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International Library of Technology 398 Sciences Pertaining to Civi. Engineering 167 ILLUSTRATIONS By EDITORIAL STAFF INTERNATIONAL CORRESPONDENCE SCHOOLS FUNDAMENTAL PRINCIPLES OF MECHANICS ANALYTIC STATICS KINEMATICS AND KINETICS HYDROSTATICS PNEUMATICS RUDIMENTS OF ANALYTIC GEOMETRY Published by INTERNATIONAL TEXTBOOK COMPANY SCRANTON, PA. 1927 GJ? Fundamental Principles of Mechanics: Copyright, 1906, by INTERNATIONAL TEXT- BOOK COMPANY. Entered at 'Stationers' Hall, London. Analytic Statics, Parts 1 and 2: Copyright, 1906, by INTERNATIONAL TEXTBOOK COMPANY. Entered at Stationers' Hall, London. Kinematics and Kinetics: Copyright, 1906, by INTERNATIONAL TEXTBOOK COMPANY. Entered at Stationers' Hall, London. Hydrostatics: Copyright, 1906, by INTERNATIONAL TEXTBOOK COMPANY. Entered at Stationers' Hall, London. Pneumatics: Copyright, 190.6, 1901, by INTERNATIONAL TEXTBOOK COMPANY. Copyright, 1901, 1897, 1895, 1893, by THE COLLIERY ENGINEER COMPANY. Rudiments of Analytic Geometry: Copyright, 1906, by INTERNATIONAL TEXTBOOK COMPANY. Entered at Stationers' Hall, London. All rights reserved Printed in U. S. A. ^BB^g, PRESS OF INTERNATIONAL TEXTBOQJK COMPANY SCRANTON, FA, 398 92109 PREFACE The volumes of the International Library of Technology are made up of Instruction Papers, or Sections, comprising the various courses of instruction for students of the International Correspondence Schools. The original manuscripts are pre- pared by persons thoroughly qualified both technically and by experience to write with authority, and in many cases they are regularly employed elsewhere in practical work as experts. The manuscripts are then carefully edited to make them suit- able for correspondence instruction. The Instruction Papers are written clearly and in the simplest language possible, so as to make them readily understood by all students. Necessary technical expressions are clearly explained' when introduced. The great majority of our students wish to prepare them- selves for advancement in their vocations or to qualify for more congenial occupations. Usually they are employed and able to devote only a few hours a day to study. Therefore every effort must be made to give them practical and accurate information in clear and concise form and to make this infor- mation include all of the essentials but none of the non- essentials. To make the text clear, illustrations are used freely. These Illustrations are especially made by our own Illustrating Department in order to adapt them fully to the requirements of the text In the table of contents that immediately follows are given the titles of the Sections included in this volume, and under each title are listed the main tppics discussed. INTERNATIONAL TEXTBOOK COMPANY B CONTENTS NOTE. Tills volume is made up of a number of separate Sections, the page numbeis of which usually begin with 1. To enable the reader to distinguish between the different Sections, each one is designated by a number preceded by a Section mark (), which appears at the top of each page, opposite the page number. In this list of contents, the Section number is given following the title of the Section, and under each title appears a full synopsis of the subjects treated. This table of contents will enable the reader to find readily any topic covered. FUNDAMENTAL PRINCIPLES OF MECHANICS, 27 Pages Motion 1-10 Introductory Definitions 1-4 Matter; Body, substance, and material; Motion and rest; Path; Initial and final position; Displacement. Velocity 5-10 Velocity in uniform motion; Formulas for uniform motion ; Velocity in variable motion ; Average or mean velocity; Instantaneous velocity; Direction and magni- tude of velocity. Force and Mass 1 1-52 Force ' 11-14 Definition of force; Balanced and unbalanced forces; Equilibrium; Weight; Measure of force; Magnitude of a force; Direction of a force; Mechanics; Dynamics; Kinematics; Statics. Fundamental Laws of Dynamics 15-35 Law of inertia ; Facts on which the law is founded ; Galileo's law of the independent effect of forces; Acceleration; Formulas for uniformly accelerated mo- tion; Initial and final velocities; Mass; Determination of the mass of a body; Law of action and reaction; Stress ; Tension, compression, and 'pressure. Immediate Consequences of the Preceding Laws 36-52 Important formulas; Space passed over in uniformly accelerated motion; Modification of the formulas when the body has an initial velocity; Retardation; Composition and resolution of forces. :v vi CONTENTS ANALYTIC STATICS, 28, 29 28 Pages Concurrent Coplanar Forces 1-33 Definitions 1-3 Coplanar and non-coplanar forces; Analytic statics; Graphic statics ; System of forces ; Equilibrants ; Inter- nal and external forces. Fundamental Principles and Formulas 4-10 Moments 11-13 Conditions of Equilibrium 14-23 General statement of the conditions of equilbrium; Equi- librium of three forces ; Selection of axes. Stresses in Framed Structures 24-33 Structure ; Machine ; Frames ; Trusses ; Supports ; Reac- tions ; Struts and ties ; Determination of stresses. Parallel Forces 34-49 Coplanar Forces 34-45 Two forces ; Theorem of moments ; Definition of a couple ; Any number of forces. Non-Coplanar Parallel Forces 46-49 29 Center of Gravity 1-28 Definitions and General Properties 1-6 Important Cases .' 7-28 Symmetrical figures ; Determination of the center of grav- ity by addition and subtraction; Center of gravity of polygons; Center of gravity of areas bounded by cir- cular arcs ; Center of gravity of a plane area bounded by an irregular curve; Center of gravity of solids. Coplanar and Non-Concurrent Forces 29-43 Couples 29-35 Effect of a circle; Equivalance and equilibrium of coax- ial couples. Equivalence and Equilibrium of Coplanar Non-Concur- rent Forces ". 36-43 Friction 44-65 Sliding Friction 44-54 Definitions and general principles; Angle and coefficient of friction. Resistance to Rolling 1 55-57 The Inclined Plane 58-65 CONTENTS vii KINEMATICS AND KINETICS, 30 Pages Composition and Resolution of Velocities 1-7 Uniform Motion in a Circle 8-22 Angular Velocity 8-12 Centripetal and Centrifugal Forces 13-17 Motion of a Train on a Curved Track 18-22 Work and Energy 23-34 Work 23-28 Energy 29-34 HYDROSTATICS, 31 Properties of Liquids : 1-2 Liquid Pressure 3-24 Pascal's Law and Its Applications 3-6 General Theory of Liquid Pressure 7-14 Pressure on an Immersed Surface 15-24 Buoyant Effort of Liquids 25-33 Immersion and Flotation 25-28 Specific Gravity 29-33 PNEUMATICS, 32 Properties of Air and Other Gases 1-16 Vacuum; Barometer; Tension of gases; Mariotte's law; Manometers and gauges ; Gay-Lussac's law. The Mixing of Gases. 14-16 Pneumatic Machines 17-41 The Air Pump 17-22 Air Compressors 23-25 The Siphon 26-27 The Locomotive Blast 28 Pumps 29-41 Suction pump; Lifting pump; Force pump; Plunger pump; Double-acting pumps; Steam pump; Centrif- ugal pumps; Hydraulic ram; Power necessary to work a pump. RUDIMENTS OF ANALYTIC GEOMETRY, 33 Graphs of Equations 1-14 Equations of Lines 15-31 viii CONTENTS RUDIMENTS OF ANALYTIC GEOMETRY (Continued') Pages The Straight Line 16-22 Equation of the straight line, Graph of any equation of the first degree, Applications, Pressure on the back of the dam The Parabola 23-31 Focus; Directrix, Vertex, Axes, Parameter, Applica- tions , Jet of water issuing from a small orifice , Moment in a beam. FUNDAMENTAL PRINCIPLES OF MECHANICS MOTION INTRODUCTORY DEFINITIONS MATTER-BODY 1. Matter may be defined as either- (a) whatever is known, or can be supposed capable of being known, through the sense of touch; or (b) that which occupies space 2. A phenomenon is whatevef happens in nature. The motions of the heavenly bodies, the rolling of a ball on the ground, the generation of steam from water, the formation of rust on the surface of iron, the circulation of the blood, the beating of the heart, thinking, speaking, walking ali these are phenomena It will be seen from the definition and examples just given that the word phenomenon, as understood and used in science, does not mean "something extraordinary or wonderful." For science, there is nothing extraordinary, and a phenomenon is simply a fact anything that happens. 3. Body, Substance, Material. A body is any limited portion of matter, as a block of wood, a coin, a stone, a piece of flesh, an animal. Another meaning of the word will be given presently. copvmoHTio BY INTRHNATIONAU TEXTBOOK COMPANY ALL RIHT MMMVBD 27 I I. T 8-3 2 FUNDAMENTAL PRINCIPLES 27 4. All bodies do not possess the same properties, or do not possess them in the same degree. This fact makes it necessary to distinguish different kinds of matter, to which different names are given 1 as, iron, water, air, flesh. The word substance is often employed in the same sense as the word matter, but in its more common use, it refers to matter of a special kind. Thus, water, steel, iron, are substances, or kinds of matter. The word body is also used, especially in physics and chemistry, to denote matter of a particular kind. In this sense, water, air, hydrogen, gold, are called bodies. In mechanics, however, the term is usually taken in the sense given to it in Art. 3. 5. Those substances that are used for works of art are called materials: steel, iron, brick, stone are examples. 6. Particle. A particle is a body so small that its dimensions may be disregarded. A particle is accoidmgly treated as if it were a geometrical point, except that it is sometimes necessary to consider some of its physical properties, such as weight. A particle is often called a material point. All bodies are treated as being com- posed of an indefinitely large number of particles. 7. It will be noticed in what follows that, in many cases, properties and principles that have been established and stated as relating to material points are extended to whole bodies. Thus, after investigating and explaining the laws governing the motion of a material point under certain circumstances, these laws are illustrated by taking as examples the motions of wheels, cars, engines, etc. The reason is that in cases of this kind the properties considered do not depend on the size of the bodies. For example, the speed of a car moving in a straight track is measured by the speed of any of its points, and what applies to the speed of a point applies to the speed of the car. So, too, the force requj v to move in a :traight track a train of a certain weigh the same as the force necessary to move a particle of the same weight; and, therefore, the 27 OF MECHANICS 3 laws obtained m the case of a particle may be applied to the motion of a tram or any othei body, bo long as the only thing to be taken into account is the weight of the body. 8. Deformation. When a body is pulled, pressed, or struck m any direction, its shape is more or less changed, either permanently or temporarily. A change of this kind is called deformation, and the body undergoing it is said to be deformed. 9. A rigid body is a body that is not deformable, that is, a body whose foim and dimensions remain the same, to whatever action it may be subjected There are no absolutely rigid bodies m nature; but it often simplifies matters to consider bodies at first as being per- fectly rigid, which is equivalent to neglecting their deform- ability, the latter being afterwards taken into account. Besides, in many cases, deformability plays so small a part that, for practical purposes, it may be neglected entirely. MOTION AND REST 10. Relative Position. The relative position of a point with respect to another is determined by the length and direction of the straight line between the two points When either the length or the direction, or both the length and direction, of this line change, the relative position of the two points is said to change The relative position of two bodies with respect to each other is determined by the relative position of the points of one body with respect to those of the other. When the rela- tive position of one or more points m one of the bodies with respect to the points of the other body changes, the relative position of the two bodies, with respect to each other, is also said to change. 11. Motion is a change iiT-rtie relative position of two bodies, and is said to be post* 1 ' M by either body with respect to the other. 4 FUNDAMENTAL PRINCIPLES 27 12. Rest is the condition of two bodies not in motion with respect to each other, and is said to belong to either body with respect to the other. Each body is said to be fixed with respect to the other body. 13. When the motion or rest of a body is referred to, without specifying 1 any other body, it is generally understood that the other body is the earth or its surface. Thus, the motion of a train, of a steamer, of a horse, as usually spoken of, means the change of position of the train, steamer, or horse with respect to the ground that is, to the surface of the earth, or to objects on that surface. 14. A body may be in motion with respect to another body, and at rest with respect to a third body. For example, the smokestack of a locomotive is at rest with respect to the boiler, since their relative position does not change; yet, both may be moving with respect to objects on the ground. Like- wise, a man standing on the deck of a moving vessel is at rest with respect to the vessel, but in motion with respect to the water, the shore, etc. 15. Path. The path, or trajectory, of a moving point is the line over which the point moves, or, in geometrical language, the line generated by the moving point. The path of a point may be straight or curved, or a combination of straight and curved lines. 16. Initial and Final Position. When the motion of a point during a certain time or over a certain portion of its path is considered, tha position occupied by the point at the beginning of the time is called the Ini- tial position of the point, Flt< ' and the position at the end of the time is called the final position. Suppose that a particle starts from A, Fig. 1, and moves so that it describes, in a certain time, the curve A M B\ this curve is the path of the particle. With respect to the time considered, A is the initial and B the final position of the particle,. , , , 27 OF MECHANICS 5 17. Displacement. The length of the straight line joining the initial and the final position of a moving point is called the displacement of the point while the point passes from the former to the latter position. In Fig 1, in which the point moves from A to B m the curved path AM B, its displacement during this motion is the length of the straight line AB 18. The space passed over, or described, m a certain time by a moving point is the length of the part of the path over which the body passes during that time. In Fig 1, the space described by the particle between A and B is the length of the curve A MB 19. Direction of Motion In a Curve. By the direc- tion of a curve at any point is meant the direction of its tangent at that point So, too, when a point is moving in a curved path, the direction of motion at any moment is the direction of the tangent to the path at the point occupied at that moment by the moving point. In Fig 1, the line PT just touches the curve AMD at P, it is, therefore, a tangent to the curve at P, and shows the direction in which the particle is moving when it reaches the position P VELOCITY VELOCITY IN UNIFORM MOTION 20. Uniform Motion. A point is said to move with uniform motion when it passes over equal distances in any and every two equal intervals of time The fact, for instance, that a point moves over 10 feet during the first second of its motion, and over an equal space during the fiftieth second, is not enough to define the motion as uni- form: that the motion may be uniform, the point must describe the same space (10 feet m this case) in every second, whatever instant is chosen in which to begin to count the time; thus, the space described by the point between the middle of the third second and the middle of 6 FUNDAMENTAL PRINCIPLES 27 the fourth must be the same as the space described in the twentieth second, or during the second in which the time of motion changes from 27 15 to 28.15 seconds, or, in short, during any and every interval of 1 second. 21. Definition of "Velocity. It is a familiar fact that bodies move more or less rapidly, more or less slowly By this is meant that they require more or less time to move from one place to another, or that they move over longer or shorter distances in a given time. In ordinary language, the term speed is used to denote the degree of quickness or rapidity of motion. The measure of the speed, or of the degree of quickness, of a motion is called velocity. How this measure is obtained will now be explained. 22. Uniform. Velocity. Let a point M> Fig. 2, move along the path A B, which may have any form, either curved or straight; and suppose the motion to be such that equal lfIG 2 ""* spaces are passed over in equal times; that is, suppose the motion to be uniform. Also, let the number of units of length passed over in a unit of time (as the number of feet passed over in 1 second) be denoted by v. It is evident that the magnitude of this number v depends on how fast the point is moving. Thus, if the point passes over 10 feet in 1 second, it is moving twice as fast as if it passed over only 5 feet in 1 second Hence, the num- ber v may be used to measure the rapidity of the motion, in which case v is the velocity of the moving point. In general terms, then, the velocity of a point moving with uniform motion is the space passed over by the point in a unit of time. When, as here assumed, the motion is uniform, the veloc- ity also is said to be uniform, or constant. What is meant by variable motion and variable velocity will be explained presently. Velocity is expressed in units of length per unit of time. For example, if a body moving uniformly passes over 7 feet in 1 second, its velocity is 7 feet per second. In 1 minute, the body passes over a space of 7 X 60 = 420 feet; hence, 27 OF MECHANICS 7 its velocity can be expressed also as 420 feet per minute. A train that moves uniformly and travels 45 miles an hour has a velocity of 45 miles per hour. 23. Formulas for Uniform Motion. Let a point or body moving uniformly pass over a space j during any time t. Then, it is obvious that the space described in a unit of time is j t. Hence, v = S - t (1) For example, if the point moves over a space of 10 feet in 2 seconds, v = 10 2 = 5 feet per second If the velocity and the time are known, the preceding formula gives the space passed over. s = vt (2) Similarly, if the velocity and the space are given, the same formula, or formula 2, gives the time. t= s - (3) v EXAMPLE 1 The wheels of a carnage moving with uniform velocity make 3,760 revolutions between two stations A and B The diameter of each wheel is 4 feet, and the time employed by the carnage to travel from one station to the other is 7 6 hours It is required to find the velocity of the carnage, m feet per second (By the velocity of the carriage is meant the velocity of any of its points ) SOLUTION During each revolution, the carriage passes over a distance equal to the circumference of the wheel, or 4 X 3 1416 ft. The total distance traveled, or space passed over, m feet, is s = 3,760 X 4 X 3.1416 The time required to travel this distance is 7 6 br. As the velocity is required in feet per second, this time must be reduced to seconds, which gives / = 75X60X60 Substituting In formula 1 the values of s and / just found, EXAMPLE 2. The distance between two ports is 400 miles What must be the velocity, in knots, of a vessel that will cover the distance in 18 hours? (A knot is a velocity of 6,080 feet per hour.) SOLUTION To find the velocity m feet per hour, we have, i - 400 mi. = (400 X 5,280) ft , t - 18 hr Therefore, by formula 1, , = *Lxpo f , per hr . . x M knote . 19 . 298 lnot , FUNDAMENTAL PRINCIPLES 27 VELOCITY IK VARIABLE MOTION 24. Variable motion is that motion in which the moving point, or body, passes over unequal spaces m equal times, or over equal spaces in unequal times 25. Average or Mean Telocity. It will be remem- bered that the distinguishing characteristic of uniform motion is the equality of all distances passed ovei m equal intervals of time, and that the quotient obtained by dividing any space 5 by the number / of units in the conesponding interval of time is a constant quantity (v, formula 1 of Art 23). Unless all these quotients are equal, the motion, and therefore the velocity, is not uniform Suppose, for example, that a tram moves over a distance of 60 miles in 45 minutes. If it is known beforehand that the motion is uniform, the velocity may be found by dividing 60 miles by 46, which gives v = - miles 45 3 per minute. In this case, the space described in any num- ber t of minutes would be, by formula 2 of Art. 23, v t = &/ miles. Suppose, on the contrary, that it is not known whether the motion has been uniform during the 45 minutes, but that the distance passed over during the first 5 minutes is known to be 5.5 miles, and that the distance passed over between the end of the thirtieth and the end of the fortieth minute is 13.98 miles Then, the quotients obtained by dividing the spaces by the numbers measuring the corre- sponding times are, respectively, 5-5 _ T in 13.98 _ , OQQ 60 _ 4 T " ' ~W ~ lf6W] 46 " 8 Since these quotients are not equal, there is no uniform motion, and therefore no uniform velocity. If, however, another train is imagined moving with a uni- form velocity of $ miles per minute, this tram will pass over the distance of 60 miles in 45 minutes. For this reason, \ miles per minute is called mean, or average, velocity of the actual train considered; it is not a velocity that 27 OF MECHANICS 9 the train really possesses, but rather the equivalent velocity with which a body moving uniformly would move over the same space in the same time Likewise, if the motion of the tram during; the first 5 minutes is considered, its 5 5 mean velocity, foi that interval, is - = 1 10 miles per 5 minute. In general, if a body moves over a space j in the time /, its mean velocity v m , during that time, is defined mathematically by the formula 26. Variable Velocity Instantaneous Velocity. When a point is moving with variable motion, its velocity at any instant is the velocity that the point would have if, at that instant, its speed ceased to change, that is, if, after that instant, the point moved neither faster nor more slowly than it is moving. Since the motion is variable, the velocity at any instant is not any velocity with which the body actually moves over any part of its path, but simply the velocity with which it would move if its speed became invariable. The velocity of a moving point or body at any given moment is called the instantaneous velocity at that moment. In variable motion, the velocity is said to be variable, because it changes or varies from instant to instant. 27. To illustrate the character of variable motion and velocity, imagine a train A to start from rest and leave a station at the same time that another train B is passing the station with a uniform velocity of 60 miles per hour, and suppose the two trains to run on two parallel tracks. At first, B will move ahead of A, since A' starts from rest. vSuppose, however, that A moves faster and faster, so that to a person in B the motion of B with respect to A will appear to become slower and slower; there may then be a time when B will be losing instead of gaining speed, with respect to A, or when A will appear to be moving past B. Thus, during the first minute, B will move over 1 mile, and 10 FUNDAMENTAL PRINCIPLES 27 A may move over only 4 mile; but, during: the following minute, A may move over li miles, and then, it will be i mile ahead of JS. Since A is at first losing space with respect to B and then is gaming, there must be an instant at which A is neither losing nor gaining; at that instant the two trains are evidently at rest with respect to each other, and if the speed of A ceased to change, A would continue to move with the same velocity as , or 60 miles per hour. The velocity of A at that instant is, therefore, 60 miles per hour. It must be understood that, as already stated, the tram A does not actually move with the velocity of 60 miles per hour during any interval of time; but this is the velocity with which A would move if, after the instant considered, its motion underwent no further change. 28. Direction and Magnitude of Velocity. By the direction of the velocity of a moving point is meant the direction in which the point is moving. 29. By the magnitude of a velocity is meant the numerical value of the velocity, expressed in units of length per unit of time, as feet per second or miles per hour. EXAMPLE A float moves down a stream from a point A to o. point D, a distance of 800 feet, in 2-J minutes The float is observed at two intermediate points B and C, whose distances from A are, respectively, 350 and 575 feet, and it Is found that it moves from A to B in 45 seconds, and from B to Cm 70 seconds. Required, the mean velocity, m feet per second, of the surface of the stream (a) between A and D, (6) between A and B t (c) between B and C; (rf) between C and D SOLUTION (a) Here s = 800 ft , / = 2| min = 150 sec ; and the formula of Art 25 gives v m = r = r^r = 5.33 ft. per sec. Ans. (6) Here s = 350 ft , t = 45 sec., and, therefore, v m = 7 = ~r=- = 7 78 ft per sec. Ans. /45 (c) Here s = 575 ft - 360 ft = 225 ft., t = 70 sec , and, therefore, 225 t ** 70 27 OF MECHANICS 11 (<f) Here s = 800 ft - 575 ft = 225 ft , t = 150 sec - (45 + 70) sec. = 35 sec , and, therefore, v m = - = -T- = 6 429 ft per sec. Ans. EXAMPLES FOB PRACTICE 1 A tram moves with uniform velocity between two stations 375 miles apart, the distance is traveled in 7i hours. What is the velocity of the train, in feet per second? Ans 73 333 ft per sec 2 A train moves from station A to station , a distance of 90 miles, in 1$ hours, from station B to station C, a distance of 17 miles, m % hour; and from station C to station D, a distance of 64 miles, in 2 hours Find its mean velocity, m feet per second (a) between stations A and >, (b) between stations B and C\ (c) between stations B and D. ( (a) 59 012 ft per sec. Ans { (d) 49 867 ft per sec [(c) 4752ft. per sec. FORCE AND MASS FORCE DEFINITIONS RELATING TO FORCE MEASURE OF FORCE 30. Definition of Force. It is known from experience that, when a body is at rest, it can be set in motion by the action of another body. Thus, if a block of iron is lying on the ground, it can be set m motion by pulling it with a rope, by pushing it with the hand, or by placing a strong magnet near it. Furthermore, the block may be pushed by one per- son in one direction, while another person pushes it m the opposite direction; or, the magnet may be placed on one side of the block, at the same time that the block is pulled by a string from the opposite side. In cases of this kind, there may be no motion, on account of the neutralizing effects of two or more actions on the same body. We say, however, that there is a tendency to motion; for the moment one of the opposing actions is removed, the other causes the body to move 12 FUNDAMENTAL PRINCIPLES 27 31. This action of one body on another, producing, 01 tending to produce, motion in the latter body, is called force, and the former body is said to exert force on the latter The characteristic of the action known as force is, then, that, when exerted on a body originally at rest, and not influenced by other bodies, it results in the motion of the first-mentioned body. The nature of the ensuing motion, and the result of the action in question, when the body acted on is already in motion or under the action of other bodies, are complicated phenomena to be either actually deter- mined by experiment or inferred from experimental data. 32. Balanced and Unbalanced Forces. If two or more forces act on a body in such a manner that they cause no motion, owing to the neutralizing effects they tend to pro- duce, each force is said to be balanced by the combined action of the others, and is referred to as a balanced force. When a force acts on a body, and there is no opposing force preventing the motion of the body, the force is called an unbalanced force. These definitions apply to moving bodies as well as to bodies at rest If two or more forces are balanced when exerted on a body at rest, they are also balanced when the body is in motion; that is, they do not affect the motion of the body. 33. Equilibrium. A body is in equilibrium when it is under the action of balanced forces. The balanced forces themselves are also said to be in equilibrium. It should be noticed that equilibrium and rest are not equivalent terms A body may move uniformly while acted on by balanced forces, in which case it is in equilibrium, but not at rest In this case, however, the motion of the body is not due to the balanced forces acting on it. This subject will be better understood after the law of inertia, presently to be explained, has been studied 34. Weight. Experience teaches that, when a body is unsupported, it falls to the ground. This fact is ascribed to a force exerted by the earth on all bodies, which force is known 27 OF MECHANICS 13 by the general name of force of attraction, or simply attraction, and also force of gravity, or simply gravity. The atti action of the earth on any particular body is called the weight of the body The methods of comparing the weights of bodies are well known. 35. Measure of Force. For engineering purposes, force is expressed in units of weight, such as pounds, kilo- grams, etc. The reason for this will be more apparent when it is considered that every force can be replaced by a weight. Thus, suppose a body P, Fig. 3, to be suspended from the extremity A of a per- fectly symmetrical A Aj B beam AB resting at its center on a knife -^ edge F. If we wish / f\ \,Q to prevent P from falling, we may pull downwaids on the string at Q, 01 attach at Q a. weight equal to P, or tie a small piece of iron at Q and place a magnet underneath. In all these cases, the weight of P balances, and is therefore equivalent to, the force at Q, by whatever means the latter may be produced. As the two forces are equivalent, the one may be meas- ured by the other, and we may say that the force acting at Q is 20 pounds, kilograms, tons, etc., according as P weighs 20 pounds, kilograms, tons, etc. Suppose, for instance, that the weight of P is 10 pounds, and that we pull at Q with sufficient force to keep P from falling, but without moving it. Then, the force with which we are pulling is 10 pounds. If P falls, the pull is less than 10 pounds; and if it rises, the pull is greater than 10 pounds. 36. Magnitude of a Force. By the magnitude of a force is meant the numerical value of the force, expressed in units of weight. If, for example, a force is equivalent to a weight of 12 pounds, its magnitude is 12 pounds. 37. Direction of a Force. The direction of a force is the direction in which the force moves, or tends to move, the body on which it acts. 14 FUNDAMENTAL PRINCIPLES 27 DEFINITION AND DIVISIONS OF MECHANICS 38. Mechanics is the science of force and motion. This science is divided into two general branches, (a) dynamics, (b) kinematics or phoronomics. 39. Dynamics is the science of force and its effects. In this science, the forces applied to bodies are given, and the resulting effects of these forces are determined, or the conditions of motion are given, and the forces necessary to produce that motion determined. Rest is considered as a special case of motion in which the velocity is zero. 40. Kinematics treats of motion alone, without refer- ence to either force or the physical or mechanical propeities of bodies. Suppose, for instance, that a body is moving in a circle with uniform velocity, and that it is required to deter- mine what force is necessary to preserve this kind of motion, this is a problem in dynamics, for it relates to force Suppose that it is required to determine what space the body travels in a certain time; this is a problem in kinematics, for it can be solved without having regard to anything but the velocity and path of the body. 41. Subdivisions of Dynamics. Dynamics is sub- divided into two branches: (a) kinetics, (b] statics. 42. Kinetics treats of unbalanced forces and their effects. 43. Statics treats of the equivalence and equilibrium of forces. NOTE The preceding divisions and definitions are of very recent origin Formerly, the term dynamics was used in the sense m which kinetics 13 now used, the latter terra being then unknown Even today, the old meaning is frequently given to the term dynamics, the best modern writers, however, use it in the sense just defined that is, to denote the science of force in general, whether it pro- duces motion or not 44. Applied mechanics is the application of mechan- ical principles to the works of human art. 27 OF MECHANICS 16 FUNDAMENTAL LAWS OF DYNAMICS I-iAW OF INKltTIA 45. Statement of l ho IJHW. In giving the definition of force (Art. 30), icference was made to the fuel, known from experience, that a body may act on anothei body, originally at rest, in such a manner as to pioduce motion in the latter body. This action was defined as foice That definition relates only to the effect of force on a body origi- nally at lest. The law presently to be stated expresses the general effect of force under any eiieumsUnces; that is, whether the body on which it is excited is at icst or in motion. Tins law, known as the lir\v of lm>rltn, or the first law of motion, may be formulated as follows: // a body is not acted on by fom\ the body, m> u tvhol<\ /.s either at test ot inovmg uniformly in a sh (tight faith. This is not as definite a statement of the law of ineilia as can be given. But the complete statement requires, in order to be understood, some familiarity with mechanical piinciples and conceptions of a high order, with which the beginner cannot be supposed to be acquainted By saying that a body as a whole is in motion, it is meant that all the points of the body are in motion. Each of the wheels of a locomotive, for instance, has a motion us a whole, and, if the track is straight, the wheel, as a whole, is said to be moving in a straight path, whose direction is that of the track. Any one would understand what was meant by saying that a billiard ball was rolling in a diiection par- allel to one of the cushions of the billiard table. In the simple cases of motion just given, the direction of the motion of the body as a whole is plainly seen to be flu* direction in which its center of figure is moving. In the OIKO of unsymmetrical bodies, the direction of their motion is the direction in which a certain point in them, called the oonlor of fifravlty, is moving, For the present, however, the student may think of the motion of symmetrical bodies 16 FUNDAMENTAL PRINCIPLES 27 only, such as balls, disks, cubes, prisms, etc., all of which have a center of figure. 46. Facts on Wlilcli the Law is Founded. That a body does not start to move by itself is a fact so familiar that it is scarcely necessary to state it That a body, once set in motion, does not come to rest unless acted on by force is not so obvious. Bodies often come to rest apparently by themselves, without any assignable cause, and this gave rise to the belief, prevalent among the ancients, and still entertained by some, that rest is the natural condition of all bodies, and that all moving bodies have a tendency to come to rest. A little consideration will show that this is an erroneous view of the nature of motion. If a block is placed on a stone pavement and struck sidewise, it will slide for a short distance and soon come to rest; if the experiment is tried on an asphalt pavement, the block will move farther and come to rest after a longer time; it will slide longer and farther on a piece of marble, and longei and faither still on a sheet of ice. These facts plainly indicate that the tendency to come to rest is not inherent in the block itself, but depends on the action of other bodies namely, the bodies on the surfaces of which the block slides, and we naturally infer that, were it not for this action (or frlc- tlonal resistance, as it is called), the block would not come to rest at all. Another common obstacle to motion is the resistance of the medium, by which is meant the action of the substance through which the body moves. Thus, if a block is placed on a horizontal suiface, immersed in mercury, and then struck sidewise, it will move through a very short distance, If the experiment is repeated m water, the block will describe a longer space before coming to rest. The space will be longer in air, and longer still in a space (called a vacuum) from which the air has been removed by means of an air pump. Here, as before, the inference is that, did the medium offer no resistance, or, more properly, were there no medium, the body would, move forever in a straight path. 27 OF MECHANICS 17 47. In the preceding examples, it will be observed that, whether, the body is brought to rest by fnctional resistance or by the resistance of the medium, the diminution of its velocity is continuous, in other words, the body comes to rest gradually, and its velocity constantly diminishes while the resisting force acts An example of the action of force in increasing the velocity of a moving body is afforded by the familiar phenomenon of a falling body Here, the body is constantly acted on by the attraction of the earth, and its velocity is observed to increase very lapidly As another example, suppose that a person pushes a car on a horizontal surface. The car moves at first with almost no velocity; but, as the person continues to push, the velocity keeps growing greater and greater. 48. The necessity of applying force to a moving body in order to change the direction of its motion is also a familiar fact. If a prismatic block is sliding on a surface in the diiec- tion of its axis, no change m the direction of its motion will be observed unless the prism is struck, or pulled, or pushed, or in some other way interfered with by some other body. Of course, all foices do not change the direction of motion, but only those exeited by bodies that are, so to speak, outside the path of the moving body. Thus, in the example just given, a pull exerted by a rope in the direction of the axis of the prism would cause no change in the direction of motion; but, if the lope were pulled at right angles to the axis of the prism, the direction of motion would evidently change. 49. Summing up, it may be said that, whenever the motion of a body has been observed to change \ cither in direction or in velocity , or in both, it has always been possible to trace the change to the influence of other bodies ikat ZA, of force. Furthermore ', bv diminishing' those influences, the changes referred to are corre- spondingly diminished; and the conclusion has been reached thai, could these influences be entirely eliminated, the said changes would not take place at all. If, then, a body is at rest, and no unbalanced force acts on it, it will continue at rest; if it is moving 1 in a straight line ILT398 3 18 FUNDAMENTAL PRINCIPLES 27 with uniform velocity, it will continue to move m the same straight line with the same velocity If a body moves under the action of a force for a certain time, its velocity will constantly change during that time; but, if the force ceases to act, the body will continue to move uniformly in the direction and with the velocity it had at the instant the foice was withdrawn. It is not necessary, however, in order that the body may continue to move uniformly, to withdraw the force or forces acting on it. the same result will be obtained if other forces are introduced equal and opposed to those already acting. An instance of this is afforded by the motion of a train- at first, the tractive force of the engine is greater than the combined resistance of friction and the air, and the velocity of the train constantly increases, but, as will be explained elsewhere, the resistance increases with the velocity, so that there is a moment when the traction of the engine is balanced by the resistance, and from that moment on, the train moves uniformly. 50. Conversely, if either the velocity or the direction of motion of a body changes^ the body must be under the action of unbalanced forces, for, according to the law of inertia, rectilinear motion with uniform velocity (in which rest must be included as a special case) is the only possible motion of a body not acted on by unbalanced forces. Thus, if a body moves in a curve, it may be concluded that, whatever its velocity may be, some unbalanced force or forces must be constantly acting on the body. The moment these forces cease to act, the body will continue to move uniformly in the direction of the tangent to the path at the point occupied by the body at that moment. For example, if a stone is tied to a string and swung around, the pull of the string will keep the stone moving in a circle; but the moment the string breaks, the stone will fly off on a tangent to this circle. A train is kept on a curve by the resistance of the rails acting against the flanges of the wheels; but if this resistance is not great enough, the wheels get off the rails, and the train, in leaving the curve, moves straight ahead along the tangent 27 OF MECHANICS 19 51. Inertia of Bodies. The fact that the motions of all bodies follow the first law of motion is often expiess>ed by saying that all bodies possess the property of Inez-tin. In this sense, it is very common in mechanics to refer to what the motion of a body is or would be, "by virtue of the inertia of the body", that is, according to the law of inertia. Sup- pose, for instance, that a -* particle is moving along the path A B, Fig. 4 It follows fiom the first law of motion that the particle must be under the action of unbalanced forces, since the path is not straight. Suppose, also, that, by some means, it is found that the veloc- ity of the particle, when at /*, is 6 feet per second. Then, we say that, if the forces were suddenly withdrawn (or bal- anced), the particle, "by virtue of its inertia," would move uniformly along the tangent P T, describing a space of 6 feet in every second GALILEO'S LAW OF THE INDEPENDENT EFFECT OF FORCES 52. Preliminary Explanation. Let A, Fig. 5, be a body acted on by a force Suppose the force to be such that, if it acts on the body for a certain time /, and the body is at rest when the force A r~~ 7- 1 *" begins to act, the body / / moves over the path A A' I /in that timej so that its / / final position is A 1 . Sup- / / pose, now, that the body, / / instead of being at rest /, /, when the force is applied, A PlGlG Al is moving along the path A A lt in such a manner that, if the force did not act, the body would move to A l in the time /. It is found from experiment that if, while the body has this motion, the force referred to above constantly acts on it during the time *, the direction of 20 FUNDAMENTAL PRINCIPLES 27 the force remaining; unchanged, the final position of the body after this time is a point A,', such that the line A^AJ is equal and parallel to A A 1 . In other words, the final position of the body, with respect to the position the body would have occupied if the force had not acted, is the same whether the force starts the body from rest or acts on the body while the latter is in motion and acted on by other foiceb Had the body been originally at rest, the effect of the force would have been to cause the displacement A A', in the direction of the force, in this case, the final position of the body, had not the force acted, would have been A. If the force had not acted and the body had been moving along A A t , its final position would have been As, the effect of the force is to cause the displacement A^AJ of the final position, which displacement is equal and parallel to A A'. 53. Statement of Galileo's Law. From the pre- ceding facts, and from others of a similar character, is derived the following general law, which is Galileo's law of the Independent effects of forces: The effect of a force on a body is the same whether the body js at rest or in motion, and, if several forces act simultaneously on a body, each force produces its effect independently of the othct forces. The meaning of this proposition is that each force produces the same amount of displacement, parallel to its direction, as if it acted alone and moved the body from a state of rest Galileo's law may be otherwise stated as follows: When a force acts on a body, the position of the body at any time relatively to the Position the body would have if the force had not acted, is independent of the motion Produced m the body by any other forces. 54. Experimental Verification. The following experiment affords an easy verification of Galileo's law In Fig. 6, a ball e is supported in a cup, the bottom of which is attached to the lever o in such a manner that a movement of o will swing the bottom horizontally and allow the ball to drop. Another ball b rests in a horizontal groove that is provided 27 OF MECHANICS 21 with a slit in the bottom. A swinging arm is actuated by the spring d m such a manner that, when drawn back, as shown, and then released, it will strike the lever o and the ball b at the same time This gives b an impulse in a horizontal direction and swings o so as to allow c to fall On trying the experi- ment, it is found that b follows a path shown by the curved dotted line, and reaches the floor at the same instant as <?, which drops vertically This shows that the force that gave the first ball its horizontal move- ment had no effect on the vertical force that compelled both balls to fall to the floor, the vertical force producing the same effect as if the horizontal force had not acted. ACCELERATION 55. Change of Velocity Caused by an Unbalanced Uni- form Force. A uni- form force is a force whose magnitude and dnection do not change. Both from the law of inertia and fiom the definition of force, it follows that the effect of an unbalanced uniform force on a body originally at rest is to set the body m motion and impart to it a continually increasing 22 FUNDAMENTAL PRINCIPLES 27 velocity Galileo's law affords the means to determine exactly how that increase takes place. Suppose that a uni- form force acts on a body originally at rest, imparting to it, in 1 second, a velocity of 10 feet per second This means that, if the force acts during 1 second and then ceases to act, and the body is not interfered with by other forces, the body will, by the law of inertia, continue to move uniformly with a velocity of 10 feet per second If the force, instead of being withdrawn at the end of the first second, acts for 1 second longer, its effect will be independent of the motion already acquired by the body; that is, during the second second, the force will impart to the body a velocity of 10 feet per second, rn addition to the velocity the body had at the end of the first second, so that, at the end of the second second, the body will have a velocity of 20 feet per second Likewise, the vel- ocity at the end of the third second will be 30 feet per second; and so on. A similar law of change applies whenever a body is acted on by an unbalanced uniform force, that is, the velocity of the body increases by a fixed amount during every unit of time; or, in other terms, the velocity increases at a constant rate. If, for example, the change of velocity per second is a feet, the change m t seconds will be at feet per second. Of course, it is not necessary to use the second as a unit of time, nor the foot as a unit of length; any other units may be used. 56. Acceleration. The acceleration of a body moving in a straight path under the action of an unbalanced force is the amount by which the velocity of the body in the direction o the force increases in a unit of time. 57. When a body moves under the action of a uniform force, its change of velocity is the same for any two equal intervals of time, as has just been explained. In this case, the acceleration is said to be uniform, and the motion is said to be uniformly accelerated. 58. In the example given in Art. 55, the acceleration is 10 feet per second per second The import of this appar- ently confusing expression is this: J!f at any moment the I V J 27 OF MECHANICS 23 felocity of the body is v feet per second, this means that, if at that moment the force ceased to act, the body would con- tinue to move uniformly with a velocity of v feet per second. If, however, the uniform force continues to act for 1 second longer and then ceases to act, the velocity will have increased to v + 10 feet per second; that is, the body, if left to itself, will continue to move uniformly with that velocity, descri- bing v + 10 feet every second, or 10 feet more than before. Instead of saying that the acceleration is 10 feet per second per second, it is customary and sufficient to say simply that the acceleration is 10 feet per second, it being understood that the velocity is also measured in feet per second. If the acceleration of a body was said to be 50 miles per hour, the velocity would be understood to be measured in miles per hour, and to be, at the end of any one hour, 50 miles per hour greater than at the beginning of that hour. 59. Formulas for Uniformly Accelerated Motion. Let a body start from rest and move under the action of a uniform force for t seconds. If the acceleration due to this force is a, the velocity at the end of 1 second will be a, at the end of 2 seconds, 2 a; and at the end of / seconds, ta, or a t. Denoting by v the velocity at the end of / seconds, we have, then, v = at (1) "i (2) If, at any instant, the body has a velocity z> , the veloc- ity v t t seconds after that instant, is given by the formula v z/ + at (3) From this formula follows "V 1)a I A\ a - - * (4) The velocity v t) taken with respect to the interval of time considered, is called the initial velocity of the body; and v is called the final velocity. It is immaterial how the body has acquired the velocity V Q ; in any case, the effect of the applied force is to produce an increase of velocity equal 24 FUNDAMENTAL PRINCIPLES 27 to at, or v TV feet per second, in t seconds The change 1) -~ i -' 7/ of velocity produced in 1 second is, therefore, - - , as expressed by formula 4, and this agrees with the general definition of acceleration In all this, it is assumed that the force acts in the direction of motion. If the body starts from rest, v = 0, and formulas 3 and 4 become identical with formulas 1 and 2, respectively. MASS 60. Force Proportional to Acceleration. Accord- ing to Galileo's law, if several forces act simultaneously on a body, each force produces its effect independently of the others Therefore, if the forces act m the same direction, they will produce an acceleration equal to the sum of the accelerations that the forces would produce if each force acted alone. Furthermore, if the forces are equal, the accel- eration will be equal to their number multiplied by the acceleration due to any one of them. It is also evident that any force may be supposed to be equivalent to any number of forces acting in the same direction as the given foice, and whose sum is equal to the given force Thus, a weight of 5 pounds is equivalent to a weight of 3 pounds and one of 2 pounds, or to a weight of 4 pounds and one of 1 pound, or to five weights of 1 pound each, etc. Let a force of 3 pounds act on a body, imparting to it an acceleration of 5 feet per second. Then, a force of 12 pounds, which is equivalent to four forces each equal to 3 pounds, will produce an acceleration four times as great, or 4 X 5 = 20 feet per second. Any other force will produce an acceleration that will be as many times 5 feet per second as the force contains 3 pounds. For instance, a force of 49 49 pounds will produce an acceleration of -- X 5 = 81.67 feet 3 per second, nearly. In general, let a force F l impart the acceleration a^ to a certain body. Let F n be another force imparting the acceleration a n to the same body. If F n is 27 OF MECHANICS 25 twice F lt then will a n be twice ,; and, generally, if F n = nF l where n is any number, fractional or integral, then a n n ,. Now, = = ; s = - = ; 7*i ^*i ! tfl that is, ~ = Therefore, /or /// &z#z body, any two forces are to each other as the accelerations they impart to the body. From the last equation follows ^ = a n ai 61. If several forces F lt F a) F,, etc are capable of pro- ducing, respectively, the accelerations a lt a a , a,, etc. in the same body, then, 5 = 5 = 5, etc. fti a, a, Each of these quotients may be considered to represent the force necessary to give the body an acceleration of 1 foot per second (or, in general, 1 unit of length per unit of time). If the value of this force, or the common value of the pre- ceding quotients, is denoted by m, then w = = = 5, etc.; a a, #o and hence, F! = ma^F, = ma,, F a = ma a , etc. In general, if Fis any force producing in a body the accel- eration a, we have m = (1) a and F = ma (2) 62. Definition of Mass. Experience teaches that it requires different forces to produce the same acceleration m different bodies. Therefore, the value of m is different for different bodies. When two equal forces produce the same acceleration in two bodies, the two bodies are said to have the same mass. If one body requires a greater force than 26 FUNDAMENTAL PRINCIPLES 27 another in order to receive a certain acceleration, the former body is said to have a greater mass than the latter. Mass, then, is that property of bodies on which alone the acceleration they receive when under the action of given forces depends. The acceleration, of course, depends on the applied force also, what the definition means is that, so long as the force remains the same, the acceleration varies with only that property of bodies called mass According to this definiton, the mass of a body may be measured by the force necessary to impart to the body an acceleration of a unit of length per unit of time For this reason, the factor m of formulas 1 and 2 of Art. 61 is called a measure of the mass of the body, or, for shortness, the mass of the body. It should be remembered that m has different values for different bodies. 63. Determination of the Mass of a Body Accel- eration Due to Gravity. The mass of a body has to be determined experimentally. A known force is applied to the body for a certain time, and the velocity at the end of the time is ascertained by some method Then the acceleration is found by dividing the velocity by the time (formula 2 of Art. 59), and the mass by dividing the applied force by the acceleration (formula 1 of Art. 61). The mass is, of course, expressed m the same units as the force, and its numerical value further depends on the unit of time and on the unit of length, these units being involved in acceleration. In this work, mass will, unless otherwise stated, be referred to the pound, the foot, and the second as units. The most convenient force to use for determining the mass of a body is the force of gravity that is, the weight of the body. It has been ascertained by actual experiment that a body falling freely in vacua follows the laws of uniformly accelerated motion. This might have been anticipated; for, while the body falls, it is under the action of a uniform force (gravity), which is measured by the weight of the body. The velocity of a falling body has been found to increase at the constant rate of about 32.16 feet per second per second; 27 OF MECHANICS 27 in other words, when acting on a falling body, gravity pro- duces an acceleration of about 32.16 feet per second. As this acceleration is due to the weight of the body, we have, deno- ting this weight by W (from formula 1 of Art. 61), = W m 32.16 64. The acceleration produced in a body by the force of gravity is called the acceleration due to gravity, or the acceleration of gravity. The magnitude of this accelera- tion decreases from the pole, where it is about 32.26 feet per second, toward the equator, where it is about 32.09 feet per second For the United States, its average value is 32.16 feet per second. This value will be used here, unless a different value is especially stated. 65. The acceleration due to gravity is usually denoted by g. If the weight of a body i& W, we have, W i-t \ m = (1) and W = mg (2) The weight vanes proportionately to g t so that the value of m, as determined from formula 1, is always the same, as it should be. It is, of course, understood that the weight W is the weight of the body at the place where the acceleration of gravity is jr.* Substituting m formula 2 of Art. 61 the value just found for m t r-i K (I ( o \ F (3) g which gives the force necessary to produce a given acceler- ation a in a body having a given weight W. Both in formula 3 of this article and in formula 2 of Art. 61, the force F is the force acting in the direction of *The exact value of g; at any particular place, is determined by careful observations with a pendulum, to which formulas derived m advanced mechanics are applied. The following formula gives a fairly approximate value of ff (feet per second) at any point whose latitude is L and whose elevation above sea lex el is h (feet): g = 32.173 - 084 cos L - .000002 h 28 FUNDAMENTAL PRINCIPLES 27 motion. If there is another force acting m an opposite direction, F must be understood to be the difference between the two. How forces acting on the same body or particle are combined will be explained further on. EXAMPLE 1 A block of any material is placed on a horizontal smooth surface (that is, a suiface supposed not to offer any factional resistance, or whose fnctional resistance may be neglected), and Is pulled by a string for 4 seconds, a registering appaiatus (called a dynamometer] placed between two portions of the string shows that the pull is 3 pounds The weight of the block being 20 pounds, it is required to find (a) the mass of the block, (6) the acceleration of its motion; (c) its velocity at the end of the fourth second SOLUTION (a} The mass of the block is found by formula 1. Taking g equal to 32 16, JQ m = sire = 622 Ans - (6) By formula 2 of Art 61, F 3 a = ~ = -jjKo = 4 823 ft per sec. Ans. in ,\j&i (c) By formula 1 of Art 59, v = 4 823 X 4 = 19 292 ft per sec Ans EXAMPLE 2 A uniform force acts on a body originally at rest for 10 seconds, at the end of which the velocity of the body is 40 feet per second, the force then ceases to act, and a force of 15 pounds is applied for 5 seconds, at the end of which the velocity of the body is 76 feet per second To find (a} the weight of the body, (/>) the force that acted during the first 10 seconds SOLUTION (a) During the second interval (5 seconds), the velocity has changed from 40 to 75 ft per sec Therefore, the acceleration, by formula 4 of Art 59, is = = 7 ft. per sec. (= a). The mass of the body is = -=-, and the weight is d I 15 X 32.16 _. _, . . mg = - ~ = 68.914 Ib. Ans. (d) The acceleration due to the fust force is (formula 2 of Art. 59), 40 T- 10 = 4 ft per sec By formula 2 of Art. 01, the force necessarv to produce this acceleration is m X 4 = x 4 - 8.5714 Ib Ans. 27 OF MECHANICS 29 EXAMPLES FOB PRACTICE 1. Find the force necessary to start a weight of 15 pounds from rest and give it a velocity of 80 feet per second in 4 seconds Ans F = 9 3284 Ib. 2 An engine and train of cars having an aggregate weight of 160 tons start from rest and move for 1 minute (60 seconds) over a level track Assuming the traction of the engine to exceed the resistances by 2 50 tons, find the velocity of the tram at the end of the minute Ans v = 32 16 ft per sec 3 A body has an initial velocity of 20 feet per second. A force of 40 pounds is applied, and after 12 seconds the velocity is found to be 85 feet per second Find (a) the acceleration due to the force, (b) the weight of the body &/() a = 5 4167 ft per sec. Ans I (d) W = 237.49 Ib 4 A body weighing 980 pounds moves for a certain time under the action of a uniform force of 50 pounds. At the end of that time, another force of 50 pounds is applied in the opposite direction After the application of the second force (the first not being removed) , the body moves over 300 feet in 15 seconds Find the time during which the first force acted alone . , v 300-15 , , , Ans / == - = - w = 12. 