A SHORT COURSE
IN
Elementary Mechanics
FOR
Engineers
CLIFFORD NEWTON MILLS, B.S., A.M.
Assistant Professor of Mathematics, South Dakota State College of Agriculture and Mechanic Arts
36 ILLUSTMATIONS
NEW YORK D. VAN NOSTRAND COMPANY
25 Park Place 1916
.^^'
i
^
^
Copyright, 1916
BY
D. VAN NOSTRAND COMPANY
PRESS OF
BRAUNWORTH & CO.
BOOKBINDERS AND PRINTERS
BROOKLYN. N. Y.
PREFACE
This course in Elementary Mechanics is arranged for students who have previously studied Trigonom- etry. It is intended as a basis for a semester's work of three hours per week.
The subject matter is divided into three parts, namely, Kinematics, Kinetics, and Statics.
Much detailed discussion is omitted, and the course is written from the " definition viewpoint." In considering the working principles of Statics, much emphasis is placed on the importance of " moment equations." The Engineer's system of units is used.
Throughout the text are many problems, a large number of these being original. In fact, it is by solving problems that a mastery of the subject is gained. Many of the problems will not be difficult to solve, but others will require the student's best effort.
The author desires to express his thanks to Pro- fessors H. C. Solberg, Head of the Department of Mechanical Engineering, H. B. Mathews, Head of the Department of Physics, and G. L. Brown, Head of the Department of Mathematics, for numerous and valuable suggestions.
Clifford N. Mills.
South Dakota State College, August, 1916
iii .
. 371695
CONTENTS
INTRODUCTION
SECTION PAGE
1. Definitions of Mechanics, Kinematics, Dynamics,
Kinetics, and Statics ix
2. Units of Measurement x
3. Motion: Translatory and Rotary x
Greek Alphabet xi
CHAPTER I
KINEMATICS
Linear and Angular Motion
4. Velocity: Uniform Velocity i
5. Average Velocity 2
6. Acceleration 2
7. Formulas for Linear Motion 3
8. Angular or Rotary Motion 6
9. Angular Measurement: Radian 6
10. Angular Velocity 7
11. Relation between Linear and Angular Velocity 8
12. Angular Acceleration 8
13. Formulas for Angular Motion 9
14. Motion Due to Gravity 11
CHAPTER II VECTORS
1 5. Definition of a Vector 15
16. Addition and Subtraction of Vectors 16
V
vi CONTENTS
SECTION PAGE
17. Resultant Velocity 17
18. Rectangular Component Velocities 17
19. Motion on a Smooth Inclined Plane 21
20. Motion of a Projectile 23
21. Velocity and Direction of Motion of a Projectile after
a Given Time has Elapsed 24
22. Velocity and Direction of Motion of a Projectile at a
Given Height 24
CHAPTER III
KINETICS
23. Mass of a Body 27
24. Weight of a Body 27
25. Definition of Force and its Measurement 27
26. Relation between Mass and Weight of a Body 28
27. Newton's Laws of Motion 28
28. Momentum and Impulse 29
29. Third Law of Motion 34
30. Motion of Two Connected Weights 34
31. Motion of Two Connected Weights, One Moving on
a Smooth Plane or Table 35
32. Work 41
33. Unit of Work 41
34. Power 41
35. Energy, Kinetic and Potential 43
CHAPTER IV
STATICS
Composition and Resolution of Forces
36. Force. Equilibrium 4q
37. Definition of a Resultant Force 49
CONTENTS vii
SECTION PAGE
38. Resultant of Two Forces Acting on a Particle 49
39. Resultant of any Number of Forces Acting on a Particle 50 4a Condition for Equilibrium 53
41. Triangle of Forces 53
42. Moment of a Force 59
43. Moment of the Resultant of Two Forces 60
44. Relation Existing between the Moments of Forces
which Produce Equilibrium 61
45. Composition of Parallel Forces 61
46. Resultant, and Point of Application, of Two Parallel
Forces having the Same Direction 62
47. Geometrical Method of Finding the Point of Apphca-
tion of the Resultant of Two Parallel Forces Having
the Same Direction 64
48. Resultant, and Point of Application, of Two Parallel
Forces Having Opposite Directions 64
49. Center of Giavity, Center of a System of Parallel
Forces 68
50. Center of Gravity of Two Particles of Given Weights
at a Given Distance Apart 69
51. Plane Moments 70
52. Theorems on Center of Gravity 71
53. Stable, Unstable, and Neutral Equilibrium 85
54. Theorems of Guldinus or Pappus 88
CHAPTER V MOTION IN A CIRCLE
55. Direction of Acceleration in Circular Motion 92
56. Centripetal and Centrifugal Forces 93
57. Motion on a Circular Inclined Track 94
58. Motion on Railway Curves 94
59. Conical Pendulum 95
60. Motion in a Vertical Circle 97
viii CONTENTS
CHAPTER VI
ENERGY OF ROTATION
Moments of Inertia
SECTION PAGE
6i. Energy of Rotation 103
62. Moment of Inertia 103
63. Radius of Gyration 104
64. Moment of Inertia of a Thin Plate about an Axis
Perpendicular to Its Plane 105
65. Moments of Inertia of a Thin Plate about Parallel
Axes in Its Plane * 106
66. Table of Moments of Inertia 107
Tables iii
Answers 115
Index 125
INTRODUCTION
1. Mechanics is a branch of Physics which treats of the motions of bodies and the causes of the changes of these motions. It is divided into two main sub- jects, namely, Kinematics and Dynamics. In this text the subject of Dynamics is divided into two parts called Kinetics and Statics.
Kinematics (Greek, kinema, motion) is the study of motion apart from matter. It considers how a body moves, and is not concerned with any of the properties of the body which moves. It does not deal with the forces which cause motion. It is strictly a geometrical science of motion.
Dynamics (Greek, dynamiSj force) treats of the effect of forces acting upon bodies whether in a state of rest or motion.
Kinetics (Greek, kinetikos, to put in motion) con- siders the forces which produce motion. It considers why a body moves, and discusses the relations between the motion produced and the forces in- volved.
Statics (Greek, statikos, to be at rest) considers exclusively the conditions under which a body under the action of forces will remain at rest. In this case the forces acting on the body are said to be in equi- librium.
is
X INTRODUCTION
2. A physical quantity is anything that can be measured. There are three fundamental units of measurement, length, mass, and time. Any physical quantity can be expressed in terms of one or more of these so-named fundamental units.
The object that moves is called matter. It has never been defined, but we do know many properties concerning it. A Umited portion of matter is called a body. A material particle is an ideal body whose dimensions, length, breadth, and thickness, are ex- tremely small in comparison with some special unit. For instance, the dimensions of the earth and planets are very small in comparison with the distance of the earth from the sun. Therefore in the study of Astronomy the earth and planets many times are considered as material particles.
The only property inherent in matter which affects the motion of bodies is called Inertia, which is defined as the reaction of matter against a change of motion or a state of rest.
3. Motion. A displacement is a change of position without reference to time. If we take into considera- tion the time taken to make the change of position then it is called Motion.
If a body moves in space, its path is a line, either straight or curved. If all material particles which make up a body describe parallel lines, not circular, the motion is said to be Translator^. Rectilinear motion is a special case of translatory motion when the lines are straight.
INTRODUCTION
XI
If all particles in a moving body describe circular paths, or concentric circles, about a fixed point called the axial point, which in this case has no motion, the motion is called Angular or Rotary.
A body is said to have relative motion with respect to another body when it is continually changing its position with respect to that body. In the case of a moving train, we always consider its motion relative to some part of the earth's surface, or objects on the earth's surface. A passenger on a train may be at rest relative to the train, but he would be in motion relative to the trees, houses, etc., which the train passes.
GREEK ALPHABET
|
Letters. |
Name. |
Letters. |
Name. |
|
A a |
Alpha |
N V |
Nu |
|
B P |
Beta |
H $ |
Xi |
|
Ty |
Gamma |
0 o |
Cmicron |
|
A 8 |
Delta |
ITtt |
Fi |
|
E c |
Epsilon |
Pp |
Rho |
|
Z ^ |
Zeta |
2 o- s |
Sigma |
|
H^ |
Eta |
T T |
Tau |
|
© e |
Theta |
Y V |
Upsilon |
|
I L |
Iota |
^ <!> |
Phi |
|
K K |
Kappa |
xx |
Chi |
|
A \ |
Lambda |
^ l/r |
Psi |
|
M fi |
Mu |
n 0, |
Omega |
ELEMENTARY MECHANICS FOR ENGINEERS
CHAPTER I
KINEMATICS Linear and Angular Motion
4. Velocity. The velocity of a moving body is the rate at which it changes its position. A velocity is completely determined when its direction of motion and the rate of change of position are given. If the rate of change of position is given, not mentioning the direction of motion, then the nimierical value of the rate of change is called the speed of the moving body. Velocity may be uniform or variable. In this text, however, we will consider only uniform and imiformly accelerated velocity.
If a body moves in the same direction and passes over equal distances in equal intervals of time, how- ever small these may be, its velocity is said to be uniform. The phrase " equal intervals of time " is very important in uniform motion. A body may describe equal distances in equal times and yet its motion may not be uniform, hence the clause " how- ever small these may be " is very necessary in the
2 ELEMENTARY MECHANICS FOR ENGINEERS
definition. For instance, a train may describe 40 miles in each of two consecutive hours, and yet its motion may have varied during each hour. There- fore uniform motion implies a constant irate of change of position.
If s is the distance described with a uniform velocity of V units of length per second in t seconds, then
s=vt (i)
5. Average Velocity. The average velocity of a moving body is the number of units of length de- scribed divided by the time taken. This is the same as the constant velocity which the moving body must have in order to describe the same distance in the same time. If a body moves with different velocities in the same equal times, and the difference between consecutive velocities is constant, then
average velocity = , {2)
n
where each v denotes a different velocity. The numerator of the fraction is an arithmetical series, and its sum is {n/2) {vi-{-Vr). Hence, if the velocity is uniformly accelerated, then
average veloaty = , ... (3;
2
where vi and Vn denote the initial and final velocities. Hereafter v, vij and ^^2, will denote average velocity, initial velocity, and final velocity.
6. Acceleration. The acceleration of a moving
KINEMATICS 8
body is the rate of change of its velocity. Accelera- tion may be positive or negative; if negative it is called retardation. Hereafter acceleration will be denoted by the letter a.
Acceleration is said to be uniform when equal changes of velocity take place in equal intervals of time, however small these may. In uniformly accelerated motion the average acceleration equals the increase in velocity divided by the time taken to increase, or
average acceleration = -^ ^. . . (4)
I
7. From the definitions of uniform and uniformly accelerated velocity one can estabhsh the following formulas:
V2=-vi-\-at, (5)
s^l{vi+V2)t=vt, ... (6)
s = vit+^af, (7)
2as=V2'^—vi^ (8)
If the initial velocity is zero then vi of the formulas becomes zero.
EXAMPLES I
1. Compare the velocities of two bodies which move uniformly, one through 5 feet in half a second, and the other through 50 yards in one minute.
2. Compare the velocities of two bodies which move uniformly^ one through 180 feet in one-quarter minute,
4 ELEMENTARY MECHANICS FOR ENGINEERS
and the other through if yards in three-eighths of s second.
3. A body has described 50 feet from rest in twc seconds with uniform acceleration; find the velocity acquired.
4. A railway train travels 162 miles in three hours find the average velocity in feet per second.
5. A particle starts from rest with a constant accelera- tion of 20 feet per second per second. Determine the distance passed over in the fourth and sixth seconds and the total distance passed over in twelve seconds.
6. A train attains a velocity of 54 miles per hour ir 4I minutes after starting from rest. Find the average acceleration.
7. A car, moving at a velocity of 28 miles per hour, i« subjected to a uniform retardation of yi feet per seconc per second by the action of the brakes. How long wil it take the car to come to rest, and how far wiU it trave during that time?
8. A particle starting from rest describes 63 feet in the fourth second. Find the acceleration.
9. A particle describes 72 feet, while its velocity in- creases from 16 to 20 feet per second. Find the whole space described from rest, and the acceleration.
10. A particle, in passing over 9 feet, has its velocity increased from 4 to 5 feet per second. Find the whole space described from rest, and the acceleration.
11. The velocity of a particle changes from 10 to 25 feet per second in three seconds. What is the accelera- tion? When will its velocity be 75 feet per second? What is the total time of motion from rest? What space will it have passed over?
12. A particle is found to be moving at the end oi
KINEMATICS 5
ten seconds with a velocity which, if continued uniformly, would carry it through 45 miles the next hour. Find the acceleration.
13. The velocity of a particle changes from 30 to 20 feet per second in passing over 25 feet. What is the retardation? How long will it be before coming to rest, and what distance will it pass over in that time?
14. With what acceleration and how far must a par- ticle move to have a velocity of 30 miles per hour in half a minute after starting from rest? What retardation would destroy this motion in ten seconds? How far will the particle have moved in the given time?
15. A particle moving with a velocity of 58! feet per second is retarded and brought to rest in 500 feet. What is the retardation?
16. A car moves with a velocity of 60 miles per hour. It is brought to rest in three minutes. Find the retarda- tion, and the space traversed before the car comes to rest.
17. In the eleventh and fifteenth seconds from the beginning a particle moves through 24 and 32 feet re- spectively. Find the initial velocity, and the accelera- tion.
18. A particle describes 81 feet in three seconds. The acceleration then ceases, and during the next three sec- onds it travels 72 feet. Find the initial velocity, and the acceleration.
19. A car moves with an acceleration of i foot per second per second. In what time will the car acquire a velocity of 30 miles per hour, if the initial velocity is 240 feet per minute?
20. A particle starting from rest describes 171 feet in the tenth second. What is the acceleration?
6 ELEMENTARY MECHANICS FOR ENGINEERS
8. Angular or Rotary Motion. Angular motion is of the greatest importance in the study of pulleys, shafts, and flywheels. There are two kinds of rotary motion which we will consider, namely: uniform and uniformly accelerated.
If any point in a rotating body, except points which are on the axis of rotation, describes equal central angles in equa] intervals of time, however small these may be,, the motion is uniform. If the point has its angular velocity increased or decreased by the same amount during consecutive intervals of time, how- ever small these may be, the motion is imiformly accelerated.
9. Angular Measurement: Radian. In angular motion the angle described by any point in a rotating body can be measured by the number of revolutions the body makes, or by a unit called the Radian.
In circular measure, any arc equals the radius of the circle multiplied by the subtended central angle, or
angle subtended = — -. — . . . . (o) radius
If the arc is equal in length to the radius, then the value of the subtended angle is unity, and in this case the angle is called a Radian. In Fig. i the angle AOB is one radian if the arc AB equals in length the radius OA or r. The circiunference of any circle
equals 2Trr, Hence it follows that there are 27r
KINEMATICS 7
radians in 360 degrees. Therefore one radian equals approximately 57.3 degrees, (tt = 3.1416.)
10. Angular Velocity. Suppose a circular plate turns about an axis which is per- pendicular to its plane and passes through its center 0 (Fig. 2). P is any point in the plate, except the axial point 0. ^The straight line OP will turn through so many degrees in a given time. Hence the angular velocity of a particle rotating in a circle equals the angle described divided by the time required, or
angular velocity = -, .... (10) t
where d is the angle described. Since the second is usually taken as the unit of time, the angular velocity is expressed in terms of so many radians per second. For instance, if a flywheel makes 480 revolutions per minute, its angular velocity is i67r radians per second. If the wheel has uniform rotation then the i6ir radians is the actual angular velocity for any second. If the wheel rotates with different speeds then the i67r radians is the average angular velocity. Angular velocity will be denoted by the letter co. If the angular velocity is uniformly accelerated then
average angular velocity = ^^^^-t^. . (n)
2
If the angular velocity of a rotating wheel for the
8 ELEMENTARY MECHANICS FOR ENGINEERS
first second is 27r radians, for the second second 47r radians, for the third second Gir radians, and so on, then at the end of the tenth second the wheel is making lo revolutions per second and its angular velocity is 207r radians per second. From the begin- ning the wheel has made a total of 55 revolutions. The average angular velocity equals |(27r+207r) radians per second, or iitt radians per second. Any point in the wheel, except the axial point, has described an angular distance of iiott radians. 11. Relation between Linear and Angular Velocity. In Fig. 3, ^ is the angle described in a circle of radius r in t seconds. Then the angular velocity of the particle moving from A to B is
Fig. 3. w = - (12)
-n arc ^^_ linear velocity times t
(13)
r^. r linear velocity . .
Therefore « = -, .... (14)
or
"^^^ C15)
12. Angular Acceleration. Angular acceleration is the rate of change of angular velocity and is ex- pressed in radians per second per second. Average angular acceleration is equal to the gain in angular
KINEMATICS
9
velocity divided by the time taken to increase or decrease, which may be expressed as follows:
£02 — COl
average angular acceleration = . . (i6)
t
Denoting angular acceleration by the Greek letter a, formula (i6) becomes
£02 — COl
a-
.... (17)
13. Similar formulas as established in section 7 for linear motion can be established for angular motion.
w2 = coi+a:/, (18)
0 = |(£oi + co2)/, .... (19)
e = oiit-\-\af, .... (20)
2aB = o)2^-03i^ (21)
If the initial angular velocity is zero, then coi of the above formulas becomes zero.
EXAMPLES II
1. A particle describes a circle of radius 5 feet with a uniform linear velocity of 8 feet per second. Find the angular velocity.
2. A wheel turns about its center, making 200 revolu- tions per minute. What is its angular velocity?
3. If the minute hand of a clock is 2 feet long, find the angular velocity and the Hnear velocity of the end of the hand.
4. The lengths of the hour, minute and second hands of a watch are .48, .8 and .24 inch respectively. Find
10 ELEMENTARY MECHANICS FOR ENGINEERS
the ratios of the angular velocities; also of the linear velocities of the ends of the hands.
5. A railway car moves with a speed of 45 miles per hour, and the diameter of the wheels is 3 feet. Find the angular velocity of the wheel.
6. The linear velocity of a point on a rotating body is 72 feet per second, and its distance from the axis of rotation is 6 feet. How long will it take the point to move 60 radians?
7. A flywheel rotates initially 3 times per second, and after twenty seconds it makes 140 revolutions per minute. How many revolutions will the wheel make, and what time will elapse before stopping, if the retardation is uniform?
8. A car is moving with a speed of J mile per minute, and the wheels are 2 J feet in diameter. Find the angular velocity of the wheels. If the car comes to rest in 300 feet under uniform retardation, find the angular retarda- tion.
9. The path of the earth in going around the sun is an ellipse. Assume the path to be a circle of radius 93,000,000 miles. If the time of revolution is taken as 365 days, find the angular and linear velocity.
10. P is a point in a body turning about a fixed axis, and PN is a line drawn from P at right angles to the axis, N being the axial point. If PN describes an angle of 375 degrees in three seconds, what is the angular velocity of the point? If PN is 6 feet in length, what is the linear velocity of the point?
11. A point is describing a circle of radius 42 feet, with a uniform velocity of 12 feet per second. Find the change in its velocity after it has described one-sixth of the circumference.
KINEMATICS 11
12. A flywheel is making x revolutions per second, and after three-fifteenths second it is making 24 revolu- tions per second. If the wheel comes to rest after making a total of 1583^ revolutions, find the value of x and the retardation.
13. A flywheel is making 12I revolutions per second, and after seventeen seconds it is making x revolutions per second. If the wheel comes to rest after ninety- seven seconds, find the value of x^ the retardation, and the total number of revolutions the wheel makes before coming to rest.
14. If a flywheel of radius 6 feet makes 30 revolutions per minute, what is its angular velocity, and what is the linear velocity of a point on its rim? What uniform retardation will bring the wheel to rest in 20 seconds?
