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MECHANICAL ENGINEERINGSCIENCE

(DYNAMICS)

ByHambeh S. T.Abikoye O. E.Ogundele S. O.



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Preface

This book is written principally for student of engineering inthe Polytechnics and Universities. It is aimed at properunderstanding of the concepts of Mechanical Engineering

Science (Dynamics) which will provide strong theoreticalbackground for future courses in engineering and thephysical sciences.

The language is simple to understand and relevant tostudent discipline. International systems of units areemployed except where stated.

Its hope that this book will provide a solid foundation for theunderstanding of advanced courses in mechanicalengineering such as: Mechanics of Machine, Strength of

Materials and Design.

Hambeh S. T.



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Table of Contents

Preface iiChapter One1.0 Dynamics 5

1.1 Linear Motion 51.2 Newton’s Laws of Motion 181.3 Angular Motion 22

Chapter Two2.0 Friction 352.1 Solid Friction 352.2 Determination of the Coefficient of Friction 392.3 Lubrication 40

Chapter Three

3.0 Work, Power and Energy 423.1 Work 423.2 Power 453.3 Energy 48

Chapter Four4.0 Principles of Simple Machines 514.1 Machines 514.2 Mechanical Advantages,

Velocity (or Movement) Ratio and Efficiency 514.3 The Lever 53

4.4 The Wheel and Axle 544.5 The Inclined Plane 554.6 The Screw and The Screw-Jack 564.7 Pulleys 584.8 Belt and Chain Drives 614.9 Gear Wheels 63

Chapter Five5.0 Stress and Strain 66



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Chapter One

1.0 DYNAMICSUnlike static, dynamics deals with motion and forces

acting on a body. In this section we shall be dealing withmotion of particles and the forces acting on them.

1.1 LINEAR MOTION1.1.1 DISPLACEMENT

Displacement from elementary physics was definedas distance measured in a specified direction. It is differentfrom distance in that, distance is a scalar quantity whiledisplacement is a vector quantity. The unit of displacement issame as that of distance, the metre (m). Distance isindependent of the path followed and of the time taken.

1.1.2 SPEEDSpeed is distance traveled per unit time. The distance

can be expressed in any convenient unit such as metre,kilometer etc, and the unit of time can also be anyconvenient value such as an hour, a minute or a second.

If a motor car travels a distance of 48km in one hour,its average speed is 48km/hr, but it is extremely unlikely thatthe car will travel at exactly this speed during the whole hour.Its speed will be at times higher and at other times lowerthan this value.

A body has constant speed only if it moves over equal

distances in equal intervals of time how-ever short theintervals.

The average speed of a body is the total distancedivided by the time; thus, if a body travels a distance xmetres in t seconds, the average speed, v metes per sec isgiven by



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V = x (meters) = x meters/second………1.1T (seconds) t

1.1.3 GRAPHS RELATING DISTANCE, TIMEAND SPEED

The relationship between distance and time andbetween speed and time can usefully be represented bysimple groups as shown below:

1.1.3.1 DISTANCE/TIME AND SPEED/TIMEGRAPHS FOR CONSTANT SPEED

Fig. 1.1 (a) is a graph showing the distance traveledduring a period of 20secs by a body moving at constantspeed. The slope of this straight line graph calculated interms of the units used for the two axes is 100m divided by20secs i.e. 5m/s and therefore represent the speed at whichthe body is moving.

100

80

Distance(m) 60 (a)

40

20

0

6

Speed(m/s) 4

2 (b)

0

5 10 15 20

Time

(S)

Fig. 1.1 Distance / Time and Speed / Time Graphs For

Constant Speed



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The horizontal line AB in Fig. 1.1 (b) is a graphrepresenting the constant speed of 5m/s derived from fig. 1.1(a).

The area shaded as shown in fig. 1.1 (b)= (speed in metre / second) x (time in seconds)

= 5 (m/s) x 20 (sec) = 100m= distance traveled.

1.1.3.2 DISTANCE/TIME AND SPEED/TIMEGRAPHS FOR VARYING SPEED

Fig. 2.2 (a) is a distance/time graph for a body whichtravels a distance of 100m in 20sec at varying speeds. Theaverage speed is 100m/ 20 sec =5m/s. The initial and thefinal speeds are zero: consequently the slopes of the graphat the beginning and the end of the period must be zero. The

100

Distance (m) 80

60 B (a)

40

20

0 A C

8

6

Speed (m) 4 (b)

2

0 5 10 15 20Time (sec)

Fig. 1.2 Distance/Time and Speed/Time Graphs for

varying speed



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slope at any immediate distance is obtained by drawing atangent to the graph at that instant: For example at 5seconds, the tangent is shown by the line AB.

Slope of AB = BC = 42 (m) = 5.6m/s AC (10 – 2.5) (sec)

Similarly, it is found that the slope of the graphbetween 7 seconds and 13 seconds is constant at about7.1m/s. Thus, by drawing tangents at various points of thedistance/time graph of fig. 1.2 (a) the speed/time graph offig. 1.2 (b) can be derived.

During the 1 second from 4.5 sec to 5.5 sec, theaverage speed is practically 5.6 m/s, so that the distancetraveled during that 1 second = 5.6m, and is represented bythe area of the shaded strip in fig. 1.2 (b). Similarly for all thethin strips into which we might divide the area under thegraph. Hence the total area enclosed by the speed /time

graph of fig. 1.2 (b) represents the total distance traveled.If the only information available about the movementof a body was the speed/time graph such as that shown infig. 1.3, the simplest method of determining the averagespeed is by means of mid-ordinates.



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s

Thus, if the base line is divided into say 6 equal lengths, asin fig. 1.3, and the mid-ordinates V1, V2, etc are drawn andmeasured, then;Average speed, V, = V1 + V2 + V3 + V4 + V5 + V6

6If t be the length of the base of the speed/time graphDistance traveled = area enclosed by graph

= average speed X time= vt

The larger the number of ordinates used, the moreaccurate the result.

1.1.4 LINEAR VELOCITYThe speed of a body can be stated without any

reference to the direction of movement of that body.Consequently, speed is a scalar quantity. If however, wespecify the direction of motion as well as the speed of thebody, the quantity is then termed the velocity of the body; if acar is traveling in a northern direction at a speed of 40

Speed

V3 V4

V2 V5

V1 V6

Time

Fig. 1.3



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kilometres / hour, the velocity is said to be 40km/hnorthwards. Since the velocity has both magnitude anddirection, it is a vector quantity and can be represented by astraight line drawn to scale in the direction of velocity. If abody travels a distance x in a constant direction in time t,

and if v is the average velocity, then,

V = x/t ……………………………….. 1.2

And the speed and velocity are numerically the same.

1.1.5 ACCELERATIONWhen the velocity of a body is increasing, the body is

said to be accelerating, whereas if the velocity is decreasing,the body is said to be retarding (or decelerating). Retardationmay be regarded as negative acceleration. Suppose that the

velocity of a train on a straight horizontal track increases by1.5m/s every second, from standstill until the train attains aspeed of 30m/s, then at the end of 1 second the speed is1.5m/s, at the end of the next second it is 3m/s, at the end ofthe third second it is 4.5m/s etc until at the end of 20sec thevelocity is 30m/s, and the variation of velocity with time canbe represented by the straight line OA in fig. 1.4 . It followsthat acceleration can be defined as the rate of change ofvelocity: and when the rate of change remains constant, asin fig. 1.4, the acceleration is said to be uniform.

1.1.5.1 EQUATIONS OF UNIFORMLY

Velocity (m/s)

30 A

20

10

0

Time(s)

Fig. 1.4 Uniform Acceleration



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ACCELERATED MOTIONSuppose the initial velocity of a body moving in a

straight line to be u, as shown in fig. 1.5, and suppose thevelocity to increase at a uniform rate to v in time t: then, if theacceleration is represented by the symbol a,

Change of velocity = v – uand acceleration, a = rate of change of velocity

= Change of VelocityTime

= v – ut

Therefore, v = u + at ………………..…………………1.3If u and v are expressed in metre/second and t is in

seconds, then the acceleration is in metres per secondsevery second, i.e. metres per second squared, the symbolbeing m/s2.

Since the velocity is assumed to vary at a uniform ratebetween u and v, it is represented by the straight line AB in

Velocity

B

at

A C

u t

0 Time D

Fig. 2.5 Uniform Accelerated



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symbol of g. owing to the radius of the earth being slightlysmaller at the north and south poles than it is at the equator,the gravitational pull is slightly higher at the poles than it is atthe equator. Consequently, the value of g is about 9.832m/s2 at the poles and about 9.780m/s2 at the equator. In London,

at sea level, the value of g is almost exactly 9.81m/s

2

.It follows from equation (1.3) that if a body, initially atrest, falls freely for time t, the final velocity v is given by

V = gt……………………………… 1.6If x is the distance traveled, then from equation (1.4)

x = average velocity X time= ½ vt = ½ gt2 ……………….. 1.7

and from equation (1.5)v = √2gx ………………………. 1.8

1.1.6 RELATIVE VELOCITY

When we speak of the velocity of a body, wegenerally mean its velocity relative to the earth, which isitself moving at a high speed; i.e. we think of the rate ofdisplacement as if the earth were at rest. Similarly wesometimes speak of the velocity of one body relative to theearth. For simplicity, we shall limit our discussion of relativevelocity to velocities in, or parallel to a single straight line.