189 sec. a f LAW OF ACTION AND REACTION 66. Action and Reaction. Force has been defined as the action of a body on another, producing, or tending to produce, a change in the motion of the latter body. When the weight of a body is spoken of as the action of the earth on the body, we are concerned only with the effects of this action on the body in question, and for this reason disregard whatever action the body may, in its turn, exert on the earth. In the same manner, we say that a magnet attracts a piece of iron; this is sufficient for us to know, if the only thing being dealt with is the condition and motion of the piece of iron. But it must not be concluded from these common and con- venient forms of expression that a body can act on another without itself being acted on by the other body. All action between bodies is mutual. Take, for instance, a magnet, hold it in the hand, and bring it near a small piece of iron- the iron will immediately move toward the magnet and 607 6100 SO FUNDAMENTAL PRINCIPLES 27 adhere to it. Now, place the magnet on the table and hold the piece of iron in the hand, bringing it near the magnet: the magnet will move toward the piece of iion and adhere to it. If both the magnet and the piece of iron are placed on corks and let float on the .surface of water, they will both move, each toward the other, and adhere to each other. Furthermore, in this motion, the acceleiation of the magnet and that of the iron are found to be in the inverse ratio of the masses of the two bodies. This shows that the two bodies are acted on by the same force (Art. (52), 01 that the action of the magnet on the iron is equal to the action of the iron on the magnet. When a body is at rest, it is evident that the rorces rtCting on it must balance among themselves (law of inertia) Take a body weighing 4 pounds lying on a flat surface If the weight of the body were not balanced, the body would move; the surface, therefore, must exert an upward force equal to 4 pounds, in order to keep the body from falling If we press against the wall with a force of 10 pounds, the wall presses on the hand with the same force, but in an opposite direction. A weight of 20 pounds hanging from a string will evidently pull the string downwards with a force of 20 pounds; but the string must pull the weight upwards with the same force, as otherwise the weight would fall. In general, whenever one body acts on another, the latter body acts on the former with the same force, but in the opposite direction. In considering the mutual action of two bodies, we are usually concerned with the condition of only one of them; the force acting on it is called the action, while the force exerted by it on the other body is called the reaction. 67. Statements of the iJaw of Action and Reaction. From the facts stated in the preceding article, the following general law has been derived; it is known as the law of action and reaction, and was first stated by Newton: To every action^ there is always an equal and opposite reaction. 27 OF MECHANICS 31 r -> 68. As the accelerations produced in two bodies by the same force are inversely proportional to the masses of the two bodies, the law of action and reaction may be stated in the following terms Whenever two bodies aft on each other, they produce > or tend to produce, in each other accelerations ni opposite directions, -and these accelerations are inversely propoi tional to the masses of the two bodies. This happens whether the two bodies act on each other by means of a string stretched between the two, or by means of a rod connecting them, or through some unknown medium, as in the case of a magnet and a piece of iron. In every case where a body moves another, the latter moves, or tends to move, the former 69. Sti-ess. When only the action of one body on another is considered, this action is called force. If the mutual action of two bodies, or two parts of a body, is Jp lOOlb.. I .20026. .lOOlb. \.100lb. '' FIG 7 considered, that action is called stress. A stress includes a pair of equal and opposite forces, an action and a reaction, a force and a cotmterforce. A stress is measured by the magnitude of either of the forces of which it consists. In applied mechanics, the term stress is generally restricted to the forces acting between two parts of a body Thus, when a string is pulled at both ends with a force of 10 pounds, every portion of the string is under a stress of 10 pounds; for, if the two portions of the string are considered as being situated on the two sides of a plane perpendicular to its direction, each portion will be pulled away from the other with a force of 10 pounds. Therefore, to keep the two por- tions from separating, each must pull the other with a force of 10 pounds. Again, consider the bar A B, Fig. 7, pushed from each end with a force of 100 pounds. Imagine the bar 32 FUNDAMENTAL PRINCIPLES 27 lo be divided by a plane PQ- when the portion AM presses on B M with a force of 100 pounds toward the right, the por- tion B M must press on AM with a force of 100 pounds toward the left. In these examples, the string is under a stress of 10 pounds, and the bar under a stress of 100 pounds. 70. Tension, Compression, Pressure. When, as in the case of the string just considered, the forces tend to pull each portion of a body from the other, or to lengthen the body on which they act, the stress is called tension. When, as in the case of the bar, the forces tend to move each por- tion toward the other, or to shorten the body on which they act, the stress is called pressure or compression. The term pressure, however, is more commonly applied to the stress between two contiguous bodies, and the term com- pression to the stress between two contiguous parts of the same body. In the example of the bar, the stress is ordi- narily called compression. In the case of a heavy body resting on a table, the force exerted by the body on the table and by the table on the body is a pressure. EXAMPLE 1 Two bodies MI and M,, Fig 8, of masses OTI and m t> respectively, rest on a smooth horizontal surface, and are connected by a string S a A force of /'' jf a j * a [ jf t | Sl - pounds is applied to Mi ' ' ' by means of a string 5, FlG 8 m the direction shown. To find the acceleration of the resulting motion, and also the tension in each string. SOLUTION In all problems of this kind, the strings are supposed to be rigid and of no mass or weight, if the mass and stretching of each string were considered, the problem would be more compli- cated. In almost all cases that occur in practice, the ma<tf of the string, rope, chain, etc may be neglected without any sensible error; and the same is true of the extensibility, or stretching capacity, of the connection For example, the mass of the coupler between two railroad cars need not be considered in determining the pull between one car and the other; nor is it necessary to take into account the amount of stretching m the coupler Since the two bodies ^/"i and M, are rigidly 1 connected, they must move with the same acceleration Let this acceleration be a, and let 27 OF MECHANICS 33 the tensions m the strings Si and S a be 7i and T a , respectively Since the body MI is pulled by the string S\ with a force equal to F, it must pull the string toward the left with the same force, so that 7~i is evidently equal to F Also, Af,. acts on M, through the string S, and imparts to it an acceleration a towaid the right, therefore, the force exerted by MI on Af a is (from foimula 2 of Art 61) in a a By the law of action and reaction, Jlf t exerts on Jlf t the hame force, m,a, but directed toward the left The tension m S, is, therefore, T, = m 9 a The net force acting on M^ is the difference between F and m a a, or Fm a a. Consequently, since the resulting acceleration is a, we have, by formula 2 of Art 01, F w, a = m t a; whence, a = (1) 7i + Wa Also, T, = m 3 a = -^-^ (2) &! + m, v ' The same results may be obtained by assuming at the outset that the two masses m l and m, are equivalent to one single mass (m l + tti) acted on by the force F From this, equation (1) follows at once as a special case of formula 2 of Art 6 1 . Knowing the acceleration of Af t , which is a, the unbalanced force T a acting on Hf, is found by for- mula 2 of Art 61: ~, itt t F T, = m a a = 7 j + m, which is the same as before. EXAMPLE 2 A body Af lt Fig 0, whose weight is W pounds, is attached to a string, the latter is passed over a smooth peg P, and then tied to a second body 7l/ a having a weight of pounds and free to move along a smooth horizontal surface To find the pull m the string and the resulting acceleration of the two bodies. Pro. 9 SOLUTION This problem does not diffei from the preceding 1 in principle, and the method of solving it is identically $he same, but it has been given in order to caution against a common mistake made by beginners The object of the peg is simply to change the direc- tion of the force, it has no effect on the magnitude of the pull, which is the same in the vertical as In the horizontal portion of the string It is to be carefully borne in mind, however, that the pull on the string is not the weight of M^ and that therefore the force pulling M t is not Wi pounds, as might appear at first sight. The weight W* is a force acting on the bodies Jtfi and Jtf t , and, like the force F m the preceding example, has to move these two bodies. If there were I LT39S-4 34 FUNDAMENTAL PRINCIPLES 27 no body M t , the unbalanced force acting on AF* would he W tt but, in the present case, this force is diminished by the leaction of M, transmitted through the string to M^ If the pull in the string is denoted by T and the common acceleration of the two bodies is denoted by a, the unbalanced force acting on Jlfi is, as before, W l - T, and the unbalanced force acting on M, is simply T The mass of Mi is - (formula 1 of Art. 65), and the mass of Jlf a is . ff Substituting in the formulas of the preceding solution, we obtain a = Wj. w,. "* Wi + w m M g g Tff * or Jjf jy TXf W = Ef _1_ TV \*J EXAMPLE 3 A body weighing W pounds lies on the surface of a horizontal table. Both the table and the body are moving upwards with an acceleration of a feet per second To find the pressure of the body on the table. SOLUTION The pressure of the body on the table is equal to the pressure of the table on the body, which is the total upward force acting on the body, call this force P The downward force acting on the body is W. Therefore, the unbalanced force producing the W a. acceleration a is P W, and, as P W => am ^~ t we find, for the downward pressure exerted by the body, /.-ZS+Wr-^ + f) (1) ff g This shows that the weight of the body is increased by the amount - . Ifthetableismovingdownwards.ais negative, andjP Ji!.lj!rfv t If in this case a, ff, then, P = 0, or there is no pressure on the table. If a is negative (downward) and greater than g t 8 ay a g + a', then, g This shows that, if the body remains in contact with the surface of the table, it will exert an upward pressure equal to -^-A; but this can only be done by placing the body under the table. In this case, it is evident that, since the table tends to move faster than the body, the latter must offer a resistance, or exert an upward pressure on the table. This problem should be carefully studied, as it is a good illustration of how the results of mathematical formulas {Ire to be Interpreted. 27 OF MECHANICS 35 EXAMPLES FOB PRACTICE 1. In example 1 of Art 70, the force .F is 500 pounds, and M* and M t weigh 8 tons and 3 tons, respectively Find (a) the resulting acceler- ation, in feet per second, (b) the tension on the rope, in pounds (1 ton = 2,000 pounds) . A ,/()= 731ft per sec Ans \(t>) T= 13636 Ib. 2 With the weights of M t and Af, as in example 1, find: (a) the force (tons) necessary to give the two bodies an acceleration of 10 feet per second, (b) the tension in the rope, m tons F = 3 4204 tons T = .9328 ton Ans 3. In Fig. 10, the three bodies M lt M,, M a , weighing 18 tons, 16 tons, and 20 tons, respectively, are connected by the ropes S L FIG 10 and S a A force F of 15 tons is, applied to M l as shown. Find, (a) the resulting acceleration a, m feet per second, (b) the tension T, in the rope S tt (c) the tension T, in the rope S 3 f a) a = 8 9333 ft per sec. Ans { b) T a = 5 5556 tons [ c) T a = 10 000 tons 4 A body that on the surface of the earth weighs 10 pounds, is weighed m an ascending balloon by means of a spring balance, and found to weigh 15 pounds With what acceleration is the balloon ascending? [See equation (1} m example 3 of Art. 70.] Ans a = 18 08 ft per sec. 5. An engine pulls a tram of The total force of the locomotive i train (the engine not included) resistance of 10 pounds per ton of during the first minute, find (a) feet per second, (b) the velocity v minute, in miles per hour, (c) the the second car to the first. seven cars weighing 80 tons each that is, the total force acting on the -is 8 tons. Assuming a constant tram (engine and tender excluded) the acceleration a of the train, in of the tram at the end of the first tension T m the coupler connecting a = 1 0643 ft per sec Ans. 4 b) v = 43.541 mi per hr. T = 6.8571 tons 86 FUNDAMENTAL PRINCIPLES 2? CMMEDIATE CONSEQUENCES OF THE PRE- CEDING 1LAW8 IMPORTANT FORMULAS 71. Space Passed Over In Uniformly Accelerated Motion. Let a body start from rest and move during: t seconds under the constant action of a force acting in the direction of motion. Let a be the acceleration of the body, and v the velocity at the end of the time t. As explained in Art. 59, v is equal to a t. It can be shown by the use of advanced mathematics that the space j desciibed by the body in the time t is the same as if the body had moved din- ing that time with a constant velocity equal to one-half the actual final velocity, or |; that is, 2i .-\* (i) If, in the second member of this formula, v is replaced by its equivalent a t, the result is ~ t = a f. Therefoie, 2 j = i/ a (2) This is one of the most important formulas in mechanics and should be memorized. 72. If, in formula 1 of Art. 71, t = 1, then, s = $ a; or, the space described during the first second is numerically equal to one-half the acceleration. 73. From the general formula v = at, we get t - . a This value substituted in formula 1 of Art. 71 gives t s = - , whence, v* = 2 as, and a v = V2 as which gives the final velocity at the end of a given space. 74. To find the space and the velocity in terms of the mov- ing force and the mass or the weight of the moving body, we 27 OF MECHANICS 37 have, by formula 2 of Art. 61, and formula 1 of Art. 65, Zp a = = g. These values in formula 2 of Art. 71 give m W s - * = -* *r/ a (1) *~ 2 2 X W*' UJ The same values in the formula of Art. 73 give (2) 75. Modification of the Formulas "WTien the Body Has an Initial Velocity. If, before the force begins to act, the bodv has an initial velocity v , or, what is the same thing, if only an interval of time during which the velocity changes from v a to v is considered, the preceding formulas must be modified as follows: As the force has no effect on the motion already acquired by the body, the latter, having a velocity z/ , will describe, by virtue of its inertia alone, during the time /, a space equal to v 1-, to this must be added the space described by the body in virtue of the action of the force, as given by formula 2 of Art. 71. Therefore, the total space is given by the formula J = v,t + \af (1) or, putting a = v ~ v (formula 4 of Art. 59), j =.* + *(-.)* = ifa + w.)/ (2) which is the mean of the initial and the final velocity multi- plied by the time. 76. The equation a = v ~ v gives / = "L=J!* f Substi- / a tuting this value in formula 2 of Art. 75, i W s = + . x - \ a- whence, v' v*' => 2 a s, and v = Vz>o" + 2 a s This gives the final velocity in terms of the initial velocity, the space, and the acceleration. To find s and v in terms of force and mass, it is only p necessary to substitute for a. m 38 FUNDAMENTAL PRINCIPLES 27 77. Falling Bodies. It has been explained (Art. 63) that, when a body falls freely under the action of gravity, its acceleration is g feet per second, the value of g being neaily constant and equal to 32 16. The preceding formulas apply to falling bodies, for a falling body is simply a body mov- ing under the action of a force equal to its own weight, with a uniform acceleration equal to g. It is, therefoie, not strictly necessary to write new formulas for the motion of falling bodies. As, however, it is customary to use the symbol g for the acceleration of a falling body, and h for the space described, or height fallen through, it is con- venient to have the general formulas expressed in terms of these symbols. By putting h for s and g for a in formula 1 of Art. 59, in formulas 2 and 1 of Art. 71, and in the for- mula given in Art. 73, the following formulas are obtained for a body falling freely from rest during / seconds: v=gt (1) k = iff (2) h = %vt (3) v = *hgh (4) For a body falling for t seconds, after having acquired a velocity z , formula 3 of Art. 59, formulas 1 and 2 of Art. 75, and the formula of Art. 76, give v = v +gt (5) h = v.t + lgf = (v. + ir/)* (6) *-*(. + )/ (7) (8) 78. It should be remembered that, theoretically, these formulas are exactly correct only for bodies falling m vacua. When bodies fall through the air, the latter offers a resist- ance that varies with the surface of the body; hence, all bodies do not fall in air with exactly the same acceleration^ but in a vacuum they do. This has been verified by the fol- lowing experiment: A feather and a ball of lead are sup- ported at the upper (inside) part of a long tube from which the air has been removed; by a convenient arrangement the support is quickly turned from the outside, so that both 27 OF MECHANICS 39 bodies will begin to fall at exactly the same time; and it is found that they reach the bottom of the tube simultaneously; nor does the ball get ahead of the feather during any portion of the time of falling. Again, if a paper disk is placed on top of a dollar, and the latter let fall, the paper will remain constantly on the dollar and fall in the same time. The reason is that in this case the paper does not encounter any more resistance than does the coin. EXAMPLE 1 What is the velocity at the end of 30 seconds of a body moving with an acceleration of 8 feet per second? SOLUTION Substituting the given values in formula 1 of Art 59, v = 8 X 30 = 240 ft per sec. Ans EXAMPLE 2 A force of 10 pounds starts a body from rest and causes it to move over 65 feet If the weight of the body is 40 pounds, what is the final velocity? SOLUTION. By formula 3 of Art 65, Fff 10 X 32 16 . ., .. a = -^ = 4Q - = 8 04 ft. per sec. Substituting the known values in the formula of Art 73, v = V2 as = -N/2 X 8 04 X 65 = 32 33 ft per sec Ans. EXAMPLE 3. A ball is thrown downwards with a velocity of 5 feet per second from a tower 200 feet high. What will be its velocity when it reaches the ground? SOLUTION Here, the initial velocity v a ~ 6 ft per sec , and the height h = 200 ft., are given. Substituting these values in formula 8 of Art 77, v - V6 a + 2 X 32 16 X 200 = 113 53 ft per sec. Ans EXAMPLE 4 If a ball dropped from a bridge reaches the water in 2 seconds, at what height is the bridge above the surface of the water? SOLUTION From formula 2 of Art. 77, the height through which the ball drops is found to be iX 32.16X2" 6432ft Ans. EXAMPLES FOR PRACTICE 1. A body starts from a state of rest and moves with an acceleration of 20 feet per second for 30 seconds. Over what distance has the body passed? Ans. 9,000 ft. 2. A body weighing 160.8 pounds is started from rest by a force of 40 pounds. Over what space has the body passed if the final velocity is 16 feet per second? Ans. 16 ft. 40 FUNDAMENTAL PRINCIPLES 27 3 A body having an initial velocity of 25 feet per second moves, with an acceleration of 8 feet pei second, over a distance of 29 feet Find the final velocity of the body. Ans 33 ft per sec 4 A body falls freely for 2 3 seconds If the initial velocity of the body is 5 feet per second and its final velocity is 79 feet per second, through what distance has the body fallen? Ans 96 6 ft 5 A body having an initial velocity of 10 feet per second falls freely for 12 seconds Through what distance does the body pass? Ans 2,435 5 ft RETARDATION 79. Uniformly Retarded Motion. When a body is moving in a certain direction and a force acts in the opposite direction, the effect of the force is to diminish the velocity by equal amounts in equal times, and the motion is said to be uniformly retai-ded. The amount by which the veloc- ity is decreased per unit of time is called the retardation due to the force. According to Galileo's law, if a force acting 1 on a. body at rest produces in it an acceleration a, the same force acting on the same body, when the latter is m motion, will pro- duce the same acceleration a and the velocity v = a t, in its own dnection, whatever the previous direction of the motion may have been Therefoie, if the force acts in a direction opposite to that in which the body is moving, and if the latter has an initial velocity v , the velocity after t seconds will be v, a t. Retardation, then, is nothing but negative acceleration, and all formulas relating to uniformly acceler- ated motion apply to uniformly retarded motion, by simply changing the sign of a. For this reason, the term accelera- tion, when taken in its most general sense, includes letarda- tion as a special case, and the latter term is seldom used in mechanics. 80. Law of Retardation. Let a. body be moving in a straight line, and let its velocity at a certain moment be TV If at that moment a force is applied in the opposite direc- tion, and allowed to act during / seconds, the final velocity v 27 OF MECHANICS 41 and the space s described are given by the following formulas (See Arts 59 and 75): v = v at (1) j = v t t %at* (2) To find the time in which the body will be brought to rest, v must be equal to zero; that is, v a t = 0, or / = . a Substituting this value in equation (2), and denoting by s 1 the space described during the time /, we have, / . &' - J a ! = '- a a 2 a If the velocity v a is supposed to have been generated by a force equal to the retarding force, the time required for the force to produce this velocity must evidently have been , a which is the same as the time required for the retarding force to destroy the velocity z/ The space described during the action of the force, up to the time the velocity v, is attained, is given by the foimula j = $af=l;a^=^ = S > a 2 a It follows, therefore, that, if a uniform force acts on a body during a certain time and carries it through a certain dis- tance, an equal but opposite foice, it applied after the first force, will destroy the velocity generated by the first force in the same time and in tlie same space (that is, after the body has described the same space] required by the first force to generate the velocity m question. SOJLDT1ON OF PROBLEMS 81. Fundamental Equations of Motion. The fol- lowing are the three fundamental equations of motion: F=ma (1) v-v a = at 1 ,v They form two groups, (1) and (2), one consisting of one equation and the other of two. In group (1), there are three quantities; and as there is only one equation, two of the 42 FUNDAMENTAL PRINCIPLES 27 quantities must be known before the other can be determined. If /'is to be determined, m and a must be known, if m is to be determined, F and a must be known, etc. In group (2) , there are five quantities, and, as there are only two equations, three of the quantities must be known in order to determine the other two; the process of finding these two is simply the process of solving two equations with two unknown quantities. If the two groups are taken together, it will be observed that a is common to both If two of the quantities m group (1) are given, a becomes known, and to solve group (2), only two more quantities of that group have to be given Or, if three of the quantities of group (2) are given, a becomes known, and only one of the quantities of gioup (1) (either /'or m) is required to find the other. The process of elimina- tion is exceedingly simple. When there is no initial velocity, v = 0. 82. General Solution of Some Important Problems. As an illustration of the use of the fundamental equations, let it be required to find F, a, and /, when ?;/, v, v ot and are given. In group (2), the quantities v, v ot and s are known. The unknown quantities are a and /. From the first equation of the group is obtained a = ^-~r^'> and fr m th e second, a = ^ s ~~ Vt -. Equating these two values of a, transpo- sing, and canceling the common factor /, (v v ) / = 2 (s v t}; whence / = -- - v + v a which is the same value that would have been obtained from formula 2 of Art 75. This value of t substituted in the equation a = ^-^-^ gives a _ - Q _ (*LrJ?o) ( v + v ) - ?. "J!s! ~" ~ " ~ Finally, F - ma * 2s 27 OF MECHANICS 43 83. Again, let m, v t v,, and t be given, and let it be required to find F, s, and a. The value of a follows at once from the relation v z/ = at, which gives a = ^^. Substituting this value in the second equation of group (2), formula 2 of Art. 75 is again found. From group (1) is obtained F=ma= (-.) t The values for F given in this and the preceding article are very important, and should be memorized. EXAMPLE 1 A body weighing 250 pounds is moving with a velocitj' of 25 feet per second, a force acting in the same direction through a space of 75 feet changes the velocity to 60 feet per second To find the magnitude of the force SOLUTION Here, m = SHTB (formula 1 of Art 85), v = 26, Aft ID v 50, and s = 75 Substituting these values in the formula given in Art 82, -(50 -26") m(v* - zO 32 16 ^ ; 260 X (2,600 - 626) 2s = 2X75 " " 8216X2X76 = 97.17 Ib. Ans EXAMPLE 2 A body whose weight is 192 96 pounds moves with a velocity of 15 feet per second. To find the magnitude of a force neces- sary to change the velocity to 45 feet per second, that force acting during 15 seconds in the direction of motion 192 96 SOLUTION Here, tn = - pn -' - =6, v 15, v = 45, and / <a 15. oZ 10 Substituting these values m the formula given in this article, f = , = - . tt EXAMPLES FOR PRACTICE 1 A body weighing 100 pounds moves for 6 seconds along a smooth horizontal surface under the action of a force of 10 pounds Find the space passed over. Ans. s = 57.888 ft. 2 A body moving with uniformly accelerated motion passes over 100 feet in 12 seconds. What is the acceleration? Ans. a 1.3880 ft. per sec, 44 FUNDAMENTAL PRINCIPLES 27 3 What force is necessary to give a body weighing 6 tons a velocity of 10 feet per second in 75 feet? W-31* Ans F = ~- = 12438 T = 248 76 Ib 4 A body starts fiom rest and moves with an acceleration of 40 feet per second Find the velocity after the body has passed ovei 75 feet Ant. v = 77 46 ft per sec 5 Find the time in which the body in the preceding example has described the space of 75 feet Ans t = 1 9365 sec. 6 A body is thrown vertically downwards from a height of 10,000 feet and reaches the ground m 6 seconds Find the velocity with which the body was projected (Use formula 6 of Ait 77.) Ans z/o = 1 ,570 2 ft per sec COMPOSITION AND RESOLUTION OF FORCES 84. Point of ApplI cation, Line of Action, and Direction of a Force. When a force is applied to a particle or material point, the particle or point is called the point of. application of the force. The line of action of a force is the straight line along which the force tends to move its point of application The direction of the force is the direction of this line (see Art. 37). 85. Graphic Representation of Forces. As a great many mechanical problems are solved by means of geometry, c ^ A it is convenient, and almost FIC, 11 necessaiy, to represent forces graphically that is, by means of lines. A force is repre- sented graphically by drawing a line parallel to the line of action of the force, and of a length proportional to the magnitude of the force, the direction of the latter being indicated by an arrowhead marked on the line. In Fig. 11, O is the point of application of a force, and the force is repre- sented by O A, whose length is as many units of length as there are pounds in the force. The arrow indicates that the force tends to move O from O toward A along the line OA, If the force is 100 pounds, OA may be made A | A inch, or mch, or 100 millimeters, etc.; the unit of length used is 27 OF MECHANICS 45 immaterial, provided that the same unit is used for all forces in the solution of any one problem It is not necessary, in ordei to represent a foice by a line, to draw the line through the point of application, or any other special point. Any line parallel to the line of action of the force may be used for the purpose 86. Vectors. Any quantity that, like force and velocity, has magnitude and direction is called a vector qumitlty; and the line, as OA, Fig 11, by which the quantity is repre- sented graphically is called a vector. A vector is, then, simply a line having a length proportional to the magnitude of a vector quantity, and an airowhead to indicate the direc- tion of that quantity Such quantities as mass and volume, which have no direction, cannot be represented by vectors. 87. That extremity of a vector to which the arrowhead points is called the end of the vector; the other extremity is the origin. In Fig. 11, O is q the origin and A the end of the vector OA. 88. Two vectors having the same origin or the same end will here be referred to as being in 11011- Pl 12 B cyclic order. If the end of one coincides with the origin of the other, they will be described as being in cyclic oi-der. In Fig 12, the vectors OA and O B have the common origin O, PIG 18 and are, therefore, in non-cyclic order; so are the vectors OA and J3A, Fig. 13, which have the common end A. In Fig. 14, the end of the vector O A coincides with the origin of the vector A B, these two vectors are in cyclic order. 46 FUNDAMENTAL PRINCIPLES 27 89. Resultant and Components. When a body is acted on by several forces, there is usually one single force that, if acting alone, would produce the same effect as the several forces combined This single force is called the resultant of the other forces. With respect to the result- ant, the combined forces are called components. 90. Composition and Resolution of Forces. The process of finding the resultant when the components are known is called the composition of foi-ces. The process of finding the components when the resultant is known is called the resolution of forces. Both the resolution and the composition of forces are of the utmost importance in dynamics, especially in statics, and will be explained more fully further on. But here the funda- mental principles must be stated and explained. 91. Composition of Colllnear Forces. Forces hav- ing the same line of action are called colllnear forces. The resultant of several colhnear fotces is equal to thein alge- braic sum. For example, if a force of 15 pounds pulls a body upwards, and a force of 10 pounds pulls the body downwards, along the same line, the effect will be the same as if the body were pulled upwards only, with a force of 15 10 = 5 pounds. If the upward direction is treated as positive, the downward direction will be negative, and the downward pull of 10 pounds will be represented by ( 10). The resultant is, therefore, 15 + ( 10) = +5. The same principle applies to any number of forces. Thus, the resultant of the forces 15 pounds, 37 pounds, 39 pounds, 13 pounds, and 78 pounds is 15 + 37 - 39 + 13 - 78 = 65 - 117 * -52 pounds. The negative sign indicates that the resultant is a downward pull of 52 pounds. This principle, which is sufficiently obvious, has already been applied in this Section. 92. Concurrent and Non-Concurrent Forces. Two or more forces not having the same line of action, but whose lines of action meet at a point, are called concurrent 27 OF MECHANICS 47 forces. If the lines of action of several forces do not meet at one point, the forces are said to be no n- concurrent. 93. Resultant of Two Concurrent Forces: Paral- lelogram of Forces. Let O t Fig 15, be a particle acted on by two concurrent forces F l and F, in the directions O A and OB, respectively. These lines are vectors representing the two forces. Com- plete the parallelogram OA CB, by drawing AC parallel to OB, and BC parallel to OA. Draw the vector C, forming the diagonal of the parallelogram. It can be proved, both Fl<< Jr> mathematically and by experiment, that this vector represents the resultant of the forces jf\ and F a ; in other words, that, if the forces fl and F; are represented in magnitude and direc- tion by the sides OA and OB of a parallelogram, the result- ant will be represented in magnitude and direction by the diagonal C. This principle is called the law of the par- allelogram of forces, and can be stated in general terms as follows: // two concurrent forces are represented in magnitude and direction by two vectors taken in non-cyclic order, their resultant will be represented^ in magnitude and direction, by another vector forming the diagonal of the parallelogram constntcted on the vectors representing the two forces. 94. Formulas Derived From the Parallelogram of Forces. When the magnitudes of the forces F,. and F at Fig. 15, and the angle between their lines of action are given, the magnitude and direction of the resultant R are very readily computed by the principles of trigonometry. In the triangle OA C, OC = 4O A* + AO~*TOA. A~Cl~wTA But OC = R, OA = F lt A C * OB = F tt and angle A = 380 . Substituting these values, and writing 48 FUNDAMENTAL PRINCIPLES U7 cos L for cos (180 Z,), the preceding equation becomes R = V/T,' + /,' + 2 /s/; cos Z, (1) The angle Mi that R makes with F^ is found from the same triangle by the principle of smes, which gives sm sin A A C = - -; or, replacing sin A by sin L (for A = 180 L), and writing F a for A C, and R for C, sin Mi _ F. m sin Z, R ' z? whence, sin Mi = -j sin Z, (2) y? r Similarly, sin M, = -~ sin Z, (3) R The values of M and yfl/ a may be checked by the i ela- tion Mi + M a = Z. EXAMPLE Two concurrent forces having the magnitudes 450 pounds (= F,.) and 675 pounds (= F,} make an angle of 60 Required (a) the magnitude jR of their resultant, (fl) the angles MI and M, made by the line of action of R with the lines of action of FI and f lt respectively SOLUTION. (a) Here, /? t = 450 Ib ,/?". = 075 Ib , Z = B0, and formula 1 gives A = -\I450 1 + 676 s + 2 X 450 X 675 X cos 60 = 980 75 Ib Ans (d) Substituting m formulas 2 and 3 the values of F tt F lt A", and L (angles are given to the nearest 10 seconds), fl7^ sin Mi = ~Y & X sin 60, whence, M* = 86 SB' 10". Ans sm M t = 935-75 X sin 60, whence, Jlf a = 23 24' 50". Ans Checking, M v + Jlf, = 36 SB'- 10" + 23 24' 50" = H0. 95. Rectangular Components. When the two forces are perpendicular to each other, they are called rectangular- components. In this case, angle L = 90, sin L = 1, cos L = 0, and the preceding formulas become = V/^'+T^ (1) sin M, = cos M, = 5 = ^^__ (Q) * V/v + /; 27 OF MECHANICS 49 For this condition, however, the angles MI and M* are more easilv determined by their tangents or cotangents. This special case is illustrated in Fig. 16, which is similar to Fig. 15, except that the angle A OB is a right ^ ^ f 2 * angle The parallelo- ^ % V^ fa gram A CB is a rect- angle, and the right ^\ triangle O A C gives R ' = OC = = ^F\* + F t ', as found A before Also, Fl0 - 16 Similarly, tan M t = = (4) 96. Resolution of a Force Into Two Components. The next thing to consider is the converse problem of find- ing the components of a foice when the force is given. This is called resolving a force into components. We shall o -Ja B v here confine ourselves p 7 >- -7 Xz fa 7 i to the particular prob- / lem of resolving a force / into two components whose directions are given. Let O C = R, Fig. 17, be a force acting at O. 17 It is required to find its components along two given directions. To do this, draw through O two indefinite lines Xi and OX ti parallel to the given directions. From C draw a parallel to OX,, meeting OXt at A, and a parallel to OX lt meeting OX, at B. Then, OA and OB are the required components Ft. and F a . To find Ft. and F t by calculation. Since the directions of Ft. and F, are given, M* and M* are known, and also L = M, + Af n . Then, from formula 2 of Art. 94, ILT398 5 FUNDAMENTAL PRINCIPLES 27 _ R sin Mi _ R sin M^ / \ sm L sin (M l -f- Similarly, f.- . sin If one of the components is to make an angle M l with the >rce R, and the other an angle of 90 yl/i, that is, if the 70 components are to be perpendicular to each other, for- ula 2 of Art. 95 gives F. = R sin M, (2) Similarly, F t = R sin M a = -ff cos M l (3) The student should again consider Fig. 16, and see how iese formulas are derived. If the figure is kept in mind id the geometrical relations of the quantities involved are imembered, no difficulty in deriving, remembering, and Dplying the formulas- will be experienced 97. By the component of a force in a certain direction is sually understood one of two rectangular components. The >rce is supposed to be resolved into two components at ght angles to each other, one of which has the direction in uestion. Such a component is also called the resolute and le resolved port of the force in the given direction. EXAMPLES JFOB PRACTICE 1. If, in Fig 17, C represents a force of 40 pounds acting; on a idy at 0, and the angles that the components Fi and F, make with ie given force are 46 and 20, respectively, what are the values of ,ese components? . f/^ 15.10 Ib. Ans 't/?; - 31.21 Ib. 2 A rectangular component F 9 of 60 pounds makes an angle of 40 ith the resultant J?, Find the other component FI and the resultant. Ana /^ " 60 ' 34 lb - An8 'lj? . 78.82 lb, 98. The Triangle of Forces, In the construction lown in Fig. 17, the complete parallelogram OACB has 2en drawn, in order that the principle on which the oonatruc- on is founded may be understood. But, in practice, it is not scessary to draw more than one-half of the 27 OF MECHANICS 51 that is, one of the equal triangles OAC, O B C. For, since OB = AC, the line CA drawn through C parallel to OX, in order to deteimme the point A, gives at once the length of the line OB representing the component F a . Also, the angle M a between F a o and R is equal to the angle OCA. Hence the following construc- tion (Fig. 18). When a force R is given, and it is required to hnd its components in two given directions, ^1 RIO is draw from one extremity, as O, of the vector representing the given force, a line parallel to one of the given directions, and from the other extremity, as C, a line parallel to the other given direction The two sides O A and AC of the triangle thus formed ate the requited components. The vectors repre- senting the two components aie in cyclic order with each other, but in non-cyclic order with the vector representing their resultant R. This construction is called the triangle of forces. Notice that, although A C gives the magnitude and direc- tion of F t , the force F* must be supposed to be really acting at O, as shown by the dotted line OB. The line C B might have been drawn through C, to meet OX, at B In this case, however, the triangle would have been as shown by the light dotted lines, but the results would have been the same as before. EXAMPLES FOR PRACTICE 1. Two forces, F,. = 1,000 pounds and F, = 8,000 pounds, act simultaneously on a particle, the lines of action of the two forces make with each other an angle of 30. Find (a.) their resultant jR; (t>) the inclination of the resultant to the two components (angles to nearest 10"). f(a) J? - 3,898.2 Ib. Ans \ IK f^/t = 22 37' 60" I (0 > Utf. = 7 22' 10" 52 FUNDAMENTAL PRINCIPLES 27 2, Given a force R = 20 tons, to resolve it into two components /", and F t making with R angles M* = 30, Jlf a = 45 Aus 1 F * = 14 41 tons Aus l/> 3 = 10 SHU tons 3 Find- (a) the resultant of two rectangular forces FI = 40 pounds and F, = 80 pounds, (d) the Inclination of the resultant to the com- ponents f (a) R = 50 Ib An s 1 ih\ JM* = JW r >-' 10" L w l^/, = f)8 7'fiO" 4 A force of 100 tons is inclined to the vertical at an angle of <H) Resolve it into a horizontal component H and a vertical component / AriB Iff => Sfi(K)3 tons. Ans I V = 50 tons. ANALYTIC STATICS (PART 1) CONCURRENT COPLANAR FORCES DEFINITIONS 1. Coplanar and Non-Coplanar Forces. When the lines of action of several forces he m the same plane, the forces are said to be coplanar. If the lines of action are not in the same plane, the forces are non-coplanar. 2. Analytic Statics and Graphic Statics. So far as the mathematical treatment of statics is concerned, this science may be considered as being mainly a branch of applied geometry. And, as geometrical problems may be solved either analytically or graphically that is, either by computation or by construction so, too, statics may be either analytic or graphic, according to the method of solu- tion used. Analytic statics treats of the equilibrium and equivalence of forces by means of the arithmetical and algebraic relations existing among the forces, which relations, in the case of forces represented by vectors, are the same as the relations existing among the vectors, and depend on the geometric properties of the figures formed by combinations of such vectors. In analytic statics, all results are found by calculation. Graphic statics treats of the equilibrium and equivalence of forces by means of geometric figures. In graphic statics, all results are found by measurement. COPYNIIIHTBD MY INTBRNATIONAI. TEXTBOOK COMPANY. ALL HlaHTI MMBRVU 28 2 ANALYTIC STATICS 28 3. System of Forces. The aggregate of all the forces acting on a body or on a group of bodies is called a system of forces. The simplest sybtem of forces is obviously that in which there is only one force. 4. Equivalent Systems. Two or more systems of forces are equivalent when they produce the same effect, or, what is the same thing, when they have the same resultant The resultant of any numbei of forces is itself equivalent to the system formed by the components. 5. Equillbrants. When a system of forces is balanced by a single force, the latter i& called the equill brant of the system Conversely, if several forces balance one single force, they are termed the equilibrants of that force. The equilibrant of several forces is evidently equal in magnitude, but opposite in direction, to the re- sultant of those forces. Hence, it is also called the antl -resultant. In Fig. 1, Ft and F a are two forces acting at O. Their resultant R is found by the principle of the parallelogram of forces. The force Q, equal to jR, but acting in the opposite direction, will evidently balance R, or the sys- tem of forces F l and F t to which R is equivalent. Therefore, Q is the equilibrant of F,. and F t . Conversely, Fi and F, are the equilibrants of Q. It is also evident that, if the forces F l} F tt and Q are in equilibrium, any of them may be considered as the equili- brant of the two others. The same principle applies to any number of forces. 6. The Kesultaiit Not an Actual Force. It should be understood that, when several forces act on a body, the resultant is not an actual force acting on the body, but an imaginary force whose effect would be the same as the com- bined effects of the components. Similarly, when a single 28 ANALYTIC STATICS 3 force acts on a body, its components are imaginary forces that would produce the same effect as the single force. The resultant may replace the components and the compo- nents may replace the resultant, but in no case are the com- ponents and the resultant supposed to act simultaneously. The equilibrant, on the contrary, is a real force applied in order to balance a system of other forces; or, if the system is balanced, any one of the forces is the equilibrant of the others. 7. Internal and External Forces. With reference to a body or system of bodies, a force is said to be external when it is exerted by a body outside of the system. The forces exerted by parts of a body, or of a system of bodies, on one another are called internal forces. 8. The terms "internal" and "external" are relative; and it is evident that a $ p force may be external _ | with respect to a body %\ I and internal with re- A 1, W 72 r spect to another | | body, or to a system >, | j Jj, of which the body in A fb) Q question forms a part. Fl 2 For instance, the forces F t and F, shown in Fig. 2 (a) are external to the body A B CD on which they act. If the body is in equilibrium, any part of it, as AB PQ, to the left of a section PQ, must push on the other part with a force equal to F lt while the other part pushes on it with a force F tt = F t . This mutual push, or stress, between AB PQ and CDQP consists of two forces, both of which are internal with regard to the whole body AB CD. But if the condition of only the part AB PQ is under consideration, the push of CDQP on it may be regarded as an external force. The portion CDQP may be removed, as shown in Fig. 2 (), and the portion AB PQ considered as an independent body acted on by the external forces Fi and F. i I 4 ANALYTIC STATICS 28 FUNDAMENTAL PRINCIPLES AND FORMULAS 9. Free Body: Principle of Separate Equilibrium. When a body is cousideied by itself as a whole disconnected from other bodies, it is called a free body. Any pait of a system or body may be treated as a free body, provided that it is assumed to be under the action of external foices equal to the internal forces that keep it con- nected with the rest of the body or system of which it forms a part. Thus, in Fig. 2 (/>), the portion ABPQ may bfe treated as a free body, after introducing the force F t) equal O to the force exerted by the rest of the body A B CD, that is, by CD QP, on the portion ABPQ. As another example, take the system formed by a string OP, Fig 3, carrying a weight W. If it is desired to study the condition of a part of the string, as O A, the part A P and the weight W may be sup- . posed to be removed; but, in order that the condition of O A may remain unaltered, it is necessary to intro- duce at A an external force equal to the sum of the weight Wand the weight of the part AP of the string. In general, when a body or a system of bodies is in equilibrium, every part of it must be in eqtiihb- Flt * - H rium and may be treated as a separate body by itself, provided that external forces are introduced to replace tht; internal forces exerted by the rest of the body or system on the pait removed. This principle is called the principle of separate equilibrium; its application, called the free- body method, is of great value in the solution of static problems. 10. TruiiHfetiiblllty of the Point of Application. Concurrent forces have already been defined as those whose lines of action intersect at a point, They may or may not be actually applied at their point of intersection. Their point of application is not essential, provided that the forces act on a rigid body. This Is expressed by the following 28 ANALYTIC STATICS 5 principle, which is usually known as the principle of the transmissibdity or transference of force, but will here be called the principle of the transferability of the point of application: When a force acts on a rigid body, the force may be supposed to be applied at any point on its line of action^ provided that this point and the body are rigidly connected. Let a force f, Fig. 4, act on the body M ' N m the direction Kio 4. OX. A string may be tied at A, or at^?,, and a pull exerted from X with the force /''and along the line OX; or it may be imagined that the string is tied to a point A, and that this point is the end of a rigid rod connected to the body at A l} A t or to any other point along OX; no matter where the string is tied, and no matter from what point on OX a pull is exerted, the effect will be the same. Again, it may be imagined that a rigid strut ,S is attached to the body and that the force is applied at A, either by pulling from X or by pushing from O. It may thus happen that several concurrent forces acting on a body do not intersect within the body. For example, in Fig fi, the points of application of the forces F^ F t) and/ 7 , actually applied at the points M^ M lt and M, of the body PQ t may be transferred to the common point of inter- section O of their lines of action, assuming this point to be Fio 5 6 ANALYTIC STATICS 28 rigidly connected to the body. Thus, the force F, may be imagined to be a pull at the end O of a string tied to M,, and the force F a as a push at the end O of a strut between O and M 3 . 11. Resultant of Any Number of Concurrent Forces. The resultant of two concurrent foices can be easily determined by either the parallelogram or the triangle of forces, as explained in Fundamental Principles of Mechanics. The resultant of more than two forces can be found by successive applications of either of these principles. Let four concurrent forces F lt F t , F a , F t be represented, respectively, by the vectors O A, OB, O C, OD, Fig. 6. The re- sultant R? of F l and F, is found from the par- allelogram OAEB constructed on the vec- tors O A and OB. This resultant is now com- bined with another of the given forces, say F t , by constructing the parallelogram OEGC, which gives the result- ant R" of R' and F tt and, therefore, of F lt F tt and TV Finally, the parallelogram G KD gives the result- ant R of R" and F tl and, therefore, of F lt F t , F and F 4 . When analytic methods are used, this process is exceed- ingly laborious, as it involves the calculation of the magni- tude and direction of each resultant, A much simpler process is afforded by the resolution of each force Into two rect- angular components, as explained in tha next article. If the problem is solved graphically, the triac&te of forces is used instead of the parallelogram: A E ia fleawn equal and parallel to OB or F t ; then, EG equal and pafalllipj to /i; then, GK Fu 6 I \ 28 ANALYTIC STATICS equal and parallel to F 4 , the vector O If, between O and K, gives the resultant This method will be further explained and illustrated in connection with graphic statics. At present, only analytic methods are under consideration. 12. Method by Rectangular Components. Let fi, F,, F a , Ft, Fig. 7, be four concurrent forces represented, r Of I 1 f* 1 1 FIG respectively, by the vectors A,, OA a , OA a , OA 4 . Take any two lines X f X, Y' Y through O and at right angles to each other. These lines are used as lines of reference; they are called coordinate axes, and may be taken in any convenient position. The axis X' X is usually referred to as the x axis, or axis of x, and the axis Y' Y t as the y axis, or axis of y. The angles made by the given forces with the axis of x are denoted by #,,-#., etc., as shown. 8 ANALYTIC STATICS 28 As explained in Fundamental Principles of Mechanics, the force /*, can be icsolved into two components in the direc- tions OX and Y, by drawing from A t a line .-/, C paral- lel to OX, and a line A,B^ parallel to OY. These lines determine a parallelogram OB^A^ C, in which OB, and O C, represent the components of F, in the directions OX and O Y, respectively. The component OB, is called the x compo- nent of F,; and O C, the y component. These components will be denoted by A' and K M respectively. Instead of con- structing the parallelogram OBiA^C,., it suffices to draw from A t the line A/?i perpendicular to OX] this determines the two components of F lt since B^A, = O C, = ]',. The components of F,, F , F< are similarly found; they will be denoted by X t and Y 3 , A' a and Y a , etc. It should be understood that, although K,, for instance, may be repre- sented by the vector B^A^ that component really ads through 0, its position being O C. The given forces aic now reduced to the system formed by the A components ( ) /?,, OB t , Z? D , #>9,, and the y components O C, OC,, O C,, OC 4 , and the resultant of the given forces is the same as thai of these components. Since the x components act in the same line, their resultant is equal to their algebraic sum (sou Fundamental Pnucifilfs of Mechanics}. The same principle applies to the resultant of the y components. Taking forces acting toward the right or upwards as positive, and those acting toward the left or downwards as negative, we luivc, X, <= OB X t = fl/?., X a = - Off,, A' t = - Oh\ Y, = 0C, Y, = OC, Y, = r)C, K 4 =r - 0(\ Denoting the resultant of the x components ant] that of the y components by X r and Y ft respectively, we hnve, X f = A' t + A", + X tt + A". = OB, + OB, - OB, - OJi, [a] Y,= Y,+ Y.+ K.+ K 4 = OC + (9C + OC - ^f, U) These resultants are represented in the figure by Oft, and OC n respectively. The given system of forces has, therefore, been reduced to the two forces X? O J7 f and K- =s C r ; and their resultant /?, found by the parallelogram or by the triangle of forces, is the resultant of the given forces, In other words, the forces X r and Y r are the rectangular 28 ANALYTIC STATICS 9 components, in the directions OX and O Y, of the required resultant. 13. The calculation of the magnitude and direction of R is accomplished as follows. The light tuangle O .?, A* gives: OB, = OA, cos Jf lt B,A, = OA t sin // or, since OB^ = X^ OA r = F lt and B l A l = Y lt X, = F t cos ff lt Y! = F l sm //i The values of X, and Y a , X e and K n , and X t and Y<. are found in the same manner It may be well to remembei the pnn- ciple that the component or resolute of a force hi any direction is equal to the force multiplied by the cosine of the acute angle that the line of action of the force makes with that direction. Thus, the x component of F, is F, cos H a For the other component, the sine should be used instead of the cosine Substituting m equations (a) and (b] of the preceding article the values of A',, K,, etc., the following expressions are obtained: X r = F, cos H, + F, cos H a F* cos H 3 F< cos H,. Y r = Fi sin H, + F* sin H, + F, sin H, - F. sm H* Having determined X r and Y r , the right triangle OB r A r gives the magnitude of the resultant R, and the inclination H f of this resultant to OX- R = OA r = ^OB r 3 + B r AS = *&' + Y r " (1) tan H - - - ; (2) 14. As explained in Plane Trigonometry, Part 2, the sym- bol of summation -V, read sigma, is often used to indicate the algebraic addition of several quantities denoted by the one letter affected by subscripts or accents. If, for example, several quantities are denoted by X^X^X^X^ etc., their algebraic sum X l + A r , + A' + A' 4 ... is denoted by the expression - A', read signia x. With this notation, the algebraic sum of the x components A', Xv, etc. of any number of forces may be denoted by - X, and the algebraic sum of the y components by 2 Y. Also, the sum of the expressions F t cos ff lt F, cos H t , etc. may be denoted by -i'^cos If, If the components of the resultant 10 ANALYTIC STATICS are, as before, denoted by A' r and )',., the following equations apply to any number of forces X r = 2X = .i'/'cob // (1) Y r = 2Y = ZF sin // (2) Having computed X r and K,, the magnitude and dnection of the resultant R are determined by formulas 1 and 2 of Art. 13. EXAMPLE It is required to hml tlie resultant of the three c<>n- the l-ll. S current foices J'\, /'\, J'\ represented in Fitf H, their magnitudes and relative positions being as shown SOLUTION Take A" A' inclined nt W to the line of m-lion of /,. The Inclinations of the other two forces to A'' V aio u-iulily determined from the given angluH. Thus, //, = 45", //. = A' ; Os1, - 1HO" - (7fi n H ifi") ** IMI" //n A'' O A* B HO" - (!()" ^ ^0" For the JT componontH, with the usual convention iw to following eqimtioliK uru obtained; A', = 1(K) )H4ri 11 * 70.711 Ih, A' =-- - JK) ros UO" -- ~ ir>.(MH) Hi, A; ~lo() cos L>(] - LlO.IUi lh. and for they corupononts: Y, KH)siu4f 11 *. 70.711 lh. r, m IH) Niu HO" 77,iu n>. K M * ~ifiHina -fii.:ttiij, Therefore, X, i 1 ^ - 70.711 - 4fi.OOO - 14JI.WV - - 115.24 Ib, K- - A 1 X 70.711 + 77.48 - ftl ,OS 7 ,ST>1 Ib. 28 ANALYTIC STATICS 11 The negative sign of X e indicates that this component of the resultant R is directed toward the left, as indicated by the vector O B r Formula 1 of Art 13 now gives, R = <\/115.24' + 97~3oT n -= 150 86 Ib. and formula 2 of the same article gives, tan H r = ?Tj^, Hr = 40 11' 20" llo U4 In finding the value of tan H r , the negative sign of X r is dis- regarded, as the only thing that is required is the numerical value of the angle The signs of X r and Y r show in what quadrant the result- ant is In the present case, X r (= O B r } is negative and Y f (B,A,} is positive. This at once shows that R lies in the angle X 1 O Y t and that H r is, therefore, the angle that R makes with O X' NOTE In this, as in many other problems In this Course, angle s aie jdyen to the nearest 10" Thus, an angle of 86 10' 37" is called Bfi 10' 40", an angle of 17 47' 28" is called 17 47' 20" MOMENTS 15. Definitions. The moment of a force about u point, or witli respect to a point, is the product obtained by multiplying the magnitude of the force by the perpendicular distance from the point to the line of action of the force. In Fig. 9, the moment of F about the point C is Fp\ about the point C, it is Fpv 16. The point to which a moment is referred, or about which a moment is taken, is called the center of moments, or origin of moments. The perpen- dicular p or p l from the origin on the line of action of the force is called the lever arm, or simply the arm, of the force with respect to the origin. 