UNIVERSITY OF TORONTO
DEPARTMENT OF CIVIL
Municipal and Structural
IN PR E PAR A TION
TEXT-BOOK OF MECHANICS
BY
LOUIS A. MARTIN, JR.
VOL. II KINEMATICS AND KINETICS
TEXT-BOOK
OF
MECHANICS
¥ BY
LOUIS A. MARTIN, JR.
(M. E., STEVENS^ A. M., COLUMBIA)
Assistant Professor of Mathematics and Mechanics in
Stevens Institute of Technology
VOL. I. STATICS
FIRST EDITION FIRST THOUSAND
NEW YORK
JOHN WILEY & SONS
LONDON: CHAPMAN & HALL, LIMITED
1906
3 A 907
V' 1
COPYRIGHT, 1906
BY
LOUIS A. MARTIN, JR.
LIBRARY
757062
UNIVERSITY OF TORONTO
ROBERT DRUMMOND, PRINTER, NEW YORK
UNIVERSITY op TORONTO
<T Of CIVIL EN6IN&6BING Municipal tqd Structural
PREFACE
Answers to. the examples given in this book will be supplied to students by the publishers, free of charge, on the receipt of the written order of the instructor.
01 statics, me stuuy ui wiiicn can auvaiuageuusiy oe begun with such knowledge of mathematics as is re- quired for admission to most Colleges and all Technical Schools. The elements of Analytical Geometry are intro- duced as the treatment of the subject requires, but no Calculus is used in the first volume.
The second volume, the MS. of which is nearly ready, will treat of Kinematics and Kinetics. In this volume extensive use will be made of the Calculus.
Thus, while studying Mechanics, the student becomes acquainted with a use to w^hich pure mathematics, as now taught in our schools, may be put. Starting with a knowledge of Algebra, Geometry, and Trigonometry,
in
907
LIBKAKY
7S7062
UNIVERSITY OF TORONTO
ROBERT DRUMMOND, PRINTER, NEW YORK
UNIVERSITY of TORONTO
.T OP CIVIL ENSINifcBING Municipal md Structural
PREFACE
THE Text-book of Mechanics, of which this is the first volume, is designed as an introductory course to Applied or Technical Mechanics for Technical Schools and Colleges. It is based upon notes prepared for the use of the Freshman and Sophomore classes at Stevens Institute of Technology.
At the same time an attempt has been made to produce a graded course in the application of mathe- matics. With this end in view the first volume treats of Statics, the study of which can advantageously be begun with such knowledge of mathematics as is re- quired for admission to most Colleges and all Technical Schools. The elements of Analytical Geometry are intro- duced as the treatment of the subject requires, but no Calculus is used in the first volume.
The second volume, the MS. of which is nearly ready, will treat of Kinematics and Kinetics. In this volume extensive use will be made of the Calculus.
Thus, while studying Mechanics, the student becomes acquainted with a use to which pure mathematics, as now taught in our schools, may be put. Starting with a knowledge of Algebra, Geometry, and Trigonometry,
iii
IV PREFACE
he at once puts them to practical use, and it is designed to have this application keep pace with his advancing knowledge of pure mathematics.
Throughout it is aimed to make the book a teachable one. As a course in Mechanics should primarily fit the student to solve its problems, numerous examples and many exercises are introduced to illustrate each principle as developed.
Of the two hundred and fifty-five problems contained in this volume many are to be found in most books on Mechanics; some have been especially prepared, while others have been selected from examinations set at Stevens.
As regards the subject-matter in a book of this nature nothing new can be expected. My only claim to originality lies in the presentation. Should any errors be found in the work, I shall esteem it a favor to be informed of them.
In concluding, I would here express my thanks to my wife, Alwynne B. Martin, for her aid in preparing the MS. and in reading the proof.
Louis A. MARTIN, Jr.
HOBOKEN, N. J., March, 1906.
CONTENTS
INTRODUCTION
PAGE
Mechanics, Kinematics, Dynamics, Statics, Kinetics 1
Motion, Velocity, Acceleration 1
Exercises 1 to 6 2
Matter, Mass 4
Momentum 5
Exercise 7 5
Force 5
Gravitational Units of Force and Mass 7
Examples 8
Exercises 8 to 1 1 .. 9
STATICS ANALYTICAL STATICS
CHAPTER I FORCES ACTING AT A SINGLE POINT
SECTION I
Triangle of Forces
Representation of a Force 13
Principle of the Triangle of Forces 14
Action of Supports 14
Examples 16
Exercises 12 to 21 .. 19
VI CONTENTS
SECTION II Components. Composition of Forces
PAGE
Resolution of Forces, Components 21
Exercises 22 to 26 22
Example 22
Exercises 27 to 30 23
Composition of Forces 24
Exercises 31 to 85. . 25
SECTION III
Conditions for Equilibrium
Conditions for Equilibrium 26
Example 27
Exercises 36 to 40 '. 29
SECTION IV
Statical Friction
Friction 30
Direction in which Friction Acts 30
Coefficient of Friction 31
Exercises 4! and 42 32
Angle of Friction 33
Total Reaction of a Surface 34
Example 34
Exercises 43 to 50 37
SECTION v Moments
Arm of a Force 38
Moment of a Force 38
Exercises 51 to 64 39
Varignon's Theorem of Moments 40
Condition for Equilibrium 41
Example 42
Exercises 56 to 58. . 42
CONTENTS Vll
CHAPTER II FORCES ACTING ON A RIGID BODY
SECTION VI
Resultant of Two Forces. Couples
PAGE
Rigid Body 43
Transmissibility of a Force 43
Resultant of Two Non-parallel Forces 44
Resultant of Two Parallel Forces 44
Exercises 59 to 6 '4 45
Couple 46
Moment of a Couple 47
Translation of a Couple 48
Rotation of a Couple 49
Equivalent Couples 50
Resultant of Couples 50
Exercises 65 and 66 50
Resultant of a Force and a Couple 51
Exercise 67 51
SECTION VII
Resultant of any Number of Forces
Resultant of Any Number of Forces 51
Exercises 68 to 70 54
SECTION VIII
Conditions for Equilibrium
Rotation and Translation 55
The Cause of Rotation and Translation 56
Conditions for Equilibrium, First Method 56
Example 57
Exercises 71 to 77 58
Conditions for Equilibrium, Second Method 59
Example 59
Exercise 78... 60
Viii CONTENTS
PAGE
Relation between Three Forces Acting on a Body 60
Example 61
Exercises 79 to 84 62
SECTION IX
Parallel Forces. Centroids or Centers of Gravity
Resultant of Parallel Forces 63
Exercises 85 and 86 65
Centroids or Centers of Gravity 66
Examples 66
Exercises 87 to 92 68
SECTION X
Summary of the Methods of Statics
Forces Acting at a Point 69
Forces Applied to a Body 70
Choice of Equations Used in the Solution of a Problem 70
Example 70
Exercises 93 to 96,. 72
CHAPTER III
APPLICATIONS OF THE PRINCIPLES OF STATICS TO THE SIMPLE MACHINES
SECTION XI
The Lever and the Wheel and Axle
Machines 73
Mechanical Advantage 73
Simple Machines 73
The Lever 74
Example 74
Exercises 97 to 101 75
The Balance 75
Exercises 102 to 104 76
The Common or Roman Steelyard 76
Example 76
Exercise 105, . 78
CONTENTS IX
PAGE
The Danish Steelyard 78
Exercise 106 78
The Wheel and Axle 79
Exercises 107 to 109 79
SECTION XII
The Pulley
The Fixed Pulley 80
Example 80
Exercises 110 and 111 81
The Movable Pulley 82
Example 82
Exercise 112 82
Systems of Pulleys 83
Example 83
Exercises 113 to 120 84
SECTION XIII
The Inclined Plane and the Wedge
The Inclined Plane 85
Exercises 121 to 126 86
The Wedge without Friction 87
Example 87
Exercises 127 to 129 88
The Wedge including Friction 88
Example 88
Exercises ISO and 131 90
SECTION XIV
Miscellaneous Machines
The Bent-lever Balance 91
Exercises 132 to 134 91
The Differential Wheel and Axle 92
Exercise 135 93
The Platform Scale 93
Exercise 136 93
l"he Knee 93
Exercises 187 to 139 94
CONTENTS
GRAPHICAL STATICS
CHAPTER IV GRAPHICAL ARITHMETIC
SECTION XV
Summation, Division, and Multiplication
PAGE
Graphical Arithmetic 95
Laws of Signs 95
Summation 96
Exercises 140 and 14! 96
Multiplication and Division 96
Examples 97
Exercises 142 to 145 97
Example 98
Exercises 146 to 148 99
SECTION XVI
Combined Multiplication and Summation
Multiplication and Summation 100
Exercises 149 to 152 102
Application to the Finding of Centroids 102
Example 102
Exercises 153 and 154 104
CHAPTER V FORCES ACTING AT A SINGLE POINT
SECTION XVII
Components. Resultants
Resultant by the Parallelogram of Forces 105
Resultant by the Triangle of Forces 106
Exercise 155 107
Components 107
Exercises 156 to 158. . . 107
CONTENTS XI
PAGE
Polygon of Forces 108
Exercises 159 to 163 109
SECTION XVIII
Conditions for Equilibrium
Conditions for Equilibrium 109
Example 110
Exercises 164 to 171 HI
Three Forces Acting on a Body 112
Exercises 172 to 174 112
CHAPTER VI FORCES ACTING ON A RIGID BODY
SECTION XIX
Resultants
Resultants by Parallelogram of Forces 113
Resultants by Triangle of Forces 114
Exercise 175 114
Resultant by Funicular Polygon 115
Exercises 176 to 179 117
SECTION XX
Conditions for Equilibrium
Conditions for Equilibrium 117
Examples 119
Exercises 180 to 188 122
CHAPTER VII APPLICATIONS TO STRUCTURES
SECTION XXI
Stresses in Members of Framed Structures
Applied or External and Internal Forces 123
Structures, Framed and Non-framed 123
Example 124
Exercises 189 to 194 .127
Xll CONTENTS
SECTION XXII
The Funicular Polygon for Parallel Forces Considered as a Moment
Diagram
PAGE
Bending Moments 127
Moment Diagram 129
Exercises 195 to 200. . .129
SECTION XXIII
Graphical Method for Finding Centroids
Graphical Method for Finding the Centroids of Laminae 130
Example 130
Exercises 201 and 202 131
APPENDIX
APPLICATION OF TWO-DIMENSIONAL METHODS TO FINDING STRESSES IN THREE-DIMENSIONAL STRUCTURES
Example 133
Exercises 203 and 204 134
Example 135
Exercises 205 to 209 136
PROBLEMS FOR REVIEW Exercises 210 to 255.. . 137 to 142
UNIVERSITY OF TORONTO
DEPARTMENT OF CIVIL ENGINEE
Municipal and Structural
TEXT-BOOK OF MECHANICS
INTRODUCTION
THAT branch of physics which is the simplest and also the oldest and which is usually treated as introductory to the other branches of this science is called Mechanics.
Mechanics treats of the motions of bodies and the equilibrium of forces.
It is divided into Kinematics and Dynamics.
Kinematics treats of the motions of bodies without reference to the forces producing these motions or to the masses moved.
Dynamics is that branch of Mechanics which treats of the equilibrium of forces and the motions of bodies under the action of forces. It is accordingly subdivided into two parts, — Statics and Kinetics.
Statics treats of the equilibrium of forces.
Kinetics treats of the motions of bodies as related to the forces producing these motions and to the masses moved.
Motion is change of position.
Velocity is the rate of change of position.
2 TEXT-BOOK OF MECHANICS
Acceleration is the rate of change of velocity.
By rate of change is meant the total change divided by the time occupied in making the change, provided the change progresses uniformly.
Thus, the population of a town in 1890 was 1500, in 1900 it was 4000; to find the rate at which the popula- tion changed we have:
Total change 4000 — 1 500
Rate of change = ^r~
Time 10
= 250 persons per year.
A body which moves uniformly over a space of 60 feet
60 in 5 seconds changes its position at the rate of — =12 feet
0
per second.
The velocity of a uniformly moving body is therefore
calculated thus: velocity = -r . In algebraic language
this is expressed, v=— , where velocity, space, and time are denoted by their initial letters.
EXERCISE i. If a body moves 144 feet in 3 seconds, what is its velocity?
EXERCISE 2. A body moves with a uniform velocity of 40 miles and 1600 yards per hour; what is its velocity in feet per second, and how many feet will it traverse in 10 seconds?
EXERCISE 3. A railway-train explodes two torpedoes which are placed on the rails 176 feet apart. Two seconds of time elapse between the explosions. Compare the velocity during that interval with the mean velocity, which is indicated by the statement that the train takes an hour and a half to perform the journey between two stations 45 miles apart.
INTRODUCTION 3
If the velocity of a moving body changes, the rate oj change of velocity is called its " acceleration." Thus, if a moving body has a velocity of 20 feet per second at one point of its path, and this velocity increases uni- formly so that 3 seconds later its velocity is 50 feet per
second, its acceleration is — — =10 feet-per-second per
o second.
This shows that
Change in velocity Acceleration —
Time required to produce change'
'U2 — 'Ui
or a = — — • , where v2 and Vi are the velocities of the
/2-/l
body at the times /2 and ti respectively.
Acceleration, being measured by the velocities im- parted per second, is stated in terms of feet-per-second per second.
EXERCISE 4. A body starts from rest with an acceleration of 2 feet-per-second per second. When will it have a velocity of 1000 feet per second ?
EXERCISE 5. The velocity of a body decreases during 10 seconds from 10 feet per second to 7 feet per second. What is its acceleration?
EXERCISE 6. A body is made to record its own velocity, which shows that at a certain instant it is moving at the rate of 112 feet per second; but after an interval of 1/20 second its velocity is 114 feet per second. What is its acceleration?
From the above it will be noticed that the unit of
velocity is obtained from the formula v=—, by placing
t
s = i foot and / = i second; thus, v = — =i foot per second.
4 /TEXT-BOOK OF MECHANICS
This may be stated thus: A body has Unit Velocity when it traverses uniformly a unit of distance in a unit of time.
Similarly, the unit of acceleration is derived from
a =-2— - by placing v2 — Vi = i foot per second and fe— *i
/2_/1 = I second; thus, a=— = i foot-per- second per
second; or, A body has Unit Acceleration when its velocity increases uniformly by a unit of velocity in a unit of time.
This method of deriving units will be found of uni- versal application.
Matter is the material of which bodies are composed.
A Body is a limited portion of matter.
Mass is the scientific name for a quantity of matter.
We can recognize in bodies a property which depends partly on their size and partly on the substance of which they are composed. Thus, if we take two balls of iron of considerably different sizes and hang them up by long strings of the same length, a very slight effort is suffi- cient to give considerable motion to the small ball, while a strong push is necessary to displace appreciably the large ball. These balls are said to differ in mass.
A heavy fly-wheel, properly mounted on ball-bearings, continues to rotate for a long time when set in motion; much effort, however, is required to either stop it or to start it when at rest. The fly-wheel is said to have mass, and the greater the mass the greater the effort required to stop it in a given time.
These and similar observations lead us to recognize
INTRODUCTION 5
that property of bodies to which the name of "Mass" has been given.
The Unit of Mass is the mass of a certain lump of platinum marked "P. S., 1844, i Ib." (P. S. being the abbreviation of Parliamentary Standard).
Momentum. — Observation leads us to recognize a quantity which depends upon both the mass of a body and its velocity. Two bodies of equal mass moving with equal and opposite velocities will on impact (collision) come to rest. If, however, the masses are unequal or their velocities different, the result of impact is not rest. The motion which a given body can communicate by impact to another body depends upon this quantity, which we call "Momentum."
The momentum of a body is the product of its mass and its velocity, or
Momentum = mXv.
The Unit of Momentum is the momentum of a mass of i pound moving with a velocity of i foot per second.
EXERCISE 7. A baseball of 5 oz. mass has a horizontal velocity of 66 feet per second. A blow causes it to travel back on the same line with a velocity of 62 feet per second. Find the change of momentum.
Momentum is a fundamental property of a moving body. If we observe the motions of bodies, we will find but few cases in nature in which the velocity is uniform: bodies generally have acceleration. This leads us to consider change in momentum. That which changes the momentum of a body is called "Force."
Force is measured by the rate of change of momentum, or
6 TEXT- BOOK OF MECHANICS
Change of momentum Force = ^r— — ; — TT— — . Time required for change
m X change of velocity m(v2 — vi)
J~f -— - M.. .,, I .... !— . — - . *
Time required for change t2 — h
V2 — Vi
but — = acceleration = a.
h-h
Therefore F = mXa.
The Force acting on any mass is equal to the product of the mass by the acceleration produced by the force.
The Unit of Force is called the "Poundal," and is that force which when acting upon a mass of i pound pro- duces an acceleration of i foot-per-second per second for F=mXa = iXi=i poundal.
m(v2 — ^i)
From the formula F = — -it will be noticed that
t2-ti
F(t2 — ti)=m(v2 — Vi), or Force multiplied by the time during which it acts equals the change of momentum produced.
The product Ffa — h) is called Impulse.
.". Impulse = change of momentum.
It has been experimentally demonstrated that any body near the earth's surface falls with a sensibly constant acceleration, usually denoted by g. This means that the velocity of the body, and therefore its momentum, is changing. In the light of what has been said we should say that a force is acting on the body.
This force is known as " Weight" and is the result of the attraction which the earth exercises on all bodies in accordance with the law of gravitation.
INTRODUCTION 7
Thus, if w represents the weight of a body whose mass is m and as
Force = mass X acceleration,
we have w = mXg.
As g is approximately equal to 32 feet-per-second per second, the weight of a body whose mass is, say, 10 Ib. equals
10X32=320 poundals.
In engineering practice it is not usual to use the abso- lute unit of force, — the poundal. Instead, forces are measured in terms of the weight of some given body, usually the weight of the standard of mass, — i pound. Thus, when in practice we speak of a force p we do not mean a force of p poundals, but a force p times as great as the weight of i pound. Such a unit is known as a gravitational unit oj force.
If we wish to use as the unit of force the weight of a mass of i pound, as before denned, and still retain the equation Iw = mg1 which is a necessary deduction from the laws of mechanics, we must change the unit of mass.
The necessity of this change becomes evident from the
w equation m=— , which gives m = i only upon making
o
w = g. The new unit of mass is then the mass of a body
possessing g times the weight of the original unit of mass.
If, then, we speak of a mass of 5^ Ib., we mean a mass
possessing 5^ times the weight of the original unit of
t (7
mass, or it contains -- ( = 5) times the new unit of mass,
o
and is therefore a mass of 5 units.
8 TEXT-BOOK OF MECHANICS
As, by definition, a poundal is that force which acting on a mass of i Ib. produces an acceleration of i foot-per- second per second, and as we observe in the case of fall- ing bodies that a force equal to the weight of i pound of mass acting upon a mass of i pound* produces an accel- eration of g feet-per-second per second, the force equal to the weight of i pound of mass must equal g poundals. This is expressed more compactly as follows: i pound = g poundals.
In the Absolute System of Units the unit of mass is fun- damental, and the unit of force is derived.
In the Gravitational System of Units we assume the unit of force and then derive the unit of mass from it.
Example. — A force of 10 Ib. acts upon a weight of 20 Ib. for 5 seconds. Find the change of momentum produced and the mass of the moving body in gravita- tional units.
Solution. — We have the law Ft = mv.
.'. The change of momentum = w^ = .F/==(io) (5)= 50 units of momentum.
The moving body has a weight of 20 Ib. ; one unit of mass has a weight of g Ib. .". The mass of the moving
20 body is — units of mass.
o
It should be observed that no specific name has been assigned to either the unit of momentum or the unit of mass.
Example. — What velocity would a mass of 64 pounds acquire in 10 seconds if acted on by a force of 4 pounds? (# = 32 ft.-per-sec. per sec.)
Solution. — Find first the acceleration produced by the
INTRODUCTION 9
force by means of the formula F = ma, wherein F = ^
pounds and m = — =2 units of mass. 32
Thus, a = — = 2 ft.-per-sec. per sec.
As the acquired velocity will be the change in velocity and as the acceleration is the rate of change of velocity, we have
v
2= — . .*. v = 20 ft. per sec. 10
The following exercises should be solved in gravita- tional units.
EXERCISE 8. A mass of 100 pounds possesses an accelera- tion of 10 feet-per-sec. per sec. What force must be acting upon it ?
EXERCISE 9. What is the weigl.t of a mass on which a force of 10 pounds produces an acceleration of 3 feet-per-sec. per sec. ?
EXERCISE 10. What force would increase the velocity of a 3<D-pound mass from — 10 feet per sec. to 30 feet per sec. in 5 minutes ?
EXERCISE n. What acceleration would a force of 6 pounds produce in a mass of 3 2 units of mass ?
STATICS
ANALYTICAL STATICS
CHAPTER I
FORCES ACTING AT A SINGLE POINT SECTION I
TRIANGLE OF FORCES
Statics treats of the equilibrium of forces.
When a body is acted on by forces which are in equi- librium the body is at rest or moves with a constant velocity.
Representation of a Force. — A force can conveniently be represented by a straight line; one end of the line will represent the point of application of the force, the direc- tion of the line gives the direction of the line of action of the force, the number of units of length in the line represents the number of units of force, while the arrow- head shows the "sense" of the force.
It can be experimentally de- monstrated that if OA and OB (Fig. i) represent two forces acting upon a particle O, the diagonal OC of the parallelo- gram constructed upon OA and OB represents in direc-
14 ANALYTICAL STATICS
tion and magnitude the resultant of the forces OA and OB, or the single force OC may replace the forces OA and OB without affecting the state of the particle O.
If we assume a force OC' (Fig. 2) equal in magnitude, but opposite in direction to OC, acting upon O together with OA and OB, it is evident. that equilibrium of the forces would result.
Let us now start with any one of the forces OC', OA, OB, as OA (Fig. 2), and in any order put the remaining
forces end to end, being careful to preserve their lengths and direction, and it will be found that a closed triangle results. (Prove this by Geometry.) Therefore,
// three forces impressed on a particle are in equilibrium, they can be represented in direction and magnitude by the sides o) any triangle drawn so as to have its sides parallel to the forces.
Before applying this principle to some examples we will call attention to the manner in which the actions of cords, planes, and supports in general may be repre- sented by their equivalent forces.
A flexible cord or string is always assumed to transmit any force along its fibres in the direction of its length. If two equal and opposite forces keep a string in equilib- rium, the string is said to be in tension and the tension is
FORCES ACTING AT A SINGLE POINT
Fir, 3
cons cant throughout its length. This tension is meas- ured by the force applied at either end. The tension in a cord passing over a smooth (frictionless) peg or surface is unaltered. If a string is knotted to another string at any point of its length, the tension will no longer be the same for both portions, in fact we may consider the two portions as two separate strings.
The action of a string can thus always be represented by a force having for direction the direction of the string, and this force will measure the tension to which the string is subjected.
A smooth (frictionless) plane always exerts a force normal to its surface. Thus in Fig. 3, representing a beam resting upon two smooth surfaces, tLe arrows NI and N2 represent the reactions of the planes. |w If a body rests upon a smooth peg, the peg exerts a force normal to the surface of the body at the point of contact. This is illustrated in Fig. | 4, representing a beam resting upon w a peg, P, and against a smooth wall, WW\ here S represents the reaction of the peg, and N the reaction of the wall.
A body which has been separated from all supports and upon which the reactions of the supports are all represented by their equivalent forces is called a jree body. Of course the action of gravity must also be shown as a force, representing the weight of the body, acting upon the free body.
In statics it is of the utmost importance always to
FIG. 4
16 ANALYTICAL STATICS
represent the body to which the principles of equilibrium are to be applied as a free body. The greatest care must be taken to represent all the forces acting upon the body by their respective arrows.
In solving problems involving the above principles the method to be pursued may be outlined as follows:
(a) Draw a diagrammatic sketch to aid in obtaining a clear conception of the problem.
(b) Find a point at which three forces act in equilibrium.
(c) Locate this point with its impressed forces, making it a "free body."
(d) Construct the triangle of forces.
(e) Apply your knowledge of mathematics (Geometry, Algebra, and Trigonometry) so as to calculate the re- quired forces.
(/) Interpret your answer so as to obtain its mechan- ical significance.
Example. — A picture weighing 3 Ib. hangs vertically from a nail by a cord passing through two rings in the frame. The parts of the string form an equilateral tri- angle. Find the tension of the string.
Solution. — (a) Fig. 5 illustrates the prob- lem.
(6) At N (Fig. 5) we have three forces acting, the tensions of the two strings and the reaction of the FlG 5 ' nail, which is equal to the weight
of the picture and acts upward. (c) In Fig. 6 NC represents the reaction of the nail, and Na( = T1) and Nb ( = T2) the tensions of the strings NA and NB respect- ively. These lines represent forces and are in no way
FORCES ACTING AT A SINGLE POINT
connected with the lines in Fig. 5, except by reason of their parallelism.
(d) Fig. 7 represents the corresponding triangle of forces.
(e) It now remains to calculate the values TI and T2 from the triangle of forces. This may be done
in many ways ; for instance, from Q drop ±QS to PR; then as PQ = QR, PS = SR = 3/2 (Geometry). From APSQ,
PS
_
3/2 _
ro 2 = —
Ib.
and
^3 . ..(Geometry.)
(/) We thus arrive at the conclusion that the tensions in the strings are equal, and that in each it equals 1.7 + Ib. This means that if the string is strong enough to support a weight of 1.7+ Ib., as indicated in Fig. 8, it will safely support the picture as shown in Fig. 5. - A
Example. — A piece of ma- chinery weighing 4000 pounds is suspended by two ropes making angles of 30° and 45° with the vertical. Find the ten- sions in the ropes. FlG- 9
Solution. — (a) Fig. 9 illustrates the problem.
(b) At C we have three forces acting.
(c) Fig. 10 shows these forces, the unknown forces due to the action of the ropes being
denoted by R and S.
FIG. 8
FIG. 10
1 8 ANALYTICAL STATICS
(d) Fig. ii represents the triangle of forces.
(e) To calculate the magnitude of the forces R and S note that the angles between the forces 4000 and R and between 4000 and S (Fig. n) are respectively equal to 30° and 45°. (Why?) Therefore the angle between the forces R and S becomes 105°. By the law of sines (Trigo- FIG. ii nometry) we have
S R 4000
sin 30° sin 45° sin 105'
/ 4000 \ \ sin75<>|
4000 sin 30°
.'. S = : o — = 2070 pounds,
sin 75°
4000 sin 45° and R = : 5 — = 2928 pounds.
sin 75
(/) These results to be interpreted as in the previous example.