If a train A is traveling, say, east at 80 km/h, thenrelative to a second train B traveling east at 50 km/h on aparallel track, the first train is moving at:

80 – 50 = 30 km/h.To an observer in the second train, A would appear to

be traveling at 30 km/h eastward; while to an observer on A,train B would appear to be traveling westward at 30 km/h.

If the second train B had been traveling westward at50 km/h, the first train A would be traveling eastward relativeto the second train at:

80 – (-50) = 130 km/hAnd would appear to be traveling eastward at 130

km/h to an observer in the second train B. To an observer in



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A, train B would appear to be traveling Westward at 130km/h.

1.1.7 RESULTANT OF TWO VELOCITIESSuppose an aero plane to be flying at a constant

velocity such that in perfectly still air it would travel northward at 500 km/h relative to the ground and suppose the airto have a constant velocity of 80 km/h in an eastern directionrelative to the ground. Under such circumstance the planewill still be traveling at 500 km/h north ward relative to theair.

We can represent these two velocities vectorially as inFig. 1.6 where vector OA represents the air velocity of 80km/h eastern relative to the ground, and vector ABrepresents the plane velocity of 500 km/h northern relative tothe air.



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The vector sum of OA and AB is OB i.e. OBrepresents the velocity of the aero plane relative to theground.

It will be noted that we started from point O with thevector OA representing a velocity component relative to theground. Consequently the vector OB gives the resultantvelocity also relative to the ground i.e. relative to the objectwe regard as stationary.From fig. 1.6 it will be seen that:OB = √(OA2 + AB2)

= √(802 + 5002)= 506Km/h

If is the angle between OB and AB,

B

500km/h

O A80km/h

Fig. 1.6 Resultant of two velocities



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Tan = OA/AB = 80/500 = 0.16

= 9.10 Hence the resultant velocity of the aero plane is 506

Km/h, in a direction 9.10 East of North.Let us now consider the general case of a body

possessing two simultaneous velocities represented byvectors OA and AB in Fig. 1.7 and suppose to be the anglebetween the directions of the two velocities. The resultantvelocity is the vector sum of OA and AB, namely OB in Fig.1.7. Double arrow heads have been inserted on OB toindicate that it represents the resultant of two velocities.

If both the magnitude and the direction of the resultantvelocity is required, the simplest method at this stage is todraw the diagram to a large scale and measure, the length of

OB and the value of angle . If, however, only themagnitude of the resultant is required, this can be calculatedfrom the relationship.

OB = √(OA2 + AB2 + 2 X OA X AB Cos)

B

O A Fig. 1.7 Resultant of two velocities



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1.1.8 RESOLUTION OF A VELOCITY INTO TWOCOMPONENT VELOCITIES

A velocity can be resolved into two components, theonly condition being that the resultant of the two vectorsrepresenting the component velocities must be the same in

magnitude and direction as the vector representing theoriginal velocity. The procedure is similar to the resolution offorces.

In practice, the directions of the components aregenerally specified; and these directions are usually at rightangles to each other, in which case they are referred to asrectangular components. Thus, if vector OB in Fig. 1.8represents the magnitude and direction of the velocity withwhich,

say a, cricket ball is thrown into the air, this velocity can beresolved into a horizontal component OA and a vertical

component AB. If be the angle between OA and thehorizontal component,

B

900

O A

Fig. 1.8 Resolution of Velocity into Rectangular Components



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Horizontal Component of the Velocity =

OA = OB Cos ………………….……….. 1.9And, Vertical Component of the Velocity =

AB = OB Sin ……………………………. 1.10

1.2 NEWTON’S LAWS OF MOTION A force may, in static be defined by the change in

configuration if produces by a standard body. In dynamics, aforce also produces changes in a body, but of far greaterimportance is the fact that a force changes the motion of abody; this change in motion is used to define a force, and thechange is accounted for in Newton’s three laws of motion:these laws are the foundations of classical dynamics andwere first stated in Newton’s famous book Principia in 1687;these laws are axioms in dynamics. The three laws can bestated as follow:

1.2.1 FIRST LAW OF MOTIONA body continues in its state of rest or of uniform

motion in a straight line unless it is compelled by an externalforce to change that state. This law may be called the ‘law of inertia’. A body will not change its state of rest or of uniformin a straight line unless compelled to do so; i.e. it resists anychange of velocity in magnitude or direction. From this law,we may define a force as any push or pull which changes ortends to change the state of rest of a body or its uniformmotion in a straight line.

1.2.3 MOMENTUMThis is the name given to the product of the Mass m

of a body and its Velocity v i.e. Momentum = mvMass has no direction and is therefore a scalar

quantity; but velocity has magnitude and direction and istherefore a vector quantity. Hence, momentum must also bea vector quantity; and a vector which represents the velocity



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of a body can also, to a different scale, represent themomentum of that body. Values of momentum can thereforebe added and resolved vectorically in the same way as forvelocity.

1.2.4 SECOND LAW OF MOTIONWhen a body is acted upon by an external force, therate of change of momentum is proportional to the force andtakes place in the direction of the force.

This law is in effect the definition of a force in respectto its effect on a body.

If a force F acts upon a body of Mass m for a Time t and causes its velocity in the direction of the force toincrease form v1 to v2,

Initial Momentum = mv1 And final Momentum = mv2

Therefore, average rate of change of Momentum =(mv2 – mv1)/t = m (v2 – v1)/t = maa = average acceleration during time t

Hence, from Newton’s Second Law, F ma = ma x a constant ………………………. 1.11The SI Unit of Force is the Newton and it is defined as

the force required to give a mass of 1kg an acceleration of1m/s2. In order to give a value of one for the constant,substituting m = 1kg, a = 1m/s2 and F = 1N in expression1.11, we have;

1[N] = 1[Kg] x 1[m/s2] x a constant

Hence the constant is unity.It follows that if F be the force, in Newton, required togive a mass m, in Kilograms, acceleration a, in metres persecond squared, then;

F = ma…………………………………….. 1.12

EXAMPLE A diesel engine pulling a train along a level track has

its oil supply cut off when the train is traveling at 60km/h. It is



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observed that the speed falls to 40km/h after the train hastraveled a distance of 1200m. The mass of the engine andcarriages is 80Mg. Assuming the retardation to be uniform;calculate the total force resisting motion.

SOLUTION 60km/h = 60,000(m) / 3600(s) = 16.67m/s40km/h = 40,000(m) / 3600(s) = 11.11m/sFrom v2 = u2 + 2ax(11.11)2 [m/s]2 = (16.67)2 [m/s]2 + 2a x 1200[m]

a = 0.0642m/s2 Mass of train = 80Mg = 80,000KgSince F = maRetarding Force = 80,000[kg] X 0.0642[m/s]2

= 5136N = 5.136KN

1.2.5 THIRD LAW OF MOTIONTo every force, there is an equal and opposite force.This law may be stated in another way. If a body, A exerts acertain force on another body B, then B exerts on A a forceof equal magnitude but in opposite direction.

This law applies to bodies whether they are at rest orin motion. Thus, if a beam or other body exerts a certaindownward force upon a support, the support exerts an equalupward force on the beam - the later force being referred toas the reaction of the support.

The fact that this relationship is equally true in regardto two bodies in motion is not so widely realized, e.g. that thebackward pull of a trailer on a motor vehicle is equal to theforward pull of the vehicle on the trailer. Let us look a littlemore closely at the state of affairs when a motion tractor ispulling a trailer along a level road. If the tractor is moving ata constant velocity, the forward pull exerted by the tractorexactly balances the backward pull of the trailer due tofriction and wind resistance.



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If, however, the forward pull exerted by the tractorexceeds the backward pull due to friction and windresistance, acceleration occurs. In this case, the forward pullexerted by the tractor may be regarded as being the sum ofthe following two components:

a. the pull F1 required to haul the trailer a constantvelocityb. the pull F2 (= ma) required to give the mass m of the

trailer an acceleration a. Hence, the total forward pullexerted by the tractor on the trailer is (F1 + F2) and isexactly equal to the total backward pull exerted by thetrailer on the tractor.

Another example of the application of Newton’s ThirdLaw is the recoil of a gun firing a bullet or a shell.

EXAMPLE A rope supports a mass of 50Kg. Calculate the pull onthe rope when the mass is being;a. Raisedb. Lowered, with an acceleration of 0.5m/s2.