17. A moment is expressed in foot-pounds, inch-tons, etc., according to the units to which the force and its arm are referred. If, for instance, the force is 10 pounds and the arm is 60 feet, the moment is 10 X 60, or 600, foot-pounds; if the force is 8 tons and the arm is 6 inches, the moment i& 12 ANALYTIC STATICS 8 X 6, or 48, inch-tons. The term foot-pound is used in kinetics m another sense; but the two meanings are so different that there is no danger of confusion 18. Sign ami Direction of u Moment. When, look- ing in the direction of the anowhead of a vector represent- ing a force, the center of moments lies on the right of the line of sight, as C and C, Fig. 10, the moment is considered positive; if the center is at the left, as C v , the moment is considered negative This may be stated otheiwise thus* Imagine the level aim p to be levnlvmg about C in the direction indicated by ../>. the arrowhead. In this , c , case, / will t evolve from ' left to nght, 01 clock- wise, and the moment is considered positive. In the case of /> on the contraiy, if this line re- volvc'd about C, follow- ing the direction of the ai low-head, its motion would be counter-clock- wise; the moment is then considered nega- tive. The direction of the motion just refcned to is called the direction of th iiioniciil, and is supposed to have the same sign as the moment itself. If M l M^ and Jlf, are the moments of /''about C, (',, and C" fl respectively, then, M - 4- Fj> - /> //, - + /'/, jffr /!/ * - />, 19, Itcin'cHentutlon of ti Momont by an Ai-ou, If the lines CO and CA, Kitf. 10, ure drawn, the area of the trianffle CO A is 10 Ax CD m |/-> * \M whence, M * 2(i OA X CD) 2 X area OA C i'io 10 28 ANALYTIC STATICS 13 The magnitude of the moment M is, therefoie, numeri- cally equal to twice the aiea of the triangle whose base is the vector representing; the force F and whose opposite vertex is the origin of moments. 20. Moment of ti Force In Terms of the Moments of Its Components. The moment of a force about any point in //\ plane is equal to the algebraic sum of the moments of r/<; (omponcnts in any diiections. Let the foice R, Fig. 11, be resolved into the two com- ponents Ft and F, in any directions, and let C be the origin a * V<*, . PIG. 11 of moments. The arms of , F l} and F a are, respectively, p r , Ai and p at as shown. The moments of these three forces about C will be denoted by M r , M^ and M t , respectively. Draw CO, and <9 A" perpend ictilai to CO. Denote the angles made by the lines of action of the three forces with OX "by If,, #i, .//., as shown, and the resolutes of the forces in the direction OA'by X,,X lt and JC. Then (Art. 12), X r = X,. + X 9 ', that is (Ait. 13), R cos H r = F! cob HI + F, cos Hj> whence, multiplying both members of this equation by CO, ft X CO cos H r = F l X CO cos ff^ + F^X CO cos ff, (a) 1 LT398- 14 ANALYTIC STATICS 2* But H f = 90 COE, and, therefore, cos /7, = sin CO& and CO cos ff, = CO sin (TO^ = C^( triangle C6>>) =A Also, CO cos // t = C(9 sin C<9 A = A CO cos // = CO sin COD, = A Substituting these values in (a), that is, jWr = ^ + ^/,- In general, if the algebraic sum of the moments of any number of concurrent forces is denoted by 2 Af t and the moment of the resultant by Af rt as above, then, Mr = XM CONDITIONS OP EQUILIBRIUM 21. The Absolute Condition. When any number of concurrent forces are in equilibrium, it is evident that they can have no rebultant; that is, if their resultant is R, we must have R 0. This is the general condition, sometimes called the absolute condition, of equilibrium Expressed in the form of R 0, however, this condition is of no value in the solution of problems. It is necessary to express R in terms of other quantities, or to derive some other relations by means of which unknown quantities can be detei mined. 22, Condition of lUtsoluton. Since the resultant of several forces in equilibrium is zero, itb resolutes m any two directions mUvSt be zero; and, as each resolute is equal to the algebraic sum of the corresponding resolutes of the given forces, it follows that the sum of the tesolutes of the forces in any direction must be zero* The resolute of a force in a direction perpendicular to itself is zero. There is, then, one direction (at right angles to the resultant) ftlongr which the sum of the resolutes of any number of unbalanced concurrent forces is zero ; but if the sum of the resolutea is zero for any two directions, the forces are in equilibrium; for both directions cannot be perpen- dicular to the line of action of the resultant. 28 ANALYTIC STATICS 15 23. Condition of Moments. Since the resultant is 0, its moment about any point must be zero; and, as the moment of the resultant is equal to the sum of the moments of the components, we must have, M + M, + M',+ . . . =0, where, M lt M,, Jlf a , etc. are the moments of the components about any point in their plane. It is necessary to specify that the sum of the moments must be zero, when taken with respect to any point, or every point in their plane, for, m a system of forces not in equilib- rium, the sum of their moments about a point on the resultant is also equal to zero, because in this case the lever arm of the resultant is zero, and its moment, therefore, is zero. But, if the forces have a resultant, the sum of their moments about a point outside of the resultant cannot be zero. Conversely, if the sum of the moments of the forces about three points not in the same straight line is zero, the forces have no resultant; for these three points cannot all lie on the line of action of the resultant, and the only case m which the sum of the moments of a system of unbalanced forces can be zero is that in which the origin of moments is a point m the line of action of the resultant. 24. Genei-al Statement of the Conditions of Equi- librium. Summing up: Any balanced system of coplanar concurrent forces must satisfy the following conditions, each of which is necessary and sufficient for equilibrium, and involves the other two: 1. The resultant of the forces must be sero. 2. T/ie sum of the resolutes of the forces in each of any two directions must be zero. 3. The sum of the moments of the forces about three points in their plane \ not in the same straight line^ must be zero. Condition 1 is expressed algebraically by the equation R 0. Condition 2 may be expressed thus (see Art. 14): EX = S Condition 3 may be expressed thus (see Art. 20): 16 ANALYTIC STATICS 1 28 If the lever arms of the forces F n F* t etc are denoted by Pi t A, etc., the algebraic sum of their moments, or 2 Af, jt> is equal to F 1 p^ 4- F a /> a + F 3 p* + , etc Denoting this sum by 2 Fp, formula 2 may be written *** vpp = (3) 25. Equilibrium of Three Forces. Problems i elating to the equilibrium of three forces are usu- ally most conveniently solved by means of the triangle of foices. It is to be observed that the eqiuhbrant of any number of forces is numer- ically equal to their resultant, but acts in the oppobite direction. Let the forces F lt F yi Fig. 12(tf), act through O. By di awing through O f , us shown at (), the vectors (Y AS and \-l Aj ) representing the given forces, and completing the triangle O 1 Aj AJ> the re- sultant R of /'I and F, is obtained. In this triangle of forces, the two compo- FIG lli nents are in cyclic order with each other, but in non-cyclic order with the resultant (see Fundamental Principles of Mechanics}. 28 ANALYTIC STATICS 17 In Fig. 13 (a), the three forces F^F,, and Q, acting at O, are supposed to be m equilibrium. The force Q is the equi- librant of F, and F t ; it is numerically equal to their result- ant R, but acts in the opposite direction. Consequently, a triangle O 1 AS A* 1 , Fig. 13 (b), equal to the ti tangle giving the resultant /?, can be formed with the vectors lepresent- ing /-" F yt and O\ but the anowheads on Q and 7? must point in opposite directions, which means that Q must be taken in cyclic order with F l and F tt . 26. Selection of Axes. When there are more than three forces in equilibimm, the most convenient method for determining any of them, when the others aie known, con- sists m finding expiessions foi the resolutes of all the forces m two rectangular dnections, and using formulas 1 and 2, Art 24. A similat method applies to the determination of the resultant, as aheady explained Theoretically these directions are entirely aibitrary; but, practically, the direc- tion of one of the foices, often one whose magnitude is not known, is almost always a very convenient one to use, as in this case one of the resolutes of that force is equal to zero, and the other is equal to the force itself. When this is done, either direction along the line of action of the force may be taken as positive and the other as negative; but this does not necessarily mean that the direction taken as positive is the direction of the force acting along that line. For example, forces acting at a point O may be resolved into components parallel and perpendicular to the line of action, say OK, of one of the forces; and, for the purpose of this resolution, the direction OK may be treated as positive and the opposite direction as negative. This simply means that, if the resolutes parallel to OK are to be added algebraically, the arithmetical difference must be taken between those resolutes whose direc- tion is OK and those whose direction is opposite; if the dif- ference is positive, the direction of the resultant resolute is from O toward K\ if negative, from A" toward O. Similarly, in taking moments, any point can be used as an origin; but it is convenient to take the origin on the line or 18 ANALYTIC STATICS 28 action of one of the forces, as in this case the moment of that force is zero, and the force is thereby eliminated from the equation of moments. These methods will be better understood by a study of the following problems: EXAMPLE 1. A weight of 500 pounds (see Fig. 14) hangs by three j t / I \ W-BOOlb. FIG. 14 ropes tied to a ring at O, the inclinations of the two slanting ropes O JS t and O E, to each other and to the horizontal being as shown. It Is required to find the tensions in the ropes OEi and O E 9 . SOLUTION BY TRIANGLE OF FORCES. The tensions F,. and F n in the ropes are represented by the vectors A v and A tt and the weight by the vector W. These three forces form a balanced system, each force being the equilibrant of the other two. Considering W an the 28 ANALYTIC STATICS 19 equilibrant of F l and F a , the triangle of forces OBC is obtained by drawing through B a line parallel to OA lt and through O a line parallel to O A* In the present case, CO is in the prolongation of O At The triangle might have been constructed anywhere else; It is not necessary to draw it m the position here shown Nor is it neces- sary to draw it accurately to scale, as its only object is to serve as a guide in the calculation In this triangle, K= AiOC = 180 - Ai OA, = 180 - 75; and, therefore, sin K = sin (180 - 76) = sm 76 Also, Y being vertical, Li = 90 - ff m = 00 - 60 = 30 L, = A, O Y = 90 - Hi = 90 - 46 = 46 The triangle O B C now gives sm L, = -^U sin 30 = 258 82 Ib. Ans. - , sin A sm 7o F t = -^-sin L 3 = -^osm 45 = 366 03 Ib. Ans. 9 ^ sin 75 SOLUTION BY RESO LUTES Let the vertical line Y and the horizontal hue OX be taken as axes of coordinates As usual, the resolutes of Fl will be denoted by A' and YI, and those of F, by X* and K,, as shown in the figure. They are here drawn for purposes of illustration, but it is not necessary to draw them in order to apply the general formulas. The vertical resolute of the weight is W\ the horizontal resolute is zero. Placing the sum of the horizontal resolutes equal to zero, we have, noticing that X l is negative, - Ft cos Hi + F t cos H t + = 0; that is, - Fi cos 45 + F a cos 60 - (a) Similarly, for the vertical resolutes, Fi sin Hi + F, sm H, - W = 0; or, Fi sm 45 + F t sin 60 - W = (*) Prom (a), cos 46 which, substituted in (b), gives etn ftn Ft sin 45 + F t cos 45 ^L^_ _ whence, cos w W _ " sm 46 + cos 45 sm ^ " 8in 46 cos ^ + cos 45 sln cos 60 W OAO ?f . OAO sin (45 + 60-) = STW Sm ^ = iSTTS 5 sin 80 as found before. The value of F t may now be found from (c) . 20 ANALYTIC STATICS 28 The horizontal and vertical resolutes have been used simply to illustrate the general method, but it will be seen that, by resolving the forces m these directions, the solution is long and tedious If however, resolutes perpendicular to the lines of action of the unknown forces FI and F a are taken, the solution will be very much simplified First, let the resolutes be taken perpendicular to O J5 y Then, we shall have, considering the direction from O toward E 3 as positive, resolute of F a = resolute of F t = f,. sm 75 resolute of W = -W^smL,= - W sm 30 Therefore (formula 1, Art 24), /r lS m75- W sin 30 = whence, -=a sm 30 sin 75 PIG 1C i&oo zfc. In a similar manner, , may be found by taking the resolutes perpendicular to FI This method should be care- fully studied, as its simplicity makes it of very great practical importance. EXAMPLE 2. A weight W of 1,500 pounds is hung from the extremity O, Fig 15, of a hori- zontal bar projecting out of a vertical wall and held by a rope ON, Inclined to the vertical at an angle of 40 Required the tension T in the rope and the compression C in the bar SOLUTION UY TRIANGLE OK FORCES. It is obvious that W is the equilibrunt of C and T. The actual lines of action of W and T are shown by the vectors O B and O A. Draw B D per- pendicular to O //, to meet A' O produced at D. Then, in the triangle O B D, the vectors D O and B D represent T and C } respec- tively. Also, BOD = 40. C - W tan 40 = 1,500 tan 40 - 1,258.0 Ib. Ans T - 1 -^so = 1,958.1 Ib Ans. cos 40 cos 40 SOLUTION BY MOMENTS. If moments are taken about N, the moment of T will evidently be zero, since //lies on the line of action 28 ANALYTIC STATICS 21 of T, the moment of W will be WxNS=lf>WXEO The moment of C whose line of action is OE, will be CX EN There- fore (Art 2O) , CXEN = i,mxo, whence, C = 1,600 J^ = 1,600 tan 40, as found before If moments aie taken about E, the moment of C, the resultant, will be 7cio, that of H' will be the same as befoie, 01 1,500 X E O, and that of ywill be TX E A negative, because the origin E is on the A* FIG 10 left of O A (Art. 18) Equating the sum of the moments of the com- ponents to the moment of the resultant, 1,500X^0- TxEK~ 0, v. T -i n EO 1.600 L500 whence, y . 3 ,600 ^. - -^y - ^ as found before The method of moments is sometimes very convenient. The center of moments should, if possible, be taken on the line of action of one of the unknown foices, as that force is thereby eliminated EXAMPLE 3 To find the equilibrant of the five forces represented by the full-line vectors in Fig. 10. SOLUTION Let OA lt the line of action of F,, be taken as the x direction, and Y, the perpendicular to it at 0, as the y direction of 22 ANALYTIC STATICS 28 resolutes. The angles of F lt F a , F 3> etc , with O X will be denoted by HI, // a , &3, as in Art 12, and the resolutes of the forces by X lt X a , y it y,, etc. Here we have H* = 0, H, = 50, ff, = A n O X 1 = 180 - (//, + 90) = 180 - 140 = 40, H< = 60 - H* = 60 - 40 = 20, H, = 30. Therefoie, JT X - F = 20 X 3 = F, cos 50 = 35 cos 50 X, = - F, cos 40 = - 25 cos 40 X< = - F t cos 20 = - 40 cos 20 X, - F t cos 30 = 45 cos 30 Denoting the x resolute of the equihbrant by X gt we must have (Art 24), X f + X,+ X, + X, + Xi + X, = whence, X 9 = -(Xi+X, + X, + X< + X*) = _ 20 - 35 cos 50 + 26 cos 40 + 40 cos 20 - 45 cos 30 - - 24.730 Ib Similarly, K x - Y, = - F t sin 50 = - 35 sin 50 Y, = - F 3 sin 40 = - 25 sin 40 K = F* sin 20 40 sin 20 y t = F, sm 30 = 45 sm 30 and y f s -(y 1 + y.+ y.+ y 4 + y 1 ) = 35 sm 50 + 26 sm 40 - 40 sin 20 - 45 sin 30 - 6.700 Ib The magnitude of the equihbrant Q may be found from the rela- tion Q = VAV + y,', but it is better to determine first the angle H q between Q and OX. For this purpose, it is not necessary to take the sign of X g into account (Art 14) and, therefore, Y a 6 700 EXAMPLES FOR PRACTICE 1 Two concurrent forces Fj. = SO tons and F t = 25 tons act at an angle of 170. Find (a) the magnitude Q of their equilibrant ; (*) the angle L made by Q with F,. Ano / (a) Q - 6.0132 tons Ans \(t>)L - 48 64' 2. Resolve a force /? 100 pounds into two equal components Fi - J" 1 , making an angle of 70. Ans. F l F, 61 039 Ib. 3. Find a general relation between a force F and either of two equal components F^ making an angle If with each other. Ans. F" 2-F, costtf 28 ANALYTIC STATICS 23 4 Given the equilibrant Q = 10 tons of the two forces F and F t , Fig 17, find the magnitudes of those two forces, the angles being as shown (The vectors F l and F, in the figure are drawn of arbitrary lengths, their purpose being to indicate direc- tions, not magnitudes, the latter being as yet unknown ) Ans IF,. = 11 154 T. Ans \F, - 2 9886 T. 6 Find the equilibrant Q of the forces represented in Fig 18, and the angle H, it makes with the line of action O X of the 20-ton force (Find resolutes parallel and perpen- dicular to O X ) Anc Ans W \Jf f 10 735 tons 5 is/ 50" 6 A weight of 15 tons is supported by two ropes, one horizontal and the other making an angle of 45 with the horizon (135 with the former rope) Find the tensions Ti and T t in the two ropes, 7i being tension In horizontal r P e< S 21.213 tons Q-lOTon* FIG 17 PIG 18 24 ANALYTIC STATICS STRESSES IN FRAMED STRUCTURES DEFINITIONS NOIE The theoiy of eoncuirent forces finds sunie of its principal applications in the determination of stiesset, in framed structuies Although a complete treatment of such a subject would be foreign to the scope of this Section, a. few simple examples will be given, in order that the utility of these principles may be seen Before giving those examples, it will be necessary to define some of the teims employed 27. Structures. A (structure is a statical combina- tion of parts designed for the transmission of foice. By saying that a structure is a statical combination, it is meant that the forces transmitted by the parts are supposed to be all balanced and produce no motion. Bridges and buildings are examples 28. MacliluoB. A machine is a combination of paits designed for the transmission of motion In any special case in which a machine produces no motion, its vanous parts being at rest, the machine is to be treated as a stiuctiue. 29. Frames*. The term frame is applied to any combi- nation of bats, strings, ropes, or other straight paits con- nected together so that their center lines form a polygon or a pait of a polygon. Eveiy one of the straight parts so con- nected is called a member of the frame, and the connection of two or more members is called a Joint. Although in some cases the center lines of the various members connecting at a joint do not meet exactly in a point, they are considered so to meet, and the forces acting along the members are treated as concurrent forces meeting at the center of the joint. 30. Trusses. A trubs is a rigid frame contesting; of triangles, such as occur in bridges and buildings, and in some branches of carpentry. 31. Supports. The place or places on which a stuiuture rests, and to which, therefore, the forces acting on the stiiic 1 - ture considered as a whole are transmitted, are called 28 ANALYTIC STATICS 25 siipports, because they support, or sustain, the structuie. Such are the piei s of a bridge and the foundation of an engine. When a stiucture, as a bridge, rests on diffeient supports, the point ot application of the resultant pressure on any sup- port is called <\ point of support. 32. Reactions. At every joint of a frame there are seveial forces acting, namely: 1 The external forces, or forces applied to the frame from the outside 2 The mutual reactions, or the forces exerted by the various members on one another at the joint. At some joints, there maybe no external foices diiectly applied, but the membeis may act on one another on account of external foices acting at other joints. As to every action there is an equal and opposite reaction, eveiy suppoit of a stiuctuie exerts on the btiucture a force equal and opposite to the foice transmitted by the structure to the support. This foice excited by the support on the struc- ture is called the reaction at the support, and is to be con- sidered as one of the external forces acting on the structure. 33. Struts ami Ties. The two kinds of stress ten- sion and compression that may occur in the members of a frame are defined in Fundamental Principles of Mechanics. A tension is often called a pull, and a compression a thrust. Those members that are designed to resist compression are called struts; those that are designed to resist tension are called tie*?. 1>KTKRM I NATION OF STRKS8ES 34. Introductory Explanations. Here those frames only will be dealt with in which the forces acting on every member aie applied al its extremities, and m which the con- ditions are such that the mutual reactions at the joints may be taken to have lines of action meeting at the common intersection of the center lines of the membeis. This being understood, lei A^A, A, A, A,A t Fig. 19, be three members of a frame, A being the point of intersection 26 ANALYTIC STATICS 28 of the center lines of the members, and A lt A t ,A a , the joints connecting the three members with other members. It will be first assumed that there is no external force acting at A. Let F^P\,F t be the resultants of the forces acting on A,. A, A, A, and A, A at A^A a , and A tt respectively. By the principle of separate equilibrium (Art. 9), every member, considered as a free body, must be in equilibrium; therefore, the resultant of the forces acting at A on A^A must bal- ance F^ and that resultant and F must be collinear and act along A As, so that the resultant at A is a force Fi equal FIG. 19 in magnitude to F lt but acting in an opposite direction. Likewise, AA t is kept in equilibrium by the forces F, and F,, both having the direction of the member, and F, being the resultant of the forces exerted by the other members on A, A at A. Similarly for A, A, The vector representing F, is drawn on one side of the member, in order to avoid confusion, although F, really acts along the member A A,. Now, transfer the forces Fi,F, t and F t to A (Art 10), where they are represented by accented letters. For convenience, as well as for the sake of uni- formity, the forces thus transferred are represented by veer FtftF, 1 , having their common origin at A that is, 28 ANALYTIC STATICS 27 with the arrowheads pointing away iromA. It is evident that the force F lt being the resultant of the foices acting directly on A* A at A, is the resultant of F, f and FJ, and, since F,.' is the equihbrant of F lt it is likewise the equili- brant of F,' and F a f ; in other words, the forces FS, FJ ', and Fa' (or their equivalents F^F t , F 3 ) form a system of concur- rent forces in equilibiium Substantially the same explanations would apply in case there weie external forces acting at A. These forces, with the forces Fi f , F,' t and F a f , would form a system of balanced concurrent forces. The problem of determining the forces acting on the members is thus reduced to the general problem of the equi- librium of concurrent forces. It is now necessary to ascer- tain what effect those forces have on the members of the frame that is, what members are in tension and what members are in compression. 35. Character of the Stress In a Member. Considui the condition of A l A It has been explained that this mem- ber is held in equilibrium by the two forces F l and F[', this pair of forces constitutes the stress in the member, and the magnitude of either force is a measure of the stress (see Fundamental Principles of Mechanics) . It will be observed that the tendency of the forces 7*1 and FI is to stretch the member, and that, therefore, the member is m tension. The stress in A,. A is, therefore, a tension of /i pounds, tons, etc., as the case may be. Similarly, the stress in A t A is a com- pression equal to F, (by which is meant that each of the two forces constituting the stress has a magnitude equal to F a ), and the stress m A a A is a tension equal to F*. 36. In determining the character of the stress in a member, however, it is not necessary to consider the two forces acting at the extremities of the member. Referring again to the joint A, where the forces FS, F t f , and F<J (or their equals F lt F F a ) form a balanced system, it is seen that FJ and F, f have their arrowheads pointing away from the ioint; these forces may be said to be pulling at the joint 28 ANALYTIC vSTATIC\S 28 in the direction of the membeis /, / and /,./, icspectively, and these two members are in tension. The force /*/, on the contrary, when supposed, as it should be, to be acting on the joint along ./M, as shown by the dotted vectoi /.y, pushes against the joint in the direction of the member . / /; hence, this member is in compulsion. 37. Summing up T/iefonfathuX'il(>ni>aHit'iHbe> Hica^itcs tension ot compiesvon atconiing "S ? '//<"'/ sttflfiw/ to mt on the joint through the membo > it pttlh at ot frtewA on the joint. It has been seen, for instance, that /'V, when supposed to be acting on the joint thiough the membei /,./ (as shown by F t ") presses on that joint, and that / . / is in compression. 38. The forces F,', /<"/, Fl (which, for shoitness, are called the stresses in the membeis) have been represented by vectors having their common unj>i at the joint. This is convenient in cornpli- 1 ezited problems in which the method of resolutes is used ^ Hut, in almost all cases, the vectors may be drawn along the lines represent- ing the members, and then the arrowhead on any force shows at once whether the force is (that is, measures) a pull or a thrust. 39, Noot'MHary Data For Deter- mining? Htrosaea, In determining the stresses in the members meeting at a joint, the external foroas or the stresses in some of the other members must be known. Suppose, for instance, that the stress in the member 28 ANALYTIC STATICS 29 Fig. 20, is known to be a compiession equal to F lt repre- sented by the vectoi A N, and that it is required to find the stresses F t and F, in the members A, A and A a A Since the forces F t ,F 3 , and F-, ai e in equilibrium, a vectoi triangle can be constiucted with the vectoi s lepresenting them Through N draw A T < I/ parallel to A* A, meeting A, A at Jlf, and mark the arrowheads on N Jlf and M 'A so that the vectors AN, NAf, and /If. 1 will be in cyclic order Then, N .If = F 3 ; MA = F n . As F, pi esses on the joint A, the member A\A is m com- piession. Similaily, F a) although given in direction by the vectoi JVM, must be supposed to be acting along A a A, and it is seen that, when thus applied, it presses on the joint A\ therefoie, 1 3 A also is in compression 40. Opposite Arrowheads 011 a Member. By refer- ring to Fig. 21, it will be seen that the member connecting the two joints M and O has two arrowheads pointing in opposite directions The student should not fall into the en or of think- ing that each arrowhead indi- cates the direction in which the part of the member on which it is marked acts on the other part. Thus, if the member is cut by a plane X K, the arrowhead F does not indicate the direction in which the part Z acts on the part MZ\ if that were the case, the member would be in compression, whereas it is in tension. The arrowhead F indicates the direction in which MO acts on O, and, as this action is transmitted from MZ to Z, the arrowhead F really indicates the direction in which M Z acts on O Z\ which shows that MZ pulls on OZ, and that, therefore, OZi& in tension. Likewise, the arrowhead F' indicates the direction in which the member acts on the joint M, or the direction in which, a^y part Q the member, as Z, acts on the part Jtf pelow it. - The forces acting in ILTJ9S-7 f ' 30 ANALYTIC vSTATICS 28 the direction of /'"and JP lire, of course, equal, as each measures the stress in the member M O\ either may be considered as the action, and the other as the equal and opposite reaction. EXAMPI.B. A tiuss consisting of tlnue horizontal membeis PC, CQ t and B f), Fig liii, and four equal inclined members/*./?, 12 C t CB t and B Q, carries a weight of ft' pounds hung from the cent IT C. The truss rests ou two piers /' untl )\ its length, lietwutni thu joints at /* and Q, IB 2/, and itn height is h. Neglecting thu weight of thts Htrwt:- ture, it IB required to find the Htreiw in eavh momlier und thts rtruc-tiniiH at the supports. SOLUTION BY TKIANOI.R OF FORCRH.- On iiwcmnt of the Hymmetry of the figure, the angle* PCJ) and Q CH MO eqtwl; tliwy ure both denoted by //. Also, /; C IH parallel to Ji Q t und D CJt f'/tQ - L, say. The anglea // and A, however, are nttt ntipptihcd to bo known and have to be exprowed in terms of / nnd h. The figure /;t '-~. A <> (i\ Niu // - tan // - m 180 In - aln 2 /^ - W 28 ANALYTIC STATICS 31 Of the forces that balance at C, it is evident that, owing to the sym- metry of the tiuss, the foices along CQanA CP are equal and opposite, and form, therefore, by themselves a balanced system. Hence, the weight Wmust be balanced by the foices F and f, acting; along the members CD and C B. The stresses in these two members are evidently pulls, and it is obvious that /<\ = F, In the triangle CMN, the vector CM represents the weight W, i\I ' N is parallel to CB and CN Is in the prolongation of D C Therefore, MN FI and NC F a . Since F^ => F tt then, also, A", = A\ Now, #1 = 180 - (90 + H) = 00 - H N = 180 - (Ki 4- A.) - 180 - 2 A', - 180 - 2 (90 - H) = 2ff W W Then, F t -F m - ^sin A', = -sm (90 - H) W rr , --- ?- r cos H = S sm H cos H " 2 sin// Coming now to the joint B t It is known that the force /*i acts on It along the member CB, hence, the two forces F a and F^ acting along DB and QB can be determined Take B M 1 = F v and draw .fl/'.A" parallel to DB, meeting BQ at A r/ Then, .rt/'JV' represents the force F* acting along D B t and N 1 B the force F+ acting along QB The stresses in these members are evidently thrusts (Art 37) As M> N' is parallel to C Q, the angles B M' N' and B JV' M f are both equal to B C Q, or H. The triangle B M 1 N 1 being isosceles, we have F* = FI . For F, t we have, _ F a = -^.smZ sin /f jq/ Putting Fi = - jr> as found above, and sin L = 2 sin H cos .#, ^. = ~ X 2 B The forces acting at Q are the thrust /?! of the member B Q, the horizontal force F t along CQ, and the reaction of the pier, which, as will be shown hereafter, acts vertically upwards Take Q M" to represent F t (notice that it is not necessary to measure the distance from Q in the direction of the arrowhead), draw M" N" horizontal and QN 1 ' vertical, and, starting with F*, whose direction is known, mark the arrowheads on the sides of the triangle M" Q N" in cyclic order Then, QN" will represent the reaction /? and N" M" the force F t in Q C. The latter force acts in the direction Q C, thus showing the stress in Q C to be a pull. The triangle M" QN" gives = ZT rr p rr W COS If W t Wl F.-F.cMjr-FtixHir- 2sln/r - T cotjsr- ^ For the reaction, , , JZ - /^ tan // ~ cot /f tan /f = ~ * 32 ANALYTIC STATICS 28 The reaction being equal and opposed to the pressure exerted by the truss on the pier, or to the part of the load transmitted by the truss to the pier, it is seen that one half of the load is transmitted to Q, and, as the truss is symmetrical, the other half is transmitted to the other pier P This subject will be more fully explained in connection with the theory of parallel forces By this theoiy, the value of the reaction can be determined first and the calculation begun at the joint through which the reaction acts, this is the method used in piactice. On account of the symmetry of the truss, the stress in DP \^ obviously the same as the stress in B O SOLUTION BY MOMENTS The solution of this pioblem by the method of moments is as follows For joint C, momenta are taken about a point on the line of action of one of the unknown forces Fj. or F, The equal and opposite forces F t need not be taken into account If the joint 13 is taken as the origin of moments, the lever arm of W is fi U = 1 /, and that of F a is BG = J&Csm L = DC shi L = ~ sin /. = ~* ~ cos H 2 COB H The moment of W\& negative, and that of F 9 is positive (Art. 18). Therefore (Art. 24), - WXBU+F,XBG = 0, or, writing the values of B U and B G just found, Wl Ft I sin L *M whence, 2 cos// W cof, H sin L This was the value found for F l (which is equal to /",) by using the ' triangle of forces, and may be transformed in the same mannei t Passing now to joint B, F t is determined by taking moments about D t \ in the line of action of F a . The lever arms of F t is DJ, and the lever f arm of F Is the perpendicular from t) on Q R produced, but this per- \ pendtcular is equal to B G 1 , because QB and CD are parallel. Therefore, F L X DJ- F t X B G = 0, i whence, F* = F* X because the triangle B CD is isosceles, arid DJ*=BG. *To find F a , take moments about Q, in the line of F t . The lever arm of F 9 is Q S = k, that of F^ is Q Z = / sin H. Then, Fi A - F l I sin H = ' h ~> W or, because FI - ^ g (see the solution by the triangle of forces given above) , 28 ANALYTIC STATICS 33 Passing now to joint Q, to find F t , take moments about S 1 in the line of R The lever arm of F t is S/ = S sm H = ~ sin H, that of F, is (PS = /; Then, whence, t I sm // = 0, sin H t . , . \ F t I sin H . , but (see above) -j = F at therefore, F = f = Wj, To hnd R, take moments about C^m the line of F, The lever arm of R is Q C = /, that of F t la C T = ~Q Z = / sm JV. Then, Fl s\n H - I? I = 0, whence, J? = FI sm /f = /?, sm // = 2~^-X sm // = -^ In practice, it is not necessary to draw the perpendiculars C U, B G, C T, etc , as their values can be at once written down by means of the fundamental trigonometric relations among the elements of a right triangle EXAMPLES FOR PRACTICE 1. A weight of 2 tons is suspended from a derrick, Fig. 23; the length of the boom A B is 40 feet; the guy rope CB is fastened at a point C, 30 feet from A, and the boom is 10 feet out of vertical. Find FIG 28 the tension FI in the guy rope and the thrust F in the boom. = .96840 tons , 2 7542 tons a 34 ANALYTIC STATICS 28 2 A trapezoidal frame having the dimensions shown in Pig 24 rests on two piers and carries two weights of 5 tons each at the joints A and B Find the stresses F tt F,, and F 3 in the members A t A C, and CD, respectively, and the reaction /? ut either pier The members A B and CD are horizontal, and the inclinations of A C and B D to CD are equal. [F^ 2 fi tons, thrust Ans \ F ' = 5 - 5002 tons > Ans F a 2 fi tons, pull 5 tons PARALLEL FORCES COPLANAR FORCES TWO FORCES 41. Resultant or Two Parallel Forces Having the Same Direction. Let two parallel forces /*i and F, Fig. 26, act on a rigid body A BCD. The lines of action of the two forces are K^L^ and K*L*, respectively, and the points of application Ei and , are any two points on those lines (see Art 10). Draw E^E t . The effect of the forces will not be changed if any two equal and opposite forces R^ T^ and fi t 7 1 ,, acting 1 along 1 the line ./?, /?,, are introduced, for these two forces will evidently balance each other. The two forces F l and F t may, therefore, be replaced by the four forces f t and E, 71, F, and /?. T.. The resultant of F, and t T, is -fi 1 , ,S lf the diagonal of the parallelogram E t A^ 5, 7*,. Similarly, the resultant of F and E* T, is ^, 5 1 ,. The lines of action 28 ANALYTIC STATICS 35 of ^ ,5*! and E, S, meet at a point E 1 . By the principle of the transferability of the point of application (Art. 10), the points of application of E^ S,. and E* S, may be transferred to ', provided that the point Ef is supposed to be rigidly connected to the body A CD. In this new position, the forces are represented by E 1 SJ and E 1 S, f . The force E 1 57 may now be resolved again into its components E' 77, equal and parallel to j\ 7\, and E 1 'A7, equal and parallel to F^\ PIG 26 and E' S, f into its components E 1 T,', equal and parallel to E t T, t and & W, equal and parallel to /".. As E' TJ and E 1 T, f balance each other, only two forces, acting along the line K L, now remain, whose resultant R is equal to F v -f- F*. The point of application of this force may be taken any- where, as at E } on the line K L, In the similar triangles E f E* and NJJS'SJ we have EEi E'E WSJ ~ & E'E X W 5' whence 36 ANALYTIC STATICS 28 In the same manner, we get, from the triangles E E' E* E 1 N,' = E 1 E X N,' S,' or, because NJ S a f = E 1 T. f = E' 77 = AV SJ t EE,XE' N.' = E 1 E X A r / S/ This, compared with (), gives - t X EN! = ^/T 3 X /;' AY, that is, ", X Fi = ; X /=;, /^ X Vi\ = 1*\ X EE* (b) , JC> AI j'a f \ whence, - = (c) i\ From this we obtain, according to the laws of piopoition (see Geometry), _ EE, , ' that is, r = , and F, X E,E* = R X E E, (d} Likewise, - 5 = ' and ^ X ^ ^ = /v> X 7 ' : ^ The preceding results may be stated as follows 1. 77w resultant of two paiallcl font's having the same direction is equal to their turn > its line of attioti ts parallel to the lines of action of the two foitt's, and its d nation is the same as the common direct ton of the fwo fanes. 2. The line of action of the tfsidtant is so silimtt'tl that it divides any line mteicepted between the components into two seg- ments niveisely proportional to those components [equation (r)]. 3. The resultant and the lomponcnts aie so t elated and situated that, if the points of applitation of thf fhiee are taken on any straight line (as JK^E^ Fig, 25) tutfi netting then lines of action^ the produtts of any two of thf three forces by the dis- tances of their respective points of application fnwi (he point of application of the third force are equal [equations {) , (</) , (<)]. The last statement applies to distances taken on any line between the lines of action of the two forces; fur it will be remembered that E l and , were taken arbitrarily on K* t 28 ANALYTIC STATICS 37 and K a L*. Any other points, as <7, and G a , might have been taken, and the same reasoning would have led to the same conclusions Let /?!/?* = /, REi = l^EE* /. Then, equations (6), (rf), and (<?) may be written K /, = /:/. i F.I = X. FJ = /?/J 42. Equilibrium of Three Parallel Forces. By leversing the direction of A*, Fig 25, we have the equili- brant Q of the forces F t and /'", Arithmetically, the value of Q is F! -\- Fj. But, if we consider forces acting in one direction ass positive and those acting in the opposite direc- tion ab negative, we have whence, Q + Ft + F a = This, with any one of equations (/) of Art. 41, gives the conditions of equilibuum of three parallel forces. The eqmhbrant must always lie between the other two forces. 43. Theorem of Moments. Let O, Fig. 25, be any point on the plane of the forces, and draw O P, perpendicular to their lines of action. Let // M^ Kf r be the moments of F lt F 9) and A' about O Then, Mi = FiXOJ\ (a} M* - F t X ( O P, + A /\) (b) Afr^RxloPi+PtP) Adding (a) and ((>) , = tf X or, because /% X A A = JtxPiP (see Ait. 41 ) , W + Af.^/tXOPi + JtxPiP^XX (OP, + I\P] = ^/ r Therefore, the moment of the ) exultant of /wo parallel fotces about any point in their plane equals the algebiait. sum of the moments of the components. The student may take the point O' and verify this prin- ciple, paying due attention to the signs of the moments, according: to the convention explained in Art. 18. 38 ANALYTIC STATICS 28 If, instead of the resultant, the equilibrant Q is taken, its moment M q is equal to M r , and M, = - Mr = -Mi-M t whence, M, + M + ^ = 44. Two Pai*allel Forces Acting in Opposite Direc- tions. When the two component forces act in opposite directions, as /'land /*" Fig. 26, the resultant is found as fol- lows. If a force Q = /*! F a is applied at a point E in , E^ produced, such that /*",/= Q / the three forces F lt 1?,, and Q will be in equilibrium, accord- ing to the principles stated in Arts. 42 and 43. Since Q balances F l and F tt , it must be FlG ^ equal and opposite to their resultant R. Therefore, in this case, the resultant is equal to the difference of the components, and the line E t E l produced is divided by the three forces F lt F*, and R m the manner indi- cated by equations (/) of Art. 41. The law of moments applies in this case also, but due attention must be paid to signs 45. The principle of moments just stated for the case of two parallel forces and their resultant (or of three balanced parallel forces) is known as the principle of the lever, or the law of the lever, as it was first discovered by Archimedes in the determination of the conditions of equilibrium of a lever. 46. Definition of a Conple. There is an apparent exception to the foregoing conclusions that must be particu- larly noticed. If F^ = F,, Fig. 26, the formulas would give R = F l F, - 0, and, fiom equations (/) of Art. 41, So, although the resultant is 0, the distance of its point of application from the points of application of the other two F I F I forces cannot be determined, since the fractions ~- and ~~ 28 ANALYTIC STATICS 89 do not represent any numbers. This simply means that, in the case here considered, it is impossible to replace the two given forces by a single force. A system of two equal parallel forces acting in opposite directions but not in the same line, is called a couple. The theory of couples is veiy important, and will be more fully treated further on. But here it may be stated, from what has just been explained, that a couple cannot be either replaced or balanced by a single force. 47. Resolution of a Force Into Two Parallel Com- ponents. Equations (/) of Ait. 41 afford a means to resolve any force into two parallel components passing through any two given points situated on a line intersecting the line of action of the given force. For, if ./?, / and /, are given, we have: whence, F, = l * i T *a Similarly, ft = ' *i T * If R and one of the distances, as / lt from the line of action of J? to the line of action of one of the components are given, and also the magnitude Ft of this component, the other component is found from the relation Its distance /, from the line of action of R is determined as follows: F.I. = FJ whence, l* Fi The distances A and /, may be either perpendicular or oblique. EXAMPLE 1. It is required to find the magnitude and line of action of the resultant R of the parallel forces ^i(* 500 pounds) and F t (= 260 pounds), the distance / being 4 feet, as shown in Fig. 27. w +0* 40 ANALYTIC STATICS 828 SOLUTION According to statement 1 of Art 41, the rebitlUnt is equal to the sura of Fj. and F a , that is, /t" = 500 + 250 = 750 Ib Ans The distance / t from the hue of action of FI to the line of action of >?, along the hue /, is given by equations (/) of Art 41 : l t , fiom which, A = ~- A F,l = From the above, F, = 250, / = -i, and A 1 = 7oO, therefore, 750 500 X 4 "750~ Ans EXAMPLE 2 In Fig 28, the forces F^= 200 pounds) and /'(= r>0 pounds) act in opposite directions; It is required to find the resultant A' when the distance I is 6 feet. _rf_ / -.--< r^ 'I ' ~~~--l- '/a- FIG 28 R Kio '."I SOLUTION The resultant is equal to the difference between the forces (Art. 44); that is, R A 1 , - A, = 200 Ib. - 50 Ib = 150 Ib. Ans. From equations (/) of Art 4 1 , Fii _ 200 X /. 160 8 ft COX 6 _ 0f TBO" " w t< Ans. EXAMPLE 3 In Fig 29, resolve the force V\'( 450 pounds) into two parallel forces FI and F a , when A = 10 feet and /, = 8 feet SOLUTION. From equations (/) of Art 41, AV. , ., A^/! /f,-- 7 ',and^- -/ In this example, A = 450 Ib , / = 10 ft., /, 8 ft., and //, + / = 10 + 8 - 18 ft. Therefore, 460 X 10 Ans, 260 Ib. 28 ANALYTIC STATECS 41 ANY NUMBER OF FOHCPS 48. Magnitude of the Remiltant. Let /*,, /^ 7s, /%, Fig 30, be parallel coplanai foices acting through points A i, A,, etc of a rigid body, their lines of action being, respectively, K* Z,,, K t L lt etc. By the pimciple of the trans- ferabihty of the point of application, each force may be sup- i. posed to be applied at any point on its line of action; but, for reasons presently to be explained, it will be assumed that the points of application are the fixed points /,, .,/,, etc. In the first place, the i exultant R /.s equal to the algebra it sum of the components ', and //A I me of action Is parallel to the lines of action of the components For, according to Arts. 41 and 44, the icsultant A" of /'I and F t is parallel to I<\ and /" and equal to /^ + J<\\ the resultant R" of R' and /". is parallel to A" and /;, and equal to A" + F* /", + / + /'" (the negative sign being implied); the resultant A? of R" and F t is parallel to R n and fl, and equal to.-^" + -ft = A + /?", + F, + /;. 42 ANALYTIC STATICS 28 The same reasoning applies to any number of forces; so that, if their algebraic sum is denoted by 2' F, then R = 2 F. 49. Line of Action of the Resultant. To locate the line of action K L of the resultant, let N be any point m the plane of the forces, and NP<. a. perpendicular to the common direction of the forces, intersecting JK' 1 L l at /*,, A' 3 L a at /> etc.; and KL, the still unknown line of action of the resultant, at P Let N P* - p NP, = A, etc., and NP = p r If moments are taken about N, the following equations obtain (Art 43). ' Moment of JR' = F^ + F.p* Moment of R" = moment of R' + F,p, Moment of R - Rp = moment of R" + F t p< = Ft pi + F t p t + F a A + F t A In general, if the algebraic sum of the moments of any number of coplanar parallel forces is denoted by ~ Fp, and the lever arm of the resultant R by p r , then Rp, = SFfr whence, p r = ^ = ^ (1) This locates the line of action of R with respect to the point N. If, instead of the perpendicular distances p t) Ai etc. of the lines of action of the forces from A'', the oblique dis- tances NTi. = a,, NT, = a, } etc., along any line NT* (as when the forces are applied at several points of the same straight line), are given, the following equations are obtained: A = NPi = -AT 71 cos H = a, cos H p a = NP t = N T, cos H = a, cos H, etc. A = Af/* = NT cos ff = a r cos .#" Substituting these values of A, A. etc. in the equation of moments, Ra r cos H = /!! cos ff -\- F t a, cos #+ . . . etc.; whence, * r = *<" + *+ = ( 2 ) 28 ANALYTIC STATICS 43 If N is taken on the line of action of one of the forces, that force is eliminated from formulas 1 and 2, since in this case both its p and its a are equal to zero Thus, for two forces, the piecedmg equations become identical with equa- tions (/) of Art. 41. It should be carefully borne in mind that, in using the inclined distances a lt a t , etc., the same rules for signs are to be observed as for the perpendicular distances A)A> etc - Thus, F^a,. is to be taken as positive and F a t as negative. 50. Conditions of Equilibrium. When the forces are in equilibrium, we must have 2 F = 0, and S(Fp) = 0; that is, the algebraic sum of the forces must be zero, a?id the algebraic sum of their moments about any point in their plane must be aero Both of these conditions are necessary for equilibrium For, in the first place, it is evident that if S F is not zero, there is a resultant, and no equilibrium is possible. In the sec- ond place, if the resultant of all the h 8 ' forces acting in one-^1 ^ J* direction is equal I JL w 100U> FIG 81 to the resultant of all the forces acting in the opposite direction, the condition -F = obtains, but, if those two resultants have not the same line of action [in which case 2(Fp) is not zero] , they form a couple (Art. 46), and the system cannot be in equilibrium It should be remembered that the arithmetical meaning of the expression 2F = is that the arithmetical mm of all the forces acting in one direction is equal to the arithmetical sum of all the forces acting in the opposite direction. EXAMPLE 1 A weight of 100 pounds is hung from the extremity A of a straight lever A JB } Fig. 31. The lever is suspended by a rope passed around a smooth peg and carrying a weight W\ and from the extremity B is hung another weight W*. The dimensions being as shown, it is required to find the magnitudes of W* and W, that the lever may remain in equilibrium. (The weight of the lever is neglected.) 44 ANALYTIC STATICS 28 SOLUTION Since JTis the eqmlibrant of H 7 , and 100 lb , the sum of the moments of the three forces about any point in their plane-must be zero If moments are taken about 5 1 , which is in the line of action of W (for W acts upwards at .S 1 ), the foice W will be eliminated, its moment about 5 being zero The moment of /f, being positive, and that of the 100-lb weight negative, we have, 6 Wi - 100 X 8 = 0, Q/V) whence, W, = = 133 S3 lb Ans To find W, we have, 100 + Wi - W=Q W = 100 + /Ft = 100 + 133 33 = 233 33 lb Ans. Notice that it Is not necessary that the lever should be horizontal From the mathematical conditions of equilibrium (Art. 5OJ, it follows that the three weights given above will hold the lever in equlhbnuiu in any position EXAMPLE 2. A rigid bar A, Fig 1 32, rests on two supports Si and S lt OOOlb. ^~~^ son. G) O la I V* I *. I ^ Jt f _ - I - //7^ _ J I ^" " V ~~'^|^ ' ' /{/ *^ \ and carries weights of 600, 300, 250, and 50 pounds placed as shown. The weight of the bai .being neglected, it is required to find the reactions J?i and A* a at .5\ and S y , respectively SOLUTION Here we have a system of balanced parallel forces, consisting of the given weights, which act downwards, and the two reactions, which act upwards. Taking moments about >S" U , and observing the rule of signs, we have, - 600 X -10 + ^ X 28 - 300 X 23 - 250 X H + 50 X 10 =0; whence, j?, = --.^ = l,ir>7.1 lb. To find J! a , we have: J?i + #, - 800 + 300 + 250 + SO = 1,200, whence, J? t = ],200 - JK V = 42.9 lb In order to check these results, the value of Jf may be found by taking moments about 5^ The moments of the weights at the right 28 ANALYTIC STATICS 45 of Si are positive, the moment of /? and the moment of the weight at the left of Si are negative. Therefore, - 600 X 12 + 300 X 5 + ,250 X 20 - ./?, X 28 + 50 X 38 = 0, whence, 1 200 = -W" = 42 -9 Ib i approximate to tenths. EXAMPLES FOR PRACTICE NOTB -lu these examples the weights of bars and ropes are neglected 1 A straight bar is supported at two points S 1 , and 5, 25 feet apart, and carries five loads between the supports as follows 100 pounds, placed 8 feet from 5,, 150 pounds, placed 8 feet Vj^ from Sn 200 pounds, | placed in the middle of the bar, 300 pounds, placed 15 feet from J?,, and 450 pounds, placed FlG M 20 feet from 5\ Find the reactions fa and fa. at Si and S a Ans. fa = 500 Ib.; fa, = 700 Ib 2. A man carries a weight of 30 pounds hung from a stick resting on his shoulder, the distance from the weight to the shoulder is 2 5 feet. Find the pressure P on the shoulder- (a,) when the man holds the stick at a distance of 1 foot from the shoulder, (b] when he holds the stick at a distance of 1.5 feet from the shoulder. a / (a) P = 105 Ib Ans -\(,5) />= 80 Ib ii and A,, Fig 33, is to will crush under a pres- sure exceeding 275 pounds, but it is desired to have the weight as far from A t as possible Find the maximum dis- tance x at which the weight can be placed from A, Ans x 13 75 ft 4 Five coplanar par- allel forces act on a body Aff, Fig 34; magnitudes and direction of forces, Find the magnitude and direction of their resultant fa and the distance a of its line of action from 0, measured along the hue X. Ans. R - 350 Ib., acting upwards; a <* 4 875 ft to the right of O. NOTB First take moments about 0, using the inclined distances instead of the lever armi then, as a oheok, take moments about P. I LT 398-8 3 A rigid bar, supported on two piers carry a weight of 500 pounds. The pier A FlO 34 and distances along OX, are as shown 46 ANALYTIC STATICS 28 E 5 In example 4, find two forces Ft and F a parallel to the given forces and equivalent to them, F l to act at a distance of 10 feet to the right of 0, and F, at a distance of 14 feet to the right of O A (F t = 798 44 Ib , acting upwards \F 3 = 448 44 Ib , acting downwards 6 A bar A 1 A 3 , Fig 35, is to carry two equal weights W, sus- B! ^B 2 pended as shown The I s bar is supported by two ^Aa ropes AI BI and A* J I of which AI J3i cannot r' -j- s' ] be subjected to a tension greater than 1,000 pounds, and A, B* can- not be subjected to a ten- FIG 85 sion greater than 600 pounds. What is the greatest value that W can have, and what are the tensions F* and F t in the two ropes when the bar carries its greatest load? f W - 720 Ib Ans IF* - 840 Ib [F, = 000 Ib. NOTE Assume first Fi 1,000, and find Wby taking momenta about As; tlion, Fa - 2 W Fi As this is greater than 600, start by ausumint ft = 000 pounds, and proceed to find W and ^"i NON-COPLANAR PARALLEL FORCES 51. Magnitude of the Resultant of Any Number of ^, A B Parallel Forces. *"""""" T ^f 1 /** /7* /P 77* ,.-'^- : ~^ -^ a ..._.^ 4 Fl ^- 36, be parallel N^I--"'" forces acting on a \ \ ^ 4 body at the points^,, Ai, etc. The forces may be either copla- nar or not. In either case, they may be combined in pairs, as in Art. 48. Hence, the resultant is equal to the algebraic sum of the components. 52. Center of *io Any System of Par- allel Forces. According to the principles stated in Art. 41, R 28 ANALYTIC STATICS 47 the resultant R 1 of F l and F,, Fig. 36, passes through a point A' P in the line A t A a , whose distance from A l is A^ A,X-~ f = A t A* p X - Since this distance does not depend on the F l + F 3 inclination of the forces to A^A^ and since the fraction p , "--rr remains the same when for F, and F, are substituted A + F n any forces proportional to the latter, such as n /\ and n F* (n being any number, integral or fractional), it follows that, so long as the points A v and A* remain fixed and the forces remain parallel and their relative magnitudes and directions unchanged, the lines of action of the forces may be revolved about A,, and A, and made parallel to any direction whatever m space, without changing the position of the point A f , through which the resultant R 1 must constantly pass. The same principle may be proved of the point A" through which the resultant R" of R' and F a must pass, and of the point A, traversed by the total resultant R. It follows that, for every system of parallel forces applied at or, more properly ', passing 1 through any points whatever^ there zs a fixed point through which the resultant must pass, whatever the (common} direction of the forces may be. Furthermore^ that point is the same for all systems of parallel forces in which the relative magnitudes of the forces are the same, whatever their absolute magnitudes may be. Such a point, as A in Fig. 36, is called the center of the system of forces considered. 53. Center of Gravity Defined. In the particular case in which the points of application A^ A,, etc. are the particles of a tyody, and the forces acting are the weights of those particlea, the .center of the system of parallel forces thus formed Is called the center of gravity of the body. The weight of tihe body, which is the resultant of the weights of Us $rarttcls> must therefore be treated as a force whose line of fldifbli 'passes through the center of gravity of the body. TN |K|^tI^^s for determining the position of the center of 48 ANALYTIC STATICS 28 54. Coordinates of Center of Parallel Forces. When all the points of application of the forces lie in one plane, the center of the forces is easily found as follows: Let the parallel forces F lt F,, F a act at the points A,, A,, A,, Fig. 37, and let <9,Yand Y be any two mutually perpen- dicular lines drawn in the plane containing A 1} A a , A a These reference lines are called coordinate axes, and the distances of any point from them are called the coordinate's of that point Distances perpendicular to O K, or parallel to OX, are usually denoted by the letter x, sometimes with an accent or subscript; distances perpendicular to OX, or parallel to OY, are denoted by the letter y. Distances above OXa.nd T JK / ,fl| **<2*f J-f& Pro 87 those to the right of O Y are treated as positive; distances below X, and those to the left of O Y, as negative. Let the coordinates of At be O HI K^ A l = x, and O A", = Jf 1 A l = y lt Similarly, let the coordinates of A, be j. a and ;/,; those of A a be x a and y a ; and those (still unknown) of the center A of the forces be x c and y e . Since the position of A is independent of the common direc- tion of the forces, the latter may be imagined as acting in the plane X Kin a direction parallel to OX. If, now, moments are taken about any point on OX, the lever arm of F l will evidently be equal to A^ //i = y lt the lever arm of Ft will be equal to y t , that of A a equal to y,, and that of R equal to y c . Then, by formula 1 of Art. 49, \-F t y t - 28 ANALYTIC STATICS 49 If, on the contrary, the forces are imagined as acting in a direction parallel to O Y t and moments are taken about any point on O K, the following formula is obtained, by pursuing the same method as before. - &*+ F >x> + F 'Xi ft + K + ft The coordinates x e and y e locate the center A. As the foregoing reasoning is evidently applicable to any number of parallel forces whose points of application all he in one plane, the following general formulas can be at once written (2) 55. Moment About a Line. The product F l y l con- sidered in the last article is called the moment of the force F t about the lino OX, or with respect to OX. In general, when a force is considered as applied to a special point, the moment of the force about any line perpendicular to the line of action of (although not necessarily in the same plane with) the force is the product of the magnitude of the force and the perpendicular distance of the point of application of the force from that line. ANALYTIC STATICS (PART 2) CENTER OF GRAVITY DEFINITIONS AND GENERAL PROPERTIES 1. The center of gravity of a. body has already been defined as the center of the parallel forces of gravity acting on the particles of the body, or the center of the weights of all the particles, each particle being taken as the point of application of its own weight (see Analytic Statzcs> Part 1). These forces are considered parallel because they are all directed toward the earth's center, whose distance from the surface is so great, compared with the dimensions of ordinary bodies, that the angle between the lines of action of the weights of any two particles of a body is practically zero. (Two terrestrial radii meeting the surface at two points distant 100 feet from each other make an angle at the center equal to about 1 second.) The abbreviation c. g. will here be used to signify center of gravity. The c. g. of a body is called also center of mass, center of inertia, and centrold. These terms, however, will not be used here. 2. Immediate Consequences of the Definition. Three important consequences follow at once from the preceding definition, namely: 1. As stated in Analytic Statics, Part 1, the weight of a body may be treated as a single force acting vertically through the o/ (fo body, , This is expressed by saying that the TBXTHOOK COMPANY, ALL MIHT MMBMVKD 529 ANALYTIC STATICS 29 weight of a body may be supposed concentrated at the c. g of the body 2 The position of the c. g. of a homogeneous body (that is, a body whose substance is the same throughout, or whose particles have all the same weight] depends on the form of the body only, not on the material of which the body is made. For the centei of a system of parallel forces depends on the relative, not on the absolute, magnitudes of the forces, and on the relative positions of their points of application. 