Ce 202 Engineering Mechanics

ENGINEERING MECHANICS

FOR THE COURSE OF

CE 202 : ENGINEERING MECHANICS

WORLD UNIVERSITY OF BANGLADESH DEPARTMENT OF CIVIL ENGINEERING

september2013

©

This copy is written for the students of 1st semester of Dept. of civil

engineering for 2 credit course of CE 202 : Engineering Mechanics. It is not

for sell or any kind of financial prifit making.All rights whatsoever in this book

are strictly reserved and no portion of it may be reproduced any process for

any purpose without the written permission of the owners.

AUTHOR

S M Tanvir Faysal Alam Chowdhoury, B.Sc.(Civil)

Lecturer in Dept. of Civil Engineering

WORLD UNIVERSITY OF BANGLADESH

SYLLABUS

MODULE 1 Importance of Mechanics in engineering; Introduction to Statics; Concept of Particle and Rigid Body; Types of forces: collinear, concurrent, parallel, concentrated, distributed; Vector and scalar quantities; Force is a vector; Transmissibility of a force (sliding vector). Introduction to Vector Algebra; Parallelogram law; Addition and subtraction of vectors; Lami’s theorem; free vector; Bound vector; Representation of forces in terms of i, j, k; Cross product and Dot product and their applications. Two dimensional force system; Resolution of forces; Moment; Varignon’s theorem; Couple; Resolution of a coplanar force by its equivalent force-couple system; Resultant of forces. MODULE 2 Concept and Equilibrium of forces in two dimensions; Free body concept and diagram; Equations of equilibrium. Concept of Friction; Laws of Coulomb friction; Angle of Repose; Coefficient of friction MODULE 3 Distributed Force: Centroid and Centre of Gravity; Centroid of a triangle, circular sector, quadrilateral, composite areas consisting of above figures. Moments of inertia: MI of plane figure with respect to an axis in its plane, MI of plane figure with respect to an axis perpendicular to the plane of the figure; Parallel axis theorem; Mass moment of inertia of symmetrical bodies’ e.g. cylinder, sphere, cone. Concept of simple stresses and strains: Normal stress, Shear stress, bearing stress, Normal strain, Shearing strain; Hooke’s law; Poisson’s ratio; Stress-strain diagram of ductile and brittle materials; Elastic limit; Ultimate stress; Yielding; Modulus of elasticity; Factor of safety. MODULE 4 Introduction to Dynamics: Kinematics and Kinetics; Newton’s laws of motion; Law of gravitation & acceleration due to gravity; Rectilinear motion of particles; determination of position, velocity and acceleration under uniform and non-uniformly accelerated rectilinear motion; construction of x-t, v-t and a-t graphs. Plane curvilinear motion of particles: Rectangular components (Projectile motion); Normal and tangential components (circular motion). Kinetics of particles: Newton’s second law; Equation of motion; D.Alembert’s principle and free body diagram; Principle of work and energy; Principle of conservation of energy; Power and efficiency.

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MODULE 5 Introduction of SI unit, force systems that produce rectilinier forces, flexible cords

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LECTURE CONTENTS

SL Lecture No. Lecture Title Page

Introduction to mechanics, Statics, 1. Lecture 1 Concept of Particle and Rigid body, 6-7

Force, Transmissibility of force

2. Lecture 2 Classification and type of forces 8-11

Introduction to Vector Algebra, 3. Lecture 3 Parallelogram Law, Addition and 12-16

Subtraction of vector Lami’s theorem, Free vector, Bound 5. Lecture 4 vector, Representation of forces in term 17-18

of I,j & k

5. Lecture 5 Cross product, Dot product, examples 19-20

6. Lecture 6 Examples 21-22

Two dimensional force system, 7. Lecture 7 Resolution of forces, Moment, 23-24

Varignon’s theorem

8. Lecture 8 Couple, Example 25-29

Resolution of a coplanar force by its equivalent force-couple system, 9. Lecture 9 Resolution of forces, Example 30-33

10. Lecture10 Example 34-38

11. Lecture11 Concept and equilibrium of forces, Free 39-40

body diagram

12. Lecture12 Free body diagram , example 41-42

13 Lecture13 Equation of equilibrium, example 43-49

Concept of friction, Columb law of 14 Lecture14 friction, Angle of repose 50-54

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15 Lecture15 Coefficient of friction, Kinetic friction 55-59

16 Lecture16 Examples 60-68

17 Lecture17 Distributed force, Centroid & Centre of 69-73

Gravity

18 Lecture18 Determination of Centroid & C.G. 74-78

19 Lecture19 Centroid of an arc, Centroid of a 79-82

triangle, Example


21 Lecture21 Moment of Inertia, Parallel axis theorem 87-90

22 Lecture22 Moment of inertia of plain figure, 91-97

Perpendicular axis theorem, Radius of gyration


24 Lecture24 Concept of stress, different types of 102-110

stresses & strains, Hook’s law

25 Lecture25 Possoin’ratio,Stress-Strain curve for ductile & brittle material, Example 111-117

26 Lecture26 Projectiles 118-120

27 Lecture27 Projectiles at inclined plane 121-126

28 Lecture28 Examples of Projectiles, Kinematics, 127-129

Rectilinear motion 29 Lecture29 Relative motion and example 130-138

30 Lecture30 Kinetics of particle, curvilinear motion, 139-141

example

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31 Lecture31 Rolling motion 142-145

32 Lecture32 Solution to kinetic problem, 146-148

33 Lecture33 Solution to problems 149-151

34 Lecture34 Example 152-157

35 Lecture35 Work, Power, Energy 158-163

36 Lecture36 Impulse and Momentum 164-169

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Lecture 37

Introduction of SI unit,

38 Lecture 38 force systems that produce rectilinier

forces, flexible cords

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Lecture 1 : Introduction to Mechanics STATICS Statics is the branch of mechanics concerned with the analysis of loads (force, torque/moment) on physical systems in static equilibrium, that is, in a state where the relative positions of subsystems do not vary over time, or where components and structures are at a constant velocity. When in static equilibrium, the system is either at rest, or its center of mass moves at constant velocity. The study of moving bodies is known as dynamics. By Newton's first law, this situation implies that the net force and net torque (also known as moment of force) on every body in the system is zero. From this constraint, such quantities as stress or pressure can be derived. The net forces equaling zero is known as the first condition for equilibrium, and the net torque equaling zero is known as the second condition for equilibrium. See statically determinate.