Example. — A heavy particle whose weight is W pounds is placed on a smooth (frictionless) inclined plane, AB. The height of the plane is a and its length is c (Fig. 12). Find the force acting parallel to AB required to sustain the particle upon the plane. Also find the pressure exerted by the particle upon the plane.
Solution. — (a) Fig. 12 illustrates the problem.
(&) Whenever a surface enters into a problem it should be remembered that there always exists a normal reaction perpendicular to the surface, which in the case of a friction-
FORCES ACTING. AT A SINGLE POINT
less plane is the only reaction. The particle then be- comes the point at which three forces act in equilibrium.
(c) The three forces are (Fig. 13): ist. The weight, W.
2d. The sustaining force, P.
3d. The normal reaction of the plane,
(d) The triangle of forces is shown in Fig. 14.
(e) In this example no angle is given. This precludes the use of Trigonometry. On due consideration it will be noticed that Axyz is similar to A ABC, (why?) and therefore
FIG. 14
=
xy AB'
Pa or more simply — =—\ .'. P = W—\
also
W
(/) This shows us that the force P necessary to sustain the particle upon the plane is less than W, the weight of the particle, for a<c,
As the reaction of the plane upon the particle is N, it follows that an equal but opposite force must be the pressure exerted by the particle upon the plane.
EXERCISE 12. A picture weighing 7 Ib. hangs by a cord passing through two hooks and over a small nail. The hooks are 18 inches apart and the cord is 4 feet long. Find the tension of the cord.
20 ANALYTICAL STATICS
EXERCISE 13, If the cord in Exercise 12 be lengthened to 6 feet, how will the tension in it be altered ?
EXERCISE 14. A cord is stretched horizontally between two posts 6 feet apart. What is the tension in it, and how much longer does it become when a 7-lb. weight suspended from its middle makes it droop 3 irches?
EXERCISE 15. A particle weig ling 10 lb. is sustained upon a smooth inclined plane by a force whose line of action makes an angle of 60° with the horizontal. If the angle of the plane is 30°, find the sustaining force.
EXERCISE 16. Same as preceding exercise with angle of plane 45° and angle between line of action of the force and horizontal 30°.
EXERCISE 17. An inclined plane has a base of 4 feet and a length of 5 feet. What will be the reaction of the plane upon a particle whose weight is 100 pounds, if the sustaining force be applied parallel to the base of the plane ?
EXERCISE 18. A weight w is supported on a smooth plane inclined at an angle a to the horizon, by means of a force in- clined at an angle 0 to the plane. Find the magnitude of the force, and the pressure on the plane. If there is no pressure on the plane, in what direction does the force act ?
EXERCISE 19. A man pushes a garden-roller weighing 80 pounds up a plank 10 feet 3 inches long and with one end 2 feet 3 inches above the ground. If the handle is horizontal, find the force applied and the pressure of the roller on the plank.
EXERCISE 20. Two sticks, AB and BC, loosely jointed at A, rest upon horizontal ground at B and C, making angles of 65° and 40° respectively with the ground. If a weight of 200 pounds be suspended from A, find the forces transmitted by the sticks to the ground.
EXERCISE 21. Draw a triangle ABC with base BC hori- zontal and its vertex A under BC. Let AB and AC be threads fastened to two fixed points B and C, and to a third thread at A. If the third thread supports a weight W and the angles
FORCES ACTING AT A SINGLE POINT
21
of the triangle are A, B, and C respectively, find the tensions in the threads.
SECTION II
COMPONENTS. COMPOSITION OF FORCES
WE have just noticed that the triangle of forces gives us an easy method for calculating the relations existing between three forces in equilibrium.
For the consideration of the equilibrium of more than three forces another method must be used. To intro- duce this, we will first consider the
Resolution of Forces
In Fig. 15 assume the force R and the lines of action
of two other, unknown forces OA and OB. Through C
draw CD\\ to OA, and CE\\ /B
to OB. Let OD and OE
then represent the forces Fy
and Fx respectively. By the
parallelogram of forces, Fx
and
sultant.
Fx and Fy are called the Components of R; and R is
said to be resolved into the component forces Fx and Fy. !B In case the lines of action
of the components OA and OB are oblique, the analyt- ical calculation of Fx and Fy is not easily accomplished. To simplify the calculation
we shall, whenever possible, assume OA and .OB at
right angles as in Fig. 16. Then if ZCO£ = 0,
Fy would have R as re-
22 ANALYTICAL STATICS
and Fy = Rsmd.
EXERCISE 22. A force of 100 pounds acts at an angle of 30° to the horizontal. Resolve same into vertical and hori- zontal components.
EXERCISE 23. Same as Exercise 22 if force is inclined at 45° to the horizontal.
For convenience we shall designate all forces acting upward and to the right as plus, and therefore all forces acting downward and to the left as minus.
EXERCISE 24. Resolve a force of 2 tons, acting at an angle of 1 60° to the horizon, into vertical and horizontal compo- nents.
EXERCISE 25. Same as Exercise 24, if the line of action of the force makes an angle of (a) 270°; (6) 210°; (c) 315° with the horizontal.
EXERCISE 26. The pull on the rope of a canal-boat is 100 pounds, and the direction of the rope makes an angle of 60° with the parallel banks. Find the force urging the boat forward.
In the propulsion of a canal-boat, is a long or a short rope more advantageous ? Why ?
Example. — Explain the action of the wind in propelling a ship.
Solution. — Let AB, Fig. 17, represent the direction of the ship's keel; CD the position of the sail, which we as- sume flat. Let the pressure of the wind be equivalent to a force P acting on the sail in the direction indicated. Resolve the force P into two components T and TV", one parallel to the sail and the other at right angles to the sail. The first component, T, produces little or no
FORCES ACTING AT A SINGLE POINT 23
effect, and we will neglect it. The component N acts on the ship through the mast. We may resolve N into the components X and Y, parallel and at right angles to the keel. The resistance offered by the water to a motion in a direction at right angles to the keel is so great that the component Y produces but little motion. The ship is so built that the water may offer only a small
FIG. 17
resistance to motion parallel to the keel, and the ship moves in this direction under the action of the force X.
EXERCISE 27. Show that a vessel may sail due east against a southeast wind.
EXERCISE 28. If F be the force of the wind, a and /? the inclinations of the wind and sail to the keel of a boat, find the headway force and the leeway force.
EXERCISE 29. A boat is sailing due west; the wind is from the northeast. The sail is set at an angle of 75° with the keel of the boat. Of the two possible positions of the sail select the most advantageous and calculate the headway force.
EXERCISE 30. A boat is towed along the centre of a canal, 25 feet wide, by mules on each bank; the length of each rope is 36 feet. Find the force exerted by each mule when the force urging the boat forward is 200 pounds,
24 ANALYTICAL STATICS
Composition of Forces
The method of resolving a force into its components leads us to a method of combining forces which is, for
purposes of calculation, much more convenient than the prin- ciple of the parallelogram of */ ^^ forces. Consider three forces
F, F", F'" all acting upon the point O and inclined to the horizontal at angles 0', 0", FIG. I8 0"', respectively (Fig. 18).
From what has preceded it
follows that each of the three forces may be replaced by a horizontal and vertical component, i.e.,
F by F'cosd' and F'sin0'; F" by F" cos 0" and F" sin 6" ; F" by F'" cos V" and F" sin 0"'.
To simplify the notation, we shall designate F cos 0' by FX and F' sin 0' by Fy, etc., and therefore call the horizontal line OA the axis of X, and OB the axis of Y.
As Fxr, Fx", FX" all have a common line of action (the axis of X), we may add them algebraically to obtain their resultant,
similarly, Ry = Fy' + Fy" + Fy'",
where Rx and Ry are the x and y components of the re- sultant, R, of the three forces F, F" , F'" .
FORCES ACTING AT A SINGLE POINT 25
Fig. 19 shows how Rx and Ry may be utilized to obtain the resultant
and the direction of its line of action
_i RV ct = tan 1 •=*.
RX
It should be noticed that the argument just applied would be in nowise affected if the X axis and the Y axis
FIG. 19
were not horizontal and vertical so long as they remained mutually perpendicular.
EXERCISE 31. Three forces of 6, 8, and 10 pounds act on a particle at angles of 120° to each other. Find the resultant in magnitude and direction:
ist. By assuming axis of x coincident with force 6;
2d. 3d-
10.
EXERCISE 32. ABCDEF is a regular hexagon. Find the magnitude of the resultant of the forces represented by AB, AC, AD, AE, and AF. (Assume AB = a).
EXERCISE 33. Three smooth pegs are driven into a vertical wall and form an equilateral triangle whose base is horizontal.
26 ANALYTICAL STATICS
Find the pressure on each peg, if a thread, having equal weights of 10 pounds attached to its ends, be hung over the pegs.
EXERCISE 34. Three men pull an iron ring. The first pulls with a force of 100 pounds in a southeasterly direction, the second pulls northeast with a force of 70 pounds, and the third pulls towards the north with a force of 50 pounds. In which direction will the ring move ?
EXERCISE 35. Two forces of 20 pounds each and one of 21 pounds act at a point. The angle between the first and second is 120°, and between the second and third 30°. Find the resultant in direction and magnitude.
SECTION III CONDITIONS FOR EQUILIBRIUM
We know that the resultant of any number of forces may be expressed by R=^Rx2 + Ry2.
If a force equal in magnitude but opposite in direction to R is applied to the body, it will balance the original forces.
If the original forces are themselves in equilibrium, no other force is necessary to produce rest or motion with constant velocity, and their resultant is therefore zero.
Whenever forces are in equilibrium we may set
R-o,
and as R2 = Rx2 + Ry2, both Rx and Ry must be o. (Why?) Therefore Fx' + Fx" + Fxm + . . . = o,
and Fv'+Fy"+Fy'"+...=o;
whence follows the theorem:
FORCES ACTING AT A SINGLE POINT 27
// any number of forces acting on a body keep it in equilibrium, the respective sums of the components oj the jorces along any two straight lines at right angles to each other are equal to zero.
In the solution of problems based upon this theorem \ve may proceed as follows:
(a) Draw the sketch illustrating the problem.
(b) Find a point at which the forces act in equilibrium.
(c) In a separate sketch show this point with all the forces acting upon it and draw "any two straight lines at right angles to each other." These lines are usually drawn so as to coincide with as many forces as possible.
(d) Write the equation expressing the fact that the respective sums of the components of the forces along these lines or axes are equal to zero.
(e) Solve the equations for the unknown quantities they may contain.
(/) Interpret the results so obtained.
Example. — A rod AB, whose weight may be neglected, is hinged at A and supports a weight W at B. It is held up by a wire BM fastened to a fixed point M vertically above A. If AB is horizontal and the angle ABM = 30°, find the tension, T, in the wire and the compression, C, in the rod.
Solution. — (a) Fig. 20 illus- trates the problem.
(b) The wire and the rod
RG. 20 both act upon the point B,
and by the nature of the problem this point is in equi- librium,
28
ANALYTICAL STATICS
(c) In Fig. 21 the forces acting at the point B are shown. BX and BY are the lines along which the components will be taken.
FIG. 21
(d) The components of the forces in the X and Y directions are:
|
Forces |
X Components |
Y Components |
|
C |
C |
0 |
|
W |
0 |
-W |
|
T |
- T cos 30° |
T sin 30° |
Therefore C + o - T cos 30° = o and o - W + T sin 30° = o.
(e) C-
or
-W+T— =o.
2
T=+2W, and C =
(/) In Fig. 21 the arrow T represents the action of the wire upon the point B. The reaction upon the wire at B is necessarily an equal but opposite one and may
FORCES ACTING AT A SINGLE POINT 2 9
be represented as in Fig. 22. As the wire is in equi-
librium, an equal force must
act upon the wire at M, this \.M
force being supplied by the
wall. These equal and op-
posite forces (T = zW) at M
and By Fig. 22, produce a
tension of T = 2W Ib. in the
FIG. 22 wire.
Explain the 'effect of the forces acting upon the rod AB.
EXERCISE 36. A man weighing 160 Ib. rests in a hammock suspended by ropes which are inclined at 30° and 45° to two vertical posts. Find the pull in each rope.
EXERCISE 37. A man weighing W Ib. is seated in a loop at the end of a rope / feet long, the other end being fastened to a point above. What horizontal force will pull him m feet from the vertical, and what will be the pull on the rope? If the rope is just strong enough to support the man when the rope is vertical, will it support the man when displaced from the vertical?
EXERCISE 38. A thread whose length is 2/ is fastened at two points, A and B, in the same horizontal, and distant / from each other. The thread carries a smooth ring of weight W. Find the tension in the thread.
EXERCISE 39. What weight can be sustained on a smooth inclined plane, the ratio of whose height to base is 5 : 12, by a horizontal force of 10 pounds and a force of 50 pounds parallel to the plane. What is the pressure on the plane?
EXERCISE 40. Five men pull upon five ropes knotted together towards the south, northeast, east, northwest, and 30° south of west respectively. If the first three exert forces of 70, 30, and 40 pounds respectively and the knot does not move, find the forces exerted by the remaining two men.
tends to take Place
30 ANALYTICAL STATICS
SECTION IV
STATICAL FRICTION
Thus far the surfaces of bodies in contact have been assumed perfectly smooth, that is, they offered no re- in which Motion > sistance to the motion of * the bodies parallel to their surfaces. In reality no A - — >• body can • be made per- fectly smooth, and thus if a body weighing W Ib. is to be moved along a '
horizontal plane, a certain force P is necessary to start ' the body, Fig. 23.
This force P overcomes an equal but opposite force F, arising from the irregularities in the surfaces of contact which fit more or less closely into one another. This is shown in Fig. 24, which represents a very highly magnified view of
the contact between th'e weight and plane. This force,, F, is called Friction. From an inspection of Fig. 24 it is evident that the irregularities will always grip in such a way as to resist any motion of the weight W. So that: The direction in which friction acts is always opposite to the direction in which motion would take place if there were no friction.
The amount of friction up to a certain limit is always just sufficient to prevent motion. But only a limiting amount of friction can be called into play. We may
FORCES ACTING AT A SINGLE POINT 3!
suppose that this limiting amount of friction is reached when either the irregularities give way and break, or when the body lifts sufficiently to allow the irregularities to clear one another.
Thus, if a body rests upon a horizontal table, the pres- sure of the table balances the weight ; and as these forces are both vertical, there is no tendency to move in the hori- zontal direction and no friction is called into play. Apply a small force parallel to the surface; the body does not move; sufficient friction is exerted to just balance the applied force; increase the applied force and still the friction increases so as to hold the force in equilibrium until the applied force reaches a certain magnitude, which the friction cannot reach, and the body moves. (Try this experiment with a book.)
Experiment shows that the amount o) this limiting friction varies as the normal pressure between surfaces in contact, or is directly proportional to the force with which they are pressed together. Therefore, if
F = limiting friction and N = normal pressure,
.Foe TV, and .*. F=/*N,*
where ^ is a constant depending upon the material of the surfaces in contact and the state of their polish, but not on their area or shape.
This constant /*, the Coefficient of Friction, is deter-
* The sign ex is read "varies as," and fi is called a proportionality factor.
ANALYTICAL STATICS
mined by experiment and for some substances is approxi- mately as follows:
Wood on wood or metal, surface dry 0.4 to 0.6
" ' ' " " ' lubricated . . o . i to o . 2
Metal on metal, surface dry 0.2
" " " lubricated 0.075
Steel on ice o .02
EXERCISE 41. Find the greatest horizontal force which can be applied to a sled shod with steel and resting on a horizontal sheet of ice without causing motion, if the weight of the sled is 50 Ib. and the load is 200 Ib.
EXERCISE 42. An iron mass of 1000 Ib. rests upon a hori- zontal wooden floor. What will be the least horizontal force necessary to move the same if the coefficient of friction is 0.5? Between what limits may the applied force vary with- out producing motion?
Consider a weight W placed upon a rough inclined plane (Fig. 25). The tendency to move is in the down-
FIG. 25
FIG. 26
ward direction and we thus have the friction acting up- ward and parallel to the plane.
This force with the other forces acting upon W are. shown in Fig. 26, and as the weight is assumed at rest we have :
FORCES ACTING AT A SINGLE POINT 33
Forces X Components Y Components
W -Ws'md -WcosO
/ / o
no n
where I (sigma, the Greek S) denotes summation. Thus IX is read, the sum of the X components. There-
fore n = W cos 6, f = W sin 0.
Thus the friction j = W sin 0, where 0 is the inclination of the plane.
Assume 0 to vary from o to — , then sin 6 varies from o
to i and the friction, to preserve equilibrium, would have to vary from o to W.
It is evident that / cannot possibly equal W or even approach this value, and therefore equilibrium cannot be preserved throughout the whole change of 6. As 6 increases from o, some limiting angle must be reached beyond which the plane cannot be inclined with- out causing the body to slide.
If we call this limiting angle
FIG. 27 a, then in Fig. 27 the weight
will be on the point of sliding and is retained in equilibrium by the limiting friction F, and the values of n and / above deduced become
N = Wcosa,
.F W sin a from which {J-=~^ = ^Tr — — = tan a can be obtained
r N [f cos a
34 ANALYTICAL STATICS
The limiting angle, a, is known as the Angle of Friction or the Angle oj Repose; therefore: The tangent oj the angle oj friction is equal to the coefficient oj friction. As has already been pointed out (page 15), a surface always exerts a normal pressure or reaction upon a body in con- tact with it. If the plane is rough we will have, under cer- tain conditions, the friction as
an additional reaction (Fig. 28).
ur ^te The resultant of these forces is
called the Total Reaction of the
surface. If <j> be the angle between the normal to the surface and the total reaction, we have
F
and as j^ = JJL = tan a,
tan (j) = tsma} .'. <f>=a. Therefore:
The greatest angle that the total reaction of a rough surface can make with the normal is equal to the angle oj friction.
Example. — A heavy body rests on a rough plane in- clined at an angle 30° to the horizontal, the coefficient
2
of friction being — ^ A horizontal force is applied to V3
the body, and is gradually increased until the body begins to move up the plane; find the magnitude of the hori- zontal force.
FORCES ACTING AT A SINGLE POINT
35
Solution. — As /£ = —=, the tangent of the limiting angle (a) to which the plane may be inclined without causing
2
the weight to slide is — ,— , or V3
tan a = — -^= 1.15 + ;
As the plane is only inclined at 30°, the body will remain at rest upon the plane with only part of the total amount of friction (acting upward) brought into play (Fig. 29).
FIG. 29
FIG. 30
If the force p, Fig. 30, be now applied, this force will tend to cause motion up the plane, thus relieving /, and as p increases it causes / to act down the plane, finally overcoming /, thus producing motion up the plane.
The value of p which we are to consider is the limiting one just suffi- cient to balance the limiting value of / acting downward. We will call these values of p and /, P and F.
Fig. 31 shows this condition of equilibrium.
36 ANALYTICAL STATICS
Fig. 32 shows us all the forces acting upon the body. We have:
Forces X Components Y Components
P P o
N —N sin 30° N cos 30°
F —F cos 30° — F sin 30°
W o ' -W
w
FIG. 32 and the conditions of equilibrium are
2X=P-N sin 30°-^ cos 3o° = o, = N cos 30°-^ sin 3o°-TF=o,
or
W = ^N--F.
2 2
(0
These equations contain three unknown quantities, TV, P, and jP. So that we must find another equation before we can effect a solution. This third equation is obtained through the law of friction. We know the coefficient of friction, and as F is the limiting value oj the friction and AT" the normal pressure between the surfaces,
we have
= —AT = - — 3 AT.
V3
(3)
FORCES ACTING AT A SINGLE POINT 37
We can now find P in terms of W from equations (i), (2), and (3). Substituting F from (3) in (i) and (2),
P=ljv+JV=:^Ar, (4)
2 2
\f 1 V 1 V3
Hr__^_ _o.V==-^2V. . . . (5) Substitute JV from (4) in (5),
EXERCISE 43. A weight W rests in equilibrium on a rough inclined plane, being just on the point of slipping down. On applying a force W parallel to the plane, the weight is just on the point of moving up. Find the angle of the plane and the coefficient of friction.
EXERCISE 44. What is the coefficient of friction when a body weighing 50 Ib. just rests on a plane inclined at 30° to the horizontal ? If the plane were horizontal, what horizontal force would be required to move the body?
EXERCISE 45. What horizontal force will be required to support a weight of 300 Ib. upon a smooth inclined plane whose height is 3/5 of its length ?
EXERCISE 46. A block of iron weighing 10 Ib. rests on a level plate. A string attached to the block passes over a pulley so placed above the plate that the string makes an angle of 45° with the vertical. After passing over this pulley the string supports a weight. Find the least value of this weight which will make the block slip, the coefficient of fric- tion being 0.2.
EXERCISE 47. Explain with diagrams two methods for experimentally determining the coefficient of friction.
38 ANALYTICAL STATICS
EXERCISE 48. A weight of 60 pounds rests on a rough level floor. Find the least horizontal force that will move it (^ = 0.5). Find the total reaction of the floor. What is its direction ?
EXERCISE 49. Find the least angle of inclination of a wooden incline that stone blocks may slide down under the
action of gravity.
Hr)
EXERCISE 50. On a hill sloping i in 50 a loaded sled weighing one ton is kept from sliding down. Show that the pull of the horses may vary from 360 to 440 pounds, the coefficient of friction between sled and snow being 0.2.
SECTION V MOMENTS
We will now consider another method of obtaining the relations existing between forces acting on a particle.
In Fig. 33 consider the force F acting on a particle at A and conceive a fixed point O to be rigidly connected to A. The force F will then tend to move the particle A about O as centre. The effect of the force in producing rotation about O depends not only upon the magnitude of the force F, but also upon the dis- tance of the line of action of F from O. This perpendicular distance, d, is called the arm oj the force. We may then state as a definition that the
Moment oj a jorce with rejerence to a point is the product oj the jorce and its arm,
wherein the word moment is used in its old-fashioned sense of importance or influence; so that the moment of
FORCES ACTING AT A SINGLE POINT
39
da
a force with reference to a point means its influence in producing rotation about this point.
The point O, from which the perpendicular is drawn, is called the origin o) moments and may be chosen arbi- trarily.
The algebraic sign of a moment is considered positive if it tends to turn the system in a direction opposite to
that of the hands of a FI
watch, and negative if in the other direction. This assumption is purely arbi- trary, but it will be noticed that it conforms with the positive and negative di- rections assumed in the measurement of angles in Trigonometry. Thus in Fig. 34 the moments of the forces with reference to the origin of moments O are — F\d\'y ~{~F2d2, -\-F^d^] F 4(0) —05 and — F^d^.
As the moment of a iorce = Fd, the unit moment is the moment due to a force of i pound whose arm is i foot. No name has been assigned to this unit.
EXERCISE 51. A force of 5 pounds acts along one side of an equilateral triangle whose side is 2 feet long. Find the mo- ment about the vertex of the opposite angle.
EXERCISE 52. A force of P pounds acts along the diagonal of a square whose side is 2n feet. Find the moments of P about each of the four vertices.
EXERCISE 53. If P is the thrust along the connecting-rod of an engine, r the crank-radius, and the connecting-rod is inclined to the crank at 150°, show that the moment of the
F. FIG. 34
4o
ANALYTICAL STATICS
FIG. 35
thrust about the crank-axis is one-half the greatest moment possible.
EXERCISE 54. At what height from the foot of a tree must one end of a rope, whose length is / feet, be fastened so that B a given force acting at the other end may have the greatest tendency to overturn it, assuming the force applied at the ground. EXERCISE 55. The post AC, Fig. 35, of a jib-crane is 10 feet long; the jib CB is inclined at 30°, and the tie AB at 60°, to the vertical. If the weight lifted is 10 tons, find the moment about C tend- ing to upset the crane.
The moment of a force may be geometrically repre- sented by twice the area of a / triangle whose base is the force and whose altitude is the arm of the force. Thus in Fig. 36, 2 (area of AOAB) =Fd = moment of F about O.
This geometrical representation will be used to prove Varignon's Theorem o) Moments:
The sum of the moments of two forces, FI and F2,
about any point O in their plane is equal to the moment of their result- ant, R, about the same point.
In Fig. 37 (moment of Fl about O) = 2AOAB, (moment of F2 about O) , (moment of R
FIG. 36
FIG. 37
about O)=2AO4C. We must prove that
FORCES ACTING AT A SINGLE POINT 41
(moment of FI) + (moment of F2) = (moment of R), or
Draw BE and DF \\ OA. Then
AOAB-0=AOAE and AOAD=C>AOAF. (Why?)
Prove A E = CF, by means of A A BE and ADCF. Then
AOAEOAOFC. (Why?)
From this the remainder of the proof may be easily deduced.
Theorem: If any number 0} forces acting upon a par- ticle are in equilibrium, the algebraic sum o] their moments about any point will be zero.
To prove this proposi- tion consider the forces FI, F2, . . . , Fig. 38, to be in x*,/ \ equilibrium, and O any ori- Y-^d8 \
gin of moments. We must
FIG. 38 now prove
F1dl+F2d2 + . . .=o.
Join O and A and find the component forces _L to OA ; then
Fi sin #! +F2 sin 62 + . . . =o. (Why?)
But sin/9!=, sin/92 = , etc.
Therefore FI c-
and Fidi+F2d2 + . . .=o;
or J moment = IM = o.
ANALYTICAL STATICS
Example. — A brace AB rests against a smooth vertical wall and upon a rough horizontal plane, and supports a weight W at its upper end. Find the compression in the brace if the angle CAB is 0.
Solution. — The forces acting at A are W, AT, and C,
Fig. 39-
As these forces are in equilibrium we may select any
point as our origin of moments and equate the sum of the
moments to zero. But as we seek a relation between C and W, we should attempt to ex- clude AT from our equation of moments. This can only be done by choosing the origin of moments upon the line of action of N, thus reducing its arm to zero. Besides, the
origin of moments should be chosen conveniently for
calculation. Vertically above B at O is a good position.
Thus,
2M=-W(OA)+C(OAcos6)=o; .'. C = Wsec6.
EXERCISE 56. Find the normal reaction of the wall in the preceding example.
EXERCISE 57. A weight W is attached to a string which is secured at A to a vertical wall and pushed from the vertical by a strut BC perpendicular to the wall; find the pressure on BC when the angle CAB is 0.