15. A pulley 5 feet in diameter is driven by a belt travelling 500 feet a minute. Neglecting the slipping of the belt, find {a) the angular velocity of the pulley, {b) its number of revolutions per minute, and (c) the number of revolutions the pulley makes in 37r minutes.
14. Motion Due to Gravity. The formulas found in section 7 for linear motion may be used to determine motion due to gravity if we take the constant of acceleration as g, where g is approximately 32 feet per second per second.
If a body falls from rest the initial velocity is zero, and the space travelled equals
s = W (22)
If a body is thrown downward with an initial velocity vi the space travelled equals
s^vit^lgi^ (23)
12 ELEMENTARY MECNANICS FOR ENGINEERS
If a body is thrown upward with an initial velocity Vi the space travelled equals
s = vit-^gf (24)
From formula (8) we get
2gS==V2^—Vi^ (25)
The time to the greatest height equals
t = '-^. . ..... (26)
The greatest he^'ght equals
h='f. . ..... (.7)
The distance travelled during any second equals
d = Vi±ig{2t''l) . . . . (28)
EXAMPLES III
1. A body falls from rest. Find (i) the distance fallen in ten seconds, (2) time to fall 10 feet, and (3) the initial velocity if the body should describe 1000 feet in five seconds.
2. A body falls from rest, and during the /th second passes over 144 feet. Find the total distance fallen in the t seconds.
3. How long will it take for a stone to drop to the bottom of a well 144 feet deep?
4. A stone is dropped from a tower 128 feet high, and at the same instant another stone is projected upward from the ground. If they meet half way up the tower, find the velocity of projection of the second stone.
KINEMATICS 13
5. A stone is dropped into a well, and after four seconds the report of its striking the water is heard. What is the depth to the water? Take the velocity of sound as 1 190 feet per second.
6. A stone is thrown downward with a velocity of 96 feet per second and reaches the bottom of a well in three seconds. What is the depth of the well?
7. The greatest height attained by a body projected upward is 400 feet. Find the initial velocity and the time the body is in the air.
8. A ball is thrown upward with a velocity of 100 feet per second. Find how high the ball will rise. At what time will the ball be 50 feet from the ground?
9. A ball is thrown upward with a velocity of 75 feet per second. When will the velocity be 22 feet per second and at what height will the ball be?
10. A stone is thrown upward and during the fifth second travels 16 feet. Find the initial velocity and the space travelled over in five seconds.
11. A bullet shot upward passes a certain point at the velocity of 400 feet per second. Find the time when the bullet will be at a point 1600 feet higher.
12. The surface of the water in a well is 256 feet below the curb. If a stone is dropped into the well, how many seconds will it be until the splash is heard?
13. A body falls from rest and during the tth second falls 176 feet. How long had it been falling before the be- ginning of the tth second?
14. A ball is thrown upward with a velocity of 160 feet per second. At what time will it be 256 feet high, and what is the velocity at this height?
15. A ball is thrown upward with a velocity of 100 feet per second. Find the time occupied by it in its
14 ELEMENTARY MECHANICS FOR ENGINEERS
ascent in describing that portion of its path which Hes between the heights of 60 and 120 feet above the point of projection.
16. A ball is thrown upward with a velocity of 128 feet per second. What is the greatest height attained, and the time required to reach the highest point?
17. Acceleration of gravity on the moon is assumed to be i the acceleration on the surface of the earth. How high will a body rise if it is thrown upward from the sur- face of the moon with a velocity of 40 feet per second?
18. The top girder of a building is x feet above the pavement. A man stands on the girder and throws a stone 144 feet upward. The stone hits the pavement seven seconds after the time of projection. Find the height of the girder above the pavement.
19. A ball is thrown downward and describes 720 feet in t seconds, and 2240 feet in 2/ seconds. Find the time and the velocity of projection.
20. A body falls freely from the top of a tower, and during the last second it falls yif of the whole distance fallen. Find the height of the tower.
21. A ball is thrown upward with a velocity of 64 V2 feet per second, and when it has attained half its greatest height another ball is thrown upward with the same velocity from the same point. Determine when and where they will meet.
CHAPTER II
VECTORS
15. A vector is a straight line having definite length, direction, and sense. The ^' sense " of a line determines whether the direction is to the right or left. Many physical quantities can be expressed by a number denoting so many units. Such quantities are called scalar quantities. But many other quan- tities need to be specified more definitely than by £0 many units. For instance, velocity and accelera- tion of a moving body, action of a push or pull; each has direction as well as magnitude. Such quantities are called vector quantities, and may be represented by straight lines defined as vectors.
If a boat moves northeast with a speed of 12 miles
/V
>£ O
145'
â– >e
Space Diagram Vector Diagram
Fig. 4.
per hour its motion can be represented by a vector (Fig. 4).
In the space diagram the line OB indicates only the direction of motion, while in the vector diagram . 15
16 ELEMENTARY MECHANICS FOR ENGINEERS
the line OA represents both the direction of motion and the magnitude of the speed.
16. Addition and Subtraction of Vectors. To find
p a p' s
Space Diagram Vector Diagram
Fig. 5.
the sum of two vectors p and q (Fig. 5) draw ah equal to vector ^, and from the end h draw he equal in length and parallel to vector q\ join ac. Then ac is the geometric or vector sum of the vectors p and q. We may write the equation
vector />+ vector g = vector r, . . . (29) or
p-^q=r (30)
The value of r may be calculated as follows: from the point c drop a perpendicular cd upon ah pro- duced. Hence ac^ = a(P-\-dc'^, or
r^ = p^+q^+2pq cos 6, ... (31)
'? <i
Space Diagram Vector Diagram
Fig. 6.
e being the angle between the directions of p and q. It is evident that the formula (31) covers all possible
VECTORS
17
in
cases arising from different values of B. Thus, when e = o'',r = p+q; ^ = 9o°,r2 = ^2^^2. e = iSo'',r = p-q.
To subtract vector q from vector p let vector q have opposite sense (Fig. 6).
17. Resultant Velocity. A single velocity which is equivalent to two simultaneous velocities is called their resultant, and these two simultaneous veloci- ties are called the components.
If two component velocities magnitude and direction, by two adjacent sides, AB, AC, of a parallelogram, then their resultant velocity will be rep- resented by the diagonal AD (Fig. 7). By the use of for- mula (31) we can find the value of the resultant.
If a man walks due northeast at a speed of 3 miles per hour across the deck of a vessel going east at 4 miles per hour, the resultant velocity with which
be represented, c
N.E
4S*
Space Diagram
4 vn/'/S A
Vector Diagram
Fig. 8.
the man is moving over the water is 6.478 miles per hour, in the direction N. 7o°53' E. (see Fig. 8).
18. Rectangular Component Velocities. A veloc- ity may be considered as the sum of any number
18 ELEMENTARY MECHANICS FOR ENGINEERS
of component velocities. Given the direction and magnitude of a velocity to find the two components of this velocity having directions at right angles to each other. Take two lines XX' and YY', at right
x\
Y
Fig. 9.
angles to each other, as the axes of reference (Fig. 9). Let V denote the given velocity, and Vx, Vy the horizontal and vertical velocities. The component velocity
F.= F cos <^, .... (32)
and the component velocity
Vy = Vsm<t> (33)
It is easily shown that the resultant velocity of any number of simultaneous velocities is expressed by the formula
/^2=X2+F^ (34)
where
X = z;i cos Q!+z;2 coSjS-f . . . -f-z;„cos7, (35) and
Y=vi sin a+z;2 sin i3+ . . . -{-Vn sin 7. (36)
VECTORS 19
Also the angle </> which the resultant R makes with the X axis can be found by the formula
Y tan<^=- (37)
EXAMPLES IV
1. Add the vectors (10°, 15); (0°, 20).
2. Subtract the vector (45°, 12) from the vector (60°, 18).
3. Add the vectors (0°, 12); (45°, 8); (135°, 10) and
(225°, 4).
4. A steamer is headed due north at 12 miles per hour. The current runs northwest at 4 miles per hour. Find the true velocity of the steamer in magnitude and direc- tion.
5. A steamer is going northeast at the rate of 24 miles per hour. Find the northern and eastern component.
6. A stone is projected at an angle of 45° to the hori- zontal with a velocity of 600 feet per second. Find the horizontal and vertical component velocities at point of projection.
7. In order to cross a river at right angles, the river flowing with a uniform velocity of 2 J miles per hour, in what direction should a swimmer head if his velocity in still water is 5 miles per hour, and how long will it take hun to cross, if the river is 200 yards wide?
8. A body is moving in a straight line with a velocity of 20 feet per second; find the component of its velocity in a direction at an angle of 60° to its direction of motion.
9. A vessel is steaming in a direction due north across a current running due west. At the end of two hours the
20 ELEMENTARY MECHANICS FOR ENGINEERS
vessel has made 22 miles in a direction 35° west of north. Find the velocity of the current, and the rate at which the vessel is steaming.
10. A man wishes to cross a river to an exactly opposite point on the other bank. He can pull the boat with three times the velocity of the current. Find the inclination to the current at which he must keep the boat pointed.
11. A certain point possesses velocities represented by 3, 19 and 9 inclined at angles of 120° to one another. Find the resultant velocity in magnitude and direction.
12. A vessel is headed due north at the rate of 12 feet per second; the current flows east with a velocity of 5 feet per second, and a sailor is climbing a vertical pole at the rate of i foot per second. Find the velocity and the direction of the sailor in space.
13. An airship has a velocity of 40 miles per hour heading due south. The wind is blowing southeast with a velocity of 30 miles per hour. Find how far the air- ship travels in one hour and its direction.
14. In problem (13) suppose the wind to be blowing northeast instead of southeast at the same velocity. Find the distance travelled in one hour and its direction.
15. Add the vectors (30°, 15); (60°,- 15); (135°, 10).
16. Subtract the vector (10°, 14) from the vector
(225°, 14).
17. A weather vane on a ship's mast points northeast when the ship is headed due east at 16 miles per hour. If the velocity of the wind is 20 miles per hour, what is the true direction of the ship?
18. Two vessels are 10 miles apart at points A and B, The vessel at A is headed N. 30° E. and is steaming with a velocity of 12 miles per hour. Find in what direction the vessel at B must head at the same moment, if it ha^
VECTORS 21
a velocity of 15 miles per hour, in order to come in collision with the first vessel. Find also at what angle it will strike the first vessel and the time of travel.
19. Motion on a Smooth Inclined Plane. Let AB (Fig. 10) be the vertical sec- tion of a smooth plane mak- ing a given angle 6 with the horizon tal. Let P be the particle which moves on the plane.
The downward accelera- tion of the moving particle
due to gravity is g. Resolve this acceleration into its components, and it is evident that the acceleration along the plane is g sin 0, Hence in the formulas for a falling body, we need only substitute g sin d iorg,
EXAMPLES V
1. Show that the velocity acquired by a body in slid- ing from rest down a smooth plane of length k is the same as that acquired by a body in falling freely through the vertical height which the body is above the horizontal.
2. A particle slides from rest down a smooth inclined plane which is 20 feet long and 14 feet high. What is its velocity when it reaches the ground, and how long does the sliding take?
3. A particle sliding down a plane 24 feet long ac- quires a velocity of 12 feet per second. Find the inclina- tion of the plane.
4. A body slides from rest down a given plane. Com-
22 ELEMENTARY MECHANICS FOR ENGINEERS
pare the times of describing the first and last thirds of its length.
5. Show that the horizontal velocity of a body mov- ing down a plane inclined at an angle 6 to the horizontal is gt sin ^ cos ^, and the vertical velocity is gt sin^ 0.
6. A body slides from rest down a plane inclined at an angle ^ = tan~^f to the horizontal. Find (i) its velocity at the end of ten seconds; (2) the horizontal and vertical components of this velocity; (3) the distance travelled.
7. A railroad track is inclined at an angle ^=tan-^ XF(y to the horizontal. A car starts down the track; find how far the car will go in four minutes, and the veloc- ity it acquires.
8. In problem 7 suppose the car has an initial velocity of 30 miles per hour. Find the distance travelled and the velocity acquired.
9. A smooth plane is inclined at an angle of 30° to the horizontal. A body is projected up the plane with an initial velocity of 160 feet per second. Find when it is 288 feet from the starting point.
10. A plane 18 feet long is inclined 15° to the hori- zontal. With what velocity must a body be projected^ in order to just reach the top?
11. A body is projected down a plane inclined 12° to the horizontal. If the body is projected with an initial velocity of 10 feet per second, how far will it go in four seconds?
12. Find the velocity with which a body should be projected down an inclined plane of length 20 feet, so that the time along the plane shall be equal to the time of falling vertically; the angle of inclination of the plane being 30°.
VECTORS
23
13. A plane 65 feet in length is inclined at an angle 0 = tan~i T2- to the horizontal. A body slides down the plane, and then passes without loss of velocity onto the horizontal plane; after how long a time from the beginning of the motion will it be 200 feet from the foot of the plane?
14. A plane is 288 feet long and 64 feet high. Show how to divide the plane into three parts so that a ball may slide over the portions in equal times, and find these times.
15. A body slides down an inclined plane, and during the fifth second after starting passes over a distance of 72 feet. Find the inclination of the plane.
• 20. Projectiles. The path of a projectile undisturbed except by the action of gravity is a parabola. The pro-
f < ty time
!//! i ^! iXi
£Jl — M. i. ^ I 1 \C
Fig. II.
jectile is subject to two component motions, the hori- zontal velocity which is constant in both magnitude and direction, and the vertical uniformly accelerated velocity which is constant in direction but not in magnitude. (Fig. 11.)
24 ELEMENTARY MECHANICS FOR ENGINEERS
In the preceding figure, 6 is the angle of projection. AC is the horizontal range, O is the highest point of the path. The time it takes to travel from ^ to C is called the time of flight.
21. Velocity and Direction of Motion after a Given Time has Elapsed. Let u be the velocity, and B the angle which the direction of motion at the end of time t makes with the horizontal. Now u cos 5 = horizontal velocity at end of time t, . (38)
= vi cos 6 J the constant horizontal velocity. . (39)
Again w sin j5 = vertical velocity at end of time /, . (40)
= vi sin d—gt (41)
Hence, and
1^2 =vi^ — 2Vigt sin 6-\-gH^j , . (42)
. o z;i sm e—gt f .
tan 5 = -^ (43)
Di cos Q ^
22. Velocity and Direction of Motion at a Given Height. The horizontal velocity is constant and equal to u cos B=vi cos 6, and the vertical component is u sin B, From (25), section 14,
Hence and
u sin B =Vvi^ sin^ d - 2gh, . . (44)
u^=vi^-2ghy U5)
r» Vvi^sin^ d — 2gh f .N
tan 5 = — ^ .... (46)
V\ cos Q
VECTORS 25
The greatest height = ^ . . (47a)
rr,, -, . , , 2^)125111^005^ 1^1^51112^ / ,s
The horizontal range = =â– . (470)
The time of flight = ^'^^^^ . . (47^) g
The maximum range Is easily seen to be — , since
o
sin 2^ is greatest when 0 = 45°.
Suppose the angle of projection is — 6. Then
2
z^i^ sin 2/ — 6] o . / \ o .
-p \2 / _vi^ Sin (tt — 2^) vi"^ sin 2S
^^^^~ g ~ g ~ g
which shows that the same horizontal range is given
for d and — 6. 2
All of the preceding discussion has been entirely theoretical, and of no practical use, since the resist- ance of the air plays a very important part in the study of projectiles.
EXAMPLES VI
1. A bullet is projected with a velocity of 500 feet per second at an angle of 40° with the horizontal. Find (i) the greatest height attained, (2) the range and time of flight, (3) the velocity and direction of motion of the bullet when it is at a height of 400 feet.
2. A ball is thrown with a velocity of 100 feet per second; find the maximum range, and the two direc-
26 ELEMENTARY MECHANICS FOR ENGINEERS
. tions in which the ball may be thrown so as to give a range of 144 feet. Also the time of flight for the maxi- mum range.
3. A ball is thrown horizontally from the top of a tower 100 feet high, with a velocity of 144 feet per second. Find (i) the time of flight, (2) distance from the foot of the tower of the point at which the ball hits the gromid, and (3) its velocity when it hits the ground.
4. From the top of a tower 96 feet high, a ball is thrown with a velocity of 96V3 feet per second, at an angle of 60° with the horizontal; find the distance from the foot of the tower of the point where the ball hits the ground.
5. What is the time of flight of a projectile, when the angle of projection is 45°, and the range is 6000 feet?
6. What is the range of a projectile when the angle of projection is 30° and the initial velocity is 200 feet per second?
7. The range of a projectile is 3250 feet and the angle of projection is 32°. What is the time of flight?
8. A bullet is projected from a gun and strikes the ground at a distance of 2000 feet. Find the initial velocity and the time of flight, if 6 is 30°.
9. When the angle of elevation is 40*^ the range is 2449 feet. Find the range when ^ = 29!°.
10. If a man can throw a ball 90 yards, how long is it in the air, and to what height does it rise?
11. The velocity of a projectile when at its greatest height is \/2/$ of its velocity when at half its greatest height. Find the angle of projection.
12. A ball is projected up an inclined plane of inclina- tion 20°. The angle 6 is 50°, and the initial velocity is 600 feet per second. Find the range on the plane and the time of flight.
CHAPTER III KINETICS
23. Mass. The mass of a body is defined as the quantity of matter in the body. The mass of any body is determined by a beam scale. Such materials as coal, wood, iron, gold, etc., represent certain amounts of mass. With the exception of gases, mass is a visible and an invariable quantity.
24. Weight. All material particles of matter are attracted toward the center of the earth by a force called gravity. In any given body the lines of action of the forces acting on the particles which make up the body are apparently parallel to each other. The resultant of all the gravitational forces acting on any body is called the weight of the body. In other words, the force with which any body is attracted toward the center of the earth is the weight of the body. Weight is determined by a spring balance. Since the weight depends upon the distance of the body from the center of the earth, it is variable. It is also invisible.
25. Force. Force may be defined as that which changes or tends to change the state of rest or motion of a body. Forces are generally represented by a push or pull. If a force produces motion then there must be an acceleration of the body moved. Also the
27
28 ELEMENTARY MECHANICS FOR ENGINEERS
amount of acceleration produced depends upon the amount of mass to be moved. Hence the force varies as the product of the mass and the acceleration, or F = kma, If proper units of force, acceleration, and mass are chosen, then k = i and
Force = mass X acceleration, or
F = mXa. (48)
The unit of force is the pound, which is that force which will accelerate a mass of one pound g feet per second per second.
Since acceleration is a vector quantity, and force is proportional to the acceleration, then force is a vector quantity, having both direction and magnitude.
26. Relation between Mass and Weight. Using the above definition of force one can write the follow- ing equation :
Weight = mass X^, .... (49)
or mass = — (i;o)
g
In engineering work, W/g is used to represent the mass of a body, W being given in pounds. For instance, a body weighing 100 pounds has a mass of 100/32 pounds.
27. Newton's Laws of Motion.
I. Every particle of matter continues in its state of rest or motion unless it is acted upon by some external force.
KINETICS 29
II. The acceleration of a body is proportional to the force appUed, and takes place in the direction of the straight line in which the force acts. III. To every action there is an equal and opposite reaction.
The first law refers to the property of matter, called inertia, to resist being put into motion, or to resist having its motion changed. The second law gives us a means to measure the accelerating force, as shown in the preceding section, that is, F = mass X accelera- tion. The third law asserts that forces always occur in pairs, which is called the two-sided nature of every force. No force can be exerted unless there is some resistance to overcome. There will be no action unless there is something to act upon which will react.
The third law is quite misleading to beginners. If the law is true, let the student explain why it is that a wagon moves if pulled by a horse. In other words, explain the reaction.