3 // the c. g. of a body acted on by its own weight is sup- ported, the body will remain in equilibrium, whatever its Posi- tion. For the weight of a body is equivalent to a single force acting through its c. g., and, if this point is supported that is, if a force equal to the weight, but acting in the opposite direc- tion, is applied at this point the two forces will balance each other. Conversely, // a body acted on by its own weight and by other forces is in equilibrium, the resultant of the other forces must be vertical, act upwards, and pass through the c. g of the body. Thus, if the triangular plate ABC, Fig 1, hangs from three strings A O, BO, CO, and is in equilibrium, its weight W, acting through the center of gravity G, must be balanced by the resultant of the reactions of the strings. This resultant must, therefore, be vertical, act upwards, and pass through G. 3. The Center of Gravity of a Body May be Outulde the Body. When several forces act on a body, the line of action of either their resultant or their equilibrant may be altogether outside the body. For example, in Fig. 2, the equilibrant Q of the forces F l and F t acting on the lever A^ A t passes through a point A, outside the lever. This means FIG ANALYTIC STATICS that the only single force that can balance Ft. and F t is the force Q acting through A 9 ', but, as the point A,, is outside the lever, that point must be imagined to be rigidly connected to the lever, in order that Q may act on the latter. In the same manner, it often happens that the c. g. of a body is outside the body. Thus, the c. g. of the curved Q FIG. 2 FIG 8 rod A /?, Fig. 3, is a point G outside the rod If it is desired to suspend the rod so that it will be in equilibrium in all positions, the point G must be connected to the jod by some means, as by another rod G H. The additional weight of the rod GH alters the position of the c. g. of the whole system; but the change may be made small, and its amount easily calculated, as will be explained presently. 4. Center of Gravity or a Line. Properly speaking, geometrical lines and surfaces have no c. g , since they have no weight. By an extension of the definition, however, every line and surface is said to have a c. g., the expression being taken in the sense now to be explained. Consider a body A fi, Fig, 4, of uny form, and a section, as PQ, containing the centers of a number of particles of the body. The resultant of the weights of these particles will pass through a fixed point 0, which is the center of the system of parallel forces constituted by those weights. Likewise, 0> is the center of the system of parallel FIG 4 4 ANALYTIC STATICS 29 forces constituted by the weights of the particles in the sec- tion P'ff. If a sufficient number of sections is taken to include all the particles of the body, and the centers thus found are joined, a line A O 'B will be obtained, which will be intersected by the resultants of the various groups of particles considered The center of the system formed by these resultants is the c. g of the body, and is said to be also the c g of the line ADO'S. If the body is homogeneous and of uniform cross-section, all the resultants referred to will be equal, and the number of them acting on any portion of the line A O' B will be proportional to the length of that portion. Now, in deter- mining the center of any system of parallel forces, the forces may be replaced by any quantities proportional to them For instance, if the x coordinates of the points of application of two forces F l and F* are # and jc tt the x coordinate x t of their center will be given by the formula (see Analytic Statics, Part 1) r* i z? Xi -)r -=;Xi F^XI-\- F, x, _ F 1 If F a and /i are proportional to two given lengths / B and / F I that is, if ' = -', we have A l\ ,/. *i + 7 x, l lQ it * ' Therefore, .-+/* = L#* +.** F> +F. /, + /. So, in finding the center of a system of parallel forces whose points of application lie in a line, and whose dis- tribution along any part of the line is proportional to the length of that part, that length may be used instead of the corresponding forces. 5. Center of Gravity of a Plane Area. Take now a homogeneous prism ACGE, Fig. 5. If all the particles 29 ANALYTIC STATICS 5 FIG. 5 enclosed by a cylindrical surface PQ are considered, the resultant of their weights will be a vertical force whose line of action will meet the upper surface of the prism at some point O. Similarly, other points may be found where the resultants of the weights of vertical rows of pai tides meet the surface ABCD, thus obtaining 1 a system of parallel forces whose points of application may be taken on the plane AB CD, and whose center is called the c. g. of the surface A B CD. This point is evidently the same as the point where the vertical line through the c g. of the prism pierces the plane A B CD. As the forces act- ing on any such space as P are proportional to the area of this space, areas may be substituted for forces when dealing 1 with the c. g of a pl?ne surface. 6. General .Definition of the Center of Gravity of Jjlnes and Surfaces. A line is said to be homogeneous, or of uniform weight, when parallel forces are distributed along the line in such a manner that the resultant of the forces acting on any part of the line is proportional to the length of that part. The center of the parallel forces thus acting is the c. g. of the Hue; and their resultant is called the weight of the line. The same terms apply to a surface when the parallel forces acting through it (that is, whose lines of action pierce it) are such that the resultant of the forces acting through any part of the surface is proportional to the area of that part. The center of the parallel forces thus acting is the c. g, of the surface; and their resultant is called the weight of the surface. 7. Static Moment. Let Wb& the weight, or any num- ber (length, area) proportional to the weight, of any figure line, surface, or solid; and let x be the distance of the center 6 ANALYTIC STATICS 29 of gravity of the figure from any point or line. Then, the product Wx is called the static moment of the figure with respect to that point or line According to the theory of parallel forces, the static moment of any figure or system of figures about any point or line is equal to the algebraic sum of the moments of the parts of which the figure or system is composed If the distances of the centers of gravity of several areas A^A^ etc from a given point or line are denoted, respectively, by AH .a,., etc , the static moments of these areas about the point or line are XiA lt x,A a , etc. If the distance of the c g of the system formed by the aggregate of these areas from the given point or line is denoted by x c) then, x e (Ai + A,+ . . . ) = x^Ai + x*A> + In general, denoting the sum of the areas by 2 A, and the algebraic sum of their static moments by SxA, the following equation may be written: x t 2 A = IxA\ V y. A whence, x c = =~^ A similar formula applies to lines and volumes. The formula shows that 1. The algebraic sum of the static moments of the patts of any figure or system of figures about the c. g. of the figure or system of figures, or about a line containing that center, is equal to zero. For, in this case, x e = 0, and, therefore, SxA (= x t !A) = Conversely, 2. // the algebraic sum of the moments of the parts of a figure or system of figures about a point is zero, that point is the c. g. of the figure or system. 29 ANALYTIC STATICS IMPORTANT CASES SYMMETRICAL FIGURES 8. Definitions. A figure is symmetrical with respect to a point O when every straight line passing through that point meets the figure in pairs of points equidistant from the point O, the two points of each pair being; on opposite sides of the point O. The circle, the ellipse, the sphere, the circular ring, are each symmetrical with respect to its center. This center is called the geometric centei', center of figure, or cen- ter of symmetry of the figure in question. A parallel- ogram is symmetrical with lespect to the point of intersection of its diagonals. 9. A figure is symmetrical with respect to a straight line, called an axis of symmetry, if every straight line perpendicular to the first-mentioned line meets the figure in pairs of points equidistant from said line or axis, the two points of each pair being on opposite sides of the axis Thus, a rectangle is symmetrical with respect to either of the lines joining the middle points of two opposite sides. An ellipse is symmetrical with respect to either of its principal axes, and n, ciicle with respect to any diameter. The axis of a right circular cylinder is an axis of symmetry, and so is the axis of a right circular cone. 10. A solid is symmetrical with respect to a plane, called a piano of symmetry, when every straight line perpendicular to the plane meets the solid in pairs of points equidistant from the plane. 11. Center of Gravity of Symmetrical Figures. It is evident that, if a figuie is symmetrical with >espect to a point, a line, or a plane, the c. g. of t/ie figure coincides with the point, or lies in the line or plane, of symmetry, as the case may be. Thus, the c. g. of a circle coincides with the center of the circle; the c. g. of a parallelogram coincides with the inter- section of the diagonals; the c, g. of an isosceles triangle 8 ANALYTIC STATICS 29 lies in the perpendicular from the vertex to the base (at what distance will be seen further on), and the c g of a icgular pyramid lies in a plane through the vertex peipendicular to the plane of the base 12. Again, if a figure has two axes of symmetry, the c. jg- m will be at the intersection of those axes, and 7? a solid has three planes of symmetry meeting at a that point is the c. g. of the solid. DETERMINATION OF THE CENTER OF GRAVITY BY ADDITION AND SUBTRACTION 13. Addition Method. If a figure (by which is meant either a line, a surface, or a solid) or system of figures can be divided into parts whose centers of gravity are known, the c. g. of the whole figure or system is easily found by the principles explained in Analytic Statics, Part 1. For example, let it be required to find the c. g of a system consisting of two homogeneous spheres C and C,, Fig. 6, whose weights are Z-J^, and whose distance FlQ fl apart, measured be- tween centers, is a. This is equivalent to finding 1 the center of two parallel forces W t and W t acting through d and C,. Let this center be G; then, whence, G C, = - ^ a W, + H7. 14. Subtraction Method. If, on the contrary, the c. g. of a system is known, and also the c. g. of all its parts but one, the c. g. of the remaining part is found by the same general principles just referred to, the only difference being that subtraction is used instead of addition. As an illustration, let it be required to find the c. g-. of a figure obtained by cutting from a rectangle B CDR, Fig. 7, a rectangular corner EFHI and a circular piece 0, the 29 ANALYTIC STATICS 9 dimensions being as shown Take EDior the axis of x and EB for the axis of y\ and let x,' and y' be the coordinates EH' and H 1 G' of the c. g of the rectangle B CDE\ x" and y, the coordinates EH" and //" O of the c. g. of the circle, x'" and y'", the coordinates //'" and H'" G 1 " of the c g. of the rectangle EFH '/; and ^ and j/, the coordinates of G, the required c. g. According to Art. 5, areas may, when dealing with surfaces, be substituted for forces in the equa- tions of moments and of centers of parallel forces. In this a j{j sense, the moment of CD with respect to ED is the same as if the whole area were concentrated at &; that is, the moment of B CD E with respect to ED is a by'. The area A of the figure whose center of gravity is required is a b cii bi TT r*. The c. g. of BCDE is at the inter- section of the two diagonals, or at the middle point of CE, and, therefore, xf = EH' = ? andj/' = H' G 1 = . 2 2 Likewise, x" EH" = c, y" and H" O = b - r, ~ ~~2 The formula of Art. 7 now gives + aibt x' 1 ' A-\- ctibi 10 whence, x e = ANALYTIC STATICS 29 (A + a, bi + n r')^ - a, b, x"' - x r> x" a b ! 61 TT r* Similarly, whence, _ y _ Ay e ff, b.y" 1 , ab , = a by 1 - a, b,y'" - TT r'y" = 1 \ab* - a, bS -_2_7r r*(b--_r) "I A 2L ab-atbi-'ni* \ 15. The subtraction method, just illustrated, is very much simplified when the c. g. of a figure consisting of two parts and the c. g. of one of the parts are known, and it is required to find the c g. of the other part. This special case is illustrated in Fig. 8, where C, the c. g. of the figure BCDJS, and the c. g. of the part C E D, are known, and it is required to find G,, the c. g. of the . .X PIG 8 part BCE. Let total area = A\ and area CDE = A,. Then (Art. 13) , Ai X G l G, A x G G: whence, GG,= A EXAMPLE 1 -To find the c g. of the channel section represented in Fig 9 L SorurioN.-Owmg to the symmetry of the figure with respect to the center line OX, its center of gravity G t must be on that line. To find the distance OG e . *-, of G e from the back of the channel, notice la " er i8 the dlfference be tween the rectangles Q D>, whose centers of gravity G and G' are at the 29 ANALYTIC STATICS 11 distances O G = 5 inches, and O G> - + 5 J 8 i + 5 = ?j 2 8 ID inches from A D Denoting the area of the section by A, and talcing moments about A /?, A x t = (AJtXAD)XOG- (A> JB> X A' D 1 ) X O G> = 150 X 8 X 16 -36 X 11 X 46 X16 " I G I 1 *- 1_ ! * S \JLZ i<if. i) The area of the section is A = A B X A D - A 1 B> X A 1 W = (M) - ^ X 11 Hence, .v c t= OG e ** ] 1,870 X 8 X W - x n x45 Km Id B x_eo_ r _3B_x 11 8 XJJO-36X 11 flT" in EXAMPLE 2. To find the c. g. of the angle section represented in Fig. 10, and having dimensions as shown. ILT3W-9 12 ANALYTIC STATICS 29 SOLUTION Produce CD to meet HF at / The section is now divided into two rectangles, BCIH and DEFI The c g of a rectangle being at the point of intersection of the diagonals, itp distance from either of two parallel sides is equal to one-half of either of the other two parallel sides Taking moments about H B, and denoting the distances of the c g of the section from H B and ffF by x e and y t , respectively, and the aiea of the section by A, whence, Xc = HP = 9 11X13 8" 1 " 16 r 16 ' 9 , 11 XJ8 T-CTT - 1.3879" Moments about H ' F\ whence, = 2 8879 In Aus. EXAMPLES FOB PRACTICE 1 Find the center of gravity G 1 of the area obtained by taking the rectangles ^and D H, Fig 11, from the rectangle A BCD The FIG 11 dimensions are as shown .in., nearly In., nearly ANALYTIC STATICS 13 2 Find the c g of the area obtained by cutting off the circle A BI C lt Fig 12, whose radius O l A is 3 inches, from the circle ABC, whose radius OA is 10 Inches Ans O G = 69231 in = H in., nearly 3 Find the c g of the angle section represented in Fig 13 = 92500 m. = ff m , nearly = 2 1750 m = 2Vfc In., nearly r L FIG 13 4 Find the c g of the Z section represented in Fig 14 NOTE In talcing moments about QM, remember tlmt the moments of areas on opposite sides have opposite sljrns Ario fOP = .41667 in = ADS \ PG = l 4167 in _ in , nearly i( nearly CENTER OF GRAVITY OF POLYGONS NOTE In some of the articles that follow, formulas and rules are given without explaining how they nre obtained This is done when- ever the processes Involved require the use of advanced mathematics, or, being elementary, are too long and complicated to be given in connection with this instruction 16. A Fundamental Principle. // a straight line divides a plane figute in such a manner that eveiy hue paral- lel to a hxed direction meets the perimeter of the figure at one point on each side of the first- mentioned line and is bisected by said hne> the c. g. of the figure lies on that line. Let OX, Fig. 15, be a line dividing the Pro. 10 figure OQXQi into two parts, in such a manner that all lines, 14 ANALYTIC STATICS 29 as QQRR etc., parallel to the fixed line /P"are bisected at their point of intersection with OX. Then, according to the principle just stated, the c. g. of the area OQXQ^ lies m the line OX. 17. Triangle. Let ABC, Fig. 16, be any triangle. DtawAA' from A to the middle point A' of the opposite side. Any line, as RS, parallel to B C is bisected at its intersection J with A A'. Therefore, according to the proposition of the preceding article, the c g. of the triangle lies on A A'. For a similar reason, the c. g. must lie on B B 1 , joining the vertex B and the middle point /}' of the opposite side A C Therefore, the c g of the triangle is at the intersection A of the lines A A 1 and B B't or of either of them with the line CO from C to the middle of An. It is shown in geometry that the distance of the point G, where A A 1 , /?/?', and CO meet, from any of the vertexes is equal to two-thirds the length of the line joining that vertex to the middle point of the opposite side, or A G = \AA' t BG = Inn 1 , CG - \ CC f . The lines A A 1 , BB>, CO are called the modlan linen.' Therefore, The c. g. of a triangle lies at the intersection of the median lines, and its distance from any vertex is egual to two- thirds the length of the median line from that vertex to the opposite side. 18. The perpendicular distance from the e.g. of a triangle to any of the sides is equal to one-third the altitude of the tri- angle, when that side is taken as the base. 29 ANALYTIC STATICS 15 Foi, drawing 1 A ff and < /if perpendicular to B C, Fig. 16, two similar triangles AA'H and GA'K are formed, which A A' GA 1 whence, bearing in mind that GA 1 = \AA' or A A' GA' A A 1 19. Sometimes, the distances of the vertexes of a tri- angle from an axis or line of reference are given, and it is required to find the distance of the c. g. from the same line. In Fig. 17, let .1,, the distances of the vertexes of the tri- angle A^AtA* from the line A' A 7 be y lt y,, y a . The c. g. of the triangle is on the median A* M, and, as already explained, x ^ GM = \AiM. Let the distance G P of the c. g. from X' X be denoted by y e . Draw MN perpen- dicular and MB parallel to A 7 A'. The figure gives y e = GP= DP + GD = MN + GD (1) In the trapezoid A,P t P a A tt the line M N joins the middle point of A a A* and P t P tt therefore, In the similar triangles A^M B and GM D, 1:1? YB' that is - -/J- fi ! whence, GD = ^AijB = ^(AiPi MN} = ijj/, i(y, + j/ a )] (3) Substituting in equation (1) the values of MN and GD given by equations (2) and (3), and reducing, This formula is perfectly general, provided that due atten- tion is paid to the signs of the coordinates. That is, distances 16 ANALYTIC STATICS 29 measured on one side of the reference line should be consid- ered positive, and those measured on the opposite side, neg- ative. Thus, if the vertex A, lies below A'' A", and y t and j/, are considered positive, y, should be considered negative. 20. Trapezoid. Let B CD , Fig 18, be any trapezoid, having the bases BE = ,, CD = b,; altitude DPI = /i, median line Mi M, = m The median line /*/, Af, (that is, a line through the middle points of the bases) bisects every line parallel to the bases; therefore, G, the c. gf of the trape- zoid, must lie on that line (Art 16). It is necessary, there- fore, only to find the distance, as GP^ of G from either base. Fin 18 Draw B D; this line divides the trapezoid into two triangles having the same altitude //, and d l and b, for their bases. The distance of the c. g. oiBCD from CD is \ h (Art. 18), and its distance from BE is h h = % h. The distance of the c. g. of JS D E from B E is h. Denoting the area of the trapezoid by A, and taking moments about /?/j, A X GP, = area B CD X I h + area BDE X \ h whence, = \ b, h X 6 X i h => (b, + 2 b,) 6' m A similar value may be found for /> by simply Inter- changing fa and b* in this formula. 29 ANALYTIC STATICS 17 Draw M t K* perpendicular to the bases. The similar triangles M, Mi A" t and GM^P* give MnM* = MjJCi m = _ h GM, ~GP\~* GM, whence, GM, = ^-~- (2) 3 \ 0i + b* / which gives the distance from the middle point of z to the c. g., measured along the median line. A similar expression may be found for GM^ by simply interchanging &i and b t . The distance B P l is found by the following formula, in which N is the angle between CB and CJ, the latter line being perpendicular to the bases: +*, + (bl + 2 ^ a) k ^ N ~ ***} (3) *i + o, J 21. There is another formula that is often useful for the determination of the c g. of a trapezoid. Let the non- parallel sides EB and D C, Fig. 19, meet at V. As before, Mi M, is the line through the middle points of the bases. From geometric princi- ples, it is known that this line passes through V. Let d be the c. g. of the triangle DVE, G, the c g. otCVB, and G that of the trapezoid BCDE. Also, let V M, VM, = a t) VG a. As explained in Art. 17, = $ a,, and V G t = s . Taking moments about , we have, area DVEXG^G = area CVBX G, G\ whence, t g area C E^g VMf VMS since the areas of two similar triangles are proportional to the squares of their homologous sides. In terms of the .it U ^ 18 ANALYTIC STATICS 29 quantities a lt a,, and a, the preceding proportion may be written: a 3 dj. _ af a "f a, 1" Solving this equation for a, the following result is finally obtained: _ 2 a," - / a 3 ,' - a,' 22. Trapezoid: Graphic Solution. Produce CD to E'> Fig. 38, making D E' = d lt draw -fi 7 (7, and produce it to its intersection d with EB produced. The similar triangles GM,E' and GMi d give- fad M.Ef. GM,' whence, M 1 d = ~XfaE' (1) GM, The value of C^/, is given by formula 2 of Art. 20, and the value of GM t is given by the same formula, by inter- changing bt and b,. Therefore, m J. + 2J. x a , , 3 ^ + ^ By construction, fa E' = M t D + D E 1 - i b t + <5 X . Sub- stituting these values in (1), fa d = I- 6 / + * X (i*. + ^) = U + ^ a = faj5 + * t ^a + 0i therefoie, B d = b, Hence, the following construction for finding the c. g. of any trapezoid. Join the middle points of the bases. Produce the bases in opposite directions, making the prolonged part of each base equal to the other base Draw a line joining the extremities of the prolonged segments The point where thts line intersects the median line is the required c. g. 23. Any Quadrilateral. Let B CDE, Fig, 20, be any Quadrilateral. Draw the diagonals BD and EC, and find the 29 ANALYTIC STATICS 19 middle point M of one of them, in this case EC. Take on D B the distance B N D /, and draw MN. The c. g. of the quadrilateral is a point G on MN, obtained by taking on M 'N a distance AfG = i M ' N. EXAMPLK 1 To find the c g. of the channel section represented in Fig 21, the dimensions being as shown NOTE The section being symmetrical, the dimensions of the lower part are the same an the corresponding: dimensions of the upper part, and need not be given SOLUTION The c, g lies on the line of symmetry O Q drawn through the middle point of, and perpendicular to, SJ The only thing to be determined is the distance O G = x e FIG 20 In complicated cases like this, it is often convenient to hnd a general formula first, and then substitute the numerical values. In order to do this, the dimensions will be represented by the letters written beside the figures on the diagram Since the channel section to the differ- ence between the rectangle B CIJ and the trapezoid DE FH> then, denoting the area of the rectangle by 1?, the area of the trapezoid by T, that of the section by A, and their centers of gravity by G>, Gt, and G, respectively, x t - G, - G G f - \b - G G r (1) Taking moments about a line YZ drawn parallel to BJ through the center G r of the rectangle B CIJ, we have, since the moment of R about G r is 0, G,_G t ~A GG, (2) 20 ANALYTIC STATICS 29 Now, T = \ (EF+DH) X PQ = i[(A - 2 j - 2/) + (A - 20] (* - ** = (A - $-20 (b -*,) and (formula 1 of Art 2O) , > X - 2 j - 20 , * iT b- t, 3A-4s-6t 3 X 2(/;-J-20 ^1 S l\ *' /.__!p__\/__^ i-v U>- * T ( 1125' I U*f J * &S> ZZlf-^'lZS'i JS J T G Ljff. A U w y tf ^ FIG 22 Substituting the values of T'tind <7 r G 1 / in (2), - j - 2 - (* - O (8 A - 4 j - 0] 29 ANALYTIC STATICS 21 Substituting the values of O G, and G G r m (1), r = J 6 _ <* -M[M*-J- 2 The dimensions, expiessed in sixteenths of an inch, are 6 = 54; // = 20S, / = (, s = 8, f t = 7 Therefore, A = b !i - (t> - t,) (// - s - 2 1) = 54 X 208 - (54 - 7) (208 - 8 - 12) ; and _ 54 _ ^T)4 - 7) [7(208 - 8 - 12) + $(54 - 7|] Xf ~ 2 2[fi4 X 208 - (54 - 7) (208 - 8 - 12)] = 27 - 47 (7 X 188 + X47\ 3 / 27 - 14 187 = 12 HHS sixte. nths 2 X 2,396 = 804 in , or, approximately, x e = f in. Ans. EXAMPLE 2 To find the c. g of the plane figure represented in Fig 22, the dimensions being as shown SOLUTION The figure may be divided into the two rectangle^ B CL M and EFIJ, the trapezoid C 'D K L, and the isosceles tri- angle FH1 Momenta will be taken about BX and B K The altitude of the trapezoid is DN ' - CJVtan 60 = (CL - DJf)ta.a 60 = 3 4641 ft The operations are given m the following table, which needs no explanation. Area Lever Arm With Respect to Moment About Figure Square Feet BX BY BX BY BCLM ao oooo i oooo 5.0000 200000 100 OOOO f DN CL+aDK CDhL \ 2 3 CL-\-DK 50000 99.4353 138.5640 = 3 5877 El'IJ 10 1250 < 4 s ' f a 25 -+DJV+CJi J + ED+CX = 7 7141 L -=5fi350 78 1053 56 PS3I FIff i 2656 I 3 5 6350 130853 7iigo I = 10 3391 59 1034 210 6l57 303 636l Having the area and the resultant moments, ,r c = BP 302.6361 _ 10rtl . _ 4 . ,or 5.1205 ft. Aus. PG 59 1034 210.6167 59.1034 3.5635ft. Ans. 22 ANALYTIC STATICS 29 EXAMPLE FOR PRACTICE Find the area and the c. g of the Irregular T section represented in Fig. 23 3 3242 sq in Ans \A : \OP: \PG-- 2 0695 in 7826 in CENTER OF GRAVITY OF AREAS BOUNDED BY CIRCULAR ARCS 24. Circular Sec- tor. Let O PS V, Fig 24, be a circular sec- tor, having the radius O P = r, and the central angle P O V = L. As the sector is symmetrical with respect to the line O S> bisect- ing the angle L t the c. g. lies on that line, and it is only necessary to determine the distance OG of the c. g. from the center of the sector This # distance is given by the for- mula p < ^ 240 r sin G = OG = In the denominator of this formula, the angle L is ex- pressed in degrees and deci- mals of a degree. 25. For the sectors whose arcs are, repectively, a semi- circumference, a quadrant, and a sextant, formula takes the following special forms: 4 r Semicircle, y e = - = .4244 r preceding (I; Quadrant, y e = 8 ^ in45 -! = .8002 r (2) Sextant, y c ^ Bin80 --.6866r (3) 7T 29 ANALYTIC STATICS 23 26. Circular Segment. Let BCD, Fig. 26, be a circular segment, having c the radius OB = r, and the central angle BOD = L, expressed m de- s \ grees and decimals of a degree. Here, G, the c. g. of the segment, lies on the line of symmetry O C. The distance y e = <O G is found by the following formula: _ Q G = 240 r sin 3 rcL 180 si If the length of the choid B D is denoted by c and the area of the segment B CD by A, the preceding formula can be reduced to (2) y. - 12 A 27. CircularTrapezoid, or Flat Ring. In Fig. 26, let B CD EH I be a circular trapezoid, or flat ring, having the radii rt. and r tt and the central angle L, expressed in degrees and decimals of a degree. The c. g. lies on the line of sym- metry O C, and its dis- tance y e from the center O of the ring is given by the formula: = 240 sin Z, r t ' -*,' Fie,. -jo nL r**-r,* EXAMPLE To find the c. g. of the area represented In Fig 27 SOLUTION Take the center line O y, bisecting the two circular arcs, for the axis of y, and the perpendicular OX, through the center of the arcs, for the axis of A' As usual, x e and y e are the coordinates of G, the required c g , and A is the area of the figure The diagram gives A = rectangle B FH T+ segment CD E segment KJ I. The 24 ANALYTIC STATICS 82ft moment of A with respect to OX\& equal to the algebraic sum of the moments of these three areas To find the area of the rectangle, KI must first be found, since TK and IH are given To find the IS? JF S Pis 27 areas of the segments and their lever arms, the angles Z, and Z, must be determined The figure gives, expressing angles to the nearest minute, K I = 1KQ = 2^lr a a O Q* = 2VJ^ fl ~(r^ = 2Vr 1 " - (r, -~4p = W 967 ft sin i Z, = , whence Z, = 112 3( sin , -^-, whence Z t = 23' Having all the required elements, the coordinates .r, nnd y r ure found by the usual methods, with which the student is now supposed to be perfectly familiar. In finding x ct however, the operations will be much shortened by observing that, if a distance K ' H' = / H is taken, and the line H' F< drawn perpendicular to TK, the figure at the right of F' H> is symmetrical with respect to O y, and its moment about Fis consequently zero Therefore, the moment of the whole area about Xis simply the moment of the rectangle B F 1 H 1 T, and hence, ' ' A The results, which should be verified by the student, are- A = 118 35 sq. ft. Ans. x t = OP= 18741ft. Ans y c - PG = 7 7607 ft. Ans. 29 ANALYTIC STATICS 25 EXAMPLES FOR PRACTICE 1. Find the c g of a circular sector whose radius is 10 feet and whose central angle is 45 y Ans y c = 6 497 ft j 2 Find the c g of a segment whose chord is 8 inches and whose radius is B 5 inches Ans. y e = 3 815 m. 3 Find the c g of a circular trapezoid whose radii aie (i inches and 3 inches, and whose shorter chord (as E /, Fig, 26) is 3 inches. Ans. y e = 4 466 m. 4. Find the c g of the half circular seg- ment BCD (Fig 28). x t = 2.839 m y e = 13.798 in. PIQ 28 CENTER OF GRAVITY OP A PLANE AREA BOUNDED BY AN IRREGULAR CURVE 28. Approximate Analytic Method. To determine the c. g. of a figure, as /? CD E, Fig. 29,, having- an irregular contour, proceed as follows: Draw two lines of reference OX and O Y perpendicular to each other in any con- Y venient positions; preferably, one of the lines, as OX, should as nearly as pos- sible bisect the area. Divide OX into a sufficient number of parts, so that, by erecting perpendic- ulars at the points of. division, the par 01 $he perimeter of the curve intercepted by two conseGUt$TO;c$rtoates, as HiH^ and /;/ may be treated Pro. 29 as a straight a trapezoid., oorceflgondlng strip .fifr.ff'././i, as .#/,//, etc., and the I i t 26 ANALYTIC STATICS ordinates ff^H^/if^ etc. Find the distances of the centers of gravity of the strips, considered as trapezoids, from the lines OX and. Y. Treat the area of the whole figure BCDE as equal to the sum of the areas of the trapezoids, and apply the method of moments, as usual That is, if the sum of the areas of the trapezoids is denoted by 2 T, and the sum of their moments about OX and O Y by - Ty_ and 2 Tx t respectively, the coordinates x t and y e of the c. g. of the whole figure are given by the formulas STx _ _ *'~ Tf t v ~, Although this method is only an approximation, it is sufficiently close for almost all practical purposes. The c. g. of each trapezoid may usually be taken at the middle point of its median line 29. Experimental Methods. A very convenient method of finding the c. g. of an irregular figure (or of any other figure) consists in drawing the figure to scale on a piece of cardboard of uniform thickness, then cutting off the remaining part of the cardboard and balancing the part thus left on a knife edge in. two positions. In Fig. 30, let B B' CO be the piece cut out of the card- board, having the form of the figure whose c. g. is lequired; let B C and B 1 C 1 be two positions of the knife edge, for each of which the piece of cardboard remains in equilibrium. Then, their intersection G is evi- FlG M dently the required c. g. This method may be used not only for irregular figures, but also for those figures the determination of whose center of gravity leads to complicated formulas. The method, however, is not very accurate, and, as a check, the piece of cardboard should be balanced in more than two positions say in six or eight. This will give an idea of the degree of accuracy attained. 29 ANALYTIC STATICS 27 30. The method just described is also very convenient for determining the distance of the c g. of a symmetrical body from a plane perpendicular to a plane or axis of sym- metry. A locomotive, for instance, is symmetrical with respect to a. plane midway between and parallel to, the axis of the cylinders. The distance of the c. g from either end of the locomotive (that is, from a plane through either end perpendicular to the axes of the cylinders) is found by balancing the locomotive on a horizontal rod perpendicular to the plane of symmetry, passed through ungs attached to the locomotive and symmetrically located with respect to that plane. The position of the rod and rings is changed until the locomotive is found to remain balanced (that is, with its two ends at the same level) when suspended. The distance of the rod, when the locomotive is thus balanced, from either end of the locomotive, is, approximately, the distance required. 31. Still another method for finding the e.g. of a plane figure forming: the faces of a thin plate of uniform thickness consists in suspending the plate by a string and marking on either face a vertical line that is the prolongation of the direction of the string; then suspending the plate in a differ- ent position (that is, tying the string to a different point In the plate) and marking a line similar to the one marked before. The intersection of the two lines is the required c. g. This method can be conveniently used for determining the c. g. of any body when that center is outside the body (as an angle section, a bent rod, etc.). Here, when the body is suspended in one position by a string, a line containing the c. g. is obtained by making two marks on the body: one at the point where the string is fastened and another directly under it, By tyio$ the string to another point of the body, another lin is( determined, whose intersection with the first gives the requfrf&jS 28 ANALYTIC STATICS 29 CENTER OF GRAVITY OF SOLIDS 32. Right Cylinder or Prism. The c g. of a homo- geneous right cylinder or prism evidently coincides with the middle point of the line joining the centers of the bases. 33. Right Cone or Regular Pyramid. The c. g. of any right cone or regular pyramid lies on the perpendicular from the vertex to the base (line joining vertex with cen- ter of base) at a distance from the ver- tex equal to three-quarters of the length of that perpendicular. 34. Conical Frustum. The c. g. of a conical frustum, Fig. 31, lies on the line joining the centers of its bases. Its distance from the lower base is given by the formula PIG 81 v - o G-X r* y e LA LJ- 7 PV. 7- v 4 (r, + rj in which r t = radius of lower base; r, = radius of upper base; A = altitude of frustum. 29 ANALYTIC STATICS 29 COPLANAR NON-CONCURRENT FORCES COUPLES DEFINITIONS-EFFECT OF A COUPLE 35. A statical couple, or simply a couple, has already been defined as a system of two equal non-collmear parallel forces having opposite direc- / tions. In Fig. 32, the forces F,. and n F, constitute a couple acting on the body BCD. Although F, and F, aie equal in magnitude, they are denoted by different letters for convenience in refer- ring to their lines of action. 36. The lever arm, or simply the arm, of a couple is FlQ 32 the perpendicular distance (A/ 5 . = p) between the lines of action of the two forces constituting the couple. 37. The plane of a couple is the plane determined by the lines of action of the two forces constituting the couple. 38. The axis of a couple is any line perpendicular to the plane of the couple. Such will be the meaning given to the term here, although some writers use it in a different sense. It follows from this definition that all couples whose planes are either coincident or parallel have the same axis. 39. Coaxial couples are couples having the same axis. Couples not having the same axis are called non-coaxial. 40. Resultant Moment of the Forces of a Couple. The resultant moment of the two forces of a couple about any 30 ANALYTIC STATICS 29 point in their plane is constant and equal to the moment of either fotce about any point on the line of action of the other fotcc. This is easily shown Paying due attention to bigns, the resultant moment of the two foices A and 1*\ about any point 0, Fig 32, is -F a Xff*0 + F l Xtf l O = -F*(ff t O- //, 0} = - F. X Hi H* = - F*p = - I\p For a point <7 between the lines of action of the forces, the resultant moment is - F, X ff. 0> - F, X Hi O 1 = -F, (// (V + //, 0} = -F t p = -Ftp 41. The constant resultant moment of the foices of a couple about any point m their plane is called the moinout of the couple, and is numerically equal to the pioduct of either force by the arm of the couple. 42. Notation. A couple is expressed cither by its moment (for all coaxial couples having the same moment are equivalent, as will be shown piesently), or by writing its two forces in parenthesis with the arm between; thus, (F^fiiFt). The latter is a more convenient form of expres- sion for some purposes, and will be often used here. 43. Effect of a Couple. The effect of a couple acting on a rigid body not acted on by other forces is to turn the body about an axis passing through its c g. A full demonstration of this proposition cannot be given here, as it would be necessary to make use of some kinetic principles that have not yet been explained. Moreover, it is to be observed that, for the A\ Q. \ IB purposes of statics, it is not i necessary to know what the * jf* ~n V.F, effect of a couple is, for all the theorems relating to the FlG 83 equilibrium and equivalence of couples can be stated and proved without any reference to what the effect would be if the couples were unbalanced, The foregoing principle, however, has been stated here, in 29 ANALYTIC STATICS 31 order to caution the student against a common error preva- lent among beginners. If the couple (F lt p t F a ), Fig 33, acts on the body A B, its tendency is to turn the body about an axis passing through its center of gravity G, not about an axis or point situated between the two forces. 44. Direction tind Sign of a Cotiple. The sign of a couple will here be treated as positive or negative according as the moment of either force about a point in the line of action of the other is positive or negative (see Analytic Statics, Part 1). In both Fig. 32 and Fig. 33, the moment of Fi about any point in the line of action of F, t or of F, about any point in the line of action of F lt is negative, and the moment of the couple is considered negative. The motion that the couple (F 1} ^,F a ), Fig 32, tends to produce is evidently the reverse in direction of the motion of the hands of a watch with its face placed upwards and its center at H,. or H t . This motion is said to be counter-clock- wise, or left-handed Motion similar to that of the hands of a watch is said to be clockwise, or right-handed. Clockwise motion is said to have a right-handed direction; counter- clockwise motion, a left-handed direction. It will be noticed that the sign of a couple is positive when the couple tends to produce clockwise motion; other- wise, however, the couple is negative. This distinction is of value only when couples are to be combined by algebraic addition. The direction and sign of a couple can be very readily determined by imagining the arm of the couple to be a line rotating about its center so that each extremity follows the direction of the force applied at it. Thus, in Fig. 32, if we conceive Hi ff* to begin to rotate about its center so that H will follow the direction of F l} and H a the direction of F a , it will be seen that the rotation will be counter-clockwise. The couple is, therefore, a left-handed couple, and its sign is negative. 32 ANALYTIC STATICS 29 EQUIVALENCE AND EQUILIBRIUM OF COAXIAL COUPLES 45. Equilibrant and Equivalent Couples. Since a couple cannot be replaced by a single force, it follows that no single force can balance a couple In order to balance a couple, another couple must be opposed to it. Either couple is called the equilibi-ant of the other. Two couples are equivalent when they can each be balanced by one and the same couple that is, when they have the same equihbrant. 46. Equilibrium of Two Coaxial Couples. Two coaxial couples balance each other if they have equal but opposite moments that is, if their moments are numerically equal but have opposite directions (or signs} DEMONSTRATION The demousttation of this principle is given below Although it is not essential that it be learned or even read, it affords a very useful and interesting exercise in the composition and resolution of forces It will be necessary to distinguish three cases, as follows Case I. When the two couples ate in the same plane and the fotces of one are parallel to the forces of the other Let the couples be (F lt p t F s ) and (F,',p', F a '), Pig. 34 It Is assumed that, with the usual notation as to signs, F^p = F^f p', or F,.p + Fj.'p' = On the other hand, the algebraic sum of the four forces Fi, F*, /V, FJ is equal to F i zero, or 2F = Therefore, the ji A a four forces, or the two couples, a satisfy the two necessary con- jp * o tf , ditions of equilibrium of parallel 3 ^--^.7 j,. -,Q| forces (Analytic Statics, Part 1). and so form a balanced system. \\ f uy It follows from this that a 1 ' couple may be replaced by an- other couple in the same plane, IG> M If the two couples have the same moment and their forces are all parallel, and that the effect of a couple is not altered by moving the couple In Its plane parallel to itself. Case 11. When the two couples are in the same plane and the lines of action of the forces of one intersect the hnes of action of the forces of the other. 29 ANALYTIC STATICS Let the couples be (F lt p, F,) and (F^p 1 , F,') , Fig 36. According to the preceding demonstration, the couple (F^f, F,') may be replaced by another couple with one of its forces acting through 0,, provided that the direction of the forces is not changed, and, moreover, the magnitude of the forces may be changed, provided that the arm is so changed as to keep the moment constant Thus, if Pi is made parallel to FJ and F t ', and equal to F lt and if, at the end of the arm t 0," = p, the force P t is applied, equal in magnitude to />,, but opposite in direction, then the two couples (F l ',p l ,F, 1 ) and (/* <9i0,", A) will be equivalent, for, by hypothesis, FJ & is numerically equal to F t p t and therefore to P t X 0i O," By trans- ferring the point of application of P, to the intersection / of the lines A "*- A. / X I/ A f i o a ^ i yi FIG 86 of action of P t and F,, and the point of application of F t to the same point, the two couples (F lt p, F m ] and (FJ.f, F,') are replaced by the forces P t and F t acting at 1( and P, and F, acting at 7 Since ^ and PI are equal, their resultant must act along the bisector of the angle />, d F lt which evidently is the same as the bisector of O, O-, O," Similarly, the resultant of P, and F, must act along the bisector of O,fO t , aud, as P a and F t are equal and parallel, respectively, to PI and FI, the resultants of the two pair of forces are numerically equal Now, owing to the equality of O^ O a and t O," t the right triangles (not fully shown) X O a f and t 0,"7 are equal, and the line 70 t (not shown) is the bisector of both O, 0i 0," and 0,70,". The two resultants have, therefore, the same line of action, and, as their algebraic sum is zero, they balance each other. It follows that a couple may be replaced by any other couple acting in the same plane, provided that the two have the same moment 34 ANALYTIC STATICS 29 Case III. When the planes of the couples do not coincide, but are parallel (as they must be, since the couples are supposed to be coaxial) . Let PQ and P' Q, Fig 36, be the planes of the two couples, and let (F^pi.F,) be the couple acting in the plane PQ Since the moment of the couple in the plane P' Q is F t p lt that couple may be replaced by another couple (F l ' l pS, F a ') having the same force and arm as (F lt p lt F a ), and the lines of action of whose forces are parallel to the lines of action of F l and F 3 This follows from the two cases pre- viously considered, for the couple (/V.A', FJ) is equivalent to any other couple acting m the plane P' Q and having the same moment. Draw 0, OJ and O OJ, meeting at / We may now compound F l Q Fit, 30 with FJ, and F t with FJ. Since d O, is equal and parallel to OJ OJ, the point /is the middle point of O l OJ and O, OS The resultant of jFi and FJ Is R = F, + F,' = 2 F,, acting through 7 The resultant of F, and FJ is ' = F t + FJ = 2 F, = 2 F lt acting through /. As these two resultants act In opposite directions along the same line (for they are both parallel to the forces of the couples) and have the same magnitude, they balance each other. Therefore, the two couples to which the two resultants are equivalent balance each other. 47. It follows from the preceding principle that the only couple are its moment and its axis. ver, that the axis is not any ion. Any line perpendicular 29 ANALYTIC STATICS 35 to the plane of a couple may be taken as its axis; and, con- versely, if the axis of a couple is given, the couple may be supposed to act in any plane perpendicular to that axis. The force and the arm of the couple may be changed atpleasuie, provided that their product, which is the moment of the couple, remains unchanged. 48. Resultant and Etinlllbrant of Any Number of Coaxial Couples. 77/6' resultant of any numhet of coaxial confilcs is a single couple having the same axis as the component couples and whose moment is the algebraic sum of the moments of the component couples. This principle is a consequence of the one stated in the last article, as can be shown by a process of mathematical reasoning that it is not necessary to give here. In general, let M^ M^ M a , etc. be the moments of any number of coaxial couples, M r the moment of their result- ant, and M q the moment of their equilibrant Then, Mr = 2M> and M t = - IM Also, if several coaxial couples are in equilibrium, the following equation must obtain: Mr = SM = EXAMPLES FOB PRACTICE 1 Find the resultant of the following couples, in which the sign before each parenthesis indicates the direction of the couple (10 Ib , 2 ft , 10 Ib ), - (7 Ib , 6 ft , 7 lb.), (25 lb., 12 ft , 25 lb.); - (8 Ib , 40 ft., 8 lb ) Aus M r = - 42 ft -lb. 2 (fl) Find the lever arm/ of a couple with a foice of 100 pounds that will balance the following couples: (30 lb , G f t , 30 lb.), (20 lb , 6ft.,201b.), - (125 lb ,10ft , 1251b ); (70 lb ,4 ft , 70 lb ), - (80 Ib., 3 ft., 80 lb.). () What is the sign of the balancing couple? Ana /()/ = 9 3 Ana \ (b) Couple positive 3. One of the forces of a couple acts through a certain point O, and is equal to 300 pounds, the moment of the couple is 2,574 foot- pounds. How far from O is the line of action of the other force? Ans R 58 ft. 4. The moment of a couple is 1,500 foot-pounds. Express it (a) as a couple having a force of 75 pounds; () as a couple having an arm of 12 feet. ._ f (a) (75 lb , 20 f t , 75 Jh ^ Ans V v ;, V I25 lb, 12 it., us& It./ 36 ANALYTIC STATICS 29 EQUIVALENCE AND EQUILIBRIUM OF COPLANAR NON-CONCURRENT FORCES 49. To Make the Line of Action of a Force Pass Through a Given Point. Let F, Fig. 37, be a force act- ing on a body, and O, a point in the body or rigidly connected with the body. Let the perpendicular distance OP of O from the line of action of the force * be denoted by x. It is obvious that, if the two forces P and F f , equal Jijl | to each other and to F, are applied jfx at 0, they will have no effect on the I j condition of the body, since they will 2>j- so !o balance each other. The single force I | Fis, therefore, equivalent to the sys- I i tern of forces F, F', F', applied as \-Jf' shown. Now, the two forces F and 1 F' form a couple whose moment is Fx Therefore, the force F acting along LK is equivalent to an equal and parallel force F 1 acting through O, together with a couple whose moment is equal to the moment of F about O. In general, The line of action of a force may be shifted parallel to itself > so that it -will pass through any chosen point, provided that a couple is introduced having a moment equal to the moment of the force about that point. 50. Resultant of a Couple and a Force In the Plane of the Couple. Let the force F, Fig. 38, and a couple (F tt p, F t ), whose plane contains the line of action of the force, act on a rigid body. The couple may be replaced (Art. 46) by another couple (F' t x t F')> whose forces are each numer- ically equal to F, provided that the lever arm x is such that F 1 x 29 ANALYTIC STATICS 37 Also, the couple may be so turned and shifted that one of its forces F 1 will act along the line of action of F, but m an opposite direction, as shown. The two forces F and F 1 balance each other, so that the system is reduced to the single force F 1 equal and parallel to F. Therefore, The resultant of a couple and a single force in the plane of the couple is a single force equal and parallel to the given force, acting along a line whose distance from the line of action of the given force is equal to the moment of the couple divided by the magnitude of the given force. The direction in which the distance x, or O P, should be measured is indicated by the character of the couple. The moment F' x of the resultant force about any point in the line of action of the given force must have the same sign as the moment of the couple. In the figure, the couple is right- handed, or positive The moment of F 1 about O must, there- fore, be rig-ht-handed, which indicates that F 1 is on the left of F. 51. Resultant of Any Number of Coplanar Non- Concurrent Forces. Let F^F^F^F^ Fig. 39, be four non-concurrent forces having their lines of action in one plane, and let O be any point in the plane whose perpen- dicular distances from the lines of action of the forces are, respectively, A, A, A, A- As explained m Art. 49, , I * kjp' S \ ' X^a 1 ' /"* F l m&y be replaced by a V I S ">X / force F-!, equal and par- \ \ allel to F lt acting through /^i O, combined with a couple >. whose moment is /^A- ^ The other forces may be similarly replaced. The whole system is thus replaced by the four concurrent forces, /^ F a ',F t f ,F t ', acting through O and equal and parallel, respectively to F it F,,F tt , /;, together with the four couples Fip^Fipt, F a p a ,F t p t . The resultant R of the concurrent forces is found as explained elsewhere; the resultant M r of the couples is a single couple whose moment is the algebraic 38 ANALYTIC STATICS 29 sum of the moments of the forces about O. Finally, the resultant of R and M r is found as explained in Ait 50. If the given forces are resolved into components in two directions perpendicular to each other, then, with the usual notation ( see Analytic Statics, Part 1 ) , X r =2X=ZFcosff (1) Y r = IY = 2Fsrnff (2) R = Vx r +">7 (3) \r V v" tan H r = ^ = ^J-- (4) A r - sL And also, M r = 2 Fp (5) 52. Conditions of Equilibrium. When the forces are in equilibrium, both R and M f must be zero. The reso- lutes of R must, therefore, be zero, and the algebraic condi- tions of equilibrium are 2X = 2Fcostf = Q (1) 2Y = 2Fsrnff = Q (2) M r =ZM=!Fp = (3) These conditions may be stated in words thus: 1 The algebraic sum of the rcsolutes of the fat CM in each of any two directions at right angles to each other must be sero. i B 2. The algebraic sum of the \ j / moments of the forces about any \ \ / and every point in their plane NO must be scio I 53. Equilibrium of I Three Forces. If three forces F t , F at F , Fig. 40, are in equilibrium, tmy one of }fi\ them is the eqtiilibrant of the other two, and, therefore, equal and opposite to their I lesultant. Thus, /'I must be FlQl 40 equal and opposite to the re- sultant R of F, and F,. As R passes through the point of 29 ANALYTIC STATICS 39 intersection of the lines of action of F, and F t and ft must balance V?, the line of action of F, must pass through O. In general, // three coplanar foices are in equilibrium, they must be con- curtent (unless they aie parallel), and if the point of nitersection of the lines of action of two of the forces is known, the line of action of the other must pass through that point. EXAMPLE 1 Four forces, F l = 100 pounds, F a = 200 pounds, F 3 = 125 pounds, F t = 150 pounds, Fig 41, act on a horizontal lever O B The inclinations of their lines of action to the horizontal are as shown, and O A-, - 4 feet, O A a = 10 feet, O A = 15 feet, O At. = 20 feet Required the magnitude, direction, and line of action of the resultant A J SOLUTION Using formulas 1 and 2 of Art 51, we have, since F, = 100, F tt - 200, F t = 125, F< = ISO, and H, = 45, H* = 50, 77, = 80, 77* = (50, X r - S F cos H - - F, cos 77 t - F, cos H, + F, cos H 3 + F*. cos ff t - - 100 cos 4f) - 200 cos 50 + 125 cos 80 + 150 cos 60 = - 102 56 Y r 2 F sin H = 7^ sin ff^. F, sm H? F, sm 7/ a + F t sin ff t - 100 sin 45 - 200 sin 50 - 125 sin 80 + 150 sin 60 - - 75.694 By formula 3 of Art 51, A' = VX r + YS = Vl02~5H a + 757694" = 127 47 Ib Ans By formula 4 of Art 51, tan H r = % - Jh.rsn; whence H, = 36 25' 40". Ans. J\f !VM OU Since p, = O A, sin 45 = 4 sin 45 - 2.8284; />, - O A, sin 50 = 10 sin 50 = 7.6604, p, = OA> sin 80 = 15 sm 80 14.772, and 40 ANALYTIC STATICS 29 p* = OAi sin 60 = 20 sin 60 - 17 321, formula 5 of Art 51 gives M r = IFp = - ^i A + F, p, + F, p* - F. p. = - 100 X 2 8284 + 200 X 7 6604 + 125 X 14 772 - 150 X 17 321 = 497 59 ft -lb The value of Y r shows that the force R acts downwards, its line of action must, therefore, be on the right of O, since its moment about this point is positive Hence, Mr 49769 Mr = Rpr, whence p r = -g 127 47 = EXAMPLE 2 Two forces, F, = 2,000 pounds and F, = 800 pounds, Fig 42, act on a horizontal beam 25 feet long resting on two supports PIQ 42 at its extremities O and B The distances O AI and O A t are, respect- ively, 8 and 16 feet. The inclinations of FI and F, are as shown. It is known that the reaction R" at B is vertical Required the magni- tudes of the two reactions R" and R 1 1 and the inclination //' of R 1 to the horizontal SOLUTION Since the moment of R' about O is zero, R" may be found by taking moments about that point anil using formula 3 of Art 62, which gives, Fipi + F* X OA, - R" X OB - 0; that is, 2,000 X 8 sin 30 + 800 X 16 - A 1 " X 2T> 0; , _.. 2,000 X 8 sin 30 + 800 X 16 M01K .. whence, R" OK ~ - = 832 lb. Ans. it) By formula 1 of Ai. 52, IF cos H = 0; that is, R' cos H' - F, cos 30 = 0; whence, R 1 cos H 1 = F, cos 30 - 2,000 cos 30 1,732 (1) Similarly, by formula 2 of Art. 52, 2 F a\n // - 0; that is, Ri sm If' - F, sin 30 - F, + R" - 0; whence, R' sin H 1 = ^ sm 80 + F, - R" 2.000 sin 30 + BOO - 832 - 968 (3) 29 ANALYTIC STATICS Dividing (2) by (1), R< sin H 1 _ . .,, _ 968 X'cosff' ~ taU ** -l7732 ; whence, H' = 29 12', nearly Ana. From (2), ff 1,984 Ib. Ans. APPLICATION'S 54. Mutual Reactions. In Fig. 43 (a) are represented two bars hinged at A,, and A at and to each other at B The bars are acted on by forces F t and F,. Through the joint B t the bar A^. B exerts \ on A,B a force R a f , which is equal and opposite to the force Rt f exerted by A, B on A* B. This force RJ R,' is the mutual reaction between the two bars at the joint B. How it is determined will be explained presently. Since the system formed by the two bars is m equilibrium under the action of the external forces F lt F t) R lt R tt these forces form a balanced sys- tem to which the general equations of equilibrium can be applied. From these Pl i 4S equations, the reactions Ri. and R, can be found when all other conditions, such as distances, etc., are known. The mutual reaction at B is determined by applying the principle of separate equilibrium (see Analytical Statics, 42 ANALYTIC STATICS 29 Part 1). The part A,B may be removed, and A^B treated as a free or separate body, provided that a foice is introduced at B equal to the force exerted by A. B on A* B t that is, a force equal to RJ. This condition is represented in Fig '43 (b), where A,B is shown as a free body acted on by the external forces R i} -ft, and RJ. By applying to the system constituted by these forces the general equations of equilibrium, RJ may be determined. 55. Method of Sections. The free-body principle finds a very useful application in the determination of stresses in framed structures by the method or sections, illustrated in Fig. 44. The truss A, B* B* A, rests on piers A, and A n and is loaded at the joints d, C, and C 91 as shown. TT, W PIG. 44 Let the reactions R* and R, first be determined. Consider- ing the truss as a whole, the external forces acting on it are the weights W lt W, and W, and the reactions JR l and R tt These forces form a balanced system, and, therefore, the algebraic sum of their moments about any point mubt be zero (Art. 52). Taking moments about A^ in order to eliminate R lt W. XA 1 C*+ WX A, C+ W t XAtCt-R.X A,A t 0, from which R, can be found. To find ^?i, moments may be taken about A a ; or, knowing ./?, R l may be found from the equation J 1 Y = 0, which in this case gives Ri + R> - W> - W* - W The stresses in the members may be found by the method explained in Analytic Statics, Part 1, proceeding joint by joint, 29 ANALYTIC STATICS 43 beginning either atA^ or at/ a , since now R* and R* are known. Or the method of sections, referred to above, may be used, as follows: Let it be required to find the stresses in the member CC a . Imagine the truss to be cut in two by a plane PQ inter- secting CC a and the members CB, and B^B*, which meet at JS 3 . The part of the truss at the right of PQ may be supposed to be removed by introducing at the points of separa- tion E, F, ff, forces 5, T, U, equal to the actions of B* E on EB^BiF on FC> and C t H on H C, respectively, which are the measures of the stresses m the three members inter- sected by the plane The part A^B^EFH may now be treated as a free body acted on by the external forces R^ W lt W, S, T, and 7, of which only the last three are unknown. If moments are taken about B t (the point of intersection of 6" and T), the forces S and T will be eliminated, and an equation will be obtained in which the only unknown quantity will be U. The same result would have been obtained if the mem- bers CC at BI G, and B t A n had been cut by the plane PQ. To find S, moments are taken about C. In every case, several members should be cut, of which all but one meet at a joint, and this joint should be, taken as the origin of moments. If, however, the stresses in some of the mem- bers cut are known, it is immaterial whether they meet at a joint or not: the only thing necessary is that, of the mem- bers whose stresses are unknown, all but one should be concurrent. ILT3D8 M 44 ANALYTIC STATICS 29 FRICTION SLIDING FRICTION DEFINITIONS AND GENERAL PRINCIPLES 66. Definition of Sliding Friction. Let a block CDE, Fig. 45, rest on a horizontal surface AB capable of resisting or balancing the weight W of the block. The block will then be in equilibrium under the action of the weight W and the reaction R, the latter being equal and opposite to W. If, now, a horizontal force F( which, for convenience, will be 1 supposed to act along ir ^ a line passing through the c. g of the block) is applied, and it is assumed that no other force than W, R, and F acts on the block, the latter being under the action of an un- Fl 45 balanced force, will move in the direction of F with an acceleration equal to g (see Fundamental Principles of Mechanics}. So long as Vv there is no other force acting on the block, motion will ensue, however small /'may be. Experience, however, shows that a small force, whether horizontal or not, often produces no effect on a body resting on a surface, and sometimes a very great force is required before the equilibrium of the body is disturbed. Thus, to drag a trunk or a box over the floor may require the efforts of several strong men. We also know that the resistance is 29 ANALYTIC STATICS 46 greater the rougher the surfaces in contact: a rough box Is not so easily dragged over the sidewalk as a polished stone block over a smooth wooden floor. There is, then, a force brought into action whenever there is a tendency of a body to slide on another. This force, whose effect is to prevent, or which tends to prevent, motion, is called sliding- friction, or simply friction. It is obviously caused by the roughness of the surfaces m con- tact. No matter how smooth a surface may appear, it always has small projections or elevations separated by small depressions. When the surfaces of two bodies are m contact, the projections of one surface go into the hollows of the other, the two surfaces thus become more or less interlocked and cannot slide freely on each other. In order to cause sliding, a force is necessary, whose magnitude depends on the roughness of the two surfaces. As this roughness varies with different bodies, it may be anticipated (and this is known from experience to be the case) that friction must be a very variable force, depending both on the nature and on the conditions of the bodies m contact. 