the loss in momentum of the other body ; i.e. (see eqs. (5) and (6), p. 65 of M. of E.), M,( F 2 - <7) = M } (C - V,). This is precisely what takes place when a spring in a state of compression (the ends tied together by a cord) is placed endwise between the two suspended balls of the apparatus of Fig. 57, and the cord is then severed without violence (burning is best). The spring in regaining its natural length exerts equal horizontal pressures in opposite directions at any instant against the two masses. This pressure is variable, but at any instant has the same value for one mass as for the other. That is, this phenomenon may be looked upon as a case of impact where the first period is lacking, the period of restitution occupying the whole time of impact. In this case M l and J/ 3 are at rest just before impact, that is, 70; while their velocities after the expansion of the spring are V^gH^ and + 1/2 0) = M t (0 [ VtyHA) ; i.e., the velocities immediately after the action of the spring are inversely propor- tional to the masses. The same method in another form consists in saying that from the principle of the conservation of momentum the total momen- tum before impact, viz., 0, here, is equal to that after impact, which is M, V, + M, F 2 , or M,( - VtyH?) + J/ 2 ( + V%glQ. Equating the latter expression to the zero momentum first men- 52 NOTES AND EXAMPLES IN MECHANICS. tioned, we have M l VZgll, Jf 2 V2gJJ^ 9 as before. This result is verified by the apparatus. 50. The Cannon as Pendulum. As the converse of the Ballis- tic Pendulum, the cannon itself may be suspended in a horizontal position from a pivot 6', Fig. 58. With the ballistic pendulum the impact was inelastic, i.e., it consisted of a period of compression only, that of restitution being lacking. Here, however, the expansion of the hot gases generated when the powder is ignited acts like the spring in the preceding |DM 2 case, causing a forward pressure at any instant ^*7* against the ball and an equal and simultaneous backward pressure against the cannon ; i.e., the impact is one having no period of compression but only one of restitution. The velocity F 2 of the ball leaving the muzzle may therefore be inferred from a knowledge of the two masses concerned and the height of recoil 77, (A being the point where the centre of gravity of the gun comes to rest for an instant). Putting the total momentum before impact, which is zero (cannon at rest, in lowest position), equal to that immediately after, the cannon having then just begun its motion from D toward A with a velocity at D of V l = V^gl^ , we have or , = As before, we here treat the masses as material points. 51. Simple Circular Pendulum of Small Amplitude (Fig. 59). B is a material point suspended from a fixed point C by an imponderable and inextensible cord. Being allowed to sink from initial rest at the point A (cord taut), it follows the circular arc ABO with increasing velocity. At any point B between A and its velocity (which of course is tangent to the curve) is v = V%g X vertical height I)F~ (see foot of p. 80, M. of E.), and the forces acting on it are its weight G, vertically downward, and the tension S, in the cord. This tension increases as the body approaches O, since 2 (norm. SIMPLE 0IKCT7X,Afi PENDULUM. 53 comps.) must Mtf -=- rad. of curv., [eq. (5), p. 76, M. of E.], i.e., #_cos0:=: .^; whence S= ^fcos 0+^-1. . (1) y " jfw-j At (9 the tension is greatest and = r 1 + -j- . If the starting point A is taken high enough, the cord may be broken before is reached. To find the tangential acceleration p t9 or rate at which the velocity is increasing, we note that from eq. (5), p. 76, M. of E., 2 (tang, comps.) must = Mp t , whence Mp t = G sin -f- S X ; or p t = g sin 0. . (1) This can be written p t = |- Z p 2 T)" gravity of actual plate and complete FlG - 61 - trapezoid, denoting those of the semicircle by a? 2 and y 2 . Re- quired a?! and y l , using the inch as linear unit. The area of the trapezoid is F = \ X 6 X (10 + 16) = 78 sq. in. ; that of the semicircle is F^ i7r(2) 2 = 6.28 sq. in. ; so that the area of the actual plate is their difference, F^ , = 71.71 sq. in. Also, for the point // we have y a = 4 X 2 ~ BTT = 0.85 in. ; and OJ) = x, * III the same, or iu a parallel, plane. -- 56 NOTES AND EXAMPLES IN MECHANICS. = 8 in. ; while (see p. 23, M. of E.) y = | .||: 1~ = 2.77 in.; and x = OD $ X 2.77 = 7.54 in. (C bisects the base MN\ conceive perpendiculars let fall from C and E upon OK, and use the similar triangles so formed in obtaining the value of a?.) From eq. (3), p. 19, M. of E., since the trapezoid is made up of the actual plate and the semicircle, we have x = (F& + X) (F> + jg ^ - Fy-F,y, and = - - By substitution, therefore, OY = [78 X 7.54 - 6.28 X 8] + 71.71 = 7.5 in. ; and y v = [78 X 2.77 - 6.28 X 0.85] -~ 71.71 = 2.94 " 55. Example 4. Stability of Two Cylinders. Fig. 62 gives an TWO cylinders, end view of two smooth and ho- Equal lengths -, * i /. mogeneous circular cylinders of equal length ( I' in feet) but of radii 1 in. and 3 in., respectively. A weighs 800 Ibs. per cub. ft., E only 100 Ibs. per cub. ft. A and B being placed, as shown, on two smooth planes at 45 with the ^2 ------- horizontal, it is required to find whether this is a stable position, or if A will crowd B out of place. Call their total weights G, and r a for the present. Since CD is one half of OC, the angle COD is 30. First, the position being supposed stable, to find the press- ure at point 2, we take A as a free body. The forces acting on it are three, shown at in Fig. 63. P l is the pressure (or reaction) of the inclined plane against A at point 1, P 2 is the pressure from the other cylinder, and G l the weight of this cylinder, A. These are all directed through (smooth surfaces), the angles being as shown. For equilibrium P 2 must be equal ai.d opposite to OK, the (ideal) resultant of G l and P l . Jn triangle OPf, OK\ G, :: sin 45 : sin 60 ; .-. OK= G[ ^2 -~ 1/3] = P % . Rad. of Cyl A is 1 inch. $ NUMEEICAL EXAMPLES STATICS. 57 In Fig. 63 we have acting through C the four forces acting on the large cylinder B (on supposition R of equilibrium). G t is its weight, P z is the pressure of the other cylinder, P 3 the pressure at point 3 of the inclined plane on the right, P 4 of that on the ^ *P* y G2 left. FIG. 63. If now the resultant of P 2 and G 9 were found to pass above the point 3, instability would be proved; since to occasion pressure at point 4 that resultant should evidently pass below 3. Or, which amounts to the same thing, assuming equilibrium at C and with P 4 as drawn, if we compute the value of P 4 by put- ting 2 (compons. -| to P 3 ) = 0, and a negative result is obtained, instability is proved ; and vice versa. Hence we write C 1 C 1 P 4 = G, cos 45 - P 2 cos 60 ; whence P 4 = ^ - L. . (1) v 2 \ 6 Now G, = |>( T V) 3 X 800Z'] Ibs. ; and G, = [n(f^ X 100Z'] Ibs., on substituting which in (1) we obtain P 4 = [^rf'( + 309.8)] Ibs.; which is positive, and thus verifies the supposition of equilib- rium. 56. Example 5. Toggle-joint (Fig. 64). The two straight links, of equal length, are pivoted to the two y blocks, as indicated. A horizontal pull of 80 Ibs., \ ftt making equal angles with the two links, is exerted on the horizontal pin of the joint B. What press- ures are thereby induced (for given position of parts) on the surfaces E and F (horizontal and vertical)? (Those on E' and F' will be the same, respectively.) Each link is evidently a straight two-force piece and hence under a com- pressive stress along its axis ; call this stress P '. The free body in Fig. 65 enables us to find P' f rom ^ (hor. comps.) = 0; i.e., 2P f cos a = P ; 58 NOTES AND EXAMPLES IN MECHANICS. Fig. 66 shows the upper block free, from which by P E = P' sin a, and from comps. P F =. P' cos a. n a= 0, hor. =4SO Ibs. ; and If, in Fig. 65, P were resolved into components along the axes of f & \ the two links, each such component / p j ' "B would be the equal and opposite of FIG. 65. FIG. 66. the corresponding P ' . Evidently, if a approached a right angle. P ', and also P E , would increase without limit. 57. Example 6. Simple Crane. The simple crane in Fig. 67 carries a load of 4 tons at (7, 12 ft. from the axis of the I ^^c vertical shaft, while its own weight is 1 ton, the centre of gravity being 3 ft. from the shaft. The socket at B is shallow so that lateral support is provided at A. Required the pressure at A and the horizontal and ver- Fl - 67 - FlG - 68 - tical pressures from side and bottom of socket B. The crane being considered free in Fig. 68, and the reacting pressures being put in, as shown, we have, from 2 (moms, about B) = (foot and ton), A X X 8 4x 12 1X3=0: whence A x = 6.37 tons. From 5" (vert, compons.), .Z^ 5 = ; or 2 (hor. compons.), B x A x = ; or By = 5 tons. B x = 6.37 tons. 58. Example 7 Door and Long Hinge-rod. The door weighs 200 Ibs., and is supported in a vertical plane in the manner indicated in Fig. 69. The continuous vertical rod ABC is con- sidered without weight, and from the nature of the mode of support of the door receives horizontal pressures at each NUMERICAL EXAMPLES STATICS. 59 of the points A,B, 6", and O. Kequired the values (J., B, = force -f- mass, the mass being = 20 -f- 32.2 = 20 x 0.0310 = 0.620 ; .-. p = ( 1.25) -=- .62 2.012 ft. per sec. per sec., and from eq. (3) of p. 54, M. of E. (in which, for present purposes, we put s = 8 1 , v 0, c = v l , and p as above), we have *, = [ (16.05) 2 ] -f. [2(- 2.012)] = 64 ft. If we inquire the pressure P between the block and curved guide just before reaching Z>, we note that that pressure must not only support the weight of the body, but must also provide a proper deviating force - to retain it on the curve ; whence P = 20 Ib, + i- Ib, = 60 Ibs. (See p. 83, M. of E.) The pressure on DE is only 20 Ibs. 68 NOTES AND EXAMPLES IN MECHANICS. Again, suppose the block to start from rest at .Z?, the angle BCD being 45 ; find v l and s l . (The acceleration on DE is the same as before.) v l = V% X 32.2 x 4(1 cos 45) = 8.68 ft. per sec. 8l = (O 2 - <) -T- [2(- . 2.012)] = 18.75 ft. 66. Example 14. Harmonic Motion of a Piston (Fig. 81). On account of the great mass, and large radius, of the rim of a fly-wheel on the same shaft, the rotation of the crank is practically uniform; at least during any one turn, i.e., the crank-pin is considered to move with a uniform velocity in a cir- cle. From the design of the piston and slot this body oscillates with harmonic motion in a horizontal path (see foot of p. 59, M. of E.) ; the left to right stroke alone is to be considered. The pressure called P is the total effective steam-pressure, i.e., the difference between the total pressure of the steam, now on the left of the piston, and the total atmos- pheric pressure on the right face. Friction on the guiding sur- faces is neglected, and since the motion is horizontal the weight of the piston has no component along its path. If P is constant throughout the whole stroke (left to right), and = 6000 Ibs., and the crank turns uniformly at the rate of (u :=) 200 revolutions per minute, r being = 8 inches; what must be the value of the pressure P 1 between crank-pin and the side of the slot just after the dead-point C is passed, i.e., at the beginning of the stroke ? Ditto, when the crank-pin is 45 from C\ and again, when it is at 0, 90 from (7? The weight of the piston and rod is 160 Ibs. Between C and 0, P' is smaller than P and is a resistance, as regards the motion of the piston. If that motion were uniform the full amount of the 6000 Ibs. would be felt at the pin, for in that case the acceleration would be zero and the horizontal forces, P and P', would be equal and oppositely directed ; but from C to the motion is accelerated (the piston has no velocity at C\ HARMONIC MOTION OF PISTON. 69 so that a certain amount, = mass X accel., of the 6000 Ibs. is absorbed in the " inertia" of the mass, so to speak, leaving only the remainder to be felt as a pressure P' at the crank-pin. This is expressed analytically by the relation 2 (hor. comps.) Mp ; i.e., P-P' = Mp. Beyond O toward D the constraint of the mechanism is such as to bring about the gradual stopping of the piston, which at O has its greatest velocity ( c, = to that which the pin has at all times), so that independently of the 6000 Ibs. on the left the piston is, as it were, thrown against the crank-pin, the pressure produced against which at any instant from to D must = Mp over and above the 6000 Ibs. due to steam action ; i.e., P f = 6000 Ibs. -j- Mp (where p is the numerical value of the acceleration, whose algebraic value is now negative), or, analytically, P P' = Mp where p has its algebraic value (negative when a number is in- serted for it). The linear velocity of the crank-pin is = %7tru, = 2 X -f - f -VA = 13.97 ft. per sec. (using the ft., lb., and sec.). As the pin ap- proaches and passes a dead-point the motion of the foot of the perpendicular let fall from it upon the horizontal diameter, along that diameter, is not only the motion of the piston, but is at this point normal to the path of the pin ; hence the normal accelera- tion of the pin is the actual acceleration of the piston at a dead- point. Therefore c 2 -f- r (see eq. (4), p. 75 ; and also section 75, and first line of p. 60, M. of E.) is the acceleration of the piston at (7; and therefore just after passing the dead-point O we have 6000 1452 = 4548 Ibs. The acceleration of the piston is proportional to the displace- ment, and hence at 45 from C we have P' = P M\ G - cos 45J = 6000 1452 X .707 = 4937 Ibs. At 0, p = and P = P' = 6000 Ibs. 70 NOTES AND EXAMPLES IN MECHANICS. Beyond at 45 P' = 6000 + 1452 X .707 = 7063 Ibs. ; while just before reaching D P' = 6000 + 1452 = 7452 Ibs. On the return-stroke steam is admitted to the right of the piston and P' occurs on the other side of the slot, with same variation during the stroke as before. 67. Example 15. Conical Pendulum, or Simple Governor-ball (Fig. 82 [a]). If the oblique part of the cord is to be 20 in. in length, what tangential velocity in a horizontal circle (centre at C) and what radius = ^, for that circle, must be given to the material point G of 10 Ibs. weight (= G) in order that motion in the circle shall be self- [a] (TJcUoK*. [bj perpetuating and the weight G' of FIG. 82. 40 Ibs. may be sustained at restl Fig. 82 [b~\ shows the moving weight as a free body, the only forces acting being a gravity-force of 10 Ibs. and an oblique cord- tension, P, which by above conditions is to be 40 Ibs. The mo- tion of G being confined to a horizontal plane, it has no vertical acceleration ; therefore 2 (vert, compons.) should balance, or p cos OL G = 0, whence cos a = -p = 0.25, and a should = 75 3 1/ . Hence r should be made = (20 in.) x sin a = 20 X 0.968 19.36 in. = 1.613 ft. Since the motion is to be in a curve, 2 (normal compons.) at any instant should M X (vel.)* G tf -r- rod. ; i.e., P sin a -[- = ; and combining this with the \J P cos a = G derived above we have tan a = c* -+- gr, whence the required velocity must be c = 1/32.2 X 1.613 X 3.871 = 14.2 ft. per sec. The proper radius being as above (r = 1.613 ft.), this implies rotation about C at the rate of c 14.2 = 1.40 revolutions per second, or 84 per minute. CONICAL PENDULUM EXAMPLE. 71 68. Example 16. The Weighted Governor or Conical Pendulum (Fig. 83). The four inextensible cords, each 16 inches long, connect the three "material points," or small //,/$,//* bodies, as shown ; the two upper cords being at- tached to a fixed support at A. G l and G, are two balls of equal weight; the weight of each = G l = 8 Ibs., while the block # 2 weighs 12 Ibs. If now the balls are caused to rotate about the vertical axis A CB at slowly increasing rate (revolutions per minute), by pressing against them FlG 83 - laterally with a vertical board whose plane contains the axis ACB, the angle a gradually increases and the block 6r 2 is lifted along the axis towards A. When the speed of rotation has reached any desired figure (rev. per min.) a has some correspond- ing value, and if the board is now removed a retains that value and the balls continue their motion (forever, if no friction) in the corresponding horizontal circle, sustaining the block 6r 2 at rest (at least its centre of mass is at rest) in some position B. Required the distance AB, 2A<7, when a speed of rotation of 120 revs, per min. has been attained ? Let I denote the cord-length of 16 in., r the unknown radius of the horizontal circle, S l the tension induced in each of the upper cords, 2 that in the lower, and c the unknown linear velocity of each ball. Let u = the number of revolutions per unit time (so that c = %7rru). Fig. 84 shows one of the balls as a free body. Since its vertical velocity is always the same (zero), i.e., its vertical acceleration = (the motion being confined to a horizontal plane), 2 (vert, comps.) = ; i.e., , cos a $, cos a G l = ; . (1) while on account of the curvilinear motion "2 (normal compons.) = Me* -f- r, or in or = - (2) 72 NOTES AND EXAMPLES IN MECHANICS. The tangent to the curve is -| to the paper and 2 (tang, comps.) is evidently = 0, whence the tangential acceleration must be zero ; i.e., c is constant, as we have as- sumed all along. With 6r 2 free, in Fig. 85, we have balanced forces ; whence 2 (vert, compons.) = 0; i.e., 2$, cos a 2 = (3) By elimination, noting that c = 2 ) tan a r - q G.-4- G 9 *i . i p A ( 1 y 'I a ' /'"y Jty , * .JllL \J + .. oo /-* ' ffr 6r, ' tan a? k.rfu G l We note, therefore, that the required distance AC is inde- pendent of the length I and is inversely proportional to the square of the number of revolutions per unit time. (A similar result was found with the simple conical pendulum ; see p. 78, M. of E.) Hence, numerically, with the foot, pound, and second (so that u = - 1 // = 2 revs, per sec.), 2 v 32 2 g i 10 AB, = 2 AC, = 4 x 987*x4 ' 8 = L018 ft< ; or 12 ' 22 in ' 69. Example 17. Cannon-ball under Gravity and Air-resistance. A round cannon-ball, weighing 24 Ibs., is at a certain point of its path moving with a velocity of v = 800 ft. t~b~ per second in a direction making an angle of 20 below the horizontal. The resistance offered to it by the air at this Fl0 - 86 - speed is 80 Ibs. and acts in the line of motion, since the body is round and has no motion of rotation. Required the amount and position of the (ideal) resultant force E. See Fig. 86. Since there are only two forces acting on the ball, P and G, E must have an amount and position determined by the diagonal of the parallelogram formed on P and G. See figure for the known angles. Hence (from formula on p. 7, M. of E.) E = VP* + CT + 2PG cos 110, i.e., E = V80 a + 24 2 + 2 x 80 X 24 X ( 0.3420) = 75.25 Ibs. NUMERICAL EXAMPLES DYNAMICS OF MATERIAL POINT. 73 To find the angle 0, note that in the triangle PRO we have 24 sin 6 : sin 70 :: G : R\ whence sin = =^-^(0.9397); i.e., 6 = < O.-ZO 17 26', and hence R is 2 34' above the horizontal. Since the action- line of R is not coincident with the line of motion of the ball at this instant, the path of the hall must he curved, the radius of curvature at this point depending on the mass, on the square of the velocity, and on the value of the normal component of R ; while the rate of retardation of the velocity (negative tangential acceleration) depends on the mass and the tangential component of R. (See next example.) 70. Example 18. Ball in Curved Path. Radius of Curvature, etc. (Fig. 87). A large ball weighing 200 Ibs. (so that its mass = M 200 -f- 32.2 200 X 0.031 = 6.2, in the foot-pound- second system of units) at a certain point of its path has a velocity of 700 ft. per sec., the resultant force R at this instant being == 300 Ibs. and making an angle of 140 with the direction (see v in figure) of motion. Required the radius of curvature, r, at this point of the path, and also the tangential acceleration. By a rectangular parallelogram of forces we resolve R along the tangent and normal, obtaining for its tang, compon., T,=R cos 140, = 300 X (- 0.76604) = - 229.812 Ibs., while N y =B sin 140, = 300 x 0.6428 = 192.84 Ibs. From eq. (5), p. 76, M. of E., 2 (norm, comps.) = Mv* ~- r, and 2 (tang, comps.) = Mp t , whence _ 6.2 X (700)- ... mgof 192.84 5>T5( " 229.812 and p t = g-g = 37.06ft. per sec. per sec. This last value means that the retarding effect of the com- ponent Tis such that if the rate of retardation remained constant 74 NOTES AND EXAMPLES IN MECHANICS. for one second, at the close of that second the velocity in the path would be 700 37.06 = 662.94 ft. per sec. 71. Example 19. Steam Working Expansively and Raising a Weight. In Prob. 4 of p. 61, M. of E., supposing the boiler-gauge to read 80 Ibs. per sq. in. (above one atmos- phere) and the total length of stroke. f-AH-G-> --HT-IR FIG. 8?a. K s n ON, to be 16 inches, with cut- off at one third stroke (so that 5, = -J of 16 in.), the diameter of piston being 10 inches ; how great a weight (} can be raised if the (circular) pis- ton is to come to rest at the end of the stroke, having started from rest at the beginning of the stroke? Required also the time occupied from to B, and the position of the piston when its velocity is a maximum. From p. 62 we have the equation . . (2) now to be solved for G. s n = f ft. ; s l = f ft., and hence the ratio s n : s, 3. The area of piston = xr* %f- X (5) 2 = 78.57 sq. in. .-. Air-pressure above piston, = A, = constant 78.57 X (15 Ibs. per sq. in.) = 1178 Ibs. ; while the steam- pressure under pis- ton while it is passing from to B, 8, , = 78.57 X 95 = 7464.15 Ibs. Noting that log e = common log X 2.302, we have from eq. (2) (using the foot, pound, and second) 7464 X t[l + 2.302 X .47712] = 1178 X f + G X f Solving, G = 4044 Ibs. (so that M = 4044 x .031 = 125.36). The acceleration from O to B is constant and p i (S l A G)-^M = 2241 -f- 125.36 = 17.88 ft. per sec. per sec. ; and [eq. (2), p. 54, M. of E.] s t = %pj? ; hence time from OtoB = t l = y ^| = V.0496 = 0.222 sec. BALL FALLING ON SPRING. 