Class _T4^^ Book._ ^■4 n Copightl^? COPYRIGHT DEPOSnV APPLIED MECHANICS FOR ENGINEERS ■>? h§?>^« THE MACMILLAN COMPANY NEW YORK • BOSTON • CHICAGO ATLANTA • SAN FRANCISCO MACMILLAN & CO., Limited LONDON • BOMBAY • CALCUTTA MELBOURNE THE MACMILLAN CO. OF CANADA, Ltd. TORONTO APPLIED MECHANICS FOR ENGINEERS A TEXT-BOOK FOR ENGINEERING STUDENTS BY E. L. HANCOCK ASSISTANT PROFESSOR OF APPLIED MECHANICS PURDUE UNIVERSITY Weto gorft THE MACMILLAN COMPANY 1909 All rights reserved ^^' A1- LIBRARY of CONGRESS Two Cooles Received JAN 12 1909 Cepyrltfttt Entry Copyright, 1909, By the MACMILLAN COMPANY. Set up and electrotyped. Published January, 1909. NorfajootJ Pregg J. 8. Cushing Co. — lieiwick & Smith Co. Norwood, Mass., U.S.A. a /<^1 PREFACE In the preparation of this book the author has had in mind the fact that thQ student finds much difficulty in seeing the applications of theory to practical problems. For this reason each new principle developed is followed by a number of applications. . In many cases these are illustrated, and they all deal with matters that directly concern the engineer. It is believed that problems in mechanics should be practical engineering work. The author has endeavored to follow out this idea in writing the present volume. Accordingly, the title "Applied Mechanics for Engineers " has been given to the book. The book is intended as a text-book for engineering students of the Junior year. The subject-matter is such as is usually covered by the work of one semester. In some chapters more material is presented than can be used in this time. With this idea in mind, the articles in these chapters have been arranged so that those coming last may be omitted without affecting the continuity of the work. The book contains more problems than can usually be given in any one semester. While it is difficult to present new material in the matter of principles, much that is new has been intro- duced in the applications of these principles. The sub- ject of Couples is treated by representing the couples by means of vectors. The author claims that the chap- ters on Moment of Inertia, Center of Gravity, Work and Energy, Friction and Impact are more complete in theory and applications than those of any other vi PREFACE American text-book on the same subject. These are matters upon which the engineer frequently needs infor- mation ; frequent reference is, therefore, given to origi- nal sources of information. It is hoped that these chapters will be especially helpful to engineers as well as to students in college, and that they will receive much benefit as a result of looking up the references cited. In general, the answers to the problems have been omitted for the reason that students who are prepared to use this book should be taught to check their results and work inde- pendently of any printed answer. The author wishes to acknowledge the helpful sugges- tions obtained from the many standard works on me- chanics. An attempt has been made to give the specific reference to the original for material taken from engi- neering works or periodical literature. He wishes, more- over, to express his thanks to Dean C. H. Benjamin and Professor L. V. Ludy for their careful reading of the manuscript, to Professor W. K. Hatt for many of the problems used, and to Dean W. F. M. Goss, whose con- tinued interest and advice have been a constant source of inspiration. It is hoped that the work may be an inspira- tion to students of engineering. E. L. HANCOCK. Purdue University, November, 1908. TABLE OF CONTENTS CHAPTER I ARTICLES 1-15 PAOS Definitions 1 Introduction — Force — Unit of Force — Unit Weight — Rigid Body — Inertia — ]\Iass — Displacement — Repre- sentation of Forces — Concurrent Forces — Resultant of Two Concurrent Forces — Resolution of Force — Force Triangle — Force Polygon — Transmissibility of Forces. CHAPTER II ARTICLES 16-19 Concurrent Forces 9 Concurrent Forces in a Plane — Concurrent Forces in Space — Moment of a Force — Varignon's Theorem of Mo- ments. CHAPTER III ARTICLES 20-21 Parallel Forces 20 Parallel Forces in a Plane — Parallel Forces in Space. CHAPTER IV ARTICLES 22-29 Center of Gravity 27 Definition of Center of Gravity — Center of Gravity de- termined by Symmetry — Center of Gravity determined by Aid of the Calcuhis — Center of Gravity of Locomotive Coun- terbalance — Simpson's Rule — Application of Simpson's Rule — Durand's Rule — Theorems of Pappus and Guldinus. vii Viil CONTENTS CHAPTER V ARTICLES 30-34 PAGE Couples 50 Couples Defined — Representation of Couples — Couples in One Plane — Couples in Parallel Planes — Couples in In- tersecting Planes. CHAPTER VI ARTICLES 35-36 Non-concurrent Forces 56 !N'on-concurrent Forces in a Plane — Non-concurrent Forces in Space. CHAPTER VII ARTICLES 37-65 Moment of Inertia 69 Definition of Moment of Inertia — Meaning of Term — Units of Moment of Inertia — Representation of Moment of Inertia — Moment of Inertia, Parallel Axes — Inclined Axes — Product of Inertia — Axes of Greatest and Least Moment of Inertia — Polar Moment of Inertia — Moment of Inertia of a Rectangle — Triangle — Circular Area — Elliptical Area — Angle Section — Moment of Inertia by Graphical Method — jMoment of Inertia by Use of Simpson's Rule — Least Moment of Inertia of Area — The Ellipse of Inertia — IMoment of Inertia of Thin Plates — Right Prism — With Respect to Geometrical Axes — Of Solid of Revo- lution — Of Right Circular Cone — IMoment of Inertia of ]\Iass — Moment of Inertia of Xon-honiogeneous Bodies — Of Mass, Inclined Axis — Principal Axes — El]i]3soid of Inertia — Moment of Inertia of Locomotive Drive Wheel. CHAPTER VIII ARTICLES 66-70 Flexible Cords Ill Introduction — Cords and Pulleys — Cord with Uniform Load Horizontally — Equilibrium of Cord due to its Own Weight — Representation by Means of Hyperbolic Function. CONTENTS ix CHAPTER IX ARTICLES 71-84 PAGE Rectilinear Motion 123 Velocity — Acceleration — Constant Acceleration — Freely Falling Bodies — Bodies Projected vertically Upward — Newton's Laws of Motion — Motion on an Inclined Plane — Variable Acceleration — Harmonic Motion — Motion with Repulsive Force Acting — Resistance varies as Dis- tance — Attractive Force varies as Square of Distance — Motion of Body falling through Atmosphere — Relative Velocity. CHAPTER X ARTICLES 85-94 Curvilinear Motion 142 Representation of Velocity and Acceleration — Tangen- tial and Normal Accelerations — Simple Circular Pendulum — Cycloidal Pendulum — JMotion of Projectile in Vacuo — Body projected up an Inclined Plane — Motion of Projec- tile in Resisting Medium — Path of Projectile, Small Angle of Elevation — Motion in Twisted Curve. CHAPTER XI ARTICLES 95-100 Rotary Motion 169 Angular Velocity — Angular Acceleration — Constant Angular Acceleration — Variable Acceleration — Combined Rotation and Translation — Rotation in General. CHAPTER XII ARTICLES 101-128 Dynamics of Machinery 177 Statement of D'Alembert's Principle — Simple Transla- tion of a Rigid Body — Simple Rotation of Rigid Body — X CONTENTS PAGE Reactions of Supports ; Rotating Body — Rotation of a Sphere — Center of Percussion — Compound Pendulum — Experimental Determination of Moment of Inertia — Deter- mination of g — The Torsion Balance — Constant Angular Velocity — Rigid Body Free to Rotate — Rotation of Sym- metrical Bodies — Rotation of Locomotive Drive Wheel — Rotation about an Axis not a Gravity Axis — Rotation of Fly Wheel of Steam Engine — Rotation and Translation — Side Rod of Locomotive — The Connecting Rod — Body rotating about an Axis, One Point Fixed — Gyroscope — The Spinning Top — Motion of Earth — Plane of Rotation — Gyroscopic Action Explained — Precessional Moment, Special Case — General Case — Car on Single Rail. CHAPTER XIII ARTICLES 129-144 Work and Energy 229 Definitions — Units of Work — Graphical Representation of Work — Power — Energy — Conservation of Energy — Energy of Body moving in Straight Line — Work under Action of Variable Force — Pile Driver — Steam Hammer — Energy of Rotation — Brake Shoe Testing Machine — Work of Combined Rotation and Translation — Kinetic Energy of Rolling Bodies — Work-Energy Relation for Any Motion — Work done when Motion is Uniform. CHAPTER XIV ARTICLES 145-168 Friction 261 Friction — Coefficient of Friction — Laws of Friction — — Friction of Lubricated Surfaces — ^lethodof Testing Lu- bricants — Rolling Friction — Friction W^heels — Resistance of Ordinary Roads — Roller Bearings — Ball Bearings — Friction Gears — Friction of Belts — Transmission Dyna- mometer — Creeping of Belts — Coefficient of Friction of Belts — Centrifugal Tension of Belts — Stiffness of Belts CONTENTS XI PAGE and Ropes — Friction of Worn Bearing — Friction of Pivots : Flat Pivot, Collar Bearing, Conical Pivot, Spherical Pivot — Absorption Dynamometer — Friction Brake — l^rony Fric- tion Brake — Friction of Brake Shoes — Train Kesistance. CHAPTER XV ARTJCLFS 109-178 Impact 315 Definitions — Direct Central Impact, Inelastic; Elastic — Elasticity of Materials — Impact of Imperfectly Elastic Bodies — Impact Tension and Impact Compression — Di- rect Eccentric Impact — Center of Percussion — Oblique Impact of Body against Smooth Plane — Impact of Rotating Bodies. Appendix I. Hyperbolic Functions, Tables. Appendix II. Logarithms of Xumbers. Appendix III. Trigonometric Functions, Tables. Appendix IV. Squares, Cubes, Square Roots, etc., of Numbers. Appendix V. Conversion Tables. Index 383 APPLIED MECHANICS FOR ENGINEERS CHAPTER I DEFINITIONS 1. Introduction. — The study of the subject of mechanics of engineering involves a study of matter^ space^ and time. The subject as presented in tliis book consists of two parts ; viz., statics, including the study of bodies under the action of systems of forces that are in equilibrium (balanced), and dynamics, including a study of the motion of bodies. 2. Force. — A body acted upon by the attraction or repulsion of another body is said to be subjected to an attractive or repulsive force, as the case may be. Forces are usually defined by the effects produced by them, as for example, we say, a force is something that produces motion or tends to produce motion, or changes or tends to change motion, or that changes the size or shape of a body. The study of relations between forces and the motions produced by them is usually designated as the study of Statics and Dynamics, Forces always occur in pairs ; for example, a book held in the outstretched hand exerts a downward pressure on the hand, and the hand exerts an equal upward pressure on the book. 3. Unit of Force. — The unit of force used by engineers in this country and England is the pound avoirdupois. It 2 APPLIED MECHANICS FOE ENGINEEBS is sometimes, however, necessary to use the absolute unit of force. This ijaay be defined as follows : The absolute unit of force is .that force which acting on a unit mass during unit time will produce in the mass, unit velocity.] This absolute unit of force is called a poundal. In France, Germany, and other countries where the centimeter-gram- second system is used, the engineer's unit of force is the kilogram. The absolute unit of force, in such countries, is the force which acting upon a mass of one gram weight (at Paris) w^ill produce a velocity of one centimeter per second, in a second. Such a unit is called a dyne. 4. Unit Weight. — The weiglit of a cubic foot of a sub- stance will be called the unit weight of the substance and will be represented by 7. Below is given a table of such weights taken at the sea level. It will be seen that the unit weight of a substance divided by the unit weight of pure water gives its specific gravity. (See Table I on opposite page.) 5. Rigid Body. — In studying the state of motion or rest of a body due to the application of forces acting upon it, it is not necessary to consider the deformation of the body itself, due to the forces. When so considered it is customary to say that the body is a rigid body. Unless otherwise stated bodies will be considered as rigid bodies in this book. 6. Inertia. — The property of a body that causes it to continue in motion, if in motion, or remain at rest, if at rest, unless acted upon by some other force, is called inertia. This is Newton's First Law of Motion. (See Art. 76.) DEFINITIONS TABLE I Unit Weights and Specific Gravity of Some Materials (Kent's *' Engineer's Pocket Book ") Material Specific Gravity Unit Weight Brass 8.2 to 8.G 511 to 536 Brick Soft 1.6 100 Common 1.79 112 Hard 2.0 125 Pressed 2.16 135 Fire 2.24- -2.4 140-150 Brickwork — mortar 1.6 100 Brickwork — cement 1.79 112 Concrete 1.92- -2.24 120-140 Copper 8.85 552 Earth — loose 1.15- -1.28 72-80 Earth — rammed 1.44- -1.76 90-110 Granite 2.56- ■2.72 160-170 Gum .92 57 Hickory .77 48 Iron — cast 7.21 450 Iron — wrought 7.7 480 Lead 11.38 709.7 Limestone 2.72- 3.2 170-200 Masonry — dressed 2.24- 2.88 140-180 Nickel 8.8 548.7 Pine— white .45 28 Pine— yellow .61 38 Poplar .48 30 Sandstone 2.24-i 2A 140-150 Steel 7.85 490 White Oak .77 48 7. Mass. —The mass of a body is the quantity of matter it contains. Mass differs from weight, in that the weight varies with the position on the surface of the 4 APPLIED MECHANICS FOB ENGINEERS earth and with the height above tlie surface, while the mass remains the same. The engineer's definition of mass, viz. that it is equal to the weight divided by the acceleration of gravity (see Art. 76), may be expressed ^= — • I^oth a and g vary for different localities, but the quotient is constant ; that is, the quantity of matter in a body is independent of its position with reference to the earth. The weight of a body may be determined by means of the spring balance. Such a balance is the only true measure of weight, since the equal-armed balance gives the same weight regardless of distance from the center of the earth. The equal-armed balance really measures mass. 8. Displacement. — By the displacement of a body is meant its change from one position to another. A dis- placement involves a movement in a definite direction. It may be represented by an arrow, the length of the arrow representing the distance moved and the direction of the arrow the direction of the motion. Thus, if a man walks due east one mile and then due north one mile, we might represent his displacement from the original posi- tion by an arrow drawn northeast of a length equal to V2 miles. Or, in Fig. 1, if P^ represents a displacement of a body in the direction indicated and Pj a subsequent dis- placement in the direction of Pj, then B represents a dis- placement equivalent to P^ and F^. It is seen that B may be determined by constructing a parallelogram on Pj and P2 as sides and drawing the diagonal. Quantities that may be represented by arrows are known as vector quantities, and tlie arrows themselves as vectors. DEFINITIONS 5 9. Representation of Forces. — Forces have a certain magnitude, act in a certain direction, and have a definite point of application. If a man, for example, attaches a rope to a log and pulls on the rope, his pull may be meas- ured in pounds ; it acts along the rope, and it has a point of application which is the same as the point of attach- ment of the rope to the log. It has been found convenient, for the purpose of analysis, to represent forces by arrows (vectors of Art. 8), the length of the arrow representing the magnitude of the force and the direction of the arrow giving the direction in which it acts. Thus, a 10-pound force, acting in a direction 30° with the horizontal, is represented by an arrow drawn in the same direction and having its point of application in the body and having a length representing 10 lb. (In this case, if 2 lb. represents 1 in., the length of the arrow is 5 in.) The line along which a force acts will be referred to as its line of action, 10. Concurrent Forces. — When two or more forces act upon the same point of a body, their lines of action are concurrent^ and the forces are known as concurrent forces, 11. Resultant of Two Concurrent Forces. — If two forces having the same point of application act on a body, there is some single force that might be applied at the same point to produce the same effect. This single force is called the resultant of the two forces, and is found as follows : construct upon the arrows representing the forces a parallelogram and draw the diagonal from the point of application. This diagonal represents the re- sultant of the two forces in macfnitude and direction 6 APPLIED MECHANICS FOR ENGINEEES Fig. 1 (Art. 8). Thus, if P^ and P^ (Fig. 1) are the forces, then R is the resultant. Algebraically ^ = ^ r^^ + r,^ -^2I>,r,cos AOB . 12. Resolution of Force. — We have just seen how two concurrent forces may be replaced by a single force called their resultant. In a similar way a single force may be resolved into two forces. These forces are the sides of a parallelogram of which the single force is a diagonal. It is clear, then, that there are an infinite number of compo- nents into which a single resultant may be resolved. It is necessary, therefore, in speaking of the components of a force, to state specifically which are intended. It will be seen in problems that follow that the components most often used are at right angles to each other, and usually the vertical and horizontal components. In such a case the components are the projections of the force on the verti- cal and horizontal lines. 13. Force Triangle. — It follows directly from the par- allelogram law of forces (Art. 11) that if we draw from any point a line parallel to and representing one of two concurrent forces, P^ say, and from the extremity of this line another line parallel to P^ and of the same length, then the remaining side of the triangle will be represented by H. This triangle is called the force triangle. In general, the resultant of two concurrent forces may be found by drawing JDEFIXinONS 7 lines parallel to the forces as above. The line necessary to complete the triangle is tlie resultant, and its arrow is always away from the point of application. The equal and opposite of this resultant would be a single force that would hold the two concurrent forces in equilibrium. 14. Force Polygon. — If more than two forces are con- current, we may find their resultant by proceeding in a way similar to that outlined above. Thus, let the forces be P-^, P^, P3, P4, etc. (Fig. 2), all passing through a point ; from any point draw a line equal and parallel to P-^, from the ex- tremity of the line draw an- other equal and parallel to P^^ from the extremity of this last line draw another equal and parallel to P3, and proceed in the same way for the other forces. The figure produced will be a polygon whose sides are equal and parallel to the forces. The resultant will be given in magnitude, direction, and point of application by the line necessary to close the polygon. The arrow, representing the direction of the resultant, will always be away from the point of applica- tion. (See Fig. 2.) If the polygon be closed, the sj^stem of forces will be in equilibrium. The single force neces- sary to produce equilibrium will, in any case, be equal and opposite to P. The student should construct force poly- gons by taking the forces in different orders and checking the resultant in each case. 8 APPLIED MECHANICS FOR ENGINEERS By drawing the lines OA^ OB^ OQ^ etc., it is easy to see that OA represents the resultant of P^ and P^, that OB represents the resultant of OA and Pg, and so of Pj, Pg, and Pg, etc. That is, it is easy to' see that the force polj^- gon follows directly from the force triangle. By means of the force polygon it is easy to find graphically the resultant of any number of concurrent forces in a plane. The work, however, must be done accurately. The student should show that the force polygon may be used for finding the resultant of concurring forces in space, by considering two forces at a time. The force polygon in this case is called a twisted polygon. 15. Transmissibility of Forces. — It is a matter of expe- rience that the point of application of a force may be changed to any point along its line of action without changing the effect of the force upon the rigid body. This, of course, is on the assumption that all the force is transmitted to the body. The law may be stated as fol- lows : The point of application of a force may be transferred anywhere along its line of action without changing its effect upon the body upon which it acts. CHAPTER II CONCURRENT FORCES 16. Concurrent Forces in a Plane. — It will often be convenient to consider forces as acting on a material point; this is equivalent to considering the body without weight and simply a point. If a material point (0) (Fig. 3) be acted up- on by a num- ber of forces in a plane, P^, P^^ -T^g, x"^^, etc., each one mak- ing angles a^, «2^ "3' ^4' ^^^-^ respectively, with the posi- tive 2:-axis, it is desirable to find the resultant of all of them in magni- tude and direction; that is, the single ideal force that could produce the same effect as the system of forces. Each force P may be resolved into components along the a> and y-axes, giving P cos a along the rc-axis, and P 9 10 APPLIED MECHANICS FOB ENGINEERS sin a along the ?/-axis. The sum of these components along the a;-axis may be expressed, 2a; = P^ cos a^ + P^ cos a^ + Pg cos a^ + etc., the proper algebraic sign being given cos a in each case. In a similar way the sum of the components along the y-axis may be written, 2y = -Pi sin a^ + P^ sin a^ + P^ sin ^3 + etc. These forces, ^x and 2y, may now replace the original system as shown in Fig. 4. And these may be combined into a single re- sultant which is the diagonal of the rectangle of which the two forces are sides (Art. 11). This gives the resultant in magnitude and direction, and this resultant force is the sin- gle force which if allowed to act upon the material point would produce the same effect as the system of forces. It should be remembered in all that follows that this resultant force has no real existence ; it sr sx ->-— z Fig. 4 CONCUmiENT FORCES 11 is used to simplify the solution of problems. Analytically the resultant may be expressed, i2 = V(2x)2+(Ey)2, and its direction a as such an angle that tan a = -^y. ^x (See Fig. 5.) If the material point be at rest or moving uniformly, this resultant force must be equal to zero; This means that (22:)^+ (2z/)2 = 0, that is, that the sum of two squares y must be zero ; but this can happen only when each one, separately, is 2y zero (since neither can be *^' negative being squared). We therefore have as the necessary and sufficient conditions for the equilibrium of a material point, acted upon by a system of concurring forces in a plane, iJ = or 2a5 = 0, and 2?/ = 0. When It is not zero, the system of forces causes accel- erated motion in the direction of R\ when li=0, the 12 APPLIED MECHANICS FOR ENGINEERS material point remains at rest, if at rest, or continues in motion with uniform velocity, if in motion. In this case the system of forces is said to be balanced. As an illustration of the foregoing, consider the case of a body of weight G- situated on an in- clined plane, making an angle 6 with the horizontal. (See Fig. 6.) There is a certain force P making an angle cf) with the plane, whose component along the plane acts upwards, and also a force of friction F upwards. The other forces acting on the body are Gr, the force of gravity acting vertically, and iV, the normal pressure of the plane. Taking the a:-axis along the plane positive upward and the ^/-axis perpendicular to it positive upward, we get, '2x=Pcos(b-^F- asm 6, and 2^/ = N+ P sin (f>— Gr cos 0. For equilibrium Pcos<^ + i"- (? sin (9=0, N+Psmcf)- G^cos(9 = 0. Therefore, N= G cos — P sin <^, asm0-F Fig. 6 P = COS (f> CONCURRENT FORCES 13 This last equation gives the magnitude of P required to preserve equilibrium, supposing that the force of friction, ^, and ^ are known. Problem 1. An angle iron whose weight is 20 lb. and angle a right angle, rests upon a circular shaft, radius 2 in. Find the normal pressure at .4 and B (Fig. 7). Problem 2. (jiven three concurring forces, 100 lb., 50 lb., and 200 lb., whose directions referred to the a;-axis are 0°, QO"", 180^ respectively ; find the resultant in magnitude and direction. Problem 3. A body (Fig. 8) whose weight is G is drawn up the inclined plane with uni- form velocity due to the action of the forces I^ and P', Find the force of friction and the IGO lb. P acts Fig. 8 normal pressure, if P = 100 lb., P' = 100 lb., G parallel to the plane and P' acts horizontally. Problem 4. A w^heel is about to roll over an obstruction. The diameter of the wheel (Fig. 0) is -V and its weight 800 lb. Find the force P necessary to start the w^heel over the obstruction. 14 APPLIED MECHANICS FOR ENGINEERS Problem 5. A weight of 10 tons is supported as shown in Fig. 10. Find the force acting in the tie A and the member B, 17. Concurrent Forces in Space. — If the material point (0) be acted upon by a system of concurrent forces not in a plane P^^ -^2' 3' 4' etc., whose direc- tion angles are etc., respec- tively, one of which is shown in Fig. 11, the resultant force may be found in magnitude and direction by an analysis similar to that used in the preceding case. The sum of the components of all the forces along the a;-axis is 2 a: = Pj cos a^ + P^ cos a^ + P^ cos a^ + P^ cos a^ + etc. Similarly, the sum of the components along the ^/-axis, 2y = Pj cos ySj + P^ cos ySg + -P3 cos ySg + P^ cos ^^ + etc., Fig. 11 CONCURRENT FORCES 15 and the sum of the components along the 2-axis, S 2 = Pj cos 7j + Pg C^S 72 + ^3 ^^^ 73 + ^4 ^^^ 74 + ^^^• The original system of forces may now be replaced by a system of three rectangular forces 2a:, 2y, and tz (Fig. 12). ^ Finally, this system may be replaced by a resultant which is the diagonal of a paral- lelopiped con- structed with ^x^ 2y, and ^z as edges. In P^'" Fig. 12 magnitude this resultant may be expressed •sz zx R = V(2;x)2 + (2^)2 + {^z)\ (See Fig. 13) and its direction given by the angles a, yS, and 7. These angles are given by the equations ^x _ 2y S2 cos a = -~^ COS /3 = ~^ cos 7 = -^. Jit ^ -/^ For equilibrium jR must be ; that is, (22:)2 + (2y)2+ (2^)2=0, ^x = 0, 2?/ = 0, ^z = 0. This gives three equations of condition from which three unknown quantities may be determined. In the preced- ing case of Art. 16 there were only two equations of condition ^x= and Sy = ; consequently, only two un- known quantities could be determined. and therefore. 16 APPLIED MECHANICS FOR ENGINEEERS Fig. 13 Problem 6. Three men (Fig. 14) are each pulling with a force P at the points a, b, and c, respectively. What weight Q can they raise with uniform motion if each man pulls 100 lb.? Each force makes an angle of 60° with the horizontal. Problem 7. Three concurring forces act upon a rigid body. Find the resultant in mag- nitude and direction. The forces are defined as follows : Pi = 75 lb. a. Pg = 80 lb ^3 = 147^2'; 73 = ? Fig. 14 63° 27' ; )8i = 48° 36' ; y^ = a, = 153° 44' ; P. = 95 lb. ; a. = 76° 14' Hint, yj, yo, and yg may be found from either of the following relations : cos (a + fS) cos («-/?)+ cos^y = 0, C0S2« + COs2/5 + COS^y = 1. Problem 8. Each leg of a pair of shears (Fig. 15) is 50 ft. long. They are spread 20 ft. at the foot. The back stay is 75 ft. long. Find the forces acting on each member when lifting a load of 20 tons at a distance of 20 ft. from the foot of the shear legs, neglecting the weight of structure. CONCURRENT FORCES 17 20 TONS Fig. 15 18. Moment of a Force. — The moment of a force with respect to any point in its plane may he defined as the prod- uct of the force and a perpendicular let fall from the point on the line of action of the force. Let P (Fig. 16) be the force and the point and a the perpen- dicular distance of the force from the point ; then Pa is the moment of the force with respect to the point 0. This moment is measured in terms of the units of both force and lengthy viz. foot-pounds or inch-pounds, and is read foot- pounds moment or inch-pounds moment to distinguish it from foot-pounds work or inch-pounds work. For convenience the algebraic sign of the moment is Fig. 16 18 APPLIED MECHANICS FOR ENGINEERS said to be positive when the moment tends to turn the body in a direction counter-clockwise^ and negative when it tends to turn the body in the clockivise direction. The moment may be represented geometrically as fol- lows : let EF represent the magnitude of P, drawn to the desired scale, and draw EO and FO, The area of the triangle OEF ^ {EFa, or EFa=2A0EF^ that is, the moment of the force with respect to a point is geometrically represented hy twice the area of the triangle^ whose lase is the line representing the magnitude of the force and ivhose vertex is the given point. 19. Varignon's Theorem of Moments. — The moment of the resultant of two concurring forces with respect to any point in their plane is equal to the algebraic sum of the moments of the two forces with respect to the same-point. The given forces P and P^ may be represented by OP and OPj and their resultant by OR. Take as the point with respect to which moments are to be taken, and construct the dotted lines as shown in Fig. 17. The rao- FiG. n' jnent of OP with respect to is twice the area of the triangle OOP (Art. 18), = 2 area of the triangle OOP, since the tri- CONCURRENT FORCES 19 angles have the same base and the same altitude. That is, the moment of CP with respect to is the same as the moment of CB with respect to = CBa^ where a is the perpendicular let fall from on OR, In a similar way, it is seen that the moment of CP^ with respect to is equal to the moment of CA with respect to 0; that is, to OA • a. Therefore, the sum of the moments of P and P^ with respect to equals (OB + OA) • a. But (OB + OA)a = (0A + AB) 'a==R ' A, since OB = AR (equal triangles OPB and AP^R), When the point is taken between the P and Pp the moment of the resultant equals the differ- ence of the moments of P and P^ Let the student show that this is true. Cor. 1. If there are any number of concurring forces in a plane, it may be shown that Varignon's theorem holds by considering the resultant of two of them with the third, and so on. The more general theorem may then be stated as follows : The moment of the resultant of any num- ber of concurring forces in a plane with respect to any point in that plane is equal to the algebraic sum of the moments of the forces with respect to the same point. Cor. 2. If the point be taken in the line of action of iJ, then a = 0^ and therefore the sum of the positive moments equals the sum of the negative moments. The moment of a force with respect to a line at right angles to the line of action of the force is the product of the force and the shortest distance between the two lines. The moment of a force with respect to a line not at right angles to the line of action of the force is the same as the moment of the component of the force in a plane perpendic- ular to the line. CHAPTER III PARALLEL FORCES 20. The Resultant of Two Parallel Forces. — In consid- ering two parallel forces in a plane three cases arise : (a) when the forces are in the same direction ; (5) when they are unequal and in opposite directions; (^) when thej- are equal and in opposite directions, but having dif- ferent lines of action. Fig. 18 Case (a). When the forces are in same direction. The two forces are P and P^, and the distance between their lines of action is a. For the sake of analysis put in two 20 PARALLEL FORCES 21 equal and opposite forces T as shown in Fig. 18. These forces will have no effect as far as the state of motion of the body is concerned. The resultant of T and P is iJ, and that of T and P^ is By Transfer It and R^ to the point of intersection of their lines of action A. Here resolve them into components parallel to their original components ; the two forces 2^ nullify each other, and there are left the two forces P and P^ acting along the same line AU. The resultant of P and P^ then, is equal to P + P^ and acts in the same direction as the forces. To determine the position of P with reference to the forces we have from similar triangles T P' irom which — i = , or a; = — . P X R That 18^ the resultant of two parallel forces in the same direc- tion divides the distance between them in the inverse ratio of the forces. Cor. For any point in the same plane, it is easy to show that the moment of the resultant is equal to the algebraic sum of the moments of the two forces with respect to this same point. Draw a line through paral- lel to T and let m be the distance from to the line of action of P^ If now Pm be added to both sides of the equation Px = Pa^ we shall have Ii(7n + x) = P{a + m) + P-^m^ and this is the relation we were to find. X T Pi and a — X AF AF Py- a — •^ n V Xz=- 22 APPLIED MECHANICS FOR ENGINEERS When (9 is a point on the line of action of B, the moment of iJ = 0, and we have the moment of P equal to the moment of P^. This is often a convenient relation to use in the solution of problems. Following out the above reasoning, let the student show that the moment of the resultant of any number of parallel forces in a plane with respect to a point in the plane is equal to the algebraic sum of the moments of the forces with respect to that same point. Case (J). When the forces are unequal and opposite in direction. In this case the analysis is exactly similar to Case (a) and leads to exactly the same conclusions. It is left as an exercise to be worked out by the student. Case (c). When the forces are equal and opposite^ but not acting along the same line, they form a couple. These will be treated later. (See Art. 30.) Problem 9. Two parallel forces, one of 20 lb. and one of 100 lb., have lines of action 24 in. apart. Find the resultant in mag- nitude, direction, and point of application : (1) When they are in the same direction. (2) When they are in opposite directions. Problem 10. A horizontal beam of length I is supported at its ends by two piers and loaded with a single load P at a distance of T from one end. Find the pressure of the piers against the beam. Problem 11. The locomotive shown in Fig. 19 is run upon a turntable whose length is 100 ft. Find the position of the engine so that the table will balance. 21. System of Parallel Forces in Space. — If the forces are all parallel, it is evident that the resultant is equal in magnitude to the algebraic sum of the forces, and that its line of action is parallel to the forces. It remains, then, to determine the point of application of this resultant. I PARALLEL FORCES 23 Fig. 19 Suppose the forces represented by Pj, P^, P3, P4, etc., and let their points of application be ^i^i^i, ^2^/2^2' ^zVz^z'^ x^y^^, etc. (Fig. 20). (In order to avoid a complicated figure only two forces are shown.) The two forces Fig. 20 24 APPLIED MECHANICS FOR ENGINEERS Pj and P2 li^ i^^ ^ plane and have a resultant R = P^ + P^ whose point of application is at a distance z^ from the xy- plane and on a line joining L^ and L^ at L^ , Draw L^A perpendicular to L^A, Then from Art. 20, R^ L^B = P^L^A^ which multiplied by sec a gives R L^L^ = P^LJLy^ L,y P. or —2 — = 1 -^2^1 -^1 + ^2 Now in the plane of z^ and z^ draw ig^ ^^^ -^'^' perpen- dicular to 2-^, and we have z^ — Zc ^2^1 -^1^ ^1 ~ % so that 2' - ^2 = p ^ p (^1 - ^2)' -^1 + ^2 A + ^2 Consider now ^' with P^\ these forces lie in a plane. Let their resultant be R^^ and its point of application x^\y^^^z^\ Following out the above reasoning for this case, the resultant is seen to be R^^ = P^ + P^ + P^ and z'^ = -^ ^ "^ -^3^3 __ -^1^1 "I" ^2^2 "^ -^3^3 . Extending this process so as to include all of the forces P4, Pg, etc., and calling the final resultant M and its point of application x, y, i, we have R= P^+P^ + P^ + Pi+ etc. and i - -Pi^i+-P2^2 + A ^3 + -P4^4 + etc. ^ 2^ Pi + P^ + A + A + etc. 2P' PABALLEL FORCES 25 and by a reasoning similar to the above y = -Pi.Vi + A .y2 + Aj/« + ^ 4?/ 4 + etc. ^ 2Pz/ ^ P^ + F^ + P^ + F^ + etc. SP _ _ PiX^ + P<>x<2, + Pg^s + -^A + etc. _ 2Px This point of application of the resultant is called the cen- ter of the system of parallel forces. As an illustration of the above, suppose P^ = 50 lb., P^ = 100 lb., ^3=300 lb., P4 = 10 1b., andP5 = -400 lb., and their points of application respectively 2, 1,— 5; — 1, -2,4; 2,1,-2; -2,1,1; 1,1,1. The resultant in this case equals 50 lb. + 100 lb. + 300 lb. + 10 lb. - 400 lb. = 60 lb. and its point of application __ 50 (2) + 100 (-1) + 300 (2) + 10 (-2)- 400(1) _^ ^- QQ -^. _ __ 50(1) + 100(- 2)+ 300(1) + 10(1)- 400(1) _ ^ ~ 60 "" "" ' __50( _5) + 100(4) + 300(-2) + 10(l)-400(l) _ "^ - 60 - "■ ^■^• As another illustration consider the problem of finding the center of the system of parallel forces P^, P^^ Pg, in Fig. 21. The figure represents a Z-iron of the same cross section throughout, and Pj, P^-, and Pg are therefore the weights of the individual parts (considering the Z-iron as divided into three parts — two legs and the connecting ver- tical portion). If the weight of a cubic inch of iron =.26 lb., Pj = .78 lb., P2 = 2.08 lb., P3 = 1.04 lb., and therefore i2 = 3.9 lb. The points of application of P^, P^^ and Pg 26 APPLIED MECHANICS FOR ENGINEERS are (- ^, - 1 91), Q, - 1-, 5), and (2, - i, i), respectively, so that _ .78(- 1)4- 2.08(1) + 1.04(2) _ '^~" 3.9 = .70 in., i/ 78(-i) + 2.08(-i) + 1.04(--i) ^ ^g^.^^^ 3.9 .78(-V-)+ 2.08(5) + 1.04(1) _ 3.9 "■ m, This point x^ y, 2J is, in this case, the center of gravity of the Z-iron. Problem 12. Parallel forces of Pp Pg' ^s' ^^^ ^4 ^^^ ^^ the corners of a rectangle 3 ft. by 2 ft. and perpendicular to its plane. Find the point of application of the resultant, if P^ = 10 lb., Pg = ^0 lb., Pg = 100 lb., P^ = 200 lb., Pj and P^ being 2 ft. apart, and Pg on same side as Pg. Problem 13. Eight parallel forces act at the corners of a one-inch cube, making an angle of 45° with one of its faces. Find the point of application of the resultant force, if Pj = 30 lb., Pg = 50 lb., P3 = 10 lb., P, = 20 lb., P, = 100 lb., Pq=5 lb., P, = 10 lb., Pg = 40 lb. The student should prove that the moment of the resultant of any system of parallel forces in space with respect to any line in space, equals the sum of the moments of the forces with respect to this same line. The solution of Problems 12 and 13 is made much shorter by using this principle. Fig. 21 CHAPTER IV CENTER OF GRAVITY 22. Definition of the Center of Gravity. — The center of gravity of a body may be defined as the point of appli- cation of the resultant attraction of the earth for that body, and the center of gravity of several bodies con- y sidered together, as the point of application of the resultant attraction of the earth for the bodies. The expressions for x, y, and z. Art. 21, may be used for locating the center of gravity, in the latter case, without change, P^, P^, Pg, etc., 'representing the weights of the individual bodies. In such cases the center of the system of parallel forces is the center of gravity of the body. The attention of the stu- dent is called to the fact that the forces acting upon the particles of a body, due to the attraction of the earth, are not parallel, but meet in the center of the earth. For all practical purposes, how- ever, they are considered parallel. ^1 B F„ Fig. 22 27 28 APPLIED MECHANICS FOR ENGINEERS If the unit weight times the volume be substituted for weight, that is, if we write instead of P^, 7^ V^ and P^^ 72 ^"2, etc., then S, y^ z become - ^ 7i^"i ^1 + 72 ^^2 ^2 + 78 ^s^3 + etc. ^ ^7^^ 7il^i + 72^'2 + 73f'^3+etc. 27r' — __ 27 Vy - __ 27Fi ^ ■" 27 r' ^ " 27 r* And if the bodies are all of the same material and so have the same heaviness, 7 is constant and may be taken outside the summation sign, where it cancels out. This gives values for S, y, and i, ^_2F^ 7/-^ZM 5-2Z^ formulae exactly similar to those of Art. 21, where the P's are replaced by F's. If the bodies are thin plates of the same material, of constant thickness 5, we may write for F^, V^, F3, etc., hF^, hF^, bF^, etc., where the P's represent the areas of the faces of the plates. Making this substitution for the F's, 5, ^, z may be written - _ IF^x^ + hF^x^^ + hF^ X r^ + etc. ^ ^Fx ^■" bF^ + bF^ + bF^ + etc. "" 2P' - 2P7/ 2P2: the J being a constant factor, cancels out. These formulae are applicable to finding the center of gravity of areas, and are much used by engineers for finding the center of gravity of sections of angles, channels, T-sections, Z-sec- CENTER OF GRAVIir 29 ^~ ~" / j_0.4" [ *;^ ^ ^. ^ 1 10 ' Fig. 23 tions, etc. As an illustration let it be re- quired to find the center of gravity of the angle section shown in Fig. 22. It is convenient to select the X- and ^/-axes as shown and to divide the area up into the two indicated areas F^ and F^. We then have F, + F^ ^ F, + F, ' x^T/^ being the center of gravity of F^, and 0:2^2 the center of gravity of F^. It is left to the student to make the numerical substitution and to calculate the values for x and ^. Second Method. The same results for x and y might be obtained from the expressions X - -^1^1 - F^H .-7 _ F^yi - F^y^ F,-F, y F,-F, where now F^ is the area formed by completing the rec- tangle whose sides are 4 in. and 3 in. and x^y^ the center of gravity of this rectangle referred to the coordinate axes, and F^ the area of the rectangle whose sides are 3 in. and 2 in. and x^y^ the coordinates of its center of gravity re- ferred to the same axes. Let the student find the center of gravity of the angle section by this method and com- pare the results with those obtained by the previous method. 30 APPLIED MECHANICS FOR ENGINEEES Problem 14. Find the center of gravity of the channel section shown in Fig. 23. Problem 15. Find the center of gravity of the T-section shown in Fig. 24. '' .. 1 . f I Fig. 24 Fig. 25 Problem 16. Find the center of gravity of the U-section shown in Fig. 25. Given the fact that the center of gravity of a semicircular 4r area is — from the diameter. (See Prob. 22.) Problem 17. Find the position of the center of gravity of a trapezoidal area, the lengths of whose parallel sides are a^ and Og? re- ^ ^ spectively, and the distance between them h. (See Fig. 26.) Hint. Draw the diaiio- nal AB and call tlie tri- angle ACB, F^, and the triangle ABD, F^. Given, the center of gravity of a triangle is \ the distance from the base to the vertex. (See Prob. 21.) Select ^jD as the a:-axis, then Fig. 26 y ^l.Vl + ^2.^2 where y^ = | A and ^2 = i ^* ^^^ center of gravity is seen to lie on a line joining the middle points of the parallel sides. CENTER OF GRAVITY 31 "^- Problem 18. A cylindrical piece of cast iron whose height is 6 in. and the radius of whose base is 2 in., has a cylindrical hole of 1 in. radius drilled in one end, the axis of which coincides wdth the axis of the cylinder. The hole was originally 3 in. deep, but has been filled with lead until it is only 1 in. deep. Find the center of gravity of the body, the unit weight of lead being 710 and of cast iron 450. (Fig. 27.) Problem 19. Find the center of gravity of a portion of a reinforced concrete beam. (See Fig. 27 Fig. 28.) The beam is reinforced with three half-inch steel rods, centers 1 in. from the bottom of the beam and 1 in. from the sides. The center of the middle rod is 4 in. from the sides. (y for steel = 490 lb. per cubic foot ; y for concrete = 125 lb. per cubic foot.) Note. It is seen that the thickness cancels out of the expression for the center of gravity, and might, therefore, have been neglected. Z ^ yf'^ <^--6^-\6^ Iz: Fig. 28 32 APPLIED MECHANICS FOR ENGINEERS 23. Center of Gravity determined by Symmetry. — In some areas and solids it is often possible to determine the center of gravity from considerations of the symmetry of the figure ; for example, the center of gravity of a paral- lelogram is at its geometrical center. This is also true of the circle, square, cylinder, sphere, etc. Whenever any axis is an axis of symmetry, that is, an axis such that for every element of area or volume on one side there is an equal area or volume on the other side, symmetrically placed, the center of gravity must be on that axis. This was found to be true in the case of the channel section, the T-section, and the U -section. In each of these cases the vertical line through the center of gravity is an axis of symmetry. The student will be able to note many more such cases, and by a little thought will often be able to find either x^ y^ or z from observation. 24. Center of Gravity determined by Aid of the Calculus. — The expressions for x^ y^ and z used to locate tlie center of gravity in Art. 22 may be put in the form of the quo- tient of two integrals, and these expressions may be inte- grated when there is no discontinuity in the expressions between the limits taken. With this understanding we may write __2Px_|^P(f) __ ^Pz _^dP{z) " 2P ~ {dP CENTER OF GRAVITY 83 As an illustration, suppose it is desired to obtain the center of gravity of a right circular cone of altitude h and radius of base r. Take the 2:-axis as the axis of the cone with the vertex at the origin. (See Fig. 29.) It is evident z Fig. 29 that ^ = and 2 = 0, so that it is only necessary to find x. The volume of any dv cut from the cone by two parallel planes, perpendicular to x and separated by a distance dx, is iry^dx^ and the weight of this dv is ^iryHx = dP. Therefore ^ ^n I xdP 1 x^iry^dx x = I dP I ^iry^dx But from similar triangles y : x gives 2 sc = r : h or y = -x. This 12 Jo x^dx A' y^T-oJ xHx T r X2 r 3 3 , h 4 34 APPLIED MECHANICS FOB ENGINEERS The expressions for x, y, and 2, involving c?P, may be changed to similar ones involving dv, and these become for homogeneous bodies, since dP = ydv^ 1 xdv j ydv j zdv J dv I dv j dv and for thin plates of constant thickness b the dv may be replaced by hdF, giving values of x, y, and 5 for area, ^xdF _^ JydF __ jzdF 7 ' ^=-p . ^ = -7 X Fig. 30 The center of gravity of thin homogeneous wires of con- stant cross section may be found by replacing the dv in the above formulae by ads, where a is the constant area of cross section and ds is a distance along the curve. The formulae then become J xds ds [yds CENTER OF GRAVITY 35 Problem 20. Find the center of gravity of a parabolic area shown in Fig. 30, the equation of the parabola being i/ = 2px. I xdF J dF Here dF = ydx, so that ^ — — a ^ ~~ " 2 ^n** ^ Ci/dx V2^CJdx ^^^]o Jo Jo It is left as a problem for the student to show that y = ^ h. Y Fig. 31 Problem 21. Find the center of gravity of a triangle whose alti- tude is h and whose base is a. Take the origin at the vertex and draw the 2:-axis perpendicular to the base. (See Fig. 31.) . Here dF = (y -\- y')dx, and from similar triangles ?/ + V' rr ??x, so that dF = - xdx, ^ ^ h h 36 APPLIED MECHANICS FOR ENGINEERS and 0? = hjo"^ 3 Jo ■2 ~h h. ?H- -I hJo 2_\ The center of gravity is f the distance from the vertex to the base, and since the median is a line of symmetry, it is a point on the median. It is, in fact, the point where the medians of the triangle intersect. Problem 22. Find the center of gravity of a section of a flat ring, outside radius R^ and inside radius i^g. (See Fig. 32.) Let the angle of the sector be 2 0. Take the origin at the center and let the X-axis bisect the angle 2 6. a xdF i dF here, dF = pdpda, and x = p cos a. so that \ i cos a da ' p'^dp r (pdpda Fig. 32 Integrating the numerator first, R i cos ada { p^df Rl_ 3 3 /^ + R 8 cos ada — — ^ ^(2sin0) CENTER OF GRAVITY Integrating the denominator, da I pdp = ^'J^ — ^ J ./(^ = (/?^2 _ /^g^) (9. Therefore, 37 If i?2 = 0, the sector becomes the sector of a circle, and x becomes 2 ^ sin e 3 ^ = ^^1-^ If the sector is a semicircle, that is, if 2 5 = ir, then, since = ^, - 2p fll Fig. 33 Problem 23. Find the center of gravity of a portion of circular wire (Fig. 33) of length L and whose chord = 2h. Take the center of the circular arc as origin and let the x-axis bisect Z. Then ( xds X = -^ ; but R\x — ds\ dy, ds= -dy, X 'B R I dy 'ft _ 2 ii6 _ radius x chord I ~ L " arc 38 APPLIED MECHANICS FOB ENGINEERS For a semicircular wire X — Diameter TT Problem 24. Find the center of gravity of a paraboloid of revo- lution. The equation of the generating curve being z/^ = 2 jox, and the greatest value of x^ is a. Note. Use the same method as that used for the right circular cone. r Fig. ai Problem 25. Find the center of gravity of a semi-ellipse (Fig. 34) whose equation is X — J xdF s dF where dF=2ydx = 2- Va^ - x^ dx, CENTER OF GRAVITY 39 therefore ciJo _ X = —. 7^^ 1 f""** 8^ ^ ^'Va-^c/x i[.Va-.'^ + a^sin-g; _ 3 _4a ¥ * 2 Problem 26. Find the center of gravity of a hemisphere, the radius of the sphere being r. Let the equation of the generating circle of the surface be x^ -^ 7/ = r^. Then CxdP X =^^ , where dP=yTry'^dx = y7r(r^ - x^)dx^ J dP and ^^y^(r^ - a:2)rfx = yw\rx - g-J,,^-^— X = JyTrr' = ?r. Problem 27. Find the center of gravity of the area between the parabola, the y-axis, and the line AB in Problem 20. Problem 28. A quadrant of a circle is taken from a square whose sides equal the radius of tlie circle. (See Fig. 35.) Find the center of gravity of the remaining area. Fig. 35 Problem 29. Suppose that the corners A and C of the angle iron in Fig. 22 are cut to tlie arc of a circle of y\^ in. radius and the angle at B is filled to the arc of a circle of | in. radius ; what would be the change in x and ^? 40 APPLIED MECHANICS FOR ENGINEERS K Problem 30. Show that the center of gravity of the segment of a circle (Fig. 36), included between the arc 2 5 and the chord 2dy is given by ^ = — — , where F is the area of the segment. 12 F Fig. 36 25. Center of Gravity of Counterbalance of Locomotive Drive Wheel. — In Fig. 37 the drive wheel is indicated by the circle and the counterbalance by the portion inclosed by the heavy lines, the point is the center of the wheel, and a is the angle subtended by the counterbalance. The point 0' is the center of the circle forming the inner boundary of the counterbalance, and yS is the angle sub- tended by the counterbalance at this point. Let F^ repre- sent the area of the segment of radius r and F^ the area CENTER OF GRAVITY 41 « of the segment of radius r^ Also let x-^ represent tlie distance of the center of gravity of F^ from 0, and x^ the distance of the center of gravity of F^ from 0^ Then, from Problem 30, „ „ o q 8 a^ i / 8 a*^ x-^— and x^' = 12F^ ' 12 F^ But 2^2, the distance of the center of gravity of F^ from (?, So that a:, the distance of the center of gravity of the F X ~ F X counterbalance from (9, equals —^^^ — ^^^~^' ^^ ^^ seen that F^ equals area of sector minus area of triangle equals ar cos -, and similarly. i<„ = ^—-^ — ar. cos ^. 