SOLUTION When the mass is stationary or moving at a uniform velocity,Pull on rope = Weight of the mass

= 50 X 9.8= 490.5N

a. If F be the upward pull, in Newton, when the mass isbeing raised with an acceleration of 0.5m/s2, the forceavailable for upward acceleration is (F – 490.5)Newton.Hence,(F – 490.5)[N] = 50[Kg] X 0.5[m/s2]so that F = 575.5N

b. If F be the upward pull in Newton, when the mass isbeing lowered with an acceleration of 0.5m/s2, the



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force available for downward acceleration is (490.5 – F) Newton.Hence,(490.5 – F)[N] = 50[Kg] X 0.5 [m/s2]so that F = 465.5N

1.3 ANGULAR MOTIONAngular motion implies motion described in a circular path.

1.3.1 ANGULAR DISPLACEMENTWhen a shaft rotates, a point on the surface of the

shaft moves in a circular path and a line joining the point tothe centre of the rotation sweeps out an angle as in Fig. 1.9.This angle is referred to as the ‘Angular Displacement’ of thepoint. The unit of Angular Displacement is the Radian, whereOne Radian is the angle suspended at the centre of a circle

by an arc equal in length to the radius, as in Fig. 1.10.

The relationship between the arc length x , the radius r

and the angular subtended , is given byx = r …………………..……………………… 1.13Thus, when the arc length x equals the radius r, we

have x = r = and so is 1 Radian or 1 rad. When the arc

Position Some

Time lake

r

Centre Angular Centre rad

Displacement r

Initial Position

of point

Fig. 1.9 Angular Displacement Fig. 1.10 The Radian

r

x



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length is a complete circle, e.g. the point on the rotating shafthaving moved through one complete revolution, then as the

circumference of a circle is 2πr we have x = 2πr = r and so

= 2π rad. As one complete revolution is a movement

through 3600 we must have 2π rad = 3600 and hence,

1 rad = 3600 2π = 1 rad = 57.3 .

It is generally easier to remember that 2π rad equalsto 3600 than 1 rad is 57.30.

EXAMPLE a. If a point on a shaft is given an angular displacement

of 1.2 rad, what is the angular displacement indegrees?

SOLUTION

Since 2π rad = 3600 1 rad = 3600

2π Then 1.2 rad = 1.2 X 3600

2π = 68.80

b. Calculate the distance moved by a point on the tyrethread of a car wheel of radius 560mm if it rotatesthrough 1.5rad.

SOLUTION

Since x = r and the distance moved by the point is the arc length,

distance moved = r = 560[mm] X 1.5[rad]= 840mm



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1.3.2 ANGULAR VELOCITYWhen, for example, a shaft is rotating then a point on

the shaft surface has an angular displacement which isvarying with time. It can be said to have an angular velocity.Angular velocity is defined as the rate of change of angular

displacement with time. It is denoted by ω (omega) and hasthe unit of radian per sec (rad/s).

If a point on a rotating shaft takes a time t to rotate

through an angle , even the average angular velocity duringthat time interval is

Average angular velocity ω = …………….. 1.14t

If there is constant angular velocity then equal angulardisplacements are covered in equal intervals of time – however short the interval.

The angular velocity is related to the frequency ofrotation. Thus if n revolutions are made per second, then,

since one revolution is an angular displacement of 2π

radians, n revolutions is an angular displacement of 2πnradians. This angular displacement occurs in 1 sec, hence,Average angular velocity

ω = angular displacement 2πn [rad]Time 1[s]

ω = 2πn ………………………………………….. 1.15

EXAMPLE A flywheel rotates at 2400 rev/min. What is its angularvelocity?

SOLUTION:

Since ω = 2πnand n = 2400 rev/min = 2400 = 40rev/s, then

60



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ω = 2π X 40 = 251 rad/s.

1.3.3 ANGULAR ACCELRATIONWhen the angular velocity of a body is changing,

there is said to be an angular acceleration. Angularacceleration is defined as the rate of change of angular

velocity with time. The symbol for angular acceleration is (alpha) and the unit is radian per second square (rad/s2).

If the angular velocity of a rotating object changes

from ω0 to ω1 in a time interval of t , then the averageangular acceleration during that time interval is

Average acceleration = (ω1 – ω0)… ………… 2.16 t

If there is constant angular acceleration, then the angularvelocity is changing by equal amounts in equal intervals oftime – however short the interval.

EXAMPLE The angular velocity of a grinding wheel changes from zeroto 150 rad/s in 30s, what is the average angularacceleration?

SOLUTION

Since = (ω1 – ω0) = (150 – 0) [rad/s]

t 30[s]= 5.0 rad/s2



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1.3.4 RELATIONSHIP BETWEEN LINEAR ANDANGULAR MOTION

B

O x

r

A Fig. 1.11

Consider a point on the rim of a wheel of radius r

moving with a constant angular velocity ω, as in Fig 1.11. Ina time t the radius OA rotates through an angle , where,

ω = and so = ωtt

But the arc length covered in this time, x, is given by

x = r

and so x = rωtThe distance covered round the circular path is x in a

time t. But the distance covered divided by the time taken isthe linear speed v for the point on the rim of the wheel.

v = x t

Hence, v = rωtt

v = rω ………………………….. 1.17



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The linear velocity of a point on the rim of the wheel isnumerically equal to the speed but is always directed alongthe tangent to the rim of the wheel.

Now consider the motion of a point on the rim of thewheel when the angular velocity is changing and there is an

angular acceleration. The point on the rim of the wheel willthen have a linear acceleration causing it to accelerate roundits circular path. (Note that this is a linear tangentialacceleration and is not the same as the radial accelerationwhich is referred to as the centripetal acceleration). If for Fig.1.11, the point on the rim has a linear velocity v0 at point Aand this has change to v1 by point B, then as this occurs in atime t, the average linear acceleration a is given by:

a = change in linear velocitytime taken

a = (v1 – v0)

tIf ω0 is the angular velocity when the point is at A and

ω1 the angular velocity when it is at B, then usingexpression, eqn. 1.17.

v0 = rω0 and v1 = rω1

Hence,

a = (rω1 – rω0)t

a = r (ω1 - ω0)t

But the angular acceleration is (ω1 – ω0)/t, hence,a = r ………………………………………….. 1.18

EXAMPLE a. A grinding wheel has radius of 100mm and rotates at

30 rev/s. What is the grinding speed at thecircumference of the wheel?SOLUTION



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Since ω = 2πn

ω = 2π X 30 [rev/s]

ω = 60π rev/s

As v = rω

v = 0.100[m] X 60π [rad/s] v = 18.8 m/sb. The angular velocity of a car wheel increases from 5

rad/s to 50 rad/s in 30s. If the wheel has a radius of350mm, what is

i. The average angular acceleration, andii. The average linear acceleration of a point on the rim

of the wheel?

SOLUTION

i. Since = ω1 – ω0 = (50 – 5) [rad/s]

t 30[s] = 1.5 rad/s2

ii. Since a = r a = 0.350(m) X 1.5 [rad/s2]a = 0.525 m/s2

EQUATIONS OF ANGULAR MOTIONFor motion with constant angular acceleration we can

write a number of equations, comparable to the equations forlinear motion. Equation 1.16 can be rewritten in the form

ω1 = ω0 + t …………………………………. 1.19

During the time interval t the average angular velocityis

Average angular velocity = ω0 + ω1 2

Since, according to equation 1.14

Average angular velocity = /t

Then /t = ω0 + ω1 2



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= ½(ω0 + ω1)

Since ω1 = ω0 + t (eqn. 1.19), then

= ½ (ω0 + ω0 + t) t

= ω0t + ½t2 ………………………….. 1.20

If we take eqn. (1.19) and square it, we haveω12 = (ω0 + t)2

ω12 = ω0

2 + 2ω0t + 2t2

= ω02 + 2 (ω0t + ½t2)

Hence using eqn. (1.20)

ω12 = ω0

2 + 2 ……………………..... 1.21

The following table shows a comparison of the linearmotion and the angular motion equations and quantities.

LINEAR MOTION ANGULAR MOTION Distance x Angle

Velocity v Angular Velocity ω

Acceleration a Angular Acceleration

v = x/t ω = /t

v = u + at ω1 = ω0 + t

x = ut + ½at2 = ω0t + ½t2

v2 = u2 +2ax ω12 = ω0

2 + 2

EXAMPLE

A wheel, initially at rest, is subjected to a constantangular acceleration of 2.0 rad/s2 for 50s, calculate theangular velocity attained and the number of revolution thewheel makes in that time.

SOLUTION

Since ω1 = ω0 + t

And ω0 = 0, then



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ω1 = 0 + 2.0 (rad/s2) X 50 (s)= 100 rad/s

Since = 0 + ½ x 2.0 [rad/s] x 502[s]2 = 2500rad

One momentum is 2πrad, therefore, the number of

revolutions n isn = 2500 [rad]

2π [rad] = 397.9

TORQUE AND ANGULAR MOTIONTorque, or the moment of a force, T about an axis is definedby

T = F.r …………………………………………. 1.22Where r is the radius of the turning circle to which the

force is tangential. This is sometimes stated as torque is theproduct of the force and the perpendicular distance of the

force from the axis of rotation.Consider a rigid body which is rotating about an axis

through 0, as in Fig. 1.12.