EXERCISE 58. A rod whose length is BC = l is secured at a point B in a horizontal plane, and the end C held up by a cord AC so that /.ABC is 0 and the distance AB = a\ re- quired the compression in BC due to a weight W applied at C,
CHAPTER II
FORCES ACTING ON A RIGID BODY SECTION VI
RESULTANT OF TWO FORCES. COUPLES
To the present our work has been confined to the investigation of forces applied to a single point or particle.
We shah1 now consider the action of forces applied at different points of a rigid body.
A body is said to be rigid when the particles of which it is composed retain their relative positions no matter what external forces may be applied to the body. Prac- tically no such thing as a rigid body is found in nature, but if the body considered " gives" under the action of the applied forces, the methods now to be described may still be employed, provided the positions of the points of application considered are assumed after all changes in the body have ceased.
If any force F (Fig. 40) be applied to a body at A,
the effect of this force will
, .P . . , FIG. 40
remain unchanged if its point
of application be moved to any other point in the line of action of the jorce, such as A'. This principle is known as the transmissibility oj a jorce<
43
44
ANALYTICAL STATICS
FIG. 41
Let the question be proposed to find the resultant of \ ! two non-parallel forces FI and
F2, Fig. 41, applied to a body, B, at the points AI and A2.
Then by the above principle the force FI may have its point of application trans- ferred to any point in its line of action; the same holds for the force F2. Let us select the point P at the intersec- tion of their lines of action as the new point of applica- tion. The transference of the forces FI and F2 to the
new point P reduces the problem to the simple case
of the parallelogram of
forces, and the resultant R
may now be applied to any
point in its line of action
PAS.
Consider now the compu- tation of the resultant oj two
parallel jorces FI and F2, Fig.
42. It is evident that the
method of transmissibility of
forces will in this case lead
to no result. (Why?) So
a special artifice must be
used.
Two equal and oppositely
directed forces F having a
common line of action AiA2 are introduced. As these
FIG. 42
FORCES ACTING ON A RIGID BODY 45
forces neutralize each other their introduction will not affect the original problem. Combine these forces with FI and F2 and obtain the forces RI and ^2. Now, instead of considering FI and F2, find the resultant of RI and R2, which we will denote by R, whose point of application is anywhere in the line PC, at C if we so choose.
By resolving R\ and R2 at P into components along PC and parallel to A^A2 the magnitude oj R will be jound to be Fi+F2, and its direction will be parallel to FI and F2. (Prove this.) To completely determine the resultant it is sufficient to locate any convenient point, say C on AiA2, on its line of action. From the similar triangles A2MN and PCA2 we find that
F2 PC Fl PC
-F-CAJ similarly j = — ;
/. -2 = , or F1(CAl)=F2(CA2).
As the angle between the forces FI, F2 and the line AiA2 does not enter into this discussion, it should be noticed that the resultant R always passes through C irrespective of the direction of the parallel forces.
This point C is therefore called the Center of the Parallel Forces.
EXERCISE 59. Show how to find the resultant of two un- equal parallel forces acting in opposite directions but not in the same straight line.
Through Ex. 59 we arrive at the result shown in Fig. 43. The resultant of the two unequal parallel forces, FI and F2j is parallel to them, but divides the line AiAz
46 ANALYTICAL STATICS
externally at C. The resultant has the same direction c as the greater of the. two forces, F2, and equals F2 — Fi. Also F2 (CA 2) =
EXERCISE 60. Three forces are repre- sented by the lines AD, BC, and DB in a parallelogram A BCD. Show that ^4C is FlG- 43 their resultant.
EXERCISE 61. Two forces acting upon a body are located with reference to a set of rectangular axes as follows: F\, point of application ( — 4, —4), with arrow representing force extending to (2, 6); F2, point of application (i, 7), arrow ex- tending to (14, 6). Find the resultant.
EXERCISE 62. Two parallel forces of 3 and 5 pounds act in the same direction at points 2 feet apart. Find their resultant in magnitude and position.
EXERCISE 63. Two parallel forces of 3 and 5 pounds act in opposite directions at points 2 feet apart. Find their re- sultant in magnitude and position.
EXERCISE 64. "We have a set of hay-scales and sometimes we have to weigh wagons that arc too long to go on them. Can we get the correct weight by weighing one end at a time and then adding the two weights ? "
Consider the case oj two equal parallel jorces acting in opposite directions (Fig. 44). If we here apply the artifice used when the forces are unequal, we find that the lines of action of R± and R\ are still parallel and the method fails.
But from, the results obtained in Ex. 59 we have R = F1-F1=o and F1(CA1)=Fl(CA2)-, .'. CA1=CA2.
But as C must divide A iA2 externally, CAi can only equal CA2, as CA2 = °°. (The sign ± is read " ap- proaches," /, CA% approaches infinity.) In other words,
FORCES ACTING ON A RIGID BODY
47
FIG. 44
the two jorces cannot be reduced to a single force in a
definite position.
This combination o] two equal parallel jorces acting in
opposite directions, but not
in the same straight line,
which cannot be replaced
by a single force, is called
a couple.
As shown a couple can- not be reduced to a single
force, and its action
therefore is not that of
a force. It thus becomes
imperative to study more
fully the properties of couples.
The tendency of a couple is to cause rotation. This
is illustrated in the operation of a screw copying-press,
or in the winding of a watch or clock. This tendency
to produce rotation is measured, as explained on page 38,
by the moments of the forces.
Select any point O, Fig. 45, in the plane of the couple
and draw trie line Oab per- pendicular to the lines of action of the forces F. The sum of the moments ste of the forces about O is ° -F(aO)+F(bO)=F(bO- aO)=F(ab). The distance ab is called the arm oj the couple, and we have shown that the moment oj the
FIG. 45
couple [F(ab)] about any point in its plane is the prod-
ANALYTICAL STATICS
uct oj one oj the forces and the arm; the sign to be determined, as on page 39, by the direction in which the body acted on would tend to turn.
Theorems Concerning Couples
In the proofs of the following theorems we shall always consider the forces of the couple applied at the ends of their arm. That a couple can always be reduced to this form is evident from Fig. 45. Here we may move the forces F along their lines of action until the tails of their arrows are respectively at a and b.
We will first prove that the action of a couple remains unaltered if it is moved anywhere in its plane in such a position that the forces remain parallel to their original directions, or
The translation 0} a couple does not alter its effect upon the body.
In Fig. 46 consider the couple F(A iA2)F. We wish to show that it is equivalent to the couple F(Alf'A2ff)F) in which the F's are the unmarked ones.
To do this first move its forces by the principle of transmissibility to the position A\A2. Then anywhere on AifA2 prolonged take A\fA%r equal to Ai'A2=AiA2. In- troduce at AI" and A2" two pairs of equal and oppositely
|
1 1 R4=2F ' , f |
: :» : |
F \\ |
|
. |
1 F V |
' l |
|
^ A2 |
FIG. 46 directed forces F.
TI.e body originally acted on by the
FORCES ACTIXG ON A RIGID BODY 49
couple F(AiA2)F will now be acted upon by six forces, but its state of rest or motion will not be altered.
Now the singly marked F's may be replaced by the singly marked R = 2F, their resultant. Similarly for the doubly marked forces. The 7?'s balance each other and their action may be neglected. Thus all forces but the unmarked ones at AI" and A2" are accounted for. These evidently form the couple F(Ai"A2")F, which is therefore equivalent to the couple F(A1A2)F.
We will now show that a couple may be rotated about any point without altering its action.
In Fig. 47 consider the couple F(AiA2)F. On any line MA2' lay off MA^ = MAi and A\A2'=A\A2. At AI and A 2 introduce the pairs of forces F J_ to AI 'A2 '. Now the singly marked F's reduce to their singly marked resultant R. Similarly for the doubly marked .F's. /M
The resultants R have a common line of action (which
bisects £AiMAi) and are of equal magnitude. (Prove this.) They thus neutralize each other and there re- mains only the couple F(A^'A2')FJ which proves the proposition.
As a corollary to the above theorems we can state that
A couple can be shijted into any position in its plane without altering its action.
Another important theorem concerning couples is the following :
ANALYTICAL STATICS
TAi
FIG. 48
Any couple may be replaced by another couple oj equal moment.
This may be demonstrated thus: In Fig. 48 consider the couple F(AiA2)F. Let A3 be any point in AiA2 prolonged. Introduce two pairs of equal and opposite forces P _L to A^3 at A 3 and A 2. Let us, however, make P of such magnitude that P(A2A3) = F(A1A2). The resultant of the singly marked P and F will be Ri=F + P acting at A2 (see page 44), and the resultant of the doubly marked P and F is R2 = F + P acting at A 2. These resultants neutralize each other and there remains only the couple P(A2A3)P', but as P(A2AS) was made equal to F(AiA2), this new couple has the same moment as the original couple. Combining this result with the previous theorem we arrive at our present theorem.
We can now state that the resultant oj any number oj couples is a couple whose moment is the sum oj the moments oj the given couples.
For the given couples may all be reduced to equivalent couples having a given arm. They may then be shifted until the lines of action of their forces coincide in pairs. Their forces may then be added and there results a couple whose moment is equal to the sum of the moments of the given couples.
EXERCISE 65. Prove the last theorem for the case of three couples whose moments are Pa, Qb, and Re and whose arms are a, b, and c respectively.
EXERCISE 66. Along the sides AD and CB of a rectangle
FORCES ACTING ON A RIGID BODY 51
A BCD, whose side AD is s feet long and whose side AB is s yards long, forces of F pounds act, and along AB and CD forces of 5-F pounds act. Find the moment of the equivalent couple. In what position should this equivalent couple be placed with reference to the rectangle A BCD?
It is important to notice that a single force F acting at C, Fig. 49, and a couple P(a)P acting on a body in the same plane cannot be in equi- librium. For, the couple P(a)P may be replaced by a couple F(b)F, if Pa = Fb, and this new couple may be placed anywhere in the plane. If placed as shown in Fig. 49, the forces at C are FlG- 49
in equilibrium, leaving the force F at D unbalanced.
EXERCISE 67. Assume a force of 100 pounds and a couple whose forces are 50 pounds and whose arm is 4 feet. Find their resultant.
SECTION VII RESULTANT OF ANY NUMBER OF FORCES
We are now in position to find the resultant of any number of forces acting upon a rigid body.
Let the forces FI, F2t ^3, ... act at the points PI(XI, yj, P2(#2> ^2), -Pa (#3) ^3) ... of a body referred to rectangular axes. Consider the force FI resolved into two components Xi and FI, Fig. 50. At the origin O apply two opposing forces each equal and parallel to Xly and similarly for FI. This will not alter the given problem. Xi acting at PI may now be replaced by Xi acting at O plus a couple
ANALYTICAL STATICS
whose moment is -Xiji, and Yl acting at P may be replaced by FI acting at O plus a couple whose moment is YiXi. Similarly for all the other forces F2, F3, . . .
0 X,
FIG. 50
The resultant of all the component forces acting at O may now be found as already explained on pages 24 and 25. We have
and
We have thus found the magnitude and direction of the resultant, but it is evident that the line of action of the resultant need not pass through the origin.
So far we have neglected the couples introduced above. By adding their moments we may obtain the moment of the resultant couple :
and tana =
= Z(Yx-Xy).
We shall assume this to be the moment of a couple whose forces are R, R, and whose arm is a. R having
FORCES ACTING ON A RIGID BODY
53
been previously determined, the equation M = Ra = I(Yx-Xy) determines a. These results are illustrated in Fig. 51. R acting at O shows the magnitude and direc- tion of the forces acting at O, Fig. 50, and RaR the couple whose moment is M, placed in the convenient position for finding their resultant (see page 51).
o R.-Sx,
FIG. 51
The forces R, R acting at O neutralize each other, and this leaves KL as the final resultant, which may be applied at any point in its line of action KL.
The intercept of the line of action of the resultant KL, Fig. 51, upon the X-axis may readily be found to be
(OX) =- — , and as sin a = -^-, we have
sm a
7-*
R
To condense this theory for actual computation we need simply fill out the following table:
|
Forces |
X |
y |
X |
Y |
Yx |
Xy |
Yx-Xy |
|
Fi |
Xl |
y\ |
xl |
Yl |
YlXl |
xiyi |
YlXl-Xiyi |
|
.f?. |
X2. |
y* |
X2 |
V2. |
2 |
X2y: |
Y*c:~Xzy* |
54 ANALYTICAL STATICS
Then obtain the IX, IY, I(Yx-Xy)=Ra. Finally,
and the intercept of the line of action of the resultant upon the X-axis is
which gives us all the data necessary for the location cf the resultant.
EXERCISE 68. Three forces FI, F<2, F3 have their points of application at (2, 4), (4, — i), (—2, — 2), and the arrows rep- resenting their ends at (6, 7), (—2, —6), and (—7, 10), respectively. Find their resultant.
EXERCISE 69. Find the resultant of the following forces, given a point on each line of action, the angle each line of action makes with the -Y-axis, and their magnitudes:
i st force: (3,2); 100°; 50 pounds. 2d force: (—1,3); 200°; 100 pounds. 3d force: (—2, —4); 30°; 60 pounds.
EXERCISE 70. Find the resultant of the following forces: PI = 40 pounds, ^2 = 30 pounds, and PS = 90 pounds, whose lines of action make angles of 60°, 280°, and 140° with the X-axis and whose intercepts on the X-axis are 9 feet, 5 feet, and o feet, respectively.
SECTION VIII
CONDITIONS FOR EQUILIBRIUM
The motion of a rigid body may always be considered as either a translation, a rotation, or a combination oj the two.
FORCES ACTING ON A RIGID BODY
55
FIG. 52
In the motion of translation the spaces described simul- taneously by the different parts of the body are parallel and equal to each other. Thus if, in Fig. 52, AI, BI, Ci represent the positions of the particles of a body at a time /i, and A2, B2, C2 represent their posi- tions at some subsequent time t2, the body is said to possess a motion of translation.
In the motion of rotation the parts of the body describe concentric arcs of circles about a certain line, called the axis oj rotation. Thus in Fig. 53 the two positions shown
are those of a body pos- sessed of a motion of rota- tion, and a perpendicular to the plane of the paper passing through O would be the axis of rotation.
Every more complex mo- tion can be considered as the combination of a trans- lation and of a rotation. Thus, in Fig. 54, if the body ^l^iCi moves to the position A2B2C2, we may first conceive it to move by translation to A'B'C' and then by rotation about O as an axis from A'B'C' to A2B2C2.
Thus, only two kinds oj motion, Translation and Rota- tion, need be considered.
Translation is always caused by the resultant oj the
FIG. 53
ANALYTICAL STATICS
forces, and Rotation by the resultant of the couples acting
on a body.
The forces shown in Fig. 50, page 52, acting on the
body have been reduced, as explained on pages 52 and
53, to the force R and a couple whose moment isM = Ra. This body, if acted on only by the forces FI, F2, F3, . . . , would possess a com- pound motion ; the translation of which \ would be due to the 1 force R, and the rota- tion of which would be due to a couple whose moment is Ra.
Conceive now that the body is at rest, possessing neither trans- lation nor rotation.
FIG. 54
This would imply that R (the cause of the translation) is zero and M = Ra (the cause of the rotation) is zero.
Therefore, if any rigid body is in equilibrium under the action of forces,
IX = o, 2T = o, and I(Yx-Xy)=o,
the X and Y axes to be assumed at pleasure.
In the application of this principle proceed as follows :
(a) Draw a sketch illustrating the problem.
(b) Draw a diagram showing the body considered as a jree body.
FORCES ACTING ON A RIGID BODY
57
(c) Into the last sketch introduce the coordinate axes and fill out a table similar to that shown on page 53, thus obtaining the IX, the ZY, and the 2(Yx-Xy).
The choice of the position of the axes must be left largely to the ingenuity of the student.
(d) Solve the equations obtained by putting the three sums just found each equal to zero for whatever un- known quantities are sought.
(e) Interpret the results.
Example. — A beam AB rests on smooth horizontal ground at A and on a smooth inclined plane at B\ a string is fast- ened at B and, passing over a smooth peg at the top of the plane,
FIG. 55
supports a weight P. If W, the weight of the beam, acts at the centre of the beam and « be the inclination
of the plane, find P and the reactions on the rod. Solution. — (a) Draw a diagram illustrating the problem (Fig. 55).
(b) Draw the rigid body under considera- tion (beam) alone (Fig. 56), showing all the reactions and other forces acting upon it.
(c) Introduce the coordinate axes and complete the table:
FIG. 56
|
F |
X |
y |
X |
y |
Y* |
Xy |
Yx-Xy |
|
% |
- 2/ COS 0 - 1 cos 0 |
- 2/ sin 0 -I sine |
o 0 |
Nj -W |
— 2Njl COS0 Wl cos 0 |
0 o |
- zN,l cos e Wl cos d |
|
P |
0 o |
0 o |
— N2 sin a + P cos a |
A/s cos a P sin a |
0 0 |
0 0 |
0 o |
58 ANALYTICAL STATICS
Then, as the beam is in equilibrium, 2X = o = — AT 2 sin a + P cos a = o, 2Y=o = Ni-W + N2 cosa + P sina=o, and I(Yx-Xy)=-2N1l cos 0 + Wl cos 6 = 0. (d) Solving the equations, we obtain
W W W
NI= — ; P = — sin a; N2 = — cos a.
22 2
EXERCISE 71. A beam 20 feet long and weighing 1000 pounds rests at an angle of 60° with one end against a smooth vertical wall and the other end on smooth horizontal ground. It is held from slipping by a rope extending horizontally from the foot of the beam to the foot of the wall. Find the tension in the string and the reaction at the ground and wall.
EXERCISE 72. Same as Exercise 71, but assuming the wall rough, with 11 = 0.2.
EXERCISE 73. Same as Exercise 71, but assuming the ground rough, ^ = 0.3.
EXERCISE 74. Same as Exercise 71, but assuming the wall rough, /* = o.2, and the ground rough, ^ = 0.3.
EXERCISE 75. A beam 20 ft. long rests with its upper end against a smooth vertical wall. Its lower end rests on a smooth horizontal plane and is prevented from slipping by a rope 1 6 feet long fastened to it and the base of the wall. The weight of the beam is W and acts at a point 5 feet from its upper end. Find the reactions of the wall and the plane and the tension in the rope.
EXERCISE 76. A cellar-door AB, hinged at its upper edge, A, rests at an angle of 45° with the horizontal when a hori- zontal force F is applied to its lower edge, B. If the weight of the door, W, be assumed to act at its midpoint, find the force F and the reaction at the hinge A .
EXERCISE 77. A uniform rod, length 20 and weight II',
FORCES ACTING ON A RIGID BODY
59
rests with one end against the inner surface of a smooth hemi- spherical bowl whose edge is horizontal and whose radius is r. The rod is also supported at some point of its length by the edge of the bowl. Find its position of equilibrium.
Another method of obtaining the equation I (Yx-Xy) =o, used in the solution of the above exercises, which is perhaps less laborious than. V that described on page 57, is the following:
From Fig. 57 it follows, by Varignon's theorem of moments, page 40, that
Yx-Xy = -Fd\
y
FIG. 57
But 2(Fd) represents the sum of the moments of all the forces about the origin O, and as the origin O may be taken in any position we have as the conditions oj equilib- rium the three equations :
Sum o) the components of all the forces along any line=o\ Sum of the components of all the forces along a line JL to
the first line = o', Sum of the moments of all the forces about any point = o.
Example. — A uniform cylindrical shell of radius r stands upon a horizontal plane; two smooth spheres of radii a and b, such that a + b>r, are placed within it. Show that the cylinder will not upset if the ratio of its weight to the weight of the upper sphere exceeds the ratio 2r — a — b:r.
Solution. — Fig. 58 illustrates the problem. Fig. 59
6o
ANALYTICAL STATICS
shows the cylindrical shell as a free body, W being its weight. From this figure we obtain
(i) (2)
^(moments about O) = 2rN-rW-zRi=o. . (3)
(We select O as origin, as this eliminates R% from equa-
tion^)).
x- — \
R
FIG. 58
-I-
FIG. 59
Consider now the equilibrium of the upper sphere, W\ being its weight, and the corresponding triangle of forces (Fig. 60).
This triangle is similar to AABC (Fig. 58).
R1 2r-a-b
2r-a-b
or
. • (4)
W
From (2), (3), and (4) we have ~^r =
EXERCISE 78. Write the equations for equilibrium for Ex- ercises 75, 76, and 77 by the method just described.
An important principle which in many cases simplifies the solution of problems is the following:
// three forces in the same plane keep a body in equilib- rium, they must be parallel or meet in a point.
FORCES ACTING ON A RIGID BODY
6l
From pages 44 and 45 it is evident that parallel forces acting on a body may be in equilibrium. If the three forces are not parallel, two of them at least intersect and their resultant acting at the point of intersection must balance the third force. Therefore the line of action of the third force must pass through the point of intersec- tion of the first two; otherwise a couple would result and the body would not be in equilibrium.
Example. — A rod AB whose weight may be neglected and which is 35 inches long carries a weight, W, at C, 20 inches from A. A thread 49 inches long is tied to the ends of the rod and slung over a smooth peg D. Find the tension in the thread and the inclination of the rod to the horizontal when it comes to rest.
Solution. — In Fig. 61, as the thread ADB passes over a smooth peg at D, the tensions in AD and DB must be equal to, say, T pounds.
FIG. 6 1
FIG. 62
FIG. 63
In Fig. 62 the lines of action of the forces T, T, W, as there are only three, must pass through Z>, and we may apply the method of the triangle of forces, Fig. 63.
As
, a=& and T W
W
sin a sin{i8o—
sm(a+/?)*
62 ANALYTICAL STATICS
From Fig. 61,
:CB = b:a = 2o:i$; (Why?)
:a = 35:i5; (Why?) 49:0 = 35:15;
49X15 a = — — = 21;
35
6 = 28.
AH 28(i\/2)
then cc,s 0= -7—7= — - .
AC 20 10
Also T = W%\/~2.
EXERCISE 79. A uniform beam weighing W pounds rests on two smooth planes inclined at 30° and 60° to the horizontal. Find the angle which the beam makes with the horizontal in the position of equilibrium, and also the pressure on the planes.
EXERCISE 80. A spherical shot weighing 100 pounds lies between two smooth planes inclined respectively at 30° and 60° to the horizontal. Find the pressure on each plane.
EXERCISE 81. A straight rod 4 inches long is placed in a smooth hemispherical cup and when in equilibrium one inch projects over the edge. Find the radius of the cup.
EXERCISE 82. A rod 3 feet long is in equilibrium resting upon a smooth pin and with one end against a smooth ver- tical wall. If the pin is one foot from the wall, show that the inclination 6 to the horizontal is given by 3 cos3 6 = 2.
EXERCISE 83. A balanced window-sash, whose height is 3 ft. and width 4 ft., weighs 100 pounds. If one of the cords is broken, what must be the least coefficient of friction in order that the remaining cord may sustain the sash ?
FORCES ACTING ON A RIGID BODY 63
EXERCISE 84. A hemisphere, whose center of gravity is at a distance of f of its radius from the centre and is on the radius perpendicular to its plane surface at its centre, is sustained by friction against a vertical wall and a horizontal plane of equal roughness. Find the greatest inclination of the plane of the base to the horizon.
SECTION IX
PARALLEL FORCES. CENTROIDS OR CENTRES OF GRAVITY
In the consideration of the conditions for the equilib- rium of forces acting on a body we meet a difficulty not involved in the consideration of the forces acting on a single particle. While considering a particle the force of gravity (the weight of the particle) was assumed to act at the point occupied by the particle itself, but in the consideration of a body assumed to consist of many particles, each having its own weight, it becomes im- perative to find the resultant jorce oj gravity before we can apply the conditions of equilibrium.
To this end we will first find the
Resultant of any Number of Parallel Forces
In Fig. 64 conceive the forces FI, F2, F^ ... to be parallel and to have their points of application at P\(x\y\\ P2(x2y2), Pz(xzyz\ .... The forces do not necessarily lie in the plane xy, but may have any direction whatsoever so long as they remain parallel.
Consider first the forces FI and F2\ their resultant, by the principle described on page 45, is F'=Fi-\-F2, and the line of action of t'.e resultant divides P\P2 in the
64
ANALYTICAL STATICS
ratio F2:Fi. The coordinates of P' may now be deter- mined for
y 1=—} = ' — — — = — — — (by similar triangles),
f
from which we obtain
FIG. 64
Find now the resultant F" of Ff and F3. Here
and — =.
/f _
similarly
FORCES ACTING ON A RIGID BODY 65
The resultant of F" and F± may now be found, etc. The final resultant will be R = FI+ F2 + F3 + . . . , and the Centre of the Parallel Forces (see page 45) will be at the point (x, y)* where
_ FiOCi + F2X2 + ^3^3 + • • • - F\y 1+^2^2 + ^3^3 + ...
which we will abbreviate to
_ 2Fx . - IFy * = and =
Note particularly that the above operations are wholly independent of the direction of the parallel forces, and the position of the point (x, y) would remain unchanged even if the direction of the forces were altered.
EXERCISE 85. The legs of a right triangle are 3 and 4 feet long; a force of one pound acts at the right angle, a negative force of 2 pounds at the greater acute angle, and a force of 3 pounds at the smaller, all forces being parallel. Find the centre of the forces.
EXERCISE 86. Four forces of 2, 3, i, and —4 pounds act perpendicularly to the plane xy at (i, — i), (o, 4), (— i, 3), and (i, 2) respectively. Find the resultant and one point on its line of action.
It can be shown that the attraction which the earth exercises on a particle is directed toward the centre of the earth.
In Fig. 65 let O represent the centre of the earth, and PI and ?2 two particles near its surface; then the arrows
*x is read " x bar."
66
ANALYTICAL STATICS
FIG. 65
Wi and W% would represent the forces of attraction. If instead of the particles PI and P2 we consider a body
AB, then each particle of the body is acted on by a force as represented. These forces are certainly not parallel, but bodies such as we ordi- narily deal with in mechanics are very small in comparison to the earth, under which circumstances the lines of ac- tion of the weights of the particles are very nearly par- allel and for practical purposes can be assumed perfectly parallel.
Under these conditions the centre o] ike parallel forces representing the weights 0} the particles of a body is called its centre of gravity, or centroid.
As this point is a point in the line of action of the resultant weight, it follows that a force equal and oppo- site to this resultant acting through the centroid will support or balance the body in any position.