2 .> 1 w7 /3 2^ I r^ coSy-— r cos Therefore, a; = — 26. Simpson's Rule. — When the algebraic equation of a curve is known, it is expressed as y =/(2:), and the area between the curve and either axis i^ always determined by integration. In Fig. 38 the area ABCD is expressed by the integral ^^ ^^ jjjdx = jj(x)dx, when the curve represented by y —f(x) is continuous between A and B. In many engineering problems the curve is such that its equation is not known, so that approximate methods of 42 APPLIED MECHANIC^ FOR ENGINEERS obtaining the areas under the curve must be resorted to. One of these methods of approximation is known as Simpson s Rule, Suppose the curve in question is the r Fig. 38 curve AB (Fig. 39) and it is desired to find the area between the portion AB and the a;-axis. Divide the length I — a into an even number of equal parts n (here 71=10). Consider the portion OBUF 2ind imagine it mag- nified as shown in Fig. 40, Pass a parabolic arc through n D ^R c ^3^ _^ A [^ /" ^ y. f, V, E 2/4 2/5 F 2/6 y-i Vs Vo 2/10 n 7 \JV ^ Fig. 39 the points 0, D, Gr\ then the area CDEF is approxi- mated by the area of the parabolic segment CGrDI plus the area of tlie trapezoid CDEF, therefore area GGrBFF = K2/2 + yd^^+ 1(2/3 - 2 [^2 + ^4])^^' since the area of CENTER OF GRAVITY 43 the parabolic segment is | tlie area of the circumscribing parallelogram. Since EII= = A a:, this area may be 71 written lA<2/2 + 4^/3 + 2/4)- In a similar way the next two strips to the right will have -— — (y^ + 4 y. + ^g), and the next two strips, an an area. A y area, (2/6 + "^ 2/7 +2/3)' ^^^ ^^ ^^^- Adding all these so o as to get the total area under the portion of the curve AB^ we get total area h — a 3 . 10 [2/0 + Kvi + y^ + yr, + yi + y^) + -0j2 + y^ + yQ + ys) + yio] > ys D or in general for n divisions, total area = -^ [^/o + -^C^i + 2/3 + ^5 + * * * ^n-i) o n + -(2/2 + ^4 + 2/6 + ••• 2/^-2) + 2/n]. and this is Simpson's formula for determining approxi- mately the area under a curve. It is easy to see that the smaller A2:, the less the approxi- mation will be. 27. Application of Simpson's Rule. — Simpson's Rule may be made use of in determining approximately not only areas, but volumes and moments. On account of its use in adding moments Simpson's formula may be employed in finding the center of gravity of areas or volumes bounded by lines or sur- faces whose equations are not known. Suppose, for example, it is desired to E n Fig. 40 44 APPLIED MECHANICS FOR ENGINEERS know the volume and position of the center of gravity of a coal bunker of a ship as shown in Fig. 41. The bunker is 80 ft. long and the areas A^, A^, A^, A^, A^, are as follows: ^, = 400 sq.ft., ^^ = 700 sq. ft., ^2=650sq. ft., ^3=600 sq. ft., A^ = 400 sq. ft. The distance between the successive areas is 20 ft. Applying Simpson's formula for volume, Qf) Summing the values A^Xq, A^x^, A^x^, etc., we obtain 'S,vx = 80 (3) (4) A, lA^o + 4(^12^1 + Agx^} + 2 A^x^ + A^x^\, A^ A^ A^ At ^ y^ y\ ^i X\ X: X\ y L^ 1 1 1 1 1 1 1 1 1 1 -—4-— / _ ...->- 1 1 1 1 1 1 J Fig. 41 where x^ = 0, a^j = 20, x^ = 40, x<^ = 60, x^ = 80. The po- sition of the center of gravity from the fore end can now be obtained from the relation ^vx x — CENTER OF GRAVITY 45 A value of x might also have been obtained from the formula Vq + ^'l 4- v^ + ^3 + ^4 by simply adding the terms in the numerator and denom- inator. Compare the value obtained by using this formula with that ob- tained by using Simpson's formula. Problem 31. A reservoir with five- foot contour lines is shown in Fig. 42. Find the volume of water and the distance of the center of gravity from the surface of the water, if the areas of the contour lines are as follows : Aq = 0, A^ = 100 sq. ft., A, = 200 sq. ft., A, = 500 sq. ft., A, = ^^^ ^^ 600 sq. ft., .45 = 1000 sq. ft., A^ = 1500 sq. ft., Aj = 2000 sq. ft., Ag = 2500 sq. ft. Making substitutions in the Simpson's formula, it becomes, for the volume, Yo\nme = -^lA,-\.i(A, + A,-\-A,+ A,)^2(A,-^A,-\-A,)+A,-]. Summing the values AqXq^ A^x^, ^2^2* ^^-j by Simpson's formula, we have '2(VX — ^ (3) (8) IAqXq-^4:(A^x^-\-A^x^-\-A^^x^-{-AjX.) 4- 2(A2X2 + A^x^ + A^^)+A^x^'], where Xq =0 ft., x^ = o ft., arg = 10 ft., etc., so that V Both numerator and denominator are computed by Simpson's formula. Compute X by means of the formula, - _ V(^X(^ + v^x^ + vnX2 4- ^V^s 4- etc. 1^0+^1 + ^2 + ^3+ ^^' and compare with previous result. Problem 32. Compute x for the parabolic area of Fig. 30, by 46 APPLIED MECHANICS FOB ENGINEERS using Simpson's Rule, and compare the result with that obtained by- integration. Problem 33. By Simpson's Rule find the area and center of gravity of the rail section shown in Fig. 43. All = 2.05 ^10 = 2.07 A,= A^ =1.89 A.= .55 .61 1.95 A^=A9 A, = m ^5 = .51 A, = .ol and the horizontal distances are as follows : A, = .82 ^0 = 2.95 = 4'' Uc 1.24' u. 1.0'' = 4.08" w-. = 1.18" = 4.24" w. = 1.0" = 2.5" u, = 1.0" Wg = 1.24" M2 = 2.23" u^ = 5.5" u, =6" Fig. 43 Problem 34. Find the center of gravity of the deck beam section shown in Fig. 44. Use the formula — _ F^xi + F2X2 + F's^s + ^'^^• Fi + F "+ F, + etc. ' and divide the bulb area into convenient areas, say F^, F^, F-,, etc. Check the re- sult thus obtained with that obtained by balancing a stiff paper model over a knife edge. 28. DuranJ's Rule. — A method of finding the area of irregular areas was published by Professor Durand in the Engmeering Neivs^ Jan. 18, 1894. The rule states that the total area of an irregular curve equals Fig. 44 CENTER OF GRAVITY 47 (^^)[(1 - 0-f>)^^0 + (1 + 0.1)^1 + ^2 + ^'3 + ^'4 + % + ...(i+o.iK_i + (i-o.6X], .^« where the i^'s have the same meaning as before. The divi- sions may be even or odd. The student is advised to make use of this rule as well as Simpson's Rule and compare the results. Fig. 45 j,tt," 29. Theorems of Pappus and Guldinus. — Let the disk in Fig. 45 be any slice cut from a solid of revolution by two parallel planes perpendi-eular to the axis of revolution and at a distance dx apart. The volume of this slice is dv = 7r(i/^^ — y^^dx^ so that the volume of the whole solid is V == IT i Qy^ — y^^dx. The generating figure of this slice, dF^ equals {^y^ — y^dx^ and the distance of its center of gravity from the ^^-axis is ^2_^1. We have seen that then _ CydF and this, considering the expression for volume, becomes 1 y = - therefore, V 2irF yF. 48 APPLIED MECHANICS FOR ENGINEERS This may be stated as a general principle as follows : The volume of any solid of revolution is equal to the area of the generating figure times the distance its center of gravity moves. Problem 35. Find the volume of a sphere, radius r, by the above method, assuming it to be generated by a semicircular area revolving about a diameter. Problem 36. Assuming the volume of the sphere known, find the center of gravity of the generating semicircular area. Problem 37. Find the volume of a right circular cone, assuming that the generating triangle has a base r and altitude h. Problem 38. Assuming the volume of the cone known, find the center of gravity of the generating triangle. Problem 39. The parabolic area of Problem 20 revolves about the a;-axis ; find the volume of the resulting solid. Problem 40. Find the vol- ume of an anchor ring, if the radius of the generating figure is a and the distance of its center from the axis of revolution is r. Let the curve AB (Fig. 46), of length Z, be the gen- ^^* crating curve of a surface of revolution. The area of the surface generated by ds will be dF= 2 nryds,, and the area of the whole surface will be F = 2 7r i yds. The center of gravity of this curve AB is given by the expression = jfyds; Cyds jds - F y =- — -., ov F^t-Kyl- lirl CENTER OF GRAVITY 49 This may be stated as follows : The area of any surface of revolution is equal to the length of the generating curve times the distance its center of gravity moves. Problem 41. Find the surface of a sphere, radius r, assuming the generating line to be a semicircular arc. Problem 42. Find the center of gravity of a quadrant of a circu- lar wire, radius of the circle r; use results obtained above. Problem 43. Find the surface of the paraboloid in Problem 39. CHAPTER V COUPLES 30. Couples Defined. — In Art. 20, Case (c), it was shown that the resultant of two parallel forces in a plane was equal to the algebraic sum of the two forces. The con- sideration of the case when the forces are equal and oppo- site in direction, that is, where the resultant is zero, will ^^ ^i^ {a) A Pi Fig, 47 ib) now be considered. It is easy to see that since the result- ant is zero, the two forces tend to produce only a rotation of the rigid body about a gravity axis perpendicular to the plane of the forces. Such a pair of equal and opposite parallel forces is called a couple. Let it be represented as in Fig. 47 (a), the two forces being P, and d the distance between the lines of action of the forces. This distance d is called the arm of the couple ; one of the forces times 50 COUPLES 51 the arm is called the moment of the couple. It was found ill Art. 20 that the algebraic sum of the moments of the resultant and the system of parallel forces with respect to any point in their plane, is zero. In this case, since the resultant is zero, the moment of the forces of the couple with respect to any point in the plane is equal to the sum of the moments of the two forces with respect to that point. Let the point be (7, Fig. 47 (a), distant x from the force P; then — Px — P(^— d — x) represents the sum of the moments of the two forces with respect to the point C (calling distance below O negative). This sum is equal to Pc?, the moment of the couple. The student should take C in different positions and show that the moment of the two forces with respect to any point in the plane is always Pd, Since the moment consists of force times distance, it is measured in terms of the units of force and distance ; that is, foot-pounds or inch-pounds, usually. If the couple tends to produce rotation in the clockwise direction, the moment is said to be negative; and if counter- clockwise, positive. 31. Representation of Couples. — The couple involves mag- nitude (moment) and direction (rotation), and may, there- fore, be represented by an arrow, the length of tlie line being proportional to the moment of the couple, and the arrow indicating the direction of rotation. In order to make the matter of direction of rotation clear, the agree- ment is made that the arrow be drawn perpendicular to the plane of the couple on that side from which the rota- tion appears counter-clockwise. This means that if we look along the arrow pointing toward us, the rotation 52 APPLIED MECHANICS FOR ENGINEERS the couple any point appears counter-clockwise. Thus, the couple of Fig. 47 (a), whose moment is Pc?, may be represented b}^ the arrow in Fig. 48 (a), where the length of line AB is propor- tional to Pd and the couple is in a plane through B and perpendicular to AB. The line AB is sometimes called the axis of the couple; it may be drawn perpendicular to the plane of at in that plane, since the mo- ment is con- stant for any point in the plane. In a similar way, the couple (i) in Fig. 47 whose moment is Bid^ is represented completely by the arrow (5), Fig. 48, the length CD being propor- tional to F^dy IN'oTE. The arrows are placed slightly away from the ends, so that the moment arrows may not be confused with force arrows. These arrows, like force arrows, may be added algebraically when parallel, resolved into components and compounded into resultants ; the prin- ciple of transmissibility holds and also the triangle and polygon laws as seen for force arrows. Several important conclusions follow easily as a result of this arrow representation. Since a moment arrow represents both force and distance and direc- tion of rotation, it is evident that it cannot be balanced by a single force arrow even though they have the same line of action and are opposite in direction. Hence, we conclude that a single force cannot balance a couple. Fig. 48 COUPLES 63 32. Couples in One Plane. — If the couples are all in the same plane, their moment arrows are all parallel, and may- be added algebraically, so that the resultant couple lies in the same plane and its moment is the alyehraic surn of the moments of the individual couples. For example, in Fig. 49, the couples P^d^, ^2^2' ^3^3' P^d^, P^d^, P^d^^ are all in the plane (aJ) ; their resultant couple must also be in this plane, and its moment must be equal to the algebraic sum of the moments of these couples. It is evident since the above is true that a couple may be transferred to any part of its plane without changing its effect upon the rigid body upon which it acts. This means, when applied to some particular rigid body, as a closed book, that the effect of a couple acting in the plane of one of the covers of the book (book remains closed) tends to produce rotation about an axis, perpendicular to the cover through the center of gravity of the book ; and that this rotation is the same no matter where the couple acts, provided it remains always in the same plane. The moment arrow of the resultant couple will be perpendicular to the cover of the book and on the side from which the rotation appears counter-clockwise. The student should endeavor to see the application of the above theorem and to see that it agrees with his observations. Fig. 49 54 APPLIED MECHANICS FOR ENGINEERS 33. Couples in Parallel Planes. — The moment arrow represents a couple in magnitude and direction of rotation and shows that the plane of the couple is perpendicular to its line. This moment arrow represents any couple of given moment and direction in any plane perpendicular to its line. It is evident, then, that a couple may he trans- ferred to any parallel plane zvithout changifig its effect upon the rigid body upon which it acts. Applied to the case of the book in the preceding article, it may be said that the effect of the couple would be unchanged if it acted in the plane of the other cover or in any of the leaves. 34. Couples in Intersecting Planes. — Suppose all the couples in a plane (1, 2) be added and let AB (a) (Fig. 48) represent the moment arrow of the resultant couple, and let the sum of all the couples in the plane (2, 3) in- tersecting (1, 2) be represented by CD (J) (Fig. 60). These moment ar- c >— i> {()) ^ rows are perpen- ^ dicular to their respective planes and may be moved about without changing the ef- fect of the couple. Move A and to some point on the line of intersection of the two planes. The resultant moment arrow is now found by the parallelogram law. The resultant couple has a moment represented by AU and acts in a plane perpendicular to AU and making an angle a with the plane (2, 3). COUPLES 55 Problem 44. Two forces, each equal 10 lb., act in a vertical plane so as to form a positive couple. The distance between the forces is 2 ft. ; another couple whose moment is equal to 20 in. -lb. acts in a horizontal plane and is negative. Required the resultant couple, its plane, and direction of rotation. Problem 45. A couple whose moment is 10 ft. -lb. acts in the a:?/-plane ; another couple whose moment is —30 in. -lb. acts in the a:2;-plane, and another couj^le whose moment is — 25 ft. ili-lb. acts in the ?/2;-plaiie. Required the amount, direction, and location of the resultant couple that w^ill hold these couples in equilibrium, (x, y, and 2-axes are at right angles with each other in this case.) / CHAPTER VI NON-CONCURRENT FORCES 35. Forces in a Plane. — The most general case of forces in a plane is that one in which the forces are non- concurrent and non-parallel. We shall now consider such a case. Let the forces be Pj, P^, Pg, P^, etc., as shown in T Fig. 51 Fig. 51, and let them have the directions shown. For the sake of analysis, introduce at the origin two equal and opposite forces Pj, parallel to Pj, two equal and oppo- 66 NON-CONCURRENT FORCES 57 site forces Pg' Parallel to P^^ and so on for each force. The introduction of these equal and opposite forces at the origin cannot change the state of motion of tlie rigid body. The original force P^ taken with one of the forces P^, introduced at the origin, forms a couple wliose moment is P^dy The same is also true of the forces P^^ Pg, P^, etc., giving respectively moments P2<^2' ^s^h' P^d^^ etc. In addition to tliese couples there is a system of con- curring forces at the origin Pj, P^, Pg, P^, etc. The resultant of this system is, as has been shown (Art. 16), R = V(S2:)2 + (S2/)2. The moments Pid^ -^2^2' ^3^3' ^^^-^ being all in one plane, may be added algebraically (see Art. 32), giving the moment of the resultant couple as SPc?. The systeyyi of non-concurrent forces in a plane may he reduced^ then^ to a single force R at the origin {arhitrarily selected^ and a single couple whose plane is the plane of the forces. For equilibrium, JK = and SfVZ = 0, or Sac = 0, ^y = 0, and SJPtZ = 0; that is, for equilibrium, the sum of the com- ponents of the forces along each of the two- axes is zero and the sum of the moments with respect to any point in the plane is zero. Considering as a special case the case Where the forces are concurring, it is seen that Pd is always zero (see Art. 16). The case of a system of parallel forces in a plane may also be considered as a special case of the above. (Art. 20 and Art. 31). Problem 46. The following forces act upon a rigid body : a 58 APPLIED MECHANICS FOB ENGINEERS 1 TON 2 TONS force of 100 lb. whose line of action makes an angle of 45° with the horizontal, and whose distance from an arbitrarily selected origin is 2 ft. ; also a force of 50 lb. whose line of action makes an angle of 120° with the horizontal, and whose distance from the origin is 3 ft.; and a force of 500 lb. whose line of action makes an angle of 300° wdth the horizontal and whose Find the resultant force and the Fig. 52 distance from the origin is 6 ft. resultant couple. Problem 47. It is required to find the stress in the members ^5, EC, CD, and CE of the bridge truss showm in Fig. 52. ^OTE. The member AB is the member between A and B, the member CD is the member between C and D, etc. This is a type of Warren bridge truss. All pieces (members) are pin-connected so that only tw-o forces act on each member. The members are, there- fore, under simple tension or compression ; that is, in each member the forces act along the piece. Usually, in such cases there are no loads on the upper pins. Solution of Problem. The reactions of the supports are found by considering all the external forces acting on the truss. Taking moments about the left support, we get the reaction at the right sup- port, equal 4500 lb. Summing the vertical forces or taking moments about the right-hand support, the reaction at the left-hand supj^ort is found to be 3500 lb. Cutting the truss along xy and putting in the forces exerted by the left-hand portion, consider the right-hand portion (see Fig. 53). The forces C and T act along the pieces, forming a system of concurring forces. For equilibrium, then, 2x = and 2^/ = 0, giving two equations, sufficient to determine the unknowns C and T. The forces in the members CD and CE may now be considered known. 4600 LB&. I Fig. 53 NON-CONCUBBENT FOBCES 59 Cutting the truss along the line ZW and putting in the forces exerted by the remaining portion of the truss, we have the portion represented in Fig. 54. This gives a system of concurring forces of which C and 2 T are known, so that from the equa- tions ^x = and 2?/=0 the remaining forces d and e may be found. Fig. 54 The truss is pin-con- 2 TONS Problem 48. Find the stress in each of the members AB and BC of the simple roof truss shown in Fig. 55. nected, and all the members are under simple tension or compression ex- cept the horizontal piece, which is under flexure. The method of cutting the truss, employed in the preceding problem, cannot be employed to advantage here, since the stress in the horizontal piece is not along the piece. The simplest method of solution for such a case is to take the whole member in question and consider all of the forces acting. In this case we have (see Fig. 56) a system of non-concur- ring forces in a plane. For equilibrium 2^ = and 2?/ = and SPrZ = 0, from which P, mined, Fig. 55 ^1 i P, and Pj may be deter- T f 1 TON 1 TON Fig. 56 Problem 49. In the crane shown in Fig. 57 (a) find the forces acting on the pins and the tension in the tie AC. Tlie method of cutting cannot be used in this case since the vertical and horizontal 60 APPLIED MECHANICS FOB ENGINEERS members are in flexure. Taking the horizontal member and consid- ering all of the forces acting upon it, we have the system of non-con- curring forces shown in Fig. 57 (6). Three unknowns are involved, Pg, Pi, and P2, and these may be determined by three equations ^x = 0, 2?/ == 0, and ^Pd = 0. It is to be remembered that the pin pressure at E is unknown in magnitude and direction. In all such cases it is usually more convenient to resolve this unknown pressure into its vertical and horizontal components, giving two un- known forces in known di- rections instead of one un- known force in an unknown direction. This will be done in all problems given here. 4 TONS In the present case the two forces Pi and P2 are the components of the unknown pin pressure. The tension in the tie A C may be found by considering the forces acting on the whole crane and taking moments about B. Thus ^Pd = gives, calling the tension in the tie T, T35sin45° = 8000 (25), y^ SOOO (25) Fig. 57 or 35 sin 45' Problem 50. In the crane shown in Fig. 58 (a) find the tension in the ties T and T' and the comj^ression in the boom. The method of cutting may be used here to determine the tension T and the com- pression in the boom, since AB is not in flexure, if we neglect its own weight. Cutting the structure about the point A and drawing the forces acting on the body, we have the system shown in Fig. 58 b. NON-CONCURRENT FORCES 61 c/ (b) Fig. 58 The forces T^ may be considered as acting at the center of the pulley. The system of forces is concurring, so that 2x == and 2^ = are suf- ficient to determine T and C. T may be found by consider- ^ -^*v, ing the forces acting on the whole crane and taking mo- ments about the lowest point B, Note. Neglecting friction, the tension W in tlie cord sup- porting the weight is trans- mitted undiminished through- out its length. Problem 51. Find the horizontal and vertical com- ponents of the forces acting on the pins of the structure shown in Fig. 59. Suggestion. First take the vertical strip and consider all the forces acting on it. Problem 52. Find the forces acting on the pins of the structure shown in Fig. 60, the weight of the members AD, BF, and CE being 600 lb., 400 lb., and 100 lb., respec- tively. Problem 53. A traction engine is passing over a bridge, and when it is in the position shown in Fig. 61 one half of the load is carried by each truss. The weight of the engine is transmitted by the floor beams to the cross beams, and these are carried at the pin connections of the truss. Find the stress in the members AB^ BC, CE, CD, and DF, for the position of the engine shown. Fig 62 APPLIED MECHANICS FOR ENGINEERS XoTE. The floor beams are supposed to extend only from one cross beam to another. Problem 54. In Problem 50, suppose the weight of the boom to be one ton ; find the tensions T and T and the pin pres- sures. Note. The boom is now under flexure, so that the method of cutting cannot be made use of. Problem 55. A dredge or steam shovel, shown in outline in Fig. 62, has a dip- per with capacity of 10 tons. When the boom and dipper are in the position shown, find the forces acting on AB, CD, and EF. Suggestion. Consider first all the forces acting on CD, then all the forces acting on AB. Fig. 60 Fig. 61 Note. The member EF has been introduced as such for the sake of analysis ; it replaces two legs, forming an A frame. The projection of the point i^ is 6 ft. from the point E, Problem 56. Suppose the members of the structure in Problem NON-CONCURRENT FORCES 63 55 to have weights as follows: AB, 15 tons, and CD, 3 tons, not in- cluding the 10 tons of dipper and load. Find the stresses as required in preceding problem. Fig, 62 Problem 57. Suppose the beam in Problem 5 to be 20 ft. long and to have a weight of 2000 lb. ; find the pin reaction and the ten- sion in the tie. Problem 58. Assume that the compression members of the War- ren bridge truss of Problem 47 have each a weight of 500 lb. ; find the stress in the members jBC and CE. 36. Forces in Space, Non-intersecting and Non-parallel. — If a rigid body is acted upon by any system of forces in space Pj, P^') P3, P4, P5, Pg, etc., making angles with the arbitrarily chosen axes, x^ y, and ^, a^, ^j, 7^ ; a^^ ySg, 72 ; ttg, ^3, 73, etc. (see Fig. 63), it can be shown that the system may he replaced hy a single force and a single couple. The single force acts at the origin, and its direction angles are a, /8, and 7. The resultant couple acts in a plane whose direction angles are X, /i, v. Introduce at the origin two equal and opposite forces Pj parallel to the line of action of P^. One of these forces Pj together with the original P^ form a couple whose moment is P^d^^ and this couple may be represented by a moment arrow at the origin, perpendicular to the 64 APPLIED MECHANICS FOR ENGINEEBS plane of the couple (see Art. 31). We thus have re- placed the force P^ by an equal and parallel force at the origin and a couple represented at the origin by a mo- ment arrow P-^dy Proceeding in the same way with P^, Pg, P4, etc., we finally have instead of the original system Fig. 63 of non-concurring, non-parallel forces in space, a sys- tem of concurring forces P^ P^, P31 P^. etc., at the origin and a system of moment arrows P^d-^, ^2^2' -^3^3' ^^^"> represented at the origin. The forces may be combined into a single resultant as in Art. 17, and we then have NON-CONCURRENT FORCES 66 whose direction cosines are cos «= --, COS /3 = -j^, cos 7 = -^- Since the moment arrows also follow the same laws as the force arrows, we may also write tlie moment of the resultant couple, where M^ = Pi^i cos \ + P^d.^ cos ^2 + etc., 3Iy = Pi'^i cos fi^ + P^d^ cos /i2 + etc., M^ = Pid^ cos v^ 4- -^2^2 ^^^ ^2 + e^e- The direction angles of ilfare X, /i, i^, and these are de- fined as follows (see Fig. 64) : cos \ = —~, M My COS M = ^ COS V = — ^ il!f This system of forces pro- duces a translation of the body in the direction of R and a rotation about a gravity axis parallel to M. li B=0 and ilf^O, the body only rotates or is translated with uniform motion, and if M= and H^ 0^ the body has only translation with possibly uniform rotation. For equilibrium both Ii= and M= ; that is, 2x = 0, % = 0, 2^ = 0, ilf, = 0, Fig. 64 66 APPLIED MECHANICS FOB ENGINEERS My = 0, and M^ =0, or expressed in words: the sum of the components of the forces along each of the three arbitrarily chosen axes is zero^ and the sum of the moments with respect to each of these axes is zero. It may be further shown that the single force and resultant of the preceding article may be replaced by a single force and a couple whose plane of rotation is per- pendicular to the line of action of the force. Suppose iHfand R both drawn at the origin and let a be the angle between them. M may be resolved into com- ponents along and perpendicular to i2, Gr cos a along jR, and Gr sin a perpendicular to R, Gr sin a may be replaced by another couple having the same moment. Let the forces he — R and + R and allow the — R to act along the line of action of the resultant force. The other force of the couple acts along a line parallel to the direction of the resultant force. The forces + R and — R along the line of action of the resultant force neutralize each other, and we have left (a) a force, R^ parallel to the original resultant, and (5) a couple, Gr cos cc, acting in a plane per- pendicular to R. That is, the system reduces to a single force and a sin- gle couple whose plane is perpendicular to the line of action of the force, or we may say, the effect of any system of forces acting on a rigid body,, at any instant,, is to cause an angular acceleration about the instantaneous axis of rota- tion and an acceleration of translation along that axis. Such a system of forces is called a screw wrench. The instanta^^us axis is called the central axis. This axis passes thrQg2 adF— 2 i XT/ cos a sin a c?l'+ j a;^ sin^ a dF = cos^ a j ?/2c?j^ — 2 sin a cos a j xydF+ sin^ a j a;^^?^ = i'^ cos^ a — sin 2 a I xydF + Ty sin^ a. In a similar way 7'^ = I^ sin2 a + 2 sin a cos a j a^yc^i^ + Fy cos^ a. These are the required formulae for obtaining the mo- ment of inertia with respect to inclined axes. It follows that Ti \ ji — jt I ji That is, the sum of the moments of inertia of an area with respect to two rectangular axes in its plane is the same as the sum of the moments of inertia with respect to any other two rectangular axes in the same plane and passing through the same point. This states that the sum of the moments of inertia for any two rectangular axes through a point is constant. It will be seen in Art. 45 that this constant is the polar moment of inertia. 43. Product of Inertia. — The integral j xydF is called a product of inertia^ for want of a better name. In case the area has an axis of symmetry, either the x- or y-axis may be taken along such an axis. The product of inertia then becomes zero, since if x is the axis of symmetry. 76 APPLIED MECHANICS FOR ENGINEERS for every + y there is a corresponding — ?/. A similar reasoning shows the product of inertia zero when 7/ is the axis of symmetry. In such cases I'^^ = I'x C0S2 a + I'y sin2 a and j'^ = l'^ sin^ a + I'y cos^ a. When j xydF is not equal to zero it is necessary to select the proper limits of integration and sum the integral over the area in question. This is illustrated in Article 63. 44. Axes of Greatest and Least Moment of Inertia. — It is often important to know for what axis through the center of gravity the moment of inertia is least or greatest ; that is, what value of a makes /^ or /^ a maximum or a mini- mum. For any area i^, i^, and I xydF are constant after the X and y axes have been selected. Using the method of the calculus for finding maxima and minima, we have, putting / xydF= z, --—■ = (/y — i^) sin 2 a — 2i cos 2 a, Equating the right-hand side to zero, the value of a that gives either a maximum or minimum is seen to be given by the equation 9 i ^y ■i: or, what is the same thing. sin 2a = 2i and cos 2a = lu-I. ± V4 t-2+ (/^_/j2 ± V4>+ (J,-/J2 MOMENT OF INERTIA 11 It is seen upon substituting these values of sin 2 a and cos 2 a in ^'^ = 2(7, - /^)cos 2 a + 4 i sin 2 a that the positive sign before the radical indicates a mini- mum and the negative sign a maximum value for i^. Investigating the values of a which give I^ maximum or minimum values, it is seen that the value of a for which 7^ is a minimum gives 7^ maximum, and the value of a for which i^ is a maximum gives I^ minimum. These axes for which the moment of inertia is greatest and least are known as the Principal Axes of the Area. This subject will be further discussed in Art. 53. It is seen from the above that when either the x- or 2/-axis is a line of symmetry, so that I xi/dF= 0, the val- ues of a which give maximum or minimum values for 7^ and 7y are all zero. This means that the x- and y-axes, themselves, are the principal axes. 45. Polar Moment of In- ertia. — The moment of inertia of an area with respect to a line perpen- dicular to its plane is called the polar moment of inertia of the area. Consider the area repre- sented in Fig. 69 and let the axis be perpendicular to the area at its center of gravity. Let dF represent an infini- Fia. 69 78 APPLIED MECHANICS FOR ENGINEERS tesimal area and let r be its distance from the axis. Representing the polar moment of inertia by i^, we have but so that r^ = x^ + ^2, or That is, the polar moment of inertia of an area is equal to the sum of the moments of inertia of any tivo rectangular axes through the same point. It has already been shown that I^ + ly^ constant (see Art. 42) for any point of an area. 46. Moment of Inertia of Rectangle. — Let it be required to find the moment of inertia of the rectangle shown in Fig. 70 (a), with respect to the axis, x. We may write / ;=/yW. Since dF=bdy^ this becomes 4 = *TW^=^' w^ 42/ -.Y {a) Fig. 70 MOMENT OF INERTIA 79 To find the moment of inertia with respect to a gravity- axis parallel to x it is only necessary to make use of the formula Ig^ = 7'^ — Fd^^ from which we have ],^ = ^bMandk%. = ^ From comparison we may write the moment of inertia with respect to a gravity line perpendicular to x^ and the polar moment of inertia for the center of gravity 47. Moment of Inertia of a Triangle. — It is required to find the moment of inertia of the triangle shown in Fig. 70 (J) with respect to the axis a:, coinciding with the base of the triangle. We have j^ = j yHF^ where dF = xdy, But x = -(h — I/), from similar triangles, giving ft The moment of inertia with respect to horizontal gravity 36 axis may now be determined. 1^^=!^ ^—Fd?=-^^^ and "" "18" 80 APPLIED MECHANICS FOR ENGINEERS It is left as an exercise for the student to find tlie moment of inertia with respect to an axis through the vei'tex paral- lel to the base, and also the polar moment of inertia for the center of gravity. 48. Moment of Inertia of a Circular Area. — The moment of inertia of a circular area with respect to a horizontal gravity axis a;, as shown in Fig. 71, may be found as fol- lows : J^^ = j 2/^dF, Changing to polar coordinates, re- membering that y = psm6^ and dF= dp(^pd6^^ the inte- gral becomes I,, = fp^ sin^ OpdOdp. Fig. 71 This integral involves two variables, p and 0. It will, therefore, be necessary to make use of a double integra- tion. For this purpose, write '-gx sin2 edej pHp = -j 1 (1 - cos 2 e')de 4L2 4 Jo ~ * MOMENT OF INERTIA 81 The corresponding radius of gyration is kg^ = ^. On account of the symmetry of the figure tliis is the moment of inertia for any line in the plane through the center of gravity. It follows that [l --^* [ ^.-"'■* OV 1 and that < ^ 2 ^P -2- Fig. 72 49. Moment of Inertia of Elliptical Area. — Let it be required to find I^^ and I^y of the elliptical area shown in Fig. 72. The equation of the bounding curve is ^ + ^=1 a^ V' and I^y =^x\lF =fx^ 2 ydx. From the equation of the bounding curve ^ a 82 APPLIED MECHANICS FOB ENGINEERS SO that Iay = — I x^'V a? —x^ dx ^y a Jo = —[^(2 x^ - a2) V^2T^ + ~ sin-i-T" a[_8 8 a Jo = — T"'^ ^^^ therefore kgy = g. In a similar way I^^=fy^dF=Jf2xdy = "/Jo^^^^^ - / ^2/ = ^^^ and therefore ^^^ = ^. Since /^ = 7^^, + J^^, the polar moment of inertia is ab TT ^ (a2 + 62)^ and k^ = ^ Va2 + 52. It is seen that when a = b = r the equations obtained for the elliptical area are the same as those obtained for the circular area, just as they should be. 50. Moment of Inertia of Angle Section. — When an area may be divided up into a number of triangles, or rec- tangles, or other simple divisions, the moment of inertia of the whole area with respect to any axis is equal to the sum of the moments of the individual parts. This method is often made use of in determining the moment of inertia of such areas as the section of the angle iron, shown in Fig. 73. We shall now determine the moment of inertia of this section with respect to the horizontal and vertical gravity axes, Ig^ and J^^, and also with respect to an axis v (see MOMENT OF INERTIA 83 Art. 53), making an angle a with the axis x. Consider the section divided into two rectangles, one 5" x f '' which we may call jF\ and the other 3|'' x f ^', which we may call Fc^. The moment of inertia of the section, with respect to X, is equal to the moment of inertia of F^ with respect to x plus the moment of inertia of F^ with respect to x, so that la. = M^Xiy + 5(|)(.808)2 + J,(2gl)3 1 + -V-(f )(1-19/ = 7.14 in. to the 4th power. Similarly I.y = iWi-Xiy + ¥(l)(l-30)^ + iV(f )(5)^ + 6(|)(.88)=' = 12.61 in. to the 4th power. Note. The problem of finding the moment of inertia of angle sections, channel sections, Z-bar sections, and the built-up sections shown in Figs. 75, 76, 77, 78, 79, is of special interest and importance to engineers, occurring as it does in the computation of the strength of all beams and columns. gy . — i- > n > i I .^- '^ -T y ?4 t . ,/ i k: — 1.62 ^9^ 9^ Fig. 73 84 APPLIED MECHANICS FOB ENGINEERS 9^ ' r -" i 1 ^^ H\ - — —^>x-— — - -f""^ 5 .^-T »n c n gx 9y Problem 62. Find the moment of inertia of the Z-bar section shown in Fig. 74 for the gravity- axes Qx and gy. Hint. Divide the area into three rectangles. Problem 63. Com- pute the moment of inertia for the channel I Fig. 74 section, shown in Fig. 23, Problem 14, for the horizontal and vertical gravity axes. Problem 64. Required the moment of inertia of the T-section (Fig. 24, Problem 15), also the moment of inertia of the U-section (Fig. 25, Problem 16) with respect to both horizontal and vertical gravity axes. Problem 65. The section shown in Fig. 75 consists of a web section and 4 angles, as shown. Find the moment of inertia of the whole section with respect to the horizontal gravity axis. Given, the moment of inertia of an angle section with respect to its own gravity axis, g'^ is 28.15 in. to the 4th power. Problem 66. Consider the section given in Problem 65 to ^^^- '^^ be so taken that it includes two rivet holes, as indicated by the posi- tion of the rivets in Fig. 75. Compute the moment of inertia of the whole section, when the moment of inertia of the rivet holes is deducted. The distance from the center of the rivet hole to the out- side of the angle section may be taken as 3 in. Compare the result with that obtained in the succeeding problem. Problem 67. The same section shown in Fig. 75 is shown in Fig. 76 with two cover plates. Find the moment of inertia of the whole beam section with respect to its horizontal gravity axis, now that the cover plates have been added. MOMENT OF INERTIA 85 Problem 68. Find the moment of inertia of the section of a box girder, shown in Fig. 77, with respect to its horizontal gravity axis. The moment of inertia of one of the angle sections with respect to its own horizontal gravity axis, is 31.92 in. to the 4th power. Problem 69. Find the mo- ment of inertia of the column section, show^n in Fig. 78, with respect to the two gravity axes gjj and cjy. The column is built up ^^* ' of one central plate, two outside plates, and four Z-bars. The legs of the Z-bars are equal, and have a length of 3^ in. The moment of inertia of each Z-bar 1.82' T -til g — ^fl" — g section with respect to its own horizontal and vertical gravity axis is 42.12 and 15.44 in. to the 4th power, respectively. Problem 70. Find the moment of in- ertia of the section shown in Fig. 79, with respect to the horizontal and verti- cal gravity axes g-c and gy. This section is made up of plates and angles. The moment of inertia of each angle section with respect to both its own horizontal and vertical gravity axes is 28.15 in. to the 4th power. 51. Moment of Inertia by Graphical Method. — It will often be necessary to lind the moment of inertia of a plane section whose bounding curve is of a complicated form, as -18- FiG. 77 86 APPLIED MECHANICS FOR ENGINEERS in the case when it is necessary to compute the strength of rails or deck beams. The graphical method given below may be used for such cases. :5 S — EEEEH ■*4 ilt V>i* 'V d ,1J. xL 9' tL 9^ -16- 99 Fig. 78 Let the section be a rail section, Fig. 80. Draw two lines, AB and CD, parallel to the required gravity axis, at any distance, Z, apart. The present section is symmet- rical with respect to the ^/-axis, so that it will only be necessary to consider the part on one side of that axis, say the part to the right. Suppose the section divided into strips parallel to AB and CD^ and let x denote the length of one of these strips, and dy its width. For each value of X there is a length x^ found, such that Then for every point P on the original section whose coordinates are x and y^ there will be a point P^ on the transformed section whose coordinates are x' and y. Sup- pose all these points, P\ constructed, and a boundary line drawn through them. Let J^ denote the area of the orig- MOMENT OF INERTIA 87 f/U ^ uJi yiG. SM -6^v I X 1.78" -^^—g' L Q'^ T Y I h28- ^L^^Vl 16 Fig. 79 inal curve, and F the area of the transformed curve, also N= ^ydF = ^yxdy = l^x'dy = IF. But \ydF=yF (^Art. 24), where j^ is the distance of the center of gravity of F from the line AB. It follows that This locates the center of gravity. The moment of inertia will be found by substituting for each x^ x\ and for each x\ x'\ such that Every point, P\ now goes over into a point P", forming a new transformed boundary. Call the area of this last y y y^ curve F^^ , Since x^^ = x^'-^ and x^ = x^^ x^' = x'^. 88 APPLIED MECHANICS FOR ENGINEERS Fig. 80 Therefore the moment of inertia r^= ^fdF= ^y^xdy = pfx''dy=PF'^, giving the moment of inertia of the original section with respect to the line AB. To determine I^^ it is simply necessary to use the for- mula Ig-^ = I' ^ — Fd^, This gives The areas of the sections are measured by means of a planimeter. 52. Moment of Inertia by Use of Simpson's Rule. — An approximate value for the moment of inertia of irregular sections, such as rail sections, may be obtained by the use of Simpson's Rule. Let the irregular area be the rail sec- MOMENT OF INERT I A 89 tion (Fig. 43) and let it be required to find the moment of inertia of tlie section with respect to the base of the rail. We may write /; = -li/A = ylA, + y\A, + ylA^ - + y^A,, where the ^'s represent the areas and the y's the distance from the center of the ^'s to the base of the rail. In this case 2/0 == 4> Hi = 1' ^2 = 4' ^^^"> ^^^^ ^0 = --95, A^ = 1.95, 7l2 = .61, etc. (see Problem 33). A more exact summation of the terms would be given by adding by means of Simpson's Formula. This gives (Art. 2G) r= 6 r -X 3(12) + 2(2/1^2 + i/4^'4 +•••) + yX I where u^, u-^^ ti^, etc., have the values given in Problem 33. The student should compare the result obtained by this method with tliat obtained by the method of direct addi- tion given above. Compare the value obtained with that resulting when Durand's Rule (Art. 28) is used. Use both methods to find the moment of inertia of the sections in Problem 33 and Problem 34. 53. Least Moment of Inertia of Area. — In considering the strength of columns and struts it is necessary to know the least moment of inertia of a cross section, since bend- ing will take place about an axis of its cross section having such least moment of inertia. It was shown in Art. 42 that, if the moment of inertia of the area with respect to two rectangular axes in its plane is known, the moment of inertia with respect to any other axis. 90 APPLIED MECHANICS FOR ENGINEERS making an angle a with one of these, could be found. It was further developed (Art. 44) that the value of a that would render the moment of inertia a minimum was given by the equation 2 (xydF tan 2 a = — ^ —• In case either of the axes x or y is an axis symmetry, the value of a given by this criterion is zero, so that, for areas having an axis of symmetry, the axis of least moment of inertia is the axis of symmetry or the one perpendicular to it. As an illustration of the problem in general let it be required to find the least moment of inertia of the angle section shown in Fig. 73 with respect to any axis in the plane of the area through the center of gravity. Let v be the gravity axis making an angle a with the rr-axis. The problem then is to find such a value of a that Ig^ will be a minimum. From Art. 42 we have Igy = Ig^ cos^ cc " Sill ^ai xi/cixdy + Igy slu^ a. In Art. 50 it was found that Ig^ = 7.14 and Igy = 12.61. We proceed now to find the value of j xydxdy for the angle section. For this purpose, suppose the section com- posed of two rectangles, F^ (5 in. x | in.), and F^ (p^^ in. X I in.), and then find the value of the integral, for the two rectangles separately. Considering first the area F^^ and using the double integration, we get r-'' rJyLdy = P1T(zlM2I^ _ (^.495)^1 •^1.62 *^-.495 ^1.62 L ^ 2 J = .505r^^M8)!_(l:62)2n 2.222. MOMENT OF INERTIA 91 In a similar way for F^, we have .025 [^M2I^-I:995)_^1 = 3.288, 2-J Therefore, j xydxdy for the whole area of the angle sec- tion is 5.51 in. to the 4th power. From this we find tan 2 a = ^1^ = 2.02. 5.47 Therefore 2 a = 63°40', a = 31° 50'. The expression for J^„ now becomes J,, = 7.14 cos2(31° 500 - 5.51 sin(63° 45') + 12.61 sin2(31°500 = 3.72 in. to the 4th power. This gives the least radius of gyration, hg,= .84 in. Problem 71. Find the least moment of inertia Igy and least radius of gyration kg^ of the Z-section shown in Fig. 71. In this case Ig^ = 15.44 in. to the 4th power and Igy = 42.12 in. to the 4th power. Ans. Least Igy = 5.66 in. to 4th power and least kgy = .81 in. Problem 72. An angle iron has equal legs. The section, similar to that in Fig. 73, is 8 in. x 8 in. wdth a thickness of I in. Find Ig^y Igy, ^ast Igy aud least A:^. Ans. Ig^ = Igy =48.65 in. to the 4th power, Ig^ = 19.59 in. to the 4th power, kgy = 1.59 in. Problem 73. Find the moment of inertia of column section, shown in Fig. 78, with respect to an axis v making an angle of 30" with gx. What value of a gives 7^„ miiiinum in this case? 92 APPLIED MECHANICS FOR ENGINEERS 54. The Ellipse of Inertia. — It is interesting to note, at this point, the relations between the moments of inertia with respect to all the lines, in the plane of the area pass- ing through a point. We have seen that for every point in an area there is always a pair of rectangular axes for which the moment of inertia is a maximum or a minimum ; that is, there is always a pair of principal axes. The cri- terion for such axes was found to be tan 2a = __^^, which means, since the tangent of an angle may have any value from to infinity, positive and negative, that for every point there is always a pair of axes such that ^ = 0, or j xydF= 0. This means that the expression for J^ may always be reduced to the form Jy = I^ cos^ (^ + ly sin2 a by properly selecting the axes of reference, where now Jp and ly represent the principal moments of inertia. If we divide through by F^ the equation becomes ^2 -- ^2 cos^ a+k^ sin2 a. Let p=Mji, sothatyi;^ = ^and^^ = ^, fC^ Ky. K y then h\ = ^ — ^cos^ a + 2 — i: sin^ a, ky fc-^ or dividing by k% -j _ p2 cos^ ci .p^ sin2 a MOMENT OF INERTIA 93 which is the equation of an ellipse referred to the princi- pal axes of inertia as axes. It may be written It is evident that k^ is inversely proportional to p, so that the major axis of tlie ellipse is along the axis of the least moment of inertia and minor axis along the axis of greatest moment of inertia. The ellipse of inertia has no physical significance, but merely shows the relation be- tween the moments of inertia with respect to the different axes through a point. If, then, the moments of inertia for all axes in a plane, through a point, be laid off on these axes, to scale, the locus of the end points will be an ellipse. The ellipse of inertia furnishes a graphical method for finding the moment of inertia for any axis through a point. 55. Moment of Inertia of Thin Plates. — Suppose the plate of constant thickness t and unit weight 7 and let x be the distance of any dM from the axis of reference, then ^x= j x^d3I= ^ I x\iV= - M x^dF. But this expression under the integral sign is the expression for the moment of inertia of the area of one of the faces of the plate, if F represents the area of a face. Therefore, the moment of inertia of a thin plate with reference to an axis in its j^^awe equals — times the moment of inertia of the area, of its face with reference to the same axis. A similar statement is seen to hold with reference to the polar moment of inertia of a thin plate by replacing x by p the distance of c?iirfrom a point. The following re- sults are deduced at once. 94 APPLIED MECHANICS FOB ENGINEERS I, for thin circular plate : J _Kx = — I, for elliptical plate (1) major axis : J- ^-/tfirab^^^Mlfi . j^ _h (2) minor axis : J- _jt/'Trb^\_Ma'^ J. _« "^"7^ 4 J T' "^"2 (3) polar axis : ■^ffp - ~ ~~^^ 4- « ; 1 — . f^pg - — 2 — (4) central axis, 2r, making angle a with major axis: ^gv _-/t( 'rraW\ _Ma%^ j. ^ ab_ ff\ir')~ 4r^ ' "" 2r MOMENT OF INERTIA 95 56. Moment of Inertia of Right Prism ; Geometrical Axis. — The moment of inertia of a right prism of height h with respect to its geometrical axis is given by the ex- That is, the moment of inertia of a rigid prism of height h is equal to — times the polar moment of inertia of its base. 9 From this result we may write : For right circular cylinder : Right parallelopiped : therefore, where d is the diagonal of the base. Hollow cylinder : r r^hfiTT\ 7rri\ M.^, ^. . Vr'^ + r'^ Elliptical cylinder : -^^2/-y-4-C^ +^)- 4 ' ^gy- 2 57. Moment of Inertia of Right Prism ; Axis Perpendicu- lar to Geometrical Axis. — Let the axis be perpendicular to the geometrical axis through the base. Consider a tliin slice cut from the prism by two parallel planes, distant dy 96 APPLIED MECHANICS EOE ENGINEERS and perpendicular to the prism. The slice so cut may be considered a thin plate, Avhose mass is "^dvF^ where F is 9 the cross section of the prism. Suppose the distance of this slice from the base is y. Then the moment of inertia of this slice with respect to the axis through the base is {ci) Fig. 81 equal to its moment of inertia about its own parallel gravity axis plus its mass times the square of the distance between the axes. (See Art. 41.) This will be made more clear by reference to the special case of the right circular cylinder of Fig. 81 (a). Adding the moments of all the slices, we liave g ''>g \ 3/ For the right circular cylinder : I. 31 ('-J^ F=a/? A2 MOMENT OF INERTIA 97 For right elliptical cyliiider : (1) major axis : Fj, = 3ll y + — )• (2) minor axis: ![,= wf- + ^. For right rectangular cijlinder : (1) parallel to h: r, = 3lf^ + -\ (2) parallel to h, : 7j = iff (^^ + | 58. Moment of Inertia of Solid of Revolution. — Con- sider tlie moment of inertia of a solid of revolution with respect to its axis of revolution. Imagine the solid cut into thin slices, all of same thickness, by parallel planes perpendicular to the axis of revolution. Each slice is a circular disk of thickness dg and radius re, and its polar moment of inertia with respect to the axis of revolution is ^di/TTx^ • — • The moment of inertia of the solid of revo- ^ . ^ lution is the sum of the moments of the small slices, so that the limits of integration and the relation between x and y depending upon the particular solid considered. For right circular cone : The right circular cone is illustrated in Fig. 81 (6). For this case "^ Jo 2a ^ 2fi ,Wo^ ^ lOrt 10 f^^^cly. cj ■" 'Igh^^o" " 10 (/ 10 since x = -y. 98 APPLIED MECHANICS FOR ENGINEERS For a sphere : since a;^ = r^ — 7/^ = ^ ro^di/ - 2 T^dy + y'dy-) =Wy - ^ + = TZ: . A r^ = I Mr^ therefore A: 2 = f ^2. ^ 15 5 ^'5 i^or an ellipsoid of revolution : (1) prolate spheroid : 2 a^-a z a a^^-a +r g^-a zg I since ^ = - v a^ — a;^, a and v = - irah\ L^ = -^^^^^ • -— a^ = | TMTft^. 