1903 liililiir ,. 3 1924 031 255 197 olin.anx ^ ' Cornell University Library The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031255197 ELEMENTARY TREATISE ANALYTICAL MECHANICS. BY WILLIAM G^^PECK, Ph.D., LL.D., PROFESSOR OP MATHEMATICS, ^MECHANICS, AND ASTRONOMY IK COLUMBIA COLL&GB. Copyright, 1887, by William G. Peck. A. S. BARNES & COMPANY, NEW YORK AND CHICAGO. PUBLISHERS' NOTICE. PECK'S ACADEMIC AND COLLEGIATE COURSE. I. MANUAL OF ALGEBRA, n. MANUAL OF GEOMETRY AND CONIC SECTIONS. III. ANALYTICAL GEOMETRY. IV. DIFFERENTIAL AND INTEGRAL CALCULUS. V. POPULAR PHYSICS [from Ganot]. VI. ELEMENTARY MECHANICS. VII. ASTRONOMY AND OPTICS. VIII. DETERMINANTS. IX. ANALYTICAL MECHANICS. PREFACE. rr^HE following treatise has been prepared for use as a text-book in the School of Mines, but it is hoped that it may find a place in other Col- leges and Schools of Science. It is intended to em- brace all the principles of Analytical Mechanics that are needefi by the student of Engineering, Archi- tecture, and Geodesy. Its order of arrangement is the result of. much practical experience, and its methods of demonstration have been thoroughly tested in the class-room. In its development the methods of the Differen- tial and Integral Calculus have been freely used, but not to the exclusion of the more elementary proc- esses of Analysis. The plan adopted has been found to be well suited to Experimental Illustration, and the numerous practical examples scattered through the book have proved to be of great utility in impressing principles on the mind of the student. IV PEEFACE. For convenience of reference, the first equation on each page has. been indicated by a number placed at the top of that page. The fundamental principles of the Science are, for the most part, based on the axiomatic laws of Newton, but the more modern ideas of "Work and Energy have been carefully considered and incorpo- rated in the text. Qoupilli^re's elegant method of treating the resistance of friction has been fully illustrated in the processes of finding the moduli of the elements of Mechanism. It is believed that the order of arrangement will be found to be logical, the definitions defer and pre- cise, and the demonstrations simple and comprehen- sive. Columbia College, JvJ/y 4, 1887. CONTENTS, I. — Defiititions and Inteoductoey Eemaeks. Definition of a Body ; Rest and Motion ; Force 9 Definition of Mechanics 10 Equilibrium ; Grarity, Weight, Mass, Density 11 Uniform and Varied Motion 12 Momentum ; "Work, Energy 13 Representation and Measure of Forces 14-15 Newtonian Laws of Motion 15-16 II. — Composition and Resolution of Foeces. Definitions; Composition of Homologous Forces 17-18 Parallelogram of Forces 18-20 Polygon and Parallelopipedon of Forces 20-22 Resolution and Composition of Rectangular Forces . . . 22-25 Resultant of Two Forces ; Relation between them 26-27 Translation ; Rotation ; Principle of Moments 28-31 Resultant and Composition of Parallel Forces 31-36 Translation ; Rotation ; Equilibrium 36-39 Quantity of Work ; Theorem of Work 39-41 III. — Centee of Geavity and Stability. Preliminary Definitions and Principles 42-44 Method of Applying the Calculus 44-45 Applications to Lines, Surfaces, and Solids 45-53 VI CONTENTS. FASEB The Centrobaric Method 53-55 /Experimental Method 55 Method by Composition 65-59 Stability and Equilibrium of Bodies 60-64 Practical Problems 64^69 IV. — Elementakt Machikes. Definitions ; Applied and Useful Work ; Modulus.. 70-71 Trains of Mechanism ; Mechanical Powers 71-73 The Cord ; the Lever 73-75 Friction 75-78 Equilibrium Bordering on Motion 78-79 Method vt Finding the Modulus 79-80 Modulus of the Lever 80-81 The Compound Lever 83 Elbow-joint Press ; its Modulus 83-85 Weighing Machines 85-89 The Inclined Plane ; its Modulus 91-95 Single Fixed Pulley ; its Modulus 95-98 Friction of Eope on an Axle 98-99 Single Movable Pulley ; Combinations of Pulleys. . . 99-103 Wheels and Axles ; Windlass ; Differential Windlass 103-108 Wheel-work 108-109 The Screw ; Differential Screw ; the Wedge 110-114 V. — Kinetics. Definition of Motion ; Fundamental Equations.,.. 115-116 Uniformly Varied Motion ; Applications 116-128 Motion of a Body in a Resisting Medium 139-133 Falling Bodies when Gravity is Variable 132-137 Motion of a Body down a Vertical Curve 137-139 Motion of Projectiles 139-145 Components of Deflecting Force 146-148 Vibration ; Angular Velocity and Acceleration 148-151 COKTENTS. vii PASES The Simple Pendulum 153-155 De I'Ambert's Principle ; the Compound Pendulum 155-160 The Keversible Pendulum ; Practical Applications. . 161-166 VI. — Cbnteifugal Force. — Mombbtt of Ineetia. Centrifugal Force ; Application to Masses ; Illustra- tions : 167-172 Surface of a Eevolving Liquid 173-173 Application to Figure of the Earth 173-176 Elevation of Outer Eail of Curved Track 176-177 Conical Pendulum ; Governor 177-180 Moment of Inertia ; Kelation to Parallel Axes 181-183 Polar Moment ; Experimental Determination 183-185 Application to Lines^ Surfaces, and Volumes 185-194 Radius of Gyration 194-195 VII. — WOEK AND EnEEGY. IMPACT. Eelation between Work and Energy 196-198 Work of Gravity when a Body Falls Down a Curve. 198-301 Work to Eaise a System of Bodies 301-303 Work in Producing Rotation 204^306 Kinetic Energy of a Eevolving Body 306-307 Fly-wheels with Applications 307-315 Elasticity ; Impact 315-331 VIII. — Mechanics of Liquids. Classification ; Transmission of Pressures 335-337 Pressure due to Weight ; Centre of Pressure 337-333 Buoyant Effort of Fluids 336-337 Floating Bodies 337-339 Specific Gravity ; Methods of Finding. 339-341 Method by Use of Balance 341-343 Hydrometers ; Alco-ometer 343-347 Vlll COKTBlSrTS. PAGXB Velocity of Jet ; Modifications of Formulas 248-356 Time for a Vessel to Empty Itself 256-357 Flow of Water in Pipes and Channels 257-264 IX. — Mechajtics of Gases and Vapoks. Gases and Vapors ; Atmospheric Pressure 365-267 Laws of Mariotte and Gay Lussac , . 268-373 Absolute Temperature .... 273-274 Manometers; Siphon Gauge; Diving Bell 274-278 The Barometer and its Uses 278-283 Work due to Expansion of Vapor ; Steam 284-288 Efflux of a Gas or Vapor 288-390 X. — Hydraulic autd Pnetjhatic MACHiiirBS. Definitions ; Water Pumps. 391-393 Lifting Pump ; Modifications 293-296 Forcing Pump ; Modifications 396-300 Fire Engine ; Eotary Pump 300-302 Hydraulic Press 302-304 Method of Storing Up Pressure 305-306 Siphons ; Intermitting Springs 306-309 The Hydraulic Eam 309-310 Archimedes' Screw 310-311 Blowers ; Blacksmith's Bellows : 313-314 Air-pump ; Fountains 314r-319 Note. — Names of Greek letters used in this book : a. Alpha. i8. Beta. ' y. Gramma. 6. Delta. e. Theta. TT. Pi. f}. Rho. 2. Sigma. T. Tau. Phi. Omega. MECHANICS. I.— DEFINITIONS AND INTRODUCTOEY PRINCIPLES. Definition of a Body. 1. A body is a collection of material particles. A body whose dimensions are exceedingly small is called a material point. In what follows the term point will generally be used in this sense. Rest and Motion. 3. A point is at rest when it remains in the same relative position with respect to certain surrounding bodies that we regard as fixed ; it is in motion when it continually changes this relative position. The terms rest and motion as used in Mechanics are purely relative ; it is probable that no point in the physical universe is absolutely at rest. Force. 3. A force is that which tends to change the state of a body with respect to rest or motion. If a body is at rest, whatever tends to set it in motion is a force ; if in motion, whatever tends to make it move faster or 10 MECHAKICS. slower, or whatever tends to change the direction of its motion, is a force. A force may act on a body for an instant and then cease, in which case it is called an impulse, or an imptilsive force ; or, it may act continuously, in which case it is called an in- cessant force. For the purposes of mathematical investigation, an inces- sant force may be regarded as made up of a succession of impulses acting at eqnal but exceedingly small intervals of ' time. This interval, in the language of the calculus, may be denoted by dt, t being an independent variable and dt its constant differential. If the successive or elementary im- pulses are all equal the force is said to be constant, other- wise it is variable. If forces act on a fixed body they produce stress, or press- ure; if they act on a body that is free to move they produce motion : in the former case they are called forces of press- ure, in the latter case they are called moving forces. The same force under different circumstances may produce either pressure or motion : thus, if gravity act on a body that is supported it produces pressure, but if it acts on a body that is not supported it produces motion. Definition of Mechanics. 4. Mechanics is the science that treats of the action of forces on bodies. It is divided into two branches : Statics, which treats of the laws of pressure ; and Kinetics (or Dynamics), which treats of the laws of motion. In kinetics it is not motion alone that is considered, but the relation of forces to motion. That branch which treats of pure motion, without reference to the bodies moved or to the forces which produce the motion, is sometimes called Kine- matics. 1.] DIIFINITIONS AND INTRODUCTORY PRINCIPLES. 11 Equilibrium. 5. If the forces acting on a body balance each other, that is, if they c&unteract each other's effects, they are said to be in equilibrium. Such a set of forces does not change the state of the body with respect to rest or motion ; if the body is at rest it will remain so, or if in motion its motion will remain unchanged, so far as these forces are concerned. When forces balance each other through the medium of a body at rest they are said to be in statical equilibrium ; when they balance each other through the medium of a mov- ing body they are in dynamical equilibrium. Gravity, Weight, Mass, and Density. 6. The earth exercises an attractive force on bodies tending to draw them towards its centre. This force, which acts on every particle of a body, is called the force of gravity. If a body is supported, the force produces a pressure that is called the "weight of the body. It is obvious then that the weight of a body varies as the quantity of matter it contains and as the force of gravity conjointly. The mass of a body is the quantity of matter it contains. If we denote the weight of a body by W, its mass by M, and the force of gravity by g, we have, from what precedes, W W = Mg and J/ = — (1) The density of a body , or the degree of compactness of its particles, is proportional to the quantity of matter in a given volume. We may take as the measure of a body's density, its mass divided by its volume ; or, denoting the density by d and the volume by F, we have 12 MECHANICS. [*• M d = — and M=Vd (2) Combining (1) and (3), we have W=:Vdg (3) The unit of weight that we have adopted is the avoir- dupois pound as determined by counterpoising it against the government standard, -and the iinit of mass is the quantity of matter in sucli a pound. The force of gravity varies froip point to point, but the weight of a body, as determined by a spring balance, varies in the same ratio ; hence, the measure of its mass, W-i- g, re- mains constant. The unit of density is the density of distilled water at 39° F. T7niform and Varied Motion. T. The velocity of a moving point is its raie of motion. If a point moves over equal spaces in equal times its velocity is constant and its motion is uniform ; otherwise its velocity is variable and its motion is varied. If the velocity continually increases the motion is accelerated ; if the velocity continually decreases the motion is retarded. The unit of velocity is assumed to be the velocity of a point which moves over one foot in owe second. This is equivalent to the assumption that the unit of length is one foot and the unit of time is one second. In uniform motion, the velocity of a body is the number of feet passed over by the body in one second ; in varied motion the velocity at any instant is the number of feet that it would pass over in one second if its velocity were to remain un- changed for that time. . In uniform motion, if we denote the velocity by v, and the space passed over in t seconds by s, we have 4.] DEFINITIONS AND INTRODUCTOKY PRINCIPLES. 13 ' = 1 (^) From equation (4), we have s = vt, and t = - (5) V Momentnm. 8. The momentum of a body is its quantity of motion. It is obvious that the quantity of motion of a body, varies conjointly with the mass of the body and with its velocity ; hence, the measure of the body's momentum is equal to mv. The unit of momentum is the momentum of a unit of mass, moving with a unit of velocity. Work, Energy. 9. A force is said to perform "work when it overcomes a resistance. Any kind of work may be assimilated to that of raising a weight, as in lifting a bucket of water from a well. In this case it is obvious that the work performed varies con- jointly as the weight raised and as the height through which it is raised. Denoting the weight in pounds by W and the height in feet by h, we have for the measure of the work per- formed Wh. The unit of work is the work required to raise_ a weight of OTie pound through a height of one foot. ■ This unit, which is termed a physical unit, is denoted by the symbol 1 ft. lb., and is called a foot pound. Thus, the work required to raise 7 It's, through 5 ft. is equal to 35 ft. lbs., that is, to 35 foot pounds. The rate of work of a force is the number of units of work it can perform in a given time. In Mechanism, the rate is usually expressed in terms of a horse power, which is technically assumed to be 33,000 ft. lbs. per minute or 550 ft. lbs. per second. Thus, an engine of 10 horse power is 14 MECHANICS. one tliat is capable of performing 5j500 units of work per second. Energy is the capacity of a body to perform work : it may be potential, or kinetic. Thus, the weight of a clock when wound up has a certain amount of potential energy which is utilized, as the weight runs down, to keep the clock train in motion ; a moying body, as the falling hammer of a pile-driver, has an amount of kinetic energy which is utilized in performing the work of driving the pile into the ground. Geometrical Blepresentation of a Force. 10. A force is said to be given when we know its intensity or magnitude j its point of application, that is, the point at which it is supposed to act ; and its line of action, that is, the direction in which it tends to move the point of appli- cation. A force is represented geometrically by a straight line whose length is proportional to its intensity ; „ _^ one extremity as 0, represents the point ^ . ^ of application and an arrow head shows the direction in which the force is supposed to act. If a force is applied to a solid body, that is, to a body whose particles are rigidly connected, the point of application may be taken at any point on the line of action of the force. Measure of Forces. 11. A force is measured by comparing it with another force of the same kind taken as a unit. Pressures are measured in pounds, the unit being one avoirdupois pound. When we speak of a pressure of n pounds we mean a force which, if directed vertically upward, would just sustain a, weight of n standard pounds. Moving forces are measured in terms of the momenta they can generate. DEFIinTIOXS AND INTRODUCTORY PRINCIPLES. 15 The Timt of an impulsive force is an impulse that can generate a unit of momentum. The unit of a constant force is a constant force which acting on a unit of mass for a unit of time can generate a unit of momentum. The measure of a variable force at any instant is the mo- mentum it could generate in a unit of time if it were to remain constant for that time. The unit of pressure may be called a statical unit, and the unit of an incessant moving force may be called a kinetic unit. Both of these units are of the same kind : for, from Art. 6 the statical unit, one poimd, is equal to a v/ait of mass multiplied by g ; but it will be shown hereafter that the force of gravity acting on a unit of mass for a unit of time will generate about 33.16 units of momentum: hence the statical unit is about 33.16 times the kinetic unit. All kinds of forces may therefore be expressed in terms the kinetic unit. The Newtonian La.ws of Motion. 13. The following three laws, commonly known as the Newtonian Laws, which are deduced from universal ex- perience, are accepted as axiomatic in treating of the motions of bodies : First Law. Every body continues in a state of rest or of uniform motion in a straight line until com- pelled by impressed forces to change that state. Second Law. Change of motion is proportional to the impressed force, and tahes place in the direction of the straight line in which the force acts. Third Law. To every action there is always an equal and contrary reaction: or, the mutual actions of any two bodies are always equal and oppositely directed. The first law is equivalent to the assertion that no body has power of itself to change its state with respect to rest or motion. In the second law the term change of motion is enuivalent 16 MECHANICS. to the expression change of momentum. This law is equivalent to the assertion that of several forces acting on ajbody each produces its own efEect as though the others did not exist, and this whether the body is at rest or in motion. If equal and opposite forces act on a body the law still holds good : thus, if a boat is moving up a river with a given velocity and a body is compelled to move from stem to stern with an equal velocity, we may regard the body as acted upon by equal and opposite forces, each of which produces its own efEect ; but these effects balance each other so that the body remains at rest with respect to objects on shore. The third law holds true whether the forces considered are statical or kinetic. If a heavy body exerts a downward press- ure on a table, the table exerts an equal upward pressure ; if the earth exercises an attraction on a stone drawing it down- ward, the stone exerts an attraction on the earth drawing it upward, the momentum generated in each case being the same ; if a moving body A impinges upon a body B which is free to move, A imparts some of its momentum to B, and the efEect on A is the same as though it were acted upon by a force equal and opposite to that exerted by A upon B. II.— COMPOSITION AND EESOLUTION OP POECES. Definitions. 13. Composition of forces, is the operation of finding a single force whose effect is the same as that of two, or more given forces. The required force is called the resultant of the given forces. Resolution of forces, is the operation of finding two or more forces whose combined effect is equivalent to that of a given force. The required forces are called components of the given force. Composition of Forces whose directions coincide. 14. Prom the rules laid down for measuring forces, it fol- lows, that the resultant of two forces applied at a point, and acting in the same direction, is equal to the sum of the forces. If two forces act in opposite directions, their result- ant is equal to their difference, and it acts in the direction of the greater. If any number of forces be applied at a point, some in one direction and others in a contrary direction, their resultant is equal to the sum of those that act in one direction, diminished by the sum of those that act in the opposite direction, or call- ing those that act in one direction plus and those that act in a contrary direction mimes, the resultant is equal to the alge- braic sum of the components. Porces that have a common line of action are called homol- ogous ; their algebraic sum may be indicated by writing the expression for one of the forces in a parenthesis and prefixing 18 MECHA.2S'ICS. [«• the symbol S. Thus, if we denote the resultant of the group by R and one of the homologous components by P, we have R = ^ (P), (6) which is read. The resultant of a set of homologous forces is equal to their algebraic sum. The forces treated of in this article and also in the following articles, are supposed to be applied at points of a solid body. "When their lines of action intersect, they are said to be con- current. Parallelogram of Forces. 15. Suppose two forces, P and Q, to be applied to a solid body at a point and let them be represented in direction and intensity by OP and OQ : complete the parallelogram PQ and draw its diagonal OR. Let the body contain a unit of mass ; then if the forces are impulses OP and OQ will represent the velocities generated by P and Q (Art. _ —yr^ 11), and inasmuch as each produces its / ^^ I own eflEect as though the other did not / ^^ exist (Art. 12), the body will be found /^ ^ at the end of one second somewhere on Kg. 2. PR by virtue of the force P and some- where on QR by virtue of the force Q ; it will therefore be at R. Had been acted on by an impulse represented by OR, it would in like manner have moved from to ^ in one second. Hence the impulse OR is equivalent in effect to the two impulses OP, and OQ ; that is. If two impulsive forces he represented, hy adjacent sides of a parallelogram, their resultant will be rep- resented by that diagonal of the parallelogram which passes through their commoV' point. If the forces are constant forces, OP and OQ will represent COMPOSITION AND EESOLUTION OF FORCES. 19 the yelocities they can generate in a unit of time, and for the same Reason as before OR will represent a constant force equivalent in effect to the forces P and Q : hence, the prin- ciple holds true for constant forces. It is also true for forces of pressure ; for, if we apply a force equal and directly op- posed to the resultant of the two moving forces it will hold them in equilibrium, converting them into forces of pressure, but it will in no manner change the relation between them and their resultant. Hence, the principle holds for all kinds of forces : it may be enunciated as follows : If two forces he represented in direction and in- tensity by adjacent sides of a parallelogram, their re- sultant will be represented by that diagonal of the par- allelogram which passes through their common point. This principle is called the parallelogram of forces. Geometrical Applicatious of the Parallelogram of Forces. 16. 1°. Given two forces ; to find their resultant. Let OP and 0§ be the given forces. Complete the parallelogram QP and draw its diagonal OR ; this will be the resultant required. 2°. Given, a force and one of its com- ponents ; to find the other. " Fig. 3. Let OR be a force and OP one of its components. Draw PR and complete the parallelogram PQ ; OQ will be the other component. 3°. Given, a force and the directions of its components ; to to find the components. Let OR be a force and OP, OQ, the directions of its components ; through R draw RQ and RP parallel to PO and "g^ QO; then will OP and Og be the re- pig. 4. quired components. 20 MECHANICS. 4°. Given, a force and the intensities of its components ; to find the directions of the components. * Let OR be a force, and let the intensities of its components be represented by lines equal to OP and OQ ; with as a centre and OP as a radius, describe an arc, then with i2 as a centre and OQ as, a. radius, describe a second arc, cutting the first at P ; draw OP, and RP, and com- plete the parallelogram PQ ; OP and OQ will be the re- quired components. mg. B. To construtjt }?-'->" Polygon of Forces. IT. Let OQ, OP, OS, and OT, be a system of forces ap- plied at a point, 0, and lying in a single plane, their resultant ; on 0^ and OP con- struct the parallelogram PQ, and draw its diagonal OR', this will be the re- sultant of OP and 0§. In like manner construct a parallelogram on OR' and OS; its diagonal OR", will be the resultant of OP, OQ, and OS. On OR" and OT construct a parallelogram, and draw its diagonal OR; then will OR be the resultant of all the given forces. This method of construction may be extended to any number of forces whatever. If we examine the diagram, we see that QR' is parallel and equal to OP, R'R" is parallel and equal to OS, R"R is parallel and equal to OT, and that OR is drawn from the point of application, 0, to the extremity of R"R. Hence, we have the following rule for constructing the resultant of several concurrent forces : Thrawgh their common point draw a line parallel Fig. 6. COMPOSITIOIT AND RESOLUTION OF FORCES. 31 and equal to the first force; through the extremity of this draw a line parallel and equal to the second force; and so on, throughout the system; finally, draw a line from the starting point to the extremity of the last line draivn, and this will be the resultant re- quired. This application of the parallelogram of forces is called the polygon of forces. The construction holds true, even when the forces are not in one plane. In this case, the lines OQ, QR', R'R", &c., form a t'wisted polygon, that is, a polygon whose sides are not in one plane. When the point R, in the construction, falls at 0, OR re- duces to 0, and the forces are in equilibrium. The simplest case of the preceding principle i^ the triangle of forces. Thus, to find the resultant of P and Q (Fig. 3), we draw OP parallel and equal to the first force, and from P we draw PR parallel and equal to the second force ; then the line OR is the required resultant. If three forces are parallel and equal to the three sides of a triangle taken in order, the forces are in equilibrium. Paxallelopipedon of Forces. 18. Let OP, OQ, and 08, be three concurrent forces not in the same plane. On these, as edges, construct the parallelopipedon J... .^ OR, and draw OR, OM, and SR. From /" "'""-^ X i the principle of Art. 15, OM is the re- f ^^^Ei—^L^p sultant of OP and 0§ ; and OR is the L/ "--..._^ I /' resultant of OM and OS; hence, OR f^' "m is the resultant of OP, OQ, and OS; Fig. 7. that is. If three forces be represented by the concurrent edges of a parallelopipedon, their resultant will be repre- 22 MECHANICS. [7- sented by the diagonal of the parallelopipedon that passes through their common point. This principle is called the parallelopipedon of forces. It is easily shown that it is a particular case of the polygon of forces; for, OP is parallel and equal to the first, PMio the second, MR to the third force, and OR is drawn from the origin, 0, to the extremity of MR. Components of a Force in the direction of Kectangulax Axes. 19. First. To find analytical expressions for the compo- nents of a force in the direction of two axes. Let ^J2 be a force in the plane of the rectangular axes OX and OY. On it as a diagonal construct a parallelogram ML, whose sides are parallelogram to OX and or. Denote AR by R, AL by X, AM, equal to LR, by Y, and the angle LAR, equal to the angle the force makes with OX, by a. From the figure, we have Pig 8. X = i2 COS a, and Y := R sma (7) In these expressions the angle a is estimated from the positive direction of the axis of X, around to the force, in accordance with the rule laid down in Trigonometry. The component X will have the same sign as cos a, and the com- ponent I^the same sign as sin «. Secondly. To find the components of a force in the direction of three rectangular axes. Let OR, denoted by R, be the given force, and OX, OY, and OZ, the given axes. On OR, as a diago- nal, construct a parallelopipedon whose edges are parallel to the axes. Fig. 9. 8] COMPOSITIOlir AND RESOLDTION OP FOKCES. 23 Then will OL, OM, and ON be the required components. Denote these by X, Y, and Z, and the angles they make with OR by a, 13, and y. Join B with L, M, and If, by straight lines. From the right-angled triangles thus formed, we have X = E cos a, Y ^:^ R cos (i, and Z = R cos y . . . (8) The angles a, (i, and y are estimated from the positive direc- tions of the corresponding axes, as in Trigonometry, and each component has the same sign as the corresponding cosine. If a force be resolved in the direction of rectangular axes, each compoiient will represent the total effect of the given force in that direction. For this reason such components are called effective components. It is plain that the compo- nent in the direction of each axis is the same as the projec- tion of the force on that axis, the projection being made by lines through the extremities of the force, and perpendicular to the axis. Hence, we may find the efEective component of a force in the direction of a given line, by multiplying the force into the cosine of its inclination of the line. It is obvious that the projection of the resultant of two or more forces on a given line is equal to the algebraic sum of the projections of its components on that line. It is also ob- vious that the resultant of two forces is equal to the sum of the projections of the forces on the direction of the resultant. Analytical Composition of Rectangular Forces. 30. First. When there are but two forces. Yj Let AL and AM he rectangular forces," denoted by X and Y, and let AR, de- noted by R, be their resultant. Denote - the angle LAR by a. Then, because ZR = Y, we have, from the triangle A LR, ' Fig. lo. 24 MECHANICS. [9. R= 'VX^+Y^; cos« ; and sm a = -5 R (9) The first of these gives the intensity, the second and third the direction of the resultant. Secondly. When there are three forces not in one plane. Let OL, OM, and ON, be rec- tangular forces denoted by X, Y, and Z, and let OR, denoted by R, be their resultant. Denote the angles which R makes with OL, OM, and ON by a, (i, and y. Then, from the figure, we have R = V-S:^+T» + Z». . . (10) X Y Z cos «=-=•; cos /3 = -j5 ; and, cos y = -^ KM Ji Pig. It (11) The first gives the intensity of the resultant, the others its direction. Examples. 1. Two pressures of 9 and 12 pounds act on a point, and at right angles to each other. Required, the resultant pressure. Ans. The resultant pressure is 15 lbs., and it makes an angle of 53° 7' 47" with the direction of the first force. 3. Two rectangular forces are to each other as 3 to 4, and their result- ant is 20 lbs. What are the intensities of the components ? Ans. X-1% lbs., and r = 16 lbs. 3. A boat fastened by a rope to a point on shore, is urged by the wind perpendicular to the current, with a force of 18 pounds, and down the current by a force of- 22 pounds. What is the tension on the rope, and what angle does it make with the current ? Ans. R = 38.425 lbs. ; a = 39° 17' 30". 4. Required the intensity and direction of the resultant of three forces at right angles to each other, having the intensities 4, 5, and 6 pounds, respectively. 12.J COMPOSITION AKD RESOLUTION- OF FORCES. 25 Ans. R = Vl6 + ^35 + 36 = 8.775 lbs. ; and a = 62° 52' 51" ; P = 55° 15' 50" ; 7 = 46° 51' 43". 5. Three forces at right angles are to each other as 2, 3, and 4, and their resultant is 60 lbs. What are the intensities of the forces ? Ans. 22.284 lbs., 33.426 lbs., and 44.568 lbs. Application to Groups of Concnrrent Forces. 31. The principles explained in the preceding articles, enable us to find the resultant of any number of concurrent forces. Let P, P', P", &c., be a group of concurrent forces. Call the angles they make with the axis of X, a, a', a", &c. ; the angles they make with the axis of Y, /3, j3', jS", &c. ; and the angles they make with the axis of Z, y, y', y", &c. Re- solve each force into rectangular components parallel to the axes, and denote the resultants of the groups parallel to the axes by X, Y, and Z. We have, (Art. 19), X=2(Pcos«), r=2(Pcos;3), Z^-S.{P cosy). If we denote the resultant by R, and the angles it makes with the axes by a, b, and c, we have, as in Article 30, R = VXa + Y^ + Z\ cos a = -S-, cosb = n", and cos c = -^ (IZ) When the given forces lie in the plane XY, Z reduces to 0, cos j3 becomes sin a, cos b becomes sin a, and the formulas reduce to X = S (P cos «), and Z= S (P sin a). j^ Y R = '/X^ + F^ and cos at = •^, and sin a = -^ . . . (13) Examples. 1. Three concurrent forces, whose intensities are 50, 40, and 70, lie in the same plane, and make with an axis, angles equal to 15°, 30°, and 45°. Re.qJ^ir^d th^ resultant. 26 MECHANICS. [14. Here, X = 50 cos 15° + 40 cos 30° + 70 cos 45° = 132.435, and r = 50 sin 15° + 40 sin 30° + 70 sin 45° = 83.44. Am. R = 156, and A = 31° 54' 13". 3. Three forces, 4, 5, and 6, lie in the same plane, and make equal angles with each other. Required the intensity of their resultant and the angle it makes with the least force. Ans. B = VS, and a = 310°. 3. Two forces, one of 5 lbs. and the other of 7 lbs., are applied at the same point, and make with eaich other an angle of 130°. What is the • intensity of their resultant. Ans. 6.34 lbs. Formula for the Resultant of Two Forces. 33. Let F and P', be two concurrent forces, and let the axis of JT be taken to coincide with F ; a will then be 0, and we shall have sin et = 0, and cos « = 1. The value of Xwill be P + P' coses', and the value of Fwill be P'sin«'. Squar- ing these values, substituting in Equation (9), and reducing by the relation sin^ a' + cos^ a = 1, we have ^e- la R=VF^+P'^ + 2PF' cos a' (14) The angle a' is the angle between the given forces. Hence, The resultant of two concurrent forces is equal to the square root of the sum of the square f of the forces, jjlus twice the product of the forces into the cosine of their included angle. If a = 0, we have E = P + P'. If a' = 180°, we have R=P — P'. Examples. 1. Two forces, P and Q, are equal to 34 and 30, and the angle between them is 105°. What is the intensity of their resultant ? Ans. R = 33.31. 16.] COMPOSITION AKD EESOLUTION OF FORCES. 27 3. Two forces, P and Q, whose intensities are 5 and 13, have a result- ant whose intensity is 13. Eequired the angle between them. Ans. Cos a = 0, or a = 90°. 3. A boat is impelled by the current at the rate of 4 miles per hour, and by the wind at the rate of 7 miles per hour. What will be her rate per hour when the direction of the wind makes an angle of 45° with that of the current ? Ans. B = 10.3m. 4. Two forces, and their resultant, are all equal. What is the angle between the two forces f Ans. 120°. 5. Two forces at right angles are to each other as 1 is to Vs, and their resultant is 10 lbs. What are the forces 1 ' Ans. 5 and 5 ^. 6. If three forces are to each other as 1, 3, and 3, and are in equilib- rium, show that the forces act along the same straight line. Relation Between Two Forces and their Resultant. 33. Let P and Q be two forces, and R their resultant. Then because QP is a parallelogram, the side PR is equal to Q. From the triangle ORP, we have, P-.q-.R-.-.wa. ORP : sin ROP : sin OPR. But, ORP=QOR, and OPR=180° — QOP ; hence, P : Q -.R :: sin QOR : sin ROP : sin QOP ; . . . . (15) That is, of two forces and their resultant, each is pro- portional to the sine of the angle between the other two. If we apply a force R' equal and directly opposed to R, the forces P, Q, and R' will be in equi- librium. The angle QOR is the sup- plement of QOR', and POR is the supplement of POR' ; hence, sin QOR= sin QOR', and sin POR = sin POR' ; we have also, R = R'. Pig. 13. Fig. 14. 28 MECHAKICS. [16- Making these substitutions in the preceding proportion, we have, P :Q:R' : : sin QOR' : sin FOR' : sin QOP .... (16) Hence, if three forces are in equilibrium, each is -pro- portional to the sine of the angle between the other two. Examples. 1. A weight of 50 lbs., suspended by a string, is drawn aside by a hori- zontal force until the string makes an angle of 30° with the vertical. Required the value of the horizontal force, and the tension of the string. Ans. 28.8675 lbs., and 57.735 lbs. 2. A point is kept in equilibrium by three forces of 6 lbs., 8 lbs., and 11 lbs. ; what are the angles between the forces? Ans. 77° 21' 53", 147° 50' 34", and 134° 47' 34". 3. A weight of 25 lbs. is attached to the ends of two strings whose lengths are 3 and 4 ft., the other ends of the strings being attached at points of a horizontal beam which are 5 ft. apart. What are the tensions of the strings? Ans. 30 lbs., and 15 lbs. Translation ; Rotation ; Principle of Moments. 34. A body is said to have a motion of translation when all of its points move in parallel straight lines : it has a motion of rotation when its points move in arcs of circles whose centres are in a straight line ; this line is called the axis of rotation, and any plane perpendicular to it is called a plane of rotation. If a body is restrained by a fixed axis, any force applied to it, whose line of action does not intersect the axis, will tend to impart a motion of rotation. Let P be a force acting on a body which is restrained by an axis OZ. -,, „ •' rig. 16. Draw a line AB, perpendicular to the force and also to the axis. Let A be taken as the point COMPOSITION AKD RESOLUTION OF FOECES. 29 of application of the force, and at this point resolve it into two components P" and P', the former parallel, and the hAA&v perpendicular to OZ. The component P" can have no tendency to produce rotation about the axis ; hence, P', which is the projection of P on a plane of rotation, is the only component that is effective in producing rotation. It is obvious that the tendency of P' to produce rotation varies conjointly with P' and with AB ; we may therefore take the product P' x ^5 as the measure of the tendency of the force P to produce rotation around OZ : this product is called a moment. Hence, the moment of a force with respect to an axis is the product of the projection of the force on a plane of rotation by the distance of the force from the axis. Let us first consider the case in which the forces lie in a plane of rotation. Let P and Q be two forces applied ^ ^_,-,yH at 0, and let R be their resultant. Suppose also that is a point of a solid body which is restrained by an axis perpendicular to the plane POQ at C ; from G draw perpendiculars to P, Q, and R, denoting them re- spectively by j», q, and r. Take the line OC as the axis of X, and the perpendicular OYas. the axis of Y; denote the angle XOP by «, XOQ'bj P, and XOR by ^. Because R is the resultant of P and Q, the projection of R on the axis of Y is equal to the sum of the projections of P and Q on the same axis (Art. 19), that is, E sin -P lever arm of the resultant will / i^^ be 0, and we shall have, from gc^ i — s-b, the principle of moments Mg.ao. (Art. 24), P X CL=Q^CN; or, P : Q:: GN: CL. But, from the similar triangles CNS and LNM, we have, ON: CL:: SN : SM. Combining the two proportions, we have, P : Q :: SN: SM. That is, the line of direction of the resultant divides the line joining the points of application of the compo- nents, inversely as the components. From the last proportion, we have, by composition, P : Q: P + Q:: SN: SM: SN+ SM; and, by division, P : Q : P— Q :: SN: SM: SN—SM. In the first case (Fig. 19), P + Q = B, ani SN + SM — MN; in the second case (Fig. 20), P — Q = R, and SN — SM = MN. Substituting in the preceding propor- tions, we have, P : Q : R:: SN: SM: MN; or, P : Q: R:: ON: CL: NL (18) That is, of two parallel forces and their resultant, each is proportional to the distance between the other two. Two equal parallel forces acting in contrary directions but 34 MECHANICS. not directly opposed constitute a couple; the distance be- tween them is the lever arm of the couple. Prom what precedes we see that the resultant of a couple is 0, and that its point of application is at an infinite distance. It is easily shown that the tendency of a couple to produce rotation around an axis perpendicular to the plane of the forces is equal to either force into the lever arm of the couple. Composition of Parallel Forces. 26. 1°. To find the resultant of two parallel forces lying in the same direc- tion : Let P and Q be the forces, M and N their points of application. Make MQ' = Q, and iVP' = P ; draw P'Q' cutting MIf in S; through S draw SR parallel to MP, and make it equal to J" + ^ ; it will be the resultant. . For, from the triangles P'SJSTand Q'SM, we have. ^^-.,4- <2' f fN B Fig. 21. P'JSr : Q'M ■.■.8N: SM ; or, P : Q:: SN: SM. 3°. To find the resultant of two parallel forces acting in opposite directions : Let P and Q be the forces, M and JV" their points of application. Prolong QN till NA = P, and make MB = Q ; draw AB, and produce it till it cuts JVM produced in S; draw SR parallel to MP, and equal to BP, it will be the resultant required. For, from the triangles SIfA and SMB, we have. M P Pig. 22. AN : BM : : SN : SM; or, P : Q :: SN: SM. F COMPOSITION XNO EESOLUTIOST OF FOKCES. 35 These constructions, which are essentially the same, suggest the methods of resolving a force into two parallel components, applied at given points. 3°. To find the resultant of any number of parallel forces : Let P, P', P", P'", be parallel forces. Find .the resultant of P and P', by the rule already given, it will be ^' = P + P' ; find the result- „^ .--'1 ant of R' and P", it will be P" = P + .-'"^'"'if f T p- P' + P" ; find the resultant of R" and T P'", it will be P = P + P' + P" + P'". p If there be a greater number of forces, the operation of composition may be continued ; the final result will be the " ' resultant of the system. If some of the j,. ^ forces act in contrary directions, com- bine all that act in one direction, as just explained, and call their resultant R' ; then combine all that act in a contrary direction, and call their resultant R" ; finally, combine R' and R" ; their resultant, R, will be the resultant of the system. If the forces P, P', etc., be turned about their points of application, their intensities remaining unchanged, and their directions remaining parallel, the forces R', R", R, will also turn about fixed points, continuing parallel to the given forces. The point through which R always passes is called the centre of parallel forces. Co-ordinates of the Centre of Parallel Forces. 37. Let P, P', P", etc., be parallel forces, applied at points that maintain fixed positions with respect to a system of rectangular axes, and let R, equal to S (P), be their re- sultant. Denote the co-ordinates of the points of application of the forces by x,y,z; x', y, z', etc. ; and those of R by K,, «/„ Zi- 36 MECHANICS. [19. Turn the forces about their points of application, till they are parallel to the axis of F, and in that position find their moments with respect to the axis of Z. In this position the lever arms of the forces are x, x', etc., and the lever arm of R is Xx. From the principle of moments (Art. 34), we have, Rxt =z Px+ P'x' + , etc. or, -IW <"' By making the forces in like manner parallel to the axis of Z, and taking their moments with respect to the axis of X, we have. yi = ^iPy) S (P) • And in like manner, we find, S(P.) 2l = X(P) (30) (31) Forces Applied at Points Invariably Connected.— Tendency to Produce Translation. 38, Let P, P', P", etc., be forces situated in any manner in space an^ applied at points that are invariably connected, as, for example, to points of the same solid body. Assume any point ^ 0, and through it draw any three lines, OX, OY, and OZ, at right angles to each other, and take these lines as axes. Denote the angles that P, P', P", etc., make with the axis of X by a, a', a", etc. ; the angles they make with the axis of YhjPjfi', fi", etc.; and the angles they make with the axis of Z by y, y', y", etc. ; also denote the co-ordinates of their points of application by x, y, z; Fig. 24. COMPOSITION AND liESOLCTION OF FOEOES. 37 «', y', z' ; x", y", z", etc. Then let each force be resolved into components parallel to the axes, as shown in Pig. 34. We have for the group parallel to the axis of X, J" cos a, P' cos a', F" coa a", etc.; for the group parallel to the axis of Y, P cos 0, P' cos P', P" COS P", etc.; and, for the group parallel to the axis of Z, P COS y, P' cos y', P" cos y", etc. Denoting the resultants of these several groups by X, Y, and Z, we have X = 2 (Poos a), F = S (Pcos (i), and Z = S (Pcos y). Now, when the system of forces does not reduce to a couple, it must have a single resultant, that is, X, Y, and Z must in- tersect at a common point whose co-ordinates may be denoted by «!, yi, and 2j, the values of which may be found as in the last article. Forces Applied at Points Invariably Connected.— Tendency to Produce Rotation. 