F

A

O

Fig. 1.12. Rigid Body

O



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A particle of that body, at A, rotates through a turningcircle of radius r. If F is the tangential force acting on thisparticle at A, then using Newton’s Second Law of Motion wehave,

F = ma

Where a is the linear acceleration in the direction of the

force. If is the angular acceleration, then

a = r

and so F = mr Hence, the torque T acting on the particle is

T = Fr = mr2 The rigid body is made up of a large number of smallparticles, each having different turning circles. The totaltorque acting on the body is the sum of the torques acting on

all small particles. Thus,Total torque = Sum of all (mr2) terms

All the particles will have the same angular acceleration.Thus,

Torque T = (Sum of mr2 terms) X The sum of all the mr2 terms is called the moment of inertia,symbol I, of the body. Hence,

T = I ………………………………………….. 1.23

This equation for angular motion can be compared

with the equation F = ma for linear motion. Torque is theangular equivalent of force, angular acceleration is theangular equivalent of linear acceleration and moment ofinertia is the angular equivalent of mass.

Mass is being defined as having a quality of inertia orreluctance of change velocity. Moment of inertia is a similarconcept of angular motion, representing the inertia orreluctance to change angular velocity.



xxxii

EXAMPLE What torque has to be applied to a flywheel, with a

moment of inertia of 30kg/m2, to give it an angularacceleration of 0.5 rad/s2?

SOLUTION Since T = I T = 30 [kg/m2] X 0.5 [rad/s2]

= 15Nm

MOMENT OF INERTIAThe moment of inertia, symbol I, of a body about a

particular axis is defined by the equation;

I = mr2 ……………………………… 1.24Where r is the distance of a particle of mass m from the axis

of rotation and the symbol is used to indicate that the

moment of inertia is the sum of all the mr2

terms for all theparticles in the body. Fig. 1.13 shows the moments of inertiaof some simple bodies.

rRadius r

axis axis

(a) For a Sphere I = 2

/ 5 mr 2

(b) For a Disc I = ½mr 2

l

Axis Radius r Axis (c) For a Ring I = Mr 2 (d) For a Slender Rod I = 1 / 12 Ml 2

Fig. 1.13 Moments of Inertia



xxxiii

The moment of inertia of any body depends on theposition of the axis about which the moment is considered. Ifthe moment of inertia I0 about a particular axis is known thenthe theorem of parallel axes enables the moment of inertia to

be calculated for any parallel axis; if I is the moment ofinertia about some parallel axis a distance d from the axisconsidered for the moment of inertia I0, then,

I = I0 + Md2 ………………………….……………. 1.25

Where M is the total mass of the body. Thus for a disc ofradius r and mass M , the moment of inertia about an axisthrough the centre is ½Mr2, the moment of inertia of thesame disc about an axis tangential to the circumference ofthe disc is ½Mr2 + Mr2 = 3 / 2Mr2.

Some objects though more complex than the objects

shown in Fig. 1.13 can be considered to be composed of anumber of such simple objects. We could have, for instancean object considering of two spheres mounted at the ends ofa slender rod, i.e. a dumb-bell. The moment of inertia of thisobject about an axis through the centre of the rod can bedetermined by adding together the moments of inertia of thecomponent parts. The moment of inertia of a sphere aboutan axis through its centre is 2 / 3Mr2.

What is required is the moment of inertia of thesphere about an axis through the centre of the rod. If the rod

has a length , i.e. the distance between the centres of the

sphere is + 2r then the distance of the axis of rotation fromthe centre of a sphere is ½ + r and so the moment of inertiaof a sphere about this axis is, using the parallel axis

theorem, 2 / 3Mr2 + M (½ + r)2.As there are two spheres the total moment of inertia

due to the spheres is twice this value. The rod has also a

moment of inertia of 1 / 12M2. Hence, the total moment ofinertia is;



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I = 2 [2 / 5Mr2 + M(½ + r)2] + ½M2

Whatever the form of an object and however its mass isdistributed, it is always possible to represent the moment ofinertia is the form I = mk2. Thus in the case of the disc where

I = ½mr2

about the central axis, K2

= ½r2

, K is called theradius of gyration. The significance of this radius is that wecan consider the body effectively to be behaving for rotationas through it had al its mass concentrated at a point adistance k from the axis of rotation.

EXAMPLE A flywheel has a mass of 300kg and a radius of gyration of1.0m. What is(a) The moment of inertia of the flywheel, and(b) The torque necessary to give it an angular

acceleration of 0.5 rad/s

2

?

SOLUTION (a) Since I = mk2

I = 300 (kg) X 1.02 (m)2 = 300 kgm2

(b) Since T = I T = 300 (kgm2) X 0.5 (rad/s2)

= 150 Nm.



xxxv

N

W

Fig. 2.1

Chapter Two 2.0 FRICTION2.1 SOLID FRICTION

When a person walks along a road, he or she isprevented from slipping by the force of friction at the ground.

In the absence of friction, for example on an icy surface theperson’s shoe would slip when placed on the ground. Thefrictional force always opposes the motion of the shoe.

Suppose a body to be resting on a horizontal table. Itis acted upon by a downward force W due to thegravitational pull of the earth upon it (its weight). There mustalso be a reaction N normal to the surface of the table. Thisreaction, exerted by the table on the body, is equal andopposite to force W as shown in Fig. 2.1, where for-clarity-the line of action of N is shown displaced slightly from that ofW . Actually, they are in direct opposition.

If the body were resting on a surface inclined to thehorizontal, the value of N would be less than that of W and

the line of action of N would not be directly opposite that ofW.

If a horizontal force P is applied to the body, pullingtowards the right as in fig. 2.2, the body will not move unlessP is large enough. This is due to the presence of a force offriction F exerted on the body by the table towards the leftand just sufficient to balance force P. This force of friction Fhas the feature of adjustment so as to be exactly equal and



xxxvi

opposite to P, i.e. if P is increased, F increases by exactlythe same amount, until motion occurs.

While the body is at rest, we may regard it as being inequilibrium under the action of the four forces shown in Fig.2.2. The table top exerts two forces, N vertically (equal andopposite to W) and F horizontally (equal and opposite) asshown in Fig. 3.3(a). From the force diagram of Fig. 2.3(b), itis seen that these two forces can be replaced by a resultant

force R acting at an angle to the left of the line of action of

force N, the value of being given by:

Tan = F/N ………………………………………. 2.1

N

P

W

Fig. 2.2: Frictional Drag

P N

N

R

F

(a) (b)

Fig. 2.3: Resultant Reaction



xxxvii

We may therefore regard the body as being in

equilibrium under the action of three forces, namely W, Pand R, shown in Fig. 2.4(a). These forces can berepresented vectorially by the triangular force diagram of Fig.2.4(b).

As the pulling force is increased, the friction force F(equal and opposite to P) also increases and the resultant R

inclines more to the left, i.e. angle increases. There is,however, a limit to this adjustment of F to resist a growingvalue of P. This limit is reached when the body is on thepoint of motion towards the right. Thus, when motion just

begins to take place, the friction operating against motion isthe maximum or limiting friction.This condition (i.e. when component F has reached itslimiting value or what is the same thing, when R has reachedits greatest inclination to the vertical) is shown in Fig. 2.5(a).The corresponding angle between R and the perpendicularor normal to the sliding surface is termed the angle of

Friction and is represented by . From equation 2.1 it followsthat:

R

W P

R

W

Fig. 2.4: Three Forces Acting on Body.



xxxviii

Tan = Sliding Friction Force F

Normal Reaction N

The ratio of the sliding friction force to the normalreaction of the supporting surface is termed the coefficient ofstatic friction when the object is just on the point of slidingand is represented by U,

i.e. U = Sliding Friction Force = tan Normal Reaction ….2.2

So far, we have referred to just the limiting frictionalforce that occurs when an object is just on the point of

sliding, such friction being referred to as static friction sincethe object is not moving. However, friction still occurs whenan object is sliding, the friction in this situation being calleddynamic friction. The frictional force in such a situation isgenerally slightly less than the limiting frictional force for thestatic condition. We thus have a coefficient of dynamicfriction, defined in the same way as for static friction which isslightly less than the static coefficient.

R P

W P

R

(a) W (b)

Fig. 2.5: Angle of Friction.



xxxix

For a body of weight W, pulled along a horizontalsurface by a horizontal force P. P and W are equal andopposite to the sliding friction force and the normal reactionrespectively; hence for this condition:

U = P/W ……………………………………. 2.3

Equation (2.3) is not applicable if the pulling for P isinclined to the surface or if the surface is not horizontal.