28. Momentum and Impulse. The momentum of a moving body is defined as the product of the mass of the body and its velocity, or
Momentum = — X^. , ... (51)
The impulse of a force is the product of the force and the time during which the force acts. The term " impulse '' is usually considered only in cases where
30 ELEMENTARY MECHANICS FOR ENGINEERS
the force acts for a very small interval of time, as in the case of blasting with powder, collisions, and the firing of guns.
^. - total impulse , .
Trnie average force = ^ —. — : . . ( c 2 )
time required ^
If a force F acts on a mass of Wig pounds for / seconds and causes its velocity to be changed from z'l to V2^ the change in momentum during that time is TF/g (2^2— ^1). If the acceleration a is constant, then the change in velocity (2^2—^1) =0^1* Hence
1
But
therefore
—{v2-n)= — at .... (53) g g
— a = forceF, (54)
g
— (zJ2— ^1)= force X time, . . (55)
which means that the change in momentum is equal to the impulse.
Example i. A car weighing W pounds is set in motion by a uniform force of Pi pounds and in h seconds it attains a velocity of V feet per second. A retarding force of P2 pounds brings the car to rest in t2 seconds. Find the relation between Pi, P2, andF.
Solution, During the time h the gain in mo-
W . mentum is — V units, and the impulse is Pi/i, hence
g
KINETICS 31
W Pih=—V (a;
g
W During the time /2 the gain in momentum is V
units, and the impulse is — P2/2, hence
P2/2 = -F (Jb)
Therefore
Pl/l=P2/2=-/^(/l+/2). . . {C)
(Multiply {a) by P2, {h) by Pi, and add.)
Example 2. A car weighing 64 tons (20chd lbs.) is acted upon by a net force of i ton. How long will it take to attain a velocity of 30 miles per ^our start- ing from rest, and how far will it have moved?
Solution, Let / be the time required in seconds, * then the impulse is 2000^ units.
Thirty miles per hour is 44 feet per second, so the gain in momentum is
or
therefore
Hence
64-2000 .. 44,
, 64 • 2000
2000^=—'^ 44.
32
^ = 88 seconds.
32 ELEMENTARY MECHANICS FOR ENGINEERS
The average speed is half the maxiraum, and the distance travelled in feet is
22-88 = 1936 feet.
Example 3. A weight of 20 pounds is acted upon by a constant force which in five seconds produces a velocity of 15 feet per second. Find the force, if the weight starts from rest.
Solution, The acceleration is 3 feet per second per second, hence -F = ft'3 = if pounds.
EXAMPLES VII
1. Of two bodies moving with constant velocities, one describes 36 miles in one hour and twenty min- utes, the other 55 feet in ij seconds; the former weighs 50 pounds, and the latter 72 pounds. Compare their momenta.
2. How far will a cannon ball weighing 10 pounds travel in one minute, supposing it to possess the same momentum as a rifle bullet of 2 ounces moving with a velocity of 1000 feet per second?
3. Calculate the momentum of a hanuner of 5 tons, let fall vertically half a foot.
4. A body of snow 28 pounds in weight falls from the roof of a house to the ground, a distance of 40 feet. What is the momentum?
5. Compare the amounts of momentum in a pillow of 20 pounds which has fallen vertically i foot and an ounce bullet moving at 200 feet per second.
6. What force will in one minute give a weight of i ton a velocity of 10 miles per hour?
KINETICS 33
7. How long must a force of 20 pounds act on a weight of 512 tons to give it a velocity of i foot per second?
8. The moving force is 3 pounds and the weight acted upon is 10 pounds. Find the distance described in ten seconds.
9. A force of 2 pounds acts on a weight of 40 pounds for 30 seconds. Find the velocity acquired and the space described.
10. A hammer weighing 3 tons falls vertically 4 feet. What is the momentum? If the force of the blow is ex- pended in .018 second, what is the average force of the blow?
11. A car weighing 100 tons has a constant resistance of 20 pounds per ton. What force will be required to give the car a velocity of 40 miles per hour in two min- utes?
12. An engine exerts a constant draw-bar pull of 14 tons on a car weighing loVioi tons up an incline of 10 per cent grade, and the resistance of the rails, etc., amounts to 12 pounds per ton. How long will it take to attain a velocity of 30 miles per hour, and what dis- tance will be travelled?
13. How long will it take the engine and car in problem 12 to go 2048 feet up the incline, starting from rest and coming to rest at the end without use of the brakes?
14. A car travelling 60 miles per hour is brought to rest by a uniform resisting force in 4096 feet. Find the total resisting force in pounds per ton.
15. A bullet weighing i ounce enters a piece of wood with a velocity of 2000 feet per second and penetrates it 6 inches. What is the average resistance of the wood in pounds to the penetration of the bullet?
16. A bullet weighing i ounce leaves the barrel of a
34 ELEMENTARY MECHANICS FOR ENGINEERS
gun 3 J feet long with a velocity of 1200 feet per second. What is the impulse of the force produced by the dis- charge? If it takes .003 second for the bullet to traverse the barrel, what is the average force exerted on it?
29. The third law of motion expresses the con- servation of energy. For instance when a projectile is fired from a cannon, the impulse or change of momentum of the shot due to the explosion is of equal ajuount to that of the recoiling cannon in the opposite direction.
30. Motion of two weights connected by a light inextensible string passing over a small smooth peg or
pulley. In Fig. 12 suppose W2 is the greater of the two weights. The total accelerating force is W2 — W1, and the total mass is W1+W2
g hence
acceleration a =
Fig. 12.
W2-W1 W1 + W2
g- (56)
Consider each weight separately and the upward
accelerating force on the right is T- tension of the string throughout; downward acceleration is W2 — T. section 25,
W2
T = — a.
Wi, T being the on the left the Hence by (48),
(57)
KINETICS
35
and
i
Hence, it follows that
2W1W2 W1+W2
Also and
^7^ V W2-W1 ^,
Velocity = z> = ^^^ , ^^^ gh •
space â– â–
W2-\-Wi
W2^Wl
2{Wl + W2)
g/^. . .
(58)
(59) (60) (61)
31. Again, suppose Wi to slide along a smooth horizontal plane or table. In Fig. 13 let W2 be
*^
0;
m
Fig. i3«
suspended by a string passing over the pulley on the edge of the table. The accelerating force is W2, a;id the total mass in
motion is -^^ — ^, therefore g
W2
a=
W1+W2
g
(62)
36 ELEMENTARY MECHANICS FOR ENGINEERS
Now, since
_ accelerating force on TFi _ T ^" mass of If 1 ~Wi^'
it follows that
.r W1W2 (. ,
What would be the acceleration if W\ were opposed by a given horizontal force F?
Example i. A hammer weighing 10 pounds strikes a nail weighing i ounce with a velocity of 20 feet per second and does not rebound. The nail is driven into a block of wood which offers a uniform resistance of 400 pounds to the penetration of the nail. How far will the nail penetrate the wood?
Solution. Momentum of hammer before impact
_ 1 0 . ^>,, _ 2_5
= -32^-20—4^. Momentum of hammer and nail after impact
= "^ • F, where F is the initial velocity after
the impact.
Hence — — —^ F. Therefore F equals feet
4 i6*32 161
per second. Let t be the time of penetration. Then
Impulse 400 t = — 20. 32
Hence / -equals -h second. The average velocity
during the penetration is | F or -— - feet per second.
KINETICS 37
The distance moved by the nail is
12 1600 I o • T.
— • • — = 1.8 inches.
I lOI 64
Example 2. A cannon weighing 40 tons fires a 1500-pound shell with a velocity of 1200 feet per second. With what initial velocity will the cannon recoil? If the recoil is overcome by an average force of 75 tons, how far will the cannon travel and how long will it take?
Solution. Momentum of projectile = momentum of the cannon.
I i;oo 40 • 2000 Tr -^ — -1200 = ^^^ F.
32 32
(F is the initial velocity of cannon.) Therefore
V = — = 22I feet per second.
40 • 2000
Let / be the time of the recoil.
Impulse of the retarding force = momentum of the shell.
, IS00-I200
75' 2000-/ = -^^ .
32
Hence
/ = f second.
Therefore the distance moved by the cannon
= -• — •^ = 4.22 feet. 228
Example 3. Two weights, W2 of 24 pounds and Wi of 18 pounds, are connected by a string passing
38 ELEMENTARY MECHANICS FOR ENGINEERS
over a smooth pulley. Find (i) tension of the string, (2) the acceleration and (3) the space travelled in 3 seconds. Solution.
(i) T = ^— — = 20y pounds.
24+18 •
(2) a = — --32=4* feet per (second)^.
24+18
(3) space = i'^^— ^- 32-32 = 2of feet.
2 24+10
EXAMPLES VIII
1. A shot whose weight is 400 pounds is projected from a gun weighing 50 tons with a velocity of 1000 feet per second. Find the initial velocity of the cannon.
2. A bullet weighing i ounce is fired from a gun which weighs 10 pounds. The velocity of the bullet is 1000 feet per second. Find (i) the initial velocity of the gun, (2) the time of recoil, and (3) the space travelled by the gun, if the recoil is overcome by a time average force of 10 lbs.
3. A cannon weighing a tons fires a 1000-pound shell with a velocity of 1000 feet per second. The cannon recoiled with an initial velocity of 20 feet per second. What is the weight of the cannon?
4. A gun fires 480 bullets per minute, each bullet weighing i ounce and having a horizontal velocity of 1600 feet per second. Find the averagd force exerted on the gun.
5. A shot of 1000 pounds is fired from a 60- ton cannon with a velocity of 1800 feet per second. The recoil is
KINETICS 39
resisted by a constant pressure of 15 tons; through how many feet will the gun recoil?
6. A shot weighing 1000 pounds is fired with a velocity of 1600 feet per second from a cannon weighing 80 tons. What steady pressure will stop the cannon after a recoil of 4 feet?
7. A weight of 10 pounds falling freely a distance of 12 feet drives a nail, weighing i ounce, 3 inches into a block of wood. Find the uniform resistance.
8. A bullet strikes and becomes imbedded in a sus- pended target. The masses of the shot and target are as I : 100, and the target moves with a velocity of 10 feet per second as soon as the shot is imbedded. What is the initial velocity of the bullet?
9. Two weights of 3 pounds and 5 pounds are sus- pended by a string passing over a pulley. Find the acceleration, and the tension of the string.
10. An elevator car weighing i ton is made to ascend with a constant acceleration of 16 feet per second per sec- ond. What is the tension of the cable?
11. A man of weight 160 pounds stands on the plat- form of an elevator cage which has an upward accelera- tion of 2 feet per second per second. What pressure is exerted on the platform?
12. The weights at the extremities of a string passing over a pulley are 500 and 502 pounds. After 3 seconds of motion, 3 pounds are removed from the greater weight. What time will elapse before the weights are at rest again?
13. Two weights of 5 pounds and 4 pounds, together pull one weight of 7 pounds over a smooth pulley by means of a connecting string. After descending a given space the 4-pound weight is instantly taken off. Find through what space the remaining 5 pounds will descend.
40 ELEMENTARY MECHANICS FOR ENGINEERS
14. A given weight of 500 pounds is placed on a smooth horizontal table and is pulled by a weight of 250 pounds which hangs vertically from the edge of the table. Find the acceleration, and the tension in the rope connecting them.
15. A weight of 60 pounds is resting on a smooth table and connected to a weight of 20 pounds hanging vertically from the edge of the table. What force must oppose the motion of the 60-pound weight in order to give an acceleration of Jg?
16. A smooth inclined plane whose height is J of its length has a small pulley at the top over which a string passes. To one end of the string is attached a weight of 12 pounds which rests on the plane, while from the other end, which hangs vertically, is suspended a weight of 8 pounds. Find the distance travelled by either weight in 5 seconds.
17. In example 16 what weight must be added to the 8-pound weight in order that a distance of 144 feet will be travelled in 6 seconds?
18. A body of weight 9 pounds is placed on a smooth table at a distance of 8 feet from its edge, and connected by a string passing over the edge, with a weight of i pound. Find (i) the acceleration, (2) the time that elapses before the body reaches the edge, and (3) its velocity upon leaving the table.
19. A body of weight 10 pounds is placed at the bottom of a plane, inclined 30° to the horizon and of length 12 feet, and is connected with a weight of 6 pounds by a string passing over the top of the plane. Find the acceleration, and the time it takes for the lo-pound weight to reach the top.
20. In problem 19 what weight attached to the vertical string will produce no motion?
KINETICS 41
32. Work. When a force acts upon a body and causes motion against resistance it is said to do work. Work is measured by the product of the force and the distance through which the body moves, or
W==Fs (64)
33. Unit of Work. Since work is measured by the product of the force and the distance, the unit is taken as that amount of work done by a unit force acting through a unit distance. The most generally used unit is the work done by a force of i pound acting through a distance of i foot, and it is called the foot-pound of work.
34. Power. Power is the rate of doing work. In most cases the horse-power* is the unit, and is equiv- alent to 33,000 foot-pounds per minute.
EXAMPLES rX
1. A man weighing 200 pounds walks to the top of a hill which is 3300 feet high. Find the work done against gravity.
2. The highest point attainable in the Washington monument is 500 feet above the base. How much work does a man do who weighs 165 pounds in going to the top?
3. A man can pump 25 gallons of water per minute to the height of 15 feet. How much work is he able to do in a lo-hour day. Assume i gallon of water to weigh 8 pounds.
4. 100 cubic feet of water is lifted to a height of 200
* Denoted by H.P.
42 ELEMENTARY MECHANICS FOR ENGINEERS
feet during each minute. What must be the H.P. of an engine to do this work, if one-fourth of the power is lost through friction?
5. What is the H.P. of an engine capable of raising 10 tons of coal through a distance of 33 feet in 2 minutes?
6. The inclination of a mountain path to the horizon is 30°. How much work is done against gravity by a man of weight 150 pounds, in walking a mile along the path?
7. A tank of 22,500 gallons at an elevation of 66 feet is to be filled with water in 2 hours by a steam pump. Find the H.P. of the pump.
8. The efficiency of a given pump is 75 per cent. 52.8 cubic feet of water is raised to a height of 300 feet in one minute. What is the H.P. of the pump.
9. What power must be expended to pull at 15 miles per hour a 170-ton car up a plane inclined tan"^ ^f to the horizon?
10. What is the H.P. of an engine which keeps a train of 180 tons weight moving at a uniform rate of 60 miles per hour, the total resistance being 10 pounds per ton?
11. Two men put a car weighing 5 tons into motion by exerting on it a force of 80 pounds. The resistance being 10 pounds per ton, how far will the car have moved in one minute?
12. A train of weight 200 tons, is ascending an incline of which the inclination is sin~^ 2^^, and the total re- sistance is 20 pounds per ton. If the engine is of 240 H.P. and working at full power, find the speed at which the train is moving.
13. A stream delivers 6000 cubic feet of water per minute to the highest point of a water wheel 66 feet in diameter. If 80 per cent of the available work is use-
KINETICS 43
fully employed, what is the horse-power developed by the wheel?
14. A car of weight 200 tons is drawn up a plane, inclined at an angle sin~^ -5^0 to the horizon, at a rate of 30 miles per hour by an engine of 400 H.P. Find the total resistance per ton.
15. A stream of water is 5 feet wide and 3 feet deep, and flows J mile per hour. The water falls 10 feet and in so doing passes over a wheel of 70 per cent efficiency. It takes I H.P. per minute of work to grind one bushel of corn. How much would one pay at 2 cents per bushel for grinding one hour?
16. A moving stairway moves at the rate of 2 feet per second. The maximum number of people who can ride on the stairway at any one time is 120. The length of the stairway is 50 feet, and its vertical rise is 30 feet. (a) Neglecting friction and assuming that the average weight of one person is 150 pounds, what is the maxi- mum power required to operate the stairway? (b) If the efficiency of the system is 60 per cent, what is the maximum power required?
35. Energy is defined as the capacity for doing work. There are two kinds of energy, called potential and kinetic.
Potential Energy is called energy of position, and it is measured by the product of the weight and the vertical distance through which the weight has been lifted, or
V.E. = Wh (65)
Kinetic Energy is called energy of motion, and it is
44 ELEMENTARY MECHANICS FOR ENGINEERS
measured by one-half the product of the mass and the square of the velocity, or
K.E.=|-z;2 (66)
g
Hence it is easily proven that the work done is equal to the change of kinetic energy; that is
K. E. = force X distance, or
K,E.^i-(v2^-vi^)^FXs, . . (67) g
It is evident that problems which are solved from the consideration of changes of momentum, might often be solved by considering the change of kinetic energy.
Example i. A weight of 3 tons has been lifted
vertically a distance of 8 feet. What is the potential
energy which the weight possesses at this height?
Solution.
T.E.^Wh,
Therefore
P. £. = 3X2000X8=48,000 ft.-lbs.
Example 2. A shell of weight 1024 pounds leaves the muzzle of a cannon with a velocity of 1500 feet per second. Find the work done by the shell, and the kinetic energy of the cannon at the instant it begins to recoil if its weight is 30 tons.
Solution, The work done by the shell is equal to its kinetic energy.
KINETICS 45
Hence
K. E. = --— .72 = -. ^t^. (1500)2=36,000,000 ft-lbs. 2 g 2 32
Smce the momentum of the gun equals the momentum of the projectile, the initial velocity of the gun is 25.6 feet per second, and the
K. E. =1.30- 2000 (25.6)2 = 614,400 ft.-lbs. 2 32
Example 3. A bullet weighing i| ounces moving at a velocity of 1024 feet per second overtakes a block of wood which weighs 8 pounds and having a velocity of 32 feet per second. Find the velocity of the wood after impact and how much kinetic energy has been lost, supposing the block does not rotate?
Solution, Let F = velocity of the bullet and block after impact.
Momentum of bullet = -^ ^ = 3 imits,
32 32
Q
Momentum of block = — ^2=8 units. 32
Hence total momentum before and after impact = 11 units.
Therefore
V = ~^~^ = 43 48 feet per second. 0+3%
K. E. of bullet=i-2-.— (1024)2 = 1536 ft.-lbs. 2 32 32
46 ELEMENTARY MECHANICS FOR ENGINEERS
K. E. of block =-. A. (32)2 =128 ft.-lbs. 2 32
Total K. E. before impact = 1664 ft.-lbs.
Total K. E. after impact = -.?±^- (4348)2
2 32
= 239.19 ft.-lbs.
Loss of K.E. after impact = 1664— 239.19
= 1424.81 ft.-lbs.
EXAMPLES X
1. The hammer of a pile driver weighs i ton. Find Its potential energy when it has been lifted 10 feet.
2. A body of weight 320 pounds is thrown up ver- tically with a velocity of 32 feet per second. What is its kinetic energy (i) after half a second, and (2) after one second?
3. What is the kinetic energy of a street car weighing 3.62 tons when it is moving at a speed of 7 J miles per hour, and is laden with 20 passengers averaging 150 pounds each in weight?
4. A ball weighing 5 ounces, and moving with a velocity of 1000 feet per second, strikes a shield, and after pierc- ing it moves on with a velocity of 400 feet per second. JIow much energy has been expended in piercing the shield?
5. What is the kinetic energy of a hammer of 5 tona which falls 4 feet?
6. A 32-pound ball is thrown vertically upward with a velocity of 20 feet per second. What is its kinetic energy when it has gone 5 feet?
7. A bullet of weight 2 ounces is fired into a target with
KINETICS 47
a velocity of 1280 feet per second. The weight of the target is 10 pounds and it is free to move. Find the loss of kinetic energy by the impact.
8. A shot of 1000 pounds moving at 1600 feet per second strikes a fixed target. How far will the shot penetrate the target, exerting upon it an average force of 12,000 tons?
9. Find the initial velocity of a shell weighing 1024 pounds if the total kinetic energy is 32,000 foot-tons.
10. An ounce bullet leaves the muzzle of a rifle with a velocity of 1024 feet per second. If the barrel is 4 feet long, calculate the pressure of the powder, neglect- ing all friction.