39. Continuing the supposition of the last article, we see that X acts with a lever arm y^, to produce negative rotation around the axis of Z, and that Y acts with a lever arm a;, to produce positive rotation around the same axis ; hence, the entire tendency to rotation around the axis of Z is Yx, - Xy,. But Yxi = 'S.{P cos |3 x) and Xy^ = S, {P coa ct y) ; hence, the preceding expression becomes S(Pcosj32;) — S(Pcos ccy). In like manner it may be shown that the tendency to pro- 38 MECHANICS. [22- duce rotation around the axis of X is measured by the ex- pression S (Pcos y y) — S (Pcos /32). Also, the tendency to rotation around the axis of Y is meas- ured by the expression, S (P cos a «) — 2 (P cos y x). Equilibrium. 30. The system will be in equilibrium when there is no tendency to motion of translation in the direction of either axis, and when there is no tendency to rotation around either axis, hence we must have (Arts. 38, 39), 2 (Pcos a) = 0, \ S (Pcos i3) = 0, 1 (22) S (Pcosy) = 0- ) , S (Pcos |8a;) — S (Pcos a y) = 0, | S (Pcos y «/) — S (Pcos i3 «) = 0, V (23) S (Pcos « «) — ^ (Pcos y a;) = 0. ) Hence, the conditions of equilibrium are, 1st. Tlve algebraic sum of the components of the forces in the direction of any three rectangular axes inust be separately equal to 0. 2d. Tfie algebraic sum of the moments of the forces, with respect to any three rectangular axes, must be separately equal to 0. Case of a Body Restrained by a Fixed Axis. 31. If a body is restrained by a fixed axis, about which it is free to revolve, we may take this line as the axis of X. Since the axis is fixed, there can be no motion of translation, neither can there be any rotation about either of the other two axes of co-ordinates. All of Equations (23), and the first 24.] COitPOSITIOK AlfD KESOLUTIOJf OF FOECES. 39 and third of Equations (23), will be satisfied by virtue of the connection of the body with the fixed axis. The second of Equations (23) is, therefore, the only one that must be satis- fied by the relation between the forces. "We must have, therefore, Z {F cos y y — Pcoa p z) = (24) That is, if a body is restrained by a fixed axis, the forces applied to it will be in equilibrium when the algebraic sum of the moments of the forces with respect to this axis is equal to 0. Quantity of "Work.— Theorem of "Work. 32. It was shown in Art. 9 that the work performed by a force whose line of action coincides with that of the resistance is equal to tJie force multiplied by the space through which it acts. It remains to consider the case in which a force acts obliquely to the resistance. Let P be a force acting on a body which is compelled to move along the line AB ; suppose the body to move over the infinites- imal distance 00 in the element of time dt. Draw €p perpendicular to P, and PP' perpendicular to AB ; ^ Pig. 23. denote Op by 6p, the angle BOP by a, and let the quantity of work performed in the time dt be called the elementary quantity of loork. Now OP', equal to P cos a, is the effective confponent of P in the direction of AB, and the work that it performs is equal to P cos « X 00, or P X 00 cos a ; but 00 cos a is equal to Sp, and consequently the elementary quantity of work of P is equal to PSp. Here Sp falls on the force and the work is assumed to be positive. If is constrained by the action of other forces to describe 40 MECHANICS. [25. the path OC, then will 6p fall on the prolongation of P and the elementary quantity of work of P will obviously be nega- tive. The distance 6p is sometimes called the virtual velocity of P, and the product P6p the virtTial moment of P. Let P and Q be two forces applied to a point of a solid body and let R be their resultant ; and suppose that is constrained to move over an infinitesimal distance OC in the time dt. Prom C draw perpendiculars Cp to P, Gq to Q, and Cr to R, denoting Op by 6p, Oq by 6q, and Or by 6r. Take the line OCas the axis of X, and the perpendicular OT as the axis of Y; denote the angle XOP by a, XOQ by /3, and XOR by 0. Then because R is the resultant of P and Q the projection of R -on the axis of X is equal to the sum of the projections of P and Q on the same axis, that is, R cos 1^ = P cos t« + § cos |3, or, multiplying through by OG, we have 72 (0(7 cos ^) = P(00'cos«) = §(00 008/?); but, OG cos = dr, 00 cos «s = rfjo, and 00 cos (i =: 6q; hence, ^ (28) ''-■^- w In using these formulas for finding the co-ordinates of the centre of gravity we take the axes of co-ordinates so as to get the simplest results (Art. 36) ; we also take the integrals be- tween such limits as to include the entire magnitude. Centre of Gravity of a Straight Line. 38. Let AB be a straight line whose length is I. Assume the direction of AB as the axis of X, A as the origin, and let any dis- tance from A, ss ADjhe denoted by x; then will dm be equal to dx. Substituting in (38) (which is the Fjg. i only formula required) and taking the integral between the limits and I, we have, I xdx ^i = '4 = ¥ (31) Hence, the centre of gramty is at the middle point of the line, as was previously shown. 46 MECHAlSriCS. [32. Centre of Gravity of a Triangle. 39. Let ABO he a plane triangle ; draw a line from the vertex A to ^the middle of the base BC, and take this as the axis of X; and let A be the origin. ^^ is a line of sym- metry, and consequently it will pass through the centre of grav- ity. At a distance from A equal to X draw BS parallel to BC, and at an additional distance dx draw a second parallel. The area between these consecutive parallels is equal to dm. Now, if we denote BG hy h and AFhy h, we have, from the similar triangle ADE and ABC, Fig. 31. h : b :: X : DE, or DE = j-x. h The area of dm is therefore equal to h X X dx sin a, a being the angle AFC. Substituting this in (38), dropping the constant factor, and integrating between the limits and li, we have / x^dx r xdx (32) Hence, the centre of gravity of a triangle is on a line drawn from the vertex to the middle of the base and two thirds of the distance from the vertex to the base. 33.] CENTRE OF GBAVITT ANT) STABILITY. 47 Centre of Gravity of a Pjrramid or Cone. 40. Let ABODE be any pyramid and let AG he & line from its vertex to the centre of gravity of its base. Take AO &a the axis of X, and denote the area of the base by h; also denote AG by A. Now, we may regard the pyramid as made up of slices formed by ^' planes parallel to the base, and whose distances apart, esti- mated along AG, are equal to dx; the actual thickness of each slice is dx x sin a, a being the angle between A G and the base. It is obvious from the principles of Geometry that A G will pass through the centre of gravity of each slice ; hence, the centre of gravity of the pyramid will be on A G. At a distance from A denoted by x pass a plane parallel to the base, and at an additional distance dx pass a second parallel plane ; the volume included by these planes will be dm. Denoting the section of the first of these planes by b', we shall have I -.V -.-.W : x\ or 5' = \a?. Hence, the magnitude of the corresponding slice is equal to jx'dx sin a. Substituting this for dm in (38), striking out the common constant factor, and integrating from to h, we have ph I x^dx / x^dx Jo Hence, the centre of gravity of a piframid is on a 48 MECHANICS. [34- line drawn from the vertex to the centre of gravity of the base, and at three fourths of the distance from the vertex to the base. The demonstration is applicable to any pyramid or cone. Centre of Gravity of an Arc of a Circle. 41, Let ABG be an arc whose radius is r, and let the origin be at the centre 0. Take the axis of X perpen- dicular to the chord A G, in which case it will be a line of symmetry. Denote the semi-angle of the arc by a and any variable angle, BOK, by 0. Now, if we give to B the constant increment dQ, Kg. 33. the difiEerential of the arc, or dm, will be equal to rdQ and its lever arm OL, or x, will be equal to r cos Q. Substituting in (38) and integrating from = — « to fl = -f a, we have fy ^03 Odd g^,^.^^ /: rdO ^^" (34) But ^r sin a = AC, and 2ra — arc ABO; hence, X, = 7-kty. or arc ABC : AC :: r : x,. arc ABC That is, the centre of gravity of an arc of a circle is on the diameter which bisects its chord, and its dis- tance from the centre is a fourth proportional to the arc, chard, and radius. When a = ^tt, or 90°, the are becomes a semi-circumfer- ence, sin a = 1, and equation (34) gives Xy = — . 35.] CENTKB or GRAVITY AND STABILITY. 49 Centre of Gravity of one Branch of a Cycloidal Arc. ' 43. Let ODX be one branch of a eycloid whose base is OX and whose axis is OD. Let the axis of X coincide with the base and let the axis of Y be paral- lel to the axis of the curve ; then, (Calculus, p. 100), will CD be a line of symmetry. Any element K of the curve will be equal to ^fS^ + dy^, or substituting for dx, its value (Gale, p. 100) and reducing, we have dm = a/^t (3r — «/)-* dy. Substituting in (29), suppressing the common constant factor and integrating from y = to y ^^ 2r, we have Vi J^i^r-yyUy |r- = f (2r-) (35) NoVe.— The numerator is reduced by Formula A, and the integration is completed by formula [29] Calculus, giving N=l(^-yfy-ir(^r-y)K The denominator is integrated by formula [39], giving i? = -3(3r-y)*. For y =r 0, iV" = - |r -v/Sr. and D = - 3 v^ ; for y = 3r both iVand D are : hence, the result given in (35). The value of yy is the distance of the centre of gravity of OKD from the axis of X, but by reason of symmetry it is also the distance of the centre of gravity of XE'D from the same axis. 50 MECHANICS. [36. Hence, the centre of gravity of the entire curve is on the axis of the curve and at a distance from the base equal to f GZ). Centre of Gravity of a Sector of a Circle. y 43. Let -40(7 be the sector; let the axis of J' bisect it and assume the same notation as in Art. 41. In this case dm is equal to the elementary sector OXL and its value is ^rMO : because this sector may be regarded as a triangle, its centre of gravity is two thirds of the distance from to JT, and consequently its lever arm x is equal to fr cos 6. Substi- tuting in (28), suppressing the com- mon constant factor, and integrating from — B to -|- «, we have Fig. 35. X, = ./—a cos 6 de |r- (36) If a = |-7r (or 90°), the sector becomes a semicircle, sin « = 1, and we have ir If we make dm equal to the second differential of a magni- tude, the position of the centre of gravity is found by double integration. Centre of Gravity of a Sector of a Circular Ring. 44. A circular ring is bounded by two arcs of concentric circles and by two radii, as BOG'B'. Take the origin at the centre of the pirplg ; ^PRote the radius of the outer circle by 37.] CENTEE OF GBAVITT AND STABILITY. 51 r, that of the inner circle by r', the angle XOGhy a' and the angle XOB by a. Draw any radius between r and r', as OL, and denote the angle which it makes with the axis of X by B. If we increase d by dO, and draw the corresponding radius OL', the part of the ring between these radii will be its first differential. Now take any distance OK greater than r' and less than r and denote it by 2 : describe two arcs, one with the radius z and the other with the radius z + dz; the part of the first differential included between these arcs will be the second differential oiBOG'B'. Eegarding this as a rectangle, its length is zdd and its breadth is dz and its area, dhn, is equal to zdzd& ; furthermore, its leyer arm x is equal to z cos d. Substituting these values for dm and x in (28), and taking the double integral between the limits indicated, we have a;, = pa. pr J a' Jt' zHz cos ddd r r i^dzdd Ja' Jr' 2r^—i 2 r^_--r^ y^ cos QdQ Jo! sm a — sin « 3 r» — r'« (37) Because the centre of gravity is on the bisecting line of the sector, the value of a^i is sufficient to determine it. If we make «' = — \'n, « = ^tt and r' = 0, we have as before, 4r ^1 - Btt" Centre of Gravity of a Semi-Ellipsoid. 45. Let the axis of X be taken to coincide with the axis of the ellipsoid, the origin being at the centre, and suppose 52 MECHANICS. [38. the volume to be generated by reyolving a semi-ellipse around its transverse axis. The elementary volume will be a slice perpendicular to the axis of X, whose radius is y and whose thickness is dx. The value of y^ taken from the equa- tion of the ellipse is —^(c? — oP) ; hence, dm, = TtyMx = TT -J (p? — a;') dx. Substituting this in (28), striking out the common constant factor, and integrating from a; = to a; = a, we have. / (a^ — a;2) x dx / (fl^ — a:^) dx X, = = *a (38) That is, the centre of gravity of a semi-prolate sphe- roid is on its axis and at a distance from the centre equal to three-eighths of the transverse axis of the me- ridian curve. Centre of Gravity of a Trirectangnlar Spherical Pyramid. 46. Let be the centre of sphere whose radius is r, and let A 00, AOB, and BOO be three rectangular planes. The volume bounded by these planes and the trirectangnlar spherical triangle, ABO, is the pyramid in question. Let A OD be any angle (p, and through OD pass a plane, DOO, cutting the base of the pyramid in the quadrant, OD ; draw OE, making the angle DOE equal to d, and through 0-E'pass Fig. 87. 39.] CENTRE OF GKAVITY AND STABILITY. 53 the plane EOC; the semi- wedge hetween these planes is the first differential of the pyramid. Draw a radius, OF, making the angle BOF equal to 0, and also the consecutive radius ON, making FON equal to dB ; then draw arcs of small circles through F and N, cutting the arc CE in K and L ; the elementary pyramid determined by the radii to F, N, K, and L, is the second differential of the pyramid. The arc FK is equal to ED cos = rd4> X cos d, and NF is equal to rdd ; hence, the area of the base of the elementary pyramid, regarded as a rectangle, is r^ cos 6 dO d, and because its altitude is r, the volume of the pyramid is given by the equation, d^ = \r^ cos 6 dO d^. The centre of gravity of the pyramid is at a distance from equal to three fourths of r ; hence, the co-ordinate x of this point is given by the equation, X = ^reos 6 cos ^. Substituting these values in (28), striking out the common constant factor, and taking the double integral between the limits of the figure, we have, / I '"' COS* d dO cos ^ d(f> / I CO8 dd d^ Note. — From example 8, p. 151, Calculus, we have at once, fco&^ edd = i (cos sin e + e). From the principle of symmetry we see that y^ and z^ are each equal to |r. The centre of gravity is on the radius that is equally inclined to OA, OB, and 00, and its distance from is equal to Va>i' + 2/1* + Zi', or to f r ^. 54 MECHAKICS. [40- The Centrobaric llethod. 47. If we clear equation (29) of fractions, and multiply both members by ^tt, we have, 27r^i fdm = /'^■nydm (a) First, let m be a plane curve ; then will dm be equal to '^dx^ H- dy'^, and (a) becomes, after limiting the curve by the ordinates a and h, 2ny^ r (da? + dy^)^ = f* '^■ny {da? + dy^)^ (40) Eeferring to the Calculus, pp. 91-92, we see that the first member of this equation is 27ryi multiplied by the length of a definite portion of a plane curve ; and that the second member is the surface generated by revolving that portion of the curve around the axis of X. Hence, the area of a surface of revolution is equal to the length of its generating arc multiplied by the cir- cumference described by its centre of gravity. Secondly, let m be a plane area ; then will dm = ydx, and the ydm of (29) will equal ^y x ydx, and we shall have, as before, 27r«/i / ydx = / "Ky^dx (41) The first member of this equation is 27ryj multiplied by a definite plane area, and the second member is the volume generated by revolving this area around the axis of X. Hence, the volume of a solid of revolution is equal to its generating area multiplied by the circumference, described by its centre of gravity. These principles may be used for finding the measure of 4a.] CBNTEE OF GRAVITY AND STABILITY. 55 surfaces and solids of reyolution, or, in certain cases, for finding the centres of gravity of their generating magnitudes. We are only concerned with the latter. To illustrate the manner of applying the rules just deduced, let us take the following examples : 1°. Centre of gravity of a semicircular arc. — The surface generated by revolving it about its diameter is ^ni^, and the length of the semicircle is wr ; hence, from (40), %r inr^ := 277^1 X T^r, or y^ = — (43) 2°. Centre of gravity of a cycloidal arc. — The surface gen- erated by revolving one branch of the arc around its base is 64 (Calc, p. 167), -5-Trr^, and the length of the branch is (Oalc, p. 158), 8r ; hence, from (40), ^TTrS = 277^1 X 8r, or yi = ^r = |(2r) (43) 3°. Centre of gravity of the area of a semicircle. — The vol- . 4 ume generated by revolving it around its diameter is ^wr*, and the area of the semicircle is ^tti^ ; hence, from (40), Irrrs = 27r«/i X ^nr^ or «^i = 3^ (44) 4°. Centre of gravity of the area of one branch of a cycloid. — The volume generated by revolving it around its base is (Calc, p. 170), 5TrV, and the area of the branch is (Calc, p. 163), Srrr*; hence, from (40), 5 5 hn^r^ = 27iyi X S^r*, or yi = -^r = -^{^r). Note.— The centre of gravity is on tlie line of symmetry in each case. 56 MECHANICS. Experimental Determination of the Centre of Gravity. 48. In certain cases the approximate position of the centre of gravity of a body may be found with sufficient accuracy as follows : Attach a string at any point G, and suspend the body by it ; when the body comes to rest, mark the direction of the string ; then suspend the body by a second point, B, and when it comes to rest, mark the direction of the string ; the point of intersection, G, p. will be the centre of gravity of the body. Instead of suspending the body by a string, it may be bal- anced on a point. In this case, the weight acts vertically downward, and is resisted by the reaction of the point ; hence, the centre of gravity lies vertically over the point. If, therefore, a body be balanced at any two points of its surface, and verticals be drawn through the points in these positions, their intersection will be the centre of gravity of the body. Centre of Gravity by Composition. 49. When we know the centres of gravity of two or more bodies, or parts of the same body, and the weights of each, the centre of gravity of the whole may be found by the method of composition ; thus, if A and B are the centres of gravity of two bodies whose weights are W and W, the weights being parallel forces, their resultant, W + W, will be applied at that point oi AB which divides it into seg- ments that are inversely as the forces. If, for example, the weight of ^ is 9 lbs., and that oi B 1 lbs., we divide AB into 16 equal parts, and lay off 7 of these from A towards B, which will determine the centre of gravity of A and B. CENTEE OF GRAVIXY AND STABILITY. 57 K the two bodies are homogeneous^ we may use their magnitudes instead of their weights. Having found the common centre of gravity of A and B, we may compound it with C in the same manner, and so on till all the bodies have been considered. As an example, let it be required to And the centre of gravity of a polygon as A, i?/ C, B, E. Divide it into triangles, and find the centre of gravity of each triangle. The weights of these triangles are proportional to their areas, and may be represented by them. Let 0, 0', 0", be the centres of gravity of the triangles into which the ^' ^' polygon is divided. Join and 0', and find a. point 0'", such that 0' 0'" : 00'" :: ABC : AOD ; then will 0'" be the centre of gravity of the triangles ABG and ^ CD. Join 0" and 0'", and find a point, G, such that 0"'G : 0"G :: ADE : ABG + AGD ; then will G be the centre of gravity of the polygon. To find the centre of gravity of a polyhedron ; if we take a point within and join it with each vertex of the polyhedron, we shall form as many pyramids as the solid has faces ; the centre of gravity of each pyramid may be found by the rule. If the centres of gravity of the first and second pyramid be joined by a straight line, the common centre of gravity of the two may be found by a process similar to that used in finding the centre of gravity of a polygon, observing that the weights of the pyramids are proportional to their volumes, and may be represented by them. Having compounded the weights 58 MECHANICS. of the first and second, and found its point of application, we may, in like manner, compound the weight of these two with that of the third, and so on ; the last point of applica- tion will be the centre of gravity of the polyhedron. Composition by the Method of Moments. 50. When we have several bodies, and it is required to find their common centre of gravity, it will often be found convenient to employ the principle of moments. To do this, we first find the centre of gravity of each body separately, by rules already given. The weight of each body is then regarded as a force, applied at the centre of gravity of the body. The weights being parallel, we have a system of parallel forces, whose points of application are known. If these points are all in the same plane, we find the lever arms of the resultant of all the weights, with respect to two lines, at right angles to each other in that plane ; and these will make known the point of application of the resultant, or, what is the same thing, the centre of gravity of the system. If the points are not in the same plane, the lever arms of the resultant are found, with respect to three axes, at right angles to each other ; these make known the point of appli- cation of the resultant weight, or the position of the centre of gravity. The methods of applying this and the preceding article are illustrated by the following : Examples. 1. Required the point of application of the resultant of three equal weights, applied at the vertices of a plane triangle. Solution. — ^Let A, B, and C be the vertices of the triangle and Z) the middle of BG. The resultant of the weights at B and C will be applied at -D. The resultant of the three weights will be applied at a point O of AI), and because the weight at D is twice that at A the distance AQ'm CENTEB OF GRAVITY AND STABILITY. 59 f?wo thirds of AD, that is, the required point is at the centre of gravity of the triangle. 2. Required the point of application of the resultant of a system of equal parallel forces, applied at the vertices of a regular polygon. Ans. At the centre of the polygon. 3. Parallel forces of 3, 4, 5, and 6 lbs. are applied at the successive vertices of a square, whose side is 13 inches. At what distance from the first vertex is the point of application of their resultant? Solution. — Take the sides of the square through the first vertex as axes ; call the side through the first and second vertex, the axis of X and that through the first and fourth, the axis of Y. We shall have, from Formulas (28, 29), 4 X 13 + 5 X 12 - x, = jg =6; , 6 X 13 + 5 X 12 22 and J,,.:. jg- =-3. Denoting the required distance by d, we have d = Vx^^ + 3/1" = 9.475 in. Ans. , 4. Seven equal forces are applied at seven of the vertices of a cube. What is the distance of the point of application of their resultant from the eighth vertex? Solution. — Take the eighth vertex as the origin of co-ordinates, and the three edges passing through it as axes. We shall have, from Equa- tions (38, 29, 30), denoting one edge of the cube by a, x^ = ^a, 2/1 = ja, and z^ = ^a. Denoting the required distance by d, we have d = Voii* + yi' + Zi* = ^a Vs. Ans. 5. Two isosceles triangles are constructed on opposite sides of the base 6, having altitudes equal to h and h', h being greater than h'. Where is the centre of gravity of the space within the two triangles? Solution. — It must lie on the altitude of the greater triangle. Take the common base as an axis of moments ; then will the moments of the triangles be ^bh x ^h, and y>h' x ^7i' ; and we have uih'^-n'^) ,,, ,,, 60 MEOHAKICS. That is, the centre of gravity is on the altitude of the greater triangle, at a distance from the base equal to one third of the difference of the two altitudes. 6. Where is the centre of gravity of the space between two circles tangent to each other internally? Solution. — Take their common tangent as an axis of moments. The centre of gravity will lie on the common normal, and its distance from the point of contact is given by the equation, Trr* — vrr" r^ +rr' + r'^ 1 n-r* — Trr" r + r' ' 7. Let there be a square, divided by its diagonals into four equal parts, one of which is removed. Required the distance of the centre of gravity of the remaining figure from the opposite side of the square. Ans. ^-g of the side of the square. 8. To construct a triangle, having given its base and centre of gravity. Solution. — Draw through the middle of the base, and the centre of gravity, a straight line ; lay off beyond the centre of gravity a distance equal to twice the distance from the middle of the base to the centre of gravity. The point thus found is the vertex. 9. Three men carry a cylindrical bar, one taking hold of one end, and the others at a common point. Required the position of this point, in order that the three may sustain equal portions of the weight. Ans. At three-fourths the length of the cylinder from the first. STABILITY A^T> EQUILIBRIUM:. Stable, TTnstable, and Indifferent Equilibrium. 51. A body is in stable equilibrium when, on being slightly disturbed from a state of rest, it has a tendency to return to that state. This will be the case when the centre of gravity of the body is at its lowest point. Let v\ A he a body suspended from an axis 0, about which it is free to turn. When the centre of gravity of A lies vertically below the axis, it is in equilibrium, for the weight of the body is exactly ^-■ counterbalanced by the resistance of the axis. Kg. 40. CBNTKE OF GRAVITY AND STABILITY. 61 e-. 5JS /P Moreover, the equilibrium is stalle ; for if the body be de- fleeted to A', its weight acts with the lever arm OP to restore it to its position of rest, A. A body is in unstable equilibrium when, on being slightly disturbed from its state of rest, it tends to depart still farther from it. This will be the case when the centre of gravity of the body occupies its highest position. Let ^ be a sphere, connected by an inflexible rod with the axis 0. "When the centre of gravity of A is verti- cally above 0, it is in unstable equilibrium ; for, if the sphere be deflected to the position A', its weight will act with the lever arm OP to increase the deflection. The motion continues till, after a few vibrations, it comes to rest below the axis. In this last position, it is in stable equilibrium. A body is in indifferent, or neutral, equilibrium when it remains at rest, wherever it may be placed. This is the case when the centre of gravity continues in the same horizontal plane on being slightly disturbed. Let ^ be a sphere, supported by a horizontal axis OP through its centre of gravity. Then, in whatever position it may be placed, it will have no tendency to change this posi- tion ; it is, therefore, in indifferent, or neutral equilibrium. oy~ Pig. 41. Kg. 48. Kg. 43. In figure 43, A, B, and G, represent a cone in positions of stable, unstable, and indifferent equilibrium. If a wheel be mounted on a horizontal axis, about which it is free to turn, the centre of gravity not lying on the axis, it 62 MECHANICS. will be in stable equilibrium, when the centre of gravity is directly below the axis ; and in unstable equilibrium when it is directly above the axis. When the axis passes through the centre of gravity, it will, in every position, be in neutral equilibrium. "We infer, from the preceding discussion, that when a body at rest is so situated that it cannot be disturbed from its posi- tion without raising its centre of gravity, it is in a state of stable equilibrium; when a slight disturbance depresses the centre of gravity, it is in a state of unstable equilibrium; when the centre of gravity remains constantly in the same horizontal plane, it is in a state of neutral equilibrium. This principle holds true in the combinations of wheels and other pieces used in machinery, and indicates the im- portance of balancing these elements, so that their centres of gravity may remain in the same horizontal planes. Stability of Bodies on a Horizontal Plane. 53. A body resting on a horizontal plane may touch it in one, or in more than one point. In the latter . case, the salient polygon, formed by joining the extreme points of contact, as abed, is called the polygon of sup- <. port. \\\ When the direction of the weight of . — / i \ V-; — 7 the body, that is, the vertical through / //"^< X/ / its centre of gravity, pierces the plane / ; — - — / within the polygon of support, the '^' body is stable, and will remain in equilibrium, unless acted upon by some other force than the weight of the body. In this case, the body will be most easily overturned about that side of the polygon of support which is nearest to the line of direction of the weight. The moment of the weight, with respect to this side, is called the moment of stability. Denoting the weight of the body by W, the distance from CEKTEE OF GRAVITY AND STABILITY. 63 its line of direction to the nearest side of the polygon of sup- port by r, and the moment of stability by S, we have, 8= Wr. The moment of stability is the moment of the least extra- neous force that is capable of oTerturning the body. The weight of a body remaining the same, its stability increases with r. If the polygon of support is a. regular polygon, the stability will be greatest, other things being equal, when the direction of the weight passes through its centre. The area of the polygon of support remaining constant, the stability will be greater as the polygon approaches a circle. When the direction of the weight passes without the poly- gon of support, the body is unstable, and unless supported by some other force than the weight, it will turn about the side nearest the direction of the weight. In this case, the product of the weight into the distance from its direction to the nearest side of the polygon, is called the moment of in- stability. The moment of instability is equal to the least moment of a force that can prevent the body from overturning. If. the direction of the weight intersect any side of the polygon of support, the body will be in a state of equilibrium bordering on rotation about that side. In what precedes, we supposed the supporting plane to be horizontal, and that the only force acting is the weight of the body. Stability when One Body Presses on Another. 53. When other forces than the weight are acting on a body, and when the plane of contact of the two bodies has any position, the conditions of stability require, first, that the resultant of all the forces shall pass within the polygon of 64 MECHANICS. %1 Fig. 45. support, to prevent rotation; and, secondly, that it shall be normal to the plane of contact, to prevent sliding. For example, let A be a movable body pressed against a fixed body B, and touching it at a single point, P'. In order that A may be in equilibrium, the resultant of all the forces acting on it, including its ■weight, must pass through the point of contact, P' ; otherwise there would be a tendency to rotation about P', which would be meas- ured by the moment of the resultant with respect to this point. Furthermore, the direction of the resultant must be normal to the surface of B at the point P', else the body A would have a tendency to slide along the body B, which tendency would be meas- ured by the tangential component. The pressure on B de- velops a force of reaction, which is equal and directly opposed to it. Practical Problems. 1. A horizontal beam AB, which sustains a load, is supported by a pivot at A, and by a cord BE, the point E being vertically over A Required the tension of DE, and the vertical pressure on A. Solution.— Denote the weight of the beam and load by W, and suppose its point of application to be G. Denote GA byp, the perpendicular distance, AF, from A to DE, by y , and the tension of the cord by t. If we take ^ as a centre of moments, we have, in case of equilibrium, Wp = tp' t: wE- y Or, denoting the angle EDA by a, and the distance AD by b, we have, Fig. 46. ' = J sin o ; W-, P b sina' CENTEE OF GRAVITY AND STABILITY. 65 To find the vertical pressure on A, resolve t into components, parallel and perpendicular to AB. We have for the latter component, denoted t' = tsma= W^ of The vertical pressure on A, plus the weight W, must be equal to t'. Denoting the vertical pressure by P, we have, p + .W=W^; or,P=l^(|.-l)=Tr(^); P= W^. When DG=0 ; or, when Z) and C coincide, the vertical pressure is 0. 2. A rope, AD, supports a pole, DO, one end of which rests on a horizontal plane, and from the 3) other is suspended a weight W. Required the ^^ tension of the rope, and the thrust, or pressure, on ^^^"^B'ir the pole, its weight being neglected. ^^ l^'' w Solution. — Denote the tension of the rope by t, ^ ' the pressure on the pole by p, the angle ADO by a, j,j ._ and the angle ODTT by /3. There are three forces acting at D, which hold each other in equilib- rium ; the weight W, acting downward, the tension of the rope, acting from D, toward A, and the reaction of the pole, acting from toward D. Lay oflE Dd, to represent the weight, and complete the parallelo- gram daoD ; then will Da represent the tension of the rope, and Do the thrust on the pole. From Eq. 16, we have, ' <: TT:: sin /?: Sinn; .: t = W^^. sm a We have, also, from the same principle, JO : W : : sin (o + /J) : sin n ; .-. p= W -. -. ^ \ I-/ J- sva.a If the rope is horizontal, we have a = 90° — (3, which gives, W i = Wtan/3, and p = -„. ^ cos p 3. A beam I'B, is suspended by ropes attached at its extremities, and fastened to pins A and //. Required the tensions of the ropes. 66 MECHANICS. Solution.— Denote the weight oi the beam and its load by W, and let c be its point of application. Denote the tension of the rope BS, by t, and that of FA by i'. The forces in equilibrium, are W, t and t'. The plane of these forces must be vertical, and further, the directions of the forces must intersect in a point. Produce AJF and BH, till they intersect in K, and draw Kc ; take Kc, to represent the TV^eight of the beam and its load, and complete the parallelogram Kbcf ; then wUl Kb represent t, and Kf wUl represent t'. Denote the angle eKB by a, and cKF by ^.• We shall hare, as in the last problem. And, TT : < : : sin (o + |8) : sin ;8 ; W : V : : sin (a + /3) : sin a ; t= Tf sin /J sin (a + /3)' V= W sin (a + |8)' 4 A gate AS, is supported at on a pivot, and at J. by a hinge, attached to a post AB. Required the pressure on the pivot, and the tension of the hinge. Solution.— Denote the weight of the gate and its load by W, and let G be its point of application. Produce the vertical through C, tiU it intersects the horizontal through A in D, and draw DO. Then wUl AD and DO be the directions of the required components of W. Lay off Dc, to represent W, and complete the parallelogram, Dcoa ; then will Do represent the pressure on 0, and aD the tension on the hinge, A. Denoting the angle oDc by a, the pressure on the pivot by ^, and on the hinge by^, we have, W P- and p' =psina. If we denote OJEJ by 6, and DE by h, we shall have, Hence, h i^V> +A« :, and sin o = ■ Vi^+h' p = T , and p' = — i— Vjs + h' CENTRE OF GEAVITY AND STABILITY. 67 B 3 Pig. 50. 5. Having two rafters, AG and BC, abutting in notches of a tie-beam AB, it is required to find the pressure, or thrust, on the rafters, and the direc- tion and intensity of the pressure on the joints at the tie-beam. Solution. — Denote the weight of the rafters and their load by 3m) ; we may regard this weight as made up of three parts — a weight w, applied at C, and two equal weights ^w, applied at A and .B respectively. De- note the half span AL by s, the rise CL by h, and the length of the rafter AG oi CB by I. Denote, also, the angle GBL by o, the thrust on each rafter by t, and the resultant pressure at each of the joints A and -Bby^. Lay ofE Go to represent the weight w, and complete the parallelogram Gboa ; then will Ga and Gh represent the thrust on the rafters ; and, since Ghoa is a rhombus, we have. t = w 2 sin a wl '2A' Conceive t to be applied at A, and there resolve it into components parallel to GL and LA ; we have, for these components, I sm a = ^, and r cos a = ■—• The latter component gives the strain on the tie-beam, AB. To find the pressure on the joint, we have, acting downward, the forces ^w and |w, or the single force w, and, acting from L toward A, the force ^ ; hence. If we denote the angle DAE by ft we have from the right-angled triangle DAE, DE ws s The joint should be perpendicular to the force p, that is, it should make with the horizon an angle whose tangent is ^. 68 MECHANICS. 6. In the Jast problem suppose the rafters to abut against the wall. Required the least thickness that must be given to it to prevent it from being overturned. Solution. — Denote the weight thrown on the wall by w, the length of wall that sustains the pressure p by V, its height by h', its thickness by X, and the weight of each cubic foot of wall by ««' ; then will the weight of this part be w'h'l'x. The force ^ acts with an arm of lever h' to overturn the wall about its lower and outer edge ; this force is resisted by the weight w+w'h I'x, acting through the centre of gravity of .the wall with a lever arm equal to \x. If there be an equilibrium, the moments of these two forces are equal, that is, *"« -L, , >i/,i >« i«8h' ,,,„ g^ X K =z(w + w hTx)g, or —=-— = wx + v/h'Vx'. Reducing, we have, x" + —rm, ^ = -ttttt ; vih'V v/l'h 7. A sustaining wall has a cross section in the form of a trapezoid, the face on which the pressure is thrown being vertical, and the opposite face having a slope of six perpendicular to om horizontal. Required the least thickness that must be given to the wall at top, that it may not be overturned by a horizontal pressure, whose point of application is at a, distance from the I bottom of the wall equal to one third its height. / D EI" G Pig. 51. Solution. — Pass a plane through the edge A parallel to BC, and consider a portion of the wall whose length is one foot. Denote the pressure on this by P, the height of the wall by 6A, its thickness at top by x, and the weight of a cubic foot by w. Let fall from the centres of gravity and 0' of the two portions, perpendiculars OG and 0'^, and take the edge Z) as an axis of moments. The weight of the portion ABGF is equal to %whx, and its lever arm, DO-, is equal to A + \x. The weight of the portion ADF is Zwh^, and its lever arm, BE, is f A. In case of equilibrium. CENTEE OF GEATITT AND STABILITY. 69 the sum of the moments of their weights must be equal to the moment of P, whose lever arm is 2h. Hence, 6whx(h + ^) + Swh' X ih = Px2h; or, 6whx + 3wx' + 2wh^ = 2P. Whence, x' + %hx : '6w ■ = -7i±\/ a(P - wK^ '6'w + h'>. 8. Required the conditions of stability of a square pillar acted on by a force oblique to the axis, and applied at the centre of gravity of the upper base. Solution. — Denote the intensity of the force by P, its inclination to the vertical by a, the breadth of the pillar by 2a, its height by x, and its weight by W. Through the centre of gravity of the pUlar draw a vertical AO, and lay off ^C equal to W ; prolong PA and lay off AS equal to P ; complete the parallelogram ABDC, and prolong the diagonal till it intersects EG- at F. If F is between H and Cr the pillar will be stable ; if at ^, it will be indiif erent ; if beyond H, it will be unstable. To find an expression for FCf, draw BE perpendicular to A(jf. From the similar triangles ADF and AFG, we have, BE-.FG; - ^"^^^ ^ // i/ / ■//a AE-.AG F& = -- AE SS. G Hs. 52. But AO = X, DE — Psin a, and AE = TT + Pcos a, fcence, we have. FG: Px sin a W + Fcosa' And, since SG equals a, we have the following conditions for sta- bility, indifference, and instability, respectively : Pc sin a a > a < W + Pcosa Px sin a W+ Pcosa' Px sin a W+ Poos a IV.— ELEMENTAEY MACHINES. Definitions and General Principles. 54. A machme is a contrivance by means of which a force applied at one point is made to produce an efEect at some other point. The applied force is called the power, and the force to be overcome the resistance ; the source of the power is called the motor. Some of the more common motors are muscular effort, as exhibited by man and beast in various kinds of work ; the weight and kinetic energy of water, as shown in various kinds of water-mills ; the expansive force of vapors and gases, as displayed in steam and caloric engines ; the force of air in motion, as exhibited in the windmill, and in the propulsion of sailing vessels ; the force of magnetic attraction and repul- sion, as shown in the magnetic telegraph and other magnetic machines ; the elastic force of springs, as shown in watches and various ^ther machines. Of these the most important are steam and water power. Applied and ITseful Work.— Modulus. 55. Machines simply transmit and modify the action of forces. They add nothing to the work of the motor ; on the contrary, they absorb and render inefficient much of the force which is impressed on them. Of the applied work, a part is expended in overcoming friction, stiffness of cords, bands or chains, resistance of the air, and adhesion of the parts. This goes ta wear out the machine. A second portion is expended in overcoming ELEMEHTAKT MACHIKES. 71 shocks, arising from the nature of the work to be accom- plished, as well as from imperfect connection of the parts, and from want of hardness and elasticity in the connecting pieces. This also goes to strain and wear out the machine, and to increase the waste already mentioned. In any machine the quotient obtained by dividing the quantity of useful, or effective work, by the quantity of ap- plied work, is called the modulus of the machine. As the resistances are diminished, the modulus increases, and the machine becomes more perfect. Trains of Mechanism. 56. A machine usually consists of an assemblage of mov- ing pieces called elements, kept in position by a con- nected system called a fram.e. Of the moving pieces, that which receives the power is called the recipient or prime mover, that which performs the work, is called the operator or tool, and the connecting pieces constitute what is called a train of mechanism. Of two consecu- tive elements, that which imparts motion is called a driver, and that which receives motion is called a follo"wer. Each piece, except the extremes, is a follower, with respect to that which precedes, and a driver, with respect to that which follows. In studying a train of mechanism we first find the relation between the power and resistance for each element neglecting hurtful resistances. We then modify these results so as to take account of all resistances, such as friction, adhesion, stiffness of cords, and atmospheric resistance. Having found the relation between the power and resistance for each piece, we begin at one extreme and combine them, recollecting, that the resistance with respect to each driver is equal to the power with respect to its follower. We might also find the modulus of each element, and take 73 MECHANICS. the product of these partial moduli as the modulus of the entire machine. The Mechanical Powers. 57. The elements to which all machines can be reduced, are sometimes called mechanical povrers. They are seven in number — viz., the cord, the lever, the inclined plane, the pulley, the wheel and axle, the screw, and the wedge. The first three are simple elements ; the pulley, and the wheel and axle, are combinations of the cord and lever ; the screw is a combination of two inclined planes twisted round an axle ; and the wedge is a simple combina- tion of two inclined planes. In finding the relation between the power and resistance we shall, in the first instance, not only neglect all hurtful resistances, but we shall also suppose cords, levers, and con- necting links, to be destitute of weight. The Cord. 58. Let AB be a cord solicited by two forces, P and E, applied at its extremities, A and B. In order that the cord may be in equilibrium, it is evident, in the first place, that the forces ^^ J J ^^ must act in the direction of the Mg. 63. cord, and in such manner as to stretch it, otherwise the cord would bend ; and in the second place, the forces must be equal, otherwise the greater would prevail, and motion would ensue. Hence, if two forces ap- plied at the extremities of a cord are in equilibrium, tJie forces are equal and directly opposed. Let ABhe a cord solicited by groups of forces applied at its extremities. In order that these forces may be in equilibrium, the resultants of the groups at A and B must be equal and directly opposed. -mg. m. ELEMEKTARY MACHINES. 