75 Above B, the steam -pressure S diminishes, and when at some point m it has become = A + #, i.e., to 5222 Ibs., the resultant or accelerating force, 8 (A + 6r), is zero ; above this point m that force is negative, i.e., the velocity diminishes, and hence the velocity is a maximum at m. Let s m be the distance of m from Oy then from Boyle's Law s, : s m : : 5222 Ibs. : S t , whence s m = I- Jfft = 0.635 feet, = T.620 in. 72. Example 20. Ball Falling on Spring (Fig. 88). A ball weighing two pounds (G) falls freely from rest, and after falling 5 ft. (= h) comes in contact with the head of a spring, which it gradually compresses during its further descent until brought to rest again momentarily (at w,). The resistance (P) of the spring is proportional to the depth of compres- sion (s) and is 60 Ibs. (P ) at the end of the first inch (s ). (Provision is made against side-buck- ling.) Required the maximum compression mjn l9 = 8, . At the end of the free fall the velocity of the balls is c VZyli (i.e., c 2 = 2^A), since so far there is but one force acting, its own weight G. At any distance s, however, below ?n (the point of first contact) the resultant downward force is G P, FIG. 88." P being the upward pressure of the head of the spring against the ball at this instant, and the acceleration is therefore variable and is p = force ~- mass, = (G P) -f- (G -f- g). Let down be positive. Substituting in vdv = pds, noting that P:P ::s:s o9 and then integrating between the points m and m, , we have i i i r /**> p r si vdv = d s -rfPds ; and / vdv = I d s -^- I sds. g G g / J Gs Q J ( ,,_o ) -g ; K:;] 1 . r ,_; = ,-^.< Numerically, with the inch, pound, and second, Finally, m,m l , = ,,=-{- 2.03 inches (and 1.96 in.). 76 NOTES AND EXAMPLES IN MECHANICS. The negative result refers to a point (call it m') 1.96 in. above m . This is the position where the ball would momentarily come to rest for the second time, if it adhered to the head of the spring after the latter had regained its natural length, supposing the lower end of the spring to be fixed. This motion of the ball while in contact with the spring is really harmonic, whose central point, from which the u displacement " would be reckoned, is -$ of an inch below m , i.e., at the point where the pressure of the spring = 2 Ibs. (the weight of ball), so that as the ball passes that point its acceleration is zero and the velocity a maximum. This point is midway between m, and in'. 72a. The Engineer's " Mass," The engineer measures the mass of a body (in case a problem connected with its motion is under treatment) by the fraction, weight -j- accel. of gravity ; or G -f- g. This is not scientific, but is so firmly rooted in engineering prac- tice that no different measure can well supplant it. It seems to imply that the amount of matter in a body depends on the existence of the attraction of gravitation; whereas, of course, such is not the case. This measure arises from the fact that a convenient way for the engineer to determine the magnitude of any force P (or resultant) acting on a body and producing an acceleration (p) of its velocity is to compare it with the force of gravity exerted on the body, whether the circumstances of the problem are affected by gravitation or not. In the phrase force = mass X accel., or P = Mp, the word mass is simply a name given to the fraction G - g, the origin of which is as follows : In the actual problem the force P produces an acceleration = p in the velocity of the body. In the ideal experiment of allowing the same body to have a free fall in vacua we know that the only force would ~be the weight 6r, and that the resulting acceleration would be g ; and since the forces must be propor- tional to the accelerations, we have ( 54, M. of E.) P : G : : actual p : ideal g\ or, P p. In other words, the engineer uses the gravitation measure of a force (p. 48, M. of E.). CHAPTER Y. MOMENT OF INERTIA OF PLANE FIGURES. 72b. Phraseology. Unless otherwise specified, we are to under- stand by " moment of inertia of a plane figure" the rectangular moment of inertia ; i.e., the axis of reference lies in the plane of the figure (and not i to it as with the " polar" moment of in- ertia). This is a useful function of the plane figure, to be used in the theory of beams under bending strain. 73. Moment of Inertia of Section of I-beam (Corners not Rounded). Fig. 89 shows the form and dimensions of the section, which is symmetrical about each of the axes X and J", and is for present pur- poses subdivided into three rectangles and four right triangles. Making use, then, of ] i, results obtained for those elementary forms, *-' ^ 4 ~ and of the transferral formula between the gravity axis of any figure and a parallel axis (see p. 94 and eq. (4), p. 93, of M. of E.), we have for the moment of inertia about axis X \ \ *l FIG. 89. -v That is, numerically, = 566 + 195.8 + 356 = 1117.8 bi. in. = Similarly, the moment of inertia about the axis Y is = 22.5 + 0.17 + 9.79 = 32.46 bi-quad. in. = I y 77 78 NOTES AND EXAMPLES IN MECHANICS. 74. Moment of Inertia of a Section of a Built Box-beam (Fig. 90). The beam is composed of two "flange-plates" (upper and lower), two vertical "stem-plates" and four "angle-bars" of equal legs, riveted together. See figure for notation and dimensions. He- quired the moment of inertia of the whole section about JT, its horizontal gravity axis of sym- metry. In Fig. 91 we have the loca- tion of the gravity axis g (parallel to X) of a single " angle" section, according to the hand-book of the New Jersey Steel and Iron Co., so that from the dis- tance 1.68 in. we compute the d' = 10.32 in. of Fig. 90, or distance of axis g from axis X. ar -*>- =-* Allplates X thick FIG. 90. From the same book we find that the I g of the j Section of Angle-Bar 6X6X.K" FIG. 91. angle-section is 20 bi. in. (very nearly), and its area F' = 5.75 sq. in. Let t = thickness of all plates = in. ; and t' = diam. of rivet-holes = f in. First, neglecting the rectangular gaps made by the rivet-holes, we have the IJs of the various component sections as follows : (four " angles") . . . 4[/ + ^^ /2 )]=4[20+5f(10.32) 2 ] =2528 ; (flange-plates) (stem-plates) ..... 2^] =11*2 ; making a total of 6681 bi. in. Treating the small rectangles left by the rivet-holes as con- centrated in their respective centres of gravity [and thus neglect- ing their local (gravity) moments of inertia], "V~ Subtractive) .,, ' I x due to V = 4[(20-^" a ]+4[j204-2J \ = 243 + 432=675. rivet-holes ) Hence, l x of actual section = 6681 675 = 6006 bi. in. MOMENT OF INERTIA BY SIMPSON'S RULE. 79 It will be noticed that the first term, -^bt 3 [or local (gravity) moment of inertia], in the I x of the section of a flange-plate, above, is very small compared with the second, or " transferral term" (bt . d*). This is due to the fact that all parts of a thin flange-plate section are very nearly at the same distance from the final axis of reference, X. In such a case it is customary to neglect the local term, as no practical error results from so doing. 75. Moment of Inertia of Irregular Curve-bounded Plane Figures by Simpson's Rule (see 93, M. of E.). If an exact result for the I x of the figure shown in Fig. 92 were desired, we might first conceive of its subdivision into narrow strips parallel to JT, of variable length v and infinitesimal width dz, then express its l x as f (small area) X 3 2 , mf(vdz)z*, =f(z*v)dz ; and finally perform the integration, if v were an algebraic function of z. If such is not the case, however, (or if such is the case but the in- tegration net practicable,) we can resort to Simpson's Rule ( 15), noting that the u 9 x, and dx of that rule correspond to the (z*v\ z, and dz, respectively, of the present problem. We divide the whole height of the figure (from the axis X] into an even number n of equal parts, and through each point of division draw a parallel to X, thus determining a series of widths, v , v l9 v 9 , etc. For example, this construction being made for the upper part of the rail-section in Fig. 93, with n~ we have its I a9 =f(z*v)dz, approximately = eths of an inch. FIG. 93. With numerical substitution, therefore, the lengths marked in the figure having been scaled in fiftieths of an inch (noting that 80 NOTES AND EXAMPLES IN MECHANICS. z = 0, s, = %z 6 , z^ = f 2 6 , s 3 = f-2 6 , etc.), we have, as the I x of the portion of the rail-section lying above the axis X, [0 + 4(1 X 40 -f 3 2 X 80 + 5* X 163) -f 2(2 2 X 50 + 4 2 X 136) + 6 X 141] bi-quad. fiftieths ; which divided by (50) 4 , or 6,250,000, gives I x for upper part = 25.27 bi. in. Similarly, the vertical height of the lower part being divided into four equal lengths, we have for the l x of that part (nearly) SIT? [ + ^ X 53 + 3 2 X 244) + 2(2 2 X 103) + 4 2 X 266] = 129,870,000 bi. fiftieths. Dividing by (50) 4 , we have 20.77 bi. in. ; so that the total I x of the complete rail-section = 20.77 + 25.27 = 46.04 bi. in. Xis a gravity axis parallel to the base of the rail-section, and has been located by cutting out the shape from card -board and balancing on a needle-point. If the plane figure is of such a form that a division into strips perpendicular to, and all terminating in, the axis of reference (X) is convenient, the exact cal- culus form for its I x is %fy*dx (see latter part of 93, M. of E.), for each strip is an elementary rectangle. See Fig. 94. "o x If Simpson's Eule is to be applied, divide the FIG. 94. base OX of the figure into an even number, n, of equal parts and scale the extreme ordinates y = AO and y n = BX, also the intermediate ordinates y l , y a , etc., at the points of divi- sion. "With Ti 6, for example, we have as an approximation TTx I x = ~ 76. Graphical Method* for the Gravity Axis and Moment of Inertia of a Plane Figure (Fig. $&).A"B"C" is the figure in question (drawn in full size). It is required to construct the special gravity axis, 7?, that is parallel to the base A"B"^ and to obtain the moment of inertia about that axis. Divide the figure into strips parallel to A" ' B' ', of small width * From Ott's Graphical Statics. MOMENT OF INERTIA GRAPHICAL METHOD. 81 (no width being more than one eighth, say, of the total width -| to A"B"\ These widths need not be equal. Through the centres of gravity of the strips (1, 2, 3, etc.) draw indefinite lines (1 . . 1', 2 . . 2', etc.) || to A" B" (with most strips it is accurate enough to take the centre of gravity midway between the sides). Along any right line 0' I, parallel to A" B" , lay off the lengths O'A, AB^ BC, etc., proportional, respectively, to the areas R Moment of Inertia Graphic Method. FIG. 95. FI , F t , etc., of the successive strips, in the order and position shown (in most instances * each such area may be assumed pro- portional to the length of the strip measured through the centre of gravity). Through 0' and / draw lines at 45 with 0'7, as shown, to determine the "pole" 0, from which the " rays" OA, OB, etc., are now drawn. Then through any convenient point Z draw ZC'" \\ to 00'. From the intersection a with 1 . . 1' draw ab \\ to OA to intersect 2 . . 2' in some point 5, then be \\ to OB, and so on ; until finally, through i, ids drawn || to IO to determine C by intersecting ZC'" . The required gravity axis R passes through C \\ to A" B" . (For proof consult 376, M. of E.) The moment of inertia about axis R may be obtained by multiplying together the area of the given figure A" B" C" (call it F) hy the area (call it F r ) of the " inertia-figure" (i.e., the area included between the two lines Ca and (7*, and the broken line, * When the strips are narrow and of equal width. 82 NOTES AND EXAMPLES IN MECHANICS. or "equilibrium-polygon," dbcdefghi ; shaded in Fig. 95). The proof of this relation (strictly true only for infinitely narrow strips) is as follows : At any vertex of the equilibrium-polygon, as at g, there are two segments meeting; prolong them to intersect the gravity axis R in some two points, as m and n. Then mng is a triangle with base m . . n (call it &,) and altitude a? 7 . But on the left of Fig. 95 the shaded triangle OF^ is evidently similar to mng\ whence the proportion # 7 : a? 7 : : F, : ^F\ i.e., F^x, \F~k, . Multiplying by a? 7 , we have F^x? = F(^xjc,). Now FJK* is the moment of inertia (about It) of strip No. 7 (considered infinitely narrow), and -J^,& 7 is the area of the triangle mng. We have therefore proved that the moment of inertia of any one strip is equal to the product of the whole area F of the given plane figure by the area of a triangle like mng and obtained in a similar manner. If all the triangles like mng were drawn, their united areas would evidently be that of the "inertia-figure" abcdefghi-C-a. Hence the sum of the moments of inertia of all the strips, i.e., the moment of inertia of the whole figure A"B"C", is j _ j area of plane \ ( area of the (figure A" "C" \-\ "inertia-figure" In practice these areas are most conveniently and accurately obtained by means of a planimeter; otherwise, subdivision into small trapezoidal strips may be resorted to. If the scale of the drawing is one half of the actual size the result must be multi- plied by 16, i.e., the fourth power of 2; and similarly for other ratios. By the use of a planimeter with this actual figure (Fig. 95) the results F= .90 sq. in., and F' = .42 sq. in., have been ob- tained ; therefore I R = 0.378 bi. in. CHAPTER YI. DYNAMICS OF A RIGID BODY. 77. Example of Rotary Motion. Axis Fixed and Horizontal (Fig. 96). The body AB consists of an irregular solid and a light drum, rigidly connected and mounted on a horizontal axle whose journals are 2 inches (= 2r) in diameter, at C\ the radius of the drum being a = 5 in. The weight of AB is 150 Ibs. = G l , and its centre of gravity is situated in the axis of rotation (hence G l has no effect as regards changing rotary velocity, having always a zero FIG. 96. moment about the axis). The weight G, = 20 Ibs., is attached to an inextensible cord which is wound on the drum and whose weight is neglected. A constant friction F, = 10.5 Ibs., acts at the circumference of the journals. It is required to compute the radius of gyration Jc c of the rotating body by the experiment of noting the time ^ occupied by G in sinking a measured distance s, , from rest. Suppose tfj 5 sec. and s, = 10 ft. At any instant during this motion, the tension in the cord being , we have for the angular acceleration of AB, which is shown free in Fig. 97, taking moments about axis C, while at the same instant the downward linear acceleration JP, = 6a, of G, enters into the rela- tion involving the sum of downward forces for G as a free body, viz.: /i S = mass. X accel. = p. (2) 83 84 NOTES AND EXAMPLES IN MECHANICS. Solving for p, we would find it constant, as also ; hence G has a uniformly accelerated rectilinear motion, and AB a uni- formly accelerated angular motion. Therefore (with foot, pound, and second), since (from eq. (2), p. 54, M. of E.) s l kpt*i an d hence p = 2s l -=- t*, we have p = 20 H- 25 = 0.80 and is the con- stant linear acceleration, not only of 6r, but for any point in the, surface of the drum, so that the angular acceleration of body AB is = p -f- a = 0.80 -i- -f% 1.92 radians per sec. per sec. From eq. (2) we now have S = 20 - (20 X .80) -=- 32.2 = 19.504 Ibs. as the constant tension in the cord ; and finally, from eq. (1), [19.504 X A ~ 10.5 X T V]32.2 * a 1.92 X 150 : 0.812 sq.tt.; i.e., the rad. of gyration Jc c 0.901 ft. = about 10.8 inches. 78. Solution of Example of Compound Pendulum. (For the statement of the example see p. 121, M. of E.) Fig. 98. (A =6", r = 1.2". Locate axis oo so that the time of o , oscillation t' shall be \ sec.) Referred to the horizontal-gravity axis cc we have for the solid cone (from 101, M. of E.) I = -f^M\f + JrA 2 ], ' sq. in. With the inch and second g = 386.4, and FIG. 98. gt'* -t-2jt*= 4.889. Hence, from eq. (1), p. 120, M. of. E., 8 _ 4,889 |/23.90 1.556 = + 9.615, and + 0.163 in. That is, with either oo or o'o' as axis of suspension, co being made = 0.163 in., and co' = 9.615 in., the duration of an oscilla- tion (of small amplitude) will be \ sec. of time. As a check we note that co + co' 9.778 in., which = 2 X 4.889 length of a simple pendulum beating half-seconds (see above and also eq. (3), p. 119, M. of E.). 79. Points of Maximum and Minimum Angular Velocities iu Motion of Crank-pin of a Steam-engine. Fig. 99 shows the results of applying the tentative graphic method mentioned in the mid- dle of p. 124, M. of E. In making trials for successive positions of the connecting-rod, advantage can be taken of the fact that if ROTATION EFFECT ON BEAKINGS. 85 the point be found where the axis of the connecting-rod (pro. longed if necessary) intei sects the vertical line (vertical in this instance), drawn through the centre C of the crank circle, the distance of that point from C represents the amount of the tangential component, T, of the force P' ', on the same scale on which the length, C . . n, of the crank would represent the force P. Hence various positions of the connecting-rod are drawn until those two are found for which the distance mentioned above is f -|, i.e., i^, of radius C . . n. In the working of most engines the steam is used expansively so that P is variable, being greatest near the beginning of the stroke. This would cause the points n and m to be nearer to A ; and similarly, on the return-stroke, n" and m" nearer to B. 80. Rapid Rotation of a Body on a Fixed Axis. Effect on Bearings. So long as a wheel or pulley in a machine is perfectly symmetrical about the axis of rotation the pressures on the bear- ings are due simply to the weight of the pulley and the pulls or thrusts of cogs, belts, cams, etc., which may be acting on the pulley or on the shaft on which it is mounted ; whether rotation is proceeding or not. But if, through any imperfection in the adjustment or mounting, the centre of gravity lies outside of the axis of rotation (its distance from that axis being p), other press- ures are brought upon the bearings, the same in amount as if the pulley were at rest and a force == oo^Mp acted through the centre of gravity, away from, and ~j to, the axis of rotation. 86 NOTES AND EXAMPLES IN MECHANICS. For example, let the pulley in Fig. 100, of weight = 644 Ibs., be out of centre by one fourth of an inch and be rotating uniformly at the rate of 210 revolutions per minute, i.e., J- rev. per second. Its angular velocity is therefore GO = 22 radians per second and the pressures on the bearings due to the eccen- FIG. 100. tricity of the pulley at this speed is the same as if the pulley were at rest and a " centrifugal force" P, =. [484(644 -r- 32.2) -^-g-], 201.6 Ibs., acted on the body, as shown in figure, where G is the centre of gravity. The pressures on the two bearings due to this ideal " centrifugal force" will be inversely proportional to their distances from the pulley-centre and parallel to P (aside from the pressures due to the weight of the pulley, action of belts, cogs, etc.). The continual change of direction of these centrifugal press- ures, as the wheel revolves, is likely to cause injurious vibrations in the framework of the machine. 81. " Centrifugal Couple." (Continuing the discussion of the last paragraph.) Even if the axis of rotation does contain the centre of gravity (or rather centre of mass in this connection) of the revolving body ; if, even then, we find that any plane containing the axis divides the body into two parts,* the right line connecting whose centres of mass is not perpendicular to the axis, the ideal centrifugal forces, Go' 2 M l p 1 and oo^M^, of these two component masses form a couple, causing pressures at the two bearings ; which two pressures, themselves, form a couple. For example, Fig. 101, where the axis of the shaft AB and the cen- tres of the two masses are in the same plane, and where the products M^ and J/ 2 P 2 are equal, the centre of mass of the whole rigid body must lie in the /'^ \ a axis of rotation (i.e., neglecting the mass M2 y b i of the shaft) ; but the line joining the ' centres of the component masses is not perpendicular to the axis AB. A and B are fixed bearings and a uniform ro- ~ FIG. 101. tation with angular velocity of GO is proceeding. Evidently the * We here suppose each of these parts to be homogeneous aud to be sym- metrical about a plane passed through its center of gravity aud perpendicular to the axis of rotation. If such is not the case, we must resort to the principles of \22-l22c iuclus. of M. of E., to determine the pressures at the bearings. PRACTICAL NOTES ON PILE-DRIVING. 87 action on the bearings is the same as if the body AB were at rest and were acted on by the two ideal centrifugal forces P l = Go t M l p l and P^ &> 2 J/ 2 /J 2 , in the positions shown in the figure (on the right). These forces are equal (since M l p l = -3/ a p a ), forming therefore a couple, so that the reactions at the two bear- ings (pressures of bearings on shaft) would form a couple of moment P'~b = P^a. For example, let M l weigh 120 Ibs. ; J/" a , 100 ; the distances ~p l and p 9 being 2 and 2.4 ft., respectively ; while the angular velocity is 22 radians per second (210 revs, per min.). Also let a = 3 f t. and I 4 ft. ; then pi = f p *= - ( 22 ) 2 x 2 - 4 = 2no ibs - So long as the rotation is uniform these pressures P' lie in the plane of the axis and the two mass-centres ; which plane, of course, is continually changing position. 