57. Limiting Equilibrium. Given the block CDE, Fig. 45, resting on the surface A B, the force F may either move it or leave its equilibrium undisturbed. In the former case, F must be greater than the friction; in the latter case, .Fmust be less than the friction Let F m be the greatest force that can be. applied to the block without moving it. Then, a force equal and opposite to F m will represent the maximum friction that can exist between the block and the surface AB, any force greater than F m will cause motion, and any force less than F m will be balanced by the friction. But it is not to be supposed that the friction is constantly equal to F m : so long as the applied force is less than F nn the friction is just equal and opposite to the applied force; the friction grows with the applied force up to the value F m , beyond which equilibrium ceases to exist and the body moves under the action of the difference between the applied force and the maximum friction F m . 46 ANALYTIC STATICS $29 When the block is acted on by a force equal to /v l} it is said to be in a condition of limiting equilibrium, or on the point of moving, for the least mciease in the applied force is sufficient to produce motion In (Ins <MSU, there exists between the two bodies A />' and C HH tin 1 greatest possible friction that under the given cucuinstancus can exist between them. This maximum force of fiietion is called limiting friction, and will heieaftcr be designated by POT. Numerically, P m = J ? m . 58. Passive Forces. Friction, like many other resist- ances, is a passive force that is, a force preventing motion, but not producing it. The reason foi this is that friction is brought into action by the application of other forces to which it is opposed; and, as it can never exceed those forces, it can never pioduce motion in the direction of its own line of action. The .same is true of this reactions of supports. A pier may be capable of exerting n reaction of 1,000 tons; but, if a stone weighing 1 pound is placed on it, the pier will exert on the stone an upward pressure of only 1 pound, or just enough to balance the weight of the stone. Forces producing, or tending to produce, motion, are called active forces. Passive forces are balancing forces and never acquire greater magnitudes than the active forces they oppose. Usually, as in the ca.se of friction and reactions, n passive force cannot exceed a certain limit; but between Kero and that limit it can have any value, and so long as the active force to which the passive force is opposed does not exceed that limit there will be equilibrium. 59. Shifting of tlio Ijlno of Action of the tteaotlou. The force of friction is a tangential force, by which is meant a force acting along the surface of contact of the two bodies between which it is exerted, In Fig, 46, the force of friction is not directly opposed to /; but acts along the surface CE, in the direction EC, How, then, can this force balance F, not being in line with it? 29 ANALYTIC STATICS 47 The reason is that, on the application of F, the reaction R is shifted so that it no longer passes through the point where the vertical through the c g. of the body meets the sup" porting surface. This is illustrated in Fig. 46. The line of action of the applied force F meets the vertical line through G at 0. The resultant of W and F, found in the usual manner, is F r , whose line of action meets AB at O 1 . Transferring F r to O 1 (where, for convenience, it is repre- sented by Fr'), and again resolving it into its components F 1 F and W = W t it is seen that, in order that there may be equilibrium, the reaction R must be equal and PIG. 40 opposite to F r ', and its components Y and P must be, respectively, equal and opposite to W'(= W} &ndF / (=s F). The vertical component Y is the resistance of A B to direct or normal pressure; and at present it will be assumed that the surfaces in contact are capable of offering this resistance, whatever the value of Y or W may be. The horizontal component P is the force of friction acting along the surface of contact. In the case illustrated in the figure, F is sup- posed less than F m that is, less than the maximum force that can be opposed by the friction. Therefore, P, which can take any value between and F m , will, in this case, be just equal to F 1 , or F, and balance the latter force. 48 ANALYTIC STATICS 29 It thus appears that the effect of the friction is to shift the point of application of the reaction from K to O', and to change the line of action of that reaction from the vertical direction to the direction O 1 L. The body CD E may be considered as being acted on by the two equal and opposite forces R and F r , or by the two couples - (W,KO f , Y) and (F, OK, P). Let the student show that the moments of these two couples are numerically equal. 60. Case In Wnlch F Is Greater Than /*,. We shall now consider the case in which F is greater than F m , or P m In Fig. 47, the block CDE is acted on by its own Fin 47 weight W and the horizontal force F. As before, F r ' F r is the resultant of W and F, transferred to the point (V on the surface of contact; its components are F 1 = F and W W< As, m this case, /MS greater than F mt the force of friction, which has its maximum value P, H = F m , is not sufficient to balance F 1 , and there will be sliding of the block on A B. The reaction R of the latter surface is the resultant of Y = W = H^and P m . This reaction is evidently less than F r ', and its line of action makes with the normal O M to AB an angle Z less than the angle / made by the line of action of F r with that normal, 29 ANALYTIC STATICS 49 If, Y (that is, W or W) remaining constant, F is made less than P m , say equal to O' S, the resultant will be O 1 T, and the reaction will be O 1 U, whose components are Y = W and MU equal to the friction />, which in this case, will be equal to OS. The conditions of equilibrium may, therefore, be stated by saying; either that /"must not be greater than P m , or that / must not be greater than Z. ANGLE AND COEFFICIENT OF FRICTION 61. Maximum Resistance. The results of the fore- going discussion may now be generalized. Let N, Fig. 48, be the normal force between two bodies whose surface of con- tact is A B. The force N is the normal component of the resultant force acting on CD E (or on the other body), when that resultant, after its point of application has been trans- "jr FIG. 48 ferred to the intersection of its line of action with the surface of contact, is resolved into two components, one parallel, and one normal, to that surface. Given, besides the normal component N (usually called the normal pressure), all other conditions such as the nature of the two bodies, the extent of their surface of contact, etc, the maximum 50 ANALYTIC STATICS 29 friction P m that can exist between them is determined by actual experiment Once P m is known, the maximum reac- tion R, which we shall call the maximum resistance that can exist between the two surfaces, is The angle made by the line of action of the reaction with the normal OS to the surface of contact is given by the, familiar expression, Li S fm -fft If, the normal pressure and all other conditions remaining constant, a force F greater than R is applied to CDE (in which case N is the normal component of F) , it is obvious that there cannot be equilibrium, since the component of F parallel to A B, which component is equal to M S, is greater than P m . If, on the contrary, F is less than R t or equal to R that is, if its line of action falls within the angle Z or coincides with L equilibrium will obtain. 62. Angle of Friction: Condition of Equilibrium. The angle Z, Fig 48, is called the angle of friction, and may be defined as the angle between the line of action of the maximum resistance and the normal to the surface of contact. The condition of equilibrium explained in the preceding articles may be thus stated: In order that there may not be sliding between two bodies in contact, it is necessary and sufficient that the resultant of the applied forces (as F r , Fig. 47} shall not make wtth the normal to the surface of contact an angle greater than the angle of friction. 63. Coefficient of Friction. Until recently, it was thought that, for any two given substances, the maximum friction P m was directly proportional to the normal pres- sure .Wand independent of all other circumstances. Accord- ing to this view, if P m ', Pa/'t /V", etc. are the maximum frictions corresponding to the normal pressures N' % N", N" f , pi p tf p tff " etc., then ^ = = ; = ^. The common value of these ratios will be denoted by c. 29 ANALYTIC STATICS 61 Having determined the ratio c of the friction to the nor- mal pressure for any particular case, the friction in any other D case could be at once found from the relation = c, N whence, P m = c N. Also, since tan Z = ^2 = c N it would follow that the angle of friction was constant for every two substances sliding on each other. That there is generally some dependence of the force of friction on the pressure is a familiar fact. Thus, referring again to Fig. 45, daily experience shows that, if a pressure Q is applied to the block, the effort required to drag the block will increase as the pressure Q increases. The relation between pressure and friction, however, is not always so simple as stated above. The law that friction is proportional to pressure is approxi- mately true only m some cases, or under certain conditions. What these conditions are, and how friction varies under different conditions, are problems to be solved by direct experiment. Although the ratio of P m to N is usually variable, it is customary and convenient to express P m as a fraction of the normal pressure, and write, P m = fN (1) The factor /, or the ratio of the maximum friction to the normal pressure, is called the coefficient of fi-Iction, or friction factor. It has been determined experimentally for various substances under various circumstances; its values have been tabulated, and the approximate laws of its variations stated. Here, the theory only of friction will be considered. In order, however, that a general idea of the values of / may be obtained, it will be remarked that, in the majority of cases coming within the practice of the mechanical engineer, / is usually much less than .5. Thus, for unlubri- cated metals sliding on one another, the average value of / is about .18; for dry and smooth wood sliding on the same, / averages about .38; and for wood sliding on metal, both 52 ANALYTIC STATICS dry and smooth, about A. Tn civil-cntfinerring wink, how- ever, values of /often oceui that e\eeeil ..", ,is in the i ,IHI nf brick sliding on hiick or on stone, masoniy on hiKk\\otk, civ. Since tan X ~ / "" 1 , which is the same .is the v.ihut *f / from foimuhi 1, \vo may write tan /. ~- f (2) The angle of friction m.iy, therefore, he defined .is .in angle whose trigoiuuneti tu tangent is uinuil ti the encHu ieut of friction, it being understood th.it one of the sides if tins angle is the normal to the surf.iee of contuet nf llic tw> bodies whose fiiction is considered. 64. titatlo Frletlon and Kliiftlc l^rlctloiu In to set a hody in motion over another hody, u fojve is neecs- sary whose component parallel to tlic snrf.iee of roni;<el i^ greater than P m . Once the body is in motion, it seem** th.it it should continue in motion, by virtue of its inert ut, if n force just equal to />, were constantly uiiplied to if; fur this force would ho sufficient to overcome the frietion. KXJHTI- ence, however, shows that the force nceessary t Ktiut n body sliding on another is almost always greater th;ui the force necessary to keep it sliding, once motion has Itegmi. In other words, the resistance of friction in greater when motion is to bo produced, than when it iff to IMJ nminlaiitct). In the former case, tjie friction IH culled fHetloii of n-nt, or static fi'totlon; in the latter case, fi'teflon of auttflon, clyiminle ri-It-tlon, or Idnettc frietion. In cither ease, the maximum frietion will bo here designated by A,, niul the coefficient of friction by /, it being undorwtood that. IM a rale, P m and /have different values for the two 65. In practice, it is necessary to bear in inlmt what the function of friction is when any particular problem in tp lie solved. In designing a nmcliine, where niotitm han to 1 produced and maintained, sufficient force should JMJ allowed to overcome the friction of rest, an othnrwive the machine could not be started. But, in calculating the c<rriolt*m<y (A term to be defined elsewhere) of the machine while ia 29 ANALYTIC STATICS 53 motion, the friction of motion should be used. In providing for the equilibrium of a structure, however, where friction is a favorable force, the friction of motion should be used; for, although under ordinary circumstances the structure, being in a state of rest, will offer a frictional resistance equal to the maximum friction of rest, the least shock is often suf- ficient to produce a disturbance of equilibrium; the structure is, so to speak, started, and then the only resistance prevent- ing it from continuing to move will be the friction of motion. This is especially the case m structures subjected to shocks, such as bridges and engine foundations. FIG. 49 EXAMPLE A block CD E, Fig 49, whose weight is W, rests on a. horizontal surface A B The coefficient of friction between the block and the surface is /. A force F, acting in a. vertical plane containing the c g of the block, IB applied at an angle H to the horizontal. Required the magnitude of F, that the block may be on the point of sliding along A JB. SOLUTION All problems similar to this may be solved in two manners, and the result found directly m terms of either / or Z, Of 64 ANALYTIC STATICS 29 course, one result can be transformed algebraically into the other from the relation tan Z = f 1 Let O be the intersection of the line of action of F with the vertical through the c g of the block The latter is held in equilibrium by the forces F, W, and the total reaction J? of A B t whose line of action O>O must pass through O (Art 53) As explained in Art 62, this, line of action must make with the normal to A 2?, or with the vertical, an angle equal to Z Therefore, F r being equal to and colllnear with R t angle OJT= WOJ = Z The triangle O JT gives TJ W W - WOJ) snZ W W = sin (90 + H - Z) sm Z = cos (H - Z) sin Z (1) 2 The line of action of the resultant F r of the forces F and W meets the surface A B at 0'. Let F and JP be transferred to this point, as shown. Resolving F into its horizontal and vertical com- ponents F cos H and F sin H, the resultant normal pressure acting at N= W-FsinH Therefore, the resistance of friction is: P m = Nf = (W-FtinH)f As this force must balance the horizontal component of F, there results- ( W - F sin H )f=F cos H\ W f whence, F = __ , , rr (2) 1 cos /f+ /sin If ^ f To reduce equation (2) to equation (1), we have Wf ^ w tan Z cos ff+fsinH cos /f + tan Z sm ^T sin Z *,* cosZ _, sin Z smZ cos(H-Z) EXAMPLES FOR PRACTICE 1. Find the angle of friction, to the nearest minute, corresponding to each of the following coefficients of friction: (a) /=*.16; (d) f** .25; (c) f - .50, (d) f = .65. Ans. 8 32' 2T 14 2' 2T 26 84- Z - 33 1' 29 ANALYTIC STATICS 55 2. Find the coefficient of friction corresponding to each of the fol- lowing angles of friction (a) Z => 12 15', (6) Z = 30, (c) Z = 8 35'; (d) Z = 3 17'. f (a) f = .217 A O J(*) f = 577 Ans lW /= .151 [(d) /= 057 3 A block of marble weighing 100 pounds is kept sliding with uniform velocity on a horizontal pine floor by a force inclined to the horizontal at an angle H = 30. If the coefficient of kinetic friction between marble and pine is 45, what must the magnitude of the force be (a) if H is an angle of depression? (6} if H is an angle of eleva- tion? & IW F= 70221b. Ans \(b) F= 41251b. 4. A force of 15 pounds, inclined to the horizontal at an angle of elevation of 30, is just enough to keep a block of cast iron, weighing 100 pounds, sliding uniformly on a horizontal cast-iron plate Find the coefficient and the angle of friction . f / = 14 Ans \Z = 7 68' 5. Taking the angle of friction of rest for brick sliding on brick as 35 30' , with what normal pressure N must a brick weighing 6 pounds be pressed against a vertical brick wall that the brick may not slide down? (Use only two decimal places for /.) Ans N = 8 45 Ib. RESISTANCE TO ROLLING 66. Cause of Resistance to Rolling. Let a homo- geneous cylinder A D, Fig. 50, of radius r, rest on a hori- zontal surface X' X. If a horizontal force F is gradually applied to the cylinder along any line NL perpendicular to the axis of the cylinder, it will be found that no motion can be produced before the force F exceeds a certain limit. Now, the line of action of the weight W of the cylinder meets the supporting surface at A, directly under the center O. Did the cylinder touch the surface X' X only at A, it is obvious that any horizontal force, however small, would cause motion; for, as the resultant of that force and the weight could not be vertical, its line of action could not pass through A, and, therefore, such a resultant could not be balanced by the reaction of the supporting surface. And, since experiment shows that it is possible to apply a horizon- tal force to the cylinder without causing motion, it follows 56 ANALYTIC STATICS 29 that not only the point A, but a part AB of the cylinder must be in contact with X' X; in which case it is easy to understand how the resultant F r of F and W may be counter- acted by the icaction R of the suiface X 1 X. That contact cannot take place at the point it A only is otherwise evident from the fact that all substances are more or less compressible, so that, while the weight of the cylmdei causes a small depression in the supporting sur- face, the reaction of the latter causes a small flattening of the cylindrical surface, as shown in the figure. 67. Coefficient of Boiling: Friction. Suppose that the cylinder is in a state of limiting equilibrium with respect to rolling; that is, that any increase of the force F will cause the cylinder to roll. Let the line of action of the resultant of Wand -Fmeet the surface of contact at Af t at a horizontal distance c from the theoretical point of contact A. Resolving the reaction R into its components F and W, it will be seen that the cylinder is in equilibrium under the action of two couples; namely, (F, Afff, -F), and -(W,NH,-W). Therefore, Fio 60 But, as the deformation of the surfaces of contact is very small, we may write, with sufficient approximation, MH = A N = h; and, therefore, Fh = Wc\ whence F = W- h 29 ANALYTIC STATICS 57 It has been ascertained by experiment that the distance c is independent of the dimensions and weight of the cylinder and depends only on the materials of the two surfaces in contact The particular value of c for any two materials rolling on each other (it is not necessary that one of the surfaces should be a plane, as X 1 X} is called the coefficient of rolling friction for those two materials This' coeffi- cient is not an abstract number, but a length, and its value, of course, depends on the unit of length used. The follow- ing are approximate values of the coefficient c. For elm rolling on oak, c = .032 inch For iron on iron and steel on steel, c .02 inch. It appears from the foregoing formula that, for any two materials, the magnitude of the force F producing limiting equilibrium depends on its lever arm h. What is necessary and sufficient in order to produce limiting equilibrium is that the moment Fh of the applied force should balance the con- stant moment We The resistance to rolling may, there- fore, be said to be expressed by a couple We rather than by a single force. This constant couple, whose moment is obtained by multi- plying the normal pressure acting between the two surfaces by the coefficient of rolling friction , is called a friction couple. 68. To Determine Whether Equilibrium Will Be Broken by Sliding or by Rolling. Referring again to Fig. 50, it must be noticed that rolling about M (practically about A} cannot occur if the cylinder slides before Fh reaches the limit We (here Wdenotes the sum of all the normal forces acting between the two surfaces) If / is the coefficient of sliding friction, ^must not be greater than Wf, or W - must not be greater than Wf\ therefore, - must not be greater /i than /, or c must not be greater than fh. If c is greater than fh, sliding will begin before rolling can take place. If c fh, sliding and rolling will begin simultaneously. 58 ANALYTIC STATICS 29 TUB INCTJNKl) / I / I /V JL Fifi.6i 69. Definitions. An hiulliuMl piano is, as its implies, a plane surface inclined to the hon/.on. In Fitf. ftl, the plane PQRS, making with the horizontal plane /' 7V/.V (or any other hori- zontal plane) an angle ' T, is an inclined plane. 70. Any vertical plane perpendicular to an inclined plane is called a principal plane, and its intersection with the inclined plane is called a lino of Uot'llvlty. A line of declivity may alho be defined as a line lyintf in the inclined plane and perpendicular to the intersection of the ItitU-r plane with any horizontal plane. In Fig, fil, PQ /", . I />' ( ', A' A 1 V t being vertical planes perpendicular to /'j^A'A*. aie principal planes, and the lines Ql\ J1 A } RS are lines of declivity. 71. The angle between an inclined plane tind the horizon- tal is called the aii^lo of the inclined plane, and is the same us the angle that any line of declivity imilcea with the horizontal. Thus, in Fig, 51, the angle of the plane is the common value of the angles QPT, BAC } .flSf/tnade by the lines of declivity QP, BA, RS with the horizontal. 72. In practice, an inclined plane is always the surface of some body, as a plank, an inclined rail, the side of a hill, etc. If, for any special purposes, it is desirable to take into account a definite extent of this surface, as /'? RS in Fig. 61, the line A n (or any other parallel to it and included between PS and QR) is called the length of the inclined plane; JB C is the height, and A C the base. 73. Equilibrium of n Body on an Inclined Plane, Let J t MJ t ', Fig. 62, be a body resting on an inclined plane AB. The view here represented IB a section made by a principal plane through the c. g. of the body. All 29 ANALYTIC STATICS 59 forces are supposed to lie in that plane. The angle of the inclined plane is B A C = H. If only the portion AB were considered, AB would be the length, BC the height, and A C the base. But these dimensions are not needed for the purpose of the present discussion. The line W is a vertical through the c. g. of the body, W is the weight of '\ Fib 62 the body, and Fa. force whose line of action meets W at O. Through O draw YY 1 and XX 1 , the former per- pendicular, the latter parallel, to A B. Let the line of action of F make an angle K with Y Y 1 . Denote the equilibrant of F and W by ?, their resultant by F rt and the angle that Q and F, make with Y Y' by L. The triangle O T U gives UT _ sin f/0 r 0tf sin C>r/ Now, UT= F-,OU= W\*\\\UOT = sin ( + #); and sin 7Y7 sin TOF - sin (180 QOF) sin _ = sin (L + A"). Substituting in the preceding equation, _^ _ sin ( + #). whence, I L T 398-U sn + 60 ANALYTIC STATICS 29 The equihbrant Q of /*" and W is the rctiction A 1 oi the inclined plane, acting through the point / whcic the line ot action of the resultant 7v meets the plane. When the body is in a position of limiting 1 equilibiium with respect to sliding, R makes with the normal J N an angle equal to the angle of faction Z (Ait. <>2). As JN is parallel to )')'', it follows that, in this case, L = xf, and the foiou /*' is the maximum foiec acting; along the line OF that the body can resist without sliding:. It is often said that this is the foico that is just enough to start the body moving ovt i r the plane; but this is not coirect. If the body is aheady moving, this force, being just enough to balance the friction, will keep the body moving with constant velocity. If the tootlv is at rest, the force will simply keep it in a condition of limiting equilib- rium; the body will not move, but the least increase in the force will be sufficient to produce motion. 74. To find the value of /"for which the body is in a condition of limiting equilibrium (or will move with con- stant velocity, once staited), it is necessary to distinguish two cases. CUHO I. The body is on the point of moving up tht plane. In this case, ATmuat be on the right (in the figure) of Y Y' t but it may be acute, right, or obtuse. If, however, A' is acute, it must be greater than //; that is, /''must act on the right of V V\ for, if /* lay on the left of V l\ the re.sult- ant F r would act either in the angle VOX' or in the angle V OX 1 : in the former case, the equilibrant would be some force directed like Q' t and, in order that the plane might furnish this equilibrant by its resistance, the reaction JR 1 should have a normal component acting downwards. This cannot take place, as the plane is not supposed to be capable of exerting any downward reaction. If the resultant fell in the angle VOX 1 , the equilibrant would be a force like Q"; this equilibrant might be furnished by the reaction R" of the plane; but in this case the friction, being: the component 1 of R" parallel to the plane, wouid act upwards, and the body could not be on the point of moving upwards, 29 ANALYTIC STATICS 61 It being, then, understood that, in the case under con- sideration, K must be greater than H, the force j^may be found from the preceding equation by writing Z instead of L\ l (1) r l ^ p _ This value may be expressed in terms of the coefficient of friction / as follows: Sin(Z + ff) _ ^y sin Z cos H + cos Z sm H sm(Z + K} sm Z cos K + cos Z sin K PIG fiB Dividing both terms of the fraction by cos 2", and writing / instead of ^4 ( = tan Z, Art. 63), cos Z / cos K -f sm K Leaving: W, ff, and / unchanged, it is seen from for- mula 1 that, if K is changed, F will have its least value 62 ANALYTIC STATICS 29 when the denominator is the greatest possiblethat is, when sin (Z+K] = 1; whence, Z + K = 90, and 90 - K = Z, or FOX = Z This very important result is expressed by saying that the best angle of t> action up an inclined plane is the angle of fnctton. Case 11. The body is on the point of moving down the plane (Fig. 53). In this case, the friction, which is the component parallel to the plane of the total reaction R, acts upwards. The line of action of R, and therefore of F ft must lie on the right (in the figuie) of the normal JN. By a process of reasoning similar to that employed in the preceding case, the following formulas are obtained for the present condi- tions: F = W !-^L~ f cos *L (4) sin K f cos K 75. Discussion of Formula 3 of Art. 74 Angle of epose. So long as H is greater than Z, formula 3 will give a positive value for F. In this case, F, being the equilibrant of Q and W> must act outside the angle WOQ, that is, K must be greater than, Z. The positive value of F indicates that, if the body is left to itself, it will slide down the plane, and that, therefore, a force is necessary to keep it from sliding. If H= Z, then sin (H - Z) = 0, and, therefore, F=Q. This means that the body, if left to itself, will rest on the plane in a state of limiting equilibrium; no force is necessary to keep the body from sliding; but if the angle of the plane is increased and no force applied, the body will slide. For this reason, the angle of friction is often called the angle of repose, and defined as the greatest angle with the horizon that the surface of contact of the two bodies to whose friction it refers can make without the bodies sliding on each other This relation is made use of in the determination of the 29 ANALYTIC STATICS 63 coefficient of friction. Suppose the plane AB t Fig. 54, to be the tipper surface of a plank hinged at A, and that it is desired to find the coefficient of friction between cast iron and marble. The plank is lined with a plate of cast iron and turned about the hinge until it is nearly horizontal. A block of marble is then laid on the plank, and the latter turned upwards until the block begins to slide. The angle of inclination of the PIG. 64 plank to the horizontal at which this takes place is the angle of friction, and its tangent is the coefficient of friction. When H is less than Z (see Fig 54), the component of W along the plane is not sufficient to overcome the fric- tion. Therefore, in order that the body may be on the point of sliding down, the force F must have a component acting down the plane; that is, K must be on the left of YY' In this case, . / T rr\ (1) or sin (Z + JK) = W -~ c ~^ H ~~- S i5-^. / cos K + sm K (2) EXAMPLE 1 A wooden box 10 feet long, 6 feet wide, and 4 feet deep is used for carrying coal up and down an inclined steel-rail track, the grade of the track being 10 in 100 (which means that the track rises 10 feet for every 100 feet of length, measured horizontally) . The weight of coal will be taken as 54 pounds per cubic foot, and the 64 ANALYTIC STATICS coefficient of kinetic friction between wood and steel as .4. The box being full, required- (a) the magnitude and inclination of the least force that will keep the box moving up the plane with constant veloc- ity; (d) the magnitude of a force parallel to the rails necessary to produce the same effect; (c) the magnitude of a force parallel to the rails that will keep the box moving downwards with constant velocity. (d) If the available force acting upwards, parallel to the track, is 4,000 pounds, to what depth can the box be filled? SOLUTION (a) For the weight W we have, neglecting the weight of the box, W = 10 X X 4 X 54 = 12,960 Ib The angle Z of friction, to the nearest minute, is the angle whose tangeut is 4, that is, Z = 21 48'. For the inclination H of the track we have tan H = Vnr, whence H = 5 43' Foi the least force for which the box will be on the point of mov- ing upwards, or that will keep the box, after the latter has been started, moving upwards with constant velocity, we must have (Art. 74) K = 90 - Z = 68 12'. These values in formula 1 of Art. 74 give f = 12,960 sin (5 43' + 21 48') = 6,988 Ib. Ans. (5) In this case, K = 90, and formula 1 of Art. 74 gives, noticing that sin (90 + Z] = cos Z, FIG. 66 (c) As here H is less than Z, formula 1 of Art. 75 is used, making (see Fig 54) K = Y O X' = 90, and sin (Z+ K] = cos Z. '-^"^.-^-.^U. An, (d) Let x be the height of the coal above the bottom of the box; then, W '= 10 X 6 X .r X 54 Formula 1 of Art 74 gives, noticing that here F= 4,000 and K = 90, W 8ln (21 A nrw 4>00a coa sin 27 81? 29 ANALYTIC STATICS 65 or, substituting the value of ZTjust given, 10X6X^X54 = 4,000 ^ ^ gp! whence, ^ = ^X^|^|~ = 2481ft =2ft 5fm, nearly. Ans. EXAMPLE 2 A piece of rock ABC, Pig 55, lying on the floor of a mine Is kept from sliding down by a prop M N Weight of A B C is 3 tons; inclination of CA to horizontal, 60; inclination of NM, 45; coefficient of friction, .75 Required the pressure on the prop. SOLUTION The pressure on the prop is equal and opposite to the reaction F of the prop The inclination of NM to the horizontal being 45, its inclination VO F to the vertical is likewise 45. Here, H = 60, theiefore, K => YOF = 60 + 45 = 105, sin K = cos 15, cos K = - sin 15 Formula 4 of Ait 74 gives sin 60 -3X cos 60 _ 866 - -f X 5 EXAMPLES FOB PRACTICE 1. What force will keep an iron block weighing 4 tons, placed on an iron plate inclined at 45 to the horizontal, in a condition of limit- ing equilibrium with respect to upward motion: (a) if the force is parallel to the plate? (b) if the force makes with the plate an angle of 30? Take / - 20. NOTE In this case, it is more convenient to use formulas Involving /, rather =i 8394T. = 3.613 T. 2 If, in the preceding example, the force is parallel to the plate and acts upwards, what must its magnitude be, that the block may be on the point of moving downwards? Ans F = 2 263 T. 3. Taking the coefficient of kinetic friction between steel and pine as .16, what is the least force that can keep a steel block having a weight of 2 tons moving with constant velocity up a pine plank inclined at 20 to the horizontal? Ans F - 1,944 Ib 4. A block of marble weighing 1,000 pounds is to be kept moving with constant velocity up an inclined white-pine plank by a force of 600 pounds; what must be the inclination of the plank, assuming that the best angle of traction is used, and that / = .45? Ans, H - 12 SSf KINEMATICS AND KINETICS COMPOSITION AND RESOLUTION OF VELOCITIES 1. Graphic Representation of Velocity. Velocity, like force, is a vector quantity that is, a quantity having both magnitude and direction and, like all vector quantities, can be represented by a straight line, called a vector (see Fundamental Principles of Mechanics] . The vector is drawn parallel to the direction of the velocity represented; its length is made, to any convenient scale, equal to the mag- nitude of that velocity, and the arrowhead on the vector is placed so that it will point in the direction of the motion under consid- eration. In Fig l.let AB be the path of a moving point, and v the velocity the point has when it occupies the position P on its path. The direction of the motion at the instant considered is that of the tangent PTto the curve A B. The velocity v may be represented by the vector OM drawn parallel to P T through any convenient point. If v is expressed in feet per second, and a scale of 5 feet per second to the inch is adopted, the Similarly for any o other scale. length of the vector O M should be \ . 5 eOPYRIOHTRD Y INTERNATIONAL TIX^BOOK COMPANY, ALL RIHT RBHRVBD 30 2 KINEMATICS AND KINETICS 30 2. Parallelogram ol Velocities. Let a body or par- ticle O, Fig. 2, be moving in the direction OX, on a flat surface A B, with a velocity v, relative to that surface, that is, in such a manner that, if the line OX, is fixed on the sur- face A B, the particle O will move in that line describing v, units of length per unit of time. Let P, be the position of the moving particle after the time /. Then, OP, = v>t (1) While the particle has this motion, let the surface A B move in the direction O A' with uniform velocity z/ l( and let A' B' be FIG 2 the position of the surface after the time t, O / being the cor- responding position of 0, and O'X a f , parallel to OX,, the corresponding position of OX t > Then, O0 f = v l t (2) The moving particle will now be at PJ t the distance O'PJ being equal to P a , or v, t. The figure O O' PJ P. is evidently a parallelogram. In the same manner, it can be shown that, after a time /', in which the moving particle has described the distance OQ t along the line OX t) while the surface has moved so 30 KINEMATICS AND KINETICS 3 that O" is the position of the starting point O, the final position Q t ' of the particle will be the end of the diagonal of the parallelogram O O" Q, 1 Q t) in which 0"Q,' = OQ a = Vm f (3) O 0" = Q a Q, f = v, f (4) Dividing (2) by (1) gives v t OP, a PS and dividing (4) by (3) gives 2i = " v, 0QJ Equating these two values of , v, oa = o o" OPJ O"Q,' According to the theory of similar triangles, the last equation shows that O, Q, f , and /V are in the same straight line. It follows, therefore, that, at every instant, the mov- ing particle is on the straight line O P a f or, what is the same thing, that the particle moves in that line. The spaces OP,', OQJ described by the particle in times t and /', respectively, are to each other as O O r is to O O" that is, as z/,/ is to Vt ^, or as t is to t'. Therefore, the motion of the particle along OP,' is a uniform motion. If O O" repre- sents v lt and OQ a represents v, t the diagonal OQJ will evidently represent the space described by the particle in a unit of time that is, the velocity v of the particle along its path O /V . Hence, the following construction: From any point P, draw two vectors PMi and PJbf,, representing, to any convenient scale, the velocities z/i and v,, respectively. Construct a parallelogram PM l MM* on those two vectors. Then will the diagonal PM represent, to the scale adopted, the velocity of the moving particle in its path. 3. It will be observed that the velocity v is determined by the same general method used for finding the resultant of two concurrent forces. With respect to the velocities v t and zr., the velocity v, which is the actual velocity of the KINEMATICS AND KINETICS 30 particle, is called the resultant velocity, and v t and v t are called the components of v in the directions OX t and OX,, respectively. Since the particle has the two velocities z^ and v t at the same time, these velocities are said to be simultaneous . The principle of the pavallelogfram of velocities, which is similar to that of the parallelogram of forces, may be stated as follows: // two simultaneous velocities of a particle are represented in magnitude and direction by two vectors drawn from the same origin, and a parallelogram is constructed on these two vectors, the resultant velocity is tepresented in magnitude and direction by that diagonal of the parallelogram that passes through the common origin of the two vectors, this diagonal being treated a? a vector having tJie same origin 4. The process of finding the resultant of two or more simultaneous velocities is called composition of velocities. As in the case of forces, any velocity may be considered to be the resultant of two velocities in any given directions. Thus, the velocity v t Fig. 3, may be consid- ered as the resultant of the simultaneous velocities v l and v t in the direction OX, and OX,, respectively; or as the resultant of the simultaneous veloci- ties vl and v tt ' in the direction O AV and OX,', respectively, etc. To resolve a PIG 8 . .^ . . velocity into its com- ponents in given directions is to find the values of these components, or to replace the given velocity with these com- ponents, the process whereby this is accomplished is called resolution of velocities. V*. 30 KINEMATICS AND KINETICS 5. Since velocities are combined and resolved in the same manner as forces, all that has been said relating to the composition and resolution of forces applies to the composi- tion and resolution of velocities. Thus, instead of the par- allelogram of velocities, the triangle of velocities is often used. In Fig. 2, for example, the resultant v may be deter- mined by drawing PM^ to represent z/,, and then MM to represent v,, and drawing PM. The magnitude of PMcan be ascertained either graphically (by constructing the par- allelogram or the triangle accurately to scale), or analytically (by applying the principles of trigonometry), as in the case of forces. EXAMPLE 1 A ship is pro- pelled by its screw in a north- east direction at the rate of 15 knots, while the current carries it due south at the rate of 5 knots Find the resultant motion of the ship (A knot is a velocity of 1 nautical mile, or 6,080 feet, per hour ) GRAPHIC SOLUTION Draw the north-and-south line NS, Fig 4 From any point A and to any convenient scale, draw A B = 15, making an angle of 45 with A N. This will represent the velocity of 15 knots toward the northeast From B draw B C due south, that is, parallel to N S, and equal to 5, using the same scale as before Join A C, and mark the arrowhead so that it will be in non-cyclic order with A and B C. The length of A C, measured to the scale used for i and v,, will give the magnitude of the resultant velocity. The direction is determined by measuring the angle NA C with a protractor. Ans. FIG 4 ANALYTIC SOLUTION The tuaugle ABC gives i) = A C = V15" + fi" - 2 X 5 X 15 cos 45 = 12 knots Also, sin M . ain 45 5 sin 45 v 12 The angle M, taken to the nearest minute, may be either 17 8' or 180 - 17 8' = 162 52' As B C, opposite M t is the shortest side of the triangle, the value 17 & must be taken. Then, NA C 6 KINEMATICS AND KINETICS 830 = 45 + 17 ' = (JU H' The ship, therefore, is moving in .1 dm c luui N (i2 K' E, at the rate of li! knots, nearly Ans. EXAMPLB 2 A river 4 miles wide h.is a turiunt velmilj of .'I milts per houi A boat whose paddle wheels can cany it thiough <"> miles per houi in still water is to cross the rivei fiom a point ,/, Kig. f>, so as to land at a point />' dneitly opposite the starting point A, It is leqmied to hud: (a) the ilutv- turn in which the hoal must lu: headed, (ft) the time lequiic-d for a trip across the i ivei KOIUTION () Snue I he veloc- ity of the cm rent and that im- parted to the boat by the paddle wheels are both uniform, the boat must move along the lim; .,-/ // with a uniform velocity v (to hi determined). Let /'be th posi- tion of the boat at any instant, and I* Q the diieetum in whieh it is headed, and let i\ he thevelonty in]]>arted to it in that direotion by the paddle wheel. In addition to ,,the boat has a velocity v t down the stieam, t-qiuil to the vdoeity of the current. The velocity v is the resultant of the. velocities f, and v t , The triangle PNN lt being right-angled at N, gives v = PN = V/ 5 7/; r - //, A/ a V, - r,' i V' - .'! = fi.1110 mi. per lir. :/, :i 1 , Also, sm M - (&) The time required is AR 4_ "" ** 5ilOO "" (5 AIIH. nearly. Ant. EXAMPLES JTOB PKACTICK 1. A point has two aimultaneous velocities, nun of 100 fct't pur second and one of 200 feet per second. The vectors represent ing these two velocities make an angle of 4fi with cueh other. Kind: (a] the resultant velocity v ; (b) its inclination M to iho velocity of 200 feet. (Angles are given to the nearest 10 2. A balloon moves upwards with a velocity of fiO feet par necund, and at the same time the wind carries it la a horizontal direction at the 30 KINEMATICS AND KINETICS 7 rate of 20 feet per second Find (a) the resultant velocity z/, (6) its inclination M to the vertical . f (a) v = &3 852 ft per sec. Ans \ (A) M = 21 48' 1C?' 6. Absolute and Kelatlve Velocity. Velocity, like motion, is always relative, it represents the rate of motion of a body with respect to another, and the same velocity can have different values according to the condition of the objects to which it is referred. Thus, when a locomotive is running, the velocity of the piston with respect to the cylinder in which it moves is the space that the piston describes in the cylinder per unit of time, the velocity of the piston with respect to the ground is equal to the velocity of the piston with respect to the cylinder added to or subtracted from the velocity that, in common with the whole engine, the cylinder has with reference to the ground added, if piston and engine are moving in the same direction; otherwise, subtracted. 7. In nearly all practical questions, it is customary to refer velocities to the surface of the earth. When the velocity of a body is given without any qualification, it is usually understood to be the velocity of the body relative to the ground. This velocity is customarily, although not properly, called absolute velocity; while the term relative velocity is restricted to indicate velocity with respect to objects that are themselves in motion with respect to the ground. Thus, in the example of the preceding article, the relative velocity of the piston is its velocity with respect to the cylinder; while the absolute velocity of the piston is its velocity with respect to the ground. In the case represented m Fig. 2, the velocity of the moving particle, relative to the surface A J3, is v,; the absolute velocity of the particle is v\ the absolute velocity of AB is /. It is obvious that, if a body moves on another with a certain relative velocity, while the latter is in motion, the absolute velocity of the former body is the resultant of its relative velocity and the absolute velocity of the other body. KINEMATICS AND KINETICS 30 UNIFORM MOTION IN A CIRCLE ANGUXAR VELOCITY 8. Angular Displacement. Let a point P, Fig. 6, be moving uniformly in a circular path of radius r. The center T of the circle is C, and the velocity of the moving point is v. If the point moves from a position P to a position />, in the time /, the angle H swept over by the radius in passing from the position CP to the position CP,. is called the angular displacement of P with respect to C. Angular displacement may be measured PlQi6 in degrees, but is usually measured in radians. As explained in Geometry, Part 2, the radian measure of the angle H is arc , or r H (in radians) = arc PP - r If the length of the arc PP l is denoted by s t r r Also, s = rff (2) 9. Angular Velocity. Since the velocity v is uniform, s = vt\ that is, rH = vt\ whence tf-H/ (1) r The angular displacement is, then, proportional to the time, and - is evidently the angular displacement per unit of time, since, when / is made equal to 1 in the preceding formula, H becomes equal to -. This displacement per unit 30 KINEMATICS AND KINETICS 9 of time is called the angular velocity of the moving 1 point. The angular velocity may be defined also as the angle described by the moving point in a unit of time. As already stated, it is customary to express angular velocity in radians per unit of time. It is also customary to represent this quantity by the Greek letter to (o-md-ga). Therefore (see formula 1), = ~ (2) 10. Halations Between Angular and Linear Teloc- ity. The velocity of a moving point in the direction of the tangent to its path that is, what in previous articles has been called the velocity of the point is often called linear or tangential velocity, in order to distinguish it from angular velocity. When the word velocity is used without any qualification, linear velocity is meant The velocity in the direction of the path, especially when the latter is a circle, is also called circumferential velocity. From Art. 9, the following' obvious and very important relation between linear and angular velocity is obtained: -* (1) r v = 1 to (2) If two points move in two circles of radii r^ and r, with the same angular velocity to, and linear velocities v and v,, then, v l = r, to, and v, = r, CD; whence, by division, = (3) V, r, that is, the linear velocities of two points moving in different circles with the same angular velocity are directly proportional to the radii of the respective paths. 11. If the linear velocities are the same, and the angular velocities are to,, and to,, we must have v = r w, = r a to,; whence, ^i = TJL to, r* that is, for the same linear velocity \ the angular velocities of two moving points are inversely as the corresponding radii of the paths. I LT 398-13 10 KINEMATICS AND KINETICS 80 12. If, in Fig. 6, the radius CP is imagined to move with P t all the points in CP will evidently have the same angular velocity, since they will all describe the same angle in the same time. If the linear velocity of the point P', situated at unit's distance from the center, is v r , then, v f = 1 X a) = of The angular velocity of P may, therefore, be defined as being the linear velocity of, or the length of the arc described in a unit of time by, a point on the radius CP sit- uated at unit's distance from the center, when the radius is considered as moving with the point P. 13. Angular Velocity In Terms of Number of Revo- lutions Per Unit of Time. In practical engineering prob- lems, it is customary to state the velocity of circular motion in terms of the number of revolutions per unit of time (usually per minute); that is, the number of times that the moving point passes over an entire circumference in a \init of time. If this number is denoted by n, and the radius of the circle by r, the space passed over by the point in a unit of time is 2 n r X n; hence, for the linear velocity of the point, we have v = 2nrn and for its angular velocity (formula 2 of Art. 10), CD = - = 2;rw r It n is the number of revolutions per minute, the veloc- ities v and oj, referred to the second, are: Conversely, if v or CD, referred to the second, is given, the number of revolutions per minute is given by the formula n = ^ = 9.5493 ^ = 9.5493 o> (3) nr r 14. An$rular Velocity of a Rotating Body. When- ever any body, as a wheel, revolves about a fixed axis, every 30 KINEMATICS AND KINETICS 11 point in the body describes a circle whose radius is equal to the distance of the point from the axis. Since all points revolve through the same angle in the same time, they all have the same angular velocity. This common angular velocity is called the angular velocity of the rotating body. EXAMPLB 1 A drum W, Fig. 7, whose radius J? is 1.5 feet is fixed to a revolving shaft 5". The motion of W is transmitted to another shaft s by means of a belt B B passing around the drum W and a drum w fixed on the shaft s. If the velocity of the belt is 10 feet per second and the drum w makes 100 revolutions per minute, it is required to deter- mine the number N of revolutions that S or W makes per minute, the radius r of the drum w, and the angular velocity of the two drums in radians per second. It is assumed that the circumferential velocity of each drum is the same as the velocity of the belt SOLUTION The circumferential velocities of the two drums, in feet per second, are, respectively, g^ and "on- (Art 13). Since each of these velocities is equal to the velocity of the belt, From (1) we get (see also formula 3 of Art. 13), replacing Jf by its value 1.5, N - 9.6493 X ~ = 63 602 A ^ s - 1.5 And from (2) (see the same formula) , r - 9.5493 X -^ - .95493 ft. Ans. For the angular velocity Wj (radians per second) of W, formula 1 of Art. 1O gives Wt *" 3? * n " 6 6667< AnSi And for the angular velocity <a, of w (formula 2 of Art. 13). ta t - 2 - 10.472. Ans. 12 KINEMATICS AND KINETICS 30 EXAJMPLB 2 Two drums D and ZX, Fig 8, of radii r and r 1 feet, are fixed on a shaft 55 A rope M N L Q is wound around the drums in opposite directions, so that when MN rises and winds around D t the FIG 8 portion descends by unwinding from ZX. The rope passes around a pulley P carrying a weight W. It is required to find the motion of the weight W when the shaft 5 5 revolves at the rate of revolutions per minute NOTE This arrangement is known by the various named of Alfferoutlal windlass, differential -wheel and axle, and Chinese wheel and axle, and is used to obtain a slow motion of IV, which, under some conditions, is n mechanical advantage SOLUTION. It is assumed that every turn of the rope around the drums is a circle, and that M N and Q L remain parallel. This is near enough to the actual conditions occurring in practice Let the shaft turn in the direction indicated by the curved arrows. Then N M will rise and Q L will descend For the angular velocity of the shaft and the drums (formula 2 of Art 13), we have ta = 10472 n radians per sec. Let v and v 1 be the velocities otNfif and QL t or of M and Q, respectively, m feet per second. Then (for- mula 2 of Art. 10), v sa r ta = 10472 n r v> s* #-/ at .10472 nr 1 30 KINEMATICS AND KINETICS 13 During any time f, the center O of the pulley, and therefore the weight W, moves through a distance OO lt which is easily determined. The length of the rope between M and Q is reduced by an amount equal to NNi + L L t = 2 OOi But, as during this time a portion v / of the rope has wound about D, and a portion v' / unwound from -Z7, 2 O O l = v t - v' f, whence O Oi = i( v')t Therefore the point O and the load W move uniformly with a velocity z/a, equal to \(v v 1 } Substituting the values of v and j/ found above, Vw = \(r -r')ta = t( 10472 n r - .10472 n r') - 05236 n(r - r 1 ) Ans EXAMPLES FOR PRACTICE 1. A flywheel 6 feet in diameter makes 76 revolutions per minute Find 1 (a) its angular velocity ru, in radians per second, (b) the veloc- ity v of a point on the rim, In feet per second A _ / (a) u> = 7 854 Ana X (d) v - 23.562 2. A point revolves in a circle 8 feet in diameter with an angular velocity of 4 5 .radians per second Find: (a) its linear velocity v, (6) the number n of revolutions it makes per minute. v ~ 18 ft< P er sec> ft - 42972 3. Two drums A and .Z? on parallel shafts are driven by a belt traveling at the rate of 1,000 feet per minute; the radius of A is 5 feet, and B makes 300 revolutions per minute. Find, (a) the number n of revolutions that A makes per minute; (d) the angular velocities io a and <y* of the two drums, m radians per second, (c) the radius r of B \ (a) n = 31.831 Ana J W/ w " 333S Ans ' d H<w, = 31.416 (c} r - .531 ft. CENTRIPETAL AND CENTRIFUGAL FORCE 15. Restatement ol the Law of Inertia. A provi- sional statement of the law of inertia was made in Funda- mental Principles of Mechanics. Now that the theory of the center of gravity has been explained, the law can be expressed in a more complete and definite manner as follows: // a body is under the action of no force t its center of gravity is either at rest or moving uniformly in a straight line, 14 KINEMATICS AND KINETICS 30 16. Centripetal Force. Let a body be- moving in such a manner that its center of gravity G, Fig. 9, travels uni- formly in a circle A B of radius r. As examples of this kind of motion may be mentioned a car moving m a curved track, and the balls of a ball governor. According to the law of inertia, such motion cannot take place unless the body is acted on by unbalanced forces, for, were the body under the action of no force, the path of its center of gravity would be a straight line instead of a circle It can be shown by the use of advanced mathemat- ics that, in order to preserve this motion, the body must be con- stantly acted on by a force P whose line of action is directed toward the center of the circle and passes through the center of gravity of the body. This force, which is exerted on the revolving body, is called centrip- etal force. If the mass of the body is denoted by m, and the linear velocity of its center of gravity by v, the magnitude of the centripetal force P is given by the formula FIG 9 P = m v W Since m = , where W denotes the weight of the body, and g the acceleration of gravity, we have also, P-Z& (1) gr In applying this formula, v should be expressed in feet per second, and r in feet, since g is referred to the foot and second as units. 30 KINEMATICS AND KINETICS 16 If, instead of v t the angular velocity a) is given, we have, from Art. 10, v = rto, v* = r* a)'. Substituting this value in formula 1, and reducing, To express P m terms of the number n of revolutions per inute, we have (Art stituting in formula 2, minute, we have (Art. 13), <a ^ , and, therefore, sub 30 If the revolving body is acted on by any system of forces, their resultant must pass through the center of gravity of the body, be directed toward the center of the circle, and have the magnitude given by any of the preceding formulas. 17. Illustrations of Centripetal Force. A familiar example of centripetal force is afforded by the circular motion of a ball tied to a string and swung around, the other end of the string being held in the hand. The ball constantly tends to move along the tangent, or "fly off on the tangent," and would do so, if it were not kept in the circular path by the pull of the string. It should be noticed that, while the centripetal force is always directed toward the center of the circle, it is not necessarily exerted from that center; in other words, the body to the action of which the centripetal force is due does not need to be at the center. In the case of the ball given above, the hand, which exerts the centripetal force, is placed at the center of the circle. In the case of a train moving on a curved track, on the contrary, the centripetal force is] the resultant of the weight of the car and the pressure of the rails on the wheels; and this force, although directed toward the center of the curve, is exerted at the circumference. 