73 Let ABCBhea. cord, at the points A, B, C, D, of which groups of forces are applied. If these forces are in equilib- rium through the intervention of the cord, there must neces- sarily be an equilibrium at each point, and this, whatever may be the lengths of AB,BO, and CD. If we make these " Pig. is. infinitely small, the equilib- rium will still subsist. But in that case the points A, B, C, and B, will coincide, 'and all the forces will be applied at a single point. Hence, we conclude, that a system, of forces applied in any manner at points of a cord will he in equilihrium, when, if applied at a single point without change of intensity or direction, they will maintain each other in equilibrium. Prom what precedes, we see that the function of a cord in mechanism is simply to transmit forces, without modifying them in any manner. The tension of a cord is the force by which two of its ad- jacent parts are urged to separate. If a cord be solicited in opposite directions by equal forces, its tension is measured by either force. If the forces are unequal, the tension is meas- ured by the less. The Lever. 59. A lever is an inflexible bar, free to turn around an axis, called the falcnim. Levers are sometimes divided into three classes, according to the relative positions of the power, resistance, and fulcrum. In the first class the fulcrum is between the power and resistance, as in the ordinary balance ; in the second class the resistance is between the power and the fulcrum, as in the ordinary nut cracker ; in the third class the power is 74 MECHANICS. [45 between the resistance and tlie fulcrum. Strictly speaking, the last two are only varieties of a single class. 1st Class. an Ci-asb, 3d Class. y Fig. 56. Fig. m. v! Fig. B8. Levers may be curved, or straight ; and the power and re- sistance may be either parallel or oblique to each other. We shall suppose the power and resistance to be perpendicular to the fulcrum ; otherwise we might conceive each to be resolved into two components, one perpendicular, and the other par- allel, to this axis. The latter would tend to make the lever slide along the axis, developing hurtful resistance, while the former alone would tend to turn the lever about the fulcrum. The perpendicular distances from the fulcrum to the lines of direction of the power and resistance, are called lever arms. In the bent lever MFN^ the perpendicular distances FA, and FB, are the lever arms of P and R. To determine the conditions of equilibrium of the lever, denote the a J^ power by P, the resistance by R, and their lever arms by p and r. We have the case of a body restrained by an axis, and if we take this as the axis of moments, we shall have for the condition of equilibrium (Art. 31), Pp = Rr; or, P : R : : r : p That is, the power is to the resistance, as the lever arm of the resistance, ig to th& hver c^rtv, q/ the power. Fig. 59. (45) 46.] ELEMENTAKT MACHIKES. 75 This relation holds good for every kind of lever. If P and B intersect, their resultant will pass through the point of intersection, and also through F ; its value may be found by the parallelogram of forces.' If several forces act upon a lever at different points, all being perpendicular to the direction of the fulcrum, they will be in equilibrium, when the algebraic sum of their moments, with respect to the fulcrum, is equal to 0. This principle enables us to take into account the weight of the lever, which is to be regarded as a vertical force through the centre of gravity of the lever. MBCHAifiCAL Advantage. — In. any machine the quotient of the resistance by the power is called the mechanical ad- vantage. Thus, in the case above considered, we have, for the mechanical advantage, |=f («) When By- F there is said to be a gaiti of mechanical advan- tage, and when R <_ F there is a loss. Because there is an equilibrium between B and P, their elementary quantities of work must be equal, from the theorem of work (Eq. 35) ; hence, if By- F, we have, 6r < Sp, and if i? < P, we have, dr > 6p. These principles apply to every case in which there is an equilibrium between the power and resistance. Friction. 60. Friction is the resistance that one body experiences in moving on another when the two bodies are pressed to- gether by a force that is normal to both. This resistance arises from inequalities in the surfaces of the bodies, the projections of the one falling into the depres- sions of the other. To overcome the resistance, a sufficient force must be applied either to break off or bend down the 76 MECHAKICS. projecting points, or else to drag the body over them ; this force must be equal and directly opposed to for,ce of friction, which acts tangentially to the two surfaces. Between certain bodies, friction is somewhat different when motion is just beginning, from what it is when motion has been established. The friction developed when a body is passing from a state of rest to a state of motion, is called friction of quiescence ; that between bodies in motion, is called friction of motion. The following laws of firiction have been established by experiment, viz. : First, friction of quiescence between the same bodies is proportional to the normal pressure, and independent of the extent of the surfaces in contact. Secondly, friction of motion between the same bodies, is proportional to the normal pressure, and independent, both of the extent of surface of contact, and of the ve- locity of the moving body. Thirdly, for compressJhle bodies, friction of quiescence is greater than friction of motion: for bodies which are incompressJMe, the difference is scarcely appreciable. Friction may be diminished by the interposition of un- guents, which fill up the cavities, and so diminish the rough- ness of the rubbing surfaces. For slow motions and great pressures, the more substantial unguents are used, such as lard, tallow, and certain mixtures ; for rapid motions, and light pressures, oils are generally employed. Methods of finding the Coefficient of Friction. 61. The quotient obtained by dividing the force of friction by the normal pressure, is called the coefficient of friction ; its value for any two substances, may be determined as follows : 4T.] ELBMENTAET MACHINES. 77 Let AB be a horizontal plane formed of one of the sub- stances, and a cubical block of the other. Attach a string, OC, to the block, so that its di- rection shall pass through the centre of gravity, and be parallel Ar— to AB ; let the string pass over a Si ■^c> fixed pulley, C, and let a weight, p F, be attached to its extremity. DP Increase F till just begins to slide along the plane, then will F be the force of friction. Denote the normal pressure by P, and the coefficient of fric- tion by/. From the definition, we have, /=! (47) In this manner, values for / may be found for different substances, and arranged in tables. The value of /, for any substance, is its coefficient of fric- tion. Hence, we may define the coefficient of friction to be the friction due to a normal pressure of one pound. Having the normal pressure in pounds, and the coefficient of friction, the entire fription may be found by multiplying these quantities together. * There is a second method of finding the value of /, as. follows : Let AB be an inclined plane, formed of one of the sub- stances, and a block, of the other. Elevate the plane till the block just begins to slide 'down by its own weight. Denote the incli- nation, at this instant, by a, and the weight of 0, by W. Eesolve W into two components, one normal to the plane, and the other parallel to it. Fig.ei. Denote the former by P, and the latter 78 MECHAKICS. [*8. by Q. Since OF" is perpendiculai' to AG, and OP to AB, the angle, WOF, is equal to a. Hencej P = Wcoa a, and Q = IF sin a. The normal pressure being equal to TFcos a, and the force of friction being Wsm a, we shall have, from the principle already explained, , TTsina . ^ f = -^^ , or, f = tan ec (48) ' PT cos fit' ' ■' ^ ' The angle es is called the angle of Motion, The yalues of /for a few of the more common cases are given in the fol- lowing TABLE. Bodies between whU^flietUm takes place. Co^gident dffiic&oa. Iron on oak 63 Cast-iron on oak 49 Oak on oak, fibres parallel 48 Do., do., greased 10 Cast-iron on oast-iron 15 Wrought-iron on wronght-iron 14 Brass on iron .* 16 Brass on brass ' ^ Wrought-iron on cast-iron 19 Cast-iron on ebn 19 Soft limestone on the same 64 Hard limestone on the same 38 EcLnilibrium Bordering on Motion. 63. Let ^ be a movable body, resting on a fixed body B ; let P be the resultant of all the forces acting on A, when it is on the eve of sliding along B in the direction from D to C, and let E, equal and directly opposed to P, be the corresponding reaction ; that there may be no tendency to 4»-] ELEMENTAET MACHIlSrES. » 79 rotation, the direction of P must in- tersect the surface, DC, at some point within the polygon of support. At A, the point of application of P, draw AN, normal to DC. If ow, let P be resolved into two components, one in the direction of the normal, and the other in the di- rection of the tangent DO; if we denote the angle between P and the normal by 0, the former will be equal to P cos 0, and the latter to P sin 0. The normal component of P multiplied by / will be the measure of the entire friction between A and B, which is directed from G towards D ; the tangential component of P exactly balances the force of friction and is directed from D towards C; hence, we have Pcos X / = P sin 0, or / = tan 0, or = tan -y . . . (49) That is, when a body is on the eve of sliding along a second body, the reaction of the second body lies on the side of the normal opposite to the direction of in- cipient motion and makes an angle with it equal to the angle of friction. Method of Finding the Modnlns. 63. The modulus of a machine is the quotient of the ele- mentary quantity of work of the resistance by the elementary quantity of work of the power. Assuming the same notation as before and denoting the modulus by M, we have ,, RSr R dr ,_.. ^=P6-p-p''6-p (^^) When friction is considered, P will be a function of/, that is, it will depend upon / for its value ; let this value of P be 80 MECHANICS. [61. denoted by Pj \ it is obyious that the yalues of R, 6r, and (Jp will be entirely independent of /. When friction is not con- sidered, P is obviously what Pj becomes when/ = ; denote this value of P by P^ When P^ and 11 are in equilibrium, we have P„(5p = R8r, or -T- = 6r 6p R' Substituting in (50), we have R ^=%xt (51) Hence, we have the following rule for finding the modulus of a machine when friction is taken into account : Find the value of P in terms of R and f; in this result make f equal to 0; then divide the latter re- sult by the former. To Find the Modulus of a Lever. 64. Let AB represent a lever and suppose its fulcrum to be a solid cylinder turning in a cylindrical box ; also, suppose both lever and fulcrum to be horizontal. Denote the power by P, the resistance by R, the weight of the lever by W, and let the distances of these forces from the centre of the ful- crum be p, r, and w. If P, R, and W are vertical the re- sultant reaction OQ will pass through the point of contact and will be directed vertically upward. When P is on the eve of overcoming all the resistances, it will be equal to P, of the preceding article, the fulcrum Fig. 68. 62.] ELEMESTTAKT MACHINES. 81 will be on the eve of sliding in the direction OT, and the normal ON, which passes through the axis of the fulcrum, must be so situated as to make the angle NOQ, denoted by ^, equal to the angle of friction (Art. 63). Take as a centre of moments and denote the radius of the fulcrum by p ; the corresponding lever arm of P, will be p — psm (p, that of R will he r + p sin , and that of W will he w — p siiKJ) ; and because the forces are in equilibrium, we shall have, from Art. 31, P/{p — p sin (j>) -{-W{w — p sin 0) = R{r + p sin (ft), from which equation we find, p = ^ (^ + psin^) — W (w — psini^) . ' p — p sin • • • • \ ) Making/ = 0, which makes sin =i 0, we have Po = ^^ (,3) Substituting the values of Pq a-iid Pf in equation 51, we haye TLr Br—Ww p~psm .... p Ii(r + psuirp) — W(w — psuKp) f Note. — ^We have sin * = — from Trigonometry. Vl+/» Example.— Let i2 = 100 IBs., 1^= 5 lbs., r = 30 in., p = BO in., w = 15 in., p = 3 in., and sin = 3 (/ = .314 nearly): What is the modu- lus? Ans. ,96 nearly, that is, about 96% of the applied work becomes effective. Note. — In many cases the value of P/ may be determined by experi- ment and the value of Po may be found from the equation of equilibrium when friction is neglected. 83 MECHANICS. [55. The Componnd Lever. 65. A compound lever is a combination of simple levers AB, BG, CD, so arranged that the resistance in one acts as a power in the next, through- out the combination. Thus, a power P produces at 5 a re- sistance R', which, in turn, ■*- "R ' * !," P I i^! P^ B," 1 produces at C a resistance R", and so on. Let us assume the notation of the figure. From the principle of the simple ^'fr ^^ lever, we have the relations, Pp = R'r", R'p' = R"r', R"p" = Rr. Multiplying these equations, member by member, and striking out common factors, we have Ppp'p" = Rrr'r" ; or, P : R :: rr'r" : pp'p" (55) And similarly for any number of levers. Hence, in the compound lever, the power is to the resist- ance as the continued product of the alternate arms of lever, commencing at the resistance, is to the con- tinued product of the alternate arms of lever, com- mencing at the power. Modulus. — If p, pf, p', are each equal to 10 ; if r, r, r'' are each equal to 2 ; and if R equals 100 lbs., we have Po = Sib. If we find by experi- ment that a force of 1 lb. applied at A will' just overcome a resistance of 100 lbs. applied at D, we have P/ = 1 Z6. Then, to find the modulus, substitute these in (51), and we have M=~ = SQ% (56) that is, 80% of the applied force becomes effective. ST.] ELEMBNTAKY MACHIKES. 83 Fig. 65. The Elbow-Joint Press. 66. Let BD and BO represent equal bars, taving hinge- points at B, 0, and D, and let BE be a bar that works back and forth through a guide be- tween E and B\ Let P be the power applied at B, and per- pendicular to DO. When P acts to depress B, the link BO turns about O, and the force transmitted through the link BB causes the piece BE to move towards F, so as to compress a body placed between E and F. This machine is called the elbow-joint press, and is used in some kinds of printing, in moulding bullets, in stamping coins, in crushing stone, and in many other like kinds of work. To find the mechanical advantage, let us resolve P into two components in the direction of the links 5 C and ^Z*; these components are equal, and if we denote one of them by Q, and the angle, DBO, by ;3, we have, from (14), whence A/a(l + cosi9f)r If we denote the angle QBB by <*, the effective component of § is ^ cos a, and this is equa?! amd directly opposed to the resistance R ; hence, R _ cos «_ P (57) ^2(1 -f-cos/J) As B approaches the line BO, cos « approaches 1, and cos fi approaches — 1 ; at the limit cos « = 1, and cos = — 1, 84 MECHANICa. • and the mechamc^l advantage becomes infinite. It is to be noted that the elementary space through which the pressure is exerted varies inversely as the mechanical advantage ; hence, at the limit it is 0. To find the modulus of the reciprdcating piece DE : as- sume the figure in ■e; y which AB represents the part within the guide, being its cen- tre. In addition to the notation already adopted, let the length of J ^ be denoted by 21, its breadth by 2b, and the distance, OD, by d. The tendency of Q is to produce rotation, pressing the piece downward at A, and upward at JB ; when Q is on the eve of overcoming E, there will be a reaction S at A, and a reaction S' at £, each of which, from Art. 63, will lie on that side of the corresponding normal which is opposite to the direction of incipient motion> and in each case the inclina- tion to the normal will be equal to , the angle of friction. At the instant in question there will be an equilibrium be- tween Q, S, 8', and B ; consequently, the algebraic sum of their components in the direction of OD, and in the direction perpendicular to OD, will be separately equal to 0, and also the algebraic sum of their moments with respect to will be equal to 0. Hence, we have, QooBa = B + {S + S')sm (b) Qsmaxd—{S+S')cos cos(t> Making = /, factoring, and clearing, we have, Q {fd sin a — hf^ sin a — I cos a) + Bl = 0. Solving with respect to Q, -whiph is the same as Q^, we have, ^'~ Zees a — /c^sina+ 5/*sina ' Making/ = 0, we have, Q _ ^^ _ _^ lf\ I cos a cos a ^"^ ' Hence, from Bq. (51), after reduction, we have, ,, I — fd tan a + ip tan «s ,.„. M=z y- y-^ — '!- (o8) Hence, the modulus varies with a, as was to have been ex- pected ; at the limit, Jf = 1. Weighing Machines. 6?. Nearly all weighing machines depend on the princi- ple of the lever ; the resistance is the weight to he determined, and the power is a counterpoising weight of known value. There are two principal classes of weighing machines : in tlie first, the lever arm of the power is constant, and the power varies ; in the second, the power is constant, and its lever arm varies. The ordinary balance is an example of the first class, and the steelyard of the second. The Common Balance. 68. The common 'balance consists of a lever, AB, called 86 MECHANICS. the beam, having a knife-edge fulcrum, F, and two scale- pans, D and E, suspended from its extremities by means of knife-edge joints at A and B. The beam is sup- ported by a standard, FK, resting on a foot-plate, L. The standard is made vertical by leveling screws passing through the foot-plate. The knife- edges and their supports are of hard- j^ ^ ened steel; and to prevent unnecessary wear, an arrangement is made for throwing them from their bearings when not in use. A needle, N, playing in front of a graduated scale, QH, shows the amount of deflection of the beam. A good balance should fulfill the following conditions : 1°, it should be true ; 2°, it should be stable — that is, when the beam is deflected it should tend to return to a horizontal position ; 3°, it should be sensitive — ^that is, it should be deflected from the horizontal by a small force. In order that a balance may be true, its lever arms must be equal in length, and when the beam is horizontal, both the beam and scale-pans must be symmietrical with respect to two planes through the centre of gravity of the beam, the first plane being perpendicular to the beam, and the second per- pendicular to the fulcrum. In order that it may be stable, the centre of gravity of the beam must be below the fulcrum, and the line joining the points of suspension of the scale-pans must not pass' above the fulcrum. In order that it may be sensitive, the line joining the points of suspension must not pass below the fulcrum, the lever arms must be as long, and the beam as light as is consistent with strength andstiffness, the knife-edges must be horizon- tal and parallel to each other, and the friction at the joints 59] ELEMENTAET MACHINES. 87 must be as small as possible. The sensitiveness of a balance diminishes as the load increases. The true weight of a body may be found by a balance whose lever arms are not equal, by means of the principle demonstrated below. Denote the length of the lever arms, by r and r, and the weight of the body, by W. When the weight W is applied at the extremity of the arm r, denote the counterpoising weight by W ; and when it is at the extremity of the arm r', denote the counterpoising weight by W". We shall have, from the principle of the lever, Wr = W'r', and Wr' — W"r. Multiplying these equations, member by member, we have, W^rr'=W"W'rr'; .-. W^VWHT'; (59) that is, the tru>e weight is equal to the square root of the product of the apparent weights. A still better method, and one that is more free from the effect of errors in construction, is to place the body to be weighed in one scale, and put weights in the other, till the beam is horizontal ; then remove the body to be weighed, and replace it by known weights, till the beam is again horizontal ; the sum of the replacing, weights will be the weight required. If, in changing the load, the positions of the knife-edges be not changed, this method is almost perfect. The Steelyard. 69. The steelyard is an instrument for weighing bodies. It consists of a lever, AB, called the beam ; a fulcrum, F; a gcale-pan, D, attached at the extremity of one arm ; and a known weight, U, movable along the other arm. We shall suppose the'Hreight of ^ to be 1 lb. This instrument is some- times more convenient than the balanoB, but it is not so accu- 88 MECHANICS. rate. The conditions of sensitiveness are essentially the same as for the balance. To graduate the instrument, place a pound weight in the pan, D, and move the counter- poise E till the beam rests horizontal — let that point be marked 1 ; next place a 10 lb. -At|^CIIg Pfii iii M i .' i Mi i iii;,'Tr|PTrwT| iB weight in the pan, and move ' \ ^ the counterpoise -£' till the beam is again horizontal, and let that ■" fi 68 point be marked 10; divide the intermediate space into nine equal parts, and mark the points of division as shown in the figure. These spaces may be subdivided at pleasure, and the scale extended to any de- sirable limit. We have supposed the centre of gravity to coincide with the fulcrum ; when this is not the case, the weight of the instrument must be taken into account as a force applied at its centre of gravity. We may then graduate the beam by experiment, or we may compute the lever arms, corresponding to difEereilt weights, by the principle of moments. To weigh a body with the steelyard, place it in the scale- pan, and move the counterpoise E along the beam till an equilibrium is established ; the mark on the beam will indi- cate the weight. The Compound Balance. 70. Compound balances are used in weighing heavy articles, as merchandise, coal, freight for shipping, and the like. A great variety of combinations have been employed, one of which is shown in the figure. ABi&a, platform on which is placed the object to be weighed; BO is a guard firmly attached to the platform ; the platform ELEMEKTAEY MACHINES. 89 Fig. 69. is supported on the knife- edge fulcrum E, and the piece B, through the me- dium of a brace CD ; OF is a lever turning about the fulcrum F, and suspended by a link from the point L ; LN\& a lever having its ful- crum at M, and sustaining the piece Z> by a link EH ; is a scale-pan suspended from the end iV^of the lever LN. The instrument is so constructed, that EF: OF-.-.KM-.LM; and we shall suppose that KM is made equal to -^^ of MN. The parts are so arranged that the beam LN shall rest hori- zontally when no weight is placed on the platform. If, now, a body Q be placed on the platform, a part of its weight will be thrown on the piece D, and, acting downward, will produce an equal pressure at K. The remaining part will be thrown on E, and, acting on the lever FO, will pro- duce a downward pressure at G, which will be transmitted to L ; but, on account of the relation given by the above -pro- portion, the effect of this pressure on the lever LN will be the same as though" the pressure thrown on E had been ap- plied directly at K. The final effect is, therefore, the same as though the weight of Q had been applied at K, and, to counterbalance it, a weight equal to ^ of Q must be placed in the scale-pan Q. To weigh a body, place it on the platform, and add weights to the scale-pan till LNis horizontal, then 10 times the sum of the weights added will be the weight required. By ap- plying the principle of the steelyard to this balance, objects may be weighed by using a constant counterpoise. 90 MECHANICS. Examples. 1. In a lever of the first class, the lever arm of the resistance Is 2f inches, that of the power, 33J, and the resistance 100 lbs. What power is necessary to, hold the resistance in equilibrium ? Ans. 8 lbs. 3. Four weights of 1, 3, 5, and 7 lbs., are suspended from points of a straight lever, eight inches apart. How far from the point of application of the first weight must the fulcrum be situated, that the weights may be in equilibrium ? Solution. — ^Let x denote the required distance. Then, from Art. (31), l>«a; + 3(a;-8) + 5(a;-16) + 7(!B-34) = 0; .•. X = 11 in. Ana. 3. A lever, of uniform thickness, and 13 feet long, is kept horizontal by a weight of 100 lbs. applied at one extremity, and a force P applied at the other extremity, so as to make an angle of 30° with the horizon. The fulcrum is 20 inches from the point of application of the weight, and the weight of the lever is 10 lbs. What is the value of P, and what is the pressure on the fulcrum ? Solution. — The lever arm of P is- equal to 134 in. x sin 30° = 63 in., and the lever arm of the weight of the lever is 53 in. Hence, 30 X 100 = 10 X 53 + Px 63 . .-. P = 34 lbs. nearly. We have, also, R = ^X' + y= = -v/aiO + 24 sin 30°)» + (24 cos SOy. and. 4. A heavy lever rests on a fulcrum 2 feet from one end, 8 feet from the other, and is kept horizontal by a weight of 100 lbs., applied at the first end, and a weight of 18 lbs., applied at the other end. What is the weight of the lever, supposed of uniform thickness throughout ? ■ Solution. — Denote the required weight by x ; its arm of lever is 3 ft. We have, f rorq the principle of the lever, 100 X 2 = a; X 3 + 18 X 8 ; .-. x — 18% lbs. Ans. R = 133.8 lbs. ; X 30.785 B 123.8 - •^^' a ■■ = 80° 30' 03". ELBMENTAKT MACHINES. 91 5. Two weights keep a IioTizontal lever at rest ; the pressure on the fulcrum is 10 lbs., the difEerence of the weights is 4 lbs., and the differ- ence of lever arms is 9 inches. What are the weights, and their lever arms ? Ans. The weights are 7 lbs. and 3 lbs. ; their lever arms are 15f in., and 6| in. 6. The apparent weight of a body weighed in one pan of a false bal- ance is 5 J lbs., and in the other pan it is 6^ lbs. What is the true weight ? W = ^/^xii = 6 lbs. Ans. The Incliued Plane. 71. An inclined plane is a plane that is inclined to the horizon. In this machine the resistance is the weight of a body act- ing vertically downward, and the power is a force applied to the body, either to prevent motion down the plane, or to pro- duce motion up the pliane. The power may be applied in any direction, but we shall suppose it to be in the vertical plane that is perpendicular to the inclined plane. To find the mechanical advantage when friction is not con- sidered. Let BA be the plane, a body resting on it whose weight is denoted by B, and let P be the force necessary to hold it in equilibrium. The resultant of P and B, denoted by Q, must pass through the polygon of support and be perpendicular to AB (Art. 53). If we denote the angle between P and B by P, and the angle between Q and B, which is equal to the inclination of the plane, by ee, the angle between P and Q will be equal to /? — «, and from equation (15) we shall have P : B :: sin «: sin ((3 — «), 92 MECHANICS. [60. or. R _ sin ()3 — a) P sin « (GO) If the power is, parallel to the plane jS — a, or 0, will be equal to 90°, or sin (/3 — a) will be equal to 1, and be- CB cause sin a = — - equation (60) be- comes Pig. 71. R P AB GB' ■ ■ (61) That is, the power is to the resistance as the height of the -plane is to its length. If the power is parallel to the base of the plane, that is, to the horizon, P — a will be equal to 90° — a, and AC because cot a = -^-5 equation (60) will become R P AG GB Kg. 18. (62) That is, the power is to the resistance as the height of the plane is to its base. If a increase, the value of P will increase, and when a be- comes 90°, P becomes infinite; that is, no finite horizontal force can sustain a body against a vertical wall, without the aid of friction. ExASfPLES. 1. A power of 1 lb., acting parallel to an inclined plane, supports a weight of 3 lbs. What is the inclination of the plane? Ans. 30°. 3. The power, resistance, and normal pressure, in the case of an inclined plane, are, respectively, 9, 13, and 6 lbs. What is the inclina- tion of the plane, and what angle does the power make with the plane? Ans. Inclination of power to plane — <^ — 90° — a — 38° 46' 54". 63.] BLEMEJS-TARY MACHINES. 93 Solution. — If we denote the angle between the power and resistance by ^, and the inclination of the plane by a, we have, from Eq. (12), 6 = VW + 9' + 2 X 9 X 13 cos 0; .•.^ = 156° 8' 20". Also, from Bq. 12, for the inclination of the plane, 6:9:: sin 156° 8' 20" : sin o ; .-. a = 37° 21' 26". 3. A body is supported on an inclined plane by a force of 10 lbs., act- ing parallel to the plane; but it requires a force of 12 lbs. to support it when the force acts parallel to the base. What is the weight of the body, and the inclination of the plane? Am. The weight is 18.09 lbs., and the inclination 33° 33' 25". Modulus of the Inclined Plane. 73. If a block is placed upon the inclined j)lane AB ■whose inclination, a, is greater than the angle of friction, it will slide down unless supported by some force; if the inclination of the plane is less than to the angle of friction it will have no tendency to move either up or down the plane unless acted upon Fig" 73 by some other force than its weight. To find the modulus when friction is considered : let us suppose the body to be on the eve of motion up the plane under the action of the force F, which makes an angle with the normal, Olf, equal to P. The reaction, Q, due to the action of P and R will, from Art. 62, be directed as shown in the figure, being equal to the angle of friction. Because there is an equilibrium between P, R, and Q, we shall have, from Eq. (16), P : R:: sm [180° — (0 + «)] : sin (0 + /?); or, P : R :: &m(p cos a + sin ec cos , remem- bering that P is now equal to Pf, and replacing tan by /, we have _ „ /coi« + sin« . ^'-^/cos/J + sin/J ^*'*^ Making/ = 0, we have P. = R'^. (65) sm jS ^ ' Substituting in (51), we have for the modulus, ^ _ sin 06 /cos/3 + sinj3 _ / cot /? + 1 " sin |3 /cos a + sin a ~/ cot a + 1 ' (66) The line of least traction is the line along which Pf is the least possible; in this case the denominator of the value of Pf must be a maximum: assuming M = / cos |8 + sin /3 and differentiating with respect to ji, we have ^ = — / sin /3 + cos ;3 = 0; . ■. /3 = cot-y. Differentiating again Hence, m is a maximum, and consequently P, is a mini- mum, when the complement of j8 is equal to the angle of fric- tion, that is, when the direction of Pf makes an angle with plane equal to the angle of friction. Let the student find the modulus when Pf is on the eve of moving the body down the plane. Work when the Traction is Parallel to the Plane. TS. If we make /3 = 90° in Eq. 64, the force of traction 67-] ELBMENTART MACHINES. 95 P, will be parallel to the plane and will be directed from A towards B. In this case, we haye P, = Rf cos a + iZ sin a. Multiplying both members by AB, and replacing ABco's.a by A G, and ^5 sin a by GB, it becomes P,y.AB = RfxAG+RxGB. . . . (67) The first member is the work of the force of traetion in drawing the body from A to B ; the first term of the second member is the work that would be required to draw the body horizontally from A to G; and the second term is the work that would be required to lift the body vertically from G to P. Formula (67) finds an application in computing the work of a locomotive in drawing a train up an inclined plane. The Pulley. 74. A ptilley is a wheel having a groove around its cir- cumference to receive a cord ; the wheel turns on an axis at right angles to its plane, and this axis is supported by a frame called a block. The pulley is said to he fixed, when the block is fixed, and movable, when the block is movable. Pulleys are used singly, or in combinations. Single Fixed Pulley. 75. In this machine the block is fixed. Denote the power by P, the resistance by R, and the radius of the pulley by r. It is plain that both the power and resistance should be at right angles to the axis. Hence, if we take the axis of the pulley as an axis of moments, we have (Art. 31), in case of equilibrium, Pr = Rr; or, P = R . . . . (68) 96 MECHANICS. [69. Stiffness of Cordage. •JG. Let be a pulley, with a cord, AB, wrapped round its circumference ; and suppose a force, P, applied at B, to overcome a resistance, R. As the rope winds on the pulley, its rigidity acts to increase the arm of lever of R, and to overcome this rigidity an additional force is required. This additional force may be represented by the expression, + IR\ i^^; Kg. 75. in which d depends on the character and size of the rope, a on its natural rigidity, bR on the rigidity due to the load, and D is the diameter of the wheel. The values of d, a, and b have been found by experiment for different kinds of rope, and tabulated. The moment of the resistance is found by multiplying the preceding expression by its lever arm, -J-Z), or r ; denoting this moment by S, it may be written. S = m + nR (69) Modulus of the Single Fi:sed Pulley. 1"7. Let the solid axle of the pulley turn in a cylindrical box as shown in the figure ; also, suppose the axis to be horizontal. Let the power, P, supposed verti- cal, be applied at one extremity of a cord passing over the pulley AB, and the resistance R at the other extremity; denote the radius of the axle by p, the radius of the pulley by r, and the weight of the Kg. 75*. 70.] ELEMENTAET MACHIKES. 97 pulley and cord (supposed to pass through the centre of the pulley) by W. When P acts to turn the pulley the axle rolls up in its box until it reaches a point 0, beyond which, if it be further dis- placed, it 117111 slide back. At this point, when F is on the eve of overcoming the resistances, the reaction, Q, will be vertical and will make with the normal at an angle equal to , the angle of friction, and we shall have, as in Art. 64, P(r — p sin (t)) = B(r -\- p sin (p) + Wpsm . . . . (70) Solving with respect to F (now P/), we have W . r + psin^+-=psm^ Pf = F ~ (71) Making/ = 0, or sin ^ = 0, we have Fo = R (72) Hence, we have ^ = - - -^^"V ■ (^^> /• + p sin ^ + ^-p sin si * f In which sin '^ vT+7 If we would take stiffness of cordage into account we must add 8, (Eq. 69), to the second member of (70), and when we make / = 0, we must also make 8=0, which gives for the modulus, M= LJUL^t (74) r + p sin ^ + -^ ( Fp sin (^ + /S') 98 MECHAKICS. [75. To find M experimentally we apply a weight P that is just on the eve of overcoming the resistances : then will the quo- tient of B by this weight be the value of the modulus. Friction of Rope on a Fixed Axle. 78. Let the circle represent the cross-section of a fixed cylindrical axlcj and let PLKR be a rope tangent to it at L and K. Sup- pose that a force P is on the eve shown in the figure : there will be a p''' \. J otion all alon? the rone. +: 5 /^ Kg. 76. Let A, B, and G, be three consecu- tive points of the arc KL, and denote the angle KOA by Q : then will the angle between AB and the prolongation of OB be equal to dd. If we denote the tension on BA which acts to resist motion by t, the tension on BC tending to produce motion will be denoted hj t -\- dt, and the reaction Q due to these forces will be situated as shown in Fig. 77, the angle being the angle of friction. Because there is an equilibrium between t, t + dt, and Q, we shall have (Eq. 16), * t:t + dt:: sin (90° + (p) : sin (90° -> + dd) (75) The third term of this proportion equals cos : sm 4> d9. (77) 78.] ELBMBNTAKY MACHINES. 99 Prom (77) we have, after reduction, J=fde (78) Integrating the first member from t = R io t =: P, and the second member from fl = to 5 = KOL (denoted by 0'), we have, Passing to numbers, (79) becomes P R OT P = Ref^' (80) In (80), e = 3.71828, / is the coefQcient of friction be- tween the rope and the cylinder, and 6' is the arc that measures the angle KOL. The unit of 0' is the radian which is very nearly equal to 57°.3. Example. — ^Let the rope be wrapped entirely around the cylinder, that is, let ff — 2w, and suppose that / = A. What is the value of P f Ans: P = J2(3.71828)'-"» = ^ x 13.44 nearly. Single Movable Pulley. 79. In this pulley the block is movable. The resistance is applied by means of a hook attached to the block ; one end of a rope, enveloping the lower part of the pulley, is attached at a fixed point, C, and the power is applied at its other , extremity. We shall suppose, in the first place, ^1 that the two branches of the rope are parallel. Assuming the notation of Art. 75, neglecting friction, and taking -4 as a centre of moments, Fig. 78. we have, in case of equilibrium. R 100 MECHANICS. [81. Px-^r = Br; P = iB. That is, when, the power and resistance are parallel, the power is one half the resistance. The tension of the cord OA is the same as that of BP. It is, therefore, equal to one half the resistance. If the resistance of the point be replaced by a force equal to P, the equilibrium will be undisturbed. In the second place, let the two branches of the enveloping cord be oblique to each other. Suppose the resistance C to be replaced by a force equal to P, and denote the angle be- tween the two branches of the rope by 20. If there is an equilibrium between P, P, and B, the horizontal compo- nents of P, P, will balance each other, and the sum of their vertical components will be equal to B ; hence, we have, 'HP cos .l ■•■•■■« In these equations, the origin of spaces is taken at the point from which the body is projected downward. Motion of Bodies Projected Vertically Upward. 95. Suppose a body projected Tertically upward from the origin of spaces, with a Telocity v', and afterward acted on by the force of gravity. In this case, the force of gravity acts to retard the motion. Making, in (97) and (98), s' = 0, f z= — g, and s = h, they become, v = v' —gt (110) h = v't — igf .... (Ill) In these equations h is positive upward, and negative down- ward. From equation (110), we see that the velocity diminishes as the time increases. The velocity is 0, when, v' v'—gi = 0, or when if =- (112) v' When t is greater than — , v is negative, and the body re- traces its path : hence, the time required for the hody to reach its highest elevation, is equal to the initial veloc- ity, divided by the force of gravity. Eliminating t, from (110) and (111), we have, ^=V (11^) Making v.= 0, in the last equation, we have, ^ = -.....(114) Hence, the greatest height to which the tody will 115] KINETICS. 121 ascend, is equal to the square of the initial velocity, divided hy twice the force of gravity. This height is that due to the initial Telocity (Eq. 108). v' It, in (110), we make t = f', we find, 9 v^gt' (115) v' If, in the same equation, we make t =z — 1- f , we find, v=-9t' (116) Hence, the velocities at equal times hefore and after reaching the highest points are equal. The difference of signs shows that the body is moving in opposite directions at the times considered. If we substitute these values of v successively, in (113), we find in both cases, ^=—4^' (^1^) hence, the points at which the velocities are equal, in ascend- ing and descending, are equally distant from the highest point; that is, they are coincident. Hence, if a body he projected vertically upward, it will ascend to a certain point, and then return upon its path, in such manner, that the velocities in ascending and descending are equal at the same points. Examples. 1. Through what distance will a body fall from rest in a vacuum, in 10 seconds, and through what space will it fall during the last second? Ans. 1608^/i;., and305J/i. .2. In what time will a body fall from rest through 1300 feet? Ans. 8.63 see. 3. A body was observed to fall through a height of 100 feet in the last second. How long was the body falling, and through what distance did it descend? Ms, < = §.9 ; s ;= 308.4 /<. 122 MECHANICS. 4. A body falls through 300 feet. Through what distance does it fall in the last two seconds ? Ans. The entire time occupied is 4.32 seconds. The distance fallen through in 2.83 sec., is 86.57 ft. Hence, the distance required is 800 //. — 86.57/*. = 313.43 /if. 5. A body is projected upward, with a velocity of 60 feet. To what height will it rise ? Ans. 55.9 ft. 6. A body is projected upward, with a velocity of 488 ft. In what time will it rise 1610 feet ? We have, from equation (111), 1610 = 483i! - 16tV<» ; .-. t=^^±i^; or, t = 26.3 see., and t = 8.83 sec. The smaller value of t gives the time required ; the larger value gives the time occupied in rising to its greatest height, and returning to the point 1610 feet from the starting-point. 7. A body is projected upward, with a velocity of 161 feet, from a point 214f feet above the earth. In what time will it reach the earth, and with what velocity will it strike ? Ans. 11.2 see. ; 199 ft. 8. Suppose a body to have fallen through 50 feet, when a second begins to fall just 100 feet below it. How far will the latter body fall before it is overtaken by the former ? Ans. 50 feet. 9. Suppose a body to descend from rest for 8| seconds, and then to move uniformly for 3J seconds with the acquired velocity ; what is the entire distance passed over ? Ans. 478 ft. nearly. 10. A body falls from a height of 400 ft. at the same instant that a body is projected upward from the earth with a velocity of 100 ft. ; at what height above the earth will they pass each other ? Ans. 142| ft. 11. With what velocity must a body be projected vertically upward, that it may rise to a height of 210 ft. in 8 seconds ? Ans. 118J ft. Restrained Vertical Motion. 96. We have seen (Eq. 93) that the measure of a moving force is equal to the acceleration due to the force multiplied by the mass moTed ; hence, conversely, the acceleration due to a moving force is equal to the force divided by the mass moved. Consequently, in case of a body falling freely, the 118.] KINETICS. 123 moTing force varies directly as the mass moved, the accelera- tion g being constant. If, however, we increase the mass moved, without changing the moving force, we shall corre- spondingly diminish the acceleration. To show how this may be done, let ^ be a fixed /^Y\ V^ pulley, mounted on a horizontal axis, and W and W', unequal weights attached to the extremities of a flexible cord passing over the pulley. If the weight, W, be greater than W, the former will [^. descend, and draw the latter up. '^ Fig. 93. In this case, the moving force is the difference of the weights, W and W ; the mass moved is the sum of the masses of W and W, together with that of the pulley and connecting cord. The different parts of the pulley move with different velocities, but the effect of its mass may be replaced by that of some other mass at the circumference of the pulley. Denoting this mass, together with the mass of the cord, by m", and the masses of W and W by m and m', we have the Sntire mass moved equal to m + m' + m", and for the moving force we have (»« — m')g ; hence, the accel- eration, denoted by g', is given by the equation, -j,g (118) m -|- jre' -|- m" This force being constant, the motion produced by it is uniformly varied, and the circumstances of that motion will be made known by substituting the above expression for/, in equations (97) and (98). Examples. 1. Two weights of 5 lbs. and 4 lis. are suspended from the extremities of a cord passing over a fixed pulley, the weight of the pulley and cord being neglected. What is the acceleration, what distance will each weight describe in the first second, and what is the tension of the cord ? 