11. Two weights of 5 and 10 pounds respectively impinge directly, moving with velocities of 8 and 10 feet per second. Find the common velocity after impact, and the loss of kinetic energy.
12. What space average force will bring to rest in 22 feet the laden car in example 3?
13. A bullet weighing 2 J ounces leaves a gun with a velocity of 1550 feet per second; the length of the barrel is 2J feet. Find the average accelerating force upon the bullet within the barrel.
14. A locomotive of weight 15 tons acquires a speed of 20 miles per hour in one mile, under the action of a constant difference of a moving and a resisting force. Find the accelerating force.
15. A train of weight 300 tons is travelling with a speed of 60 miles per hour. Find the force which will stop it in 500 yards.
16. An engine with all brakes on and steam shut off reduced its speed from 60 to 20 miles per hour in 800 yards. At the instant the speed was 20 miles per hour the engine
48 ELEMENTARY MECHANICS FOR ENGINEERS
collides with a car. How much additional space would have saved the collision?
17. A train of weight 100 tons, is travelling at the uniform velocity of 30 miles per hour on a level track, and the total resistance is 20 pounds per ton. Part of the train of weight 20 tons becomes detached. Find how much the train will have gained on the detached part after one minute, and the velocity of the train when the detached part comes to rest.
18. A jet of water issues in a parallel stream, at 33 feet per second, from a round nozzle of radius 2 inches. What is the horse-power of the jet, and what thrust should a fireman give to support the jet?
19. A weight of 3 tons falls a distance of 3 feet, and in so doing lifts a weight of 168 pounds to a height of 100 feet. Find the work done by the fall of the 3 -ton weight, and what per cent of the work is effective?
CHAPTER IV
STATICS
COMPOSITION AND RESOLUTION OF FORCES
36. A force is completely determined when its direction, magnitude, and point of application are specified.
Equilibrium exists when two or more forces act upon a body, or a particle, and the body is kept in a state of rest.
Since force has magnitude and direction it can be represented by a vector. Hence the composition of forces may be found by the addition of vectors.
37. A resultant force is a single force which if appHed to the particle will have the same effect as any number of concurrent forces. _
38. Resultant of two forces acting at a point but inclined at an angle d to each other.
p 5
Space Diagram Vector Diagram
Fig. 14. Fig. 15.
Represent the forces in direction as in Fig. 14, and in magnitude and direction as in Fig. 15.
49
50 ELEMENTARY MECHANICS FOR ENGINEERS
Let AB represent the force P in magnitude and direction. Let BC represent the force Q in magnitude and direction. Now vector ^jB-|- vector 5C = vector AC J the resultant. Since in the triangle ABC the two sides AB, BC, and the angle d are given, the numerical value of ^C may be found by use of trigonometry. Hence
i?2=p2^Q2+2P2cos0. . . . (68)
Also, using the same scale of units as for P and Q, R may be measured in the same manner. This gives a check upon the computation.
39. Resultant of any number of forces in one plane acting upon a particle.
From trigonometry, the sum of the projections of a broken line is equal to the projection of the straight line joining the ends of the broken line.
Since a force may be broken up into its com- ponents, in the same manner as velocity, the pro- jection upon a given Hne of the vector representing the resultant is the sum of the projections upon the same line of the component forces.
If three forces P, Q, and R, incHned at angles of a, /3, and y respectively, to the X-axis, act upon a particle the resultant is easily found by means of the previous theorem.
Let the forces P, Q, and 5 act upon a particle at Oj as represented in Fig. i6.
Now __
R^=OD^+CD^,
STATICS
51
but OD=Oa+Ah+Bc
= F cos a+Q cos /3+5 cos y=Xy (69) and CD = aA+bB+cC
=P sin a-\-Q sin ^-\-S sin 7 = F. (70) Hence
i?2=X2+F2, (71)
and
^ ^ CD Y
tan rf>= — = — ,
. (72)
where <^ is the angle between the resultant and the X axis.
space Diagram
Vector Diagram
Fig. 16.
If there are n forces acting on the particle, then
F = SP„sin0«
(70a)
X=IlPn COS On, ... . (6ga) 1
^ ^ F
tan <^ = Y-
(72a)
52 ELEMENTARY MECHANICS FOR ENGINEERS
The letter sigma, S, stands for " the sum of all such
EXAMPLES XI
1. Two forces, P and Q, act on a particle. What is their resultant, (i) when they act in the same direc- tion, and (2) in opposite directions?
2. In the following examples, P and 0 denote forces, 0 the angle between the forces.
(a) P = 2o; Q = i5; ^ = 90°; ^.AR
(b) P=i2- R = iyy 61 = 90°; finde {c) P = io; <3 = 4o; 0 = 6o°; find i?
(d) P= 8; g= 4; 0=120°; findi?
(e) P = io; Q= 6; i? = i4; find <9.
3. Two equal forces act on a particle; find the angle between them when the square of the resultant equals their product.
4. Two forces of 84 and 187 pounds have a resultant of 205 pounds. Find the angle between the forces.
5. Let ABC be a triangle and O the middle point of the side AC. If the three forces, represented in mag- nitude and direction by BA^ BO, and BC, act upon the point J5, find the magnitude and direction of the resultant.
6. Two forces, P and Q, act on a particle. P is three times Q, and the square of the resultant is 13 times P. Find the forces if they act at an angle of 60°.
7. Three forces, equal to 10, 20, and 25 pounds, making angles of 10°, 75°, and 225° with the X-axis, act on a particle. Find the magnitude and direction of the resultant.
8. Four forces, equal to 12, 18, 20, and 30 pounds
STATICS 53
making angles of 15°, 30°, 120°, and 210° with the X-axis, act on a particle. Find the magnitude and direction of the resultant.
9. Four men pull on ropes inclined to the X-axis at angles of 0°, 60°, 180°, and 330°, with forces of 100, 50, 200, and 50V3 pounds. Find the resultant.
10. Four forces of 24, 10, 16, and 16 pounds act on a particle. The angle between the first and second is 30°; between the second and third, 90°; between the third and fourth, 120°. What is the resultant?
11. Three forces, proportional to i, 2, and 3 respect- ively, act on a particle. The angle between the first and second is 60°, between the second and third is 30°. What angle does the resultant make with the first?
40. Equilibrium. If a number of forces act on a particle and keep it at rest, then the vector dia- gram will be a closed polygon. Therefore, formulae (69) and (70) of Section 39 become
X = 0, and Y = 0.
Therefore R^=X^+Y^= 0, (73)
which expresses the condition that must exist when the action of forces produces equilibrium.
41. Triangle of Forces. When three forces act on a particle and produce equilibrium, then the following relation (known as Lame's Theorem) exists.
force P _ force Q _ force R , v
sin Z ACB ~ sin Z BAC~ sin Z ABC ^^"^^
where ABC is the vector triangle.
54 ELEMENTARY MECHANICS FOR ENGINEERS
Example i. Four forces of P, Q, i?, and S pounds acting at angles of a, jS, y, and 5, with the Z-axis, Y
Space Diagram Vector Diagram
Fig. 17.
keep a particle in equilibrium. Draw the space and vector diagrams. See Fig. 17.
In the above diagram one sees at once that the figure is closed, and that the resultant is zero.
Example 2. A body of weight W pounds is sus- pended by two cords of length a and b inches. The ends of the cords are fastened to two supports on
Space Diagram
Vector Diagram
Fig. 18.
the same horizontal line, and 2c inches apart. Find the tensions in each cord.
STATICS 55
Solution. Let the cord a be inclined to the hori- zontal 6° and the cord b be inclined 0°. Draw the space and vector diagrams. Fig. i8.
There are three forces acting on the body, W pounds acting downward due to gravity, and the two tensions.
By triangle of forces
Ti sin (90° -0) cos^
and
W sin {6 -\-(t>) sin {d-\-<t>)
T2 _ cos <f) W~sm {d+cf>y
|
sin |
(9 = 0 = |
Va2- |
-c\ |
cos COS |
e-- Cf>: |
c |
|
a |
f |
a' |
||||
|
sin |
b |
-C2 |
c ~b' |
Now, if the values of W, a, b, and c are known, then the tensions Ti and T2 can be found, since
and
Example 3. Three forces, P, Q, and R, produce equilibrium. If R = t, pounds, Q = 4 pounds, find P when the angle between R and Q is 90°.
Solution. Suppose Q to act along the horizontal, and R to act vertically. Fig. 19.
By triangle of forces
o
P _ sin 90
R sin a sin a
56 ELEMENTARY MECHANICS FOR ENGINEERS but
Therefore,
AB 3 sin a = ^T— =-.
OB s
•P=3*~ = 5 pounds.
/?.J'
.90'
:r
■*<?"'• 0
Space Diagram
/?-J
Fig. 19.
Also, resolving vertically,
F sma^R Therefore
P
;^--=i=SPoxmds. sin a f
Example 4. Solve example 2 b) resolution of forces,
SduHan, Resolving the forces along lines of action of the forces Ti and T2 we get
Ti sin(^+^)=TFcos^, and
rasin(d+^) = IFcos0. Therefore,
COS0
ri-lF-
sin («+♦)•
STATICS 67
and j,^^^ cos<t>
sin {d+<t>Y agreeing with results found in Example 2.
EXAMPLES Xn
1. A wheel has six equally spaced radial spokes, all in tension. If the tensions of four consecutive spokes are 200, 100, no, and 180 pounds respectively, find the tensions in the other two. Work by vector polygon, and also by resolution of forces.
2. A body of 100 pounds weight is suspended by two strings of length 3 and 4 feet attached to two points in the same horizontal line whose distance apart is 5 feet. Find the tensions of the strings.
3. Two men carry a body of weight 100 pounds be- tween them by means of two ropes fastened to the weight. One rope is inclined 30° to the vertical and the other at 45°. Find the tensions in each rope.
4. Foiu- forces of 3, 4, 6, and 5 pounds keep a body at rest. The first three forces make angles of 0°, 90°, and 180° with the X-axis. Find what angle the fourth force makes with the horizontal.
5. Four forces of i, 4, 5V3, and P pounds are in equilibrium. The first three forces make angles of 60**, 0°, and 150° with the X-axis. Find F and its direc- tion.
6. A body of weight W pounds is fastened to a fixed point by means of a string of length loV^ feet. It is acted on by a horizontal force of 100 pounds. Find the value of W and the tension of the string when the weight has been pulled 10 feet from the vertical.
68 ELEMENTARY MECHANICS FOR ENGINEERS
7. A body of weight W pounds is suspended by two strings of length 6 and 8 feet respectively, from two points in the same horizontal line 10 feet apart. If the tension of the longer string is 42 pounds, find the ten- sion in the other and the value of W.
8. Two forces of 4 and 9 pounds respectively act on a particle, and their lines of action are inclined to each other at an angle of 60°. Find the direction and mag- nitude of the third force, which will produce equilibrium.
9. A pole rests vertically with its base on the ground and is held in position by three ropes all in the same horizontal plane and drawn tight. From the pole the first rope runs northeast, the second 10° west of north. Find the direction and tension of the third rope, when the pull on the first rope is 20 pounds, and the pull on the second rope is 30 pounds.
10. Two rafters, AB and AC, making an angle of 120°, support a weight of 500 pounds placed at A where they meet. What is the pressure on each rafter?
11. Three ropes are tied together, and a man pulls at each. If, when their efforts are in equilibrium, the angle between the first and second rope is 90°, between the first and the third is 150°; what are the relative strengths of each man as regards pulling?
12. On a smooth plane of length 5 and height 2, a weight of 10 pounds is kept at rest by a force along the plane. Determine the pressure on the plane and the force along the plane.
13. Three cords are tied together at a point. One of these is pulled due north with a force of 6 pounds, another due east with a force of 8 pounds. What pull on the third cord will produce equilibrium?
14. A weight of 10 tons is hanging by a chain 20 feet
STATICS
59
long. Neglecting the weight of the chain, how much is the tension in the chain increased by the weight being pulled out by a horizontal force F to a distance of 12 feet from the vertical through the point of support? What is the horizontal force?
15. A body of weight 10 pounds, is supported on a smooth plane by a force of 2 pounds acting along the plane and a horizontal force of 5 pounds. What is the inclination of the plane?
42. Moment of a Force. The moment of a force about a point is the measure of its turning effect about that point, and is equal to the product of the force and the length of the perpendicular drawn from the given point to the line of action of the force.
The value of a moment may be represented in a vector diagram. Let the force P, Fig. 20, be repre-
O
Fig. 20.
sen ted in direction and magnitude by the line AB, and let 0 be the given point, OD the perpendicular from 0 upon AB, or AB produced.
Join OA and OB. The moment of the force P about the point 0 is POD. But ABOD is twice the area of the triangle AOB. Hence the moment of a force about a point is represented by twice the
60 ELEMENTARY MECHANICS FOR ENGINEERS
area of the triangle whose base is the line represent- ing the force, and whose vertex is the point about which the moment is taken. The point 0 is called the origin of moments.
43. Moment of the Resultant of Two Forces. Let P and Q, Fig. 21, be the two forces applied at the
|
* / 1 / Ja |
|
Fig. 21.
point 0, and let R be their resultant. Suppose we take B as the origin of moments. Then draw per- pendiculars ^, r, and q to the lines of action of P, i?, and Q.
Denote the angles XOA, XOC, and XOB, by a, 0, and jS respectively. By the principle of projec- tion
i? sin 0 = P sina-fQsin jS, . . . (75)
or
R^OD sin e = P'OD sin a+Q-OD sin /3. (76) Hence
R'r = P'P-\-Q'q. ..... (77)
Therefore the moment of the resultant of two forces
STATICS 61
is equal to the algebraic sum of the moments of the forces taken separately.
The principles used in determining the preceding theorem may be applied in finding the resultant of three or more forces.
Moments of forces which tend to cause rotation counterclockwise will be considered as positive and moments of forces- which tend to cause rotation clockwise will be considered as negative.
44. A system of coplanar forces is in equilibrium when no translatory or rotary motion is produced. Hence, Section 40,
X = JlFn cos en = 0,
F = 2F„sin 0„=O. 1
[so tne ajgeoraic sum 01 tne moments 01 ine lorces
with respect to a given point or axis is equal to zero.
Fidi+F2d2+ . . . +Fndn = 0, . . (78)
where the d represents the distance of the line of action of each force from the origin of moments. In each case the sign of the moment will be deter- mined by the kind of rotation the force would pro- duce.
45. Composition of Parallel Forces. Let P and Q, Fig. 22, be two forces in the same plane and applied at points A and -B of a body. Let P be the greater force.
The lines of action, when prolonged, will meet
62 ELEMENTARY MECHANICS FOR ENGINEERS
in some point 0. Now suppose the points of applica- tion to be transferred to 0, the resultant may be determined by the parallelogram of forces. The
resultant R makes angles of a and 0 respectively, with the forces P and Q. Resolving forces along the line of action of R we get
jR = P cosa+Q cos jS. . . . (79)
Now, suppose Q to turn around B until it becomes parallel with the force P and having the same direc- tion. The point 0 recedes to the left to infinity. Hence, a and /8 approach zero, and
R = P+Q. ..... (80)
Again, let Q turn around B until it is parallel to force P, but having opposite direction. By construct- ing this case of rotation we can draw the conclusion that a becomes zero, and /3 becomes 180°. Hence
R=P-Q (81)
46. To find the resultant, and its point of applica- tion, of two parallel forces havmg the same direc-
STATICS
63
R'Ptq Fig. 23.
tion. Let P be greater than Q, and A and B their points of appHcation, Fig. 23.
Produce QB to P\ making /?
5P' equal to the vector length of the force P. Measure on AP the length AQ' so that AQ' is the vector length of q- the force Q. Draw P'Q' cut- ting AB at 0. Through O draw OR parallel to the direc- tion of P and Q, and make
OR equal to P-\-Q, which is the resultant. From the triangles P'BO and Q^AO we have
P'B:Q'A::BO:OA.
Since P^B and Q'^ represent the forces P and Q respectively, we have
P:Q::BO:OA .... (82)
If ^-B = a and BO=x, we can easily show that
aP
P+Q'
(83)
which is the distance of O from B, the point of application of force Q. Also, by the principle of moments we get
P'AO = Q'OB,
and, if OB equals x and AB equals a, we have
P'{a-x)=Q-x, which agrees with formula (82).
64 ELEMENTARY MECHANICS FOR ENGINEERS
47. The following method of finding the resultant of parallel forces having the same direction is inter- esting, and easily applied. Let P and Q (Fig. 24)
be two parallel forces having the same direction and equal in magnitude to ^P and BQ.
At A and B, introduce two equal and opposite forces, M, acting in the line AB, one acting at A and the other acting at B. The resultant of M and Q is Ru and the resultant of M and P \?> R2. Now Ri and R2 intersect at some point O. By con- structing the parallelogram of the forces Ri and i?2, their resultant is R and its line of action cuts AB 2X the point C. The distance of the point C from either ^ or J5 agrees with the results of sec- tion 46.
48. To find the resultant of two parallel forces acting in opposite directions. P>Q. Let P and Q be the forces; A and B their points of application.
STATICS 65
Fig. 25. Prolong QB until BP' equals P, and draw AQ^ equal to Q. Draw P'Q' and produce it until
Q
\
e\ 4- o
...
Fig. 25.
it cuts AB produced at point 0; draw OR parallel to AP and equal to Q'P. Now R = P-Q. From the triangles OP^B and OQ^A we have
or
P:Q::BO:AO (84)
If ^^ = a and ^O = x, then
.=^. ...... (85)
which is the distance of 0 from B, the point of applica- tion of force Q. Also by principle of moments we
get
Q'BO=PAO ..... (86) Hence
Q-x=P-{x-a). Therefore
"rk <^"
By the same method as used in section (47) one can
66 ELEMENTARY MECHANICS FOR ENGINEERS
find the resultant of two parallel forces having opposite direction,, and its line of action.
EXAMPLES XIII
1. A light horizontal beam ^JB of 12-foot span carries loads of 700, 600, and 900 pounds at distances of i foot, 5 feet, and 10 feet respectively from A. Find the sup- porting forces.
2. Two forces, P = ioo pounds and Q = 6o pounds, are applied at the points A and B, and act in the same direc- tion. Find the magnitude and line of action of the resultant ii AB is 16 feet.
3. Find the magnitude and line of action of the result- ant when (the forces acting in the same direction)
(i) P= 9; 2=27; ^5 = 12 inches;
(2) P = i8; 2 = 24; ^5 = 28 inches;
(3) P= 8; Q = io; AB= 9 inches.
4. Find the magnitude and line of action of the result- ant when (the forces acting in opposite directions)
(i) P = 2o; Q = io; ^P = i2 inches;
(2) P = io; Q = so; ^5 = 12 inches;
(3) P = i8; Q = i2; ^5 = 15 inches.
5. The forces having the same direction, if Q = ii, AO = j inches, AB = Sj inches; find P and R.
6. The forces having opposite directions, if Q=ii, O is 15! inches to the left oi B, AB is 8f inches; find P and R.
7. Two men carry a stone which weighs 240 pounds,
STATICS 67
by means of a plank 5 feet long. One man carries double what the other carries. How must the stone be placed on the plank?
8. A uniform rod, AB, 20 feet long weighing 2 pounds per foot, can turn freely about a point in it, and the rod is in equilibrium when a weight of 10 pounds is placed at the end A. How far from the end A is the point about which it can turn?
N. B. The weight of a uniform rod may be taken to act at its middle point.
9. A valve 4 inches in diameter is held down by a lever and a weight, the length of the lever being 20 inches and the valve spindle being 4 inches from the fulcrum. Find the pressure per square inch which will lift the valve when the weight hung on the end of the lever is 40 pounds and the lever itself weighs 10 pounds.