To find the position of the centroid we may use the formulae above derived for parallel forces, if we remem- ber that the forces FI, F2, F%, . . . under considera- tion are now the weights W\, W2, Ws, .... Thus the coordinates of the centroid are
IWx
IWy
and =
Example. — Find the centre of gravity of three parti- cles each weighing W pounds placed upon the circum-
FORCES ACTING ON A RIGID BODY 67
ference of a circle, radius r, so that the first and third are 90° from the second.
Solution. — Draw diagram illustrating problem (Fig. 66). Assume two rectangular axes. Then the position of the C. of G. is found as follows:
IWx
X~^W"
W+W + W
= IWy
w+w+w
the C. of G. is reaUy at C, Fig. 66.
FIG. 66
FIG. 67
Example. — Find the centroid of a circular plate or lamina, radius r inches, in which a circular hole tangent to
the circumference and of radius — has been drilled (Fig.
67).
6S ANALYTICAL STATICS
Solution. — :In this case consider by reason of symmetry that the C. of G. of the whole plate is at its centre, at which point its weight may be considered as concen- trated. Similarly for the hole, with the exception that the weight is taken with the negative sign, signifying that this weight has been removed from the whole plate.
Assume the density of the plate to be w pounds per square inch; then the weight of the whole disc is xr2w,
and of the removed portion —x-w.
4
8 sr.
x=~iw=~ ~1 ^~\~ ~T=~6m-
xr2w+[ — K-W] i
4 and
I r2 \ -n-w)
\ 4 /
( - \
47 which places the centroid at C, Fig. 67.
EXERCISE 87. Weights of i, 2, and 3 ounces are placed at the vertices of an equilateral triangle whose sides are 6 inches long. Find the distance of their C. of G. from the vertices. EXERCISE 88. A common form of a cross-section of a reser- voir wall or embankment wall is a trapezoid whose top and bottom sides are parallel (Fig. 68). If the top side equals 4 feet, bottom side 8 feet, and height 9 feet, find s Ft. the centre of gravity of the section with
FIG. 68 reference to the left and lower sides. (Divide
the figure into a rectangle and a triangle, or into two triangles. C. of G. of rectangle is at intersection of diagonals and of triangle at intersection of medians.)
FORCES ACTING ON A RIGID BODY
69
EXERCISE 89. A lamina has the form of a square with an isosceles triangle attached to one side. The side of the square is a, the height of the triangle is h. Find the position of the centre of gravity of the figure, and determine the value of h, if the C. of G. lies in the base of the triangle.
EXERCISE 90. Where must a circular hole of one foot radius be punched out of a circular disc of 3 foot radius, so that the centre of gravity of the remainder may be 2 inches from the centre of the disc?
EXERCISE 91. A circular disc, 8 inches in diameter, has a hole 2 inches in diameter punched out of it; the centre of the hole is 3 inches from the circumference of the disc. Find the C. of G. of the remaining portion.
EXERCISE 92. Find the centroids of the plates shown in Fig. 69%
|
1 • 1 A |
<-!->• i |
L_ 2I'£ 4 -i |
H^ |
=d |
FIG. 69
SECTION X
SUMMARY OF THE METHODS OF STATICS
As may be observed from the preceding sections, we may divide the methods employed in statics into two large classes, based upon the treatment of forces acting — I. At a single point.
II. On a rigid body.
If the forces acting at a single point are only three in number, the "Triangle of Forces" may be used advan- tageously.
70 ANALYTICAL STATICS
If more than three forces act at a single point, we may best proceed by putting
2(X- components of the forces) =o and 2(Y- components of the forces) =o.
If the forces considered are applied to a rigid body, then they will be in equilibrium if
2(X- components of the forces) =o, J?(F-components of the forces) =o, and ^(Moments about any point) = 2( Yx — Xy) = o.
In dealing with rigid bodies the weight of the body considered must always be assumed as concentrated at the centre of gravity, or centroid, of the body.
The method of finding the centroid as developed in Section IX, although universally true, can be applied only to a system of concrete particles or to a system of bodies the positions of whose centroids are known until the student has familiarized himself with the methods of the calculus.
// is not always necessary, in dealing with the equilib- rium of the forces acting on a rigid body, to write out all the equations for equilibrium as explained in Section VIII. It lies with the ingenuity of the student to select that one of the equations SX = o, IY = o, I (Moment) =o which will involve only the one unknown force sought. This is illustrated in the following
Example. — A straight rod 10 feet long, when un- weighted, balances about a point 4 feet from one end;
FORCES ACTING ON A RIGID BODY
71
--- ---- 10-Ft; ------- »
but when loaded with 20 pounds at this end and 4 pounds at the other it balances about a point 3 feet from the end. Find the weight of the rod.
Solution. — Draw two diagrams representing the condi- tions of the problem; and also a diagram showing all the forces acting on the rod (Fig. 70).
From the first state- ment, and from the definition of the cen- troid, we can say that the centroid of the rod is 4 feet from the left- hand end, and that the weight W may be assumed there con- centrated.
For the solution of this example it will be most convenient to write the equation for the moments about the point of support when loaded, for the reaction of the support (an unknown force) will then not enter our equation of moments.
-" "10"Ft'- ------- -*
X
FIG. 70
Thus
(2o)(3)-(TF)(i)-(4)(7)=o; .'. W = 32 pounds.
If the reaction (R) of the support were required, the moments should have been taken about the centroid, as then the unknown weight would not have entered
72 ANALYTICAL STATICS
our equation. Thus
.'. R = $6 pounds.
If the weight were known and the reaction of tl.e support were required,
IY = — 20-}- R — 32— 4 = 0 would give the result
^ = 56 pounds.
Solve each of the following exercises by the use of a single equation.
EXERCISE 93. A weightless rod ten feet long is supported at its centre. 100 pounds is hung at one end and 200 pounds 3 feet from this end. Find the pressure on the support if enough force is applied at the other end to hold the rod in equilibrium.
EXERCISE 94. A heavy rod 5 feet long has a weight of 200 pounds attached to one end and is supported 2 feet from that end. What force must be applied at the other end to produce a pressure of 150 pounds on the support?
EXERCISE 95. — A rod weighing 25 pounds and loaded at each end with 200 pounds is supported so as to be in equilib- rium. If the unloaded rod balances on a knife-edge placed i foot from one end and is 6 feet long, find the pressure on the support.
EXERCISE 96. A man carries a load of 40 pounds attached to the end of a stick resting on his shoulder. If the man exerts a force of 20 pounds at the other end of the stick, what is the pressure on his shoulder
(a) if the stick is horizontal ?
(b) if the forward end of the stick dips 30° ?
CHAPTER III
APPLICATIONS OF THE PRINCIPLES OF STATICS TO THE SIMPLE MACHINES
SECTION XI THE LEVER AND THE WHEEL AND AXLE
THE principles of statics as discussed in Chapters I and II will now be applied to what are usually termed simple machines.
Any contrivance for making a force applied to a body at a given point and in a given direction available at some other point or in some other direction is called a machine.
The machines considered will all be supposed in equilibrium under the applied force, the available force, and the reactions of the supports of the machine. If motion is desired, the applied force must be slightly increased.
The ratio of the available force to the applied force is called the mechanical advantage o) the machine.
The simple machines usually include the lever, the wheel and axle, the pulley, the inclined plane, the wedge, and the screw.
In the solution of problems involving simple machines no special formulae will be deduced or required; each case is to be analyzed, all the forces acting are to be
73
74
ANALYTICAL STATICS
shown, and the required forces calculated in terms of the known forces and the dimensions of the machines by the direct application of the principles of equilibrium already studied.
The Lever
The lever is a rod or bar, either straight or curved, supported at one point. This point is called the ful- crum.
Example. — A straight rod is loaded so that its centroid
^centroid js one third of its length from
JL X one end. When weights of
(5) (1(M i i
^-^ \^J 5 and 10 pounds are sup-
ported from the ends, the rod
balances about the middle point. Find the weight of the rod.
Solution. — (i) Draw a diagram illustrating the prob- lem (Fig. 71).
(2) Draw a sketch showing all forces acting on the rod (Fig. 72); these |y
include the reaction of the fulcrum and the c !.
weight.
Assume the origin of moments so as to ex- clude from the moment equation (2M = o) the
|
•1 |
f |
R |
I" |
|
FIG. |
72 |
||
|
requ |
lired. Then |
= I5 pounds.
APPLICATIONS TO SIMPLE MACHINES
75
EXERCISE 97. A lever is to be cut from a bar weighing 3 pounds per foot. What must be its length that it may bal- ance about a point 2 feet from one end, when weighted at this end with 50 pounds ?
EXERCISE 98. A straight lever 6 feet long wreighs 10 pounds, and its centre of gravity is 4 feet from one end. What weight at this end will support 20 pounds at the other when the lever is supported at one foot from the latter, and what is the pressure on the fulcrum ?
EXERCISE 99. A lever is supported at its centroid, which is nearer to one end than the other. A weight P at the end of the shorter arm is balanced by 2 pounds at the end of the longer; and the same weight P at the longer arm is balanced by 1 8 pounds at the shorter. Find P.
EXERCISE 100. In a pair of nut-crackers the nut is placed one inch from" the hinge; the hand is applied at a distance of 6 inches from the hinge. How much pressure must be applied by the hand if the nut requires a pressure of 20 pounds to break it, and what will be the force acting on the hinge ?
EXERCISE 101. The oar of a boat is io£ feet long; the row- lock is 2j feet from the end. A man applies a force of 70 pounds at 2 feet from the rowlock; the average pressure of the water is exerted at 6 inches from the other end of the oar. Find the force urging the boat forward, and the total pressure of the water on the blade of the oar.
The balance (Fig. 73) in its simplest form is a lever which turns very freely about a fulcrum F. It is used for comparing the weights of two bodies. A balance is true if the beam is horizontal when-
FIG. 7.3
ever equal masses are placed in the sc^le-pans PPt \
76 ANALYTICAL STATICS
EXERCISE 102. Find the condition that a balance may be true.
EXERCISE 103. A false balance rests with the beam hori- zontal when unloaded, but the arms are not of equal lengths. A weight W when hung at the end of the shorter arm, b, bal- ances a weight P, and when hung at the end of the longer arm, a, it balances a third weight, Q. Find the correct weight of W.
Can you suggest another way of ascertaining correctly the weight of W ?
EXERCISE 104. The arms of a false balance are in the ratio of 20 to 21. What will be the loss to a tradesman who places articles to be weighed at the end of the shorter arm if he is asked for 4 pounds of goods priced at $1.50 per pound?
The Common, or Roman, Steelyard (Fig. 74) con- sists of a lever support- ed on knife-edges at D. The object to be weighed is placed at W, and the distance of Q from the fulcrum is adjusted until the lever is horizontal.
The centroid, C, of the lever and scale-pan does not usually coincide with the point of suspension D. Example. — To graduate the steelyard. Assume the weight of the lever to be w. With no load at W adjust Q until the lever is horizontal; then x = a.
Thus I(M)=wy-Qa = o.
Now assume a load W in the pan; then
APPLICATIONS TO SIMPLE MACHINES
77
In this equation d, <2, and a are known quantities to be determined by experiment; the variable quantities are W and x.
Assume
then
c/ = 3 inches, Q = 6 pounds, a = 2 inches; 6# — 12,
This is the equation of the straight line plotted in Fig. 75. Assuming any value of W we obtain a certain value of x.
FIG. 75
If W = o, then 00 = 2, so that when the steelyard is un- loaded Q must be placed 2 inches from D.
If W = i> then x = $/2 inches. If W = 2, then # = 3 inches, etc.
The zero mark would then be placed at 2 inches to the right of D, the one-pound mark 1/2 inch farther to tie right, the two-pound mark 1/2 inch still farther, etc.
78 ANALYTICAL STATICS
If now, instead of measuring x from D, x be measured from E, the zero mark, the equation W = 2X — 4 would have to be transformed to a new set of axes Wf, x' (Fig. 75), so that W = Wf and x = x* +2; whence
-4, or If now W = W' =
If, in the equation W' = 2oc*, x' is given a minus value, that is, is measured to the left of the zero mark, the value W' = W becomes minus (Fig. 75).
The physical interpretation of this is that instead of a weight an upward push must be supplied at W to pro- duce equilibrium.
EXERCISE 105. In a common steelyard the pan is supported 3 inches to one side of the fulcrum, and the center of gravity is i inch to the other side. The weight of the lever is 3 pounds, that of the movable constant weight is 2 pounds. What is the smallest load that can be weighed ?
The Danish Steelyard (Fig-. 76) consists of a bar A A I ^/
•p w
FIG. 76
terminating in a ball. The load is placed at W, and the fulcrum F is moved until equilibrium is established.
EXERCISE 106. Graduate a Danish steelyard, assuming the distance from the load to the centroid to be 12 inches and the weight of bar and scale-pan 20 pounds.
APPLICATIONS TO SIMPLE MACHINES
79
00=
The Wheel and Axle
consists of a wheel or drum of considerable diameter to which is rigidly attached a drum or axle of smaller diameter. Both drums turn freely upon the same axis
(Fig. 77)- A rope is coiled around
each drum, one clockwise, and
the other counter-clockwise. The forces are applied to
these ropes. An end view
of this machine is shown in Fig. 78, from which it
may be seen that the wheel and axle is simply a lever with fulcrum F at the axis.
The wheel and axle may be called a continuous lever; for, if motion results, new radii take the place of the ones previously in use and the machine operates as before.
FIG. 77
FIG. 78
1 EXERCISE 107. In a wheel
and axle the diameter of the wheel is 10 feet and the radius of the axle is 18 inches. A force of 100 pounds is applied to the rope coiled about the axle. Find the available force at the end of the rope coiled about the wheel if this rope leaves on a horizontal tangent. Find, also, the mechanical advantage. What pressure do the bearings sustain?
EXERCISE 108. In a wheel and axle the radius of the wheel is 3 feet. The axle is of square section, the side of the square
30 ANALYTICAL STATICS
being 6 inches long. Find (a) the greatest and (b) the least vertical power that must be exerted to slowly lift a weight of 252 pounds attached to the rope coiled around the wheel. If the wheel and axle weighs 150 pounds, what is the pressure on the bearings ?
EXERCISE 109. The circumference of a wheel is 60 inches, the diameter of the axle is 5 inches. What force must be applied to the circumference of the axle to support 100 pounds at the circumference of the wheel ?
SECTION XII THE PULLEY
The pulley is a small wheel with a groove cut in its outer edge. The pulley can turn on an axis through its centre; the ends of this axis are carried by a block within which the pulley turns. If the block is so fastened as to be immovable, the pulley is said to be "fixed"; otherwise the pulley is designated as "movable."
Fig. 79 shows a fixed pulley; the applied force is F pounds, the available force is W pounds. Example. — Calculate F and the
reaction of the staple supporting the FIG. 79
block.
Solution. — Fig. 80 shows the forces acting upon the pulley. R is the force transmitted by the block from the staple to the axis of the pulley.
Since three forces acting in equilibrium upon a rigid body (pulley) have their lines of action meeting in a point (page 60), Fig. 81 may be used to represent the forces.
APPLICATIONS TO SIMPLE MACHINES 8 1
Using Cj Fig. 80, as the origin of moments, we have
2M = Fr— Wr = o', .'. F = W. In Fig. 81, assuming the x and y axes as shown,
as, by Geometry,
FIG. 80
FIG. 81
EXERCISE no. Find the force necessary to sustain a weight of 100 pounds by means of a fixed pulley; find also the mag- nitude and direction of the force on the staple to which the pulley is fastened, if the applied force makes an angle of (a) o°, (b) 30°, (c) 90°, (d) 150° with the vertical.
From the above it follows that a fixed pulley has a mechanical advantage of unity (it does not increase the available over the applied force) : it simply changes the direction of the applied force.
EXERCISE in. Design a system of fixed pulleys by means of which a horse walking on horizontal ground may raise a bale of hay weighing 200 pounds from the ground to a ver-
82 ANALYTICAL STATICS
tical height of 20 feet. Show by diagrams all the forces acting on the various pulleys.
In Fig. 82 a movable pulley is shown. Example. — Assuming the pulley, Fig. 82, to weigh w pounds, find the force F necessary to sustain the weight W.
FIG. 82
FIG. 84
Solution. — Fig. 83 shows all forces acting on the pul- ley, T representing the tension in the rope. Taking moments about C,
From Fig. 84,
=o; /. F=T.
= F cos d+F cos d-W-iv=o;
2 COS 6'
EXERCISE 112. Find the force necessary to sustain a weight of 100 pounds by means of a single movable pulley; also, find direction and 'magnitude of the force acting on the staple sup- porting one end of the rope if the force makes an angle of (a) 90°, (b) 35°, (c) o° with the horizon.
APPLICATIONS TO SIMPLE MACHINES
System of Pulleys
Various combinations of pulleys are in use. For con- venience we shall designate three important systems as the first, second, and third; these systems are illustrated in Figs. 85, 86, and 87, respectively.
i w I
FIG. 85
FIG. 86
FIG. 87
In finding the various forces acting in a system of pulleys, it should be remembered that ike tension in any one string is the same throughout its length, and that each portion of the system may be considered in equilibrium by itself.
Example. — Find the force necessary to sustain a weight W by means of the third system if one movable pulley (weighing w pounds) is used. Also find the pull on the staple if the fixed pulley weighs w{ pounds.
84
ANALYTICAL STATICS
Solution. — The machine is shown in Fig. 88.
Consider first the equilibrium of the movable pulley, Fig. 89, where /, ti represent the tensions in the several strings.
FIG. 88 FIG. 89 FIG. 90 FIG. 91
The equilibrium of the weight gives (Fig. 90)
, and
W-w
Consider now the equilibrium of the fixed pulley, Fig.
91
or
2Y=R-ti-w1-ti=o; /. ^ =
^W-qw 3
EXERCISE 113. What force is necessary to sustain a weight of 1000 pounds by means of the first system of pulleys, using two movable pulleys and regarding them as weightless? What is the pull upon each point of support?
APPLICATIONS TO SIMPLE MACHINES
EXERCISE 114. Same as Exercise 113, considering each pul- ley to weigh 20 pounds.
EXERCISE 115. What weight will a force of 10 pounds applied to the first system, containing five weightless pulleys, support ?
EXERCISE 116. A force of 20 pounds applied to the second system, containing two pulleys in the movable block with the rope fastened to the fixed block, will sustain what weight ? What is the pull on the point of support ?
EXERCISE 117. Same as Exercise 116, but having the rope fastened to the movable block.
EXERCISE 118. A system of the third class has three weightless movable pulleys. What weight will a force of 10 pounds sustain, and what is the tension in the staple ?
EXERCISE 119. Same as Exercise 118, if each pulley weighs 2 pounds.
EXERCISE 120. In the system illustrated in Fig. 92 find the force, necessary to sustain 100 pounds, and the pull on the staples, if each pulley weighs 4 pounds.
FIG. 92
SECTION XIII
THE INCLINED PLANE AND THE WEDGE
Any plane inclined to the horizon is an inclined plane.
In solving problems involving inclined planes, it is simply necessary to consider the equilibrium of the weight, considered as a particle, upon the plane. Care should be taken not to neglect the normal reaction of the plane, and, if friction is involved, the friction jorce.
The inclination of a plane is usually denoted by the angle it makes with the horizon. When referring to
86 ANALYTICAL STATICS
roads or railways the inclination is generally expressed as the grade of the incline, which means the ratio of the height to the length. Thus on a "five per cent grade" a rise of 5 feet would be obtained on walking 100 feet up the incline.
Less frequently the term pitch is used to denote the ratio of the height to the base.
EXERCISE 121. What force acting parallel to the plane is needed to support two tons on a smooth incline, if the grade is 8 per cent?
EXERCISE 122. Same as Exercise 121, if the coefficient of friction is 0.3.
EXERCISE 123. The pitch of a plane is 0.25, and the coeffi- cient of friction between a certain body and the plane is 0.35. Will the body, if placed on the plane, slide down or remain at rest ? What will be the friction force exerted ?
EXERCISE 124. A weight rests on a smooth inclined plane. Show that the least force which will keep it in equilibrium must act along the plane. (Assume a force acting at an angle 6 with the plane and show that 6 is zero for a minimum force.)
EXERCISE 125. Two unequal weights W\ and W2 on a rough inclined plane are connected by a string which passes over a smooth pulley in the plane. Find the greatest inclina- tion of the plane consistent with equilibrium, in terms of Wi, W2, and the coefficients of friction /*i and /*2.
EXERCISE 126. Two rough bodies W\ and W2 rest upon an inclined plane and are connected by a string parallel to the plane. If the coefficient of friction is not the same for both, determine the greatest inclination and the tension of the string, consistent with equilibrium, in terms of W1} W2, and the co- efficients of friction //i and j*2.
APPLICATIONS TO SIMPLE MACHINES
The Wedge
This machine consists of a double inclined plane made of some hard material, such as iron or steel. It is used for splitting wood, or overcoming great resistances over short dis- tances. In Fig. 93, ABC repre- sents a wedge. The applied
force, F, acts normally to J5C; the available force will be a force equal and opposite to R, normal to AB.
Example. — Assuming the angle shown in Fig. 93, and all surfaces to be frictionless, find the relation existing between R and F.
Solution. — Fig. 94 shows all forces acting on the wedge, N being the normal resultant of the reactions of the plane MN on AC.
FIG 94
FIG. 95
P FIG. 96
Three forces to be in equilibrium must always act through one point. Therefore Fig. 95 truly represents the forces, and by the principle of the triangle of forces Fig. 96 is obtained.
From which we have F
or
F=R sin a.
88
ANALYTICAL STATICS
EXERCISE 127. In the wedge shown in Fig. 97 find the relation between F and P in terms of the angles a, /?, and 0. All surfaces to be considered smooth, and GG being guides preventing lateral motion of the rod transmitting the force P.
Hint. — Find a force R normal to AB balancing F and then consider the rod as a free body with a force "^__ e(lual and opposite to R acting upon it together with the reactions of the guides, etc.
EXERCISE 128. A wedge in the form of a right triangle, whose sides are 5, 4, and 3, is driven by a horizontal force of 300 pounds, applied nor- mally to side 3, along a horizontal plane. What weight, supported so as to prevent its moving horizontally, may be lifted if applied to the side 5, friction neglected?
EXERCISE 129. In Fig. 97, assuming a: = 45°, #=30°, /?=i2o°) and P=5oo pounds, find P.
Action of the Wedge, including Friction
In Fig. 98 let A BCD be the wedge driven by a for^e P along the plane MN, HE a block sliding upon AB, and GG a fixed guide preventing the horizontal motion of HE.
Let 6 be the inclination of AB, and /*, /*i, //2 the co- efficients of friction at the surfaces DC, AB, and GG, respectively. FIG. 98
The problem consists in finding the relation between P and Q, assuming the wedge and block HE as weightless.
APPLICATIONS TO SIMPLE MACHINES
89
In Figs. 99 and 100 are shown the wedge and block respectively, with all forces acting on the same. From Fig. 99,
cos 6 = 0, . (i)
= 0. . . . (2)
\y i
uLt
s
Ff/AaS
-J7**?
FIG. roo From Fig. 100,
ri sin 6 + fiiNi cos 6—S=Oj . . ri cos 0 — /£; ATi sin 6 — Q — jj>2S = o. From (i) and (2), by elimination of R,
sin 6 + (u+ «i) cos 01 JVi.
(3) (4)
From (3) and (4), by elimination of 5, Q = [(i - & p2) cos 6 - O + 1*2) sin
P (i — /*/*i ) sin 0 + (p. + fti ) cos
" Q (J ~ ftP*) cos ^- (/«i + j"2) sin which is the required solution.
90 ANALYTICAL STATICS
If the coefficients of friction are equal, or jj. = /*i = /*2 = tan a, where a is the angle of friction (see page 34),
sin 0 2fi P (i — //2) sin 6 -\-2fj. cos 6 cos 6 i — fi2
dn d / 2fji \ sin 0
~ \i-//2/ cos 0
tan # + tan 2a = — — -0 = tan
i —tan 20. tan 0
by the trigonometric relations
2 tan
tan 2X =
tan # + tan and tan (
i — tan x tan y
EXERCISE 130. Find the relation of P to Q in the machine shown in Fig. 98, if Q is applied normally to AB by means of a block restrained from moving parallel to AB.
FIG. 101
EXERCISE 131. Find P, if Q = iooo pounds, and the coeffi- cient of friction ^ = 0.2, and ^1 = 0.4, from the wedge shown in Fig. 101, where A ABC is isosceles and /.ABC- 60°.
APPLICATIONS TO SIMPLE MACHINES
91
SECTION XIV
MISCELLANEOUS MACHINES
The Bent-lever Balance. — This balance is shown in Fig. 102, where ABC is the bent lever turning about a pivot'at B. D is the index Nf which points to some division on the graduated arc MDN. C is the centroid of the lever CBA.
EXERCISE 132. Find the rela- tion existing between <£ and W, if W is the load in the pan and <j> is the angle through which BC is displaced from its position, when W=o, by the load W. Assume as constants of the balance BC = a, BA=b, w= weight of pan, Wi= weight of lever CBA, and that when W=o the end A is on a horizontal through B while BC makes an angle (3 to the vertical.
EXERCISE 133. Find the relation between F and W in the combi- nation of levers shown in Fig. 103.
6-'—-
FIG. 102
FIG. 103
EXERCISE 134. Find the weight, W, that can be sustained by a force of 100 pounds and the pull on each point of sup- port in the combination of pulleys shown in Fig. 104, if the
9 2 ANALYTICAL STATICS
movable pulleys weigh 5 pounds each and the fixed pulleys 3 pounds each.
The Differential Wheel and Axle. — This machine is shown in Fig. 105. It is similar to the wheel and axle, but instead of one axle it has two of different diameters with the same rope coiled in opposite directions around them. In the loop of this rope hangs a single movable pulby to which the weight is attached.
FIG. 105
FIG. 106
FIG. 107 FIG. 108
An end view of the machine is shown in Fig. 106, and the forces acting on the separate parts in Figs. 107 and 1 08, where T represents the tension in the rope.
Taking moments about C, Fig. 107, we have
From Fig. 108,
Thus,
= 2T-W = o', /. T=-.
2
w
= T(R-r)=—(R-r)9
P =
W(R-r)
2(1
APPLICATIONS TO SIMPLE MACHINES 93
From this equation it should be noted that the nearer r = R the smaller does P become until R = r when P = o.
EXERCISE 135. In a differential wheel and axle the radius of the wheel is 3 feet, and the radii of the axles 20 inches and 18 inches respectively; find the weight sustained, and the pressure on the bearings, if applied force is 200 pounds.