134 MECHANICS. Arts, gi = 3.574: ft. ; s = 1.787 /«. To find the tension of the cord denoted \>j t : we have the moving force acting on the heavier body equal to (5 — i) g, and the acceleration due to this force is ( — — 1 g ; the moving force acting on the lighter body is {t — 4) g, and the corre- sponding acceleration is | — j- j g ; equating these accelerations, and solving, we have, < = 4f Iba. 3. A weight of 1 lb., hanging over a pulley, descends and drags a second weight of 5 lbs. along a horizontal plane. Neglecting the mass of the pulley and all hurtful resistances, to what will the acceleration be equal, and through what space will the descending body move in the first second ? Ans. g' = 5.3632 ft.; s = 3.6811 ft. 3. Two bodies, each weighing 5 lbs., are attached to a string passing over a fixed pulley. What distance will each body move in 10 seconds, when a pound weight is added to one of them, and what velocity wUl have been generated at the end of that time, all hurtful resistances being neglected ? Ans: s = 146.3 ft. ; v = 29.24: ft. 4. Two weights, of 16 oz. each, are attached to the ends of a string passing over a fixed pulley. What weight must be added to one of them, that it may descend through a foot in two seconds, hurtful resistances being neglected ? ' w Ans. The required weight being w, we have, g' = =^ g ; but s = \g'l*, which for s = 1, and t = 2, gives, w = 0.505 oz. Atwood's Machine. 97. Atwood's machine is a contrivance to illustrate the laws of falling bodies. It consists of a vertical post, AB, about 10 feet in height, supporting, at its upper extremity, a fixed pulley, A. To obviate, as far as possible, the resistance of friction, the axle is made to turn on friction rollers. A silk string passes over the pulley, and at its extremities are fastened two equal weights, C and D. To impart motion to the weights, a small weight, G, in the form of a bar, is laid on O, and by diminishing its mass, the acceleration may be rendered as small as desirable. The rod, A B, graduated to feet and decimals, is provided KINETICS. 135 with sliding stages, E and F; the upper one is in the form of a ring, vhich will permit G to pass, but not G; the lower one is in the form of a plate, which is intended to intercept the weight C. Connected with the instrument is a seconds pendulum for measuring time. Suppose the weights, C and D, each equal to 150 grains, the weight of the bar 24 grains, and let a weight of 62 grains at the circumference of the pulley, produce the same resistance by its inertia as that actually produced by the pulley ™. t and cord. Then will the fraction ■ m -\- m' + m" become equal to -^^ ; and this, multiplied by 32^, gives g' ■— 2. This value, substituted for g, in (104) and (105), gives, V = %t, and h = tK Fig. 94. If, in these equations, we make t =zl sec, we have h = 1, and « = 2. If we make ^ = 2 sec, we, in like manner, have A = 4, and v = 4. If we make / = 3 sec, we have A = 9, and v = &, and so on. To verify these results ex- perimentally, commencing with the first : — The weight, 0, is drawn up till it comes opposite the of the graduated scale, and the bar is placed on it. The weight thus set is held in its place by a spring. The ring, E, is set at 1 foot, and the stage, F, at 3 feet from.the 0. When the pendulum reaches one of its extreme limits, the spring is pressed back, the weight, CO, descends, and as the pendulum completes its vibration, the bar O strikes the ring, and is retained. The acceleration then becomes 0, and C moves on uniformly, with the velocity acquired, during the first second ; and it will be observed that G strikes the second stage just as the pendulum completes its second vibration. 126 MECHANICS. Had F been set at 5 feet from the 0, C would have reached it at the end of the third vibration of the pendulum. Had it been 7 feet from the 0, it would have reached it at the end of the fourth vibration, and so on. To verify the next result, we set the ring, E, at 4 feet from the 0, and the stage, F, at 8 feet from the 0, and proceed as before. The ring will intercept the bar at the end of the second vibration, and the weight will strike the stage at the end of the third vibration, and so on. By making the weight of the bar less than 34 grains, the acceleration is diminished, and, consequently, the spaces and velocities, correspondingly diminished. The results may be verified as before. Motion of Bodies on Inclined Planes. 98. If a body is placed on an inclined plane, and aban- doned to the action of its own weight, it will either slide or roll down the plane, provided there be no friction between it and the plane. If the body is spherical, it will roll, and in this case friction may be disregarded. Let the weight of the body be resolved into two components, one perpendicular to the plane, and the other parallel to it : the plane of these components will be vertical, and also perpendicular to the given plane. The effect of the first component will be counteracted by the resistance of the plane, whilst the second will act as a constant force, urging the body down the plane. The force being constant, the body will have a uniformly varied motion, and equations (97) and (98) will be applicable. The acceleration may be found by pro- jecting the acceleration due to gravity on the inclined plane. Let AB represent the inclined plane, and P the centre of gravity of a body jj^ resting on it. Let PQ represent the Fig. gs. 119-] KINETICS. • 127 force of gravity, denoted by g, and PR its component, parallel to AB, PS being the normal component. Denote PR by g', and the angle ABO by a. Then, since PQ is perpendicular to BC, and QR to AB, the angle, RQP, is equal to ABO, or to «. Prom the right-angled triangle, PQR, we have, g' = g sin a. But the triangle, ABO, is right-angled, and, if we denote its height, A 0, by h, and its length, AB, by ?, we shall have sin a = y, which, being substituted above, gives, /=f (119) This value of g' is the acceleration due to the moving force. Substituting it for/, in equations (97) and (98), we have, • v = v'-\-^t, (119) s^s' + v't + ^f^ (130) If the body start from rest at A, taken as the origin of spaces, then wiU v' — 0, and s' = 0, giv^g, v = ^t (131) s = §t' (133) To find the time required for a body to move from the top to the bottom of the plane, make s=l, in (133) ; there will result. Hence, the time varies directly as the length, and in- versely as the square root of the height. 128 MECHANICS. [134. For planes having the same height, but different lengths, the radical factor of the value of t remains constant. Hence, the times required for a tody to move down planes having the same height, are to each other as their lengths. To determine the velocity with which a body reaches the bottom of the plane, substitute for t, in equation (121), its value taken from equation (123). We have, after reduction, V = V^gh. But this is the velocity due to the height h, Eq. 107. Hence, the velocity generated in a hody whilst m,oving down an inclined plane, is equal to that generated in falling through the height of the plane. Let be the centre of a circle in a vertical plane, and let AB'bQ its vertical diameter. Through A draw any chord A C and regard it as an inclined plane, ; also draw CD perpendic- ular to AB. Denote J C by I, AD by h, and AB by 2r. The time required for a body to roll down AO is, from Eq. (123). =VI- But from a property of the circle, we have I = VA X ir. Substituting this in the preceding equation, we have after reduction, V^ 2 {AB) (124) From (106) we see that this is the time required for a body to fall through the height AB. Hence, the time required for KIM-BTICS. 129 a body to roll down any chord through A is equal to the time required for the body to fall through the vertical diameter ; in other words, the time required for a body to roll down any chord passing through the point A is constant. Examples. 1. An inclined plane is 10 feet long and 1 foafc high. How long will it take a body to roll from top to bottom, and what velocity will it ac- quire. Ans. t = 2^ sec. nearly; v — S.02 ft. _ 3. How far will a body descend from rest in 4 seconds, on an inclined plane whose length is 400 feet, and whose height is 300 feet? Ans. 193 ft. 3. How long will it take a body to descend 100 feet on a plane whose length is 150 feet, and whose height is 60? Ans. 3.9 sec. 4. There is a track, 2^ miles in length, whose inclination is 1 in 35. What velocity will a car attain, in running the length of the road, by its own weight, hurtful resistances being neglected? Ans. 155.15 ft., or, 106.3 m. per hour. 5. A railway train, having a velocity of 45 miles per hour, is detached from the locomotive on an ascending grade of 1 in 300. How far, and for what time, will the train continue to ascend the inclined plane ? Ans. Distance = 13,540 ft.; t = 6m. 50.3 see. 6. A body weighing 6 lbs. descends vertically, and draws a weight of 6 lbs. up an ineliiied plane of 45°. How far wiU the first body descend in 10 seconds? Ans. 3.44 ft, Body Falling in a Resisting Medium. 99. In the cases hitherto considered, we have supposed the motion to take place in vacuo: if a body falls through a resisting medium, like the atmosphere, it will experience a resistance which is generally assumed to vary as the square of the velocity. This resistance acts to diminish the acceler- ation due to gravity ; hence, the acceleration thus diminished is expressed by the equation,

hence (135), becomes, s = ^l{vM + 1), (136) Which shows the relation between the space s and the time t. Falling Bodies when Gravity is Variable. 101. In accordance with the Newtonian law, the attraction exerted by the earth on a body at different distances varies inversely as the squares of those distances. Denoting the radius of the earth, supposed a sphere, by r, the force of gravity at the surface by g, any distance from the centre greater than r, by s, and the force of gravity at that distance by ^, we have Substituting this value of 4> in (93), and at the same time making it negative, because it acts in the direction of s nega- tive, we have gj.2 dt» (137) 138.] KINETICS. 133 Multiplying by 2ds, and integrating both members, we have, %='-^^a: o..^ = ^Jf^a (m) If we make v = 0, when s — h, we have 0-^ + C- ■ o- ^^^' 0_ ^ +0, ..6_--^, and, i;2 = 3^r3/-_^^ (139) Equation (139) gives the velocity generated whilst the body is falling from the height h to the height s. If we make A =: cx), and s = r, (139) becomes, v^ = 2gr; .: v =V2gr (140) In this equation the resistance of the air is not considered. If we make ff — 32.088 ft., and r = 30,933,596 ft., their equatorial values, we find, V — 36,664: ft., or nearly 7 miles. That is, if a tody were to fall from an infinite dis- tance to the surface of the earth, its terminal velocity would he nearly 7 miles per second. Equation (140) enables us to compute the terminal velocity* of a body falling from an infinite distance to the sun. In this case g = 890.16 ft., and r = 430,854.5 miles, and the corresponding terminal velocity is, V = 381 miles per second. To find the time required for a body to fall through any ds space, substitute -r for v in (139), and solve the result with respect to dt ; this gives. 134 MECHAKICS. [141. 7j / A ds'/s I li sds ., _. ^^ = -V 2-^ • V^iil = -y 3^ • V^7=7^ • ■ ■ ^''') The negatiTe sign is taken because s decreases as t increases. Reducing (141) by Formula E, and integrating by Formula [36], Calculus, we find. If ^ = when s = A, we have, „ / h Tzh This, in (143), gives, ^=\/3^[(^*-*)*-l^^"'''^^"¥+^^ — (^*^) Making s = r in (143), we find. Which gives the time required for a body to fall from a dis- tance 7i to the surface of the earth or sun. Body- Falling TJuder the Action of a Force that Varies as the Distance. 103. If the earth were homogeneous, a point within its surface would be attracted downward by- a force that would be directly proportional to the distance from the centre (Calculus, p. 198). Let us then suppose the earth to be homogeneous and that an opening is made along a diameter from surface to surface. Denote the acceleration at the surface by ff, the radius by r, and the acceleration at a 146.] KINETICS. 135 distance s from the centre by ^. From the principle stated, we have, g:tt>::r:s; .: ^ = ^-s (145) Substituting this in (92) and at the same time giving it the minus sign because it acts in the direction of s negative, we have, d^s gs Multiplying by 2ds and integrating, we have, -rs=— ^— +C, ort;8=(7 — ^— (146) dt^ r r ^ ' Making v = when s = r, we have C= -r^, and this in (146) gives v» =^{r'> — s>) (147) If we make s = 0, we find v = Vgr, which is the velocity at the centre ; it is also the maximum velocity. If the body moves on beyond the centre, s becomes negative, and we find that the velocities are equal at equal distances from the centre, whether s be positive or negative. When s = — r, v reduces to 0, ahd the body falls toward the centre again, and so on continually. To find the time required for the body to fall from surface ds^ to surface, replace v^ by its value, ^ in (147) and solve with respect to df, and making the result negative because t is a. decreasing function of s, we have. ^^=-^/^V^f=S (''') 136 MECHANICS. [149- Integrating (148) from r to —r and denoting the corre- sponding time by t, we have. i = "Y^J (149) which we shall see hereafter is the time; of vibration of a simple pendulum whose length is r. The kind of motion considered in this article is called hannonic, being the same as that of the particles of a vibra- ting cord, or spring. When the body sets out from one extremity A' of the diameter A'O' to fall through that diameter, suppose a second body to start from the same point and to travel around the semicircle A'MG' with a constant velocity Vgr, the maximum velocity of the first body. The time required for the second body to reach '^' Cwill be -nr -i- Vgr or it\/ -, that is, the two bodies will reach C together. It is easily shown that the velocity of the first body is everywhere equal to the projection of the velocity of the second body. For let B'H' be denoted by s ; then will the velocity of the first body at H' be equal to A /-('■*— «')• Draw IT' Jlf perpendicular to A'O', and at if draw MT io represent the velocity '^/gr , and let MN be its projection parallel to A'O'. We shall have MN= MTcos NMT; but cos NMT ^/r^ — : : cos IT MB' = ^ ' ^ ° ; hence, MN=^\J^- (f-^), . that is, the velocityof the first point at H' is the projection of the corresponding velocity of the second point. KINETICS. 137 From what precedes, we may regatd harmonic motion as the projection of uniform circular motion, that is, if a point revolves uniformly in a circle, the projection of the point on a diameter will move harmonically.. 2°. Plane Curvilikeab Motion. Principle Employed. 103. If a material point moves in a plane curve, we may regard its path as resulting from two simultaneous motions, which are respectively parallel to two right lines lying in the plane of the curve. Thus, we may conceive the point to be moving parallel to the axis of X in accordance with a certain law, and parallel to the axis of Y in accordance with some other law ; or we may conceive it to be moving in the direc- tion of the tangent according to some law, and in the direc- tion of the normal in accordance with some other law. In accordance with Newton's second law we may study the motion of a body in any direction as though it had no motion in any other direction. Motion of a Point Down a Cnrve in a Vertical Plane. 104. Let a body fall down a curve, situated in a vertical plane, under the action of gravity regarded as constant ; and let the axis of Y be vertical, distances downward being positive. At any point of the curve whose ordinate is y, the ac- celeration due to gravity being denoted by g, we have for the tangential component of this acceleration g sin 6, or (Calc. in (92), we have T X "\^ ^ r y y" ^ p e ^"/^ Y / \ i / 12) g~-. Substituting this 138 MECHANICS. [150. ' dy ds-d^s , ■■^ds'''^^-=^^^- Taking the integral between the limits y' and y we have, after reduction, v' = 25- (y - y'), oTv = V2g {y - y') (150) The second member is the velocity due to the height y—y'. Hence, the velocity generated in a body rolling down a curve, gravity being constant, is equal to that generated in falling freely through the same height. ds .Replacing v in (150) by -^ and solving, we have. dt = ds ds dy V2g{y-y') dy^2g{y-y') (151) Integrating (151) between the proper limits, we have the time of falling through the height, y — y'. Time of Descent on an Inverted Cycloid. 105. Let APB represent one half of a branch of an in- verted cycloid, the origin being at A, and the values of y being positive downward. Its differ- ential equation (Oalc, p. 100) is , ydy dx . (152) V^ry—y^ From (153) we find, by reduction. y y Q p V ~-v^j^ IB B K~"~-^^^~__ dy Fig. 99. 153.] KINETICS. 139 Substituting in (151), and reducing, we have, dt=\F-. ^- ^ = (153) To integrate the second member of (153), we make the dis- tance y — y' = z; whence, "ilr — y = 'Hr — y' — z, and dy = dz. Substituting in (153), it becomes dt = A A . ^^ _^ (154) V (/ V{2r —y')z — z> ^ ' Applying formula [26], Calculus, we have, ^ = Y/r.^ersin-^-?^,+^ (l^S) If we suppose the body to fall from P to B, we have at P, = 0, and at B we have, 2 = 3r — y'. Hence, the value of t between those limits is =v? which is entirely independent of y' ; that is, the time re- quired for a body to fall from any point of an inverted cycloid to its vertex is constant. Motion of Projectiles. 106. If a body be projected obliquely upward in a vacuum, and then abandoned to the force of gravity, it will be con- tinually deflected from a rectilinear path, and, after describ- ing a curvilinear path, called its trajectory, will finally reach the horizontal plane from which it started. The starting-point is the point of projection ; the dis- tance from the point of projection to the point at which the 140 MECHANICS. [16T. Fig. 100. projectile again reaches the same horizontal plane is the range, and the time occupied is the time of flight. The only forces to be considered, are the initial impulse and the force of gravity. Hence, the trajectory will lie in a vertical plane through the direction of the initial impulse. Let CAB be this plane, A the point of projection, AB the range, and AG & vertical through A. Take AB and ^ C as co-ordinate axes ; denote the angle of projection, DAB, by a, and the velocity due to the initial impulse by v. Resolve v into two components, one in the direction AG, and the other in the direction AB. We have, for the former, v sin a, and, for the latter, v cos a. The velocities, and, consequently, the spaces described in the direction of the co-ordinate axes, will (Art. 103) be en- tirely independent of each other. Denote the space described in the direction A G, in any time t, by y. The circumstances of motion in this "direction, are thos3 of a body projected vertically upward with an initial velocity, v&ma, and then continually acted on by the force of gravity. Hence, Equation (111) is applicable. Making, in that equation h ■=y, and «' = « sin a, we have. y ■= V sin at — ^gt* (157) Denote the space described in the direction of the axis, AB, in the time t, by x. The only force in this direction is the component of the initial impulse. Hence, the motion will be uniform, and the iirst of iequations (5) is applicable. Making s = a, and v = vcoa a, we have, X = V cos at (158) If we suppose ^ to be the same in equations (157) and (158), 159.] KIKETICS. 141 they will be simultaneous, and taken together, will make known the position of the projectile at any instant. Finding the Talue of t from (158), and substituting in (157), we have, sin « qx^ ,, „„. y = X — —f — 5-, (159) ^ cos a iiV^ cos^ a ^ ' which expresses the relation between x and y for all values of t ; hence, it is the equation of the trajectory. Denoting the height due to v by Ti, we. have, v^ = 2^A, which reduces (159) to the form sin es 1 „ y = X — - , , x^ (160) •' cos a 4:11 cos^ a ^ ' This equation satisfies the test b^ = iac (An. Geom., p. 158) ; hence, we conclude that the trajectory is a parabola. To find the range, make y = in (160) ; this gives, a; = 0, and x— ih sin « cos a ; the first value of x corresponds to the point of projection, and the second is the value of the range. Denoting the range by r, and making 2 sin a cos a =: sin 3a, we have, r — 2hsm2a (161) The range will be a maximum when a — 4:5°, in which case we have, r = 2Ji, (162) If in (161) we replace tc by 90° — a, the value of r will not be changed ; hence, there are two angles of projection, complements of each other, that give the same range. The trajectories in the two cases are not the same, as may be shown by substituting the values of a, and 90° — a, in equa- tion (160). The greater angle of projection gives a higher elevation, and consequently, the projectile descends in a line lia MECHANICS. [163. nearer the vertical. It is for this reason that the gunner selects the greater of the two, when he desires to crush an object, and the less when he desires to batter,. or overturn the object. If « = 90°, the value of r is 0. , That is, if a body be projected vertically upward, it will return to the point of projection. To find the time of flight, make x = r in (158), and deduce the value of t ; this gives, t = — ^ (163) V cos a ^ ' If the range and initial velocity are constant, the time of flight will be greatest when « is greatest. 107. To find the highest point, JT, we have simply to find the point where the tangent is horizontal. To do this, we find the differential coefiBcient of y from equation (160), and make it equal to ; this gives, dy sin a X dx cos « 3A cos^ a = 0; from which we find x = %h sin a cos «, or half the range ; substituting this in (160), we find y = h sin^ « ; these values of X and y are the co-ordinates of K. To find the equation of the trajectory referred to parallel axes through K, we first change the sign of y in (160), so that ordinates may be positive downward ; this gives, sin a 1 „ cos a 4ji cos^ a In this equation we make, y ^= y' — h 8va? a, and x = x' ■\- 2h sin a cos a, and reduce: the resulting equation, after dropping the dashes," is, x^ = 4Jicos^ay (164) 165.] KINETICS. 143 In this parabola the parameter is 4A cos^ a, and conse- quently the distance from K to the directrix is h cos* a ; add- ing this to the ordinate of K, which is h sin* a, we see that the directrix of the curve is at a distance from AB equal to A. The distance from K to the focus is equal to h cos* « : hence, if ct < 45°, the focus is below AB ; if « = 45°, the focus \5 GO. AB; if a > 45°, the focus is above AB. 108. To find an angle of projection such that the trajec- tory may pass through a given point, we substitute in (160), for the reciprocal of cos* a, its value 1 -f tan* « ; that equa- tion may then be written, 4:hy = 4:h tan ecx — (1 -|- tan* ee) xK Denoting the co-ordinates of the given point by x', and y', the equation of condition that the curve shall pass through it is, ihy' = 4A tan ax' — (1 + tan* «)«'*, which can be written, tan* a r tan a = ^—j^ (165) Solving (165), we have. tan a = r — " (166) X ' This shows that there are two angles of projection, under either of which, the point may be struck. If we suppose, ' a;'* = 4/%* — 47j?/', (167) the quantity under the radical sign will be 0, and the two angles of projection will become one. If x' and y' be regarded as variables, equation (16?) repre- sents a parabola whose axis is a vertical line, through the 144 MECHANICS. point of projection. Its vertex is at a distance, Ji, above the point. A, its focus is at A, and its parameter is AJi, or twice the maximum range. If we suppose, ^'2 < iJii _ /Uiy', the point {x', y'), will lie within the parabola just described, the quantity under the radical sign will be positive, and there will be two real values of tan «, and, consequently, two angles of projection, under either of which the point may be struck. If we suppose, a;'2 > 4/^2 — Ahy', the point {x', ?/'), will be without this parabola, the values of tan a will both be im- aginary, and there will be no angle under which the point can be struck. Let the parabola B'LB represent the curve whose equation is x'» = 4^2 — 47««/'. donceive it to be revolved about AL, as an axis, generating a paraboloid of revolufion. Then, from what precedes, we conclude, first, that every point within the surface may be reached from A, under two different angles of projection; secondly, that every point on the surface can be reached, but only by a single angle of projection; thirdly, that no point without the surface can be reached at all. 109. If a body is projected horizontally from an elevated point, A, its trajectory will be made known from equation (160) by Kg. 102. 168-] KINETICS. 145 simply making « = ; whence, sin a = 0, and cos « = 1. Substituting and reducing, we have, x^ z=: —4Ji,y (168) For every value of x, y is negative, which shows that the trajectory lies below the horizontal through the point of pro- jection. If we suppose ordinates to be positive downward, we have, a? = 4Jiy (169) To find the point at which the trajectory will reach any horizontal plane whose distance below A is h', we make «/ = A' in (169), whence, X — 2Vhh' (170) On account of the resistance of the air, the results of the preceding discussion must be greatly modified. They ap- proach more nearly to the observed phenomena, as the velocity is diminished and the density of the projectile in- creased. The atmospheric resistance increases as the square of the velocity, and as the cross section of the projectile ex- posed to the action of the resistance. In the air, it is found, under ordinary circumstances, that the maximum range is obtained by an angle of projection, not far from 34°. Note. — In the following examples atmospheric resistance is neglected. Examples. 1. What is the time of flight of a projectile in vacuum, when the angle of projection is 45°, and the range 6000 feet ? Ans. 19.3 sec. 2. What is the range of a projectile, when the angle of projection is 30°, and the initial velocity 200 feet ? Ans. 1076.9 ft. 3. The angle of projection under which a shell is thrown is 33°, and the range 3350 feet. What is the time of flight ? Ans. 11.35 ?ec., nearly. 146 MEOHAKICS. 4 Find the angle of projection and velocity of projection of a shell, so that its trajectory shall pass through two points, the co-ordinates of the first being x = 1700 ft, j^ = 10 ft, and of the second, x = 1800 ft, y = 10 ft Ans. a = 39'19" ; v = 2218.3 ft. 5. At what elevation must a shell be projected with a velocity of 400 feet, that it may range 7500 feet on a plane which descends at an angle of 30° ? SoIjUtion. — The co-ordinates of the point at which the shell strikes are, xf — 7500 cos 30° = 6495 ; and y' = - 7500 sin 30° = - 3750. And denoting the height due to the velocity 400 ft, by h, we have, 7i = ^ = 2486 /if. Substituting these values in the formula. . _2h± VW-Thy' - a;'" x' and reducing, we have, Ans. a = 4:° 34' 10" ; and a = 55° 25' 41". Tangential and Normal Components. 110. A point cannot move in a curve except under the action of an incessant force, whose direction is inclined to the direction of the motion. This force is called the deflecting force, and can be resolved into two components, one in the direction of the motion, and the other at right angles to it. The former acts simply to increase or diminish the velocity, and is called the tangential force : the latter acts to turn the point from its rectilinear direction, and being directed toward the centre of curvature is called the centripetal force. The normal reaction, which is equal and directly opposed to the centripetal force, is called the centrifugal force. To find expressions for the tangential and centripetal forces, let the acceleration due to the deflecting force at any ITl.] KINETICS. 147 point, P, of the curve be resolved into components parallel to the axes OX and OY, and denote these components re- spectively by <^' and 0". Let these components be again resolved into components acting tangentially along PT, and normally along PN. Denote the algebraic sum of the tangential components by T, and the algebraic sum of the normal components by N. Assuming the notation of the figure, we have. T n\ /C\ 'P / A^ S. / X Kg. 103. r=0'cos6l + ^"sin6i (171) JV=^' sin — 0" cos 61 (173) But from Art. 92, 0' = ^, and 0"=-^/, and from Calculus, p. 12, we have, „ dx -, . . dt/ cos ^ -j- , and sm 6 = -r-. ds ds Substituting these in (171), and reducing, on the supposi- tion that t is the independent variable, we have, „_^ dx d^ dy dP ' ds'^ dP ' ds _d(d^j±_df) _ d (ds^) _ d^s ~ dp = %ds ~ dP • 2ds " dt^ (173) Substituting the same quantities in (173), we have, d^'x - dy — d^y • dx _ ds^ dx • d^y — dy • d^x N = dP • ds dp 148 MECHANICS. [IT*. But the second factor of the last member is equal to the reciprocal of the radius of curvature denoted by R (Calculus, p. 65) ; substituting this in the preceding equation, and re- ducing, we have, ^=-J, (174) ■ which is equal and directly opposed to the acceleration due to the centrifugal force. Denoting the centrifugal force, when the mass of the body is M, by F, we have, ^ = qf . . . . . (175) Hence, the centrifugal force of a body, whose mass is M, that is, the force that it exerts normally to the curve which it is compelled to describe, varices directly as its mass into the square of its velocity, and inversely as the radius of curvature of the curve. The subject of centriiiigal force will be further considered hereafter. 3°. Pbbiodic MoTioiT. Rectilinear and Curvilinear Vibration. 111. Periodic motion is a kind of variable motion, in which the spaces described in certain equal periods of time are equal. This kind of motion is exemplified in the phe- nomena of vibration, of which there are two cases. 1st. Rectilinear vibration. Theory indicates, and experi- ment confirms the fact, that if a particle of an elastic fluid be slightly disturbed from its place of rest, and then aban- doned, it will be urged back by a force, varying directly as its distance from the position of equilibrium ; on reaching this position, the particle will, by virtue of its inertia, pass to the other side, again to be urged back, and so on. 176.] KINETICS. 149 To find the time required for the particle to pass from one extreme position to the opposite one and back, let us denote the displacement at any time t by s, and the acceleration due to the restoring force by <^ ; then, from the law of the force, we shall haye = w's, in which n is constant for the same fluid at the same temperature. Substituting for its value. Equation (92), and recollecting that acts in a direction con- trary to that in which s is estimated, we have, • -w = -'^ (1^^) Multiplying both members of (176) by Ms, we have. df whence, by integration. = %n^sds ; -^ = nh^+ G=- tfl. dv The velocity v will be when s is greatest possible ; denot- ing this value of s by a, we shall have, n^c? + C = ; whence, C = — r^a^. Substituting this value of G in the preceding equation, it becomes, j,2 = ^=^2(flS_43)j whence, ndt= ^-= ...(iyy) dr V a* _ ^ Integrating the last equation, we have, nt-^ <7=sm-'| (1'^^) Taking the integral between the limits s = -\- a and 150 MECHANICS. s = — a, and denoting the corresponding time by ^r, t being the time of a double vibration, we have. \nT = TT ; whence, t = 277 The value of t is independent of the extent of the excur- sion, and dependent only upon n. Hence, in the same medium, and at the same temperature, the time of vibration is constant. 2ndly. Curvilinear vibration. Let ABG be a vertical plane curve, symmetrical with respect to BB. Let ^ C be a horizontal line, and denote the distance SB by h. If a body were placed at A and aban- doned to the action of its own weight, being constrained to remain on the curve, it would, in accordance with preceding principles, move toward B with an accelerated motion, and, on arriving at B, would possess a velocity due to the height h. By virtue of its inertia it would ascend the branch BO with retarded motion, and would finally reach O, where its velocity would be 0. The body would then be in the same condition that it was at A, and would, consequently, descend to B and again ascend to A, whence it would again descend, and so on. Were there no retarding causes, the motion would continue for ever. From what has preceded, it is obvious that the time occupied by the body in passing from ^ to 5 is equal to that in passing from B to C, and also the time in passing from C to 5 is equal to that in passing from B to A. Further, the velocities of the body when at G and H, any two points lying on the same horizontal, are equal, either being that due to the height BJ^. I'S] KIKETICS. 151 Angular Velocity.— Angiilar Acceleration. 113. When a body revolves about an axis, its points, being at different distances from the azis, will have different veloc- ities. The angular velocity is the velocity of a point whose distance from the axis is equal to 1. To obtain the velocity of any other point, we multiply its distance from the axis by the angular velocity. To find a general expression for the velocity of any point of a revolving body, let us denote the angular velocity by co, the space passed over by a point at the unit's distance from the axis in the time dt, by dd. The quantity dO is an infinitely small arc, having a radius equal to 1 ; and, as in Art. 93, it is plain that we may regard the angular motion as uniform, during the infinitely small time dt. Hence, as in Article 93, we have, d0 If we denote the distance of any point from the axis by I, and its velocity by v, we shall have, « = Zw; or, « = Z^ (179) The angular acceleration due to a force is the rate at which the force can impart angular velocity. If we denote it hy 01, we have, as in Art. 93, _do) _^ 'P'-'dt- dfi- The corresponding acceleration of a point whose distance from the axis is I, will be, % = Z^ (180) The corresponding measure of the moving force is found by multiplying (180) by the mass m. 153 MBCHAN-ICS. [181 The Simple Pendulum. 113. A pendulum is a heavy body suspended from a horizontal axis, about which it is free to yibrate. In order to investigate the circumstances of vibration, let us first con- sider the hypothetical case of a single material point vibrating about an axis, to which it is attached by a rod destitute of weight. Such a pendulum is called a simple pendulum. The laws of vibration, in this case, will be identical with those explained in Art. Ill, the arc ABC being the arc of a circle. The motion is, therefore, periodic. Let ABC 'bQ the arc through which the vibration takes place, and denote its radius by I. The angle CD A is called the amplitude of vibration ; half of this angle ADB, denoted by «, is called the angle of deviation ; and I is called the length of the pendulum. If the point starts from rest, 2A, A, it will, on reaching any point li, of its path, have a velocity v, due to the height UK, denoted by h. Hence, V = Vap" (181) If we denote the variable angle HDB by 6, we shall have DE = loosO ; we shall also have I)£! = I cos a; and since h is equal to DK — DU, we shall have, h = I (cos — cos a). Which, being substituted in the preceding formula, gives, V = V^gl (cos — cos «). From the preceding article, we have, ,d0 at 1 82.] KINETICS. 153 Equating these two values of v, we have. 'dt- VUgl (cos 6 — cos«). Whence, by solving with respect to dt. dt v: de ^9 -v/cos 6 — cos « (182) If we develop cos 6 and cos a into series, by McLaukin's theorem, we shall have, cose = l--+j^-^-etc.; cos«=l--+^^^-etc. When tc is very small, say two or three degrees, 6 being still smaller, we may neglect all the terms after the second as inappreciable, giving cosS — cos a = ^(a' — 6^). Substituting in equation (182), it becomes, dt^\/-.—^^^= (183) _d9_ Integrating equation (183), we have. t =z\/- sin-' + C. 9 « Taking the integral between the limits 6 z=a and 6-= — a, t will denote the time of one vibration, and we shall have. V 9 (184) 154 MECHANICS. [185. Hence, the time of vibration of a simple pendulum, is equal to the numAer 3.1416, m^ultiplied into the square root of the quotient obtained by dividing the length of the pendulum, by the force of gravity. For a pendulum, whose length is V, we shall have, 34) and (185), we have, or, t : t' :: VT : VV (186) From equations (184) and (185), we have, by division, t__Vl That is, the times of vibration of two simple pendu- lums, are to each other as the square roots of their lengths. If we suppose the lengths of two pendulums to be the same, but the force of gravity to vary, as it does slightly in difEerent latitudes, and at different elevations, we shall have, t = ^\J^-, and r = 7rY^,. Whence, by division, or, t : t" :: Vf ■■ Vg (187) 7'-y g' That is, the times of vibration of the same simple pendulum, at two different places, are to each other inversely as the square roots of the forces of gravity at the two places, If we suppose the times of vibration to be the same, and the force of gravity to vary, the lengths will vary also, and we shall have. 7r\/-, and t ■= tt\ / —.• \ g ■■ \ g' 188.] KINETICS. 155 Equating these values and squaring, we have, \ = y, or, l:l'::g:g' (188) That is, the lengths of simple pendulums which vi- brate in equal times at different places, are to each other as the forces of gravity at those places. Vibrations of equal duration are called isochronal. De TAmbert's Principle. 114. When several bodies are rigidly connected, it often happens that they are constrained to move in a different manner from what they would, if free. Some move faster and some slower than they would, were it not for the connec- tion. In the former case there is a gain, and in the latter a loss, of moving force, in consequence of the connection. It is obvious that the resultant of all the impressed forces is equal to that of all the effective forces, for if the latter were reversed, they would hold the former in equilibrio. Hence, all the moving forces lost and gained in con- sequence of the connection are in equilibrium. This is known as De I'Ambert's principle. The Compound Pendulum. 115. A compound pendulum is a body free to vibrate about a horizontal axis, called the axis of suspension. The straight line drawn from the centre of gravity of the pendulum perpendicular to the axis of suspension is called the axis of the pendulum. In practical applications, the pendulum is so shaped that the plane through the axis of suspension and the centre of gravity divides it symmetrically. Were the particles of the pendulum entirfely disconnected. 156 • MECHANICS. but constrained to remain at invariable distances from the axis of suspension, we should have a collection of simple pendulums. Those at equal distances from the axis would vibrate in equal times, and those unequally distant would vibrate in unequal times. The particles nearest the axis would vibrate more rapidly than the compound pendulum, and those most remote would vibrate slower; hence, there must be intermediate points that would vibrate in the same time as the pendulum. These points lie on the surface of a circular cylinder whose axis is that of suspension ; the point in which this cylinder cuts the axis of the pendulum is called the centre of oscillation. If the entire mass of the pen- dulum were concentrated at this point, the time of its vibra- tion would be unchanged. Hence, the centre of oscillation of a compound pen- dultim is a point of its axis, at which, if the mass of the pen- dulum were concentrated, its time of vibration would be un- changed. A line drawn through this point, parallel to the axis of suspension, is called the axis of oscillation. The distance between the axis of oscillation and the axis of sus- pension is the length of an equivalent simple pendulum, that is, of a simple pendulum, whose time of vibration is the same as that of the compound pendulum. Angnlar Acceleration of a Compound Pendulum. 116. Let GK be a compound pendulum, C its axis of sus- pension, G its centre of gravity, and suppose the plane of the paper to pass through the centre of gravity, G, and perpen- dicular to the axis, G. We may regard the pendulum as made up of infinitely small filaments, parallel to the axis of suspension, and consequently perpendicular to the paper. The circumstances of vibration will be unchanged if we sup- pose each element to be concentrated in. the point where it meets the plane of the paper. Denote the mass of any such KINETICS. 157 Fig. 103. element, as S, by m, ijs distance from G, by r, and tbe mass of the entire pendulum by M. Through Cdraw a horizontal line, OB, and draw SB, OA, and PB, perpendicular to it. On HS prolonged, take SE to represent the moving force impressed on 8. Then will 8E be equal to mg, (Bq. 93), and its moment with respect to G will be mg x HO. Denote the angular accelerationhj (p^; then will the actual acceleration of S, in a direction perpendicular to SO, be equal to r(pi (Eq. 180), and the effective moving force to inri ; because this force acts at right angles to SO, its moment is equal to mr^^i. Because mg is the moving force impressed on S, and 'mri, and a gain when mg is less than mrcpy The moment of this force with respect to O is equal to mg x Off — mr^^. Similar expressions may be found for each of the elementary particles of the pendulum. By de I'Ambert's principle, the moving forces lost and gained, in consequence of the connection of the parts, are in equilibrium ; hence, the algebraic sum of their moments with respect to the axis, 0, is equal to 0, that is, S{mg xOff)—:s. (mr^cPi) = 0. But 01 and g are the same for each particle ; hence, _ S(ot xOH) fi- s (mr«) ^' Prom the principle of moments, we have, S (ot X Off) = Mx GA. 158 MECHANICS. [189. Substituting above, ve have, finally, M xCA . . That is, the angular aoceleration varies as, CA,'the lever arm of the weight of the pendulum. The expression X (mr^) is called the moment of inertia of the body with respect to the axis of suspension, Mg is the weight of the body, and Mg x CA is the moment of the weight, with respect to the same axis. Hence, the angular acceleration is equal to the mo- ment of the weight, divided by the moment of inertia, both tahen with respect to the axis of suspension. Length of an Eqnivalent Simple Pendulum. 117. To find the length of a simple pendulum that will vibrate in the same time as the given compound pendulum, let be the centre of oscillation, and draw OB perpendicular to CB. Denote GO by I, and GO by k. Were the entire mass concentrated at 0, each value of mr would become equal to ml, and we should have, for its moment of inertia, J/f, and for the moment of the mass, M x GB, and for the angu- lar acceleration, M xCB But the pendulum is to vibrate in the same time, whether it exist as a compound pendulum, or as a simple pendulum, its mass being concentrated at its centre of oscillation ; the value of 01 must, therefore, be the same in both cases. Placing the value just deduced equal to that in equation (189), we have, MxCB MxGA MP ^ ~ £(w3)^' 190.] KINETICS. 159 whence, by reduction. Jf ^8 = 2 (jwr2) X GB CA From the similar triangles, OGA and COB, we have, GB CA h Substituting, and reducing, we have, ' = ^ a») That is, fhe length of the equivalent simple jjendu- lum is equal to moment of inertia of the pendulum divided by the moment of its mass, both taken with respect to the axis of suspension. Beciprocity of Axes of Suspension and Oscillation. 118. Let G be the axis of suspension, the centre of oscillation, and let a line be drawn through parallel to the axis of suspension. This line is called the axis of oscillation. Let the planfe of the paper be taken as before, and suppose the elements projected on it, as in Article 116. Let 8 be any element, and denote its distance frofti the axis of suspension by r, and from the axis of oscillation by t ; denote OC by I, and the angle OGShj(l>. If the axis of oscillation be taken as an axis of suspension, and the length of the corresponding simple pendulum, de- noted by r, we have, from the preceding article, Fig. 107. I'. M{1 — k) (191) 100 MECHANICS. [193. In the triangle, OSC, we have, hence, J.{m&) — I,{mr^) + J,{mP) — 2J.{mr gos0)Z. But, from equation (190), we have, X{mr^) = MM ; and because I is invariable, we have, l.{niF) = S(m)Z3 = MP ; if we suppose CO horizontal, rcos^, the projection of r on CO, will be the lever-arm of m, and the expression, S [mr cos )?