Solids

Statics is used in the analysis of structures, for instance in architectural and structural engineering. Strength of materials is a related field of mechanics that relies heavily on the application of static equilibrium. A key concept is the centre of gravity of a body at rest: it represents an imaginary point at which all the mass of a body resides. The position of the point relative to the foundations on which a body lies determines its stability towards small movements. If the center of gravity falls outside the foundations, then the body is unstable because there is a torque acting: any small disturbance will cause the body to fall or topple. If the centre of gravity falls within the foundations, the body is stable since no net torque acts on the body. If the center of gravity coincides with the foundations, then the body is said to be met stable.

Particle:

A body of infinitely small volume i.e. negligible dimensions but having mass concentrated at a point is called particle. Such a body cannot exist theoretically.

Rigid Body:

A rigid body may be defined as a body in which the relative positions of any two particles do not change under the action of forces. Nobody is perfectly rigid. A body when subjected to external forces, it must undergo some form of deformation, however small it may be.

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Force

It is the agent which changes or tends to change the state of rest or motion of a body. The force is a vector quantity. The characteristics of force are magnitude, direction and point of application. The unit of force is ‘N’, ‘KN’ ‘Kgf’ etc.

Principle of transmissibility of Forces

The state of rest or of a motion of a rigid body is unaltered; if a force acting on a body is replaced by another force of the same magnitude and direction but acting anywhere on the body along the line of action of the replaced force. If deformation of the body is to be considered, the law of transmissibility will not hold good. By transmission of force, only the state of the body is unaltered, but not the internal stresses which may develop in the body. This force is a sliding vector as this has a unique line of action in space but not a unique point of application.

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Lecture 2 : Classification & Types Of Forces

1.1.1 CLASSIFICATION OF FORCE

The general classification of force is presented below:

Force

Gravitational force Electromagnetic force Strong force Weak force

But engineering mechanics point of view, a system of forces can be classified as:

System of Forces

Coplanar Forces Non-Coplanar Forces

(lying on the same plane) (do not lie on the same) plane)

Coplanar forces may be Non-Coplanar forces may be

■ Collinear ■ Concurrent

■ Concurrent ■ Non-Concurrent

■ Non-Concurrent

■ Parallel

■ Parallel

■ Non-Parallel 8

■ Non-Parallel

Table 1.2 TYPES & CHARACTERISTICS OF FORCE SYSTEM

Force System Line Diagram Characteristics

Coplanar forces Y Lines of action of all forces

lie on the same plane

x

Non-Coplanar forces y Lines of action of all forces

do not lie on the same F1

plane

x

F fF

z F2

Collinear Forces Lines of action of all forces

lie on the same st. line

Concurrent Forces Lines of action of all forces

pass thru the same point

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Parallel Forces Lines of action all forces

F1 are parallel to each other

F2

F3

According to the effect produced by the force

a) External Force

This force is applied to externally to a body.

b) Internal Force

This force is developed into the body to resist the deformation or change of shape of a body.

c) Active Force

This force causes a body to move or to change its shape.

d) Passive Force

This force represents the motion of a body.

According to the nature of force

a) Action or Reaction Force

When two bodies come into contact with each other, each body will exert a force on the other

body. Out of these forces one is known as action and other is known as reaction. It is seen that

action and reaction are equal.

b) Attraction and Repulsion

The non-contact force exerted by one body on the another body without any visible medium are

known as attraction or repulsion force e.g. magnetic force.

c) Tension or Compression

When a pull is applied on a structural member, it tends to elongate/increase in length, pull is

known as tensile force. If a structure member is subjected to two equal and opposite pushes and

the member tends to shorten/decrease in length, member is said to be in compression.

According to the force applied to a point or over an area

a) Concentrated Force: The point of application of this force is considered to be a point.

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b) Distributed Force: The force which is distributed over an area, that force is known as

distributed force.

Multiple Choice Questions 1. Forces are called concurrent when their lines of action meet in

a) One plane, b) One point, c) Different planes , d) Two points

2. Forces are called coplanar when all the forces acting on a body

a) One plane, b) One point, c) Different planes, d) Different points

3. A force is completely defined when we specify

a) Its magnitude, b) Its direction, c) Its point of application, d) All of the above

4. The amount of matter contained in a body is called its

a) Mass, b) Volume, c) Weight, d) None of the above

5. Two non collinear parallel equal forces acting in opposite direction

a) Balances each other, b) Constitute a couple, c) Constitute a moment

d) Constitute a moment of couple

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Lecture 3 : Introduction To Vector Algebra

Introduction to Vector Algebra

We use two kinds of quantities in mechanics-scalars and vectors. Scalar quantities are those with which only a magnitude is associated. Examples of scalar quantities are time, volume, density, speed energy and mass. Vector quantities, on the other hand, possess direction as well

as magnitude, and must obey the parallelogram law of addition. Examples of vector quantities are displacement, velocity, acceleration, force, moment and momentum. Velocity is specified by a direction as well as speed. However, the mass of a body is completely specified by a magnitude, and hence mass is a scalar quantity. Since the weight of a body is the force with which it is attracted by the earth, weight has a downward direction and thus is a vector quantity. Since weight and mass are different physical concepts, they should not be measured in the same units. The gram is a unit of mass. The force with which the earth attracts a one-gram mass at a standard location sometimes is called a “gram-weight” of force. * Since weight is proportional to mass in any given locality, this experiment is not affected by the slight variations consequent to laboratory conditions

Parallelogram Law of forces

If two forces acting at a point be represented in magnitude, direction and sense by the two adjacent sides of a parallelogram, their resultant is represented in magnitude, direction and sense by the diagonal of the parallelogram passing through that point .