82. Piles. (See p. 140, M. of E.) As a practical matter it should be understood that as the head of the pile becomes broomed, or splintered, during the driving, the penetration occasioned by the blow may be much smaller than would otherwise be the case (only one quarter as much in some cases) ; hence it is economical of power that the head be adzed off occasionally, and especially should this be done just previously to the last few blows when the measurement of penetration is made, to be used in a formula for safe bearing load. A formula for the safe load has been proposed (see p. 185, Eng. News, Feb., 1891) in which, besides adopting the divisor 6 of eq. (2), p. 140, M. of E., the value of s' is assumed one inch greater than that actually observed. In cases where the actual s' is small (as an inch or less) this allows a very wide margin of safety. If s' is large, the margin of safety is practically little more than that afforded by the divisor 6. This formula may be written : 7 , _ 1 _ &h _ ( or, for the ) 1 Gh ~ 6 V + one inch ' ( Ib. and ft., i ~ 6 ' + T V ' 88 NOTES AND EXAMPLES IN MECHANICS. while if G is in Ibs., h in feet, and s f in inches, it takes the form * 7 J ' 71 safe load ^n Ibs. = , , -. . (See Baker's Masonry Construction, and Trautwine's Pocket Book, for many practical details in the matter of piles.) In experiments in the laboratory of the College of Civil Engineering at Cornell University, with round oak stakes 2 in. in diameter and a 4-lb. hammer falling four feet, the actual bear- ing loads have been found to be from about one half to seven eighths of that given by the expression GJi -~ s'. 83. Kinetic Energy of Rotary Motion. Numerical Example. The grindstone in Fig. 102 has an initial angular velocity, GO O , corresponding to 180 rev. per min., about its horizontal geometrical axis, which is fixed. How many turns will it make before it is brought to rest by the two frictions, at A and at 6"? The pressure of the plank OD on the stone at A is due to the weight of 180 Ibs. suspended at rest at D. Assume that the friction, F') or tangential action of the plank against the stone, is always one third of the normal pressure, which is vertical and is evidently 360 Ibs. (since the plank is a body at rest and therefore under a balanced system of forces, so that by moments about O we find the normal pressure to be double the 180 Ibs.). Hence the friction F' is = 120 Ibs., and with regard to the plank is a force directed toward the right ; but toward the left, as regards the revolving stone. The stone is a homogeneous right cylinder weighing 600 Ibs., and of radius = 2', its geometrical axis being the axis of rotation. Radius of journals at C is 1 in., and the mass of the projecting axle is neglected. The journal friction, F", tangent to the circumference, is to be taken as constant and as being ^V of the normal pressure, N"^ on the journal. FIG. 102. WORK AND ENERGY IN ROTARY MOTION. 89 We now consider the stone free in Fig. 103. acting are as shown, and the rotation clockwise with retarding angular velocity. Though the system is not a balanced one as regards rotation, since the moment-sum about C is not zero but = OMk*, it is such as regards motion of the whole mass vertically ; for the centre of mass has a zero vertical acceleration, being at rest at all times. Hence we may put 2 (vert, comps.) = 0, and thus obtain, noting that the pressure N" is practically vertical, N" = G + N 1 ; and hence F" = 960 -=- 20 = 48 Ibs. To apply here the principle of Work and Energy, as expressed in eq. (2) of p. 144, M. of E., the range of motion considered being that occupied by the stone in coming to rest, we note, Fig. 103, that G, N', and N" are neutral; that F' and F" (both assumed constant) are resistances ; and that there are no working forces. Let u denote the unknown number of turns made in coming to rest, ds' an element of the path of the point A, and ds" an element of the path of a point in circumference of the journal. With the foot, pound, and second, we have, therefore : Work ) /* / done \ = / F'ds' = F' / ds' = [120 X u X 2* x 2] ft. Ibs.; m F' ) J J on Work ) rn /*n done \ = \ F"ds" = F" / ds" = [48 X u X 2* X T V] ft. Ibs. on F" } J Jo Since Jc c * for the cylinder (see p. 99, M. of E.) is = %r* = 2 sq. ft, and the initial angular velocity is G? O = 27r(180 ~ 60) = 18.85 radians per .second, we have 1 } = ^V - [ 18 - 85 )'' X 2] ft. Ibs. The final kinetic energy is to be zero and the work of the work- ing forces is zero ; hence from eq. (2), p. 144, M. of E., = 4807TW -f STTU + [0 (18.85) 3 (18.6)]. . . (3) Solving this, we obtain u = 4.33 turns, for the stone to be 90 NOTES AND EXAMPLES IN MECHANICS. brought to rest. Eq. (3) might be read : the initial K. E. is entirely absorbed in the work of overcoming friction. (How would the time of coming to rest be determined ?) 84. Work and Kinetic Energy. Motion of Rigid Body Paral- lel to a Plane. Numerical Example. Conceive of a rigid body, or hoop, Fig. 104, consisting of two material points, M' and M ", so connected with a rigid, but imponderable, framework that the centre of gravity of the two material points lies at the centre of the circle formed by the outer edge of the frame. This outer circle, or wire, is to roll without FIG. 104. slipping from a state of rest (cen- tre at o) to a position 6 ft. lower, vertically (centre at n) ; , beyond which the point of insertion will follow the arc DA', the portion of cord between it and C being straight. Beqnired the (final) angular velocity Go n of the rotating body when A reaches A. Considering free this collection of rigid bodies (two masses and cord), knowing that all mutual actions between them, if normal to surfaces, are neutral, and assuming no internal friction, 92 NOTES AND EXAMPLES IN MECHANICS. we note that the only forces external to the collection are the gravity-force of 48.3 Ibs. (neutral because its point of application does not move) ; the normal reactions of the bearings against the sides of the journals (these are neutral because the path of the point of application is always ~| to the action-line of the force) ; the pressure of the peg against the side of the cord (neutral, for the same reason as that just mentioned) ; and the gravity-force of 10 Ibs., which is the only working force. The path of the point of application of this last force is vertical and coincident with the action-line of the force, and the length of this path (which is equal to that of its own projection on the action-line) is equal to the total length of cord, s, that runs over the peg. Evi- dently s = n x 2 f t. = 6.28 ft. (while A is passing to E\ increased by the difference between the distance EC (considered straight ; a is very small in this case) and the straight distance A'C\ _ i.e., s = 6.28 + [ Vl6 a + 2 9 14] = 8.406 feet. As both the initial and the final Kinetic Energy of the 10 -lb. mass are zero, and the initial K. E. of the rotating body is zero, the work 10 X 8.406 84.06 ft. -Ibs., of the one working force, is entirely expended in generating the final K. E., ^co^Mk\ of the rotating body, .-. 84.06 = Jco B f X 1.5 X (3) 2 ; and oo n = 3.53 rad. per sec., which corresponds to o? B -r- 2?r = 0.562 rev. per sec. As to the tension, S', induced in the chord as the 10-lb. weight reaches its lowest point and begins to ascend, which is when the point of insertion of the chord in the drum passes through the point A (in its circular path), we note that at that instant, since the chord cannot stretch, the horizontal acceleration of the inser- tion-point, which is its normal acceleration at A ', must be the same* as the vertical acceleration of the 10-lb. weight, and this is upward, since the downward velocity is slackening. Hence the resultant force, S' G, on G at this instant is upward, and (ft -f. g) x (normal accel. of a point moving with linear velocity = os n X 2 ft. in a curve whose rad. of curv. is 2 ft.), i.e., - (# + g ) x (< B X 2 ft.) 2 -T- 2 ft. = 7.72 Ibs. .-. S' = 10 + 7.72 = 17.72 Ibs. * Approximate; the error is less the longer the distance A' C, compared with OA' , strictly true only when the peg is at an infinite distance toward the right (A somewhat similar approximation was made on p. 59, M. of E., in consider- ing the connecting-rod of a steam-engine as infinitely long.) INTERNAL FRICTION LOST WORK. 93 At other parts of the motion the cord-tension is smaller. Let the student trace the remainder of the motion. 86. Work of Internal Friction. The student should note well that the work spent on the friction between two rubbing parts of the collection of rigid bodies under consideration is formed by multiplying the value of the friction by the distance rubbed through by one of the parts on the surface of the other, inde- pendently of the actual path in space of any point of either body ; for instance, in the examples of 144 and 145, M. of E., the friction (F" = 400 Ibs.) between the crank-pin and its bearing (in the end of the connecting-rod), the range of motion of the mechanism being that corresponding to a half-turn of the crank, is multiplied by the length of a half-circumference of the crank- pin itself, viz., ttr" , = 0.314 ft. (less than 4 inches), whereas the centre of the crank-pin describes a distance of nr 3.14 ft. at the same time. 87. Locomotive at Uniform Speed. Required the necessary total mean effective pressure P Q in the cylinder of a locomotive on a level track to maintain constant the speed of a train whose resistance, J?, at that speed is given. This resistance is due entirely to frictions of various kinds in the train and to the resist- ance of the air, and, the track being level, is equal to the tension in the draw-bar immediately behind the locomotive. Fig. 106 shows the collection of rigid parts forming the locomotive \ G <- 2g -> alone and the forces external to them, all mutual frictions being disregarded. For sim- plicity consider that there is ' '' : V:^^V^^-:- ; ^T^^ only one cylinder and piston ; FlG - 106 - and take, as the range of motion of the parts, that corresponding to one backward stroke of the piston, i.e., to a half-turn of the driving-wheels. The external forces are : P Q on the piston ; an equal P Q on the front cylinder-head ; Y, Y, the pressures of the rails against the truck- wheel s ; $, S, those of the track against the driving-wheels, which are supposed not to slip on the track; G, the weight of the locomotive ; and 7?, the tension in the draw- 94 NOTES AND EXAMPLES IN MECHANICS. bar. Of these, the I r 's and the S's are neutral for reasons given in the example of 84 (perfect rolling); G also is neutral, the elements of its path being always ~| to the action-line of the force ; R is a resistance, overcome through a distance = nr ; the P on the front cylinder-head is a working force, its point of application describing a horizontal path of length = nr ; while the P Q on the piston is a resistance whose application-point moves forward in absolute space a distance = nr %a. The kinetic energy of the mechanism being the same at the end as at the beginning of the stroke, the K. E. terms cancel out from the equation of work and K. E., leaving P**r - P Q (7tr - 2a) - Rnr = ; .-. P = itt> or P = the draw-bar resistance multiplied by the half-cir- cumference of the driver and divided by the length of one stroke. For example, if a 200-ton train offers a resistance of 10 Ibs. per ton (in the draw-bar) at a speed of 20 miles per hour, then with 2.5 ft. radius for the drivers and 1 ft. radius for the crank-pin circle, we have P 7857 Ibs. ; i.e., with two cylinders, 3928 Ibs. for each piston. Suppose each piston to have an area of 100 sq. in., then the average amount by which the steam-pressure on one side should exceed the atmospheric pressure on the other is 39.28 Ibs. per sq. in. As steam is ordi- narily used expansively when the train is under full headway, this means an initial steam-pressure at the beginning of stroke of per- haps 80 or 90 Ibs. per sq. in. above the atmosphere (as a roughly approximate illustration). Above, we have assumed the drivers not to slip on the rails. Slipping will not occur ordinarily if H is less than about one fifth of the total weight resting on the drivers. 88. Locomotive Tractive Effort at Starting. Track Level. "When the train is once in motion the steam-pressure on the piston in the latter part of the stroke is much smaller than the average, so that the speed slackens temporarily, the great mass of the train acting as a fly-wheel. To start, however, this steam-pressure must reach a certain minimum amount which, we MECHANICS OF THE LOCOMOTIVE. 95 FIG. 107. FIG. 109. call P. For instance, with one engine at its dead-centre, the duty of starting devolves on the other alone, which is then in the mid-stroke position (nearly), its crank being vertically over (or tinder) the driver-axle, and the horizontal component of the pull on the crank-pin is = P. Fig. 107 shows the driving-wheel as a free body, the vertical components of the acting forces being omitted, as having no moments about D, the point of contact with the track. (Steam is now pressing on the left face of the piston, not shown, and on the hinder (left end) end of cylinder-head (see Fig. 106 for direction of motion, etc.). The action of the driver is that of a lever whose fulcrum is at D (no slip). P' is the horizontal press- ure of the locomotive-frame against the driver-axle, as induced by the pull P\ and by moments about D we havePV .P(r-\-a)' 9 Considering now the locomotive-frame in Fig. 108, we find a forward force P' at A, the bearing of the driver-axle ; the draw-bar resist- ance R, backward ; and a back- ward steam-pressure = P on the hinder cylinder-head at B ; FIG. 108. whence, for equilibrium (i.e., more strictly, for a very small (r\ acceleration), P' P = It, and finally P \~)R, as the neces- sary total effective steam- pressure on the one piston to start the train when the draw-bar resistance is R (neglecting the resist- ances in the locomotive itself). (No vertical forces shown.) Similarly, if the crank-pin were vertically under the axle, as in Fig. 109, we obtain for the pressure at axle (induced by P) the value P' = P f-j P, [since, from moments about D, P'r = P(r a)"]. Hence, as regards the locomotive-frame, we now find a 'backward force P' at the axle-bearing ; but there is 96 NOTES AND EXAMPLES IN MECHANICS. now a forward steam-pressure, P, against the front cylinder- head, so that P P' = R, which after substitution gives us P = f-7?, the same as before. \aJ If the wheels are not to slip, the horizontal action of the rail on the driver, viz., P", Figs. 107 and 109, must not exceed thje ultimate friction, F, or about one fifth the weight on the drivers. But P" = ^P, so that P" = R-, hence R should not exceed F. 89. Appold Automatic Friction-brake, or Dynamometer of Absorption. Fig. 110 shows the Appold friction-brake, which is automatic by reason of its *' compensating lever," BC. The brake is formed of a ring of wooden blocks, con- nected by an iron hoop, whose two ends are pivoted to the lever, as shown, at FIG. no. FIG. in. C. The pulley W revolves within. If the friction is insufficient to keep the weight G raised, it sinks and the end B of the lever presses against the fixed stop BK, thus tightening the hoop, increasing the friction, and raising G (and vice versa when the friction is too great). Fig. Ill shows the Appold brake as a free body. The normal pressures of the pulley against the wooden blocks have no mo- ments about (7, while the tangential actions (i.e., the frictions) have a common lever-arm, = , about 0. Hence, by equilibrium of moments, Fa = Gb PC, where F is the sum of the frictions. Evidently the pressure P at the end of the lever should be known, for accuracy. (See correspondence in the London Engineer, from Novem- ber, 1887, to March, 1888.) Hence if v is the velocity (ft. per sec. for instance) of a point in the pulley-rim, the power absorbed is Z = Fv == f Jv, 90. The " Carpen tier" Dynamometer of Absorption. (Exhibited in Paris in 1880.) This is automatic, like the Appold brake, but THE " CARPENTIER " DYNAMOMETER. 97 FIG. 112. instead of automatically altering the tension in the strap to keep the weight " floating," it changes the arc of contact A B until the friction is so altered ( F'} as to equilibrate the weight G' with help of the smaller weight G. See Fig. 112. N is a pulley keyed upon the shaft of the motor to be tested, and therefore moving with it. A weight G is fastened to a strap BA C against which the pulley JV^rubs, but the upper end of this strap is attached to a block C, which is a rigid part of another pulley, or disk, J/, beyond the pulley N. The disk M is loose on the shaft, the block C projecting over the face of the pulley JV, attached to the pulley M at D and carries a weight G' which we at first assume to be supported by a fixed platform & . When the shaft begins to turn (see arrow), and the tension in CA occasioned by the friction due to the arc AB and weight G is greater* than G', then G' will begin to rise, the disk M turning. But, as M turns, the uppermost point of contact of the strap AB on the pulley N moves to the right ; i.e., the arc of contact AB becomes smaller, with a consequent reduction of the friction between N and the strap, so that after a little the tension in CA, pulling on C, is just sufficient ( $') to keep the weight G' at rest at some point k'. When this state of equilibrium is reached, we have, by moments, for the equilibrium of the strap, S f r=F'r-\-Gr', The strap DG' is so ___ and from that of the disk, S f r = G'r' ; i.e., F' = that if v = velocity of the pulley-rim, the power = Z, = Fv, == -- G \v 9 (ft.-lbs. per second, for instance.) 91. Numerical Example in Boat-rowing. As an illustration of the relations of the quantities concerned in a simple, but typical, case of propulsion on the water, let us suppose, in the problem of Fig. 166 of p. 161, M. of E., that the distances from the oar-handle, A, and oar-lock, (?, to the middle of the blade are 9 ft. and 6 ft., respectively ; and that a pull of 20 Ibs. is exerted on A. The pressure at the oar-lock is then 30 Ibs. ; and that of tne blade * Or, rather, its moment greater than that of G'. 98 NOTES AND EXAMPLES IN MECHANICS. against the water, 10 Ibs. Hence the boat is subjected to two forward oar-lock pressures of 30 Ibs. each ; to two backward pressures against the foot-rest, of 20 Ibs. each ; and to some back- ward resistance R of the water against the hull (R depending on the square of the velocity ; R = zero if there is no velocity). The difference between the oar-lock and foot-rest pressures is a forward force of 20 Ibs. ; and if the velocity of the boat at the beginning of the stroke is such that R is = 20 Ibs., the effect is to barely maintain that particular speed while the 20 Ibs. pressure is acting on each oar-handle, (and R remains constant also.) If R is smaller than 20 Ibs. the velocity will be accelerated, and R will increase ; if larger, the velocity diminishes (and R also), and of course will diminish at a much more rapid rate when the oar is lifted from the water. For an ordinary small skiff (with pointed stern as well as bow) containing one person, the water-resistance R is (roughly) about one half pound for each sq. foot of the wetted surface of the hull, when the velocity is 10 it. per sec. / for other velocities, as the square of the velocity. If in above case each oar-handle, while under the 20 Ibs. press- ure, passes through a distance AE '= 3 ft., measured on the boat, and the blade slips backward in the water an absolute distance s, of 6 inches (say), = J- ft., the absolute distance passed through by the boat will be (from the geometry of the figure) 5.5 ft. The work spent^oii siip is 2 X 10 X -J- = 10 ft.-lbs. ; so that, of the work, 2P .AE= 2 X 20 X 3 = 120 ft.-lbs., exerted by the oar-handle pressures, relatively to the boat (see fourth line of p. 161, M. of E.), 110 ft.-lbs. remain for the work of over- coming the resistance, R, and increasing the K. E. of the mass of boat. R is overcome through the distance s 3 CD, 5.5 ft., so that if all of the 110 ft.-lbs. are spent on R (i.e., if the velocity is to be maintained constant), we must have 110 = R X 5.5, i.e., R = 20 Ibs., as proved above. [In order that R may have this value with a small skiff the velocity must be (roughly) about 11 or 12 ft. per sec., at the beginning of stroke.] Note that the absolute distance through which the two 20-lb. oar-handle pressures work is 5.5 -f- 3 = 8.5 ft. But, of the abso- lute work done by them, viz., 2 X 20 X 8.5 = 340 ft.-lbs., in the NUMERICAL EXAMPLE IN BOAT-ROWING. 99 stroke, the amount 2 X 20 X 5.5, = 220 ft.-lbs., is absorbed in overcoming the foot-rest pressures through 5.5 ft., the remainder being an amount 2 X 20 X 3, = 120 ft.-lbs., = 2P . AE] to be spent on the work of slip, of liquid-resistance, and change of K. E. (if any). 92. Remarks on the Examples of 155, M. of E. In the first example the work to be computed is that done by the tension P, in the draw-bar, considered as a working force acting on the train behind the locomotive. If the velocity were uniform, a value 10 X 200 = 2000 Ibs. would be sufficient for P; but as the velocity is to be increased, the " inertia" of the train is brought into play and the amount required is three times as great in this instance. In Example 2 P has the same significance. As to Example 3, multiplying the 15,000 Ibs. resistance by the speed reduced to ft. per sec. will give the power in ft.-lbs. per sec. Dividing this number by 550, we obtain 461 H. P. (see p. 136, M. of E.). In Example 4 the resistance is greater than before, in the ratio of (10) 2 to (15)', i.e., it is 2 times 15,000 Ibs. The distance through which it is overcome per second being 1J times its former value, we find the power spent on water-resistance to be (in ft.-lbs. per sec.) 2J X 1| X 15000 X -Q^~QQ- ' wllicl1 divided b 7 55 gives 1556 H. P. (That is, the respective powers are as the cubes of the speeds.) Example 5. Since 80 per cent of the power, Z, of the work- ing force (steam-pressure) is to be 461 H. P., we write 0.80 X L = 461, and obtain L = 576 H. P. Example 6. If the thrust or pull of the connecting-rod of the engine on the crank-pin be resolved at every point into a tangen- tial and a normal component, 7 7 and N (Fig. 113), we note that T is a working force and N neutral. Hence at this point in the line of transmission of power we can ascribe all the power to the force T. T is variable, and by its average value, T m , we mean a value whose product by the length of the circumference described by the crank-pin shall be the same amount of work as that actually done by the variable 100 NOTES AND EXAMPLES IN MECHANICS. T per revolution. Now 461 H. P. means 253,000 ft.-lbs. of work per second, which divided by 1.0, the number of turns per sec., gives the work done by T in each turn. Hence T m %7t X 1.5 should = 253,000; or T m = 26,890 Ibs. Example 7. Fig. 114. At we have the sphere in its initial position, the forces acting on it being its own weight G (a resistance), and the two components, ^Vand T, of the pressure of the inclined plane against it. Since there is no slipping (i.e., perfect rolling), both N '///////////, and T are neutral (see 84). Let v = initial velocity of the centre FIG. in. of gravity, and GO O the initial angular ve- locity. s f sin 30 is the unknown height through which the centre will rise before the velocity becomes zero. G v 2 The initial K. E. of translation is - ~ ; and of rotation, Now v 9 = ay and & 3 f /* (p. 102). Hence by eq. (XY), p. 147, or ^ = 0.39 ft 93. Efficiency of a Wedge. Block on Inclined Plane. In the numerical example 3 of p. 172, M. of E., it is to be noted that the " efficiency" of the mechanism, or ratio of the work usefully employed in overcoming Q, to that exerted by the one working force P, is 57.6 -=- 153.6 = 37.5 per cent. Fig. 115 shows as a free body the block mentioned in prob- lem 6, p. 172, M. of E. We are to find the force P in the given action-line, such that a uniform motion (equilibrium) can be maintained up the plane. The action of the plane on the block is represented by the normal pressure N and the tangential action, or friction,/^. [The FIG. 115. student should not assume thoughtlessly that N must = G cos /? (the normal component of G), for the unknown P 9 not being ROLLING RESISTANCE. , , 101 parallel to the plane, has an influence in determining N\ here it tends to relieve the pressure on the plane.], />% forces balanced, 2X, and 2Y, = ; whence P cos a fN G sin /3 = 0; and P sin tf -f- N G cos /? = ; [sin/?+/cos/?] and finally P= - . 1 . ....... (1) COS <* --r Sin a Problem 7 calls for the value of P if the motion is to be down the plane; other things as before. Fig. o/ / ,x 116 shows the change. The friction acts in a Y contrary direction ; P will be smaller ; JV, larger. Solving as before, we finally have 6 ? [sinyg -/cos/?] cos a f sin a We note here [eq. (2)] that for ft =