A form of Platform Scales is shown in Fig. 109. The levers are so arranged that x pounds at A balance the
ABC
|
W |
||
|
P |
j |
|
|
AM || |
||
|
D |
G FIG. 109 |
N |
same load W wherever it may be put on the platform. This result is secured by making the horizontal arms EM and NH equal, and also EF equal to GH.
In this machine there are three fulcrums.at B, E, and H, while the platform rests on the levers EF and DH at M and N respectively.
EXERCISE 136. Assume the distances (Fig. 109) as follows: EF = GH=io ft., EM = NH = 2 ft., HD = 2$ ft., BC=i inch, AB = $o inches. Find x, if W = $ tons and is placed (a) at the centre of the platform, (b) one foot from the left edge of the platform.
The combination of levers shown in Fig. no is some- times called a knee. This machine can be used to advantage where very great pressure is required to act through very small space only, as in coining money, in the printing-press, etc.
At A and D are fixed pivots, and the levers AB and DC are joined by pins to BC at B and C.
94
ANALYTICAL STATICS
As EC transmits force only in the direction EC, the forces acting upon the levers AE and DC can be repre- sented as in Fig. in.
EXERCISE 137. In Fig. no, assuming Z CA B = Z EC A = 30°, AH = 10 inches, AC = $ feet, DC = 2 feet, and
FIG. no
feet, find pressure at K produced by a horizontal force of 100 pounds acting at H.
EXERCISE 138. Draw diagrams showing the forces acting on the fixed points A and D and on the bar EC.
EXERCISE 139. Calculate the direction and magnitude of the forces acting at A and D due to the force given in Exercise 137.
GRAPHICAL STATICS
CHAPTER IV GRAPHICAL ARITHMETIC
SECTION XV SUMMATION, DIVISION, AND MULTIPLICATION
THE theory of statics and the solution of statical prob- lems has already been considered. The results, however, were obtained by the use of analytical methods. Methods for solving statical problems graphically will now be developed.
As an introduction, some problems in Graphical Arith- metic will first be considered.
Numbers may be represented by the lengths of lines drawn to some scale. Thus, . • • i • repre- sents 5.
If plus and minus numbers are to be used, some as- sumption as to signs becomes necessary. It will be as- sumed that plus numbers are to be shown by lines drawn to the right, and minus numbers by lines drawn to the
Q5
g 6 GRAPHICAL STATICS
lejt. Thus, — • - • — > represents +3, and
represents —3, the arrow-head being used to show direc-
tion.
Summation (addition and subtraction) is performed graphically by placing the arrows representing the num- bers to be summed tail to head, and measuring the dis- tance between the tail of the first and the head of the last arrow, care being taken to denote the direction of this line by the appropriate sign. Thus 3 + 2 — 4 is repre-
sented graphically by -- ' ° .' s; \ N,and equals — ,
EXERCISE 140. Sum graphically:
(b) -6+4-5 + 7-2.
(c) 12-3-4 + 2-10+5.
EXERCISE 141. Rewrite each sum in Exercise 140, changing the order of the terms, and then again sum graphically.
Multiplication, when performed graphically, must always be preceded by Division.
In order to perform division, it is necessary to use lines drawn vertically, and another assumption as to signs must be made. Assume lines drawn upward as plus and downward as minus.
Then 2-^-3 is shown graphically in Fig. 112. The dividend is placed vertically, the divisor horizontally, ^S always tail to head. The tail of the dividend is then joined to the head of 2-1 /"' the divisor. The quotient, although
not drawn in Fig. 112, is the tangent of the angle 6, for tan 0 = §. This quotient may be obtained graphically by multiplying by
GRAPHICAL ARITHMETIC
97
one. In Fig. 113 this is done by making the multiplier a continuation of the divisor. Then x, the required quotient, is drawn vertically to meet the sloping line.
EXERCISE 142. Prove geometrically that in Fig. 113 x=— .
Example. — Find graphically x = .
o
' -4 -
|
\ \ \ \ x \ 1 |
...-*.. t* |
|
•5 ' *p* >" FIG. 115 |
|
|
+3' \1x X ' \ s FIG. 114 |
. o . , .'fo
FIG. 113
Solution in Fig. 114 shows —4-5-3 to obtain the dotted line, then the multiplication by unity to obtain x.
EXERCISE 143. Find graphically:
(a) x=; (J) *-^; (c) x-=±.
Example. — Find graphically x =
s
The solution in Fig. 115 is obtained by dividing 3 by 5, thus obtaining the direction of the sloping line, then multiplying by ( — 6), as above explained, to find x.
EXERCISE 144. Find x graphically:
~3
GRAPHICAL STATICS
It is important to note that the triangles used in mul- tiplication and division need not be right-angled. Thus,
x = may be found from each of the Figs. 116 to 118.
EXERCISE 145. Give a geometric proof that x=-^-- for the construction shown in each of the figures from 116 to 118.
2 /
r'lG. 117
FIG. 118
Example.— Find graphically * = /_ w_ y To find x the form of the expression should first be changed to
x
(-3)
Then find y = -r — \, as shown in Fig. 119, and then
( — $)
# =— - as shown in Fig. 120. Or, better, we may solve
the entire problem in one diagram by combining Figs. 119 and 1 20, as in Fig. 121.
GRAPHICAL ARITHMETIC 99
From the preceding work it becomes evident that, in performing multiplication and division graphically, the problem must be put in a fractional form, in which the numerator always contains one more factor than the denominator. This may be accomplished by the intro-
duction of unit factors. Thus, x= should be
put in the form *-
>-"f
2&TLT
FIG. 119 FIG. 120 FIG. 121
Again, in x = - - an inconveniently large diagram would result if the form be changed to x= . x . This may be avoided by factoring; thus x= /2\/L\ •
EXERCISE 146. Find x graphically:
,.t -_(3)(4)(S). ,n «_MM. (,) X=M.
4
(/) *-
EXERCISE 147. Find x graphically:
.
— -jp^j-: - (-3)(-5) •
«> « ^
100 GRAPHICAL STATICS
EXERCISE 148. By means of cross-section paper find x, y, and z. In each case select an appropriate scale.
. _
(6375)(3957) = (.087) (.0035) (.01975) (-.029)000995)
SECTION XVI
COMBINED MULTIPLICATION AND SUMMATION
Consider now a problem such as u = — — - — — .
It is found most convenient to change the given expres- sion into one composed of parts similar to those already
solved; thus, u=—— + ~-+-^-. Instead of solving o o o
each term separately and then summing the results, it is more convenient to perform all the divisions in one diagram (Fig. 122).
Here P (called the Pole) is selected anywhere at a distance 8 from the line along which 2, 4, P and 6 are set off, and the rays ' drawn from P to the extremities
of 2, 4, 6. The division is then performed as in the left-hand
portion of Fig. 118. FIG. 122 .....
To perform the multiplication
the rays in Fig. 122 could be prolonged (Fig. 123) and the distance 3, 5, 7 laid down horizontally and lines x,
GRAPHICAL ARITHMETIC
101
y, z located between their proper rays and at their proper distance from P, as in Fig. 118.
FIG. 123
By this construction the lines x, y, z will not fall in the same, straight line, and therefore their sum is not found directly. A more con- venient construction is shown in Fig. 124. Here at any point, O, in a vertical line OM, construct a triangle similar to the one with base 2 in Fig. 123, but with altitude 3, thus obtaining x. This is most conveniently done by drawing the sides of the triangle parallel to the corresponding rays of Fig. 122. Next construct a triangle similar to the one with base 4, but with altitude 5, etc. Thus we obtain x, y, z all on the
102
GRAPHICAL STATICS
line OM and following one another head to tail. Then OM=u is the required result.
The diagram in Fig. 124 can appropriately be called a Summation Polygon or Diagram.
EXERCISE 149. Find graphically:
(fl) ^
2-4+30.
(b) x-
7-8+5-6+3-4+I-2 5
EXERCISE 150. Find x graphically:
(b) x-
-2) (3) + (3) (-5) -(6) (-7)
EXERCISE 151. A horizontal weightless rod 10 feet long has weights of 2, 3, and 4 pounds attached at £, J, and } of its bngth. Write down the value of the x of its centroid accord-
2Wx ing to the formula x=—=-— , and find x graphically.
EXERCISE 152. Same as Exercise 151, with additional weights of i and 5 pounds at the ends of the rod.
The construction just studied lends itself well to find- ing the centroids of plates.
Example. — Find the centroid of a quadrant of a circle, Fig. 125.
Solution. — Consider the quadrant divided into sections of equal width as shown. Let the plate weigh w pounds per square foot and let the width of each section be a feet. Each section will have a mean altitude, as shown by the dotted lines; denote these by hi, h2, . . . . Assume each section to be a trapezoid, and assume their
f.
x a x a
FIG.
I25
GRAPHICAL ARITHMETIC 103
centroids to be at the midpoints of their medians. Neither of these assumptions is strictly true, but by decreasing the width of the sections, and thus increasing their num- ber, the assumptions approach nearer and nearer the truth. It follows that
the area of any section is ah, 11 weight " " " " wah,
which weight is assumed to act at the midpoint of the median.
Placing the y-axis as shown in Fig. 125 and putting #=the abscissa of the centroid, we have
ZW
which is an expression precisely suiting our graphical methods.
The graphical construction for x is shown in Fig. 126. At (a) is shown the quadrant with the mean altitudes (the sections are not shown, as these are unnecessary for purposes of construction). At (b) the divisions are per- formed. Here, instead of using the h's as indicated in the formula, half h's are used throughout. Why does this not affect the result? At (c) we have the sum- mation polygon, and Iv, the required abscissa, is here
io4
GRAPHICAL STATICS
found, x is then plotted as shown in (a); the centroid must lie on the dotted line. Where is the final position of the centroid?
(b)
FIG. 126
EXERCISE 153. Construct a parabola whose parameter is 8 and find the centroid of a plate of this shape between the vertex and the latus rectum. Divide the plate into four sec- tions.
EXERCISE 154. Same as Ex. 153, but divide the plate into 8 sections and compare the result with that of Ex. 153.
CHAPTER V
FORCES ACTING AT A SINGLE POINT SECTION XVII
COMPONENTS. RESULTANTS
WE already know that the diagonal of a parallelogram constructed upon two forces as sides, correctly represents the resultant if the tails of all forces are at the one ver- tex of the parallelogram. Thus, in Fig. 127, if AB and AC represent forces acting on a c, particle at A, AD represents their resultant. If now the arrows
representing the forces A B and AC
, . ., FIG. 127
be drawn to a scale, i.e., each unit
of their length being accurately made to represent a unit of their force, and the parallelogram be accurately drawn, then, by applying the same scale of length to AD, the number of units of length in AD would correctly indicate the number of units of force in the resultant of AB and AC. This is, then, a graphical method of obtaining the resultant of two forces acting at a point.
In many constructions it is inconvenient to complete the parallelogram as above, so the triangle of forces is used. This is illustrated in Fig. 128. It is here required to find the resultant of the forces represented by AB and
106 GRAPHICAL STATICS
AC. In a detached diagram, ab is drawn || and = to AB and be \\ and = to AC, and the triangle then com- pleted by the line ac. It will be remembered that the force ca represents the force producing equilibrium with AB and AC and thus ac, acting in the opposite direc-
. A ^B o
FIG. 128
tion to the other forces about the triangle, represents the resultant of AB and AC in magnitude.
This resultant is now, however, not in its proper posi- tion. A force parallel and equal to ac acting at A is the resultant in both magnitude and position.
Notice the difference between the parallelogram and the triangle of forces, the first gives the complete result- ant in position, magnitude, and direction, while the second gives only the magnitude and direction.
Instead of the notation employed in Fig. 128 it will be found more convenient to use the notation of Fig. 129.
Here the forces are denoted by naming the letters on each side of the forces; thus, the horizontal force would be ab. In the triangle of forces these letters are placed at the ends of the arrow representing the force.
FORCES ACTING AT A SINGLE POINT 107
EXERCISE 155. Find graphically the resultant of
(a) 3 lb. and 5 lb., included angle 90°;
(b) 5 " "7 " " " 60°;
(c) 4 " "9 " " " 120°;
both by the parallelogram and by the triangle of forces.
Components
To find the components of a force P, Fig. 130 (a), in the direction of the lines i and 2, proceed as indicated in Fig. 130 (b). Draw p \\ and = to P, and through its ends
-4
^x
(a) ~N-^- -___ (&) _
a
FIG. 130
draw lines || to i and 2 respectively; then ab and be rep- resent the required components of P, in magnitude and direction.
EXERCISE 156. Demonstrate the correctness of the above construction by means of the triangle of forces.
EXERCISE 157. Find the components of P=io lb. inclined at 45° to the horizon along lines inclined at
(a) o° and 90°;
(b) 30° " 60°;
(c) -30° " 120° to the horizon.
EXERCISE 158. Find graphically the components of a force whose magnitude^ direction, and point of application are 20 lb., 210°, and (2, 6) respectively, along lines inclined at angles of 45° and 60° to the axis of X,
io8
GRAPHICAL STATICS
Resultants
If it is desired to find the resultant of more than two forces acting upon a particle, the principle of the parallel- ogram of forces may still be applied, provided the result- ant of any two forces be found, and this resultant be then combined with another of the given forces, etc.
This is illustrated in Fig. 131, where the given forces are OA, OB, OC. The result- ant of OA and OB is OX, and the resultant cf OX and OC is OF. Therefore the resultant of OA, OB, OC is OF.
As the number of forces increases it is evident that this construction becomes more and more complicated.
In its place a construction based upon the polygon o) forces is used. This is shown in Fig. 132. The forces whose resultant is required are shown in Fig. 132 (a). In
d
the detached diagram, Fig. 132 (b), an open polygon is drawn having for its sides lines parallel and equal to the forces ab, be, and cd arranged tail to head. The arrow ad required to close the polygon and taken in a direction around the polygon opposed to that of the given forces
FORCES ACTING AT A SINGLE POINT IOQ
represents the required resultant in magnitude and direction but not in position. A force parallel and equal to ad acting at o completely represents the resultant.
EXERCISE 159. Find the resultant of the forces shown in Fig. 133, by means of the polygon of forces, and demonstrate the dependence of this construction upon that of the triangle of forces.
EXERCISE 160. Find the resultant of the forces shown in Fig. 133 by means of the polygon of forces, and demonstrate the correctness of the construction by resolving each force into vertical and horizontal components and showing that
Resultant =V(2X)2+ (IY)2..
EXERCISE 161. Assume three sets of forces of 3, 4, and 5 forces respectively, and find the resultant of each set.
EXERCISE 162. Six men pull upon six " FlG cords knotted together. They exert forces of 50, 60, 40, 100, 80, and 125 pounds and pull towards the N., N. 30° E., E. 60° S., W., NW., and S. 30° W., respect- ively. Find the direction in which the knot will move.
EXERCISE 163. If O is any point in the plane of the tri- angle ABC, and D, E, F are the midpoints of the sides of the triangle, show that the system of forces OA, OB, OC is equivalent to the system OD, OE, OF.
SECTION XVIII
CONDITIONS FOR EQUILIBRIUM
For the equilibrium of forces acting on a particle it is evident that the resultant must be zero. Therefore from the above discussion of the polygon of forces it follows that if the forces acting upon a particle are known to pro- duce equilibrium, the polygon formed of them must close,
no
GRAPHICAL STATICS
Example. — A particle weighing 5 pounds is supported upon a rough inclined plane by forces of 3 and 2 pounds acting horizontally and vertically respectively. Find the normal reaction of the plane and the friction force, if the inclination of the plane is 30°.
Solution. — Fig. 134 illustrates the problem. To solve graphically all forces acting on the particle should be shown, as in Fig. 135. As the reaction of the plane, de, and the friction force, ea, are unknown, they are repre- sented by their lines of action only.
31bs.
51bs.
FIG. 135
FIG. 136
In Fig. 136 the polygon of forces is shown. To con- struct this start with the known forces and then close the polygon by lines parallel to the lines of action of the unknown forces.
The arrows de and ea represent the normal reaction of the plane and the friction force respectively. If measured with the same scale as was used in plotting the known forces, their magnitudes are directly obtained. The total reaction of the plane, being defined as the sum of the normal reaction and friction, is represented by da. (Why ?)
FORCES ACTING AT A SINGLE POINT
III
D B
FIG. 137
In the following exercises diagrams illustrating the frames should first be drawn to scale so as to obtain the relative inclination of the members.
EXERCISE 164. Fig. 137 shows a derrick, ABC, and a chain, DCE, supporting a load of 10 tons. Assume AC = 25 feet, AB=g feet, BC = 20 feet, and D the midpoint of A B.
Find graphically the stresses in AC and BC if the chain is fastened at C. (This releases DC of stress.)
EXERCISE 165. Find stresses in AC and BC, Fig. 137, if the chain passes freely over a pin at C. (This causes a stress of 10 tons in DC.)
EXERCISE 166. Find the stress in AB and BC of the tri- angular frame, Fig. 138, if 2 tons are suspended from B.
EXERCISE 167. What will be the stress in AC, Fig. 138, if the frame is loaded as in Exercise 166?
EXERCISE 168. If in Fig. 138 two cables inclined to the horizon at 45° and 90° and stretched with forces of 2 and 5 tons respectively be attached to B, find the stress in AB and BC.
EXERCISE 169. Find stresses in AB and BC, Fig. 139, if the dotted line represents a rope passing without friction over B and fastened at E.
EXERCISE 170. If a kite in equi- librium in the air makes an angle of 45° with a horizontal plane, and if the pressure of the air on the kite is equal to three times
20ft.
FIG. 138
FIG. 139
the weight of the kite, find the direction of the string at its point of attachment to the kite, and the magnitude of the tension in the string in terms of the weight of the kite.
112
GRAPHICAL STATICS
EXERCISE 171. A string A BCD is fastened to supports at A and D which are on the same level. Two weights of 10 pounds and x pounds are tied to the string at B and C. If AD is 30 feet and BC is horizontal and AB = BC = CD=i2 feet, find x and the tensions in AB, BC, and CD.
When considering the relation between forces acting on a body the important principle concerning the equilibrium of three forces stated on page 60 should be carefully remembered; for by it the problem can sometimes be reduced to the case considered in this section.
EXERCISE 172. Find the reactions at the points of support A and B when the crane, shown in Fig. 140, is loaded at C with 4000 pounds. (The support at A consists of a collar, and at B of a footstep.)
EXERCISE 173. Make the rod DC (Fig. 140) a "free body " and thus find the stress in the rod EF.
FIG. 141
EXERCISE 174. If in Fig. 141 the frame ABC is hinged to the wall at A and rests against the wall at B, find the support- ing forces at A and B.
"U^VERS.TY 6f- TORONTO
' r.vY, £NGIN£«|*
NT OF V* IV'V *•
'^~l' Alld St
CHAPTER VI
FORCES ACTING ON A RIGID BODY SECTION XIX
RESULTANTS
THE finding of the resultant of forces acting on a rigid body is next to be considered. Assume the forces shown in Fig. 142 and let their resultant be required. It will be remembered (see p. 43) that the point of applica- tion of a force may be placed anywhere upon its line of action and thus the forces i and 2 may be applied at the point of intersection of their lines of action and the parallelogram completed to obtain th e resultant X. X, considered as replacing i and 2, may now be simi- larly combined with 3, and and 3, obtained in both magnitude and position. This resultant, F, may have its point of application anywhere along its line of action.
"3
FIG. 142 F, the resultant of i, 2.
114 GRAPHICAL STATICS
This use of the parallelogram o] jorces is cumbersome when a larger number of forces is to be dealt with. The triangle or polygon of forces is again, as in Section XVII, used to simplify the construction.
In Fig. 143 (a) the forces are shown, and in Fig. 143 (b) the forces are combined to obtain the resultant. First ab and be give ac, then ac and cd give the final resultant
al, Fig. 143 (b), in magnitude and direction but not in position.
Notice that from no portion of the work already com- pleted can we infer at what point of the body this result- ant should act.
To find the position of the resultant, return to Fig. 143 (a) and find the intersection of the lines of action ab and be at O. Through this point we know that their resultant must act; therefore through O draw a line parallel to ac of Fig. 143 (b). This partial resultant is now combined with cd, and the intersection of their lines of action is found at P. Finally through this point the resultant of ac and cd (or of ab, be, and cd), that is the force ad, must pass. The problem is thus com- pletely solved.
EXERCISE 175. Find the resultant of the forces shown in Fig. 144
FORCES ACTING ON A RIGID BODY
(a) by means of the parallelogram of forces;
(b) by means of the triangle of forces.
If it should happen that the given forces are parallel or even nearly so, it is evicjent that the method for find- ing their resultant outlined above is inapplicable, as the intersection of the lines of action of any two such forces cannot be found at all, or is inconveniently distant. It is therefore necessary to develop some modification of the preceding method.
Consider the forces ab,
be, cd, and de, shown in Fig.
145. Instead of finding the
resultant of the given forces
directly, introduce into the problem a force oa, taken
entirely at random or to suit convenience, and then pro-
ceed to find the resultant of this new set of forces, as
already explained on page 114.
This is done in Figs. 145 and 146, and oe is found to be the resultant of oa, ab, be, cd, and de. Thus oe equals the sum of the given forces plus oa. But as oa was introduced for convenience of solution only, it should now be removed, i.e., subtracted from oe, to obtain the sum of the given forces or the required resultant. This is done by drawing the line ae\ for, if the direction of the arrows be noted, we see that
oa-\-ae=oe',
n6
GRAPHICAL STATICS
This is equivalent to reversing the direction of the force oa and adding it to oe.
Thus a force parallel and equal to ae, having its line of action passing through P, the intersection of the lines of action of oa and oe, is the required resultant.
The polygon formed of the forces ab, be, cd, de, and ae, Fig. 146, is called the polygon oj forces, or more dis-
10
FIG. 145
FIG. 146
tinctively the magnitude polygon. It depends only upon the forces irrespective of their positions. The point o is called the pole, and the lines oa, ob, . . . , oe are rays.
The polygon formed by the lines of action of the forces oa, ob, oc, od, and oe, Fig. 145, is called the funicular polygon. This is the Latin equivalent of string polygon, and it is so named because it gives the form assumed by a weightless string fastened at the points M and N and subjected to the given forces ab, be, . . . , de.
In applying this method to a set of given forces, it will be found convenient to first letter all the forces, then draw the magnitude polygon, thus determining the mag-
FORCES ACTING ON A RIGID BODY 117
nitude and the direction of the resultant, as in Fig. 146. Next, instead of assuming the force oa as in the above demonstration, place the pole in any convenient position and draw the rays. Then start the funicular polygon by drawing the line of action of oa anywhere in the diagram, showing the positions of the given forces, parallel to the ray oa. Through the intersection of the first and last sides of the funicular polygon draw the resultant parallel and equal to the arrow representing the resultant in the magnitude polygon.
Notice particularly the system of lettering employed and follow it closely in the exercises.
EXERCISE 176. Determine graphically the resultant of the following forces: (20 lb., 45°> o, o), (*5 lb-> 9°°> 13. °)> (3° lb-> 60°, 7, o), and (25 lb., 315°, 30, o). Each force is given by its magnitude, the direction of its line of action with reference to the X-axis, and one point on its line of action.
EXERCISE 177. Find the resultant of (20 lb., 105°, 5, o), (25 lb., 263°, n, o), (30 lb., 98°, 15, o), (10 lb., 80°, 17,0), (15 lb., 280°, 2, o).
(Note. — The angles in this exercise can be conveniently plotted by the use of a table of natural tangents.)
EXERCISE 178. Find the resultant of forces of 10, 20, 30, 40, 50 lb. acting vertically upward, with successive intervals between their lines of action of 3, 2, 6, and 4 feet.
EXERCISE 179. Same as Ex. 178, but with 10- and 3o-lb. forces acting vertically downward.
SECTION XX
CONDITIONS FOR EQUILIBRIUM
The magnitude polygon shown in Fig. 146 is said to be "open" because, in addition to the given forces, ab, . . . , de, an extra force, ae, the resultant, is required to
n8
GRAPHICAL STATICS
close the polygon. If the given forces placed tail to head do of themselves form a " closed" magnitude poly- gon, no closing line is required and the given forces have no resultant.
This, however, does not mean that the body on which these forces act is in equilibrium, for the absence of a resultant simply precludes translation.
It yet remains to be shown that no couple acts upon the body. To do this we turn to the funicular polygon. From Fig. 145 it will be noted that the funicular poly- gon has all but one of its vertices on the lines of action of the given forces, and this one vertex P lies on the line of action of the resultant. The polygon is said to be "open" because all of its vertices do not lie on the lines of action of the given forces. If, however, the lines of action of ao and oe should coincide, then the funicular polygon would be "closed," i.e., have each of its ver- tices upon one of the lines of action of the given forces, and the forces ao and oe would neutralize each other and produce equilibrium. This can only happen if the mag- nitude polygon is also closed. Study carefully Fig. 147.
aoreuZ
FIG. 147
It is, however, possible that the magnitude polygon be closed, and that the first and last sides of the funicular
FORCES ACTING ON A RIGID BODY
119
polygon do not coincide but are parallel. Under these conditions the funicular polygon is open and a couple, whose forces are ao and oe, results. Study carefully Fig. 148.
Thus it is seen that for equilibrium oj translation the graphical requirement is a closed magnitude polygon, and for equilibrium o] rotation a closed funicular polygon is
FIG. 148
necessary. These, then, are the graphical conditions of equilibrium.
The following examples illustrate the method of pro- cedure in the application of the above principles.
Example. — Upon a uniform beam, AB, 10 feet long, weighing 150 pounds, a force of 100 pounds acts at an angle of 315° at a point 6 feet from A : find the reactions of the supports if the beam is horizontal, and its end A is hinged to a vertical wall while the beam rests upon a smooth knife-edge one foot from B.
Solution. — Fig. 149 (a) illustrates the problem. Fig. 149 (b) shows the beam as a "free body." There is an
120
GRAPHICAL STATICS
(6)
P=?
unknown vertical reaction P at C, and a reaction Q, unknown in magnitude and direction, at A.
Letter the forces, known and unknown, and start the magnitude polygon, Fig. 149 (c), using any convenient
scale. Proceed to the line of action of cd, the point d being undetermined.
Assume a pole, o, and draw the rays, ao, bo, and co. Now start the funicular polygon, Fig. 149 (b), at A, the only known point in the line of action of Q. This is essential, as otherwise the funicular polygon can- not be closed. Complete the funicular polygon, clos- ing it by the line od, and c through o, Fig. 149 (c), draw
od parallel to od, Fig. 149 (b), and its intersection with cd determines the point d.