18. Centrifugal Force. By the law of action and reaction, a particle (or body) moving in a circle must exert on the body to whose action the centripetal force is due a reaction equal and opposite to that force. This reaction is 16 KINEMATICS AND KINETICS 30 called centrifugal force. Thus, in the case of the ball considered in the preceding article, the pull exerted on the hand through the string is the centrifugal force exerted by the ball on the hand. Suppose the source of the centripetal force P, Fig. 9, to be at the center C of the circle. The force P that acts on the body G may be represented by the vector G M, while the centrifugal force Q, which is exerted by the body G, may be represented by the equal and opposite vector C N. 19* Caution Against Some Common Misconcep- tions Regarding Centrifugal Force. The following facts must be clearly understood, as this subject is one about which students (and even teachers) often fall into very gross errors: In studying' the motion of a body in a circular Path, the centripetal, not the centrifugal, force must be considered as the only force acting on the body. In Fig. 9, the body G is under the action of the force P, not under the action of the force Q. A body moving in a circular path has no tendency to move along the radius, its tendency being to move along the tangent. In Fig. 9, the body G, if left to itself, would move along G T, not along G C'. A body moving in a circle is at every instant under the action of an unbalanced force (the centripetal force}; as, otherwise, the center of gravity of the body would move in a straight line. Although the centripetal and the centrifugal force are collinear, equal in magnitude and opposite in direction, they do not act on the same body, and cannot, therefore, balance each other. It should be kept in mind that, when it is said that two or more forces balance, the meaning of the state- ment is that they produce no motion on the body on which they all act; and, if several forces act on different independent bodies, it cannot be said that they form either a balanced or an unbalanced system, as, under such circumstances, the 30 KINEMATICS AND KINETICS 17 expressions balanced system and unbalanced system have absolutely no meaning. EXAMPLE A ball J3 t Pig 10, whose weight is W, is suspended by a string of length / from a point O, and made to revolve uniformly in a horizontal circle MN, at " the rate of n revolutions per minute It is required to determine (a) the radius r of the circle M N, (b) the centripetal force P acting on the ball, (c) the tension T m the string NOTE This contrivance is called a conical pendulum, and its operation Is very similar to tbat of the ball erovomor of an engine SOLUTION (a) The two forces acting on the ball are the weight Wai the ball and the tension 7"of the string. According to the principles stated in the preceding articles, the resultant P of these two forces must be directed toward the center C of the circle, and have the magnitude given by formula 3 of Art that is, In the triangle BED, 900 jr P - W tan V, or. since tan V = tan B O C (2) Equating the second members of (1) and (2), Wjjt'v^ = _Wr_ . ~900> " <>lP-~r*' whence V/^ Squaring and solving for r, 18 KINEMATICS AND KINETICS 30 (6) Substituting in equation (2) the values ]ust found for V/ 1 r* and r, D __ 90Q.g- ~ 900 , . ~ * Ans In the triangle BDE, T= ED EXAMPLES FOR PRACTICE 1 A ball weighing 50 pounds revolves uniformly on a smooth horizontal surface, about an axis to which the ball is connected by a steel rod (a) If the rod is 10 feet long and the ball makes 00 revolu- tions per minute, what is the tension in the rod? (&) If the greatest tension that the rod can stand is 2,500 pounds, what is the greatest number of revolutions per minute that the ball can make without breaking the rod? / t a \ 1,381 Ib. Ans I W 121 2 The height h (= CO, Pig 10) of a conical pendulum Is 2 feet. How many revolutions per minute can the pendulum make? (Observe that the number of revolutions is independent of the weight of the pendulum and of the length of the suspending string.) Ans. 38.3 3 If, in example 2, the weight of the pendulum is 10 pounds and the length of the string 16 feet, find (a) the centripetal force P\ (5) the tension T in the string. . / (a) 70 4 Ib Ans.< ) 801b< MOTION OF A TRAIN ON A CURVED TRACK 20. General Theory. In Fig. 11 are represented the rails Ri and R, of a curved track, and the flanges F^ and F t of two directly opposite wheels of a car. The inclination H, or ^i R t N, of the track to the horizontal is shown very much exaggerated for the sake of clearness. The weight W/that comes on the two wheels acts through G, at a height a above 30 KINEMATICS AND KINETICS 19 the track equal to the height of the center of gravity of the car. The difference NR^. in elevation between the outer rail Rt and the inner rail R t is called the superelevation of the outer rail When the car is moving uniformly in the curve, G moves m a horizontal circle whose radius is practically the same as the radius of the center line of the track. According to the theory of centripetal force, the resultant P of all the forces acting on that part of the car whose weight is carried by the two wheels here considered must pass through G and be directed toward the center of the circle in which G moves; that is, P must be horizontal. For convenience all forces will be resolved into components parallel and perpendicular to R^R t . The components of P in these directions are denoted by X and Y t as shown in the tri- angle G DE. The forces acting on the car (or on the part here con- sidered) are W, which is resolved into the component X, and y., and the pressures of the rails. The components of these 20 KINEMA ^ICS AN ) KINETICS 30 pressures perpendiculs r 'to R l R, e re denoted by Y l and K,; the component paralle to or alon^ RI R, is denoted by X'. If, as shown m the figure, X 1 acts toward R a , the pressure is exerted by the rail R! on the flange ft; if X' acts toward R lt the pressure is> exerted by R, on the flange F,. In the gen- eral derivation of the formulas, X' will be treated as acting toward R a ; if, in any particular case, the value of X 1 is found to be negative, this means that X 1 acts in the oppo- site direction, and that it is, therefore, exerted by the lower rail R,. Let v = linear velocity of car, in feet per second, e = superelevation of outer rail, in feet; b = R^R, = distance between centers of rails, m feet; r = radius of center line of track, in feet. According to the principles of statics, we have, since P is the resultant of the forces acting on the car, X' + X = X (1) Y* + y> - Y, = Y (2) From (1), X' = X-X, (3) Now, X = P cos H = cos H (Art. 16) = X gr gr gr b R l R a b These values in (3) give X ' = ~(^F-e*-e} (D b \gr ] To find Y lt moments are taken about R a . Since the moments of X' and Y, about R, are each zero, the algebraic sum of the moments of Y 1 and W must be equal to the moment of /*; that is, Y*b- WXKR, = P^GK (4) The figure gives, MI being horizontal and MS vertical, KR* = JR t cos H = (MR, - MJ) cos H = (i6- MG tan H) cos H = i b cos H- a sin H 30 KINEMATICS AND KINETICS 21 Substituting these values in (4), writing -^-^- for P, and gr solving for Y 1} (2) The value of Y, is found from equation (2) by substi- tuting the expressions for the values of Y, Y^ and Y a . The result is Y, = ** - *- cos H+ iA + a sin tf (3) b L\ "/ \ gr I J 21. Car Moving Without Exerting Lateral Pres- sure. The pressure A'' exerted between the rail and the flange increases with the velocity v, as is evident from formula 1 of the preceding article. As this pressure is very injurious both to the rail and to the wheels, the track is so constructed as to eliminate it, that is, the superelevation e is so made that, for the greatest velocity of trams moving on the track, the pressure X' shall be zero. Putting the second member of formula 1 of the last article equal to zero, whence, squaring and solving for e, b^, (1) Usually, (---) is a very small fraction. If this fraction is \ffr/ neglected, the following approximate formula is obtained: e = b ?*- (2) gr 22. When A'' is zero, the algebraic sum of the moments of Yi and Y, about G is zero, since the moments of both W and P are zero; and, as the lever arms of Ki and Y, are equal, it follows that Y l == F a . Also, since the resultant P of Yi, K B> and W is horizontal, the sum of the vertical compo- nents of yi and X, must be numerically equal to W, and 22 KINEMATICS AND KINETICS 30 the sum of their horizontal components must be equal to 1 If, therefore, a horizontal line A Q is drawn through A, the vector Q G will represent Xi + Y,, and the vector O si will represent the horizontal component of Y l + y a , this component being equal to P It is thus seen that, in this case, the centripetal force P is obtained by resolving the weight W into two components one, GQ, peipendicular to the track, and one, GJS, horizontal: GQ simply bal- ances the pressure Y l + Y, of the rails, while GE is the centripetal force. 23. Car on the Point of Upsetting About the Outer Wheel. If the velocity increases sufficiently, the pressure X' will become so great that the car will upset by turning about the rail R-,.. When the car is on the point of upsetting, the wheel F, is on, the point of being lifted from the rail J? at and there is, consequently, no pressure between the wheel and the rail at J? 3 . The relation that exists between e and v for this condition is obtained by making Y a = in formula 3 of Art. 20, expressing cos // and sm H in terms of e and b, and solving for v or e. The pioe- ess is comparatively complicated, and will not be given here. There is, however, an approximate solution that is sufficiently close for practical purposes. In railroad work, the angle H is always very small, and its cosine can, with- out any appreciable error, be taken equal to 1. Putting the second member of formula 3 of Art. 20, equal to zero, and writing 1 for cos H and 7 for sin H (see Fig, 11), b and . gr \ gr I b d(2a EXAMPLE 1. A curved track is to be constructed for a maximum train velocity of 45 miles per hour. If the radius of the curve is 2,600 feet, and the distance between centers of rails 6 feet, what must 30 KINEMATICS AND .KINETICS 23 be the superelevation of the outer rail, that there may be no lateral pressure? SOLUTION Formula 2 of Art. 21 will be used Here, b 5; r = 2,500, g 32 16, and v = ' = 66 ft per sec. Therefore, 66 a e = 5 X Q4> , A _ Knn = .271 ft = 3i in , nearly Ans . 32 16 X 2,500 (_ EXAMPLE 2 With the track constructed as m example 1, what is the maximum velocity a train can have without upsetting, the height of the center of gravity of each car above the track being 6 feet? SOLUTION The formula m Art. 23 gives .-f *" + **-** persec . = 134 mi. per hr., nearly. Ans. EXAMPLES FOR PRACTICE 1. A curved track of 1,600 feet radius is to be built for a maximum tram velocity of 40 miles per hour, the distance between rails is 5 feet What must the superelevation be, that there may be no lateral pressure? Ans 535 ft 2 The radius of a curved track being 1,500 feet, tne superelevation 3| inches, and the distance between, the centers of rails 5 feet, what is the maximum velocity a tram can Jiav> without exerting any lateral pressure? Ans. 36.17 mi. per hr. WORK AND ENERGY WORK 24, Definition of Work. Let a force F, supposed to remain constant in magnitude and direction, act on a body while the point of application of the force undergoes a certain displacement. This displacement does not neces- sarily have the direction of the force, nor is the force F necessarily the only force acting on the body. Let the pro- jection of the displacement on a line parallel to the direc- tion of the force be denoted by s. Then, the product Fs is called the work of the force F for the given displacement. Thus, in Pig. 12, if the point of application of the force F moves from A to A f , its displacement is A A'\ the projection of this displacement on a line parallel to the line of action 24 KINEMATICS AND KINETICS 30 of the force is AS, A' S being perpendicular to the direc- tion of the force. In this case, then, s = A S, and the t work of the force, while B / the point of application /^X S moves from A to A 1 , *\^ fr, is FX AS. The pro- / jection A S is called the MT component of the dis- f ,--*'* placement A A' parallel to the line of action of the force When, as in Fig. 12, the projection Pl - 12 of A' on A S moves in the same direction as the force, the force is said to do work on the body on which it acts When, as in Fig 13, the pro- jection of A 1 on A S moves in a direction opposite to that of the force, the body is said to do work against the force, or the force is said to do the negative work FX AS on the body. If the line of action of the s ^ force F, Figs. 12 and 13, coincide / """"--^^ with the path A A' of the point / "~"^'7^ of application, then s = A A'. If / /' \ the point of application moves / in the direction of the force, the / work of the force is F X A A 1 } / otherwise, Fx A A'. / > If the line of action of F is / /' perpendicular to A A f , the projec- tion A S reduces to the point A, and F X A S = F X = 0; that is, the work done by the force is zero. This shows that, when ' Fr0tW the point of application of a force moves in a direction per- pendicular to that of the force, the force does no work, 25. Formulas for Work. Denoting the work of the force F, Fig. 12, by U, we have U=FXAS (1) 30 KINEMATICS AND KINETICS 25 If Fis resolved into two components Tand N, the former along A A', and the latter perpendicular to A A', then, T T = /''cos M, and F - - This value in (1) gives cos y// cos M This formula shows that the work of a force can be obtained by multiplying the component of the force in the direction of the displacement of its point of application, by the length of that displacement. 26. Work of the Resultant of Several Forces. When several forces act on a body, the algebraic sum of their works is equal to the woik of their resultant. This is an immediate consequence of the principle stated in the preceding article. Let F he the resultant of several forces, say F^ J 7 ,, F a . Let the body on which these forces act move in any direction through a distance x, and let the correspond- ing work of the forces be U, /i, /, /. Denoting the com- ponents of the forces in the direction of the displacement by X, X lt A' a , X 3 , we have, according to the preceding article, U = Xx, U, = X t x, U* = A',*, U, = A'.*. Therefore, #++/. = (X t + Xn + X.)x According 1 to the principles of statics, Ji 4- X, + X a = X\ therefore, 7 X + /, + U, = Xx = U, as stated^ It should be remembered that the resultant F'\& a force that can replace the forces F^F*, F,, not a force acting simultaneously with them. Since, when seveial forces are in equilibrium, their result- ant is zero, it follows that, if a body is moving under the action of balanced forces (in which case the body must be moving with uniform velocity), the algebraic sum of the works of the forces for any displacement of the body is equal to zero. 27. Effort and Resistances. Usually, forces are applied to bodies in order to produce or to prevent motion. The forces thus applied are called efforts, and those that I L T 398-14 26 KINEMATICS AND KINETICS 30 oppose them are called resistances. Thus, when a train is moved by a locomotive, the pull of the locomotive is the effort; the friction of the wheels and the resistance of the air are the resistances. Generally, when the work done on a body is referred to, the work done by the effort is meant, although, in some cases, its value is determined by calcula- ting the work of the resistance. Thus, if a weight W is raised through a height h, the work of the resistance W is numerically Wh> which, if the body comes to rest at the end of the distance h } is also the work of the effort. 28. Unit of "Work. As work is measured by the prod- uct of a force and a distance, its numerical value depends on the units of force and length employed. The unit of work may be defined as the work done by a force equal to the unit of force acting through a distance equal to the unit of length. If the unit of force is the pound, and the unit of length is the foot, work is expressed in foot-pounds; if the unit of force is the ton, and the unit of length is the inch, work is expressed in inch-tons; etc. 29. Work Done In Raising a System of Bodies. It can be shown by the use of advanced mathematics that the work performed in raising a system of bodies is equal to the aggregate weight of all the bodies multiplied by the height through which their center of gravity is raised. Thus, if stones piled in any manner at the bottom of a tunnel shaft are raised and formed into a pile on the surface, the work per- formed is equal to the combined weight of all the stones multiplied by the vertical distance between the centers of gravity of the two piles. The stones may be raised one or more at a time and deposited on the surface in any manner whatever; the work will always be the same. EXAMPLE 1. The average pull of a locomotive on the train being 7 5 tons, what is the work done by the locomotive in hauling the train 1 mile? SOLUTION. Here F = 7 5 T., J = 1 mi. = 5,280 ft Therefore, V - 7.6 x 6,280 = 39,600 ft. -tons. Ana. 30 KINEMATICS AND KINETICS 27 EXAMI'LB ->. fiiven an inclined plane CD, Pig 14, 500 feet long and inclined to the horizontal at an tingle //of .SO", it is reqtmed to determine the work necessary tn just pull up from C to D a block whose weight W is SiO pounds The coefficient of friction / between the block and the plane is 2fi SOIUTION The re- sistances opposing the motion of the block are the weight W and the friction P The work U of the effort is numeri- cally equal to the sum of the works of those re- sistances Resolving Jf into two components A r and T, the foimer noinial to the plane and the lattei along the plane, \ve have (.see Analytic Statics}, T = Wsln H, N = H r cos //, /' = fN = fW cos //. The work done against I\ in foot-pounds, is /' X CD fiOO / J , and that done against W is 7'X C/) - . r >(H) T. Therefore, U -= f)(K) T + r>(H) /' = fiOO (T + P) = 5(M) ( W sin // + / W cos //) = fi(K) W (sin U + /cos //) = fi()0 X SX) (iln + ,2T> X cos M ) = HH.rtM ft -Ib. Ans. EXAMPI.K U. A cylmdnctil cistern 10 feet in diameter is filled with water to u distance of fiO feet above the bottom. What work must be done in pumping all the water into a tank l> r > feet in diameter, whose bottom is 1(X) foot above the bottom of the cistern? The weight of water is taken ocjual to (12.5 pounds per cubic foot. SorimoN. The- volume F"o the water to be raised iu (* x 4 10 'x>) cu. ft. I'or the wuSght of thiw volume wo have IT V 1(1" W = l r X (JB.fi , X 50 X 02.5 i If y is the height of the water In thu tank after pumping;, r , it X ir> V*> 4 y 4 X * X J ' X nO 4 ^ * x 4 X W 10" X 50 whence * - x 1B . - ff x 1B - ' - lfi Tne rtiHtancu of the center of gravity of the water In the clBtern from the bottom of the cistern is 1'5 ft. The distance of the center y >f gravity of the water in the tauk from the bottom of the tank ia ~. 28 KINEMATICS AND KINETICS '30 *or the distance h between the two centers of gravity we have, therefore, Then (Art 29), U=Wh = 1L ^~ X 60 X 62 5 (?5 + ^^) = 21,135,000 ft -Ib 4 N J X 15 I Ans. NOTE In practical problems, it Is seldom necessary to use moie than five significant figures POWER 30. Work, as defined in Art. 24, has no reference to time. The work performed in raising a given weight through a given height is the same whether the time occupied is a second or an hour. However, when the capacities of diffei- ent agents for doing work are to be compared, time must be considered; and to indicate the amount of work performed in a given time, or the rate of doing work, the term power is used. If an engine does in an hour twice as much work as another engine, it is said to have twice the power of the second engine. 31. Horsepower. Power may be expressed in tenns of any convenient units, as foot-pounds per second, foot- tons per minute, etc. Thus, if an engine woiking uniformly raises 40,000 pounds through a distance of 10 feet in 1 min- ute, the work it performs per minute is 400,000 foot-pounds, and its power may be stated as 400,000 foot-pounds per min- ute. The foot-pound per minute is too small a unit for many purposes, and a larger unit is more generally used by engi- neers. This unit is called the horsepower, and is equal to 33,000 foot-pounds per minute, or 550 foot-pounds per second. The abbreviation for horsepower is H. P. Let F = effort (or resistance), in pounds; s = distance, in feet, that point of application is moved; / = time occupied, in minutes; t, = time occupied, in seconds; ff = horsepower. 30 KINEMATICS AND KINETICS 29 TT PS Fs 8 EXAMPLE 1 What horsepower is required to pull a tram weighing 400 tons at a speed of a mile per minute, if the resistance at this speed is 12 pounds per ton? SOLUTION The force F necessary to overcome the resistance is 400 X 12 = 4,800 Ib s = 6,280 ft Substituting In the foimula, 4,800X5,280 _ H= 33,000 X~r = 768H P ' AnS EXAMPLE 2 A crane hoists a load of 5 tons a height of 22 feet m 20 seconds What horsepower is developed? SOLUTION The resistance F is 6 X T= 10,000 Ib.; 20 sec. = i mm Substituting in the formula, 10,000 X_22 H H ~ 33,000 X * ** H P 10,000 X 22 _ H = ~~- = 2 H p ENERGY 32. Kinetic Energy. Let a body of mass m and weight W be moving 1 with a velocity v a . If a force F is applied to the body in a direction opposite to the direc- tion of motion, that force will bring the body to rest in a distance s, such that F s = a 1 m V* ( 1 ) This formula, which should be committed to memory, is obtained from the following, derived in Fundamental Prin- ciples of Mechanics F^'Jii^-J^l (2) 2s In the present case, v = 0, and, therefore, F = ~~, Fs = - i m v? 2s The negative sign simply indicates that F and v have opposite directions. When only numerical values are con- sidered, the sign is disregarded, and the equation written as in formula 1 . Notice that Fs is the work done by the body against the force in the space j. Whatever the force may be, this work 30 KINEMATICS AND KINETICS 30 is always the same, since it is always equal to i m v a ' It follows, then, that the body can, on account of its velocity and mass, perform an amount of work numerically equal to a" m v,". 33. The capacity that an agent has for performing work is called enei-gy. The energy that a moving body has, on account of its velocity and mass, is called the kinetic energy of the body; and, as ]ust explained, is measured by the product -3 m v a *. Since this quantity is equivalent to work, it is expressed in units of work, such as foot-pounds or foot-tons Denoting the kinetic energy of a moving body by jR" t and replacing m by (see Fundamental Principles of Mechanics] , g formula 1 of Art. 32 may be written, Also, Fs = ^p- (2) EXAMPLE Find the kinetic energy, m foot-tons, of a car weighing 25,000 pounds, and moving at the rate of 30 miles an hour. SOLUTION Here, W = 25,000 Ib = -^ tons. The velocity of the tram, in feet per second, is 30X5,280 , . 60 X 60 ^ ^ V ' Therefore (formula 1), K = \ X |^- X 44* 376 24 ft -tons. Ans. NOTU The velocity has been reduced to feet per necond, because jf In expressed in feet per second We might also find v In feet per hour and reduce jf to feet per hour In all cases, it is necessary to refer v andjr to the same units, 34. Equation of Energy. From formula 2 of Art. 32, we have, leplacing m by , Fs = (z>* zJo") (1) 2.P" This formula gives the work performed by an unbalanced force in changing the velocity of a body from z> to v, or its 30 KINEMATICS AND KINETICS 31 kinetic energy from v a " to v". If the force F is the difference between an effort P and a resistance R, then, (P R} s = (v* v ") *f whence Ps = R s + -^ ( - v, a ) (2) 2.T This formula is called the equation of energy, and may be stated thus: The work done by the effort zs equal to the work done against the resistance plus the increase of kinetic energy. EXAMPLE In hoisting coal from a mine, the load to be hoisted, including cage and car, is 12,000 pounds, the load starts from rest, and when it is 50 feet from the bottom, it is moving with a velocity of 30 feet per second What is the pull in the hoisting rope? SOLUTION. The resistance R is the weight, 12,000 Ib. The effort P is the unknown pull in the rope, v, = 0, v = 30 ft per sec., and s = 50 ft Substituting in the formula, PS = tfj + |(- -.-)_ 12,000 x so + 2 x 32i6 (30> " QI) = 767,910 ft.-lb. 767,910 50 15,358 2 Ib. Ans. NOTE In the solution of this example It is assumed that, when the load has reached a height of 60 feet, it is still being accelerated at the average rate of accelera- tion necessary to fflve it a velocity of 30 feot per second in a space of 60 feet When the velocity becomes uniform, the pull in the rope is that due to the load only, since there la then no Increase in the kinetic energy. 35. Potential Energy. As has been explained, a body may have a capacity for doing 1 work by reason of its velocity. There are, however, other states or conditions than that of motion that give a body a capacity for work: (1) Suppose that a body of weight W is raised through a height h\ in thus raising the body, the work Wh is dons against the force W. If, now, the body is permitted to descend through the same vertical distance h, the weight W becomes the acting force, and the work Wh is done by it. The body in its highest position has the work Wh stored in it, and thus possesses a stock of energy equal also to W h. (2) Suppose that a spring is extended or compressed; work is done against the resistance of the spring to change of lengtH. 32 KINEMATICS AND KINETICS 30 If the spring is released, it will return to its original form, and in so doing can do precisely the amount of woik that was expended on it. Thus, in the extended or compressed state, there is an amount of work stored in the spimg, and the spring possesses energy because of its sti etched 01 compressed condition. Similarly, compressed air possesses energy merely because it is compressed and can do work in returning to its original state. The energy that a body possesses by reason of its position, state, or condition is called potential energy. 36. It is to be remarked that the potential energy that a body has, due to its position, is relative to a certain plane below the body. Thus, if the body weighs W and its height above the ground is h, its potential energy with respect to the ground is Wh With lespect to a plane whose distance below the body is h lt the potential energy is 37. Conservation of Enerjyy. The law of the con- servation of eneigy asserts that energy cannot be destroyed. When energy apparently disappears, it is found that an equal amount appears somewhere, though perhaps m another form. Frequently, potential energy changes to kinetic energy or vice versa. To illustrate, take the case of a steam hammer or a pile driver. The ram is raised to a height h above the pile, and, on being released, strikes the head of the pile and comes to rest. In its highest position, the ram has the potential energy Wh In falling, it attains a velocity v that, just at the instant of striking, has the magnitude ^%ffh, hence, at this instant, the ram has lost all its potential energy, but h'as a kinetic energy -- = IV h. ** The loss of one kind is therefore just balanced by the gain of the other kind. After the blow, the ram comes to rest and has neither potential nor kinetic energy. Apparently there is a loss of energy, but really the energy has for the most part been expended in doing the work of driving the pile a certain distance into the earth. A small part has 30 KINEMATICS AND KINETICS 33 been expended in heating the ram and head of the pile, and another part in producing sound. Heat is a form of energy due to the rapid vibration of the molecules of bodies. In many cases where energy appar- ently disappears, it reappears in the foim of heat Thus, the work clone against frictional resistances appears as heat, which is usually dissipated into the air as fast as produced Conversely, there aie examples of heat energy transformed into other forms. A pound of coal has potential energy, which, when the coal is burned, is liberated and changed into heat energy. The heat applied to water produces steam, and the steam has potential energy. Finally, the steam m giving up its energy does work in a steam engine, and this woik is expended in overcoming the friction of shafts, belts, and machine parts, and ultimately reappears m the form of heat. However, in all these transformations, the total amount of energy the sum of the kinetic and potential energy remains always the same. Such is the law of the conservation of energy. EFFICIENCY 38. In every machine, there is an effort or driving force that produces motion in the machine, and a resistance against which the machine does work. Take, for example, the rais- ing of a load by a crane. The effoit is exerted by the work- man on the crank of the windlass, and the resistance is the weight of the load lifted. In the case of a machine tool for cutting metal, the effort is the pull of the belt on the driving pulley, and the resistance is the resistance of the metal to the cutting tool. In a given time, say 1 minute, a quantity of work is done by the effort and another quantity of work is done against the resistance. Were it not for frictional resist- ances, these two works would be just equal. In actual cases, however, only a part of the woik of the effort is usefully expended in doing work against the resistance. The remain- der is used in overcoming the friction between the various sliding 1 surfaces, journals and bearings, etc. The work thus 34 KINEMATICS AND KINETICS 30 expended appears in the form of heat energy, and serves no useful purpose. The ratio ful work Qr useful work total work of effort total energy supplied the efficiency of the machine. Representing the efficiency by E, the useful work by /, and the total work (= energy supplied) by /, we have, EXAMPLE A water motor receives 150 cubic feet of water per second, with a velocity of 45 feet per second If the maximum power that can be obtained from the motor is 459 H P , what is its efficiency? SOLUTION The weight of the water delivered per second is 150 X 62 5 Ib , the kmet:c energy of which (Art 33) is 150X626X45' .. , , n 2 ft -Ib per sec ( = U) Since 1 H P = 650 ft -Ib per sec , the useful work per second done by the motor is (459 X 550) ft -Ib ( = /) . Therefore, 150 X65X 45' - 8513 ' r ffi 5 per Centl Ans ' HYDROSTATICS DEFINITIONS PROPERTIES OF LIQUIDS 1. Hydromechanics is that branch of mechanics that deals with liquids, their properties, and their applications to engineering. It may be divided into two chief branches: hydrostatics and hydraulics. 2. Hydrostatics treats of liquids in a state of rest. 3. Hydraulics treats of liquids in motion. 4. In the first of these divisions are included such prob- lems as the pressure of water on enclosing vessels and sub- merged surfaces; in the second are included problems relating to the flow of liquids through orifices, in pipes, and m channels. 5. !Llquid Bodies. A liquid body, or simply a liquid, is a body whose molecules change their relative positions easily, being, however, held in such a state of aggregation that, although the body can freely change its shape, it retains a definite and invariable volume, provided the pressure and temperature are not changed. Water and alcohol are exam- ples of liquid bodies. 6. A perfect liquid is a liquid without internal friction; that is, one whose particles can move on one another with absolute freedom. On account of this characteristic property, a perfect liquid offers no resistance to a change of form. 7. A viscous liquid is a liquid that offers resistance to rapid change of form on account of internal friction, or vis- cosity. Tar, molasses, and glycerine are examples of viscous liquids. COPYRIaHTBD BY INTIBNATIONAL TEXTBOOK COMPANY ALU KIOHTB 31 2 HYDROSTATICS 31 All liquids are more or less viscous. For the purposes of hydrostatics, however, water, which is the liquid mainly dealt with in this work, may be treated as a perfect liquid, its vis- cosity at ordinary temperatures being too small to be taken into account. 8. Compressibility. All liquids offer great resistance to change m volume; that is, they can be compressed but little. Under the pressure of one atmosphere (about 14 7 pounds per squaie inch), water is compressed about 20067 of its original volume. Foi engineering purposes, it may be assumed that water is practically incompressible. 9. Density. The density of a homogeneous body is the mass of the body per unit of volume, and may be obtained by dividing the total mass of the body by its volume. If V, W, ;, and D are, respectively, the volume, weight, mass, and density of a body, then, or, since m (see Fundamental Principles of Mechanics), Both from the definition and from this formula, it follows that the density of any one body varies inversely as the volume of the body; in other words, when the mass of the body remains the same, the density is greater the less the volume of the body, or the space that the body occupies. This agrees with the ordinary use of the words density and dense, which are employed to denote the degree of com- pactness, a body being said to be more or less dense accord- ing as its molecules are supposed to be more or less close to one another. 10. Weiglit of "Water. The weight of 1 cubic foot of water varies with the temperature: at 39.2 F., which is the temperature of maximum density, it is 62.425 pounds. For nearly all engineering purposes, 62.5 pounds is used as a 31 HYDROSTATICS 3 convenient and sufficiently approximate value. This value will be used throughout this work Since a. column of water 1 square inch in cross-section and 1 foot high is -j4r cubic foot, its weight is 62 5 -f- 144 = 434 pound. This value is of very frequent application, and should be memorized. LIQUID PRESSURE PASCAL'S LAW AND ITS APPLICATIONS 11. Statement of Pascal's Law. A perfect liquid transmit* pressure equally in all directions This principle is Pusc-aPs lu\v, and follows di- 7 A rectly fiom the definition of a per- "* feet liquid. i The difference between a liquid H and n solid as regards transmis- \ sion of pressure may be illus- trated as follows: In Fig. 1 are shown two cylindrical vessels of the same size. The vessel a is fitted loosely with a wooden block of the same size as the vessel. The vessel b is filled with water, whose depth is the same as the length of the wooden block in a. m * l a Both vessels are fitted with air- FlGl tight pistons P. For convenience, let the weight of the block and that of the water be neglected, and suppose that a force of 100 pounds is applied to each piston. Assume the piston area to be 10 square inches; then the pressure per square inch is 100 -T- 10 <= 10 pounds. In the vessel a, this pressure will be transmitted, undiminished, to the bottom of the vessel; it is easy to see that there will be no pressure on the sides. In the vessel b, an entirely different result is obtained. The pressure on the bottom will be the same as in the other HYDROSTATICS 31 case, that is, 10 pounds per square inch; but, owing to the fact that the molecules of water are perfectly free to move, this pressure of 10 pounds per square inch is transmitted in every direction with the same intensity; that is to say, the pressure at all points, such as c, d, e, f,g,h, due to the external force of 100 pounds, is exactly the same, namely, 10 pounds per square inch. 12. Verification of Pascal's Law. An experimental proof of Pascal's law may be effected with the apparatus shown in Fig. 2 The vessel is filled with water, and is fitted with pistons of different diameters. Let a force be applied to the small piston c , as a result, the water in con- tact with c is subjected to a certain pressure According to Pascal's law, this pressure must be transmitted with undimimshed intensity in all directions. For the sake 16 of convenience, let the area of the piston c be 1 square inch, and let a force of 6 pounds be applied to it. The pressure exerted by the piston on the water is, there- fore, 5 pounds per square inch. If the area of the piston a is 40 square inches, and Pascal's law is true, the water must exert on the face of a a total pressure of 40 X 5 = 200 pounds. It is found by experiment that a force of 200 pounds must be applied to the piston a to prevent it from moving outwards. This shows that the fluid pressure against a is 200 pounds, or 5 pounds per square inch, which is the same as that against c. Similar results are obtained with the pistons b, d, <?, and /. If their areas are, respectively, 7, 6, 8, and 4 square inches, the pressures exerted on them are found to be, respectively, 35, 30, 40, and 20 pounds. 31 HYDROSTATICS 6 Pascal's law may be formally stated as follows: The pres- sure i>er unit of area exerted anywhere on a mass of liquid is transmitted undimimshed in all directions, and any surface in contact with the liquid will be subjected to this pressure in a direction at right angles to the surface. 13. Application of Pascal's Law. In Fig. 3, let the area of the piston a be 1 square inch, and that of b, 40 square inches. According to Pascal's law, 1 pound placed on a will balance 40 pounds placed on b. If a moves down- wards 10 inches, then 10 cubic inches of water will be forced into "the tube b. This will be distributed in the tube b in the form of a cylinder whose cubical contents must be 10 cubic inches, whose base has an area of 40 square inches, and whose altitude must therefore be 10 - 40 = } inch. This shows that a movement of 10 inches of the piston a causes a movement of i inch of the piston b. This is an example of the familiar principle of work that the force, PlQ 3 or effort, multiplied by the distance through which it moves, is equal to the resistance multiplied by the distance through which it moves. 14. The Hydraulic Press. The principle just stated finds an important application in the hydraulic press, which is shown in Fig. 4. As the lever a is depressed, the piston b is forced down on the water in the cylinder c. The water is forced through the bent tube d into a cylinder fitted with a large piston e> and causes that piston to rise, the plat- form / is thus lifted, and the bales placed between it and an upper fixed platform are compressed. Assume the area of the piston b to be 1 square inch, and that of the piston e to be 100 square inches. Also, assume the length of the lever between the hand and the fulcrum to be ten times the length between the fulcrum and the piston rod b. If the end of the lever is depressed, say 10 inches, the piston b is 6 HYDROSTATICS 31 depressed one-tenth of 10 inches, or 1 inch, and the piston e is raised rihr inch, since 1 cubic inch of water is displaced. If P denotes the force applied and Q the piessure on the platform /, then P X 10 inches = Q X Toir inch, hence, Q = 1,000 P, or P = ToW Q. A force of 40 pounds applied PIG by hand thus produces a pressure of 40 X 1,000 = 40,000 pounds. But, if the average movement of the hand per stroke is 10 inches, it will require 1,000 -r 10 = 100 strokes to raise the platform 1 inch, which again shows that what is gained in power is lost in speed. 31 HYDROSTATICS GENERAL THEORY OF LIQUID PRESSURE 15. Downward Pressure. The weight of a mass of liquid is a force, and will produce a piessure on any surface in contact with the liquid independently of the pressure pro- duced by other forces. Thus, in the case of the vessel shown in Fig. 2, if the vessel is vertical so that the piston d is at the bottom, d is subjected, in addition to the pressure of 5 pounds per squaie inch transmitted from the piston c, to an additional pressure due to the weight of the water above it. Evidently, the piston e will be under an additional pressure due to the weight of the water, but the pistons a, b, and /, being above the liquid, will be subjected only to transmitted pressuie. To arrive at the laws govern- ing the pressure due to the weight of a liquid, let us con- sider first the pressure on the bottom of a vessel containing the liquid. In Fig. 5 are shown two vessels of different shape but with the water level in each at the same height 24 feet. Assume the area of the bottom of a to be 500 square inches, or $M square feet; then the vol- ume of the water contained in a is -fH X 24 = 83i cubic feet, and the weight is 83i X 62.5 = 6,208.3 pounds. The pressure per square inch is 5,208.3 -5- 500 = 10.42 pounds, nearly. An easier solution, howevei, is effected as follows: The weight of a column of water 1 square inch in cross-section and 1 foot long is .484 pound (Art. 10). Now, above each square inch of the bottom there is a column of water 24 feet high, whose weight is, therefore, 24 X .434 = 10.42 pounds, and this weight is the pressure per square inch on the bottom of the vessel, In vessel 6, assume the cross-section of the lower part to be 600 square inches and that of the upper part to be ILT39&-1S PIG 5 8 HYDROSTATICS 31 100 square inches. It is evident that the volume of water contained is smaller than before, and it might be concluded that the pressure on the bottom is, consequently, smaller; but this conclusion would not be true, as will appear from the following reasoning: Let water be poured into the empty vessel until the lower part is filled; the pressure on the bottom is, according to the method of calculation just given, 10 X .434 = 4.34 pounds per square inch. Imagine, now, a thin piston or disk fitted tightly into the smaller part and resting on the surfa'ce of the water below, and then let water be poured in until the smaller part is full. This piston acts as the bottom of a vessel 14 feet high and with a uniform cross-section of 100 square inches. Hence, the pressure on this piston, due to the weight of the water above, is 14 X .434 = 6.08 pounds per square inch. According to Pascal's law, this pressure is transmitted in all directions, and therefore acts on the bottom of the vessel; hence, the total pressure on the bottom is 4.34 + 6.08 = 10.42 pounds per square inch, the same as in the first case. It is evident that the result will be the same if the piston is left out. 16. From the reasoning used in the preceding paragraph, it appears that the pressure per unit of area on the bottom of a vessel depends only on the vertical distance between the bottom and the liquid surface, and not at all on the shape of the vessel. This principle is one of great importance. 17. Intensity of Pressure. When a surface is sub- jected to a uniformly distributed pressure, the pressure per unit of area is called the intensity of pressure, or unit pressure. Usually, the unit of force is the pound, and that of area is either the square inch or the square foot; and so the intensity of pressure is expressed either in pounds per square inch or in pounds per square foot. In general, if P is the total pressure uniformly distributed over a surface, and A is the area of the surface, the intensity of pressure p is given by the formula 31 HYDROSTATICS 9 18. Head. The distance from any horizontal layer of a liquid body to the surface of the liquid is termed the head on that layer. Let h = head, in feet, on any horizontal layer; p = pressure per square inch on the layer, in pounds; w = weight of a column of liquid 1 foot long and 1 square inch in cross-section. Then, p = wh (1) For water, w = .434, and, therefore, p = .4347* (2) Also, h => -f~-r; that is, .4o4 h = 2.304^ (3) EXAMPLE 1 The depth of water in a stand pipe is 80 feet (a) What is the pressure per square inch on the bottom ? (b} What is the pressure per square inch on a horizontal layer 65 feet from the surface? SOLUTION. (a) Substituting the value of h in formula 2, p = 434 X 80 = 34 72 Ib per sq in. Ans (b) Here // = 65, hence, substituting in formula 2, p = 434 X 65 = 28 21 Ib. per sq. in. Ans EXAMPLE 2. What must be the height of water in a stand pipe to give a pressure of 80 pounds per square inch on the bottom? SOLUTION. Substituting in formula 3, h - 2.304 X 80 = 184 32 ft. Ans. 19. Upward and Lateral Pressure. So far, only downward pressure has been discussed. It is necessary now to consider upward pressure, as well as lateral (sidewise) pressure on vertical surfaces. Let the vessel shown in Fig. 6 (a) be filled with a liquid to the level a. The part of the liquid in a b rests on the layer at 6, and produces over that surface an intensity of pressure of w// t pounds per square inch, where A x is the head, in feet, on the layer at b. According to Pascal's law, this intensity of pressure is trans- mitted to all the bounding surfaces below the level b\ hence, there is a pressure of wh^ pounds per square inch, due to the liquid in ad, exerted at c,d,e,f t g, and k } at right angles to 10 HYDROSTATICS 31 the surfaces. At any point below the level b there is, how- ever, additional pressure due to the weight of the liquid between the point and the level b. At c, which is at the same level as b, the total upward pressure per unit of area is w h lt the same as the downward pressure on the layer at b. Consider now the layer at de at a distance fi a below the surface a. The pressure on this layer, due to the weight of the liquid above, is wh t pounds per square inch, and by Pascal's law this pressure is transmitted to all parts of the bounding surface below the level de, just as if the layer dc were a solid piston. The liquid below de can exert no pres- sure at the points d and e\ hence, at these points th2 pressure per unit of area is the same as the downward pressure on the to layer de, namely, w h, pounds per square inch. The same reasoning shows that the lateral pressure per unit of area at the points / and g is w A 3 , where h, is the head on the layer fg. The following important law, which is a direct consequence of that of Pascal, may now be stated: The intensity of pressure at any point of a surface enclosing a liquid is normal to that surface; it depends only on the depth of the point below the surface of the liquid^ and is equal to the intensity of downward pressure on a horizontal layer of the liquid having' the same head as the point in question. The formula p = w h is, therefore, general, and expresses the intensity of pressure at any point whose distance from the surface of the liquid is A. 31 HYDROSTATICS 11 EXAMPLE In Pig 6, suppose the depth of the various layers below the level a to be as follows: depth of t>, 10 feet, of de, 17 feet, of fg t 26 feet, and of k, 30 feet. The liquid being water, what are the pressures per square inch at the points c, d, <?, /, g , kt SOLUTION Using formula 2 of Art 18 in each case, the pressures at the various points are found to be as follows c = 434 X 10 = 4 34 Ib per sq m. Ans. d = .434 X 17 = 7.38 Ib. per sq. in Ans. Pressure at, * = <4S4 X 1? = 7 88 lb P er ** in " AnS Pressure at, f _ 434 x 25 10 85 lb per sq in. Ans. g = .434 X 25 = 10 85 lb. per sq in. Ans k = .434 X 30 = 13 02 lb. per sq in Ans. 20. Graphic Determination of Pressure. The varying intensity of pressure for different points below the surface of a liquid may be determined graphically as follows: Let OM, Fig. 6 (), represent, to any convenient scale, the head h t on the bottom k of the vessel. Draw MN perpen- dicular to MO, and representing, to any convenient scale, the intensity of pressure w h* on the bottom of the vessel. It should be observed that OM represents a distance and MN a pressure, and that there is no necessary relation between the two scales. For instance, OMma.j represent h* to a scale of 2 feet to the inch; and MN } the pressure w 7i 4 to a scale of tsV inch to the pound. Draw ON. Then, any horizontal line limited by OM and ON will represent the intensity of pressure at any depth equal to the vertical distance of that line below O. Thus, CJ3 t whose distance below is A,, represents w k^ Likewise, D 2? represents w h t . If the pressure scale is -sV inch to the pound, it will be found that the line FG measures about H, which represents 11 pounds per square inch. For very accurate work, the scale should be as large as possible. It should be observed that the tangent of the angle MONis numerically equal to w, since MO hi 21. Pressure Due to Matternal Load, If the surface of a liquid is subjected to a pressure, this pressure, accord- ing to Pascal's lav?, is transmitted undiminished to all parts 2 HYDROSTATICS 81 f the enclosing vessel, and must be added to the pressure ue to the weight of the liquid. Assume the surface a of 'ig. 6 to be subjected to a pressure of 30 pounds per square ich. In the example of Art. 19, the pressure at c, due to the ead of water, is 4.34 pounds per square inch. Adding the D pounds per square inch due to the external force applied t a, the result is 4.34 + 30 = 34 34 pounds per square inch .ikewise, the pressure at d and e is 7.38 + 30 = 37 38 pounds er square inch; the pressure at / and g is 10 85 + 30 = 40.85 ounds per square inch; and the pressure at k is 13.02 + 30 = 43.02 pounds per square inch. Let G = total load on surface of liquid, in pounds; A = area of surface loaded, in square inches; /^ pt = = intensity of pressure on surface, in pounds A per square inch; P = intensity of pressure, in pounds per square inch, at a point h feet below liquid surface. Then, p = wh+p* = wh + (1) When the liquid is water, p = .434^+A = .434>% + (2) A EXAMPLE A vessel filled with salt water weighing 1 03 times as uch as fresh water has a circular bottom 13 inches in diameter The p of the vessel is fitted with a piston 3 inches in diameter, on which laid a weight of 75 pounds. What is the intensity of pressure on the ttom, if the depth of the vessel is 18 inches? SOLUTION In this case, w = 1 03 X 434 = .447 lb.; h = 18 in. 1 6 ft , and A = 3 1 X 7864 sq. m By formula 1, p = .447 X 1 5 + 3 , x ?6 7854 - 11.28 lb per sq. in. Ans, 22. Atmospheric Pressure. When a liquid is exposed the air, as in the cases represented in Figs. 5 and 6, there an external pressure, due to the weight of the air, on the rface of the liquid. This pressure is called atmospheric essure. It varies according to locality and atmospheri nditions, but, for nearly all practical purposes, it may taken as 14.7 pounds per square inch. This pressure is 31 HYDROSTATICS 13 transmitted by the liquid and should be treated as the pres- sure p n considered in the preceding article. It is customary, however, in hydrostatic and hydraulic calculations to take the atmospheric pressure as the zero of reference. When the pressure of a liquid is referred to, the pressure in excess of the atmospheric pressure is meant. This is called gauge pi-essure. When the atmospheric pressure is added to the gauge pressure, the resultant pressure, which is really the total pressure on the liquid or surface considered, is called absolute pressure. Thus, a gauge pressure of 20 pounds per square inch is equivalent to an absolute pressure of 20 + 14.7 = 34.7 pounds per square inch. 23. Equilibrium of Liquids at Best. Since the pressure on a horizontal layer due to the weight of a liquid PIG 7 is dependent only on the height of the liquid, and not on the shape of the vessel, it follows that, if a vessel has a number of radiating tubes, as in Fig. 7, the water in each tube will be at the same level, no matter what may be the shape of the tubes. For, if the water is higher in one tube than in the others, the downward pressure at the level , due to the height of water m this tube, is greater than that due to the height of the water in the other tubes. This excess of pres- sure will cause a flow toward the other tubes, which will continue until there is no further excess that is, until the 14 HYDROSTATICS 31 free surfaces are at the same level. Then the liquid will come to rest, and will be in equilibrium. The principle here stated is embodied in the familial saying, Wafer seeks tts level. This principle explains why city reservoirs are located on high elevations, and why water, when leaving the hose nozzles, spouts so high. If there were no resist- ance by friction and air, the water would rise to a height equal to the level of the reservoir If a long pipe with a length equal to the vertical distance between the nozzle and the level of the water in the reservoir were attached to the nozzle and held vertically, the water would just reach the end of the pipe. If the pipe were lowered slightly, the water would trickle out. Fountains, canal locks, and artesian wells are examples of the application of this principle. EXAMPLE 1 The water in a city reservoir is 150 feet above a hydrant What is the pressure per square inch at the hydrant? SOLUTION By formula 2, Art 18, p = 434 X 150 = 65 1 Ib per sq in Ans EXAMPLE 2 The pressure on a water mam, when the water is not flowing, is shown by a gauge to be 72 pounds per square inch What is the elevation of the reservoir above the mam? SOLUTION. By formula 3, Art 18, h = 2.304 X 72 = 165.89 ft Ans. EXAMPUJ8 FOR PRACTICE 1. What is the intensity of pressure on the bottom of a stand pipe 90 feet high? Ans 39 06 Ib, per sq In. 2. A cylindrical vessel 15 inches m diameter is filled with water. The top of the vessel is fitted with a piston on which is laid a weight of 300 pounds. The depth of the vessel being 24 inches, determine the intensity of pressure on the bottom, of the vessel Ans 2 566 Ib per sq. in. 3. If the intensity of pressure, due to an external load, on a vessel filled with water is 30 pounds per square inch, what is the intensity of pressure at a point 12 inches below the surface of the water? Ans. 30.434 Ib. per sq. in 31 HYDROSTATICS 15 PRESSURE ON AN IMMERSED SURFACE 24. Total Pressure on a Flat Immersed Surface. Let M N, Fig. 8, be a plate immersed in water. It is pro- posed to determine the total pressure acting on the upper surface of this plate. In the first place, the total piessure />, being the resultant of the pressures acting at all the points of M N, is perpendicular to MN. The magnitude of P is determined by means of the following principle, which is derived by advanced mathematics: The total pressure on any plane surface immersed in a liquid is equal to the product of the following quantities ( 1 ) the area of the surface, (2) the dis- tance of the center of gravity of the surface ftom the level of the liquid, (5) the weight of the liquid per unit of -volume. In applying this princi- ple, length, area, and vol- ume should be referred to the same unit. Thus, if dis- tances are expressed in feet, areas should be ex- pressed in square feet and volumes in cubic feet. The princi- ple obviously applies to the pressure on the bounding sur- faces of the vessel or receptacle containing the liquid as well as to the pressure on the immersed surface. Let A be the area of the surface M N> Fig 8; G, the center of gravity of that surface, its distance from the level of the liquid being h a \ and W, the weight of the liquid per unit of volume. Then, the foregoing principle may be stated in symbols thus: P = A h e W It should be particularly noticed that the total pressure P does not act through the center of gravity G. The deter- mination of the point of application of P will be dealt with farther on. Pro. 8 16 HYDROSTATICS 31 EXAMPLB To find the total pressure on the side ML N K oi the rectangular tank shown in Fig 9, the dimensions being as indicated. SOLUTION Since ML N K 'is a rectangle, its center of gravity G is at a distance from the upper side L N equal to 5, or 6 25 ft 2 LI Also, A = MK X KN = 3 2 X 6.25, = 3 125 ft.; hence, h r = 3 125 ft and W = 62 5 Ib Substituting these values in the formula for P, P = 3 2 X 6.25 X 3 125 X 62 5 = 3,906 Ib , nearly Ans. 25. Center ol Pressure. The point where the line of action of the total or resultant pressure acting on an immersed surface meets that surface is called the center of pressure. This point does not coincide with the center of gravity of the surface unless the latter is horizontal In all other cases, the center of pressure lies below the center of gravity. Thus, in Fig. 8, the center of pressure C lies below the center of gravity G. A general for- mula for determining the position of the center of pressure cannot be either derived or applied without the use of advanced mathematics. Special for- mulas applying to some important cases are given in the fol- lowing articles. The distance of the center of pressure from the level of the liquid will in all cases be denoted by A as shown in Fig. 8. 26. Center of Pressure ol a Rectangular Surface. In Fig. 10 is represented a vessel V containing a liquid. The bounding surface RSTU is shown inclined to the ver- tical; but the formulas derived in this and the following two articles apply both to an inclined and to a vertical surface. Let K L M N be a rectangular part of the inclined surface, the sides K L and M N being horizontal The line EFYis an axis passing through the middle points of K L and M N. The heads on the points E and F are denoted by hi and Jt t , respectively. It should be observed that the formulas given 31 HYDROSTATICS 17 below apply whether the rectangular surface considered forms part of one of the bounding surfaces of the vessel or whether it is the surface of any immersed body. The center of pressure Cis on the line E F, and its depth h c below the level of the liquid is given by the formula CD To find the distance FC, draw FHQ perpendicular to the JZ FlQ 10 verticals through F t C t and ., and denote the length EF, or f* T NK, by 5. The similar triangles FCH and FEQ give ^ / y jD ~^r] whence, the following equa- = -\ that is, EQ b tion obtains: PC ' X (2) If the value of h e found by formula 1 is substituted in formula 2, the result, after several transformations have been made, is X* (3) 18 HYDROSTATICS 31 27. If the edge K ' L is flush with the surface of the water, h* = 0, and formulas 1 and 3 of the preceding article become, respectively, A. -iXv- 1 ! = *A, (1) FC= ' (2) 28. Center of Pressure of Triangular Surface. Only the case of an isosceles triangle with its base horizontal will be treated in this Section. And first, a triangle MFN t FIG 11 Fig. 11, with its vertex F above its base will be considered The line EFYis perpendicular to the base MN The alti- tude FE is denoted by a The rest of the notation is similar to that in Fig. 10. The center of pressure is on the line FE, and its depth below the surface of the liquid is given by the formula h t = 3/ '' a +-' (1) The distance PC is determined in the same way as for the rectangle treated in Art 26, the result being FC = fit. A, (2) 31 HYDROSTATICS 19 29. If the vertex is flush with the surface of the liquid, //, = 0, and the two formulas of the preceding article become 7 3 til n i / t X h c - = i /it ( 1 ) 4 fli 30. Let, now, the vertex be below the base, as shown in Fig 12. With the same notation as before, _ R .. i- , -.~ T ~" ?s " rm /\ .J'..i _A_____^__ f __/.__\ 1 i ' 1 i ' ! i '/ , * ^ / ! /rJ y S PIG. 12 3 A. 1 , i 2 i h t (1) (2) 31. If the base is on the surface of the liquid, h t 0, and the two formulas of the preceding article become h e = %hi (1) B *xi (2) 32. Center of Pressure of Circular Surface. Let M ' N, Fig. 13, be a circular surface. The line EFY con- tains the lowest point E and the highest point F of the circle. S30 HYDROSTATICS 31 The diameter of the circle is denoted by d. The rest of the notation is the same as in previous articles. The center of pressure is on the diameter EF, and the distances k e and FC are given by the formulas 7B '. T-T (D (2) FC = ' ~ ' X kt h. /js'^T^r,^-'^" -^ ,.""' A 1 ,*: V . " .' x T-.F- -/ -\ r> * .* ,-. i i / / ' - i / / i \\ !/ / )tf I H/fc, / /'l^ PlQ 18 33. If the circle just touches the surface of the liquid, h t = 0, and formulas 1 and 2 of the preceding article become +& (1) h c = EXAMPLE. The vertical circular plate a, Fig. 14, 8 inches In diam- eter, covers an opening of equal size. The center G is 24 inches below the surface of the -water, and 20 inches below the hinge b, What horizontal force applied at c is sufficient to move the plate ? SOLUTION. Area of plate is 50 266 sq. in Head above center of gravity is 2 ft. Then (Art. 24), the total pressure on the plate is X 62.5 X 2 = 43.63 Ib. 144 31 HYDROSTATICS 21 Here, A, = 24 + f = 28 in , and A, = 24 - f = 20 m The depth of the center of pressure below the surface is, by formula 1, Art 32, h e = 24 167 in , nearly The center of pressure is 167 in below the center of gravity G, and, therefore, 20 167 m. below the hinge b The ' ;_ _y moment of P, the total pressure, about r the hinge must be equal to the moment of .P 1 about the hinge, hence, FX 40 = P X 20 167 = 43 63 X 20.167 = 879 89 in -Ib ; whence, F = 879 89 -*- 40 = 22 Ib , nearly Ans. 5 $ 34. Plates 01- Gates With Liquid Press xi i-e on Both Sides. The plate A B, Fig. 15 is subjected to water pressure with the level a' on one side and a!' on the other. The total pres- _ _ B ~ -^ FIG 14 FIG. 15 sure on one side is /", acting through the center of pres- sure C'\ the total pressure on the other side is P' r , acting through the center of pressure C". The heads on the centers of pressures O and C" are denoted by h e ' and h e ", respect- ively; and the difference between the levels a 1 and a" is denoted by e. The two horizontal forces P' and P" have a resultant P whose magnitude is P' P". Let h t be the depth, below a f , of the point of application C of this result- ant. Taking moments about the point E> we have, Ph e = /"^'-/>"U," + *); P' h f P" ( h it 4- e\ p /l < ( + e whence, h t = 22 HYDROSTATICS 31 EXAMPLE Let the gate in Fig 15 be rectangular and 6 feet wide; let the depth BE be 9 feet and the difference *, 2 4 feet. Required the pressures /*, P", P, and the distance h e SOLUTION From formula of Art 24, the total pressure on the right side is / J/ = 0X6x625Xo = 15,188 Ib. Ans The total pressure on the left side is P" = 6X66x625X-fj- = 8,167.5 Ib Ans. The resultant force is P = 15,188 - 8,167 5 = 7,020 5 Ib From formula 1 of Art 27, h c ' - % X 9 = 6, and h = X 6 6 = 44, also, e = 9 - 6.6 = 2 4. Substituting in formula of Art 34, . 