= Mkl. Substituting for these expressions their values given above, and putting the value of S (mfi), thus found, in (191), we have, , _ MJcl + MH^ — ^Mkl _ M{P — M) . "~ M{1 -k) ~ M{1 — h) ' or, I' = 1 (192) Hence, the axis of suspension and oscillation are convertible; that is, if either be taken as an axis of suspension, the other will be the axis of oscUlation. This property of the compound pendulum is employed to determine the length of the seconds' pendulum, and the value of the force of gravity at different places on the sur- face of the earth. KIKETICS. 161 The Beversible Pendulum. ^ 119. A reversible pendulum sufficiently accurate for purposes of illustration may be constructed as follows. A rectangular bar of steel, CD, about four feet long, is provided witb two knife-edge axes, A and B, having ^ their edges turned toward each other ; these axes are [uV] attached to sliding sleeves with clamp screws, so that they may be set at different points of the bar CD ; the axes should be mounted so that they shall be perpen- dicular to the plane of the bar, and so that their plane shall pass through the axis of the bar. The pendulum thus constructed is suspended on horizontal plates of hard steel, so placed that the pen- dulum may vibrate freely between them, and around p. ^ either axis at pleasure. To adjust the axes A and B so the times of oscillation around them shall be equal, we proceed by the method of approximation. Set the axis A in some convenient position, and allow the pendulum to vibrate ; note the time required to make, say 100 vibrations ; then, dividing this by 100, we have the time of a single vibration. Substitute this value for t, in (184), and deduce the corresponding value of I. Set the axis 5 at a distance from A equal to I, and we have a first approximation. Eeversing the pendulum, and repeating the operation, we find a second approximation. By continuing this process of approximation, we ultimately find two positions of the axes around which the times of vibration are equal. The distance between the axes is then the length of the equivalent simple pendulum. Knowing the length of the equivalent simple pendulum and the corresponding time of vibration, we may determine the acceleration due to the force of gravity from formula (184); 162 MECHANICS. [193. we may also find the length of a seconds' pendTilTim; that is, of a pendulum, that vibrates once in a second, from the proportion (186), by making t' = 1, and substituting the known values of t and I. The pendulum used for scientific purposes is far more complex in its modes of adjustment and use than the one above described, and far greater precautions are taken to avoid errors. Pendulums of this kind have been used for determining the figure of the earth, and for various other scientific purposes. By a series of carefully conducted experiments, it has been found that the length of a seconds' pendulum in the Tower of London is 3.2616 ft., or 39.13921 inches. Experiments made in different latitudes show that the force of gravity continually increases from the equator to- ward either pole. According to Plantamour's modification of Bessel's barometric formula, the value of g is given for any latitude by the formula 5r = 5''(l — .0026257 cos 3Z), (193) in which g' is its value in latitude 45°, and L is the latitude of the given place. According to Airy, the value of g' is about 32.17/^. This, in (193), gives g = 32.17(1 — .0026257 cos2Z). For the latitude of New York (40° 45'), we have, ^=33.16/^. nearly. Formula (193) takes account of the variation of g, due to latitude only. It may be still further modified to take ac- count of variation in height above the level of the sea, as fol- lows : let g be the value of gravity at the level of the sea in 19*-1 KINETICS. 163 the latitude L, and let g" be its value in the same latitude at an eleyation above the sea level denoted by « ; if we denote the radius of the earth by R, we shall then have, by the Newtonian law, g-.f-.'.^R^zf-.m, or, g" =g.^^^. Developing the denominator, and dividing R^ by it, neglect- ing all terms of the quotient after the second, as insignificant, we have, Substituting in (193), we have, g" = g'{l — .0026257 cos 2^) (l - S (194) From which we may find the value of gravity in any latitude, and at any elevation. If z is given in feet, R must also be given in feet : it is assumed that, R = 20,886,860 ft. Practical Application of the Pendulum. 130. One of the most important uses of the pendulum is to regulate the motion of clocks. A clock consists of a train of wheelwork, the last wheel of the train connecting with a pendulum-rod by a piece of mechanism called an escape- ment. The wheelwork is kept in motion by a descending weight, or by the elastic force of a spring, and the wheels are so arranged that one tooth of the last wheel in the train escapes from the pendulum-rod at each vibration of the pendulum, or at each beat. The number of beats is rendered visible on a dial-plate by indices, called hands. 164 MECHANICS. On account of expansion and contraction, the length of the pendulum is liable to variation, which gives rise to irregu- larity in the times of vibration. To obviate' this, and to render the times of vibration uniform, several devices have been resorted to, giving rise to what are called compensat- ing pendulums. "We shall indicate two of the most im- portant of these, observing that the remaining ones are nearly the same in principle, differing only in mode of application. Graham's Mercurial Pendulum. 131. Graham's mercurial pendulum consists of a rod of steel about 43 inches long, branched toward its lower end, to embrace a cylindrical glass vessel 7 or 8 inches deep, and having between 6 and 7 inches of this depth filled with mercury. The exact quantity of mercury, being dependent on the weight and expansibility of the other parts of the pen- dulum, may be determined by experiment in each case. When the temperature increases, the steel rod is lengthened, and, at the same time, the mercury rises in the cylinder. When the temperature decreases, the steel bar is shortened, and the mercury falls in the cylinder. By a proper adjust- ment of the quantity of mercury, the effect of the lengthening or shortening of the rod is exactly counterbalanced by the rising or falling of the centre of gravity of the mercury, and the axis of oscillation is kept at an invariable distance from the axis of suspension. Harrison's Gridiron Pendulum. 123. Harrison's gridiron pendulum consists of five rods of steel and four of brass, placed alternately with each other, the middle rod, or that from which the bob is sus- pended, being of steel. These rods are connected by cross- pieces in such a manner that, whilst the expansion of the steel rods tends to elongate the pendulum, or lower the KINETICS. 165 boby the expansion of the brass rods tends to shorten the pendulum, or raise the bob. By duly propor- « tioning the sizes and lengths of the bars, the axis of oscillation may be maintained at an invariable distance from the axis of suspension. From what has preceded, it follows that whenever the dis- tance from the axis of oscillation to the axis of suspension remains invariable, the times of vibra- tion must be absolutely equal at the same place. The pendulums just described are principally used for astronomical clocks, where great accu- , ., racy and uniformity in the measure of time are indispensable. ^'s- ''^• Basis of a System of Weights and Measures. 123. The pendulum is of further importance, in a practi- cal point of view, in furnishing the standard that has been made the basis of the English system of weights and measures. It was enacted by Parliament, in 1834, that the distance between the centres of two gold studs in a certain described brass bar, the bar being at a temperature of 63° F., should be an " imperial standard yard." To be able to restore it in case of its destruction, it was enacted that the yard should be considered as bearing to the length of the seconds' pendulum in the latitude of London, in vacuum, and at the level of the sea, the ratio of 36 to 39.1.393. From the yard, every other unit of linear measure may be derived, and thence all meas- ures of area and volume. It was also enacted that a certain described brass weight, made in 1758, and called 3 lbs. Troy, should be regarded as authentic, and that a weight equal to one half that should be "the imperial standard Troy pound." The -jV^^th part of the Troy pound was called a grain, of which 7000 con- stituted a pound avoirdupois. To provide for the con- 166 MECHANICS. tingeney of a loss of the standard, it was connected with the system of measures, by enacting, that if lost, it should be restored by allowing 252.724 grains for the weight of a cubic inch of distilled water, at 63° P., the water being weighed in vacuum and by brass weights. From the grain thus estab- lished, all other units of weight may be derived. Our own system of weights and measures is the same as that of the English. Examples. 1. The lengtli of a seconds' pendulum is 39.13931 in. If it be shortened 0.130464 in., how many vibrations wUl be gained in a day of 24 hours? Ans. 144 nearly. 2. A seconds' pendulum on being carried to the top of a mountain, was observed to lose 5 vibrations per day of 86400 seconds. Required the height of the mountain, reckoning the radius of the earth at 4000 miles. Ans. h = 0.2815 mi. = 1322 ft. 3. "What is the time of vibration of a pendulum whose length is 60 in., when the force of gravity is 32 J ft. ? Ans. 1.3387 see. 4. How many vibrations will a pendulum 36 inches in length make in one minute, the force of gravity being the same as before ? Ans. 63.58. 5. A pendulum makes 43170 vibrations in 12 hours. How much must it be shortened that it may beat seconds ? Ans. .0544 in. 6. In a certain latitude, the length of a pendulum vibrating seconds is 39 inches. What is the length of a pendulum vibrating seconds, in the same latitude, at the height of 21000 feet above the first station, the radius of the earth being 3960 miles ? Ans. 38.9318 in. 7. If a pendulum make 40000 vibrations in 6 hours, at the level of the sea, how many vibrations will it make in the same time, at an eleva- tion of 10560 feet, the radius of the earth being 3960 miles ? Ans. 39979.8. 8. What is the length of a pendulum that will beat sidereal seconds, the length of the sidereal day being 23 firs. 56 min. 4 sec. ? Ans. 39.0334 inches. 9. What is the length of a pendulum that makes as many vibrations per minute as it is laches long ? Ans. 53.03 inches. VI.— OENTEIPUGAL FO EC E.— MOMENT OP IN"EETIA. Centrifogal Force in Terms of Ang^ilar Velocity. 124. When a material point is constrained to move in a curve it offers a resistance to the force that deflects it from its rectilineal path. This resistance, as stated in Article 110, is called the centrifugal force. It is shown, in the article referred to, that the value of the centrifugal force is given by the Equation, F=m~, (195) in which m is the mass of the particle, v its lineal velocity, and r the radius of curvature of its path at the instant in question. If we denote the angular velocity of the point around the centre of curvature by w, its linear velocity will be equal to rw, and this substituted in (195), gives F=mr(^ (196) This equation is often more convenient than (195). If we suppose the point to be restrained by a rigid curve, the centrif- ugal force is equal, and directly opposed to the reaction of the curve. If the material point m is whirled around a fixed point, being retained by a string, the centrifugal force is the measure of the tension of the string. 168 MECHANICS. Centrifugal Force of an Extended Mass. 135. "We have supposed, in what precedes, that the dimen- sions of the body under consideration are extremely small ; let us next examine the case of a body, of any dimensions whatever, constrained to revolve about a fixed axis. If the body be divided into infinitely small elements, whose direc- tions are parallel to the axis, the centrifugal force of each element will be equal to the mass of the element into the square of its velbcity, divided by its distance from the axis. If a plane be passed through the centre of gravity of the body, perpendicular to the axis, we may, without impairing the generality, of the result, suppose the mass of each element concentrated at the point in which this plane cuts. the line of direction of the element. Let XOY be the plane through the centre of gravity per- pendicular to the axis of revolution, AB the projection of the body on the plane, and C the point in which it cuts the axis. Take G as the . ^ ..^ origin of a system of rectangular co-ordi- nates ; let GX be the axis of X, GY the axis of Y, and m be the point at which the mass of one filament is concentrated, and denote that mass by m. Denote the co-ordinates of to by « and y, its distance ^' ' from G by r, and its velocity by v. The centrifugal force of the mass, m (Eq. 196), is equal to mr (t)^. Let this force be resolved into components parallel to GX and GY. We have, for these components, inr u? cos m GX, and mr to^ sin m GX. ■ 197.] CENTBIFUGAL FORCE. — MOMENT OE IKEETIA. 169 But, from the figure, cos m OX = - , and sin m GX = - . r r Substituting these in the preceding expressions, and reduc- ing, we have, for the components, ~ mx o)', and my <^. Similar erpressions may be deduced for each of the other filaments. If we denote the resultant of the components parallel to OX by X, and of those parallel to GY by Y, we haye, X = 2 {mx) Gj8, and F = S {my) u\ If we denote the mass of the body by M, and suppose it concentrated at its centre of gravity, 0, whose co-ordinates are x^, and yi, and whose distance from G is Vi, we shall have, from the principle of the centre of gravity (Art. 37), S (mx) = Jfoj, and S (my) = Myi. Substituting above, we have, X = M<^Xi, and Y = Mufiji. If we denote the resultant force by R, we have. B = VX^ + Y^ — Mofi Vxf+Y^ = Mu?r^ (197) The direction of the resultant R, Eq. (9), is given by the equations, X Xt, . Y yi ,^nQ, cosa=-g- = -; .sma = -^ = J^; (198) that is, the resultant passes through G. 1 IC 170 MECHANICS. But equations (197) and (198) are expressions for the intensity of the centrifugal force of the mass M concentrated at 0. Hence, the centrifugal force of an extended mass, constrained to revolve about a fixed axis, is the same as though the mass were concentrated at its centre of gravity. Experimental lUustratious. 136> The principles relating to centrifugal force admit of experimental illustration. The instrument represented in the figure may be employed to show the value of the centrifugal force. „ E D ^ is a vertical axle, on which is mounted a wheel, F, communicating with a train of wheelwork, by means of which the axle may be made to revolve with any angular velocity. Fig. ni. At the upper end of the axle is a forked branch, BO, sustaining a stretched wire. D and E are balls pierced by the wire, and free to move along it. Be- tween B and ^ is a spiral spring, whose axis coincides with the wire. Immediately below the spring, on the horizontal part of the fork, is a scale for determining the distance of the ball, E, from the axis, and for measuring the degree of compres- sion of the spring. Before using the instrument, the force required to produce any degree of compression of the spring is determined experimentally, and marked on the scale. If a motion of rotation be communicated to the axis, the ball D will at once recede to G, but the ball E will be re- strained by the spring. As the velocity of rotation increases, the spring is compressed more and more, and the ball E ap- proaches B. By a sujta|3le arrangement of wheelwork, the 3CF CENTEIFUGAL FOECE. — MOMENT OF INERTIA. 171 angular velocity of the axis corresponding to any compression may be ascertained. "We have, therefore, all the data neces- sary to verify the la^c of centrifugal force. If a circular hoop of flexible material be mounted on one of its diameters, its lower point being fastened to the horizontal beam, and a motion of rotation imparted, the portions of the hoop farthest from the axis will be most affected by centrif- ugal force, and the hoop will assume an elliptical form. If a sponge, filled with water, be attached to one of the arms of a whirling machine, and motion of rotation im- parted, the water will be thrown from the sponge. This principle has been used for drying clothes. An annular trough of copper is mounted on an axis by radial arms, and- the axis connected with a train of wheelwork, by means of which it may be put in motion. The outer wall of the trough is pierced with holes for the escape of water, and a lid confines the articles to be dried. To use this instrument, the linen, after being washed, is placed in the annular space, and a rapid rotation imparted to the machine. The linen is thrown against the outer wall of the instrument, and the water, urged by the centrifugal force, escapes through the holes. Sometimes as many as 1,500 revolutions per minute are given to the drying machine, in which case, the drying process is very rapid and very perfect. If a body revolve with sufiicient velocity, it may happen that the centrifugal force generated will be greater than the force of cohesion that binds the particles together, and the body be torn asunder. It is a common occurrence for large grindstones, when put into rapid rotation, to burst, the fragments being thrown away from the axis, and often pro- ducing much destruction. When a wagon, or carriage, is driven round a corner, or is forced to run on a circular track, the centrifugal force is often suflBcient to throw loose articles from the vehicle, and 173 MECHANICS. even to overthrow the vehicle itself. When a car on a rail- road track is forced to turn a sharp curve, the centrifugal force throws the cars against the rail, producing a great amount of friction. To obviate this diflEiculty, it is customary to raise the outer rail, so that the resultant of the centrifugal force, and the force of gravity, shall be perpendicular to the plane of the rails. Form of the Surface of a Revolving Liquid. IST. If a vessel of water be made to revolve about a verti- cal axis, the inner particles recede from the axis on account of the centrifugal force, and are heaped up about the sides of the vessel, imparting a concave form to the upper surface. The concavity becomes greater as the angular velocity is in- creased. To determine the form of the concave surface, we assume the principle yet to be demonstrated, viz. : that the resultant action on any point of the free surface is normal to that surface. Let the figure represent a section made by a plane through the axis of rotation sup- posed vertical. Let BA G be the section of the upper surface of the water, A its lowest point, and P any other point. Take A as the origin of co-ordinates, the axis of Y being vertical, and the axis of X horizontal, and denote the co-ordinates of P by a; and y. The material point P is urged horizontally by the centrif- ugal force Pf, whose acceleration (Eq. 196) is equal to kw^ and it is urged downward by the force of gravity Pg, whose acceleration is g. The resultant of these forces, PR, is, from what precedes, normal to the curve APG. Because Pf is perpendicular to A Y, and PR to the tangent PT, the angle fPR is equal to YTP ; but the tangent of fPR is equal to 199.] CENTRIFUGAL FORCE. — MOMENT OF INERTIA. 173 -^, and the tangent of VTP is (Calc, p. 13) equal to -T-- Equating these values and clearing of fractions, we have. u)^x dz = g dy. (199) Integrating (199), observing that the arbitrary constant under the given conditions is 0, we have. u'x ^9. j-=gy, or x-^ = -^,x, (200) which is the equation of a parabola whose axis coincides with the axis of revolution. Hence, the surface is a paraboloid of revolution whose axis is the axis of revolution. Centrifugal Force at Points of the Earth's Surface. 138. Let it be required to determine the centrifugal force at different points of the earth's surface, due to rotation on its axis. Suppose the earth spherical. Let ^ be a point on the sur- face, PQP' a meridian section through A, PP' the axis, RQ the equator, and AB, perpen- dicular to PP', the radius of the parallel of latitude through nf. A. Denote the radius of the earth by r, the radius of the par- allel through A by r', and the latitude of A, or the angle QGA, by I. The time of revolution being the same for every point on the earth's surface, the velocities of Q and A will be to each other as their distances 174 MECHANICS. [801. from the axis. Denoting these velocities by v and v', we have, V •.v' :: r : r', whence, , _ w' ~~ r But from the right-angled triangle, GAB, since the angle at A is equal to I, we have, r' ^ r cos I. Substituting this value of r' in the value of v', and re- ducing, we have, ■ v' = V cos I. If we denote the centrifugal force at the equator by/, we have, f=- (301) In like manner, if we denote the centrifugal force at A by /', we have, f -"Hi J - r'' Substituting for v' and r' their values, previously deduced, we get, r = t^ (202) Combining equations (201) and (202), we find, f:f':: 1 : cos I, .: /' =/cos I (203) That is, the centrifugal force at any point on the earth's surface, is equal to the centrifugal force at the equator, multiplied by the cosine of the latitude. 204.] CEUTEIFUGAL FOECE. — MOMENT OF IKEETIA. 175 Let AE, perpendicular to PP' , represent /', and resolve it into two components, one tangential, and the other normal to the meridian section. Prolong GA, and draw AT) perpen- dicular to it at ^. Complete the rectangle FD on ^^ as a diagonal. Then will AD be the tangential and AF the nor- mal component. In the right-angled triangle, AFE, the angle at A is equal to I. Hence, FE= AD = f sinl = f Gosl sinl = ^^^ (304) AF = /' cosl =/ cos2 1 (205) From (304), we see that the tangential component is at the equator, goes on increasing till I = 45°, where it is a maximum, and then goes on decreasing till the latitude is 90% when it again becomes 0. The effect of the tangential component is to heap up the particles of the earth about the equator, and, were the earth in a fluid state, this process would go on till the effect of the tangential component was counterbalanced by the component of gravity acting down the inclined plane thus formed, when the particles would be in equilibrium. The higher analysis shows that the form of equilibrium is that of an oblate spheroid, differing but slightly from that which our globe is found to possess by actual measurement. Prom equation (205), we see that the normal component of the centrifugal force varies as the square of the cosine of the latitude. This component is directly opposed to gravity, and, conse- quently, tends to diminish the apparent weight of all bodies on the surface of the earth. The value of this com- ponent is greatest at the equator, and diminishes toward the poles, where it is 0. Prom the action of the normal compo- nent of the centrifugal force, and because the flattened form 176 MECHASriCS. of the earth due to the tangential component brings the polar regions nearer the centre of the earth, the measured force of gravity ought to increase in passing from the equator toward the poles. This is found to be the case. The radius of the earth at the equator is about 3962.8 miles, which, multiplied by Stt, will give the entire circumfer- ence of the equator. If this be divided by the number of seconds in a day, 86,400, we find the value of v. Substituting this value of v and that of r just given, in equation (301), we find, / = 0.1113/i!., for the centrifugal force at the equator. If this be multi- plied by the square of the cosine of the latitude of any place, we have the value of the normal component of the centrifugal force at that place. If the earth were to revolve 17 times as rapidly as it now does, the centrifugal force at the equator would be equal to 0. 1112 /i. X 289, or to 32.1368/^., that is, the centrifugal force at the equator would be very nearly equal to cf. In that case, the apparent weight of a body at the equator would be equal to 0. Elevation of the Outer Rail of a Curved Track. 1S9. To find the elevation of the outer rail, so that the resultant of the weight and cen- trifugal force shall be perpendic- "fflT ular to the line joining the rails, /"^^N, assume a cross section through f 1 the centre of gravity, G. Take 1 ga \(h the horizontal, GA, to represent riu |I-:^\\ the centrifugal force, and GB \y '\ % to represent gravity. Construct ^. g j their resultant, GO. Then must Fig. ii4. DU be perpendicular to GC. 308.] CENTRIFUGAL FORCE. — MOMENT OF INERTIA. 177 Denote the velocity of the car by v, the radius of the curved track by r, the force of gravity by g, and the angle, DBF, or its equal, £GO,hj «. From the right-angled triangle, GBC, we have, BO tan« = -^. But BO, -ffhieh is equivalent to GA, is equal to — , and GB is equal to g ; hence, . If' tan cc ^ —. gr Denoting the distance between the rails, by d, and the elevation of the outer rail above the inner one, by h, we have, tana == -^, very nearly. Equating the two values of tan a, we have, 1 = ^, ,.h = d^ (206) a gr gr ^ ' Hence, the elevation of the outer rail varies as the square of the velocity directly, and as the radius of the curve in- versely. It is obvious that the elevation ought to be different for different velocities, which, from the nature of the case, is im- possible. The correction is, therefore, made for some assumed velocity, and then such a form is given to the tire of the wheels as to partially correct for other velocities. The Conical Pendulum. 130. A conical pendulum consists of a solid ball at- tached to one end of a rod, the other end of which is con- nected, by means of a hinge-joint, with a vertical axle. 178 MECHANICS. [aOT- When the axle is put in motion, the centrifugal force gener- ated in the ball causes it to recede from the axis, until an equilibrium is established between the weight of the ball, the centrifugal force, and the tension of the connecting rod. When the Telocity is constant, the centrifugal force will be constant, and the centre of the ball will describe a horizontal circle, whose radius will depend upon the velocity. Let it be required to determine the time of revolution. Let BD be the vertical axis, A the ball, B the hinge-joint, and AB the connecting rod, whose mass is so small, that it may be neglected, in compar- ison with that of the ball. Denote the required time of revolution, by t, the length of the arm, by I, the accelera- tion due to the centrifugal force, by /, and the angle ABG, by (p. Draw AG perpen- dicular to BD, and denote AG, bv r, and '' ' ■' ' Ym 115 BG, by h. ^ From the triangle, ABG, we have, r ^h tan )= sin (/>• We have, therefore, g : f :: cos : sin 0, .•. f — g tan 0. Equating these values of/, we have, Arr% tan Hence, Solving with respect to t, =: g tan 0. (208) That is, the time of revolution, is equal to the time of a double vibration of a pendulum, whose length is h. The Governor. 131. The principle of the conical pendulum is employed in the governor, a machine attached to engines, to regulate the supply of motive force. AB is a vertical axis connected with the machine near its working-point, and and revolving with a velocity propor- tional to that of the working-point; FU and GB are arms turning about AB, and bearing heavy balls, D and F, at their extremities ; these bars are united by hinge- joints with two other bars at G and F, and also to a ring at H, that is free to slide up and down the shaft. 180 MECHANICS. The ring, H, is connected with a lever, HE^, that acts on the throttle valve in the pipe that admits steam to the cylinder. "When the shaft revolves, the centrifugal force causes the halls to recede from the axis, and the ring, H, is depressed ; and when the velocity has become sufficiently great, the lever closes the valve. If the velocity slackens, the balls approach the axis, and the ring, H, ascends, opening the valve. In any given case, if we know the velocity required at the work- ing-point, we can compute the required angular velocity of the shaft, and, consequently, the value of t. This value of t, substituted in equation (308), gives the value of h. We may, therefore, properly adjust the ring, and the lever, HK. Examples. 1. A ball weighing 10 lbs. is ■whirled round in a circle whose radius is 10 feet, with a velocity of 30 feet. What is the acceleration due to centrifugal force? Ana. 90 ft. 3. In the preceding example, what is the tension on the cord that re- strains the ball? Ans. < = 38 lbs., nearly. 3. A body is whirled round in a circular path whose radius is 5 feet, and the centrifugal force is equal to the weight of the body. What is the velocity of the moving body? Ans. v = 12.7 ft. 4. In how many seconds must the earth revolve that the centrifugal force at the equator may counterbalance the force of gravity, the radius of the equator being 8963.8 miles? Ans. t = 5,068 sees. 5. A body is placed on a horizontal plane, which is made to revolve about a vertical axis, with an angular velocity of 3 feet. How far must the body be situated from the axis that it may be on the point of sliding outward, the coefficient of friction between the body and plane being equal to .6? Ans. r = 4.835 ft. 6. What must be the elevation of the outer rail of a track, the radius being 3960 ft., the distance between the rails 5 feet, and the velocity of the car 30 miles per hour, that there may be no lateral thrust? Ans. 0.076 ft., or O.d in., nearly. 7. The distance between the rails is 5 feet, the radius of the curve 600 feet, and the height of the centre of gravity of the car 5 feet. What 209.] CENTRIFUGAL FORCE. — MOMENT OF INERTIA. 181 velocity must the car have that it may be on the point of being over- turned by the centrifugal force, the rails being on the same level? Ans. v = 9S ft., or 66f miles per hour. 8. A body revolves uniformly in a circle whose radius is 5 ft., and with such a velocity as to complete a revolution in 5 seconds. What is the acceleration due to the centripetal force? Ans. Jtt'^ 9. A body weighing 1 lb., is whirled around horizontally, being re- tained by a string, whose length is 6 feet. What is the time of revolu- tion when the tension of the string is 3 lbs. ? Ans. Sir Vf seconds. Moment of Inertia. 133. The moment of inertia of a body with respect to an axis, is the algebraic sum of the products obtained by multiplying the mass of each element of the body by the square of its distance from the axis. Denoting the moment of inertia with respect to any axis by K, the mass of any ele- ment of the body by m, and its distance from the axis by r, we hare, from the definition, X = S (mr^) (309) If we denote the mass of any element by m and its distance from the axis by r, the velocity of m when the angular veloc- ity of the body is 1 will be equal to r and the momentum of m will be mr : but the particle is moving in a direction that is perpendicular to r ; hence the moment of the momentum with respect to the axis is mr^. The algebraic sum of the moments of the momenta of all the elements of the body will be equal to S (mr^), which is the expression for the moment of inertia of the body, that is, the moment of inertia of a body with respect to any axis is the algebraic sum of moments of the jnomenta of all its elements, with re- spect to the same axis, when the body is revolving with an angular velocity equai to 1. 183 MECHANTCS. 210.] If we denote the algebraic sum of the moments of the mo- menta of all the elements of a body by L, the angular velocity being u>, it is obvious that we shall have, i = 2 (mr^) Id — Aw (210) Moment of Inertia with, respect to Parallel Axes. 133. The moment of inertia of a body varies with the po- sition of the axis with respect to which it is taken. To investigate the law of variation, let AB represent any section of the body by a plane perpendicular to the axis ; C, the point in which this plane cuts the axis ; and Q, the c't'a point in which it cuts a parallel axis through *' the centre of gravity. Let P be any element of the body, whose mass is m, and denote PO hj r, PO 'hj s, and CG by h. From the triangle GPG, according to a principle of Trig- onometry, we have, r-3 = s^ + ^2 — 2sk cos CGP. Substituting in (309) and separating the terms, we have, E —1. (ms^) + -2 (m;fc3) _ 22 {msk cos OGP). Or, since k is constant, and 2 (m) = M, the mass of the body, we have, X = 2 (ms^) + MB — 3^2 {ms cos CGP), But s cos OGP = GH, the lever arm of the mass m, with respect to the axis through the centre of gravity. Hence, S {ms cos COP) is the algebraic sum of the moments of all the particles of the body with respect to the axis through the 311.] CElfTEIFUGAL FOSCE. — MOMENT OF INERTIA. 183 centre of gravity ; but, from the principle of moments, this is equal to 0. Hence, K=J. (ms^) + Mh^ (311). The first term of the second member is the 'moment of inertia, with respect to the axis through the centre of gravity. Hence, the moment of inertia of a body with respect to any axis, is equal to the moment of inertia with re- spect to a parallel axis through the centre of gravity, plus the m,ass of the body into the square of the distance between the two axes. The moment of inertia for any system of parallel axes is least possible when the axis passes through the centre of gravity. If any number of parallel axes be taken at equal distances from the centre of gravity, the moment of inertia with respect to each will be the same. Equation (311) enables us to find the moment of inertia with respect to any axis when we know the moment of iner- tia with respect to any parallel axis. If the first axis does not pass through the centre of gravity of th6 body, we pass to a parallel axis by diniinishing the given moment of inertia by Mk^ ; we then pass to the second axis by increasing the last result by Mk'^; k and k' are the distances from the first and second axes respectively, to the parallel axis through the centre of gravity of the body. Polar Moment of a Plane Surface. 134. The polar moment of inertia is the moment of inertia of a plane surface with respect to an axis perpendic- ular to the plane. 184 MECHANICS. [212. Let A GBD be any plane area, and the point in which it cuts an axis per- pendicular to it : through draw any two rectangular axes, OX and OY. Let m be an infinitesimal element whose co-ordinates are x and y, and whose dis- tance from is r. Then will the polar moment of inertia of the area be given by the equation, A' = 2 {mr'^'). . . . Kg. 118. (212) But, r^ = 2/2 + a;^ : substituting in (213) and separating the terms, we hare, /r = S [mf) + S (mx^). (213) The first term of the second member of (213) is the mo- ment of inertia of the given area with respect to the axis of X, and the second term is its moment of inertia with respect to the axis of T; denoting the former by Z, and the latter by /„, we have. K=I, + ly. (314) Hence, the polar moment of inertia of a plane area is equal to the sum of the moments of inertia of the area with respect tn any two rectangular axes in the plane of the area passing through the foot of the polar axis. Experimental Determiuation of Homent of Inertia. 135. When a body can be handled conveniently, its mo- ment of inertia may be found experimentally as follows : Make the axis horizontal, and allow the body to vibrate about it, as a compound pendulum. Find the time of a single vi- bration, and denote it by t. This value of t, in equation 215.] CENTRIFUGAL FOKCB. — MOMENT OF INEETIA. 185 (184), makes known the value of I. Determine the centre of gravity, and denote its distance from the axis by h. Find the mass of the body, and denote it by M. We have, from equation (190), Mkl — S (mr^) — K. (315) Substitute for M, I, and h, the values already found, and the value of K will be the moment of inertia, with respect to the assumed axis. Subtract from this the value of Mk"^, and the remainder will be the moment of inertia with respect to a parallel axis through the centre of gravity. When a body is homogeneous and of regular figure, its moment of inertia is most readily found by means of the cal- culus. To make formula (209) suitable to the application of the calculus, we have simply to change the sign of summation, S, to that of integration, /, and to replace m by dM and r by x. This gives, E = fxHM (216) Moment of Inertia of a Straight Line. 136. Let AB represent a physical straight line, G its centre of gravity, and E any ele- ment limited by planes at right angles to its length and infinitely near to each other. Denote the mass of the line by M, -its length by 11, the distance GE by x, and the length of the element by dx ; also denote the mass of the element by dM: we shall then have, Mdx M : dM :: 21 : dx, or, dM = -^^— ; D n? R 1 G E Fig. C 119. 186 MECHAKICS. [317. Substituting this in (216), and taking the integral between the limits — I and -\-l, we have. That is, the moment of inertia of a right line with respect to a perpendicular axis through its centre of gravity is equal to the mass of the line multiplied by one third of the square of half its length. For a parallel axis whose distance from G is d, the moment of inertia being denoted by K', we have, E' ^ Jf^lV^a) (218) For a parallel axis through the end of the line we have d^^l, and (318) becomes, E' = m{^^ + A='^MP (319) Formulas (317), (318), and (319) are entirely independent of the breadth of the filament AB : they will therefore hold good when the filament is replaced by the rectangle EF. In this case M represents the mass of the rectangle, 21 is its length, and d is the distance of the centre of gravity of the rectangle from an axis parallel to one of its ends. Moment of Inertia of a Thin Circular Plate. 137. In the_^rs^ place, let us find the polar moment of in- ertia, the axis being perpendicular to the plane of the circle at its centre. From the centre with a radius x describe a circle, and again with the radius x + dx describe a concentric circle; all points of the included ring are equally distant aaO.] CBNTBIFUGAL FOECE.— MOMENT OF IKEKTIA. 187 from the axis, and consequently its moment of inertia is equal to its mass multiplied by the square of X. Denoting the mass of the circle by /-"^ M and the mass of the elementary ring by dM, we have, i ,,• y M : dM : : -nr^ : 2-nx dx, from -which we deduce, ^' Mg. ISO. ,,, M ^ , %M , dM = — 5 x2TTxdx = — =-« dx. Substituting in (316), and integrating between the limits X = and a; = r, we have, 2M /»'■ Mr^ For a parallel axis at a distance d from the axis through the centre, we have, X' = m(^ + d^) (331) In the second place, to find the moment of inertia of a circle with respect to any diameter, we observe that the moment of inertia is the same for all diameters ; hence, from (314), we have, Mt^ K=^ (333) And for a parallel axis at the distance d, K' = if(J + ^') (333) 188 MECHANICS. [224. To find the moment of inertia of a cir- cular ring with reference to an axis perpen- dicular to the plane of the ring through its centre, we integrate (220) between the limits r' and r, r' being the inner radius of the ring and r the outer one. In this case, we have, 2if/r* r\ M Fig. 121. K-. r^ \4 ) = ^,ir^-r"){r^ + r'') [a) in which M is the mass of the outer circle. If we denote the mass of the ring by M', we have, M : M' V^ . T'™ M or — s = M' Substituting in (a), and reducing, we have. For a parallel axis, we have. (334) Kz= M i- 3 d^\ (335) It is to be noted that the preceding expressions are entirely independent of the thickness of the circular plate ; hence, they are- applicable to cylinders of any length (either solid or hollow), when their axes are parallel to the axis of rotation. Let the student solve the same problem, using polar co-ordinates. Moment of Inertia of any Solid of Revolution. 138. Let PA Q be the meridian section of a solid of revolu- tion, and take the axis OX of the solid as the axis of X. Let 226.] CENTEIFUGAL FORCE. — MOMENT OF INEETIA. 189 OZ, perpendicular to OX, be the axis^ with respect to which the moment of inertia is to be found. At a distance x from the origin pass a plane PQ perpendicular to OX, and at a distance X + dz pass a parallel plane. These planes will include an elementary slice of the solid whose volume is ny^dx, y being the ordinate of the meridian curve corresponding to the abscissa x. If we denote the entire volume by V, its mass by M, and the mass of the slice by dM, we have, ■nM M : dM : : F : mfdx, or dM -. V yMx. The moment of inertia of the slice with respect to OZ, may be found from Equation (223), by making M = dM, r = y, and d=-x: calling this moment of inertia dK, we have, dK . nM jty Whence, by integration. dx. (226) the limits being taken to include the entire body. Moment of Inertia of a Cylinder. 139. Let the axis of the cylinder coincide with the axis of X, and let the axis OZ be taken through the centre of gravity of the cylinder ; de- note the length of the cylinder by 2Z, and the radius of its cross section by r. Then will y = r, and V = 2-iTrH : sub- stituting in (226), and integrating between the limits — I and -|- I, we have. ID Fig. 128. 190 MECHANICS. [aa?- Fot an axis parallel to CD, we have, X- = M^l + l+d^ (328) Moment of Inertia of a Cone. 140. Let the axis of the cone coincide with the axis of X, and let OZ be perpendicular to it at the vertex of the cone ; denote the height of the cone by h, the radius of its base by r, and its mass by M. T 1 . . We shall have y ■= jx, and V = -nr% : substituting, in (226), and taking the integral between the limits and /*, we have. To find the moment of inertia when OZ passes through the centre of gravity of the cone, we subtract from (239) the /3 \' quantity M x yjhj , whence, "^ = ^S'^ + ¥ - re^^^ = lA' - 't^ ^'''^ Moment of Inertia of a Sphere. 141. Let the axis of X be taken to coincide with a diame- ter of the sphere, and let the axis of Z be perpendicular to it at its left hand extremity ; denote the radius of the sphere by r, and its mass by M. 4 "We shall then have, y^ = %rx — «', and F = ^-nr^ ; substi- 231.] CENTRIFUGAL FORCE. — MOMENT OF INERTIA. 191 tuting, in (226), and taking the integral between the limits and %r, we have, 3Jf r^l^iM — ^rofi + x^ „ , A , or. ZM r^l 3 \ 7 ^= Wo V'''' + '■'^ - -^y"" = r^'" (^^1) Por a parallel axis through the centre we subtract, from (231), the quantity, Mr"^, which gives, K' =^-Mr^ (232) The origin might have been taken at the centre in the first place, in which case we should at once have deduced equa- tion (232), from which we readily deduce equation (231). Moment of Inertia of a Solid with Respect to Its Axis. 143. Recurring to Article 138, let us suppose the circular slice, PQ, to revolve around the axis OX: its moment of inertia, which is the differential of the moment of inertia of the solid with respect to OX, may be. found, from (220), by making r z:^ y, and M equal to the mass of the slice. Art. 138 ; making these substitutions, and calling the result dK, we have, dK=-Y ■ ^dx. Whence, by integration, K^'^fy^dx (233) ■ In which y is the ordinate of the meridian curve. 192 MECHANICS. [234. Moment of Inertia of a Cylinder. 143. To find the moment of inertia of a cylinder with respect to its axis, we assume the co-ordinate axes, as in Art. 139. "We then have, V = ttt^ x 2Z and y ■=r. Substituting, in (233), and integrating from — Z to + Z, we haye, ^=WJ-1 ^^- = ^^' (234) a result that corresponds with (220). Moment of Inertia of a Cone. 144. Assume the position of the axis, and the notation of 1 r Art. 140. "We then have V = gWr^A, and y — ^x: substitut- ing, in (233), and taking the integral between the limits and li, we have, 2 r% />A ^4 3 Moment of Inertia of a Sphere. 145. Let the axis OX be a diameter of the sphere, and let OZ be perpendicular to it at its left hand extremity. If we 4 assume the notation already employed, we have, V = ^Trr'j O and 2/2 = 2rx — x^ ; substituting, in (233), taking the inte- gral between the limits and 2r, we have, 3 M r^ 2 ^ = 8 ^/o (^'■'^ ~ ^'■^ + oi^)dxz= ~Mr% (236) which corresponds to equation (232). 237.] CENTRIFtTGAL POECE. MOMENT OE INEETIA. 193 Moment of Inertia with Bespect to Axis of Symmetry. 