10. Three men carry a log which is uniform and has a length of 8 feet, and weighs 240 pounds. If one of the men Hfts at an end, and the other two lift by means of a cross bar, how ought the cross bar be applied in order that each man may carry one-third the weight, neglecting the weight of the cross bar?
11. A rod, ABj of length 5 feet and weight 10 pounds, is found to balance itself if supported on a fulcrum 3 feet from A. If this rod were placed horizontally on two points, one under A and the other under B, what pressure would it exert on each point?
12. A uniform bar, AB, 18 feet long rests in equilib- rium upon a fulcrum 2 feet from A. A weight of 5 pounds is hung at B and no pounds at A. What is the weight of the bar?
13. A horizontal uniform beam, AB, of length 21 feet, weighs 10 tons, and rests on two supports at the ends.
68 ELEMENTARY MECHANICS FOR ENGINEERS
What is the pressure on each support when a weight of 4 tons is (i) placed at the middle and (2) 7 feet from A?
14. A horizontal beam, AB, of length 16 feet, weighs 8 tons and rests on two supports. At points 2, 4, 6, 8, 10, 12, and 14 feet from A, weights of i, 2, 3, 4, 5, 6, and 7 tons respectively, are placed. Find the supporting forces.
15. Forces equal to the weights of i pound, 2 pounds, 3 pounds, and 4 pounds act along the sides of length a of a square, ABCDj taken in order. Determine the resultant.
16. A ladder whose weight is 168 pounds, and whose length is 25 feet, rests with one end against a smooth vertical wall, and with the other end upon a smooth horizontal floor; if it is prevented from slipping by a peg at its lower end, and if the upper end is 24 feet above the floor, find the reaction of the peg, ground, and the wall.
17. A bar, AB which weighs 20 pounds, 30 inches long, is hinged at A so as to be free to move in a vertical plane. The end B is supported by a cord, BCy so placed that the angle ABC is 150° and AB is horizontal. A weight of 20 pounds is hung on the bar at a point D, between A and B and 12 inches from A, Fuid the tension in the cord and the pressure of the rod on the hinge. Also find what angle the fine of action of the pressure on the hinge makes with AB.
49. Center of a System of Parallel Forces, Center of Gravity. LeL A, B, and C (Fig. 26) be the points at which parallel forces Fi, F2, and F3, respectively, act. The line of action of the resultant force R may be found by applying successively the results of Sec- tion 46.
STATICS
69
Fig. 26.
Now, Fi and F2 may be replaced by a forceFi+F2, at a point H in AB, such that AH/BH = F2/Fi. Also the force acting at H, and the force F3 act- ing at C may be replaced byaforceFi+Fs+Fa at the point K in EC, such that
Any system of pai allel forces, acting in the same
direction, may be replaced by a force equal to the algebraic sum of the given forces. If G is the point where the resultant force acts, then this point is called the center of the parallel forces.
Center of Gravity. Gravitational forces have a tendency to pull all bodies toward the center of the earth. Since the bodies are small in comparison with the size of the earth, the lines of action of the forces acting on the particles which make up the bodies may be assumed to be parallel. The point of application of the resultant of the parallel forces which act on the particles which make up a body is called the center of gravity of the body.
The resultant of these gravitational forces acting on a body is called its weight, and the line of action of the resultant passes through the center of gravity.
50. Center of gravity of two particles of given weights at a given distance apart.
Let A and B, Fig. 27, be the positions of the centers
70 ELEMENTARY MECHANICS FOR ENGINEERS
of gravity of the two bodies of weights wi and W2. Let G be the center of gravity of the system. Taking
Fig. 27.
moments about the point B we get
{wi -{-W2) • GB = wi • AB, Taking moments about the point A we get
(•wi-\-W2)'AG = W2'AB.
Hence
and
GB^-^^ AB, .... (88)
101-^1)1)2
AG = -^^'AB. .... (89)
Wi -\-W2
If 'Wi=W2 the center of gravity is midway between
them, i.e.
Afi — rm
AG = GB = UB.
51. Plane and Line Moments. The product of any element by its distance from a given plane, or line, is called a plane, or line, moment, where element means a unit length, unit area, or unit volume.
The products aidi, ^2^2, etc., may be called moments of areas, where a is the area of the plane figure, and d is the distance of the center of gravity of the plane figure from a point, line, or plane.
The products Vidi^ V2d2, etc., may be called
STATICS 71
moments of volumes, where V is the volume of the geometric solid, and d is the distance of the center of gravity of the solid from a point, line, or plane.
52. Theorems on Center of Gravity :
Theorem I. If a system of particles, whose weights are wi, W2, . . • Wn, be in the same plane, and if OX and OF be two fixed straight lines in the plane at right angles, and if the distances of the particles from OX and OF be yi, y2, . . . Jn, and 0:1, :r2, . . . Xn, respectively, the distance of their center of gravity from OX and OY being X and F, then
^1+^2+ . . . -i-Wn ' ' '
and
1Vi-\-W2+ . . . -JrWn
Theorem IL The moment of an entire line, area, or volume, is equal to the algebraic sum of the moments of its parts.
The lever arm of a line, area, or volume, is the distance of the center of gravity from the given point, line, or plane, about which the moments are taken.
Theorem III. The center of gravity of a uniform straight rod is at its middle point.
Theorem IV. The center of gravity of a uniform triangular plate is at the point of intersection of the medians.
72 ELEMENTARY MECHANICS FOR ENGINEERS
Theorem V. The center of gravity of a uniform plate having the shape of a parallelogram is at the point of intersection of the diagonals.
Theorem VL The center of gravity of a uniform plate having the shape of a polygon may be found by dividing the polygon into triangles and applying Theorem II.
Theorem VII. The center of gravity of a right cone, or tetrahedron, of uniform material is at a dis- tance of J the altitude from the base, and on the line joining the center of the base to the vertex.
Theorem VIII. The center of gravity of a cir- cular arc of length /, which subtends an angle 2d
at the center, is at a distance of r from the
center of the circle and on the bisector of the angle subtended.
Theorem IX. The center of gravity of a cir- cular sector which subtends an angle 2^ at the center is on the bisector of the subtended angle at a dis- tance of -r from the center. Hence the center
3 ^ _ of gravity of a semicircular area is at a distance of
— from its straight-line boundary. 3^
The center of gravity of a circular segment may
be found by applying Theorem II.
Theorem X. The distance of the center of gravity
of a hemispherical shell from the plane of its rim is
half the radius of the shell.
STATICS 73
Theorem XI. The center of gravity of a sector
2 ft
of a sphere is at a distance of - r cos^ - from the center
42
of the sphere on the bisector of the angle subtended
at the center.
The center of gravity of a soHd hemisphere is at a distance of f r from the plane boundary and on the perpendicular through the center of the plane base.
The proofs of these theorems are easily found by employing methods of Calculus, but elementary proofs will be given here.
Theorem I can be developed by using the prin- ciples employed in Section 50.
Theorem II is deduced readily from Theorem I.
Proof of Theorem III.
Let AB he a, uniform rod, and G its mid-point. Divide the rod into pairs of parti- cles of equal weight situated at ^ — - — ^--- ^
equal distances from G. The center of gravity of each pair, such as the particles wi and W2, is midway between them, Section 50, that is, at the mid-point of the rod.
Proof of Theorem IV.
Given a triangular plate ABC. Divide the plate into very narrow strips par- allel to AB. Each of the strips may be considered as a rod whose center of gravity is at its mid-point. But the mid-points of all these parallel strips are on the
74 ELEMENTAKY MECHANICS FOR ENGINEERS
median CD, of the triangle. If we had divided the plate into narrow strips parallel to BC, the mid- points of the strips would be on the median AE. Therefore the center of gravity of a uniform tri- angular plate must be at the point of intersection of the medians.
Proof of Theorem V.
Given a plate A BCD having the shape of a parallel- ogram. The diagonals of any parallelogram bisect
each other. The center of gravity of the triangle ABC is in the line BD, and the center of gravity of the tri- angle ADC is in the line BD. Likewise the e.g. (center of gravity) of the triangles ABD and BDC are in the line AC, Hence the e.g. of the plate is at the point G, the intersection of the diagonals.
Proof of Theorem VII.
Given a tetrahe- dron 0-ABC. The e.g. of the base ABC is gi, and the e.g. of the face BOC is ^2.
Let G be the e.g. of the pyramid, or tetrahedron. Divide the pyramid into a large number of very thin plates parallel to the base ABC. The e.g. of the pyra-
STATICS
75
mid is on the line Ogi. If we divide the pyramid into plates parallel to the face BOC, then the e.g. of the pyramid is on the line Ag2. Hence Ag2 and Ogi must intersect at some point which is G, We know that Hgi = iAH, and Hg2 = \0H, By similar tri- angles gig2 = \0A, and Ggi=\OG, Therefore Ggi^'iOgi. Extending this analysis it is readily proved that the e.g. of any pyramid, or cone, is \ the distance from the e.g. of the base to the vertex.
Proof of Theorem VIII.
Let ABC be a portion of a circular arc which sub- tends an angle 2d at the center, r being the radius, and B being the mid-point of the arc. Divide the arc into a large number of very short lengths and let MN be one of these parts. Let a be the angle which OS makes with the radius OA, S being the mid-point of the arc MN, Since MN is very short then the distance of MN from OA is r sin a. Apply- ing Theorem I, the distance of the e.g. of the arc ABC from OA, y , is
y^^ir-sm a
MN)
Hmn
HiMN' sin a)
^MN
MN- sin a = WW,
76 ELEMENTARY MECHANICS FOR ENGINEERS the projection of MN on OA . Therefore
^{MN- ?>ma)=Ac, and
2;ijZiv=arc^c.
Hence
-_r'Ac_r{r—r cos 2B) _r(i— cos 2B) ~AC~ AC IC
r
_r(i— cos 20) _r-s>{v? d Vd ~d *
By symmetry it is evident that G, the e.g. of the arc ABC, must be on the radius OB. Hence
0G = ^ rsin^
sine
Proof of Theorem IX.
In the figure for the proof of Theorem VIII we may consider OMN as a triangle, since MN is so short that it is assumed as a straight Hne. The e.g. of any triangle is on the median and J the length of the median from the vertex. Hence the e.g. of each triangle which is a part of the sector 0-ABC is on the arc of a circle of radius %r. Therefore, by Theorem VIII, the e.g. of the sector 0-ABC is on the radius OB and at a distance of fr- (sin d)/e from the center.
STATICS
77
Proof of Theorem X.
Given ABC a. cross-section of hemispherical shell. Let ACNM be a cross- section of a zone. Divide the zone ACNM into a large number of equal zones, such as xyrs. The e.g. of each zone is on the axial line OB. Therefore, if G is the e.g. of the zomqACNM,
Hence for a hemispherical shell
OG = ^{o+OB)=ir,
Proof of Theorem XL
Given 0—^^C a cross-section of a spherical sector. Divide the sector into equal small pyramids having a common vertex 0, the bases of the pyramids being the curved surface of the sphere. The e.g. of each pyramid is at a distance of |r from the center 0. Hence the e.g. of the sector is the same as the e.g. of a shell of radius fr. Let 2^ be the angle subtended by the arc ABC,
78 ELEMENTARY MECHANICS FOR ENGINEERS
Then
^^ fr cos d+lr 3 / , ^x
0G = - '-±- = p{i-{- cos e),
2
3 9^
= 1^ COS^-. 2
EXAMPLES XIV
1. Two bodies are attached to the end of a rod 5 feet long. One weighs 8 pounds and the other 14 pounds. Find the center of gravity of the system.
2. Three bodies are attached to the corners of a hori- zontal equilateral triangle 14 inches on a side. The bodies weigh 10, 15, and 20 pounds. Find the position of the center of gravity of the system.
3. A horizontal rod, AB, is 4 feet in length, and of weight 10 pounds. It is trisected in the points C and D, and at the points A, B, C, and D, are placed bodies of 2, 4, 6, and 8 pounds weight respectively; find the center of gravity of the system.
4. If, in (3) the body at A is removed, and another body is substituted, find the weight of this unknown body so that the new center of gravity may be at the middle point of the rod.
5. To the end of a horizontal rod 4 feet long, and of weight 8 pounds, is attached a weight of 12 pounds. Find the center of gravity of the system.
6. A wheel of weight 100 pounds is attached to the end of a straight uniform axle 4 feet long and of weight 50 pounds. Find the center of gravity of the system.
7. If a horizontal square tin plate weighs 5 ounces, and a small body weighing 2 ounces is placed at one corner of the plate, where is the center of gravity?
8. Find the center of gravity of the figure A when the
STATICS 79
three pieces are of uniform material, and the central piece is half a side in length, and is joined at the mid- points of the sides.
9. Find the center of gravity of a T-square, the two pieces being of the same dimensions, 2 feet in length and 2 inches wide.
10. A board consists of a square of side 2 feet, and another square of side 6 inches, having the center of its side the same as the center of the side of the larger square. Find the distance of the center of gravity from the center of the base of the larger square.
11. In example 10 suppose an equilateral triangle is described on one side of the square, instead of the smaller square. Find the distance of the center of gravity from the center of the square.
12. A piece of cardboard consists of a rectangle of length a and width b, and a triangle of altitude h de- scribed on the side a. Find the distance of the center of gravity from the base.
13. A trapezoid with upper base b and lower base a has an altitude h. Find the position of the center of gravity and its distance from the base a.
14. From a triangular board is cut off, by a line parallel to its base, ^ of its area. Find the position of the center of gravity of the remainder, if the base of the triangle is 40 inches and its altitude is 36 inches.
15. An isosceles right triangle ABC has squares described externally on its sides. The lengths oi AB and BC are each 27 V2 inches, the right angle being at B. Show that the center of gravity is on the line joining B to the middle of AC, and 26 inches from B.
16. A quadrilateral consists of two isosceles triangles on opposite sides of a base 12 inches long. The side of
80 ELEMENTARY MECHANICS FOR ENGINEERS
the larger triangle is lo inches and the side of the smaller triangle is 8 inches. Find the distance of the center of gravity from the common base.
17. The middle points of two adjacent sides of a square are joined, and the triangle formed by this straight line and the edges is cut off. Find the distance of the center of gravity of the remainder from the center of the square.
18. From a circular disc of radius 12 inches is cut out a circular area whose diameter is 3 inches and center ij inches from the circumference of the disc. Find the position of the center of gravity of the remainder, and its distance from the center of the disc.
19. From an equilateral triangle, of side 20 inches, is cut out a circular area of radius 5 inches, the circle being tangent to the middle of a side. Find the position of the center of gravity of the remainder, and its distance from the center of the triangle.
20. The lever arm of a testing machine .weighs 2000 pounds, and is placed horizontally on a knife edge. It sustains a downward pull of 8 tons placed 5 inches from the knife edge, and carries a load of i ton on the opposite side of the knife edge, and placed 30 inches from it. Find the distance of the center of gravity of the lever arm from the knife edge.
21. The radius of a circle is 20 inches. Find the center of the circular arc which subtends an angle of 50 degrees at the center.
22. Find the center of gravity of a semicircular arc of a circle of radius 25 inches.
23. The radius of a circle is 20 inches and the central angle of a sector is 60 degrees. Find the center of gravity of the sector. Find center of gravity of a semicircular area.
STATICS 81
24. In example 23 find the center of gravity of the seg- ment bounded by the chord and the arc.
25. Let AOB be a given sector of a circle of radius 42 inches, angle AOB being 120 degrees. Through 0\ the center of the triangle AOB, draw lines O'H and O'K parallel to OA and OB, H and K being on the chord AB. If the area O'HK is removed, find the distance of the center of gravity of the remainder from the center of the circle.
26. A cast-iron eccentric consists of a short cylinder 12 inches in diameter, and has through it a cylindrical hole 3 inches in diameter, the axis of the hole being parallel to that of the eccentric and 4 inches from it. Find the distance of the center of gravity from the center of the eccentric.
27. Out of a square plate, of side 12 inches, show how to cut a triangle having one side of the square for a base, so that the remainder may have its center of gravity at the vertex of the triangle.
28. Out of a square plate, of side 12 inches, is cut an equilateral triangle of side 6 inches, the center of the base of the triangle coinciding with the center of the square, and the base parallel to the side of the square. Find the distance of the center of gravity from the center of the square plate.
29. Where must a circular hole, of 2 -foot radius, be punched out of a circular disc, of 5 foot radius, so that the center of gravity of the remainder may be 5 inches from the center of the disc.
30. A square hole is punched out of a circular disc, the diagonal of the square being the radius of the circle. If the radius of the circle is 20 feet, find the distance of the center of gravity of the remainder from the center of the circle.
82 ELEMENTARY MECHANICS FOR ENGINEERS
31. An I-section of a girder is made up of three rect- angles, two flanges having their sides horizontal, and one web connecting them having its long side vertical. The top flange section is 8 inches by 2 inches, and the bottom flange is 16 inches by 4 inches. The web section is 10 inches by 2 inches. Find the distance of the center of gravity of the area of the cross-section from the bot- tom of the flange.
32. A piece of iron consists of an isosceles trapezoid with semicircular areas described upon the ends. The altitude of the trapezoid is 4 feet and the ends measure 2 feet and 2 J feet. Find the distance of the center of gravity from the center of the 2j-foot base.
;^2,' A circular board 10 inches in diameter has two square holes cut out of it, one of side 2 inches, and the other of side i inch, the centers of the squares being on diameters mutually perpendicular, and 2 inches from the center of the circular plate. Find the center of gravity of the remainder.
34. A counterpoise on a wheel of an engine has the shape of a crescent and fits inside the rim of the wheel 8 feet in diameter, and the crescent subtends an angle of 120° at the center of the wheel. The center of the circle determined by the inner rim of the crescent is on the rim of the wheel. Find the distance of the center of gravity of the counterpoise from the center of the wheel.
35. A solid consists of a right cylinder of radius r and 4 feet long and a right cone of altitude 2^ feet, the base coinciding with one end of the cylinder. If the cylinder and cone are made of the same uniform material, find the distance of the center of gravity of the solid from the vertex of the cone.
STATICS 83
36. A solid consists of a right triangular prism hav- ing an equilateral base of side 12 inches and a triangular pyramid with its base coinciding with the base of the prism. The altitude of the prism is 2 feet, and the height of the pyramid is 18 inches. The axis of the prism and pyramid being in the same line, find the distance of the center of gravity of the solid from the common base.
37. A solid consists of a right cylinder and a hemi- sphere described on one end. If the altitude of the cylinder is 4 feet, and its radius is i foot, find the dis- tance of the center of gravity from the center of the cylinder.
38. A piece of solid iron consists of two cylinders, one placed on the other and their axes being in the same line, the altitude and radius of the lower cylinder being 10 inches and 4 inches, and the upper being 3 inches and I inch. Find the distance of the center of gravity from the bottom of the solid.
39. The upper and lower bases of a frustum of a right cone have diameters 5 and 20 inches respectively. The altitude is 15 inches. Find the distance of the center of gravity from the lower base.
40. The longest element of a cone makes an angle of 60° with the diameter of the base. If this element is 20 feet in length, find the distance of the center of gravity from the plane of the base.
41. A concrete dam is 10 feet high, 2 feet thick at the top, and 6 feet thick at the bottom. The front face being vertical, find the distance of the center of gravity from the vertical face.
42. A solid right cone stands on a base 24 inches in diameter, and its altitude is 15 inches. The upper part of the cone is removed by passing a plane parallel to the
84 ELEMENTARY MECHANICS FOR ENGINEERS
base and 5 inches from the vertex. Find the distance of the center of gravity from the lower base.