In the figure vectors P and Q are represent in B C Magnitude, direction and sense by OA and OB respectively. The resultant R is represented by OC in magnitude, Q R and direction. α We now find magnitude of resultant R. Ø P α O A D From C we draw CD perpendicular to OA produced. Let α = angle between two forces P and Q = <AOB Now <DAC = <AOB = α (corresponding angle) In parallelogram OACB, AC is parallel and equal to OB So, AC = Q

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2 2 2

In ∆OCD, OC = OD +DC 2 2 2

Or, R = (OA+AD) +DC 2 2 2 2

Or, R = OA +2. OA.AD + AD +DC 2 2 2 2 2 2

Or, R = P +2PQcosα +Q cos α+ Q sin α 2 2

Or, R = (P + Q +2PQcosα) ½ Direction of resultant vector R Let Ø = angle made by resultant with OA Then from ∆OCD, tan Ø = CD/OD = Qsinα/ P+Qcosα

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So, Ø = tan¯ [Qsinα/ P+Qcosα]

Composition and Resolution of Concurrent Forces by Vector Methods

In order to add scalar quantities, one has merely to make the algebraic addition. When one wishes to add two vector quantities, the process is more difficult because their directions must be considered. The vector sum of two vector quantities is the single vector quantity that would produce the same result as the original pair.

The addition of vector quantities is greatly simplified by representing the vector quantity graphically. A vector is the line segment whose length represents the magnitude of a vector quantity and whose direction is that of the vector quantity. The sense along the line is indicated by an arrow. For example, a force of 100lb. acting at an angle of 30° above the horizontal may be represented by the line OA. Fig. 1, which is 5 units long and has the correct direction. Each unit of length thus represents 20lb. When vectors do not have the same line of action, their vector sum is not their algebraic sum but a geometric sum.

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This geometric sum may be determined by either graphical or analytical methods. Graphical methods are simple and direct but are limited in precision to that obtainable by drawing

instruments. Analytical methods have no such inherent limitations. In this experiment both

graphical and analytical methods will he applied to forces as examples of vector quantities, but

the same methods apply to all vector quantities.

The vector sum, or resultant, of a set of forces is the single force that will have the same effect,

Vector Summation by Graphical Methods: As an example of vector addition let us consider

the case of two forces acting on a body in such a direction that the forces are concurrent, that is

their lines of action, if projected would intersect at a point. The vectors OA and OB representing

two such forces are shown in Fig. 2. Their vector sum or resultant R, is found by constructing a

parallelogram having the two vectors as sides and drawing the concurrent diagonal, as shown in

Fig. 3. This diagonal vector R represents in magnitude and direction the single force that is

equivalent to the origina1 pair that is their vector sum. When the resultant of more than two

vectors is to be obtained graphically a polygon method is used. This is illustrated in Fig. 4. The

vector A is first constructed by the use of a chosen

scale and reference direction. Then, from the head

of A, the vector B is drawn. It is clear that the vector

M is the resultant of vectors A and B, since M

would be the concurrent diagonal of a

parallelogram if such a parallelogram had been

drawn, as was done in Fig. 3. Similarly, it follows

that the vector R is the resultant of M and C or of A,

B, and C. When the resultant of several forces is

required this method is simpler than the

parallelogram method. It should be noted that when

the parallelogram method is used, the arrows, with

their tails together, all radiate from a common point.

But in the polygon method the tail of the second arrow coincides with the head of the first, etc.

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Summation of Vectors by Analytical Methods: The resultant of two vectors may be

determined analytically by the use of the trigonometric laws of sines and cosines. Consider the vectors A and B in Fig. 5. The magnitude of the resultant R can be obtained by the application of the law of cosines:

2 2

R = A + B2 + 2ABcosβ (1) The direction of R can then be obtained from the law of sines: Sinφ/Sinβ = B/R Since sinβ = sinθ, Sinφ = (B/R) sinθ (2)

Components of Vectors: Any single force may be replaced by two or more forces whose joint action will produce the same effect as the single force. These various forces are said to be

components of the single force. The most useful set of components is usually a pair at right angles to each other, as shown in Fig. 6. The force B is the resultant of Forces Bx and By. Therefore conditions are unchanged by replacing the single force B by forces Bx and By, called their X- and Y- components. It is obvious from Fig. 6 that Bx = B cosβ and By = B sinβ. Component Method for Addition of Vectors: Fig. 7 illustrates the component method of computing the resultant of A,

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B, and C. The X- axis is so chosen that it coincides with the vector A, and the vectors B and C are resolved into X- and Y-components. The three forces A, B, and C have been replaced by five forces (A has no Y- component). The slim of the component along either axis may be computed by algebraic addition. Calling the sum of the X-components Fx and the sum of the Y-

components Fy, it follows that the resultant R is given by the equation 2 2 2

R = (FX) + (FY) (3) and that the angle φ- the angle that R makes with the X-

axis may be determined from the equation tanφ = Fy/Fx (4)