Close the magnitude polygon by drawing da. Then cd = P = i4o pounds is the vertical reaction at C and da=Q = no pounds is the reaction at A in both direction and magnitude.
These are the forces which represent the actions of the knife-edge and hinge upon the beam. What forces do the wall and knife-edge have to resist?
Example. — The weightless bar, Fig. 150 (a), supported by smooth pegs at A and B and by the cord CD, is in equilibrium. Find the reactions of the supports. Solution, — Fig. 150 (b) shows the bar as a "free body."
(O
FORCES ACTING ON A RIGID BODY
121
5 lib.
200 Ib.
Letter the. forces (known ones first), and start the mag- nitude polygon, Fig. 150 (c). Having carried the magni- tude polygon as far as possible, start the funicular polygon at E, the intersection of the lines of action of the two last- named unknown forces; tliis, again, is essential for the closing of the funic- ular polygon.
Close first the funicu- lar polygon and then the magnitude .polygon by drawing de and ea in Fig. 150 (c) through d and a parallel to the corre- sponding lines of action in Fig. 150 (b).
The reactions are found to be a/ = 730 Ib., de = 725 Ib., and ea = 200 Ib.
The application of the graphical principles of equilibrium offers no diffi- culties with the exception of the two- illustrated in
|
d f. |
c rf _ |
|
|
ix^ |
o —2-] |
yc c |
|
(» .'I' " |
b b |
tod
FIG. 150
the above examples. These ^should therefore be care- fully noted.
Notice, also, as a check upon the correctness of a fu- nicular polygon, that any two of its adjacent sides must meet upon the line of action of that force designated by the same letters as the sides after omitting the 0's. Thus the sides oa and ob meet on the line of action' of the force ab, etc.
122 GRAPHICAL STATICS
EXERCISE 180. Assume four forces and find a fifth wholly unknown force necessary to produce equilibrium.
EXERCISE 181. Assume four non-parallel forces and the lines of action of two other forces. Find the magnitudes of these forces for equilibrium.
Note. — This system of forces requires the lines of action of all forces involved to pass through one point. This may be shown by replacing the known forces by their resultant and noting that this resultant and the two unknown forces can only be in equilibrium under the above conditions. (Why?)
EXERCISE 182. Same as Exercise 181, but all forces parallel.
EXERCISE 183. Assume two forces and the line of action of another. Find a fourth wholly unknown force and the mag- nitude of the third force for equilibrium.
EXERCISE 184. Assume two forces and the lines of action of three other forces. Find the magnitudes of these forces for equilibrium.
T
V 5'
10 ft. ^f_ 10Jlb. 50o!lb.
1000
3000
1000
FIGS. 151 TO 154
EXERCISE 185. Find the reactions of the supports of tl.c beam shown in Fig. 151.
EXERCISE 186. Find the reactions of the supports of the beam shown in Fig. 152.
EXERCISE 187. Find the reactions of the abutments of the frame shown in Fig. 153.
EXERCISE 188. Find the reactions of the abutments of the frame shown in Fig. 154. The end A rests on frictionless rollers, and B is hinged to the foundations,
CHAPTER VII
APPLICATIONS TO STRUCTURES SECTION XXI
STRESSES IN MEMBERS OF FRAMED STRUCTURES
UP to the present only the forces acting on a rigid body have been considered. These forces are called the applied forces or external forces. The external forces act- ing upon a body cause a strained condition of the body due to the interaction of the various parts of the body in withstanding the action of the applied forces. This interaction of the parts of the body allows a transmission of forces from one part of the body to another. The forces so transmitted are known as internal forces.
Structures are contrivances for resisting forces. They may be divided into two types, framed and non-framed. A framed structure, or frame, is one composed of a system of straight bars fastened together, at their ends only, by pins so as to allow a hinge-like motion. Since the tri- angle is the only geometric figure in which a change of shape is impossible without a change in the length of its sides, the triangle is necessarily the basis of the arrange- ment of the bars in a frame.
Non-framed structures consist of one continuous mem-
123
124
GRAPHICAL STATICS
her, or a number of members so fastened together through- out their lengths as to make one solid piece.
Only the forces acting in framed structures will be considered.
Consider the simple frame shown in Fig. 155. A load of 1000 pounds acts at the apex M, and the frame rests upon two abutments N and P.
The first step in finding the forces acting on and trans- mitted by the frame is to show the frame as a "free body," Fig. 156. In this figure, in order to readily refer
\
FIG. 156
to the various members of the frame and the applied forces, designate each portion into which the plane is divided by a letter as shown. The load of looo pounds would then be referred to by naming the areas on each side, as CD ; similarly the members of the frame are EA, AD, AB, etc. Find now the reactions of the abutments DE and EC by means of the magnitude and funicular polygons as in Figs. 157 and 156. The reac- tions are found in Fig. 157 to be ce and ed. Thus a complete knowledge of the external forces is obtained.
To find the internal forces, consider the pin i joining the bars AD and EA, The forces acting on this pin are
\
c FIG.
APPLICATIONS TO STRUCTURES
125
FIG. 158
shown in Fig. 158 (a). DE is the known reaction, and
AD and EA are the unknown forces transmitted by the
members of the same name.
These forces must be in equi-
librium, therefore the triangle
of forces must close, Fig.
1 58 (b), and ae and da are the
required internal forces trans-
mitted by AE and AD respectively.
Consider now pin 2. The forces acting on it arc shown in Fig. 159 (a). Here, in addition to DC, the given load, AD, is known. For, as the member A D itself is in equilibrium and presses upon pin i in the direction da, Fig. 158 (b), it must press with an equal and opposite force on pin 2.
The polygon of forces for pin 2 is shown in Fig. 159 (b).
FIG. 159
c
FIG. i 60
This polygon shows that the member AB transmits no force and is therefore useless for this particular loading of the frame.
The process begun above may be continued until as many force polygons as there are pins have been con- structed. Instead, however, it is much shorter to com- bine all these polygons into one diagram based upon the magnitude polygon, Fig. 157, as shown in Fig. 160. The reason for this is evident if we consider that each polygon
126 GRAPHICAL STATICS
has for its sides one or more sides already belonging to preceding polygons.
In order to facilitate the construction of the polygons of internal forces, the names of all the forces acting at each pin, always taken in one direction (clockwise) and always starting with known forces, are set down thus:
For pin i, EDAE; pin 3, EABE;
pin 2, ADCBA] pin 4, CEBC.
Then, as in Fig. 160, draw lines parallel to the corre- sponding members transmitting the forces, through the proper points, until all the schemes have been com- pleted.
Consider now very carefully the information conveyed by these polygons of internal forces (Fig. 160). If we consider the action of the member AD upon pin i we need the scheme EDAE. Follow this in Fig. 160 and note that DA presses pin i downward and to the left. For equilibrium it is necessary for pin i to react with an equal and opposite force, pushing member AD upward and to the right. Now regard pin 2; here the scheme is ADCBA. This, with the assistance of Fig. 160, shows that AD pushes pin 2 upward and to the right. Thus pin 2 reacts on AD by pushing it downward and to the lejt. Thus AD is pushed upward and to the right at i, and downward and to the left at 2, by equal forces. These equal and opposite forces produce in AD a stress, and in this case the stress is known as a compression, and its magnitude is given by the length ad, Fig. 160. The member AD must be made of sufficient strength to with- stand tiiis compression.
APPLICATIONS to STRUCTURES
127
1000 Ibs.
500 Ibs.'
EXERCISE 189. Explain the stress in member AE, Fig. 156.
EXERCISE 190. Explain the stress in member BC, Fig. 156.
EXERCISE 191. Find the stresses in the various members of the frame shown in Fig. 155 if P is rigidly fastened to the abutment and N rests on frictionless rollers, and the load of 1000 Ibs. at the apex is inclined at 45° to the horizon.
EXERCISE 192. Find the stresses in the members of the frame shown in Fig. 161.
10 ft.
10 ft.
FIG. 161
FIG. 163
loft. " 15ft.
FIG. 162
EXERCISE 193. Same as Ex. 192 for Fig. 162. EXERCISE 194. Same as Ex. 192 for Fig. 163.
SECTION XXII
THE FUNICULAR POLYGON FOR PARALLEL FORCES CONSIDERED AS A MOMENT DIAGRAM
Fig. 164 shows a horizontal bar supported at the ends and supporting the weights P and Q. The magnitude polygon is shown at the right, and the funicular polygon directly under the bar. From these polygons we derive the magnitudes of the reactions cd and da.
128
GRAPHICAL STATICS
As this bar is in equilibrium the sum of the moments of all the forces acting on the bar about any point, A, must be zero. Thus, the sum of the moments of the forces to the left of A must be equal and opposite in sign to the sum of the moments of the forces to the right of A. At A, then, there is a tendency to bend the bar by rotating the left-hand end clockwise and the right-
FIG. 164
hand end counter-clockwise about A. The measure of this tendency to bend the beam at A is called the bend- ing moment, or simply the moment, at A , and it is equal to the sum of the moments of the forces, either to the left or to the right of A , about A . In Fig. 164 this moment would be
To obtain M graphically we must remember that in graphics division must always precede multiplication. From Fig. 164 (b) we see that RI =da and P = ab are both divided by D (see page 100). It only remains to mul-
APPLICATIONS TO STRUCTURES 1 29
tiply by x and y respectively. This is done in Fig. 164 (a),
7? P
where u:*=jrX and rv = ~f)y (see page 101);
or
.'. mD=-M.
As the sign is immaterial, depending only on which side of A we consider the moment as acting, we see that
m-D is the bending moment at A.
m being a vertical distance bounded by the funicular polygon, we see that the funicular polygon can be con- sidered as a moment diagram.
EXERCISE 195. Construct a diagram similar to Fig. 164 and show that (a) nD is the bending moment at B; (b) pD is the bending moment at C.
In the application of the above principle care must be taken to measure the various distances in the proper units. A good rule to follow is to measure the forces and other vertical distances with the scale of force (in pounds) and all horizontal distances with the scale of space (in feet). Thus, m, p, ab, cd, etc. (Fig. 164) should be measured in pounds and x, y, D, etc. in feet.
EXERCISE 196. Find graphically and check by calculation the bending moments at B, C, Z>, E, and F in the following horizontal beam: The beam, AG, is 30 feet long, AB=BC= etc. = 5 feet; abutments are at A andG; loads of 3000, 1000, 2000 Ibs. are hung at B, C, and E respectively.
EXERCISE 197. A horizontal beam AE, 20 feet long, is di-
130 GRAPHICAL STATICS
vided into four equal parts by B, C, and D. The beam is supported at A and B and loaded at C, D, and E with 1000, 3000, and 2000 Ibs. respectively. Find graphically the bend- ing moments at B, C, D, and E.
EXERCISE 198. Check the results of Ex. 197 analytically.
EXERCISE 199. A beam similar to that of Ex. 197 is sup- ported at A and C and loaded with 3000, 2000, and 3000 Ibs. at B, D, and E respectively. Find the moments at B, C, and£>.
EXERCISE 200. A horizontal beam AF, 25 feet long, sup- ports loads of 1000, 500, and 300 pounds at one end and at 10 and 25 feet from this end respectively. The supports are 5 and 15 feet from this same end. Find graphically the moments at 5 -foot intervals along the beam.
SECTION XXIII
GRAPHICAL METHOD FOR FINDING CENTROIDS
To find the centroid of a lamina, divide the lamina into portions the position of whose centroids are known, and consider the weights of these portions to act at their respective centroids in any convenient direction. By means of the funicular polygon find the resultant of these weights. The centroid must He somewhere on the line of action of this resultant. Now assume the weights to act in any other direction, always through their respect- ive centroids, and again find their resultant. The cen- troid must also lie upon the line of action of this resultant. The centroid is thus located at the intersection of the lines of action of these resultants.
This is illustrated in the following
Example. — Find the centroid of the plate shown in Fig. 165.
APPLICATIONS TO STRUCTURES 13!
Divide the plate into rectangles. The weight of each rectangle is proportional to its area; thus the weights of the rectangles can be represented by 20X6, 20X4, 8X4 respectively.
Consider first the weights ab, be, and dc as acting ver- tically downward; their resultant is found to be ad.
Next assume the weights to act horizontally. Repre-
FIG. 165
sent them by ef, )g, and gh\ their resultant is now eh. Therefore the centroid of the plate is at C, the intersec- tion of the lines of action of the resultants.
EXERCISE 201. Find graphically the centroid of the plates shown in Fig. 69.
EXERCISE 202. Find graphically the centroid of the lamina described in Exercise 153.
APPENDIX
APPLICATION OF TWO-DIMENSIONAL METHODS TO FINDING STRESSES IN THREE-DIMEN- SIONAL STRUCTURES
IN this volume only two-dimensional statics has been considered. A few examples will illustrate how the prin- ciples already studied may be applied to three-dimen- sional structures.
Example. — Find the stresses in the members of the structure represented pictorially in Fig. 166 (a). Here AD,
FIG. 1 66
BD, and CD represent three rods fastened at A, B, and C to a vertical wall MN. A and B are on the same level and 8 feet apart; EC is perpendicular to AB at its midpoint. The rods are each 5 feet long.
133
134 APPENDIX
Solution. — Pass a plane through C, D, and E and find the stresses in the member CD, and in an imaginary mem- ber ED, by considering the equilibrium of the point D under the action of the forces transmitted by these mem- bers and the weight of 100 pounds, Fig. 166 (b).
By the similarity of triangles, we have
- = — , . '. -R = 7 5 pounds ; 100 4'
also — = -, .*. 5 = 125 pounds.
100 4
The tension in CD is thus found to be 5 = 125 pounds.
Now pass a plane through A, B, and D. This plane contains the rods AD and BD and the imaginary rod DE. As AD and BD are to replace ED, a force equal and opposite to R, Fig. 166 (b), together with the forces transmitted by AD and BD} must be in equilibrium. These forces are shown in their true relative positions, together with their triangle of forces, in Fig. 166 (c).
By comparing Aoczw with AABD we can show that
P=Q, .'. wy=yz = —. As Aocyz is similar to AEBD we have
P S — =— , or P=62.5 pounds.
/ o o
Thus the compressions in AD and BD are equal each to 62.5 pounds.
EXERCISE 203. A pair of " sheer-legs" is formed of t\to equal spars fastened together at the top so as to form an inverted V. The spars are inclined at 60° to each other, and
STRESSES IN THREE-DIMEXSIOXAL STRUCTURES 135
their plane is inclined at 60° to the ground. A rope attached to the top of the sheers and inclined at 30° to the horizon lies in a vertical plane bisecting the sheers and holds them in posi- tion. Find the stresses in the spars and in the rope when a weight of 10 tons is lifted.
EXERCISE 204. A pair of sheers, such as described in Ex. 203 but of the following dimensions, supports a load of 122 tons. The legs of the sheers are 116 feet long; they are 45 feet apart at the bottom; the supporting guy is 146 feet long, and the sheers afford a horizontal reach of 35 feet. Find the stresses in the legs.
Example. — The vertical post of a crane is 10 feet long. The jib is 30 feet long and the stay is 24 feet long. There
FIG. 167
are two back-stays, making angles of 45° with the hori- zontal; these lie in planes due south and due east of the post. A weight of 4000 pounds is sustained by the crane. Find the forces transmitted by the jib- and back- stays when the jib lies to the northwest of the post. Solution,— Fig. 167 (a) illustrates the problem. Consider
136 APPENDIX
the equilibrium of the forces acting at C and lying in the plane ABC. If their triangle offerees, Fig. 167 (b), is drawn to .scale, we obtain the forces transmitted by AC and BC graphically. Represent these by 5 and R.
Now consider the point A in equilibrium under the action of AC, AB, and an imaginary stay AD. These forces all lie in the plane ACBD. From the triangle shown in Fig. 167 (c) find the forces transmitted by AB and AD, let these be Q and P.
Then consider A in equilibrium under the action of ASj AE, and a force equal and opposite to P, Fig. 167 (c). These forces are shown in their true relative positions in Fig. 167 (d). From this figure we obtain Y=X, the stresses in the stays AS and AE.
EXERCISE 205. Calculate the stresses found graphically in the preceding example.
EXERCISE 206. Find the stresses in the structure described in the preceding example when the jib lies to the north of che post.
EXERCISE 207. A weight of 100 pounds is sustained by a tripod having equal legs so arranged that the distance between each pair of feet is equal to the legs. Find the compression in each leg.
EXERCISE 208. Find the compression in each leg of a tri- pod the feet of whose legs rest at the vertices of an equilateral triangle whose sides are V3 feet long, if the legs are 5 feet long and the tripod sustains a weight of 1000 pounds.
EXERCISE 209. A, B, and C are the vertices of an isosceles triangle drawn upon a horizontal ceiling. AB, the base of the triangle, is 6 feet long; the altitude CD is 5 feet long. To A and B two strings, each 5 feet long, are attached; at C a string 3 feet long is fastened. The strings are joined at their lower ends and support a weight of 100 pounds. Find the tension in each string,
PROBLEMS FOR REVIEW
210. Two weights of P and Q pounds are attached to a string 21 inches long at points 8 and 6 inches respectively from the ends. If the ends of the string are fastened to two points on the same level 14 inches apart, and the central portion of the string is horizontal when equilibrium results, find the ratio of P to Q.
211. Draw an equilateral triangle ABC with the base AB horizontal and C downward. Let a weight at C be tied by threads AC and BC to the fixed points A and B: if the thread BC is cut, by how much does the tension in AC increase?
212. A load, W, of 2000 pounds is hung from a pin, P, at which pieces AP and BP meet like the tie-rod and jib of a crane. The angles WPB and WPA are respectively 30° and 60°. Find the forces transmitted by AP and BP, and state which piece acts as a strut and which as a tie.
213. Draw an equilateral triangle ABC. Let BC repre- sent a weightless lever acted on at B by a force of 13 pounds, acting from A to B, and at C by a force of 9 pounds, acting from A to C. Find graphically the pressure on, and the position of, the fulcrum.
214. An incline, the ratio of the height to the base being i to 10, supports a body weighing 100 pounds. If the coeffi- cient of friction is .2, what force inclined at 30° to the incline would move the body up the plane ?
215. A body rests upon a smooth plane inclined at 35° to the horizon. If the body weighs 10 pounds, find the tension in the supporting string when the string is inclined to the plane at an angle of (a) 25°, (b) — 10°.
138 PROBLEMS FOR REVIEW
2 1 6. A dead-weight safety-valve is 4 inches in diameter. The weight of the valve is 20 pounds. What additional weight must be added so that steam should blow off when the pressure reaches 80 pounds per square inch?
217. A square lamina is divided into four equal squares by lines parallel to the sides; a circle is inscribed in one of these squares, and the portion of the lamina within the circle is removed. Find the centroid of the remainder.
218. A uniform beam, whose length is 4 feet and whose weight is 20 pounds, has weights of 4 and 8 pounds suspended from its extremities. Where must a single support be placed to produce equilibrium ?
219. A, B, and C are three smooth pegs in a vertical wall, A being the highest. AB and AC make angles of 30° and 60° respectively with the vertical through A on opposite sides. A string, carrying two weights of 1 2 pounds each, passes over the pegs and the weights hang freely. Find the pressure on each peg.
220. A sphere, diameter i foot, hangs against a smooth vertical wall by a string 6 inches long fastened to its surface and to the wall; find the tension of the string and the pres- sure of the sphere against the wall.
221. A uniform wire AD, 15 inches long, is bent upwards at right angles 4 inches from A and 6 inches from D. Prove that if it be suspended from A it will rest with the second bend vertically below A.
222. A step-ladder has the form of the letter A. The semi- angle at the vertex is 6. If it rests on a smooth horizontal plane, and its legs are kept from slipping by a cord connecting them together half-way up, how much greater does the ten- sion in the cord become when a weight W is placed on top of the ladder?
223. The post BA of a crane is 10 feet high; the jib BC is 24 feet long and is movable about B. The tie is shortened so as to drop the load of 6000 pounds at a point, D, 20 feet from & Find graphically the stresses in the tie and jib,
PROBLEMS FOR REVIEW 139
224. A rod of length b is supported horizontally, and to its extremities are attached the ends of a string of length s. If a heavy ring of weight W is slung on the string, find the com- pression in the rod.
225. Draw a square A BCD; a force of 8 pounds acts from A to D, and two forces of 12 pounds each act from A to B and from C to D; find their resultant.
226. A beam balances about the midpoint of its axis when weights of 20 and 40 pounds are suspended one from each end, and it balances about a point one-third of the length from one end when the weights are interchanged. Find the weight of the beam and the distance of its centroid from one end.
227. A round table is supported on three legs A, B, and C. AB = AC=$ feet, BC=4 feet. A weight of 100 pounds is placed at P; the distances of P from AB 'and AC are re- spectively J and i foot. What is the compression in each leg?
228. A sheer-leg is formed by two sheer-poles, BC and DC, each 25 feet in length and secured to a base-plate in the ground at B and D. The wire guy, AC, is attached to the ground at a point A which is 60 feet from ED. The vertical from the top, C, of the poles meets the ground at a distance of 10 feet from the center of BD, which is 15 feet long. Find the stresses in the guy (and the compression in the poles) when a weight of 20 tons is suspended from C.
229. A uniform beam 12 feet long and weighing 56 pounds rests on and is fastened to two props 5 feet apart, one of which is 3 feet from one end of the beam. A load of 35 pounds is placed (a) at the end farthest from a prop, (b) at the middle of the beam, (c) at the end nearest a prop. Calculate the pressure on each prop in each case.
230. A crane, whose post, tie-rod, and jib measure 15, 20, and 30 feet respectively, supports a load of 10 tons suspended by a chain passing over a pulley at the jib-head. Find the stresses in each member (i) when the lifting-chain passes
140 PROBLEMS FOR REVIEW
from the pulley to the drum parallel to the jib, (2) when the drum is placed so that the chain passes from the jib-head parallel to the tie-rod.
231. In a pair of pincers the jaws meet one inch from the pin forming the joint. If the handles are grasped with a force of 30 pounds on each handle at a distance. 6 inches from the pin, find the compression exerted on an object held be- tween the jaws, and also the force resisted by the pin.
232. A painter's scaffold 20 feet long and weighing 150 pounds is supported by vertical ropes attached 2 feet from each end; if two painters weighing 125 and 175 pounds are at 4 feet and 9 feet from one end of the scaffold respectively, and pots of paint weighing 30 pounds are at 6 feet from the same end, find the tensions in the ropes.
233. A uniform bar projects 6 inches beyond the edge of a table, and when 2 ounces is placed one inch from the project- ing end the bar topples over; wrhen it is pushed out so as to project 8 inches beyond the edge, one ounce at the end makes it topple over. Find the weight of the bar and its length.
234. In a common steelyard the weight of the beam is 10 pounds, and acts at a distance of 2 inches from the fulcrum. Where must a weight of 4 pounds be applied to balance it ?
235. The weight of a window-sash 3 feet wide is 5 pounds; each of the weights acting on the cords is 2 pounds. If one cf the cords be broken, find at what distance from the middle of the sash the hand must be placed to raise it with the least effort. What pressure must the hand exert?
236. A piece of lead placed in one pan, A, of a balance is balanced by 10 pounds in the other pan, B. When the same piece of lead is placed in the pan B it required n pounds in the pan A to balance it. Find the ratio of the lengths of the arms of the balance.
237. Two equal uniform spheres of weight W and radius a rest in a smooth spherical cup of radius r. Find the pres- sure between either sphere and cup and the pressure between Uie spheres.
PROBLEMS FOR REVIEW 141
238". A uniform beam weight W is hinged to a horizontal plane and rests against a vertical wall. Find the reaction of the hinge and the pressure on the wall if the inclination of the beam is a.
239. A smooth uniform beam rests against two smooth hori- zontal rods, and its lower end rests against a smooth horizontal plane. The beam is 2.1 feet long, and the rods touch it at points a and b feet from its lower end. If the inclination of the beam is a, find the reactions of the supports.
240. A beam rests between two rough horizontal rods. The beam lies in a vertical plane. Assume all necessary data and write the equations for equilibrium.
241. ABC is a rigid equilateral triangle; the weight is not considered; the vertex B is fastened by a hinge to a vertical wall, while the vertex C rests against the wall under B. If 100 pounds is hung from A , find the reactions at B and C graphi- cally.
242. Three forces of 10, 15, and 50 pounds, making angles of 30°, 90°, and —135° with the horizon, act upon a particle. Find the resultant force acting upon the particle. In what direction wrill the particle move ?
243. A uniform beam weighing 10 pounds is supported at its ends by two props. If the length of the beam is 5 feet, find where a weight of 30 pounds must be attached so that the pressures on the props may be 15 and 25 pounds respect- ively.
244. A carriage- wheel, whose weight is W and whose radius is r, rests upon a level road. Find the least horizontal force applied at the axle necessary to draw the wheel over an ob- stacle whose height is h.
245. A mass whose weight is 750 pounds rests on a hori- zontal plane and is pulled by a force, P, inclined at 15° to the horizon. Find the value of P, which will just start the mass if the coefficient of friction is .62.
246. Find the total resistance of the plane in Ex. 245.
247. A body whose weight is 10 pounds is supported on a
142 PROBLEMS FOR REVIEW
smooth inclined plane by a force of two pounds acting along the plane and a horizontal force of 5 pounds. Find the in- clination of the plane.
248. Two equal rafters / feet long support a weight, W, at their upper ends. Find the stress in the tie-rod a feet long connecting their lower ends.
249. A davit is supported by a foot-step at A and by a collar at B placed 6 feet apart. A boat weighing 2 tons is supported by two such davits and is about to be lowered. Assuming that the boat hangs 5 feet from the vertical through the foot-step and collar, and that each davit supports one- half the weight of the boat, determine the forces at A and B.
250. A man, sitting upon a board suspended from a single movable pulley, pulls downward at one end of the rope which passes under the movable pulley and over a pulley fixed to a beam overhead, the other end of the rope being fixed to the same beam. If the man weighs 180 pounds, what force must he exert so as to maintain equilibrium ?
251. Make sketches of (a) a system of weightless pulleys in which one pound balances 32 pounds, (b) a system of weightless pulleys in which one pound balances 15 pounds.
252. Given four weightless pulleys, three movable and one fixed, around each pulley passes a separate rope; the load is a man weighing 160 pounds. Find the pull exerted by the man on the free end of the last rope in order to maintain equi- librium.