146. Let APBQ be an ellipse, AB its transverse, and CD its conjugate axis. Let the origin of co-ordinates be taken at tbe centre 0; at a distance from equal to x, draw a double ordinate, PQ, and at a distance, x + dx, draw a parallel double ordinate : the area of the included filament is equal to 'iydx ; denoting the mass of this filament by dM, that of the ellipse being denoted by M, and calling the semi-axes of the curye a and h, we hare. M : dM : : nab : 2ydx, or 2Mydx nab (237) The moment of inertia ot PQ with respect to the axis AB, which is the differential of the entire moment of inertia, is given by Equation (217), by making M equal to the value of dM, Eq. 337, and I = y. Substituting these values, and integrating, we have. K^ %M Znab I y^dx (238) From the equation of the ellipse, referred to its centre, we have, y^Ua^-x^)^. Substituting in (338), reducing, and taking the integral between the limits — a and -|- a, we have. 3 Mb^ /»+ " 1 3 nd^J -a (239) 194 MECHANICS. [240. Applying formula B twice, and completing the integration by the sine formula, Calc, p. 107, we finally obtain E=:^Mb^ (240) In like manner, with respect to the conjugate axis, we find • the moment of inertia to be K-:^Ma^ (241) By a siinilar course of investigation we can find the moment of inertia of any plane area with respect to an axis of sym- metry. Centre and Radius of Gyration. 147. The centre of gyration of a body with reference to any axis is a point lying on a perpendicular to this axis through the centre of gravity, such that if the entire mass of the body were concentrated at it, its moment of inertia would not be changed. The distance from the centre of gy- ration to the axis is called the radius of gyration. Let M denote the mass of the body, and Ji its radius of gy- ration ; then will the moment of inertia of the concentrated mass, with respect to the axis, be equal to M.W ; but this must, by definition, be equal to the moment of inertia of the body with respect to the same axis, or to S {mr^) ; hence. MJfi = 2 {mr% or A = \/^-^- (342) Hence, to find the radius of gyration with respect to an axis, we divide its moment of inertia with respect to that axis, by the mass, and then extract the square root of the result. Since M is constant for the same body, it follows that the radius of gyration will he the least possible when the mOPient 343.1 CENTRIFUGAL POECE. MOMENT OF INERTIA. 195 of inertia is the least possible, tliat is, when the axis passes through the centre of gravity. This minimum radius is called the principal radius of gyration. If we denote the principal radius of gyration by h^, we shall hare, for the straight line. 1hz=~, and A = \/| + d^, (343) 'I V3' """^ '""V3 and in like manner we may treat the other magnitudes that have been considered. The straight line through the centre of gyration parallel to the axis with reference to which the centre of inertia is taken, is sometimes called the axis of gyration. If this line be revolved around the axis of rotation of the body, it will gener- ate a cylinder ; if the mass of the body be concentrated in any manner on the surface of this cylinder, its moment of inertia will not be changed. In a compound pendulum, the centre of gravity, the centre of oscillation, and the centre of gyration, all lie on the same straight line ; the last of these points lies between the other two. Equation (190) may be written 1 = ^=^, or U = h\ or h=VM (344) Hence, A is a mean proportional between Jc and I. When any two of these quantities are given, the other one may be found by geometrical construction. VII.— WOEK AND ENEEGY.— IMPACT. Relation Between Work and Energy. 148. The terms -work and energy are defined in Article 9. From that article and from Article 33 we see that the elementary quantity of work of a force is measured by the intensity of the force multiplied by the projection of the elementary path of its point of application on the direction of the force. Let a force P be exerted on a free body whose mass is m, and suppose that the point of application passes over a dis- tance k in the element of time dt. Let the distance 1c be projected on F, and denote the projection by ds. If we de- note the elementary quantity of work by dQ, we shall have, dQ = Fds. But in this case we have, from Art. 92, ^^""dt^- Substituting this value of F in the preceding equation, we have, d^ d' , Q is positive and work is stored up ; if 6" < 6', Q is negative, and the wheel gives out work. If the angular velocity increase from 6' to 6", and then de- crease to 6', and so on, alternately, the work accumulated during the first part of each cycle is given out during the second part, and any device that will make 6' and 6" more nearly equal, will contribute toward equalizing the motion of the machine. By suitably increasing the mass and radius of gyration, their difference may be made as small as desirable. Let the half-sum of the greatest and least angular velocities be called the mean angular velocity, and denote it by 6'". all 1 Qi We shall have — ~ — = 6'", and by factoring the second member of (258), we have, Q = iMk^ ((9" + 6') {e" - e') ; whence, by substituting the value of 6" -f 6', Q = Mk^ {9" — e') 6'" (259) Let us suppose the difference between the greatest and 260.] WORK AND ENERGY. — IMPACT. 209 least velocity, equal to the n^ part of their mean, that is, that ft"' 6" _ e' = — . n This, in (359), gives « = ^-', or, «. = !«,....« From this equation the moment of inertia of the wheel may be found, when we know n, Q, and 0'". The value of n may be assumed ; for most kinds of work a value of from 6 to 10 will be found to give sufficient .uniformity ; the value of 0'" depends on the character of the work to be performed, and Q is made known by the character of the motion to be regu- lated. To find the value of Q in any practical case we must find the quantity of work that is stored up in the first part of the cycle, which is also the quantity given out in the second part of the cycle. The method of proceeding is best illustrated by a practical example. The Crank and Crank Motion. 158. A crank is a device for converting reciprocating motion into rotary motion, or the reverse. In the diagram, DG represents a reciprocating piece, that is, a piece that moves back- ward and forward being re- .^'^^~^ strained by suitable guides, as Pn n 9,g g^fx--^--^-F is the case with the piston rod ~° ° ^\^ ■*■ J in a locomotive ; OB is a con- p. j^_ " ' necting rod having a hinge Joint at G and connected with AB loy means of a short axle called the crank pin ; AB is the crank arm solidly con- nected with an axle;tvhich is perpendicular to the plane GBA, and which is called the crank axis. 210 MECHANICS. When the crank pin is at E the reciprocating piece is at one limit of its play, and when the crank pin is at F the re- ciprocating piece is at the other limit of its play. The dis- tance EF, which is equal to twice the length of the crank arm, is called the throTV of the crank ; it measures the dis- tance through which the reciprocating motion of OD takes place. If we suppose a force P to act along DC, either pushing or pulling, it will have no tendency to produce rotation of the crank when the crank pin is either at ^or i?'; these points are therefore called dead points. "When the crank pin is at any other point, P will tend to produce rotation, the meas- ure of the tendency being equal to the component of P in the direction of BG, multiplied ly the perpendicular distance from A to this direction. If we suppose P to be constant, its component in the direc- tion of -BC will be variable, and it is obvious that the varia- tion will increase with the obliquity oi BG to GD. For this reason the connecting rod should be made as long as the nature of the mechanism will permit ; when it is 10 or 13 times as long as the crank arm, we may, in. most practical cases, regard the component along BG as constant, and as acting parallel to GD. We shall so regard it in the follow- ing discussion. We may suppose the force P to act in one direction, say in the direction from Gto I) ; or we may suppose it to act alter- nately in the directions from G to D and from D to G. In the former case the crank is said to be a crank of single action, and in the latter case it is a crank of double ac- tion. We have an example of the former in the ordinary lathe where the force applied to the treadle only acts downward ; we have an example of the latter in the locomotive where the force of the steam acts alternately in both directions. 261.] WOEK AND ENEEGT. — IMPACT. 211 Fly-Wheels to Regulate Crank Motion. 159. To compute the dimensions of a fly-wheel to regulate crank motion, let us first consider the case of a crank of sin- gle action. Let us take the connecting rod so long that it may, in all positions, he regarded as sensibly parallel to the re- ciprocating piece, which is supposed to lie in the direction ITK, and sup- pose the force P to act downward only, and through a distance equal to ILK, the diameter of the circle HGK de- scribed by the crank pin. Suppose the resistance overcome to be constant and equal in effect to a force Q acting with a lever arm 1. Denote the length of the crank arm by r. The work of P in one revolution is equal to P x 3r, and the work of Q in the same period is ^ x ^tt ; these are, by hypothesis, equal to each other, that is. ZnQ = %Pr, or -£ = - Pr n (261) When the crank pin is at IT, the moment of P is ; as the crank pin advances in its path, RGB, the moment of P in- creases, being equal io P •>(. AM. As long 2& P y. AM <, § X 1, the work of the power is less than the work of the resistance ; when P x AM = § x 1, the work of the power is equal to the work of the resistance ; and when P x -3/^ > § X 1, the work of the power is greater than the work of the resistance. In the last case the excess of work of the power is being stored up in the fly-wheel which is supposed to be mounted on the crank axis. 213 MECHANICS. [362. To find the position of the crank arm when work begins to be stored up, make AM O P X AM= Qxl, whence ^- = ;^^ (262) Denoting the corresponding angle EAC by (/>, the first member of (262) is equal to cos i^, and from (261) the second member is the reciprocal of n ; hence, we have, cos = -, or ^ = 71° 27' (263) Whilst the crank pin is moving over the arc GED, equal to 142° 54', work is being stored up. To find the amount of work stored up, we observe that the corresponding work of P is P X OD, or to P X 2?- sin 71° 27', and denoting this work by W, we have, W=^P X 2r sin 71° 27' = 2Pr x 0.9480 (264) The work of Q in the same period, denoted by W, is given by the equation, „, „ _ 142° 54' W =Qx2nx ^gQ^. Or, because, § x 27r = P x 2r, we have, W = 2P»- x ^Q^' = 2Pr x 0.3968 ..... (265) Subtracting (263) from (264), we have, for the amount of work stored up, W—W'=z 2Pr [.9480 — .3968] — Pr x 1.1024 .... (266) Substituting this for Q, in Equation (260), we have, jf;fc. = !L!iZy^i^4 ^^g^^ u If we denote the number of revolutions of the crank per second by iV, the value of 9'" will be 27riV"; this reduces (367) to the form. 268.1 WOKK AND ENEEGT. — IMPACT. 213 MB: n X Pr X 1.1024 : (268) Poncelet says that a value of n between 7.5 and 10 will secure a sufl&cient degree of regularity. Making n = 10, and substituting for n^ its value 9.87, we have finally. Pr X 11.024 (269) The first member of (269) is the moment of inertia of the required fly-wheel, and the second member is its value ; this latter is completely known from the conditions of the prob- lem. We may now assume Jc, the radius of gyration, and com- pute M; or, we may assume M, or, what is the same thing, we may assume the weight of the wheel, and compute &. In either case the problem is completely solved. Crank of Double Action. 160. Let us next consider a crank of double action, em- ploying the same notation as before. Let the force P act downward through the distance, 2r, and then upward through same dis- tance ; the quantity of work of P in an entire revolution of the crank will then be equal to 4:Pr, and we shall have, as before, Pr iPr=2^Q, or ^ = - (270) When the moment of P is equal to the moment of Q, we have, P X AM= Q, or — Pr' 214 MECHANICS. [271. Whence, by substitution, as before, we have, cos ^ = - , or ^ = 50° 28' (271) There will be four points, C, D, Q, and L, at which there will be an equilibrium between P and Q. In passing from to D the velocity will increase ; from D to G the velocity will decrease ; from G to L the velocity will increase again ; and from L to Cthe velocity will decrease. From Cto D work will be stored up to be given in passing from D to G, and in passing from G to L &u equal amount of work will be stored, to be given out in passing from H to G; that is, there are two cycles of change in each revolution. The work stored up in passing from OtoDis given by the equation, W— W' = P X2r sin 50° 28' - Q x 2rr ^J^'- Or, by reduction, W— W = Pr(1.54 — 1.12) = Pr X 0.42. Substituting this for Q in (260), and making n = 10, we have, Pr X 4:2 MV = ^^-^- (272) Denoting the number of revolutions per second by N, the value of 6»"' is 2-nN, which reduces (272) to Pr X 42 which gives the moment of inertia of the required fly-wheel. To compare the results in (269) and (273), let us suppose the work of the power during one revolution to be the same in both. In this case the value of P in (273) is one half that 274.] WOEK AND ENEEQT.— IMPACT. 315 of P in (269) ; consequently the moment of inertia of the fly- wheel to regulate single crank action is 11.034: -4- 2.1, or b'^ times that of the fly-wheel to regulate double crank action. Elasticity. — Impact. 161. Elasticity is that property by virtue of which a body tends to resume its original form after having been dis- torted by the action of some force. A body may be distorted iy compression, by extension, or iy twisting. The force that acts to change the form of a body may be called the force of distortion, and the force that tends to restore the body to its original shape may be called the force of restitution. If the force of restitution is equal to the force of distortion, the body is perfectly elastic ; if the force of restitution is 0, the body is perfectly inelastic. No body is either perfectly elastic or perfectly inelastic. Ivory, glass, and steel are among the more elastic bodies ; soft putty and tempered clay are examples of bodies that are nearly inelastic. If we denote the force of distortion by d, the force of resti- tution by r, and their ratio by e, we have, « =^ (274) The value of e, which depends upon the molecular action of the particles of a body, can be taken as a measure of the body's elasticity ; it may then be called the coefficient of elasticity. If one ball impinges upon another a succession of effects takes place which are dependent upon the molecular constitu- tion of the two bodies, and in general these effects take place in an exceedingly short space of time, so that the force ex- erted may be called impulsive. The first effect of the im- pact is to produce a compression of both bodies, and this 316 MECHANICS. [27S. continues until the molecular forces called into action are sufficient to resist further distortion. The balls, being en- dowed with a certain degree of elasticity, then tend to recover their spherical form, and in this effort they exert a further pressure upon each other, the force of restitution being equal to e times the force of distortion, (Bq. 274). Momentum and Velocity. 163. In the case of impact of two spherical balls, A and B, let us denote the mass of -4 by m and the mass of B by m', and furthermore let us denote the velocity of A before impact by v ^/--^^ mjv' and that of -5 by y', v being greater than v'. Before collision the aggre- gate momentum of the two bodies '^^ — ^ ^^ will be equal to mv -f- m'v'. When collision takes place a mutual pressure P will be set up, the effect of which will be to diminish the velocity of A and to increase that of B, and because action and reaction are equal, the momentum of A will be diminished as much as that of B is increased, and the same law will continue during the period of restitution. Hence, the aggregate momentum is not changed during the period of impact. Let us assume, as shown in the figure, that the centres of the balls are moving along the same right line, the velocity of A being v and that of B being v'. At any time t, before or after impact, denote the distance of A from a fixed plane, at right angles to the given direction, by x, and that of B from the same plane by x': the distance of the centre of gravity of A and B from the same plane will be given by the equation, mx + m'x' /„™e\ aJi = ^ — - (375) m + m ^ At the time t + dt the distance of A from the fixed plane 270-] WORK AND ENEEGT. — IMPACT. 217 will hex + vdt and the distance of B from the same plane will be x' + v'dt, and the distance of the centre of gravity of A and B from that plane will be given by the equation, m + m ^ ' Subtracting (375) from (376), and dividing by dt, we have, dxi mv + m'v' dt in + m! (377) It has just been shown that the second member of (377), which represents the aggregate momentum of the two bodies, is unchanged by the collision; hence, the first member, which represents the velocity of the centre of gravity, is also unchanged by the collision. Direct Central Impact. 163. If the centres of the two balls (Fig. 130) are moving along the same straight line when the collision takes place, the impact is said to be direct and central. Let us consider the direct and central impact of two spheres of the same material, and let e denote their common coefficient of elas- ticity. Denote the mass of the ball A by m, its velocity before im- pact by V, and its velocity after impact by u ; denote the mass of the ball B by m', its velocity before impact by v', and its velocity after impact by u' ; also, denote the common velocity of the two balls at the instant of greatest compression by w. The aggregate momentum, or simply the momentum of the two bodies before impact, is mv -{- m'v', and their momentum at the instant of maximum compression is {m -f m') w ; but from Art. 162, these two expressions are equal ; hence, / , n , ' r mv + m'v' ,„„_, (m -f- m) w = mv + mv' or w = ; ,— ■ • ■ • (378) ^^ ' m + m ^ 318 MECHANICS. [279- To find the velocities lost by A and gained by B during the period of compression, we subtract w from v and v' from w, giving after reduction, v — 'w= '^' , (v — v'), (279) m + m ^ ' «, _ ^,' = _ J!L_ („ _ ^,') (380) The velocities lost by A and gained by B during the period of restitution will be equal respectively to the second mem- bers of (279) and (280), each multiplied by e (Art. 161) : hence, the total loss of velocity by A, and gain of velocity by B during the entire impact, will be given by the equation, v-u={l + 6) —^—,{v - v'), (281) M - «' = (1 + e) _J?L_ (v _ J,') (282) The velocities of A and B after collision, found from the preceding equations by simple transposition, are m' « = ^ _ (1 + e) --^,(v - v'), (283) u' = 4;' + = wzdm cos 0. But dm cos ^ is equal to the horizontal projection of dm, in other words it is a horizontal section of a vertical prism of which dm is the base ; denoting this cross section by dm', we have, p' = w%dm' (302) 330 MECHANICS. If ^ < 90°, p' will be positive, and the pressure will be downward; if > 90°, p' will be negative, and the pressure will be upward. Let figure 134 represent a submerged vertical filament : from what precedes, we see that it will be pressed downward by a force that is equal to the weight of a column of the fluid whose base is the cross-section of the filament and whose alti- tude is the distance at AB from the free surface ; and it will be pressed upward by a force that is equal to the weight of a column of the fluid whose base is the cross-section of the fila- ment and whose altitude is the distance of UF from the free surface. This principle is used hereafter in finding an expression for the buoyant effect of a fiuid. Examples. 1. A hollow sphere is filled with a liquid. How does the entire press- ure, on the interior surface, compare with the weight of the liquid ? Ans. The pressure is three times the weight of the liquid. 3. A hollow cylinder, with a circular base, is filled with a liquid. How does the pressure on the interior surface compare with the weight of the liquid ? Ans. The pressure is equal to — ; — times the weight of the liquid ; here h is the height of the cylinder, and r is the radius of its base. 3. A right cone, with a circular base, stands on its base, and is filled with a liquid. How does the pressure on the internal surface compare with the weight of the liquid ? Zr + 2 Vh^ 4- r^ Ans. The pressure equals times the weight of the liquid ; here h is the altitude, and r the radius of the base. 4. Required the relation between the pressure and the weight in the preceding case, when the cone stands on its vertex. Vh^ + r* Ans. The pressure is times the weight of tht liquid. MECHANICS OF LIQUIDS. 231 5. What is the pressure on the lateral faces of a cubical vessel filled with water, the edges of the cube being 4 feet, and the weight of the water 63^ lbs. per cubic foot ? A.na. 8,000 lbs. 6. A cylindrical vessel is filled with water. The height of the vessel is 4 feet, and the radius of the base 6 feet. What is the pressure on the lateral surface ? Ans. 18,850 lbs., nearly. 7. A sluice-gate 10 feet square is placed vertically in the water, its upper side coinciding with the free surface; what are the respective pressures on the upper and lower halves, the weight of water being 62^ lbs. per cubic foot ? Ans. 7,812.5 lbs. and 23,437.5 lbs. ■ 8. What are the pressures on the two triangles formed by drawing one diagonal of the gate ? Ans. 10,416f lbs. and 20,838J lbs. Centre of Pressure. 169. The centre of pressure on a surface is the point at which the resultant pressure intersects the surface. Let us consider the case of a submerged plane surface, ABED, whose prolongation inter- sects the free surface, XOF, of the liquid in OX, and whose inclina- tion to the free surface is equal to a. Let OJTbe taken as the axis of X, and let OY, perpendicular to OX, and lying in the plane of ABED, be taken as the axis of Y, the value of y being positive downward. Let ABED be divided into infinitesimal elements by lines parallel to OX ; let DB, denoted by dm, be one of these ele- ments, and denote the co-ordinates of its centre of gravity, P, by X and y. The resultant pressure on DB will pass through P, and be- cause the distance of P below the free surface is equal to y sin a, the value of this pressure will, be, from (399), w • y sin a • dm, 232 MECHANICS. [303. and because it is perpendicular to ABED, its moment with respect to the axis of X will be wy^ smadm, and the sum of the moments of all of the elements with re- spect to OX will be w sin « fy^dm (303) Denote the co-ordinates of the centre of gravity of ABED by Xi and yi, and the co-ordinates of the centre of pressure, G, by x' and y'. Then, from Art. 168, the entire pressure on ABED will be wyi sin « fdm, and its moment with respect to OX will be y' y. w sin ay^ fdm (304) But the expressions (303) and (304) are equal ; equating them, and finding the value of y', we have, y' = ^ (305) In like manner, the sum of the moments of the elementary pressures with respect to the axis of Y is w sin a fxydm, and that of the entire pressure with respect to the same axis is x'w sin axi fdm. ' Equating these expressions, and solving, we have. 30r.] MECHANICS OF LIQUIDS. 233 If the surface has a line of symmetry perpendicular to OX, which is generally the case in practice, we may take that line as the axis of T, in which case x' will be equal to 0, and the centre of pressure will be determined by equation (305). Equation (305) shows that tJie distance of the centre of pressure on the plane surface from the intersection of its plane with the free surface, is equal to the moment of inertia of the surface divided hy the moment of the area of the surface, hath taken with respect to the line of intersection. If we denote the moment of inertia of the surface by M (P + y^), the moment of the mass will be equal to My^, and from equation (305), we have, y'-| +yr> (307) in which h is the principal radius of gyration parallel to OT, and yi is the distance of the centre of gravity from the same line OX. Because equation (305) is entirely independent of a, the position of the centre of pressure with respect to OX is independent of the inclination of the plane. Examples. 1. Where is the centre of pressure on a rectangular flood-gate, the upper line of the gate coinciding with the surface of the water. Solution. — If we denote the length of the gate by %l, we have from (217), h^ = P'; we also have y^ =1. Substituting them in (307), we hiivo, Hence, the centre of pressure is on the line of symmetry and two thirds of the distance from the top to the bottom of the gate. 2. Let it be required to find the pressure on a submerged rectangular 334 MECHANICS. flood-gate ABGD, the plane of the. gate being vertical. Also, the distance of the centre of pressure below the surface of the water. E & J iC 1C" D Fig. la B Solution. — Let EF be the intersection of the plane with the surface of the water, and suppose the rectangle AO to be prolonged till it reaches ^^. Let C, C, and C", be the centres of pressure of the rectangles EC, EB, and AC, respectively. Denote the distance OC", by z, the distance ED, by a, and the distance EA, by a'. Denote the breadth of the gate, by 6, and the weight, a unit of volume of the water, by w. The pressure on EC will be equal to ^a'hw, and the pressure on EB will be equal to ^a'^hw ; hence, the pressure on AG will be equal to which is the pressure required. Prom the principle of moments, the moment of the pressure on AG, is equal to the moment of the pressure on EC, minus the moment of the pressure on EB. Hence, from the last problem, \hw (ffl= — a'^) X z= ibwa' x fas — ^bwa'^ x |a', which is the required distance from the surface of the water. This distance may be found directly from Equation (305). 3. Let it be required to find the pressure on a rectangular flood-gate, when both sides are pressed, the water being at different levels on the two sides. Also, to find the centre of pressure. Solution. — Denote the depth of water on one jg side by a, and on the other side by a', the other elements being the same as before. |c The total pressure wUl, as before, be equal to, ibwia" -a">). Estimating z from C upward, we have Fig; 137.' Am. MECHANICS OB LIQUIDS. 235 4. A sluice-gate, 10 feet square, is placed vertically, its upper edge coinciding with the surface of the water. What is the pressure on the Tipper and lower halves of the gate, respectively, the weight of a cubic foot of water being taken equal to 62^ lbs. ? Ans. 7812.5 lbs., and 33437.5 lbs. 5. What must be the thickness of a rectangular dam of granite, that it may neither rotate about its outer angular edge nor slide along its base, the weight of a cubic foot of granite being 160 lbs., and the co- efficient of friction between it and the soil being .6 ? Solution. — First, to find the thickness necessary to prevent rotation outward. Denote the height of the wall, by h, and suppose the water to extend from the bottom to the top. Denote the thickness, by t, and the length of the wall, or dam, by I. The weight of the wall in pounds, will be equal to IM X 160 ; and this being exerted through its centre of gravity, the moment of the weight with respect to the outer edge, as an axis, will be equal to ^tHh X 160 = SOlht^. The pressure of the water against the inner face, in pounds, is equal to This pressure is applied at the centre of pressure, which. Example 1, is at a distance from the bottom of the wall equal to ^h ; hence, its moment with respect to the outer edge of the wall, is equal to Ih" X 10.4166. The pressure of the water tends to produce rotation outward, and the weight of the wall acts to prevent this rotation. In order that these forces may be in equilibrium, their moments must be equal ; or mhfi = W X 10.4166. Whence, we find, t - A\/.130a = .36 x h. Next, to find the thickness necessary to prevent sliding along the base. The entire force of friction due to the weight of the wall, is equal to 160Z^X, and x represents KE. If we describe a semicircle on EL, and through D draw an ordinate, DH, we have, from a property of the circle, DII= ^/DK ■ DL = VW. Hence we have, by substitution, KE — 2DS. 252 MEcnANics. Since there are two points on KL at which the ordinates are equal, there must be two orifices through which the fluid will spout to the same distance on the horizontal plane KE; one of these is as far above the centre, 0, as the other is below it. If the orifice be at 0, midway between ^and L, the ordinate, OS, will be greatest possible, and the range, KE', will bo a maximum. The range in this case will be equal to the diam- eter of the circle, LHK, or to the distance from the surface of the water in the vessel to the horizontal plane. Let a semi-parabola, LE', be described, having its axis, LK, vertical, its focus at K, and its vertex at L ; then if we suppose a jet to issue from K, being directed upward at dif- ferent angles by a short pipe, it may be shown, as in Art. 108, that every point within the curve may be reached by two jets, every point on the curve may be reached by one jet, and that points lying without the curve cannot be reached at all. A jet thrown obliquely upward, as shown at A in Fig. 147, is an arc of a parabola, since each particle may be regarded as a projectile ; the circumstances of motion of the particles will be made known by Equation (160). In like manner, the same equation will make known the circumstances of motion when a jet is projected obliquely downward. Coefficients of Efflux and Velocity. 185. When a vessel empties itself through a small orifice at its bottom, it is observed that the particles of fluid near the top descend in vertical lines ; when they approach the bottom they incline toward the oriflce, the converging lines of fluid particles tending to cross fsach other as they emerge from the vessel. The result is, that the stream grows nar- rower, after leaving the vessel, until it reaches a point at a distance from the vessel equal to about the radius of the MECHANICS OF LIQUIDS. 253 orifice, when the contraction becomes a minimum, and below that point the vein again spreads out. This phenomenon is called the contraction of the vein. The cross-section at the most contracted part of the vein, is not far from -^^ of the area of the orifice, when the vessel is very thin. If we denote the area of the orifice, by a, and the area of the least cross-section of the vein, by a', we shall have, a' = ka, in which ^ is a number to be determined by experiment. This number is called the coefficient of contraction. To find the quantity of water discharged through an orifice at the bottom of the containing vessel, in one second, we have only to multiply the area of the smallest cross-section of the vein, by the velocity. Denoting the quantity discharged in one second, by Q', we shall have. Q' = hav'^gh. This formula is only true on the supposition that the actual velocity is equal to the theoretical velocity, which is not the case, as has been shown by experiment. The theoretical velocity has been shown to be equal to V2^A, and if we de- note the actual velocity, by v', we shall have, v' = I V^gh, in which I is to be determined by experiment ; this value of I is slightly less than 1, and is called the coefficient of velocity. In order to get the'actual discharge, we must replace V'^gh by I V^gh, in the preceding equation. Doing so, and denot- ing the actual discharge per second, by Q, we have, Q = Ha V2gh. The product Id, is called the coefficient of efflux. It 254 MECHANICS. [316. has been shown by experiment, that this coefficient for orifices in thin plates, is not quite constant. It decreases slightly, as the area of the orifice and the velocity are in- creased ; and it is further found to be greater for circular orifices than for those of any other shape. If we denote the coefficient of effiux, by m, we hare, Q = ma ^/2gh (316) In this equation, h is called the head of water. Hence, we may define the head of loater to be the distance from the orifice to the plane of the upper surface of the fluid. The mean value of m corresponding to orifices of from \ an inch to 6 inches in diameter, with from 4 to 20 feet head of water, has been found to be about .615. If we take the value of A = .64, we shall have. That is, the actual velocity is only -^^ of the theoretical velocity. This diminution is due to friction, viscosity, etc. Efflux ThrouglL Short Tabes. 186. It is found that the discharge from a given orifice is increased, when the thickness of the plate through which the flow takes place, is increased ; also, when a short tube is introduced. When a tube AB, is employed which is not more than four times as long as the diameter of the orifice, the value of m becomes, on an average, equal to .813 ; that is, the discharge j,. ^^ per second is 1.325 times greater when the tube is used, than without it. In using the cylindrical tube, the contraction takes place at the outlet of the vessel, and not at the outlet of the tube. MECHANICS OF LIQUIDS. 355 Compound mouth-pieces are sometimes used formed of two conic frustums, as shown in the figure, having the form of the vein. It has been shown by Btelwein, that the most effective tubes of this form should have the diameter of the cross-section CD, equal to .833 of the diameter AB. The angle made by the sides Kg. iso. CF and DE, should be about 5° 9', and the length of this portion should be three times that of the other. Examples. 1. With what theoretical velocity will water issue from a small orifice 16^ feet below the surface of the fluid ? Ans. 33J ft. 3. If the area of the orifice, in the last example, is ^ of a square foot, and the ooeffleient of eflux .615, how many cubic feet of water will be discharged per minute? i Ans. 118.695 cw. y?. 3. A vessel, constantly filled with water, is 4 feet high, with a cross- section of one square foot; an orifice in the bottom has an area of one square inch. In what time will three fourths of the water be drawn ofE, the coefficient of efflux being .6 ? Ans. | of a minute, nearly. 4. A vessel is kept constantly full of water. How many cubic feet of water will be discharged per minute from an orifice 9 feet below the upper surface, having an area of 1 square inch, the coefficient of efflux being .6 ? Ans. 6 cubic feet, about. 6. In the last example, what will be the discharge per minute, if we suppose each square foot of the upper surface to be pressed by a force of 645 lbs. ? Ans. 8f cubic feet, about. 6. The head of water is 16 feet, and the orifice is -j^ of a square foot. What quantity of water will be discharged per second, when the orifice is through a thin plate? Solution. — In this case, we have, Q = .615 X .01 V2 X 33J x 16 = .197 cubic feet. When a short cylindrical tube is used, we have, § = .197 X 1.325 =: .?61 cubic feet. In Etelwein's compound mouth-piece, if we take the smallest cross- 35G MECHANICS. [317- section as the orifice, and denote it by a, it is found that the discharge is 2J times that through an orifice of the same size in a thin plate. In this case, we have, supposing a = -j^ of a square foot, Q = .197 X 2J = .49 cubic feet. Time Required for a Vessel to Empty Itself. ISt. Let it be required to find the time in which a Tessel will empty itself through a small orifice in its bottom. As- sume a horizontal plane of reference through the orifice, and denote the distance from this plane to the free surface by z ; denote the area of the free surface by S, and suppose this surface to be depressed by the outflow through a distance dz in the time dt ; the entire volume of the outflow in the time dt will then be equal to Sdz. But the effective velocity of outflow is equal to m\^2gz;, if we denote the area of the orifice by a, we shall therefore have for the volume of the outflow ma^/^gz dt. Equating these values, we have ma'\/"lgzdt =8dz,oxdt =L := • -r, (317) ma'\/%g 2* in which the second member is made negative because z di- minishes as t increases. Integrating (317) from li the original height of the liquid above the orifice to 0, we have t = — ^ _ /'" ^^ (318) which gives the required time. Examples. 1. Let S be constant and equal to A : in this case we have from (318), for the time required for the vessel to empty itself. 319-1 MECHANICS OF LIQUIDS. 257 t = — = —4/— 319) ma y-^g mar g ^ ' in which m may be taken equal to .615. 3. Let the surface be a surface of revolution whose axis is vertical, and let the equation of its meridian curve be a;* =:p^z. In this ease we have S=vx^ — irp'z^ ; substituting in (318), we have t = ^-= f°irp^dz (320) ^V2gJh Performing the integration from h to z, we have t = -^= (h-z) (331) From which we see that t varies as the height through which the free surface falls. Making z = 0, in (330), we have for the time required for the vessel to empty itself, t=^^^ (332) ma 4/3^ Other examples may be solved in a similar manner. Motion of Water in Open Channels. 188. When i^ater flows through an open channel, as in a river, canal, or opeli aqueduct, the form of the channel being always the same, and the supply of water being con- stant, it is a matter of observation that the flow becomes steady ; that is, the quantity of water that flows through any cross-section, in a given time, is constant. On account of adhesion, friction, etc., the particles of water next the sides and bottom of the channel have their motion retarded. This retardation is imparted to the next layer of particles, but in a less degree, and so on, till a line of particles is reached 358 MECHANICS. whose velocity is greater than that of any other filament. This line, or filament of particles, is called the axis of the stream. In the case of cylindrical pipes, the axis coincides sensibly with the axis of the pipe ; in straight, open chan- nels, it coincides with that line of the upper surface which is midway between the sides. A section at right angles to the axis is called a cross-sec- tion, and, from what has been shown, the velocities of the fiuid particles will be different at different points of the same cross-section. The mean velocity corresponding to any cross-section, is the average velocity of the particles at every point of that section. The mean velocity may be found by dividing the volume which flows through the section in one second, by the area of the cross-section. Since the same vol- ume flows through each cross-section per second, after the flow has become uniform, it follows that, in channels of vary- ing width, the mean velocity, at any section, will be inversely as the area of the section. The intersection of the plane of cross-section with the sides and bottom of the channel, is called the perimeter of the section. In the case of a pipe which is constantly filled, the perimeter is the entire line of intersection of the plane of cross-section, with the interior surface of the pipe. The -mean velocity of water in an open channel depends, in the first place, upon its inclination to the horizon. As the inclination becomes greater, the component of gravity in the direction of the channel increases, and, consequently, the velocity becomes greater. Denoting the inclination by I, and resolving the -force of gravity into two components, one at right angles to the upper surface, and the other parallel to it, we shall have for the latter component, g sin /. This is the only force that acts to increase the velocity. 332.] MECHANICS OF LIQUIDS. 359 The Telocity will be diminished by frictioiij, adhesion, etc. The total effect of these resistances will depend upon the ratio of the perimeter to the area of the cross-section, and also upon the velocity. The cross-section being the same, the resistances will increase as the perimeter increases ; conse- quently, for the same cross-section, the resistance of friction will be the least possible when the perimeter is the least pos- sible. The retardation of the flow will also diminish as the area of the cross-section is increased, other things remaining unchanged. If we denote the area of the cross-section by a, the perim- eter by P, and the velocity by v, we shall have, " ^^^ = /W, in which f(v) denotes some function of v. Since the inclination is very small in all practical cases, we may place the inclination itself for the sine of the inclination, and doing so, it has been shown by Peont, that the function of V may be expressed by two terms, one of which is of the first, and the other of the second degree, with respect to v ; or, 2-p- = mv -\- nv\ Denoting -^ by J2, — by k, and - by I, we have, finally, kv + Iv^ = RI, in which h and I are constants, to be determined by experi- ment. According to Etelweik, we have, h = .0000342651, and I = .0001114155. 260 MECHANICS. [323. Substituting these values, and solving with respect to v, we have. v= — 0.1088941604 + V.0118580490 + 8975.414385^/, from which the velocity can he found when E and I are known. The values of 7c and I, and consequently that of v, were found by Pkont to be somewhat different from those given above. Those of Etelwein are selected for the reason that they were based upon a much larger number of experi- ments than those of Peojsty. Having the mean velocity and the area of the cross-section, the quantity of water delivered in any time can be computed. Denoting the quantity delivered in n seconds by Q, and re- taining the preceding notation, we have, Q = nav (323) The quantity of water to be delivered is generally one of the data in all practical problems involving the distribution of water. The difference of level of the point of supply and delivery is also known. The preceding principles enable us to give such a form to the cross-section of the canal, or aque- duct, as will insure the requisite supply. Were it required to apply the results just deduced, to the case of irregular channels, or to those in which there were many curves, a considerable modification would be required. The theory of these modifications does not come within the limits assigned to this treatise. For a complete discussion of the whole subject of hydraulics in a popular form, the reader is referred to the TraiU d' Hydraulique D'AuBissoN. Motion of Water in Pipes. 189. The circumstances of the motion of water in pipes, are closely analogous to those of -its motion in open channels. MECHANICS OF LIQUIDS. 261 The forces which tend to impart motion are dependent upon the weight of the water in the pipe, and upon the height of the water in the upper reservoir. Those which tend to prevent motion ^Je depend upon -the depth of water in the ML --— ^ j V ■ lower reservoir, friction in the pipe, ^^^^^ ^B*^ adhesion, and shocks arising from ir- i^^^ regularities in the bore of the pipe. The retardation due to shocks will, for the present, be neglected. Let AB represent a straight cylindrical pipe, connecting two reservoirs R and R'. Suppose the water to maintain its level at E, in the upper, and at G, in the lower reservoir. Denote AH by h, and BC hy h'. Denote the length of the pipe by I, its circumference by c, its cross-section by a, its inclination by 4>, and the weight of a unit of volume of water by w^ Experience shows that, under the circumstances above in- dicated; the flow soon becomes uniform. We may then re- gard the entire mass of fluid in the pipe as a coherent solid, moving with a mean uniform velocity down the inclined plane AB. The weight of the water in the pipe will be equal to wal. If we resolve this weight into two components, one perpen- dicular to, and the other coinciding with the axis of the tube, we shall have for the latter component, wal sin 0. But I sin is equal to DB. Denoting this distance by h", we shall have for the pressure in the direction of the axis, due to the weight of the water in the pipe, the expression wah". This pressure acts from A toward B. The pressure due to the weight of the water in R, and acting in the same direction, is wah. The forces acting from B toward A, are, first, that due to the weight of the water in R', which is equal to wah' ; and, secondly, the resistance due to friction and adhesion. This resistance depends upon the length of the pipe, its circumfer- 262 MECHAKICS. [324. ence and the velocity. It has heen shown, by experiment, that this force may be expressed by the term, cl {kv + k'l^). Since the velocity has been supposed unif orin, the forces acting in the direction of the axis must be in equilibrium. Hence, wah + wdh" = wah' + cl {kv + k'v^) ; whence, by reduction, k w k' , a/k + h"-h'\ w c\ I • I The factor - is equal to one fourth of the diameter of the pipe. Denoting this by d, we shall have, - =. ^d; denoting k , k\ h + h" — li' . — by m, — by n, and 7 by s, we have, mv + niy^ = \sd (3^4) The values of m and n, as determined experimentally by Peony, are, m = 0.00017, and n = 0.000106. Hence, by substitution, .00017W + .000106?;a = ^sd. If V is not very small, the first term may be neglected, which will give, V = 48.56 Vsd. If we denote the quantity of water delivered in n seconds by Q, we shall have, Q = nav = 48.56 naVsd. (325) MECHANICS OF LIQUIDS. 