/'#, for axle-friction. The stone being considered free, we note that its weight and the normal component of the reac- FIG. 119. tion of the bearing are neutral forces, the friction, or tangential EXAMPLES DYNAMICS. 105 component,/' #, is a resistance ; and that there are no working forces. The initial K. E.* of the stone is >*&? (o? being the initial i7 angular velocity) ; its final K. E., zero. Hence the initial K. E. is all absorbed in the work of friction. Since the number of turns in coming to rest is 160, the total distance through which the friction at the circumference of the journal is overcome is 160 X 7r[1.5 -r- 12] = 62.85 ft. "With the foot and second, GO Q = 27T x -W - = 4?r radians, and g = 32.2. Therefore, from the Principle of Work and Kinetic Energy [eq. (XV), p. 147, M. of E.], P-? Example 4. To move A horizontally with an acceleration = 15 (foot and sec.) would require a horizontal force = mass X 2 ace. = oo~o X 15 = .93 Ibs. But the horiz. compon. (friction) of the action of E on A is only fN 0.3 X 2 = 0.6 Ibs., which is < .93. Hence A will riot keep abreast of B, but will gradually fall behind B. 97. Brake-strap, Lever, and Descending Weight, Numerical Example (Fig. 120). The weight Q of 600 Ibs. is to be let down without acceleration, the rope un- winding from a drum of 1 ft. radius. On the shaft of the drum is keyed a pulley of 2 ft. radius, the friction on whose rim, due to its rubbing under the encircling stationary strap, can be varied in amount by a force P exerted on the lever AB. It is required to compute a proper value for the pressure P in the present instance to prevent acceleration of the 600-lb. weight, the coefficient of friction of the strap on the pulley being assumed to bef = 0.30. * Kinetic Energy. FIG. 120. 106 NOTES AND EXAMPLES IN MECHANICS. The strap covers three quarters of the pulley-rim (i.e., a = f TT). See figure for other dimensions. If Q sinks without acceleration, the tension in the vertical part of the rope must be 600 Ibs., and the rotation of pulley be uniform ; hence moments must balance for the pulley, drum, and shaft ; so that (with the foot, pound, and second) 600 X V =< F X 2', where ./'"is the sum of the requisite frictions (tangential forces) of strap on pulley ; i.e., F = 300 Ibs. But from the equilibrium of the curved portion of strap, by moments about centre of the curve, S x ^' S n X 2'-|-^X 2'= 0. 8 n is the tension in the vertical straight part of the strap; /$, that in the horizontal. S n is greater than S and bears to it the / o \ relation S n S ef ; or, \fit = log e (~J. That is v ^e p. 184, M. of E.), (S n : S ) = the number whose common logarithm is (.45)* X 0.43429 = 4.12. Combining this with F = S n S (see above), we have (4.12 l)S = F 300 Ibs.; whence o = 300 -j- 3.12 = 96.15 Ibs. ; and ^ n , = 4.12^ , = 396.13 Ibs. The requisite force P is then found by noting that for the equilibrium of the lever, the moments of the three forces S n , , and P, must balance about the fulcrum A ; i.e., PX^^^xi + ^Xi; or P = (& + #.) = 61.5 Ibs a pressure easily applied by one man. CHAPTER VII. MECHANICS OF MATERIALS AND GRAPHICAL STATICS. 98. Intensity of the Shearing, and of the Normal, Stress on an Oblicme Section of a Prism under Tension. (This treatment may be more readily understood than that now given in 182 of M. of E.) From the prism under tension in Fig. 193, M. of E., con- sider by itself a portion shown in Fig. 121, between the right section JN &\\& any oblique section, ML. The area of the right section being F and the intensity stress per unit area of that section being p, Fp expresses the total stress on the FlG - right section. The total stress on ML is also of course = Fp. Its component P" normal to the plane ML is evidently P" = Fp sin of. This is the total normal stress on this oblique plane ML. But the area ML over which this normal stress is dis- tributed is not = F, that of the right section of the prism, but = (F -T- sin a). Hence to obtain the normal stress on ML per unit of area, i.e., the intensity of normal stress on ML^ we must divide P" by (F ~ sin a) and thus obtain : Normal stress, per unit of \ c T) sin OL (I i area, on oblique section \ ^ " * ' Similarly, the other component, P' (of the total stress on ML), which is tangential to ML, is in amount Fp cos a, which is the total shearing stress on ML. To obtain the intensity of 107 108 NOTES AND EXAMPLES IN MECHANICS. this shearing stress, we divide P f by the area (F r -s- sin ) of ML, over which P' is distributed, and obtain : Shearing stress, per unit \ '^ \ = p. sin a cos a. . (2V of area, on oblique section ) In these two equations p is an abbreviation for P -~- F, and a is the angle that the oblique plane ML makes with the axis of the prism. The reason for ascertaining the stress per unit area in any case is, of course, that the safety of the material depends upon it and not simply on the total stress. 99. Spacing of Rivets in a Built Beam. The statement in the middle of p. 293 of M. of E. that " The riveting connecting the angles with the flanges (or the web with the angles), in any locality of a built beam, must safely sustain a shear equal to J (the total shear of the section) on a horizontal length equal to the height of the web" may be most directly utilized as follows : Imagine the horizontal continuity of the web to be broken and consequently a vertical seam ren- dered necessary, as shown in Fig. 122. Then, whatever spacing of rivets would be necessary in this ideal seam can be adopted in the real horizontal FlG - 182 - seam made by riveting together the web and angles, the rivets being considered to be in double shear (or single) in the ideal case, if so in the actual. For example, taking the data of the example on p. 294, M. of E., there must be enough rivets in the vertical seam, of length = h = the height of web, to safely stand the total shear of J = 40,000 Ibs. Since each rivet can safely endure a shear of 9000 Ibs. in double shear (see p. 294), the number of rivets required in the height of web would be 40000 -^- 9000 = 4.44 ; i.e., they should be spaced 4.5 in. apart since the height of web is 20 in., and 20" -f- 4.44 = 4.5". But since the pressure on the side of each hole is limited to i>470 Ibs. (see p. 294), on this basis the number of rivets in height MECHANICS OF MATERIALS. 109 of web should be 40,000 -f- 5470 = 7.2, which implies putting them at a distance apart, centre to centre, of 20 -r- Y.2 = 2.7 in. This spacing, therefore, should be adopted in the horizontal seam between web and angles at this part of the beam (near abutment). 100. I-beams treated without the Use of the Moment of Inertia, The assumption is so frequently made (for simplicity in treating the web) that the web carries all the shear, that the corresponding as- sumption that the two flanges carry all the tension and compression is also sometimes used to simplify the analysis. With thin webs the re- sults obtained are accurate enough for practical purposes. For ex- ample, suppose a horizontal I-beam to rest upon supports at its extremities and to bear several detached loads. To find the tensile or cornpressive stress in the flanges at any right section as m, consider free the part of beam on the left of that section (see Fig. 123). R is the abutment reaction ; P l and P 9 the loads between m and A. Considering the total compression P' to be uniformly distributed over the area of the upper flange, and hence (for geometrical purposes) to be applied in the centre of gravity of that flange, and similarly the total tension P" in the centre of gravity of the lower flange (h! =. vertical distance be- tween), while the web carries the shear alone ; we obtain (by taking moments about the intersection of <7and P"\ from which P' is found. Evidently P"= P ', since they are the only forces having horizontal components. Let F' = the area of either flange, then the stress per unit area in either flange is P' jp' = -p-,. If this result is too large for safety the flange-area must be increased. By this method, the proper amount of sectional area can be computed for the flanges at each of several sections of a beam to 110 NOTES AND EXAMPLES IN MECHANICS. carry a specified loading, and a beam of " uniform strength" be designed. In such a case, however, the weight of the beam itself cannot be estimated in advance, but after a gradual tentative adjustment can be taken into account in a satisfactory manner. (The upper flange being in compression may need to be braced or latticed with those of the accompanying parallel beams to prevent horizontal buckling.) Plate girders of variable section ape designed on this general principle, the flange-area being increased toward the middle of the span by riveting on additional plates. 101. Load of " Incipient Flexure" of Long Slender Columns, The result brought out in eq. (8) of p. 366, M. of E., that in the case of a long, slender, round-ended, prismatic column, no flexure at all takes place until a definite value for the load is reached, and that, with that value, any small deflection whatever can be main- tained, may be arrived at quite rationally as follows, without intricate analysis : Let the horizontal bed-plates of a testing machine, Fig. 124, be advanced toward each other until the slender round- m ended rod AB between them is deflected a small amount = a at the middle. If p is the radius of curvature of the elastic curve at R, we have (from free body 7? . . A) T> V - Pa ; or P = . p ap (This radius of curvature, p, is nearly = to that of the circle determined by the three points B, It, and A ( T %- of it). Now let the plates be separated until the deflection is only = a' ; call the new radius of curvature, p'. Then the EI FIG. 124. new value of the force or pressure at each end is P' = -7,. CL fJ But, for this very flat curve, as the deflection is diminished, the radius of curvature changes in an inverse ratio ;* i.e., p' : p : : a : a' ; or, ap = a' p'. Therefore P = P' ; i.e., the pressure induced ly the elasticity of the rod against the plates does not diminish with a diminishing deflection, hut remains constant. Or, conversely, if a less force than this is applied to the column, no deflection * Strictly, the relation is a(2p a) = [^Jj?] 3 , = practically a constant. TESTS OF WOODEN POSTS. Ill takes place ; while if an actual load, whose weight is greater than the above force, be placed on the column, the upward pressure against it due to the elasticity of the column (as the latter bends) being always less than the weight (unless perhaps when the de- flection becomes extreme), the load sinks with an accelerated motion and the column finally breaks (since with increased deflec- tion the stress in the outer fibre is augmented). In most cases in practice, columns are not sufficiently slender to bear out all the above-described phenomena, but enough has been said to show that while with a horizontal beam the deflec- tion is nearly proportional to the load applied (within elastic limit), such is far from being the case with a column. If a piece of card-board cut from a visiting card, and about half an inch wide by three or four inches long, be pressed end- wise with the finger on the scale-pan of a letter-scale, the value of the pressure corresponding to different deflections can be easily noted and the above claims roughly verified. 102. Recent Tests of Large Wooden Posts Prof. Lanza, of the Massachusetts Institute of Technology, Boston, has made tests on yellow-pine flat-ended posts, giving results as follows : Highest breaking stress 5400 Ibs. per sq. in. ; lowest, 3600 ; average, 4544. These yellow-pine posts were nearly cylindrical in form and almost all of them 12 ft. in length (a few 2 ft. long); diameters, from 6 to 10 in. With white oak posts, flat-ended, and of about the same sizes as the former, the highest breaking stress was 4600, the lowest, 3000 Ibs. per sq. in. ; with exception of two which reached 6000. In all these cases of pine and oak posts failure occurred by direct crushing, lateral deflections being inconsiderable, showing that all were practically " short Hocks" Eight separate tests were made with the load applied eccen- trically, about two inches off the centre, the result being to diminish the strength by about one third. All these posts had been in use for years and were well seasoned. Each had a hole about 2 in. in diam. along the axis, from end to end. Other tests have been made at Watertown, Mass., with the 112 NOTES AND EXAMPLES IN MECHANICS. Government testing machine on timber columns, of rectangular sections, mostly about 5 by 5 in., and 7.5 by 9.7 in. ; with a num- ber 5 by 15 in., and 8 by 16 in. Their lengths ranged progress- ively from 15 in. to 27 ft. Flat-ended supports. From these tests Mr. Ely concludes that if the breaking load in pounds be expressed as P = FC, where F is the sectional area in square inches, one of the following values for C should be taken accord-' ing to the kind of timber and the ratio of the length I to the least side, >, of the rectangular section ; thus : For white pine : For I H- b = to 10 ; 10 to 35 ; 35 to 45 ; 45 to 60, Make C= 2500; 2000; 1500; 1000. For yellow pine : For Z -^- 5 = to 15; 15 to 30; 30 to 40; 40 to 45; 45 to 50; 50 to 60, Make G= 4000; 3500; 3000; 2500; 2000; 1500. 103. The Pencoyd Tests of Full-size Rolled-iron I-beams, Channels, Angles, Tees, etc., used as Columns. These were made in 1883 by Mr. Christie, at the Pencoyd Iron Works of Phila- delphia, Pa., and were very careful and extensive. The following table is based on them. By "fixed ends" it is here meant that the ends are so securely attached to the contiguous supports that the fastenings would not be ruptured if the column were sub- jected to a breaking load; by "flat ends," that the ends are squared off and bear on a fiat, firm surface. " Hinged ends " refers to the ends being fitted with pins, or ball-and-socket joints, of proper size, with centres practically in the axis of the column (this axis being the line containing the centres of gravity of all sections of the column, which is prismatic); while "round ends" implies that the ends have only points of contact such as balls or pins bearing on a flat plate, the point of contact being in the axis of the column. The first column of the table contains the ratio of the length, , of the column to #, the least radius of gyra- tion of the section (except that for hinged ends the pin must be at right angles to the least radius of gyration). Factors of safety are recommended as given, being different for fixed and flat ends THE PENCOYD EXPERIMENTS WITH COLUMNS. 113 Ratio I k Fixed Ends. Flat Ends. Factor of Safety. Hinged Ends. Round Ends. Factor of Safety. 20 46,000 46,000 3.2 46,000 44,000 3.3 40 40,000 40,000 3.4 40,000 36,500 3.6 60 36,000 36,000 3.6 36,000 30,500 3.9 80 32,000 32,000 3.8 31,500 25,000 4.2 100 30,000 29,800 4.0 28,000 20,500 4.5 120 28,000 26,300 4.2 24,300 16,500 4.8 140 25,500 23,500 4.4 21,000 12,800 5.1 160 23,000 20,000 4.6 16,500 9,500 5.4 180 20,000 16,800 4.8 12,800 7,500 5.7 200 17,500 14,500 5.0 10.800 6,000 6.0 220 15,000 12,700 5.2 8,800 5,000 6.3 240 13,000 11,200 5.4 7,500 4,300 6.6 260 11,000 9,800 5.6 6,500 3,800 6.9 280 10,000 8,500 5.8 5,700 3,200 7.2 300 9,000 7,200 6.0 5,000 2,800 7.5 320 8,000 6,000 6.2 4,500 2,500 7.8 340 7,000 5,100 6.4 4,000 2,100 8.1 360 6,500 4,300 6.6 3,500 1,900 8.4 380 5,800 3,500 6.8 3,000 1,700 8.7 400 5,200 3,000 7.0 2,500 1,500 9.0 420 4,800 2,500 2,300 1,300 440 4,300 2,200 2,100 460 3,800 2,000 1,900 480 1,900 1,800 from those proposed for hinged and round ends. In the other columns the numbers given are the respective breaking stresses in Ibs. per sq. in. of sectional area, which number must be multi- plied by that area in sq. in., for the actual breaking load ; divid- ing which by the proper factor of safety we obtain the safe load in pounds. For example, required the breaking load of a 9-in. light iron I-beam of the N. J. Steel and Iron Co., 14 ft. long and used as a column with flat ends. From p. 40 of the hand-book we find : least / = 4.92 and the area of section = 7 sq. in. Hence the least k 2 is I ~ F, = 0.703, and k itself = 0.838 in. ; so that I -+- k = 168 in. -f- 0.838, = 200. The breaking load, then, is FC = 7.00 X 14,500 = 101,500 Ibs. ; and the safe load would be 101,500 -r- 5 20,300 Ibs. (If Rankine's formula were applied to this same case a fair agreement with this result would be found.) 114 NOTES AND EXAMPLES IN MECHANICS. 104. Cast-iron Channel as a Column. In the case of the ,^ channel-shaped section in Fig. 125, com- posed of three rectangles, all of the same width, t, it is desirable, for economy of material if the prismatic body is to be used as a column, that the moments of inertia about the two gravity axes X and T should be equal (see 311, M. of E.). The problem, therefore, presents itself in this form : Given the width 5 of the base DC of the section, and the thickness of metal t, what value should be given to the length c, of the projecting sides AD and BG, of the channel to secure this result? The algebraic statement of the conditions involved leads to an equation of high degree for the unknown quantity c. But, ~by numerical trial, a few reliable values have been found, given below, by simple interpolation between which all ordinary cases in practice may be satisfied. X 1 . , c >: "f | I I i ! i 1 SL I D -> [ ~> c FIG. 125. When t = (infinitely thin), " t 0.0835 ^5, " t = 0.1665 = 5, j solid I rectangle make c = 1.375 ; = 1.305; t 0.5005 = r G = 1.005. In the construction of such a column the edges A and B of the projecting sides should be tied or braced together at inter- vals ; or occasional transverse webs may be introduced. It is remarkable in this problem that the distance u of the centre of gravity C from the base DC is almost exactly equal to one half of 5 in every instance, and may be so assumed in locat- ing the axis of the column that the load may be applied in that axis. 105. Vertical Reactions of Horizontal-faced Piers bearing a Beam and Loads. Graphical Method. The construction on p. 404 of M. of E. is the one usually given and is the most convenient; still, the proof is a little puzzling to the student because the PIER- REACTIONS GRAPHICS. 