2.2 DETERMINATION OF THE COEFFICIENT OFFRICTION

Fig. 2.6 shows a simple apparatus by which experiments onthe amount of friction between two surfaces can beperformed. A wooding board A is supported horizontally on atable or bench; and on A rests of rectangular block B of

known weigh. A spring balance S, calibrated in Newton isattached to B. the horizontal pull P applied to S is increaseduntil B begins to move, and the reading on S is noted:

(a). Just before the movement begins, and(b). While B is being pulled along at a steady speed.

The force to overcome dynamic friction is slightly lessthan that required to overcome static friction. The test is

B S

P

A

W

Fig. 2.6: Experiment on Friction



xl

repeated with various values of load W by placing metalblocks of known weight on top of block B.

Experiments between pairs of solids with dry surfaceshow that if the surfaces are of uniform character throughoutin regard to finish and condition, the force P required to

maintain steady slow motion is approximately proportional tothe perpendicular force W between the two sliding surfaces,i.e. the ratio P/W, namely the coefficient of friction, isapproximately constant.

It is also found from other experimental results that fordry solid surfaces the frictional force;a. depends upon the nature of the surface in contact;b. is independent of the area of the surface in contact;c. is independent of the speed of sliding.

The above must be regarded as only rough empiricalrules applicable in regard to relative slow motion and

moderate pressures and not necessarily correct at highspeeds or high pressures or in regard to any but drysurfaces. Actually, it is very difficult to exclude some form oflubrication, however slight its amount may be. Consequently,experiments on friction are difficult and the results are notsimple.

2.3 LUBRICATIONSometimes friction force is desirable as for instance,

in the grip of a tyre in a road, or a belt on a pulley or a brakeblock on a drum. But in machinery, where one piece hassliding or turning motion on or in another, any friction forceagainst the motion results in loss of energy which isconverted into heat. It is therefore desirable to reducefriction, and this is done by interposing a lubricant betweenthe solid surfaces, thereby keeping them out of direct contactwith each other.

Oils of various kinds are the best known lubricants;but for heavy loads, greases of various degrees of viscosity,which can resist pressure and avoid being squeezed out, are



xli

used. The lubricant employed depends upon manycircumstances, such as the pressure in a bearing, and thespeed as well as the cost. For high speeds, forcedlubrication is used in which Oil is supplied to bearings underpressure from a pump, the ideal condition being that the

metal surfaces shall be completely separated by a film or oilin which a rotating shaft floats.The resistance offered to lateral motion by a liquid is

quite unlike that between solid surfaces. It increases withspeed and with area of contact and pressure has little or noeffect on it. In practice, friction resistance is often acomplicated matter of forces exerted by solids andintervening fluids, in which not very much light is thrown bysimple experiments on the sliding resistance between solidbodies, and even this is affected by variable amounts of filmsof grease, moisture or even air partially separating the

surfaces.

EXAMPLES: 1. A block of metal having a mass of 60kg requires a

horizontal force of 140N to drag it at a constant speedalong a horizontal floor. Calculate

a. The coefficient of friction, andb. The angle of friction

SOLUTION: Since mass of block = 60kgTherefore, Weight of block = 60 x 9.8 = 588.6NFriction Force = 140NTherefore,Coefficient of Friction = 140 N = 0.238

588.6 (N)

= tan From trigonometrical tables, = 13.4o



xliii

Force [N]

50

40

30

20

10

0

2 4 6 8 10 12

Distance[m]

Fig. 3.1 Work Represented by an area

These sides of the rectangle are 50mm by 60mm so that thetotal area is 3000mm2.

Hence the work done is represented by an area of3000mm2 on a scale of 1mm2 to 0.2J.

Therefore, Work done = 0.2 [J/mm2] x 3000 [mm2] =600J.

For calculating the work done by a force of constantvalue, the diagram is of no advantage: for instance, in thecase just considered, we could have calculated the workdone by merely multiplying the force by the distance, thus:

Work done = 50(N) x 12(m) = 600JWhen the force varies in some known but perhaps

rather complicated manner, calculation of the area of thework diagram may be convenient way of determining thework done.

3.1.2 WORK DONE BY AN OBLIQUE FORCE



xliv

In section 1.2.6, it was shown that when a force F,

acting on a body, is inclined at an angle to the direction ofmotion (Fig. 3.2). The force can be resolved into two

components, namely FCos acting in the direction of motion

and FSin acting at right angles to the direction of motion.

Since there is no movement of the body in the

direction of component FSin, the later does not do anywork. Hence the work done by the Oblique Force F is the

product of the first component, FCos, and the distance xthrough which the body moves.

i.e. Work done = (FCos) X x

3.1.3 WORK DONE IN ROTATIONLet us consider the case of a Crank handle or Pulley

attached to a shaft as in Fig. 3.3, and suppose a force of70N to be exerted at the right angles to the Crank arm,200mm long, or at the circumference of the pulley of 200mmradius, in order to turn the Shaft against some resistance.Also, suppose the force to be always in the direction of themotion of the point to which it is applied.

F F Sin

F Cos

Motion Fig. 3.2 Work done by an Oblique Force



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200mm 200mm

Fig. 3.3 Work done in Rotation

The distance through which the point of application ofthe force travels in 1 revolution is

2π X 0.2(m), namely 1.257m,Therefore work done in 1 revolution= 70[N] x 1.257[m] = 88J.

In general, if a force F Newton acts at a Radius r metres,

Work done in 1 Revolution =F x 2πr Joules ……………………………………………. 3.2

The product of a turning force and the radius of thecircle at which it acts is termed the ‘torgue’ or ‘turningmoment about the axis of rotation’. Hence, for a force FNewton acting at a Radius r metres: Torque (or TurningMoment) = T = Fr Newton Metres and from equation 3.2.Work done in JoulesIn n revolutions = Fr x 2πn

= torque in Newton metres x angle in radius= TØ…………………………..………….. 3.3.

3.2 POWERPower is the rate of doing work and the S.I Unit of

Power is the Watt, namely 1 Joule per second. In practice,the Watt is often found to be inconveniently small;consequently the Kilowatt (KW) is frequently used, theKilowatt being 1000 Watts. For still larger powers, theMegawatt (MW) is used where:

1 MW = 1000Kw = 1 000 000 W



xlvi

Similarly, when we are dealing with large amount ofwork (or energy), it is often convenient to express the latterin Kilowatt hours.

1Kwh = 1000Watt hours= 1000 x 3600 Watts Seconds or Joules

= 3 600 000 J = 3.6 MJ.

3.2.1 POWER REQUIRED FOR ROTATIONIf T be the torque or turning moment, in Newton

metres, and if n be the speed, in revolutions per second,then from equation (3.3) we have:Work done per second = Torque in Newton metres x

Speed in radians per second= T (Newtons metres) x 2πn (radians/second) = 2πnT Joules/Second or Watts

Power = 2πnT

= ωt Watts ……………………………………. 3.4where ω = angular velocity in radians/second

= 2πn radians/second If the rotational speed by N revolutions per minutePower = 2 πTN/60 watts ……………………………… 3.5



xlvii

W W

B

S2 P S1

r

Fig. 3.4 Brake Test

3.2.2 DETERMINATION OF THE OUTPUT POWEROF A MACHINE BY MEANS OF A BRAKE

In the case of comparatively small machines, theoutput power can be measured by some form of mechanicalbrake such as that shown in Fig. 3.4. where a belt (or rope)on an air or water-cooled pulley has its ends attached tospring balances S

1and S

2, calibrated in Newton. The

balances are supported by a rigid horizontal beam B, andthe tension on the belt can be controlled by wing-nuts W.

Suppose the brake pulley to be rotating clockwise andthe tension on the belt to be adjusted to give readings of Pand Q Newton on S1 and S2 respectively. The pull P exertedby S1 has to balance the pull exerted by S2 and the frictionforce F between the belt and the pulley,

i.e. P = Q + F



xlviii

F = (P - Q) NewtonIf r be the effective radius of the brake, in metres,Torque due to brake friction = Fr

= (P - Q)r Newton metresIf n be the speed of the pulley in revolutions/minute,

then from equation 3.5.

Output Power = 2π(P - Q) rN / 60 ………….….……. 3.6

The output power of the machine is converted intoheat at the brake, and the size of the machine that can betested by this method is limited by the difficulty of dissipatingthis heat.

3.2.3 EFFICIENCY OF A MACHINEIn all machines, some of the power supplied to the

machine is lost in over coming friction, etc., so that the usefulpower available is less than the input power. The ratio of theoutput power to the input power is termed ‘The efficiency of the machine’.

i.e. Efficiency = Output PowerInput Power

And is expressed as a ‘per unit’ value or as apercentage;

Thus if the Input and Output Powers are 75KW and60KW respectively,

Efficiency = 60/75 = 0.8 per unit= 0.8 x 100 = 80 per cent.

The Input and Output Powers must obviously beexpressed in the same unit.