363 The velocity will be greatly dimiDished if the tube is curved to any considerable extent, or if its diameter is not uniform throughout. It is not intended to enter into a dis- cussion of these cases ; their complete development would require more space than has been allotted to this branch of Mechanics. General Remarks on the Distribution and Flow of Water in Pipes. 190. Whenever an obstacle occurs in the course of an open channel or pipe, a change of velocity must take place. In passing the obstacle, the velocity of the water will increase, and then, impinging upon that which has already passed, a shock will take place. This shock consumes a certain amount of kinetic energy, and thus diminishes the velocity of the stream. All obstacles should be avoided ; or, if any are un- avoidable, the stream should be diminished, and again en- larged gradually, so as to avoid, as much as possible, the necessary shock incidient to sudden changes of velocity. For a like reason, when a branch enters the main channel, it should be made to enter as nearly in the direction of the current as possible. All changes of direction give rise to mutual impacts amongst the particles, and the more, as the change is more abrupt. Hence, when a change of direction is necessary, the straight branches should be made tangential to the curved portion. The entrance to, and outlet from a pipe or channel, should be enlarged, in order to diminish, as much as possible, the coefiQcients of ingress and egress. When a pipe passes over uneven ground, sometimes ascend- ing, and sometimes descending, there is a tendency to a col- lection of bubbles of air, at the highest points, which may finally come to act as an impeding. cause to the flow. There 264 MEOHAKICS. should, therefore, be suitable pipes inserted at the highest points, to permit the confined air to escape. Finally, attention should be given to the form of the cross- section of the channel. If the channel is a pipe, it should be made cylindrical. If it is a canal or open aqueduct, that form should be given to the perimeter which would give the greatest cross-section, and, at the same time, conform to the necessary conditions of the structure. The perimeter in open channels is generally trapezoidal, from the necessity of the case ; and it should be remembered, that the nearer the form approaches a semicircle, the greater will be the flow. IX.— MECHANICS OF GASES AND VAPOKS. Gases and Vapors. 191. G-ases and vapors are distinguished from other fluids by their great compressibility and correspondingly great expansibility. They continually tend to expand, and if left free the expansion will go on till counteracted by some ex- traneous force, as that of gravity, or the resistance offered by a containing vessel. The force with which a gas or vapor tends to expand is called its tension, or its elastic force. When the pressure exerted by a gas or vapor on a surface is uniform, that is, when it is the same at all points of the surface, we take as its unit of measure the pressure on a square inch of the surface. If we denote this unit by^, the area pressed by a, and the total pressure by P, we then have, P^ap (326) When the pressure is variable, we take for the unit of measure at any point the pressure that would be exerted on a square inch, if the pressure were the same at every point of the square inch, as at the point in question. But we may regard the pressure on any infinitesimal element, dm, as con- stant throughout the element ; hence, in this case we have, as before, P — dmxp (337) Many of the principles already demonstrated for liquids hold good for gases and vapors, but there are certain prop- erties arising from elasticity which are peculiar to aeriform bodies, some of which if is now proposed to investigate. 266 MECHAKICS. Atmospheric Air. 193. The gaseous fluid that envelops our globe, and ex- tends on all sides to a distance of many miles, is called the atmosphere. It consists principally of nitrogen and oxy- gen, together with small but variable portions of watery vapor and carbonic acid, all in a state of mixture. On an average, it is found that 1000 parts by volume of atmospheric air, taken near the surface of the earth, consist of about, 788 parts of nitrogen, 197 parts of oxygen, 14 parts of watery vapor, 1 part of carbonic acid. The atmosphere may be taken as a type of gases, for it is found by experiment that the laws regulating density, expan- sibility, and elasticity, are the same for all gases and vapors, so long as they maintain a purely gaseous form. "It is found, however, in the case of vapors, and of those gases which have been reduced to a liquid form, that the laws change just before actual liquefaction. This change appears to be somewhat analogous to that ob- served when water passes from the liquid to the solid form. Although water does not actually freeze till reduced to a temperature of 32° Fah., it is found that it reaches its maxi- mum density at about 39°, at which temperature the particles seem to commence arranging themselves according to some new law, preparatory to taking on the solid form. Atmospheric Pressure. 193. If a tube, 35 or 36 inches long, open at one end and closed at the other, be filled with pure mercury, and inverted in a basin of the same, it is observed that the mercury B JMECHAinCS OF GASES AND VAPOBS. 367 at the level of the sea will fall in the tube until the vertical distance from the surface of the mercury in the tube to that in the basin is about 30 inches. This c column of mercury is sustained by the pressure of the atmosphere exerted upon the surface of the mercury in the basin, and transmitted through the fluid, accord- ing to the general law of trans?)iission of pressures. The column of mercury sustained by the elasticity of the atmosphere is called the barometric column, be- cause it is generally measured by an instrument called a barometer. . In fact, the instrument just described. Fig. 15?. when provided with a suitable scale for measuring the altitude' of the column, is a barometer. The height of the barometric column fluctuates somewhat, even at the same place, on account of changes of temperature, and other causes yet to be considered. Observation has shown, that the average height of the barometric column at the level of the sea, is a little less than 30 inches. But the weight of a column of mercury 30 inches in height is nearly 15 lbs. ; hence, a pressure of 15 lbs. on'a square inch is adopted as a tmit of measure, and is technically called an atmosphere. This unit is often employed in measuring the pressure of elastic fluids, particularly in the case of steam. Thus, when we say that the pressure of steam in a boiler is two atmos- pheres, we are to understand that there is a pressure of 30 lbs. on each square inch of the interior of the boiler. In general, when we say that the tension of a gas or vapor is equal to n atmospheres, we mean that it is capable of exerting a pressure of n times 15 lbs. on each square inch of the surface with which it may be in contact. The specific gravity of mercury being nearly 13.6, a mer- curial column of 30 inches is equivalent to a water column of 34 feet. 268 MECHANICS. Mariotte's Law. 194. When a given mass of any gas or vapor is com- pressed so as to occupy a smaller space, other things being equal, its elastic force is increased ; on the contrary, if its volume is increased, its elastic force is diminished. The law of increase and diminution of elastic force, first discovered by Maeiotte, and bearing his name, may be enunciated as follows : The elastic force of a given, mass of any gas, whose temperature remains the same, varies inversely as the volume which it occupies. As long as the mass remains the same, the density must vary inversely as the volume occupied. Hence, from Mari- otte's law, it follows, that. The elastic force of any gas, whose temperature re- mains the same, varies as its density, and conversely, the density vanes as the elastic force. Mariotte's law may be verified for atmospheric air, by an instrument called Mabiotte's tube. This is a tube, A BCD, of uniform bore, bent so that its two branches are parallel to each other. The shorter branch, AB, is closed at its upper extremity, whilst the longer one is open. Between the two branches, and attached to the frame, is a scale of equal parts. To use the instrument, place it in a vertical po- sition, and pour mercury into the tube, until it just cuts off communication between the two branches. The mercury will then stand at the same level, BC, in both branches, and the tension of the air in AB, will be exactly equal to that of the ex- Kg. 153, temal atmosphere. If an additional quantity of mercury be poured into the longer branch, the air in the shorter branch will be compressed, and the mercury will rise 328.] MECHAliriCS OF GASES AKD VAPORS. 269 in both branches, but higher in the longer, than in the shorter one. Suppose the mercury to have risen in the shorter branch, to K, and in the longer one, to P. There will be an equilibrium in the mercury lying below the hori- zontal plane, KK ; there will also be an equilibrium between the tension of the air in AK, and the forces which give rise to that tension. /These forces are, the pressure of the ex- ternal atmosphere, transmitted through the mercury, and the weight of a column of mercury whose base is the cross-section of the tube, and whose altitude is PK. If we denote the height of the column of mercury sustained by the pressure of the external atmosphere, by h, the tension of the air in AK, will be measured by the weight of a column of mercury, whose base is the cross-section of the tube, and whose height is h -|- PK. Since the weight is proportional to the height, the tension of the confined air is proportional to h + PK. Now, whatever may be the value of PK, we have, from the assumed law, AK : AB :: h : h + PK; whence, ^^=;rri4 (^^«) If PK = h, we have, AK=^AB; if PK = 2h, we have, AK=iA£; if PK = nh, n being any positive number, AB entire or fractional, we have, AK=: -. Formula (338), deduced from Maeiotte's law, was verified by Duiong and Aeago for all values of n, up to w = 37. The law may also be verified when the pressure is less than an atmosphere, by the following apparatus : ^^ is a tube of uniform bore, closed at its upper and open at its lower- ex- tremity ; CD is a deep cistern of mercury. The tube, AK, is either graduated into equal parts, commencing at A, or has attached to it a graduated scale of brass or ivory. 270 MECHANICS. [329- To use the instrument, pour mercury into the tube till it is nearly full ; place the finger over the open end, invert it in the cistern, and depress it till the mercury stands at the same level without and within the tube, and sup- n a pose the surface of the mercury in this case to be at Hit B. Then will the tension of the air, in AB, be equal to that of the external atmosphere. If the tube be raised vertically, the air va. AB will expand, its tension will diminish, and the mercury will fall in the tube, to maintain the equilibrium. Suppose the level of the mercury in the tube, to have reached K. In this position of the instrument the tension dI of the air in AE, added to the weight of the column of mercury, KE, will be equal to the tension of the external air. Now, it is found, whatever may be the value of KE, that whence. AE: AB :: 7i : h — EE; ^^ = ^E (^^^) If EE = \Th, we have, AE = 2AB ; if EE = f A, we have AE = SAB ; in general, if EE= -7i, we have, AE = AB {n + 1). Formula (339) has been verified, for all values of n, up to n = 111. It is a law of Physics that, when a gas is suddenly com- pressed, heat is evolved, and when a gas is suddenly expanded, heat is absorbed ; hence, in making the experiment, care must be taken that the temperature be kept uniform. More recent experiments have shown that Mariotte's law is not strictly true, especially for high tensions, yet its variation , is so small that the error committed in regarding it as true is not appreciable in practical mechanics. 330.] MECHASriCS OF GASES AND VAPORS. 271 Gay Lnssac's Law. 195. II the volume of any gas or vapor remain the same, and its temperature be increased, its tension is increased also. If the pressure remain the same, the volume of the gas in- creases as the temperature is raised. Let us assume that a given mass of gas or vapor occupies a certain volume at the temperature 33° F. ; then the law dis- covered by Gay Ldssac, and which bears his name, may be enunciated as follows : If the volume remains the sam,e, its increase of tension varies as its increase of temperature; if its tension remains the same, its increase of volume paries as its increase of temperature. According to Regnadlt, if a given mass of air be heated from 32° Fahrenheit to 312°, the tension remaining constant, its volume will be increased by the .3665th part of its volume at 32°. Hence, the increase for each degree of temperature is the .00204th part of its volume at 32°. If we denote the volume at 32° by v, and the volume at the temperature t' de- grees, by v', we have, v' = «;[1*4- .00204 (;;' — 32)] (330) _ Solving with reference to v, we have, v' " = 1 + . 00204 (^'-32) <^^^^) Formula (331) enables us to compute the volume of a mass of air at 32°, when we know its volume at the temperature f degrees, the pressure remaining constant. To find the volume at the temperature t" degrees, we haye simply to substitute t" for t' in (330). Denoting this volume by v", we have, v" =zv[l+ .00304 {t" — 32)]. 275 MECHAKICS. [332. Substituting for v its value, from (331), we get, , 1 + .00204(^-32) > " = *' 1 + . 00204 (^'-32) ^^^^' This formula enables us to compute the volume of a mass of air, at a temperature t", when we know its volume at the temperature f ; and, since the density varies inversely as the volume, we may also, by means of the same formula, find the density of any mass of air, at the temperature t", when we have given its density at the temperature t'. * Absolute Temperature. 196. In the practical applications of the laws of Maeiottb and G-AT Ltjssac it is often found convenient to reckon tem- peratures from a zero point, so taken that the volume of a gas or vapor of given tension shall always b^ proportional to its temperature. To find the position of such a zero point on the Fahrenheit scale, let ns consider the case of an ideal air ther- mometer. Let ^^ be a tube of uniform cross- ^ section closed at the bottom and open at -the top ; ' let P'be an air-tight piston without weight, which moves freely and without friction up and down the tube ; and suppose a given mass of air to be con- fined between the bottom of the tube and the piston. When the temperature of the confined air is 32° F. suppose the piston to be at O, and when its tem- ^ perature is 212° suppose the piston to be at D. ^^ jgg Divide CD into 180 equal parts and continue the scale to the bottom of the tube ; then if the scale thus de- termined is numbered from the bottom upward, the lower division being numbered 0, it is obvious that the numbers thus found will be proportional to the corresponding volumes 333.] MECHANICS OF GASES AND VAPORS. 273 of the confined air. To find the number of divisions of the scale between G and A. Denote ^C by 1 ; then, from the law of expansion as deduced by Eegnault, CD will be equal to .3665 ; and if we denote the number of divisions in GA by x, we shall have, GD : AG wW) : x; or .3665 : 1 : : 180 : a; ; whence, X = 491, nearly (333) If, therefore, the Fahrenheit scale be extended downward 491 divisions below the freezing point of water, and if the lowest point of the scale be taken as the zero point, the volume of the confined air will always be proportional to its temperature. On the scale thus determined the freezing point of water is at 491°, and the boiling point is 671° ; at the former tempera- ture the volume of the confined air is 1, and at the latter temperature it is 1.3665 ; and we obviously have, 491° : 671° : : 1 : 1.3665. Temperatures counted from the zero point thus determined are called absolute temperatures. The zero point of the absolute scale is 4^1° — 32°, or 459° below the zero point of the Fahrenheit scale ; hence, we may convert any ordinary temperature into the corresponding absolute temperature by adding to it 459°. Thus, 35° F. (ordinary temperature) is equivalent to 484° F. absolute tem- perature. The absolute zero point of the centigrade scale is 491°-=-1.8, or, 373°, nearly, below the centigrade zero point. Hence, any ordinary temperature, as given by the centigrade ther- mometer, may be converted into absolute centigrade temper- ature by adding to it 373°. Thus, 15° C. (ordinary temper- ature) is equivalent to 388° C. absolute temperature. If we denote the volume of a given mass of gas or vapor by 2U MECHANICS. [334. V, the pressure per square inch on its surface by F, and its absolute temperature by T, the relation between these quanti- ties may be expressed by the equation. FV = a constant (334) If Tis constant, we see that P varies inversely as V; which is Makiotte's law. If F is constant, P varies as T ; if P is constant, V varies as T; which is Gat Lussao's law. Hence, we see that Makiotte's and Gut Lussac's law are both particular cases of a more general law which is given by equation (334). Manometers. 197. A manometer is an instrument for measuring the tension of gases and vapors, particularly of steam. Two principal varieties of manometers are used for measuring the tension of steam, the open, and the closed manometer. J^m The Open Manometer. 198. The* open manometer consists of an open glass tube, AB, terminating near the bottom of a cistern BF. The cistern is of wrought-iron, steam-tight, and filled with mercury. Its dimensions are such, that the upper surface of the mercury will not be materially lowered, when a portion of the mercury is forced up the tube. BB is a tube, by means of which, steam may be ad- mitted from the boiler to the surface of the mercury in the cistern. This tube is some- times filled with water, through which the pressure of the steam is transmitted to the E mercury. Fig. 156. MECHANICS OF GASES AND VAPORS. 275 To graduate the instrument. All communication with the boiler is cut off, by closing the stop-cock, E, and communi- cation with the external air is made by opening the stop-cock, D. The point H of the tube at which the mercury stands is then marked 1. From the point H, distances equal to 30, 60, 90, etc., inches are laid off upward, and the correspond- ing points numbered 2, 3, 4, etc. These divisions corre- spond to atmospheres, and may be subdivided into tenths and hundredths. To use the instrument, the stop-cock, D, is closed, and communication made with the boiler, by opening the stop- cock, E. The height to which the mercury rises in the tube indicates the tension of the steam in the boiler, which may be read from the scale in terms of atmospheres and decimals of an atmosphere. If the pressure in pounds is wished, it may at once be found by multiplying the reading of the in- strument by 15. The pressure, when the surface of the mercury stands at 1 is the actual pressure of the atmosphere at the time of ob- servation ; hence, if great accuracy is required, the reading of the instrument must be corrected for the excess of 30 in. over the barometric column. Thus, if the reading of the manometer is 3 and the barometric column is 29 in., the true reading is 3 — -gV = 2||. The principal objection to this kind of manometer is its want of portability, and the great length of tube required, when high tensions are to be measured. The Closed Manometer. 199. The general construction of the closed manometer is the same as that of the open manometer, except that the tube, AB, is closed at the top. The air confined in the tube, is compressed in the same way as in Maeiottb's tube. To graduate this instrument. Find by experiment the 276 MECHANICS. [^35. point H, that is, the point at which the mercury stands in the tube when the pressure is an atmosphere of 30 inches ; this we mark 1, and then "we determine the other divisions by- means of the following formula. Denote the distance in inches, from H to the top of the tube, by I ; the pressure on the mercury, in atmospheres, by n, and the distance in inches, from H to the upper surface of the mercury in the tube, by x. The tension of the air in the tube is equal to that on the mercury in the cistern, diminished by the weight of a column of mejcury whose altitude is x. Hence, in. atmospheres, it is X '^-30- The bore of the tube being uniform, the volume of the compressed air is proportional to its height. When the press- ure is 1 atmosphere, the height is I ; when the pressure is (« — q7^) atmospheres, the height is / — x. Hence, from Mariotte's law, l:n-^^::l-x:l. Whence, by reduction, x^ — (30re ■\-t)z= — 30Z(ra — 1), Solving, with respect to x, we have, 30»» '+ I x = 2 ?±A/-ao<(»-i)H.(55i+J)! The upper sign of the radical is not used, as it would give a value for x, greater than /. Taking the lower sign, and assuming 1 = 30 in., we hare, x=15n + 15~ V — 900(w — 1) + (15w + 15)» . , . (335) MECHANICS OF GASES AND VAPOES. 377 Making n=2, 3, 4, etc., in succession, we find for x, the values, 11.46 in., 17.58 in., 20.93 in., etc. These distances being set ofE from IT, upward, and marked 2, 3, 4, etc., indi- cate atmospheres. The intermediate spaces may be subdi- vided by the same formula. In making the graduation, we have supposed the tempera- ture to remain the same. If, however, it does not remain the same, the reading of the instrument must be corrected by means of a table computed for the purpose. The instrument is used in the same manner as that already described. ITeither can be used for measuring tensions less than 1 atmosphere. The Siphon Gauge. 300. The siphon gauge is used to measure tensions of gases and vapors, less than an atmosphere. It consists of a tube, ABO, bent so that its two branches are parallel. The branch, BC, is closed at the top, and filled with mercury, which is retained by the pressure of the atmosphere ; the branch, AB, is open at the top. If the air be rarefied in any manner, or, if the mouth of the tube be exposed to the action of a gas whose tension is sufficiently small, the mercury will no longer j-ITTot. be supported in BC, but will fall in it, and rise in BA. The distance between the surfaces of the mercury in the two branches, given by a scale between them, indicates the tension of the gas. If this distance is expressed in inches, the tension can be found, in atmospheres, by dividing by 30, or, in pounds, by dividing by 3. The Diving-Bell. 201. The diving-bell is a bell-shaped vessel, open at the bottom, used for descending into the water. The bell is ^^ ■31 278 MECHANICS. placed with its mouth horizontal, and let down by a rope, AB, the whole apparatus being sunk by weights properly adjusted. The air contained in the bell is compressed by the pressure of the water, but its increased elasticity prerents the water from rising to pig. isg. the top of the bell, which is provided with seats for the accommodation of those within the bell. The air, constantly contaminated by breathing, is continually re- placed by fresh air, pumped in through a tube, FG. Were there no additional air introduced, the volume of the com- pressed air, at any depth, might be computed by Maeiottb's law. The unit of the compressing force, in this case, is the weight of a column of water whose cross-section is a square inch, and whose height is the distance from DC to the sur- face of the water. The principle of the diving-bell is used in the diving- dress. The diver is surrounded by a water-tight envelope, fitted with a helmet, into which air is pumped from above, as in the diving-bell. There is, as in the diving-bell, an escape valve, by means of which circulation of the air is maintained. The diving-bell is used in constructing foundations and other submarine works ; the diving-dress is principally used in sub- marine explorations. The Barometer. 303. The barometer is an instrument for measuring the pressure of the atmosphere. It consists of a glass tube, her- metically sealed at one extremity, filled with mercury, and inverted in a basin of that fluid. The pressure of the air is indicated by the height of mercury that it supports. A variety of forms of mercurial barometers have been devised, all involving the same mechanical principle. The most impor- tant of these are the siphon and the cistern barometers. A MECHANICS OF GASES AND VAPOKS. 279 The Siphon Barometer. 203. The siphon barometer consists of a tube, GDE, bent so that its two branches, CD and DE, are parallel to each other. A scale is placed between them, and attached to the same frame as the tube. The longer branch, -rfiC CD, is hermetically sealed at the top, and filled vith mercury ; the shorter one is open to the air. "When the instrument is placed vertically, the mercury sinks in the longer branch and rises in the shorter one. The distance between the surface of the mercury in the two branches indicates the pressure of the atmos- phere. To prevent shocks when handling the instru- ^ ment, the tube is drawn to a narrow neck in the neigh- Fig. 159. borhood of the point marked B in the diagram. The Cistern Barometer. 304. The cistern barometer consists of a glass tube, filled with mercury, and inverted in a cistern of the same. The tube is surrounded by a frame of metal, attached to the cistern. Two longitudinal openings, near the upper part of the frame, permit the upper surface of the mer- "^ cury to be seen. A slide, moved up and down by a rack and pinion, may be brought exactly to the upper level of the mercury. The height of the column is then read from a scale, whose is at the surface of the mercury in the cistern. The scale is graduated to inches and tenths, and smaller divisions are read by means of a vernier. The figure shows the essential parts of a com- plete cistern barometer ; KK represents the frame ; *-^ HH, the cistern, the upper part of which is of -^-^^x^, glass, that the mercury in the cistern may be seen through it ; L, & thermometer, to show the temperature of s 380 MECHANICS. m w D Fig. 161. the mercury; JV, a sliding-ring bearing the vernier, and moved up and down by the pinion, M. The cistern is shown on an enlarged scale in Fig. 161 ; A is the barometer tube, terminating in a small opening, to prevent sudden shocks when the instrument is moved from place to place ; H, the frame of the cistern ; B, the upper portion of the cistern, made of glass, that the mercury may be seen ; B, a piece of ivory, projecting from the upper surface of the cistern, whose point corre- sponds to the of the scale ; 00, the lower part of the cistern, made of leather, or other flexible material, and firmly attached to the glass part ; D, a screw, working through the frame, and against the bottom of the bag, CO, by means of a plate, F. The screw, D, serves to bring the surface of the mercury to the point of ivory, B, and also to force the mercury to the top of the tube, when it is desired to transport the barometer from place to place. To use this barometer; it is suspended vertically, and the level of the mercury in the cistern brought to the point of ivory, B, by the screw, D ; a smart rap on the frame will detach the mercury from the glass, to which it tends to ad- here. The ring, JV, is run up or down till its lower edge appears tangent to the surface of the mercury in the tube, aud the altitude is read from the scale. The height of the at- tached thermometer should also be noted. The requirements of a good barometer are, sufficient width of tube, perfect purity of mercury, and a scale with an accu- rately graduated vernier. The bore of the tube should be as large as practicable, to diminish the effect of capillary action. On account of the repulsion between the glass and mercury, the latter is de- pressed in the tube, and this depression increases as the diam- eter of the tube diminishes. MECHAKICS OF GASES AND VAPOKS. 381 In all cases, this depression should be allowed for, and the reading corrected hy a table computed for the purpose. To secure purity of the mercury, it should be carefully dis- tilled, and after the tube is filled, it should be boiled to drive off any bubbles of air that may adhere to the tube. Uses of the Barometer. 205. The primary object of the barometer is, to measure the pressure of the atmosphere. It is used by mariners as a weather-glass. It is also employed for determining the heights of points on the earth's surface, above the level of the sea. The principle on which it is employed for the latter purpose is, that the pressure of the atmosphere at any place depends on the weight of a column of air reaching from the place to the upper limit of the atmosphere. As we ascend above the level of the ocean, the weight of the column diminishes; consequently, the pressure becomes less, a fact which is shown by the mercury falling in the tube. Difference of Level 'Between Two Stations. 206. Let us suppose the atmosphere to be in a state of equilibrium, and the temperature of the atmos- phere and of the mercury to be 32° F., both at A',— ,p,'d' the upper and at the lower station ; and further- more, suppose that the force of gravtity does not j^. =1?, <5 vary in passing from one station to the other. Conceive a cylindrical colufan of air whose cross-section is 1 square inch to extend from the bottom of the atmosphere to the top. Then, be- cause the atmosphere is supposed to be in equi- ^ j | |p j, librium, the lateral pressures in each horizontal Kg. lea. layer will balance each other, and we may tr^t the column as though it were inclosed in a vertical tube. Let A A' he a portion of such a column, A being in the horizontal plane passing through the lower station, and A' in 383 MECHANICS. [336. the horizontal plane passing through the upper station ; de- note the pressure of the atmosphere at A by P, and at ^' by P' ; denote the density of the air at A by D, and at ^' by D' ; at any point a between A and A' and at a distance from A equal to z, let the pressure be denoted by p and the density by (J. The change of pressure in passing from the height z to the height z + dz, denoted by dp, will obviously be equal to the weight of the infinitesimal layer whose height is dz ; but the weight of this layer, equal to its volume into its density inio gravity, is equal to dz x ^ X g ; and because p decreases as z increases, we have, dp= — dz X S X g (336) From Maeiotte's law we have, p:P::8:D, or, P = ^^ (337) Dividing (336) by (337), we find, , f=-?^^ (338) Integrating (338) from the bottom of the column to the top, and denoting the height AA' hj h, we have, IP' -IP = - ^h, or, IP - IP'.= ^h, . . . (339) in which I denotes the Naperian logarithm. Multiplying both members of (339) by M, the modulus of the common system, and reducing, we have, %(J) = ^'», and solving, we find. 340.] MECHANICS OF GASES AND VAPOKS. 283 A = log (p), (340) MDg in which log denotes the common logarithm. For the case assumed, the pressures at A and A' are measured by the heights of the mercurial columns at these stations; denoting these heights by H and H', we have finally, '' = rng'^a (|> (341) which is the fundamental barometrical formula. In applying formula (341) to practice, a great number of cor- rections must be made. It must be corrected for the tem- perature of the stratum of air between the stations A and A' ; it must be corrected for difEerence of temperature of the columns of mercury at the two stations; it must be cor- rected for the variation of gravity between the level of the sea and each of the two stations ; and it must be corrected for the variation in bhe amount of aqueous vapor at the two stations. The method of making these corrections does not fall within the scope of this treatise. For the complete formula, as used in geodesy, the reader is referred to the exhaustive treatise, " On the Use of the Barometer in Surveys and Reconnaissances," by Colonel R. S. Williamson, U. S. Engineers, published by the Engineer Department for the use of the army engineers ; and also to the admirable tables of Gutot, published by the Smithsonian Institution. In both of these works elaborate tables are given, by means of which the process of barometrical leveling is made comparatively simple. In "Professional Papers, No. 12," published for the use of the Corps of U. S. Engi- neers, pp. 146-180, may be found Plantamoue's modi- fication of Bessel's formula, arranged and adapted to En- glish measures, with a complete set of tables for its use, by Colonel Williamson. 284 MECHANICS. [343. Work Due to the Expansion of a Gas or Vapor. 307. Let tlie gas or vapor be confined in a cylinder closed at is lower end, and having a piston working air-tight. When the gas occupies a portion of the cylinder whose height is h, denote the pressure on each square inch of the piston by p ; when the gas expands, so that the alti- tude of the column becomes x, denote the pressure on a square inch by p'. Since the volumes of the gas, under these suppositions, are proportional to their altitudes, we shall have, from Mariottb's law. — . -iW Pig. 168. whence p : p :: X : h; xp' =ph (343) If we suppose^ and h to be constant, and x and^' to vary, the above equation will be that of an equilateral hyperbola, whose asymptotes are ^Cand AM. From (343) we have, , ph ^ X The elementary quantity of work performed on each square inch of the piston whilst the gas is expanding from the height X to the height x -f - dx is equal to p' dx ; denoting this by dQ', we have, after substituting for p' its value from the preceding equation, dQ' = ph — . Integrating between the limits h and x, we have. 343.] MECHAS-ICS OF GASES AND VAPORS. 285 Q' = ph {Ix — IJi) = ph . li^^, (343) in which I denotes the Xaperian logarithm. In (343), Q' is the quantity of work performed on each square inch of the piston ; if we denote the area of the piston in square inches by A and the total quantity of work by Q, we have, after substituting for ^ its value taken from (342), ih Q = Aph . Z (J,) (344) If we denote by c the number of cubic feet of gas when the pressure on a square inch of the piston is p, and sup- pose it to expand till the pressure is p', we shall have c z= Ah, or, if A is expressed in square feet, we shall have Ah , 144c „,. . ,„... . c — jj-^, or A = -J- . This, m (344), gives Q = Uicp.l{^), (345) which gives the quantity of work of c cubic feet of gas whilst expanding from a pressure of p pounds on a square inch to a pressure of p' pounds. Steam. 308. If water be exposed to the atmosphere, at ordinary temperatures, a portion is converted into vapor, which mixes with the atmosphere, constituting one of the permanent ele- ments of the aerial ocean. The tension of watery vapor thus formed, is very slight, and the atmosphere soon ceases to absorb any more. If the temperature of the water be raised, an additional amount of vapor is evolved, and of greater tension. When the temperature is raised to that point at which the tension of the vapor is equal to that of the atmos- 286 MECHASriCS. phere, ebullition commences, and the vaporization goes on with great rapidity. If heat be added beyond the point of ebullition, neither the water nor the vapor will increase in temperature till all of the water is converted into steam. When water is converted into steam under a pressure of one atmosphere, each cubic inch is expanded into about 1700 cubic inches of steam, of the temperature of 212° ; or, since a cubic foot contains 1728 cubic inches, we may say, in round numbers, that a cubic inch of water is converted into a cubic foot of steam. If water is converted into steam under a greater or less pressure than one atmosphere, the density will be increased or diminished, and, consequently, the volume will be dimin- ished or increased. The temperature being also increased or diminished, the increaso of density or decrease of volume will not be exactly proportional to the increase of pressure ; but, for purposes of approximation, we may consider the densities as directly and the volumes as inversely proportional to the pressures under which the steam is generated. Under this hypothesis, if a cubic inch of water be evaporated under a pressure of a half atmosphere, it will afford two cubic feet of steam ; if generated under a pressure of two atmospheres, it will only afford a half . cubic foot of steam. "Work of Steam. 309. When water is converted into steam, a certain amount of work is generated, and, from what has been shown, this amount of work is very nearly the same, whatever may be the temperature at which the water is evaporated. Suppose a cylinder, •whose cross-section is one square inch, to contain a cubic inch of water, above which is an air-tight piston, that may be loaded with weights at pleasure. In the first place, if the piston is pressed down by a weight of 15 pounds, and the inch of water converted into steam, the MECHANICS OF GASES AKD VAPOES. 287 weight will be raised to the height of 1738 inches, or 144 feet. Hence, the quantity of work is 144 x 15, or, 2160 units. Again, if the piston be loaded with a weight of 30 pounds, the conversion of water into steam will give but 864 cubic inches, and the weight will be raised through 72 feet. In this ease, the quantity of work will be 72 x 30, or 2160 units, as before. We conclude, therefore, that the quantity of work is the same, or nearly so, whatever may be the press- ure under which the steam is generated. We also conclude, that the quantity of work is nearly proportional to the fuel consumed. Besides the quantity of work developed by simply convert- ing an amount of water into steam, a further quantity of work is developed by allowing the steam to expand after entering the cylinder. This principle is made use of in steam-engines working expansively. To find the quantity of work developed by steam acting ex- pansively. Let AB represent a cylinder, closed at A, and having an air-tight piston B. Suppose the steam to enter at the bottom of the cylinder, and to push the piston upward to C, and then suppose the opening at which the steam enters to be closed. If the piston is not too heavily loaded, • the steam will continue to expand, and the piston will be raised Kg. ]64. to some position, B. The expansive force of the steam wiU obey Makiotte's law, and the quantity of work due to expansion will be given by equation (344). Denote the area of the piston in square inches, by A ; the pressure of the steam on each square inch, up to the moment when the communication is cut off, by p ; the distance A O, through which the piston moves before the steam is cut off, by h ; and the distance AB, by nJi. If we denote the pressure on each square inch, when the piston arrives at B, by p', we shall have, by Maeiotte's law. 288 MECHANICS. t^*®- P p : p' : : nh : h, .•. ^ = n an expression which gives the limiting value of the load of the piston. The quantity of work due to expansion being denoted by Q', we shall have, from equation (344), Q' = ApTi X l[^) = Aph.l{n) (346) If we denote the quantity of work of the steam, whilst the piston is rising to G, by Q", we shall have, Q" = Aph. Denoting the total quantity of work during the entire stroke of the piston, by Q, we shall have, Q' +Q''^Q = Aph[l + l{n)] (347) Efflux of a Gas or Vapor. 310. Suppose the gas to escape from a small orifice, and denote its velocity by v. Denote the weight of a cubic foot of the gas by w, and the number of cubic feet discharged in one second by c, then will the mass escaping in one second, CZO CZD be equal to — , and its kinetic energy will be equal to ^v^. But, from Art. 148, the kinetic energy is equal to the accu- mulated work. If, therefore, we denote the accumulated work by Q, we shall have. But the accumulated work is due to the expansion of the 348.] MECHANICS OF GASES AND VAPORS. 289 gas, and if we denote the pressure within the orifice by p, and without, by^', we shall have, from Art. 207, Equating the second members, we have, whence. Substituting for g, its value, 32^ ft., we have, after reduc- tion. When the difference between p and p' is small, the pre- ceding formula can be simplified. p p p' Since ^ = 1 + - — ^ , we have, from the logarithmic for- mula. When p — p' is very small, the second, and all succeeding terms of the- development, may be neglected, in comparison with the first term. Hence, SubstitutJTig, in the formula above deduced, we have. V w v = OH/^xl^-, 290 MECHANICS. r349. or, since — , is, iinder the supposition just made, equal to 1, we have, finally, ^=^V^ ^'^'^ CoefBcient of Efflnx. 311. When air issues from an orifice, the section of the current undergoes a change of form, analogous to the con- traction of the vein in liquids, and for similar reasons. If we denote the coefficient of efflux by Ic, the area of the orifice by A, and the quantity of air delivered in n seconds by Q, we shall have, from equation (348), Vi'fF) e=96i.^y^i^|,j (360) In which w is to be expressed in the same unit as p and p'. According to Koch, the value of h is equal to .58 when the orifice is in a thin plate; equal to .74 when the gas issues through a tube 6 times as long as it is wide ; and equal to .85 when it issues through a conical nozzle 5 times as long as the diameter of the orifice, and whose elements make an angle of 6° with the axis. X.— HYDRAULIC AND PNEUMATIC MACHINES. Defiuitions. 313. Hydraulic machines are machines for raising and distributing water, as pumps, siphons, hydraulic rams, and the like. The name is also applied to machines in which water power is the motor, or in which water is employed to transmit pressures, as water-wheels, hydraulic presses, and the like. Pneumatic machines are machines to rarefy and con- dense air, or to impart motion to air, as air-pumps, ventilating blowers, and the like. The name is also applied to those machines in which the kinetic energy of air is the motive power, such as windwills, and the like. Water Pumps. 313. A water pump is a machine for raising water from a lower to a higher leyel, by the aid of atmospheric pressure. Three separate principles are employed in pumps : the suck- ing, the lifting, and the forcing principle. Pumps are named according to the principles employed. Sucking and Lifting. Pump. 314. This pump consists of a barrel. A, to the lower ex- tremity of which is attached a sucking-pipe, B, leading to a reservoir. An air-tight piston, C, is worked up and down in the barrel by a lever, E, attached to a piston-rod, Z* ; P is a valve opening upward, which, when the pump is at rest, closes by its own weight. This valve is called the piston- 293 MECHANICS. Pig. 165. valve. A second valve, G, also opening upward, is placed at the junction of the pipe with the barrel; this is called the sleeping-valve. The space, LM, through which the piston moves up and down, is tlie play of the piston. To explain the action of the pump : suppose the piston to be in its lowest position, and everything in equilibrium. If the extremity of the lever, E, be de- pressed, and the piston raised, the air in the lower part of the barrel is rarefied, and that in the pipe, B, by virtue of its greater tension, opens the valve, and a portion escapes into the barrel. The air in the pipe', thus rarefied, exerts less pressure on the water in the reservoir than the external air, and, consequently, the water rises in the pipe, until the tension of the internal air, plus the weight of the column of water raised, is equal to the tension of the ex- ternal air ; the valve, G, then closes by its own weight. If the pistol} be again depressed to the lowest limit, the air in the lower part of the barrel is compressed, its tension be- comes greater than that of the external air, the valve, P, is forced open, and a portion of the air escapes. If the piston be raised once more, the water, for the same reason as before, rises still higher in the pipe, and after a fex double strokes of the piston, the air is completely exhausted from beneath the piston, the water passes through the piston-valve, and finally escapes at the spout, F. The water is raised to the piston by the pressure of the air on the surface of the water in the reservoir ; hence, the piston should not be placed at a greater distance above the water in the reservoir, than the height at which the pressure of the air will sustain a column of water. In fact, it should be placed a little lower than this limit. The specific gravity of mer- HTDEAULIC AND PNEUMATIC MACHINES. 