3 ga'^ 15 » (2) oblate spheroid : since ^ = t ^^^ ~" 3/^ ^^^ ^ = o ^r^^^^- 6 3 59. Moment of Inertia of Right Circular Cone. — When the moment of inertia of a right circular cone with respect to an axis through its vertex parallel to the base is to be found we may proceed as in Art. 58. MOMENT OF INERTIA 99 Imagine the cone to be cut by parallel planes into slices as shown in Fig. 81 (J). The moment of inertia of the whole cone with respect to x is equal to the sum of the moments of the small slices with respect to x. Then but x=--y h'' so that J^:=l!r!i^ + l!r?1^3^^/^3 2 + 3^2 4^ 5 g b V20 . 60. Moment of Inertia of Mass ; Parallel Axis. — If the moment of inertia of a mass with respect to an axis in space is known, it is im- portant to be able to determine its moment with respect to any par- allel axis. Let dM be the mass of a particle of the body, a' and a (Fig. 82) the parallel axes dis- taiit cZ, and r' and r the distances of the mass from the two axes. From the triangle r^ = by dM and integrating over the body, we have ^rHM=-^ ^r^HM+ ^dHM - J2 r^d cos edM, Fig. 82 r'2 -f d'^— 2 r'd cos 6, multiplying 100 APPLIED MECHANICS FOR ENGINEERS But r' COS 0=d\ the distance of dM from a plane through a\ perpcndicuhxr to the plane a^ and a, so that ^r\UI= ^r^d3I+ d^M- 2 d^d'dM, or la = la' + d'^M - 2 d3Id^, where d^ represents the distance of the center of gravity of the mass from the plane through a' perpendicular to the plane of a' and a. From this relation, if the position of the body is known and its moment of inertia with respect to an axis in space, its moment of inertia with respect to any other parallel axis may be found. In particular, if d^ = 0, that is, if the center of gravity of the body lies in a plane through a^ perpendicular to the plane of a' and a, the relation reduces to That is, the moment of inertia of a mass with respect to any axis in space is equal to its moment of inertia ivith respect to ^^ m mm^m Fig. 83 a parallel axis, lyiyig in a gravity plane ^ perpendicular to the line joining the two axes^ pins the mass times the square of the distance between the bodies. The student will notice that this relation is very similar to the one developed for the moment of inertia of plane areas with respect to parallel axes, in Art. 41. MOMENT OF INERTIA 101 Problem 74. Show tliat the moment of inertia of a right circular cylinder, altitude h, and radius of base ?-, with respect tg a gravity axis parallel to the base is Igx = mI — h "/ ). J^^id find Wui moiucnt of inertia with respect to an axis parallel to this and at a distance d from the base. Problem 75. Show that the moment of inertia of a right circular cone, altitude h, and radius of base r, with respect to a gravity axis parallel to the base is Igx = ^^ M (r- + i A-). Problem 76. Find the moment of inertia of a slender rod of length I with respect to an axis through one end and perpen- dicular to the rod. Let the cross section be F and the mass M: then Problem 77. It is required to find the moment of inertia of the cast-iron disk fly wheel shown in Fig. 83 with respect to its geomet- rical axis. Hint. The wheel may be regarded as made up of three hollow cylinders, the moment of the whole wheel being equal to the sum of the moments of the three parts. The dimensions are as follows : diameter of wheel 2 ft., width of rim and hub 4 in., thickness of rim and web 2 in., thickness of hub 1\ in., and diameter of shaft 2 in. All distances must be in feet. '^ ss JL. Wm v3><- /i>6 15 Fig. 84 102 APPLIED MECHANICS FOB ENGINEERS Problem 78. Find the moment of inertia of the cast-iron fly- wheel shown in Fig. 84 with respect to its axis of rotation. There are six elliptical spokes, and these may be regarded as of the same cross section throughout their entire length. 61. Moment of Inertia of Non-homogeneous Bodies. — When the bodies are not homogeneous, the expressions for the moment of inertia given in this chapter do not hold, since in that case 7 is no longer constant. In case the law of variation of 7 is known, as for example if 7 varies as the distance from the line, then this variable value of 7 may be used and the moment of inertia found. Let it be required to find the moment of inertia of a right circular cylinder with respect to an axis through its base, if the density varies as the distance from the base, in such a way that 7 = 2/' where ?/ is a distance measured from the base (Art. 57). Then /;= Ci'^Fdykl + '^FdyyA =[ I4F y^ F /■ FJif^.oh . ¥ 9 = £±[T,f-^^'^ If 7 varies in some other way, the proper value must be used in the integral. In most cases, however, 7 is a constant. 62. Moment of Inertia of a Mass ; Inclined Axis. — We shall now study the problem of finding the moment of inertia of a solid with respect to an axis inclined to the coordinate axes. Suppose the moments of inertia of the MOMENT OF INERTIA 103 body with respect to the three coordinate axes known from the expressions: and let it be required to find the moment of inertia of the body with respect to any other axis OA making angles a, yS, 7 with the coordinate axes. (See Fig. 85.) Let dM equal the mass of an infinitesimal portion of the body and d its distance from the axis OA. Since r^ = a;^ _|_ ^2 _(_ ^2^ q^ _ ^ ^^^ ol + y cos /3 + 2 cos 7 and (p = 7^— OA^ = (x^+ ^2 -f 22) — (x cos a+ y cos /3 + 2J COS7)2, z dM t Fig. 85 we may write = / [ (^ + ^^ + ^0 — (^ cos a + y cos /3 + 2 cos 7)^] dM, 104 APPLIED MECHANICS FOR ENGINEERS This reduces, since cos^ a + cos^yS + cos^ 7 = 1^ to J^ , = j(f+ ^2) cos2 adM+ J (^2 + ^2) cos2 /3dM + f (2j2 + 2/2) cos2 ydM— 2 cos a COS yS (xydM — 2 cos y8 cos ^KyzdM— 2 cos 7 cos a (xzdM, or io^i = -^ cos2 oc + J^/ cos2 ^ + Iz cos2 7 — 2 COS a COS ^ (xydM — 2 cos yS COS 7 r^/^c^iHf — 2 cos 7 cos a KxzdM^ which gives the moment of inertia of the body with respect to an inclined axis in terms of the moments of inertia with respect to the coordinate axes and the prod- ucts of inertia \xydM^ \yzdM^ 'diiAxxzdM. 63. Principal Axes. — If the three products of inertia KxydM^ \yzdM^ and \xzdM are each equal to zero, the expression for Iq^ reduces to the form Iqa = Ix C0S2 Ci + ly C0s2 /3 + I^ C0s2 7. In this case the coordinate axes x^ y, and z are called the principal axes for the point and the moments i^, Z^, and I^ the principal moments of inertia. If the point is the center of gravity of the body and the products of inertia are each equal to zero, the prin- cipal axes are called the principal axes of the body. It can be shown that it is always possible to select the" coordinate axes x, y, and z so that the products of inertia given in the expression for Iqj^ will each be zero. It fol- lows that for every point of a body there exists a set of rectangular axes that are principal axes. MOMENT OF INERTIA 105 Problem 79. Find the moment of inertia of the ellipsoid whose surface is given by the equation with respect to the axes a, i, and c, and with respect to an inclined axis making angles a, /3, y with a, h, and c, respectively. The volume of an ellipsoid is o o o and loA — la cos2 a + /j COS^ ^ -\- Ic cos^ y. 64. Ellipsoid of Inertia. — It is always possible to re- duce the expression for Iqj^ to the form J^^ = I^ COS^ OL + ly cos^ 13 + I^ cos^ 7, by selecting the axes x^ y, and z so that the products of inertia are zero. Dividing this equation through by M^ we have ^Ia = ^1 cos2 a + kl cos2 /3 + k^ cos^ 7. Let /) = ^f^% so that "'X so that Z. -^rzM. h — P^OA r.j.A -L ^ h"^OA ^0 A^y ^^ ^^ ^x^y which when divided through by ^^^ becomes 1 — (£_52?_^4. (p c^^-"^ yQ)^ Cp cos 7)' 106 APPLIED MECHANICS FOR ENGINEERS This is seen to be the equation of an ellipsoid whose semi-axes are k^ky^ kjc^^ and kjcy\ the equation may be written 1 = X' k^ ky + 2/' 2 7.2 + 2 7_2 f^x '^z '^x f^y where x\ y\ and z^ represent the coordinates of a point on the line OA at a distance p from 0. It is evident that if we draw all the lines through and then locate all points x^^ y^^ and z^ on these lines, such that p = ^ the locus of all the points will be the ellip- k soid of inertia for that point. For the position of the co- ordinate axes selected, the principal axes of the ellipsoid of inertia coincide with the principal axes of the body. Problem 80. Write the equation and construct the inertia ellip- soid for the center of gravity of a right circular cylinder, altitude h and radius r. Problem 81. Construct the inertia ellipsoid for the center of a solid sphere of radius r. Problem 82. Show that the moment of inertia of the seg- ment of the circle F\ (Fig. 86) with respect to the axis OZ is r-V^ + lsin4 4\2 2 / the moment of the sector OBSA, minus J ah^, the mo- ment of the triangle OAB or Fig. 8(3 ^oz = Y^(2a-sin2aj, MOMENT OF INERTIA 107 and the moment of inertia of Fi with respect to 05 is — ( - — - sin a ) , the moment of inertia of the sector, minus \ ha^, the moment of inertia of the triangle ^Oi? or Iqs = — Ta - sin a M + ^ sin^^ j ~| . Problem 83. Show that the moment of inertia of the counter- balance, Fig. 37, with respect to a line through 0, perpendicular to OS, and in the plane of the wheel, is Ioz=[f^{2a-sin2a) -!^^(^2fi-sin2l3y'-^^ - F,(00r,]^ where Fo = ^ ~ar. cos ^, 00' — r. cos ^ - r cos -, and t is the thick- 2 2 2 2 ness, as explained in Art. 25. Problem 84. Find the moment of inertia of the counterbalance. Fig. 37, with respect to a line through perpendicular to the plane of the wheel. It may be written /o = j y^ [4 a - sin 2 cc -2 sin a ^1 + ?sin2^]1 - ^ fd ^ - sin 2 )8- 2 sin ^(1 + I sin^ |)] -f i^^^ - F,(00')4^^ f if 65. Moment of Inertia of Locomotive Drive Wheel. — The drive wheel may be represented as in Fig. 87, and may be considered as made up of a tire, rim, twenty elliptical spokes, counterbalance, and equivalent weight on opposite side of center, and hub. The tire, rim, and hub may each be considered as hollow cylinders whose moments may be found as in Art. 56. The moment of the spokes is easily found by considering them elliptic cylinders (see Art. 57), with the short axis of the ellipse in the plane of the wheel. The moment of the counterbalance is equal to the moment of the weights carried by the crank pin, radius of counterbalance times radius of crank 108 APPLIED MECHANICS FOR ENGINEERS The dimensions of the wheel are as follows : radius of tread 40'^, radius of inside of tire 36'^, width of tread 5'', outside radius of rim 36'', inside radius of rim 34'', width of rim 4y . There are twenty elliptical spokes 24" long, 31" X 21". The hub is 10" outside radius, 4|" inside radius, and 8" thick, radius of crank pin circle 18". The counterbalance has an outside radius of 34", an inside Fig. 87 radius of 7' 11.5", thickness 7^", mass 20.2, and distance of its center of gravity from the center 28.8", a = 94° 40', yS = 30° 20', 7 = 490 (Art. 64). We shall neglect the moment of inertia of the flange and shall consider the spokes cylindrical throughout their length, and that 10% of the moment of inertia of the spokes is included in that of the counterbalance and MOMENT OF INERTIA 109 boss. The moment of inertia of the wheel with respect to its axis of rotation will first be found. The moment of inertia consists of: I^ for tire = 415, I^ for rim = l-JU, /q for spokes = 246, I^ for hub = T, I^ for counterbalance = 344, and for boss = 73. The total moment is 1224. With respect to a gravity axis OZ, Fig. 86, in the plane of the wheel when the counterbalance is in a position where the line joining its center of gravity to the center is perpendicular to OZ, we get for the moment of the vari- ous parts: loz for tire = 207, loz for rim = 69, loz for spokes = 123, loz for hub = 3, loz for counterbalance = 279, and for boss = 73. The total moment is 755. In computing the moment of the spokes in this case, it was necessarj^ to consider that it differs for each spoke. The value 48 was obtained by computing the moment of inertia of a spoke perpendicular to OZ^ multiplying by 20, deducting 10% for the part of spokes in counterbalance and boss, and then dividing the remainder by 2. This, of course, is only a reasonable approximation. Problem 85. Compute the moment of inertia of a pair of drivers and their axle with respect to their axis of rotation. Use the data given above and assume the axle as cylindrical, the diameter being 9^' and the length 68'^ Ans, 2451. Problem 86. Compute the moment of inertia of the pair of drivers and their axle, given in the preceding problem, with respect to an axis midway between the wheels and perpendicular to the axle. Consider the counterbalance of both wheels in such a position as to give a maximum moment of inertia and the distance between the centers of the wheels 60". Ans. 314."). Problem 87. Find the moment of inertia of two cast-iron car wheels and their connecting steel axle with respect to (a) their axis of roti^tion, (b) an axis midway between the wheels and perpendicular 110 APPLIED MECHANICS FOB ENGINEERS to the axle. Consider the car wheels as composed of an outside tread, a circular web, and a hub ; each part may be considered a hollow cyl- inder with the following dimensions : tread, outside radius 16'^ inside radius 14", width o^" ; web, outside radius 14'', inside radius 5^", thickness 1.5" ; hub, outside radius 5 J", inside radius 2|", width 8" ; axle (considered cylindrical), 5" diameter and 7' 3" long. Distance between centers of wheels 60". According to the assumption made above, the flange has been neglected, the web is considered a hollow disk, and the axle of uniform diameter throughout its length. The results will be approximately as follows : (a) 40, (b) 320. Problem 88. The value 755 is the greatest value for the moment of inertia of a drive wheel with respect to a gravity axis in its plane. The least value will be with respect to an axis at right angles to this through the centers of gravity of the counterbalance and wheel. The student should compute this least moment of inertia. Problem 89. In Problem 86 the drivers have been considered as having their cranks in the same plane. In practice they are 90° apart. Find the moment of inertia with respect to the axis stated when the wheels are so placed. CHAPTER VIII FLEXIBLE CORDS 66. Introduction. — A cord under tension due to any load may be considered as a rigid body. In the analysis of problems in which such cords are consid- ered, the method of cutting or section may be used. Since the cord is flexible (requiring no force to bend it), it is easy to see that, no matter what forces are acting upon it, it must have at any point the direction of the result- ant force at that point, and so must be under simple tension. If the ^'^- ^^ cord is curved, as is the case where it is wrapped around a pulley, the resultant force is in the direction of the tangent. Consider, as the simplest case, a weight TF suspended by a cord, as shown in Fig. 88 (a). The forces acting on W are shown in (6) of the same figure. The cord has been considered cut and the force T^ acting vertically upward, 111 112 APPLIED MECHANICS FOE ENGINEERS has been used to represent the tension. Summation of vertical forces = 0, gives T = TF". When the weight W^ is supported by two cords as in Fig. 88 (c), the cords A and B are under tension and may be cut. The system of forces acting on the point is shown in (c?), where T^ and T^ represent the tensions in the cords A and 5, respec- tively. ^x=0 and Sy = give and 2^ cos a =2^^^ cos /9 These two equations are sufficient to determine the unknown tensions F and T^^. If two weights TTj and W^ are attached to the cord, as shown in the case of the cord ^5 (7i) (Fig. 89), each portion is under tension. Con- sider the cord cut at A and D and repre- sent the tensions by Fig. 89 T^ and T^ respectively, From ^x=0 and 2^ = we have an( T^ cos 7 = ^2 cos a T^ sin 7+^2 sin a^W^+ W^, A consideration of the forces .acting at 5, if we call the tension in the portion BC^ T^^ gives, when the summation of the X and y forces are each put equal to zero. and y^cos 7= T3COS/3 2\sin7- 2^3sinyS= W^ FLEXIBLE CORDS 113. In a similar way, consider the forces acting on the point c, and we have T^ cos /3 = 2^2 ^^^ ^ and T^ sin ^ + T^ sin a = W<^, Of the six equations given above only four are independ- ent; consequently, of the six quantities T^^ T^, T^, a, /3, and 7, two must be known in order to determine the other four. In general, if there are n knots such as B and O of Fig. 89, with the weights TF^ IFg, TPg, TF^, etc., attached, it will be possible to get n + 2 independ- ent equations. These will be sufficient to determine the tension in each portion of the cord and its direction, provided the tension at A, say, and its direction are known. If the weights are close together, the curve takes more nearly the form of a smooth curve. Two special cases of this kind are discussed in this chapter in Art. 68 and Art. 69. 67. Cords and Pulleys. — When a cord passes over a pulley, without friction, the tension is transmitted along its length undiminished. A weight W attached to a cord which passes vertically over a pulley is raised by a direct downward pull P on the other end of the rope. If there is no friction, P is equal to TF. In the case of a system of pulleys, as shown in Fig. 90, the cord may be considered as under the same Fig. 90 114 APPLIED MECHANICS FOB ENGINEEBS tension throughout and parallel to itself in passing from one sheave to the other. It is then possible to cut across the cords, just as was done in the case of the bridge truss, Problem 47, where the stress was along the member in each case. Cutting all the cords at and considering all the forces acting on the sheave 5, we get, calling the tension in the cord P, 6 P = TF or the tension in the cord is TF/6. A consideration of the upper sheave gives y= 7P = 7/6 (TT). The various cases of cords and pulleys that come up in engineering work may be taken up in a similar way, but in any case of cutting cords, it must be remembered that all cords attaching one part to another must be cut and the tension acting along the cords inserted before the principles of equilibrium can be applied. The consideration of the friction between cords and pulleys will be taken up in Chapter XIV. 68. Cord with Uniform Load Horizontally. — When a cord is suspended from two points A and P, Fig. 91 (a), and A. [B loaded with a uniform load horizontally in such a way that the points of attachment of the load to the cable are very close together, the cable takes Fig. 91 the form of a [a) T ib) ;rrn J wwij FLEXIBLE CORDS 115 continuous curve. The resultant tension in the cable in this case is in the direction of the cable at any point. Suppose the cable cut at the points and (7, where is the middle point and Q any point between and jB, and consider the forces acting on the cut portion. At Q there is a tension T making an angle a with the horizontal, Fig. 91 (6). At 0, the lowest point on the curve, the tension (P) is horizontal. The curve is supposed loaded with a uniform load of W pounds per linear foot, so that Wx represents the total horizontal loading. Writing down the equations Sa: = and 2y = 0, we obtain P= ^cosa, Wx = T sin a, Wx and by division tan a = — — - . But tan a = ^, giving ^ = ^. ax dx P Therefore, y = or x^ =4^^' which is the equation of the curve taken by the cable under the assumed loading. This is a parabola. The deflection of the curve at any point can be found by putting in the value of X for that point and solving for y. If I equals length of span and d the maximum deflection at the center, then d = — — . 8P The length of the cable may be found by considering the formula ^^ 116 APPLIED MECHANICS FOR ENGINEERS where ds is measured along the curve. The length OB^ or semi-length of the cable for a span Z, is, since ds^ = dx^ + (7^2, op From the equation of the curve x^ = v. we have dji_ _ Wx dx'~'~P' Therefore, 1^2 + ^ Expanding the two terms of the above expression into infinite series and adding like terms, we may express the total length of the parabola in terms of Z, TF, and P : Total length - 1 + ^^ - ^^-\ + • • 24 P*^ 708 P* or in terms of I and c? Total length = I -f ^^^ - '^^^ + 70 8^ 32^* 3^ ht In general, the convergence of either of the above series will be sufficiently rapid that only the first two terms need be used. In such cases they will be found con- venient for computation. FLEXIBLE CORDS 117 The point of maximum tension in the cable is determined by considering the equation P = T cos a ov T = cos a It is easy to see that T is greater, the smaller the value of cos a, that is, the larger the value of «, and this is greatest at the points of attachment A and B, Tliis is the problem of the suspension bridge where tlie weight of the cable is neglected. 69. Equilibrium of a Flexible Cord Due to its Own Weight. — Where the loading along a flexible cord or cable is uni- form, as in the case where the cable bends due to its own weight, the shape of the curve taken is no longer a parabola, as will be shown in what follows. Suppose a portion of the cable, 00, Fig. 92, with its load, cut free, and let the tensions in and C ( is the middle point and O is any point between and jB) be P and T respectively. Then, 2x = and 2^/ = give P = T cos a and W8= T sin a, and from these two equations, by division, we have tan a = Ws dy dx or -^ = Ws ~P' where 8 represents distance along the curve, and is re- lated to X and y in such a way that ds'^ = do^ + dy'^. Eliminating dy between this and the previous equation, we obtain ds P ds W ' dx = >|i + p2 4 p2 jy2 + s ,2 118 APPLIED MECHANICS FOR ENGINEERS Therefore, ^ = w" ^^Se ( ^ + \ "ttT^ + "^ =w''^^ s + 4 F2 + s^ W («) This gives a relation between x and s. A relation be- tween y and s may be obtained by eliminating dx between the equations -^ = -— - and d8^ = dx^ + dy^. This gives dx P dy=^ sds + Therefore, y =V^+4=V- h S^ — ■ W^ w (5) Eliminating s between (a) and (6), we get the equation of the curve taken by the cable to be 4- — = — (WX VVX\ (0 This is the equation of the catenary with the origin at the vertex and the y-axis the axis of symmetry. If the length of span is Z, the maximum deflection of the cable at the center may be determined by substituting 2: = - in (c). The semi-length of the cable may be found I from (a) by substituting x = ~ and solving for s. For this purpose (a) may be written in the exponential form ; FLEXIBLE CORDS 119 remembering that a"^ = n may be written m = log^ n, we then have p Since P = T cos a or T= , it is evident that the maxi- COS a mum tension occurs at the supports A and B. 70. Representations by Means of Hyperbolic Functions. — Equations (c) and (cZ) may be expressed in a simpler form by using the hyperbolic functions, remembering that the hyperbolic sine and cosine are expressed . . Wx sinh — , Wx cosh — P 2 We have for equation ((?) , P P , Wx '"-w^w'^'^'-p-^ and for equation (c?) p . ^ W^ '"If 7^- It is evident that for rapidity of computation of s and ^, tables giving the values of the hyperbolic sine and cosine, for various values of — ^, would be convenient. For this reason a table is given in Appendix I, and the student is requested to use this table in solution of the problems. e Wx p — e Wx P 2 9 e Wx p + e~ Wx P , 120 APPLIED MECHANICS FOR ENGINEERS Problem 90. A suspension bridge as shown in Fig. 93 has a span of 1200 ft. and the cable a maximum deflection at the center d = 120 ft. The weight of the floor is 2 tons per linear foot. Find the Fig. 93 equation of the cable and the tension at and at B. If the safe strength of cable is 75,000 lb. per square inch, find the area of wire section of cable necessary to support the floor. Problem 91. Find the length of the cable in the preceding problem. Problem 92. A flexible wire weighing ^ lb. per foot is sup- ported by two posts 200 ft. apart. The horizontal pull on the wire is 500 lb. Find the deflection at the center and the length of the wire. Problem 93. What pull will be necessary in Problem 92 so that the greatest deflection will not be greater than 6 in.? What is the length of the wire for this case? Problem 94. Find the tension in the wire of Problem 92 at the supports. In practice use is often made of the fact that the exponential func- tion may be expanded into an infinite series, so that P P .Wx V H = — cosh ■ ^ W W P FLEXIBLE CORDS 121 Wx may be written, remembering the meaning of cosh ■). ^ 2r2S^r3^ In a similar way we may write P ( 2 Wx WV W^x^ 2W\ P 3P3 3.4.5P6 "^ W2x3 Wl Since tan a at the supports may be written tan a = sinh , we may write it as the following series : ir 48 i^3 2 3 4 5 321^5 When the series are rapidly convergent, only the first terms need be used, so that TF.r2 , W^x^ s = x -\ , 2.3P2' and tana = I^+^^-^ P 6P3 When y = c?, x = -, so that d —- —— + 8P 2.3.4.I6P8 or approximately _ JVli . or, 7^ - Wl'^ d-^-^,andsoJ>_-^. 122 APPLIED MECHANICS FOR ENGINEERS Also at the supports tan a is approximately jn 4 €i When X — -t ^"^^ = 21> I s — - A = - H . 2 48P^ 2 6Z The total length of the cable or wire may then be expressed as Total length = Z + 3« The student should make use of these formulae in solving the pre- ceding problems. CHAPTER IX MOTION IN A STRAIGHT LINE (RECTILINEAR MOTION) 71. Velocity. — The velocity of a body is its rate of motion. If the velocity is constant (uniform), it may be defined as the ratio of the distance passed over to the time spent in passing over that distance. If the velocity is variable, the velocity at any instant is the velocity that the body would have if at that instant the motion should become uniform. Speed is sometimes used instead of velocity, especially in speaking of the motion of machines or parts of machines. Speed, however, involves only the rate of motion without reference to the direction of motion, while velocity involves both rate of motion and the direction in which the motion takes place. Since con- stant velocity is the ratio of distance to time, it may be represented as 8 t The units for measuring velocity are those of distance and time, usually feet and seconds. Thus a body has a velocity of k feet per p second or a train has ^ 9 7 a velocity of k miles F^ i' -A*— H 1 Fig. ^ per hour. A formula for expressing the relation between velocity, distance, and time for variable velocity may be derived by 123 124 APPLIED MECHANICS FOR ENGINEERS referring to Fig. 94. Suppose a body to have moved from to P over the distance s with variable velocity. Let V be its velocity at P and t the time. In moving to another position P^ distant As, the velocity changes by an amount Av and the time by an amount Af, so that at P^ V + Av = — ; A^ as Av^ As, and At approach zero as a limit, that is, the velocity is the first derivative of the distance with respect to time. 72. Acceleration. — Acceleration may be defined as the rate of change of velocity. If the velocity changes by equal amounts in equal times, tlie acceleration is said to be constant or uniform^ otherwise it is variable. Constant acceleration, then, is the ratio of the velocity to time; rep- resenting this acceleration by a^, we have The units used are those of velocity and time, and since velocity is usually expressed in terms of feet and seconds or feet per second, acceleration is usually expressed in terms of feet per second per second. This is sometimes expressed as feet per square second or simply as feet per second, it being understood that the time must enter twice. Since the acceleration equals the rate of change of velocity at any instant, we may write MOTION IN A STRAIGHT LINE 125 ^ dv d^s d = — = ? dt dt2 where a represents the variable acceleration. Since v = — and a = — , vdv = ads, by eliminatine^ dt. dt dt J & ( ^- ^^-- 73. Constant Acceleration. — When the acceleration is constant, we have the relation dv = a^dt^ a^ representing the constant value of a, and therefore . %/ Vq c/0 V = a^« + Vo ; and since or ds V = —i dt £ds = ajjtdt + In a similar way the relation vdv = a^ds gives £vdv = a,j^ds; therefore v^ vl 2 2 ~^«*' or These equations of motion give the velocity in terms of time, the distance in terms of time, and the distance in terms of velocity. 126 APPLIED MECHANICS FOR ENGINEERS 74. Freely Falling Bodies. — Bodies falling toward the earth near its surface have a constant acceleration. It is usually represented by ^ and equals approximately 32.2 ft. per second squared. The value of g varies slightly with the height above the sea level and the latitude, but for the purposes of engineering it may usually be taken as 32.2. The equations of motion for such bodies are, then, v=gt + VQ, 8 = ^gt^ + VQt, V^ — Vq If the body falls from rest, v^ = 0, and the equations of motion become v=gt, This latter is often written v^ = 2gh^ where h= s. 75. Body Projected vertically Upward. — When a body is projected vertically upward from the earth, the accelera- tion is constant and equals —g. If the velocity of pro- jection is ^Q, the equations of motion are V = — gt + v^ , 2 2 Problem 95. A body is projected vertically downward with an initial velocity of 30 ft. per second from a height of 100 ft. Find the time of descent and the velocity with which it strikes the ground. MOTION IN A STRAIGHT LINE 127 Problem 96. A body falls from rest and reaches the ground in 6 sec. From what height does it fall, and with what velocity does it strike the ground ? Problem 97. A body is projected vertically upw\ard and rises to the height of 200 ft. Find the velocity of projection v^ and the time of ascent. Also find the time of descent and the velocity with which the body strikes the ground. Problem 98. A stone is dropped into a well, and after 2 sec. the sound of the splash is heard. Find the distance to the surface of the water, the velocity of sound being 1127 ft. per second. Problem 99. A man descending in an elevator whose velocity is 10 ft. per second drops a ball from a height above the elevator floor of 6 ft. How far will the elevator descend before the ball strikes the floor of the elevator ? Problem 100. In the preceding problem, suppose the elevator going up wdth the same velocity, find the distance the elevator goes before the ball strikes the floor of the elevator. 76. Newton's Laws of Motion. — Three fundamental laws may be laid down which embody all the principles in accordance with which motion takes place. These are the result of observation and experiment and are known as Newton's Laws of Motion, First Law, Every body remains in a state of rest or of uniform motion in a straight line unless acted upon by some unbalanced force. Second Law. When a body is acted upon by an unbal- anced force, motion takes place along the line of action of the force, and the acceleration is proportional to the force applied. Third Law. To every action of a force there is always an equal and opposite reaction. 128 APPLIED MECHANICS FOR ENGINEERS The first law has already been made use of, and also the third law — see articles of Chapter II. The second law states that in case the system of forces acting on the body is unbalanced, the motion is accelerated. Motion takes place in the direction of the resultant force with an acceleration proportional to the force. It also implies that each force of the system produces or tends to produce an acceleration in its own direction propor- tional to the force. That is to say, each force produces its own effect, regardless of the action of the other forces. As a result of this latter fact, if a body is acted upon by a force P and the earth's attraction Gr^ we have P: a = a:g, where Gr is the weight of the body, g the acceleration of gravity, and a the acceleration due to the force P. From this it follows that g that is, the accelerating force equals the mass (see Art. 7) times the acceleration. 77. Motion on an Inclined Plane. — A body (see Fig. 95), of weight (r, moves down an inclined plane, with- out friction under the action of a force Gi sin 6. The acceleration down the plane equals the accelerating force Fig. 95 divided by the mass (see Art. 76) = j^ — = g sin 6, The acceleration is constant. 1 MOTION IN A STRAIGHT LINE 129 The equations of motion for such a case, then, are (see Art. 73) v = {g sin 6) f + v^^ 8 = \g (sin 0) f3 + v^t, s = ?. 2 gr sm 6 If the body starts from rest down the plane, Vq = 0. If it be projected up the plane with an initial velocity v^, the acceleration equals —g sin 6. Problem 101. A body is projected up an inclined plane which makes an angle of 60° with the horizontal with an initial velocity of 12 ft. per second. Neglecting friction of the plane, how far up the plane will the body go? Find the time of going up and of coming down. Problem 102. A body is projected down the plane given in the preceding problem with a velocity of 20 ft. per second. How far will it go during the third second? Problem 103. Suppose the body in the preceding problem meets a constant force of friction F = 10 lb. What will be the acceleration down the plane? How far will it go during the second second ? Problem 104. A boy who has coasted down hill on a sled has a velocity of 10 mi. per hour when he reaches the foot of the hill. He now goes on a horizontal, meeting a constant resistance of 25 lb. If the combined weight of the boy and sled is 75 lb., how far will he go before coming to rest ? Problem 105. Suppose that in the preceding problem the boy weighs 65 lb. and the sled 10 lb., and that the boy can exert a force of 20 lb. horizontally to keep him on the sled. Will the boy remain on the sled when the latter stops, or will he be thrown forward ? 130 APPLIED MECHANICS FOR ENGINEERS G^ = 20 LBS. Fig. 96 Problem 106. A body whose weight G = D lb. is being drawn up an inclined plane as shown in Fig. 96 by the action of the weight G = 20 lb. Suppose the resistance offered by the plane F = 10 lb., and that G starts from rest. How far up the plane will G go in 6 sec? Problem 107. Two weights, Gi = 6 lb. and G2 attached to an inextensible cord w^hich runs over a pulley, are acted upon by gravity ; no friction ; mo- tion takes place. Find the tension in the cord, and the acceleration. Consider G2 and Gi separately with the forces acting upon them, and call the tension in the cord T. Then apply the principle " acceler- ating force equals mass times acceler- ation, '^ 10 lb., Fig. 97, T G. Problem 108. Fig. 98 ing problems, descending? An elevator. Fig. 98, whose weight is 2000 lb. is descending with a velocity, at one instant, of 2 ft. per second, and at the next second it has a velocity of 18.1 ft. per second. Find the tension T in the cable that supports the elevator. Problem 109. Suppose the elevator in preced- ing problem going up with the same acceleration. Find the tension in the cable if the elevator starts from rest and attains its acceleration in 3 sec. Problem 110. A man can just lift 200 lb. when standing on the ground. How much could he lift when in the moving elevator of the preced- (a) when the elevator was ascending? (b) when MOTION IN A STRAIGHT LINE 131 Problem 111. Two weights, G and G\ are connected by an inextensible flexible cord that passes over a friction- less pulley, as shown in Fig. 99. G = 20 lb., G' = 100 lb., and there is no friction on the plane. Find the tension in the cord and the acceleration of the two bodies. Fig. m Problem 112. A 30-ton car is moving with a velocity of 30 mi. per hour on a level track. The brakes refuse to work. Ho^y far will the car go after the power is turned off before coming to rest, if the friction is .01 of the weight of the car? 78. Variable Acceleration. — It has already been shown fl Q Cm 1) Cr^ that V = — . a = — =— ^, and vdv = ads. These relations dt dt dt^ hold true no matter whether the acceleration is constant or variable. If the acceleration is constant^ the equations of motion are those that have already been worked out (see Art. 73), and by simple substitution in these equa- tions it is possible to find the velocity in terms of the time, the distance in terms of time, and the distance in terms of velocity. If the acceleration is variable, it is necessary to work out the equations of motion for each case. This may be done, when it is known how a varies, by means of either of the equations. a = or d^ dfi' vdv = ads. The latter equation will usually give the beginner less difficulty. 132 APPLIED MECHANICS FOR ENGINEERS 79. Harmonic Motion. — Let it be supposed that a body is moved by an attractive force which varies as the dis- tance. That is, the attractive force is proportional to the distance. Then the acceleration is also proportional to the distance. Let the acceleration = — ks. Then vdv = — ksds, and I vdv = —Jc \ sds ; therefore v^ — v^^ — ks\ where Vq is the initial velocity when s equals zero and k is the factor of proportionality, determinable in any special case. This equation gives the relation between the velocity and distance. Since v = Vv^ — ks^^ it is evident that v = when V^ - s= v^. This means that the body comes to rest when s has reached a certain value, viz. — ^ • From the original assumption, a= — ks^ it is ■y/k seen that the acceleration is greatest when s is greatest, Vk when s= 0. To get the relation between distance and time, the equation v = Vt^g — ks^ may be put in the form ds V » that is, when s = — % ; and is least when s is least, that is. Vv2 __ J^g2 = dt, from which, — =. sin"^ '— = f, ■Vk ^0 or -4. sin -y/kt = s. ^k MOTION IN A STRAIGHT LINE 133 This relation between the distance and time shows that as t increases s chang^es in value from — ^_ to ^, as- ^k Vk suming all values between these limits, but never exceed- ing them, since sin ^kt can never be greater than + 1 or less than — 1. The motion is, therefore, vibratory or periodic, and is known as harmonic motion. The complete period m tins case is — z. • Vyfc The relation between velocity and time may be found for this case by differentiating the last equation with respect to time. Then, V = Vq cos Vkt This shows that v^ is the greatest value of v. This motion is usually illustrated by imagining a ball attached by means of two rubber bands or springs, since the force exerted by either ^ ^ of these is proportional to L ^ _J the elongation, to two pins, IS as shown in Fig. 100. As- ^'^- ^^^ suming that there is no friction and that the ball is dis- placed to a position B by stretching one of the rubber bands, when released it continues to move backward and forward with harmonic motion. Problem 113. Suppose the ball in Fig. 100, held by two helical springs, to have a weight of 10 lb. and that it is displaced 1 in. from O. The two springs are free from load when the body is at O. The springs are just alike, and each requires a force of 10 lb. to compress or elongate it 1 in. Find the time of vibration of the body and its velocity and position after \ sec. from the time when it is released. It has been found by experiment that the force neces- 134 APPLIED MECHANICS FOR ENGINEERS sary to compress or elongate a helical spring is proportional to the compression or elongation. 80. Motion with Repulsive Force Acting. — Suppose the force to be one of repulsion and to vary as the distance ; then a = ks^ and vdv = ksds^ so that s = -\(e^~''^ - e-^''A = -^ sinh ^/kf 2Vk\ J Vyfc These equations show that as t increases s also increases and the body moves farther and farther away from the center of force. The motion is not oscillatory. 81. Motion where Resistance varies as Distance, — If a ©body w^hose weight is 644 lb. falls freely from rest through 60 ft. and strikes a -. . resisting medium (a shaft where friction on the sides equals 2 ^=10 times the distance; see Fig. 101), since accelerating force equals mass times acceleration, a-2F a-ios a = M a 9 = 9- 2* It is required to find (a) the distance the body goes down the shaft before coming to rest ; (J) the distance at which the velocity is a maximum; ((?) the total time of fall ; (c?) the velocity at a distance of 10 ft. down the shaft. After striking the shaft the relation between velocity and distance is as follows : Fig. 101 MOTION IN A STRAIGHT LINE 135 Jvdv = I 9 ds. The remainder of the problem is left as an exercise for the student. Problem 114. A ball whose weight is 32.2 lb. falls freely from rest through a distance of 10 ft. and strikes a 400-lb. spring, Fig. 102. Find the compression in the spring. It is to be under- stood that a 400-lb. spring is such a spring that 400 lb. resting upon it compresses it one inch, and 4800 lb. resting on it compresses it one foot, if such compression is possible. After the ball strikes the spring it is acted upon by the attraction of the earth and the resistance of the spring. The ac- O _ 4800 5 celeration a is then M where s is measured in feet. The relation between velocity and dis- tance is then obtained from the relation, 4800 s) ds. Fig. 102 Problem 115. A 20-ton freight car, Fig. 103, moving with a velocity of 4 mi. per hour strikes a bumping post. The 60,000-lb. spring of the draft rigging of the car is compressed. Find the com- pression s. Assume that the bumping post absorbs none of the shock. Problem 116. Suppose the car in the preceding problem to be moving with a velocity of 4 mi. per hour, what should be the 136 APPLIED MECHANICS FOR ENGINEERS strength of the spring in the draft rigging so that the compression cannot exceed 2 in. ? Problem 117. After the spring in Problem 114 has been com- pressed so that the ball comes to rest, it begins to regain its original 4 Ml. PER HR. 3wwww\/\D Fig. 103 form. Find the time required to do this and the velocity with which the ball is thrown from the spring. 82. Motion when Attractive Force varies inversely as Square Distance. — This is the case I of motion, Fig. 104, when two bodies G in space are considered, since in such cases the attractive force varies di- rectly as the product of their masses and inversely as the square of the dis- tance between them. The same at- traction holds between two opposite poles of magnets or between two bodies charged oppositely with elec- tricity. — k Suppose the acceleration = — -- and that the velocity is Fig. 104 zero. MOTION IN A STRAIGHT LINE 137 vdv= — I .,^7s, so that and VSqS — S^ -^vers 1 ^ 9 ^ -> The time required to reach the center of attraction O from the position of rest is obtained by putting s = 0. This gives ^ = —f^\2. It is seen that when s = the velocity is infinite, and therefore the body approaches the center of attraction with increasing velocity and passes through the center, to be retarded on the other side until it reaches a distance — «Q. The motion will be oscillatory. If one of the bodies is the earth, of radius r, and the other is a body of weight Gr falling toward it, the equa- tions just derived hold true. In this case it is possible to determine k. The attraction on the body at the sur- face of the earth is (7, and at a distance s is F^ so that F= (xf — ). The acceleration is therefore = — J — This gives k, then, equal to r^g. Substituting these values in the above equation, we find .=# gr^ 'VsnS — s^ 'SI h « When s = r vt the earth's surface, 138 APPLIED MECHANICS FOR ENGINEERS If Sq= cc, v= V2 gr. But this is a value of v that cannot be obtained, since Sq cannot be infinite. So that the velocity is always less than V2 gr. It is interesting to notice here that if a body were projected from the earth with a velocity greater than V2 gr^ it w^ould never return, provided there were no atmospheric resistance. Substituting ^= 32.2 and r = 3963 mi., V = 6.7 mi. per second. This is the greatest velocity that a body could possibly acquire in falling to the earth, and a body projected upward with a greater velocity would never return (neglecting resistance). If the body falls to the earth from a height A, the veloc- ity acquired may be obtained from the foregoing by put- ting s = r and SQ = h + r ; then . 2 grh ^r + h If h is small compared to r, this may be written, without serious error, /- — - V = yzgh, which is the formula derived for a freely falling body in Art. 74. 83. Motion of a Body through the Atmosphere. — When a body such as a raindrop moves through the air, the resist- ance varies approximately as the square of the velocity. Suppose a body of weight Gr projected vertically upward in such a medium and let the resistance be Ii = Ma = a=-g[l-:^v^ e-- -/'.. H U. >• 7 MOTION IN A STRAIGHT LINE 139 And the relation between velocity and distance is ex- pressed by the equation vdv= — g\l + -^vAds, or ^ a a 1 2k ° 1 + ^ 21 (7^ 1 + This gives the relation between velocity and distance. It is left as a problem for the student to determine the relation between distance and time for this case. Find the greatest height to which a body will rise and the velocity with which it strikes the ground upon re- turning. Compare this velocity with the velocity of projection. 84. Relative Velocity. — When we speak of the velocity of a body, it is understood that we mean the velocity of the body relative to the earth, more particularly the point on the earth from which the motion is observed. Since the earth is in motion, it is evident that velocity as gen- erally spoken of is not absolute velocity, and since there is nothing in the universe that is at rest, all velocities must be relative. In everyday life, however, we think of velocities referred to any point on the surface of the earth as being absolute. A person walking on the deck of a boat, for example, has a velocity relative to the earth and a velocity relative 140 APPLIED MECHANICS FOR ENGINEERS to the boat ; the former is usually spoken of as the absolute velocity, and the latter the relative velocity. Or, suppose the case of a man standing on the deck of a boat moving south with a velocity v while the wind blows from the east with a velocity Vy It is required to find the velocity of the wind with respect to the man, or, in other words, the apparent direction and velocity of the Avind as ob- served by the man. Referring to Fig. 105, we represent the velocity (see Art. 85) of the boat with respect to the earth by v^ and the velocity of the wind with respect to the earth by i^j, then V represents the velocity of the wind with respect to the man ; that is, the wind appears to the man to be coming from the southeast. The velocity V was obtained by reversing the arrow rep- resenting the velocity of the boat and find- ing the resultant of this reversed velocity and the velocity v-^ of the wind. If v-^^ be considered as the velocity with respect to the earth of a man walking- across the deck of a steamer mov- ing with a velocity v^ then F' represents the velocity of the man with respect to the boat. Fig. 105 Problem 118. An ice boat is moving due north at a speed of 60 mi. per hour, the wind blows from the southwest with a velocity of 20 mi. per hour. What is the apparent direction and velocity of the wind as observed by a man on the boat ? Problem 119. A man walks in the rain with a velocity of 4 mi. per hour. The rain drops have a velocity of 20 ft. per second in a direction making 60° with the horizontal. How much must the man incline his umbrella from the vertical in order to keep off the MOTION IN A STRAIGHT LINE 141 rain : (a) when going against the rain, (J>) when going away from the rain V If he doubles his speed, what change is necessary in the inclination of his umbrella in ( p— ^, andp — - are the direction cosines of ds^ ds^ ds^ the principal normal. From the above equations it will be seen that the result- ant acceleration a may be resolved into tangential and normal components at = —- and a^ = —-> dt p just as was done in the case of motion in a plane curve. In this case the normal is the principal normal and the radius p is the radius of absolute curvature. As an illustration of motion in a twisted curve consider the motion in a helix. The helix may be considered as CURVILINEAR MOTION 167 generated by the end of a line that moves with uniform velocity along the line OZ (Fig. 118). The edge of the thread of a screw is such a twisted curve. Let the curve Fig. 118 be given by Fig. 118, and letP be any point having coor- dinates x^ y, and z. Then x = r cos c^, ^=rsin^, 168 APPLIED MECHANICS FOB ENGINEERS will represent the curve. It follows that t; = — r sin d) -^ = — ro) sin ; V =r cos 6-^ = rco cos (f>; ^ at _ k cl(f> _cok -2 is the angular velocity of the point with respect to z; ^^ I p" represent it by co (see Art. 95), so that v = o)^ r^ -] = constant (^k is an arbitrary constant that determines the pitch of the helix). The velocity of a point moving in such a curve is constant since co is constant. The accel- eration a^ is therefore zero. We may also write a^== r cosrf)-^ = —rco cos 6; ^ ^ dt ay — r sin -^ = — rco sin ^. a. = 0. Therefore a = V a| + a^ + a^ = ray. That is, the acceleration in the direction of the resultant force is equal to (or and the accelerating force is equal to The fact of zero tangential accelerations has made this curve very useful. In many cases the helical surface formed by the revolving line has been made use of to send packages from upper floors of commercial establish- ments to the lower floors. Since a^ = 0, the packages move down with uniform motion. The helicoid is in- closed in a tube with convenient openings for the insertion of packages. CHAPTER XI ROTARY MOTION 95. Angular Velocity. — In Art. 71 linear velocity was defined as the rate of motion, and it was stated that it might be expressed as the ratio of distance to time or the rate of change of linear distance to time. The simplest case of rotating bodies is seen in uniform rotation about a fixed axis. The angular velocity is defined as the ratio of angular distance to time. Let the angular distance (measured in radians) be represented by a and the angular velocity by co. Then for uniform velocity » a and for variable velocity dt Angular velocity involves a magnitude and a direction, and may, therefore, be represented by an arrow (see Fig. 119), the length of the arrow representing the magni- tude and drawn perpendicular to the plane of motion such that if you look along the arrow, from its point, the motion appears positive or negative ; positive if coun- ter-clockwise and negative if clockwise. Velocity arrows may be compounded into a resultant or resolved into components in the same wa}' tliat force arrows were treated. For example, in Fig. 119, the 169 170 APPLIED MECHANICS FOR ENGINEERS angular velocity of a body at any instant is represented by CO. Then the angular velocity with respect to two rec- •33— Fig. 119 tangular axes x and y will be represented hj co^ = co cos \ and (Oy = co sin X, so that co^ = toj + co^. In a similar way if a body has an angular velocity CO about an axis making angles X, /x, and v with the x^ y^ and z axes, respectively, the component velocities along the axes will be given by ft)^ = ft) cos X, o)^ == ft) cos /i, ft)^ = ft) cos V so that a)2=z:«2 +0,2 +0)2. 