3.3 ENERGYWhen a body is capable of doing work, it is said to

possess energy which may take various forms such asmechanical energy, thermal energy, chemical energy andelectrical energy. In mechanics, we are concerned only with



xlix

mechanical energy which is of two kinds, namely kineticenergy and potential energy.

The Kinetic Energy (K.E) of a body is the energy itpossesses by virtue of its motion. Thus a body, set in motionby a force doing work upon it, acquires Kinetic Energy, which

enables it to do work against resisting forces. An importantengineering application is flywheel. Work is done on theflywheel as its speed is increased; and later, when themachine to which it is attached slows down, some of thestored Kinetic Energy is given out b y the flywheel and helpsto drive the machine. This is the reason why a machinedriving a fluctuating load, such as a stamping press, isusually fitted with a flywheel to maintain a more constantspeed than would otherwise be the case.

If a force is exerted on a body and there is noopposing resistance except the inertia of the body, the whole

of the work done becomes the Kinetic Energy of the body.Thus, if a force F, acting on a mass m, gives it a uniformacceleration a, then

F = maIf x is the distance traveled by the body while it

accelerates from standstill to a velocity V, then x = ½V2 /aTherefore, work done = F (Newton) X x (Metres)

= ma x ½V2 /a= ½mV2 Joules ………….……………… 3.7= Kinetic Energy of body.

POTENTIAL ENERGY (P.E) The Potential Energy of a body is the energy it

possesses by virtue of its position or state of strain. Forinstance, a body raised to a height above the ground haspotential energy since its weight can do work as the bodyreturns to the ground.

If a body having a mass m, in Kilograms, is liftedvertically through a height h, in metres, and if g is thegravitational acceleration in metres/seconds2 at that point,



l

Force required = Weight of the body= mg Newton

and Work done = Weight of the body X height= mgh Joules ……………………… 3.8= Potential Energy of the body

The Pendulum of a clock is an example of energybeing charged backwards and forwards between the Kineticform and the Potential form. Thus the oscillating mass hasits potential energy being then zero. On the other hand, thepotential energy of the pendulum is a maximum at the end ofeach swing, its speed and therefore its Kinetic Energy beingthen zero. The small loss of energy due to friction is suppliedby the impulse given regularly through the escapementmechanism from the main spring.

One of the natural sources of Potential Energy iswater lifted by evaporation from Sea – level to Lakes and

Rivers at higher levels into which it is deposited as rain orsnow. With the aid of pipes supplying water turbines at areservoir at a high level can be converted into Kinetic Energyand thereby used to drive the turbines which in turn driveelectrical generators or machinery. Thus the potential energyof the water in the reservoir is converted into useful work.

3.3.1 PRINCIPLE OF THE CONSERVATIONOF ENERGY

This important principle states that whenever energyis converted from one form to another, no energy is lost. Allthe energy involved in the conversion can be accounted forin some form or another. This constancy of the total energyis referred to as the principle of the conservation of energy.

In some situations we can have, to a reasonableapproximation, a conservation of mechanical energy. Thuswith the swinging pendulum of the clock potential energy atthe end of each swing is converted to kinetic energy at thelowest point of it’s travel. The sum of the potential and kineticenergies is a constant.



li

Chapter Four

4.0 PRINCIPLES OF SIMPLE MACHINES4.1 MACHINES

A machine is a mechanical device for transmittingmotion, force and energy. Lifting machines are usuallyarranged to enable a small effort (or driving force) to raise amuch larger load (or resisting force). To achieve this, theeffort must move through a much greater distance than thatthrough which the load is raised, the work done by the effortbeing equal to the useful work in lifting the load together withthe work required to overcome friction. This relation is knownas the principle of conservation of energy.

4.2 MECHANICAL ADVANTAGES, VELOCITY (ORMOVEMENT) RATIO AND EFFICIENCY

It was mentioned in section 4.1 that the purpose ofmost lifting machines is to enable a large load W to bemoved by the application of a relatively small effort F. Theratio of the load to the effort is termed the MechanicalAdvantage of the machine,i.e.Mechanical Advantage = Load = W

Effort f …………………. 4.1

It is evident that for a given load W, the smaller the value ofthe effort F, the greater the Mechanical Advantage.

The ratio of the distance moved by the effort to thatmoved by the load is termed, the movement ratio of morecommonly the velocity ratio,

i.e.V. R. = Distance Moved by Effort ……………… 4.2

Distance Moved by Load



lii

The value of the velocity ratio is dependent upon thearrangement of the machine and is constant for a givenmachine, whereas the value of the mechanical advantage,decreases as the load is reduced, being zero at no load.

From the principle of the conservation of energy,Work done by Effort = Work done on Load + Work done toovercome friction

The Work done in overcoming friction is convertedinto heat and is thus wasted as far as the machine isconcerned. The ratio of the Work done on the load to thatdone by the effort is termed the efficiency of the machine,i.e. Efficiency = Useful Work done

Work done by Effort ……………… 4.3

Therefore Efficiency = W X distance moved by Load

F X distance moved by Effort= W X distance moved by LoadF distance moved by Effort

= Mechanical AdvantageVelocity Ratio …………………… 4.4

For an ideal machine, i.e. a machine having nofriction, the efficiency is unity ( = 100 percent), so that;

Work done by Effort = Work done by Load,Therefore, Ideal Effort X distance moved by Effort =

W X distance moved by LoadSo that,Ideal Effort = W

Velocity Ratio ……………… 4.5andIdeal Mechanical Advantage = W

Ideal Effort= Velocity Ratio ……………………… 4.6

Let us now consider some of the simplest types ofmachines and determine the velocity ratio of each type.



liii

4.3 THE LEVERThe lever was probably the earliest device used by

man to enable him to move a large load by means of thelimited physical effort he could exert.

Fig. 4.1 shows a straight lever, pivoted at C. A mass,having a weight W, is suspended at B and a downward effortF is applied at A to balance weight W.

Suppose distances AC and BC to be a and b respectively, and the weight of the Lever to be negligible. If

the Lever is tilted anticlockwise through an angle ,Distance through which effort F moves in its own direction =

aSin So that, Work done by Effort = F X asin Similarly,Distance through which load W is moved in its own direction

= bsin

So that, Work done by Load = W X bsin Hence,

Velocity Ratio = asin = a

bsin b

B

a bA b Sin

B

a Sin C

F

W

A

F W

Fig. 4.1 A Lever



liv

A

B

A B C C C

E

D a b D

F W F

W

Fig. 4.2 The Wheel and Axle

4.4 THE WHEEL AND AXLEThe wheel and axle may be regarded as an

adaptation of the Lever to allow continuous rotation of thedevice about the pivot.

Fig. 4.2 shows a Wheel A, of radius a , and an Axle B,of radius b , carried by a shaft c . A body having weight W isattached to a cord E , wound around Axle B , and a downwardeffort F is applied to cord D wound around the rim of WheelA.

When the Wheel and Axle makes one revolution in ananticlockwise direction, a length, 2πb, of cord E is wound on

Axle B, thereby lifting the load a distance 2πb. At the sametime, a length 2πa, of cord D is unwounded from Wheel A,enabling the effort to move a distance 2πa.

Hence,Velocity Ratio = Distance moved by Effort

Distance moved by Load



lv

= 2πa = a2πb b

4.5 THE INCLINED PLANE

The principle of the incline plane was known at least5000 years ago when it was employed in the construction ofthe pyramids.

Suppose a body having weight W to be hauled up thewhole of the inclined surface (Fig. 4.3) by a force F actingparallel to the plane.

Hence, Work done by Effort = F X ABWhile the load has been hauled a distance AB up the

inclined plane, it has also been lifted through the verticaldistance BC,

So that Work done on Load = W X BCVelocity Ration = AB = 1

BC Sin

Where is the angle between the plane and thehorizontal.

B

F

W

A C

Fig. 4.3 The Inclined Plan



lvi

4.6 THE SCREW AND THE SCREW-JACKThe screw is really an inclined plane converted into a

helical inclined path around the bolt on which the screw hasbeen cut, thereby enabling the circular motion of a nut to be

converted into a linear motion along the screwed portion ofthe bolt. This means that a small effort applied tangentially atthe end of, say a Spanner moves through a distance equalto (2π X radius of Eff ort), while the nut travels a distanceequal to the lead of the thread, i.e. the distance between thecentres of adjacent threads for single-start thread.

Hence,Velocity Ratio = 2π X Radius of Effort

Load of Threade.g. if the radius at which the effort is applied is, say 200mmand the lead of the thread is 1.8mm,

Velocity Ratio = 2π X 200 (mm) 1.8 (mm)= 698

Allowing for an efficiency of, say 15 percent, we have fromequation (4.4)Mechanical Advantage = Velocity Ratio X Efficiency

= 698 X 0.15= 105

A device, using the principle of the screw, that iscommonly employed for lifting heavy objects is the screw-

jack shown in Fig. 4.4.



lvii

W r

F

Fig. 4.4 The Screw-Jack

An effort F is applied tangentially at the end of an armof radius r. for a screw having a right-hand thread, one turnof the effort in an anticlockwise direction, viewed from abovethe jack, lifts the load W through a distance equal to the

load, of the Screw. (in a single-start screw, the load isequal to the pitch f the thread).