15,188 X 6 - 8,167 5 X (4 4 + 2 4) ? 7,020.5 35. Assume the plate B, Fig 16, to be entirely below the water levels a and b, and, therefore, subjected to pressure on both sides. Let the triangle OM ' N be drawn to represent the variation in pressure due to the weight of the water on the left-hand side (see Art. 20), and let the tri- angle 0' M f N' be drawn to represent the variation of pressure due to the weight of the water on the right-hand side Since the angles M O N and M' O 1 N' are equal, the tangent of each being w (Art. 20), it follows that O 1 N' and ON are parallel. Take M0 n = M 1 1 , and through O" draw O"N" parallel to O ' N'\ then, the triangles O 1 M' N 1 and O" MN" are equal. Now, the intensity of pressure on the left of the gate, at any level, as x t below the surface, is given by the intercept K ' E, and that on the right by the intercept DE = D'E'. The resultant intensity of pressure at that point is the difference KE DE KD. Since ON and O" N" are parallel, the horizontal intercepts between them are equal, which means that the intensity of pressure is the _ _ N' M FIG 16 31 HYDROSTATICS 23 same at all points of the plate B It follows that the center of pressure for such a submerged plate coincides with its center of gravity. The total resultant pressure P on the plate is found as follows: The intercept SO" represents the intensity of pres- sure on the left side at the level /?, that is, for the head e, which is the difference between the levels a and b. Hence, since the intensity of pressure on B is everywhere equal to KD, or SO", or we, it follows that, if the area of B is denoted by A, the total pies&ure is P = Awe EXAMPLE In the example of Art 33 (see Fig 14) , suppose there is water on the right-hand side of the partition If the level of this water surface is 8 inches lower than that of the level d, find the total resultant pressure on the plate, and also the force F SOLUTION The head e = 8 in = f ft The total pressure is A w e - 50 206 X 434 X ij = 14 54 Ib Ans This pressure acts through the center of gravity, 20 ill below the hinge, hence, taking moments about the hinge, F X 40 = P X 20, whence, F=\IP=\P= ^~ = 7.27 Ib Ans. 36. Pressure on Surfaces That Are Not Plane. If the surface sustaining liquid pressure is not plane, but is (b) Fio 17 curved or irregular, the total pressure on it in any direction, neglecting the pressure due to the weight of the liquid, is the same as the total pressure would be on the projection of the surface on a plane at light angles to the given direction. To illustrate this statement, consider the three pistons in Fig. 17. At (a) is shown a piston with a curved end, at (b), a piston with an irregular end; and at (f), a piston with a flat end. In each case, the projection of the surface sustaining pressure, I L T 398-W 24 HYDROSTATICS 31 on a plane perpendicular to the piston rod, is the circular cross-section of the cylinder, and if the pressures per unit of area in the cylinders are the same, the pressure on the face of the piston is the same in each case. It is assumed that the pressure per unit of area is the same at all points of the surface; that is, the change of pressure due to the varying depth of the liquid is neglected With this restnc- tion, the law applies to all fluids, both liquid and gaseous. If the curved or irregular surface forms a part of the wall of a vessel, or is submerged, so that the pressure on it is due to the weight of the liquid, the law of Art. 24 is not true except for a projection on a vertical plane. 37. The whole subject of pressure in any direction on a curved or irregular surface may be summarized as follows: 1. If the external pressure is so great that the pressure due merely to the weight of the liquid may be neglected, the total pressure in any direction is equal .to the product of the projection of the surface on a plane perpendicular to that direction, and the pressure per unit of area. 2. When the pressure is due wholly, or in part, to the weight of the liquid, as m the case of submerged surfaces, the total pressure in any direction cannot, in general, be determined except for regularly curved surfaces, such as spheres, cylinders, cones, etc., and for these the computa- tions are difficult and must be made by advanced mathematics. 3. In one direction, however the horizontal the total pressure on a surface is easily found. It is precisely the same as the horizontal pressure on the projection of the given surface on a vertical plane. EXAMPLES FOR PRACTICE \ What is the pressure on a layer of water 33 feet below tne surface? Ana 14.32 Ib. per sq. In. 2 What elevation of a reservoir is required to make the pressure in a. water main 40 pounds per square inch? Ans. 92.16 ft 3. A vertical triangular plate forms the side of a vessel that con- tains water The base of the tnangle is 4 feet long and lies at the 31 HYDROSTATICS 25 water level; the vertex is 4^ feet below the water level (a) What is the total pressure on the triangle? (0) What is the distance of the center of pressure below the water surface? A \ (a) 843 75 Ib S U*) 2J ft 4. A circular plate 2 feet in diameter is held vertically so that its upper edge is 5 feet below the liquid surface Find the depth of the center of pressure Ans ft ^ in 5 In the gate shown in Fig 15, the depth of the head-water is 7 feet, and that of the tail-water is 6 feet (a) What is the resultant pressure on the gate per foot of length? (f>) How far above the bottom B is the line of action of the resultant? An8 /() 760 Ib Ans \ (6) 3 03 ft 6 The diameter of the plunger of a hydraulic press used m an engineering establishment is 12 inches. Water is forced into the cylinder of the press by means of a small pump having a plunger whose diameter is f inch What pressure is exerted by the large plunger when the force acting on the small plunger is 125 pounds? Ans 32,000 Ib. BUOYANT EFFORT OF LIQUIDS IMMERSION AND FLOTATION 38. Principle of Archimedes. In a mass of liquid at rest, suppose a part M N, Fig. 18, of the liquid to become solid without changing its form or density. This solid part, having the same density as before, will be held in equilibrium, or remain at rest. Let B denote the weight of the solid part; D, the total downward pressure; and U, the total upward pressure on this part. The oidinary static condi- tions of equilibrium require that the upward force U should balance the down- ward forces D and B\ that is, - U=D + B\ F <o-w whence, /- D = B (1) The difference U L) between the upward and the down- ward pressure is called the buoyant effort of the body MN. Suppose, now, that the imaginary solid part of the water is replaced by an actual solid body having precisely the same 26 HYDROSTATICS 31 shape and volume, and let the weight of this body equal W. The solid will be subjected to the same vertical pressures D and /as was the part M N of the liquid. Consider, now, the vertical forces acting on it. The weight \V and the pressure D are downwards, and tend to cause the body to sink; the pressure U is upwards, and tends to cause the body to rise. The resultant downward force is D+ W- U = W-(U-D) = W-B (2) The weight B is the weight of a mass of liquid whose volume is equal to that of the solid, or, what is the same thing, B is the weight of the liquid displaced by the solid. The following principle may therefore be stated- When a solid body is immersed in a liquid, a buoyant effort equal to the weight of the hqmd displaced acts upwanh and opposes the action of gravity. The weight of a body, as shown by a scale, is decreased by an amount equal to the buoyant effot t, that is, by an amount equal to the -weight of liquid displaicd. This principle is called the principle of Archimedes, from the name of its discoverer. EXAMPLE In Fig 18 is shown a cube immersed in water Ihe edge is 6 inches, the sides are vertical, and the lateral pressures me balanced On the upper face, there is a downward pressure I) due to the head of water of 15 inches, and on the lower face there is an upward pressure U due to the head of 21 inches. To fancl the buoyant effort J SOLUTION. The downward pressure due to the head of 1C in. or H ft , is '' 6' X 434 X It = 19 53 Ib The upward pressure on the lower face due to a head of 21 in., or 1-f ft , is 6 X 434 X 1* = 27 34 Ib. The buoyant effort of the body is, therefore, 27 34 - 19 53 = 7 81 Ib. Ans. Also, the buoyant effort is equal to the weight of the water dis- placed by the body The volume of the cube is (j-) or } cu. ft The weight of the water displaced is i X 62 5 7 81 Ib. Ans 39. The principle of Archimedes may be experimentally verified with a beam balance, as shown in Fig. 19. From one scale pan, suspend a hollow metallic cylinder l> and 31 HYDROSTATICS 27 below that a solid cylinder a of the same size as the hollow part of the upper cylinder. Put weights m the other scale pan until they exactly balance the two cylinders If a is immersed in water, the scale pan containing the weights will descend, showing that a has lost some of its weight. Now fill / with water; the volume of water that can be poured into t is obviously equal to that displaced by a. The scale pan containing the weights will rise gradu- ally until / is filled, when the scales will balance again 40. If a body im- mersed in a liquid has the same weight as the liquid it displaces, then IV = B, the resultant vertical force IV B is zero, and the body will remain at rest at any depth below the surface. If the body is heavier than the liquid it displaces, W is greater than .5, and the resultant vertical force W B is downwards, hence, the body will sink to the bottom. If the body is lighter than the liquid it displaces, B is greater than W, the resultant vertical force B IV is upwards, and the body will rise to the surface. 41. An interesting experiment in confirmation of the conclusions just derived may be performed as follows: Drop an egg into a glass jar filled with fresh water. The mean density of the egg being a little greater than that of fresh water, the egg will fall to the bottom. Now dissolve salt in the water, stirring the mixture; as soon as the salt water becomes denser than the egg the latter begins to rise. If fresh water is then ponied in until the egg and water have the same density, the egg will remain stationary in any position it may be placed below the surface of the water. PIG 19 28 HYDROSTATICS ' 31 42. Floating Bodies. A body lighter than a liquid, bulk for bulk, rises to the surface, when immersed, and floats. For equilibrium, the buoyant effect B must be just equal to the weight W of the body. But, since B is the weight of liquid displaced, it follows that the weight of the liquid displaced by a floating body is equal to the weight of the body. The depth of a floating body m a liquid depends on the relative weights of equal volumes of the body and the liquid. If the body is nearly as heavy as the liquid, it will sink until it displaces nearly its whole volume; if very light, compared with the liquid, the larger part of the body will be above the liquid surface. For example, the density of ice being about nine-tenths that of water, about one-tenth of an iceberg appears above the surface and nine-tenths is submerged. The density of pine is about one-half that of water; hence, about one-half of a floating pine log is submerged, and one-half is above water. EXAMPLE 1. Water-tight canvas air bags are used for raising sunken ships These bags are sunk when collapsed, attached to the ship by divers and then filled with air from the pumps above, (a) If the capacity of a bag is 200 cubic feet, what Is the buoyant effort? (b) How many bags will be required to lift 600 tons? SOLUTION. (a) The weight of the bag and the enclosed air may be neglected. The buoyant effort is the weight of the water displaced by the full bag, that is, 200 X 62.5 - 12,500 Ib. Ans. (b) To raise 600 tons, 600 X 2,000 __ . 12 500 = ^ are necessarv ' Ans. EXAMPLE 2 A cast-iron cylinder 14 inches long and 8 inches in diameter is closed at the ends, and the metal is i inch thick through- out. Will the cylinder float or sink in water? SOLUTION. The volume of the entire cylinder is 7854 X 8 1 X 14 = 703.72 cu. in. The hollow portion has a length of 18 in. and a diameter of 7 in.; tts volume is, therefore, .7854 X 7* X 13 - 600.80 cu. !n. The volume of metal is 703.72 - 600.30 = 203.42 cu. in. 31 HYDROSTATICS 29 Taking the weight of cast iron as 450 Ib per cu. ft., the weight of the cylinder is 450 X ^|| = 52.97 Ib If Immersed, the cylinder displaces 703.72 cu. in of water, which weighs 703 T> 62 5 X j~~ = 25.45 Ib The buoyant effort being less than the weight, the cylinder will sink Ans. SPECIFIC GRAVITY 43. Definition. The specific gravity of a solid or of a liquid substance is the ratio of the weight of any volume of that substance to the weight of an equal volume of water. This ratio varies slightly with temperature, and the values given by physicists are exactly correct only for a temperature of about 39 F , at which water has its maximum density. For practical purposes, however, it is not necessary to take changes of temperature into account. The abbreviation Sp. Gr is often used for specific gravity. 44. Specific Gravity of Solids Not Soluble in Water. The principle of Archimedes affords a very easy manner of determining 1 the specific gravity of a solid not soluble in water. The body is first weighed in air; it is then attached to a scale pan and weighed in water. The difference between the two weights will be the weight of an equal volume of water. The ratio of the weight in air to the difference thus found will be the specific gravity. Let W be the weight of the body in air and W the weight in water; then W W is the weight of a volume of water equal to the volume of the solid, and Sp. Gr. ** w- W EXAMPLE. A body weighs in air 364- ounces and in water 30 ounces. What is its specific gravity? SOLUTION. Here W 36}, and W 30. Substituting In the formula, Sp. *. - ** - f - 6.8. AM. 30 HYDROSTATICS 3J 45. If the body is lighter than water, a piece of iron or other heavy substance must be attached to it, sufficiently heavy to sink it. Then the two bodies are weighed together both in air and in water, both are weighed separately in air, and the heavier body in water. Subtracting the combined weight of the bodies in water from then combined weight in air, the result will be the weight of a volume of water equal to the volume of the two bodies. The difference between the weight of the heavy body m air and in water gives the weight of a volume of water equal to the volume of the heavy body. Subtracting this last result from the former, the result will be the weight of a volume of water equal to the volume of the light body. The weight of the light body in air divided by the weight of an equal volume of water, as just determined, is the specific gravity of the light body. Let W = weight of both bodies in air; W = weight of both bodies in water; w = weight of light body in air; Wi = weight of heavy body in air; W, = weight of heavy body in water. Then, the specific gravity of the light body is given by the formula Sp. Gr = w (W- W}-(W l - W,) EXAMPLE. A piece of cork weighs, in air, 4 8 ounces To it is attached a piece of cast iron weighing 36 ounces in air and 31 ounces m water The weight of the iron and cork together, in water, is 15 8 ounces, what is the specific gravity (a) of cork? (<J) of cast iron? SOLUTION. (a) Here w = 4 8, W = 40 8, W = 15 8; W v - 30, W t = 81 Substituting in the formula, Spec,fic gravity is (4Q g _ ^*_ (M __ _ ^ ^ (b) To apply formula of Art 44, W = 36, W - 31. no Specific gravity of cast iron is Q . ., => 7 2. Ans. oo SI 46. Specific Gravity of a Liquid. To determine the specific gravity of a liquid, proceed as follows: Weigh an empty flask; fill it with water, then weigh it and find the 31 HYDROSTATICS 31 difference between the two results; this will be equal to the weight of the water Then weigh the flask filled with the liquid, and subtract the weight of the flask; the result is the weight of a volume of the liquid equal to the volume of the water. The weight of the liquid divided by the weight of the water is the specific gravity of the liquid. Let w = weight of empty flask, W = weight of flask when filled with the liquid; W 1 = weight of flask when filled with water Then, Sp.Gr. = *-' W 1 w EXAMPLE A flask when empty weighs 8 ounces; when filled with water, 33 ounces, and when filled with alcohol, 28 ounces. What is the specific gravity of the alcohol? SOLUTION Here W = 28, w = 8, W = 33 Substituting in the formula, no __ Q Sp Gi. = ^__| = 8 Ans. 47. Nicholson's Hydrometer. Instruments called hydrometers are in general use for detei- mining quickly and accurately the specific gravity of liquids and of some solids. One of the principal forms is Nicholson's hydrometer, which is shown in Fig. 20. It consists of a hollow cylmdei, carrying at its lower end a basket //, heavy enough to keep the apparatus upright when placed in water. At the top of the cylinder is a veitical rod, to which is attached a shallow pan a The cylinder is made of such size and weight that the apparatus is somewhat lighter than water, and a certain weight W must be placed in the pan to sink Fl it to a given point c on the rod. The body whose specific gravity it is desired to find must weigh less than W. It is placed in the pan a, and enough weight w is added to sink the point c to the water level. It is evident that the weight In air of the given body is W w. The body is now removed from the pan a and placed in the basket d, an 32 HYDROSTATICS 31 additional weight being added to sink the point e to the water level. Represent the weight now m the pan by W 1 . The difference W w is the weight of a volume of water equal to the volume of the body. Hence, EXAMPLE The weight necessary to sink a hydrometer to the point c is 16 ounces, the weight necessary when the body is In the pan a is 7 3 ounces; and when the body is in the basket d, 10 ounces. What is the specific gravity of the body? SOLUTION Here W = 16, w = 7 3, W = 10 Substituting in the formula, I Q IT Q Specific gravity = ^ ~ = 3.22 Ans. 1U " i o 48. Volume of an Irregular Solid. The principle of Archimedes affords a very easy and accurate method of find- ing the volume of an irregularly shaped body. Let the weight of the body in air be W, and in water W. The difference W W is, according to the principle of Archi- medes, the weight of a volume of water equal to the volume of the body. If this volume is denoted by F, and the weight of water per unit of volume is denoted by W , the weight of the volume V is W, V, and, therefore, W V = W- W'\ whence, F= W ~ W> (1) W, If the volume is expressed in cubic feet, W,is 62.5 pounds, and, therefore, V= W ^^~ = -016(^- W) (2) If the volume is expressed in cubic inches, W* = .03617 pound, and, therefore, v = "~ = 27<647( w " w ' } EXAMPLE The weight of a body in air is 96 pounds, and m water, 48 8 pounds What is the volume of the body? SOLUTION To apply formula 2, we have W = 96 and W * 48.8. Substituting in the formula, V .016(98 - 48.8) - .76 cu. ft. Ans. 31 HYDROSTATICS 33 49. If the specific gravity of a body is known, the cubical contents of the body can be found by dividing its weight by its specific gravity, and then dividing again by either .08617 or 62 5, according as the volume is desired in cubic inches or in cubic feet. EXAMPLE A certain body has a specific gravity of 4 38 and weighs 76 pounds What is the volume of the body m cubic Inches? 76 SOLUTION.- __-_ . 479.7 en. In. Ans. EXAMPLES FOB PRACTICE 1 If a certain quantity of red lead weighs 5 pounds in air and 4 441 pounds in water, what is its specific gravity? Ans. 8.94 2. A piece of iron weighing 1 pound in air and 861 pound in water is attached to a piece of wood weighing 1 pound in air. When both bodies are placed in water they weigh 2 pound. What is the specific gravity: (a) of the iron? (6) of the wood? Aric /() 7 194 Ans> \() 602 3. An empty flask weighs 13 ounces; when filled with water, it weighs 22 ounces, and when filled with sulphuric acid, 29.56 ounces. What is the specific gravity of the acid? Ans. 1 84 4. How many cubic feet of brick having a specific gravity of 1.9 are required to make a total weight of 260 pounds? Ans. 2.39 cu. ft. PNEUMATICS PROPERTIES OF AIR AND OTHER GASES 1. Pneumatics is that branch of science that treats of the mechanical properties of gases. 2. The distinguishing property of a gas is that, no matter how small the quantity may be, the gas -will always till the vessel or vessels that contain it. If a bladder is partly filled with air and placed under a glass jar (called a receiver) from which the air has been exhausted, the bladder will immediately expand, as shown in Fig. 1. The force that a gas exerts when con- fined in a limited space is called tension. In this case, the word tension means pressure, and is only used in this sense when referring to gases. Fm. i 3. As water is the most common type of fluids, so air is the most common type of gases. It was supposed by the ancients that .air was "imponderable," by which was meant that it weighed nothing, and it was not until about the year 1650 that it was proved that air really has weight. A cubic inch of air, under ordinary conditions, weighs about .31 gram. The ratio of the weight of air to that of water is about 1 . 774; that is, air is only T^T as heavy as water. OOPYMiaHTID BY INTERNATIONAL TKXTMQOK COMPANY. AU. RIOHTft RMKHVtO 532 2 PNEUMATICS 32 It was shown in Hydrostatics that if a body is immersed in water, and weighs less than the volume of water it dis- places, the body will rise and project partly out of the water. The same principle, which is the principle of Archimedes, applies to gases. If a vessel made of light material is filled with a gas lighter than air, so that the total weight of the '. vessel and gas is less than the weight i of the volume of air they displace, the \ vessel will nse. It is on this pnnci- ! pie that balloons are made. ! 4. Since air has weight, it is evi- , dent that the enormous quantity of air that constitutes the atmosphere must exert a considerable pressure on the earth. This is easily proved by taking a long glass tube, closed at one end, and filling it with mercury. If the finger is placed over the open end, so as to keep the mercury from running out, and the tube is inverted and placed in a glass partly filled with the same liquid, as shown in Fig. 2, the mercury in the tube will fall, then rise, and after a few oscillations will come to rest at a height above the top of the mercury in the glass equal to about 30 inches This height will always be practically the same under PlQ 2 the same atmospheric conditions Now, since the atmosphere has weight, it presses on the upper surface of the mercury in the glass with equal force on every square unit, except on that part of the surface occupied by the tube. According to Pascal's law (see Hydrostatics), this pressure is transmitted in all directions. There being nothing in the tube, except the mercury, to counterbalance the upward pressure of the air, the mercury falls in the tube until it exerts an upward pressure on the upoer surface of 32 PNEUMATICS 3 the mercury in the glass sufficiently great to counterbalance the downward pressure produced by the atmosphere. In order that there may be equilibrium, the pie&sure of the air per unit of area on the upper surface of the mercury in the glass must equal the pressure (weight) exerted per unit of area by the mercury inside of the tube. Suppose that the area of the inside of the tube is 1 square inch, then, since mercury is 13 6 times as heavy as water, and 1 cubic inch ot water weighs about 03617 pound, the weight of the mer- curial column is .03617 X 13.6 X 30 = 14.7574 pounds The actual height of the mercury is a little less than 30 inches, and the actual weight of a cubic inch of distilled water is a little less than .03617 pound. When these considerations are taken into account, the average weight of the mercmial column at the level of the sea is 14.696 pounds, or, as it is usually expressed, 14.7 pounds. Since this weight, exerted on 1 square inch of the liquid in the glass, just produces equilibrium, it is plain that the pressure of the outside air is 14 7 pounds on every square inch of suiface. This pres- sure is often referred to as one atmosphere. A pressure of two atmospheres is a pressure of 2 X 14.7 = 29.4 pounds per square inch, a pressure of three atmospheres is a pres- sure of 3 X 14.7 = 44.1 pounds per square inch; etc. 5. Vacuum. Referring to Fig. 2, the space between the upper end of the tube and the upper surface of the mercury in the tube is called a vacuum, or empty space. If this space contained a gas of some kind, no matter how small the quantity might be, the gas would expand and fill the space, and its tension would, according to the amount present, cause the column of mercury to fall and become shorter; the space would then be called a partial vacuum. If the mer- cury fell 1 inch, so that the column was only 29 inches high, this would be expressed by saying that there were 29 inches of vacuum; a fall of 8 inches would be referred to as 22 inches of vacuum; a fall of 16 inches, as 14 inches of vacuum, etc. Hence, when the vacuum gauge of a condensing engine shows 26 inches of vacuum, there i$ enough air in the PNEUMATICS 32 condenser to produce a pressure of _ no X 14.7 = 30 X 14 7 = 1.96 pounds per square inch. In all cases where the mer- ^ __ cunal column is used to measure a vacuum, the height of the column, in inches, gives the number of inches of vacuum Thus, if the column were 5 inches high, or the vacuum gauge showed 5 inches, the vacuum would be 5 inches If the tube had been filled with water instead of mercury, the height of the column of water to balance the pressure of the atmosphere would have been about 30 X 13.6 = 408 inches = 34 feet This means that if a tube is filled with water, and is inverted and placed in a dish of water in a manner similar to that shown m Fig 2, the result- ing height of the column of water will be about 34 feet 6. The barometer is an instrument used for measuring the pressure of the atmosphere There are two kinds in general use the mercurial barometer and the aneroid barometer. The latter was described in Leveling The mercurinl barometer is shown in Fig 3. The principle is the same as in the case of the inverted tube shown in Fig 2 The tube and cup at the bottom are protected by a brass or an iron casing At the top of the tube is a graduated scale that can be read to rsW inch by means of a vernier. Attached to the casing is an accurate thermometer for deter- mining the temperature of the outside an at the time the barometric observation is taken. This is necessary, since mercury expands when the temperature rises, and contracts when the temper- ature falls; for this reason a standard temperature is assumed, and all barometer readings are reduced to this temperature. This standard temperature is usually taken as 32 F., at which temperature the height of the mercurial PlG 3 32 PNEUMATICS 5 column is 30 inches. Another correction is made for the alti- tude of the place above sea level, and a third correction for the effects of capillary attraction. It is not necessary here to go into details regarding these corrections. 7. The pressure of the atmosphere varies with the alti- tude above sea level, being greater in low than in high places. At the level of the sea, the height of the mercurial column is about 30 inches, at 5,000 feet above the sea, it is 24.7 inches; at 10,000 feet above the sea, it is 20 5 inches; at 15,000 feet above the sea, it is 16.9 inches; at 3 miles, it is 16 4 inches; and at 6 miles above the sea level, it is 8 9 inches. The density also varies with the altitude; that is, a cubic foot of air at an elevation of 5,000 feet above the sea level does not weigh as much as a cubic foot at sea level. This is proved conclusively by the fact that at a height of 83" miles the mercurial column measures but 15 inches, indicating that half the weight of the entire atmosphere is below that height It is known that the height of the earth's atmos- phere is at least 50 miles; hence, the air just before reach- ing the limit must be in an exceedingly rarefied state. It is by means of barometers that great heights are measured. The aneroid barometer has the heights marked on the dial, so that it can be read directly. With the mercurial barom- eter, the heights must be calculated from the reading. (See Leveling.) 8. The atmospheric pressure is everywhere present, and presses all objects in all directions with equal force. If a book is laid on the table, the air presses on it in every direction with an equal average force of about 14.7 pounds per square inch. It would seem as though it would take considerable force to raise a book from the table, since, if the size of the book were 8 inches by 5 inches, the pressure on it would be 8 X 6 X 14 7 = 588 pounds But there is an equal pressure beneath the book to counteract the pressure on the top. It would now seem, as though it would require a great force to open the book, since there are two pressures I LT 398-17 6 PNEUMATICS 32 of 588 pounds each, acting in opposite directions, and tending to crush the book, and so it would but for the fact that there is a layer of air between each two leaves acting upwards and downwards with a pressure of 14.7 pounds per square inch. If two metal plates are made as perfectly smooth and flat as it is possible to make them, and the edge of one is laid on the edge of the other, so that one may be slid on the othei, and the air thus excluded, it will take an immense force, compared with the weight of the plates, to separate them. This is because the full pressure of 14 7 pounds per square inch is then exerted on each plate with no counteracting equal pressure between the plates If a piece of flat glass is laid on a flat surface that has been previously moistened with water, it will require considerable force to lift it off the surface. This is due to the fact that the water helps to fill up the pores m the flat surface and glass, and thus creates a partial vacuum between the glass and the surface, thereby reducing the counterpressure beneath the glass. 9. Tension of Gases. In Art. 5, it was said that the space above the column of mercury in Fig. 2 was a vacuum, and that if any gas or air were present it would expand, its tension forcing the column of mercury downwards. If enough gas is admitted to cause the mercury to stand at 15 inches, the tension of the gas is evidently 14,7 2 = 7 35 pounds per square inch, since the pressure of the out- side air of 14.7 pounds per square inch only balances 15 inches, instead of 30 inches, of mercury, that is, it balances only half as much as it would if there were no gas in the tube; hence, the pressure (tension) of the gas in the tube is 7.35 pounds. If more gas is admitted until the top of the mercurial column is just level with the mercury in the cup, the gas m the tube has then a tension equal to the outside pressure of the atmosphere. Suppose that the bottom of the tube is fitted with a piston, and that the total length of the inside of the tube is 36 inches. If the piston is shoved upwards so that the space occupied by the gas is 18 inches long, instead 32 PNEUMATICS 7 of 36 inches, the temperature remaining the same as before, it will be found that the tension of the gas within the tube is 29.4 pounds per square inch. It will be noticed that the volume occupied by the gas is only half that in the tube before the piston was moved, while the pressure is twice as great, since 14.7 X 2 = 29.4 pounds. If the piston is shoved up so that the space occupied by the gas is only 9 inches, instead of 18 inches, the temperature still remaining the same, the pressure will be found to be 58.8 pounds per square inch. The volume has again been reduced one-half, and the pressure increased two times, since 29.4 X 2 = 58.8 pounds. The space now occupied by the gas is 9 inches long, whereas, before the piston was moved, it was 36 inches long, as the tube was assumed to be of uniform diameter throughout its length, the volume is now TUT = i of its original volume, and CO O its pressure is -"- = 4 times its original pressure. More- over, if the temperature of the confined gas remains the same, the pressure and volume will always vary in a similar way. The law that states these effects is called Marwtte*s law, and is as follows: 10. Marietta's Law. The temperature remaining the same, the volume of a. given quantity of gas varies inversely as the pressure. The meaning of this is: If the volume of the gas is dimin- ished to one-half, one-third, one-fifth, etc. of its former volume, the tension will be increased two, three, five, etc. times; or, if the outside pressure is increased two, three, five, etc. times, the volume of the gas will be diminished to one- hnlf, one-third, one-fifth, etc. of its original volume, the tem- perature remaining constant. Suppose 3 cubic feet of air to be under a pressure of 60 pounds per square inch in a cylinder fitted with a mov- able piston; then the product of the volume and pressure is 3 X 60 = 180. Let the volume be increased to 6 cubic feet; then the pressure will be 80 pounds per square inch, and 30 X 6 SB 180, as before, Let the volume be increased to 8 PNEUMATICS 32 24 cubic feet; it is then 24 3 = 8 times the original volume, and the pressure is one-eighth of the original pressure, or 60 X i = 71 pounds, and 24 X 7i = 180, as in the two pre- ceding cases. It will now be noticed that, if a gas is allowed to expand without change of temperature, the pioduct of any Pressure and the corresponding volume is the same as for any other pressure and the corresponding volume If the air were com- pressed, the same result would be obtained. Let p = pressure corresponding to volume v; pi = pressure corresponding to volume z^. Then, p v = p v Knowing the volume and the pressure for any position of the piston, and the volume for any other position, the pres- sure may be calculated; or, if the pressure is known for any other position, the volume may be calculated. EXAMPLE 1 If 1 875 cubic feet of air is under a pressure of 72 pounds per square inch, what will be the pressure when the volume is increased (a) to 2 cubic feet? (b) to 3 cubic feet? (c) to 9 cubic feet? SOLUTION Solving the last equation for/,, the unknown pressure gives . . . p v 72 X 1 875 _, ., (a) p! = r -- = - = --- = 67 Ib per sq m Ans. v\ & ... . 72 X 1 875 ._ ,. . (b) pi = = 45 Ib per sq in Ans. , . . 72 X 1 875 .,_ ,. (c) pi = g =15 Ib. per sq. in Ans EXAMPLE 2 Ten cubic feet of air has a tension of 5.6 pounds per square inch, what is the volume when the tension is (a) 4 pounds? (b) 8 pounds? (c) 25 pounds? (d) 100 pounds? SOLUTION Solving the same equation for v,. gives (a) , l =^ = 56 * 10 = 14cu ft Ans pi 4 7 cu ft Ans , (c) Vl = 5 6 * 10 = 2 24 cu ft Ans (<0 i = ^ = 56 cu ft Ans. 11. For the same quantity of gas, the weight per unit of volume varies inversely as the volume. For example, if 32 PNEUMATICS 9 1 pound of gas occupies n volume of 4 cubic feet, its weight per cubic foot will be i pound. If it occupies a volume twice as large, or 8 cubic feet, its weight per cubic foot will be i pound, or only one-half of what it was before. In general, it, the temperature of a fixed quantity of gas remaining the same, the weights per unit of volume when the gas occupies the volumes v and z>,, respectively, are denoted by w and w lt then, w _ v\ Wj. V and wv z0, v, (1) Also, since = v i and w p l = iv ip (2) EXAMPLE 1 The weight of 1 cubic foot of air at a temperature of 60 F , and under a pressure of 1 atmosphere (14.7 pounds per square inch), is 0763 pound, what would be the weight per cubic foot if the volume were compressed until the tension was 5 atmospheres, the temperature still being 60 F ? SOLUTION. Applying formula 2, 1 X w\ = 6 X .0763. Hence, a/, = .3815 Ib. per cu. ft. Ans. EXAMPLE 2. If in the last example the air had expanded until the tension was 5 pounds per square inch, what would have been its weight per cubic foot? SOLUTION. Here, p 14 7, p l = 5, and w = 0703 Hence, apply- ing formula 2, 14.7 X w, = 5 X .0763, whence a/, = ~~y = 02695 Ib. per cu. ft. Ans. EXAMPLE 3 If 6.75 cubic feet of air, at a temperature of 60 F. and a pressure of 1 atmosphere, is compressed to 2 25 cubic feet (the temperature still remaining 60 P.), what is the weight of a cubic foot of the compressed air? SOLUTION Applying formula 1, 6.75 X .0763 = 2 25 X w t . Hence, w, m - 22 2289 Ib. per cu. ft. An 8 . 12. Manometers and Gauges, There are two ways of measuring the pressure of a gas: by means -of an instru- ment called a manometer, and by means of a gauge. 10 PNEUMATICS 32 The manometer generally used is practically the same as a mercurial barometer, except that the tube is much longer, so that pressures equal to several atmospheres may be meas- ured, and is enlarged and bent into a U shape at the lower end, both the lower and the upper ends are open, the lower end being connected to the vessel containing the gas whose pressure it is desired to measure. The gauge is so com- mon that no description of it will be given here. With both the manometer and the gauge, the pressure recorded is the amount by which the piessure being measured exceeds the atmospheric pressure, and is called gauge pressure. To find the total pressure, or absolute pressure, the atmos- pheric must be added to the gauge pressure. In all for- mulas in which the pressure of a gas or steam occurs, the absolute pressure must be used, unless the gauge pressure is distinctly specified as the proper one to use. For conve- nience, all pressures given in this Section and in the ques-- tions referring to it will be absolute pressures, and the word "absolute" will be omitted, to avoid repetition. 13. In all that has been said, it has been stated that the temperature was constant; the reason for this will now be explained. Suppose 5 cubic feet of air to be confined in a cylinder placed in a vacuum, so that there will be no pressure due to the atmosphere, and suppose the cylinder to be fitted with a piston weighing, say, 100 pounds, and having an area of 10 square inches. The tension of the air will be 1 r q u a = 10 pounds per square inch. Suppose, now, that the air, originally at a temperature of 32 F., is heated until its temperature is 33 F that is, until its temperature is raised 1. It will be found that the piston has risen a certain amount, and, consequently, the volume has increased, while the pressure remains the same as befoie, 01 10 pounds per square inch. If more heat is applied, until the temperature of the gas is 34 F , it will be found that the piston has again risen, and the volume again increased, while the pressure still remains the same. It will be found that for every increase in temperature there will be a corresponding increase 32 PNEUMATICS 11 of volume. The law that expresses this change is called Gay-Lussac 's law^ and is as follows: 14. Gay-IJussac's Liaw. // the pressure remains con- stant, every increase tn temperature of 1 F. produces in a given quantity of gas an expansion 0/To of its volume at 32 f. If the pressure remains constant, it will also be found that every decrease of temperature of 1 F. will cause a decrease of TU of the volume at 32 F. Let z/ volume of any quantity of gas at 32 F.; v = volume of gas at temperature t; z^ =* volume of gas at temperature /,. It is assumed that the pressure of the gas remains unchanged. Then, m passing from the temperature 32 to the temperature /, the volume of the gas will be increased J _ QO algebraically by the amount TUYZ/O (t 32) = jnn v*. 492 Therefore, or, reducing, v = flo T~- (D 492 Likewise, v, = v, --' (a) 492 Dividing formula 1 by equation (a), v _ 460 + t_ v\ 460 + t, whence v - V *M (2) EXAMPLE If 5 cubic feet of air at a temperature of 45 is heated under constant pressure up to 177, what is the final volume of the air? SOLUTION To apply formula 2, we have v^ = 5; , 46; t 177. Therefore, 460 + 177 v * 5 x 07 cu ' ft AnSi 15. Suppose that a certain volume of gas is confined in a vessel so that it cannot expand; in other words, sup- pose that the piston of the cylinder before mentioned is fastened so that it cannot move. Let a gauge be placed on ihl 12 PNEUMATICS 32 the cylinder so that the tension of the confined gas can be registered If the gas is heated, it will be found that, for every increase of temperature of 1 F , there will be a cor- responding increase of TUT of the tension That is, the volume remaining: constant, the tension increases Tn of the original tension for eveiy degree rise of temperature. Let p tension of gas at temperature /, pi = tension of gas at temperature ti. Then, as in the preceding article, = y ^460 + ^ EXAMPLE If a certain quantity of air is heated under constant volume from 45 to 177, what is the resulting tension, the original tension being 14 7 pounds per square inch? SOLUTION Applying the formula, p = 14 7 X t' 7 - 18 542 lb per sq in Ans 16. According to the modern and now generally accepted theory of heat, the atoms and molecules of all bodies are in an incessant state of vibration. The vibratory movement in liquids is faster than in solids, and in gases faster than in either of the other two Any increase of heat increases the vibrations, and a decrease of heat decreases them From experiments and advanced mathematical investigations, it has been concluded that at 460 below zero, on the Fahrenheit scale, all these vibrations cease. This point is called the absolute zero, and all temperatures reckoned from this point are called absolute temperatures. The point of absolute zero has never been reached, nevertheless, it has a meaning, and is used in many formulas, absolute tem- peratuies being usually denoted by T. Ordinary temperatures are denoted by /. When the word temperature alone is used, it refers to the ordinary way of measuring temperatures, but when absolute temperature is specified 460 F. must be added to the ordinary temperature The absolute temperature cor- responding to 212 F. is 460 -f 212 = 672 F, If the abso- lute temperature is given, the ordinary temperature may be found by subtracting 460 from the absolute temperature. 32 PNEUMATICS 13 Thus, if the absolute temperature is 520 P., the ordinary tempeiature is 520 - 460 = 60. 17. Let p = pressure, in pounds per square inch, of W pounds of air; V = volume of air, in cubic feet, T absolute tempeiature of air. It is shown in advanced works on the theory of heat that these quantities are related by the following- equation: pV = .37 WT EXAMPLE 1 The pressure on 9 cubic feet of air weighing 1 pound is 20 pounds per squaie inch, what is the ordinary temperature of the air? SOLUTION Here, W= 1. Substituting the other values in the 180 formula gives 20 X 9 = 37 T, hence, T = ~ = 486.6, nearly. .37 40 5 - 4(50 = 26 6, the ordinary temperature. Ans EXAMPLE 2 What is the volume of 1 pound of air whose tempera- ture is 00 F under a pressure of 1 atmosphere? SOLUTION Here, W - 1, and T = 460 -f 60 - 520, therefore, 14 7 07 \f Knn X V = .87 X 520, whence V = ~^~ = 18 088 cu. ft Ans. EXAMPLE 3 If 3 cubic feet of air weighing 35 pound Is under a pressure of 48 pounds per square inch, what Is the ordinary tem- perature of the air? SOLUTION Applying the formula, 48 X 8 - .37 X .35 X T, JO v ^ whence, T= --%. - 1,112 Then, 1,112 - 460 - 652. A I X <50 Ans. EXAMPLK 4. What is the weight of 1 cubic foot of air at a tempera- ture of 32, and under a pressure of 1 atmosphere? SOLUTION Here T - 400 + 32 = 492; V = 1; and p = 14 7 (Art. 4). Substituting these values in the formula gives 14 7 X 1 - .37 X 492 X W, whence, Ans - If the pressure is taken as 14.696 Ib. per sq. in , the weight of 1 cu. ft. of air, at 32 and atmospheric pressure, is found to be - - 08078 lb 14 PNEUMATICS 32 EXAMPLES FOR PRACTICE 1 A vessel contains 25 cubic feet of gas at a pressure of 18 pounds per square inch, if 125 cubic feet of gas having the same pressure Is forced into the vessel, what will be the resulting pressure? Ans 108 Ib. per sq. in. 2 A pound of air has a temperature of 126, and a pressure of 1 atmosphere, what volume does it occupy? Ans. 14 75 cu. ft. 3 A certain quantity of air has a volume of 26 7 cubic feet, a pres- sure of 19.3 pounds per square inch, and a temperature of 42; what is its weight? Ans. 2 77 Ib 4 A receiver contains 180 cubic feet of gas at a pressure of 20 pounds per square inch, if a vessel holding 12 cubic feet is filled from the receiver until its pressure is 20 pounds per square Inch, what will be the pressure m the receiver? Ans 18$ Ib per sq in. 5 Ten cubic feet of air, having a pressure of 22 pounds per square inch and a temperature of 75, is heated until the temperature is 300, the volume remaining the same, what is the new pressure? Ans 31 25 Ib per sq in. THE MIXING OF GASES 18. If two liquids that do not act chemically on each other are mixed together and allowed to stand, it will be found that after a time the two liquids have separated, and that the heavier has fallen to the bottom. If two equal ves- sels containing: gases of different densities are put in com- munication with each other, it will be found that after a short time the gases have become mixed in equal proportions. If one vessel is higher than the other, and the heavier gas is in the lower vessel, the same result will occur. The greater the difference of the densities of the two gases, the faster they will mix. It is assumed that no chemical action takes place between the two gases. When the two gases have the same temperature and pressure, the pressure of the mixture will be the same. This is evident, since the total volume has not been changed, and, unless the volume or temperature changes, the pressure cannot change This property of the mixing of gases is a very valuable one, since, if gases acted like liquids, carbonic-acid gas (the result 32 PNEUMATICS 15 of combustion), which is li times as heavy as air, would remain next to the earth, instead of dispersing into the atmosphere, the result being that no animal life could exist. 19. Mixture of Equal Volumes of Gases Unequal Pressures. // two gases having equal volumes and temperatures, but different Pressures, ate mixed in a vessel vu hose volume equals one of the equal volumes of the gases, the pressure of the mixture will be equal to the sum of the two pressures, provided that the temperature remains the same as before. 20. Mixture of Two Gases Having; Unequal Vol- umes and Pressures. Let v and p be the volume and pressure, respectively, of one of the gases. Let v 1 and PI be the volume and pressure, respectively, of the other gas. Let f-'and P be the volume and pressure, respectively, of the mixture. Then, if the temperature remains the same, yp= vfi + v^ That is, if the temperature is constant, the volume after mix- ture, multiplied by the resulting pressure, equals the volume of one gas before mixture multiplied by its pressure, plus the volume of the other gas multiplied by its pressure* EXAMPJ.B Two gases at the same temperature, having volumes of 7 cubic feet and 4J cubic feet, and pressures of 27 pounds and 18 pounds per square inch, respectively, are mixed together In o vessel whose volume is 10 cubic feet. What is the resulting pressure? SOLUTION. Applying the preceding formula, J V p v + p^ v tl or />X 10 - 27 X 7 + 4t X 18. Hence, P - i?-il - 27 Ib. per sq. in. Aim. 21. Mixture of Two Volumes of Air Having Unequal Pressures, Volumes, and Temperatures. If a body of air having a temperature /, a pressure p^ and a vol- ume v l is mixed with a volume of air having 1 a temperature * a pressure /> and a volume z>,, to form a volume f having a pressure /* and a temperature t, then, either the new temper- ature /, the new volume V, or the new pressure P may be found, if the other two quantities are known, by the following 16 PNEUMATICS 32 formula, in which T lt T at and Tare the absolute temperatures corresponding to / / a , and /: PV= /A^i + A \Ti T EXAMPLE Five cubic feet of air having a tension of 30 pounds per square inch and a temperature of 80 F are compressed, together with 11 cubic feet of air having a tension of 21 pounds per square inch and a temperature of 45 F , in a vessel whose cubical contents are 8 cubic feet The new pressure is required to be 45 pounds per square inch What must be the temperature of the mixture? SOLUTION Substituting in the formula, 45 X 8 = Qflft X T, or 360 = .7352 T Hence, T = ~^ = 489 66, nearly, and / = 29 66 Ans EXAMPLES FOB PRACTICE 1 Two vessels contain air at pressures of 60 and 83 pounds per square inch The volume of each vessel is 8 47 cubic feet If all the air in both vessels is removed to another vessel, and the new pressure is 100 pounds per square inch, what is the volume of the vessel, the temperature remaining unchanged? Ans. 12 11 cu ft 2 A vessel contains 11 83 cubic feet of air at a pressure of 33 3 pounds per square inch It is desired to increase the pressure to 40 pounds per square inch by supplying air from a second vessel that contains 19 6 cubic feet of air at a pressure of 60 pounds per square inch What will be the pressure in the second vessel after the pressure in the first has been raised to 40 pounds per square inch? Ans. 55.96 Ib. per sq m. 3 If 4 8 cubic feet of air having a tension of 52 pounds per square inch and a temperature of 170 is mixed with 13 cubic feet having a tension of 78 pounds per square inch and a temperature of 265, what must be the volume of the vessel containing the mixture in older that the tension of the mixture may be 30 pounds per square inch and the temperature 80? Ans. 32 31 cu ft. PNEUMATICS 17 PNEUMATIC MACHINES THE AIR PUMP 22. The air pump is an instrument for removing air from an enclosed space. A section of the principal parts is shown in Fig. 4, and a view of the complete instrument is given in Fig. 5. The closed vessel R is called the receiver, FIG 4 and the space it encloses is that from which it is desired to remove the air. The receiver is usually made of glass, and the edges are ground so as to be perfectly air-tight. When made in the form shown, it is called a boll-Jar receiver, The receiver rests on a horizontal plate, in the center of which is an opening communicating with the pump cylin- der C by means of a bent tube /. The pump piston fits the cylinder accurately, and has a valve V 1 opening upwards. At the junction of the tube with the cylinder is another valve V, also opening upwards, When the piston is raised, 1 PNEUMATICS the valve V closes, and, sinct 1 no .111 can m't into tin- i \liu dcr from above, the piston leave's a \aamm brliind it. Tin pressure on top of rbemjj now u-ninvi'd, tin 1 ti'iismu il tli ,ur in llu- rm-iMx IM A' iMllsi's / * tn MSI-, lh' .111 in tlu- ii'i'i'ivvi thru t'\ pands and uivupu-s thr sp.u't* U'tt rmpty by the pistnn, as wi-ll as tin- spaiv in the- tube / and tin* IT or i VIM A*. Tbtj piston is iiuw pnslu-tl down, tin 1 vahc- /"i-losfs, tlk 1 \,dvt* / ' opens, and ihr an nt (' csoapcs. Tilt* valv /" is snni suppiirli'tl, as shown in Kitf. 1, by a nu-tal nxl passing tlimuuh llu- piston and itttinu it somewhat tightly. When the piston is raised or lowered, this rod moves with it. A button near the upper end of the rod confines its motion to within very narrow limits, the piston sliding on the rod during the greater part of the journey. S3. JUetfi'ooH und JLlmltH of Kxlvauwtlon, Suppose that the volume o /? and / together is four times that of ( \ and that there are, ny, 200 grains of nir in 'A* and /, and 50 grains in C, when the piston is at the top of the i-ylimU-i. At the end of the first stroke, when the piston is a^ain at the top, 60 grains of air in the cylinder ( ' will have been removed, and the 200 grains in /? and / will omipy the spaces jR t /, and C. The ratio between the sum of the spaces .# and / and the total space jV + / f Tis ^; hence, j 200 X - 160 grains - the weight of air in / and t after 32 PNEUMATICS 19 the first stroke. After the second stroke, the weight of the air in R and / will be 200 X X = 200 X (*)' = 200 X iif = 128 grams. At the end of the third stroke, the weight will be [200 X ()'] X = 200 X (*)' = 200 X rW = 102 4 grains. At the end of n strokes, the weight will be 200 X (&)" It is evident that it Zi impossible by this method to remove all the air contained in R and t. It requires an exceedingly good air pump to reduce the tension of the air in R to -gV inch of mercury. When the an has become so rarefied as this, the valve V will not lift, and, consequently, no more air can he exhausted 24. Spreiigel's Air Pump. In Fig 6, cd is a glass tube longer than 30 inches, open at both ends, and connected by means of India-rubber tubing with a funnel A filled with mer- cury and supported by a stand. Mercury is allowed to fall into this tube at a rate regulated by a clamp at c. The lower end of the tube c d fits in the flask ./?, which has a spout at the side a little higher than the lower end of c d\ the upper part has a branch at x to which a receiver R can be tightly fixed. When the clamp a.t c is opened, the first portions of the mer- cury that run out close the tube and prevent air from enterirg Fia. 6 20 PNEUMATICS 32 from below. These drops of mercury act like little pistons, carrying the air in front of them and forcing it out through the bottom of the tube. The air in R expands to fill the tube every time that a drop of mercury falls, thus creating a partial vacuum in R, which becomes more nearly complete as the process goes on. The escaping mercury falls into the dish H, from which it can be poured back into the funnel from time to time. As the exhaustion from R goes on, the meicury rises in the tubecd until, when the exhaustion is complete, it forms a continuous column 30 inches high, in other words, it is a barometer whose vacuum is the receiver R. This instrument necessarily requires a great deal of time for its operation, but the results are very complete, a vacuum of ^aiUa inch of mercury being sometimes obtained. By use of chemicals in addition to the above, a vacuum of TBTMMTO inch of mercury has been obtained. 25. Magdeburg: Hemispheres. The pressure of the atmosphere can be made manifest by means of two hollow hemispheres, such as are shown in Fig. 7. This contrivance was devised by Otto Von Guencke, of Magdeburg, and is known as the Magdeburg hemispheres. One of the hemispheres is provided with a stop- cock, by which it can be screwed on to an air pump. The edges fit accurately and are well greased, so as to be air-tight. When the hemispheres contain air, they can be separated easily; when the air is pumped out by an air pump, they can be separated only with great difficulty. The force required to separate them will be equal to the area of the largest circle of the hemisphere (projected area) m square inches, multiplied by 14.7 pounds. This force will be the same in whatever position the hemispheres may be held, .which proves that the pressure of air on them is the same m all directions. FIG 7 32 PNEUMATICS 21 26. The Weight Liifter. The pressure of the atmos phere is shown by means of the apparatus illustrated in Fig 8. Here, a cylinder fitted with a piston is held in suspension by a chain. At the top of the cylinder is a plug a, which can be taken out. This plug is removed and the piston is pushed up until it touches the cylinder head. If the plug is then screwed in, the piston will remain at the top until a weight has been hung on the rod equal to the area of the piston multiplied by 14.7 pounds, less the weight of the pibton and rod. If a force is applied to the rod suffi- ciently great to push the piston down- wards, the piston will, on the removal of the force, raise to the top of the cylinder any weight that is less than the one men- tioned. Suppose the weight to be removed, and the piston to be supported midway between the top and bottom of the cylinder. Let the plug be removed, air admitted above the piston, and the plug screwed back into its place; if the piston is shoved upwards, the farther up it goes, the greater will be the force necessary to push it, on account of the compression of the air. If the piston is of large diameter, it will also require a great force to pull it out of the cylinder, as a little consideration will show. For example, let the diameter of the piston be 20 inches, the length of the cylinder 86 inches, plus the thickness of the piston, and the weight of the piston and rod 100 pounds. If the piston is in the middle of the cylinder, there will be 18 inches of space above it, and 18 inches of space below it. The area of the piston is 20' X .7864 =* 314.16 square inches, and the atmospheric pressure on it is 814.16 X 14.7 4,618 pounds, nearly. In order to shove the piston upwards 9 inches, the pressure on it must be twice as great, or 9,286 pounds, and I LT 398-18 Pro 8 22 PNEUMATICS 32 to this must be added 100 pounds, the weight of the piston and rod, which gives 9,236 + 100 = 9,336 pounds. The force necessary to cause the piston to move upwards 9 inches will then be 9,336 4,618 = 4,718 pounds. Now, suppose the piston to be moved downwards until it is just on the point of being pulled out of the cylinder The volume above it will then be twice as great as before, and the pressure one-half as great, or 4,618 2 = 2,309 pounds. The total upward pies- sure will be the pressure of the atmosphere less the weight of the piston and rod, or 4,618 100 = 4,518 pounds, and the force necessary to pull it downwards to this point will be 4,518 - 2,309 = 2,209 pounds. 