96. Angular Acceleration. — Angular acceleration, which we shall represent by ^, may be defined as the rate of change of angular velocity, so that if ROTAIIY MOTION 171 From the preceding article and the definition of angular acceleration we may write '~~dt' ^~ dt ' '"'dt' The linear velocity and linear acceleration of a point of a rotating body may be determined in terms of the angular velocity and angular acceleration. Assume that for the instant under consideration the point is moving in the arc of a circle of radius p, over an arc ap, then pda 1 pd'^a i; = ^ = cop and at = ^^ = Op. dt dfi It will be seen that v in this case is the velocity along a tangent to the path at the point P. /- 97. Angular Acceleration Constant. — In Art. 73 we found that when the linear acceleration was constant, the equa- tions of motion reduced to a simple form. In a similar way if is constant, and the axis of rotation fixed, we have « = «o + 0^ ; (lidia = Met ; <^= — ^— ^; 20 where (o^ is the constant initial angular velocity. The expression for linear velocity and linear accelera- tion of any point P of the body becomes in this case V = a^r and at = ^r, where r is the distance of P from the axis of rotation. 172 APPLIED MECHANICS FOR ENGINEERS Problem 150. A fly wheel making 100 revolutions per minute is brought to rest in 2 min. Find the angular acceleration and the angular distance a passed over before coming to rest. Problem 151. A fly wheel is at rest, and it is desired to bring it to a velocity of 300 radians per minute in J min. Find the accelera- tion 6 necessary and the number of revolutions required. What is the velocity w at the end of 10 sec. ? Problem 152. Suppose the fly wheel in Problem 151 to be 6 ft. in diameter. After arriving at the desired angular velocity, what is the tangential velocity of a point on the rim? What has been the tangential acceleration of this point, if constant ? 98. Variable Acceleration. — In case is not constant, its law of variation must be given so that the equations of motion may be worked out. As an illustration suppose that a body moves in such a way that the angular accelera- tion 6 varies as the angular distance a. Let ^ = — - ^cc, then from the equation codco = dda^ we get (odco = — kada. Taking the limits of (o as (o^ and c», and the limits of a, and a, and of t^ and t^ we have Jcodco = — k ) ada ; therefore co = Va)2 — ka^. Integrating again, da which gives J"^"^ da r Va)2 — ka^ *^Q 1 . _iV^a — - sin ^ = t^ ■\/k «o or -^ sin -^u = <^- -\k ROTARY MOTION 173 This last equation sliows tliat a is a periodic function of the time; the motion is vibratory. Referring to Art. 79, it is seen that the motion is harmonic. In fact, if we substitute co^ = ij^p and a =^/o, we have exactly the same equation as was obtained in Art. 79. This example ap- plies to the motion of a simple pendulum, considering it as rotating about the point of support. Problem 153. The balance wheel of a watch goes backward and forward in | sec. The angle through which it turns is 1S(F; find the greatest angular acceleration and the greatest angular velocity. Problem 154. Assume the angular acceleration varies inversely as the square of the angular distance; find the relation between oj and a, and t and a. 99. Combined Rotation and Translation. — The angular acceleration of a body may be resolved into its tan- gential and normal com- ponents at = 6p and «^ = — = ofip. If now the body has in addition to its rota- tion, a translation, the total acceleration of any point P will be given by the components 6p^ od^p^ and %. In Fig. 120 the body is sup- posed to have an axis of ro- tation perpendicular to the paper, and a translation parallel to ox. Let the angle that PO makes with x be yS, so that cos (3 = - and sin ^ = --• P P ' ar^ Fig. 120 174 APPLIED MECHANICS FOR ENGINEERS Writing the x and 7/ components of the acceleration, we have a^ = — co^x Oy -aj ay= — ofiy + Ox. The tangejitial and normal components of a are, from ^'^' ^^^' fv\ fx\ at = 6p + aJ^\ a^ = co^p + aJ-V As an illustration of combined rotation and translation consider the case of a wheel of radius r rolling in a straight horizontal track. Let the acceleration of transla- tion of the center be a^, and the angular velocity of the wheel, about the center, be w, and the corresponding angular acceleration 6. Tl^en are the tangential and normal accelerations of a point on the rim situated at the top of the wheel. Problem 155. A locomotive drive wheel 6 ft. in diameter rolls along a level track. Find the greatest tangential acceleration and the greatest normal acceleration of any point on the tread, (a) when the velocity v with which the w^heel moves along the track is 60 mi. per hour, (b) when the engine is slowing down uniformly and has a velocity of 30 mi. per hour at the end of 3 min., (c) when the engine is starting up uniformly and has a velocity of 30 mi. per hour at the end of 5 min. Problem 156. A cylinder, diameter d, rolls from rest down an inclined plane, inclined at an angle <^ wdth the horizontal. What is the greatest normal and greatest tangential acceleration of any point on its circumference? I^eglect friction. BOTAllY MOTION 175 100. Rotation in General. — It lias been shown in Art. 36 that any system of forces acting upon a rigid body may be reduced to a single force and a single couple whose plane is perpendicular to the line of action of the single force. That is, the most complicated cases of rota- tion consist of an instantaneous translation combined with an instantaneous rotation at right angles to the translation. Bodies projected into the air while rotating have been mentioned in Art. 92. The projectile rotating about an axis is projected in the direction of the axis. If no forces acted upon it after leaving the gun, it could move in a straight line. It is, however, acted upon by gravity, which causes it to take a somewhat parabolic path. The resist- ance of the air causes the projectile to drift. This action of the projectile will probably be most easily explained by a consideration of the motion of a base- ball. The modern pitcher when he throws the ball gives it also a motion of rotation. The force of gravity causes the ball to take a path somewhat parabolic and the resist- ance of the air, due to the rotation, causes the ball to de- :^ — > Fig. 121 176 APPLIED MECHANICS FOR ENGINEERS fleet from the plane in which it started. The combination of the two deflecting forces makes the path of the ball a twisted curve. Different speeds and directions of rota- tion and different speeds of translation give great variety to the curves produced. The action of the baseball will be best understood by referring to Fig. 121. Let the baseball have an initial angular velocity co and an initial linear velocity v in the directions shown. The rotation of the ball causes the air to be more dense at J. than at jB, so that the ball is pushed constantly from A to B, This action causes it to deviate from the plane in which it initially moved and to take the path indicated by c. As stated above, this action in the case of a projectile is known as drift. CHAPTER XII DYNAMICS OF MACHINERY 101. Statement of D'Alembert's Principle. — A body may be considered as made up of a collection of individual particles held together by forces acting between them. The motion of a body concerns the motion of its individ- ual particles. We have seen that in dealing with such problems as the motion of a pendulum it was necessary to consider the body as concentrated at its center of gravity; that is, to consider it as a material point. The principle due to D'Alembert makes the consideration of the motion of bodies an easy matter. Consider a body in motion due to the application of certain external forces or impressed forces. Instead of thinking of the motion as being pro- duced by such impressed forces, imagine the body divided into its individual particles and imagine each of the parti- cles acted upon by such a force as would give it the same motion it has due to the impressed forces. These forces acting upon the individual particles are called tlie effective forces, D'Alembert's Principle, then, states that the im- pressed forces ivill he in equilibrium with the reversed effec- tive forces. It must be seen by the student that the principle does not deal with the forces acting between the particles of a body; these are considered as being in equilibrium among themselves. We shall see in what follows that this N 177 178 APPLIED MECHANICS FOR ENGINEERS principle, by assuming a system of effective forces acting upon the particle, enables us in many cases to apply the principles of equilibrium as developed and used in the subject of statics. 102. Simple Translation of a Rigid Body. — The principle of D'Alembert will be best understood by applying it to •' the consideration of the simpler motions of a rigid body. Let us consider the body, Fig. 121 a, and let us assume that it has simple translation parallel to x due to the action of certain impressed forces P., Fig. 121a ^ P^, Pg, etc., making angles oci, aoj ^3' ^^<^-' with X, It is seen at once that only the components of Pj, P2, P3, etc., parallel to x have any part in produc- ing motion in that direction. We may say, then, that the impressed forces are Pj cos a^, P^ cos a^-, P3 cos ccg, etc., and that these produce an acceleration a in the direction indicated. Imagine the body now divided into small particles each of mass dM^ and assume that the system of forces produc- ing the motion of the body consists of a small force dM . a acting on each particle. D'Alembert's Principle then states that these forces reversed are in equilibrium with the impressed forces. We have, then, ^x =0, DYNAMICS OF MACHINERY 179 or PjCOS ai + 1^2^^^ «2 + ^3C0S a^ + etc. — ^tlM • a = 0, or P-^cos ai + P^^os a^ + -Ps^^os a^ + etc. = ^dM • a. But since motion is parallel to x^ it is evident that Pjcos ai + P'f'^^ a^ + Pz^o^ a^ + etc. = Resultant Force = 72. Therefore, for continuous bodies, since a is the same for every particle of the body. Con- sider each particle at a distance y from x and let d be the distance of R from x ; then taking moments with respect to an axis through x and perpendicular to it, we have Rd — a \ydM = ayM^ where y is the distance of the center of gravity of the body from X (Art. 22). Dividing through by R^ we find, that is, the resultant force passes through the center of gravity of the body. 103. Simple Rotation of a Rigid Body. — We shall now apply D'Alembert's Principle to the case of a rigid body rotatinof about a fixed axis. Let B in Fisr. 122 be the body, and imagine it rotating in the direction indicated about an axis through perpendicular to the paper. Sup- pose the rotation due to the action of forces P^, P^^ Pg, P^, etc., making angles a^, ^^, 7^ a^. ^^^ j^' ^^^"> ^^'i^l^ ^ ^^^ of axes x^ 2/, ^, with origin at 0. It is evident that only the components of the forces P^, P^^ P3, P^, etc., parallel to 180 APPLIED MECHANICS FOB ENGINEERS the xz-'plaue^ Avill have any part in producing rotation. Call these projections P^, P^^ ^s^ P['> ^^^-5 ^l^^J ^^'^ ^^^ impressed forces for the motion con- sidered. The dis- tances of the lines of action of these forces from may be represented by (Jv-ttj Ctey^ ^Q^ ^49 etc. Now consider the effective forces. Imagine the body made up of individual par- ticles c^iltf situated at a distance p from 0. Eachc^iHf is acted upon by a force d3Iat=d3I6p, These are the effective forces. Equating the moments of these forces, reversed, to the moments of the impressed forces, we have 2 {Pi'di + r2.'d2 + jPs'^s + etc.) = (cUI-ep'p = 6 Cp'^d3I= 07, since j p'^dM gives the moment of inertia of B with respect to 0. (See Art. 37.) That is, when a body rotates about a fixed axis, the sum of the moments of the impressed forces in the jjlane of rotation equals 61. It is evident that any one of the forces P[^ P^, P^, F^, Fig. 122 DYNAMICS OF MACHINERY 181 etc., may be such as to offer a resistance to the indicated motion of the body. In such a case the sign of its mo- ment would be changed. 104. Reactions of Supports ; Rotating Body. — It has just been shown that one equation is sufficient to give the motion of a rigid body about u fixed axis. It is necessary, how- ever, in order to determine the re- actions of the supports, to use other equations. Consider the body B with its axis vertical, as shown in Fig. 123, and let the rotation take place as indi- cated due to the action of the forces P^, P. -£3, x'^^, etc. 2' Let Fig. 123 P^, Py^ and P^ be the reaction of the supports on the axis at 0, and P^ and P^ the reactions of the support at A on the axis. The effective forces acting upon a particle of the body B are shown in Fig. 124. Let dMhe its mass and replace 182 APPLIED MECHANICS FOR ENGINEERS the resultant force acting on d3f by its tangential and normal components and call them cZ^ and JiV respectively. It has been shown (Art. 86) that the normal force equals 3Ii and the tan- gential force equals Ma^^ and that v = cop and a^ = Op (Art. 96). We have then dN = fOJa)2p and d T= eZ JX0p, X and it is seen that both dJV and dT may be re- solved along each of the axes x^ y, and z for every dM of the body. From what has been said it is evident that the sum of the impressed forces along the a;-axis equals the sum of the reversed effective forces along the same axis. Calling the impressed forces i and the effective forces ^, we may write 2.x i — ^x^^ X mom^^J. = 2 mom^^, S mom.,, = S mom w ey^ S niom^v^ = S mom^ That is, the components of the impressed forces along each of the three axes x^ y, and z equal the components of the reversed effective forces along these axes and the moment of the impressed forces with respect to the three DYNAMICS OF MACHINERY 183 axes x^ ?/, and z equals the moment of the reversed effective forces with respect to these axes. Writing down these six conditions, we have ^Xi = — (dlVcos (f> + (dTiiin <^, 2y^- = — (dN'sm (f) — \dTcos(f>^ 2 mom^-^ ~ "" J ^^^^'^'^ • ^ + J dT cos cf) • 2, S mom,-^ = \dJVcos (f) - z+ idTsin (f) • ^, Smom^2= J dT' p. Substituting the values of tZiVand dTin the expressions on the right-hand side, we have ^dJVcos (f) = w2 Jp COS (j)d3f= afi^xdM= - ay^Mx, ^dNsin (t> =co^^ydM= - co'^M^, Jc?iV^sin ^ . 2 = ofl^ijzdM, JrfiV^cos (/) • ^ = &)2 Ja:^t7il!f, f cZrsin ^ = ^epdMsm cf> = e^ydM=e3Iy, (dTco^ (\> = e^xdM= OMx, jdTsm(f>'Z = 6^f/zdM, ^dTcos'Z = e^xzdM. 184 APPLIED MECHANICS FOR ENGINEERS The six general equations therefore reduce to the form ^Xi = QMy - 0)2 JX^, (^1) ^y. = -r 0315c - a)2ilf ^ (2) S^i=0; (3) 2 mom^^ = - «2 (yzdM + (xzdM, (4) S mom ,.^ = 0)2 fxi^i^JJr -f ^(yzdM, (5) 2 mom^^ =: r p'^d3I= 01z. (6) These equations hold true at any instant during the motion of the body. It will be seen, since x and ^ are the coordinates of the center of gravity of the body, that when the axis of rotation passes through the center of gravity, the right-hand sides of (1) and (2) become zero. It is further seen from (4) and (5) that if the body JS has a plane of symmetry as the 2:^-plane, the right-hand sides of these equations reduce to zero, since for every I xQ+ z^dM there is a corresponding I x(^ — z^dM 2iiid for every \ 7/Q+z)dM there is a corresponding \ y(^— z^dM. Therefore, ivhen the axis of rotation passes through the center of gravity and the plane xy is a plane of symmetry^ the six equations become : ^x, = 0, , (7) 2y, = 0, (8) 22, = ; (9) 2mom,-^ = 0, (1*^) 2 mom, J,— 0, (H) ^mom,,=^dI,. (12) DYNAMICS OF^MACniNERY 185 This case is the one that usually comes up in engineer- ing problems, and so these simplified equations are more often used than the six more general equations. It will be noticed that these equations are exactly the same as the conditions for equilibrium as determined in Art. 35 except that 2 mom^-^ is not zero. 105. Rotation of a Sphere. — Suppose the body B, Fig. 125, to be a cast-iron sphere, radius 2 in., con- nected to the axis by a weightless arm whose length is 6 in. Let the body be rotated by a cord running over a pulley of radius 1 in. situated 2 in. below P^. Call the constant ten- sion in the cord 10 lb. and suppose it acts in the y^-plane. Suppose a=l ft. and 6 = 6 in. Consider the motion when the sphere is in the a:^-plane. Take the xy-j^lane through the center of the sphere perpendicular to z, then ( xzdM and ( yzdM 'avq both zero. Using the foot-pound- second system of units, we have, x = ^ = -^, ilif = |, i^ = .065, (? = 8.05 1b., and Fig. 125 186 APPLIED MECHANICS FOR ENGINEERS ^- = ^'^-^^--8-?' (a) 2y,= -10-P',-P,= -^-«g^ (^) 23, = P,- 8.05 = 0; (0 2 mom,, = -P',- 10(f) + P,(|) = 0, (^) 2mom,, = P',-(8.05)(i)-P.^ = 0, (0 2momi, =lf = 6'(.065). (/) From equation (/) we get ^=12.81 radians per (second)^. Suppose the body begins to rotate from rest and that at the time under consideration it has been rotating one second. From the relation co = 00^+ 6t (Art. 97) we get €0 = z= 12.81 radians per second. Solving the remaining equations, we get P'^= - 3.623 lb.; P'^ = -1.507 lb.; P,= 8.05 lb.; P,=- 15.281 lb.; P^ = 13.616 lb. The negative signs indicate that the arrows in the figure have been assumed in the wrong direction. Problem 157. A fly wheel 3 ft. in diameter rotates about a verti- cal axis. The cross section of the rim is 3 in. x 3 in. and is made of cast-iron. Neglect the weight of the spokes. This wheel is placed on the axis in the preceding problem instead of the sphere. If the other conditions are the same, find the reactions of the supports. Problem 158. The sphere in Fig. 125 is replaced by a right circular cast-iron cone of height one foot and diameter of base one foot. The vertex is placed at the point of attachment of the sphere. If the other conditions are the same, find the reactions of the sup- ports. Problem 159. In the problem of the sphere, Art. 105, find the angular velocity at the end of 30 sec. and the reactions of the sup- ports for such speed. i^* I -1. ^ C I 1 5^ ,-u DYNAMICS OF MACHINERY 106. Center of Percussion. — It will be interesting now to consider the motion of a slender homoge- neous rod due to an impulsive force P when free to swing about a horizontal axis through one end. Let the rod be given in Fig. 126 and let be the axis perpendicu- lar to the paper about which rota- tion takes place. The length of the rod is Z, and P is the impressed force tending to produce rotation. The effective forces are repre- sented in the figure as acting on each individual particle, equal in each case to dM • a^ = dMOp, where p is the distance of dM from 0, D'Alembert's Principle for the horizontal forces gives y- ^ r 187 -- -i — X Y" c; -^ > -f{0> — p. e Fig. 126 P - P^= jdM0p = e^pdM= 0pM; similarly moments about the axis through (?, Pd = 0^pHM== 61,. But Z for a slender rod has been shown to be - Ml^ and /o = -, so that p^ = p-eMp = p-^Mp=p(i-^^^ 188 APPLIED MECHANICS FOR ENGINEERS H If Fj, = 0, then d = ^l. Under such conditions, if the rod were struck with a blow P, there would be no hori- zontal reaction at 0. This point distant 1 1 from 0, for which Pj, = 0, is called the center of percussion. The general problem of center of percussion is not quite within the scope of the present work. A right circular cylinder of height h and diameter of base d is made of cast iron. Locate its center of percus- sion, when supported as the rod in Fig. 126. 107. Compound Pendulum. — When a body rotates about a horizontal axis due to the action of gravity, it is called a compound pen- dulum. We have seen how to find the time of vibration of a simple pen- dulum and can investigate its motion completely, for small oscillations. We shall now study the motion of the compound pendulum. Let the pendulum be represented by Fig. 127 and suppose the axis of rotation is through perpendicular to the paper. Taking moments about the axis of rotation, we have Fig. 127 — Grd sin a= 6L 0' e= - Grd sin a _ Grd sin a _ gd sin a L Mkl hi It is seen that varies with sin a. We wish now to find the length of a simple pendulum that will have the same period of vibration as this compound pendulum. DYNAMICS OF MACHINERY 189 It was found, Art. 88, thtat the tangential acceleration for a simple pendulum, a^ = — ^sina, and hence its angular acceleration = — ^ -. Equating this value to the c value of found for the compound pendulum, we get as the length of a simple pendulum having the same period of vibration. This length I is the length of the compound pendulum. That is, the length of a compound pendulum is the length of a simple pendulum having the same period of vibration. The student may find this length experimentally by taking a piece of thread with a small lead ball attached to one end, and holding the other end at adjust the length of the thread until the compound pendulum and the simple pendulum vibrate simultaneously. The length of the thread is the length Z. Since I = — ^, and k^ = k§, + (P^ we may write d(l -d} = kL or 00' ' aa^ = constant, Evidently the relation is not changed if 0' and 0^' be interchanged; we may, therefore, say that the compound pendulum will vibrate with the same period when sus- pended about 0^' as an axis. This point is called the center of oscillation. The point is known as the point of sus- pension. The result may be stated as follows : m a 190 APPLIED MECHANICS FOR ENGINEERS compound pendulum the ]joiyit of suspension and the center of oscillation are interchangeable. The time of vibration of a simple pendulum was found tobei^ = 7r^/- (Art. 88). This gives for the time of vibration of a compound pendulum, gd Problem 160. A cast-iron sphere whose radius is 6 in. vibrates as a pendulum about a tangent hne as an axis. Find the period of vibration and the length of a simple pendulum having the same period. Locate the center of oscillation. Problem 161. A steel rod one inch in diameter and 3 ft. long is free to turn about a horizontal axis through one end. AVhile hanging from this axis it is suddenly acted upon by a 10-lb. force perpendicular to its length in such a way as to cause the hori- zontal component of the reaction of the support to be zero. How far from the support does the force act? Problem 162. Two drums whose radii are ?'i = 16 in. and r^ = 12 in. are mounted as shown in Fig. 128. Their combined weight is 200 lb. and Jc = U", The forces Gi and 6^2 3,ct upon the drum, as well as journal friction amounting to 16 lb. The radius of the shaft is 1 in. Find the velocities of Gi and G'2 and the drums when the point of attachment of the cord on 20 LB& Fig. 128. DYNAMICS OF MACHINERY 191 the small drum has traveled from rest at A to a point A '. Neglect the friction at B, 108. Experimental Determination of Moment of Inertia. — The computation of the moment of inertia of many bodies is a difficult matter. It is often convenient, therefore, to use an experimental method in dealing witli such bodies. The compound pendulum furnishes a means whereby such determinations may be made. From Art. 107, we find that the time of vibration of a compound pendulum is This may be written gd ^ = 5^^ ' multiplying both sides by M^ the mass of the body, we have It thus appears that if c?, the distance from to the center of gravity, is known (the center of gravity may be located by balancing over a knife edge) and also the weight Gr^ and the body be allowed to swing as a pendu- lum about as an axis, t may be determined, giving J^. If I^, be desired, it may be determined from the formula (see Art. 41), Problem 163. The connecting rod of a high-speed engine tapers regularly from the cross-head end to the crank-pin end. Its length is 10 ft., its cross section at the large end 5.59 " x 12.58'' and at the 192 APPLIED MECHANICS FOR ENGINEERS cross-head end 5.59'^ x 8.39'^ Neglecting the holes at the ends, the center of gravity is 6J: in. from the cross-head end. The rod is made of steel and vibrates as a pendulum about the cross-head end in 1.3 sec. Compute its moment of inertia. Problem 164. The student should take such a connecting rod as the one in the preceding problem and by swinging it as a pendulum find its period of vibration. Compute the moment of inertia. 109. Determination of g. — From the preceding article we see that 12 2 r *> this relation enables us to determine gr, as soon as we know Jq, i^f, and c?, by determining the time of vibration about the r 2 point 0. It is evident that __o is a constant for the ^ Md body, when the axis is through 0, and that when once determined accurately the pendulum might be used to determine g for any locality. 7-2 This constant,— , is known as the pendulum constant Md^ ^ Problem 165. A round rod of steel 6 ft. long is made to swing as a pendulum about an axis tangent to one end and perpendicular to its length. The rod is 1 in. in diameter. Determine the pendu- lum constant. Problem 166. The center of gravity of a connecting rod 5 ft. long is 3 ft. from the cross-head end. The rod is vibrated as a pendulum about the cross-head end. It is found that 50 vibrations are made in a minute. Find the radius of gyration with respect to the cross-head end. 110. The Torsion Balance. — A torsion balance consists of a body such as ABQ^ Fig. 129, suspended by means of a slender rod or wire rigidly clamped at both ends. Suppose the wire clamped at 0, and let the body ABC DYNAMICS OF MACHINERY 193 /'UU/AW//////' I) be a cast-iron disk of radius r and thickness t. The point of support is the center of gravity. The mark OA on the body is shown in the neutral position. The application of a certain torque in the plane of the disk causes it to turn through a certain angular distance so that OA assumes period- ically the positions OB and 00^ due to the resistance of the wire. It is well known that a circular rod or wire when twisted offers a resistance to the twist, such that the resisting torque varies as the angle of displacement. As the line OA moves to OB the resisting torque offered by the wire steadily increases. After the body has been given a twist ^ig. 129 the only forces tending to produce rotation are the forces in the wire. So that if we call the moment of the couple m we may write m = ca^ since the moment of resistance varies with a. If, for the particular wire in question, when a = a^ that m = m^ we may write, since c is constant, ^= -^• Taking moments about the axis of rotation, we have m = eL zi or la a. L codco da since (odco = Oda-t and the resisting torque is negative. Multiplying through by da and integrating, we get, if when a = a, (O '0' 194 APPLIED MECHANICS FOR ENGINEERS or co = ^JP^^/al-^a^. But ft) = -— , so that cZ^ c?^ =^/5 c?a ^1 Va2_^2 If we integrate, taking ^ = 0, when a = a^, we have where ^ is the time taken in turning from OB through any angle a. Suppose the upper limit zero (a = 0), then sin~i -^ = 0, TT, etc. Suppose sin~^ — = tt, then « = |V^- The value of t given represents one fourth of a com- plete backward and forward swing, so that for a complete period t = 2.^lI^. . It is seen that this time of vibration is independeyit of the initial angular displacement a^. DYNAMICS OF MACHINERY 195 It is also seen that the moment of inertia of a body might be determined by suspending it as the body ABC m is suspended. The constant (? = -— i is a constant of the wire or rod and depends upon the material and diameter. Knowing this constant, it would only be necessary to determine the period of vibration in order to find /. For practical purposes, however, it is desirable to m For this eliminate from consideration the value purpose suppose the disk provid.ed with a suspended platform rigidly attached as shown in cross section in Fig. 130. Let t be its time of vibration and I its moment of inertia about the axis of suspension. Now place on the disk two equal cylinders IT in such a way that their center of gravity is the axis of suspension. Let t^ be the period of vibration of the cylinders and support and 7j their moment of inertia. t^ I Then — = — The moment of inertia of the two cyl- ti I, inders with respect to the axis of rotation is known; Then I^ = 1+ 1^, Fig. 130 call it I^, SO that J=Z ^2 2^2_^2 This gives the moment of inertia of the torsion balance, which, of course, is a constant. The moment of inertia of any body L may now be 196 APPLIED MECHANICS FOR ENGINEERS determined by placing the bod}^ on the suspended plat- form with its center of gravity in the axis of rotation and noting the time of vibration. Calling the time of vibra- tion of the body L and the balance t^ and their moment of inertia io, we have 3 "^ Let the moment of inertia of L itself be /^, so that Then ^^ = ^^- This method may be used in finding the moment of inertia of non-homogeneous bodies, provided the center of gravity be placed in the axis of rotation. Problem 167. The moment of inertia of a torsion balance is 6300 and its time of vibration 20 sec. The body L consists of a homogeneous cast-iron disk 3 in. in diameter and 1 in. thick. Find the time of vibration of the balance when L is in place. Assume the moment of inertia of the disk. Problem 168. The same balance as that used in the preceding problem is loaded with a body L, and the time of vibration is found to be 30 sec. Determine the moment of inertia of Z. 111. Constant Angular Velocity. — The six general equa- tions may be written, PJ + P^ + '2x = OMy - oflMx, PJ + P^ + S^ = - BMx - ofiMy, PJ + P, + ^z=0, DYNAMICS OF MACHINERY 197 FJa + P^b + 2(2X- xZ-) = oy^-jxzd3I+ O^yzdM, where the P's represent the action on the bearings and 2X, Ey, and EZ represent the components of the forces producing the rotation and ^(^yZ—zY}. etc., the moments of these forces. Now if JT, y, and Z are each zero, the last equation shows that ^ = 0, and therefore the angular velocity co is constant. The condition, however, that A"^, y, and Z be each equal zero, means that the forces tend- ing to produce rotation no longer exist. Such an axis is sometimes called a permanent axis. It is a principal, axis of the body for the point. 112. Rigid Body Free to Rotate. — If in addition to the conditions that X, Y, Z, \xzdM eind \yzdMhe each zero, we impose the condition that both x and y be zero, that is, that the 2:-axis pass through the center of gravity, the equations of motion become PJ + P. = 0, P>+P,5 = 0, ^ = 0. The body is in equilibrium under the action of the reactions of tlie supports and continues to rotate with uniform velocity co about the original axis of rotation. 198 APPLIED MECHANICS FOR ENGINEERS The axis of rotation is now a principal axis of the body through the center of gravity. It is often called an axis of free rotation. Since there are three principal axes of the body through the center of gravity, there are three free axes of rotation. 113. Rotation of Symmetrical Bodies. — When a homo- geneous body having a plane of symmetry rotates with constant angular velocity cd about an axis perpendicular to that plane, the only forces acting on the body reduce to where P is the centripetal force acting through the center of gravity and p is the distance of the center of gravity of the body from the axis of rotation. The force of gravity is assumed to produce no rotation. Let the xy- plane be the plane of symmetry and suppose the axes to rotate with the body and the ^-axis be the axis of rota- tion. It is seen that equations (1) and (2) of Art. 104 are now identical and each expresses the fact 1.x = ofiMp, and the other four equations are satisfied by this condition. This will be more easily understood by applying it especially to the sphere in Fig. 125. The only force act- ing, if o) is constant and the rry-plane is the plane of symmetry, will be a centripetal force P=co^Mp^ if we neglect the tendency to rotate about the x and y axes on account of the weight of the sphere. If the body is also symmetrical with respect to the axis of rotation, we may consider for each half, P= oy^Mp^ DYNAMICS OF MACHINERY 199 where M is the mass of | of the body and p is the distance of that one half from the axis of rotation. As an illustration consider the ro- tation of a fly wheel. Suppose all the centrifugal force is carried by the rim and neglect the spokes. If the mean diameter of the wheel is r (Fig. 131), its mass M, and it rotates with constant angular velocity co about its axis, then 2F=ay'^Mp. Fig. 131 2r If the rim be considered as a thin wire, p = — (Prob. 23) TT and M= ^ irrF^ so that 9 It is seen that the tension in the rim varies with the square of the angular velocity. In particular, suppose the wheel made of cast iron and let r= 6 ft. and #= 10^' x4'^ Then P = 0)2(140.1). If the speed of rotation is six revolutions per second, we ^^^^ 0)2=1421.29 and P= 199,122 lb. Dividing by ^=40 sq. in., tlie area of cross section, we get the stress on tlie material in pounds per square inch as 4978, 200 APPLIED MECHANICS FOR ENGINEERS Problem 169. If the wheel just described should "run wild," what speed would be attained before the bursting of the rim occurred, supposing the rim to carry all the centrifugal forces? Assume the tensile strength of cast iron as 25,000 lb. per square inch. 114. Rotation of a Locomotive Drive Wheel. — The drive wheel of a locomotive, Fig. 132, may be con- sidered for the present as rotating about a fixed axis. We shall con- sider the effect of the weight of the counterbalance on the tire due to ro- tation only, on the assumption that the tire carries all the weight of the l^ 132 counterbalance. Note. It is to be understood that the wheel center carries part of the weight of the counterbalance, but a complete solution of the problem of the drive wheel is beyond the scope of this book. The above assumption is therefore made. Let M be the mass of ^ of tire and p the distance of its center of gravity from the center of wheel. Let M^ be the mass of the counterbalance, and p-^ the distance of its center of gravity from the center of wheel. Then 2 P = 0)2 (Mp + M^p^) . In particular suppose the diameter of the tread of the tire to be 80 in. ; distance of the center of gravity of ^ of tire from center 27 in., and mass of | of tire 21. The mass of the counterbalance is 20, and the distance of its center of gravity from the center of the wheel 29 in. Substitut- ing these values, we get 2 P =0)2 [ 21(f J) + 20 (fl)] = 95.5o)2. DYNAMICS OF MACHINERY 201 If now we know the speed of rotation of the wheel so that CO is known, we may determine P. Let us take co corresponding to a speed of train of 60 mi. per hour. This gives co = 2QA radians per second and F = 33,380 lb. Problem 170. If the area of a cross section of tire is 20 sq. in. under the assumption given above, the stress on the metal due to rotation about the axis would be P divided by 20, or 1669 lb. per square inch. Problem 171. If the allowable stress on the metal is 20,000 lb. per square inch, the value of o) necessary to develop such a stress is given by 0) = M 20,000 X 20 47.7 = 91.5 radians per second. This corresponds to a speed of train of 207 m. per hour. Problem 172. Consider the rotation to take place about a point on the track. Find P for a speed of train of 60 rni. per hour. Find the corresponding stress in pounds per square inch. What speed of train would develop a stress in the tire of 20,000 lb. per square inch? 115. Rotation about an Axis not a Gravity Axis, the center of gravity of a rotating part of a machine is not on the axis of rota- tion, there is a force tend- ing to bend the shaft equal to ofiMp. The distance p is the distance of the center of gravity from the axis. Suppose the body be a disk of steel, Ficy. 133, Fig. 133 •When 202 APPLIED MECHANICS FOR ENGINEERS radius 6 in., and tliickness 1 in., and let to == 60 tt radians per second. If the disk is off center -| in., the force perpendicuhir to the sliaft due to the unbalanced mass is given by the equation = (60 7r)Va)2 J,- II 0^ (^,) =1241 lb. Such a disk might be balanced by the addition of a proper Aveight placed with its center of gravity diametri- cally opposite the center of gravity of the disk and in the plane of the disk. For static balancing it would not be necessary for the added weight to have its center of gravity in the plane of the disk, but for rotation this is necessary, as will be shown in what follows. Let the shaft AB^ Fig. 134, . . carry weie^hts G- ']' ¥L G, Fig. 134 and (7^, as shown. When Gr is up, it tends to lift the end A of the shaft, due to its centrif- ugal force. When Gti is up, the end B of the shaft is lifted. Thus for each revolution, the end ^ is lifted, and then the end B ; that is, the shaft wobbles about the point (7. What really happens is something more than the mere lifting of the shaft in the bearing. Each end describes a cone with as the apex. One wheel, improperly balanced, when rotated on a shaft, causes the shaft to wobble about one of the bearings DYNAMICS OF MACHINERY 203 or to lift bodily. Two wheels improperly balanced on the same shaft, cause the shaft to wobble about some point C. The principle involved in the balancing of rotating parts is made clear by considering two masses, M^ and M^, distant r-^ and r^-, respectively, from the axis of rotation. Suppose these bodies to be in the same plane. For static balance it is necessary that M^r^ = M^r^, but for dynamic balance, in addition to this, we must have M-^r^ = M^r^. It follows, therefore, that for static balance an equivalence of moment is required^ while for dynamic balance an equivalence of both masses and distances is re- quired. If the counterbalance on the locomotive drive wheel (see Fig. 87) does not balance perfectly the parts on the opposite side of the center and the reciprocating parts, the unbalanced mass will cause the locomotive to lift up when it is above the center. When it is below the center, the weight comes down on the rail with what is known as a "hammer blow." This becomes very destructive to both rail and wheel at high speeds when the unbalanced mass is at all large. It is much the same in effect as the dropping of the weight on the driver through a given distance. Suppose the counterbalance is too heavy by 64.4 lb., and that its center of gravity is 30 in. from the center of the wheel. If the locomotive is making 60 mi. per hour, and the drivers are 80 in. in diameter, the approx- imate lifting force when the counterbalance is up is, from P= (omp, 3780 lb. For a speed of 100 mi. per hour, this lifting force would be about 10,980 lb. In 204 APPLIED MECHANICS FOR ENGINEERS either case this weight applied suddenly to the rail must be very destructive to both wheel and rail. Problem 173. A steel disk 3 ft. in diameter and 1 in. thick is not perpendicular to the axis of rotation, but is out of true by yJo of its radius. Find the twisting couple introduced tending to make the shaft wobble. Problem 174. If the unbalanced weight in a drive wheel in the above illustration is 200 lb., find the centrifugal force for a speed of train of 60 mi. per hour. 116. Rotation of the Fly Wheel of Steam Engine. — Let the fly w^heel be given as in Fig. 135 w4th radius r, and suppose that a belt runs over it horizontally as shown by — -. n ^- -2a- FiG. 135 Pj and P^. The effective steam pressure is P; iV^ is the pressure of the guides on the crosshead. (It is normal if friction is neglected.) The pressure on the crank pin is resolved into tangential and radial components 2^ and Ny The relation between P^ and P^ is shown by the expres- sion Pj =: (const.) P2 = C'A (se^ ^i^t. 156). The six general equations give, considering only the fly wheel, DYNAMICS OF MACHINERY 205 ^x = -I\- I\^ - N^ cos a - Tsin a = 0, ^y = Tcosa — iVj sin a — (? = 0, 2^ = 0, Smom^= 0, 2momj,= 0, 2 mom^ = F^r - P^r + Ta= 61,. It is reasonable to assume that the resistance of the machinery, as shown by P^ and P^^ is constant. The last equation, then, states that jT, the tangential force on the crank pin, varies with ^, the angular acceleration. From this equation we have, remembering that P^ = CT^g' C' < 1 or P^ > Pj, Ta-(P,^P,)r ^^^ L Since P^ > Pi, it is evident that the numerator will be zero when Ta = QP^ — P-^r^ so that 6 will be zero for such a case. This makes the angular velocity to either a maxi- mum or a minimum at such a point. At the dead point B^ T is zero; as a increases, 2^ increases until at a certain point B^ it equals (^P^ — P^-, At this point ^ = and o) is a minimum. Beyond Pj, o) increases and 6 increases, T has a maximum value and so does 6 at some point beyond B^^ after which ^ decreases, since it is again zero at the dead point A. Thus in passing to zero there is a value such that T= (P^ — P\)~'> so that is again zero at some point Ay At this point Ay, 6 changes from positive to negative, so that 0) is a maximum. It is evident that there are two corresponding points A-^ and B^ below the line AB. 206 APPLIED MECHANICS FOR ENGINEERS The above equation may be written codco = Y ) ^^^^ "" T I ^^^^ where the subscript zero indicates some initial value ; now ada = ds^ distance in the crank-pin circle and rda = c?s', distance in the fly-wheel circle. We may, accordingly, write = Crds - (Po - A) arc of fly wheelT- Since work done (see Art. 135) on one end of connecting rod equals the work done on the other, i Tds = I Pdx^ where x^ x^^ and dx are distances in the cylinder corre- sponding to s, Sq, and ds in the crank-pin circle. Assum- ing that this has already been shown, we may write (0)2 -«2) ^x Y I^- — 2~^ = j -^^^ ■" ^^2 - Pi) arc of fly wheel . ^0 The approximate value of I Pdx may be found by reading from the indicator card the values of P for successive values of x between the limits x and x^, A more exact treatment of the above equation may be obtained by considering that the expansion of steam in the cylinder is constant and equal to P' up to the point of cut-off and that beyond this point the pressure varies inversely as the volume. If we assume P constant and equal to P^ to the cut-off, then the limits of integration will be regarded accordingly, and we may write DYNAMICS OF MACHINERY 207 i;^!l__^ ^ J^^f% - (P2 - i^i) arc of fly wheel ?i = F'(x^-Xq) - {F^-F^) arc of fly wheel Beyond the cut-off F varies inversely as the volume of steam in the cylinder, and so F = q[ -— ) = 2i. Then \ ITV^X J X from the point of cut-off to the end of the stroke -.s' I,- —^ = ?i (A - ^1) ^^I'c of fly wheel If the pressure be regarded as constant throughout, that is, if the mean effective pressure F^' be substituted for P, we have, considering the motion from B to J., (^2 -col) I. ^ 2 = ^i2 a - (P, - P^vrr. The two first equations of this article in SX and E Y give ^ ^ (jr ^Qg ^ _ ^p^ ^ p^>^ gjj^ ^^ and ■^ \ since / "^ Problem 175. Suppose the mean effective steam pressure is 16,000 lb., the radius of the crank-pin circle 18 in., and the radius of the fly wheel 3 ft. If {P^ ~ ^\) — ^^^ ^h. and w^ = 2 tt radians per second, find iDA\ if 4=2000. Problem 176. The fly wheel in the above problem has a velocity 0)^ = 6 TT radians per second. What constant resistance (P^ — P^) will change this to 2 tt radians per second in 100 revolutions? Problem 177. Find the values of a for which co and are maxi- mum and minimum in the above problem. 208 APPLIED MECHANICS FOR ENGINEERS 117. Rotation and Translation. — In this work only the simple case of rotation and translation in a straight line will be considered, since the engineer is not usually con- cerned with more complicated motions. Let us consider the motion of a body rotating about an axis that moves parallel to itself. Suppose the body in Fig. 136 rotates about 0, and at the same time has a motion of trans- lation along X. Take the origin at 0, and allow it to be translated with the body. Any elementary mass dM of the body may be con- sidered subjected to a force dM- a parallel to X, due to the translation, a tangential force dT= 6Mp^ and a normal force dN = co^Mp, These are the effective forces. Writing down the equations of equilibrium between these forces and the impressed forces, we have at any instant 2a; = J dMa+ ydTsin a- JdJ^icosa =Ma+ 0My -co^Mx ^y = - f d]Sr sin a- fdTcos a= - ay'^My - OMx Fig. 136 23=0 2mom^ = 2inomj, = 2 mom^ ^ '^ J^'^P~~ J ^^^P sin cc = — 61^ - a3Iy. DYNAMICS OF MACHINEBY 209 It is seen at once that if the x and 7/ axes pass through the center of gravity, so that x and y are each zero, the right-hand side of the second equation becomes zero, and the first and last equations may be written ^x = 3Ia, 2 mom^ = — 61^, As an illustration let us assume that a cast-iron cylinder rolls on a straight horizontal track, due to the application Fig. 137 of certain impressed forces. Suppose the radius of the cylinder is 18 in. and its thickness 2 in., and that at the time of observation it is making 10 revolutions per second. From this time there is only a constant tangen- 210 APPLIED MECHANICS FOR ENGINEERS tial force of friction F acting parallel to X and the normal pressure iV. The cylinder comes to rest in one minute. Find F; the distance passed over, in coming to rest ; the angular velocity at the end of 10 sec, and the linear velocity at the end of 30 sec. Take the origin at the center and X horizontal. From Fig. 137 it is seen that the general equations for this case become F=:Ma, N= a, Fr=0Z. Since F is constant, a and are constant, so we have in addition to the above equations CO = COq + 6t^ 9. 2 20 ' az= ^0fi -{- (Oot. These equations are sufficient to determine the unknown quantities. Problem 178. The same cylinder given in the above illustration rolls, without slipping, down a rough inclined plane, inclined at an angle 8 to the horizontal. In addition to the forces acting as given above there is a component of gravity (see Fig. 138). Let a; be parallel to the plane, then 2a; = F4- Gsm8 = May ^y = N- GcosS = 0, Smom, = Fr = 6h, so that a is constant and equal to g sin 8 1 + ^ DYNAMICS OF MACHINERY 211 If the cylinder has the same velocity as in the above illustration, find the constant force of friction and the distance passed over in coming to rest, 8 = 10°. Fig. 138 Problem 179. A cast-iron cylinder, radius 3 in. and 6 in. long, rolls down a rough inclined plane, inclined at an angle of 30° with the horizontal. Find the acceleration dow^n the plane ; the angular acceleration ; the force of friction jp; and the normal pressure N. 118. Side Rod of Locomotive. — The side rod of a loco- motive furnishes an interesting study of a case of com- bined rotation dN" P □cLV and translation. Assume the ve- locity of the loco- motive uniform so that each dM of the side rod revolves uni- formly in the arc of a circle of radius r. See Fig. 139 (referred to the locomotive). Writing the equation of equilibrium after Q Fig. 139 212 APPLIED MECHANICS FOE ENGINEERS neglecting the thrust due to the pressure of steam on the piston, we have where P is the pressure on a crank pin due to the rotation alone and v is the tangential velocity of any dM relative to the locomotive. If v^ is the velocity of the train and r' the radius of the drive wheel, then r so that 2P-a ^/2 Problem 180. Suppose the locomotive to have a velocity of 90 mi. per hour, the radius of the crank-pin circle 20 in., the radius of the drive wheel 40 in., and the weight of the parallel rod 400 lb. Find the pressure on the crank pins due to the centrifugal force. 119. The Connecting Rod. — The connecting rod of an engine has a circular motion at one end while the other end moves backward and forward in a straight line. We shall consider the motion relative to the engine and shall assume that the fly wheel is of sufficient weight to give the crank a motion sensibly uniform. It will be con- venient to regard the motion of the connecting rod as consisting of a rotation about the crosshead end while that end is moving in a straight line. In Fig. 140 let A be the crosshead and the center of the crank-pin circle. Let I be the length of the connect- ing rod, and r the radius of the crank-pin circle. If we neglect friction, the only forces acting on the connecting DYNAMICS OF MACHINERY 213 rod at A are iV^', the pressure of the guides, and P\ the pressure exerted by the piston rod. The force exerted on the connecting rod by the crank pin has been resolved into its normal and tangential components N^ and T^ re- spectively. Suppose ft)i to be the constant angular velocity of the crank, and suppose the angular velocity of the rod about A to be represented by cd and the angular accelera- tion by 6. If we consider any element of mass dM oi the rod, it is seen that the forces acting upon it consist of a force dP = Fig. 140 dM' a parallel to OX, a normal force dN= ofiMp and a tan- gential force dT=6Mp, so that at any instant we have the same formulae as those developed in Art. 104. These may be written in this case. "Ixi = P' - T^m « - iVj cos a = Ma - co'^3Ix - OMy, ^yi = - iV^' + iV^ sin a - ^Tcos a = - cifiMy + QMx, 2 mom^i = N^ sin (cc + (/>) - Tl cos (« +)= 6L - aMy. The axis of rotation cannot pass through the center of gravity, so no further reduction is possible. 