115 amounts and direction of the two auxiliary, mutually annulling, forces P and P' are not known at the outset. Hence we here present a construction in which those two forces are completely assumed and known at the outset. Assume 1 and 6 as the auxiliary forces (equal, opposite, and in the same line), there being three loads in this instance, 2, 3, and 4; Fig. 126, Number the forces of the system as in figure and draw a FIG. 126. portion of the force-polygon with the known forces 1, 2, 3, and 4, beginning at O, and the first three rays, I, II, and III (dotted). The first segment of the corresponding equilibrium polygon should begin at #, the intersection of the action-lines of forces 1 and 2, and finally the third segment cuts the action-line of 5 in the point m. Now the fourth segment is the last in this case [of three loads, 2, 3, 4], and must pass through the intersection of the last two forces, 6 and 7, i.e., through A. Hence draw mA and a parallel to it through the pole O, this latter line being the fourth ray desired, whose intersection with the " load-line" at n f cuts off the proper length of the right-hand reaction 5. The forces 6 and 7 are now easily added to the force-diagram in an obvious manner and the latter is complete ; the magnitude of the other reaction, 7, being thus determined. Of course, the value of force 7 is also given by the rid, and the force-polygon could also be closed by running from n f to d and then from d to 0. instead of in the manner shown. Note the order of numbering in the above. The two assumed forces are made the first and the last but one, respectively ; the unknown reactions are made the last ~but two, and the last, respectively ; while the given loads are assigned to the interven- ing numbers, in any order. 116 NOTES AND EXAMPLES IN MECHANICS. (in.) 106. Construction for Use with the Treatment in 390, M. of E. (Graphical Statics). For the particular case involved in the treatment of the straight horizontal girder, built in, of 390, M. of E., the following is given, to replace the more general construction of 377, M. of E. At (I.) in Fig. 127, we have a curve, or broken line (equi- librium polygon, for instance), - 1> v*. / I FKWMG, connecting points F \^>^ m and G in two vertical lines. Across this curve we wish to draw a right line v . . m, in such a way that the area K WM above v . .m shall be equal to the sum of the areas vFK and mGM below v . . m, and also that the centre of gravity of the upper area shall be in the same vertical line as that of the two lower, or negative areas, taken together. Only one position of v . . m will do this, the algebraic expression for which is that 2(s*) = 0, and that 2(%z') = ; (the areas in question being divided into vertical strips of equal horizontal widths = 4x 9 the distance of any strip from the vertical line vF being called a?.) Annex the figure FGMK (having joined F . . G) to both the positive and negative figures above mentioned and the con- dition now becomes [see (II.) in figure] that the area of the figure FKWMG..F (the right line FG being its lower boundary) must equal that of the two triangles vFG and mGv, and that the centre of gravity of the former must lie in the same vertical as that of the two triangles combined. In other words, if the area of the curvilinear figure, FKWMG-F, with FG as base, be considered as a weight R acting through its centre of gravity, then the areas of the two triangles must represent the two upward reactions T&ud T' of two piers [see (III.) in figure] supporting a horizontal beam on which R rests. These imagi- nary piers are evidently at distances of one third the span from | FIG. 127. EQUILIBRIUM POLYGON THROUGH THREE POINTS. 117 the verticals through F and G. Adopt, therefore, the following construction : By dividing into vertical strips find the area of the curvi- linear figure with base FG. Draw the pier verticals at the one- third points. Find the vertical containing the centre of gravity of the curvilinear figure by p. 415. Compute or construct the values of T and T' on the conception of the known area R being a weight supported on the beam with T and T' as reactions. (T and T' are most easily obtained, perhaps, by scaling the distances s and t (see figure) and writing T(s + t) = Rt, and T' = R - T.) If the area of the triangle FGv must be = T, one of the values just found, knowing that this area = \ (altitude 1} X base vF, the proper length of vF is easily computed ; and similarly, using T' and the triangle mvGr, we calculate mCr. Joining v and m, the required right line vm is determined. 107. Three-point Construction. Equilibrium Polygon for Non- vertical Forces. Preliminary Step. In the construction and proof of 378a, M. of E., it is supposed at the outset that the action- line of the resultant (/?,) of all the forces acting between points A and p has been found ; similarly, that of the resultant (/2 a ) of the forces acting between p and B ; and that of the resultant (R) of the two partial resultants. It is here intended to give the detail of finding these three lines and to make clearer the scope and intent of the problem. Fig. 128. Let A^p^ and B be the three points through which the equilibrium polygon is to pass, and 1, 2, etc. (to 6 inclusive) (on left of figure), the given forces in magnitude and position ; 1, 2, and 3 acting between A and JP; and 4, 5, and 6, between p and B. Lay off the " load-line" 8TU, on some convenient scale of force, and select any pole as 0'". Draw the " rays," 0'"8^ etc. ; and also lines parallel to them, in proper order, beginning at a convenient point A" (not ^4, necessarily), so as to form an equilib- rium polygon, A" B" , as shown. The first segment is parallel to 0"'S ; the last, to 0'" U ; and the segment connecting forces 3 and 4 (in this particular instance), parallel to O'"T. The inter- section of the first and last segments gives M fl ', a point in the 118 NOTES AND EXAMPLES IN MECHANICS. action-line of R, the resultant of all the given forces, 1 . . 6 ; and similarly, the other intersections N" and 0" are points in the action-lines of R l and R^, respectively. Right lines through these three points, parallel respectively to SU, ST, and TU, M' Preliminary Step; Equil., Polygon thro' Three Points; Forces not Vertical. FIG. 128. should meet in a common point ' ', and are the respective action- lines desired. The further steps are those given on pp. 460 and 461, M. of E. Our present equilibrium polygon, A" B" , is of no further use ; but the load-line STU will still serve, after the lines M f A and M ' B have been located according to 378a. We can then draw through S and U parallels to M f A and MB, respectively, and by the intersection of these parallels with each other deter- mine the pole corresponding to the final equilibrium polygon which is to pass through A 9 p, and B. It is to be noted that the problem of the " Shear-legs" of 59 of these Notes, and also Problem 2 (of the two links), p. 35, M. of E., are cases of a three-hinged arch-rib, and can be treated graphically in the same manner ; and thus the " special" equilib- rium polygon and its corresponding pole and rays (i.e., force- diagram) determinedo The ray parallel to the segment passing through the inter- mediate joint (p) gives the amount and direction of the pressure on the hinge of that joint ; and corresponding statements may be made for the two extreme joints. CHAPTER VIII. MISCELLANEOUS NOTES. 108. Co-ordinates of Centre of Gravity Fuller Explanation. Assume the various small particles of a rigid body to be num- bered 1, 2, 3, etc., and call their respective volumes d V l , d F, , d F, , etc. (cubic feet), and their #-co-ordinates x l , a? 2 , x z , etc. (feet). If the body is heterogeneous, the particles may be of dif- ferent densities; for example, particle 1 may be of such a density that a cubic foot of material of that density would weigh 100 Ibs., while a cubic foot of the material of which particle 2 is composed would weigh 110 Ibs. ; and so on. Or, in symbols, the " heavi- ness," or rate of weight, of particle 1 is y l = 100 Ibs. per cubic foot, while that of particle 2 is y 9 = 110 Ibs. per cubic foot. A similar notation would apply to all the other particles. The respective weights, then, of the particles (or force of the earth's attraction on them) are y^ F,, 7^F 2 , y s dV 3 , etc. (pounds), and if we substitute these for the forces P t , P a , P 3 etc., in the expression for the x of the centre of parallel forces (foot of p. 16, M. of E.), we obtain _ w& V* + wA V ** . . . . r,dV, + r ,d^ + r3 dV a +... which in the compact notation of calculus, the particles being taken " infinitely small " and, therefore, " infinite " in number, can be written fxydV or m^-mfdV . . (b) JydV where C is the total weight of the body, = / yd V. If the body is homogeneous, all the particles have a common " heaviness," which we may call.y and factor out, thus obtaining 119 120 NOTES AND EXAMPLES IN MECHANICS. y ' m r "- Ym from which the y m can be cancelled, leaving _ 1 /* x= r l xdV, . . . . '.' * (c) T/t/ ^ where V denotes the total volume of the body. (Note that the factoring out of a common multiplier from a parenthesis corre- sponds to taking a constant outside of the integral sign.) 109. The Time-velocity Curve and its Use. From eq. (I.), p. 50, M. of E., we have ds -r- dt = v, the velocity of a moving point at any instant. Hence, also, ds = vdt, or the element of dis- tance equals the product of the velocity at that instant by the element of time. If now, whatever the character of the recti- linear motion, we conceive a curve to be plotted, in which the time (from some initial instant) is laid off as an abscissa, and the velocity of the moving point as an ordinate, this curve may be called a u time-velocity" curve for the particular kind of motion, being different for different kinds of motion. (Of course, proper scales must be selected in laying off distances on the paper to represent the quantities time and velocity.) The equation expressing the relation between the two vari- ables, time and velocity, may be regarded as the equation to the curve. Thus, in uniform motion we have the velocity v con- stant, so that the curve is a right line parallel to the horizontal axis, or axis of time, as AL in Fig. 130, where OA represents the constant velocity. If the motion is uniformly accelerated, that is, has a constant acceleration, we have v = v -\-pt [where p is the acceleration, or rate of change of the velocity, and v is the initial velocity (for t = 0)], and note that the quantity pt, or total gain in velocity over the initial velocity v , is directly pro- portional to the time, so that the curve obtained is a straight line inclined to the axis of abscissas, as BH in Fig. 130, BO repre- senting the initial velocity V Q . In general, let KVM, Fig. 129, be the time- velocity curve MISCELLANEOUS NOTES. 121 M/ l = for any rectilinear motion. Here we note that the product v . dt at any instant during the motion is represented by the area of the vertical strip "FT?, whose width is dt and length v (by scale). Hence the element of distance ds (feet) is proportional to that area, and the whole distance (s) described from the beginning is represented by the sum of the areas of all such strips from OK to FT?, i.e., by the area OKVR\ while the distance (s n ) de- scribed between t = and t = t n will be represented by the complete area OKVMN. If now ON be regarded as the base of the figure OKMN, we may conceive of a rectangular figure OALN having the same area and same base as OKVMN, and having some altitude A which can be looked upon as representing the "average velocity "of the motion between t and t = t n . By " aver- age velocity " would be meant that constant velocity necessary in a uniform, motion to enable a moving point to describe the dis- tance s n in the same time t n as in the ac- tual motion. In other words, the " aver- age velocity " is the result obtained by di- viding the whole distance by the whole time, and is represented by the altitude AO, since the area of a rectangle is equal to the product of its base, ON(= t n \ by its altitude, AO. As a useful instance consider again a uniformly acceler- ated motion. Its time- velocity curve is a straight line, BH, Fig. 130, the initial velocity v 9 being rep- resented by QB, and the final, v n (for t = i n ), by HN. The distance described is represented by the area of the trapezoid OBHN. This area is equal to that of a rectangle of the same base ON (i.e., t n ) and of an altitude OA = half the sum of OB and NH\ i.e., s n = %(v + v n )t n . In other words, the average velocity between t = and t = t n for the uniformly accelerated motion is J(-y Q + v n ). time FlG - v n FIG. 130. 122 NOTES AND EXAMPLES IN MECHANICS. FIG. 131. If the initial velocity of the uniformly accelerated motion is zero, the time-velocity curve becomes a straight line passing / through the origin, 0, viz., OH, Fig. 131 ; and in that case the distance s n is repre- sented by the area of the triangle OHN, and the average velocity OA is one half the final, or the final is double the average. Therefore to obtain the whole distance de- scribed we must multiply the whole time by one half the final velocity ; i.e., s n = %v n t n . If, then, in this case of uniformly ac- celerated motion with initial velocity = zero, we divide the whole distance by the whole time, we do not obtain the final velocity (a common error with students), but only the " average velocity," in the sense defined above. This result must then be doubled to obtain the final velocity. As an instance where the average velocity is one third of the final (the initial velocity being zero), consider the case of variably accelerated motion represented by the relation v = qf, where q is a constant. The time-velocity curve will have the form OPH, Fig. 131, a parabola with vertex at 0. The area OPEN will be one third of that of the circumscribing rectangle, and hence is equal to O X one third of NH, i.e., to ON X Hence OB, or one third of the final velocity, is the average velocity. Or, mathematically, in detail, ds, = vdt, = qfdt ; therefore s n = But for t t n we have v, = v n , = eft*, and hence (1) That is, the average velocity is equal to one third of the final. 110. Reduction- Formulae for Moment of Inertia of a Plane Figure. (To replace 88, M. of E., as regards the Moment of Inertia of a plane figure.) MISCELLANEOUS NOTES. 123 Definition. Any right line containing the centre of gravity ot a plane figure is called a " gravity axis " of that figure. Theorem. The moment of inertia of a plane figure about a given axis in its own plane is equal to its moment of inertia about a gravity axis parallel to the given axis, augmented by the product of the area of the figure by the square of the distance between the two axes. Proof. Fig. 132. Let dF be the area of any element of the plane figure, and the distance of that element from any axis X in the plane of the figure ; while z is its distance from a " gravity axis," g, parallel ~ to axis X. By definition we have I x = Fia axes. \ but z' = z + Hence being the distance between the two dF. Now, from the theory of the centre of gravity, we have (see eq. (4), p. 19, M. of E.)fsdF= F~z, where "z is the distance of the centre of gravity C from the axis g. But g is a gravity axis, so that -5 ; and hence fadF = 0. Also fdF = F, the whole area; %ud.fzdF=Ig. Whence, finally, I x = I g + Fd\ ... (4) Q. E. D. It also follows, by transposition, that I g = l.-F#\ . ... . . (4a) which shows that the moment of inertia of a plane figure about a gravity axis is smaller than that about any other axis parallel to that gravity axis. The moment of inertia of a plane figure plays an important part in the theory of beams subjected to bending action, the transverse section of the beam forming the plane figure in ques- tion ; somewhat as the mere area of the section does when the beam or rod is subjected to a straight pull. 124 NOTES AND EXAMPLES IN MECHANICS. 111. Miscellaneous Examples. (See opposite pages for figures.) 1. Fig. A. Given the data of the figure, find the stress in every two-force piece, and the three pressures exerted on the pin at J5. 2. Wind from the S.W., 30 miles per hour. Ship going toward the N.W. at 10 miles per hour. At what angle with ship's course should a vertical sail be placed that the air- par tides may strike it at an angle of 30 on the hinder side ? 3. Locate, by calculus, the centre of gravit}' of the plane figure in Fig. B, the equation to the upper bounding curve being xy = 20 sq. ft. 4. In the rectilinear motion of a material point weighing 12 Ibs., and mov- ing horizontally on a rough surface, friction from which is the only horizon- tal force, we note that positions A, B, C, and D are passed at the following times (by the clock), respectively : 3 h. 4 m. 8.1239 sec.; 3 h. 4 m. 8.2350 sec.; 3 h. 4 m. 8.3490 sec.; 3 h. 4 m. 8.4658 sec. The distance from A to B is 10.00 ft. ; from B to G, 10.02 ft. ; and C to D, 10.05 ft. Find (approx'ly) the acceleration of the motion as the material point passes the position B ; also for C; and what must be the value, in pounds, of the friction at B ? 5. Fig. C. Find the stress in each two-force piece, and pressure on pin at A ; also pressure of pin B on bar BD. 6. Of the solid right cylinder in Fig. D, the internal conical portion is of lead, whose spec. grav. is 11.3, while the remainder is of cast iron. Find the centre of gravity of the whole solid. 7. In Fig. 80 on p. 67, Notes, compute the pressure between the block and the guide when the former, having started from rest at A, is passing position B, 45 from A. 8. Fig. E. The block weighs 40 Ibs. The cord d . . . e is attached to it at angle 15 with the plane, which is smooth. If the block is to be permitted to slide from rest down the plane, under action of gravity, the cord, and the plane, what constant tension (Ibs.) must be maintained in the cord that a velocity of 12 ft. per sec. may be generated in 2 seconds ? Afterwards, what new value must be given to this tension if the velocity is to continue at that figure (12 ft. per sec.) ? 9. In the vertical fall of a material point, certain consecutive small space- intervals are given, = a, b, c, and d ; and the corresponding time-intervals, /, t', t", t" r . Derive approximate formulae for the velocities at mid-points of these time-intervals ; the accelerations near the end of some of these intervals ; and the corresponding resultant force that must be acting, the weight of the body being O. 10. Fig. F. Find the amount and position of the pressure between the bar AB and each of the four pins passing through it ; also the stress in DB. t ^Xl< I VH C -< J I 400 Z&s. B 1200 Z&s. Pi ~ Ui-- - J_ 12" 20 Ibs. 10 * R t L. \12lbs. [A is inside 30' N.B. The small pulley is free on the axle of wheel B . chain 2400 Ibs. 20' JV '-friction * 12000 Ibs. 126 NOTES AND EXAMPLES IN MECHANICS. 11. Fig. G. The two pulleys run on fixed bearings, without friction. If the given three weights be allowed to come to a position of equilibrium, the knot A being a fixed knot, find the angles the two oblique cords make with the vertical, respectively. 12. Find the moment of inertia about the axis X of the symmetrical plane figure shown in Fig. H. The equation to the bounding curve AB is (x a'} -r- (a a') = z* -+- h*. 13. Fig. J. The block weighs 12 Ibs., and in sliding down the rough in- clined plane encounters a variable friction, which in Ibs. 3 times the dis- tance, s, from the starting point, in feet. It starts from rest at A. Find the velocity acquired on its reaching a position 2 ft. vertically below A. 14. Compute the moment of inertia of the plane figure in Fig. K about the axis X ; also the corresponding radius of gyration. 15. Compute the y co-ord. of the centre of gravity of the plane figure shown in Fig. L. The equation to bounding curve is a; 3 = 4y, with the foot as linear unit. 16. A block of 50 Ibs. weight is started along a rough horizontal table with an (initial) velocity of 40 ft. per sec. If the friction met with is variable, and, in Ibs., equal to 700 times the body's weight -s- (1200 + the square of the veloc. in ft. per sec.), find the time and distance in which the body comes to rest. 17. Fig. M. Find the tensions in all chains and the pressure on pin A and under wheel B. 18. Fig. N. Find the moment of inertia of the plane figure about the grav- ity-axis X which is perpendicular to the axis of symmetry. 19. Fig. P. The ram A, of 1200 Ibs. weight, falls from rest through 20 ft., and has then an inelastic impact with the pile B, of weight 400 Ibs. Com- pute their common velocity after the impact, the Kinetic Energy lost in the impact, and the distance the two bodies will sink after the impact, overcom- ing the constant frictional resistance of 12000 Ibs. on side of hole. 20. Fig. Q. Block of 10 Ibs. weight on inclined plane. It starts from rest at A. If friction on AB is 2 Ibs., while on EG it is 3 Ibs., compute the time of reaching position G, and its velocity at that instant. 21. Fig. R. A block of 12 Ibs. weight falls from rest, freely through the first 7 ft., but then strikes the head of spring A, which opposes a resisting force at rate of 100 Ibs. per inch of shortening ; and 3 inches further down Strikes spring B, offering 160 Ibs. of resistance per inch of shortening. Where is the block when (momentarily) brought to rest (supposing the elastic limit of the springs not passed) ? 22. A bod} r weighing 12 Ibs. is given an initial upward vertical velocity of 10 ft. per sec., being thereafter acted on only by gravity and a variable hori- zontal force = [ T V of time in sec.] Ibs. Find the equation to its path. B f 5^ 1200 Ibs. 10-- U- ' ~"--^. 22 Ibs. R m 12 Ibs. U> ' b The small pulley is free on the axle of wheel B. 2400 tes. 20' '// 12000 Ibs. 128 NOTES AND EXAMPLES IN MECHANICS. 112. Answers to Preceding Problems. 1. 339.1 Ibs. and 240 Ibs. ; on pin A, 240 Ibs. vertically, 288.3 Ibs. at 33 42' with hor- izontal, and 466.2 Ibs. at 30 56' with vertical. 2. 41 34' ; or sail pointing 3 26' K of West. 3. so = 3.28 ft. ; y = 3.28 ft. ; area of figure = 18.31 sq. ft. 4. In passing B, ace. = 18.72 ft. per sec. per sec. ; C, 16.05; friction as the material point passes B, 6.98 Ibs. 5. 154 Ibs. compression, and 1200 Ibs. tension ; at A, pressure = 738 Ibs. at 25 with horiz. ; pressure of pin B on bar BD = 1132 Ibs. at 36 45' with vertical. 6. x = 5.39 inches from left-hand base. 7. Pressure is 42.44 Ibs. 8. First value of tension, 18.92 Ibs. ; second, 26.63 Ibs. 9. Near the end of the first interval the acceleration = at'} , , , . , . , , 2(X W that near end of second interval, + . 