For one revolution of the effort,Distance moved by Effort = 2πr

and, Distance moved by Load = Therefore, Velocity Ratio = 2πr



lviii

4.7 PULLEYSA SINGLE-PULLEY SYSTEM

A single-pulley system. Fig. 4.5 shows a pulley block fittedwith one pulley (or sheave), with a rope passing over thepulley and supporting at one end a body having weight W.This load can be lifted by applying an effort F to the otherend of the rope. With this simple arrangement, the velocityratio is unit since the distance through which the effort isapplied is exactly the same as that through which the load israised. The effort F has to be greater, than the load W toallow for friction at the pulley, so that the mechanicaladvantage (= W/F) is less than unity. The only advantage ofthe single-pulley system is that a person, when having arope downwards, is able to make use of his own weightwhen exerting the pull and thus finds it easier than to lift theload directly.

F

W

Fig. 4.5



lix

A TWO-PULLEY SYSTEM

The velocity ratio can be increased by using morepulleys. For example in Fig. 4.6 there are two pulley-blocks,A and B, each with one pulley. A rope has one end attachedto pulley-block A and passes round the pulleys of B and A,as shown. The body to be lifted is attached to pulley-block B.

A

F

B B Fig. 4.6

W



lx

If the load W were raised 1m by means of effort F,then each of the length of the rope between the pulley-blocks would be shortened by 1m and the effort F wouldmove 2m. Hence,Velocity Ratio = Distance moved by Effort

Distance moved by Load= 2 (m)1 (m) = 2

A THREE-PULLEY SYSTEM

Fig. 4.7 shows Pulley-block A fitted with two pulleys. Threepulleys are usually on the same spindle, but for convenience

A

F

B

Fig. 4.7

W



lxi

F 2

Slack Side

A B

Tight SideDriven Pulley

Driver Pulley F 1

Fig. 4.8 A Belt Drive

of explanation, they are shown one above the other. Withthis arrangement, effort F moves a distance of 3m when theload is lifted 1m, so that Velocity Ratio is now 3.

In general, if n be the total number of pulleys (orsheaves) on the two Pulley-blocks,

Velocity Ratio = n.

4.8 BELT AND CHAIN DRIVESA belt or a chain is used when a shaft has to be

driven from a parallel shaft that is too far away from the useof gear wheels. Fig. 4.8 shows a belt drive in which A is thedriver pulley and B the driven pulley. The transfer of motionfrom Pulley A to the belt and again from the belt to the PulleyB is dependent upon friction at each area of contact betweenbelt and pulley.

If dA and dB = Diameters of Pulleys A and Brespectively. And nA and nB = Speeds, inrevolutions/seconds, of A and B respectively,then, linear speed of rim of Pulley A = πdAnA and linear speed of rim of Pulley B = πdBnB



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If there is no slipping, the linear speed of the rim ofeach pulley is the same as the speed of the belt, hence,

πdAnA = πdBnB therefore,

speed of driver Pulley A = diameter of driven Pulley Bspeed of driven pulley B = diameter of driver Pulley A

i.e. the speeds of the pulleys are inversely proportional totheir diameters.When a driving torque is applied to driver Pulley A, the sideof the belt approaching the Pulley tightens and that leavingthe pulley slackens. Thus, with the driver pulley rotatingclockwise as in fig. 4.8, the lower side of the belt has a largertension than the upper side.

If F1 and F2 be the tensions in the tight and slack

sides respectively of the belt,Effective force due to friction = F1 – F2 and

Power Transmitted = Net force [Newton] X Speed of belt[meter/second

= (F1 – F2) X πdAnA watts ………………………… 4.8or (F1 – F2) X πdBnB watts ………………………… 4.9

One of the main disadvantages of belt is its liability toslip. This disadvantage can be avoided by the use of chainwhose links engage with teeth on the driver and drivenwheel one of the best known example is the use of a chainto transmit power from the pedals to the rear wheel of abicycle.

Since the number of teeth on each wheel isproportional to the diameter of the wheel,Speed of driver wheel = diameter of driven wheelSpeed of driven wheel diameter of driver wheel



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Pitch Circles

C D

A B

Driver Follower

Pitch

Fig. 4.9

= number of teeth on driven wheelnumber of teeth on driver wheel

4.9 GEAR WHEELSGear wheels are used to transmit motion and power

from one shaft to a parallel shaft in close proximity, and toenable the speed of rotation to be stepped up and down tosuit specific requirements, e.g. to reduce the high speed ofan electric motor to the relatively low speed of a lathe.

A gear wheel has a number of specially-shaped teetharound its periphery. These teeth mesh with similar teeth ona second wheel as shown in Fig. 4.9, where wheel A (thedriver) drives wheel B (the follower).

In order that the teeth may mesh correctly, the pitch ofthe teeth, i.e. the peripheral distance between the centres ofadjacent teeth, measured on the pitch circles C and D, mustbe the same for the two wheels.



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Hence, for a given pitch, the number of teeth isproportional to the diameter of the pitch circle.Therefore,

number of teeth on Wheel A. = Circumference of Wheel Anumber of teeth on Wheel B = Circumference of Wheel B

If nA and nB be the speeds of wheels A and B respectively, inrevolutions per second,Peripheral speed of driver A, measured at pitch circle

= Circumference of A X nA and, Peripheral speed of follower B, measured at pitch circle

= Circumference of B X nB Since there can be no slip, the peripheral speed at the pitchcircles is the same for the two wheels, i.e. Circumference ofA X nA = Circumference of B X nB therefore,Speed of driver A = Circumference of followerSpeed of follower B = Circumference of driver

= number of teeth on followernumber of teeth on driver ………. 4.10.

i.e. the speeds of the gear wheels are inversely proportionalto the number of teeth on the wheels.

It will be noted that gear wheels, A and B in Fig. 4.9rotate in opposite directions. The direction of rotation of Aand B can be arranged to be the same by introducing an idle

wheel C between A and B as shown in Fig. 4.10. If idler Crotates at nC revolutions per second,

nA = number of teeth on CnC number of teeth on A



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A C B

Pitch Circles

Fig. 4.10 Effect of an Idle Wheel

Similarly,nC = number of teeth on BnB number of teeth on C

therefore,

nAXnC = number of teeth onC X number of teeth on BnC nB number of teeth on A number of teeth on C

hence,nA = number of teeth on BnB number of teeth on A

i.e. the gear ratio of the gear train ABC is independent on thenumber of teeth on the idler.



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Chapter Five

5.0 STRESS AND STRAIN5.1 STRESS

When a material has a force exerted on it, thematerial is said to be stressed or in a state of stress. If a rodis subjected to a tension, the force per unit area of cross-section of the rod is referred to as tensile stress. If the rod issubjected to compression, the force per unit area of cross-section is termed a compressive stress. Hence in general:Stress = Force

Cross-Sectional Area …………….. 5.1Tensile and Compressive Stresses are sometimes

referred to as normal stresses since they act at right anglesto the cross-sectional area which is used for calculating thevalue of the stress.

The SI Unit of stress is the Newton per square meter(Symbol: N/m2). Another unit the Pascal adopted in memoryof the French Philosopher Blaise Pascal (1623 – 1662) whocarried out many brilliant experiments in hydrostatics andpneumatics is widely used.

It is often more convenient to express Stress inKilonewton per square metre (KN/m2) or Kilopascals (Kpa) orin Meganewtons per square metre (MN/m2) or Mega Pascal(Mpa) where,1KN/m2 or 1Kpa = 103 N/m2 or 103paand 1MN/m2 or 1 Mpa = 106 N/m2 or 106pa.

5.2 STRAINWhen a rod or a wire is pulled, it stretches; and the

total stretch or elongation, expressed as a fraction of the un-stretched length, is termed direct Strain or merely Strainwhen it is obvious that the change in length is due to tensionor compression. Thus, if the pull on a rod having a un-stretched length of 1m produces an extension of 2.5mm,



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Strain = ExtensionOriginal Length …………………..… 5.2

= 2.5 (mm)(1 X 1000) (mm)

= 0.0025

It will be noticed that the extension and the originallength must be expressed in the same unit. In the aboveexpression, they are both expressed in milimetres, but theresult would be exactly the same if they we both expressedin, say metres, thus:Strain = (2.5/1000) (m) = 0.0025

1.0 (m)Hence, Strain is merely a ratio and possesses no units.

5.3 TENSILE TEST ON A STEEL WIREThe behaviour of various materials, when stretchedby gradually increasing forces, may be studiedexperimentally in the laboratory.

The results of a test performed on a steel wire aregiven in the following table. The initial length of the wire was907mm and the initial cross-sectional area was 0.34mm2.