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Lecture 4 : Lami’s Theorem

Lami’s Theorem Q If a body is in equilibrium under the action of γ three forces, each force is proportional to the P sine of the angle between other two forces. α β Suppose the three forces P, Q & R are acting at a point O and they are in equilibrium as shown in figure. According to the Lami’s theorem, R P/sinα = Q/sinβ = R/sinγ α Proof of Lami’s theorem The three forces acting on a point are in (Π- α) Equilibrium and hence they can be R Q represented by the three sides of a γ triangle taken in the same order. In Π - γ figure .applying sine rule, we get Π –β P/ sin (Π -α) = Q/sin (Π -β) = R/sin (Π - γ) β P Or P/sinα = Q/sinβ = R/sinγ Free Vector Situation in which vector may be positioned anywhere in space without loss or change of meaning provided that magnitude and direction are kept intact. It is not constrained to any particular direction and location. It can be moved anywhere in space without rotation. Example 1. The velocity of water particle (having turbulent motion).

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2. Couple (the change in location of application couple in a plane does not change its effect) Bound Vector A bound or fixed vector has a definite point of application. It is specified by magnitude, direction and its point of application. Change in the point of application of force will alter its effect. y Representation of forces in terms of I,j & k FY β Many problems in mechanics require analysis in three dimensions and for such problem it is α often necessary to resolve a force into its three FX x mutually perpendicular components. The force O F acting at point O has the rectangular components FZ FX, FY, FZ. γ z Fx = F cosα, Fy = F cosβ, Fz = F cosγ

2 2 2)

Also F = √ (Fx + FY + FZ The cosines of α, β and γ are known as the direction cosine of vector F and denoted by l= cosα, m= cosβ, n= cosγ

2 2 2 2 2 2

The three angles are related by, cos α +cos β +cos γ = 1 or l +m + n = 1 Let i= vector of unit length in the +tive x direction j= vector of unit length in the +tive y direction k= vector of unit length in the +tive z direction The force F is represented by, F = Fxi + FY j+ Fzk

2 2 2

Magnitude of unit vector F = l F l = √ (Fx + FY + FZ ) But Fx = F cosα, Fy = F cosβ, Fz = F cosγ Substituting these values in earlier equation, we get F = F cosα i + F cosβ j+ F cosγ k

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Lecture 5 : Cross Product & Dot Product Cross Product or Vector Product The cross product of vectors A & B is a vector quantity and is defined as the product of the magnitude of the vector A, the magnitude of the vector B and the sine of the smaller angle between the two vectors. Its direction is perpendicular to the plane containing the vector. Now if n is the unit vector which gives the n B direction of the resultant vector R, θ R = A X B = lA l IB l sinθ. n = AB sinθ. n A Dot Product or Scalar Product Dot product of A and B is a scalar quantity and is defined as the product of the magnitude the vectors and cosines of their included angle. A. B = lA l IB l cosθ = AB cosθ Following points is to be noted

0

(i) When θ = 0 , and vectors A & B are along same direction, A. B = AB cos0 = AB

0

(ii) When θ = 90 , and vectors A & B are perpendicular to each other, A. B = AB cos90 = 0 (iii) A. B = A times projection of B on A projection of A On B = B times projection of A on B (iv) The angle between the vector A and B is O Cos θ =A. B / lAl lBl Projection of B on A = A. B/AB

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The dot product is used to find the component of a vector along an arbitrary direction and to define the term work. Example.1 Add a 20N force in the positive x direction to a 50N force at an angle 45° to axis in the first quadrant and directed away from origin. F1 = 20N, F2 = 50N B C We add vectorally, F1 + F2 = F 50N To get the sum, we may use the law of cosines 45° F for the triangular portions of the parallelogram by using the ∆OCA, α A O 20N

2 2

lFl = √ (20 + 50 + 2.20.50.cos45) = 65.68267N (Ans) The direction of the vector may be described by giving angle and sense. The angle is determined by employing the law of sine. For ∆OBA, 50/sinα = 65, 68267/sin135 Or α = 32.566° (Ans) Example.2 Force A (given as a horizontal 10N force) and B (vertical) and add up to a force C that has magnitude of 20N. What is the magnitude of force B? C y 20N Let OA represents force A (FA) = 10N B A Let OB “ “ B (FB) =? 10N Let OC “ “ C (FC) = 20N x B 20N

2 2 2

OC = OA + OB B C 2- 2

Or OB= √ (20 10 ) FB 10N = 17.32N O A Tanθ = 17.32/10 or θ = 60°

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Lecture 6 : Solved Examples Example3. A force vector of magnitude 100N has a line of action with direction cosines, l= 0.7, m=0.2 & n=0.59 relative to reference xyz. The vector points away from the origin. What is the component of the force vector along a direction ‘a’ having direction cosines, l=-0.3, m=0.1 & n=0.95 for the xyz reference? F = 100 (li +mj+nk) = 100 (0.7i +0.2j+0.59k) = 70i+20j+59k Unit vector along ‘a’ = li+mj+nk = -0.3i+0.1j+0.55k R Component of F along a is Fcosθ = [F] [a] cosθ = F. a θ = (70i+20J+59k) (-0.3i+0.1j+0.55k) N = 37.05N Fcosθ Example.4 Making use of the cross product, give the unit vector n normal to the inclined surface ABC. Given OB=8cm, OC=10cm, <ABY=150° From figure we can write, <OBA=180-150=30°, <BOA= 90° So AO = OB tan30 = 8 x tan30m =4.618m Coordinate of the points A,B &C are A(4.618,0,0), (0,8,0) , (0,0,10) respectively. z AB = -4.618i+8j+0k and AC = 4.618i+0j+10k C Area = ½.AB.AC = ½(-4.618i+8j+0k) (4.618i+0j+10k) O B y =(40i+23.09j+18k A x 150°