10. Pressures are 1600, 45T ? 2263, 1143 Ibs., respectively. There is no stress in piece DB. 11. On the left, 87 43' ; on the right, 65 IT. 12. The moment of inertia, about X, is -f [aA 3 -f- \a'h ']. 13. The velocity acquired is 7.13 ft. per second. 14. I x = 330.66 biquad. inches; rad. gyr. = 3.02 inches. 15. The area is 80 sq. ft. and y = 15.61 ft. 16. Time 3.075 sec., to come to rest. Distance 70.9 ft. 17. Tension in upper chain, 5838 Ibs. ; in the lower oblique chain, 2770 Ibs. Pressure under wheel B, 1385 Ibs. Pressure on pin at A is 7503 Ibs., and is directed toward the right at an angle of 6 37' above the horizontal. 18. Distance of centre of gravity from the horizontal line drawn through the upper corners is 2.20 inches. Moment of inertia about the gravity axis X is 8.12 biquad. inches. 19. 26.91 ft. per sec. ; 6000 ft.-lbs. lost; 1.73 ft. sinking after impact. 20. Time, A to (7, = 2.06 sec.; velocity at #=18.05 ft. per sec. 21. The block is 7 ft. 4.297 inches below its starting-point, i.e., has shortened spring A an amount of 4.297 inches. PKOBLEMS AND EXAMPLES. 129 (< F H FIG. 150. 113. Examples. 1. Fig. 149. The hollow shaft A is to be twice as strong torsionally (i. e. , as to tor- ^v ^ sional moment) as the solid shaft B ; while the material of A is only half as strong as that of B. Given the lengths Z and Z", O i " (, ~ ~~-n the radius r" of j5, and the outer radius, FIG. 149. r, of ^1, determine a proper value for r', the inner radius of A, for above conditions. 2. Fig. 150. Homogeneous prismatic beam; rectangular section ; width = h and height = h ; placed with h horizontal, and loaded uniformly over its whole length at rate of w Ibs. per running inch. Given the whole length, Z, and position of support A (at left end), where (i.e., distance a = ?) shall we place support B that the moment of stress-couple in section over B shall equal (without regard to sign) the greatest moment of stress- couple occurring between A and the point of inflection, F, of the elastic curve ? Also, after a is found, find the maximum shear, (b and h are given, aiid'also w\ elastic limit supposed not passed.) 3. Fig. 151. By the principles of the graphical statics of mechanism, having the resisting force Q given in amount and position, given also all friction angles con- cerned, and considering all kinds of frictional action, find the value of P for forward motion ; also for back- ward motion ; the efficiency of the mechanism. 4. Fig. 152. The load, of weight P = 60 Ibs., is gradu- ally applied at lower end of compound vertical round wire (of wrought iron) whose upper end is fixed. Neglecting the weight of the wire, compute the total elongation of the wire and the work done on the wire 130 NOTES AND EXAMPLES IN MECHANICS. during the gradual stretching. (Lengths are 100' and 120'; diameters J in. and T ^- in., respectively.) 5. Fig. 153. The horizontal prismatic beam is of rectangular section ; width J, horizontal. Its height h is to be three tim-es its width. Beam of timber. The length and character and amount of loading are given in the figure, the weight of the beam itself being neglected. VSSSS%%S;;S/S/A There being three (local) maxi- A IHP HHc fxx^xjxxxxxxxx ~. mum moments (of stress-couple), "* m"~n viz -> at ^' B * and ^ locate sec ~ ^ ku2*oii ^ on -B by determining distance i FIG. 153. u #"; compute the moments at A^ B, and (7; and then determine the minimum safe dimen- sions to be given to the section of the beam. (The load of five tons is uniformly distributed along the ten feet.) 6. Fig. 154. The short vertical cylindrical body abed is fixed at the upper end while sustaining at its lower end a weight of 4000 Ibs., whose line of action prolonged upward passes through the extreme edge, &, of the section db. The horizontal section of the body is a circle of radius = 0. 5 inch. It is required to compute the stress per square inch at point a of the section db, and also that at point b. Also, what would these stresses be if the line of action of the 4000 Ibs. load passed through the centre of the section ab ? 114. Answers to Problems in 113. (For 6, see below.*) 1. r f = r 4 - 4rr"\ 2. a = (1 4/i)Z, = 0.293?. The maximum shear is just on the left of support B and is J m ( V2 V)wl = 0.414i0Z. 3. (Solved graphically.) 4. Total elongation = 1.21 inches. Work done = 36.3 in. -Ibs. 5. Distance a = 4.4 ft. Moments are 1.00, 3.84, and 4.00 * Answer to Ex. 6 : At a, 12.7 tons per sq. in. tension ; at &, 7.63 tons per sq. in. compression. THE " IMAGINARY SYSTEM" IN ROTARY MOTION. 131 ft. -tons, respectively. Taking B r = 1000 Ibs. per sq in., we have h = 12, and b = 4, in. 115. The "Imaginary System." In conceiving of the imagi- nary equivalent system in 108, M. of E., applied to the material pointo supposed destitute of mutual action, and not exposed to gravitation, we employ the simplest system of forces that is capable, by the Mechanics of a Material Point, of pro- ducing the motion which the particles actually have. If now the mutual actions, coherence, etc., were suddenly re- established, there would evidently be no change in the motion of the assem- blage of particles ; that is, in what is now a rigid body again ; hence the imaginary system is equivalent to the actual system. In applying this logic to the motion of translation of a rigid body (see 109 and Fig. 122, M. of E.) we reason as follows: If the particles or elementary masses did not cohere together, being altogether without mutual action and not subjected to grav- itation, their actual rectilinear motion in parallel lines, each hav- ing at a given instant the same velocity and also the same accel- eration, p, as any other, could be maintained only by the appli- cation, to each particle, of a force having a value == its mass X p, directed in the line of motion. In this way system (II.) is con- ceived to be formed and is evidently composed of parallel forces all pointing one way, whose resultant must be equal to their sum, viz. / dM X p> But since at this instant p is common to the motion of all the particles, this sum can be written p I dM, = the whole mass M X p- If now the mutual coherence of contiguous particles were sud- denly to be restored, system (II.) still acting, the motion of the assemblage of particles would not be affected (precisely as the fall- ing motion in vacua of two wooden blocks in contact is just the same whether they are glued together or not) and consequently we argue that the imaginary system (II.), is the equivalent of what- ever system of forces the body is actually subjected to, viz. sys- tem (I.), (in which the body's own weight belongs) producing the actual motion. 132 NOTES AND EXAMPLES IN MECHANICS. Since the resultant of system (II.) is a single force, = parallel to the direction of the acceleration, it follows that the resultant of the actual system is the same. 116. Angular Quantities Rotary Motion about a Fixed Axis. In Fig. 155 let the body MLN rotate about the fixed axis C (perpendicular to paper), the initial posi- tion of the arm CL having been CY. Now the angular motion of the whole rigid body is the same as that of the arm CL. If in a small time- interval, dt, the arm passes from position CA to position CB, and thus describes the small angle ACB, whose value in ?r-measure is da radians, the angular velocity at about the mid-point of angle A CB is GO = da -r- dt. In the next and equal time-interval a slightly different angle, da' radians, is described ; and if in the figure we lay off angle B CE equal to A CB, the angle ECB, or difference between da and da may be called d*a. And so on, for any number of consecutive (dor)'s, described in equal times, each = dt. If the motion is uniform, all the (dcfy's are equal ; that is, the angular velocity is constant, each d*a being = zero. If all the (d*ay& are equal, the motion is uniformly accel- erated, the angular acceleration, 0, being thus determined : The gain, or change, of angular velocity occurring between the mid- point of ACB and that of BCD is -j- - -^, or GO' oo\ and if this be divided by the time, dt, occupied in acquiring the gain, we have for the rate of change of angular velocity, that GO 1 GO d(&> is, for the angular acceleration, the value 6 - ^- - ^-. Another form is this: if the gain of angular velocity, -^, be divided by dt, we have dt dt da' -da If the successive (^V)'s are unequal (the successive (da)^ ANGULAR QUANTITIES. ROTAEY MOTION. 133 being described in equal time-intervals, it must be remembered), the motion is some kind of variably accelerated angular motion ; e.g., in a harmonic (oscillatory) rotary motion the angular acceleration at any instant is directly proportional to the angular space between the ' ' arm ' ' or reference line of the body and the middle of its oscillation, and is of contrary sign; i.e., 6= Aa, where A is a constant. Example 1. At a certain part of its revolution a fly-wheel is found to describe just 1 in 0.01 second. Here we have da= 0.01745 radian, and dividing this by the dt,= 0.01 sec., we obtain GO = 1.745 radians per second as the angular velocity of the wheel at this part of its progress, as nearly as the data permit. (Strictly, this value of GJ is only the average value of the angular velocity for this small but finite portion of the motion. The data are insufficient to determine whether the velocity is variable or constant.) Example 2. The same wheel, besides describing 1 in 0.010 sec., is found to describe 1 2' in the next 0.01 second. Com- pute the angular acceleration, as near as may be from these data, for this part of the motion. As before, da = 0.01745 radian, and we now have the additional fact that da' = 0.01803 radian, each of the time-intervals being dt = 0.010 sec. If we substi- tute directly in eq. (VII), there results 0.000581 radian 6 = - = 5. 81 rad. per sec. per sec. (0.01 sec.)' That is to say, at this part of the wheel's motion its angular velocity is increasing at the rate of 5.81 velocity-units per sec- ond, i.e., 5.81 radians per second per second. Another method is this: The velocity at the mid-point of the da is, as before, 0.01745 -r- 0.01, = 1.74 radians per sec- ond, while at the middle of the da' it is 0.01803 -f- 0.01,= 1.803 radians per sec. ; taking the difference of which we find that 0.058 rad. per sec. of angular velocity has been gained while the wheel was passing between these two mid-points. Hence, dividing this gain by the time of passage, 0.01 sec., we obtain = 0.058 -r- 0.01 = 5.81 rad. per sec. per sec. APPENDIX. NOTES ON THE GRAPHICAL STATICS OF MECHANISM. PREPARED BY I. P. CHURCH, CORNELL UNIVERSITY. These notes are based mainly on the work of Prof. Herrmann of Aix-la- Chapelle, the earlier form of which was presented as an appendix to Vol. Ill, of Weisbach's Mechanics. It is thought that the form of presentation adopted in the following pages is that best adapted for students already familiar with the Graphical Statics of quiescent structures, which Prof. Herrmann assumes is not the case with the readers of his book. For greater clearness, all force polygons have been placed on separate parts of the paper from the mechanism itself, instead of being superposed on the latter as in Prof. Herrmann's book. The figures referred to in the text will be found in the back of this pam- phlet; while the paragraphs () referred to (of a higher number than "30") will be found in the writer's "Mechanics of Engineering." For table of contents see p. 28. 1. Assumptions. The forces acting on each part of any mech- anism .here treated will be considered to be in the same plane and to form a balanced system, i.e., to be in equilibrium ; in other words, the motions of the pieces take place without sen- sible acceleration (or the effect of inertia is disregarded). (See p. 440 of Prof. Uri win's Machine Design for consideration of the inertia of a piston.) Also, the weights of the pieces will be neglected unless specially mentioned. * 2. Graphical Treatment. A " two- force " piece, or a two-force " member of a mechanism is one acted on by only two forces ; which forces, therefore, must be equal and opposite and have a 2 NOTES ON THE GRAPHICAL STATICS OF MECHANISM. common line of action, for equilibrium. No force polygon need be drawn for such a piece. A Three-Force Piece. Here, for equilibrium, the three lines of action must meet in a common point and the force polygon is a triangle (see 325); for example, see Fig. A (bell-crank), (p. 28). For the equilibrium of & four-force piece (Fig. B), the result- ant of any two of the forces must be equal and opposite to that of the other two forces. The common line of action of these resultants is the line joining the intersections a and &, while their common amount is given by the ray 00, which is parallel to a . . b and is a diagonal of the closed force polygon (here a quad- rilateral). If the four forces act in parallel lines, and two are given, we determine the other two by an "equilibrium poly- gon," etc., by 329 ; i.e., we treat the two unknown forces as pier-reactions, even if their action-lines (one or both) are between the action-lines of the known forces. (For example, see [A] in Fig. IT, Plate Y, where from the known forces S l and $, we construct the two unknown, P and R, in given action -lines, to balance them. The equilibrium polygon begins at c in the left-hand abutment- vertical and consists of segments c . . . d, d. .. e, and e .../*, terminating in the right-hand abutment- verti- cal, at/*. We then draw c . . .f as the abutment-line, or " closing line.") No more than four forces will act on any piece. 3. Efficiency. In each of the following problems there will be but one working force or driving force, P, and but one useful resistance, Q ; all other resistances being due to friction. For present purposes we are to understand by efficiency the ratio of the value P , which would be sufficient for the driving force if there were no friction of any kind, in order to overcome Q (without acceleration), to the value P, which it must actually have, to overcome Q under actual conditions, i.e., with friction. p Hence efficiency = rj = -7^. [Strictly, the efficiency involves the distances traversed ; see 5.] 4. Backward Motion. If the useful resistance Q is a load due to gravity, i.e., a weight, the value to which the working force 3 NOTES ON THE GRAPHICAL STATICS OF MECHANISM. (now such no longer) must be diminished in order to allow the mechanism to run backward without acceleration will be called P '; and if in any case P' is found to be negative, we recognize the mechanism to be self-locking that is, it will not run back- ward (or " overhaul ") when the working force is zero, but, on the contrary, a force must be applied in the action-line of the working force in the opposite direction to cause a backward motion. For instance, in Fig. C,* the force P is necessary for the uniform downward motion of the handle A, to raise Q, and overcome friction at all points. With no friction, P would be P Q sufficient to raise Q, and the efficiency = -77; whereas, to enable Q to sink uniformly, a still smaller force, viz. P\ must be applied at A (but still positive in this case, so that the machine is not self-locking). 5. Efficiency in General. . . , with one working force P and one useful resistance Q, is the ratio of the work Qs' to the work Ps, where s r and s are the respective distances worked through in forward motion by Q and P (simultaneously); i.e., 77, or efficiency, = -~ ....... (1) Now by 142, if 2(Fs") denotes the sum of the amounts of work lost in friction at various points where rubbing occurs in the mechanism, we have Ps = Qs' + 2(Fs") ....... (a) But in the ideal case of no friction (or perfect efficiency), taking the same range of motion as before, and letting P denote the new (and smaller) value of the working force which is now sufficient to overcome , we have Hence from (#), (#), and (1) it is plain that the efficiency may be written thus : the form proposed in 3 above. * On p. 28 of this appendix. 4 NOTES OK THE GRAPHICAL STATICS OF MECHANISM. 6. Condition of being Self-Locking. If the efficiency is less than 0.50 and the lost work of friction be assumed to be the same in forward as in backward motion (in same range of motion, of course), the mechanism is self-locking. Proof. For forward motion (no acceleration), letting 2(Fs") denote the sum of the amounts of work lost in friction at the various points of rubbing, we have, for a definite range of motion, (see 142,) Ps = Qs f + 2(1*8"), (2) and for backward motion (same range), similarly, Q being a a working force and P' a resistance, (see 142,) Qs f = P's + 2(Fa") (3) As implied in the notation, assume that the friction-work is the same in backward as in forward motion of the mechanism. From (1) and (2) we deduce l' = Qs' + 2(Fs") ; or, 2(W) = Qs' g - l]. . . (4) Substituting from (4) in (3) we obtain, finally, P* = ^'[2.00 - -1 (5) Si T}_] From (5) it is evident that, when the efficiency is less than 0.50, P' is negative ; that is, the mechanism is self-locking. The assumption made above as to equality of lost work in forward and backward motion is not exactly true for any machine, per- haps. In most mechanisms the friction work is somewhat less in backward motion, and the proposition is then true if for 0.50 we put a smaller value for r/. Machines of peculiar design may be constructed with rj greater than 0.50, which nevertheless are self -locking ; but in these we find that the lost work is greater in backward than in forward motion. 7. Sliding Friction. By 156 we know that if two rough sur- faces slide on each other the mutual action or force between them is not normal to the plane of contact, but inclined to the 5 NOTES ON THE GRAPHICAL STATICS OF MECHANISM. normal at an angle 0, the " angle of friction," on that side of the normal opposed to the direction (of relative motion) of the body under consideration. Thus, Fig. 1, Plate I, if the block A is sliding toward the right relatively to B (no matter which one, if either, is absolutely at rest) the mutual force between them acts in the line o . . . b ; if toward the left, then it acts in the line o' . . . V. The pressure of B upon A is from m toward o (or o')\ that of A upon B from m toward b (or &'). 8. Example! Mill Elevator. (Plate I, Fig. 2.) The single rigid body Ac consists of a platform and vertical side-piece, which rubs at &, the left side of the fixed vertical guide 6 y , and also at #, the right side of the same. Let Q be the combined weight of the load on the platform and the platform itself, and P the required pull or tension to be applied vertically at A, to maintain a uniform vertical upward motion (forward motion, here). Besides the forces P and Q, the body Ac, considered free, is acted on by the forces J? x and 7? 2 at the rubbing surfaces a and , acting at the proper incli- nation (0) from tneir respective normals ; note which side. Evidently A . . c is a four-force piece. Four action-lines are known, but only one amount, that of Q. To find the amounts of P, jft t , and R we join d (intersection, or "co-point" of P and J2,) with c that of R^ and Q. On the right of the figure we begin the force-polygon by laying off m . . o parallel and equal to Q (by scale). After drawing through m a line parallel to 7? 2 , and through o a line parallel to c . . d ( the known action-line of the resultant of Q and 7? 2 ), by the intersection k we determine R^ = in . . k, and the resultant, o. . k, of Q and 7? a . But this resultant should balance P and 7?,, and therefore close a triangle with them in the force-polygon ; hence parallels to P and ^, . . through Ic and o respectively, finally complete the force quadrilateral m . . n, and fix the values of P and R^ which can now be scaled off. In this figure, for clearness, a large value, about 20, has been given to 0, so that/*, tan 0,= about 0.36 ; and from the force- polygon we find that for Q = 110 Ibs., P is about 140 Ibs. If there were no friction R l and R^ would be horizontal, and a 10 6 NOTES ON THE GRAPHICAL STATICS OF MECHANISM. force P =Q = 110 Ibs. would be sufficient, vertical and upward at Ay to maintain a uniform upward motion (or to permit a uni- form downward motion, once started). Hence the efficiency for upward (forward) motion, with friction, is r) = -p = - = 0.78 or about 78 per cent. In uniform backward motion with friction, R l and 7? 8 (now call them R{ and Rj ) will change their action-lines to the other sides of their normals ; i.e., they will now act along a. .d and b . . c', respectively, at an angle with the normals. Hence the line G' . . d\ in backward motion, takes the place of c . . d, in getting the new force-polygon (see on left in Fig. 2); i.e., o'. . k' is drawn parallel to G'. . d', and m'. . k r parallel to G'. , b, Q being known. Then a vertical through k f and a parallel to d'. . a through o' complete the polygon; which fixes .P'as well as R{ and RJ. For Q = 110 Ibs., P' is roughly about 80 Ibs. Since P' is upward (i.e., not negative, being in the same direction as P) the machine is not self-locking. To alter the design, however, so that it shall be self-locking, we need only place the rubbing point a near enough to b to cause a..d'to pass through, or below, G'\ for then o'. . Tc! will either coincide with o' '. . n' or pass below it, thus making P' either (zero) or negative. But .the pressures R^ and R^ will be enormously increased. Also, P for upward motion will be larger, and the efficiency smaller than before. 9. Example II. Wedge, (Plate I, Fig. 3.) Required the neces- sary horizontal force P, at the head of the wedge AB, to raise the load Q and overcome the friction at the three points of rubbing, a, c, and e. For the upward motion of the block E (wedge mov- ing to the right) the lines of pressure at these points of rubbing are a ..b, c. .e, and c..d; there being no pressure at E. The block Ee is a three-force piece, under the action of the known Q and the unknown R^ and R l (i.e., the R l which acts from G toward d) all three action-lines being given. Hence the force- triangle is immediately drawn, viz., m . . n . . 