214 APPLIED MECHANICS FOB ENGINEERS We know sin a __ Z sin (f) r' so that (f> = sin~i f - sin a j ; differentiating with respect to t, sm a J _ = «= V^2 ' l--sm2a therefore a, = ^^i'^^^'^ . -\/P —r^sin^a da since — -= — ft). dt ^ and n _ — ^\^ sin a(Z2 _ ^2^ (P — r^sin^ a)% From these last two equations, for any values of (o^ and a we may obtain o) and 6. The linear acceleration, a, has been taken the same for each dM of the rod, and equal to the acceleration of the piston. If the quantities M^ ^, ^, and I^ are known ; we might find for any steam pressure P' and the corresponding a, the forces N\ iVj, and T, In other words, we could determine the forces acting on the guides and crank pin. Problem 181. The connecting rod given in Problem 163, Art. 108, is in use on an engine whose crank has a constant angular ve- locity of 26 radians per second. The length of the crank is 2 ft., the effective steam pressure on the piston is 16,000 lb. Let a be taken as 30"^, then <^ = 5° 45' ; w = — 4.52 radians per second, and = — 69 radians per second squared. From Problem 163, / = 2172, M — 61.1, X = 5.3 ft., y = .533 ft. To determine a, consider the relation be- tween the velocity v of the piston and the velocity t\ of the crank DYNAMICS OF MACniNERY 215 pin, Fig. 140. Since the velocity of every point of the rod in the direction of its length is the same, the projections of v and vi on the rod are equal. The rela- tions will not be altered '''^X G if the figure vi\ED be turned through 90° and E be placed at and ri be made to coincide with OF and drawn to such a scale as to equal r. Then v will fall on ^^^ 1 "K " ^ '^\f^^ 1 ^'N Fig. 141 OG and will be of a length equal to that cut off on OG hj AF produced (see Fig. 141). V sin (a + d)) . , , , — _ V — L^r/ _ gjj^ ^ _^ (>Qg ^ ^^^-^ ^^ vi cos (^ V = vi (sin a + cos a tan <^) . ?^p-$. = -, For small values of sin a I by sin <^, so that V = ?;i(sin a + cos a -sin a ) = n (sin a + — sin 2 a). Then But The acceleration a = vi (cos a 4- - cos 2 a)(Di. But since vi = > t^ kA ^' Fig. 148 was so successful in operation that it maintained itself in an upright position even when loaded eccentrically. The action of the fly wheels is such as to place the center of gravity of the car and load directly over the rails. (Note. See Engineering^ June 7, 1907.) CHAPTER XIII WORK AND ENERGY 129. Definitions. — When the forces acting upon a body cause a motion of that body, work is done. We define the work done by a force as the magnitude of the force times the distance through which the hody^ upon which it acts^ moves along its line of action. This definition may be less exactly stated by saying that a force acting on a body that moves through a dis- tance does work. This brings to mind the forces consid- ered as acting in Chapters II, III, and VI, where no mo- tion was produced; that is, where the point of application did not move. Such forces produce lio work according to our definition. ■ To make the idea 1^ of work clearer, suppose the body (Fig. 149) to be acted upon by a force P, and that the body is moved until the point at A is finally at B, The work done by P is P times AB. Suppose the plane upon which C^ moves is rough, so that it offers a resistance F. In passing over the distance AB, the force F does a work of resistance equal to F times AB. 229 Fig. 149 230 APPLIED MECHANICS FOP ENGINEERS The upward force iV, which is the pressure of the support- ing surface, does no work, since no motion takes phice along its line of action. The idea of work is related to that of the motion of the body in the direction of the acting force, but is independent of time. 130. Units of Work. — Since work involves force times distance, we express it in terms of the units of force and distance; that is, in inch-pounds or foot-pounds. These are the units used by engineers in this country and England, and are the units that will be used in this book. We might say, then, that the unit of work, the foot-pound^ is the work done hy a force whose magnitude is one pound when the body upon which it acts moves through a distance of one foot. In countries where the metric system is used, the erg is used when a small unit of w^ork is convenient. The erg is the work done by a force of one dyne when the body upon which the force acts moves a distance of one centi- meter in the direction of the force. A larger unit of work, the joule, is often used; the joule is 10" ergs. En- gineers often use the kilogram-meter as a unit of work. It is the work done by a force whose magnitude is one kilogram, while the body upon which the force acts moves one meter in the direction of the force. 131. Graphical Representation of Work. — Work has been defined as the product of a force and a distance. If the force be uniform and equal to P, and the body upon which it acts be moved through a distance a, the graph- ical representation of the work done by P is given by the WOBK AND ENERGY 231 airea of a rectangle, Fig. 150, constructed on F and a as sides, since P W= Fa, Fig. 150 If the force F varies as the dis- tance through which the body is moved along its line of action, we may represent the work by the area of the triangle as shown in Fig. 151. Let the force be zero when the motion begins, and let it be F^ when the distance passed over along its line of action is OA. Then since the force varies as the distance, it is equal to F for any intermediate distance 00. The total work done, then, in moving the body through a distance OA by the variable force P, which varies as the distance, is equal to — J^^- L It is seen that p this is the same as the work done by the average force —^ acting through the distance OA. The resistance of a helical spring varies with the elongation or compression. The same law of variation holds for all elastic bodies. Another variation of force with distance with which the engineer is frequently concerned, is the case where the force varies inversely as the distance through which the body is moved. If F is the force and S the distance, the relation between force and distance may be expressed, F = ^^H^, or FS= const. But this represents the equi- FiG. 151 232 APPLIED MECHANICS FOR ENGINEERS J ^ s p^^ 1 E C 1 D 1 1 Fig. 152 lateral hyperbola. This will be made clearer by reference to the specific example of the expansion of steam in a steam cylinder. (See Fig. 152.) Up to the point of cut- off 0^ the steam pressure is the same as that in the boiler (prac- tically), and is constant while the piston moves from to C. At this point, the entering steam is cut off and the work done must be done by the expansion of the steam now in the cylinder. According to Mariotte's Law, the pressure varies inversely with the volume of steam; but since the cross section of the cylinder is constant, we may say that the pressure varies inversely as the distance. From the properties of the curve, it is easy to see that the area under the curve represents the work done. The curve ob- tained in practice representing the re- lation between the force and distance is shown in Fig. 153. The curve after cut-off is not a true hyperbola, and its area is determined by means of a planimeter or by Simpson's Rule. Fig. 153 WOBK AND ENERGY 233 132. Power. — The idea of work is independent of time. But for economical reasons it is necessary to take into consideration this element of time. We must know whether certain work has been done in an hour or ten hours. For such information a unit of the rate at which work is done has been adopted. This unit is called power. Power is the rate of doing work. It is the ratio of the work done to the time spent in doing that work. The unit of power is the horse poiver. This has been taken as 550 ft. -lb. per second, or 33,000 ft. -lb. per minute. Originally the idea of the rate of work was connected with the rate at which a good draft horse could do work. This value as used by Watt was 550 ft. -lb. per second. The horse power of a steam engine is mean effective pressure times distance traveled by the piston per second, divided by 550. 133. Energy. — Energy is the capacity for doing work; it is stored-up work. Bodies that are capable of doing work due to their position are said to possess potential energy. Bodies that are capable of doing work due to their motion are said to possess kinetic energy, A familiar example of potential energy is the energy possessed by a brick as it is in position on the top of a chimney. If the brick should fall, its energy at any instant would be called kinetic. When the brick strikes the ground, work is done in deforming the ground and brick, or perhaps even breaking the brick and even generating heat. The work done by the brick when it strikes is sufficient to use up all the energy that the brick had when it struck. 23J: APPLIED MECHANICS FOR ENGINEERS 134. Conservation of Energy. — The kinetic energy of the brick spoken of in the last article was used up in doing work on the ground and air, and upon the brick itself, so that the kinetic energy that the brick possessed when it struck was used up. It was not, however, de- stroyed, but was transferred to other bodies, or into heat. Such transference is in accord with the well-known prin- ciple of the conservation of energy. This principle may be stated as follows : energy cannot he created or destroyed. The amount of energy in the universe is constant. This means that the energy given up by one body or system of bodies is transferred to some other body or bodies. It may be that the energy changes its form into light, heat, or electrical energy. Energy cannot be created or destroyed; it is, therefore, evident that such a thing as perpetual motion is impossible. Such a motion would involve the getting of just a little more energy from a system of bodies than was put into them. i./i / 135. Energy of a Body moving in a Straight Line. — Sup- pose the body, Fig. 154, moving in a straight line as in- ^^-^ n >^ n Fig. 154 dicated with a variable velocity v. Let P be a constant working force, so that the resulting force acting on the body will be P — R. From the relation that the accelerat- WORK AND ENERGY 235 ing force equals the mass times the acceleration, we may = if~ Integrating between the limits v^ and t', and and s, we have — -^=(P-R}s, The quantity --^ is the kinetic energy of the body when it has a velocity v^ and the quantity -^ the kinetic energy of the body when it has a velocity v^. The left- hand side of the equation, therefore, represents the change in kinetic energy. The equation shows that tlie work done hy the tvorhing force equals the ivorh done hy the re- sisting force plus the change in the kinetic energy. The weight of the body is 64.4 lb., and P is a constant force, say 100 lb., and R a constant resistance = 84 lb. If the body starts from rest, what will be the velocity when it has moved a distance of 16 ft.? Substituting in the above equation, w^e have z; = 16 ft. per second. Problem 193. A car whose weight is 20 tons, and having a veloc- ity of 60 mi. per hour, is brought to rest by means of brake friction after the power has been shut off. If the tangential force of friction of 200 lb. acts on each of the 8 wheels, how far will the car go before coming to rest? Solution. Here M = -^ — ^ v^ — 88 ft. per second, i; = 0, P = 0, 236 APPLIED 3IECHANICS FOR ENGINEERS since there are no working forces, it = 1600 lb., so that 40,000 (88)2 = 1600 5. 2(32.2) ^ ^ Therefore s =3009 ft. Problem 194. Suppose the car in the preceding problem to be moving at the rate of 60 mi. per hour when the power is shut ofP, what tangential force on each of the 8 wheels will bring the car to rest in one half a mile ? Problem 195. What is the kinetic energy of a river 200 ft. wide and 15 ft. deep, if it flows at the rate of 4 mi. per hour, the weight of a cubic foot of water being 62.5 lb. ? What horse power might be developed by nsing all the water in the river ? Problem 196. The flow of water in Niagara River is approxi- mately 270,000 cu. ft. per second. What is its kinetic energy? What horse power could be developed by using all the water ? Problem 197. This amount of water, 270,000 cu. ft., goes over the falls of Niagara every second. The height of the falls including rapids above and below is 216 ft. What horse power could be developed by using all the water? What horse power could be developed by using the water, considering the height of the falls to be 165 ft., the height of free fall ? Note. It is estimated that the total horse power of Niagara Falls, considering the fall as 216 ft., is 7,500,000. The Niagara Falls Power Company diverts a part of the volume of water above the rapids into their power plants, where it passes through a tunnel into the river below the falls. The turbines are 140 ft. below the water level, and each one is acted upon by a column of water 7 ft. in diameter. The estimated power utilized in this way is 220,000 horse power. The student should estimate the horse power of each turbine, assuming the w^ater to fall from rest through 140 ft. For a full account of the power at Niagara Falls, the student is referred to Proc. Inst. C. E., Vol. CXXIV, p. 223. When the motion of the body is not along the line of the force, as is the case in Fig. 155, where the body is WOBK AND ENERGY 237 supposed moving up the plane under the system of forces shown, we resolve the force into components along and perpendicular to the direc- tion of motion. It is evi- dent that the component perpendicular to the direc- tion of motion can do no work in moving the body ^ ' Fig~155 up or down the plane. The same might be said of the component of Gr perpen- dicular to the plane and of iV, so that the work-energy equation in such a case includes only the components of the forces along the line of motion. The accelerating force in this case is P cos a, and the resisting forces are F and Gr sin a. The work-energy equation becomes ^-.^^^Fs + Ca sin a)8= (P cos «)5, 2 2 ^ where s is a distance measured along the plane. Problem 198. A body whose weight is 32.2 lb. is pulled up an inclined plane, inclined at an angle of 30^ w^ith the horizontal, by a horizontal force of 250 lb. The motion is resisted by a constant force of friction of 10 lb. acting along the plane. If it starts from rest, what will be its velocity after it has gone up a distance of 100 ft. ? Problem 199. The same body as that in the preceding problem is projected down the plane with a velocity of 5 ft. per second. How far will it go before coming to rest? Hint. In this case there is no working force acting, and the final kinetic energy is zero, so that the work-energy equation reduces to the expression : the work of resistance equals the initial kinetic energy. Problem 200. The student should solve Problem 112 by using the principle of work and energy. 238 APPLIED MECHANICS FOR ENGINEERS 136. Work under the Action of a Variable Force. — When the forces are not constant, the work-energy equation already derived does not hold true. In such a case the equation vdv = ads becomes, by integrating, 2 2 ^0 ^0 The integrals expressed cannot be determined until it is known how H and P vary. In any case, however, we see that the quantity under the integral sign represents work, and so we may say: the work done equals the resistance overcome plus the change in kinetic energy. Let us suppose that the resistance H varies as the dis- tance, and also that the force P varies as the distance. Then R = const, x (6-) = C-^s and P = const, x (s) = O^s. The work-energy equation becomes, upon substitution, In Art. 81 the case of a body of 644-lb. weight falling in a resisting medium was discussed. We may now dis- cuss this same problem by means of the principle of work and energy. In this case, E = 10s,F= a,M= 20, v^ = V2 (/h = 62. 1 ft. per second. Then 20.^-2^62^ 2 2 ^0 ^0 ^2^ 3864 --§2+ 64.4 s. 2 This gives a relation between velocity and distance. The student should complete the problem, as outlined in Art. 81. WORK AND ENERGY 239 Problem 201. A body whose weight is G4.4 lb. falls freely from rest from a height of 5 ft. upon a 200-lb. helical spring. Find the compression in the spring. It will be recalled from Problem 113 that a 200-lb. spring is such a spring as would be compressed 1 in. by a weight of 200 lb. resting upon it. It will also be recalled that in compressing such a spring, the resistance of the spring is at first zero, and that it increases in pro- portion to the compression. So in the present case we may write -rr- = — , where the distance of one inch is expressed in feet : since 1 s "12" s is expressed in feet and R is the resistance of the spring in pounds, R = 2400 s. In this case P= G,v = 0, vq = ^2 gh = ^"^-^ ^^' P^^ second, and AI= 2. Then the work-energy equation gives - ^ (^^'^y ^ 2400 — = 64.4 5, 2 2 .s=.544ft., or 6.52 in. Problem 202. A weight of 500 lb. is to fall freely from rest through a distance of 6 ft. The kinetic energy is to be absorbed by a helical spring. Specifications require that the spring shall not be compressed more than 2 in. Find the strength of the spring required. Problem 203. It requires 2000 lb. to press a certain sized nail into a board a distance of 2 in. The same size nail is to be driven to the same depth by a 5-lb. hammer (Fig. 156) in 4 blows. With what velocity must the hammer strike the nail each time? Assume that all the energy of the ham- mer is absorbed by the nail and that the resist- ance offered by the timber in question varies as the distance of penetration of the nail. Neglect the weight of the hammer as a working force. Under the assumptions made, the penetration of the nail will be the same for each blow. 3 S~? Fig. 156 \' V'^^VOJ 240 APPLIED MECHANICS FOE ENGINEERS Fig. 157 Problem 204. Specifications state that it shall require 32,000 lb. to compress a helical spring 1^ in. What weight falling freely from rest through a height of 10 ft. will compress it one inch? Problem 205. The draft rigging of a freight car shown in Fig. 157 is provided with two helical springs, one inside the other. The outside spring is a 10,000- Ib. spring, and the inside a 5000-lb. spring. A car weighing 60,000 lb. is provided with such a draft rigging. While go- ing at the rate of 1 mi. per hour it collides with a bumping post. How much will the springs be compressed ? Problem 206. The draft rigging in the preceding problem is attached to the first car of a freight train, consisting of 30 cars, each weighing 60,000 lb. How much will the springs of the first car be elongated if there is 10 lb. pull for each ton of weight when the speed is 40 mi. per hour? The speed is increased to 45 mi. per hour. How much will the springs be elongated if the resistance per ton at this speed is 12 lb. ? Problem 207. The Mallet compound locomotive (Railway Age, Aug. 9, 1907) is capable of exerting a draw-bar pull of 94,800 lb. According to the preceding problem, how many 60,000-lb. cars can be pulled at 45 mi. per hour? What strength of spring would be necessary for the first car, if the allowable compression is 11 in.? Problem 208. An automobile going at the speed of 30 mi. per hour comes to the foot of a hill. The power is then shut off and the machine allowed to "coast" up hill. If the slope of the hill is 1 ft. in 50, how far up the hill will it go, if friction acting down the plane is .06 Gy where G is the weight of the machine ? 137. Pile Driver. — A pile driver consists essentially of a hammer of weight Gr so mounted that it may have a free WORE AND ENERGY 241 . fall from rest upon the pile (Fig. 158). The safe load to be placed upon a pile after it has been driven is the problem that interests the engineer. This is usually determined by- driving the pile until it sinks only a certain fraction of an inch under each blow, then the safe load is a fraction of the resistance offered by the earth to these last blows. This resistance is very small when the pile begins to penetrate the earth, but increases as penetration proceeds, until finally due to the last blows it is nearly con- stant. If we regard the hammer Gr as a freely falling body, and consider the hammer and pile as rigid bodies, and further assume, as is usually done, that It for the last few blows is constant, we may write the work-energy equation. Fig. 158 ^Mv 1= r Rds= Rs^, since the final kinetic energy is zero and the weight of the hammer as a working force is negligible. The distance s^ is the amount of penetration of the pile for the blow in question. But v'^=2 gh^ so that lMvl=Gh. We have then as the value for the supporting power of a pile, 11 = Gh 2-i2 APPLIED MECHANICS FOR ENGIXEEBS A safe value, H^ from l to |- of i2, is taken as the safe supporting power of piles. The factor of safety and the value of s-^ for the last bloAv are usually matters of specifi- cation in any particular work. This is the formula given by Weisbach and Molesworth. Other authorities give formulae as follows : Trautwine, Ii=QO G -Vh^ if s^ is small, , ^5 aVh and U = — ; ^1 + 1 Wellington, H = — , where h is in feet and s-^ in inches; McAlpine, i? = 80 [ (7 + (. 228 VA - 1)2240] ; Goodrich, jB = -— -. For other formulae and a general discussion of the subject of the bearing power of piles, the reader is referred to Transactions of Am. Soc. C. Eng., Vol. 48, p. 180. The great number of formulae for the supporting power of piles is due to the various assumptions that are made in deriving them. In deriving the Molesworth formula, the hammer and pile were considered as rigid bodies. It will be seen that the hammer and pile are both elastic bodies, both are compressed by the blow; there is friction of the hammer with the guides, and the cable attached to the hammer runs back over a hoisting drum. There is, in most cases, a loss of energy due to brooming of the head of the pile. This broomed portion must be cut off before noting the penetration due to the last few blows. Results of tests have also been taken into consideration WORK AND ENERGY 243 and have modified the formula}. The Wellington formula differs from the Molesworth formula only in the denomi- nator, where s^ + 1 is used instead of s^. This has been done to guard against the very large values of R given when §1 is very small. According to the Wellington formula, M can never be greater than (7A, the total energy of the hammer, and this is perhaps the safer formula to use for small values of s^. Engineers have come to believe that it will be extremely difficult to get a general formula that will give very ex- act information as to the bearing power of piles, since soil conditions are so varied. The more simple formulae with a proper factor of safety are used. As an illustration of the use of the formula, let us consider the problem of providing a pile to support 75 tons. If the weight of the hammer is 3000 lb., and the height of fall 15 ft., the pile will be considered down when Gh 3000 X 15 o p^ o r • s. = -— - = = .o it. = o.b m. ^ E 1^0,000 Using a factor of safety of i, we have s^ = .6 in. Problem 209. Compute the value of s^ for the pile in the above illustration by using the various formulae given in this article. Compare the results. Problem 210. A pile is driven by a 4000-lh. hammer falling freely 20 ft. What will be the safe load that the pile will carry if at the last blow the amouut of penetration was | in.? Use a factor of safety of J. Compute by the Molesworth and the Wellington formulae, and compare. Problem 211. A pile was driven by a steam hammer. The last twenty blows showed a penetration of one inch. If two blows of the steam hammer cause the same penetration as one blow from 244 APPLIED MECHANICS FOB ENGINEERS -2r- B r a 2000-lb. hammer falling 20 ft., what weight in tons will the pile support? Assume the penetration for each of the last few blows the same. 138. Steam Hammer. — The steam hammer consists essentially of a steam cylinder mounted vertically and having a weight or hammer attached to one end of the piston rod. Let AB (Fig. 159) be the steam cylinder, D the piston, and jETthe anvil, upon which a piece of metal is shown un- der the hammer (7. The steam pres- sure in the cylinder is constant and equal to P, while the piston passes over a distance a to cut-off, and varies inversely as the volume over the remaining distance h, (r, the weight of the hammer and piston, is also a working force. The resistance, I d -B^ of the metal varies I during any blow with amount of compression. It is ^55 Q t K the Fig. 159 zero just as the hammer touches the metal, and increases up to a maxi- mum when the compression is great- est. Let R^ be the average resistance of the metal, and iZj, the exhaust pressure. Then the work-energy equa- tion for the hammer before striking the metal becomes Ml.M^R^ p. = P p, + Cp'as + a Cds, or 2 2 + E^s = Pa+ Gs +j^P'ds. WORK AND ENERGY 245 Since P' varies inversely as the volume of the cylinder, ., jy, const. we may write, I^' = — =-• Then the work-energy equation gives ^ + E.s = Pa+ Gs + c\ogt. 2 a The term — ^ ^^ zero, since the motion has been consid- ered from rest at the top of the cylinder to a distance s. The quantity c may be computed by reading from the indicator card the value of P' at s. It will be seen that 8 has been taken greater than a ; that is, the piston is beyond the point of cut-off. When the hammer finally comes to the face of the metal, the work-energy equation may be written ^ + R^h^ = Pa+ ah^ + c^ log -, where the distance h^ represents the value of s when the hammer just touches the metal, and v^^ 6', and c^ are the corresponding values of ^;, 5, and c. This equation gives the kinetic energy of the hammer when it strikes the metal. The work-energy equation for the hammer during the compression of the piece may now be written where d is the amount of compression of the metal due to the blow. This is shown by the small figure to the right, where the piece of metal has been drawn to a somewhat larger scale. 246 APPLIED MECHANICS FOR ENGINEEBS After the hammer strikes the metal, the steam pressure and the weight of the hammer as working forces, and the exhaust pressure as a resisting force, have been neglected. The work done by these pressures is small, since the dis- tance d is small. Approximately, then, the work done on the metal equals the kinetic energy at the time of first contact. Instead of using the value P^ = -, and computing the s integral j P'ds as indicated in the formulae, values of P^ and s might be read from the indicator diagram (see Art. 131) and added by means of Simpson's formula (see Art. 26). As an illustration of the foregoing, let us suppose the steam cylinder 25 in. long and 14 in. in diameter; the steam pressure P = 18,000 lb.; the exhaust pressure i?i = 2300 lb.; a=7.2 in.; d = l in.; (^=644 lb.; c^ = 10,800 lb. ; A' = 24 in. Substituting in the work- energy equation, we have for the kinetic energy of the hammer at the time of striking the iron, :^ = 11,475 ft. -lb. This gives a value for v^ = 33.8 ft. per second as compared with 11.3 ft. per second for the same weight freely falling through the same distance. Investigating now the resistance of the metal, we have, under the assumption already made, i2^c^= 11,475 ft.-lb., so that E' = 550,800 lb. WORK AND ENERGY 247 In the above discussion we have neglected the compres- sion of the anvil and hammer due to the blow, and also the friction of the piston. Problem 212. Find the kinetic energy of the hammer when k' = 18 in. Find also v and R', using the same value of d. Problem 213. A steam hammer exactly similar to the one given in the illustration above is used with the same steam pressure. It is only necessary, however, for the work for which it is intended, that the kinetic energy of the hammer for a stroke of 2 ft. be 6000 ft.-lb. What weight of hammer should be used? Problem 214. Compute the kinetic energy and velocity of the hammer in the illustration (G = 644 lb.) when the piston has moved the full length of the cylinder (h' = 25 in.). Assume that there is nothing on the anvil. Problem 215. "What value of h' in the above problem would give the hammer the same velocity as it would have if it fell freely from rest through the height h ? Compute the kinetic energy for this velocity. Problem 216. In the illustration given above, what would be the value of R' if the steam pressure and G be included as working forces, and R^ as a resisting force, during the compression of the piece ? Problem 217. In the illustration given above, suppose that, in addition to the compression of the piece J in., the anvil is compressed .02 in. Find the value of R\ 139. Energy of Rotation about Fixed Axis. — In Art. 103, where the subject of the rotation of a rigid body about a fixed axis was discussed, the following equation was derived : ^CP\d,+P^^d^+ P'^d^ + etc.} = 61 where the P's represent forces tending to rotate or retard the rotation of a rigid body about a fixed axis, the 6?'s, 248 APPLIED MECHANICS FOR ENGINEERS the distances of the lines of action of these forces from 0, 6 the angular acceleration, and I the moment of inertia of the body with respect to the axis of rotation. This equation may be put in the form ^ ^ 2 (F\d, + P^^, + P^A + etc.) Now let us suppose the moment P\di is made up of a working moment and a resisting moment, such that P\d^ = P\d^-R\d^\,P\d^=P^^^d^-R'^d"^, etc. Re- membering that (odco= 0da^\ we may write, after clearing of fractions, I(od(o = 'E(P'\d^da" + P^^^d^da^^ + P^^^d^da^^ + etc. - ^(iR'\d^\dci'^+ R^\d!\du!^ + R^\d^\dci!^ + etc.). Let the angles which P'\, P^\'^ P^-) ^tc, make with the r2;-axis be called d\^ a^^^, a^^g, etc. Then d^da^' = dsi^ d^daf^ = ds^, d^dod^ = ds^^ and d^\da'^ = ds^\^ etc. Then if con- tinuity exists so that we may integrate, we may write I jl)da) =fp^\ds^ +Jp'^^ds^ + etc. ^CE'\ds^\^ CB'^ds'^^^eiG., o)= Cp'\ds^+ r>Vs2 + etc. - CR^\ds^\- fp^^^ds^'^^ etc. Since i 0)2/= 1 (ifljdMp^ = Ji dM(copy = ^\ dMv\ ft)^— ft). WOBK AND ENERGY 249 the kinetic energy of the body, where co is the angular velocity, the left-hand side represents the difference in the kinetic energy of rotation of the body when its initial velocity is co^ and its final velocity (o. On the right-hand side, Fds represents work, since ds is measured along the line of action of P in each case. A similar statement could be made for the E's, so that the right-hand side represents the work of the working forces minus the work of the resisting forces. Here, then, as in simple transla- tion, between any two positions of a rigid body, the work done ly the working forces equals the work done by the resist- ing forces plus the change in kinetic energy. As an illustration, let us consider the case of two weights (see Fig. 160), (?i=20 lb. and a^-=10 lb., suspended from drums rigidly at- tached to each other and of radii 3 ft. and 2 ft. respectively. Let the weight of the two drums and shaft be 644 lb., and the radius of gyration 2 ft. The radius of the axle is one inch and the axle friction 30 lb. The friction acts tangentially to the axle. Assume that the initial velocity co^ is one radian per sec- ond, and the final velocity 18 radians per second, how many revolutions will the drums make ? The work-energy equation gives 1 (18)2 (80) - \ (1)2 (80) + (^2 2 irrn + 30 ^^/i = G-^2 Trr^n, Fig. 160 250 APPLIED MECHANICS FOR ENGINEERS where r^ is the radius of the large drum, r that of the small drum, r^ that of the axle, and n the number of revolu- tions. Making the substitutions, n becomes 7^ = 54.8 revolutions. Problem 218. In the above illustration, what is the velocity of _ Gi and G2 when co has its initial and final values ? In what time do the drums make the 54.8 revolutions? Problem 219. The drum in Fig. 161 is solid and has a radius r and a thickness h. Initially, it is rotating, mak- ing 0)0 radians per second, but it is brought to rest by the action of a brake. The brake is applied from below by a force P acting at the end of the beam. The force of friction between the drum and brake is — , where P' is 4 the normal pressure exerted by the beam on the drum. The radius of the axle is ri, and the axle friction (.05) P'', where P'' is the pres- sure of the axle on the bearing (neglecting the lifting caused by P)^ Required the work-energy equation. Since the drum comes to rest, the final kinetic energy is zero, so that -- coo^/ + ^' 2 Trrn + (.05) P'' 2 Trnn = 0. There are no working forces, so we find the equation reducing to the form: the initial kinetic energy equals the work of resistance. The normal pressure exerted by the beam on the drum may be found by taking moments about the hinge of the beam. Then Fig. 161 P' a + b P. WORK AXD ENERGY 251 The number of revolutions turned through in coming to rest is designated by n. The equation then becomes 1 ^ nrn (a + b) P ^ . ^^^ p„ o ^,^„. 2 2 /; Problem 220. Suppose the drum in the preceding problem to be 3 ft. in diameter, 1^ in. thick, and made of cast iron. It is mak- ing 4 revolutions per second when the force P = 100 lb. is applied to the beam. The length of the drum is 6 ft., and the rim weighs twice as much as the spokes and hub. If ^ = 1.25 ft., a -h b = S ft., and r =1 in., find the number of revolutions ni that the drum will make before coming to rest. Assume the friction of the brake on the drum to be I the normal pressure, and the friction of the axle (.05) P". Problem 221. The drum in the preceding problem is making 3 revolutions per second ; what force will be required to bring it to rest in 100 revolutions ? Problem 222. If the brake in Problem 220 is above instead of below the drum, how will the results in Problems 220 and 221 be changed? Problem 223. A square prism as shown in Fig. 162 is mounted so as to rotate due to the weight G. The elastic cord runs over the pulley B and meets the square at P'. The mechanism is such that motion begins when P is in the position shown, and ceases when the prism has made a quarter turn; that is, when P reaches P'. The diameter of the journal is 2 in., and the weight on the same is 600 lb. The force of friction on the journals is 60 lb., and on the pulley at B 10 lb. Find the tension in the cord when P reaches P'. The cord is elastic, and is made of such material that it elongates, due to a pull of 100 lb., .02 in. in each inch of length. What is the elongation per inch due to the fall of G as stated ? B Q Q^ 100 LB8. Fig. 162 252 APPLIED MECHANICS FOB ENGINEERS 140. Brake Shoe Testing Machine. — The brake shoe test- ing machine owned by the Master Car Builders' Asso- ciation has been established at Purdue University. It consists of a heavy fly wheel attached to the same axle as the car wheel. These are connected with the engine, and may be given any desired rotation. When this has been obtained, they may be disconnected and allowed to rotate. The dimensions and weight of the parts are known so that the kinetic energy of the fly wheel and rotating parts may be computed by noting the Fig. 163 angular velocity. When the desired velocity has been attained, the brake shoe is brought down on the car wheel. The required normal pressure on the shoe at A (see Fig. 163) is obtained by applying suitable weights at B, The system of levers is such that one pound at B gives a nor- mal pressure of 24 lb. on the brake shoe. The weight of the levers themselves gives a normal pressure of 1233 lb. Provision is also made for measuring the tangential pull of the brake friction ; this, however, is not shown in the figure. WORK AND ENERGY 253 The weight of the fly wheel, car wheel, and shaft and all rotating parts is 12,600 lb., and the radius of gyra- tion is V2.I6. The weight of 12,600 lb. is supposed to be the greatest weight that any bearing in passenger or freight service will be called upon to carry. The diame- ter of the fly wheel is 48 in., its thickness 30 in., diame- ter of shaft 7 in., and the diameter of the car wheel is 33 in. The brake-shoe friction is ^ the normal pressure of the brake shoe on the wheel, and the journal friction may be assumed as (.002) of the pressure of the axle on the bearing. The work-energy equation for the rotating parts after being disconnected from the engine becomes 2\32.2 J 2V32.2 y o^^ ^ ^4 24 n + (1233 + 12,600 + 24 a)(.002)2 tt ^ « = 0, since there are no working forces. Problem 224. The speed is such as to correspond to a speed of train of a mile a minute when brakes are applied. What must be the weight G so that a stop may be made in a thousand feet? What is the corresponding normal pressure on the brake shoe? Problem 225. If the speed corresponds to the speed of a train of 100 mi. per hour, what weight G would be necessary to reduce the speed to 60 mi. per hour in one mile? What is the normal pres- sure on the brake shoe necessary? Problem 226. If the velocity corresponds to a train velocity of 60 mi. per hour, and the apparatus is brought to rest in 220 revolu- tions, the weight G is 100 lb. Find the tangential force of friction acting on the face of the wheel. What relation does this bear to the normal brake-shoe pressure ? Note. In the preceding problems, the ratio (the coefficient of friction, see Art. 146) has been taken as J. One of the important 254 APPLIED MECHANICS FOB ENGINEERS uses of this testing machine is to determine the coefficient of friction for different types of brake shoes. Experiment shows that it varies generally from ^ to i, sometimes going as high as y^. 141. Work of Combined Rotation and Translation. — The relation between work and energy of simple translation and the work and energy of rotation about a fixed axis have been discussed. We shall now determine the rela- tion for combined rotation and translation when the axis remains parallel to itself. Let the body of mass M Fig. 164 (Fig. 164) be in rotation with angular velocity co about an axis at 0, and at the same time let this axis move parallel to itself with a linear velocity v. At any instant the elementary mass dM has a linear velocity of translation v-^ and a tangential velocity v = cop. Its resultant velocity is expressed as the diagonal of a parallelogram con- structed upon the two velocity arrows as sides, so that WORK AND ENERGY 255 ^12 -_ -y2 _|_ ^2 _(_ 2 VV^ COS (f>. dM Multiplying both sides of tliis equation by — --, we have J -2- = J -2- + J -Y^ + j 2—.., cos<^, or J -^— = J — wy + J -^+J ciitfwpvicosc^; but /J cos (f>= t/, and | c?il!i?/ = My=^0, since OX is a gravity line. At any instant w and v-^^ are constant. Therefore, At any instant, then, the kinetic energy of combined rota- tion and translation is equal to the kinetic energy of trans- lation of the center of gravity plus the kinetic energy of rotation. As an illustration, consider a disk of radius r and thick- ness h rolling without slip- ping down an inclined plane, inclined at an angle a with the horizontal (see Fig. 165). There is a working force (? sin a and a resisting force F=^ (.06) Gr cos a. Now the kinetic energy of the disk is made up of the sum of its kinetic energy rotation and transla- tion. If ft)Q and v^ be the respective initial angular and linear velocities, we have lcu2j- l,,2r+ 1 3j^2 -\mvI ■\-Fs=G sin a • s. Fig. 165 256 APPLIED MECHANICS FOR ENGINEERS Problem 227. Suppose the disk in the above illustration to be made of cast iron, and let r = 2 ft., 7^ = J ft., and a = 30°. At a cer- tain instant it is making 2 revolutions per second. What will be the linear and angular velocities after the disk has gone 20 ft.? Would the disk finally come to rest? 142. Kinetic Energy of Rolling Bodies. — It is convenient to express the kinetic energy of combined rotation and translation of such bodies as rolling wheels in a different form from that given in the preceding article. There is some mass M^ that will have the same kinetic energy when translated with a velocity v-^ as the kinetic energy of trans- lation plus the kinetic energy of rotation of the body of mass M; that is, 2 ""22' for a wheel rolling on a straight track cor = Vj, where r is the radius. Then M. = M+ 4/ This has been called the equivalent mass. Applying this to the disk in the preceding article, we find the work-energy equation to be, ^ ^ ^ + Fs= Grsina -s A ''A for the disk, since J= \Mr\ 31^ = f 7l!f. Problem 228. A sphere of radius r rolls without slipping down an inclined plane, inclined at an angle a to the horizontal, with an initial velocity r^. Show that its kinetic energy is the same as that of a sphere whose mass is | larger translated with a velocity Vy work: axd EyEBGY 257 Problem 229. The sphere in the preceding problem is made of steel, 12 in. in diameter, and a = :^0^. If r^ = 10 ft. per second, wliat will be the velocity 10 ft. down the plane ? There is a force of fric- tion acting up the plane = (.03) times the normal pressure of the sphere on the plane. 143. Work-Energy Relation for Any Motion. — The rela- tion between work and energy for the motions considered in this chapter holds for more complicated motions and for motions in general. The limits of the present work will not admit the proof of the general theorem. It may be said, however, that for any motion the w^ork done by the working forces equals the work done by the resisting forces plus the change in kinetic energy. In the case of the motion of a complicated machine, the total work done equals the total resistance overcome plus the change in kinetic energy of the various parts of the machine. 144. Work done when Motion is "Uniform. — AVhen the motion is uniform, the change in kinetic energy is zero, and the work-energy equation reduces to the form : tvork done equals the resistance overcome. As an illustration, let us consider the case of a loco- motive moving at uniform speed and represented in Fig. 166. Suppose P the mean effective steam pressure (see Art. 131), F the friction of the piston, F^ the friction of the crosshead, F'^ the journal friction, F^^' the crank-pin friction, T the friction on the rail. It the draw bar resist- ance, G the weight of the locomotive, and N^ and iV the normal reactions of tlie rails on the wheels. Consider only one side of the locomotive and write the work-energy equation for a distance s, equal to a half turn of the driver 258 APPLIED MECHANICS FOR ENGINEERS (from dead center A to dead center ^), that the loco- motive travels. This becomes, for the frame, Fira = F(7ra + 2 r) + i^'(7ra + 2 r) + E^ira - Rira, where R^ is the pressure of the driver axle on the frame. If we neglect friction, this becomes JP=J? -JK. Considering the rotating and oscillating parts, we obtain P(7ra+ 2r)=F(7ra + 2r) + F^ (ira + 2 r) + Tira + F'^irr^ + F'^^irr^ + B^ira, where r^ and r^ are the radii of the driver axle and the crank pin, respectively. If we neglect friction, this equa- tion reduces to the form, PQira + 2r) = Tira + Wira. Fig. 166 Taking all the parts represented in Fig. 166, we may disregard the work of friction of the cylinder and cross- head, since the sum would be zero. That is, the work done by the piston on the cylinder, due to friction, equals the work done by the cylinder on the piston due to friction. We have, then, for the work-energy equation. WOliK AND ENEUGY 259 P(ira + 2 r) = Prra + Rira + Tira + F'^ttt^ + F'^'irr^. It is seen that the pressure of the steam on the head of the cylinder, for the half of the stroke considered, is a resistance. If Ave neglect friction and assume perfect roll- ing, this equation becomes P Qjra + 2 r) = Pira + Bira, or 2r or considering both cylinders, P = ira •1 r R. This is the formula usually given for the tractive power of a locomotive having single expansion engines. This may be expressed in terms of the dimen- sions of the cylin- ders and the unit steam pressure. Let p be the unit steam pressure in pounds per square inch, I the length of the cylinder in inches, d the diameter of the cylinder in inches, and d-^ the diameters of the drivers in inches; then ■" d^ ' Considering the forces acting on the driver, disregarding ()()(• EH Fig. 167 260 APPLIED MECHANICS FOR ENGINEERS friction, and taking moments about the center of the wheel (see Fig. 167), we have, for uniform motions, Pr = Ta, or T=-P, a Taking moments about the point of contact of the wheel and rail, we have, for the position shown, P(a + r) = R'a, and since P = R — R^ we have R= —P. a It follows that T = R\ that is, the train resistance can- not be greater than the adhesion of the drivers to the rails. This adhesion in American practice is usually taken as \ to \ the load on the drivers. Problem 230. What resistance R may be overcome by a locomo- tive moving at uniform speed, diameter of drivers 62 in., cylinders 16 X 24 in., and a steam pressure on the piston of 160 lb. per square inch ? What should be the weight of the locomotive on the drivers ? Problem 231. If the diameter of the drivers of a locomotive is 68 in., and the size of the cylinder is 20 x 21 in., what train resistance may be overcome by a steam pressure of 160 lb. per square inch ? Problem 232. A locomotive has a weight of 155 tons on the drivers, if the adhesion is taken as J, this allows 31 tons for the draw- bar pull. The train resistance per ton of 2000 lb., for a speed of 60 mi. per hour, is 20 lb. Find the weight of the train that can be pulled by the locomotive at the speed of 60 mi. per hour. Problem 233. An 80-car freight train is to be pulled by a single expansion locomotive at the rate of 30 mi. per hour. The weight of each car is 60,000 lb., and the resistance for this speed is 10 lb. per ton. What must be the weight on the drivers, if the adhesion is J? CHAPTER XIV FRICTION 145. Friction. — When one body is made to slide over another, there is considerable resistance offered because of the roughness of the two bodies. A book drawn across the top of a table is resisted by the roughness of the two bodies. The rough parts of the book sink into the rough parts of the table so tliat when one of the bodies tends to move over the other, the projections interfere and tend to stop the motion. The bearings of machines are made very smooth, and usually we do not think of such surfaces as having projections. Nevertheless they are not perfectly smooth, and when one surface is rubbed over the other, re- sistance must be overcome. This resisting force to the motion of one body over another is known as friction. When the bodies are at rest relative to each otlier, the friction is known as the friction of rest^ or static friction. When they are in motion with respect to each otlier, tlie friction is known as the friction of motion^ or kinetic friction. 146. Coefficient of Fric- tion. — If the body repre- sented in P'ig. 1G8 be pulled along tlie horizon- tal plane by the force P, the following forces will 2G1 262 APPLIED MECHANICS FOR ENGINEERS be acting on it ; the downward force Gr (not shown) and the reaction li inclined back of the vertical through the angle 0. The reaction li of the plane on the body may be re- solved into two components, one horizontal and one vertical. The horizontal force is known as the force of friction, and the normal force, the normal pressure. The tangent of the XT angle ^, or — , is called the coefficient of friction. This coefficient of friction, which we shall represent by/, may be defined as the ratio of the force of friction to the normal pressure ; it is an absolute number. The coefficient of friction is usually determined by al- lowing a body to slide down an inclined plane as shown in Fig. 169. The angle 6 is increased until the force of friction F will just keep the body from sliding down the plane. The angle is then called the angle of repose^ and the tangent of is the coeffi- cient of friction. It is possible with such an apparatus to determine the coefficient of friction for various materials. It has been found that after motion begins the friction is less, that is, the friction of motion is less than the friction of rest. This is an important law for engineers. Fig. 169 147. Laws of Friction for Dry Surfaces. — Very little was known of the laws of friction until within the last seventy- five years. About 1820 experiments were made that FRICTION 203 seemed to show that, for such materials as wood, metals, etc., friction varies with tlie pressure, and is independent of the extent of the rubbing surfaces, the time of contact, and the velocity. A little later (1831) Morin published the following three laws as a result of liis experiments on friction : (1) The friction between two bodies is directly propor- tional to the pressure; that is^ the coefficient of friction is constant for all pressures. (2) The coefficient and amount of friction for any given pressure is independent of the area of contact. (3) The coefficient of friction is independent of the velocity., although static friction is greater than kinetic friction. These laws of Morin hold approximately for dry unlu- bricated surfaces, although it has been found that an in- crease in speed lowers the coefficient of friction. The coefficient of friction is a little greater for light pressures upon large areas than for great pressures on small areas. The following is a table of some of the coefficients of friction as determined by Morin : Coefficients of Friction, due to Morin Material Condition of Surface e 5 E F S5 O U. Si III Brick on limestone Dry .67 3r)° .30' Cast iron on cast iron Slightly greased .16 9° 6' Cast iron on oak Wet .^h 30° 2' Copper on oak .17 9° 38' Copper on oak Greased .11 6° 17' Leather on cast iron .28 15° 39' 2Gi APPLIED MECHANICS FOR ENGINEEBS Coefficients of Friction, due to Morin — Continued Material Condition of Surface 1 1 Leather on cast iron Wet .38 20° 49' Leather on cast iron Oiled .12 6° 51' Leather on oak Fibers parallel .74 36° 30' Leather on oak Fibers crossed .47 25° 11' Oak on oak Fibers parallel, dry .62 31° 48' Oak on oak Fibers crossed, dry .54 28° 22' Oak on oak Fibers parallel, soaped .44 23° 45' Oak on oak Fibers crossed, wet .71 35° 23' Oak on oak Fibers end to side, dry .43 23° 16' Oak on oak Fibers parallel, greased .07 4° 6' Oak on oak Heavily loaded, greased .15 8° 45' Oak on pine Fibers parallel .67 33° 50' Oak on limestone Fibers on end .63 32° 15' Oak on hemp cord Fibers parallel .80 38° 40' Pine on pine Fibers parallel .56 29° 15' Pine on oak Fibers parallel .53 27° 56' Wrought iron on oak Wet .62 31° 48' Wrought iron on oak .65 33° 2' Wrought iron on wrought iron .28 15° 39' Wrought iron on cast iron _ .19 10° 46' Wrought iron on limestone .49 26° 7' Wood on metal Greased .10 6° 0' Wood on smooth stone Dry .58 30° 7' AVood on smooth earth Dry .33 18° 16' 148. Friction of Lubricated Surfaces. — The laws of fric- tion as given by Morin and stated in the preceding article hold approximately for rubbing surfaces, when the sur- faces are dry or nearly so ; that is, for poorly lubricated surfaces. If, however, the surfaces are well lubricated so FRICTION 2G5 that tlie projections of one do not fit into the otlier, but are kept apart by a fihn or hiyer of the lubricant, tlie laws of Morin are not even approximately true. The study of the friction of lubricated surfaces, then, may be divided into two parts: (1) the study of poorly lubricated bear- ings, and (2) the study of well-lubricated bearings, the friction of which varies from | to -^^ that of dry or poorly lubricated bearings. Since the friction of poorly lubricated bearings is about the same as that of dry surfaces, we shall consider that the laws of oNIorin hold, and shall confine our attention to the friction of well-lubricated bearings. If tlie lubricant is an oil, the friction of the bearing is no longer due to one surface rubbing over the other, but to the friction between the bearing and the oil, and to tlie internal fric- tion of the oil. That is, the oil adheres to the two sur- faces and its own particles attract each other, and the motion of one of the surfaces with respect to the other changes the positions of the oil particles. It is to be expected then that the friction of an oiled bearing will depend upon the viscosity of the oil, upon the thickness of the layer interposed hetiveen the surfaces, and upon the velocity and form of the bearing. The coefficient of friction is no longer constant, but varies with the temperature, velocity, and pressure. The variation of the coefficient of friction of a paraffine oil with temperature is shown in Fig. 170 when the pressure on the bearing is 33 lb. per square inch and a velocity of rub- bing of 296 ft. per minute. It is seen that the coefficient of friction decreases with increase of temperature until a temperature of 80° F. is reached, when it increases rapidly. 