2 2 2

Unit Vector = (40i+23.09j+18k)/ √ (40 + 23.09 +18 ) = 0.804i+.464j+.371k) (ans) Assignment 1. Subtract the 20N force in example.1 from50N force. 400N 40° 500N

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2. What is the sum of forces transmitted by the structural rods to the pin at A? 3. Given the vectors A= 6i+3j+10k, B=2i-5j+5k, C=5i-2i+7k What vector D gives the following results D.A=20, D.B=5, D.C=10 Multiple Choice Questions 1. Moment of a force is a a) Scalar quantity, b) Vector quantity, c) Either a or b, d) None of the above 2. A vector having a unit magnitude is known as a) Null vector, b)Free vector, c)Unit vector,d)None of the above 3. The dot product of two orthogonal vectors is a) 1, b) o, c) No definite value, c) None of the above 4. Two vectors are said to be equivalent if they produces a) Same magnitudes, b) Same directions, c) Same effect in a certain respect d) None of these .

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Lecture 7 : Two Dimensional Forces Two dimensional force system & Resolution of forces When several forces of different magnitudes and directions act upon a body, they constitute a system of forces. If all the forces in a system lie in a single plane is known as coplanar force system. If all the forces in a system lie on the same plane and the lines of action of all forces do not pass through a single point, the system is known as coplanar non-concurrent force system. The most common two-dimensional y resolution of a force vector is into j rectangular components. It follows from the parallelogram rule that the F vector F (in figure) may be written as FY F=FX+Fy θ x

FX I

Where Fx & Fy are vector components of F in the x and y directions. Each of the two vector components may be written as a scalar times the appropriate unit vector. In terms of the unit vector I and j, FX=Fxi & FY =Fyj and thus we may write F = Fxi +Fyj Where the scalars FX and FY are the x and y scalar components of the vector F. The scalar components can be positive or negative, depending on the quadrant into which F points. For the force vector shown above the x and y scalar components are both positive are related to the magnitude and direction of F by

2 2 1

FX = Fcosθ & FY = Fsinθ F = √ (FX + Fy ), θ = tan¯ (FX/ FY) Moment of a force

The product of the magnitude of a force and the perpendicular distance of the line .O of action the force from appoint is known d as moment of a force about that point. moment of a force about a point is the measure of its rotational effect. The rotational effect of a force becomes very P important when we deal with non concurrent force system.

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Moment ‘M’ of a force P about point O as shown in figure is given by, M = P x d where d is the perpendicular distance between the line of action of the force and the moment centre. The tendency of this moment is to rotate the body in anticlockwise direction about O.

Varignon's theorem

One of the most useful principles of mechanics is Varignon’s theorem. The algebraic sum of the moments of a system of coplanar forces about a moment centre in their plane is equal to the moment of their resultant force about the same moment centre. Proof According to the figure, let F be the Y Resultant of the forces F1 and F2 acting at A. Let us consider any point B lying in the plane of the forces, as a moment centre. Let I, l1 and l2 be the moment arms of the forces F, F1 and F2 respectively F from the moment centre B. F2 θ2 We join AB and consider it as y-axis F1 And draw x-axis at right angle to it θ θ1 at A as shown in figure. O X FX2 FX1 FX Now moment of the force F about B = F X l = F (BAcosθ) = BA (Fcosθ) = BAFX

Moment of the force F1 about B = F1 X l1 = F1 (BAcosθ1) = BA (F1 cosθ1) = BAFX1….1 Moment of the force F2 about B = F2 X l2 = F1 (BAcosθ2) = BA (F2 cosθ2) = BAFX2.....2

Now adding equation 1 and 2, F1 X l1 + F2 X l2 = BA (FX1 + FX2) But the sum of the x components of forces F1 and F2 = x components of the resultant F So FX = FX1 + FX2 or BAFx = BA (FX1 + FX2) So from Fx l = F1 X l1 + F2 X l2 hence proved If a system of forces consists of more than two forces, the above proof can be extended by taking into consideration all forces.

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Lecture 8 : Couples Couple Two parallel forces of equal magnitude 1 P d P but opposite in direction are separated . by a finite distance to form a couple. d1

The algebraic sum of the forces forming a couple is zero, which means the translatory effect is zero. The algebraic d2 sum of the moments of the two forces of a couple is independent of the position in the plane of the couple, of the moment centre P d3 2 d4 and is equal to the moment centre and is always equal to the product of the magnitude of the forces and the perpendicular distance d between the two forces. Let the magnitude of the forces forming a couple P is P and the perpendicular distance between the two forces is d. Considering the moment of the two forces forming a couple about point 1as shown. Le t the moment be M1. So M1 = Pd2 –Pd1 =P (d2 –d1) = P d Now we consider the moment of the forces about point 2 as shown. Let M2 is the moment, then M2 = P (d3+d4) = P. d The moment of a couple about any point is the same. Since the only effect of a couple is a moment (rotational effect) and this moment is the same about any point, the action of a couple is unchanged if (a) The perpendicular distance of the forces i.e. arm is turned in the plane of couple through any angle about one of its end. (b) The magnitude of the forces and the arm of the couple both are changed in such a way that the moment of the couple remains unchanged. (c) The couple is shifted to any other position.

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Example 1: A person is holding a 100N weight (that is roughly a 10kg mass) by a light weight

(negligible mass) rod AB. The rod is 1.5m long and weight is hanging at a distance of 1m from the end A, which is on a table (see figure 6). How much force should the person apply to hold the weight?