293 cury being about 13.5, the height of a column of water that will counterbalance the pressure of the atmosphere may be found by multiplying the height of the barometric column by ISf At the level of the sea the average height of the barometric column is 2^ feet ; hence, the theoretical height to which water can be raised by the principle of suction alone, is a little less than 34 feet. The water having passed through the piston-valve, may be raised to any height by the lifting principle, the only limita- tion being the strength of the pump. There are certain relations that must exist between the play of the piston and its height above the water in the reservoir, in order that the water may be raised to the piston ; if the play is too small, it will happen after a few strokes of the piston, that the air in the barrel is not suflBciently compressed to open the piston- valve ; when this state of affairs takes place, the water ceases to rise. To investigate the relation that should exist between the play and the height of the piston above the water. Denote the play of the piston by p, the distance from the surface of the water in the reservoir to the highest position of the piston by a, and the height at which the water ceases to rise by x. The distance from the water in the pump to the highest po- sition of the piston will be a — x, and the distance to the lowest position of the piston a — p — x. Denote the height at which the atmospheric pressure sustains a column of water in a vacuum, by h, and the weight of a column of water, whose base is the cross-section of the pump, and altitude is 1, by w ; then will wh denote the pressure of the atmosphere exerted upward through the water in the reservoir and pump. When the piston is at its lowest position, the pressure of the confined air must be equal to that of the external atmos- phere ; that is, to wh. When the piston is at its highest 294 MECHANICS. [351. position, the confined air will be rarefied, the volume occupied being proportional to its height. Denoting the pressure of this rarefied air by wh', we shall have, from Mariotte's law, wh : wh' :: a — x : a — p — x. As the water does not rise when the piston is in its highest position, the pressure of the rarefied air plus the weight of the column already raised, will be equal to the pressure of the external atmosphere ; or, ivli — f- wx = wh. a — x Solving this with respect to to x, we have, a ± Va^ — 4pA (351) Or If, 4ph > a« ; or, i' > j^ , the value of x is imaginary, and there is no point at which the water ceases to rise. Hence, the above inequality ex- presses the relation that must exist, in order that the pump may be efEective. This condition, expressed in words, gives the following rule : The -play of the piston must be greater than the square of the distance from the loaier in the reservoir, to the highest position of the piston, divided by four times the height at which the atmosphere will support a column of water in a vacuum. Let it be required to find the least play of the piston, when its highest position is 16 feet above the water in the reservoir, and the barometer at 28 inches. HTDEAULIG AND PNEFMATIC MACHINES. 295 In this case, a = 16 fl., and A = 28 in. x 13J = 378 in. =Z\\ ft. Hence, P>mfl-< or> P>Zv\fi- To find the quantity of work required to make a double stroke of the piston, after the- water has reached the spout. In depressing the piston, no force is required, except that necessary to overcome inertia and friction. Neglecting these for the present, the quantity of work in the downward stroke, may be regarded as 0. In raising the piston, its upper sur- face is pressed downward, by the pressure of the atmosphere, wh, plus the weight of the column of water from the piston to the spout ; and it is pressed upward, by the pressure of the atmosphere, transmitted through the pump, minus the weight of a column of water, whose cross-section is that of the barrel, and whose altitude is th« .distance from the piston, to the water in the reservoir. If we subtract the latter from the former, the difEerence will be the downward pressure. This difEerence is equal to the weight of a column of water, whose base is the cross-section of the barrel, and whose height is the distance of the spout above the reservoir. Denoting this height by H, the pressure is equal to wH. The path through which the pressure is exerted during the ascent of the piston, is the play of the piston, or p. Denoting the quantity of work required, by Q, we shall have, Q = wpH. But wp is the weight of a volume of water, whose base is the cross-section of the barrel, and whose altitude is the play of the piston. Hence, Q is equal to the quantity of work necessary to raise this volume of water from the level of the reservoir to the spout. This volume is evidently equal to that actually delivered at each double stroke of the piston. Hence, the quantity of work expended in pumping, with the sucking 296 MECHAinCS. and lifting pump, hurtful resistances being neglected, is equal to the quantity of work necessary to lift the amount of water, actually delivered, from the level of the reservoir to the spout. In addition to this, a sufficient amount of power must be ex- erted to overcome hui-tful resistances. The disadvantage of this pump, is the irregularity with which the force acts, being in depressing the piston, and a maximum jn raising it. This is an important objection when machinery is employed in pumping ; but it may be partially overcome, by using two pumps, so arranged, that one piston ascends as the other descends. Another objection to the use of this pump, is the irregularity of flow, the inertia of the col- umn of water having to be overcome at each upward stroke. Modification of tke Lifting Pump. 315. To correct the irregularity of flow, it is customary to attach to the piston-rod a cylindrical piece, P, called a plunger. This piece, which moves up and down with the piston, is so adjusted as to displace in its descent one half the quantity of water that is delivered at each double stroke of the piston. Then, during the up stroke of the piston, an amount of water will be lifted up to the spout equal to that which \% forced, up by the plunger in the down stroke. In this manner the flow is rendered nearly uniform, and the rate of work of the power is made nearly constant. Pig. los. Sucking' and Forcing Pump. 31.6. This pump consists of a barrel. A, with a sucking pipe, B, and a sleeping-valve, G, as in the pump just dis- HYDEAtTLIO AND PKBUMATIC MACHINES. 297 cussed. The piston, G, is solid, and is worked up and down by a lever, E, and a piston-rod, D. At the bottom of the bar- rel, a pipe leads to an air-vessel, K, through a second sleep- ing-valve, F, which opens upward, and closes by its own weight. A delivery-pipe, H, enters the air-vessel at the top, and terminates near the bottom. To explain the action of this pump, suppose the piston, C, to be in its lowest position. If the piston be raised to its highest position, the air in the barrel is rarefied, its tension is diminished, the air in the tube, B, thrusts open the valve, and a portion escapes into the barrel. The pressure of the external air then forces water up the pipe, B, until the tension of the rarefied air, plus the weight of the water raised, is equal to the tension of the external air. An equilibrium being produced, the valve, G, closes by its own weight. If the piston be depressed, the air in the barrel is condensed, its tension in- creases till it becomes greater than that o'f the external air, when the valve, F, is thrust open, and a portion escapes through the delivery-pipe, H. After a few double strokes of the piston, the water rises through the valve, G, and, as the piston descends, is forced into the air-vessel, the air is con- densed in the upper part of the vessel, and, acting by its elastic force, urges a portion of the water up the delivery- pipe and out at the spout, P. The object of the air-vessel is, to keep up a continued stream through the pipe, H, otherwise it would be necessary to overcome the inertia of the entire column of water in the pipe at every double stroke. The flow having commenced, a volume of water is delivered from the spout, at each double stroke, equal to that of a cylinder whose Ks;. 167. 298 MECHANICS. base is the area of the piston, and whose altitude is the play of the piston. The same condition as to the play of the piston must exist as in the sucking and lifting pump. To find the quantity of work consumed at each double stroke, after the flow has become regular, hurtful resistance being neglected : When the piston is descending, it is pressed downward by the tension of the air on its upper surface, and upward by the tension of the atmosphere, transmitted through the delivery- pipe, plus the weight of a column of water whose base is the area of the piston, and whose altitude is the distance of the spout above the piston. This distance is variable during the stroke, but its mean value is the distance of the middle of the play below the spout . The difference between these pressures is exerted upward, and is equal to the weight of a column of water whose base is the area of the piston, and whose altitude is the distance from the middle of the play to tie spout. The distance through which the force is exerted, is the play of the piston. Denoting the quantity of work during the descending stroke, by Q', the weight of a column of water, whose base is the area of the piston, and altitude is 1, by w, and the height of the spout above the middle of the play, by h', we have, Q' — wh' X p. When the piston is ascending, it is pressed downward by the tension of the atmosphere on its upper surface, and up- ward by the tension of the atmosphere, transmitted through the water in the reservoir and pump, minus the weight of a column of water whose base is the area of the piston, and whose altitude is the height of the piston above the reservoir. This height is variable, but its mean value is the height of the middle of the play above the reservoir. The distance through which this force is exerted, is the play of the piston. 352.] HYDRAULIC AND PNEUMATIC MACHINES. 399 Denoting the quantity of work during the ascending stroke, by '§", and the height of the middle of the play above the reservoir, by h", we have, Q" = wJi" X p. Denoting the entire quantity of work during a double stroke, by Q, we have, Q=Q' +Q" = wp {h' + h") (352) But wp is the weight of a volume of water, whose base is the piston, and whose altitude is the play ; that is, it is the weight of the volume delivered at each double stroke. The quantity, h' + h", is the height of the spout above the reservoir. Hence, the work expended is equal to that required to raise the volume delivered from the level of the reservoir to the spout. To this must be added the work necessary to over- come hurtful resistances, as friction, etc. If Ji' = h", we have, Q' = Q" ; that is, the quantity of work during the ascending stroke, is equal to that during the descending stroke. Hence, the work of the motor is more nearly uniform, when the middle of the play is at equal dis- tances from the reservoir and spout. Forcing Pnmp with Plunger. 217. A forcing pump may be constructed as shown in Fig. 168. The barrel of the pump. A, is extended into a lateral cylinder, JT, in which is a plunger that is worked back and forth through a packing-box, B, by means of the lever, LN. The sucking-pipe, S, and the delivery- pipe, D, are separated from the barrel by sleeping valves, G and F, opening up- ward and closing by their own weight. Fig. les. 300 MECHANICS. The method of filling the pump is as follows : When the plunger is thrust toward A in the figure, the air in the barrel is condensed, forces open the valve, F, and a portion escapes into the delivery-pipe ; when the plunger is drawn back the air in the barrel is rarefied, the greater tension of the air in the sucking-pipe opens the valve, G, and the water rises to a certain height in D. A few double strokes fill the pump with water, after which there will be delivered at each double stroke a quantity of water whose volume is equal to that of a cylinder whose diameter is equal to that of the plunger, and whose altitude is equal to the 'play of the plunger. In the pump just described there is no air-chamber, and as a consequence 'the flow is irregular. To prevent binding, the lever, ZiV, is connected with the plunger, P, by means of a short link, having a hinge joint at each extremity. A similar arrangement is made for like purpose in the pumps previously described. Fire-Engine. 318. The fire-engine is a double sucking and forcing pump, the piston-rods being so connected, that when one piston ascends, the other descends. The sucking and delivery pipes are made of leather, and attached to the machine by metallic screw- joints. The figure exhibits a cross- section of the essential parts of an ordinary fire-engine. A, A', are the barrels, the pistons are connected by rods with the lever, JH, E'; B is the sucking-pipe, ter- minating in a reservoir from which the water may enter jig ^g HTDEAULIC AND PNEUMATIC MACHINES. 301 either barrel through the valTee, G, 0'; K is the air-vessel, common to both pumps, and communicating with them by valyes, F, I"; H is the delivery-pipe. It is mounted on wheels for convenience of locomotion. The lever, E, E', is worked by rods at right angles to the lever, so arranged that several men can apply their strength at once. The action of the pump differs in no respect from that of the forcing pump ; but when the instrument is worked vigorously, a large quantity of water is forced into the air-vessel, the tension of the air is much augmented, and its elastic force, thus brought into play, propels the water to a considerable distance from the delivery-pipe. It is this capacity of throwing a jet of water to a great distance, that gives to the engine its value for extinguishing fires. A pump similar to the fire-engine, is often used, under the name of the double-action forcing-pump, for other pur- poses. The Rotary Pump. 319. The rotary pump is a modification of the sucking and forcing pump. Its construc- tion will be understood from the drawing, which represents a sec- tion through the axis of the suck- ing-pipe, at right angles to the axis of the rotating portion. J is a ring of metal, revolving about an axis ; D, D, is a second ring of metal, concentric with the first, and forming with it an intermediate annular space. This space communicates with the suck- ing-pipe, K, and the delivery-pipe, L. Four radial paddles, C, are so disposed as to slide backward and forward through suitable openings in the ring A, and are moved around with Kg. 170. 303 MECHANICS. it. G* is a guide, fastened to the end of the cylinder inclos- ing the revolving apparatus, and cut as represented in the figure ; E, E, are springs, attached to the ring, Z>, and acting, by their elastic force, to press the paddles firmly against the guide. These springs do not impede the flow of water from the pipe, K, into the pipe, L. When the axis, 0, revolves, each paddle, as it passes the partition, is pressed against the guide, but is forced out again, by the form of the guide, against the wall, D. Each paddle drives the air in front of it in the direction of the arrow-head, and finally expels it through the pipe, L. The air behind the paddle is rarefied, and the pressure of the external air forces a column of water up the pipe. After a few revolu- tions, the air is entirely exhausted from the pipe, K. The water enters the channel, C, C, and is forced up the pipe, L, from which it escapes by a spout. The work expended in raising a volume of water to the spout, by this pump, is equal to that required to lift it from the level of the cistern to the spout. This may be shown in the same manner as was ex- plained under the head of the sucking and forcing pump. To this quantity of work, must be added the work necessary to overcome hurtful resistances. A machine, similar to the rotary pump, is constructed for exhausting foul air from a mine ; or, by reversing the direc- tion of rotation, to force fresh air to the bottom of the mine. The Hydraulic Press. 330. The hydraulic press is a machine for exerting a great pressure through a small space. It is used in compress- ing seeds to obtain oil, in packing hay and other goods, also in raising great weights. Its. construction, though requiring the use of a forcing-pump, depends upon the principle of equal pressures, (Art. 167). 353.] HYDRAULIC AND PNEUMATIC MACHINES. 303 u H Fig. 171. By It consists of two cylinders, A and B, each provided with a solid piston. The cylinders communicate by a pipe, C, whose entrance to the larger cylinder is closed by a sleep- ing valve, U. The smaller cylinder communicates with the reservoir JS^, by a sucking-pipe 11, whose upper extremity is closed by a sleeping valve, Z>. The piston £, is worked by the lever, G. raising and depressing the lever G, the water is raised from the reservoir and forced into the cylinder, A ; and when the space below the piston, F, is filled, a force is exerted up- ward, as many times greater than that applied to B, as the area of F is greater than B (Art. 167). This force may be utilized in compressing a body, L, between the piston and the frame of the press. Denote the area of the liarger piston, by P, of the smaller, by p, the pressure applied to B, by /, and that exerted on Fj^yF; and we shall have, . F=f^ P If we denote the longer arm of the lever G, by L, the shorter arm by I, and the force applied at the extremity of the longer arm by K, we have, from the principle of the lever, KL F:f:: P : p, (353) E:f::l:L, ■f--T- (354) Substituting above, we have, PEL F^ pi (355) To illustrate, let the area of the larger piston be 100 square inches, that of the smaller piston 1 square inch, the longer arm of the lever 304 MECHANICS. 30 inches, the shorter arm 2 inches, and let a force of 100 pounds be applied at the end of the longer arm of the lever ; to find the press- ure on F. From the conditions, P = 100, K= 100, L = %0,p = 1, and I = 3. Hence, _ 100 X 100 X 30 -.-nnnn ii. F = = 150,000 lbs. a We have not taken into account the hurtful resistances, lience the pressure of 150,000 pounds must be somewhat diminished. The volume of water forced from the smaller to the larger cylinder, during a single descent of the piston, B, will occupy, in the two cylinders, spaces whose heights are in- versely as the areas of the pistons. Hence, the path, over which / is exerted, is to the path over which F is exerted, as P is to p. Or, denoting these paths by s and S, we have, s: S :: P : p; or, since P : p :: F : f, we shall have, s : S :: F : f, .: /.s = i?W. That is, the work of the power and resistance are equal, a principle that holds good in all machines. Examples. 1. The cross-section of a sucking and forcing pump is 6 square feet, the play of the piston 3 feet, and the. height of the spout, above the reservoir, 50 feet. What must be the effective horse-power of an engine to impart 30 double strokes per minute, hurtful resistances being neglected ? Solution. — The number of units of work required to be performed each minute, is equal to 6 X 3 X 50 X 62J X 30 = 1,687,500 Ib.ft. Hence, HYDRAULIC AND PlfEUMATIC MACHINES. 305 3. In a hydrostatic press, the areas of the pistons are, 3 and 400 square inches, and the arms of the leyer are, 1 and 30 inches. Required the pressure on the larger piston for each pound of pressure on the longer arm of lever. Am. 4,000 lbs. 3. The areas of the pistons of a hydrostatic press are, 3 and 300 square inches, and the shorter arm of the lever is one inch. "What must be the length of the longer arm, that a force of 1 lb. may produce a pressure of 1000 Z6s.? Ans. 10 inches. Storing Up the Work of Hydraulic Pressure. 231. In certain branches of industry it becomes necessary from time to time to handle objects of enormous weight. Work of this kind is most advantageously performed by means of the hydraulic crane. The description of the crane itself does not fall within the scope of this work, but the manner in which the motive force is generated and stored up is an immediate application of the principle of the hydraulic press just described. The diagram shows the essential parts of what is called the accumulator. The barrel. A, rests firmly on a solid bed-plate, B, which also supports the frame FOE. A heavy ram, R, is free to move up and down in the barrel, being guided by a cross-piece, K, which is notched on to the posts E and F, This cross-piece carries a heavy weight, W, usually of scrap iron, which may be placed upon K, or suspended from it in an annular basket. At P is a pipe lead- ing into A, from a powerful forcing pump, and provided with a valve, V, opening inward. At D is a pipe lead- ing /t-owj A to the machine that oper- ates the crane ; it may be opened and closed by a stop-cock, ;S'. The stop-cock, 8, being closed, the pump forces water into Fig ira. 306 MECHAITICS. A, raising up the ram and its superposed weight. In some cases the pressure on the water in A is carried as high as 700 lbs. on a square inch. When the stored up work is to be ap- plied, the cock 8 is opened, and the water enters the cylinder, where its enormous pressure is utilized in working the crane. The principles here explained may be applied to any nla- chine where great force is required for a comparatively short period of time. The Siphon. 333. The siphon is a bent tube, fof transferring a liquid from a higher to a lower level, over an intermediate elevation. The siphon consists of two branches, AB and BG, of which the outer one is the longer. To use the instrument, the tube is filled with the liquid, the end of the longer branch being stopped with the finger, or a stop-cock ; in which case, the pressure of the atmosphere prevents the liquid from escaping at the other end. The instrument is then inverted, the end, G, being submerged in the liquid, and the stop removed from A. The liquid will flow through the tube, and the flow will continue till the level of the liquid in the reservoir reaches the mouth of the tube, G. To find the velocity with which water will issue from the siphon, let us consider an infinitely small layer at the orifice, A. This layer is pressed downward, by the tension of the atmosphere exerted on the surface of the reservoir, minus the weight of the water in the branch, BD, plus the weight of the water in the branch, BA. It is pressed upward by the tension of the atmosphere. The difference of these forces, is the weight of the water in DA, and the velocity of the stratum will be due to that height. Denoting the vertical height of DA, by h, we shall have, for the velocity, V = V2ffh. HTDBAULIC AND PNEUMATIC MACHINES. 307 This is the theoretical Telocity, but it is neyer quite realized in practice, on account of resistances, that hare been neglected in the preceding investigation. The siphon may be filled by applying the mouth to the end. A, and exhausting the air by suction. '' The tension of the atmosphere, on the upper surface of the reservoir, presses the water up the tube, and fills it, after which the flow goes on as before. Sometimes, a sucking-tube, AD, is inserted near the opening, A, rising to the bend of the siphon. In this case, the opening. A, is closed, and the air exhausted through the sucking-tube, AD, after which the flow goes on as before. The Wurtemburg Siphon. 333. In the "Wurtemburg siphon, the ends of the tube are bent twice, at right angles, as shown in the figure. The advantage of this is, that the tube, (^ once filled, remains so, as long as the plane of its axis is kept vertical. The siphon may be lifted out and replaced at pleasure, thereby stopping and re- vsJ Iti/ producing the flow at will. '^' It is to be observed that the siphon is only effective when the distance from the highest point of the tube to the level of the water in the reservoir is less than the height at which the atmospheric pressure sustains a column of water in a vacuum. This will, generally, be less than 34 feet. The Intennittin^ Siphon. 334. The intermitting siphon is represented in the figure. AB is a curved tube issuing from the bottom of a reservoir. The reservoir is supplied with water by a tube, E, having a smaller bore than the siphon. 308 MECHANICS. To explain its action, suppose the reser- b Toir to be empty, and the tube, E, to be open ; as soon as the reservoir is filled to the level, CD, the water begins to flow from the opening, B, and the flow once commenced, continues till the level of the reservoir is reduced to CD', through the opening, A. The flow then ceases till the cistern is again filled to GD, and so on as before. The instrument just described is often called Tantalus' Cup. Intermitting Springs. 335. Let A represent a subterranean cavity, communi- cating with the surface of the earth by a channel, ABC, bent like a siphon. Suppose the reservoir to be fed by percolation through the crev- ices, or by a small channel, D. When the water in the reservoir rises to the horizontal plane, BD, the flow com- mences at G, and, if the channel is sufiiciently large, the flow continues till the water is reduced to the level plane through G. An intermission then occurs till the reservoir is again filled ; and so on, intermittingly. Kg. 177. Siphon of Constant Flow. 336. We have seen that the velocity of efilux depends on the height of water in the reservoir above the external open- ing of the siphon. When the water is drawn from the reser- voir, the surface sinks, this height diminishes, and, conse- quently, the velocity continually diminishes. If, however, the shorter branch, GD, be passed through a cork large enough to float the siphon, the instrument will sink I HYDEAULIC AKD PNEUMATIC MACHINES. 309 as the upper surface is depressed, the height of DA will remain constant, and, consequently, the flow will be uniform till the siphon comes in contact with the upper edge of the reservoir. By suitably adjusting the siphon in the cork, the velocity of efflux can be increased or decreased within certain limits. In this manner, any desired quantity of the fluid can be drawn off in a given time. The siphon is used in the arts, for decanting liquids. It is also employed to draw a portion of a liquid from the interior of a vessel when that liquid is overlaid by one of less specific gravity. The Hydraulic Ram. 327. The hydraulic ram is a machine for raising water by means of shocks caused by the sudden stoppage of a stream of water. It consists of a reservoir, £, supplied by an inclined pipe, A ; at the upper surface of the reservoir, is an orifice closed by a valve, D ; this valve is kept in place by a metallic basket im- mediately below the orifice ; (? is an air-vessel comfnunicating with the reservoir by an opening F, with a spherical valve, U ; this valve ,^^ closes the orifice F, except when ^^5>^!Te u: 3 forced upward, in which case its motion is restrained by a frame- Fig. its. work or cage ; ^ is a delivery-pipe entering the air-vessel at its upper part, and terminating near the bottom. To explain the action of the instrument, suppose it empty, and the parts in equilibrium. If a current of water be ad- mitted to the reservoir, through the pipe, A, the reservoir is soon filled, and the water commences rushing out at D ; the impulse of the water forces the valve, D, upward, and closes 310 MECHANICS. the opening ; the velocity of the water in the reservoir is checked ; the reaction forces open the valve, U, and a por- tion of the water enters the air-chamber, G ; the force of the shock having been expended, the valves both fall by their own weight ; a second shock takes place, as before ; an ad- ditional quantity of water is forced into the air-vessel, and so on continuously. As the water is forced into the air-vessel, the air becomes compressed ; and acting by its elastic force, urges a stream of water up the pipe, U. The shocks occur in rapid succession, and thus a constant stream is kept up. To explain the use of the valve, P, it may be remarked that water absorbs more air under a greater, than under a smaller pressure. Hence, as it passes through the air- chamber, a portion of the contained air is taken up by the water and carried out through the pipe, II. But each time that the valve, B, falls, there is a tendency to a vacuum in the upper part of the reservoir, in consequence of the rush of the fluid to escape through the opening. The pressure of the external air then forces open the valve P, a portion of air enters, and is afterward forced up with the water into the vessel, G, to keep up the supply. The hydraulic ram is only used to raise sm^l quantities of water, as for the supply of a house, or garden. Only a small fraction of the fluid that enters the supply-pipe actually passes out through the delivery-pipe ; but if the head of water is pretty large, a column may be raised to a great height. Water is often raised, in this manner, to the highest parts of lofty buildings. Sometimes, an additional air-vessel is introduced over the valve, JS, to deaden the shock of the valve in its play. Archimedes' Screw. 338. This is a machine for raising water through small heights, and, in its simplest form, it consists of a tube wound HYDRAULIC AND PNEUMATIC MACHINES. 311 spirally around a cylinder. The cylinder is mounted so that its axis is oblique to the horizon, the lower end dipping into the reservoir. When the cylinder is turned on its axis, the lower end of the tube describes the circumference of a circle, whose plane is perpendicular to the axis. When the mouth of the tube comes to the level of the axis and begins to ascend, there is a certain quantity of water in the tube, which con- tinues to occupy the lowest part of the spire ; and, if the cyl- inder is properly inclined to the horizon, this flow is toward the upper end of the tube. At each revolution, a quantity of water enters the tube, and that already in the tube is raised, higher and higher, till, at last, it flows from the upper end of the tube. The Chain Pump. 229. The chain pump is an instrument for raising water through small elevations. It consists of an endless chain passing over wheels, A and B, having their axes horizontal, one be- low the surface of the water, and the other above the spout of the pump. Attached to this chain, and at right angles to it, are circular disks, fitting the tube, CD. If the cylinder A, be turned in the proper direction, the buckets or disks rise through the tube CD, driving the water before them, un- til it reaches the spout C, and escapes. One great objection to this machine is, the difficulty of making the disks fit the tube. Hence, there is a constant leakage, requiring great additional expenditure of force. .Sometimes the body of the pump is inclined, in which case it does not differ much in principle from a wheel with flat buckets, which has also been used for raising water. Fig. 179. 312 MECHANICS. Blowers. 330. A blower is a machine for creating and keeping up a constant flow of air, either for promoting combustion, or for the purposes of ventilation. One of the simplest, and not the least important, of this class of machines, is the black- smith's bellows, a section of which is shown in Fig. 180. The Blacksmith's Bellows. 331. The blacksmith's bellows is a machine that re- sembles the accumulator described in Art. 231, air being used instead of water. It con- sists essentially of three plates, usually of wood, A, B, and 0. The mid- dle one. A, is fixed in position, whilst the up- per and lower ones, 6' and B, are free to turn about Kg. 183. hinge-joints at B and Z> ; the upper and lower plates are at- tached to the middle one by flexible sheets of leather, nailed to their respective edges ; the joints at B and D are also cov- ered by pieces of leather attached to the plates in the same manner. By this arrangement the whole machine is sepa- rated into two air-tight compartments, U and L, the capacity of each being capable of a certain amount of contraction and expansion. The middle plate. A, has a valve, F, opening upward and closing by its own weight ; a similar valve, G, exists in the lower plate. The upper compartrdent terminates in a nozzle through which the air contained in the compartment U, is forced by the action of the weight, W. The supply of air, in the upper compartment is kept up by means of the movable plate B, which is raised by means of a link, MJf, worked by HYDRAULIC AND PNEUMATIC MACHINES. 313 a lever not shown in the diagram, and which falls by its own weight. When the plate B, is raised all the air in the com- partment L, is forced into the compartment JJ, where it is kept from returning by the closing of the valve F; whilst the plate B, is descending the valve G, is kept open by the pressure of the external air and the compartment L, is filled. When the plate B, is raised again a new supply of air is forced into the compartment TJ, and so on indefinitely. The strength of the current through the nozzle may be in- creased by increasing the weight, W. -& - -a-.P The Yentilating Blower. 232. One form of ventilating blower is shown in Fig. 181. A solid piston, P, is moved up and down in the barrel. A, by a piston-rod, R, set in motion by a steam-engine. The barrel. A, is provided with four valves, B, G, D, E, the first two of which open inward and the others open outward, as indicated in the diagram. Air passing through either D ov E is driven through the pipe, F, to the place where it is to be utilized. To explain the action of this machine, let us suppose the piston to be, say, at the upper limit of its play. Then, when the piston is depressed, the valve, G, will be closed by the tension of the compressed air, the valve, D, will be opened, the air which is forced into the pipe will close the valve, E, and all the air below the piston will be forced into the pipe, F; meantime the valve, B, is opened by the pressure of the external air and the space behind the piston is filled with air from without. When the piston is raised the valve, B, is closed by the tension of the compriessed air, the valve, E, is Pig. 181. 314 MBCHANTICS. opened, the air which is forced into the pipe closes the valve, D, and all the air above the piston is forced into the pipe, F; meantime the valve, G, is opened by the pressure of the ex- ternal air and the space behind the piston is filled with air from without. In this manner a continuous stream of air is kept flowing through the pipe, F, which may be utilized in any desirable manner. 8 f I 1 r E ^ ^^ si i±DK Fig. 182. The Air Pump. 333. The air ptunp is a machine for rarefying air. It consists of a barrel. A, in which a piston, B, is worked up and down by a lever, G, attached to a piston-rod, D. The barrel communicates with a ves- „ sel, E, called a receiver, by a narrow pipe. The receiver is usually of glass, ground to fit air-tight on a smooth bed-plate, KK. The joint between the receiver and plate may be ren- dered more perfectly air-tight by interposing a layer of tallow. A stop-cock, H, permits communication to be made at pleas- ure between the barrel and receiver, or between the barrel and external air. When the stop-cock is turned in a particu- lar direction, the barrel and receiver communicate ; but on turning it through 90 degrees, the communication with the receiver is cut off, and a communication is opened between the barrel and external air. Instead of the stop-cock, valves are often used, that are opened and closed by the elastic force of the air, or by the force that works the pump. The com- municating pipe should be exceedingly small, and the piston, B, when at its lowest point, should fit accurately into the bottom of the barrel. To explain the action of the air pump, suppose the piston to be at its lowest position. The stop-cock, H, is turned so HTDKAULIC AND PNEUMATIC MACHINES. 315 as to open a coniniunication between the barrel and receiver, and the piston is raised to its highest point by a force applied to the lever, C. The air, which before occupied the receiver and pipe, expands so as to fill the barrel, receiver, and pipe. The stop-cock is then turned to cut ofE communication between the barrel and receiver, and open the barrel to the external air, and the piston again depressed to its lowest position. The air in the barrel is expelled by the depression of the piston. The air in the receiver is now more rare than at the beginning, and by a continued repetition of the process, any degree of rarefaction may be attained. To measure the rarefaction of the air in the receiver, a siphon-gauge may be used, or a glass tube, 30 inches long, may be made to communicate at its upper extremity with the receiver, whilst its lower extremity dips into a cistern of mercury. As the air is rarefied in the receiver, the pressure on the mercury in the tube becomes less than on that in the cistern, and the mercury rises in the tube. The tension of the air in the receiver is indicated by the differenpe between the height of the barometric column and that of the mercury in the tube. To investigate a formula for the tension of the air in the receiver, after any number of double strokes, let us denote the capacity of the receiver by r, that of the connecting-pipe by p, and that of the space between the bottom of the barrel and the highest position of the piston by 5. Denote the original tension of the air by t ; its tension after the first upward stroke of the piston by t' ; after the second, third, . . .»'*, upward strokes, by t", t'", . . .t"'. The air which occupied the receiver and pipe, after the , first upward stroke, fills the receiver, pipe,\and barrel: ac- cording to Mariottb's law, its tension in the two cases varies inversely as the volumes occupied ; hence. 316 MECHANICS. In like manner, we shall have, after the second upward stroke, t':f::p + r + b:p + r, .: t" =t'j^^^. Substituting for i' its value, deduced from the preceding equation, we have. In like manner, we find, and, in general, after the n'* stroke, ^ -H^ + j + j-r If the pipe is exceedingly smaU, its capacity may be neg- lected in comparison with that of the receiver, and we then have. Let it be required, for example, to determine the tension of the air after 5 upward strokes, when the capacity of the barrel is one third that of the receiver. T In this case, ■= = #, and w = 5, whence, + r Hence, the tehsion is less than a fourth part of that of the external air. HYDRAULIC AND PNEUMATIC MACHINES. 317 Instead of the receiver, the pipe may be connected by a screw-joint with any closed vessel, as a hollow globe, or glass flask. In this case, by reversing the direction of. the stop- cock, in the up and down motion of the piston, the instru- ment may be used as a condenser. When so used, the tension after n downward strokes of the piston, is given by the formula. Taking the same case as that before considered, with the exception that the instrument is used as a condenser instead of a rarefier, we have, after 5 downward strokes. That is, the tension is eight thirds that of the external air. Artificial Fouutaias. 334. An artificial fountain is an instrument by which a liquid is forced upward in the form of a jet, by the tension of condensed air. The simplest form of artificial fountain is called Heko's ball. Hero's Ball. 235. This consists of a globe, A, into the top of which is inserted a tube, B, reaching nearly to the bottom of the globe. This tube is provided with a stop- cock, *(7, by which it may be closed, or opened at pleasure. A second tube, D, enters the globe near the top, which is also provided with a stop- cock, E. To use the instrument, close the stop-cock, C, ^^, and fill the lower portion of the globe with water through D ; then connect Z» with a condenser, and piimp air 318 MECHANICS. into the upper part of the globe, and confine it there by clos- ing the stop-cock, E. If, now, the stop-cock, G, be opened, the pressure of the confined air on the surface of the water in the globe forces a jet through the tube, B. This jet rises to a greater or less height, according to the greater or less quantity of air that was forced into the globe. The water will continue to flow through the tube as long as the tension of the confined air is greater than that of the external atmos- phere, or else till the level of the water in the globe reaches the lower end of the tube. Instead of using the condenser, air may be introduced by blowing with the mouth through the tube, D, and confined by turning the stop-cock, E. The piinciple of Hero's ball is the same as that of the air-chamber in the forcing-pump and fire-engine, already ex- plained. Hero's Fountain. 236. Hero's fountain is constructed on the same prin- ciple as Heko's ball, except that the compression of the air is efEeoted by the weight of a column of water, instead of by a condenser. -4 is a cistern, similar to Hero's ball, with a tube, B, extending nearly to the bottom of the cistern. C is a second cistern placed at some dis- tance below A. This cistern is connected with a basin, D, by a bent tube, E, and also with the upper part of the cistern, ^, by a tube, F. When the fountain is to be used, A is nearly filled with water, G being empty, A quantity of water is then poured into the basin, D, which, acting by its weight, sinks into C, compressing the air in the upper portion of it into a smaller space, thus increasing its tension. This increase of tension acting on the surface of the water in A, forces a jet through B, which rises to a greater or less Pig. 184 HYDKAULIC AND PNEUMATIC MACHINES. 319 height according to the greater or less tension. The flow will continue till the level of the water in A reaches the bottom of the tube, B. The measure of the compressing force on a unit of surface of the water in G, is the weight of a column of water, whose base is that unit, and whose altitude is the difEerence of level between the water in B and in G. If Heko's ball be partially filled with -water and placed under the receiver of an air-pump, the water will be observed to rise in the tube, forming a fountain, as the air in the re- ceiver is exhausted. The principle is the same as before ; the flow is due to an excess of pressure on the vater within the globe over that without. In both cases, the flow is re- sisted by the tension of the air without, and is urged on by the tension within. Wine-Taster and Dropping-Bottle. 237. The wine-taster is used to bring up a small portion of wine or other liquid from a cask. It consists of a tube, open at the top, and terminating below in a narrow tube, also open. When it is to be used, it is inserted to any depth in the liquid, which rises in the tube to the level of the liquid without. The finger is then placed so as to close the upper end of the tube, and the instrument raised out of the cask. The fluid ' escapes from the lower end, until the pressure of the rarefied air in the tube, plus the weight of a column ^s- iss. of liquid, whose cross-section is that of the tube, and whose altitude is that of the fluid retained, is equal to the pressure of the external air. If the tube be placed over a tumbler, and the finger removed from the upper orifice, the fluid brought up flows into the tumbler. If the lower orifice is very small, a few drops may be allowed to escape, by taking off the finger and immediately replacing it. The instrument then constitutes the dropping-bottle.- m wBh. m