27. Tlie Baroscope. The buoyant effect of air is very clearly shown by means of an instrument called the baro- scope, shown in Fig. 9. It con- sists of a scale beam, from one extremity of which is suspended a small weight, and from the other a hollow copper sphere. In air, they exactly balance each other, but when they are placed under the receiver of an air pump and the air is exhausted the sphere sinks, showing that it is really heavier than the small weight. Before the air is exhausted, each body is buoyed up by the weight of the air it displaces, and, since the sphere displaces more air, it loses more weight by reason of this displacement than the small weight. Suppose that the volume of the sphere exceeds that of the weight by 10 cubic inches; the weight of this volume of air is 3.1 grains. If this weight is added to the small weight, it will overbalance the sphere in air, but will exactly balance it m a vacuum. FIG 9 32 PNEUMATICS 23 AIR COMPRESSORS 28. For many purposes, compressed air is preferable to steam or other gases for use as a motive power; m such cases, air compressors are used to compiess the air. These are made in many forms, but the most common one consists of a cylinder, called the air cylinder, placed in front of the crosshead of a steam engine, so that the piston of the air cylinder can be driven by attaching its piston lod to the crosshead, in a manner similar to a steam pump. A cross- section of the air cylinder of a compressor of this kind is shown in Fig 10, in which a is the piston and b is the piston rod, driven by the crosshead of a steam engine not shown in the figure. Both ends of the lower half of the cylinder are fitted with inlet valves d and d', which allow the air to enter the cylinder, and both ends of the upper half aie fitted with discharge valves / and f, which allow the air to escape fiom the cylinder after it has been compressed to the required pressure. Suppose the piston a to be moving in the direction of the arrow; then the inlet valves d in the left-hand end of the cylinder from which the piston is moving will be forced inwards by the pressure of the atmosphere, which over- comes the resistance of the light spring: c t thus allowing the air to flow in and fill the cylinder. On the other side of the piston, the air is being compressed, and, consequently, it acts with the springs ^ to force the inlet valves d' in the right-hand end of the cylinder to their seats. In the right-hand end of the cylinder, the discharge valves f are opened when the pressure of the air in the cylinder is great enough to overcome the resistance of the light springs f and the tension of the air in the passages leading to the discharge pipe h, and the discharge valves / are pressed against their seats by the springs c and the tension of the air in the passages. Suppose it is desired to compress the air to 59 pounds per square inch, and to find at what point of the stroke the discharge valves will open. Now, a pressure of 59 pounds per square inch equals a pressure of 4 atmospheres, 24 PNEUMATICS 32 very nearly, hence, when the pressure in the cylinder becomes great enough to force air out through the discharge valves, the volume must be one-quarter of the volume at atmos- pheric pressure, or the valves will open when the piston has traveled three-quarters of its stroke, provided that the air is compressed at constant temperature. The air, after being discharged from the cylinder, passes out through the delivery pipe h, and from there is con- FIG 10 veyed to its destination. It has been shown that when air or any other gas is compressed its temperature is increased. For high pressures, this increase of temperature becomes a serious consideration, for two reasons: (1) When the air is discharged at a high temperature, the pressure falls con- siderably when the air has cooled down to its normal temper- ature, and this represents a serious loss in the economical working of the machine. (2) The alternate heating and 32 PNEUMATICS 25 cooling of the compressor cylinder by the hot and cold air is very destructive to it, and increases the wear to a great extent. To prevent the air from heating, cooling devices are lesorted to, the most common one being 1 the so-called water-Jacket. This is effected in the following manner The cylinder walls are hollow, as shown in Fifi 10; the cold water enters this hollow space in the cylinder wall through the pipe k k, and flows around the cylinder, finally passing out through the discharge pipe /. The water keeps cold the cyl- inder walls, which cool the air as it is compressed. 29. Hei-o's Fountain. Hero's fountain derives its name from its inventor, Hero, who lived at Alexandria about 120 B. C. This fountain, which is shown in Fig. 11, consists of a brass dish A and two glass globes B and C, and depends for its operation on the elastic prop- erties of air The dish com- municates with the lower part of the globe C by a long tube D, and another tube E connects the two globes. A third tube passes through the dish A to ^ the lower part of the globe B. _ This last tube being taken out, the globe B is partially filled with water; the tube is then replaced and water is poured into the dish. The water flows through the tube D into the lower globe, and expels the air, which is forced into the upper globe. The air thus com- pressed acts on the water and makes it jet out through the shortest tube, as represented in the figure. Were it not for 33 Fie. 11 26 PNEUMATICS .12 the resistance offered by the atmosphere and by friction, the issuing water would rise to a height above the water in the dish equal to the difference of the level of the water m the two globes. THE SIPHON 30. The action of the siphon illustrates the effect of atmospheric pressure. A siphon is simply a bent tube with unequal branches, open at both ends, and is used to convey a. liquid from a higher point to a lower, over an intermediate point higher than either of the other two. In Fig 12, a and b are two vessels, b being lower than a, and a c b is the bent tube or siphon. Suppose this tube to be filled with water and placed in the vessels, as shown, with the short branch a c in the vessel a. The watei will flow fiom the vessel a into , so long as the level of the water in b is below the level of the water in a and the level of the water in a is above the lower end of the tube ac. Tne atmospheric pressure on the surfaces of a and b tends to force the water up the tubes ac and be. When the siphon is filled with water, each of these pressures is counteracted in part by the pressure of the water in that branch of the siphon that is immersed in the water on which the pres- sure is exerted. The atmospheric pressure opposed to the weight of the longer column of water will, therefore, be more resisted than that opposed to the weight of the shorter column; consequently, the pressure exerted on the shorter column will be greater than that on the longer column, and this excess of pressure will produce motion, The action of the siphon is of great importance, and should be thoroughly understood. The following considera- tions and computations will make the subject clear: PNEUMATICS 27 Let a = area of tube, in square inches; // = dc = vertical distance, in inches, between the surface of water in b and highest point of the center line of tube, //, = e c = distance, in inches, between the surface of water in. a and highest point of center line of tube. The weight of the water in the short column is .03617 a hi, and the resultant atmospheric pressuie, tending to force the water up the short column, is 14 7 X a .03617 a //,. The weight of the water in the long column is 03617 a/i t and the icsultant atmospheric pressure, tending to force the water up the long column, is 14.7 a .03617 a k. The differ- ence between these two is (14.7 a .03617 ah^} (14.7 a - 03617 ah} = .03617 a (A - k,) But h - //, = ed = dif- ference between the levels of the water in the two vessels. It will be noticed that the short column must not be higher than 34 feet for water, or the siphon will not work, since the ,r\ PIG 18 pressure of the atmosphere will not support a column of water that is higher than 84 feet; 28 feet is considered to be the greatest height for which a siphon will work well. 31. Intermittent Springs.- Sometimes a spring is observed to flow for a time and then cease; then, after an 28 PNEUMATICS 32 interval, to flow again for a time The generally accepted explanation of this is that there is an underground reservoir fed with water through fissures in the earth, as shown in Fig. 13 The outlet for the water is shaped like a siphon, as shown. When the water in the reservoir reaches the same height as the highest point of. outlet, it flows out until the level of the water in the reservoir falls below the mouth of the siphon, if the flow of water is greater than the supply to the reservoir, in which case the flow ceases until the water in the reservoir again reaches the level of the highest point of the siphon. THE LOCOMOTIVE BLAST 32. Fig. 14 shows the front end of a locomotive: B is the exhaust pipe, the center of which is directly in line with PIG 14 the center of the smokestack S, T, T are the tubes through which the hot furnace gases are discharged. The exhaust 32 PNEUMATICS 29 steam has a pressure of about 2 pounds above the atmos- phere, and rushes through the exhaust pipe E and up the smokestack 5 1 with a very high velocity, taking the air out with it, and producing 1 a partial vacuum in the space in front of the tubes. No air can get in this space except through the grates of the firebox, consequently, the partial vacuum created in front of the tubes causes an influx of air through the grate, and produces the forced draft, or blast. The faster the engine runs, the greater is the quantity of air drawn through the grate. PUMPS 33. The Suction Pump. A section of an ordinary suction pump is shown in Fig. 15. Suppose the piston to be at the bottom of the cylinder and to be just on the point of moving upwards in the direction of the arrow. As the piston rises, it leaves a vacuum behind it. The air in P then raises the valve P, and expands in the cylinder .#, whereby its pressure is dimin- ished below that of the atmosphere. The atmospheric pressure on the surface of the water in the well causes the water to rise in the pipe P. When the piston descends, the air in B escapes through the valves . After a few strokes, the water fills completely the space under the piston in cylinder JB, so that, when the piston reaches the end of its stroke, the water entirely fills the space between the 30 PNEUMATICS 32 bottom of the piston and the bottom of the cylinder and also the pipe P. The instant that the piston begins its down stroke, the water in the chamber B tends to fall back into the well, and its weight forces the valve Fto its seat, thus preventing any downward flow of the water. The piston now tends to compress the water in the chamber B, but this is prevented through the opening of the valves u, u in the piston. When the piston has reached the end of its down- ward stroke, the weight of the water above closes the valves u, u. All the water resting on the top of the piston is then lifted with the piston on its upward stroke, and discharged through the spout A, the valve V again opening, and the water filling the space below the piston as before It is evident that the distance between the valve V and the surface of the water m the well must not exceed 34 feet, the highest column of water that the pressure of the atmos- phere will sustain, since otherwise the water in the pipe 3 would not reach to the height of the valve V In practice, this distance should not exceed 28 feet This is due to the fact that there is a little air left between the bottom of the piston and the bottom of the cylinder, a little air leaks through the valves, which are not perfectly air-tight, and a pressure is needed to raise the valve against its weight, which, of course, acts down wards. There are many vari- eties of the suction pump, differing principally in the valves and piston, but the principle is the same in all. 34. The Lifting Pump. A section of a lifting pump is shown in Fig 16 These pumps are used FIG IB when water is to be raised to greater heights than can be done with the ordinary suction pump. As will be perceived, it is essentially the same as the suction 32 PNEUMATICS 31 PlO. 17 pump, except that the spout is fitted with a cock and has a pipe attached to it, leading to the point of discharge. If it is desired to dis- charge the water at the spout, the cock may be opened, otherwise, the cock is closed, and the water is lifted by the piston up through the pipe P' to the point of discharge, the valve c preventing it from falling back into the pump, and the valve V preventing the water in the pump from falling back into the well. It is not necessary that there should be a second pipe P f , as shown in the figure, for the pipe P may be continued straight upwards, as shown in Fig 17. This figure shows a section of a lifting pump for raising water from great depths, as from the bottom of mines to the surface. The pump consists of a series of pipes connected together, of which the lower end only is shown in the figure. That part of the pipe included between the letters A and B forms the pump cylinder, in which the piston P works. That part of the pipe above the highest point of the piston travel, thiough which the water is discharged, is called the delivery pipe, and the part below the lowest point of the piston travel is called the suction pipe. The lower end of the suction pipe is expanded, and has a number of small holes m it, to keep out solid matter. C is a plate covering an opening, and may be removed to allow the suction valve to be repaired. D is a plate covering a similar opening, through which the piston and piston valves may be repaired. The pibton rod, or rather the piston stem, is made of wrought iron, inserted with wood, and connected with the piston. The only limit to the height to which a pump of this 32 PNEUMATICS 32 kind can raise water is the strength of the piston rod Lifting pumps of this kmd are used to raise water from great depths to the earth's surface; hence, a very long piston rod is neces- sary. In the lifting pump shown in Fig. 16, the water is raised from a point a few feet below the earth's surface to a point considerably higher. This requires the piston rod to move through a stuffingbox, as shown at 5 1 , and also necessitates the rod being round, in order that the water may not leak out. 35. Force Pumps. The force pump differs from the lifting pump in several important particulars, but chiefly in the fact that the piston is solid, that is, it has no valves A section of a suction and force pump is shown in Fig. 18. The water is drawn up the suction pipe as be- H - -T ^ F fore, when the piston | 5 rises, but when the piston reverses the pressure on the water caused by the descent of the piston opens the valve V and forces the water up the delivery pipe P'. When the piston again begins its up- ward movement, the valve V is closed by Fro I8 the pressure of the water above it, and the valve V is opened by the pressure of the atmosphere on the water below it, as in the previous cases. For an arrangement of this kind, it is not necessary to have a stuffingbox. The water may be forced to almost any desired height. The force pump differs again from the lifting pump in respect to its piston rod, which should not be longer than is absolutely necessary in order to prevent it from buckling, while, in the lifting pump, the length of the piston rod is a matter of indifference. 34 PNEUMATICS .".2 pressure below the plunger being less than the pressure of the atmosphere above, the air would rush in instead of being expelled. 37. Double-Acting- Pumps. In the pumps previously described, the discharge was intermittent, that is, the pump could only discharge when the piston was moving in one direction. In some cases, it is necessary that there should PIG 20 U a continuous discharge; in all cases, it takes more power to run the pump with an intermittent discharge, as a little consideration will show. If the height that the water is to be raised is considerable, its weight will be very great, and the entire mass must be put in motion during one stroke of the piston. In order to obtain the advantage of a more continuous discharge, double-acting pumps are used. Fig 20 shows a part sectional view of such a pump. Two pistons a and 6 32 PNEUMATICS 35 are used, which are operated by one handle c in the manner shown. The pump has one suction pipe s and one discharge pipe d The cylinders e and / are separated by a diaphragm , so that they cannot communicate with each other above the pistons In the figure, the handle c is moving to the right, the piston a upwards, and the piston b downwaids In moving upwards, the piston a lifts the water above it, causing it to flow through the delivery valve h into the discharge pipe d. This upward movement of the piston creates a. partial vacuum below it m the cylinder <?, and causes the water to rush up the suction pipe j into the cylinder, as shown by the arrows In the cylinder /, the downwaid movement of the piston b raises the piston valve v, and the weight of the water on the suction valve z keeps it closed. When the handle c has completed its movement to the right and begins its return, all the valves on the right-hand side open except v, and those on the left-hand side close except /; watei is then discharged into the delivery pipe by the cylin- der /, and only at the instant of reversal is the flow into the delivery pipe d stopped. 38. Air Chambers. In order to obtain a continuous flow of water in the delivery pipe, with as nearly a unifoim velocity as possible, an air cluimber is usually placed on the delivery pipe of force pumps as near to the pump cylinder as the construction of the machine will allow. The air chambeis aie usually pear-shaped, with the small end con- nected to the pipe. They are filled with air, which the water compresses during the discharge. During the suction, the air thus compressed expands and acts as an accelerating force on the moving column of water, a force that diminishes with the expansion of the air, and helps to keep the velocity of the moving column more nearly uniform. An air chamber IH sometimes placed on the suction pipe. These air chambers not only tend to promote a uniform discharge, but they also equalize the stresses on the pump, and prevent shocks due to the incompressibility of water. They serve the same pur- pose in pumps that flywheels do m steam engines. Unless 36 PNEUMATICS 32 the pump moves very slowly, it is absolutely necessary to have an air chamber on the delivery pipe. 39. Steam Pumps. Steam pumps are force pumps operated by steam acting on the piston of a steam engine, directly connected to the pump, and in many cases cast with the pump. A section of a double-acting steam pump showing 1 the steam and water cylinders, with other details, is illustrated in Fig. 21. Here G is a steam piston, and R the piston rod, which is secured at its other end to the plunger P. Fis a partition cast with the cylinder, which prevents the water in FIG 21 the left-hand half from communicating with that m the right- hand half of the cylinder. Suppose the piston to be moving in the direction of the arrow The volume of the left-hand half of the pump cylinder will be increased by an amount equal to the area of the circumference of the plunger multi- plied by the length of the stroke, and the volume of the right- hand half of the cylinder will be diminished by a like amount In consequence of this, a volume of water in the right-hand half of the cylinder equal to the volume displaced by the plunger m its forward motion will be forced through the valves V. V into the air chamber A, through the orifice D, 32 PNEUMATICS 37 and then dischaiged through the delivery pipe H. By reason of the partial vacuum in the left-hand half of the pump cylin- der, owing to this movement of the plunger, the water will be drawn from the reservoir through the suction pipe C into the chamber K, A", lifting the valves S', S', and filling the space displaced by the plunger. During the return stroke, the watei will be drawn through the valves 5, 5 into the right-hand half of the pump cylinder, and discharged through the valves V, V in the left-hand half. Each of the four suction and four discharge valves is kept to its seat, when not working, by light springs, as shown. There ate many varieties and makes of steam pumps, the majority of which are double-acting. In many cases, two steam pumps are placed side by side, having a common dehveiy pipe. This arrangement is called a duplex pump. It is usual so to set the steam pistons of duplex pumps that when one is completing the stroke the other is in the middle of its stroke A double-acting duplex pump made to run in this manner, and having an air chamber of sufficient size, will deliver water with a nearly uniform velocity In mine pumps for forcing water to great heights, the plungers are made solid, and in most cases are extended through the pump cylinder. In many steam pumps, pistons are used instead of plungeis, but when very heavy duty is icquired plungers are preferred. 40. Centrifugal Pumps. Next to the direct-acting steam pump, the centrifugal pump is the most valuable instrument for raising water to great heights. As the name implies, the effects produced by centrifugal force are made use of. Fig. 22 represents a centrifugal pump with half of the casing removed. The hub 5* is hollow, and is connected directly to the suction pipe. The curved arms a, called vanes or wliiffs, are revolved with a high velocity in the direction of the arrow, and the air enclosed between them is driven out through the discharge passage and delivery pipe DD* This creates a partial vacuum in the casing 1 and suction pipe, and causes the water to flow in through S, This water is also I LT 398-19 58 PNEUMATICS 82 nade to revolve with the vanes, and, of course, with the same velocity. The centrifugal force of the revolving water causes t to fly outwards toward the end of the vanes, and becomes jreater the farther away the water gets from the center This :auses the water to leave the vanes, and finally to leave he pump by means of the discharge passage and delivery npe D D. The height to which the water can be forced depends on the velocity of the revolving vanes. In the construction of a centrifugal pump, particular care is re- quired in giving the correct form to the vanes, for the effi- ciency of the machine depends greatly on this feature. What is required is to raise the water, and the energy used to drive the pump hould be devoted as much as possible to this one purpose. "*he water, when it is raised, should be delivered with as ittle velocity as possible, for any velocity that the water hen possesses has been secured at the expense of the energy sed to drive the pump. The form of the vanes is such nat the water is delivered at the desired height with the 33 st expenditure of energy. The number of vanes depends on the size and capacity of lie pump. It will be noticed that, in the pump shown m lie figure, the vanes have sharp edges near the hub. The bject of this is to provide for a free ingress of the water, nd also to cut any foreign substance that may enter the ump and prevent it from working properly. Almost any liquid can be raised with these pumps, but, rhen they are intended for pumping chemicals, the casing nd vanes should be made of materials that will not be acted n by the chemicals. FIQ 22 32 PNEUMATICS 39 41. Tlie Hydraulic Ram. The construction of a hydraulic ram is shown in Fig. 23. This machine is used for raising water from a point below the level of the water in a spring or reservoir to a point considerably higher, with no power other than that afforded by the inertia of a moving column of water. In the figure, a is a pipe called the drive pipe, connecting the ram with the reservoir; the valve b slides freely in a guide, and is provided with locknuts to legulate the distance that it can fall below its seat. When the water is first turned on by opening the valve , the valve b is already opened, and the water flows out through c, as shown. PlO. 28 As the discharge continues, the velocity of the water in the drive pipe will increase until the upward pressure against the valve b is sufficient to force the valve to its seat. The actual closing of the valve takes place very suddenly, and the momentum of the column of water, which was moving with an increasing velocity through the drive pipe a, will very rapidly force some water through the valve d into the air chamber /. Immediately after this, a rebound takes place, and for a short interval of time the water flows back up the drive pipe a and tends to form a vacuum under the air-chamber valve d\ this opens the snifter valve g and admits a little air, which accumulates under the valve d and is forced into the air chamber with the next shock. This air keeps the 40 PNEUMATICS 32 air chamber constantly charged, otherwise, the water, being under a greater pressure in the air chamber than in the reser- voir, would soon absorb the air in the chamber and the ram would cease to work until the chamber was recharged with air The rebound also takes the pressure off the under side of the valve b and causes it to drop, and the above-described operations are repeated. The delivery pipe is shown at c; a steady flow of water is maintained through it by the pressure of the air in the chamber /; this air also acts as a cushion when valve b suddenly closes, and prevents undue shock to the parts of the ram. The height to which water can be raised by the hydraulic ram depends on the weight of the valve b and the velocity of the water in a. 42. Power Necessary to "Work a Pump. Principle I. In all pumps, whether lifting, force> steam single- or double-acting, or centrifugal, the number of foot- Pounds of power needed to work the pump is equal to the weight of the water in pounds, multiplied by the vertical distance, in feet, between the level of the water in the well, or sowcc, and the point of discharge, plus the work necessary to ovej come the fric- tion and other resistances. Principle II. The work done in one stroke of a pump zs equal to the weight of a volume of water equal to the volume displaced by the piston during the stroke, multiplied by the total vertical distance, in feet, through which the water is to be raised, plus the work necessary to overcome the resistances. A little consideration will make Principle II evident. Suppose that the height of the suction is 25 feet; that the vertical distance between the suction valve and the point of discharge is 100 feet; that the stroke of the piston is 15 inches, and that its diameter is 10 inches. Let the diam- eters of the suction pipe and delivery pipe be 4 inches each. The volume displaced by the pump piston or plunger in one stroke equals 10 X ; 7 _ 8 5 4 X *- = .68177 cubic foot. The 1,728 weight of an equal volume of water is .68177 X 62.6 32 PNEUMATICS 41 = 42 611 pounds Now, in order to discharge this water, all the water in the suction and delivery pipes has to be moved through a certain distance, in feet, equal to 68177 divided by the area of the pipes, in square feet. 4 inches = foot, (i)" X .7854 = ^f- = .0812$ square y foot. .68177 -^ .08723- = 7 8126 feet. The weight of water in the delivery pipe is (^) a X .7854 X 100 X 62.5 = 545 42 pounds. The weight of water in the suction pipe is (i) X .7854 X 25 X 62 5 = 136.35 pounds. 545 42 + 136 35 = 681 77 pounds, which is the total weight of water moved in one stroke. The distance that the water is moved m one stroke is 7.8125 feet; hence, the number of foot-pounds necessary for one stroke is 681.77 X 7.8125 = 5,326 3 foot-pounds. Had this result been obtained by Principle II, the process would have been as follows: The weight of the water displaced by the piston in one stroke was found to be 42.611 pounds. 42.611 X 125 = 5,326.4 pounds, which is practically the same as the result obtained by the previous method, and is a great deal shorter. The slight difference between the two results is due to neglected decimals. EXAMPLE. What must be the necessary horsepower of a, double- acting steam pump if the vertical distance between the point of dis- charge and the point of suction is 96 feet? The diameter of the pump cylinder is 8 inches, the stroke is 10 inches, and the number of strokes per minute is 120. Allow 25 per cent, for friction and other resistances. SOLUTION. Since the pump IB double-acting, it raises ft quantity of water equal to the volume displaced by the plunger at every stroke. The weight of the volume of water displaced at one stroke la (A)* X .7854 X itf X 02.5 = 18 18 lb., nearly. 18.18 X 90 X 120 - 209,430 ft.-lb. per minute. Since 25 per cent. Is to be allowed for friction, the actual number of foot-pounds per minute is 209,430 + .75 - 279,240. 1 H. P. = 33,000 ft.-lb. per min,; hence, 8.462 H. P., nearly. Ans. 2 RUDIMENTS OF ANALYTIC GEOMETRY 33 If the acceleration a is supposed to have a fixed value, as 8 feet per second, and different values are assigned to /, different values will be obtained for s. Here, too, / and .v are variables /, which is varied at pleasure, is the independent variable; and s, whose values depend on those of t, is the dependent variable, or a function of /. The acceleration a, although represented by a letter, is assumed to be fixed or invariable, and is therefore a constant. 2. In general, when, for any particular purpose, some of the quantities represented by letters in an equation are made to take (or are considered as being such that they can take) different values, they are called variables. Those to which values are assigned arbitrarily are called Independent variables; those whose values depend on the values of the independent variables are called dependent variables, or functions of the independent variables. Those quantities that are supposed to remain fixed are called constants. A function may also be defined as a quantity whose value depends on the value or values of one or more other quan- tities; the very word "depends" indicates that the function can have different values (that is, can vaiy) according to the values assigned to other quantities. Thus, the area of a triangle is a function of the base and altitude; if the base is assumed to be fixed, it becomes a constant, and the area is a function of only the altitude. The velocity of a body moving under the action of an unbalanced force is a function of the magnitude of the force and the mass of the body; if the mass is assumed to be fixed, it becomes a constant, and the velocity is a function of only the force. 3. Graph of an Equation. It was stated in Art. 1 that from the equation A = xi*, a table can be made giving the values of the function A corresponding to different values of the independent variable r. Instead of a table, a diagram may be constructed to represent fte relation between the values of A and r. Such a diagram; w&ick is the graphic representation of the equation just givei^, is palled the graph of that equation, and is constructed as foj I \ 4 RUDIMENTS OF ANALYTIC GEOMETRY $33 value of ns laid off along OX from O to, say, A/, a per- pendicular is erected at M, intersecting the graph at P. Then MP will represent the value of A corresponding to the value OM of r. 4. The perpendicular distances of any point of the graph from the axes are called the coordinates of that point. Thus, the coordinates of P are MP (= OM') and M 1 P (= OM}. The horizontal coordinate OM, or, more gen- erally, the coordinate representing the independent variable, is usually called the abscissa, and the other coordinate OM, the ordinate. It is customary to reckon the abscissa along the axjs OX, called the axis of x, and the ordinate on a line parallel to O Y t through the foot of the abscissa. Thus, the coordinates of P are stated as OM and MP instead of M'P and MP. 6 RUDIMENTS OF ANALYTIC GEOMETRY 33 connects with the crank OJfby the connecting-! od ff K It is shown m mechanics that when the crank has described an angle X from the position OA, which is in line with the axis of the cylinder, the velocity u of the piston is given approxi- mately by the formula / . v , a sm 2 X\ u = v ( sm X -\ - 1 \ 2/ / in which v = linear velocity of crankpm K; I = length of connecting-rod; a = length of crank. If the ratio - is represented by <:, the formula may be If written, - = sinAr+sin2AT (1) v 2 When c is given, the graph of equation (1) can be con- structed by taking X as the independent variable, and u as v L r PlQ 3 the function Then, the value of - for any value of X can v be found from the graph, and, when v is given, u can be determined by multiplying by v the value found for --. v EXAMPLE 1 To construct the graph of equation (1) when c = $, for values of X varying from to 180, that Is, for one-half a revolu- tion of the crank, from the position OA to the position O A', Fig. 3. SOLUTION Writing, for shortness, y for -, and substituting the given value of c, equation (1) becomes y = em X + -rV sm 2 X By giving to lvalues from to 180, at intervals of 6, the follow- ing table is obtained: RUDIMENTS OF ANALYTIC GEOMETRY EXAMPLES FOR PRACTICE 1 Plat the following equations for values of x varying from 3 to 8 (a) y = X j* - 7 x* + 4 x - 6 (6) v = t x 9 - 1<> x + 4 a) The general form of the guiph is shown in Fig f) 6) The general form of the graph Is shown in Pig c) The general form of the graph is shown in Fig 7 PIG 5 10 RUDIMENTS OF ANALYTIC GEOMETRY 33 have been proposed to determine the relation between these two quantities, no exact formula has yet been found. Sup- pose that a series of experiments on wrought-iron columns gives the following results: RATIOS OF LENGTH BREAKING LOAD TO DIAMETER POUNDS PER SQUARE INCH 6 51,200 7 47,400 8 44,600 9 42,200 10 40,200 11 38,700 12 37,900 13 37,100 14 36,900 These results may be represented graphically, as shown in Fig 9. Having drawn two coordinate axes OX and O Y, r\ V, 5 Batio of Length to Diamter JUT | FIG. the values of the ratio of length to diameter, which is the independent variable, are laid off from O, along OX, to any convenient scale. Thus, OM* represents the ratio 6; OJIf represents 7; OM, t 8; etc. The corresponding values 12 RUDIMENTS OF ANALYTIC GEOMETRY 33 values written on the respective projections. Thus, the point P, is projected at MJ, on which is written 31,500,000, the population in 1860. The probable population m any intermediate year can be readily obtained from the curve. For instance, if it is desired to find the population in 1878, it will be observed that 1878 lies between 1870 and 1880, and that the interval between 1870 and 1878 is .8 of the interval between 1870 and 1880. Therefore, the required population is found by laying off M t M equal to .8 of M*M t , and drawing the ordmate MP, which JT' P-r represents the approx- / imate population in Pa/ 1878. P B / 8. In some cases, a *?' diagram is constructed m yf | y rather for the purpose _3isoopoo_ J:*/ of presenting to the o -"a p *s e y e > m a striking man- Pl ^ ner, the variations ol certain quantities, than ~ ^r f or the purpose of de- 1840 Bo70 Year values, the variations Pl 10 m the latter being too irregular. In the examples so far given, the curves are fairly regular, which shows that the variations in the functions (ordinates) are not too abrupt, and that the curves may be depended on to give tolerably approximate values of the function corresponding to intermediate values of the inde- pendent variable Example 1 of the following Examples for Practice is a case in which the curve, on account of its too great irregularity, could not be depended on to give inter- mediate values. In such cases, the extremities of the ordi- nates are joined by straight lines, instead of by a curve. 33 RUDIMENTS OF ANALYTIC GEOMETRY 13 EXAMPLES FOR PRACTICE 1 Draw a. graph representing the average daily water consumption in the city of Brooklyn between January, 1H71, and December, 1873, from the following data. YE.R MO.TH January 21,000,000 s. April . . . 17,500,000 oo rH July . . ... . . 19,600,000 , October . . . 18,500,000 23,600,000 2 April . 20,500,000 SB' I-H July . 22,500,000 October . . . 23,000,000 January . . . . . 29,000,000 CO April . . . 22,500,000 t- , 00 July ... 26,500,000 rH October ... . 22,500,000 December ... 24,500,000 Ans. The curve is shown in Pig. 11 o PIG. 11 2. (a) Taking the hours as abscissas, and the discharges as ordi- nates, construct a graph for the water discharged by a pipe from the following observed values for 1 day, (b] determine, by means of the curve, the probable discharge at 11:30 A. M.; (c] determine the prob- able discharge at 5:80 F. M. . The scale of ordlnatoi should be chosen sufficiently large to show the differences In discharge say 4 Inches to 1 cable loot. RUDIMENTS OF ANALYTIC GEOMETRY HOUR A. M. DISCHARGE CUBIC FEBT PER SKCOND HOUR A M DISCHARGE CUBIC FEET PER SECOND 1 2.45 11 2.54 o 2 67 M. 3 2 69 12 233 4 2.62 f M. 5 2.36 1 264 2.30 2 215 7 2.23 3 1 19 8 2.19 4 119 2 2O 5 226 10 2.OO 8 237 Aus. FIG 12 The general form of the graph is shown in Fig. 12 2 43 cu. ft. per sec. 2 31 cu. ft per sec. 33 RUDIMENTS OF ANALYTIC GEOMETRY 15 EQUATIONS OF LINES INTRODUCTION 9. Equation of a Line. As already explained, a graph is constructed from an equation expressing a relation between two variables, or giving the value of one variable as a function of the other. In V the graph, these variables are the coordinates of points on a line, usually curved. Sometimes, on the contrary, a line is given, and it is required to find the equation of which the line is the graph. That equa- tion is called the equation of the given line, and is a general expression of the relation between the two coordinates of any point of the line, with reference to two coordinate axes conveniently chosen. Take, for instance, a circle of radius r, Fig. 13, and two rectangular axes OX, OY, passing through its center. Let P be any point on the circumference, and x and y its coordi- nates, as shown. The right triangle OMP gives OM* + MP* = OP 1 -, that is, x'+y* - r' (1) This is the equation of the circle referred to two rectangu- lar axes through the center. In that equation, x and. y are the coordinates of any point on the circumference. No 16 RUDIMENTS OF ANALYTIC GEOMETRY 33 matter where the point is located, its coordinates satisfy equation (1). Thus, for the point P lt x = -OM lt y = -M 1 P 1 , and = OP,' = S 10. The equation of a line is useful in the study of the geometric properties of the line, and it often serves to recognize the form of a graph corresponding to a given equation If, for example, it is found in the solution of a \f mechanical problem that the path AB, Fig. 14, of a moving point is such that the coordinates x and y of any of its points, with re- gard to the axes OX and v O y, are related by the equation x" + y = a' o = - ' -- x tne quantity a being con- 1 Pl 14 stant, it can be at once concluded that the path of the point is a circle whose center is and whose radius is a. 11. Analytic geometry is that branch of mathematics m which geometric figures are studied by means of their equations. Surfaces, as well as plane lines, have equations; but in this Course only a few plane lines will be treated. THE STRAIGHT LINE 12. Equation of the Straight Line. Let X' X and y y, Fig. 15, be two axes of coordinates, and A B a straight line making with X'X an angle H, and intersecting Y' Y at 7, the distance b (= 01} being known. It should be understood that the angle H is always measured from the axis X> X upwards. Thus, if the line were 4iB the angled would be XJ^B,. Also, b, like the ordinates, is 53 RUDIMENTS OF ANALYTIC GEOMETRY 17 positive upwards and negative downwards. Thus, for the line A t Bi, the value of b is Of, This being understood, the equation now to be derived is entirely general, and, with the symbols interpreted as just explained, applies to all cases. The right triangle PMJ gives PM = JM tan H\ y = (x 4- JO] tan H. L j. r T that is, Now, therefore, y =[x - , tan H -- -} tan H = x tan H + b tan HI This is the equa- tion of the line A B. It is customary to denote tan H by a, and write the equa- tion in the form y = ax + b The distances 01 and J are called the Intercepts of AB on the axes of y and*, respectively . 1 1 i s evident that, at 7, the abscissa x is 0, and the ordinate is 01] and at J the ordi- nate y is 0, and the abscissa is OJ.t 13. Graph of Any Equation of the First Degree. Any equation of the first degree between two variables can be represented by a straight line; in other words, the graph of any equation of the first degree between two variables is a straight line, Let mx+ ny = P (1) be any equation of the first degree between the variables 18 RUDIMENTS OF ANALYTIC GEOMETRY 38 x and y, the other quantities m,n,p being constants, either positive or negative. Solving the equation for y, n n Since the tangents of the angles between and 180 con- tain all possible numbers, both positive and negative, it is always possible to find an angle whose tangent is . Let that angle be denoted by H, and denote * by b. Then, n = tan H, and equation (2) becomes n y = jirtan/f + b (3) If is positive, n H is less than 90; if negative, H is greater than 90. If on the y axis the distance b is laid off from O t upwards if b x is positive, down- wards if b is negative, and from the extrem- ity of b a line is drawn making with the x axis an angle equal to H, that line is the graph of equation (3) , r and, therefore, also FIG 16 of equation (1). If b and tan H are positive, the line will have such a position as A,B^ Fig 16, in which O A = b, and XJ,B, = H. If & is positive and tan H negative, the graph will have a position like A t B t > in which OI t = b, and XJ,B t = H. If b is neg- ative and tan H positive, the graph will be like A, JB,, in which OI> = b, and XJ,B* = H. If both b and tan H are negative, the graph will have such a position as A^ B,.. 33 RUDIMENTS OF ANALYTIC GEOMETRY 19 14. Because any equation of the first degree between two variables can be represented by a straight line, formulas in which the value of a quantity is given in terms of the first power of another are called straight-line formulas. For example, the formula p = 10,000 - 45 - r which expresses the intensity p of pressure that a column can stand, m terms of the ratio - of length to radius, is a straight- r line formula. If y is written instead of p, and x instead of-, the r formula becomes y = 10,000-45* = _45*+ 10,000 which is the standard form of the equation of a straight line. 15. The simplest way to draw the straight line corre- sponding to an equa- tion of the first de- gree is as follows: Let the equation be mx -\-ny-\-p = 0. Select the axes in any convenient position, as X 1 X and V K, Fig. 17. Making x = in the equation, and solving for y, the intercept (say <?/,) on the y axis is obtained, and the point 7, where the line intersects that axis is determined. Making y = 0, and solving for x, the intercept 07, and the point ,/i are obtained. The line is then drawn through the points 7, and /,. EXAMPLE 1. To draw the graph of the equation 8# l&y + 20-0. SOLUTION^ Draw the axes X'X< Y> Y, Fig. 17 Making x In the equation, we have -16^ + 20 = 0; whence y - 0/ 1.26 20 RUDIMENTS OF ANALYTIC GEOMETRY 33 Making y = in the equation, we have 8x + 20 = 0, whence x = OJ= -26 Laying off Of = 1 25, and OJ =25, the line A B drawn through 1 and J is the required graph Ans EXAMPLE 2 To draw the graph of the equation 10 x + 8y + 40 = J. SOLUTION Making x 0, and solving for y, Pig. 17, = 01 = - = -5 Making y = 0, and solving for x, * - n T 40 _ 4 * ~ l = ~ 10 ~ ~ 4 Laying off OJ* = 4, 07; = 5, and joining/! and J lt the required graph A! B* is obtained Ans. EXAMPLE 3 To draw the graph of the equation 4j/ 8x = 0. SOLUTION It will be noticed that this equation has no term independent of x and y, This shows that the line passes through the origin of coordinates, or that its intercepts on the two axes are zero This follows at once from Art 13, for it was there shown that >-, and, as n in this case,/ = 0, it follows that b = In this case, the graph cannot be con- structed as in the two pre- ceding examples. Since the line passes through the origin O, Pig. 17, it is only necessary to de- termine another point. This is done by assu- ming any convenient value for x and solving for y Making*- = |, the equation becomes 4j/ - 4 = 0, whence y = 1 Laying off OM^ i, and the ordmate MP = 1, the line A, t , drawn through and JP, is the required graph FOR PRACTICE Draw the graphs of the equations (a) 8 (c) 5x+4y = -20. rf I2y - 10 . Ans The are shown, respectively, at tt A, B, s AtJBt in Pig. 18 33 RUDIMENTS OF ANALYTIC GEOMETRY 21 2 Find the angle that the graph of the equation x + &y 4 = makes with the x axis. Ans. 141 20> 20" 3 Given the equation x-\- 4y = 20, find (a) the intercept of the graph on the y axis, (b) the angle that the graph makes with the x axis (Give seconds in angle to nearest multiple of 10.) 165 57' 60" Ans . {(?) APPLICATIONS 16. Reactions on a Beam. Let a beam A B t Fig. 19, resting on the supports A and B> carry a movable load W. Let the distance of the load from the left support at any instant be x, and let the length of the beam be denoted by /. For this position of the load, the reaction Ri at A is obtained by taking moments about B\ thus, RJ- W(l-x) = 0; whence I (1) Since the load W moves, x is a variable, and so is Equation ( 1 ) gives R l as a function of x. As that equation is of the tirst degree, it can be represented by a straight line, Ri being used instead of the y used in previous arti- cles. A convenient and usual way of drawing the graph is as follows: The ori- gin is taken direct- ly under A, and the axis OX is drawn parallel to A B\ the axis O Y is drawn through O. Equa- tion (1) shows that the y intercept is W. Therefore, laying off, to any convenient scale, along O Y the distance Of to represent W, one point / of the graph is obtained. The equation also shows that the x intercept (fotind by making Pio, 19 22 RUDIMENTS OF ANALYTIC GEOMETRY 33 J?, = 0) is /. Therefore, projecting the point B on OX at /, the point J of the graph is found, and the graph is the straight line IJ. The ordmate M P represents the left leac- tion when the load is at W\ the ordmate M' P' represents the left reaction when the load is at W; etc. If 1 ' O is drawn parallel to OX, the ordinates M,P, AfJP', etc. will represent the corresponding values of the right reaction, since jR, + R* = W= Of. If, for example, the load is 2,000 pounds, and a scale of 1,000 pounds to the inch is used, Of should be made 2 inches; and, if the ordinate M'P' is found to measure li inches, the reaction R^ when the load is at W>, is 1,000 X 1* = 1,500 pounds. 17. Pressure on the Back of a Dam. Let Fig. 20, be the back or inner face, supposed to be vertical, o of a dam, the water reach- ing to the top From hydrostatics it is known that the pressure p per square foot, at any point M whose depth below the sur- face is x feet, is given by the formula p = wx -* m which w is the weight of 1 cubic foot of water. Fl M As this is an equation of the first degree, it can be represented by a straight line. Taking O M,. as the axis of x and as the origin, the latter point is a point in the graph, since there is no intercept. Making* = h, the formula gives p = w h t which is the pres- sure at Mr Laying off, horizontally, M r P t = wh, the line O P l is the required graph. The intensity of pressure at any point M is given by the ordinate M P. If, for instance, the height k is 24 feet, and w is taken equal to 62.5 pounds, wh = 62 5 X 24 = 1,600 pounds per square foot. If a scale of 500 pounds per square foot to the inch is used, M^P^ should be made 1,500 -T- 500 = 3 inches; 33 RUDIMENTS OP ANALYTIC GEOMETRY 23 and, if M ' P measures li inches, the pressure at M is 500 X H = 625 pounds per square foot. It will be observed that here p is used instead of y, that the x axis is vertical; and that positive values of x are counted downwards, and positive values of p toward the left It is often necessary to make such changes in notation, so as to adapt the construction to given conditions. The general principles and methods, however, remain the same. THK PARABOLA 18. Definitions. A parabola, Fig. 21, is a curve such that, if any point P on it is taken, the distance P F of that point from a fixed point F is s r equal to its distance PN fiom a fixed line DE. The fixed point F is called the focn&j the fixed line DE, the directrix. The line N, X passing through the focus and perpendiculai to the di- rectrix is called the axis of the curve, and the point O where the curve crosses the axis is called the vertex. Twice the distance FN<> from the focus to the di- rectrix is called the par- ameter, and is denoted by 2/>; so that 2/ = 2 N.F, and p = N,F. The curve is symmetrical with respect to the axis; that is, to every point P on one side of the axis there corresponds another point P 1 on the other side, at the same distance from, and on the same perpendicular to, the axis. Any line, as JPP', perpendicular to the axis and bounded at its two ends by the curve is called a double ordluate. FIG. 21 24 RUDIMENTS OF ANALYTIC GEOMETRY 33 19. Equation of the Parabola Referred to the Axis and Vertex. Let P, Fig. 21, be any point on the curve, and O M = x and M ' P y its coordinates, the axes being the axis OX of the parabola and the perpendicular Y at O Let the parameter be 2/>. Then, the distance FN from the focus F to the directrix DE is equal to p, and OF = vfl, since ON a OF, the point O of the curve being, according to the definition of a parabola, equally distant from the directrix DE and the focus F. According to the same definition, we have PF= PN and, therefore, PF' = PN' (1) Now, PF' = PM' + FM' = y 9 + (OM- OF)' Also, = N M' = (OM+N t O)' = (x + p)* = A -\ Substituting in equation (1) these values of PF' and PN', 4 4 whence y' = %px which is the required equation. 20. Let y^ and y tt be two ordinates corresponding to the abscissas x* and x, Then, since the equation applies to all points, y* = 2px lt y' = ZpXi, whence That is, in any parabola, any two coordinates parallel to the axis of the curve are to each other as the squares of the corre- sponding coordinates perpendicular to the axis. 21. It is important that the equation of the parabola should be so mastered that it can be applied to parabolas in different positions, and when the notation is different from the one here used. It should be borne in mind that in the S33 RUDIMENTS OF ANALYTIC GEOMETRY 25 general equation, the coordinate to be squared is that per- .pendicular to the axis of the parabola. In Fig. 22 is repre- sented a parabola with the axis vertical, coinciding with the y axis. In this case, the equation should be written Fig. 23 represents a vertical parabola with its axib down- T x t o Fio. 22 Pro 23 wards The coordinate axes are denoted by O S and O T, as shown, and the coordinates of any point P by s and /, 5 being positive downwards, and t positive toward the left Under such conditions, the equation of the curve should be written 22. Problem I. To find the parameter and equation of a parabola, when a double ordinate and the corresponding abscissa are given. This is the usual way in which the parabola occurs in practice. Thus, in road construction, in which the cross- section of a road is often made parabolic, the width a of the road, Fig. 24, and the height h of the center above the ends A and B are given. In order to determine the fall of the cross-section at different distances from the center, it is necessary, or at least advisable* to determine the equation and parameter of the curve A OB. 26 RUDIMENTS OF ANALYTIC GEOMETRY 33 As usual, the parameter will be denoted by 2 p. At the point M , for which x = -, and y = A, we have, since here L the x coordinate is perpendicular to the axis, whence e lp = ^~ 4:h The general equation of the parabola is, therefore, * In practice, points on the curve are determined by assu- ming values for x, and computing the corresponding values of y. The equation may, therefore, be more conveniently written in the form a' \i aj Let OM t be divided into any convenient number of equal Jf, >F FIG 25 parts, say n, and give to x successive values correspond- ing to the points of division; thus, x = ", x = 2 -^-^, n n x = 3 -, etc.; or x = , x - 2 , # = 3 , etc. Then, n n n n the corresponding values of y beccme Thus, if OM, is divided into ten equal parts, and OM con- tains six of those parts, -Axr-ixH 33 RUDIMENTS OF ANALYTIC GEOMETRY 27 EXAMPLE Given the double ordinate A B = 40 feet, Fig 25, and the corresponding abscissa OK = 15 feet, it is desired to hnd points on the curve at intervals of 5 feet on each side of O SOLUTION Draw O M t parallel to A B, and B MI parallel to KO Since the points are to be located every 6 ft from O, and O M t = 20 ft , the latter line should be divided into four equal parts O MI, Jlf, Mt, etc Here, i0=jx*0 = 20, = ^=4, and h = 1 5. Therefore, - = ^ = 094, nearly, and the values of y are SLtM lt Mi PI = 094 X 1 at M., M, /> = 004 X 2 at Af a , M a /> = .094 X 3 1 at Jtf t , Jlf t n = 1 5 ft 09 ft , nearly, 094 X 4 = 38 ft , nearly, .094 X9 = .85 ft., nearly, 23. Problem II. To construct a parabola when a double ordinate and the con espotidmg abscissa are given. In practice, what is usually required is to locate points of the curve on the % ft ground, as in the case of road and street construction, and in railro ad curves. Under such circum- stances, the ordinates are calculated as ex- plained in Art. 22; if desirable, they may be platted, and the curve drawn through the points thus deter- mined. A purely graphic method of constructing the curve is explained in Geometrical Drawing* Another method, which is often convenient in the drafting room, is as follows: Let AB> Fig. 26, be the given double ordinate, and OK the corresponding abscissa. Bisect ATX at Cand draw CI parallel to KO and OI perpendicular to K O. Draw /A", ILT39&-21 Fro 26 28 RUDIMENTS OF ANALYTIC GEOMETRY 33 and IN a perpendicular to it, meeting the axis produced at N . The line D E, drawn through N perpendicular to OK pro- duced, is the directrix It is shown here for the purpose of explanation, but it is not necessary to draw it. Lay off OF = ON.. The point/' is the focus. From 7V , lay off along the axis any convenient distances, as N M, N M^ etc , and at the points M, M 1} etc. draw indefinite perpendiculars 8Q f > QiQS> to the axis. From F as a center, and with a radius equal to NM, describe an arc, cutting QQ' at P and P'\ these are points of the curve. Likewise, the points P! and P/ are the intersections of ?, Q l / with an arc described from Fwiih a radius equal to N M* Other points may be determined in a similar manner, and the curve drawn through them. It is usually more convenient to find the points N 9 and F by calculation, instead of by the geometrical construction described above. The parameter 2^ is computed as explained in Art. 22, and then N and OF are laid off each equal NOTE The correctness of the preceding construction will now be shown It is not necessary for the student to study the following demonstration, but he is advised to read it carefully, as it is a good exercise As explained In Art 22, 2/ = ~ In Fig 26, fl = 2 A K and A = OK. Therefore, P AK* In the right triangle N, IK, the perpendicular Of, which is equal to \A K t is a mean proportional between O K and O N a . that is, (IAK)*= OJV.XOJf, whence ON, = ^frjj = | [by equation (1)] Therefore, O N a Is the distance from the vertex to the directrix (Art 18) The point P was so determined that FP = N a M = PN. That point is, therefore, at the same distance from the focus as from the directrix, and, according to the definition of a parabola, must be a point of the curve The same reasoning applies to the points P 1 , A, etc. 33 RUDIMENTS OF ANALYTIC GEOMETRY 29 APPLICATIONS 24. Projectiles. In the most general sense of the term, a projectile is any body thrown into the air The velocity with which a projectile is thrown is called the initial velocity, or velocity of projection. Here, only pro- jectiles thrown horizontally will be considered, and the resistance of the air will be neglected. Let a projectile be thrown horizontally in the direc- tion OX from a point O, Fig 27, with a velocity v. It is required to determine the path OA of the projectile. The horizontal line OX and the vertical line O Y will be taken as axes of coordinates, y being positive downwards. Were the projectile not acted on by gravity, it would describe, in any time /, a space OM = vt, which o will be denoted by x. If it were not thrown at all, and were allowed to fall freely from O during the time t, it would fall through a distance O G = %gf, denoting, as usual, by g the accelera- tion due to gravity. This distance O G will be denoted by y. When the pro- PIG. 27 jectile, after being thrown along OX with the velocity v, is acted on by giavity, its position P, at the end of the time t, will be such that M P will be equal and parallel to O G. We have, therefore, x = vt and y = i g t*. yjt From the first of these two equations is found /* = -. v Substituting this value in the second equation, there results 2 w" whence x' = y, # which is the equation of a parabola with its vertex at O and axis vertical. This parabola is, therefore, the path followed 2 if by the projectile. Its parameter is . 30 RUDIMENTS OF ANALYTIC GEOMETRY 33 25. Jet of Water Issuing 1 From & Small Orifice. One of the most important applications of the theory of pro- jectiles is to the determination of the velocity of water flowing from a small orifice. This determination is of much importance in hydraulics. In Fig. 28 is represented a tank T from which water flows through a small orifice O. The water issues horizontally with a velocity v (to be deter- mined), and each particle, being under the same conditions as a projectile thrown horizontally with the same velocity, describes a parabola. As the jet is narrow, it may be treated as a whole as a parabolic arc OJ. A horizon- tal string OX is stretched from O, and any distance OM^ is measured. From M* Pl another string carry- ing a heavy weight is suspended, and the distance M l P t is measured. Let OM l = x lt Af l P l = y lm Then, whence EXAMPLE. What was the velocity of a jet for which the measured distances x^. and jy, were, respectively, 6 and 5 feet? SOLUTION Substituting m the formula the given values of .*-, and y lt and 32 16 for^-, x g - = V3 216 X 36 = 10 76 ft. per sec Ans 26. Moment In a Beam. Let a beam A B, Fig. 29, of length /, carry a movable load W. When the load is at a distance x from the center of the beam, the left reaction R 1 is found by taking moments about B\ thus, RJ = whence R^ = 33 RUDIMENTS OF ANALYTIC GEOMETRY 31 The moment of .#, about I^is called the bending moment at W. Denoting it by M, we have The bending moment at the center, which will be denoted by M^ is obtained by making x = 0, which gives ". - \ L Equation (1) may p 1 be written M M whence If M - M is de- noted by y, equation (1) may be written / x which is the equa- tion of a parabola whose axis is perpen- dicular to A B, and whose parameter is The curve is shown at A 1 OB 1 . It may be constructed by points, assuming values of x and finding the corresponding values of M and M t M\ or by the method of Art. 23, noticing that to the double ordinate A' #', or /, corresponds a value of y (= fCO) equal to M^ since, when x = J/, M 0, and y = M, M M,. Having drawn the curve, the bending moment, when the weight is at any point W, is found by projecting W on A' /?', and measuring the distance QP, which is equal to the moment at W> to the scale by which KO represents M,, For -tfie ordinate NP represents y, or M, M, and, therefore, M.-M= NP, M = M.-NP = Qfi-NP= QP