43. The barrel of a cannon is shaped like a frustum of a cone, being 12 feet long, 18 inches in diameter at one end and 12 inches at the other. Through this frus- tum is bored a cylindrical hole 6 inches in diameter, the axis of the cylinder and frustum being the same. Find the distance of the center of gravity from the larger end of the barrel.
44. A solid right cone and a solid hemisphere have a common base. The altitude of the cone is 12 inches, and the diameter of the base is 12 inches. Find the dis- tance of the center of gravity from the center of the common base.
45. A hemispherical shell of uniform material has an internal radius of 12 inches, and a thickness of 2 inches. Find the distance of the center of gravity from the center of the sphere.
46. A soHd monument consists of three parts, the lowest part is a cylindrical stone of height 10 feet, and radius of base 3 feet; the middle part is a stone the shape of a frustum of a cone, the altitude is 3 feet, the lower base coinciding with the upper base of the cylinder, and the upper base has a radius of 2 feet; the third part is a stone hemisphere with its base the same as the upper base of the frustum. Find the distance of the center of gravity from the base of the monument.
47. A solid right cone is 12 feet in height and the radius of the base is 2 feet. The greatest possible sphere is cut out from the cone. Find the distance of the center of gravity of the remaining volume from the vertex of the cone.
48. A solid sphere of radius 18 inches consists of two
STATICS 85
parts; the density of one half is three times that of the other. How far is the center of gravity from the center of the sphere?
49. From a solid sphere of radius 13 feet is cut off a segment by passing a plane 5 feet from the center of the sphere, and perpendicular to a diameter. Find the distance of the center of gravity of the smaller segment from the center of the sphere, and the distance of the center of gravity of the remaining volume from the center of the sphere, if the smaller segment is removed.
53. Stable, Unstable, and Neutral Equilibrium.
When a body is moved from its position of support and left, it will in some cases return to it, pass by and return again, and continue thus to vibrate until it settles in its place of support by friction and other resistances. This condition is called stable equilibrium.
In other cases, when a body is moved from its position of support and left, it will depart further from it, and never recover that position again. This is called unstable equilibrium.
In other cases still, the body, when moved from its place of support and left, will remain, neither returning to it nor departing further from it. This is called neutral equilibrium.
In genera], there is stable equilibrium when the center of gravity, upon being disturbed in either direction, begins to rise; unstable when if disturbed either way, it begins to descend, and neutral when the disturbance neither raises nor lowers the center
86 ELEMENTARY MECHANICS FOR ENGINEERS
of gravity. A body will remain at rest on a plane surface if the vertical line through its center of gravity falls within the base of support.
EXAMPLES XV
1. A rectangular block 20 feet high, 4 feet wide and 3 feet thick, is racked into an oblique form until it is on the point of falling. What is the inclination to the horizontal?
2. If the block in example i were tilted instead of racked, until it is about to fall, what is its inclination to the horizontal?
3. A cube lies on an inclined plane and is prevented by friction from sliding down; to what inclination must the plane be tipped so that the cube will be ready to turn over the lower edge?
4. What is the greatest height which a right cyUnder of 10 inches diameter may have in order that it may rest with one end on a plane inclined 45° to the horizontal?
5. A cylinder of height a and radius of base r, is placed on an inclined plane and prevented from sliding; if the inclination 0 of the plane is gradually increased, find when the cylinder will topple over.
6. A body, consisting of a cyUnder and a hemisphere joined at their bases, is placed with the spherical end upon a horizontal plane. If the radius of the sphere is 10 inches, what must be the length of the cylinder in order to have neutral equilibrium?
7. A cone and a hemisphere have a common base of radius 2 feet. Find the height of the cone when neutral equilibrium exists, when the body rests with the hemi- spherical surface on a horizontal plane.
STATICS 87
8. The altitude of a cone is 24 inches and the diameter of the base is 15 inches. What is the greatest inclina- tion on which it may stand in equilibrium on its base?
9. A right cone contains 288V3 r cubic inches of material. What is the greatest height it may have in order to rest with its base on a plane inclined 30° to the horizontal, and what is the radius of the base?
10. A right pyramid of stone is 12 feet high and has a pentagonal base of side 8 feet. How much may the pyramid be turned about an edge of the base before it will topple?
11. A rectangular stone 8 feet long, 4 feet wide and 3 feet in thickness, is laid with one-quarter of its length projecting over the edge of a concrete base, and resting on the 4 by 8 face; a stone and a quarter is then laid on the first stone with one-quarter projecting over the first stone and so on. How many courses will cause the pile to topple?
12.* A cubical block of material resting on a hori- zontal plane weighs 1000 pounds and its edge is 2 feet in length. Find the work in foot-pounds necessary to turn the block over one edge.
13. How much work is expended in turning over the cone of example 8, if it rests on a horizontal plane, and its weight is 60 pounds per cubic foot?
14. How much work is expended in turning over the pyramid of example 10, if it rests on a horizontal plane, and its weight is 4 tons?
15. What amount of work is expended in turning over a right cone, with a circular base, whose altitude
* The total work done in lifting a body is equal to the weight of the body multiplied by the vertical distance through which its center of gravity has been raised.
88 ELEMENTAKY MECHANICS FOR ENGINEERS
is 12 feet and radius of base 4 feet, the weight of the material being 100 pounds per cubic foot?
16. A cylindrical well is 150 feet deep and 10 feet in diameter. Supposing the well to be filled with water to a depth of 50 feet, how much work is expended in rais- ing the water to the top?
17. A rectangular tank 7.854 feet long, 6 feet wide, and 4 feet deep is filled with water from a cylindrical tank 78.54 square feet horizontal cross-sectional area. The level of the water, before filling begins, stands 16.8 feet below the bottom of the rectangular tank. How much work is expended in filling the rectangular tank?
18. A hollow cylindrical boiler shell, 8 feet internal diameter and 20 feet long, is placed with its axis hori- zontal. It is to be half filled with water from a reser- voir, the level of which remains constantly 6 feet below the axis of the boiler. How much work is required to lift the water?
19. A uniform chain, 60 feet long and weighing 15 pounds per foot, hangs vertically. How much work is expended in winding up the chain?
20. A chain 144 feet long hangs vertically. Its weight at the top end is 20 pounds per foot and at the lower end 10 pounds per foot, the weight per foot varying uniformly from top to the lower end. How much work is expended in winding up the chain?
21. A regular polygon just tumbles down an inclined plane whose inclination is 10°. How many sides has the polygon?
54. Theorems of Guldinus or Pappus, (a) If a plane curve revolves around any axis in its plane, the area of the surface generated is equal to the
STATICS
length of the revolving curve multiplied by the length of the path described by its center of gravity.
(b) If a plane area revolves around any axis in its plane the volume generated is equal to the revolving area multiplied by the length of the path described by its center of gravity.
Proof of Theorem {a).
Let A B he Si part of a curve which revolves about the axis 00\ Let L be the length of the curve between A and B. Divide L into a large number of small parts having lengths of h, h, h, . . . In, each part being Xi, X2, X3, ... Xn, from the axis of revolu- tion. Any part / of the curve gen- erates a surface of area 2txXL Hence the total area generated is
n n
2 (2TrXnln) = 27r2 (x„/„) . 1 1
n _
But 2 {xnQ =^Lx (Theorem 1
II, Section 52). Therefore the total surface generated is
2'kX'L.
Proof of Theorem (6).
Let A be the area which
revolves about the axis 00' ,
Divide the surface A into a
large number of small parts, ai, ^2, as, . . . an, each
being X\^ X2, x^, . . . Xn, from the axis of revolution.
90 ELEMENTARY MECHANICS FOR ENGINEERS
Any part a will generate a volume of 2TrxXa, Hence the total volume is
n n
S ( 2TrXnan) = 27rS (Xnan) . 1 1
n
But ^(xnan) =A'X (Theorem II, Section 52). There- fore the total volume generated is 2Tr x-A.
EXAMPLES XVI
1. Find the volume of a sphere by revolving a semi- circle.
2. Show that the volume of a cone may be found by revolving a right triangle.
3. Show that the surface of a sphere may be found by revolving a semi-circumference.
4. A circle, 10 inches in radius, has its center on the F-axis and 18 inches from the origin. Find the surface and volume of the torus ring by revolving the circle about the X-axis.
5. Find the volume generated by revolving a circle 10 feet in diameter about a tangent. Also the surface.
6. Find the volume generated by revolving a semi- circle 20 inches in diameter about a tangent parallel to its diameter.
7. A circle of 10 inches radius with an inscribed regular hexagon, revolves about an axis of rotation 20 inches distant from its center and parallel to a side of the hexa-' gon. Find the difference in area of the generated sur- faces. Also the difference in volumes generated.
8. An equilateral triangle rotates about an axis with- out it, parallel to, and at a distance of 10 inches from one
STATICS 91
of its sides. Find the surface and volume thus generated, a side of the triangle being 4 inches.
9. Find the surface and volume of a square ring de- scribed by a square of side i foot revolving around an axis parallel to one of its sides, and 4 feet distant from the side.
10. The center of gravity of a semi-ellipse is at a dis- tance of 4&/37r from the center of its straight line bound- ary. Its area is ^irab, where a is half the longer axis and b is half the shorter axis. Find the volume gener- ated by revolving the semi-ellipse about the axis a, if a = 3 feet and b=2 feet.
11. Find the surface and volume of a cylindric ring, the diameter of the inner circumference being 12 inches and the diameter of the cross-section 16 inches.
12. A groove of semicircular section 3 inches in radius is cut in a cylinder whose radius is 12 inches. Find (i) the area of the curved surface of the groove, and (2) the volume of the material removed.
13. The arc of a circle of 30 inches radius subtends an angle of 120° at the center. Find the area of the sur- face generated when this arc revolves about its chord; also find the volume of the solid generated by the revolu- tion of the segment about its chord.
14. A groove is cut in a cylinder casting 48 inches in diameter. The cross-section of the groove is an isosceles trapezoid with outer base 10 inches, the inner base 6 inches, and 6 inches deep. Find the volume of the material removed.
15. Find (i) the surface and (2) the volume of a ring with a circular section whose internal diameter is 12 inches and thickness is 3 inches.
CHAPTER V
MOTION IN A CIRCLE
55. In Section ii we considered the uniform circular motion of a particle, and found that the linear velocity is equal to the radius multipHed by the angular velocity. Since the average accelera- tion is equal to the change in velocity divided by the time, we can find by the vector diagram the direc-
FlG. 28.
tion of the acceleration. In Fig. 28, let the vector ab, parallel to the tangent BT2, represent the linear velocity at B, and the vector be, parallel to the tangent ATi, represent the linear velocity at A. Since the motion is uniform, vector ab equals vector
92
MOTION IN A CIRCLE ^ 93
be. Then to find the change of velocity between A and B, we must subtract the vector representing the velocity at A from the vector representing the velocity at B. Therefore vector ac represents the change of velocity between A and B. But ac = 2ah
sin -. Hence ac = 2v sin -. The time equals
2 2
^/oj (Section II, formula 12). Therefore the average acceleration equals
a a
2V sin - sin - 2 2
— — — = (oz; — — '.
i i
0) 2
The limiting value of — r-^ is unity, and the
average acceleration becomes oiv, oy^r, or v'^/r.
Now, as the angle B decreases, the angle hac approaches a right angle, and the acceleration is toward the center of the circle.
56. Centripetal and Centrifugal Forces. We have just seen that if a particle is describing a circle of radius r about a center O with angular velocity of CO radians per second, it has an acceleration of oy^r toward the center of the circle. The force is equal to the mass times the acceleration; and in this case it is W/g (jp-r pounds. This force which tends to pull an object inward is called the centripetal force. According to the third law of motion, there must be a reaction in the opposite direction to the cen-
94 ELEMENTARY MECHANICS FOR ENGINEERS
tripetal force. This opposite force is called the centrifugal force. In other words, the particle exerts a pull of W/g oy^r pounds away from the center of the circle. In some cases the centripetal force may be a thrust instead of a tension, as in the case of a car going around a curve.
57. Motion on a Circular Inclined Track. Suppose
a particle P (Fig. 29) moves with uniform velocity F, around a smooth circular track of radius OP equal to R feet. To find the inclina- tion of the track in order that the particle keeps its circular path. Now, there are only two forces acting on the par- ticle, its own weight, and the reaction Q of the track which is perpendicular to the plane of the track. The resultant of these forces, W/g- V^/R, is directed toward the center O of the horizontal circle. From the vector diagram we get at once
— ^=s <-'
which is the required angle.
58. Motion on Railway Curves. The track must be inclined so that the reaction Q is normal to the plane of the track. Suppose G is the gauge of the track in inches, V the velocity of the car in feet per
MOTION IN A CIRCLE
95
second, and R the radius of the curve which the car
is rounding. Let AB (Fig. 30) represent the gauge.
^C is the height in inches
which the outer rail is
above the inner rail, and d
is the inclination of the
plane of the track. Now,
AC =AB sine. When d is
very small, sin 6 = tan 6.
Fig. 30.
Hence AC = AB tam 9.
From Section 57, formula (92),
AC = Gtsind = G'
(93)
72
gR' • • •
69. The Conical Pendulum. If a particle is tied by a string to a fixed point O, and moves so that it describes a circle in a hori- zontal plane, the string de- scribing a cone whose axis is the vertical line through 0, then the string and particle together are called the conical pendulum. Consider the case when the circular motion is uniform. 31 represents a conical pendulum, where a particle P, tied by a string to the fixed point 0, describes the horizontal circle with uniform motion about the center H vertically under 0.
Let r = the tension of the string OP in pounds; CO = angular velocity of P about H in radians per second;
Fig. 31.
Fig.
96 ELEMENTARY MECHANICS FOR ENGINEERS
i? = radius HP of the horizontal circle; PF = weight of the particle P in pounds; a = length of the string OP in feet; 0 = angle which string makes with a vertical
line through O; /? = height of OH in feet; z; = linear velocity of the particle in feet per second.
The two forces, the tension T of the string, and W, the weight of the particle, have a resultant W /g oj^R toward the center of the horizontal circle. In the vector diagram we see that
Hence
^=^=i •••••• (95)
This shows that h, the height of the conical pendulum, is independent of the length of the string, and is dependent upon the angular velocity about H, the center of the horizontal circle. Resolving vertically
rcos6> = IF, (96)
which will give the value of T after 6 has been found. Resolving horizontally
rsine=^ (97)
MOTION IN A CIRCLE 97
since T sin 6 is the only force in the direction PH, From (96) and (97) we get
a sin^ d cos 6^ and
cose = ~-f—, .... (98)
as
V = 2Tna sin ^, .... (99)
where n is the number of revolutions per second. Hence by (96) and (98), we find that
Since
W
T = 4 — TT^w^a pounds. . . . (100)
^=i' "=Vl ^'°'^
The time of one complete revolution of the particle is
2ir radians 27r Ih , s
time = -— - = _ = 27r^ - . (102)
angular velocity co \ g
Hence the time is dependent upon h only.
60. Motion in a Vertical Circle. Suppose a particle P moves in a vertical circular path and is kept in the circular path by a string fixed at some point, or by an inward pressure of a circular track. To find the tension of the string, or inward pressure of the track, and the velocity at any point, and the condition that
98 ELEMENTARY MECHANICS FOR ENGINEERS
the particle will make a complete revolution. In Fig. 32 let
R = the normal pressure of the track or the tension
of a string; IF = the weight of the rotating particle; z^=its velocity in feet per second in any position? B, such that OB makes an angle 6 to the vertical OA , A being the lowest point on the circumference; Fa = velocity at A; r = radius of the circle in feet.
Suppose the particle is projected from the position C A with an initial velocity of
Va, then the velocity at any point B is (Section 14, formula
(25)),
v^ = VA'-2g-AH. (103)
f^ Also, if the particle is moving clockwise, the kinetic energy at B is ^ W/g'V^. At A the kinetic energy is hW/g-V^, and the potential energy Sit B is W- AH. Since, by the principle of work, no energy is lost, the total mechanical energy at B is equal to that at A. Hence
. (104)
Fig. 32.
1 W^ 9 I TI7 ATT I ^ T/ 2
2 g 2 g
or
v^+2g'AH = VA^
(105)
MOTION IN A CIRCLE 99
which is the same as (103). It is evident that the normal reaction at B is
R = '±y-^W co^B. . . . (106)
g r
The least value of R will occur when the particle is at C, the highest point of the circle. One can readily find the condition which must exist so that the par- ticle describes a complete circle, which condition is
VA>^igt^. .... (107)
EXAMPLES XVII
1. A baU weighing 10 pounds is whirled around on a smooth horizontal circular track of radius 10 feet, with a velocity of 30 feet per second. What is the ten- sion in the cord?
2. In example (i) suppose the string will only sustain a tension of 1600 ir'^/g pounds. Find the greatest num- ber of revolutions per second that the ball can make without breaking the string.
3. With what velocity must a body revolve on a smooth horizontal circular track of 5 feet radius in order that the centrifugal force may equal the weight of the body?
4. A weight of 5 pounds is attached to the end of a cord 3 feet long just capable of sustaining a pull of 100 pounds. How many revolutions per second must the body make in order that the cord may be on the point of breaking?
5. A railway car, weighing 7 tons, moving at the velocity of 30 miles per hour, describes an arc whose
100 ELEMENTARY MECHANICS FOR ENGINEERS
radius is 400 yards. What is the outward pressure on the track?
6. At what speed will an engine going around a curve of 960 yards radius, exert a horizontal thrust on the rails equal to 9V of its own weight?
7. What is the least radius of the curve around which a car may run on a level at 30 miles per hour without producing a side thrust of more than xo 0 of its own weight?
8. What is the tension of a string 4 feet long, which has a weight of 10 pounds attached to it, describing a horizontal circle once in half a second.
9. A car is going around a curve of 500 feet radius at 30 miles per hour. Find the centrifugal force if the car weighs 6 tons.
10. A skater describes a circle of 100 feet radius with a velocity of 18 feet per second. What is his inclination to the ice?
11. Calculate the tension of an endless chain, of i pound per foot, and 30 feet long, when made to rotate as a horizontal circle once a second.
12. What weight will exert a tension of 16 tt^ pounds, when rotated twice per second around a circle of radius 10 feet?
13. A uniform disc rotates 300 times per minute about an axis through its center and perpendicular to its plane. It has attached to it two weights, one of 6 pounds and the other of 8 pounds, at an angular distance of 90° apart, the first being 2 feet from the axis and the second 4 feet from the axis. Find the magnitude and direction of the resultant centrifugal force on the axis. Find also where a weight of 16 pounds must be placed on the disc to make the resultant centrifugal force zero.
MOTION IN A CIRCLE 101
14. A railway car is going around a curve of 500 feet radius at 30 miles per hour. Find how much a plumb bob, hung from the roof of the car by a thread 6 feet long, would be deflected from the vertical.
15. At what angle should the plane of the rails of a car track be inclined to the horizontal, when the gauge is 56J inches, the radius of the curve is 300 yards, and the highest velocity attainable is 45 miles per hour?
16. How much must the outer rail of a car track of 4 feet 8| inches gauge be elevated on a curve of 750 feet radius in order that a train may exert a thrust nor- mal to the track when travelling at 45 miles per hour?
17. What is the maximum speed of a car rounding a curve of 1000 feet radius, when the outer rail is 3 inches above the inner one, and the gauge is 5 feet?
18. To what angle should a circular auto course of radius J mile be banked in order to allow a speed of 90 miles per hour, supposing no friction?
19. In a high school gymnasium is a circular track of 20 laps to the mile. How much must the track be banked in order to allow a maximum speed of 15 miles per hour?
20. A string 4 feet long, fixed at one end has attached, to its other end a ball of weight i pound which describes a horizontal circle, making 60 circuits per minute. What is the inclination of the string to the vertical and the tension in the string?
21. Find in inches the change in height of the conical pendulum of example 20, if its speed is increased to 100 revolutions per minute.