Let the normal reaction of the table on the rod be N and the force by the point be F1. Then the two equilibrium conditions give

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(Note: If the brick did not provide friction, the force applied cannot be only in the vertical direction as that would not be sufficient to cancel the horizontal component of N). Let us see what happens if the brick offered no friction and we applied a force in the vertical direction. The fulcrum applies a force N perpendicular to the rod so if we apply only a vertical force, the rod will tend to slip to the left because of the component of N in that direction. Try it out on a smooth corner and see that it does happen. However, if the friction is there then the rod will not slip. Let us apply the equilibrium conditions in such a situation. The balance of forces gives

Let us choose the fulcrum as the point about which we balance the torque. It gives

Then

The normal force and the frictional force can now be calculated with the other two equations obtained above by the force balance equation.

Example2: To balance a heavy weight of 5000 N, two persons dig a hole in the ground and put a pole of length l in it so that the hole acts as a socket. The pole makes an angle of 30° from the ground. The weight is tied at the mid point of the pole and the pole is pulled by two horizontal ropes tied at its ends as shown in figure 2. Find the tension in the two ropes and the reaction forces of the ground on the pole.

27

To solve this problem, let me first choose a co-ordinate system. I choose it so that the pole is over the y-axis in the (y-z) plane (see figure 2). The ropes are in (x.y) direction with tension T in each one of them so that tension in each is written as

You may be wondering why I have taken the tension to be the same in the two ropes. Actually it

arises from the torque balance equation; if the tensions were not equal; their component in the x-

direction will give a nonzero torque. Let the normal reaction of the ground be (Nx, Ny, Nz). Then the force balance equation gives

Taking torque about point O and equating it to zero, we get

which gives

28

Lecture 9 : Resolution Of Coplanar Forces

Resolution of a coplanar force by its equivalent force-couple system The effect of force acting on a body is the tendency to push or pull the body in the direction of the force, and to rotate the body about any fixed axis which does not intersect the line of force. We can represent this dual effect more easily by replacing the given force by an equal parallel force and a couple to compensate for the change in the moment of the force. This replacement is illustrated in figure where the given force F acting at point A is replaced by an equal force F and at some point B and the counterclockwise couple M= F d. B F . B F F A. A d F Resultant of forces The most common type of force system occurs when the forces all act in a single plane, say, the x-y plane. We obtain the magnitude and direction of the resultant force R by forming the force polygon. Thus for any system of coplanar forces we may write R = F1 +F2 + F3 + ……………. =∑F Rx = ∑FX RY = ∑FY where R = √ [(∑FX) ² + (∑FY) ²]

θ = tan ̄ RY/ Rx = tan ̄ ∑FY / ∑FX

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Example 3:

The picture below shows the forces acting on a parked car. If the weight of the car acts exactly

halfway between the two wheels and the weight is 1000lbs how much force is exerted on the

rear wheel? What about the front wheel?

Writing the force equations

There are no forces in the x direction

Writing moment equation about front wheel

Subbing RB back into the sum Fy

Please note that Rf and RB are distributed over two wheels. Each front wheel supports half of Rf and each back wheel supports half of RB.

Example 4. A weight of 15kN hangs from a point C, by two strings BC and AC (Fig. to

Exmpl.5.1). Determine the tensions in the strings.

B

B

C

B

T1 T2

75°

135° 150°

A

15KN

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The load 15kN force acts downwards inducing tensions T1 and T2 in the strings CB & CA (Fig to

Exmpl.5)

Applying Lami’s theorem:

15kN T1 T2

o = o = o

sin75 sin150 sin135

o o

Or, T1 = 15kN (sin150 ) / sin75 = 7.76kN Ans. o o

& T2 = 15kN (sin 135 ) / sin 75 = 10.98kN Ans.

Example 5. ABCD is a square of 1m side. It is being acted upon by a number of forces as

shown (Fig. to Exampl.4.3). The system is at equilibrium. Find the magnitude of

i) P & Q ii)The resultant couple. P

D C

Q 200kN

A B

100KN

1m

Fig. to Example. 5

Resolving the system of forces horizontally and computing Σ H = 0

o

100kN –100kN (cos45 ) – P = 0

Or, P = 29.3kN ns.

Resolving the forces vertically and computing Σ V = 0

o

200kN –100kN (sin45 ) –Q = 0

Or, Q = 129.3kN Ans.

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ii) Couple

Moment of a couple = Σ M of constituents forces about any point

Σ MA = + 200kN x 1m + P x 1m = 200 + 29.3 kN = 229.3kN Ans.

Since the moment is + tive, the couple is counterclockwise.

Example 6. At what point on the beam a weight of 2kN is to be placed so that one of the strings

may just snap?

The weight of the beam 4kN acts halfway between A and B.

TA TB

3m

x

A D C B

AC = BC

2kN

4kN

Fig to Example 6

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TA and TB are the tensions in the two strings, which balance the net downward force.

When either TA or, TB becomes 3.5kN, the corresponding string snaps. Let string at end A

snaps.

Taking moment about B and for balance: ΣMB = 0

3.5kN x 3m = 2kN x (3 – x) + 4kN x 1.5m

Or, x = 0.75m Ans.

33

Lecture 10 : Solved Examples

Example7.The screw eye in Fig. is subjected to two forces. Determine the magnitude and direction of the resultant force.

F2=150N

10°

10° F1= 100N

125° 65°

15° 150N FR

θ

100N 15°

90°-25°=65°

Ø

Parallelogram Law. The parallelogram law of addition is shown in Figure Two unknowns are the magnitude of FR and the angle θ(theta).