0, and the amounts of J?? 2 and R l become known by scale. The wedge AB is also a three- force piece, acted on by the R l 7 NOTES ON THE GRAPHICAL STATICS OF MECHANISM. (now found) pointing from c toward Z>, and the unknown 7? 3 and the required working-force P. Its force triangle n . . o . . k is then drawn, since we have just found 7?, and have only to make o . . k horizontal ( i.e., parallel to P) and n . . ~k parallel to a . . 5, to close the triangle and thus determine P = o . . k ; (for- ward motion with friction). With no friction, J? 3 is vertical, 7? 2 horizontal, and K l follows the normal to the plane A . . B ; and the corresponding force- triangles, beginning with the known Q, are m . . n 9 . . o and n . . o . . k ; and thus P is found. From the drawing we have, with Q = 102 Ibs., P = 98 Ibs. 30 and P = about 30 Ibs.; so that the efficiency i? = ^ ~ 0. 31. i/O The angle used is about 20 or/ = 0. 36. For backward motion, 7?/ would act parallel to c . . d', R^ parallel to e . . c' ', and l? s f parallel to a . . #'; and the corre- sponding force-triangles are m '. . n'. . o' and n'. . o'. . k f , the result- ing value of P' being negative. That is, P' must act from right to left, for backward motion. Hence the mechanism is self- locking (for these particular data, in Fig. 3). Note. It can easily be shown that if the angle of the wedge (i.e., the angle between its two sides) is less than twice the angle of friction (supposed the same for all three rubbing contacts) the mechanism is self-locking. (This is best shown graphically.) 10. Example III. The Jack-Screw. ( Plate I, Fig. 4.) Kequired the value of each force P of a couple, in a horizontal plane, applied to the cross-bar a J , in order to raise the load Q and overcome the friction between the surfaces of the square-threaded screw and nut. The screw-shaft is vertical. Assume no pressure at the edges of the threads. The pressures on the helical surfaces may be considered concentrated at two points d l and d^ diametri- cally opposite, in the middle of the width of the thread. The pressure at d^ viz. J?,, will lie in a vertical plane ~| to a . . b , and make an angle -f- OL (on the left) with the vertical, while R^ on the other side makes an equal angle with the vertical (but on the NOTES ON THE GRAPHICAL STATICS OF MECHANISM. right), a is the angle which the helix of screw surface makes with the horizontal. Projecting the five forces on a plane ~| to the bar a- . . & , we find that Q must balance the vertical components of R l and 7? 2 , which justifies the force-triangle drawn at [#], where, with Q given, we easily fix R^ and 7? 2 by drawing w . . m and k. . m at an angle of -f~ a w ^h Qi as shown. Therefore the horizontal component of J^ l isp . . m, that of 7? 2 is m . .p. Projecting all the five forces on a horizontal plane, we note that the two unknown P's must balance the couple formed by the horizontal components of R l and 7? a , as if these components (which we will now call R z and 7? 4 ) were applied directly to the bar a ..b at points vertically above d t and d^ Hence (see \_C~\ in Fig. 4) we treat the known E z and 7? 4 (each = m . .p) as par- allel forces applied perpendicularly to a straight beam a . . b supported at a and 5 on || smooth surfaces whose normals are ~| to a. .l> and || to R z and 7? 4 . The reactions of these supports will be || to 7? 3 and 7? 4 and are the forces P and P required. At \_G] in Fig. 4, d l . . d^ = the distance d t . . d^ of [yl], and a. .b = distance a . . b . Hence, using the construction of 329, we make x . . y = 7? 3 , y . . x = 7? 4 , select a pole O at convenience, draw the three rays O..x, O..y^ and 0.. x, and a corresponding equilibrium poly- gon u . . t . . s . . r, beginning at any point u in the action-line u . . b of the lower P. Join u . . r, the abutment-line, and draw a line || to it through the pole to fix n'\ then x . . n' = P re- quired. With no friction, = 0, and 7? 3 and 7? 4 are each equal to m Q . .p, instead of m..p; and P is proportionally smaller than P. In this figure P is about two thirds of P\ i.e., the efficien- cy is about 0.66. For backward motion, R^ and R^ must act on the other sides of their respective normals; i.e., in \_H\ we should use a instead of a -f- 0, and if in that case a were less than 0, the point m would fall on the right of p, and P' would be negative; i.e., the screw would not run backward (would not " overhaul ") when there was no force on the cross-bar ; ( self-locking). NOTES ON THE GKAPHICAL STATICS OF MECHANISM. 11. Pivot Friction. (Plate II, Fig. 5.) Since the frictions at the base of a flat-ended pivot (see 168) are equivalent to a couple in which eacli force is \fR, or \fQ with present notation, and whose arm is f of the radius r (so that the moment of the couple is f fQ^t we may suppose the pressure concentrated at two opposite points, in the circumference having a radius = f r, as< in Fig. 5. The pressures at these points are inclined at an angle =

friction-circle on its under side. Drawing, then, these tangents || to the respective known di- rections of P and Q, we have A . . a and c . . a as their action-lines, meeting at #, through which (since the bell-crank is a three-force piece) the third force R must pass (the reaction at (7), and this must be tangent to the friction-circle at C on its upper side.' Hence a line through a and tangent to the friction-circle at C on upper side, is the action line of It, viz., a . . b. The force-trian- gle ~k . . m . . n is now easily drawn, Q being the known force and laid off first, and P = m . . n is thus determined. With no friction, P, arid Q (directions unchanged) must pass through the centres of the pins at A and B, and intersect at a . R then passes through a and the centre of C, i.e., acts along ajb Q . P = mn Q in the new force-triangle, and the efficiency = m..n -=- m..n. For backward motion we make the action-lines tangent to their re- spective friction-circles on the other side in each case ; then P' = m. .n'. 14, Example V. The Slider-crank. (Plate II, Fig. Y.) The wheel TFand the crank B form a single rigid body turning on a fixed bearing C. The connecting-rod or link, BD, is pivoted about the crank-pin at one end and to the cross-head pin at the other. The cross-head block slides in a right line between guides, and receives the pull (or push) of the piston-rod in the axial line of that rod, i.e., through the centre of the pin at D. The resist- ance Q being given, applied in line a..~b to wheel W, required the necessary value of P for uniform motion m ike given position of the mechanism. The connecting-rod is a two-force piece, and in its present position (B on the right of a vertical through C) the pressure at the cross-head pin is tangent to the friction-circle there on its upper side ; that at the crank-pin on the lower side of the friction -circle there. Hence a line drawn so as to be tan- gent to the two circles in the manner stated is the action-line of R^ the crank-pin pressure (as also that at the cross-head pin); i.e., draw e . . h. The three-force piece, WCB^ is acted on by the known Q, by RV and a pressure at the bearing (7, viz., 7? 2 , which is tangent to 11 NOTES ON THE GRAPHICAL STATICS OF MECHANISM. the friction-circle at C on the upper side ; therefore 7? 2 must act in the line b . ./"drawn from the intersection b, of Q and R l tan- gent to friction-circle at C on upper side. Hence the force-tri- angle for WCB is easily completed by making o..m||and = Q, and drawing m . . n and o . . n \\ to c . . h and f. . b, respectively, thus determining It l and R^. Now the cross-head block, Dk, is a three-force piece acted on by R l (pointing toward the left), now known ; by the unknown P and by the unknown R, which is the pressure coming from the upper guide, and must act in a line d . . e through the intersection of the action-lines of P and R^ viz., e, and makes angle = (in direction as shown) with the normal to the guide surface. Its force-triangle, then, is rks, k . . r being = and opposite to m .. n, s..r being drawn || to P and k..s\\to d. .e. Thus P is found, being = rs. As for P (i.e., with no friction), R^ would act through the pin centres, thus raising b ; R z would act through ~b and the centre of C\ while J? 3 would be normal to the guide surface. With new force-triangles, on this basis, we find P = r . . s ; whence the ef- ficiency, = P -f- P, is found. In any other position of the mechanism a similar method is available. 15, Example VI. Beam-engine with Evans's straight-line motion. (Plate II, Fig. 8.) WKM is a single rigid body (wheel and shaft) turning on a fixed bearing at K. M is the crank-pin, MB the connecting-rod, DC & link turning in a fixed bearing at D, and pivoted at C to the beam ACE, one end of which, E, is guided in a horizontal right line by the block 7^ and straight guides. The pin C being mid-way (in a straight line) between E and A, and the length CD being made equal to AC, which also = CE (between centres; D and E at the same level), A will be compelled to move in a vertical straight line, as if it were itself guided by a straight edge. The vertical piston-rod is linked to the beam at A. Required the necessary steam pressure P on the piston-head, for upward -motion, the resistance Q being applied to the wheel W in the line x . . w, and the parts having the position shown in the figure. The link DC is a two-force piece, subjected at this instant to 12 NOTES ON THE GRAPHICAL STATICS OF MECHANISM. some thrust, R^ whose action-line must be tangent (below) to the friction-circle at C, and also (above) to that at D. Similarly the link or connecting-rod BM is a two-force piece, under a tension 7?,, whose action-line is tangent to the friction-circle both at M and B. At M. thistangency is on the right ; at B it may be either on the right or the left according as the link has not yet reached, or has passed, a certain critical position which is very nearly the position chosen for the figure ; in which the tan- gency has been drawn on the left at B. The block J^is a two-force piece, under a thrust 7? 4 , directed, as shown, at an angle with the normal to the guide-surface below it. Construction. We first draw the force-polygon for TPjOf, a three-force piece, the forces being the known Q in line w. .a?, the unknown JS l in the known line c. . A, and the unknown reaction R^ at the bearing K. R z must act in a line x . . y through the point a?, and tangent, on left, to the friction-circle at I. Hence by laying off o . . m = and || to Q, and then drawing m . . n \\ to h . . c and o . . n \\ to x . . y, we determine R l and J? 2 . The beam ABCE\ a four-force piece under the forces 7? 1? 7? 3 , 7? 4 , and RV their action-lines being all known and R l now known in amount, = m . .n. The direction of JR 3 becomes known from a consideration of the piston and rod which together form a three-force piece, being acted on by jP, by the stuffing-box pres- sure 7? 6 , and by 7? 3 reversed. Since P and R 6 intersect at #, and since JK 3 must pass through a and be tangent on the right to the friction-circle at A, its action-line is easily drawn. Resuming the body ABCE, we find the intersection, e, of R z and R l (N.B. e is off the paper) and join e with/*, the intersection of 7? 4 and 7? 6 . Making n 1 ..m l = and || to R^ (now pointing down), and draw- ing n l . . r || to f . . e and m, . . r || to R 3 ; also r . . u \\ to 7? 4 and ft,, . . u || to RS : we finally determine 7? 3 , R^ and R^ Having found 7? 3 , the force-triangle for the three-force piece A . . a (piston, etc.) is easily drawn (see [Z] in Fig. 8) and P finally determined. With no friction we find P by taking friction-circle centres 13 NOTES ON THE GRAPHICAL STATICS OF MECHANISM. instead of tangencies and making R 6 and R t ~] to their respective sliding surfaces. 16. Example VII. Oscillating Engine. (Plate III, Fig. 9.) Here the connecting-rod is dispensed with, the piston acting directly on the crank-pin A, while the cylinder EE oscillates on trunnions turning in fixed bearings. The crank AB and wheel W constitute a single rigid body turning in a fixed bearing B. The resistance Q being applied to W in line d . . c, we wish to find the proper steam pressure P on the left of the piston to overcome Q and all intervening frictions, for motion (with in- sensible acceleration), of the mechanism in the position shown. P acts centrally along the axis of the piston-rod. The pressure R is that of the stuffing-box against the piston-rod, and R z that of the cylinder against the edge of the piston. Remembering that the piston and cylinder always have a common axis as they " tele- scope" in and out, we see that there are only two forces external to these two pieces when considered together, viz., the pressures at the crank-pin and at the trunnion, which two pressures (J? 2 ) must therefore be equal and opposite, both of them acting in the line h . . g tangent to the two friction-circles (on the lower side at both A and 0). We have thus found the action-line of 7? 2 . Now consider the three-force piece WBA. The forces are the known Q in line d. . c 9 the unknown 7? 2 in the line h . . g, and the reaction or pressure R \ in a line a . . e which we draw through 0, the intersection of R and (), and make tangent (on the upper side) to the friction-circle at the bearing B. Hence, making m . . n = and i| to Q, and m . . r and n . . r || to e . . h and e..a 9 respect- ively, we fix the amounts of R^ and R^ by this force-triangle m . . n . . r. * The piston is a four-force piece, under the action of 7? 2 , now known, acting toward the left in line h . . e ; by the unknown P acting along the centre line of the piston-rod ; and by the two unknown reactions, R z and R^ inclined at angle (p to their re- spective normals, and acting at known points. Hence join y, the intersection of 7? 2 and JR 3 , with a?, that of P and R^ make o , . p = and || to RV draw^? . . s \\ to R^ and o . . s \\ to y . . a?, to deter- IT 14 NOTES ON THE GRAPHICAL STATICS OF MECHANISM. mine s ; s . . t \\ to P and o . . t \\ to R^ are then drawn to fix t. P can now be scaled off, as also R z and R^. With no friction, R^ would act in the axis of the piston through the centres of A and C. In place of e we would have e (not shown in figure), and R would act through e and the centre of B. J? 3 and R would be ~] to their respective rubbing surfaces. We would then obtain R^ = m . . r and P = R z = m. . r ; whence the efficiency is = P -r- P = m . . r -f- s . , t. 16. Example VII. The Blake Ore-crusher. (Plate III, Fig. 10.) JTand L are fixed walls, HD oscillates about a fixed bearing _Z> 1? WFa is a wheel and crank rotating on a journal in a fixed bear- ing F. The connecting-rod a..c communicates motion to the two links or struts, DE and CB^ forming a toggle-joint and causing IfD 1 to oscillate. For the given position of the parts, Q being the resistance offered by a piece of ore, A, to any further motion of H toward the left, required the value of P, applied in the line W. . b to wheel WF to overcome Q and the pressures at the seven sockets or bearings. The two links Z>J^and CB are evidently two-force pieces, DE being subjected to some thrust R z in line h . . c, CB to some thrust R z in line c . . B ; these lines being drawn tangent to the various friction-circles in the manner shown. R z and Q intersect in A, hence the line f . . A, drawn through h and tangent to the friction-circle at Z\, must be the action-line of RV the reaction at the bearing D r That is, Q, R^ and 7? 2 , are the forces acting on the three-force piece IIJ) l ; therefore, Q and the three action-lines being known, we easily complete the corresponding force-triangle o . . m . . n and thus determine R, and R z . Passing to the three-force piece a . . b we have the action-lines of 7? 2 and 7? 3 already intersecting at c, and the amount of 7? 2 = o . . n. The third force 7? 4 has an action-line passing through c and tangent (on left) to the friction-circle at the crank-pin. Hence draw c. . a accordingly, and complete the force-triangle by making ~k . . i \\ and = to o . . n, (i.e., to 7? 2 ), and i . . r and 15 NOTES ON THE GRAPHICAL STATICS OF MECHANISM. Jc. .r || to a. . c and B . . c respectively ; which determines R^ and J? 4 . Finally, the three-force piece a . . F. . TFis seen to be acted on by RV now known ; by the unknown P in line P . . & ; and by the bearing-reaction R acting through b and tangent (on right) to the friction-circle at F. Draw b . . F^ then, as indicated, and the force-triangle is readily formed, t . . s . . u, in an obvious manner, whence s . . u, = P, is scaled. Without friction we would as usual draw the action-lines of the divers jft's through the centres of the various journals, or sockets, instead of tangent to their friction-circles. With a new set of force-triangles on this basis (see broken lines on the right in Fig. 10), we obtain jP , considerably smaller than P. In this exaggerated figure the efficiency is only about 0.50 ; but in Prof. Hermann's drawing, withy 7 for journal friction = 0.10, he ob- tains a value of 0.80 for the efficiency. 17. Rolling Friction (so called). This kind of resistance is due to the fact that the point of application of the force acting be- tween a wheel and the rail or surface on which it rolls is not at the foot of the perpendicular dropped from the centre of the wheel upon the rail, but a little in front (in direction of rolling) by an amount, or distance, ~b = about 0.02 in. for iron wheels on an iron rail, and about 0.50 in. for a wagon-wheel on a dry macadamized road (b = from 2."00 to 3/'00 on soft ground). See 172; (also Prof. Reynoids's article in the Philos. Transac., vol. 166.) As to the direction of the pressure of rail on wheel it is somewhere within the cone of friction so long as perfect rolling (i.e., no slipping) proceeds, being the equal and opposite of the resultant of all the other forces acting on the wheel. Thus, Fig. 11, Plate III, if Q is the weight of the roller, applied in the centre o', the necessary force P, horizontal and acting through the centre of the roller (to maintain a uniform rolling motion), is determined by the fact that the resultant of P and Q must act through c, and therefore along the line o' '. . c, where c . . d = the distance 1) just mentioned. Hence, making the line Q through o equal by scale in length and || to Q at o', 16 NOTES ON THE GRAPHICAL STATICS OF MECHANISM. and drawing o . . m \\ to o' . . c, as well as a horizontal through the lower extremity of Q, we determine both P and the rail pressure. Again, Fig. 12, if the rolling action occurs on both sides of the roller, as at c and e when a weighted plank is moved horizon- tally (to left, here) on loose rollers, we note that it is a two-force piece (neglecting the weight of the roller), the action-line of the compressive forces being e . . c, both e and c having been located at distance b from the perpendicular through the centre (in the proper directions ; and then, we may have different Z's at the two points of rolling contact). Here the upper plank is moving and the horizontal force (toward left) which must be applied to it to maintain this motion uniformly must be equal to the sum of the horizontal components of the various inclined 72's, one from each roller, the sum of the vertical components of the latter being equal to the total load on the plank. In the uniform motion of a car-wheel, EA, Fig. 13 (brakes not on), the (double) wheel and its axle constitute a single, rigid, two-force piece, acted on by the rail pressure, or reaction, at c (c . . d being made 5) and by the pressure of the bearing against the journal or axle at R\ and this pressure must be tangent to the friction-circle at I? (ou the right for motion here shown, car moving to the left). The horizontal component of the equal and opposite of this R is the tractive resistance, on a straight level track with uniform motion (besides the resistance of the air), and is continually overcome by the tension in the draw-bar of the locomotive. R is practically equal to its own vertical component which equals the portion of the car's weight borne by this wheel, and this horizontal component of R is = G tan 6, where 6 denotes the complement of the angle e . . c . . d. If the brake is in action, Fig. D* exerting a pressure P' at the point 0, this pressure must act along the line a . . o at an angle =