22. What percentage change of angular speed in a conical pendulum will correspond to the decrease in height of 2 per cent?
102 EI EMENTAHY MfiOHANICS FOR ENGINEERS
23. A string of length 4 feet and having one end attached to a fixed point, and the other end to a weight of 40 pounds, revolves as a conical pendulum 30 times per minute. Find the tension in the string and its inclination to the vertical.
24. The revolving ball of a conical pendulum weighs 10 pounds, and the height of the pendulum is 6 inches. What is the speed?
25. A ball of weight 2 pounds, attached to a string, is moving in a vertical circle of 4 feet diameter. If its velocity when passing through the lowest point is 30 feet per second, find its velocity and the tension of the string when it is 2 feet and 3 feet above the lowest point, and at the highest point.
26. A ball weighing 2 pounds is whirling in a vertical circle at the extremity of a string 4 feet long. Find the velocity of the ball and tension of the string (i) at the highest position, (2) at the lowest, (3) at a point 2 feet above the lowest point, (4) at a point 5 feet above the lowest point, if the velocity at the lowest point is just sufficient to cause the ball to make a complete revolution.
CHAPTER VI ENERGY OF ROTATION
Moment of Inertia
61. Energy of Rotation. If a particle of mass wig is moving in a circle of radius r, with angular velocity CO, then the linear velocity of the particle is cor; hence the kinetic energy is
K. E.=-^(cor)2 .... (io8) 2 g
We may suppose that a rotating body is made up of a number of particles of mass wi/g, ^2/^, ^3/^, etc., at distances ri, ^2, rs, etc., from the axis of rotation. If CO is the angular velocity of the body, then the linear velocities of the particles are cori, co/'2, cors, etc., since CO is the same for all. Hence the kinetic energy of a rotating body, being the sum of the energies of all the particles, is expressed by the equation
K. E.=i^ (cori)2+i?^^ (^r2)^+^^^ (0,^3)^+, etc., 2 g 2 g 2 g
or
K.E.=i co^nf-A .... (109)
62. Moment of Inertia. If the mass of each par- ticle which makes up a body, is multiplied by the
103
104 ELEMENTAKY MECHANICS FOR ENGINEERS
square of its distance from a given axis, the sum of the products is called the moment of inertia of the body about that axis.
Let / = the moment of inertia of the body about a given axis; w/g = the mass of any particle of the body;
r = distance of the particle from the given axis.
Then
^-^{y') (^^°)
The value of the kinetic energy is given by the equation
K. E.=i/a)2 (hi)
63. Radius of Gjrration. If the whole mass W/g of a body is concentrated at a single point which moves in a circle of radius k with the same angular velocity as the body, then the kinetic energy is I W/g oo^P. If k is so chosen that the energy is same as that of the rotating body, then
Hence
or
2 2 g
W
I = -k^ (II2)
, o Moment of mertia , .
^ ="■Ti^ • • • (113)
Mass ^ '^^
ENERGY OF ROTATION
105
Then (109) may be written
1 W
2 g
(114)
64. Moment of Inertia of a Thin Plate about an Axis Perpendicular to Its Plane.
Suppose two lines, OX and OYj perpendicular to each other and lying in the plane of the plate, meet at the point 0 where the axis pierces the plate (Fig. 33). ^ ^^^
Any constituent particle wi ' ^^'
is at a distance of xi from the Y axis, and at a dis- tance of yi from the X axis. The moments of inertia of the plate about the X and Y axes are
and
Adding these two equations we get
g
g Denoting the value of 2w/g r^ by /o we have
Io=Ix+Iy (115)
Also since x^-]-y^ = r^
ko^ = kJ^-{-ky^ (116)
106 ELEMENTARY MECHANICS FOR ENGINEERS
65. Moments of Inertia of a Thin Plate about Paral- lel Axes in Its Plane. Given the moment of inertia of the plate about an axis in the plane of the plate which passes through the center of gravity of the plate (Fig. 34), to find the moment of inertia about a new axis ZZ' in the plane of the plate, which is parallel to the axis 00' and at a distance a from it. Let /oo' be the moment of inertia of the plate about 00' \ and let Izz> be the moment of inertia of the plate about ZZ' , Then
Fig. 34.
g g
The sum 'Zwx is, by Theorem I, Section 52, equal to Xw times the distance of the center of gravity of the plate from 00', But x is zero, hence
T .^ 2 g
(117)
where W is the total weight of the plate. Dividing each term of this equation by W/g we get
k^zz' = Poo'+a^ (118)
ENERGY OF ROTATION 66. Table of Moments of Inertia
107
|
Name of Body and Dimensions |
Position of Axis about which Rotation Takes Place |
Moment of Inertia |
|
Uniform circular rod, length /, radius r |
Traverse through end |
iy/r2 /2\ g \4 3/ |
|
Uniform circular rod, length /, radius r |
Traverse through e.g. |
TF/r2 /2\ g \4 12/ |
|
Solid circular cylinder, radius r |
About its ov/n axis |
g 2 |
|
Hollow circular cyl- inder, radii, R and r |
About its own axis |
rm |
|
Solid sphere of ra- dius r Solid cone, radius of baser |
About a diameter About its own axis |
i 5 g lo |
|
Thin rectangular plate, length a, width h |
About an axis coin- ciding with side h |
W 1 ^ g 3 |
|
Thin rectangular plate, length a, width h |
Axis through e.g. parallel to side h |
W I , a^ g 12 |
|
Thin rectangular plate, length a, width h |
Axis through e.g. perpendicular to plane of plate |
W a2+62 g 12 |
|
Thin triangular plate, altitude h, base h |
Through e.g. parallel to base |
g i8 |
|
Thin circular disc, ra- dius r |
Through center, per- pendicular to disc |
EI,. g 2 |
|
Thin circular disc, ra- dius r |
About a diameter |
g 4 |
108 ELEMENTARY MECHANICS FOR EI^GINEERS
If we consider the cross-section of a uniform cir- cular rod as being very small, then W Ig becomes the length of the rod. Also if the thickness of a uniform plate is very small, then W jg represents the area of the plate.
Example i. A solid disc of iron is 20 inches in diameter and 4 inches thick. If the disc rotates 300 times per minute, what is its kinetic energy? Iron weighs .32 pound per cubic inch.
Solution. The weight of the disc is 128^ pounds. Therefore the kinetic energy equals
I287rXlOOXl007r2 .^ .,
= 2153.2 ft.-lbs.
.2X32X2X144
Example 2. A solid circular disc of radius 2 feet and thickness 4 inches, weighs 3200 pounds. How much work is required to increase its speed from 90 to 120 rotations per minute?
Solution. Change of kinetic energy = Work done (Section 35). Therefore the work required equals
3200X4X(i6x2-9^) ^ ^^ ^^ _j^g^ 2X2X32
Example 3. A thin rectangular plate 12 inches by 8 inches has a rectangular part 8 inches by 5 inches removed, the diagonals of the two parts being in the same straight line. Find the moment of inertia of the part remaining about one of the shorter outer sides.
.. _.j5
a i
ENERGY OF ROTATION 109
Solution. Let Tab be the moment of inertia of the part remaining about the side AB (Fig. ^ 35). The moment of inertia of the area abed about ^5 equals tVX 40X8^+40 X62. The moment of inertia of ABCD about ^5 equals |X8X 12X122. The difference between these two moments of inertia equals Iab- Hence Fig. 35.
Iab = 4608 — ^^-^ = 2961 ^ (inches)^. 3
EXAMPLES XVIII
1. A circular disc of radius 4 feet and weight 100 pounds makes 60 revolutions per minute. Find the kinetic energy of the disc.
2. Find the moment of inertia of a cylinder of weight 12 pounds and 4 inches radius, about an axis parallel to its own, and at a distance of 10 feet from it.
3. The radius of gyration of a flywheel is 4.VI feet. The wheel stores 20,000 tt^ foot-pounds of kinetic energy when rotating 100 times per minute. Find the weight of the wheel.
4. A flywheel requires 20,000 foot-pounds of work to be done upon it to increase its speed from 68 to 70 rotations per minute. What is the moment of inertia of the wheel? If the wheel weighs 4096 pounds, what is its radius?
5. A flywheel has a weight of 30 tons, which may be supposed to be distributed along the circumference of a circle 8 feet in radius; the wheel rotates 20 times per minute. Find its kinetic energy.
no ELEMENTARY MECHANICS FOR ENGINEERS
6. What is the kinetic energy of a circular saw having a diameter of 2 feet, and | inch thick, if a point on the circumference has a linear velocity of 6000 feet per minute? Density of steel being considered as 500 pounds per cubic foot.
7. A girder of I-shaped cross-section has two horizontal flanges 10 inches broad and 2 inches thick, connected by a vertical web 12 inches high and 2 inches thick. Find the moment of inertia of the area of the section (i) about a horizontal axis through its center of gravity, (2) about the lower edge as an axis.
8. Find the moment of inertia of the area enclosed between two concentric circles of 20 inches and 12 inches diameter, about a diameter as an axis.
9. Find the radius of gyration of a flywheel rim 6 feet in external diameter and 6 inches thick, about its axis.
.. „ If the rim is 10 inches broad,
and of cast iron, what is its moment of inertia about its axis? Cast iron weighs .26 pound per cubic inch. If the wheel rotates 200 times per minute, what is its kinetic energy?
jTjQ^^^^ 10. Fig. 36 represents the
cross-section of a girder. MN = 6 inches, KN=-2 inches, GH = 2 inches, GE = % inches, AC = 2l inches, and AB = 10 inches. Find the moment of inertia of the section (i) about AB {2) about an axis parallel to ^^ and passing through the center of grav- ity of section.
LOGARITHMS OF NUMBERS
111
LOGARITHMS OF NUMBERS
|
No. |
0 |
I |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
|
10 II 12 13 14 |
00000 04139 07918 II392 I46I3 |
00432 04532 08279 11727 14922 |
00860 04922 08636 12057 15229 |
01284 05308 08991 1238s I5S34 |
01703 05690 09342 12710 158.36 |
02119 06070 09691 13033 16137 |
02531 06446 10037 13354 16435 |
02938 06819 10380 13672 16732 |
03342 07188 10721 13988 17026 |
03743 07555 11059 14301 17319 |
|
IS i6 17 i8 19 |
17609 20412 2304s 25527 27875 |
17898 20683 23300 25768 28103 |
I8184 20952 23553 26007 28330 |
18469 21219 23805 26245 28556 |
18752 21484 240SS 26482 28780 |
19033 21748 24304 26717 29003 |
19312 22011 24551 26951 29226 |
19590 22272 24797 27184 29447 |
19866 22531 25042 27416 29667 |
20140 22789 25285 27646 2988S |
|
20 21 22 23 24 |
30103 32222 34242 36173 38021 |
30320 32428 34439 36361 38202 |
3053s 32634 34635 36549 38382 |
30750 32838 34830 36736 38561 |
30963 33041 35025 36922 38739 |
31175 33244 35218 37107 38917 |
31387 33445 3S4II 37291 39094 |
31597 33646 35603 37475 39270 |
31806 33846 35793 37658 39445 |
3201S 34044 35984 37840 39620 |
|
27 28 29 |
39794 41497 43136 44716 46240 |
39967 41664 43297 44871 46389 |
40140 41830 43457 4502s 46538 |
40312 41996 43616 45 1 79 46687 |
40483 42160 43775 45332 46835 |
40654 42325 43933 45484 46982 |
40824 42488 44091 45637 47129 |
40993 42651 44248 45788 47276 |
41162 42813 44404 45939 47422 |
41330 4297s 44560 46090 47567 |
|
30 31 32 33 34 |
47712 49136 50515 51851 53148 |
47857 49276 50651 51983 53275 |
48001 49415 50786 52114 53403 |
48144 495S4 50920 52244 53529 |
48287 49693 51055 52375 53656 |
48430 49831 SI188 52504 53782 |
48572 49969 51322 52634 53908 |
48714 50106 514SS 52763 54033 |
48855 50243 S1587 52892 54158 |
48996 SO379 S1720 S302O 54283 |
|
35 36 11 39 |
54407 55630 56820 57978 59106 |
54531 55751 56937 58092 592 1 8 |
54654 55871 58206 59329 |
54777 55991 57171 S8320 59439 |
54900 56110 57287 58433 59550 |
55023 56229 57403 58546 59660 |
SS145 56348 S7SI9 S8659 59770 |
55267 56467 57634 S8771 59879 |
55388 5658s 57749 58883 59988 |
SS509 56703 57864 58995 60097 |
|
40 41 42 43 44 |
60206 61278 62325 63347 64345 |
60314 61384 62428 63448 64444 |
60423 61490 62531 63548 64542 |
60531 6x595 62634 63649 64640 |
60638 61700 62737 63749 64738 |
60746 61805 62839 63849 64836 |
60853 61909 62941 63949 64933 |
60959 62014 63043 64048 65031 |
61066 62118 63144 64147 65128 |
61172 62221 63246 64246 6522s |
|
49 |
65321 66276 67210 68124 69020 |
65418 66370 67302 68215 69108 |
66464 67394 68305 69197 |
65610 66558 67486 6839s 69285 |
65706 66652 67578 6848s 69373 |
65801 66745 67669 68574 69461 |
65896 66839 67761 68664 69548 |
65992 66932 67852 68753 69636 |
66086 67025 67943 68842 69723 |
66181 67117 68034 68931 69810 |
|
SO SI 52 S3 54 |
69897 70757 71600 72428 73239 |
69984 70842 71684 72509 73320 |
70070 70927 71767 72591 73400 |
70157 71012 71850 72673 73480 |
70243 71096 71933 72754 73560 |
70329 71181 72016 72835 73640 |
70415 71265 72099 72916 73719 |
70501 71349 72181 72997 73799 |
70586 71433 72263 73078 73878 |
70672 71517 72346 73159 73957 |
|
No. |
0 |
I |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
112 ELEMENTARY MECHANICS FOR ENGINEERS
LOGARITHMS OF NUMBERS
|
No. |
0 |
I |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
|
55 56 57 S8 59 |
74036 74819 75587 76343 77085 |
^4"5 74896 75664 J6418 77159 |
74194 74974 75740 76492 77232 |
74273 75051 75815 76567 77305 |
74351 75128 75891 76641 77379 |
74429 75205 75967 76716 77452 |
74507 75282 76042 76790 77525 |
74586 746631 74741 75358|75435J755il 7611817619376268 76864 76938177012 77597 77670^77743 |
||
|
60 61 62 63 64 |
77815 78533 79239 79934 80618 |
77887 78604 79309 80003 80686 |
77960 78675 79379 80072 80754 |
78032 78746 79449 80140 80821 |
78104 78817 79SI8 80209 80889 |
78176 78888 79588 80277 80956 |
78247 78958 79657 80346 81023 |
78319 79029 79727 80414 81090 |
78390 79099 79796 80482 81158 |
79169 7986s 80550 81224 |
|
65 66 67 68 69 |
81291 81954 82607 83251 8388s |
81358 82020 82672 83315 83948 |
8142s 82086 82737 83378 8401 1 |
81491 821SI .82802 83442 84073 |
81558 82217 82866 83506 84136 |
81624 82282 82930 83569 84198 |
81690I81757 81823I81889 82347182413 82478J82543 82995 8305083 123 83187 83632 83696 83759 83822 84261J84323 84386:84448 |
|||
|
70 71 72 73 74 |
84510 85126 85733 86332 86923 |
84572 85187 85794 86392 86982 |
84634 8S248 85854 86451 87040 |
84696 85309 85914 86510 87099 |
84757 85370 85974 86570 87157 |
84819 85431 86034 86629 87216 |
84880 84942 85491185552 86094 '86 1 53 86688186747 8727487332 |
85003185065 85612J85673 862I3'86273 86806:86864 87390 87448 |
||
|
75 76 77 78 79 |
87506 88081 88649 89209 89763 |
87564 88138 88705 89265 89818 |
87622 88195 88762 89321 89873 |
87679 88252 88818 89376 89927 |
87737 88309 88874 89432 89982 |
87795 88366 88930 8e .87 90037 |
87852 87910 88423 88480 88986 89042 89542:89597 90091 90146 |
87967 88024 88536,88593 89098I891S4 89653(89708 90200 9025s |
||
|
80 81 82 83 84 |
90309 90849 91381 91908 92428 |
90363 90902 91434 91960 92480 |
90417 90956 91487 92012 92531 |
90472 91009 91540 92065 92583 |
90526 91062 91593 92117 93634 |
90580 91116 91645 92169 92686 |
90634 90687 9116991222 91698I91751 92221 92273 92737 92788 |
90741 90795 9127591328 91803 9185s 92324.92376 92840 92891 |
||
|
85 86 87 88 89 |
92942 93450 93952 94448 94939 |
92993 93500 94002 94498 94988 |
93044 93551 94052 94547 95036 |
93095 93601 94101 94596 95085 |
93146 93651 94151 94645 95134 |
93197 93702 94201 94694 95182 |
93247 93298 93752|93802 9425094300 9474394792 95231)95279 |
93349 !93399 93852 193902 94349 94399 94841 94890 95329 95376 |
||
|
90 91 92 93 94 |
95424 95904 96379 96848 97313 |
95472 96426 96895 97359 |
95521 95999 96473 96942 97405 |
95569 96047 96520 96988 97451 |
95617 96095 96567 97035 97497 |
95665 96142 96614 97081 97543 |
95713 96190 96661 97128 97589 |
95761 96237 96708 97174 9763s |
95809 962»4 96755 97220 97681 |
95856 96332 96802 97267 97727 |
|
95 96 98 99 |
97772 98227 98677 99123 99564 |
97818 98272 98722 99167 99607 |
97864 98318 98767 992 1 1 99651 |
97909 98363 988 1 1 99255 99695 |
97955 98408 98856 99300 99739 |
98000 98453 98900 99344 99782 |
98046 98091 98498:98543 98945 198989 99388:99432 99826 99870 |
98137 98588 99034 99476 99913 |
98182 98632 99078 99520 999S7 |
|
|
No. |
0 |
I |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
TABLE OF NATURAL VALUES
113
TABLE OP NATURAL VALUES OP THE TRIGONOMETRIC FUNCTIONS
|
Angle |
Angle |
||||
|
in |
sin |
tan |
cot |
cos |
in |
|
Degrees. |
Degrees. |
||||
|
0 |
oooo |
0000 |
Infinite |
I. 0000 |
90 |
|
I |
0175 |
0175 |
57.2900 |
9998 |
89 |
|
2 |
0349 |
0349 |
28.6363 |
9994 |
88 |
|
3 |
0S23 |
0524 |
19.0811 |
9986 |
87 |
|
4 |
0698 |
0699 |
14-3007 |
9976 |
86 |
|
5 |
0872 |
0875 |
II. 4301 |
9962 |
85 |
|
6- |
1045 |
105 1 |
9.5144 |
9945 |
84 |
|
7 |
1219 |
1228 |
8.1443 |
9925 |
83 |
|
8 |
1392 |
1405 |
7.1154 |
9903 |
82 |
|
9 |
1564 |
1584 |
6.3138 |
9877 |
81 |
|
10 |
1736 |
1763 |
5.6713- |
9848 |
80 |
|
II |
1908 |
1944 |
5 . 1446 |
9816 |
79 |
|
12 |
2079 |
2126 |
4.7046 |
9781 |
78 |
|
13 |
2250 |
2309 |
4.3315 |
9744 |
77 |
|
14 |
2419 |
2493 |
4.0108 |
9703 |
76 |
|
IS |
2588 |
2679 |
3.7321 |
9659 |
75 |
|
l6 |
2756 |
2867 |
3.4874 |
9613 |
74 |
|
17 |
2924 |
3057 |
3.2709 |
9563 |
73 |
|
l8 |
3090 |
3249 |
3.0777 |
9511 |
72 |
|
19 |
3256 |
3443 |
2.9042 |
9455 |
71 |
|
20 |
3420 |
3640 |
2.7475 |
9397 |
70 |
|
21 |
3584 |
3839 |
2.6051 |
9336 |
69 |
|
22 |
3746 |
4040 |
2.4751 |
9272 |
68 |
|
23 |
3907 |
4245 |
2.3559 |
9205 |
67 |
|
24 |
4067 |
4452 |
2 . 2460 |
9135 |
66 |
|
25 |
4226 |
4663 |
2.1445 |
9063 |
6S |
|
26 |
4384 |
4877 |
2.0503 |
8988 |
64 |
|
27 |
4540 |
5095 |
1.9626 |
8910 |
63 |
|
28 |
4695 |
5317 |
1.8807 |
8829 |
62 |
|
29 |
4848 |
5543 |
I . 8040 |
8746 |
61 |
|
30 |
5 000 |
5774 |
I. 7321 |
8660 |
60 |
|
31 |
5150 |
6009 |
1.6643 |
8572 |
59 |
|
32 |
5299 |
6249 |
I . 6003 |
8480 |
58 |
|
33 |
5 446 |
6494 |
1.5399 |
8387 |
57 |
|
34 |
5592 |
6745 |
1.4826 |
8290 |
56 |
|
35 |
5736 |
7002 |
I. 4281 |
8192 |
55 |
|
36 |
5878 |
7265 |
1.3764 |
8090 |
54 |
|
37 |
6018 |
7539 |
1.3270 |
7986 |
53 |
|
38 |
6157 |
7813 |
1.2799 |
7880 |
52 |
|
39 |
6293 |
8098 |
I . 2349 |
7771 |
51 |
|
40 |
6428 |
8391 |
1.1918 |
7660 |
SO |
|
41 |
6561 |
8693 |
I. 1504 |
7547 |
49 |
|
42 |
6691 |
9004 |
I.1106 |
7431 |
48 |
|
43 |
6820 |
932s |
1.0724 |
7314 |
47 |
|
44 |
6947 |
9657 |
I. 0355 |
7193 |
46 |
|
45 |
7071 |
I. 0000 |
I . 0000 |
7071 |
45 |
|
Angle |
Angle |
||||
|
in |
cos |
cot |
tan |
sin |
in |
|
Degrees, |
Degrees. |
ANSWERS
EXAMPLES I
1. 4 : I.