266 APPLIED MECHANICS FOR ENGINEERS This means that above this temperature the oil is so thin that it is squeezed out of the bearing, and the conditions of dry bearing are approached. The temperature at which oils show an increasing coefficient of friction is dif- .02 .03 COEFFICIENT OF FRICTION Fig. 170 ferent for different oils, even at the same pressure and velocity. The curve in Fig. 170, however, may be re- garded as typical of all oils when the pressure and velocity are constant. The following table, due to Professor Thurston, shows FRICTION 267 the relation between tlie coefficient of friction and tem- perature for a sperm oil in steel bearings when the veloc- ity of rubbing is 30 ft. per minute. Pressure, lb. Temfekatiue. Coefficient PKESSirUE, LH. TeMI'KUATUKE, Coefficient PER SQ. IN. Degrees F. OF Friction PER SQ. IN. Degrees F. OF Friction 200 150 .0500 100 110 .0025 200 140 .0250 50 110 .0035 200 130 .0160 4 110 .0500 200 120 .0110 200 90 .0040 200 110 .0100 150 90 .0025 200 100 .0075 100 90 .0025 200 95 .0060 50 90 .0035 200 90 .0056 4 90 .0400 150 110 .0035 It is seen that for a pressure of 200 lb. per square inch as the temperature increases from 90° F. the coefficient in- creases, indicating that the temperature of 90°, for the given pressure and velocity, was above the temperature at which the oil became so thin as to be squeezed out and the bearing to approach the condition of a dry bearing. For a constant temperature 110° F. and 90° F. the coefficient is seen to decrease with increase of pressure up to a certain point and then to increase. This is a typical behavior of oils when the temperature is constant and the pressure varies. At speeds exceeding 100 ft. per minute, the same authority found '' that the heating of the bearings within the above range of temperatures decreases the resistance due to friction, rapidly at first and then slowly, and gradually a temperature is reached at which increase 268 APPLIED MECHANICS FOR ENGINEERS takes place and progresses at a rapidly accelerating rate." The relation between the coefficients of rest and of motion as determined by Professor Thurston for three oils is given below. The journals were cast iron, in steel boxes; velocity of rubbing 150 ft. per minute and a tem- perature 115° F. g PERM Oil West Virginia Oil Lard Pressure, LB. PER. SQ. IN. At 150 ft. per min. At start- ing At stop- ping At 150 ft. per min. At start- ing At stop- ping- At 150 ft. per min. At start- ing At stop- ping 50 .013 .07 .03 .0213 .11 .025 .02 .07 .01 100 .008 .135 .025 .015 .135 .025 .0137 .11 .0225 250 ,005 .14 .04 .009 .14 .026 .0085 .11 .016 500 .004 .15 .03 .00515 .15 .018 .00525 .10 .016 750 .0043 .185 .03 .005 .185 .0147 .0066 .12 .020 1000 .009 .18 .03 .010 .18 .017 .0125 .12 .019 Steel Journals an d Brass Boxes. 500 .0025 .004 1000 .008 .009 It is seen that the coefficient of friction at starting is much greater than at stopping, and that these are both much greater than the value at a speed of 150 ft. per minute. For an intermittent feed such as is given by one oil hole, without a cup, oiled occasionally. Professor Thurston found for steel shaft in bronze bearings, with a speed of rubbing of 720 ft. per minute, the following coefficients of friction: FRICTION 269 Pressuke, lb. i'er sq. in. Oil 8 16 32 48 Sperm and lard Olive and cotton seed Mineral oils .150-.25 .1G0-.283 .154-.2()1 .l;38-.192 .107-.245 .145-.233 .086-.141 .101-.108 .086-. 178 .077-.144 .079-.131 .094-.222 The results show that the coefficient decreases with the pressure within the range reported, but that the results are considerably higher than those for well-lubricated bearings. He also found in connection with the same tests that with continuous lubrication sperm oil gave the following coefficients; Pressure, lb. per sq. in. 50 200 300 Coefficient OF Frictioh. .0034 .0051 .0057 The results of tests of the friction of well-lubricated bearings are summarized by Goodman (^Engineering News^ April 7 and 14, 1888) as follows: (a) The coefficient of friction of ivell-luhricated surfaces is from \ to Jq that of dry or poorly lubricated surfaces. (6) The coefficient of friction for moderate pressures and speeds varies approximately inversely as the normal pressure; the frictional resistance varies as the area in contact^ the nor- mal pressure remaining the same, (c) For low speeds the coefficient of friction is abnormally high^ but as the speed of rubbing increases from about AO to 100 ft, per miiiute, the coefficient of friction diminishes^ and again rises when that speed is exceeded^ varyiiig ajyproxi- ynately as the square root of the speed. 270 APPLIED MECHANICS FOR ENGINEERS (c?) The coefficient of friction varies approximately in- versely as the temperature^ within certain Ihnits; namely^ just hefore abrasion takes place, 149. Method of Testing Lubricants. — To make the matter of the tests of the friction of lubricants clear, it will be convenient to make use of the description of a testing cTBlJl Fig. 171 machine used by Dean W. F. M. Goss at Purdue University on graphite, and a mixture of graphite and sperm oil. In making the tests the apparatus shown in Figs. 171 and 172 was used. (See "A Study in Graph- ite," Joseph Dixon Crucible Co.) This apparatus represents, in principle, the machines generally used for testing lubricants. It is therefore shown in some detail. The weight Gr is hung from the shaft upon which it is suspended by the form of box to be tested. The desired speed of rubbing is obtained by means of the cone of pulleys, and the pressure on the bear- FRICTION 271 R-\-G ing is adjusted by the spring. The temperature of the bearing is read from the thermometer inserted in the bear- ing. When rotation takes place, the weight Gr is rotated a certain distance dependent upon the friction. This dis- tance is measured on the scale A, The forces acting upon the pendulum Gr are shown in Fig. 172, where R represents the resistance of the spring, ^the force of friction, I the distance of the center of gravity of Gr from the axis of rota- tion, (/) the angle through which Gr is deflected, and r the radius of the shaft. Taking moments about the center of the shaft, we have, when Gi is held in the position shown, due to friction, F-^r = Gri sin (/>, where F^ is the total friction on the bearing 2F = F^; Fig. 172 but so that ^_ Grl sin (f> It is customary to take G small compared with i?, so that the pressure on both sides of the bearing may be considered equal to 72, the resistance of the spring. The spring is easily calibrated so that R may be made any- thing desired by compressing the spring through the appropriate distance as indicated on the scale F(Fig. 171). The quantities (r, Z, r, and It are known, and (j> can be read so that/ can be calculated. 272 APPLIED MECHANICS FOR ENGINEERS If G is not small compared to i2, then /= so that F, 'XF, F, average pressure .XR + G-) + E Crl sin ^ B + a /= r[R^^]R The results of tests made upon a mixture of graphite and oil as a lubricant are given in the pamphlet. The tests were run under 200 lb. per square inch pressure, at a speed of rubbing of 145 ft. per minute. Oil was dropped into the bearing at the rate of about 12 drops per minute, showing a coefficient of friction of \, Problem 234. If the weight of the pendulum is 360 lb., the diameter of the shaft 4J in., distance of the center of gravity of G from the center of shaft 2 ft., the angle <^ 5 degrees, and the average resistance of the spring 1000 lb., find the coefficient of friction. The weight G should not be neglected in this case. 150. Rolling Friction. — The resistance offered to the rolling of one body over another is known as rolling fric- tion. It is entirely different from sliding friction, and its laws are not so well understood. When a wheel or cylinder (Fig. 173) rolls over a track the track is depressed \ and the wheel dis- v;?//;;;??????v??^??????/ 7777r7 Fig. 173 torted. The force P necessary to overcome this depression and distortion is known as rolling friction. FRICTION 273 The forces acting on the wheel are seen from Fig. 173 to be: P the working force, W the weight on tlie wheel, and M the resistance of the track or roadway to the rolling. This upward pressure i2 is not quite vertical, but has its point of application a short distance K' from the vertical. Its line of action passes through the center of the wheel. The distance jfiT' depends chiefly upon the roadway; it is called the coefficient of rolling friction. It is measured in inches and is not a coefficient of friction in the strict sense that / is the coefficient of sliding friction. Taking moments about the point of application of M^ we have, approximately, WK^ = Pr, SO that K' = ^, or r= When the track or roadway is elastic or nearly so, we have a condition something like that represented in Fig. 174. The wheel sinks into the ma- terial and pushes it ahead, at the same time it comes up behind the wheel. For a portion of the wheel on each side of the point the roadway is simply compressed; over the remainder of the surface in contact, however, slipping occurs, as indicated by the arrows. The resultant resistance, however, is in front of the vertical through the center. 274 APPLIED MECHANICS FOR ENGINEERS and we have, as in the case of imperfectly elastic roadways, P = K^W It has been found hj Reynolds (see Phil. Trans. Royal Soc, Vol. 166, Part 1) that when a cast-iron roller rolls on a rubber track, the slippage, due to the elasticity of the track, may amount to as much as .84 in. in 34 in. An elastic roller rolling on a hard track will roll less than the geometrical distance traveled by a point on the circumference. When the roller and tracks are of the same material, the roller rolls through less than its geo- metrical distance. 151. Friction Wheels. — The friction of bearings is often made much less by the use of friction wheels. The ar- rangement is usually something like that shown in Fig. Fig. 175 FRICTION 275 175. The two friction wheels A^ carry the shaft of the mechanism A, The friction of the shaft A is thus changed from sliding to rolling friction. Let P be the normal pressure on the shaft, and let the two equal forces P' act- ing through the centers of the friction wheels be the com- ponents of P acting on the friction wheels. The forces acting on each friction wheel are, then, the pressure P\ the friction jP, and the friction of the bearing F'. Since P, P\ and P' form a balanced system of forces, when the forces acting on the shaft are considered, 2cos/3' where 2/8 is the angle between the forces P' . The value of the friction F' of the friction wheel bearings is cos p where/ is the coefficient of sliding friction, and the moment of this friction is cosp Taking moments about the center of a friction wheel, we have Fr^ = PV3, so that cosp ^=(?^-^- It is seen that if the ratio -^ is constant, the friction may be made less by taking yS small, so that cos/3 is large. If r^ is large as compared with rg, the friction is reduced. 276 APPLIED MECHANICS FOB ENGINEERS Tlie work lost due to friction per revolution of A is 2 Trr^ F, or ^ \7\y J COS P The friction of A when resting in an ordinary bearing would be /P. In order that the friction of the friction wheels may be less than that of a plain bearing, we must have r, ^-— <1, or -^ !3 ^2 If the angle /3 is zero, that is, if there is only one fric- tion wheel, so that the center of A is vertically over J.', the friction is ^2 This is always less than the friction of a plain bearing, since -^ is always less than unity. ^2 Problem 235. li P— ^ tons and the radius of the shaft is 2 in. and the coefficient of friction is .07, what work is lost per revolu- tion ? If the shaft makes 3 revolutions per second, what horse power per revolution is lost in friction? Given also /3 = 45°, rg = | in., and Tg = 4 in. FRICTION 277 Problem 236. In the case of the shaft nieutioiied in the preced- ing problem, how much more horse power per revohition would it take if the bearing was plain ? What value of jB would give the same loss due to friction in both the plain bearing and the one pro- vided with friction wheels ? 152. Resistance of Ordinary Roads. — Resistance to trac- tion consists of axle friction, rolling friction, and grade resistance. Axle friction varies from .012 to .02 of the load, for good lubrication, according to Baker. The tractive power necessary to overcome axle friction for ordinary American carriages has been found to be from 3 lb. to 3| lb. per ton, and for wagons with medium-sized wheels and axles from 3| lb. to 4 J lb. per ton. The total tractive force per ton of load, for wheels 50 in., 30 in., and 26 in., in diameter, respectively, is, according to Baker (^Engineering Netvs^ March 6, 1902) : Tractive Force IN Pounds On macadam roads On timothy and blue grass sod, dry, grass cut On timothy and blue grass sod, wet and springy . On plowed ground, not harrowed, dry and cloddy 57 1:32 173 252 61 145 203 303 70 179 288 374 Rolling resistance is influenced by the width of the tire. According to Baker, poor macadam, poor gravel, compres- sible earth roads, and, on agricultural lands, narrow tires, usually give less traction. On earth roads composed of dry loam with 2 to 3 in. of loose dust, traction with l|-in. tires was 90 lb. per ton, and with 6-in. tires 106 lb. per ton. On tlie same road when it was liard and dry, with no dust, that is, wlien it was compressible, the 278 APPLIED MECHANICS FOR ENGINEERS traction was found to be 149 lb. per ton with l|^-in. tires and 109 lb. per ton with 6-in. tires. On broken stone roads, hard and smooth, with no dust or loose stones, the traction per ton was 121 lb. with l|-in. tires, and 98 lb. with 6-in. tires. Moisture on the surface or mud in- creases the traction. Morin found that with 44-in. front and 54-in. rear wheels on hard dry roads the traction per ton was 114 lb. with either l|-in. or 3-in. tires. On wood-block pave- ments the traction per ton was 28 lb. with 1^-in. tires, and 88 lb. with 6-in. tires. On asphalt, bricks, granite, macadam, and steel-road sur- faces, investigated by Baker, the traction per ton varied from 17 lb. to 70 lb., the average being 38 lb. Morin gives the coefficient of rolling friction for wagons on soft soil as .065 in., and on hard roads .02 in. Accord- ing to Kent ("Pocket-Book"), tests made upon a loaded omnibus gave the following results : Pavement Speed, Miles PER Hour Coefficient, Inches Eesistance, per Ton, in Lb. Granite Asphalt Wood Macadam, graveled . . Macadam, granite, new . 2.87 3.56 3.34 3.45 3.51 .007 .0121 .0185 .0199 .0451 17.41 27.14 41.60 44.48 101.09 Problem 237. Compare tlie resistance offered to a load of two tons pulled over asphalt, macadam, good earth roads, or wood-block pavement. Width of tires, 6 in. Problem 238. Compare the resistances in the above problem with that of steel rails to the same load. FRICTION 279 153. Roller Bearings. — In the roller bearings the sliaft rolls on hardened steel rollers as shown in cross section in Fig. 176. The roll- ers are kept in place in some way similar to that shown in the journal of Fig. 177. Such bearings are used where heavy loads are to be car- ried. Tests of roller bearings have been made by Dean C. H. Benjamin (^Machinery^ October, 1905), who determined the follow- ing values for the coefficient of friction. Speed 480 revolutions per minute. Fig. 176 Diameter of Roller Bearing Plain Cast-Iron Bearing Journal, in Inches Max. Min. Average Max. Min. Avera«^e 2tV 211 .036 .052 .041 .053 .019 .034 .025 .049 .026 .040 .030 .051 .160 .129 .143 .138 .099 .071 .076 .091 .117 .094 .104 .104 It was found that the coefficient of friction of roller bearings is from ^ to |- that of plain bearings at moderate speeds and loads. As the load on the bearing increased, the coefficient of friction decreased. Tightening the bear- ing was found to increase the friction considerably. Tests of the friction of steel rollers 1, 2, 3, and 4 in. in diameter are reported in the Transactions Am. Soc. C. E., August, 1894. The rollers were tested between 280 APPLIED MECHANICS FOR ENGINEEBS plates IJ in. thick and 5 in. wide, arranged as shown in Fig. 178. Tests were made with the plates and rollers of Fig. 177 ^P' cast iron, wrought iron, and steel. The friction P' for ..1^7. ^ A ^ I. -0063 unit load P was round to be Vr for cast-iron rollers .0073 and plates, ^^^ for wrought iron, and '——- for steel, Vr vr when r represents the radius of the roller in inches. The rollers were turned and the plates planed, but neither were polished. 154. Ball Bearings. — For high speeds and light or moderate loads the friction is much re- duced by the use of hardened Fig. 179 steel balls instead of the steel rollers. These bearings are now used on all classes of FRICTION 281 machinery, giving a much greater efficiency except for heavy loads. The principal objection to the ball bearing seems to be clue to the fact that there is so little area of contact between the balls and bearing plates. This gives rise to very high stresses over these areas, and consequently a considerable deformation of the balls. When the ball has been changed from its spherical form it is no longer free to roll, and the friction increases rapidly. Some authorities consider a load of from 60 to 150 lbs. sufficient for balls varying in size from | to -| inch in diameter. Figure 179 illustrates a type of bearing used for shafts, and Fig. 180 a type used for thrust blocks. The conclusions reached by Goodman from a series of tests on bicycle bearings (Proc. Inst. C. E., Vol. 89) are as follows: (1) The coefficient of friction of ball hearings is constant for varying loads, hence the frictional resistance varies directly as the load. (2) The friction is unaffected hy a change of temperature. The bearings were oiled before starting the tests. The coefficient of friction for ball bearings Avas found to be rather higher than for plain bearings with bath lubrica- tion, but lower than for ordinary lubrication. Ball bear- ings will also run easily with a less supply of oil. The following table gives the resvilts of tests of ball bearings. The bearings were oiled before starting, and the tests Avere run at a temperature of 68° F. Fig. 180 282 APPLIED MECHANICS FOB ENGINEERS 19 is: r 350 Load on Bearing Retolutions PER MiN. Revolutions PER MiN. Revolutions per Min. IN Lb. Coetf. friction Friction, lb. Coeff. friction Friction, lb. Coeff. friction Friction, lb. 10 .0060 .06 .0105 .10 .0105 .10 20 .0045 .09 .0067 .13 .0120 .24 30 .0050 .15 .0050 .15 .0110 .33 40 .0052 .21 .0052 .21 .0097 .39 50 .0054 .27 .0054 .27 .0090 .45 60 .0050 .30 .0055 .33 .0075 .45 70 .0049 .34 .0054 .38 .0068 .47 80 .0048 .38 .0062 .49 .0060 .48 90 .0050 .45 .0068 .61 .0060 .54 100 .0058 .58 .0069 .69 .0057 .57 110 .0054 .59 .0065 .71 .0060 .66 120 .0055 .66 .0075 .90 .0057 .68 130 .0058 .75 .0078 1.01 .0062 .81 140 .0056 .78 .0077 1.08 .0060 .84 150 .0060 .90 .0083 1.24 .0062 .93 160 .0075 1.20 .0081 1.29 .0058 .93 170 .0079 1.34 .0078 1.33 .0055 .93 180 .0079 1.42 .0078 1.40 .0053 .95 190 .0087 1.65 .0076 1.44 .0054 1.03 200 .0090 1.80 .0081 1.62 .0060 1.20 Another series of tests, run with a constant load on the bearing of 200 lb. and a temperature of 86^ F., shows the variation of the coefficient of friction with the speed. It is seen that as the speed increased the coefficient and the friction decreased. The preceding table, however, shows, for loads below 175 Zi., an increase in the coefficient with increase in speed. In particular, this table shows that for loads below 80 lbs. the coefficient increased with increase of speed ; for loads between 90 and 175 lbs. it increased when the speed was 150 r.p.m. and decreased when it was 350 R.P.M. Beyond 175 lbs. the coefficient increased. FRICTION 283 Revolutions per Minitk Coefficient Friction Friction Pounds 15 .00735 1.47 93 .004(35 .93 175 .00375 .75 204 .00345 .69 280 .00300 .60 It seems from the data given that the first conclusion of Goodman's should be changed to read : the coefficient of friction of hall hearings is constant for vari/ing loads^ up to a certain limits heyond which it increases with increase of load. This limit is about 150 lb. in the tests reported. Tests on ball bearings designed for machinery subjected to heavy pressures have been made in Germany (see Zeit- schrift des Vereins deutsche Ingenieure^ 1901, p. 73). It was found that at speeds varying from 65 to 780 revolu- tions per minute, vt^here the bearing was under pressures varying from 2200 lb. to 6600 lb., the coefficient of friction varied little and averaged .0015. Tests of ball bearings made by Stribeck and reported by Hess (Trans. Am. Soc. M. E., Vol. 28, 1907) give rise to the following conclusions : (a) the load that may be put upon a bearing is given by the formula r = cd-n where P is the load in pounds on a bearing, consisting of one row of balls, (? is a constant dependent upon the mate- rial of the balls and supporting surfaces and determined experimentally, d the diameter of the balls, the unit being ^ of an inch, and n the number of balls. For modern 284 APPLIED MECHANICS FOR ENGINEERS materials c varies from 5 to 7.5. (5) The coefficient of friction varied from .0011 to .0095. It was independ- ent of speed, '' within wide limits," and approximated .0015; this was increased to .003 when the load was about one tenth the maximum. The following values for the coefficient of friction for heavy loads are reported, from observation, with the state- ment that the real values are probably somewhat less: Revolutions per minute 65 100 190 380 580 780 1150 Coefficient of friction for load 840 lb. .0095 .0095 .0093 .0088 .0085 .0074 Coefficient of friction for ♦ load 2400 lb. .0065 .0062 .0058 .0053 .0050 .0049 .0047 Coefficient of friction for load 4000 to 9250 lb. .0055 .0054 .0050 .0050 .0041 .0041 .0040 It should be remembered that the friction of a ball bearing is due to both sliding and rolling friction, the sliding friction being due to the elasticity of the balls and the bearing (see Art. 150). Rolling friction is most nearly approached when the balls are hard and not easily changed from their spherical shape. All materials, how- ever, are deformed under pressure so that perfect rolling friction is impossible. On account of the sliding friction present in roller and ball bearings, it is necessary to use a lubricant to prevent wear. FRICTION 285 Problem 239. ITow many J-iii. balls will be necessary in a ball bearing designed to carry 4000 lb., if c = 7.5? If /= .0015, what work is lost per revolution, the distance from the axis of rotation to the center of balls being one inch? 155. Friction Gears. — In the friction gears the driver is usually the smaller wheel, and when there is any differ- ence in the materials of which tlie wheels are made, the driver is made of the softer material. This latter arrange- ment is resorted to, to prevent flat places being worn on either wheel in case of slipping. These gears have been used for transmitting light loads at high speeds, where toothed gears would be very noisy, or in cases where it is necessary to change the speed or direction of the motion quickly. IRON FOLLOWER Fig. 181 The use of paper drivers has made possible the trans- mission of much heavier loads by means of such gears. 286 APPLIED MECHANICS FOB ENGINEERS A series of tests, made by W. F. M. Goss, and reported ill Trans. Am. Soc. M. E., Vol. 18, on the friction be- tween paper drivers and cast-iron followers, is of inter- est in this connection. The apparatus used is shown in Fig. 181. The pressure between the wheels was obtained by a mechanism that forced the two wheels together w^itli a pressure P. A brake w^heel shown in the figure ab- sorbed the power transmitted. The coefficient of friction was regarded as the ratio of F to P, as in sliding friction. While this is customary, it is not entirely true, since we have the rolling of one body over the other. We shall, however, assume that we may call the coefficient of friction /=—• It was found that the coefficient of friction varied with the slippage, but was fairly constant for all pressures up to some point between 150 to 200 lb. per inch of width of wheel face. '' Variations in the peripheral speed between 400 and 2800 ft, per minute do not affect the coefficient of friction.''^ If the allowable coefficient of friction be taken as .20, the horse power transmitted per inch of width of face of the wheel is H.P. = 150 X .2x^\7rd xwx ^^M0288dwl^, 33,000 where d is the diameter of the friction wheel in inches, w the width of its face in inches, and iV the revolutions per minute. Using this formula, the following table is given in the article in question : FRICTION 287 Horse Power which may be Transmitted by Means of Paper Friction Wheel of One Inch Face, when run UNDER A Pressure of 150 lb. DiAMETEK Kf.volutions per M [NUTE r\v Pitt t w IN Inches 25 50 75 100 150 200 600 1000 8 .0476 .0952 .1428 .1904 .2856 .3808 1.1424 1.904 10 .0595 .1190 .1785 .2380 .3570 .4760 1.4280 2.380 14 .0833 .1666 .2499 .3332 .4998 .6664 1.9992 3.332 16 .0952 .1904 .2856 .3808 .5712 .7616 2.2848 3.808 18 .1071 .2142 .3213 .4284 .6426 .8568 2.5704 4.288 24 .1428 .2856 .4284 .5712 .8568 1.1424 3.4272 5.712 30 .1785 .3570 .5355 .7140 1.0710 1.4280 4.2840 7.140 36 .2142 .4284 .6426 .8568 1.2852 1.7136 5.1408 8.560 42 .2499 .4998 .7497 .9996 1.4994 1.9992 5.9976 9.996 48 .2856 .5712 .8568 1.1424 1.7136 2.2848 6.8544 11.420 The value of the coefficient of friction for friction gears, (Kent, ''Pocket Book") may betaken from .15 to .20 for metal on metal; .25 to .30 for wood on metal; .20 for wood on compressed paper. Problem 240. If the friction wheels are grooved as shown in Fig. 182, both of cast iron, and the small driver fits into the groove of the larger follower, we may take/= .18. Then F = 2fN = 2fP cos a = .36 P cos a. Fig. 182 Problem 241. The speed of the rim of two grooved friction wheels is 400 ft. per minute. If a = 45°, /= .18, what must be the pressure P to transmit 100 horse power ? 288 APPLIED MECHANICS FOR ENGINEERS Problem 242. What horse power may be transmitted by the gearing in the preceding problem, ii P = 6000 lb. and the peripheral velocity is 12 ft. per second? 156. Friction of Belts. — When a belt or cord passes over a pulley and is acted upon by tensions T^ and T^^ the ten- sions are unequal, due to the friction of the pulley on the belt. We shall determine the relation between T^ and T^. Let the pulley be represented in Fig. 183. The belt covers Fig. 183 an arc of the pulley whose angle is a. Consider the forces acting upon the belt and suppose 2\ and T^ to be the ten- sions in the belt on the tight and slack sides, respectively, and T the tension in the belt at any point of the arc of contact. Let F be the total friction between the pulley and belt and dF the friction on an elementary arc ds. If dp is the noi'mal pressure on an elementary arc, then dF=fdP and T^- T.^ = F, where / is the coefficient of friction. Represent as in Fig. 183 an elementary arc of the belt FRICTION 289 of length ds. The forces acting on this elementary part are, the tensions T + dT and T^ the friction dF^ and the normal pressure dP. Taking moments about the center of the pulley, we have (^T+dTy = dFr+Tr, or dT=dF. Of the forces acting upon this elementary portion of the belt, dT and dF are in equilibrium, so that T^ dP^ and T must also be in equi- librium. Since this is true, these latter forces must form a closed triangle when drawn to scale (Art. 13). We have, then, from Fig. 184, approximately, dP= TJ/3, dT=dF=fTdl3, Fig. 184 SO that or This gives or *^T2 I *^0 T or, writing it in the exponential form, This is the relation desired. The quantity e = 2.72+ is the base of the system of natural logarithms. The log 10 .4343. F=T,^T,^=T,(l u -fo. ) T./y'^ 1) 290 APPLIED MECHANICS FOB ENGINEEES When the band is used to resist the motion of a pulley- as in some types of brakes (see Fig. 185), it is known as a friction strap, see Art. 165. Fig. 185 Problem 243. A rope is wrapped four times around a post and a man exerts a pull of 50 lb. on one end. If the coefficient of friction is .3, what force can be exerted upon a boat attached to the other end of the rope ? Problem 244. A pulley 4 ft. in diameter, making 200 revolutions per minute, drives a belt that absorbs 20 horse power. What must be the width of the belt in order that the tension may not exceed 70 lb. per inch of width ? Problem 245. What should be the width of a belt J of an inch thick to transmit 10 horse power ? The belt covers .3 the smaller pulley and has a velocity of 500 ft. per minute. The coefficient of fric- tion is .27 and the strength of the material 300 lb. per square inch. Note. The power that can be transmitted by a belt depends upon the friction between the belt and pulley. So that H.P. Fv _ (T,-T,)v 33,000 33,000 FRICTION 291 'V 157. Transmission Dynamometer. — It has been shown, Art. 156, that the tension in I of a belt on the tight side is greater than the tension on the slack side. The transmission dy- namometer (the Fronde dynamom- eter), illustrated in Fig. 186, is designed to measure the difference in these tensions. Let the pulley D be the driver and the pulley E T, the follower, so that 7\ represents the tight side of the belt and T^ the slack side. The pulleys B^ B run loose on the T-shaped frame CBB, This frame is pivoted at A. If we neglect the friction due to the loose pulleys, we have the fol- lowing forces acting on the T- frame, two forces T^ at the center of the right-hand pulley B^ two forces T^ at the center of the left- hand pulley jB, a measurable re- action P at (7, and the reaction of the pin at A, Taking moments about the pin, we have P((7^) = 2 T^iBA) - 2 T^(^BA} = 2BAiT,-T,\ Fig. 18(3 so that ^ - 2 BJL The distances CA and BA are known, and P may be measured ; the difference, then, T^ — T^, may always be 292 APPLIED MECHANICS FOR ENGINEERS obtained. The value T^ — T^ is then known and the horse power determined by the rehition (see Problem 253) ' ' ' 33,000 where n is the number of revolutions, r is the radius of the machine pulley in feet. 158. Creeping or Slip of Belts. — A belt that transmits power between two pulleys is tighter on the driving side than it is on the foUoiving side. On account of this differ- ence in tension and the elasticity of the material, the tight side is stretched more than the slack side. To compen- sate for this greater stretch on one side than on the other, the belt creeps or slips over the pulleys. This slip has been found for ordinary conditions to vary from 3 to 12 ft. per minute. The coefficient of friction when the slip is considered is about .27 (Lanza). It has also been found that the loss in horse power in well-designed belt drives, due to slip, does not exceed 3 or 4 per cent of the gross power transmitted, and that ropes are practically as efficient as belts in this respect. For an account of the experimental investigations on this subject the student is referred to Institution of Mechanical Engineers, 1895, Vols. 3-4, p. 599, and Transactions Am. Soc. M. E., Vol. 26, 1905, p. 584. 159. CoeiScient of Friction of Belting. — The value of the coefficient of friction of belting depends, not only on the slip but also upon the condition and material of tlie rubbing surfaces. Morin found for leather belts on iron pulleys the coefficient of friction /= .50 when dry, .30 FRICTION 293 when wet, .23 when greasy, and .15 when oily (Kent, ''Hand Book"). Most investigators, however, including Morin, took no account of slip, so that the best value of/, everything considered, is that given in the preceding article (.27). 160. Centrifugal Tension of Belts. — When a belt runs at a high rate of speed over a pulley there is considerable tension introduced in the belt due to the centrifugal force. We have seen (Art. 86) that the centrifugal force equals ■ , where 31 is the mass and v tlie tangential velocity. Let the centrifugal force be represented by P^ and the tension in the belt due to this force by T^.. We know 3Iv^ that Pc = • Now if we consider a section of belt one r foot long and of one square inch cross section, we may consider the tensions Tc, at either end of this len^^th, in Fig. 187 equilibrium with P^ (see Fig. 187 (^) ). From Fig. 187 (5) we have approximately P^ = T^.6, but from Fig. 187 (a), 6=-, so that P^ = —^ Since Pc = — -, r r. r 294 APPLIED MECHANICS FOR ENGINEERS T = 31 v'- = 9 where W is the weight of a portion one foot long and one square inch in cross section. If 7 for leather is 56 lb., Tr= .388 lb. and T,= -~v'- = .012 v^. Hence, in designing belts, the total tension must be T^+T,^ ^i + .012z;2= T^e^-+,012v\ Problem 246. A belt runs at a velocity of 4000 ft. per minute. What tension is introduced by the centrifugal force of the belt in passing over the wheel ? Problem 247. What additional width of belt must be provided for in Problems 244 and 245 if the centrifugal force of the belt is considered ? 161. Stiffness of Belts and Eopes. — Belts and ropes used in the transmission of power are not perfectly flexible, so that some force is necessary to bend them around the pulleys. We desire to know the magni- tude of this force. Let T (Fig. 188) be the tension in the on-side of the belt and T + T^ the tension on the off-side. Neglecting the effect of the friction of the pulley, T^ represents the force necessary to overcome the stiffness of tlie rope. In the analysis here Fig. 188 * ' ^' given, it is assumed that while r+r. FRICTION 295 it takes a certain force T^ to cause the rope to wind around the pulley, it does not require any force to straighten it. This assumption is nearly true for steel wire rope, but not nearly so true for hemp rope. All rope requires some force to straighten it when coming off the pulley. Taking moments about the center of the pulley and neglecting the friction of the bearing, we have or T^ = y(^^ ~ ^i\ where d^= r + -, and d^ = r + ai+- + a^- The distance a^ is due to the stiffness of the rope, and the distance a^ the distance of the point of application of T from the center of the rope. That T does not act at the center of the rope, but at a distance a^ toward the outside, is due to the fact that the outside of the roj)e is under greater tension than the inside. Now the distance a^ for inelastic ropes decreases as T increases, and so we may write a^ = ^, where c^ is a constant, determined experimentally. For wire rope, a^ increases with in- creased radius of the pulley, and decreases with increased tension, so that we may write .(-f) «i = Y making these substitutions in the above equation, we have e^ + d^T 1\ = — -1 — tor hemp rope, 296 APPLIED MECHANICS FOR ENGINEERS and a^T , 2\ = c^-i ^— y tor wire rope. For tarred hemp ropes, c-^ has been found (see Du Bois, ''Mechanics of Engineering") to be 100, and a^^ .222, so that ^ 100 + .222y . ^^ = \ — pounds. , Cv For new hemp ropes^ T^ = —ZLl — _ pounds. For wire ropes^ m i Ao .0937 7 , y=1.08H -J- pounds. ^ + 2 In each case T is expressed in pounds and r and d in inches. Problem 248. — A new hemp rope, one inch in diameter, passes over a pulley 13 in. in diameter, under a tension of 500 lb. What is the force necessary to overcome the stiffness of the rope? What per cent is this of the total tension in the rope ? Problem 249. — A wire rope, one inch in diameter, passes over a pulley 2 ft. in diameter, under a tension of 1000 lb. What force is necessary to overcome the stiffness of the rope? Compare this force with that necessary to overcome the stiffness of the same rope under the same tension, when it passes over a pulley 12 in. in diameter. What per cent of the total tension is it in each case ? Problem 250. — A new hemp rope, 2^ in. in diameter, passes over a grooved pulley 31 in. in diameter, under a tension of 1000 lb. What force is necessary to overcome the stiffness of the rope? Allow an increase of 6 per cent for the grooved pulley. FRICTION 297 It will be seen from the formula and the problems tliat the force necessary to overcome the stiffness of ropes is greater for small pulleys than for large ones. The following empirical formula will be found useful (see Memoirs et Compte rendu de la Societe des Ingenieurs Civile^ December. 1893, p. 558, or Proc. Inst. C. E., Vol. 116, p. 455): for ropes, where c?and r are expressed in millimeters and 2\ and T in kilograms. A formula for belting is also given, t, = ^^^m-^ut']. where w is the width of the belt and t its thickness. T^ and T are expressed in kilograms and w^ t^ and r in milli- meters. The student should solve Problems 248 and 249, using the empirical formula given above. Problem 251. A belt 12 in. wide and -| in. thick passes over a pulley 18 in. in diameter under a tension of 1000 lb. What force is necessary to overcome the stiffness of the belt? In using the empirical fornmla just given it will be necessary to change pounds to kilograms and distances to millimeters. 162. Friction of a Worn Bearing. — Tlie friction of a bear- ing that fits perfectly is the friction of one surface sliding over another and is given by the equation 298 APPLIED MECHANICS FOR ENGINEERS where F is the force of friction, / the coefficient of friction, and iV'is the total normal pressure on the bearing. When, however, the bearing is worn, as is shown much exaggerated in Fig. 189, the friction may be somewhat Fig. 189 different. When motion begins, the shaft will roll up on the bearing until it reaches a point A where slipping be- gins. If motion continues, slipping will continue along a line of contact through A. Let P be a force that causes the rotation, R a force tending to resist the rotation, and M^ the reaction of the bearing on the shaft. There are only three forces acting on the shaft, so that P, i2, and R^ must meet in the point B. The direction of R^ is accord- ingly determined. The normal pressure is iV^= i^^ cos ^, and the force of friction is It is seen that 6 is the angle of friction. The moment of the friction with respect to the center of the axle is Fr m^r sin 6. FRICTION 299 If the axle is well lubricated so that 6 is small and sin 6 may be replaced by tan 6 = f , the friction is and the moment Fr = fM^r. The circle tangent to AB radius r sin is called the friction circle. Since r and 6 are known generally, this circle may be made use of in locating the point A, In other words, the shaft will continue to rotate in the bear- ing so long as the reaction B^ falls within the friction circle, and slipping will begin as soon as the direction of B^ becomes tangent to the friction circle. Problem 252. If the radius of the shaft is 1 in., = 4°, P = 500 lb., ai = 3 ft., a-z = 2 ft., angle between ai and ao is 100° and P and R are at right angles to ai and a2, what resistance R may be overcome by P when slipping occurs ? Problem 253. The radius of a shaft is 1 in., R = 20 lb., P = 20 lb., ai = 3 ft., and a2 = 2 ft. What force of friction will be acting at the point A, when the angles between P and ai and R and ao are right angles ? What must be the value of the coefficient of friction ? 163. Friction of Pivots. — The friction of pivots presents a case of sliding friction, so that the force of friction F equals the coefficient of friction times the normal pressure. That is, (a) Flat- End Pivot. — The friction on a flat-end pivot is greatest on the outside and varies linearly to zero at the center as shown in Fig. 190. The resultant force of friction 300 APPLIED MECHANICS FOR ENGINEERS R has its point of application |r from the center. We may write F=R=fP, and the moment of the friction with respect to the center is Moment = | rfP. p /^ — > ^ ^ Fig. 190 Fig. 191 The work lost per revolution is (6) Collar Bearing or Hollow Pivot. Let the outside radius be r^ and the inside radius r^ (Fig. 191); then, F=B=.fP, FRICTION 301 and the moment of friction is Moment = 2rg- fP :i rl - rl (For the position of the center of gravity, see Art. 24.) The work lost, due to friction, per revolution is 3L.2 W •» 1 r/P. ( "^'^^^i^ If r^ = 0, this reduces to the work lost per revolution for the solid flat-end pivot. ((?) Conical Pivot. The conical pivots, such as are illustrated in Fig. 192, do not usually fit into the step the entire depth of the cone. Let the radius of the cone at the top of the step be /, a half the angle of the cone, and P^ the resultant normal re- action of the bearing on the pivot. Then Fig. 192 z Sin a and the total friction F= li — fP i sin a The moment of friction in this case is fp Moment = 2 sin a 302 APPLIED MECHANICS FOR ENGINEERS since the resultant friction may be regarded as applied at I r'. The work lost due to friction per revolution may be written - ^-o 3 sin a If cc = — , this value for work lost reduces to the work lost per revolution, in the case of the flat-end solid pivot. It is easily seen, since sin a is less than unity, that if r' is nearly equal to r, the friction of the conical bearing is greater than the friction of the flat-end bearing. This might have been expected from the wedgelike action of the pivot on the step. It is also easily seen that r' may be taken small enough so that the friction will be less than the friction of the flat pivot. The work lost due to friction in the case of the conical pivot will be equal to, greater, or less than, the work lost, due to friction in the case of the flat-end pivot, p according as r'-^r sin a. (c?) Spherical Pivot, Suppose the end of the pivot is a spherical sur- face, as shown in Fig. 193. Let r be the radius of the shaft and r^ the radius of the spherical surface; then the load per unit of area of P horizontal surface is Fig. 193 irr' FRICTION 303 The horizontal projection of any elementary circle of the bearing, of radius x^ is 2 irxdx. The load on this area is \7rrv ^" and the corresponding normal pressure is dF, = ?Zf^ sec /3. X. ^ r, OD Vr^-a^ ^\.^;jT> ^Pxdxf r, \ But cos/3= — = — ^ , so that dF.= — ( — — ^ — ]• ^1 ^1 r^ Wrf-x^J The corresponding friction is/dP^, Integrating between the values x = and x = r^ the value for tlie total friction is given by F=E= n-Pnf( ^dx \ Jo r2 V Vrf - xy Since r = r^ sin a and Vr^ — r^= r^ cos a, the expression for the friction may be written 1 + cos CI IT If « = ^9 that is, if the bearing is hemispherical, Li F=^fPy and if «=0, that is, if the bearing is flat, F=fP. The moment of the friction is given by adding all the tQvm^fdP^x by means of integration; this gives Moment =:M^f'l sin"!- - ?! V/f=^\ r^ \'l r^ 2 / or in terms of a. Moment = fBr ( — V " ^^^ ^) ' 304 APPLIED MECHANICS FOR ENGINEERS The work lost due to friction, for each revolution, is found by adding the work lost by friction on each ele- mentary area of the bearing ; that is, by finding the sum of such terms as 2 irxfdP-^ by means of integration ; this gives TF= 2 ir/Fr i-^ - cot (i\ . If the bearing is hemispherical, a = - , and the moment becomes Moment ->^^^ 2 and the work lost per revolution The friction of flat pivots is often made much less by forcing oil into the bearing, so that the shaft runs on a film of oil. In the case of the turbine shafts of the Niag- ara Falls Power Company (see Art. 135) the downward pressure is counteracted by an upward water pressure. In some cases the end of a flat pivot has been floated on a mercury bath. This reduces the friction to a minimum (see Engineering^ July 4, 1893). The Schiele pivot is a pivot designed to wear uniformly all over its surface. The surface is a tractrix of revolu- tion; that is, the surface formed by revolving a tractrix about its asymptote. Its value as a thrust bearing is not as great as was first anticipated (see American Ma- chinist, April 19, 1891). The coefficient of friction for well-lubricated bearings of flat-end pivots has been found to vary from .0041 to .0221 (see Proc. Inst. M. E., 1891). For poorly lubri- FRICTION 305 cated bearings the coefficient may be as high as .10 or .25 for dry bearings. Problem 254. Show that the work lost per revolution for the hemispherical pivot is 2.35 times the work lost per revolution for the flat pivot. Problem 255. The entire weight of the shaft and rotating parts of the turbines of the Niagara Falls power plant is 152,000 lb., tlie diameter of the shaft 11 in. If the coefficient of friction is con- sidered as .02 and the bearing a flat-end pivot, what work would be lost per revolution due to friction ? Problem 256. A vertical shaft carrying 20 tons revolves at a speed of 50 revolutions per minute. The shaft is 8 in. in diameter and the coefficient of friction, considering medium lubrication, is .08. What work is lost per revolution if the pivot is flat? AVhat horse power? What horse power is lost if the pivot is hemispherical ? Problem 257. What horse power would be lost if the shaft in the 2^receding problem was provided with a collar bearing 18 in. out- side diameter instead of a flat-end block? Compare results. Problem 258. A vertical shaft making 200 revolutions per minute carries a load of 20 tons. The shaft is 6 in. in diameter and is provided with a flat-end bearing, well lubricated. If the coefficient of friction is .004, what horse power is lost due to friction? 164. Absorption Dynamometer. — The friction brake shown in Fig. 181 is used to absorb the energy of the mechanism. It may be used as a means of measuring the energy, and when so used it may be called an absorption dynamometer. The weight TF, attached to one end of the friction band, corresponds to the tension in the tight side of an ordinary belt (see Art. 156), while the force measured by the spring S corresponds to the tension on slack side of a belt. Let W= T^ and S = T^\ then 2\ = 7^2^-^% just as was found in 306 APPLIED MECHANICS FOB ENGINEERS the case of belt tension. The work absorbed per revolu- tion is work =z (T^— T^}2 7rr^, where r^ is the radius of the brake wheel. The horse power absorbed is HP ^ (T,-T,r^^r, 33,000 vv^>^\ . .^..v^ ^ ^ In many cases the friction band is a hemp rope, and in such cases it is possible to wrap the rope one or more times around the pulley, making it possible to make T^ — T^ large while T^ is small. The surface of the brake wheel may be kept cool by allowing water to flow over the inside surface of the rim, which should be provided with inside flanges for that purpose. 165. Friction Brake. — The friction brake shown in Fig. 185 consists of the lever EQ^ the friction band, and the friction wheel. Such brakes are used on many types of hoisting drums, automobiles, etc. Let the band ten- sions be T^ and T^^ and let W be the force causing the motion, that is, the working force, and P the force applied at the end of the lever EC in such a way as to retard the rotation of the drum. We have here as before ^i = T^e^°- and the work per revolution C^i~ ^2)^ '^^1 + ^^ '^^s- By taking moments about A we have T^ — T^ = — 2 a, where r^ is the radius of the shaft and F is the force of friction acting on the shaft. Taking moments about (7, we have p^(^EC^ = T^d^ sin S + T^d^ sin /8. Problem 259. A weight of one ton is being lowered into a mine by means of a friction brake. The radius of the drum is 1^ ft., FRICTION 307 radius of the friction wlieel 2 ft., coefficient of brake friction .30, 8 = 45^ ^ = 15^ r/j = cl^ = l ft, EC = G ft., radius of shaft 1 in., coefficient of axle friction .04, and the weight of the drum and brake wheel is 600 lb. Find T^, T,^, and P in order that the weight W may be lowered with uniform velocity. Problem 260. Suppose the weight in the above problem is being lowered with a velocity of 10 ft. per second when it is discovered that the velocity must be reduced one half while it is being lowered the next 10 ft., what pressure P will it be necessary to apply to the lever at E to make the change ? What will be the tension in the friction bands and the tension in the rope that supports TF? Problem 261. If the weight in the above problem has a velocity of 10 ft. per second, and it is required that the mechanism be so con- structed that it could be stopped in a distance of 6 ft., what pressure P on the lever and tensions T^ and T^g would it require ? What would be the tension in the rope caused by the sudden stop ? Com- pare this tension with IF, the tension when the motion is uniform. 166. Prony Friction Brake. — The Prony friction brake may be used as an absorption dynamometer as shown in Fig. 194 principle in Fig. 194. Let W be a working force acting on the wheel of radius r and suppose the brake wheel to 308 APPLIED MECHANICS FOR ENGINEERS be of radius r^ The brake consists of a series of blocks of wood attached to the inner side of a metal band in such a way that it may be tightened around the brake wheel as desired by a screw at B, This band is kept from turning by a lever (7AZ>, held in the position shown by an upward pressure P, at A, Considering the forces acting on the brake and taking moments about the center, we have the couple due to friction, jFV^, equal to the mo- ment FQOA), or Fr^ = F(OA}. Considering the forces acting on the wheel, and neglecting axle friction, we get Fr-^ = Wr, The energy absorbed is used in heating the brake wheel. The wheel is kept cool by water on the inside of the rim. The work absorbed is 2 7rr^Fn= 2irF( 0A^7i^ where n is the number of revolutions. The force F may be measured by allowing a projection of the arm at A to press upon a platform scales. The horse power absorbed is where OA is expressed in feet, and n is the number of revolutions per minute. 