" with the vertical (track level), on the right, here. Since P' and P" intersect at o, the only remaining force acting on the rigid body JEW (which now is a three-force piece) must pass through o. This force is P" ', the pressure of the bearing on the journal, and it must also be tangent to the friction-circle, at E. Its action-line, therefore, is easily drawn and is e . . o. A corresponding force-triangle being now constructed (not shown in the figure) with sides || to . . 0, a . . 0, and e . . o, respectively, in which the vertical projection of P" is made 6r, the portion of weight of car coming on this wheel (double), we determine the value of each force. Since ) and also w' . . o f = r (a +5). Draw a vertical through each of the points o. w, and the ratio of the given Q to the unknown P' thus fixed ; and P' would be 7J/ O found from the relation : --= l Q 21. Example X. The Differential Pulley. (Plate Y, Fig. 17.) This is a tackle in which the upper block carries only one pulley, which, however, has two grooves in || planes, but with slightly different radii. Also, since friction in the grooves is not sufficient for the purpose, projecting pegs or ridges (or some similar device) are provided to prevent the chain from slipping in either groove. The lower block D carries an ordinary pulley of one groove, the weight Q being suspended from the axle of this lower block. The chain is endless, passing twice over the pulley A (once in each groove) and once under the lower pulley, while a portion hangs freely, as shown. Given the load Q, required the vertical force (or load perhaps) P, to be applied to the chain where it unwinds from the outside groove (see figure) in order to raise Q and overcome all friction. The lower pulley is then acted on by the vertical force Q, and the two vertical and upward tensions S 1 and $, each of these three forces being tangent to its own friction-circle, as shown, on the proper side (note that $i is on the unwinding side). We find S l and S 9 by treating Q as a load ( 329) resting on a horizontal beam supported in verticals a and n. At \B\ is the force-diagram, while a . . b . . n is the equilibrium polygon. (See 329.) /Si and S 9 having thus been determined, we consider the upper pulley, which is acted on by four || forces, viz., the known $, and $ 2 , the unknown .Pand the unknown R or reaction at the journal of the pulley, The four action-lines are known, being vertical and tangent to the respective friction-circles in the manner shown. Note the direction of the uniform motion (to raise the load Q). Again we employ 329, regarding A as a horizontal beam or lever with supports in the verticals,/ and c, and loaded with S t 12 22 NOTES ON THE GRAPHICAL STATICS OF MECHANISM. and S M both known. We lay off r . . s = S a and s . . t = S and take any pole 0, drawing rays from O to r, s, and t. From any point G in the action-line of R (the left-hand reaction, or support- ing force) we draw a line || to . . rto find d, then^Z . .e\\toO..s to find e, and a line parallel to O . . t through e, to find f in the action-line of the Tight-hand supporting force, P. Drawing c . .f, a line parallel to it through fixes n' on the load-line (produced), giving in' P, and n'w = -Z? ; i.e., \A\ is the force-diagram. Without friction, the vertical action-lines would be drawn through the centres of the friction-circles, and anew construction on this basis would give P , whence the efficiency P ~ P can be found. For backward motion each force-vertical shifts over to the opposite side of the friction-circle from that shown in Fig. 17, and the result of a third construction is P' . If P' is found to be negative, that is, if n occurs above t in diagram A, the mechanism is self-locking^ as should be the case in the practical machine itself. 22. Rigidity of Heinp Eopes. Here, as with chains, the effect of the rigidity is to cause the tension where the rope is winding on to have a lever-arm about the centre of the pulley r -f- a, where r = radius of circle formed by the axis of the rope when wound on the pulley, and a = a small distance which from Eytelwein's formula for rigidity of hemp ropes may be put 0.0093<$ 2 , where d is the diameter of the rope in millimetres, and whence a will be obtained in millimetres. The tension on the unwinding side has a lever-arm of r a. Hence, having computed a, we deal with hemp ropes as with chains. The phenomena observed with wire ropes are different. (See 176.) 23. Tooth Friction in Spur Gearing. (Plate Y, Fig. 18.) This figure shows one gear-wheel driving another, both provided with " involute teeth " by which we are to understand that the normal a^..o, or o . . a at the point of contact always passes through o, the intersection of the line of centres with the pitch- circle, as motion proceeds. We assume here that two pairs of teeth are always in contact. 23 KOTES ON THE GRAPHICAL STATICS OF MECHANISM. Just now these points of contact are at a v and a,, and have there- fore a common normal a,0 2 . Rubbing occurs both at 0;, and & 2 , and evidently in such direc- tions that the pressure at a has a^ X as action line ; and that at a has 2 & 2 as action-line, making the angle of friction with the respective normals (or common normal, rather). This common, normal a^oa^ will be assumed as making an angle of 75 with the line of centres at all times (property of the kind of teeth used). Hence the resultant action of the two driving teeth upon those driven is represented by an ideal force R, the resultant of the pressures at a l and a a , and acting through 0', making an angle of 75 with the line of centres. Notice the position of this angle with reference to the direction of motion and to the driven wheel ; also that the effect of friction is to cause the action-line of J?, which without friction would act along a^oa v to be shifted || to itself a distance oo f farther from the centre of the driving wheel. This distance 00', can ensily be determined by drawing the parts concerned on a convenient scale, and will be called C in the next paragraph. 24. Example XI. Pinion Spur-Wheel, Drum and Weight. (Plate Y, Fig. 19.) The weight Q hangs by a chain or rope from the drum B which forms a rigid body with the spur-wheel //, with which the pinion A gears. A drives //, and it is required to find what force P, applied to a crank d (forming one piece with the pinion) and acting (at this instant) in the line ~b . . 03, will main- tain uniform motion ; i.e., overcome all frictions and raise Q without acceleration. Since A drives ZTwith tooth-gearing (involute and of same design as in preceding paragraph), the line of action of the result- ant pressure 7? 2 between them is 00'0:, making an angle of 75 with the line of centres, as shown (note on which side), and is drawn through the point o' on the line of centres but at a distance = C farther from the centre of the driving pinion A than a point in the pitch-circle of the latter. Call this force ^? 2 . The reaction at the bearing s is some force R l whose action- line must pass through S, the intersection of the action-lines of the other two forces, 7? 2 and P (since A is a three-force piece), 24 NOTES ON THE GRAPHICAL STATICS OF MECHANISM, and be tangent (on the right) to the friction-circle at s. The ac- tion-line of Q is vertical, and is tangent (on the left) to -a friction- circle at a ( just as in Fig. 15 a similar relation holds at K) ; it cuts g . . 1} at <7, and therefore a line drawn through g, and tangent (on right) to the friction circle at K^ is the action-line of R^ the reaction at the bearing K. We thus have the action-lines of all three forces acting on' each of the three-force pieces, A and H, while the force Q is given. Hence, the force-triangle r..m..n is easily drawn for piece H, and determines 7? 2 and R y With n"m" = and || to m . . n as a known side, we then complete the force-triangle for piece A 9 from which m" . . r" = P is scaled off. Without friction, j?? 2 would shift to the position g . . J , R l would pass through & and the centre of the circle at s. R % would pass through the centre of the circle at B and the point gr , in the new vertical action-line of Q (through centre of friction -circle mentioned above). g . . t> is parallel to g . . b and passes through the intersection o of the line of centres c . . s, and the pitch-circle of the pinion A. Drawing the dotted force-triangles on this basis, Q being given, we finally obtain P = m" . . r ". The efficiency can now be obtained, =P -r- P. 25. Belt Gearing. In Fig. 20, Plate VI, we have a pulley turning in a fixed bearing and driven by a force P. By belt connection this pulley drives another, not shown in the figure. The tension S w on the driving side is greater than that, $ , on the following side. If Z is the (ideal) resultant of S n and 8 0) then the reaction R of the bearing must act in a line through #, the intersection of P and Z and tangent (above) to the friction-circle at the bearing ; i.e., it acts along a. . &.* If we assume that the belt is on the point of slipping on the smaller of the two pulleys, we have the relation S n = 8.e ( 170) where/" = coefficient of friction, e is the Naperian Base, and a = arc of contact on the smaller pulley in n measure, or in radians. Although S n and $ are both unknown at the outset, we have * By mistake, R has been drawn in Fig. 20 along a' . . &, instead of a . . b. 25 NOTES ON THE GRAPHICAL STATICS OF MECHANISM. their ratio from the above equation, and hence can construct the action-line of Z(iov impending slip only, it must be remembered), their resultant, thus determining the point b in Fig. 20 and ultimately R and Z, as will be seen. With Z found, we can obtain 8 n and S . The value of this ratio, e fa , having been computed for a range of values off and of a, the results may be. embodied graphically in the spirals shown in Fig. 21, Plate VI, these being drawn in such a way, all starting from the point A in the circumference of the circle A . . (7, that if OA, the radius, represent the smaller tension, S Q , and the special value a AOO in any case be laid off and the radius OG produced till it inter- sects the spiral corresponding to the coefficient f proper to the case in hand, B being this intersection ; then OB = S n9 and GB S n S (which multiplied by the velocity of the belt gives the power transmitted). Or, whatever & may be, the ratio BO : AO = the ratio S n : , and may be obtained from the diagram if f and a are given. [N.B. Note carefully that o the relation -~- = e fa only holds when the belt is actually slipping on the pulley -rim (and then/* is the coefficient of fric- tion of motion) or is on the point of slipping (and theny = co- efficient of friction of rest), and is never to be used except for those conditions. Of course, in most machinery impending slip is to be avoided, and the only use of the above formula in such cases is to find the ideal maximum value, e fa , for the ratio &n ' > which the actual value should not approach if slipping is not to occur. For the uniform motion of an "idle pulley,"' ignoring axle friction, S n : S is always equal to 1.00. 26. Example XII. Brake Strap and Drum. (Plate VI, Fig. 22.) A is a lever with a fixed fulcrum or bearing at B and has attached to it both ends of the belt or strap which passes over a pulley and serves as a brake to prevent the acceleration of the descending weight Q. The chain sustaining Q unwinds from the drum (7, rigidly attached to the pulley, which turns on a fixed bearing B. Required the proper force P, in a given action-line r . . P, to be applied to the lever A, to preserve a uniform motion for Q (downward). 26 NOTES ON THE GKAPHICAL STATICS OF MECHANISM. Evidently, from the direction of motion, the tension in the strap at d is the greater, = $, and that in the portion x . . e is the smaller, = S . Since in this case there is actual slipping of the drum under the strap, the ratio S n : ol) at convenience. This gives o . .i as the action-line of Z, the (ideal) resultant of S n and $ ; of course, that there may be no slip or any approach to it, actual values must be secured for these tensions greater than those to be found by this construction, which are for impending slip c/n> small pulley (and this means the assumption of a less value /or the ratio S n : & . (See p. 186.) A is a three-force piece (so considered here) under the action of Z (ideal), P 9 and J?, the bearing reaction. Pulley B may also be treated as a three-force piece under action of Q, of Z (reversed) and R, the bearing reaction at B. P and Z intersect at Z>, Q and Z at q. Hence R^ acts through ~b and tangent (above) to its friction-circle ; while R acts through q and is tangent (on right) to friction -circle at B. Beginning, then, with pulley B, since the force Q is given, we close the triangle v . . w . . x in an obvious manner, obtaining R and Z. For pulley A, now that Z is found, we complete the force- triangle y . . z . . d, and determine R^ and P. Without friction at the bearings^ R and R l would pass through the centres of their friction-circles and the dotted force-triangles O would result, whence we have P = z . . d f . To find the belt tensions (for impending slip on smaller pul- ley) we resolve the force Z\\ to their directions; see lower part of the figure. Note. As far as finding the value of P alone is concerned, having Q given, any line whatever could be taken through the point o, as the action -line of the resultant of the two tensions, if the friction at the bearings were disregarded, and the construc- tion would result in the same value of jP, whatever the belt- tensions, provided the belt did not slip. 28. Final Remark. From an inspection of the preceding ex^ amples involving the effect of friction in the working of machines, it becomes apparent, as should be expected, of course, that in every case this effect is to put the working force at the greatest possible disadvantage, thus exacting as large a value as possible NOTES ON THE GKAPHICAL STATICS OF MECHANISM. 28 for it; and from this general principle we may often decide quickly in the matter of tangencies to friction-circles, inclination of a pressure on one side or the other from the normal, etc. Fig. A. pole Fig. D, Am.Dk. Xote Co. N. Y. CONTENTS. PAGE Assumptions 1 Efficiency 2, 3 " Overhauling" 3, 4 Sliding Friction 4 Mill Elevator 5 Wedge 6 Jack-screw 7 Pivot and Journal Friction 9 Bell-crank...? 9 Slider-crank 10 Beam-engine 11 Oscillating Engine 13 Ore-crusher 14 Rolling Friction 15 Crane (rollers of) 17 Chain Friction 18 Tackle 19 Differential Pulley 21 Rigidity of Ropes 22 Spur Gearing 22 Belt Gearing 24 Brake-strap and Drum 25 PLATE I. Figures 1 to 4 . Am. Ek. Note Co. N. T. ' ' ' ' '< J ' 'r ' 1 ' ' PLATE II. Figures 5 to 8 Fig, 8. THE "EVANS" STRAIGHT-LINE MOTION. PLATE m. Figures 9 to 14. Fig. 9. OSCILLATING ENGINE, WLU FRICTION. _ Horizontal Section itf $w\nqing Qrane and Fixed Mast, about to/itc/V it it^nis.'., yhis sectwrj, is at the base Yin the Korteontdl pRinz XY. J The pivot frictio^i al Tisnqt considered. PLATE IV. Figures 15 and 16. Am. Bk. Note Co. N. Y. PLATE V. Figures 17 to 19. Fig. 17. P DIFFERENTIAL.PULLEY PLATE VI. Figures.20 to 23, Fig,2l, BELT-FRICTION SPIRAL Fig. 2O. BELT-FRICTION LOGARITHMS (BRIGGS*). N 01234 5 G 7 8 9 Dif. 1O 11 12 13 14 0000 0043 0086 0128 0170 0414 0453 0492 0531 0569 0792 0828 0864 0899 0934 1139 1173 1206 1239 1271 1461 1492 1523 1553 1584 0212 0253 0294 0334 0374 0607 0645 0682 0719 0755 0969 1004 1038 1072 1106 1303 1335 1367 1399 1430 1614 1644 1673 1703 1732 42 38 35 32 30 15 16 17 18 19 1761 1790 1818 1847 1875 2041 2068 2095 2122 2148 2304 2330 2355 2380 2405 2553 2577 2601 2625 2648 2788 2810 2833 2856 2878 1903 1931 1959 1987 2014 2175 2201 2227 2253 2279 2430 2455 2480 2504 2529 2672 2695 2718 2742 2765 2900 2923 2945 2967 2989 23 26 25 24 22 2O 21 22 23 24 3010 3032 3054 3075 3096 3222 3243 3263 3284 3304 3424 3444 3464 3483 3502 3617 3636 3655 3674 3692 3802 3820 3838 3856 3874 3118 3139 3160 3181 3201 3324 3345 3365 3385 3404 3522 3541 3560 3579 3598 3711 3729 3747 3766 3784 3892 3909 3927 3945 3962 21 20 19 19 18 25 26 27 28 29 3979 3997 4014 4031 4048 4150 4166 4183 4200 4216 4314 4330 4346 4362 4378 4472 4487 4502 4518 4533 4624 4639 4654 4669 4683 4065 4082 4099 4116 4133 4232 4249 4265 4281 4298 4393 4409 4425 4440 4456 4548 4564 4579 4594 4609 4698 4713 4728 4742 4757 17 16 16 15 15 3O 31 32 33 34 4771 4786 4800 4814 4829 4914 4928 4942 4955 4969 5051 5065 5079 5092 5105 5185 5198 5211 5224 5237 5315 5328 5340 5353 5366 4843 4857 4871 4886 4900 4983 4997 5011 5024 5038 5119 5132 5145 5159 5172 5250 5263 5276 5289 5302 5378 5391 5403 5416 5428 14 14 13 13 13 35 36 37 38 39 5441 5453 5465 5478 5490 5563 5575 5587 5599 5611 5682- 5694 5705 5717 5729 5798 5809 5821 5832 5843 5911 5922 5933 5944 5955 5502 5514 5527 5539 5551 5623 5635 5647 5658 5670 5740 5752 5763 5775 5786 5855 5866 5877 5888 5899 5966 5977 5988 5999 6010 12 12 12 11 11 40 41 42 43 44 6021 6031 6042 6053 6064 6128 6138 6149 6160 6170 6232 6243 6253 6263 6274 6335 6345 6355 6365 6375 6435 6444 6454 6464 6474 6075 6085 6096 6107 6117 6180 6191 6201 6212 6222 6284 6294 6304 6314 6325 6385 6395 6405 6415 6425 6484 6493 6503 6513 6522 11 10 10 10 10 45 46 47 48 49 6532 6542 6551 6561 6571 6628 6637 6646 6656 6665 6721 6730 6739 6749 6758 6812 6821 6830 6839 6848 6902 6911 6920 6928 6937 6580 6590 6599 6609 6618 6675 6684 6693 6702 6712 6767 6776 6785 6794 6803 6857 6866 6875 6884 6893 6946 6955 6964 6972 6981 10 9 9 9 9 5O 51 52 53 54 6990 6998 7007 7016 7024 7076 7084 7093 7101 7110 7160 7168 7177 7185 7193 7243 7251 7259 7267 7275 7324 7332 7340 7348 7356 7033 7042 7050 7059 7067 7118 7126 7135 7143 7152 7202 7210 7218 7226 7235 7284 7292 7300 7308 7316 7364 7372 7380 7388 7396 9 9 8 8 8 N. B. Naperian log = Briggs' log x 2.302. Base of Naperian system = e = 2.71828. LOGARITHMS (BRIGGS'). N 01234 56789 Dif. 55 56 57 58 59 7404 7412 7419 7427 7435 7482 7490 7497 7505 7513 7559 7566 7574 7582 7589 7634 7642 7649 7657 7664 7709 7716 7723 7731 7738 7443 7451 7459 7466 7474 7520 7528 7536 7543 7551 7597 7604 7612 7619 7627 7672 7679 7686 7694 7701 7745 7752 7760 7767 7774 8 8 8 7 7 6O 61 62 63 64 7782 7789 7796 7803 7810 7853 7860 7868 7875 7882 7924 7931 7938 7945 7952 7993 8000 8007 8014 8021 8062 8069 8075 8082 8089 7818 7825 7832 7839 7846 7889 7896 7903 7910 7917 7959 7966 7973 7980 7987 8028 8035 8041 8048 8055 8096 8102 8109 8116 8122 7 7 7 7 7 65 66 67 68 69 8129 8136 8142 8149 8156 8195 8202 8209 8215 8222 8261 8267 8274 8280 8287 8325 8331 8338 8344 8351 8388 8395 8401 8407 8414 8162 8169 8176 8182 8189 8228 8235 8241 8248 8254 8293 8299 8306 8312 8319 8357 8363 8370 8376 8382 8420 8426 8432 8439 8445 7 7 6 6 6 70 71 72 73 74 8451 8457 8463 8470 8476 8513 8519 8525 8531 8537 8573 8579 8585 8591 8597 8633 8639 8645 8651 8657 8692 8698 8704 8710 8716 8482 8488 8494 8500 8506 8543 8549 8555 8561 8567 8603 8609 8615 8621 8627 8663 8669 8675 8681 8686 8722 8727 8733 8739 8745 6 6 6 6 6 75 76 77 78 79 8751 8756 8762 8768 8774 8808 8814 8820 8825 8831 8865 8871 8876 8882 8887 8921 8927 8932 8938 8843 8976 8982 8987 8993 8998 8779 8785 8791 8797 8802 8837 8842 8848 8854 8859 8893 8899 8904 8910 8915 8949 8954 8960 8965 8971 9004 9009 9015 9020 9025 6 6 6 6 5 80 81 82 83 84 9031 9036 9042 9047 9053 9085 9090 9096 9101 9106 9138 9143 9149 9154 9159 9191 9196 9201 9206 9212 9243 9248 9253 9258 9263 9058 9063 9069 9074 9079 9112 9117 9122 9128 9133 9165 9170 9175 9180 9186 9217 9222 9227 9232 9238 9269 9274 9279 9284 9289 5 5 5 5 5 85 86 87 88 89 9294 9299 9304 9309 9315 9345 9350 9355 9360 9365 9395 9400 9405 9410 9415 9445 9450 9455 9460 9465 9494 9499 9504 9509 9513 9320 9325 9330 9335 9340 9370 9375 9380 9385 9390 9420 9425 9430 9435 9440 9469 9474 9479 9484 9489 9518 9523 9528 9533 9538 5 5 5 5 5 9O 91 92 93 94 9542 9547 9552 9557 9562 9590 9595 9600 9605 9609 9638 9643 9647 9652 9657 9685 9689 9694 9699 9703 9731 9736 9741 9745 9750 9566 9571 9576 9581 9586 9614 9619 9624 9628 9633 9661 9666 9671 9675 9680 9708 9713 9717 9722 9727 9754 9759 9763 9768 9773 5 5 5 5 5 95 96 97 98 99 9777 9782 9786 9791 9795 9823 9827 9832 9836 9841 9868 9872 9877 9881 9886 9912 9917 9921 9926 9930 9956 9961 9965 9969 9974 9800 9805 9809 9814 9818 9845 9850 9854 9859 9863 9890 9894 9899 9903 9908 9934 9939 9943 9948 9952 9978 9983 9987 9991 9996 5 4 4 4 4 N. B. Naperian log = Briggs' log x 2.302. Base of Naperian System = e 2.71828. TRIGONOMETRIC RATIOS (Natural); including " arc," by which is meant the ' it- measure" or "circular measure" of the angle; e.g., arc 100 = 1.7453293, = j-f$of ie. arc degr. sin esc tan cot sec cos O.OOO O o.ooo inf. o.ooo inf. I.OOO I.OOO 9 L57I O.OI7 I 0-017 57-3 0.017 57-3 I.OOO I.OOO 89 T .553 0.035 2 0.035 28.7 0.035 28.6 i. ooi 0.999 88 1.536 0.052 3 0.052 19.1 0.052 19.1 i.ooi 0.999 87 1.518 0.070 4 0.070 14.3 0.070 14.3 I.OO2 0.998 86 1.501 0.087 5 0.087 IJ 5 0.087 IX 4 1.004 0.996 85 1.484 0.105 6 0.105 96 0.105 9.5 1. 006 0.995 84 1.466 0.122 7 O.I 22 8.2 0.123 8.1 1.008 0.993 83 1.449 0.139 8 0.139 7-2 0.141 7.1 i.oio 0.990 82 1.432 0.157 9 0.156 6.4 0.158 6.3 1. 012 0.988 81 1.414 0.174 10 0.174 5.8 0.176 5.7 I.OI5 0.985 80 1.396 0.192 ii O.I9I 5.24 0.194 5.14 I.OI9 0.982 79 1.379 0.209 12 0.208 4.81 0.213 4.70 I.O22 0.978 78 1.361 O.227 13 0.225 4-45 0.231 4-33 1.026 0.974 77 1.344 0.244 14 0.242 4.13 0.249 4 01 I.03I 0.970 76 1.326 O.262 *5 0.259 3-86 0.268 3.73 1.035 0.966 75 1.309 0.279 1 6 0.276 3.63 0.287 3.49 1.040 0.961 74 1.291 0.297 17 0.292 3.42 0.306 3.27 1.046 0.956 73 1.274 0.314 18 0.309 3.24 0.325 3.08 I.05I 0.951 72 1.257 0.33 2 I 9 0.326 3.07 0.344 2.90 1.058 0.946 7i 1.239 0-349 20 0.342 2.92 0.364 2.75 1.064 0.940 70 1.222 0.366 21 0.358 2.790 0.384 2.605 I.07I 0.934 69 I.2O4 0.384 22 0.375 2.669 0.404 2.475 1.079 0.927 68 I.I87 0.4OI 2 3 -39i 2.559 0.424 2.356 1. 086 0.921 67 1.169 0.419 24 0.407 2.459 0.445 2.246 1.095 9 I 4 66 1.152 0.436 2 5 0.423 2.366 0.466 2.145 1.103 0.906 65 LI34 0-454 26 0.438 2.281 0.488 2.050 1.113 0.899 64 I.II7 0.471 2 7 0.454 2.203 0.510 1.963 1. 122 0.891 63 1.099 0.489 28 0.469 2.130 0.532 i.88r I.I33 0.883 62 1.082 0.506 2 9 0.485 2.063 0.554 1.804 T.I43 0.875 61 1.064 0.523 30 0.500 2.000 0.577 1-732 1.155 0.866 60 1.047 0.541 31 0.515 .942 0.601 1.664 1.167 0.857 59 1.030 0.558 32 0.530 .887 0.625 i. 600 1.179 0.848 58 1. 012 0.576 33 0-545 -836 0.649 I -54 1.192 0.839 57 0-995 0-593 34 0.559 .788 0.675 1.483 1.206 0.829 56 0.977 0.611 35 0-574 -743 0.700 1.428 1. 221 0.8l9 55 0.960 0.628 36 0.588 .701 0.727 1.376 1.236 0.809 54 0.942 0.646 37 0.602 .662 0-754 1-327 1.252 0.799 53 0.925 0.663 38 0.616 .624 0.781 1.280 1.269 0.788 S 2 0.908 0.68 1 39 0.629 -589 0.810 1.235 1.287 0.777 5 1 0.890 0.698 , 40 - 6 43 -55 6 0.839 1-192 1.305 0.766 5 0.873 0.716 41 0.656 .524 0.869 1-150 1.325 0*755 49 0.855 0-733 42 0.669 -494 0.900 i. in 1.346 0.743 48 0.838 0.750 43 0.682 .466 0.933 1-072 1.367 0.731 47 0.820 0.768 44 0.695 -44 0.966 1.036 1.390 0.719 46 0.803 0-785 45 0.707 .414 I.OOO 1. 000 I.4I4 0.707 45 0.785 cos sec cot tan csc sin degr. arc '67 -S PM FEJ ocr AUQ 18 1947 LD U. C. BERKELEY LIBRARIES 995884 O 3 THE UNIVERSITY OF CALIFORNIA LIBRARY