2. 6 : 7.
3. 50 ft. per sec.
4. 79I ft. per sec.
5. 70 ft.; no ft.; 1440 ft.
6. .293 ft. per sec. per sec.
7. 5itf sec.;' 112.44ft.
8. 18 ft. per sec. per sec.
9. 200 ft.; I ft. per sec. per sec.
10. 25 ft.; ^ ft. per sec. per sec.
11. 5 ft. per (sec.)2; 10 sec; 15 sec; 562.5 ft.
12. 6f ft. per (sec) 2.
13. Retardation 10 ft. per (sec) 2; 2 sec; 20 ft.
14. 1.467 ft. per (sec.)2; I- mile; 4.4 ft. per (sec.)^; -^ mile.
15. 3.44 ft. per (sec)2.
16. If ft. per (sec.)2; 7920 ft.
17. 3 ft. per sec; 2 ft. per (sec.)^.
18. 30 ft. per sec; 2 ft. per (sec.)^.
19. 40 sec.
20. 18 ft. per (sec.)\
EXAMPLES II
1. o) = f radians per sec
2. 20 x/3 radians per sec.
3. 7r/i8oo radians per sec.
4. Linear, i : 20 : 360; Angular, 1:12: 720
5. 44 radians per sec.
6. 5 sec
7* 135 revolutions; 90 sec. 8. 35.2 radians per sec; 2.581 radians per sec.
115
116 ANSWERS
Q. —. r — radians per sec; 18.52 miles per sec.
365.24-1800 f ) O f
10. 2.19 radians per sec; 13.14 ft. per sec
11. 42 ft. per sec, at 120° with first direction.
12. a;=24f rev. per sec; -^ rev. per sec
13. x=\o rev. per sec; \ rev. per sec; 5883^ rev.
14. TT radians; 18.84 ft. per sec; .157 radian per sec. per sec. 15- (^) 33 radians per sec; (&) 31.83 rev.; (c) 300 rev.
EXAMPLES III
1. 1600 ft.; .79 sec; 120 ft. per sec.
2. 400 ft.
3. 3.06 sec.
4. 64 ft. per sec.
5. 231.72 ft.
6. 432 ft.
7. Fi= 160 ft. per sec; total time is 10 sec.
8. 1565 ft.; .548 sec. and 5.7 sec. after time of the projection.
9. 8o|-j ft. above point of projection; ff sec and |^ sec after projection.
10. 160 ft. per sec; 400 ft.
11. 5 sec and 20 sec. later.
12. 4-mT sec
13. 5 sec
14. 2 sec. or 8 sec; 96 ft. per sec.
15. .9463 sec.
16. 256 ft.; 4 sec
17. 150 ft.
18. 112 ft.
19. / = 5 sec; 64 ft. per sec.
20. 42I ft.
EXAMPLES IV
1. (4'' 17', 34.87).
2. (85° 50', 7-12).
3. (51° 55', 12.55).
4. 15.095 miles per hour directed N. 10° 48' W.
5. North i2\/2 miles per hour; East 12V7 miles per hour.
6. Horizontal Velocity = Vertical Velocity = 300 VJ ft. per sec.
7. Upstream 30°; 1.57 minutes.
ANSWERS 117
8. lo ft. per sec.
9. Current 6.3096 miles per hour; steamer 9.01 12 miles per hour.
10. 70° 32'.
11. 14 ft. per sec; cos~^ If with greatest velocity.
12. Viyo ft. per sec; tan~^ ^ with line making tan~^ ^ with the east direction.
13. 64.78 miles directed S. 19" 7' E.
14. 28.3 miles directed S. 48'' 28' E.
15. (I35^ 2.234).
16. (2c7''3o', 26.7).
17. N. 10° 33' E.
18. N. 46" 9' W.; 76° 9'; 35 min, 40 sec.
EXAMPLES V
2. 29.93 ft. per sec; 1.336 sec.
3. 5° 23'.
4. (V3+V7) : I.
6. (i) 192 ft. per sec; (2) 153! ft. per sec. and iis| ft. per sec; (3) 960 ft.
7. 9211 ft.; 76.8 ft. per sec.
8. 19,771 ft.; 120.8 ft. per sec.
9. 2 sec. or 18 sec.
10. 17.26 ft. per sec.
11. 93.22 ft.
12. 18.97 ft. per sec.
13. 8j sec.
14. 32 ft., 96 ft., 160 ft. respectively. from top; 3 sec
15. 30°.
EXAMPLES VI
1. (i) 1614 ft.; (2) 7694 ft. and 20 sec; (3) 473.7 ft. per sec. and 5 = 36° 3' or 143° 57'.
2. 312^ ft.; 13° 43' or 76° 17'; 4.42 sec.
3. /=! sec; 360 ft.; 165 ft. per sec.
4. 799-8 ft.
5. 19.3s sec.
6. 1082.5 ft.
7. 1 1.2 sec
8. 271.2 ft. per sec; 8.47 sec
118 ANSWERS
9. 2131.S ft.
10. 67I ft.; 4.1 1 sec.
11. 60°.
12. 8189 ft.; 19.9s sec.
EXAMPLES Vn
1. 5 : 8.
2. 750 ft.
3. 1768 units.
4. 44.2 units.
5. 64 : 5.
6. i5i^lbs.
7. 1600 seconds.
8. 480 ft.
9. 48 ft. per sec; 720 ft. 10. 3000 units; i66,666| lbs. II- 5055! lbs.
12. 41 sec; 902 ft.
13. 71 sec.
14. 59.8 lbs. per ton. •
15. 7812I lbs.
16. 2,34 units; 780 lbs.
EXAMPLES VIII
1. 4 ft. per sec.
2. (i) 6i ft. per sec; (2) /^^ sec; (3) ^^ ft.
3. 25 tons.
4. 25 lbs.
5. i4iV ft.
6. 3ij tons.
7. 477 lbs.
8. loio ft.
9. 8 ft. per sec; si lbs.
10. 3000 lbs.
11. 170 lbs.
12. 5.982 sec.
13- f given distance.
14. I g.; i66f lbs.
15. 10 lbs.
ANSWERS 119
i6. 40 ft.
17. 4 lbs.
18. T^ g.; Vs sec.; ^ ft. per sec.
19- Ts g-; 2V7 sec. 20. s lbs.
EXAMPLES IX
1. 660,000 ft.-lbs.
2. 82,500 ft.-lbs.
3. 1,800,000 ft.-lbs.
4. SI H.P.
5. 10 H.P.
6. 396,000 ft.-lbs.
7. 3 H.P.
8. 40 H.P.
9. 2080 H.P.
10. 288 H.P.
11. 172^ ft.
12. 15 miles per hour.
13. 600 H.P.
14. 13 lbs.
15. $5.25.
16. (a) 39.27 H.P.; (b) 65.45 H.P.
EXAMPLES X
1. 20,000 ft.-lbs.
2. (i) 1280 ft.-lbs.; (2)0 ft.-lbs.
3. 19,360 ft.-lbs.
4. 4102 ft.-lbs.
5. 40,000 ft.-lbs.
6. 40 ft.-lbs.
7. 3160I ft.-lbs.
8. if ft.
9. 2000 ft. per sec.
10. 256 lbs.
11. 9^ ft. per sec; ^ ft.-lbs.
12. 880 lbs.
13. 2346 lbs.
14. 76.4 lbs.
15. 48,400 lbs.
120 ANSWERS
1 6. loo yards.
17. Gain 500 ft.; 37I mi. per hour.
18. 5.57 H.P. and 185.6 lbs.
19. 18,000 ft.-lbs.; 933%.
EXAMPLES XI
1. iP-\-Q) in same direction.
2. (a) 25; (b) 5; (c) loViT; (d) 4VJ; (e) 60°.
3. I2o^ 4. 90°.
5. Direction is along BO and resultant equals sBO.
7. 4.294 lbs., making an angle of 128° 9', with X-axis.
8. 16.87 lbs. making an angle of 121° 23' with X-axis. Q. Zero lbs.; hence, in equilibrium.
10. 17.4 lbs.
11. 67° nearly.
EXAMPLES XII
1. 120 lbs. and 90 lbs.
2. 80 lbs. and 60 lbs.
3. 52 lbs. and 73 lbs.
4. cos-i|.
5. 6 lbs.; 120° with X-axis.
6. PF=ioolbs; r=i4i.4lbs.
7. r=s6 1bs.; W^=7olbs.
8. 162° 31' inclined to the force of 9 lbs.; V133 ^^s.
9. 44.58 lbs. directed S. 11° 44' W.
10. P = 5oo lbs.
11. VJ : I : 2.
12. P=9.i7 lbs. and F = 4 lbs.
13. ID lbs.
14. 2^ tons; 7I tons.
15. cos-i|.
EXAMPLES XIII
1. 22b=io58| lbs.; JRa = ii4I3 lbs.
2. 160 lbs. at a distance of 6 ft. from A.
3. (i) 36 lbs., 3 inches from B; (2) 42 lbs., 12 inches from B'f (3) 18 lbs., 4 inches from B.
ANSWERS 121
4. (i) 10 lbs., having the same direction as P and applied at a distance of 12 inches to the left of A] (2) 20 lbs., having the same direction as Q and applied 6 inches to the right of B\ (3) 6 lbs., having the same direction as P and applied 30 inches to the left of A.
5. P= 2f lbs. and R= 13I lbs.
6. P=24f lbs. and i?=i3f lbs.
7. 2>\ ft. from the lesser force.
8. 8 ft.
9. 17.9 lbs.
10. 6 ft. from end man,
11. 4 lbs. and 6 lbs.
12. 20 lbs.
13. (i) 7 tons; (2) 7f tons and 6\ tons.
14. 14^ tons and 21^ tons.
15. 2V2 lbs. inclined 45° to side ^5, and cutting AB produced I a to the left of A.
16. 24^ lbs.; 168 lbs.; 24^ lbs.
, II
17. Tension 36 lbs.; pressure on hinge 4V 91 lbs.; g=tan~^ /-
downward.
EXAMPLES XIV
1. 3^ ft. from 8 lbs. weight.
2. 4.04 in. from side joining the 10 and 20 lb. weights.
|
3. |
2^ ft. from A. |
|
4. |
81 lbs. |
|
5. |
i\ ft. from center of the rod. |
|
6. |
f ft. from center of the wheel. |
|
7. |
\ av'2 from center of the square. |
|
8. |
At the center of the cross-bar. |
|
9- |
18^ in. from end of the long arm. |
|
10. |
12.88 in. |
|
II. |
S.72 in. |
|
h^+bh-^W |
|
|
12. |
2b-\-h ' |
|
h a+2b |
|
|
13- |
3 a-\-b' |
14. 10. in. from the base.
122 ANSWERS
1 6. .9 in. from the common base.
17. ^ aVl.
18. .166 in.
19. .637 in.
20. 10 in.
21. 19.37 in.
22. 15.91 in.
23. 12.73 in. from center; 8.486 in. from center.
24. 18.39 in.
25. 23.37 in.
26. .267 in.
27. A=6(3-v7)in.
28. .21 in.
29. 26J in.
30. 1.88 ft
31. 5.48 in.
32. 1.76 ft.
33. .113 in.
34. 38.22 in. . . .
35. 4.0s ft.
36. 8.7 in.
37. 4.08 in.
38. 5.1 19 in.
39. 4-82 in.
40. 4.33 ft.
41. 2| ft.
42. 3t3 in.
43. SiV ft.
44. f in.
45. 6.52 in.
46. 6.44 ft. 47- 8.1 1 ft.
48. 3I in.
49. 7-838 ft.; 2.286 ft.
EXAMPLES XV
1. 78° 28'.
2. 78° 41'.
3. 45".
4. 10 inches.
ANSWERS 123
5. tan0=-.
6. h^sVl inches.
7. h = 2VJ ft.
8. si'' 21'.
9. A=24V3 inches and radius is 6 inches.
10. 61° 25'.
11. Equilibrium when there are 4 courses; will topple when there are 5 courses.
12. 414 ft.-lbs.
13. 97,33S7r ft.-lbs.
14. 26,158 ft.-lbs.
15. 40,212.48 ft.-lbs.
16. 30,679,687.5 ft.-lbs. 17- 235,620 ft.-lbs.
18. 43,0307r ft.-lbs.
19. 27,000 ft.-lbs.
20. 138,240 ft.-lbs.
21. 18.
EXAMPLES XVI
4. 7207r2 sq. in.; 3600^'^ cu. in.
5. 2507r^ cu. in.; ioott'^ sq. ft.
6. 5680 cu. in.
7. 355-85 sq. in.; 6830.66 cu. in.
8. 841.06 sq. in.; 485.5 cu. in.
9. 113. 1 sq. in.; 28.27 cu. in.
10. i67r cu. ft.
11. 4421.6 sq. in.; 17,686.4 cu. in.
12. 597.5 sq. in.; 952.8 cu. in. 13- 3873 sq. in.; 21,360.5 cu. in.
14. 6408.8 cu. in.
15. 444.1 sq. in.; 333.1 cu. in.
EXAMPLES XVII
1. 28i lbs.
2. n=2.
3. i)=i2.6 ft. per sec.
4. «=2.32.
124
ANSWERS
5. 706 lbs.
6. » = 21^ mi. per hour.
7. i? = 6o5oft.
8. 197.36 lbs.
9. 1452 lbs.
10. About 84".
11. 6o7r/glbs. per ft.
12. 3.2 lbs.
13. io6.77r2 lbs. inclined 20^33' to the 8-lb. force; :»; = 2.134 ft. from center on opposite side of resultant.
14. 6° 54'. IS- 8° 36'.
16. io| inches.
17. 27^ mi. per hour.
18. II" 39'-
19. 19° 48'.
20. 78° 19'; 4.934 lbs. ^ •
21. 6.23 inches.
22. 0)2=IOI%(0i.
23. r=49.3lbs.; ^=35" 51'.
24. 76,4 rev. per min.
25. Fi=27.78 ft. per sec; 72 = 26.61 ft. per sec; ^3 = 25 38 ft. per sec; 71 = 24! lbs.; 72=21^ lbs.; r3=i8i lbs.
26. Fi = ii.3i ft. per sec; 72 = 25.3 ft. per sec; 73=22.11 ft. per sec; 74=17.84 ft. per sec; ^1 = 0 lbs.; r,= i2 lbs.: T, = q lbs.; 7^4=41 lbs. ^
EXAMPLES XVIII
1. 49348 ft.-lbs.
2. 37.5 units.
3. 3600 lbs.
4. 13,216 units; 14.37 ft.
5. 263,188 ft.-lbs.
6. 1278 ft.-lbs.
7. (i) 226ii (inches)*; (2) 6357! (inches)*.
8. 34,8i67r (inches)^
9. 770-7 units; 169,040 ft.-lbs.
10. (i) 2404^ (inches)*; (2) 995.63 (inches)*.
INDEX
The numbers refer to sections
Acceleration, 6
average, 6 angular, 12 Angular measurement, 9 motion, 8 velocity, 10 Average force (time), 28
Center of gravity, 49
of parallel forces, 49 Centrifugal force, 56 Centripetal force, 56 Circle, motion in a, 55 Circular arc, 9
motion, 3 Circular inclined track, motion
on, 57 Composition of parallel forces,
45 Conditions of equilibrium, 40,
44 Conical pendulum, 59 Curve, motion on railway, 58
Displacement, 3 Dynamics, i
Energy, 35
of rotation, 61
kinetic, 35
potential, 35 Engineer's unit of mass, 26 Equilibrium, conditions of, 40,
44 stability of, 53
First law of motion, 27 Force, 25
unit of, 25 moment of a, 42 time average, 28 Forces, parallel, 45
resolution and composi- tion of, 37, 38 triangle of, 41 Foot-poimd, 33
Gravity, theorems on center of, 52
acceleration of, 14
center of, 49 Gyration, radius of, 63 Guldinus, 54
125
126
INDEX
H
Horse-power, 34
I
Impulse, 28
Inclined plane, smooth, 19
Inertia, 2
moment of, 62
table of moments of, 66
K
Kinematics, i Kinetic energy, 35
of rotation, 61
Lame's theorem, 41 Laws of motion, 27
M Mass, 23 Matter, 2 Mechanics, i Moment, of a force, 42 of an area, 51 of inertia, 62 Momentum, 28 Motion, 3
due to gravity, 14
angular, 8
first law of, 27
of connected weights,
30,31 second law of, 27 third law of, 27, 29 translatory and rotary, 3
N Neutral equilibrium, 53 Newton's laws of motion, 27
Pappus, 54
Parallel axes, moments of inertia
about, 65 Parallel forces, resolution and
composition of, 45-48 Particle, material, 2 Pendulum, conical, 59 Plane, smooth, inclined, 19 Plane moments, 51 Potential energy, 35 Power, 34 Projectile, motion of a, 20
R
Radian, 9
Radius of gyration, 63 Railway curve, 58 Resultant force, 37
velocity, 17
Second law of motion, 27 Space diagram, 15, 38 Speed, 4
Stable equilibrium, 53 Statics, I
Theorems on center of gravity,
52 Third law of motion, 27 Time average force, 28 Triangle of forces, 41
of velocities, 17
U
Uniform circular motion, 8 Unstable equilibrium, 53
INDEX
127
Vector, IS
Vectors, addition and subtrac- tion of, i6 diagram, iS, 3^ Velocity, 4
average, $ angular, 10
Velocity component, 18
resultant, 17 Vertical circle, motion in, 60
W
Weight, 24 Work, 32
in lifting, 35
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