33 If OAhQ taken as - — , a convenient length, the formula 2 TT reduces to IT P -_^ A dynamometer slightly different from the Prony dynamometer is shown in Fig. 195. It differs only in the means of measurincr F. In this case the force F is meas- FRICTION 309 ured by the angular displacement of a lieavy pendulum Wy Taking moments about the axis of W^ and calling Fig. 195 r^ the distance from that axis to its center of gravity and yS the angular displacement, we have Fr^ = W^Tr^ sin ^8, so that the horse power absorbed may be written 7-4 33,000 ' where OA^ r^, and r^ are expressed in feet. If OA be taken as — ^, this becomes 27r H.P.= W^r^ sin Pn r^lOOO The student should understand that the rotation of the mechanism at is not in every case due to a weight W being acted upon by gravity. In fact, in most cases, tlie motion will be due to the action of some kind of engine. This, however, will not change the expressions for horse power. 310 APPLIED MECHANICS FOB ENGINEERS Problem 262. If ]]\ = 100 lb., OA = — , r^ = 2 ft., and r^ = 6 27r in., what horse power is absorbed by the brake if /3 is 30°, and n is 300 revolutions per minute ? 167. Friction of Brake Shoes. — The application of the brake shoe to the wheel of an ordinary railway car is shown in Fig. 196, where F' is the axle friction, F the brake-shoe friction, N the normal pressure of the brake shoe, Gr the weight on the axle, and F^ and iV^j the reaction of the rail on the wheel. The brakes on a rail- way car when applied should be capable of absorbing all the en- ergy of the car in a very short time. The high speeds of modern trains require a system of perfectly working brakes, capable of stopping the car when running at its maximum speed in a very short distance. The coefficient of friction between the shoes and wheel for cast-iron wheels at a speed of 40 mi. per hour is about ;|, while at a point 15 ft. from stopping the coefficient of friction is increased 7 per cent, or it is about .27. The coefficient for steel-tired wheels at a speed of 65 mi. per hour is .15, and at a point 15 ft. from stopping it is .10. (See Proc. M. C. B. Assoc, Vol. 39, 1905, p. 431.) The brake shoes act most efficiently when the force of friction F is as large as it can be made without causing a Fig. 196 FRICTION 311 slipping of the wheel on the rail (skidding). The normal pressure iV, corresponding to the values of the coefficient of friction given above, varies in brake-shoe tests from 2800 lb. to 6800 lb., sometimes being as high as 10,000 lb. Problem 263. A 20-ton car moving on a level track with a velocity of a mile a minute is subjected to a normal brake-shoe pres- sure of 6000 lb. on each of the 8 wheels. If the coefficient of brake friction is .15, how far will the car move before coming to rest? Problem 264. In the above problem the kinetic energy of rota- tion of the wheels, the axle friction, and the rolling friction have been neglected. The coefficient of friction for the journals is .002, that for rolling friction is .02. Each pair of wheels and axle has a mass of 45 and a moment of inertia with respect to the axis of rotation of 37. The diameter of the wheels is 32 in. and the radius of the axles is 2i in. Compute the distance the car in the preceding problem will go before coming to rest. Compare the results. Problem 265. A '30-ton car is running at the rate of 70 mi. per hour on a level track when the power is turned off and brakes ap- plied so that the wheels are just about to slip on the rails. If the coefficient of friction of sliding between w^heels and rails is .20, how far will the car go before coming to rest? Problem 266. A 75-ton locomotive going at the rate of 50 mi. per hour is to be stopped by brake friction within 2000 ft. If the coefficient of friction is .25, what must be the normal brake-shoe pressure ? Problem 267. A 75-ton locomotive has its entire weight carried by five pairs of drivers (radius 3 ft.). The mass of one pair of drivers is 271 and the moment of inertia is 1830. If, when moving with a velocity of 50 mi. per hour, brakes are' applied so that slipping on the rails is impending, how far will it go before being stopped? The coefficient of friction between the wheels and rails is .20. 312 APPLIED MECHANICS FOR ENGINEERS 168. Train Resistance. — The resistance offered by a train depends upon a number of conditions, such as velocity, acceleration, the condition of track, number of cars, curves, resistance of the air, and grades. No law of resistance can be worked out from a theoretical consideration, be- cause of the uncertainty of the influence of the various factors involved. Formulae have been developed from the results of tests; the most important of these are given below. Let R represent the resistance in pounds and v the ve- locity in miles per hour. W. F. M. Goss has found that the resistance may be expressed as i2 = . 0003(^4- 347) ^;^ where L is the length of the train in feet (see Engineering Record, May 25, 1907). The Baldwin Locomotive Works have derived the formula b as the relation between the resistance and velocity. When all factors are considered, this becomes R = Z + ^+ .3788 (0 + .5682 {c) + .01265 (a)2, where t = grade in feet per mile, c the degree of curvature of the track, and a the rate of increase of speed in miles per hour in a run of one mile. To get the total resistance it is necessary to include, in addition to the above factors, tlie friction of the locomo- FRICTION 313 tive and tender. Tliis is given by Holmes (see Kent's "Hand Book") as R^= [12 + .3(i;- 10) W], where W is the weight of the engine and tender in pounds and R^ the resistance in pounds due to friction. 70 1 60 oc LU °- 50 CO o z 2 40 ^ 30 U^^W-^U::^!!-: ! .ltUl}!!4i^:-4!!.l- -^ :|.---!---H!.!-l U, -1; n FOR RAILROAD TRAINS -"d^ 10 20 30 40 oO 60 70 bO 90 100 VELOCITY IN MILES PER HOUR Fig. 197 Other formulae derived as the result of experiments are shown graphically in Fig. 197. 314 APPLIED MECHANICS FOR ENGINEERS The formulae themselves are as follows (see Engineering, July 26, 1907): Curve Number Formula Authority 1 ''='-^+;44 Clark 2 '== « + m Clark 3 *=*-"+i';2 Wellington 4 «=»+2» Deeley 5 i? = .2497 V Laboriette 6 R = 3.36 + .1867 v Baldwin Company 7 72 = 4.48 + 284:1; Lundie 8 J? = 2 + .24 i; Sinclair 9 72 = 2.5+ ""' 65.82 Aspinal It is evident that these formulae do not agree as closely as one would wish. The difference must be due chiefly to the different conditions under which the tests were made. These conditions should be taken into account in any application of the formulae to special cases. CHAPTER XV IMPACT 169. Definitions. — ^When two bodies that are approach- ing each other collide, they are said to be subjected to impact. If their motion is along the line joining their centers of gravity, the collision is designated as direct central impact. If they are moving along parallel lines, not the common gravity line, the impact is known as direct eccentric impact. When the collision differs from either of the above forms, the impact is known as oblique impact. The phenomena of impact may be best studied by con- sidering the two bodies somewhat elastic. Suppose for simplicity that they are two spheres, M^ and M^^ Fig. 198, and that they are moving in opposite directions with velocities v^ and v^ and that the impact is central. In Fig. 198 (a) they are shown at the instant when contact first takes place, and in Fig. 198 (J) they are shown some time after first contact when each has been deformed somewhat by the pressure of the other. The dotted lines indicate the original spherical form and the full lines the actual form of the deformed spheres. When the spheres first touch, the pressure P between them is zero, but as each one compresses the other, the pressure P increases until it becomes a maximum. The compression of the spheres is indicated in the figure by d^ and d^. We shall 316 APPLIED MECHANICS FOR ENGINEERS designate the time during which the bodies are being com- pressed as the period of deformation. After the compression has reached its maximum value the bodies, if they be partially elastic, begin to separate and to regain their original _!j ^ form. The common pressure P decreases and becomes zero, if the bodies are sufficiently elastic so that they finally separate. We shall designate this pe- riod of separation as the period of restitution^ and the velocities of separa- tion as v-^^ and v^^ If the bodies are en- tirely inelastic^ there will be no restitution. They will, in that case, remain in contact just as they are when the pressure be- tween them is a maximum and will move on with a common velocity V. (a) 170. Direct Central Impact, Inelastic. — When the bodies meet in direct central impact, separation will take place along the line joining tlie centers of gravity. Let T be the time from the first contact up to the time of maximum pressure, that is, the time of deformation^ and T^ the time IMPACT 317 from first contact up to the time of separation. Then T^— T represents the time of restitution. We have, from Art. 72, dv = a ' dt. Considering the motion of M^ during the period of deformation, we have dv^ = a^dt anda,= -| so that I dv. = — -— : j Pdt^ or iffiC r - t^i) = - fpdt. Jo In a similar way, remembering that if v^ is positive v^ is negative, Jo The two integrals f Pdt on the right-hand side of the preceding equations cannot be determined since we do not know in general how the pressure P varies with the time; we do know, however, that they are equal term for term, so that we may eliminate them. We have, then, or If the bodies were moving in the direction of Jfj and Vj > v^, we should have both Vi and v^^ positive, a^ negative, and ^2 positive. Then M^ v^ + M^v^ This is also the value for P^if both bodies are moving in the direction v^ and v^ > v^. 318 APPLIED MECHANICS FOR ENGINEERS If the bodies are inelastic, they will both move with the velocity V^ and there will be no separation. Suppose the bodies to be two lead balls, and let (7^ = 10 lb., Gr^ = 25 lb., Vi = 10 ft. per second, and v^ = 60 ft. per second. Then V = 51.28 ft. per second if the bodies are moving in the same direction, and F^= 33.3 ft. per second if they move in opposite directions. The energy lost in direct central impact of inelastic bodies may be found by subtracting the kinetic energy of 31-^ and 711^2, when the common velocity is F^ from the kinetic energy of the two bodies at the time of first con- tact. The kinetic energy of M^ before impact is E^ = I M^vl, and that of M^ is -£"2=2 ^2^% ^^ ^^^^^ ^^^ total energy before impact is ^M^v\ + ^M^v\, The kinetic energy after impact is ^(^M^ + M^ F"^, so that the loss of kinetic energy is and this equals ""Xlt^Mf if the bodies move in the same direction, or if they move in opposite directions. This energy is used up in deforming the bodies and in raising their temperatures. The kinetic energy remain- ing may be written when the bodies move in the same direction. IMPACT 319 In the above example of the lead balls, we find that the kinetic energy lost due to impact is 40.3 ft. -lb. when the bodies move in the same direction, and 1109 ft. -lb. when they move in opposite directions. When the bodies move toward each other and M-i^i\ = M^v^^ the final velocity Vis zero and the kinetic energy lost is ^ M^v'^ + 1^ M^v'^, If the masses are equal, V= ^^ ~ ^2 and the kinetic energy lost is MOh + v^ 2 2 If M^ is infinite as compared with 3f^^ and v^ is zero, the final velocity V is zero, and the kinetic energy lost is 2 Problem 268. A lead sphere whose radius is 2 in. strikes a large mass of cast iron after falling freely from rest through a distance of 100 ft. What is its final velocity? What is the loss of kinetic energy? Problem 269. A 10-lb. lead sphere is at rest when it is acted upon by another lead sphere, whose radius is 3 in., in direct central impact. The velocity of the latter sphere is 20 ft. per second. What is the common velocity of the two spheres and what is the loss of kinetic energy due to impact ? 171. Direct Central Impact, Elastic. — If the impact is not too severe, elastic or partially elastic bodies tend to regain their original shape after the deformation has reached a maximum and finally separate if they possess sufficient elasticity. Using the notation of Art. 169, we have, for the period of restitution, if jR is the force of restitution. 320 APPLIED MECHANICS FOR ENGINEERS M^\ dv^ = — J Rdt, and for M^, so that Mi(v[ - V) = -JRdt, and M^(v'^ - F) = Clidt The value of the integral j Pdt during deformation will not in general be the same as its value during restitution. Call the ratio of I Bdt to I Pdt, e. This value, which, is called the coefficient of restitution, is constant for a given material. It is unity for perfectly elastic substances, zero for non-elastic substances, and some intermediate value for the imperfectly elastic materials with which the engineer is usually concerned. The following values of e have been determined: for steel, ^ = .55, for cast iron, ^ = 1, nearly; for wood, ^ = 0, nearly. From the above definition of e, it is seen at once that we may write v^ -V v^-V e-i = —, , and Co = ~ ' V- v{ 2 Y_ ^ and these equations enable one to determine e experi- mentally. Rdt = e ) Pdt, we may write IMPACT 321 so that v[ — V{1 + gj) — e-^v^, and ^2 = V(l + ^2) - ^2^2' where ^=-E^r^' if the bodies are moving in the same direction, and M^ + M^ if they are moving in opposite directions. If the bodies are of the same material, e-^ = e^ = e. Then from the above equation it is seen that and the energy lost in impact is If the bodies are perfectly elastic, so that ^ = 1, the loss of energy is zero. Problem 270. The student should show that for any impact M^vi + M2V2 = M^v[ + il/g?;^ ; that is, the sum of the momenta before impact equals the sum of the momenta after impact. Problem 271. Two perfectly elastic bodies, having equal veloci- ties in opposite directions, meet in direct central impact. What must be the relation of their masses so that they will be reduced to rest? Problem 272. If the bodies are perfectly elastic and M^= M^, show that v[ = v^ and vl^ — Vy T 322 APPLIED MECHANICS FOR ENGINEERS Problem 273. If a ball M^ of a certain material falls upon a large mass M2 of the same material from a height h and rebounds to a height /?,, show that e =^-^. ^ h In this case il/g = 00 , rg = 0, and V = 0. Problem 274. A 20-ton car having a velocity of 40 mi. per hour collides with a 30-ton car having a velocity of 60 mi. per hour in the opposite direction. Both are destroyed. What is the loss of kinetic energy ? 172. Elasticity of Material. — All materials of engineer- ing are imperfectly elastic. Some, however, show almost perfect elasticity for stresses that are rather low. This has been expressed by saying that all materials have a limit (elastic limit) beyond which if the stress be increased the material will be imperfectly elastic. Within the limit of elasticity^ stress is proportional to the deformation pro- duced. Let the total stress in tension or compression be P, and the stress, in pounds per square inch of cross sec- tion, be/, also let d be the deformation caused by JP and X the deformation per inch of length. Within the limit of elasticity of the material the ratio ^ is a constant, and A since /= — -, and X = -, when F is the area of cross sec- F I tion and I is the length of the material, it may be written Fl . . — — . This constant is called the modulus of elasticity of Ji d the material; it is usually represented by E^ so that for tension or compression. For steel F has been found to be 30,000,000 lb. per square inch. IMPACT 323 173. Impact of Imperfectly Elastic Bodies. — Let d^ be the compression of M^ due to the impact, iincl c?2 the com- pression of ilfg* ^f -Pyji i^ tl^6 pressure between the two bodies when the compression is greatest, the average force p acting maybe represented by --^, if the limit of elasticity of the material is not passed. The work done on the two p bodies is -~ Qd^ + d.^^ and this should equal the energy lost during compression ; then p I From the preceding article we know that d. = ''L^ F I ' ' and c:?2= xT^^ where Zj and Zg are the lengths of the masses itf^ and M^ (considered prismatic), F^ and F^ the areas of cross section, and F^ and E^ the moduli of elas- ticity. The sum d^ + d^ may then be represented by ^^J^l.^'l ^ J^2-^2 F F F 7^ For convenience let ^ ^ = ^^ and ^ ^ = ^2 (iTj and ^1 ^2 ^2 may be considered as representing the hardness). Then d, + d,==pj^j±fl and i>.. = (.^1- V,) ^3^^^^^^ l^r+ffJ 324 APPLIED MECHANICS FOB ENGINEERS Problem 275. Suppose a 100-lb. steel hammer strikes an immov- able cast-iron plate with a velocity of 25 ft. per second. The hammer has a face of 3 sq. in. area and a length of 6 in. ; the plate has the same area and a thickness of 2 in. If the modulus of elas- ticity of steel is 30,000,000 and of cast iron is 15,000,000, find the greatest pressure between the two bodies due to the impact. Under the assumptions v^ = and Mo = oo , M^ = M., ij^ = 22,500,000, H^ = 15,000,000. o'2»2 Then P^ = 132,750 lb. Problem 276. Let the mass of the ram of a pile driver be M^ and let h be the height of fall. Let il/^ be the mass of the pile and s its penetration under a blow. If c?^ is the compression of the ham- mer" and c?2 the compression of the pile, the work equation becomes Pr.s+^(d, + d,)= G,h, and P. = ,y , ■ It is required to find the load that the pile will carry. Since v^ = V2 gh and V2 = 0, we may write and then Pm = ' ^ iV/i + M, V H,H^ J ' Gih 8 + The load Pm that the pile will carry is found by measuring the penetration s for the last blow. It is customary to use a factor of safety, as was explained in Art. 137. Problem 277. A wooden pile whose cross section is 1.5 sq. ft., and whose length is 30 ft., is driven by a steel hammer of 2000 lb. weight falling a distance of 20 ft. The penetration at the last blow is observed to be \ in. What load will the pile carry, using a factor of safety of 6 ? IMPACT 325 Assume the weight of the pile as 1800 lb., the modulus of elas- ticity of timber as 1,500,000, and of steel as 30,000,000, the face of the hammer 2 sq. ft., and its length 2 ft. An approximate formula may be obtained by noting that lo is large as compared with li and E^ is small as compared with Ei, so that Hi is large compared with Ho, thus making H. ^Ih Ih^ 1 . • ^ 1 N -Hjrr = -Hr=m (^Woximately). Substituting in the expression for Pm, we have ■t^m. m _^ / MxM^gli (.Ml + M,)i/2 Problem 278. The student should solve Problem 277, using this approximate formula and compare the result with that already obtained. Problem 279. Compare the results obtained in problems 277 and 278 with those obtained by using formulas of Art. 137. 174. Impact Tension and Impact Compression. — Figure 199 (a) represents a massi!i2 subjected to impact from the mass M^ falling from rest through a height li. The mass M^ is compressed by the impact. Figure 199 (5) repre- sents the body M^ as subjected to impact in tension, the mass in this case being a rod having Gr^ attached to one end and the other end attached to a crosshead A, Tlie rod, crosshead, and weight fall freely together through the distance li until A strikes the stops at B. when one end of the rod suddenly comes to rest aud tlie weight Gr^ causes tension in the rod due to impact. Suppose v^ represents the velocity of M^ when impact occurs and l^ its length, whether it be a tension or compres- 326 APPLIED MECHANICS FOR ENGINEERS sion piece. Let V be the common velocity of the bodies at the time of greatest pressure. Then V= -^ ^ ^^ (see Art. 170), and the kinetic energy necessary to bring the two bodies to rest is given by the expression, where h is the height of fall. ^1 A B 3/. M. G, {a) ib) Fig. 199 This energy is used in stretching or compressing M2, if we neglect the work done on (7^, which is supposed small in comparison. Let c^g be the deformation of M^ when the pressure between the two bodies is greatest, and let the average pressure between them be -^, as before. Then the IMPACT 327 work done on the bar may also be expressed as —^ x d^ 2 and P^d,=M^ = ^ML, 2 ' 2^2 M^ + M^ where F^ is the cross section, l^ the length, and E^ the modulus of elasticity of M^. We may, therefore, write ^-=\ir. Ml 2 ? } 4— -- '-. M, c Fig. 200 IMPACT 329 be considered a rotation about its center of gravity A com- bined with a translation of that center of gravity. Introduce at a point (, the direc- tion of motion making an angle a with the vertical to the plane. After im- Fig. 202 J s © ^^ ^.^> '-V / ' ^ \/^ /• r^^^ / 332 APPLIED MECHANICS FOB ENGINEERS pact the body iff rebounds with a velocity v-^ in the direc- tion, making an angle /3 with the vertical AB, Since the plane is considered smooth, the effect of the impact will be all in the direction of AB and the impulsive force after impact will be e times what it was before impact. Summing horizontal and vertical components of the velocities, we have v-^ sin /3 = V sin a ; v^ cos ^ = ev cos a. Dividing, we have tan yS = - tan a, e and squaring and adding. So that if a and e are known, v^ and yS may be determined. If the body is perfectly elastic, ^ == 1, /3 = «, and v-^ — v. If the body is inelastic, so that ^ = 0, /3 TT 9' "1 V sm a, it then moves along the plane with a velocity v sin a. 178. Impact of Rotating^ Bodies. — Suppose two bodies M^ ^^^d. 31^ revolve about two parallel axes and 0^ (Fig- 203) in such a way that impact occurs at a point along the line BU, Let the point at which impact occurs be distant r^ from and rj from 0^ It is evi- dent that the kinetic energy oi M^ is I- Iq^I and that this is equal to the kinetic Fig. 203 IMPACT 333 energy of an equivalent mass 3f situated at a distance r^ from O.suice IM^v- = l3f7'lco?^. Equating these values for kinetic energy, we have for the equivalent mass 31' AT J ^ the value— 2fo. Likewise the equivalent mass of 31^ at 9 71 T- 7 2 the point of impact is —IJh, The impact of the two ,2 rotating bodies 31^ and iHf^ then, may be considered by considering the impact of their equivalent masses along the line DU, From Art. 171, we have and vl = F(l + e,) - e^v^ ^2 = T^l + ^2) - ^2^'2' where V= 1^ 1 + 2^2 ^ if the bodies are moving in the same direction, and a similar expression with v^^ say, nega- tive, if they are moving in opposite directions. Let 0)2 be the angular velocity of 71^2 before impact. Let coi be the angular velocity of 31^ before impact. Let co^ be the angular velocity of 31^ after impact. Let col ^^ ^'^^ angular velocity of 31^ after impact. Then we may write co^r^ = t'2, (o^7\ = v^, co,^r^ = v.^, co[r^ = vl, so that G)i= r^ 0)2 C 3Lk^7'^ + 3J,k^i^~ ■^ Oi 2 ^01 3I,klrl + 3I,klrl 2—1 (1 + 0-^i«r (1 + ^2) -^2^2' 334 APPLIED MECHANICS FOR ENGINEERS These equations may be put in the form 1 2 2 1 a>2 = (o^-r^ (rjft)! - r^to^yO- + e^} Ix 1 2 ' 2 1 Problem 290. Suppose the moment of inertia of M^ is 3000 and its angular velocity before impact one radian per second ; that of ilfg Fig. 204 IMPACT 335 15,000 and its angular velocity zero. Let r^ = 2 ft. and rg = 3 ft. and = ^2 = 0. Then w^= .311 radian per second. 0) '= — .207 radian per second. The kinetic energy lost due to the impact is 1034 ft.-lb. Problem 291. A well drill is shown in principle in Fig. 204. The drill is supported by a cable that passes over a pulley C and is attached to a friction drum A. When A is held, the drill is raised by the operation of M^ and M^- Suppose that I is 300 and w^ = 3 radians per second ; /g = 200 and w,^ = ] r^ = 2 ft. and rg = 6 ft. Assume e^ = e^^ J. Find o)[ and w'^. What kinetic energy is lost due to each impact ? Problem 292. The moment of inertia of the trip hammer ilfg, illustrated in principle in Fig. 205, is 100,000 ; that of J/j is 00,000. If Tj = 3 ft., r2 = 10 ft., (Oj = 2 radians per second, (Og = 0, and e^^ — e^ = i, find 0)1 and wL What is the kinetic energy lost due to each impact? What is the kinetic energy of the hammer? APPENDIX I HYPEEBOLIC FUNCTIONS cosh X sinh a; = tanh X = 2 sinh X e^ — e~^ cosh X e-^ + e ^ HYPERBOLIC FUNCTIONS 339 X Cosh X Sinha: X Cosh X Sinhx 0.01 1.0000,500 0.0100002 0.51 1.13289:^ 0.5323978 .02 .0002000 .0200013 .52 .1382741 .54375;^) .03 .0004500 .0300045 .53 .1437686 .5551(;.37 .04 .0008000 .0400107 .54 .1493776 .5(;6()292 .05 .0012503 .0500208 .55 .1551014 .5781 51 (; .0() .0018006 .06003()0 .56 .1609408 .5897;n7 .07 .0024510 .0700572 .57 .1668fK)2 .()013708 .08 .0032017 .0800854 .58 .1729()85 .6130701 .09 .0040527 .0901215 .59 .1791579 .6248:)()5 .10 .0050042 .1001668 .60 .1854652 .636653(5 .11 .0060561 .1102220 .61 .1918912 .6485402 .12 .0072086 .1202882 .62 .1984363 .6604917 .13 .0084618 .1303664 .63 .2051013 .6725093 .14 .0098161 .1404578 .64 .2118867 .6845942 .15 .0112711 .1505631 .(>5 .2187933 .mmiry .1() .0128274 .1606835 .()() .2258219 .7089704 .17 .0144849 .1708200 .67 .2329730 .7212()43 .18 .0162438 .1809735 .68 .2402474 .73:^)303 .19 .0181044 .1911452 .69 .247()458 .7460()97 .20 .0200668 .2013360 .70 .2551690 .7585837 .21 .0221311 .2115469 .71 .2628178 .7711735 .22 .0242977 .22177i)0 .72 .2705927 .7838405 .23 .02( )56( 58 .2320333 .73 .2784948 .7^5858 .24 .0289384 .2423107 .74 .2865248 .8094107 .25 .0314132 .2526122 .75 .2946833 .82231()7 .26 .0339<)08 .2629393 .76 .3029713 .835:>049 .27 .0366720 .2732925 .77 .311385)6 .84837()6 .28 .0394568 .2836731 .78 .3199392 .8()15330 .29 .0423456 .2940819 .79 .3286206 .8747758 .30 .0453385 .3045203 .80 .3374349 .8881060 .31 .0484361 .3149891 .81 .3463831 .9015249 .32 .0516384 .3254894 .82 ,3554658 .9150342 .33 .05494()0 .33()0222 .83 .3(>46840 .928();U7 .34 .05835<)0 .34(35886 .84 .3740388 .9423282 .35 .0618778 .3571898 .85 .3835309 .9561 1(W .36 .0(>55029 .3(578265 .86 .3931614 .9699993 .37 .0()92345 .3785001 .87 .4029312 .9839796) .38 .07307:^ .3892116 .88 .4128413 0.9980584 .39 .0770189 .3999619 .89 .4228927 l.()1223()9 .40 .0810724 .4107523 .90 .4330864 .02()5167 .41 .0852341 .4215838 .91 .44.34234 .0408991 .42 .0895042 .4324574 .92 .4539048 .055.3cS5() .43 .0938888 .4433742 .93 .4()45315 .06)99777 .44 .0983718 .4543354 M .4753046 .0 .4972947 .1144018 .47 .1124983 .4874959 .97 .50851.37 .1294;i<)7 .48 .1174289 .498()455 .98 .5198837 .144:.72() .49 .1224712 .5098450 .i)9 .5314057 .1598288 0.50 1.1276260 0.5210953 1.00 1.54.30806 1.1752012 340 HYPERBOLIC FUNCTIONS X Cosh X Sinli X X Cosh X Sinhx 1.01 1.5549100 1.190(3910 1.51 .2338704 2.1529104 1.02 .5668948 .2062999 1.52 .3954(586 .17675(36 1.03 .5790365 .2220294 1.53 .41735()3 .2008206 l.Olr .5913358 .2378812 1.54 .4394857 .2251046 1.05 .6037945 .2538567 1.55 .4618591 .2496111 1.0(3 .6164134 .2()99576 1.56 .4844787 .2743426 1.07 .6291940 .2861855 1.57 .5073467 .2993014 1.08 .6421375 .3025420 1.58 .5304654 .3244903 1.09 .6552453 .3190288 1.59 .5538373 .3499117 1.10 .6685186 .3356474 1.60 .5774645 .3755679 1.11 .6819587 .3523997 1.61 .()013494 .4014618 1.12 .7005670 .3(342872 1.62 .6254945 .4275958 1.13 .7093449 .3863116 1.63 .6499(^21 .4539726 1.14 .7232938 .4034746 1.64 .6745748 .4805947 1.15 .7374148 .4207781 1.65 .6995149 .5074(350 1.16 .7517098 .4382235 1.66 .7247249 .5345859 1.17 .7661798 .4558128 1.67 .7502074 .5619()03 1.18 .7808265 .4735477 1.68 .7759(550 .5895910 1.19 .7956513 .4914299 1.69 .8020001 .6174806 1.20 .8106556 .5094613 1.70 .8283154 .6456319 1.21 .8258410 .5276436 1.71 .8549136 .6740479 1.22 .8412089 .5459788 1.72 .8817974 .7027311 l!23 .8567610 .5644685 1.73 .9089692 .7316847 1.24 .8724988 .5831146 1.74 .9364319 .7(509115 1.25 .8884239 .(3019191 1.75 .9641884 .7904143 1.26 .9045378 .6208837 1.76 2.9922411 .8201962 1.27 .9208421 .6400105 1.77 3.0205932 .8502601 1.28 .9373385 .6593012 1.78 .0492473 .8806091 1.29 .9540287 .6787578 1.79 .0782063 .9112461 1.30 .9709143 .6983824 1.80 .1074732 .9421742 1.31 1.9879969 .7181768 1.81 .1370508 2.9733966 1.32 2.0052783 .7381431 1.82 .16()9421 3.(W49163 1.33 .0227603 .7582830 1.83 .1971501 .03673(55 1.34 .0404446 .7785989 1.84 .2276799 .0(588(303 1.35 .0583329 .7990926 1.85 .2585283 .1012911 1.36 .0764271 .8197662 1.86 .2897047 .1340321 1.37 .0947288 .840()219 1.87 .3212100 .1670863 1.38 .1132401 .8616615 1.88 .3530475 .2004573 1.39 .1319627 .8828874 1.89 .3852202 .2341484 1.40 .1508985 .9043015 1.90 .4177315 .2681629 1.41 .1700494 .9259060 1.91 .450584(3 .3025041 1.42 .1894172 .9477032 1.92 .4837827 .3371758 1.43 .2090041 .9(39()951 1.93 .5173293 .3721810 1.44 .2288118 1.9918840 1.94 .5512275 .4075235 1.45 .2488424 2.0142721 1.95 .5854808 .44:52067 1.46 .2690979 .0:>68616 1.9() .6200927 .4792343 1.47 .2895803 .0591)549 1.97 .6550()(;7 .515(5097 1.48 .3102917 .082()540 1.98 .69040(31 .55233(58 1.49 .3312341 .1058614 1.99 .7261146 .5894191 1.50 2.352409(3 2.1292794 2.00 3.7()21957 3.6268604 HYPERBOLIC FUNCTIONS 341 X Cosh X Sinh X X Cosh X Siuh X 2.01 3.7986528 3.6(54()642 2.51 6.19:!0993 6.111S311 2.02 .8354899 .7028:U5 2.52 .2545281 .174()(5S5 2.03 .8727101 .7413746 2.53 .31(55827 .2:5692:57 2.0-i .9103184 .7802896 2.54 .3792(587 .:3()(J402:3 2.05 .9483518 .819(5198 2.55 .4425928 .3(545111 2.06 .98()7111 .8592571 2.56 .50(55(511 .4292.~)(5:i 2.07 4.02550:'>8 .8993179 2.57 .5711800 .494(5444 2.08 .0()47395 .9:598093 2.58 .(5:3(545(50 .5(50(5820 2.09 .1043012 .980(5140 2.59 .702:5958 .(527:5758 2.10 .1443131 4.0218567 2.(50 .7690059 .6947:323 2.11 .1847398 .06:^5018 2.61 .8362940 .7627595 2 12 .2255846 .1055530 2.(52 .9042(544 .8314(515 2.13 .2668523 .1480149 2.63 .9729254 . 2.50 6.1322895 6.0502045 3.00 10.0(37(3620 10.017S750 342 n YPEliB OLIC FUN CTIONS X Coslia: Sinha: X Cosh X Sinha: 3.01 10.1683456 10.1190539 3.51 16.7390823 16.7091854 3.02 .2700464 .2212451 3.52 .9070139 .8774144 3.03 .3727741 .3244585 3.53 17.0766361 17.0473312 3.04 .4765391 .4287042 3.54 .2479662 .2189529 3.05 .5813518 .5339929 3.55 .4210213 .39229()6 3.06 .6872224 .6403347 3.56 .5958178 .567:3790 3.07 .7941620 .7477408 3.57 .7723744 .7442186 3.08 .9021809 .85()2217 3.58 .9507082 .9228:325 3.09 11.0112900 .9657881 3.59 18.1308371 18.1032388 3.10 .1215004 11.0764511 3.60 .3127790 .2854552 3.11 .2328226 .1882217 3.61 .4965523 .4695004 3.12 .3452684 .3011112 3.62 .6821753 .6553927 3.13 .4588488 .4151309 3.63 .8696665 .8431503 3.14 .5735748 .5302919 3.64 19.05^)0447 19.0327924 3.15 .6894584 .6466062 3.()5 .2503288 .2243376 3.16 .8065107 .7640850 3.66 .4435377 .4178052 3.17 .9247440 .8827403 3.67 .6386909 .61:32145 3.18 12.0441695 12.0025838 3.68 .8358083 .8105854 3.19 .1647998 .123(5279 3.69 20.0349094 20.0099373 3.20 .2866462 .2458839 3.70 .2360140 .2112905 3.21 .4097213 .3693646 3.71 .4391421 .4146645 3.22 .5340375 .4940825 3.72 .6443142 .6200802 3.23 .6596073 .6200497 3.73 .8515505 .8275577 3.24 .7864428 .7472790 3.74 21.0608720 21.0371178 3.25 .9145572 .8757829 3.75 .2722997 .2487819 3.26 13.0439629 13.0055744 3.76 .4858548 .4625710 3.27 .1746730 .136{j665 3.77 .7015584 .6785064 3.28 .30()7006 .2690723 3.78 .9194324 .89(J609() 3.29 .4400587 .4028048 3.79 22.1394981 22.1169025 3.30 .5747611 .5378780 3.80 .3617777 .3394069 3.31 .7108208 .6743046 3.81 .5862933 .5641452 3.32 .8482516 .8120988 3.82 .8130681 .7911403 3.33 .9870673 .9512741 3.83 23.0421239 23.0204143 3.34 14.1272820 14.0918450 3.84 .2734843 .2519907 3.35 .2689091 .2338247 3.85 .5071715 .4858917 3.36 .411 (KiO .3772277 3.86 .7432095 .7221415 3.37 .5564583 .5220686 3.87 .9816222 .9607()38 3.38 .7024094 .()(i83()19 3.88 24.2224327 24.2017819 3.39 .8498306 .8161219 3.89 .465()658 .4452205 3.40 .9987366 .9653634 3.90 .7113454 .6911034 3.41 15.1491429 15.1161016 3.91 .9594963 .9394557 3.42 .3010()37 .2683513 3.92 25.2101431 25.1W3020 3.43 .4545147 .4221278 3.93 .4633109 .44:^6673 3.44 .6095114 .5774468 3.94 .7190247 .6995765 3.45 .7()60()88 .7343232 3.95 .9773109 .9580561 3.46 .9242033 .8927735 3.2 32428 32(533 32838 33041 33243 3:3445 33(546 33845 :34044 22 34242 34439 34(535 34830 35024 :35218 :35410 35(502 :35793 :35983 23 36173 363()1 3(5548 36735 36921 :^7106 37291 37474 37657 37839 24 38021 38201 38381 38560 38739 38916 39093 39269 39445 39619 25 39794 39967 40140 40312 40483 40654 40824 40993 41162 41330 26 41497 416(i4 41830 41995 421(>0 42:524 42488 42651 42813 42975 27 43136 43296 4345(5 4361(5 43775 4:3933 44090 44248 44404 44560 28 44716 44870 45024 45178 45331 45484 45(536 45788 459:3i) 4(5089 29 46240 46389 46538 46686 46834 46982 47129 47275 47421 47567 30 47712 47856 48000 48144 48287 48430 48572 48713 48855 48995 31 49136 49276 49415 49554 49693 49831 49968 50105 50242 50:379 32 50515 50650 50785 50920 51054 51188 5i:321 51454 51587 51719 33 51851 51982 52113 52244 52374 52504 526:53 52763 52891 53020 34 53148 53275 53402 5352i) 53655 53781 53907 54033 54157 54282 35 54407 54530 54654 54777 54900 55022 55145 55266 55388 55509 36 55630 55750 55870 55990 5(5110 56229 56:i48 5(5466 5(5584 56702 37 56820 56937 57054 57170 57287 57403 57518 57(5:34 57749 57863 38 57978 58092 5820(5 58319 58433 5854(5 58(558 58771 58883 58995 39 59106 59217 59328 59439 59549 59659 597(59 59879 59988 (50097 40 60206 60314 (50422 60530 60638 60745 60852 60959 61066 61172 41 61278 61384 (51489 61595 61700 61804 61909 (520i:5 (52118 (52221 42 62325 (52428 (52531 62634 (52736 62838 (52941 (5:5042 (53144 (5:5245 43 63347 ();5447 63548 63(548 63749 6:3848 (5:3948 (54048 64147 (54246 44 64345 64443 64542 (>4t540 (54738 (548:36 649:53 (550:30 65127 (55224 45 65321 65417 65513 65()09 (55705 65801 6589(5 (55991 (5(508(5 (56181 46 66276 (56370 (5(5464 (5(5558 (5(5(551 (5(5745 6(58:58 (5(1931 (57024 (57117 47 67210 67302 67394 (57486 (57577 (57(5(59 (57760 (57851 (57942 (580:33 48 68124 68214 68:^4 (58394 (58484 (58574 (58(5(53 (5S752 (58842 (589:50 49 69020 69108 69196 (59284 (59372 (59460 (59548 (59(5: 55 (59722 (59810 50 69897 69983 70070 70156 70243 70:^29 70415 70500 7058(5 70(571 51 70757 70842 70927 71011 7109(5 71180 712(55 71:549 1 714:53 71516 52 71600 71683 717(57 71850 7193:5 72015 72098 72181 722(53 72:545 53 72428 72509 72591 72(572 72754 728:55 72! 116 72997 7:5078 73158 54 73239 73319 73399 73480 73559 73639 73719 73798 73878 73957 346 LOGARITHMS OF NUMBERS LOGARITHMS OF NUMBERS, FROM TO 1000 (Contl/tued) No. 1 2 3 4 5 6 7 8 9 55 74036 74115 74193 74272 74351 74429 74507 74585 74(5()3 74741 56 74818 74896 74973 75050 75127 75204 75281 ' 75358 75434 75511 57 75587 75663 75739 75815 75891 75966 7()042 76117 76192 76267 58 76342 76417 76492 76566 76641 76715 76789 1 76863 76937 77011 59 77085 77158 77232 77305 77378 77451 77524 77597 77670 77742 60 77815 77887 77959 78031 78103 78175 78247 78318 78390 78461 61 78533 78604 78()75 78746 78816 78887 78958 79028 79098 791(59 62 79239 79309 79379 79448 79518 79588 79()57 79726 79796 798(55 63 79934 80002 80071 80140 80208 80277 80345 80413 80482 80550 64 80618 80685 80753 80821 80888 80956 81023 81090 81157 81224 65 81291 81358 81424 81491 81557 81624 81690 81756 81822 81888 6() 81954 82020 82085 82151 82216 82282 82347 82412 82477 82542 67 82607 82672 82736 82801 82866 82930 82994 83058 83123 83187 68 83250 83314 83378 83442 83505 83569 83632 83(595 83758 83821 69 83884 83947 84010 84073 84136 84198 84260 84323 84385 84447 70 84509 84571 84633 84695 84757 84818 84880 84941 85003 85064 71 85125 85187 85248 85309 85369 85430 85491 85551 85612 85672 72 85733 85793 85853 85913 85973 86033 86093 8(5153 8(5213 86272 73 86332 86391 86451 86510 86569 86628 86687 86746 86805 86864 74 86923 86981 87040 87098 87157 87215 87273 87332 87390 87448 75 87506 87564 87621 87679 87737 87794 87852 87909 87966 88024 76 88081 88138 88195 88252 88309 88366 88422 88479 88536 88592 77 88649 88705 88761 88818 88874 88930 88986 89042 8fK)98 89153 78 89209 89265 89320 89376 89431 89487 89542 89597 89(552 89707 79 89762 89817 89872 89927 89982 90036 90091 90145 90200 90254 80 90309 90363 90417 90471 90525 90579 90633 90687 90741 90794 81 90848 90902 90955 91009 91062 91115 91169 91222 91275 91328 82 91381 91434 91487 91540 91592 91645 91698 91750 91803 91855 83 91907 91960 92012 92064 92116 92168 92220 92272 92324 92376 84 92427 92479 92531 92582 92634 92685 92737 92788 92839 92890 85 92941 92993 93044 93095 93146 93196 93247 93298 93348 93399 86 93449 93500 93550 9;^01 93651 93701 93751 93802 93852 93^X)2 87 93951 94001 94051 94101 94151 94200 1M250 94300 94349 94398 88 94448 94497 94546 94596 94645 94(594 94743 94792 94841 94890 89 94939 94987 95036 95085 95133 95182 95230 95279 95327 95376 90 95424 95472 95520 95568 95616 95664 95712 95760 95808 95856 91 95904 95951 95999 96047 96094 96142 9(5189 9(5236 96284 96331 92 %378 96426 9(^73 9()520 9()567 9()614 9(5661 9(5708 96754 9(5801 93 96848 96895 9()941 9()988 970;U 97081 97127 97174 97220 972(56 94 97312 97359 97405 97451 97497 97543 97589 97635 97680 97726 95 97772 97818 97863 97^K)9 97954 98000 98045 98091 98136 98181 96 98227 98272 98317 983()2 98407 98452 98497 98542 98587 98632 97 98677 98721 9876(j 98811 98855 98900 98945 98989 99033 99078 98 f)9122 991()6 99211 99255 99299 99343 99387 99431 99475 99519 99 99563 99607 99651 99694 99738 99782 99825 99869 99913 99956 APPENDIX III TRIGONOMETRIC FUNCTIONS TRIGONOMETIilC FUNCTIONS 349 NATURAL SINES, COSINES, TANGENTS, ETC. O / Sine Cosecant Tangent Cotangent Secant Cosine / .000000 Infinite .000000 Infinite 1.00000 1.000000 90 10 .002* 109 343.77516 .0(>29()9 343.77371 1.0()()(K) .99991 K3 50 20 .oonsis 171.8S831 .005818 171.88540 1.00(X)2 .999983 40 30 .008727 114.59301 .008727 114.588(55 1.00004 .999962 30 40 .011()35 85.945609 .011()3() 85.939791 1.00007 .999932 20 50 .014544 68.757360 .014545 68.750087 1.00011 .999894 10 1 .017452 57.298688 .017455 57.289962 1.00015 .999848 89 10 .0203()1 49.114062 .020365 49.103881 1.00021 .999793 50 20 .0232()9 42.975713 .023275 42.9(54077 1.00027 .999729 40 30 .021)177 38.201550 .02()186 38.188459 i.ooo:^ .999657 30 40 .029085 34.382316 .01^9097 34.3(57771 1.00042 .999577 20 50 .031992 31.257577 .032009 31.241577 1.00051 .999488 10 2 .034899 28.653708 .034921 28.636253 1.00061 .999391 88 10 .03780() 2().450510 .037834 2(5.431(500 1.00072 .999285 50 20 .040713 24.562123 .040747 24.541758 1.00083 .999171 40 30 .043(;i9 22.925586 .0436(51 22.9037(36 1.00095 .999048 30 40 .04(5525 21.49:5676 .046576 21.470401 1.00108 .998917 20 50 .049431 20.230284 .049491 20.205553 1.00122 .998778 10 3 .052336 19.107323 .052408 19.0811.37 1.00137 .998630 87 10 .055241 18.102619 .055325 18.074977 1.00153 .998473 50 20 .0.58145 17.1984:i4 .058243 17.169337 1.001(59 .998:308 40 30 .0()1049 16.380408 .0611(53 16.:^9855 1.00187 .998135 30 40 .C>63952 15.6;i6793 .0(54083 15.(504784 1.00205 .997:357 20 50 .006854 14.957882 .0(37004 14.924417 1.00224 .997763 10 4 .069756 14.335587 .069927 14.300(366 1.00244 .997564 86 10 .072658 13.763115 .072851 13.72(5738 1.002(35 .997:357 50 20 .075559 13.234717 .075776 13.196888 1.00287 .997141 40 'SO .078459 12.745495 .078702 12.70(3205 1.00309 .996917 30 40 .081359 12.291252 .081629 12.250505 1.00333 .99(3685 20 50 .084258 11.868370 .084558 11.826167 1.00357 .9iK3444 10 5 .087156 11.473713 .087489 11 430052 1.00382 .996195 85 10 .090053 11.104549 .090421 11.059431 1.00408 .9959:37 50 20 .092950 10.758488 .093:^)4 10.711913 1.00435 .995671 40 30 .095846 10.43;'.431 .096289 10.385397 1.004(33 .995396 30 40 .098741 10.127522 .09922(5 10.078031 1.00491 .995113 20 50 .101635 9.8391227 .102164 9.7881732 1.00521 .91M822 10 6 .104528 9.5667722 .105104 9.5143645 1.00551 .994522 84 10 .107421 9.3091()9<) .10804(5 9.255;i035 1.00582 .994214 50 20 .110313 9.0651512 .110990 9.0098261 1.00614 .993897 40 83 o / Cosine Secant Cotangent Tangent Cosecant Sine f Fit func tions from S3* ' 40' to 90° read from bott om of tab e upward. 350 TRIGONOMETRIC FUNCTIONS NATURAL SINES , COSINES, TANGENTS, ETC. (Co7itinued) o / Sine Cosecant Tangent Cotangent Secant Cosine / o 6 30 .113203 8.8336715 .113936 8.7768874 1.00647 .993572 30 40 .11()093 8.6137901 .116883 8.5555468 1.00681 .993238 20 50 .118982 8.4045586 .119833 8.3449558 1.00715 .992896 10 7 .121869 8.2055090 .122785 8.1443464 1.00751 .992546 83 10 .124756 8.0156450 .125738 7.9530224 1.00787 .992187 50 20 .127642 7.8344335 .128694 7.7703506 1.00825 .991820 40 30 .130526 7.6612976 .131653 7.5957541 1.00863 .991445 30 40 .133410 7.4957100 .134613 7.4287064 1.00902 .991061 20 50 .136292 7.3371909 .137576 7.2687255 1.00942 .990669 10 8 .139173 7.1852965 .140541 7.1153697 1.00983 .990268 82 10 .142053 7.0396220 .143508 6.9682335 1.01024 .989859 50 20 .144932 6.8997942 .14()478 6.8269437 1.01067 .989442 40 30 .147809 6.7654691 .149451 6.6911562 1.01111 .989016 30 40 .150686 6.6363293 .152426 6.5605538 1.01155 .988582 20 50 .153561 6.5120812 .155404 6.4348428 1.01200 .988139 10 9 .156434 6.3924532 .158384 6.3137515 1.01247 .987688 81 10 .159307 6.2771933 .161368 6.1970279 1.01294 .987229 50 20 .162178 6.1(360674 .164354 6.0844381 1.01342 .986762 40 30 .165048 6.0588980 .167343 5.9757644 1.01391 .986286 30 40 .167916 5.9553625 .170334 5.8708042 1.01440 .985801 20 50 .170783 5.8553921 .173329 5.7693688 1.01491 .985309 10 10 .173648 5.7587705 .176327 5.6712818 1.01543 .984808 80 10 .176512 5.6653331 .179328 5.5763786 1.01595 .984298 50 20 .179375 5.5749258 .182332 5.4845052 1.01649 .983781 40 30 .182236 5.4874043 .185339 5.3955172 1.01703 .983255 30 40 .185095 5.4026333 .188359 5.3092793 1.01758 .982721 20 50 .187953 5.3204860 .191363 5.2256647 1.01815 .982178 10 11 .190809 5.2408431 .194380 5.1445540 1.01872 .981627 79 10 .193(i()4 5.1635924 .197401 5.0658352 1.01930 .981068 50 20 .196517 5.0886284 .200425 4.9894027 1.01989 .980500 40 30 .199368 5.0158317 .203452 4.9151570 1.02049 .979925 30 40 .202218 4.9451687 .206483 4.8430045 1.02110 .979341 20 50 .205065 4.8764907 .209518 4.7728568 1.02171 .978748 10 12 .207912 4.8097343 .212557 4.7046e301 1.02234 .978148 78 10 .210756 4.7448206 .215599 4.6382457 1.02298 .977539 50 20 .213599 4.6816748 .218645 4.5736287 1.02362 .976921 40 30 .21(>440 4.()202263 .221695 4.5107085 1.02428 .97(>296 30 40 .219279 4.5604080 .224748 4.4494181 1.02494 .975(562 20 50 .222116 4.5021565 .227806 4.3896940 1.02562 .975020 10 77 o f Cosine Secant Cotangent Tangent Cosecant Sine f o F or functions from 77° K ' to S3° 30' read from bottom of ta ible upward. TRIGONOMETRIC FUNCTIONS 351 NATURAL SINES, COSINES, TANGENTS, ETC. (Continued) 13 14 15 16 17 18 19 r Sine .224951 10 .227784 20 .23061(3 30 .23;U45 40 .230273 50 .239098 .241922 10 .244743 20 .247r)()3 30 .250380 40 .253195 50 .250008 .258819 10 .201028 20 .2()44:34 30 .207238 40 .270040 50 .272840 .275037 10 .278432 20 .281225 30 .284015 40 .280803 50 .289589 .292372 10 .295152 20 .297930 30 .30070() 40 .30:3479 50 .300249 .309017 10 .311782 20 .314545 30 .317305 40 .3200()2 50 .322810 .325508 10 .328317 20 .331003 1 Cosine Cosecant 4.4454115 4.3, COSINES, TANGENTS, ETC. (Continued) O / Sine Cosecant Tangent Cotangent Secant Cosine / 39 .629320 1.5890157 .809784 1.2348972 1.28676 .77714(5 51 10 .(531578 1.5833318 .814612 1.2275786 1.28980 .775312 50 20 .633831 1.5777077 .8194(53 1.2203121 1.29287 .77:3472 40 30 .636078 1.5721337 .82433(5 1.21:30970 1 .29597 .771625 30 40 .638320 1.5666121 .8292:U 1.2059:527 1.29909 .7(59771 20 50 .640557 1.5611424 .834155 1.1988184 1.30223 .767911 10 40 .642788 1.5557238 .839100 1.1917536 1.30541 .766044 50 10 .645013 1.5503558 .8440(59 1.1847:576 l.:50861 .7(54171 50 20 .647233 1.5450378 .849062 1.1777698 1.31183 .7(52292 40 30 .649448 1.5397(590 .854081 1.1708496 1.31509 .7(50406 30 40 .651657 1.5345491 .859124 1.1(539763 1.31837 .758514 20 50 .653861 1.5293773 .864193 1.1571495 1.32168 .756615 10 41 .656059 1.5242531 .869287 1.1503684 1.32501 .754710 49 10 .658252 1.5191759 .874407 1.143(5326 1.328:38 .752798 50 20 .6(50439 1.5141452 .879553 1.1:369414 1.33177 .750880 40 30 .(362620 1.5091(505 .884725 1.1302944 l.:3:3519 .748956 30 40 .()6479() 1.5042211 .889924 1.1236909 1.3:38(54 .747025 20 50 .666966 1.4993267 .895151 1.1171:305 1.34212 .745088 10 42 .669131 1.4944765 .900404 1.1106125 1.34563 .743145 48 10 .671289 1.489(5703 .W5(585 1.1041:5(55 1.34917 .741195 50 20 .67^43 1.4849073 .910994 1.0977020 l.:35274 .739239 40 30 .675590 1.4801872 .916331 1.091:5085 l.:356:54 .737277 30 40 .677732 1.4755095 .921(597 1.0849554 1.35997 .735309 20 50 .679868 1.4708736 .927091 1.0786423 1.36363 .733335 10 43 .681998 1.4662792 .932515 1.0723(587 1.36733 .73ia54 47 10 .684123 1.4(517257 .937iH)8 i.066i:ui 1.37105 .7293(57 50 20 .686242 1.4572127 .943451 1.0599:381 l.:37481 .727374 40 30 .688355 1.4527397 .9481K55 1.0537801 l.:378t50 .725374 30 40 .6904()2 1.448:^(53 .954508 1.0476598 1.38242 .72336<) 20 50 .692563 1.4439120 .960083 1.0415767 1.38628 .721357 10 44 .694658 1.4395565 .9(55689 1.0355:503 1.39016 .719340 46 10 .(i9()748 1.4352393 .97132(5 1.0295203 l.:594()9 .717:316 50 20 .()98832 1.4:509()02 .97(59i)(5 1.02:354(51 l.:39804 .715286 40 30 .700{K)9 1.42(57182 .982<5i>7 1.01 76074 1.4020:5 .71:5251 30 40 .702981 1.4225134 .988432 1.0117088 1.40(50(5 .711209 20 50 .705047 1.4183454 .994199 1.0058:348 1.41012 .709161 10 45 .707107 1.4142136 1.000000 1.0000000 1.41421 .707107 45 o / Cosine Secant Cotangent Tangent Cosecant Sine / For functit )ns from 45°-( y to 51°-0' reiul from bottom of tab le upward. APPENDIX IV SQUARES, CUBES, SQUARE ROOTS, ETC. SQUARES, CUBES, SQUARE ROOTS, ETC. 359 SQUARES, CUBES , SQUARE ROOTS, CUBE BOOTS, AND RECIPROCALS No. Squares Cubes Square Roots Cube Roots Reciprocals 1 1 1 1.0000000 1.0000000 1.000000000 2 4 8 1.414213() 1.2599210 .500000000 3 9 27 1.7320508 1.4422496 .;3:3:5:5:3:5:3:53 4 16 64 2.0000000 1.5874011 .250000000 5 25 125 2.23(J0(i80 1.7099759 .200000000 6 36 216 2.4494897 1.8171206 .l(i(;6( 56(567 7 49 343 2.()457513 1.9129312 .142.S.57143 8 64 512 2.8284271 2.00000(J0 .] 25000000 9 81 729 3.0000000 2.0800837 .111111111 10 100 1000 3.1622777 2.1544;M7 .100000000 11 121 1331 3.316()248 2.2239801 .090iK)454545 23 529 12167 4.7958315 2.8438670 .04:347S2(51 24 576 13824 4.8989795 2.8844991 .041(5(56(5(57 25 625 15625 5.0000000 2.9240177 .040000000 26 676 17576 5.091K)195 2.9()24960 .0:38461. "):38 27 729 11M383 5.1')889 .0:357142S(5 29 841 24389 6.3851648 3.07231(58 .034482759 30 900 27000 5.4772256 3.1072325 .0»5:):).j>). 5:3.3 31 961 29791 5.5(577644 3.1413806 .0:52258065 32 1024 32768 5.6568542 3.1748021 .0312.50000 33 1089 35937 5.7445()2() 3.2075^U3 .0:30:30:50:30 34 1156 39304 5.8:509519 3.2:306118 .0294117(55 35 1225 42875 5.9160798 3.2710663 .028571429 36 1296 46656 6.0000000 3.:5019272 .027777778 37 1369 50653 6.0827(525 3.:3:322218 .027027(V27 38 1444 54872 6.1()44140 3.:5(519754 .026:515789 39 1521 59319 6.2449^)80 3.3912114 .025641026 40 1600 64000 6.3245553 3.41<9519 .02.")0noooo 41 1681 68921 6.4031242 3.44S2172 .024:3'. K)244 42 lliyi 74088 6.4807407 3.47(i02(3(3 .02:3S(MI524 43 1849 79507 6.5574:585 3.50:5:5981 .02:52.-)5S14 44 1936 85184 ().():^:>249() 3.530:5483 .022727273 45 2025 91125 6.7())»)0*)'>')02 4(5 2116 973:^ 6.782:5:i()0 3.58:30479 !02173rtir3(") 47 2209 103823 ().(S55().')4(> 3.(50882(51 .021 27t 5596 48 2304 110592 6.92S20:V2 3.6:342411 .0208:5:5:3:33 49 2401 ii7(;4i) 7.000()()00 3.659:3057 .O2O40Sl(53 360 SQUARES, CUBES, SQUARE ROOTS, ETC. 2 SQUARES, CUBES , SQUARE ROOTS, CUBE EOOTS, AND RECIPROCALS Jfo. Squares Cubes Square Roots Cube Roots Reciprocals 50 2500 125000 7.0710678 3.6840314 .020000000 51 2601 132651 7.1414284 3.7084298 .019(J07843 52 2701 140608 7.2111026 3.7325111 .019230769 53 2809 148877 7.2801099 3.7562858 .018867925 54 2916 157464 7.3484692 3.7797631 .018518519 55 3025 166375 7.4161985 3.8029525 .018181818 5(5 3136 175616 7.48.33148 3.8258624 .017857143 57 3249 185193 7.5498344 3.8485011 .017543860 58 3364 195112 7.6157731 3.870876() .017241379 59 3181 205379 7.6811457 3.8929965 .016949153 60 3600 216000 7.7459667 3.9148676 .016666667 61 3721 226981 7.8102497 3.9364972 .016393443 62 3814 238328 7.8740079 3.9578915 .016129032 63 3969 250047 7.9372539 3.9790571 .015873016 64 4096 262144 8.0000000 4.0000000 .015625000 65 4225 274625 8.0622577 4.0207256 .015384615 66 4356 287496 8.1240384 4.0412401 .015151515 67 4489 300763 8.1853528 4.0615480 .014925373 68 4624 314432 8.2462113 4.0816551 .014705882 69 4761 328509 8.3066239 4.1015661 .014492754 70 4900 343000 8.3G66003 4.1212853 .014285714 71 5041 357911 8.4261498 4.1408178 .014084507 72 5184 373248 8.4852814 4.1601676 .013888889 73 5329 389017 8.5440037 4.1793390 .013698630 7i 5476 405224 8.6023253 4.1983364 .013513514 75 5625 421875 8.6602540 4.2171633 .013333333 76 5776 438976 8.7177979 4.2358236 .013157895 77 5929 456533 8.7749644 4.2543210 .012987013 78 6084 474552 8.8317609 4.2726586 .012820513 79 6241 493039 8.8881944 4.2908404 .012658228 80 6400 512000 8.9442719 4.3088695 .012500000 81 6561 531441 9.0000000 4.3267487 .012345679 82 6724 551368 9.0553851 43444815 .012195122 83 6889 571787 9.1104336 4.3620707 .012048193 84 7056 592704 9.1651514 4.3795191 .0119047()2 85 7225 614125 9.2195445 4.396821K) .011764706 86 7396 636056 9.2736185 4.4140049 .011(327907 87 7569 658503 9.3273791 4.4310476 .011494253 88 7744 681472 9.3808315 4.4479602 .011363636 89 7921 7049()9 9.4339811 4.4647451 .011235955 90 8100 72^)000 9.4868330 4.4814047 .011111111 91 8281 753571 9.5393920 4.4979414 .010989011 92 8464 778688 9.5f)16630 4.5143574 .010869565 93 8649 804357 9.6436508 4.5.306549 .010752688 91 8836 8305S4 9.695.3.^)97 4.54()8359 .010(538298 95 9025 857;'.75 9.74()7943 4..")(;29026 .01052(5316 96 9216 88473() 9.7979.'")90 4.57S8.570 .010416(5(57 97 9409 912673 9.848S578 4.5947009 .010309278 98 9()04 941192 9.8994949 4.61043()3 .010204082 99 9801 970299 9.9498744 4.6260650 .010101010 8QUABES, CUBES, SQUARE SOOTS, ETC. 3 361 SQUARES, CUBES , SQUARE ROOTS, CUBE ROOTS, AND RECIPROCALS No. Squares Cubes Square Roots Cube Roots Reriprorals 100 10000 100(^000 10.0000000 4.(5415888 .oiooooooo 101 10201 1030301 10.049875(5 4.(5570095 .009900990 102 10404 1061208 10.0995049 4.^)72:3287 .00i)8():i922 103 10G09 1092727 10.1488916 4.