2 2

FR = √ (100 N) + (150 N) – 2x100 Nx150 N cos 115° =213N (Ans) The angle is determined by applying the law of sines, using the computed value of FR. 150 N 212.6 N --------- = ------------ sin θ sin 115° 150 N sin θ = ------------- (0.9063) 212.6 N

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θ = 39.8° Thus, the direction Ø (phi) of FR, measured from the horizontal, is Ø = 39.8° + 15.0° = 54.8° (ans) Example8.The force F acting on the frame shown in Figure has a magnitude of 500 N and is to be resolved into two components acting along members AB and AC. Determine the angle measured below the horizontal, so that the component is directed from A toward C and has a magnitude of 400 N.

FAC=400N FAC=400

θ B θ 60°

60° 500N Ø

500N FAB

FAB 30° 30°

FIG A FIG B

A C

θ

500N By using the parallelogram law, the vector addition of the two components yielding the resultant is shown in Figure A. Note carefully how the resultant force is resolved into the two components FAB and FAC which have specified lines of action. The angle Ø can be determined by using the law of sines. The corresponding vector triangle is shown in Fig. 400 N 500N ---------- = ------- sin Ø sin 60° 400 N sin Ø = ---------- x sin 60 = 0.6928 or Ø = 43.9° 500 N Hence, θ = 180° - 60° - 43.9° = 76.1° (ans) Using this value for θ we apply the law of cosines or the law of sines to find the value of FAB

which has a magnitude of 561N .

35

Example9. The system in Figure shown below is in equilibrium with the string in the center

exactly horizontal. Find (a) tension T1, (b) tension T2, (c) tension T3 and (d) angle θ. 35° θ T1 T3

T2 40N 50N T3

T1 T2 T2 (b) 40N (a) 50N

(a) Forces at the left junction of the strings. (b) Forces acting at the right junction of the strings.. We have to solve for four unknowns (T1, T2, T3 and θ). We consider the points where the strings meet; the left junction is shown in Figure (a). Since a string under tension pulls inward along its length with a force given by the string tension, the forces acting at this point are as shown. Since this junction in the strings is in static equilibrium, the (vector) sum of the forces acting on it must give zero. Thus the sum of the x components of the forces is zero: −T1 sin 35_ + T2 = 0 ………1 The sum of the y components of the forces is zero: +T1 cos 35- 40N = 0 ………2 Now we look at the right junction of the strings; the forces acting here are shown in Figure (b). Again, the sum of the x components of the forces is zero: −T2 + T3 sin θ = 0 …………3 The sum of the y components of the forces is zero: +T3 cos θ − 50N = 0……….4 And at this point we are done with the physics because we have four equations for four

36

unknowns. We will do algebra to solve for them. From Eq. 2, we get, T1 = 40N/cos 35 = 48.8N and then Eq. 1 gives us T2, 2 = T1 sin 35 = (48.8N) sin 35 = 28.0N . We now rewrite Eq. 3. As: T3 sinθ = T2 = 28.0N……..5 And Eq. 4 as: T3 cosθ = 50.0N……….6 Now if we divide the left and right sides of 5 by the left and right sides of 6 we get: tanθ = (28.0N)/(50.0N) = 0.560 And then θ = tan− 1(0.560) = 29.3 Finally, we get T3 from Eq. 3.9: T3 = 50.0Ncos 29.3 = 57.3N Summarizing, we have found: T1 = 48.8N, T2 = 28.0N, T3 = 57.3N, θ= 29.3 B

Assignment 1A. The vertical force F acts downward at A on the two-membered frame. Determine the magnitudes of the two A components of F directed along the axes of AB and AC. Set F=500N, <ABC=45°, <ACB=60° 1B. Solve Prob. 1A with F = 350 lb.

C 2. For the stepladder shown in Figure, sides AC and CE are C each 8.0 ft long and hinged at C. Bar BD is a tie–rod 2.5 ft

long, halfway up. A man weighing 192 lb climbs 6.0 ft along the ladder. Assuming that the floor is frictionless and neglecting the weight of the ladder, find (a) the tension B D in

the tie–rod and the forces exerted on the ladder by the

floor at (b) A and (c) E. Hint: It will help to isolate parts A E

of the ladder in applying the equilibrium conditions.

Multiple Choice Questions

1. A force can be replaced by a) A force of same magnitude and couple b) A force-couple combination so that equilibrium is maintained c) A force of different magnitude and couple d) None of the above

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2. Two non-collinear parallel equal forces acting in opposite direction a) Balance each other b) Constitute a couple c) Constitute a moment d) None of the above 3. The force polygon for coplanar concurrent forces a) Must Close b) Must not close c) Anyone of the above d) None of the above 4. Moment of a force a) Varies directly with its distance from the pivot b) Varies inversely with its distance from the pivot c) Is independent of its distance from the pivot d) None of the above 5. A couple consists of a) Two like parallel forces of same magnitude b) Two like parallel forces of different magnitude c) Two unlike parallel forces of same magnitude d) Two unlike parallel forces of different magnitude

38

Lecture 11 : Concept Of Equillibrium Equilibrium: Many problems that concern the physicist and engineer involve several forces acting on a body under circumstances in which they produce no change in the motion of the body. This condition is referred to as equilibrium. The body does not necessarily have to be at rest, but its motion must retain the same velocity; hence both magnitude and direction of motion are unchanged. First Condition for Equilibrium: Insofar as linear motion is concerned, a body is in equilibrium if there is no resultant force acting upon it that is if the vector sum of all the forces is zero. This statement is called the first condition for equilibrium. This condition is satisfied if the vector polygon representing all the external forces acting on the body is a closed figure. Analytically

this condition is satisfied if each set of rectangular components of the forces separately adds to zero, or Rx = ΣFx = 0 (5) Ry = ΣFy = 0 (6)

Mechanical equilibrium

A standard definition of static equilibrium is:

A system of particles is in static equilibrium when all the particles of the system are at rest [1]

and the total force on each particle is permanently zero.