253. A rod 10 inches long can turn freely about one of its ends; a body weighing 4 pounds is hung at a point 3 inches from this end. If the free end of the rod is supported by a string inclined to it at an angle of 120°, find the tension in the string and the reaction at the fixed end of the rod.
254. Find the height of a cylinder which can just rest on an inclined plane the angle of which is 60°, the radius of the cylinder being r.
255. What is the minimum coefficient of friction necessary to prevent the sliding of the cylinder in Ex. 254?
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* Elements of the Art of War Svo, 4 oo
Metcalf's Cost of Manufactures — And the Administration of Workshops. .Svo, 5 oo
* Ordnance and Gunnery. 2 vols i2mo, 5 oo
Murray's Infantry Drill Regulations iSmo, paper, 10
Nixon's Adjutants' Manual 24010, i oo
Peabody's Naval Architecture Svo, 7 50
* Phelps's Practical Marine Surveying .8vo, 2 50
Powell's Army Officer's Examiner i2mo, 4 oo
Sharped Art of Subsisting Armies in War iSrno, morocco, i 50
* Walke's Lectures on Explosives, 8vo, 4 oo
* Wheeler's Siege Operations and Military Mining 8vo, 2 oo
Winthrop's Abridgment of Military Law, i2mo, 2 50
Woodhull's Notes on Military Hygiene i6mo, i 50
Young's Simple Elements of Navigation i6mo, morocco, i oo
Second Edition, Enlarged and Revised i6mo, morocco, 2 oo>
ASSAYING.
Fletcher's Practical Instructions in Quantitative Assaying with the Blowpipe.
1 2mo, morocco, i 50
Yunnan's Manual of Practical Assaying 8vo, 3 oo
Lodge's Notes on Assaying and Metallurgical Laboratory Experiments. . . .8vo, 3 oo
Miller's Manual of Assaying i2mo, i oo
O'Driscoll's Notes on the Treatment of Gold Ores 8vo, 2 oo
Ricketts and Miller's Notes on Assaying 8vo, 3 oo
Ulke's Modern Electrolytic Copper Refining 8vo, 3 oo
Wilson's Cyanide Processes i2mo, i 50
Chlorination Process i2mo, i 50
ASTRONOMY.
Comstock's Field Astronomy for Engineers 8vo, 2 50
Craig's Azimuth 4to, 3 50
Doolittle's Treatise on Practical Astronomy 8vo, 4 oo
Gore's Elements of Geodesy 8vo, 2 50
Hayford's Text-book of Geodetic Astronomy 8vo, 3 oo-
Merriman's Elements of Precise Surveying and Geodesy 8vo, 2 50
* Michie and Harlow's Practical Astronomy 8vo, 3 oo
* White's Elements of Theoretical and Descriptive Astronomy i2mo, 2 oo
BOTANY.
Davenport's Statistical Methods, with Special Reference to Biological Variation.
i6mo, morocco, i 25
Thome' and Bennett's Structural and Physiological Botany i6mo, 2 25
Westermaier's Compendium of General Botany. (Schneider.) 8vo, 2 oo
CHEMISTRY.
Adriance's Laboratory Calculations and Specific Gravity Tables . .i2mo, i 25
Allen's Tables for Iron Analysis 8vo, 3 oo
Arnold's Compendium of Chemistry. (Mandel.) Small 8vo, 3 50
Austen's Notes for Chemical Students i2mo, i 50
Bernadou's Smokeless Powder. — Nitro-cellulose, and Theory of the Cellulose
Molecule i2mo, 2 50
Bolton's Quantitative Analysis 8vo, i 50
* Browning's Introduction to the Rarer Elements 8vo, i 50
Brush and Penfield's Manual of Determinative Mineralogy 8vo, 4 oo
Classen's Quantitative Chemical Analysis by Electrolysis. (Boltwood. ). .8vo, 3 oo
Cohn's Indicators and Test-papers i2mo, 2 oo
Tests and Reagents 8vo, 3 oo
Crafts's Short Course in Qualitative Chemical Analysis. (Schaeffer. ). . .i2mo, i so> Dolezalek's Theory of the Lead Accumulator (Storage Battery). (Von
Ende.) 1 2mo, 2 50-
Drechsel's Chemical Reactions. (Merrill.) I2mo, i 25
Duhem's Thermodynamics and Chemistry. (Burgess.) 8vo, 4 oo
Eissler's Modern High Explosives 8vo, 4 oo
Eff rent's Enzymes and their Applications. ( Fresco tt.) 8vo, 3 oo-
Erdmann's Introduction to Chemical Preparations. (Dunlap.) i2mo, i aj>
3
Fletcher's Practical Instructions in Quantitative Assay irg with the Bleu pipe.
12010, morocco, i 50
Fowler's Sewage Works Analyses iimo, 2 oo
Fresenius's Manual of Qualitative Chemical Analysis. (Wells.) 8vo, 5 oo
Manual of Qualitative Chemical Analysis. Part I. Descriptive. (Weils.) 8vo, 3 oo System of Instruction in Quantitative Chemical Araiysis. (Cohn.)
2 vols 8vo, 1 2 50
Fuertes's Water and Public Health i2mo, i 50
Furman's Manual of Practical Assaying 8vo, 3 oo
* Getman's Exercises in Physical Chemistry i2mo, 2 oo
Gill's Gas and Fuel Analysis for Engineers i2mo, i 25
Grotemeu's Principles of Modern Dairy Practice. (Woll.) i2mo, 2 oo
Hammarsten's Text-book of Physiological Chemistry. (Mandel.) Svo, 4 oo
Helm's Principles of Mathematical Chemistry. (Morgan.) i2mo, i 50
Bering's Ready Reference Tables (Conversion Factors ) 1611:0, rr.crocco, 2 50
Hind's Inorganic Chemistry 8vo, 3 oo
* Laboratory Manual for Students i2mo, 75
Holleman's Text-book of Inorganic Chemistry. (Cooper.) 8vo, 2 50
Text-book of Organic Chemistry. (Walker and Mott.) 8vo, 2 50
* Laboratory Manual of Organic Chemistry. (Walker.) i2mo, i oo
Hopkins's Oil-chemists' Handbook 8vo, 3 oo
Jackson's Directions for Laboratory Work in Physiological Chen^&>ry. .8vo, i 25
Keep's Cast Iron 8vo, 2 50
Ladd's Manual of Quantitative Chemical Analysis i2mo, i oo
Landauer's Spectrum Analysis. (Tingle.) 8vo, 3 oo
* Langworthy and Austen. The Occurrence of Aluminium in Vege able
Products, Animal Products, and Natural Waters 8vo, 2 oo
Lassar-Cohn's Practical Urinary Analysis. (Lorenz.) i2mo, i oo
Application of Some General Reactions to Investigations in Organic
Chemistry. (Tingle.) i2mo, i oo
Leach's The Inspection and Analysis of Food with Special Reference to State
Control. 8vo, 7 50
Lob's Electrolysis and Electrosynthesis of Organic Compounds. (Lorenz. ).i2iro, i co
Lodge's Notes on Assaying and Metallurgical Laboratory Experiments. .. .8vo, 3 co
Lunge's Techno-chemical Analysis. (Cohn.) i2mo, i oo
Mandel's Handbook for Bio-chemical Laboratory i2mo, i 50
* Martin's Laboratory Guide to Qualitative Analysis with the Blowpipe . i2mo, 60 Mason's Water-supply. (Considered Principally from a Sanitary Standpoint.)
3d Edition, Rewritten 8vo, 4 oo
Examination of Water. (Chemical and Bacteriological.) i2rro, i 25
Matthew's The Textile Fibres Svo, 3 50
Meyer's Determination of Radicles in Carbon Compounds. (Tingle.). .i2mo, i oo
Miller's Manual of Assaying i2mo, i oo
Mixter's Elementary Text-book of Chemistry i2mo, i 50
Morgan's Outline of Theory of Solution and its Results i2ir_o, i oc
Elements of Physical Chemistry i2mo, 2 oo
Morse's Calculations used in Cane-sugar Factories i6mo, morocco, i 50
Mulliken's General Method for the Identification of Pure Organic Compounds.
Vol. I Large Svo, 5 oo
O'Brine's Laboratory Guide in Chemical Analysis Svo, 2 oo
O'Driscoll's Notes on the Treatment of Gold Ores Svo, 2 oo
Ostwald's Conversations on Chemistry. Part One. (Ramsey.) i2mo, i 50
Ostwald's Conversations on Chemistry. Part Two. (Turnbull.). (In Press.)
* Penfield's Notes on Determinative Mineralogy and Record of Mineral Tests.
Svo, paper, 50
Pictet's The Alkaloids and their Chemical Constitution. (Biddle.) Svo, 5 oo
Pinner's Introduction to Organic Chemistry. (Austen.) i2mo, i 50
Poole's Calorific Power of Fuels Svo, 3 oo
Prescott and Winslow's Elements of Water Bacteriology, with Special Refer- ence to Sanitary Water Analysis i2mo, i 25
4
* Reisig's Guide to Piece-dyeing 8vo, 25 oo
Richards and Woodman's Air, Water, and Food from a Sanitary Standpoint 8vo, 2 oo
Richards's Cost of Living as Modified by Sanitary Science i2mo, i oo
Cost of Food, a Study in Dietaries i2mo, i oo
* Richards and Williams's The Dietary Computer 8vo, i 50
Ricketts and Russell's Skeleton Notes upon Inorganic Chemistry. (Part I.
Non-metallic Elements.) 8vo, morocco, 75
Ricketts and Miller's Notes on Assaying .8vo, 3 oo
Rideal's Sewage and the Bacterial Purificat on of Sewage £vo, 3 50
Disinfection and the Preservation of Food 8vo, 4 oo
Rigg's Elementary Manual for the Chemical Laboratory. . . 8vo, i 23
Rostoski's Serum Diagnosis. (Bolduan.) 1200, i oo
Ruddiman's Incompatibilities in Prescriptions Svo, 2 oo
Sabin's Industrial and Artistic Technology of Paints and Varnish £vo, 3 oo
Salkowski's Physiological and Pathological Chemistry. (Orndorff.) Svo, 2 50
Schimpf's Text-book of Volumetric Analysis i2mo, 2 50
Essentials of Volumetric Analysis i2mo, i 25
Spencer's Handbook for Chemists of Beet-sugar Houses i6mo, morocco, 3 oo
Handbook for Sugar Manufacturers and their Chemists. i6mo, morocco, 2 oo
Stockbridge's Rocks and Soils Svo, 2 50
* Tillman's Elementary Lessons in Heat Svo, i 50
Descriptive General Chemistry Svo, 3 oo
Treadwell's Qualitative Analysis. (Hall.) Svo, 3 oo
Quantitative Analysis. (Hall.) Svo, 4 oo
lurneaure and Russell's Public Water-supplies Svo, 5 oo
Van Deventer's Physical Chemistry for Beginners. (Boltwood.) i2mo, i 50
* Walke's Lectures on Explosives Svo, 4 oo
Washington's Manual of the Chemical Analysis of Rocks Svo, 2 oo
Wassermann's Immune Sera : Haemolysins, Cytotoxins, and Precipitirs. (Bol- duan.) i2mo, i oo
Well's Laboratory Guide in Qualitative Chemical Analysis.. Svo, i 50
Short Course in inorganic Qualitative Chemical Analysis for Engineering
Students I2mo, i 50
Text-book of Chemical Arithmetic. (In press.)
Whipple's Microscopy of Drinking-water Svo, 3 50
Wilson's Cyanide Processes i2mo, i 50
Chlorination Process i2mo, i 50
Wulling's Elementary Course in Inorganic, Pharmaceutical, and Medical
Chemistry i2mo, 2 oo
CIVIL ENGINEERING.
BRIDGES AND ROOFS. HYDRAULICS. MATERIALS OF ENGINEERING. RAILWAY ENGINEERING.
Baker's Engineers' Surveying Instruments i2mo, 3 oo
Btxby's Graphical Computing Table Paper IQ£ X 24! inches. 25
** Burr's Ancient and Modern Engineering and the Isthmian Canal. (Postage,
27 cents additional.) Svo, 3 50
Comstock's Field Astronomy for Engineers Svo, 2 50
Davis's Elevation and Stadia Tables Svo, i oo
Elliott's Engineering for Land Drainage i2mo, i 50
Practical Farm Drainage I2mo, i oo
Fiebeger's Treatise on Civil Engineering. (In press.)
Folwell's Sewerage. (Designing and Maintenance.) Svo, 3 oo
Freitag's Architectural Engineering. 2d Edition , Rewritten Svo, 3 50
French and Ives's Stereotomy Svo, 2 50
Goodhue's Municipal Improvements i2mo, i 75
Goodrich's Economic Disposal of Towns' Refuse Svo, 3 ?o
Gore's Elements of Geodesy Svo, 2 50
Hayford's Text-book of Geodetic Astronorry Svo, 3 oo
Bering's Ready Reference Tables (Conversion Factors) i6mo, morocco, 2 50
5
Howe's Retaining Walls for Earth i2mo, i 23
Johnson's (J. B.) Theory and Practice of Surveying Small 8vo, 4 oo
Johnson's (L. J.) Statics by Algebraic and Graphic Methods 8vo,
Laplace's Philosophical Essay on Probabilities. (Truscott and Emory.) . i2mo, Mahan's Treatise on Civil Engineering. (1873.) (Wood.) 8vo,
* Descriptive Geometry 8vo,
Merriman's Elements of Precise Surveying and Geodesy 8vo,
Elements of Sanitary Engineering 8vo,
Merriman and Brooks's Handbook for Surveyors i6mo, morocco,
Nugent's Plane Surveying 8vo, 3 50
Ogden's Sewer Design i2mo, 2 oo
Patton's Treatise on Civil Engineering 8vo half leather, 7 50
Reed's Topographical Drawing and Sketching 4to, 5 oo
Rideal's Sewage and the Bacterial Purification of Sewage 8vo, 3 50
Siebert and Biggin's Modern Stone-cutting and Masonry 8vo, i 50
Smith's Manual of Topographical Drawing. (McMillan.) 8vo, 2 50
Sondericker's Graphic Statics, with Applications to Trusses, Beams, and Arches.
8vo, 2 oo
Taylor and Thompson's Treatise on Concrete, Plain and Reinforced 8vo, 5 oo
* Trautwine's Civil Engineer's Pocket-book i6mo, morocco, 5 oo
Wait's Engineering and Architectural Jurisprudence 8vo, 6 oo
Sheep, 6 50
Law of Operations Preliminary to Construction in Engineering and Archi- tecture 8vo, 5 oo
Sheep, 5 50
Law of Contracts 8vo, 3 oo
Warren's Stereotomy — Problems in Stone-cutting 8vo, 2 50
Webb's Problems in the Use and Adjustment of Engineering Instruments.
i6mo, morocco, i 25
* Wheeler's Elementary Course of Civil Engineering 8vo, 4 oo
Wilson's Topographic Surveying 8vo, 3 50
BRIDGES AND ROOFS.
Boiler's Practical Treatise on the Construction of Iron Highway Bridges. .8vo, 2 oo
* Thames River Bridge 410, paper, 5 oo
Burr's Course on the Stresses in Bridges and Roof Trusses, Arched Ribs, and
Suspension Bridges 8vo, 3 50
Burr and Falk's Influence Lines for Bridge and Roof Computations. . . .8vo, 3 oo
.Du Bois's Mechanics of Engineering. Vol. II Small 4to, 10 oo
Tester's Treatise on Wooden Trestle Bridges 4to, 5 oo
Towler's Ordinary Foundations 8vo, 3 50
Greene's Roof Trusses 8vo, i 25
Bridge Trusses 8vo, 2 50
Arches in Wood, Iron, and Stone 8vo, 2 50
Howe's Treatise on Arches 8vo, 4 oo
Design of Simple Roof-trusses in Wood and Steel 8vo, 2 oo
Johnson, Bryan, and Turneaure's Theory and Practice in the Ccsiprirg of
Modern Framed Structures Small 4to, 10 oo
Merriman and Jacoby's Text-book on Roofs and Bridges:
Part I. Stresses in Simple Trusses 8vo, 2 50
Part II. Graphic Statics 8vo, 2 50
Part III. Bridge Design 8vo, 2 50
Part IV. Higher Structures 8vo, 2 50
US. orison's Memphis Bridge 4to, 10 oo
Waddell's De Pontibus, a Pocket-book for Bridge Engineers. . i6mo, morocco, 3 oo
Specifications for Steel Bridges i2mo, i 25
Wood's Treatise on the Theory of the Construction of Bridges and Roofs . . 8vo, 2 oo Wright's Designing of Draw-spans:
Part I. Plate-girder Draws 8vo, 2 50
Part II. Riveted-truss and Pin-connected Long-span Draws 8vo, 2 50
Two parts in one volume 8vo, 3 50
6
HYDRAULICS.
Bazin's Experiments upon the Contraction of the Liquid Vein Issuing from
an Orifice. (Trautwine.) 8vo, 2 oo
Bovey's Treatise on Hydraulics 8vo, 5 oo
Church's Mechanics of Engineering 8vo, 6 oo
Diagrams of Mean Velocity of Water in Open Channels payer, i 50
Coffin's Graphical Solution of Hydraulic Problems i6mo, morocco, 2 50
Flather's Dynamometers, and the Measurement of Power izmo, 3 oo
FolwelTs Water-supply Engineering 8vo, 4 oo
Frizell's Water-power 8vo, 5 oo
Fuertes's Water and Public Health i2mo, i 50
Water-filtration Works ramo, 2 50
Ganguillet and Kutter's General Formula for the Uniform Flow of Water in
Rivers and Other Channels. (Hering and Trautwine.) 8vo 4 oo
Hazen's Filtration of Public Water-supply 8vo, 3 oo
Hazlehurst's Towers and Tanks for Water- works 8vo, 2 50
Herschel's 115 Experiments on the Carrying Capacity of Large, Riveted, Metal
Conduits 8vo, 2 oo
Mason's Water-supply. (Considered Principally from a Sanitary Standpoint.)
8vo, 4 oo
Merriman's Treatise on Hydraulics 8vo, 5 oo
* Michie's Elements of Analytical Mechanics 8vo, 4 oo
Schuyler's Reservoirs for Irrigation, Water-power, and Domestic Water- supply Liuse 8vo, 5 oo
** Thomas and Watt's Improvement of Rivers. (Post., 440. additional.) 4to, 6 oo
Turneaure and Russell's Public Water-supplies ?vo, 5 oo
Wegmann's Design and Construction of Dams 4to, 5 oo
Water-supply of the City of New York from 1658 to 1895 4to, 10 oo
Wilson's Irrigation Engineering . .Small 8vo, 4 oo
Wolff's Windmill as a Prime Mover 8vo, 3 oo
Wood's Turbines 8vo, 2 50
Elements of Analytical Mechanics , 8vo, 3 oo
MATERIALS OF ENGINEERING.
Baker's Treatise on Masonry Construction . .8vo, 5 oo
Roads and Pavements 3vo, 5 oo
Black's United States Public Works ObJonji 4to 5 oo
Bovey's Strength of Materials and Theory of Structures Svc. 7 50
Burr's Elasticity and Resistance of the Materials of Engineering 8vo, 7 50
Byrne's Highway Construction 8vo, 5 oo
Inspection of the Materials and Workmanship Employed in Construction.
i6mo, 3 oo
Church's Mechanics of Engineering 8vo, 6 oo
Du Bois's Mechanics of Engineering. VoL I Small 4to, 7 50
Johnson's Materials of Construction Large 8vo, 6 oo
Fowler's Ordinary Foundations 8vo, 3 50
Keep's Cast Iron 8vo, 2 50
Lanza's Applied Mechanics 8vo, 7 50
Marten's Handbook on Testing Materials. (Henning.) 2 vols 8vo, 7 50
Merrill's Stones for Building and Decoration 8vo, 5 oo
Merriman's Text-book on the Mechanics of Materials 8vo, 4 oo
Strength of Materials i2mo, i oe
Metcalf's Steel. A Manual for Steel-users i2mo, 2 oo
Patton's Practical Treatise on Foundations 8vo, 5 oo
Richardson's Modern Asphalt Pavements. (In press.)
Richey's Handbook for Superintendents of Construction i6mo, mor., 4 oo
Rockwell's Roads and Pavements in France i2mo, i 25
Sabin's Industrial and Artistic Technology of Paints and Varnish 8vo, 3 oo
Smith's Materials of Machines i2mo, i oo
Snow's Principal Species of Wood 8vo, 3 50
Spalding's Hydraulic Cement. . i2mo, 2 oo
Text-book on Roads and Pavements i2mo, 2 oo
Taylor and Thompson's Treatise on Concrete, Plain and Reinforced 8vo, 5 oo
Thurston's Materials of Engineering. 3 Paris 8vo, 8 oo
Part I. Non-metallic Materials of Engineering and Metailurcy 8vo, 2 oo
Part II. Iron and Steel 8vo, 3 50
Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their
Constituents 8vo, 2 50
Thurston's Text-book of the Materials of Construction 8vo, 5 oo
Tillson's Street Pavements and Paving Materials 8vo, 4 oo
WaddelFs De Pontibus. (A Pocket-book for Bridge Engineers.). . i6mo, mor., 3 oo
Specifications for Steel Bridges i2mo, i 25
Wood's (De V.) Treatise on the Resistance of Materials, and an Appendix on
the Preservation of Timber 8vo, 2 oo
Wood's (De V.) Elements of Analytical Mechanics 8vo, 3 oo
Wood's (M. P.) Rustless Coatings: Corrosion and Electrolysis of Iron and
Steel 8vo, 4 oo
RAILWAY ENGINEERING.
Andrew's Handbook for Street Railway Engineers 3x5 inches, morocco, i 25
Berg's Buildings and Structures of American Railroads 4to, 5 oo
Brook's Handbook of Street Railroad Location i6mo, morocco, i 50
Butt's Civil Engineer's Field-book . i6mo, morocco, 2 50
Crandall's Transition Curve i6mo, morocco, i 50
Railway and Other Earthwork Tables .8vo, i 50
Dawson's "Engineering" and Electric Traction Pocket-book. . i6mo, morocco, 5 oo
Dredge's History of the Pennsylvania Railroad: (1879) Paper, 5 oo
* Drinker's Tunnelling, Explosive Compounds, and Rock Drills. 4to, half mor., 25 oo
Fisher's Table of Cubic Yards Cardboard, 25
Godwin's Railroad Engineers' Field-book and Explorers' Guide. . . i6mo, mor., 2 50
Howard's Transition Curve Field-book i6mo, morocco, i 50
Hudson's Tables for Calculating the Cubic Contents of Excavations and Em- bankments 8vo, i oo
Molitor and Beard's Manual for Resident Engineers i6mo, i oo
Nagle's Field Manual for Railroad Engineers i6mo, ncrocco, 3 oo
Phiibrick's Field Manual for Engineers i6mo, morocco, 3 oo
Searles's Field Engineering i6mo, morocco, 3 oo
Railroad Spiral i6mo, morocco, i 50
Taylor's Prismoidal Formulae and Earthwork 8vo, i 50
* Trautwine's Method of Calculating the Cube Contents of Excavations and
Embankments by the Aid of Diagrams 8vo, 2 oo
The Field Practice of Laying Out Circular Curves for Railroads.
i2mo, morocco, 2 50
Cross-section Sheet Paper, 25
Webb's Railroad Construction i6mo, morocco, 5 oo
Wellington's Economic Theory of the Location of Railways. ....'. . Small 8vo, 5 oo
DRAWING.
Barr's Kinematics of Machinery 8vo, 2 50
* Bartlett's Mechanical Drawing 8vo, 3 oo
* " " " Abridged Ed 8vo, 150
Coolidge's Manual of Drawing 8vo, paper i oo
Coolidge and Freeman's Elements cf Gcr.or?! Drafting for Mechanical Engi- neers ".'.' Oblong 4to, 2 50
Durley's Kinematics of Machines 8vo, 4 oo
Emch's Introduction to Projective Geometry and its Applications 8vo. 2 50
8
Hill's Text-book on Shad&s and Shadows, and Perspective 8vo, 2 oo
Jamison's Elements of Mechanical Drawing 8vo, 2 50
Jones's Machine Design:
Part I. Kinematics of Machinery 8vo, i 50
Part El. Form, Strength, and Proportions of Parts 8vo, 3 oo
MacCord's Elements of Descriptive Geometry 8vo, 3 oo
Kinematics; or, Practical Mechanism 8vo, 5 oo
Mechanical Drawing 4to, 4 oo
Velocity Diagrams 8vo, i 50
* Mahan's Descriptive Geometry and Stone-cutting 8vo, i 50
Industrial Drawing. (Thompson.) 8vo, 3 50
Moyer's Descriptive Geometry. (In press.)
Reed's Topographical Drawing and Sketching 4to, 5 oo
Reid's Course in Mechanical Drawing 8vo, 2 oo
Text-book of Mechanical Drawing and Elementary Machine Design. 8vo, 3 oo
Robinson's Principles of Mechanism 8vo, 3 oo
Schwamb and Merrill's Elements of Mechanism 8vo, 3 oo
Smith's Manual of Topographical Drawing. (McMillan.) 8vo, 50
Warren's Elements of Plane and Solid Free-hand Geometrical Drawing. i2mo, oo
Drafting Instruments and Operations i2mo, 25
Manual of Elementary Projection Drawing i2mo, 5*
Manual of Elementary Problems in the Linear Perspective of Form and
Shadow i2mo, oo
Plane Problems in Elementary Geometry i2mo, 25
Primary Geometry i2mo, 75
Elements of Descriptive Geometry, Shadows, and Perspective 8vo, 3 50
General Problems of Shades and Shadows 8vo, 3 oo
Elements of Machine Construction and Drawing 8vo, 7 50
Problems, Theorems, and Examples in Descriptive Geometry 8vo, 2 50
Weisbach's Kinematics and Power of Transmission. (Hermann and Klein)8vo, 5 oo
Whelpley's Practical Instruction in the Art of Letter Engraving i2mo, 2 oo
Wilson's (H. M.) Topographic Surveying 8vo, 3 50
Wilson's (V. T.) Free-hand Perspective 8vo, 2 50
Wilson's (V. T.) Free-hand Lettering 8vo, i oo
Woolf's Elementary Course in Descriptive Geometry Large 8vo, 3 oo
ELECTRICITY AND PHYSICS.