(3875482 .00970S7:}8 104 1081(3 1124864 10.1980390 4.702(5(594 .009(515:W5 105 11025 1157(325 10.24(59508 4.717(5940 .00952:5810 10() 11236 1191016 10.295(5301 4.732(52:35 .0094:5:59(52 107 11449 1225043 10.:5440804 4.7474594 .0(^)9:145794 108 1161)4 1259712 10.3923048 4.7(5220:52 .0(J9259259 109 11881 1295029 10.4403065 4.7768562 .009174312 110 12100 1331000 10.4880885 4.7914199 .009090909 111 12321 13()7631 10.535(5538 4.8058955 .009009009 112 12544 1404928 10.5830052 4.8202845 .008928571 113 127()9 1442897 10.6301458 4.8:345881 .008849558 114 1299() 1481544 10.6770783 4.8488076 .008771!):30 115 13225 1520875 10.7238053 4.8(529442 .008695(552 IK) 13456 1560896 10.7703296 4.87(59990 .008(520(590 117 13689 1601(313 10.8166538 4.89097:32 .008547009 118 13924 1(343032 10.8(527805 4.9048(581 .008474576 119 14161 1685159 10.9087121 4.9186847 .008403361 120 14400 1728000 10.9544512 4.9324242 .0083333:33 121 14(541 1771561 11.0000000 4.94(50874 .0082(544(53 122 14884 1815848 11.0453610 4.9596757 .00819(5721 123 15129 18608()7 11.09053(55 4.9731898 .0081:30081 124 15:'>7() 1906624 11.1355287 4.98(5(5310 .0080(5451(5 125 15()25 1953125 11.1803:599 5.000(X)00 .008000000 12(i 15876 2000376 11.2249722 5.01:32979 .0079:3(3508 127 16129 2048383 11.2(594277 5.02(55257 .00787401(5 128 16384 2097152 11.3137085 5.0:39(5842 .007812500 121) 1(3()41 2146689 11.3578167 5.0527743 .007751938 130 16900 2197000 11.4017543 5.0657970 .007692:308 131 171()1 2248091 11.4455231 5.07875:51 .0076:53588 132 17424 22i)99()8 11.4891253 5.091(54:34 .007575758 133 17689 2352(i37 11.5325(526 5.1044(587 .007518797 134 1795() 240(5104 11.5758369 5.1172299 .0074(52(587 l.T) 18225 21(5():'>75 11.6189500 5.129H278 .007407407 i:;(j 1849() 2515456 11.(5(5190:58 5.1425(5:32 .007:352941 137 18769 2571353 11.704(5999 5.1551:3(57 .007299270 138 11H)44 2(52.S072 11.747:3401 5.1(57(5493 .00724(5:577 139 19321 2685(319 11.7898261 5.1801015 .007194245 140 19(;00 2744000 11.8321596 5.1924941 .0071428.57 141 19881 2803221 11.874:3421 5.2()4S279 .007092 11 »9 142 20164 28(5.".288 ll.i)l(5.3753 5.21710:34 .0070422.54 143 20449 2!)24207 11.9582(507 5.229:3215 .00(599:5007 144 207;5() 2985984 12.(X)()00()0 5.2414828 .006944444 145 21025 3048(525 12.041594(5 5.25:i5879 .0O(589(55.")2 146 21316 3112136 12.0.S:;04(50 5.2(55(5.374 .00(5,S49315 147 21()()9 317(5523 12.124:5557 5.277(5:321 .()0(5SO2721 148 21904 3241792 12.1(55.-)251 5.2895725 .00(575(5757 149 22201 's^mm 12.20(5.")55() 5.:5()14592 .00(5711409 362 SQUARES, CUBES, SQUARE ROOTS, ETC. 4 SQUARES, CUBES , SQUARE ROOTS, CUBE ROOTS, AND RECIPROCALS Ao. Squares Cul)('s Square Roots Cube Roots Reciprocals 150 22500 3375000 12.2474487 5.3132928 .0066()(5(i(57 151 22801 3442951 12.2882057 5.3250740 .00(5(522517 152 23104 3511808 12.3288280 5.3368033 .00657S947 153 23409 3581577 12.3693169 5.3484812 .006535948 154 23716 3652264 12.4096736 5.3(301084 .00(3493506 155 24025 3723875 12.4498996 5.3716854 .006451613 156 24336 3796416 12.4899960 5.3832126 .006410256 157 24649 3869893 12.5299641 5.3946907 .006369427 158 24964 3944312 12.5698051 5.4061202 .006329114 159 25281 4019679 12.6095202 5.4175015 .006289308 ino 25600 4096000 12.6491106 5.4288352 .006250000 161 25921 4173281 12.6885775 5.4401218 .006211180 162 26244 4251528 12.7279221 5.4513618 .006172840 163 26569 4330747 12.7671453 5.4625556 .0061349(59 164 26896 4410944 12.8062485 5.4737037 .006097561 165 27225 4492125 12.8452326 5.4848066 .00(5060(506 166 27556 4574296 12.8840987 5.4958647 .006024096 167 27889 4(357463 12.9228480 5.5068784 .005988024 168 28224 4741632 12.9614814 5.5178484 .005952381 169 28561 4826809 13.0000000 5.5287748 .005917160 170 28900 4913000 13.0384048 5.5396583 .005882353 171 29241 5000211 13.0766968 5.5504991 .005847953 172 29584 5088448 13.1148770 5.5612978 .005813953 173 29929 5177717 13.1529464 5.5720546 .005780347 174 30276 5268024 13.1909060 5.5827702 .005747126 175 30625 5359375 13.2287566 5.5934447 .005714286 176 30976 5451776 13.2(564992 5.(3040787 .005681818 177 31329 5545233 13.3041347 5.6146724 .005649718 178 31684 5639752 13.341(3(341 5.6252263 .005617978 179 32041 5735339 13.3790882 5.6357408 .005586592 180 32400 5832000 13.4164079 5.(i4(321(32 .005555556 181 32761 5929741 13.4536240 5.6566528 .0055248(52 182 33124 6028568 13.4907376 5.6(570511 .005494505 183 33489 6128487 13.5277493 5.6774114 .005464481 184 33856 6229504 13.564(3(300 5.6877340 .005434783 185 34225 6331625 13.(3014705 5.6980192 .005405405 186 34596 6434856 13.6381817 5.7082675 .00537(5344 187 34969 6539203 13.(3747943 5.7184791 .005347594 188 35344 6644()72 13.7113092 5.728(5543 .005319149 189 35721 6751269 13.7477271 5.7387936 .005291005 190 36100 6859000 13.7840488 5-7488971 .005263158 191 36481 69()7871 13.8202750 5.7589652 .005235602 192 3()864 7077888 13.85()40(35 5.7(589982 .005208333 193 37249 7189057 13.8924440 5.77899(36 .005181347 194 37636 7301384 13.9283883 5.7889(304 .005154639 195 38025 7414875 13.9642400 5.7988900 .005128205 196 38416 7529536 14.0000000 5.8087857 .005102041 197 38809 7645373 14.035(3(388 5.818(3479 .00507(5142 198 39204 7762392 14.0712473 5.8284767 .005050505 199 39601 7880599 14.10(373(30 5.8382725 .005025126 SQUAIiES, CUBES, SQUAHE HOOTS, ETC. 3G3 SQUARES, CUBES , SQUARE ROOTS, CUBE ROOTS, AND RECIPROCALS No. Squares Cubes Square Roots Cube Roots Reciprorals 200 40000 8000000 14.1421356 5.8480355 .005(K)(K)(K) 201 40401 8120601 14.17744()9 5.85776()0 .004975121 202 40804 8242408 14.212()704 5.8674643 .0049.50195 203 41209 8365427 14.2478068 5.8771307 .004926108 204 41()16 848VU;64 14.28285()9 5.8867653 .00490 19()1 205 42025 8(U5125 14.3178211 5.89()3685 .004S7.S049 206 4243() 8741816 14.3527001 5.90.-)9406 .0048.54.369 207 42849 88()9743 14.3874946 5.9154817 .0048.30918 208 43264 8998912 14.4222051 5.9249921 .004807692 209 43681 9129329 14.4568323 5.9344721 .004784(389 210 44100 9261000 14.491.3767 5.9439220 .004761905 211 44521 9393931 14.525835)0 5.9533418 .0047:39:33f) 212 44944 9528128 14.5(i02198 5.9627320 .00471()981 213 453()9 9663597 14.5945195 5.9720926 .004()1K18:36 214 4579() 9800344 14.()287388 5.9814240 .004(572897 215 4()225 9938375 14.6628783 5.99072(>4 .004(5511(53 216 4()6r)6 10077()96 14.()969385 6.0000000 .004629(5:30 217 47089 10218313 14.7309199 6.0092450 .004(508295 218 47524 10360232 14.7648231 6.0184()17 .004.5871.56 219 471K51 10503459 14.7986486 6.0276502 .00456(5210 220 48400 10648000 14.8323970 6.0368107 .(X)4545455 221 48S41 10793861 14.8660687 6.0459435 .004524887 222 492.S4 10941048 14.8996644 6.0550489 .004.504505 223 49729 11089567 14.9331845 6.0()41270 .004484.305 224 50176 11239424 14.966()295 6.0731779 .0044(54286 225 50()25 1132964 6.0911994 .004424779 227 51529 11697083 15.06()5192 6.1001702 .004405286 228 51984 11852352 15.0996()89 6.1091147 .004:3859(55 229 52441 12008989 15.1327460 6.1180332 .004:3(56812 230 52900 12167000 15.1657509 6.1269257 .004.347826 231 53361 1232()391 15.1986842 6.1357924 .004329004 232 53824 12487168 15.2315462 6.14463.37 .004310:U5 233 54289 12649337 15.2(543.375 6.1.5.34495 .004291845 2.34 54756 12812904 15.2970585 6.1622401 .00427:3.504 235 55225 12977875 15.32970i)7 6.1710058 .004255319 236 55696 13144256 15.3622915 6.1797466 .0042:>72S8 237 56169 13312053 15.3948043 6.1884628 .004219409 238 56644 13481272 15.4272486 6.1971544 .004201(581 239 57121 13()51919 15.4r)96248 6.20.-).S218 .004 1841 (Ml 240 57600 13824000 15.4919.3.34 6.2144650 .004 1(3(5(5(37 241 58081 13997521 15.5241747 6.22:30843 .004149:^78 242 58564 14172488 15.5."56:U92 6.2316797 .0041:^2231 243 51K)49 14:H8907 15.5884573 6.2402515 .00411.5226 244 59536 14526784 15.()204994 6.2487998 .004098:3(51 245 60025 1470(;i25 15.6.")24758 6.2.')7:'»248 .004081(5:^3 246 6051() 14() 28()52r)16 17.4928557 6.738(5(541 .0032(57974 307 94249 28934443 17.5214155 6.7459967 .003257329 308 948(54 29218112 17.5499288 6.7533134 .00324675:5 309 95481 29503629 17.5783958 6.7606143 .00323624(5 310 96100 29791000 17.(50(58169 6.7(578995 .00322580(5 311 9()721 300802;U 17.6351921 6.7751(590 .0032154:^ 312 97344 30371328 17.6635217 6.7824229 .003205128 313 97969 30664297 17.69180(50 6.789(5613 .003194888 314 98596 30959144 17.7200451 6.7968844 .003184713 315 99225 31255875 17.7482393 (5.8040921 .003174(503 316 9985() 31554496 17.77(53888 6.8112847 .003164557 317 100489 31855013 17.8044938 6.8184620 .003154574 318 101124 32157432 17.8325545 6.825(5242 .003144r,.54 319 101761 32461759 17.8605711 6.8327714 .003134796 320 102400 32768000 17.8885438 6.8399037 .003125000 321 103041 33076161 17.9164729 6.8470213 .0031152(55 322 103684 33386248 17.9443584 6.8541240 .0031055^)0 323 104329 33698267 17.9722008 6.8612120 .003095975 324 104976 34012224 18.0000000 6.8682855 .00308(5420 325 105625 34328125 18.0277564 6.875:5443 .00:307(5923 326 10(L>76 34645976 18.0554701 6.8823888 .00:50(57485 327 106929 349(35783 18.0831413 6.8894188 .00:5058104 328 107584 35287552 18.1107703 6.89(54:545 .003048780 329 108241 35611289 18.1383571 6.9034359 .003039514 330 108900 35937000 18.1659021 6.9104232 .0030:50:503 331 1095()1 362(54691 18.1934054 6.917:3964 .00:3021148 332 110224 3()594;^)8 18.2208(572 6.9243556 .003012048 333 110889 36926037 18.248287(5 6.931:5008 .00:500:500:5 334 111556 37259704 18.275(5(5(59 6.9:582:321 .002994012 3:35 112225 37595375 18.3030052 6.9451496 .002985075 336 112896 37933056 18.330:5028 6.9520533 .00297(5147581 7.0:5:58497 .00287:5.-)(53 349 121801 42508549 18.(5815417 7.0405806 .0028653:30 366 SQUABE8, CUBES, SQUARE ROOTS, ETC. 8 SQUARES, CUBES , SQUARE ROOTS, CUBE ROOTS, AND RECIPROCALS No. Squares Cubes Square Roots Cube Roots Reciprocals 350 122500 42875000 18.7082869 7.0472987 .002857143 351 123201 43243551 18.734V)910 7.0540041 .002849003 352 123904 43614208 18.761(5630 7.0006967 .002840909 353 124609 43986977 18.7882942 7.0673767 .002832861 354 125316 443()1864 18.8148877 7.0740440 .002824859 355 126025 44738875 18.8414437 7.0806988 .002816901 356 126736 45118016 18.8679623 7.0873411 .002808989 357 127449 45499293 18.8944436 7.0939709 .002801120 358 128164 45882712 18.92088V9 7.1005885 .002793296 359 128881 46268279 18.9472953 7.1071937 .002785515 360 129600 46656000 18.9736660 7.1137866 .002777778 361 130321 47045881 19.0000000 7.1203674 .002770083 362 131044 47437928 19.0262976 7.1269360 .002762431 363 131769 47832147 19.0525589 7.1334925 .002754821 364 132496 48228544 19.0787840 7.1400370 .002747253 365 133225 48627125 19.1049732 7.1465695 .002739726 366 133956 49027896 19.1311265 7.1530901 .002732240 367 134689 49430863 19.1572441 7.1595988 .002724796 368 135424 49836032 19.1833261 7.1660957 .002717391 369 136161 50243409 19.2093727 7.1725809 .002710027 370 136900 50653000 19.2353841 7.1790544 .002702703 371 137641 51064811 19.2613603 7.1855162 .002695418 372 138384 51478848 19.2873015 7.1919663 .002688172 373 139129 51895117 19.3132079 7.1984050 .002680965 374 139876 52313624 19.3390796 7.2048322 .002673797 375 140625 52734375 19.3649167 7.2112479 .002666667 376 141376 53157376 19.3907194 7.2176522 .002659574 377 142129 53582633 19.4164878 7.2240450 .002652520 378 142884 54010152 19.4422221 7.2304268 .0026)45503 379 143641 54439939 19.4679223 7.2367972 .002638522 380 144400 54872000 19.4935887 7.2431565 .002631579 381 145161 55306341 19.5192213 7.2495045 .002624672 382 145924 55742968 19.5448203 7.2558415 .002617801 383 146689 56181887 19.5703858 7.2621675 .002()10966 384 147456 56623104 19.5959179 7.2684824 .002604167 385 148225 57066625 19.6214169 7.2747864 .002597403 386 148996 57512456 19.6468827 7.2810794 .002590674 387 149769 57960603 19.6723156 7.2873()17 .002583979 388 150544 58411072 19.6977156 7.2936330 .002577320 389 151321 58863869 19.7230829 7.2998936 .002570694 390 152100 59319000 19.7484177 7.3061436 .002564103 391 152881 59776471 19.7737199 7.3123828 .002557545 392 153(J64 60236288 19.7989S99 7.3186114 .002551020 393 154149 60()98457 19.8242276 7.3248295 .002544529 394 155236 611()2984 19.8494332 7.33103()9 .002538071 395 156025 61()29875 19.874()069 7.3372339 .002531646 396 156816 62099136 19.8997487 7.3434205 .002525253 397 157609 62570773 19.9248588 7.3495966 .002518892 398 158404 63044792 19.9499373 7.35.57()24 .002512563 399 159201 63521199 19.9749844 7.3619178 .0025062()() SQUARES, CUBES, SQUARE ROOTS, ETC. 9 367 SQUARES, CUBES , SQUARE ROOTS, CUBE ROOTS, AND RECIPROCALS Itfo. Squares Cabes Square Roots Cube Roots Reriprorals 400 160000 (;40ouooo 20.0000000 7.3680()30 .0()2r)()()0()o 401 1()0S01 64481201 20.0249844 7.3741979 .00249:576(5 402 161604 649(>4808 20.0499377 7.3803227 .0024875(52 403 162409 65450827 20.0748599 7.3864373 .002481:590 404 163216 65939264 20.0997512 7.3925418 .002475248 405 1()4025 664:^125 20.1246118 7.398()362 .0024(591:5(5 406 16483() 66923416 20.1494417 7.4047206 .0024(5:5054 407 165()49 67419143 20.1742410 7.4107950 .002457002 408 16()464 67917312 20.19rK)099 7.4168595 .0024.-)0980 409 167281 68417929 20.2237484 7.4229142 .002444988 410 168100 68921000 20.24S45()7 7.4289589 .0024:59024 411 168921 69426531 20.2731349 7.4349938 .0024:5:5090 412 169744 69934528 20.2977831 7.4410189 .002427184 413 170569 70444997 20.3224014 7.4470:542 .002421:508 414 171396 70957944 20.34()9899 7.45:>0:399 .002415459 415 172225 71473375 20.3715488 7.4590359 .002409(5:59 416 17305() 71991296 20.3900781 7.4650223 .00240:5846 417 173889 72511713 20.4205779 7.4709991 .002398082 418 174724 73034632 20.4450483 7.4769(564 .002392:344 419 175561 73560059 20.4694895 7.4829242 .002:5866:55 420 176400 74088000 20.4939015 7.4888724 .002380952 421 177241 74618461 20.5182845 7.494S113 .002375297 422 178084 75151448 20.542()386 7.5007406 .002:5(59(5(58 423 178929 7568()iK57 20.5669()38 7.50(5(5(507 .002:5640(56 424 179776 76225024 20.5912603 7.5125715 .002:558491 4'^ 180625 767()5625 20.()155281 7.51847:30 .002:552941 426 181476 77308776 20.6397()74 7.524:3(552 .002:547418 427 182329 77854483 20.6639783 7.5:302482 .002341920 428 183184 78402752 20.()881609 7.5:561221 .0023:5(5449 429 184041 78953589 20.7123152 7.5419867 .002331002 430 184900 79507000 20.7364414 7.5478423 .002:525581 431 185761 80062991 20.7605395 7.55:36888 .002:520186 432 186624 806215()8 20.7846097 7.55952(53 .002314815 4:53 187489 81182737 20.8()8()520 7.565:3548 .002:5(n>4(59 434 18835() 81746504 20.8326(;()7 7.5711743 .002:5(^4147 435 189225 82312875 20.85665:36 7.57(59849 .002298851 43() 1900i)() 82881856 20.880(5130 7.58278(55 .00229:5578 437 1075 7.(52888:57 .002252252 445 198025 88121125 21.()9.~)()231 7.6:54(50(57 .002247191 446 198<)1() 8871()536 21.1187121 7.(540:3213 .002242 1.")2 447 199809 89314()23 21.1423745 7.(^(50272 .0022:571:5(5 448 200704 89915392 21.1660105 7.6517247 .0022:52143 449 201601 90518849 21.1896201 7.(55741:58 .002227171 368 SQUARES, CUBES, SQUARE BOOTS, ETC. 10 SQUARES, CUBES , SQUARE ROOTS, CUBE ROOTS, AND RECIPROCALS No. Squares Cubes Square Roots Cube Roots Reciprocals 450 202500 91125000 21.2132034 7.6630943 .002222222 451 20;i401 91733851 21.23()7606 7.()687()()5 ;002217295 452 204304 92345408 21.2602916 7.6744303 .002212389 453 205209 92959677 21.28379(;7 7.()800857 .002207506 454 206116 93576664 21.3072758 7.6857328 .002202643 455 207025 94196375 21.3307290 7.691.3717 .002197802 456 207936 94818816 21.35415()5 7.()970023 .002192982 457 208849 95443993 21.3775583 7.7026246 .002188184 458 209764 96071912 21.4009346 7.7082388 .002183406 459 210681 96702579 21.4242853 7.7138448 .002178649 460 211600 97336000 21.4476106 7.7194426 .002173913 461 212521 97972181 21.4709106 7.7250325 .002169197 462 213444 98611128 21.4941853 7.7306141 .0021(14502 463 214369 99252847 21.5174348 7.7361877 .002159827 464 215296 99897344 21.5406592 7.7417532 .002155172 465 216225 100544625 21.5638587 7.7473109 .0021.50538 466 217156 101194696 21.5870331 7.7528606 .002145923 467 218089 101847563 21.6101828 7.7584023 .002141328 468 219024 102503232 21.6333077 7.7639361 .002i;5()752 469 219961 103161709 21.6564078 7.7694620 .002132196 470 220900 103823000 21.6794834 7.7749801 .002127660 471 221841 104487111 21.7025344 7.7804904 .002123142 472 222784 105154048 21.7255610 7.7859928 .002118644 473 223729 105823817 21.7485632 7.7914875 .002114165 474 224676 106496424 21.7715411 7.7969745 .002109705 475 225(525 107171875 21.7944947 7.8024538 .0021()52()3 476 226576 107850176 21.8174242 7.8079254 .002100840 477 227529 108531333 21.8403297 7.8133892 .0020!i()43() 478 228484 109215352 21.8632111 7.8188456 .002092050 479 229441 109902239 21.8860686 7.8242942 .002087683 480 230400 110592000 21.9089023 7.8297353 .002083333 481 231361 in284()41 21.9317122 7.8351688 .002079002 482 232324 111980168 21.9544984 7.8405949 .002074689 483 233289 112678587 21.9772610 7.8460134 .002070393 484 2:^256 113379904 22.0000000 7.8514244 .0020(5(5116 485 235225 114084125 22.0227155 7.8.5{)8281 .002061856 486 236196 114791256 22.0454077 7.8()22242 .002057(513 487 237169 115501303 22.0680765 7.8676130 .002053388 488 238144 116214272 22.0^)07220 7.8729944 .002049180 489 239121 116930169 22.1133444 7.8783684 .002044990 490 240100 117649000 22.13594.36 7.8837352 .002040816 491 241081 118370771 22.1585198 7.8890946 .002036(5(50 492 2420()4 119095488 22.1810730 7.8944468 .002032520 4<)3 24:^)049 119823157 22.2036033 7.8997917 .002028398 494 24403() 120553784 22.2261108 7.9051294 .002024291 495 245025 121287375 22.248.5955 7.9104599 .002020202 496 246016 122023936 22.2710575 7.9157832 .002016129 497 247009 12276;M73 22.29IU9()8 7.92109()4 .002012072 498 248004 123505992 22.31591 :U) 7.9264085 .0020080.32 499 249001 124251499 22.338:3079 7.9317104 .002004008 SQUARES, CUBES, SQUARE ROOTS, ETC. 3G9 11 SQUARES, CUBES , SQUARE ROOTS, CUBE ROOTS, AND RECIPROCALS 5o. Squares Cubes Square Roots Cube Roots Reciprocals 500 250000 125000000 22.360(;798 7.9:370053 .002000000 501 251001 125751501 22.3830293 7.9422!>31 .00199(5008 502 252004 12()5()()008 22.4053565 7.94757:39 .0019920:52 503 25:3009 127263527 22.427()615 7.9528477 .001988072 504 2:401() 128024061 22.4499443 7.9581144 .001984127 505 255025 128787625 22.4722051 7.9(5:3:3743 .001980198 50() 25()036 12i)554216 22.4944438 7.9(58(5271 .00197(5285 507 257049 130323843 22.51()6605 7.97:38731 .001972.387 508 2580()4 131096512 22.5388553 7.9791122 .0019(585(4 509 251K)81 131872229 22.5610283 7.984:3444 .0019(346:37 510 260100 132()51000 22.5831796 7.9895(597 .0019(50784 511 261121 133432831 22.(3053091 7.9917883 .00195(3147 512 262144 134217728 22.(3274170 8.0000000 .001953125 513 2()3169 135005697 22.(495033 8.0052049 .001949318 514 264196 1357i)6744 22.(3715(381 8.01(40:32 .001945525 515 265225 13(3590875 22.(5936114 8.015594(5 .001941748 51(3 266256 137388096 22.71563134 8.0207794 .0019:37984 517 2(57289 138188413 22.737(3340 8.0259574 .0019342:36 518 268324 138991832 22.75fK3134 8.0:311287 .0019.30502 519 269361 139798359 22.7815715 8.0:362935 .001926782 520 270400 140608000 22.8035085 8.0414515 .00192:3077 521 271441 141420761 22.8254244 8.04(5(30:30 .001919386 522 272484 14223()648 22.8473193 8.0517479 .00191.5709 523 273529 143055(367 22.8691933 8.0568862 .001912046 524 274576 143877824 22.8910463 8.0(520180 .0O11KK397 525 275()25 144703125 22.9128785 8.0(371432 .0019047(52 52() 27(>()76 145531576 22.934(5899 8.0722(520 .0011K)1141 527 277729 14(33(33183 22.9564806 8.0773743 .0018975:33 528 278784 147197952 22.9782506 8.0824800 .00189:3939 529 279841 148035889 23.0000000 8.0875794 .001890:359 530 280! )00 148877000 23.0217289 8.092(5723 .00188(5792 531 281961 149721291 23.04:4372 8.0<)77589 .00188:32.39 532 283024 1505687(J8 23.0(351252 8. 1028; 590 .001879(599 533 281089 151419437 23.0867928 8.1079128 .00187(5173 5;u 2851 5() 152273304 23.1084400 8.112i)S03 .001872(559 535 286225 153130375 23.1:300(570 8.1180414 .0018(59159 536 28729() 1539<)0656 23.151(3738 8.12:309(32 .0(M 8(55(572 537 2883()9 154854153 23.17:52(305 8.1281447 .0018(52197 538 289444 155720872 23.1948270 8. 1:3:51 870 .0018587:3(3 539 290521 156590819 23.21(53735 8.1:3822:30 .001855288 540 291(300 157464000 23.2:579001 8.1432529 .001851852 541 292681 158340421 23.25140(37 8.14827(55 .001.S4S429 542 2937()4 159220088 23.28089:35 8.15:329:39 .00184.1018 543 294849 16010; ;oo7 23.:302:5(501 8.158:3051 .001841(521 544 2!>59;'.6 1()0989184 23.:32:38076 8.16:33102 .(^) 18:382.35 545 2971)25 I(;i878()25 23.:452:\51 8.1(58:3092 .0018:48(52 546 298116 162771336 23.:3( 5(3(5429 8.17:3;5020 .fH)1831.n02 547 299209 1().3()67323 23.. 38803 11 8.17S28S8 .001S28J.14 .548 300304 1(4566592 23.409:3998 8.18:32(595 .001824818 -49 301401 1()54(J<)149 23.4:30741K) 8.1882441 .001821494 2 b 870 SQUABES, CUBES, SQUARE BOOTS, ETC. 12 SQUARES, CUBES , SQUARE ROOTS, CUBE ROOTS, AND RECIPROCALS No. Squares Cubes 166375000 Square Roots Cube Roots Reciprocals 550 302500 23.4520788 8.1932127 .001818182 551 303601 167284151 23.4733892 8.1981753 .001814882 552 304704 168196608 23.4946802 8.2031319 .001811594 553 305809 169112377 23.5159520 8.2080825 .001808318 554 306916 170031464 23.5372046 8.2130271 .001805054 555 308025 170953875 23.5584380 8.2179(357 .001801802 556 309136 171879616 23.5796522 8.2228985 .001798561 557 310249 172808693 23.6008474 8.2278254 .001795332 558 311364 173741112 23.6220236 8.2327463 .001792115 559 312481 174676879 23.6431808 8.2376614 .001788909 560 313600 175616000 23.6643191 8.2425706 .001785714 561 314721 176558481 23.6854386 8.2474740 .061782531 5()2 315844 177504328 23.7065392 8.2523715 .001779359 563 316969 178453547 23.7276210 8.2572633 .001776199 564 318096 179406144 23.7486842 8.2621492 .001773050 565 319225 180362125 23.7697286 8.2670294 .001769912 566 320356 181321496 23.7907545 8.2719039 .00176()784 567 321489 182284263 23.8117618 8.2767726 .001763668 568 322624 183250432 23.8327506 8.2816355 .001760563 569 323761 184220009 23.8537209 8.2864928 .0017574(^9 570 324900 185193000 23.8746728 8.2913444 .001754386 571 326041 186169411 23.895(^063 8.296^)03 .001751313 572 327184 187149248 23.91(55215 8.3010304 .001748252 573 328329 188132517 23.9374184 8.3058651 .001745201 574 329476 189119224 23.9582971 8.3106941 .001742160 575 330625 190109375 23.9791576 8.3155175 .001739130 576 331776 191102976 24.0000000 8.3203353 .001736111 577 332929 192100033 24.0208243 8.3251475 .001733102 578 334084 193100552 24.0416306 8.3299542 .001730104 579 335241 194104539 24.0624188 8.3347553 .001727116 580 336400 195112000 24.0831891 8.3395509 .001724138 581 337561 196122941 24.1039416 8.3443410 .001721170 582 338724 197137368 24.12467(^2 8.3491256 .001718213 583 339889 198155287 24.1453929 8.3539047 .001715266 584 ^41056 199176704 24.1(^(50919 8.358(5784 .001712329 585 342225 200201625 24.1867732 8.3634466 .001709402 586 M339f) 201230056 24.2074369 8.3682095 .001706485 587 344569 202262003 24.2280829 8.3729(568 .001703578 588 345744 203297472 24.2487113 8.3777188 .001700680 589 346921 204336469 24.2693222 8.3824(553 .001697793 590 348100 205379000 24.2899156 8.38720(55 .001(594915 591 349281 206425071 24.3104916 8.3919423 .001692047 592 350464 207474688 24.3310501 8.39(5(5729 .001689189 593 351649 208527S57 24.351.5913 8.4013981 .001686341 594 352836 209584584 24.3721152 8.4061180 .001683502 595 354025 210(544875 24.3926218 8.410832(5 .001(580672 596 355216 211708736 24.4131112 8.4155419 .001677852 597 356409 21277(5173 24.4335834 8.42024(50 .001675042 598 357604 213847192 24.4540385 8.4249448 .001()72241 599 358801 214921799 24.4744705 8.429(5383 .001(5(59449 SQUARES, CUBES, SQUARE ROOTS, ETC. 13 371 SQUARES, CUBES SQUARE ROOTS, CUBE ROOTS, AND RECIPROCALS No. Squares Cubes Square Roots Cube Roots Reciprocals 600 360000 216000000 24.4948974 8.4343267 .001(5(5(3(3(57 601 361201 217081801 24.5153013 8.4390098 .001(5(5:3894 602 362404 218167208 24.53515883 8.443(5877 .(X)l (5(51 1.30 603 363609 219256227 24.55(50583 8.4483(305 .001(558375 604 364816 220348864 24.5764115 8.4530281 .001655(529 605 36(5025 221445125 24.5967478 8.457(5906 .001(552893 606 367236 222545016 24.6170(373 8.462:3479 .001(5501(35 607 3()8449 223648543 24.6373700 8.4670001 .001(54744(5 608 36^)664 224755712 24.(i576560 8.4716471 .001644737 609 370881 225866529 24.6779254 8.4762892 .001(542036 610 372100 226981000 24.()981781 8.4809261 .001639:344 611 373321 228099131 24.7184142 8.4855579 .001(3:3(3(5(51 612 374544 229220928 24.738(3338 8.4f)01848 .001(333987 613 375769 230346397 24.7588368 8.4948065 .001631321 614 376996 231475544 24.7790234 8.4994233 .001628(3(54 615 378225 232608375 24.7991935 8.5040350 .001(52(3016 616 379456 233744896 24.8193473 8.508(5417 .001623377 617 380689 234885113 24.8394847 8.5132435 .001620746 618 381924 23()029032 24.8596058 8.5178403 .001618123 619 383161 237176659 24.8797106 8.5224321 .001615509 620 384400 238328000 24.8997992 8.5270189 .001612903 621 385641 239483061 24.9198716 8.531(3009 .001610:306 622 386884 240641848 24.9399278 8.5361780 .001(307717 623 388129 241804367 24.9599679 8.5407501 .001(5051:36 624 389376 242970624 24.9799920 8.5453173 .001(3025(34 625 390625 244140625 25.0000000 8.5498797 .001(500000 62() 391876 245314376 25.0199920 8.5544372 .001597444 627 393129 246491883 25.0399681 8.5589899 .00159489(5 628 394384 247673152 25.0599282 8.5(535377 .001592:357 629 395641 248858189 25.0798724 8.5680807 .001589825 630 smm 250047000 25.0998008 8.5726189 .001587:302 631 398161 251239591 25.1197134 8.5771523 .00158478(5 632 399424 2524359()8 25.1396102 8.581(3809 .001582278 633 400689 253636137 25.1594913 8.58(52047 .001579779 ()34 401956 254840104 25.17935(36 8.5907238 .001577287 635 403225 256047875 25.19920(53 8.5952380 .00157480:5 636 404496 257259456 25.2190404 8.5997476 .001572:527 637 405769 258474853 25.2388589 8.6042525 .00 !.")( 59859 638 407044 259694072 25.2586(519 8.608752(5 .0015(57:39S 639 408321 260917119 25.2784493 8.6132480 .0015(34945 640 409600 262144000 25.2982213 8.6177388 .(X)15(52500 641 410881 263374721 25.3179778 8.(5222248 .0015600(52 642 412164 264609288 25.3377189 8.(52(570(53 .001557(5:32 643 41.3449 265847707 25.3.-.74447 8.(5311830 .001555210 (^t4 4147:3(^ 267089984 25..3771551 8.(535(5551 .0015527i)5 645 416025 268336125 25..3iH5S5()2 8.(340122(3 .001550:388 646 417316 26958()136 25.41(5."k>()1 8.(544.")8.'')5 .001547988 647 418009 270840023 25.43(51947 8.(541K)437 .001545595 648 419904 272097792 25.4.V)S441 8.r).^);U974 .00154:3210 649 421201 273359449 25.4754784 8.(557941.5 .0015408:32 372 SQUARES, CUBES, SQUARE ROOTS, ETC. 14 SQUARES, CUBES , SQUARE ROOTS, CUBE ROOTS, AND RECIPROCALS Xo. Squares Cubes Square Roots Cube Roots Reciprocals 650 422500 274625000 25.4950976 8.6623911 .001538462 651 423801 275894451 25.5147016 8.6()()8310 .00153()098 652 425104 277167808 25.5342907 8.6712665 .001533742 653 426409 278445077 25.5538647 8.6756974 .001531394 654 427716 279726264 25.5734237 8.6801237 .001529052 655 429025 281011375 25.5929678 8.6845456 .001526718 656 430336 282300416 25.6124969 8.6889()30 .001524390 657 431649 283593393 25.6320112 8.6933759 .001522070 658 432964 284890312 25.6515107 8.6977843 .001519757 659 434281 286191179 25.6709953 8.7021882 .001517451 660 435G0O 287496000 25.6904652 8.7065877 .001515152 661 436921 288804781 25.7099203 8.7109827 .001512859 662 438244 290117528 25.7293607 8.7153734 .001510574 663 439569 291434247 25.7487864 8.7197596 .001508296 664 440896 292754944 25.7681975 8.7241414 .00150()024 665 442225 294079625 25.7875939 8.7285187 .001503759 666 443556 295408296 25.8069758 8.7328918 .001501502 667 444889 296740963 25.8263431 8.7372604 .001499250 668 446224 298077632 25.8456960 8.7416246 .001497006 669 447561 299418309 25.8650343 8.7459846 .001494768 670 448900 300763000 25.8843582 8.7503401 .001492537 671 450241 302111711 25.9036677 8.7546913 .001490313 672 451584 303464448 25.9229628 8.7590383 .001488095 673 452929 304821217 25.9422435 8.7633809 .001485884 674 454276 306182024 25.9615100 8.7677192 .001483680 675 455625 307546875 25.9807621 8.7720532 .001481481 676 456976 308915776 26.0000000 8.7763830 .0014792^)0 677 458329 310288733 26.0192237 8.7807084 .001477105 678 459684 311665752 26.0384331 8.7850296 .001474926 679 461041 31304-6839 26.0576284 8.7893466 .001472754 680 462400 314432000 26.0768096 8.7936593 .001470588 681 463761 315821241 26.0959767 8.7979679 .001468429 682 465124 317214568 26.1151297 8.8022721 .001466276 683 466489 318611987 26.1342687 8.8065722 .001464129 684 467856 320013504 26.1533937 8.8108681 .001461988 685 469225 321419125 26.172.-047 8.8151598 .001459854 686 470596 322828856 26.1916017 8.8194474 .001457726 687 471969 324242703 26.2106848 8.8237307 .001455604 688 473:^44 325660672 26.2297541 8.8280099 .001453488 689 474721 327082769 26.2488095 8.8322850 .001451379 690 476100 328509000 26.2678511 8.8365559 .001449275 691 477481 329939371 26.2868789 8.8408227 .001447178 692 478864 331373888 26.3058929 8.8450854 .001445087 693 480249 332812557 26.3248932 8.8493440 .001443001 694 481636 334255384 26.3438797 8.8535985 .001440922 695 483025 335702375 26.3628527 8.8578489 .001438849 696 484416 33715:^536 26.3818119 8.8(;20952 .00143()782 697 485809 338608873 26.400757() 8.8()63375 .001434720 698 487204 34()0(;8392 26.419()896 8.8705757 .001432(;()5 699 488(501 3415320i)9 26.4386081 8.8748099 .001430615 SQUARES, CUBES, SQUARE ROOTS, ETC. 15 373 SQUARES, CUBES , SQUARE ROOTS, CUBE ROOTS, AND RECIPROCALS No. Squares Cubes Square Roots Cube Roots Reciprocals .001428571 700 490000 343000000 20.45751:51 8.8790400 701 4!)1401 344472101 2(5.47(5404(5 8.88:52(561 .0014205:54 702 492804 ^45948408 26.4952826 8.8874882 .001424.-)01 703 494209 347428927 26.5141472 8.89170(53 .001422475 704 495(}16 348913(5(54 26.5329983 8.895il2(J4 .001420455 705 497025 350402(525 26.5518:5(51 8.9001:504 .(H)1418440 706 49843() 351895810 2(5.570(5(505 8.iM)4:5:5(5() .00141(5431 707 499849 353393243 2(5.5894710 8.9085:587 .001414427 708 501204 354894912 26.(5082(594 8.9127:5(59 .001412429 709 502081 350400829 20.0270539 8.9169311 .001410437 710 504100 357911000 26.6458252 8.9211214 .001408451 711 505521 359425431 26.6(5458:53 8.925:5078 .00140(5470 712 50(5944 3(50944128 26.(58:5:5281 8.9294902 .001404494 713 5083(59 3024(57097 2(5.7020598 8.9:3:5(5(587 .001402525 7U 509790 3(515994344 26.7207784 8.9:5784:53 .0014(XJ5(50 715 511225 305525875 2(5.73948:59 8.9420140 .001:598(501 710 512(356 3670(51(590 1:6.75817(53 8.94(51809 .001:59(5(548 717 514089 308001813 26.7768557 8.9503438 .001:594700 718 515524 37014(5232 26.7955220 8.9545029 .a)i:592758 719 510901 371094959 20.8141754 8.958(5581 .001390821 720 518400 373248000 26.8328157 8.9(528095 .001:588889 721 519841 3748053(51 26.85144:52 8. i)( 5(59570 .OOi:58(51K53 722 521284 3703(57048 26.8700577 8.9711007 .001:585042 723 522729 37793:50(57 20.888(5593 8.9752406 .001383126 724 524170 379503424 26.9072481 8.979:57(56 .001:581215 725 525(525 381078125 26.9258240 8.98:55089 .001:579310 72(5 527070 382(557170 20.i)44:5872 8.987(5:573 .001:577410 727 528529 384240583 2(5.9(52<):;75 8.9917(520 .001:575516 728 529984 385828352 20.9814751 8.9958829 .001:57:5(526 729 531441 387420489 27.0000000 9.0000000 .001371742 730 532900 389017000 27.0185122 9.00411:54 .001:5(598(53 731 5;U3(;i 390617891 27.0:570117 9.008222i) .001:5(57989 732 535824 392223108 27.0554985 9.0123288 .001:5(5(5120 733 537289 393832837 27.07:59727 9.01(54309 .001:5(5425(5 lU 53875() 39544(5904 27.0924:U4 9.0205293 .001:5(52:598 735 540225 3970(55375 27.11088:54 9.024(52:59 .001:5(50.544 73() 541(59(5 398(588256 27.1293199 9.0287149 .0()i:558()^H) 737 5431(59 400315553 27.14774:59 9.0:528021 .00i:55(58,"')2 738 544(^44 401947272 27.1(5(51554 9.0:5(58857 .001:555014 739 540121 40358:1419 27.1845544 9.0401H)55 .00K553180 740 547(500 405224000 27.202!)410 9.0450417 .001:551:5,51 741 549081 40(58(59021 27.2213152 9.0491142 .001:549528 742 5505(54 408518488 27. 2:5 745 555025 41:549:5(525 27. 2! M( 5881 9.(X 55:5(577 .001:542282 74() 55(5510 4151 (509:5(5 27.31:5000(5 9. (M 594220 .001:540483 747 558009 41(58:52723 27.:53i:5(M)7 9.07:5472(5 .001:5:58(588 748 559504 418508992 27.:549.~)887 9.0775197 .001:5:5(5898 749 501001 420189749 27.:5(578()44 - 9.(3815(5:51 , .001:5:55113 374 SQUARES, CUBES, SQUARE ROOTS, ETC. 16 X SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS, AND RECIPROCALS JVo. Squares Cubes Square Roots Cube Roots Reciprocals 7r.o 562500 421875000 27.3861279 9.0856030 .001333333 751 564001 423564751 27.4043792 9.0896392 .001331558 752 565504 425259008 27.4226184 9.0936719 .001329787 753 567009 426957777 27.4408455 9.0977010 .001328021 754 568516 428661064 27.4590604 9.1017265 .001326260 755 570025 430368875 27.4772633 9.1057485 .001324503 756 571536 432081216 27.4954542 9.1097669 .001322751 757 573049 433798093 27.5136330 9.1137818 .001321004 758 574564 435519512 27.5317998 9.1177931 .001319261 759 576081 437245479 27.5499546 9.1218010 .001317523 760 577600 438976000 27.5680975 9.1258053 .001315789 761 579121 440711081 27.5862284 9.1298061 .001314060 762 580644 442450728 27.6043475 9.1338034 .001312336 7(53 582169 444194947 27.6224546 9.1377971 .001310616 764 583696 445943744 27.6405499 9.1417874 .001308901 765 585225 447697125 27.6586334 9.1457742 .001307190 766 586756 449455096 27.6767050 9.1497576 .001305483 767 588289 451217663 27.6947648 9.1537375 .001303781 768 589824 452984832 27.7128129 9.1577139 .001302083 769 591361 454756609 27.7308492 9.1616869 .001300390 770 592900 456533000 27.7488739 9.1656565 .001298701 771 594441 458314011 27.7668868 9.1696225 .001297017 772 595984 4(J0099()48 27.7848880 9.1735852 .001295337 773 597529 461889917 27.8028775 9.1775445 .001293661 774 599076 463684824 27.8208555 9.1815003 .001291990 775 600025 465484375 27.8388218 9.1854527 .001290323 776 602176 467288576 27.8567766 9.1894018 .001288660 777 603729 469097433 27.8747197 9.1933474 .001287001 778 605284 470910952 27.8926514 9.1972897 .001285347 779 60()841 472729139 27.9105715 9.2012286 .001283697 780 608400 474552000 27.9284801 9.2051641 .001282051 781 601M!61 476379541 27.94()3772 9.2090962 .001280410 782 611524 478211768 27.9()42()29 9.2130250 .001278772 783 613089 480048687 27.9821372 9.2169505 .001277139 784 614656 481890304 28.0000000 9.2208726 .001275510 785 616225 4837:5()(;25 28.0178515 9.2247914 .001273885 786 617796 485587656 28.0356915 9.2287068 .0012722(55 787 619369 487443403 28.0535203 9.2326189 .001270648 788 620944 489303872 28.0713377 9.2365277 .00126^)036 789 622521 491169069 28.0891438 9.2404333 .001267427 790 624100 493039000 28.1069386 9.2443355 .001265823 791 625681 494913()71 28.1247222 9.2482344 .001264223 792 6272f)4 49()793088 28.1424946 9.2521300 .001262626 793 ()28849 498()77257 28.1602,557 9.2560224 .001261034 794 6304:^.6 50056()184 28.17800.56 9.2599114 .001259446 795 632025 502459875 28.1957444 9.2637973 .001257862 79f) 633()16 504358^,36 28.2134720 6.2(576798 .001256281 797 635209 50()261573 28.2311884 9.2715592 .001254705 798 63>6804 5081()9592 28.2488938 9.27.54352 .001253133 799 638401 510082399 28.2()65881 9.2793081 .001251564 SQUARES, CUBES, SQUARE ROOTS, ETC, 17 (O SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS, AND RECIPROCALS Ko. Squares Cubes Square Roots Cube Roots Reciprocals 800 ()40000 512000000 28.2842712 9.2831777 .0()i2r)()(X)() 801 641()()1 513922401 28.:5019434 9.2870440 .00124s };;<) 802 ()43204 515849()08 28.31 9()045 9.2i)0i)072 .0O124(5S,S3 803 644809 517781627 28.:5:;7254(j 9.2947(571 .001245:5:50 804 646416 51971S464 28.:55489:58 9.298(52::59 .00124:5781 805 648025 5216()()125 28.:3725219 9.3024775 .0012422:5(5 806 649()36 523()0()()16 28.390L391 9.30(5:5278 .001240(595 807 651249 525557^)43 28.4077454 9.3101750 .001 2:59 1.-)7 808 652864 527514112 28.425:5408 9.:3140190 .0012:57(524 809 654481 529475129 28.4429253 9.3178599 .0012:5(5094 810 656100 531441000 28.4(;04989 9.321(5975 .0012:345(58 811 657721 533411731 28.4780(517 9.3255:520 .0012:5:504(5 812 659344 535387328 28.4i)5()i:57 9.;329:5(5:34 .001231.V27 813 660969 5373()7797 28.5131549 9.3:331916 .0O12:;00I2 814 6625i)6 539353144 28.5:50(5852 9.:3:5701(57 .001228501 815 6(J4225 54134:5375 28.5482048 9.:5408:586 .00122(5994 816 &mm 54333849() 28.5(>57L37 9.344(5575 .00122541K) 817 667489 54533S513 28.58:52119 9.3484731 .00122:^990 818 669124 54734:^132 28. 6001)99:5 9.3522857 .001222494 819 670761 549:353259 28.61817(50 9.3560952 .001221001 820 672400 55i:')()S000 28.():55i)421 9.3599016 .001219512 821 674041 55:3;)87661 28.65:5097(5 9.:36:57049 .00121.S()27 822 675()84 555412248 28.(^705424 9.:^(575051 .00121(5.")45 823 677329 557441767 28.6S797()6 9.:37i:^022 .00121.-.0(57 824 678976 559476224 28.7054002 9.:^750iK)3 .00121:5592 825 680625 561515625 28.72281:^2 9.: 5788873 .001212121 826 682276 5(i:;55997l> 28.7402157 9.:5826752 .001210(5.")4 827 683i)29 5(i5()09283 28.757(5077 9.:5864(;00 .001209190 823 ()8r)584 5()7()'>:r)52 28.7749891 9.:5902419 .001207729 829 687241 569722789 28.792:5(501 9.3940206 .00120(5273 830 688900 571787000 28.809720(5 9.:59779(54 .001204819 831 6905()1 57:^856191 28.827070(5 9.4015(591 .00120:5:^(59 832 692224 5759:](K3()8 28.8444102 9.405:5:587 .001201023 833 693889 5780095:^7 28.8(517:594 9.4091054 .0012OO4S0 834 695556 58009:5704 28.8790582 9.4128()t)0 .0011!M)041 835 697225 582182875 28.8776 788881K)24 30.397:5(583 9.7:>99(534 .0010.S2251 925 855(525 791453125 30.41:58127 9.74:54758 .001 (IS 1 081 926 85747() 794022776 30.4:502481 9.74(59857 .001079914 927 8593,29 71K)597983 30.44(5(5747 9.75049:50 .001078749 928 861184 799178752 30.4(5:50924 9.75:59979 .00107758(5 929 8():i041 80176r;089 30.4795013 9.7575002 .00107(5426 930 8(54900 804357000 30.4959014 9.7610001 .(X)10752(59 931 8()()761 806954491 30.5122926 9.7(544974 .001074114 932 8()8624 8095575(58 30.5286750 9.7(579922 .()O1072!i(51 fl33 870489 8121(5()237 30.5450487 9.7714845 .001071811 934 872;}5() 814780504 30.5(5141:56 9.7749743 .001070(5(54 93,5 874225 817400375 30.5777(597 9.7784(516 .001()(5<)519 93(3 87(;09() 820025856 30.5941171 9.78194(56 .0010(58:576 937 877969 822(55(5953 30.6104557 9.7854288 .0010(572:56 938 879844 825293(572 30.62(57857 9.788<)()87 .0010(5(5098 939 881721 82793(5019 30.(54:51069 9.7il2:5861 .0010(549(53 940 883600 830584000 :50.6594194 9.7958611 .0010(5:58:50 941 885481 8:5323.7(521 :30.(-;757233 9.799:5:5:5() .O01(^(52()<><) 942 8873()4 83,589(5888 30.(5920185 9.802S0:5() .0010(51571 943 8S9249 8385(51807 30.708:5051 9.80(52711 .0010(50445 944 891 13() 8412:523,84 30.72458:50 9.8(V.)7:5(52 .0010.'')9:522 945 893025 84:5908(525 30.740S.V2;) 9.8131989 .(HI 1058201 940 894911; 84( 5590.1:56 :50.757li:50 9.81(5(5591 .(M)10.")7082 947 89()S09 849278123 :50.77:5:5(551 9.S2011()9 .0010559(5(5 948 898704 851971:592 30.789(5086 9.82:5.-)72:i .001054S52 949 900601 854(570:549 30.80584:56 9.8270252 .00105:5741 378 SQUARES, CUBES, SQUARE ROOTS, ETC. 20 SQUARES, CUBES , SQUARE ROOTS, CUBE ROOTS, AND RECIPROCALS Jfo. Squares Cubes Square Roots Cube Roots Reciprocals 950 902500 857375000 30.8220700 9.8304757 .001052632 951 904401 860085351 30.8382879 9.8339288 .001051525 952 906304 862801408 30.8544972 9.8373695 .001050420 953 908209 865523177 30.8706981 9.8408127 .001049318 954 910116 868250664 30.8868904 9.8442536 .001048218 955 912025 870983875 30.9030743 9.8476920 .001047120 956 913936 873722816 30.9192497 9.8511280 .00104(5025 957 915849 876467493 30.9354166 9.8545617 .001044932 958 917764 879217912 30.9515751 9.8579929 .001043841 959 919681 881974079 30.9677251 9.8614218 .001042753 960 921600 884736000 30.9838668 9.8648483 .001041667 961 923521 887503681 31.0000000 9.8682724 .001040583 962 925444 890277128 31.0161248 9.8716941 .001039501 963 927369 893056347 31.0322413 9.8751135 .001038422 964 929296 895841344 31.0483494 9.8785305 .001037344 965 931225 898632125 31.0644491 9.8819451 .001036269 966 933156 901428(596 31.0805405 9.8853574 .001035197 967 935089 904231063 31.0966236 9.8887673 .001034126 968 937024 907039232 31.1126984 9.8921749 .001033058 969 938961 909853209 31.1287648 9.8955801 .001031992 970 940900 912673000 31 .1448230 9.8989830 .001030928 971 942841 915498611 31.1608729 9.9023835 .001029866 972 944784 918330048 31.1769145 9.9057817 .001028807 973 946729 921167317 31.1929479 9.9091776 .001027749 974 948676 924010124 31.2089731 9.9125712 .001026694 975 950625 926859375 31.2249900 9.9159(524 .001025(541 976 952576 929714176 31.2409987 9.9193513 .001024590 977 954529 932574833 31.2569992 9.9227379 .001023541 978 956484 935441352 31.2729915 9.9261222 .001022495 979 958441 938313739 31.2889757 9.9295042 .001021450 980 960400 941192000 31.3049517 9.9328839 .001020408 981 962361 944076141 31.3209195 9.9362613 .0010193(58 982 964324 94696()168 31.33(38792 9.9396363 .001018330 983 96()289 949862087 31.3528308 9.9430092 .001017294 984 968256 952763<)04 31.3687743 9.94(i3797 .00101()2(50 985 970225 955(571625 31.3847097 9.9497479 .001015228 986 972196 958585256 3i.400(;;i(;9 9.95:51138 .001014199 987 9741()9 961504803 31.41()5561 9.9;5()4775 .001013171 988 976144 964430272 31. 43241 ;73 9.9598389 .001012146 989 978121 967361()69 31.4483704 9.9(531981 .001011122 990 980100 970299000 31.4f)42654 9.9665549 .001010101 991 982081 973242271 31.4801525 9.9(;9{K)95 .0010(^9082 9i)2 984064 976191488 31.49(50315 9.9732()19 .0010080()5 993 98(5019 97914()()57 31.5119025 9.9766120 .001007049 994 988036 982107784 31.5277(555 9.9799599 .001006036 995 99(J()25 985074875 31.5436206 9.983;'.055 .001005025 996 992016 988047936 31.5."')9-1()77 9.98(5(5488 .001004016 997 994009 9i)102()973 3i.575:;()()8 9.9899<)00 .()()10();5009 998 99(;001 99401 19i)2 31.5911380 i). 9933289 .001002004 999 998001 997(X)2999 31.(5(J(5i)(513 i).99(;()(55() .001001001 APPENDIX V CONVERSION TABLES CONVERSION TABLES 381 TABLES FOR CONVERTING UNITED STATES WEIGHTS AND MEASURES METRIC TO CUSTOMARY WEIGHTS Milli^n-iuiis (^raius Grams Kilograms Tonnes to Tonnes to No. to to to Avoirdii})ois to Avoirdupois Net Tons of (Jross Tons of 1 Grains Troy Ounces Ounces Pounds 2000 Pounds 2240 Pounds .01543 .03215 .03527 2.20462 1.10231 .98421 2 .0308G .0()4;30 .07055 4.40924 2.20462 1.96841 3 .046:30 .09645 .10582 6.()1387 3.:5()693 2.952(;2 4 .06173 .12860 .14110 8.81849 4.40924 3.9:J682 5 .07716 .1()075 .176:57 11.02311 5.51156 4.92103 6 .09259 .192^)0 .21164 13.22773 6.6i:387 5.90.-)24 7 .10803 .22.-)06 .24()92 15.4:52:36 7.71618 (5.88944 8 .121346 .25721 .28219 17.63698 8.81849 7.87:565 9 .13889 .2893(5 .31747 19.84160 9.92080 8.85785 1 Kilogram = 1543 2.356:39 Grains LINEAR Ml EASURE Miliimetrrs Centimeters Meters liletcrs Kilometers Kilometers No. to m\\s of ail to to to to to Inch Inches Feet Yards Statute Miles Nautical Miles 1 2.51968 .39370 3.280833 1.093611 .62137 .53959 2 5.03936 .78740 e.oinmi 2.187222 1.24274 1.07919 3 7.55904 1.18110 9.842.500 3.2808:13 1.86411 1.(51878 4 10.07872 1.57480 13.12.33:53 4.:i74444 2.4S548 2.158:57 5 12.59840 1.96850 16.4041()7 5.468056 3.10685 2.6979(5 6 15.11808 2.36220 lO-fiS-^OOO 6.561()(;7 3.72822 3.2:575(5 7 17.63776 2.75.^)<>0 22.96.~)S:;:i 7.65.-)278 4.:U959 3.77715 8 20.15744 3.141K)0 26.24()(;67 8.748S89 4.97096 4.:51(574 9 22.67712 3.54330 29.527500 9.842500 5.59233 4.85633 • 382 CONVERSION TABLES TABLES FOR CONVERTING UNITED STATES WEIGHTS AND MEASURES CUSTOMARY TO METRIC WEIGHTS Grains : Troy Ounces Avoirdupois Avoirdupois Net Tons of Gross Tons of No. to to Ounces Pounds to 2000 Pounds 2240 Pounds Milligrams Grams to Grams Kilograms to Tonnes to Tonnes 1 64.79892 31.10348 28.34953 .45359 .90718 1.01605 2 129.59784 62.20696 56.69f)05 .^)0718 1.81437 2.03209 3 194.39675 93.31044 85.04858 1.36078 2.72155 3.04814 4 259.19567 124.41392 113.39811 1.81437 3.62874 4.06419 5 323.99459 155.51740 141.74763 2.26796 4.53592 5.08024 6 388.79351 186.62088 170.09716 2.721.55 5.44311 6.09628 7 453.59243 217.72437 198.44669 3.17515 6.35029 7.11233 8 518.39135 248.82785 226.79621 3.62874 7.25748 8.12838 9 583.19026 279.93133 255.14574 4.08233 8.16466 9.14442 1 AvoirduiD L ois Pound = -INEAR ME 453.5924277 Grams EASURE 64tlis of an Inches Feet Yards Statute Miles Nautical Miles M Inch to to to to to to Millimeters Centimeters Meters Meters Kilometers Kilometers 1 .39688 2.54001 .304801 .914402 1.60935 1.85325 2 .79375 5.08001 .609601 1.828804 3.21869 3.70650 3 1.19063 7.62002 .914402 2.743205 4.82804 5.55975 4 1.58750 10.16002 1.219202 3.657607 6.43739 7.41300 5 1.98438 12.70003 1.524003 4.572009 8.04674 9.26625 6 2.38125 15.24003 1.828804 5.486411 9.65608 11.11950 7 2.77813 17.78004 2.133604 6.400813 11.26543 12.97275 8 3.17501 20.32004 2.438405 7.315215 12.87478 14.82600 9 3.57188 22.86005 2.743205 8.229()16 14.48412 16.67925 1 Nautic ialMile = 1853.25 Meters 1 Gunte r's Chain = 20.1 1()8 Meters 1 Fathoi m = 1.829 Meters INDEX Absorption dynamometer, 305. Acceleration, 124, 125, 131, 143, 170, 171, 172. angular, 170. normal, 144. taniijential, 144. Angular velocity, 169, 196. Appendix I, Hyperbolic Functions, 339. Appendix II, Logarithms of Numbers, 345. Appendix III, Trigonometric Func- tions, 349. Appendix IV, Squares, Cubes, etc., 359. Appendix V, Conversion Tables, 380. Attractive force, 132, 136. Ball bearings, 280. Bearings, ball, 280. roller, 279. Belts, centrifugal tension, 293. coefficient of friction, 292. creeping of, 292. friction of, 288. stiffness of, 2i)4. Body, freely falling, 126. projected up inclined plane, 158. projected upward, 126. through atmosphere, motion of, 138. Brake friction, 306, 307. Brake shoes, friction of, 310. Brake shoe testing machine, 252. Car on single rail, 227. Catenary, 118. Center of gravity, 27, 32. of cone, 33. of locomotive counterbalance, 40. of rail section, 46. of triangle, 35. of T-section, 30. of U-section, 30. Center of percussion, 187, 330. Centrifugal force, 146. Centrifugal tension of belts, 293. Circular pendulum, 148. Coefficient of friction, 261. Combined rotation and translation 173. Compound pendulum, 188. Concurrent forces, 5. in plane, 9. in space, 14. Conical pivot, 301. Connecting rod, 212. Conservation of energy, 234. Conversion Tables, 381. Cords, and pulleys, 113. flexible, 111. uniform load along cord, 117. uniform load horizontally, 114. Couples, 50, 53, 54. Creeping of belts, 292. Cubes, Cube Roots, etc., 359. Curvilinear motion, 142. Cycloidal pendulum, 154. D'Alembert's principle, 177. Determination of //, 192. Direct central impact, 316, 319. Direct eccentric impact, 328. Displacement, 4. Dry surfaces, friction of, 262. Durand's rule, 46. Dynamometer, absorption, 305. transmission, 291. Eccentric impact, 328. Elasticity of materials, 322. Ellipse of inertia, 92. Ellipsoid of inertia, 105. Energy, 2:i3. and work, 229. conservation of, 234. of body moving in straight line, 234. 383 38-i INDEX Experimental determination of mo- ment of inertia, 191. Falling bodies, 126. Flat pivot, 299. Flexible cords, 111. Force, 1, 8, 56, 63. moment of, 17, 18. parallel, 20, 22. polygon of, 7. representation of, 5. tangential and normal, 146. transmissibility of, 8. triangle of, 6. units of, 1. Friction, 261. coefficient of, 261. laws of, dry surfaces, 262. laws of, lubricated surfaces, 264. of belts, 288, 292. of brake shoes, 310. of pivots, 299. of worn bearing, 297. rolling, 272. Friction brake, 306, 307. Friction gears, 285. Friction wheels, 274. Gears, friction, 285. Gravity, center of (see Center of grav- ity). Gyroscope, 216. Gyroscopic action explained, 221. Harmonic motion, 132. Hyperbolic Functions, 119, 339. Impact, 315. direct central, elastic, 319. direct central, inelastic, 316. imperfectly elastic bodies, 323. oblique, 331. rotating bodies, 332. tension and compression, 325. Inclined plane, motion on, 128. Inertia, 2. ellipse of, 92. ellipsoid of, 105. moment of, 69, 71. (see Moment of inertia). Introduction, 1. Kinetic energy of rolling bodies, 256. Laws of friction, 262, 264. of motion, 127. Length of cord, 116, 122. Locomotive counterbalance, 40. Locomotive side rod, 211. Logarithms of Numbers, 345. Lubricants, testing of, 270. Lubricated surfaces, friction of, 264. Mass, 3. Moment of force, 17, 18. Moment of inertia, 69, 71. experimental determination of, 191. graphical method, 85. greatest and least, 76, 89. inclined axis, 74, 102. non-homogeneous bodies, 102. of angle section, 82. of circular area, 80. of circular cone, 98. of elliptical area, 81. of locomotive drive wheel, 107. of rectangle, 78. of triangle, 79. parallel axes, 72, 99. principal, 104. polar, 77. right prism, 95. Simpson's rule, 88. solid of revolution, 97. thin plates, 93. Motion, curvilinear, 142. body through atmosphere, 138. due to repulsive force, 134. earth, 219. harmonic, 132. in circle, 146. in straight line, 123. Newton's laws of, 127. on inclined plane, 128. resistance varies as distance, 134. tAvisted curve, 165. Newton's laws of motion, 127. Non-concurrent forces, 56, 63. Parallel forces, 20, 22. Pendulum, compound, cycloid al, ir)4. simple circular, 148. 188. INDEX 385 Percussion, center of, 187, 330. Pile driver, 240. Pivots, friction of, 299. Plane of rotation, 220. Polar moment of inertia, 77. Power, 233. Precessional moment, 222, 225. Principal axes, 104. Principal moment of inertia, 104. ProcUict of inertia, 75. Projectile, 156, 160, 162. Pulleys and cords, 113. Peciprocals of numbers, 359. Pectilinear motion, 123. Ridative velocity, 139. liepresentation of force, 5. of couples, 51. of moment of inertia, 72. Repulsive force, 134. Resistance, of roads, 277. train, 312. varies as distance, 134. Ri.2:id body, 2. free to rotate, 197. Roller beariuij^s, 279. Rolling friction, 272. Ropes and belts, stiffness of, 294. Rotating body, reactions of supports, 181. Rotation, about axis, one point fixed, 216. axis fixed, 247. axis not a gravity axis, 201. and translation, 173, 208. fly wheel, 204. in general, 175. locomotive drive wheel, 200. rigid body, 179. sphere, 185. symmetrical bodies, 198. Side rod of locomotive, 211. Simple circular pendulum, 148. Simpson's rule, 41, 43. Specific gravity, 3. Spherical pivot, 302. Spinning top, 218. Squares, square roots, etc., 359. Steam hammer, 244. Stiffness of belts, 294. Suspension bridge, 114, 120. Tangential and normal acceleration. 144. Tangential and normal force, 146. Testing of lubricants, 270. Theorems of Pai)pus and Guldinus, 47. Top, spinning, 218. Torsion balance, 192. Train resistance, 312. Translation and rotation, 173, 208. Translation of rigid body, 178. Transmissibility of force, 8. Transmission dynamometer, 291. Trigonometric Functions, 349. Twisted curve, motion in, 165. Uniform motion in circle, 146. Unit of force, 1. of moment of inertia, 71. of power, 233. of weight, 2, 3. of work, 230. Variable acceleration, 131, 172. Varignon's Theorem of Moments, 18. Velocity, 123, 142, 169. relative, 139. Work, combined rotation and transla- tion, 254. graphical representation, 230. motion uniform, 257. units of, 230. variable force, 238. Work and energy, 229. AVork-energy relation for any motion, 257. Worn bearing, friction of, 297. 2c Text-Books on Mechanics FRANKLIN and MACNUTT — The Elements of Mechanics. A Text-Book for Colleges and Technical Schools. By W. S. FRANKLIN and Barry Macnutt of Lehigh University. Cloth, Svo, xi + 2Sj pages, $1.50 ?iet. Its special aim is to relate the teaching of mechanics to the immediately practical things of life, to cultivate suggestiveness w ithout loss of exactitude. 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