LOAD(N) 0 10 20 30 40 50 60 70 80 90 100Extension 0 0.10 0.22 0.35 0.50 0.62 0.75 1.1 2.1 4.0 7.5(mm)LOAD(N) 105 110 118 129 136 145 150 154Extension 10.4 12.9 17.8 21.8 29.1 42.7 54 70 (broke)

(mm)

An examination of the tabulated figures shows that there aretwo stages in the stretching of the wire:

(a). THE ELASTIC STAGE : Up to a Load of about 60N,the extension is small and is proportional to the load, asshown in the fig. 5.1 below. During this stage of the test, it isfound that if the load is removed, the stretch disappears and



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the wire returns to its original length. The fact that during thiselastic stage the extension is proportional to the load ischaracteristics of the elastic straining of materials.

(b). THE PLASTIC STAGE : The results for Loads of 80Nupwards are plotted in fig. 5.2, where the extension is shownto a much smaller scale than in fig. 5.1. It is evident from thetable and from Fig. 5.2 that for loads exceeding about 60N,the extension for a given increase in load is much greaterthan for loads below 60N. For instance, when the load is

increased for 60N to 150N, the extension increases from0.75mm to 54mm, i.e. the load is increased about 2.5 times,the extension increases more than 70 times.

During this plastic stage of the test, it is found thatwhen the load is increased from 60N to 150N, the extensionincreases from the said to have a plastic deformation orpermanent set. A metal that is able to undergo cold plastic

Load(N)60

50

40

30

20

10

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

Extension (mm) Fig. 5.1 Elastic Extension of a wire



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deformation, usually as the result of tension is said to beductile or to possess ductility.

5.4 HOOKE’S LAW Robert Hooke (1635 – 1703), a famous English

Physicist, Architect and Inventor, carried out a series ofexperiments with springs and wires and was the first topublish, in 1676, a clear statement to the effect that when amaterial is worked within its elastic range, the extension isproportional to the force.

5.5 MODULUS OF ELASTICITY (DIRECT) OR YOUNG’S MODULUS

For a material worked within its elastic range, wehave from Hooke’s law that the extension is proportional tothe force:

Load (N)160

140

120

100

80

60 A

40

20

0

20 40 60 80 Extension (mm)

Fig. 5.2 Tensile test on a steel wire



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Strain Stress

i.e. Stress = a constant.Strain

This constant is termed Modulus of Elasticity (Direct)

or Young’s Modulus, after Thomas Young who was the firstto determine this constant for some materials.The symbol for Modulus of Elasticity (Direct) or

Young’s Modulus is E,i.e. E = Tensile or Compressive Stress

Strain ..………………… 5.3= Force/Original Cross-Sectional Area

Change in Length/Original Length

Young’s Modulus may be regarded as a measure of the resistance which a material offers to extension or

compression, i.e. the larger the value of E, the smaller is theextension or compression produced by a given stress.When the value of Young’s Modulus is being

determined from experimental results such as those shownin fig. 5.1, it is usually better not to use figures for load andextension at any one point on the graph but to take thedifference in extension and the corresponding difference inload between two points, such as B and C in Fig. 5.1, on astraight line drawn through the points determinedexperimentally. This is due to the difficulty in determiningaccurately the point representing zero load and zero

extension.Let us now proceed to calculate the value of E from thegraph of Fig. 5.1

Load at B = 10NLoad at C = 55N

Therefore, Increase in Load = 45NInitial Cross-sectional Area = 0.34mm2

= 0.34 X 10-6 m2 Therefore, Increase in Stress =



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45(N) / (0.34X10-6) (m2)= 132.4 X 106 N/m2 or pa

Extension at B = 0.10mmand Extension at C = 0.69mmTherefore, Increase in Extension =

0.69 – 0.10 = 0.59mmInitial length of Wire = 907mmTherefore, Increase in Strain =

0.59(mm) / 907(mm)= 0.65 X 10-3

Hence, E = Increase in Stress = 132.4 X 106(N/m2)Increase in Strain 0.65 X 10-3

= 204 X 109 N/m2 or pa= 204 000 MN/m2 or Mpa= 204 GigaNewton/Metre2 (or GN/m2 or Gpa)

5.6 STRESS-STRAIN GRAPH FOR MILD STEEL

Stress

C D

B

A

Strain

Fig. 5.3 Stress-Strain Graph for Mild Steel



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Fig. 5.3 above shows a typical graph for a tensile teston a mild-steel specimen. The stresses have beencalculated by dividing the loads by the original Cross-Sectional Area the text piece, and the strains by dividing theextensions by the original length selected for the test. The

graph differs from a load-extension graph, such as that inFig. 5.2, only in the matter of scales.Point A in Fig. 5.1 and Fig. 5.3 marks approximately

the limit of proportionality of stress to strain, and is practicallythe elastic limit, i.e. the highest stress that can be appliedwithout producing permanent deformation. For stresses upto the elastic limit, the material returns to its original lengthon removal of the load.

Portion B of the graph gives the yield stress, namelythe stress at which, in a tensile test, elongation of the testpiece first occurs without increase of load. This plastic

yielding is not found in all ductile materials but is markedfeature of the softer irons and steels.The maximum stress, represented by C on Fig. 5.3 is

termed the tensile strength (replaces ultimate tensile stressand is obtained by dividing the maximum load applied duringa tensile test by the original Cross-Sectional Area of thespecimen and not by the reduced area of section afterplastic extension. At about point C, the Cross-Sectional Areaat some point along the length of the test specimen begins todecrease rapidly and the load required to increase theextension beyond C consequently decreases until thespecimen ultimately fractures at a load corresponding topoint D.

5.7 PERCENTAGE ELONGATIONA useful measure of the degree of ductility of a

material is the percentage elongation. This involves ameasurement of the gauge length of the test piece beforethe tensile test and after the test is completed and the piece



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broken, the two broken pieces being put into contact for themeasurement.Percentage Elongation =

Final Length – Initial Length X 100Initial Length …………….… 5.4

Another quantity which is sometimes measured is thePercentage reduction in area. The initial cross-sectional areaof the test piece is measured and then the smallest cross-sectional area when the test piece has been broken.Percentage reduction in area =

Initial Area – Final Area X 100Initial Area

5.8 PROOF STRESS

The graph of Fig. 5.4 is typical of the stress-strainrelationship for many materials such as copper, etc. In thisdiagram, the strain has been plotted as a percentage of theoriginal length, i.e.

Stress C B

D

A

0.1 0.2 0.3 0.4 0.5 0.6

Fig. 5.4 Proof Stress



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Percentage Strain = Elongation X 100Original Length

Point A represents the limit of proportionality for thisspecimen had the strain remained proportional to the stress,the stress-strain relationship would have been represented

by the dotted line OB, which is OA extended beyond point A.If we draw a horizontal line CD at a Stress OC, then CBrepresents the percentage strain if the strain had remainedproportional to the stress and CD represents the actualpercentage strain. Consequently, BD represents the amountby which the percentage strain has departed fromproportionality, i.e. BD is the non-proportional elongation ofthe specimen, expressed as a percentage of the originallength.

Draw the dotted line DE parallel to CB, then:OE = BD

If OE is, say 0.1 per cent, then, OC is termed theproof stress to give a non-proportional elongation equal to0.1 percent of the original length of the specimen.

It is usual to express the proof stress as the stressrequired to give a non-proportional elongation of 0.1, 0.2 and0.5 per cent of the original length.

EXAMPLEA specimen has an initial gauge length of 55mm and aCross-Sectional Area of 150mm2. A test on the specimengave the following results:

Load (KN) 0 10 20 30 35 38 40Extension (mm) 0 0.075 0.15 0.23 0.30 0.38 0.6

Determine the 0.2 percent Proof Stress in MegaNewton persquare meter or MegaPascals.



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SOLUTION The graph representing the above table is given in the figurebelow:

For 0.2 percent non-proportional extension,Actual non-proportional extension =

55(mm) X 0.2 / 100= 0.11mm

Hence, in Fig. 5.5, from a point E corresponding to anextension of 0.11mm, draw ED parallel to OA; and from point

Load(N)

40

B C D

30 A

20

10

E 0

0.1 0.2 0.3 0.4 0.5 0.6 Extension (mm)

Fig. 5.5 Load-Extension Graph



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D, draw a horizontal line to cut the vertical axis at C. Then,the proof load to give 0.2 percent non-proportional extensionis OC.From the graph in Fig. 5.5, OC = 38KN = 38 000NCross-Sectional Area = 150mm2 = 150 X 10-6 m2

and

Corresponding Proof Stress = 38 000(N)150 X 10-6 (m2)

= 253 X 106 N/m2 = 253 MN/m2 or Mpa



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References

Edwards and Christopher Hughes: Engineering Science 4th Edition ( Longman Publisher)

Carl T. Fross: Dynamics of Mechanical System 2nd Edition

(Horwood Engineering Science Series)R. S. Khurmi and J. K. Gupta: Theory of Machines (Eurasia

Publishing House Ltd)

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