Anthony and Brackett's Text-book of Physics. (Magie.) Small 8vo, 3 oo
Anthony's Lecture-notes on the Theory of Electrical Measurements. . . . i2mo, i oo
Benjamin's History of Electricity 8vo, 3 oo
Voltaic Cell 8vo, 3 oo
Classen's Quantitative Chemical Analysis by Electrolysis. (Boltwood.).8vo, 3 oo
Crehore and Squier's Polarizing Photo-chronograph 8vo, 3 oo
Dawson's "Engineering" and Electric Traction Pocket-book. i6mo, morocco, 5 oo Dolezalek's Theory of the Lead Accumulator (Storage Battery). (Von
Ende.) iamo, 2 50
Duhem's Thermodynamics and Chemistry. (Burgess.) 8vo, 4 oo
Flather's Dynamometers, and the Measurement of Power I2mo, 3 oo
Gilbert's De Magnete. (Mottelay.) 8vo, 2 50
Hanchett's Alternating Currents Explained i2mo, i oo
Bering's Ready Reference Tables (Conversion Factors) i6mo, morocco, 2 50
Holman's Precision of Measurements 8vo, 2 oo
Telescopic Mirror-scale Method, Adjustments, and Tests. . . . Large 8vo, 75
Kinzbrunner's Testing of Continuous-Current Machines 8vo, 2 oo
Landauer's Spectrum Analysis. (Tingle.) 8vo, 3 oo
Le Chatelien's High-temperature Measurements. (Boudouard — Burgess.) i2mo, 3 oo
Lob's Electrolysis and Electrosynthesis of Organic Compounds. (Lorenz.) i2mo, i oo
* Lyons's Treatise on Electromagnetic Phenomena. Vols. I. and II. 8vo, each, 6 oo
* Michie's Elements of Wave Motion Relating to Sound and Light 8vo, 4 oo
Niaudet's Elementary Treatise on Electric Batteries. (Fishback.) lamo, 2 50
* Rosenberg's Electrical Engineering. (Haldane Gee — Kinzbrunner.). . ,8vo, i 50
Ryan, Norris, and Hoxie's Electrical Machinery. Vol. 1 8vo, 2 50
Thurston's Stationary Steam-engines 8vo, 2 50
* Tillman's Elementary Lessons in Heat 8vo, i 50
Tory and Pitcher's Manual of Laboratory Physics Small 8vo, 2 oo
Ulke's Modern Electrolytic Copper Refining 8vo, 3 oo
LAW.
* Davis's Elements of Law 8vo, 2 50
* Treatise on the Military Law of United States 8vo, 7 oo
Sheep, 7 So
Manual for Courts-martial i6mo, morocco, i 50
Wait's Engineering and Architectural Jurisprudence 8vo, 6 oo
Sheep, 6 50
Law of Operations Preliminary to Construction in Engineering and Archi- tecture 8vo, 5 oo
Sheep, 5 50
Law of Contracts 8vo, 3 oo
Winthrop's Abridgment of Military Law I2mo, 2 50
MANUFACTURES.
Bernadou's Smokeless Powder— Nitro-cellulose and Theory of the Cellulose
Molecule i2mo, 2 50
Bolland's Iron Founder I2mo, 2 50
" The Iron Founder," Supplement I2mo, 2 50
Encyclopedia of Founding and Dictionary of Foundry Terms Used in the
Practice of Moulding I2mo, 3 oo
Eissler's Modern High Explosives 8vo, 4 oo
Effront's Enzymes and their Applications. (Prescott.) 8vo, 3 oo
Fitzgerald's Boston Machinist i2mo, i oo
Ford's Boiler Making for Boiler Makers i8mo, i oo
Hopkin's Oil-chemists' Handbook 8vo, 3 oo
Keep's Cast Iron 8vo, 2 50
Leach's The Inspection and Analysis of Food with Special Reference to State
ControL Large 8vo, 7 50
Matthews's The Textile Fibres 8vo, 3 50
Metcalf's Steel. A Manual for Steel-users i2mo, 2 oo
Metcalfe's Cost of Manufactures — And the Administration of Workshops. 8vo, 5 oo
Meyer's Modern Locomotive Construction 4to, 10 oo
Morse's Calculations used in Cane-sugar Factories i6mo, morocco, i 50
* Reisig's Guide to Piece-dyeing 8vo, 25 oo
Sabin's Industrial and Artistic Technology of Paints and Varnish 8vo, 3 oo
Smith's Press-working of Metals 8vo, 3 oo
Spalding's Hydraulic Cement i2mo, 2 oo
Spencer's Handbook for Chemists of Beet-sugar Houses. . . . i6mo, morocco, 3 oo
Handbook for Sugar Manufacturers and their Chemists . i6mo, morocco, 2 oo
Taylor and Thompson's Treatise on Concrete, Plain and Reinforced 8vo, 5 oo
Thurston's Manual of Steam-boilers, their Designs, Construction and Opera- tion 8vo, 5 oo
* Walke's Lectures on Explosives 8vo, 4 oo
Ware's Manufacture of Sugar. (In press.)
West's American Foundry Practice i2mo, 2 50
Moulder's Text-book i2mo, 2 50
10
Wolff's Windmill as a Prime Mover 8vo, 3 oo
Wood's Rustless Coatings: Corrosion and Electrolysis of Iron and Steel. .8vo, 4 oo
MATHEMATICS.
Baker's Elliptic Functions 8vo, i 50
* Bass's Elements of Differential Calculus i2mo, 4 oo
Briggs's Elements of Plane Analytic Geometry . . 121110, oo
.Compton's Manual of Logarithmic Computations iimo, 50
Davis's Introduction to the Logic of Algebra 8vo, 50
* Dickson's College Algebra Large i2mo, 50
* Introduction to the Theory of Algebraic Equations Large i2mo, 25
Emch's Introduction to Projective Geometry and its Applications 8vo, 50
Halsted's Elements of Geometry 8vo, 75
Elementary Synthetic Geometry 8vo, 50
Rational Geometry i2mo, 75
* Johnson's (J. B.) Three-place Logarithmic Tables: Vest-pocket size. paper, 15
100 copies for 5 oo
* Mounted on heavy cardboard, 8X 10 inches, 25
10 copies for 2 oo
Johnson's (W. W.) Elementary Treatise on Differential Calculus. .Small 8vo, 3 oo
Johnson's (W. W.) Elementary Treatise on the Integral Calculus. Small 8vo, i 50
Johnson's (W. W.) Curve Tracing in Cartesian Co-ordinates i2mo, i oo
Johnson's (W. W.) Treatise on Ordinary and Partial Differential Equations.
Small 8vo, 3 50
Johnson's (W. W.) Theory of Errors and the Method of Least Squares. i2mo, i 50
* Johnson's (W. W.) Theoretical Mechanics i2mo, 3 oo
Laplace's Philosophical Essay on Probabilities. (Truscott and Emory.). i2mo, 2 oo
* Ludlow and Bass. Elements of Trigonometry and Logarithmic and Other
Tables 8vo, 3 oo
Trigonometry and Tables published separately Each, 2 oo
* Ludlow's Logarithmic and Trigonometric Tables 8vo, i oo
Maurer's Technical Mechanics • S» , , 4 oo
Merriman and Woodward's Higher Mathematics 8vo, 5 oo
Merriman's Method of Least Squares 8vo, 2 oo
Rice and Johnson's Elementary Treatise on the Differential Calculus. . Sm. 8vo, 3 oo
Differential and Integral Calculus. 2 vols. in one Small 8vo, 2 50
Wood's Elements of Co-ordinate Geometry 8vo, 2 oo
Trigonometry: Analytical, Plane, and Spherical i2mo, i oo
MECHANICAL ENGINEERING.
MATERIALS OF ENGINEERING, STEAM-ENGINES AND BOILERS.
Bacon's Forge Practice i2mo, i 50
Baldwin's Steam Heating for Buildings I2mo, 2 50
Barr's Kinematics of Machinery 8vo, 2 50
* Bartlett's Mechanical Drawing 8vo, 3 oo
* " " " Abridged Ed 8vo, i 50
Benjamin's Wrinkles and Recipes i2mo, 2 oo
Carpenter's Experimental Engineering 8vo, 6 oo
Heating and Ventilating Buildings 8vo, 4 oo
Cary's Smoke Suppression in Plants using Bituminous Coal. (In Prepara- tion.)
Clerk's Gas and Oil Engine Small 8vo, 4 oo
Coolidge's Manual of Drawing 8vo, paper, i oo
Coolidge and Freeman's Elements of General Drafting for Mechanical En- gineers Oblong 4to, 2 50
11
Cromwell's Treatise on Toothed Gearing i2mo,
Treatise on Belts and Pulleys i2mo,
Durley's Kinematics of Machines 8vo,
Flather's Dynamometers and the Measurement of Power iimo,
Rope Driving i2mo,
Gill's Gas and Fuel Analysis for Engineers i2mo,
Hall's Car Lubrication i2mo,
Bering's Ready Reference Tables (Conversion Factors) i6mo, morocco,
Mutton's The Gas Engine 8vo, 5 oo
Jamison's Mechanical Drawing 8vo, 2 50
Jones's Machine Design:
Part I. Kinematics of Machinery 8vo, i 50
Part II. Form, Strength, and Proportions of Parts 8vo, 3 oo
Kent's Mechanical Engineers' Pocket-book i6mo, morocco, 5 oo
Kerr's Power and Power Transmission 8vo, 2 oo
Leonard's Machine Shop, Tools, and Methods. (In press.)
Lorenz's Modern Refrigerating Machinery. (Pope, Haven, and Dean.) (In press.)
MacCord's Kinematics; or, Practical Mechanism 8vo, 5 oo
Mechanical Drawing 4to, 4 oo
Velocity Diagrams 8vo, i 50
Mahan's Industrial Drawing. (Thompson.) 8vo, 3 50
Poole's Calorific Power of Fuels 8vo, 3 oo
Reid's Course in Mechanical Drawing 8vo, 2 oo
Text-book of Mechanical Drawing and Elementary Machine Design. 8vo, 3 oo
Richard's Compressed Air i2mo, i 50
Robinson's Principles of Mechanism 8vo, 3 oo
Schwamb and Merrill's Elements of Mechanism 8vo, 3 oo
Smith's Press-working of Metals 8vo, 3 oo
Thurston's Treatise on Friction and Lost Work in Machinery and Mill
Work : . . 8vo, 3 oo
Animal as a Machine and Prime Motor, and the Laws of Energetics . i2mo, i oo
Warren's Elements of Machine Construction and Drawing 8vo, 7 50
Weisbach's Kinematics and the Power of Transmission. (Herrmann —
Klein.) • 8vo, 5 oo
Machinery of Transmission and Governors. (Herrmann — Klein.). . 8vo, 5 oo
Wolff's Windmill as a Prime Mover 8vo, 3 oo
Wood's Turbines .' 8vo, 2 50
MATERIALS OF ENGINEERING.
Bovey's Strength of Materials and Theory of Structures 8vo, 7 So
Burr's Elasticity and Resistance of the Materials of Engineering. 6th Edition.
Reset 8vo, 7 50
Church's Mechanics of Engineering 8vo, 6 oo
Johnson's Materials of Construction 8vo, 6 oo
Keep's Cast Iron 8vo, 2 50
Lanza's Applied Mechanics . .8vo, 7 50
Martens's Handbook on Testing Materials. (Henning.) 8vo, 7 50
Merriman's Text-book on the Mechanics of Materials 8vo, 4 oo
Strength of Materials I2mo, i oo
Metcalf's Steel. A manual for Steel-users i2mo. 2 oo
Sabin's Industrial and Artistic Technology of Paints and Varnish 8vo, 3 oo
Smith's Materials of Machines I2mo, i oo
Thurston's Materials of Engineering 3 vols., 8vo, 8 oo
Part II. Iron and Steel 8vo, 3 50
Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their
Constituents 8vo, 2 50
Text-book of the Materials of Construction 8vo, 5 oo
Wood's (De V.) Treatise on the Resistance of Materials a., -n Appendix on
the Preseivation of Timber 8vo, 2 oo
Wood's (De V.) Elements of Analytical Mechanics 8vo, 3 oo
Wood's (M. P.) Rustless Coatings: Corrosion and Electrolysis of Iron and
Steel 8vo, 4 oo
STEAM-ENGINES AND BOILERS.
Berry's Temperature-entropy Diagram i2mo, i 25
Carnot's Reflections on the Motive Power of Heat. (Thurston.) i2mo, i 50
Dawson's "Engineering" and Electric Traction Pocket-book. . . .i6mo, mor., 5 oo
Ford's Boiler Making for Boiler Makers i8mo, i oo
Goss's Locomotive Sparks 8vo, 2 oo
Hemenway's Indicator Practice and Steam-engine Economy i2mo, 2 oo
Button's Mechanical Engineering of Power Plants 8vo, 5 oo
Heat and Heat-engines 8vo, 5 oo
Kent's Steam boiler Economy 8vo, 4 oo
Kneass's Practice and Theory of the Injector 8vo, i 50
MacCord's Slide-valves 8vo, 2 oo
Meyer's Modern Locomotive Construction 4to, 10 oo
Peabody's Manual of the Steam-engine Indicator i2mo. i 50
Tables of the Properties of Saturated Steam and Other Vapors 8vo, i oo
Thermodynamics of the Steam-engine and Other Heat-engines 8vo, 5 oo
Valve-gears for Steam-engines 8vo, 2 50
Peabody and Miller's Steam-boilers 8vo, 4 oo
Pray's Twenty Years with the Indicator Large 8vo, 2 50
Pupin's Thermodynamics of Reversible Cycles in Gases and Saturated Vapors.
(Osterberg. ) i2mo, i 25
Reagan's Locomotives: Simple Compound, and Electric i2mo, 2 50
Rontgen's Principles of Thermodynamics. (Du Bois.) 8vo, 5 oo
Sinclair's Locomotive Engine Running and Management i2mo, 2 oo
Smart's Handbook of Engineering Laboratory Practice i2mo, 2 50
Snow's Steam-boiler Practice 8vo, 3 oo
Spangier's Valve-gears 8vo, 2 50
Notes on Thermodynamics i2mo, i oo
Spangler, Greene, and Marshall's Elements of Steam-engineering 8vo, 3 oo
Thurston's Handy Tables 8vo. i 50
Manual of the Steam-engine 2 vols., 8vo, 10 oo
Part I. History, Structure, and Theory 8vo, 6 oo
Part II. Design, Construction, and Operation 8vo, 6 oo
Handbook of Engine and Boiler Trials, and the Use of the Indicator and
the Prony Brake 8vo, 5 oo
Stationary Steam-engines. 8vo, 2 50
Steam-boiler Explosions in Theory and in Practice i2mo, i 50
Manual of Steam-boilers, their Designs, Construction, and Cperaticn £vo, 5 oo
Weisbach's Heat; Steam, and Steam-engines. (Du Bois.) 8vo, 5 oo
Whitham's Steam-engine Design Svo, 5 oo
Wilson's Treatise on Steam-boilers. (Flather.) i6mo, 2 50
Wood's Thermodynamics, Heat Motors, and Refrigerating Machines. . .8vo, 4 oo
MECHANICS AND MACHINERY.
Barr's Kinematics of Machinery 8vo, 2 50
Bovey's Strength of Materials and Theory of Structures Svo, 7 50
Chase's The Art of Pattern-making I2mo, 2 50
Church's Mechanics of Engineering 8vo, 6 oo
13
Church's Notes and Examples in Mechanics 8vo, oo
Compton's First Lessons in Metal-working i2mo, 50
Compton and De Groodt's The Speed Lathe i2mo, 50
Cromwell's Treatise on Toothed Gearing i2mo, 50
Treatise on Belts and Pulleys i2mo, 50
Dana's Text-book of Elementary Mechanics for Colleges and Schools. . i2mo, 50
Dingey's Machinery Pattern Making i2mo, oo
Dredge's Record of the Transportation Exhibits Building of the World's
Columbian Exposition of 1893 4to half morocco, 5 oo
Du Bois's Elementary Principles of Mechanics :
Vol. I. Kinematics 8vo, 3 50
Vol. II. Statics 8vo, 4 oo
Vol. III. Kinetics 8vo, 3 50
Mechanics of Engineering. Vol. I Small 4to, 7 50
Vol. II Small 4to, 10 oo
Durley's Kinematics of Machines 8vo, 4 oo
Fitzgerald's Boston Machinist i6mo, i oo
Flather's Dynamometers, and the Measurement of Power i2mo, 3 oo
Rope Driving i2mo, 2 oo
Goss's Locomotive Sparks 8vo, 2 oo
Hall's Car Lubrication i2mo, i oo
Holly's Art of Saw Filing iSrao, 75
James's Kinematics of a Point and the Rational Mechanics of a Particle. (In press.)
* Johnson's (W. W.) Theoretical Mechanics 12010, 3 oo
Johnson's (L. J.) Statics by Graphic and Algebraic Methods 8vo, 2 oo
Jones's Machine Design: •
Part I. Kinematics of Machinery 8vo, i 50
Part II. Form, Strength, and Proportions of Parts 8vo, 3 oo
Kerr's Power and Power Transmission 8vo, 2 oo
Lanza's Applied Mechanics 8vo, 7 50
Leonard's Machine Shop, Tools, and Methods. (In press.)
Lorenz's Modern Refrigerating Machinery. (Pope, Haven, and Dean.) (In press.)
MacCord's Kinematics; or, Practical Mechanism. 8vo, 5 oo
Velocity Diagrams 8vo, i 50
Maurer's Technical Mechanics 8vo, 4 bo
Merriman's Text-book on the Mechanics of Materials 8vo, 4 oo
* Elements of Mechanics 12010, i oo
* Michie's Elements of Analytical Mechanics 8vo, 4 oo
Reagan's Locomotives: Simple, Compound, and Electric i2mo, 2 50
Reid's Course in Mechanical Drawing 8vo. 2 oo
Text-book of Mechanical Drawing and Elementary Machine Design. 8vo, 3 oo
Richards's Compressed Air i2mo, i 50
Robinson's Principles of Mechanism 8vo , 3 oo
Ryan, Norris, and Hoxie's Electrical Machinery. Vol. 1 8vo, 2 50
Schwamb and Merrill's Elements of Mechanism 8vo, 3 oo
Sinclair's Locomotive-engine Running and Management i2tno, 2 oo
Smith's (O.) Press-working of Metals 8vo, 3 oo
Smith's (A. W.) Materials of Machines i2mo, i oo
Spangler, Greene, and Marshall's Elements of Steam-engineering 8vo, 3 oo
Thurston's Treatise on Friction and Lost Y/brk in Machinery and Mill
Work 8vor 3 oo
Animal as a Machine and Prime Motor, and the Laws of Energetics
i2mo, i oo
Warren's Elements of Machine Construction and Drawing 8vo, 7 50
Weisbach's Kinematics and Power of Transmission. (Herrmann— Klein. ).8vo, 5 oo
Machinery of Transmission and Governors. (Herrmann — Klein. ).8vo, 5 oo
Wood's Elements of Analytical Mechanics 8vo, 3 oo
Principles of Elementary Mechanics i2mo, i 25
Turbines 8vo . 2 50
The World's Columbian Exposition of 1893 4to, i oo
14
METALLURGY.
Egleston's Metallurgy of Silver, Gold, and Mercury:
VoL L Silver 8vo, 750
VoL II. Gold and Mercury 8vo, 7 50
** Iles's Lead-smelting. (Postage 9 cents additional.) 12010, 2 50
Keep's Cast Iron 8vo, 2 50
Kunhardt's Practice of Ore Dressing in Europe .8vo, i gc
Le Chatelier's High-temperature Measurements. (Boudouard — Burgess. )i2mo, 3 oo
Metcalf's SteeL A Manual for Steel-user& i2mo, 2 oo
Smith's Materials of Machines i2mo, i oo
Thurston's Materials of Engineering. In Three Parts 8vo, 8 oo
Part II. Iron and SteeL 8vo. 3 50
Part TTT. A Treatise on Brasses, Bronzes, and Other Alloys and their
Constituents 8vo, 2 50
Ulke's Modern Electrolytic Copper Refining 8vo, 3 oo
MINERALOGY.
Barringer's Description of Minerals of Commercial Value. Oblong, morocco, 2 50
Boyd's Resources of Southwest Virginia 8vo, 3 oo
Map of Southwest Virignia Pocket-book form. 2 oo
Brush's Manual of Determinative Mineralogy. (Penfield.) 8vo, 4 oo
Chester's Catalogue of Minerals 8vo, paper, i oo
Cloth, i 25
Dictionary of the Names of Minerals 8vo, 3 50
Dana's System of Mineralogy Large 8vo, half leather, 12 50
First Appendix to Dana's New " System of Mineralogy." Large 8vo, i oo
Text-book of Mineralogy 8vo, 4 oo
Minerals and How to Study Them I2mo, i 50
Catalogue of American Localities of Minerals Large 8vo, i oo
Manual of Mineralogy and Petrography i2mo 2 oo
Douglas's Untechnical Addresses on Technical Subjects i2mo, i oo
Eakle's Mineral Tables 8vo, i 25
Egleston's Catalogue of Minerals and Synonyms 8vo, 2 50
Hussak's The Determination of Rock-forming Minerals. ( Smith.). Small 8vo, 2 oo
Merrill's Non-metallic Minerals: Their Occurrence and Uses 8vo, 4 oo
* Penfield's Notes on Determinative Mineralogy and Record of Mineral Tests.
8vo. paper, o 50 Rosenbusch's Microscopical Physiography ot the Rock-maki«g Minerals
(Iddings.) 8vo. 5 oo
* Tollman's Text-book of Important Minerals and Rocks ... .8vo. 2 oo
Williams's Manual of Lithology 8vo, 3 oo
MINING.
Beard's Ventilation of Mines I2mo, 2 50
Boyd's Resources of Southwest Virginia 8vo, 3 oo
Map of Southwest Virginia Pocket book form, 2 oo
Douglas's Untechnical Addresses on Technical Subjects i2mo. i oo
* Drinker's Tunneling, Explosive Compounds, and Rock Drills 4to.hf. mor. . 25 oo
Eissler's Modern High Explosives 8vo 4 oo
Fowler's Sewage Works Analyses 12010 2 oo
Goodyear's Coal-mines of the Western Coast of the United States. . . i2mo. 2 50
Ihlseng's Manual of Mining 8vo. 5 oo
** Iles's Lead-smelting. (Postage oc. additional.) 12010. 2 50
Kunhardt's Practice of Ore Dressing in Europe 8vo, i 50
O'Driscoll's Notes on the Treatment of Gold Ores 8vo. 2 oo
* Walke's Lectures on Explosives 8vo. 4 oo
Wilson's Cyanide Processes i2mo, i 50
Chlorination Process I2mo, i 50
15
Wilson's Hydraulic and Placer Mining i2mo, 2 oo
Treatise on Practical and Theoretical Mine Ventilation t2mo. i 25
SANITARY SCIENCE.
FolwelPs Sewerage. (Designing, Construction, and Maintenance.) 8vo, 3 oo
• Water-supply Engineering 8vo, 4 oo
Fuertes's Water and Public Health i2mo, i 50
Water-filtration Works i2mo, 2 50
Gerhard's Guide to Sanitary House-inspection i6mo, i oo
Goodrich's Economic Disposal of Town's Refuse Demy 8vo, 3 50
Hazen's Filtration of Public Water-supplies 8vo, 3 oo
Leach's The Inspection and Analysis of Food with Special Reference to State
Control 8vo, 7 50
Mason's Water-supply. (Considered principally from a Sanitary Standpoint) 8vo, 4 oo
Examination of Water. (Chemical and Bacteriological.) i2mo, i 25
Merriman's Elements of Sanitary Engineering 8vo, 2 oo
Ogden's Sewer Design i2mo, 2 oo
Prescott and Winslow's Elements of Water Bacteriology, with Special Refer- ence to Sanitary Water Analysis I2mo,
* Price's Handbook on Sanitation I2mo,
Richards's Cost of Food. A Study in Dietaries i2mo,
Cost of Living as Modified by Sanitary Science i2mp,
Richards and Woodman's Air, Water, and Food from a Sanitary Stand- point 8vo,
* Richards and Williams's The Dietary Computer 8vo,
25 So oo
00
oo
50
Rideal's Sewage and Bacterial Purification of Sewage 8vo, 3 50
Turneaure and Russell's Public Water-supplies 8vo, 5 oo
Von Behring's Suppression of Tuberculosis. (Bolduan.) i2mo, i oo
Whipple's Microscopy of Drinking-water 8vo, 3 50
Woodhull's Notes on Military Hygiene i6mo, i 50
MISCELLANEOUS.
De Fursac's Manual of Psychiatry. (Rosanoff and Collins.). . . .Large i2mo, 2 50 Emmons's Geological Guide-book of the Rocky Mountain Excursion of the
International Congress of Geologists Large 8vo, i 50
Ferrel's Popular Treatise on the Winds 8vo. 4 oo
Haines's American Railway Management i2mo, 2 50
ITott's Composition, Digestibility, and Nutritive Value of Food. Mounted chart, i 25
Fallacy of the Present Theory of Sound i6mo, i oo
Ricketts's History of Rensselaer Polytechnic Institute, 1824-1894. .Small 8vo, 3 oo
Rostoski's Serum Diagnosis. (Bolduan.) i2mo, i oo
Rotherham's Emphasized New Testament Large 8vo, 2 oo
Steel's Treatise on the Diseases of the Dog • 8vo, 3 50
Totten's Important Question in Metrology 8vo, 2 50
The World's Columbian Exposition of 1893 4*0, i oo
Von Behring's Suppression of Tuberculosis. (Bolduan.) i2mo, i oo
Winslow's Elements of Applied Microscopy i2mo, i 50
Worcester and Atkinson. Small Hospitals, Establishment and Maintenance;
Suggestions for Hospital Architecture : Plans for Small Hospital i2mo, i 25
HEBREW AND CHALDEE TEXT-BOOKS.
Green's Elementary Hebrew Grammar i2mo, i 25
Hebrew Chrestomathy 8vo, 2 oo
Gesenius's Hebrew and Chaldee Lexicon to the Qld Testament Scriptures.
(Tregelles.) Small 4to, half morocco, 5 oo
Lettews's Hebrew Bible 8vo» 2 25
16
UNIVERSITY OF TORONTO
DEPARTMENT OP CIVIL ENGINEERING
Mnuicipr.l and Structural
QA Martin, Louis Adolphe
807 Text-book of mechanics. 1st ed<
M38 v.l
v.l
P&AScLt
PLEASE DO NOT REMOVE CARDS OR SLIPS FROM THIS POCKET
UNIVERSITY OF TORONTO LIBRARY