This is a strict definition, and often the term "static equilibrium" is used in a more relaxed manner [2]

interchangeably with "mechanical equilibrium", as defined next.

A standard definition of mechanical equilibrium for a particle is:

The necessary and sufficient conditions for a particle to be in mechanical equilibrium are [3]

that the net force acting upon the particle is zero.

The necessary conditions for mechanical equilibrium for a system of particles are:

(i)The vector sum of all external forces is zero; [3]

(ii) The sum of the moments of all external forces about any line is zero.

As applied to a rigid body, the necessary and sufficient conditions become:

A rigid body is in mechanical equilibrium when the sum of all forces on all particles of the system is zero, and also the sum of all torques on all particles of the system is zero.

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A rigid body in mechanical equilibrium is undergoing neither linear nor rotational acceleration; however it could be translating or rotating at a constant velocity.

However, this definition is of little use in continuum mechanics, for which the idea of a particle is foreign. In addition, this definition gives no information as to one of the most important and interesting aspects of equilibrium states – their stability.

An alternative definition of equilibrium that applies to conservative systems and often proves more useful is:

A system is in mechanical equilibrium if its position in configuration space is a point at which the gradient with respect to the generalized coordinates of the potential energy is zero.

40

Lecture 12 : Free Body Diagram

Free body diagram

Block on a ramp (top) and corresponding free body diagram of just the block (bottom).A free

body diagram is a pictorial representation often used by physicists and engineers to analyze the

forces acting on a free body. A free body diagram shows all contact and non-contact forces

acting on the body. Drawing such a diagram can aid in solving for the unknown forces or the

equations of motion of the body. Creating a free body diagram can make it easier to understand

the forces, and moments, in relation to one another and suggest the proper concepts to apply in

order to find the solution to a problem. The diagrams are also used as a conceptual device to

help identify the internal forces—for example, shear forces and bending moments in beams—

which are developed within structures.

Construction

A free body diagram consists primarily of a sketch of the body in question and arrows representing the forces applied to it. The selection of the body to sketch may be the first important decision in the problem solving process. For example, to find the forces on the pivot joint of a simple pair of pliers, it is helpful to draw a free body diagram of just one of the two pieces, not the entire system, replacing the second half with the forces it would apply to the first half.

What is included

The sketch of the free body need include only as much detail as necessary. Often a simple outline is sufficient. Depending on the analysis to be performed and the model being employed, just a single point may be the most appropriate. All external contacts, constraints, and body forces are indicated by vector arrows labeled with appropriate descriptions. The arrows show the direction and magnitude of the various forces. To the extent possible or practical, the arrows should indicate the point of application of the force they represent. Only the forces acting on the object are included. These may include forces such as friction, gravity, normal force, drag, or simply contact force due to pushing. When in a non-inertial reference frame, fictitious forces, such as centrifugal force may be appropriate. A coordinate system is usually included, according to convenience. This may make defining the vectors simpler when writing the equations of motion. The x direction might be chosen to point down the ramp in an inclined plane problem, for example. In that case the friction force only has an x component, and the normal force only has a y component. The force of gravity will still have components in both the x and y direction: mgsinθ in the x and mgcosθ in the y, where theta is the angle between the ramp and the horizontal.

41

What is excluded

All external contacts and constraints are left out and replaced with force arrows as described above. Forces which the free body applies to other objects are not included. For example, if a ball rests on a table, the ball applies a force to the table, and the table applies an equal and opposite force to the ball. The FBD of the ball only includes the force that the table causes on the ball. Internal forces, forces between various parts that make up the system that is being treated as a single body, are omitted. For example, if an entire truss is being analyzed to find the reaction forces at the supports, the forces between the individual truss members are not included. Any velocity or acceleration is left out. These may be indicated instead on a companion diagram, called "Kinetic diagrams", "Inertial response diagrams", or the equivalent.

Assumptions

The free body diagram reflects the assumption and simplifications made in order to analyze the system. If the body in question is a satellite in orbit for example, and all that is required is to find its velocity, then a single point may be the best representation. On the other hand, the brake dive of a motorcycle cannot be found from a single point, and a sketch with finite dimensions is required. Force vectors must be carefully located and labeled to avoid assumptions that presuppose a result. For example, in the accompanying diagram of a block on a ramp, the exact location of the resulting normal force of the ramp on the block can only be found after analyzing the motion or by assuming equilibrium. Other simplifying assumptions that may be considered include two-force members and three-

force members.

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Lecture 13 : Equation Of Equilibrium

Equation of equilibrium

When a body is in equilibrium, resultant of all forces and moments acting on that body is zero. Stated in another way, a body is in equilibrium if all forces and moments applied to it are in balance. These requirements are contained in the vector equations of equilibrium which in two

dimensions may be written in scalar from as ∑FX = 0, ∑FY = 0, ∑MO =0 The third one represents the zero sums of the moments of all forces about any point O on or off the body.

Let's look at a truss

P Q S

C D

A B

FBD

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P Q S

C D

W

Ax

A By y

M A0 By

Fx0 Ax

Fy0 Ay

Additional equations could be written.

MB0 Does not provide any new info. This is not an independent equation.

We can use MB0to replace one of the above 3.

Example 1.Given

2 m

2 kN

1.5 m

2 kN

1.5 m

A B

Find: Reactions

44 F0 x

22R0 Fy0 Ax

R4 kN 3RAyRB0

Ax

FBD

Substituting RB into y-equation: RAy = -1.5 KN

2 m

2 kN

1.5 m

2 kN

1.5 m

A B RAx

RAy RB

Example2.Given 2000 lbs

3000 ft lbs A B C

3 ft 2 ft

Find: Reactions at A

Fy0 2000 lbs

M 3000 ft lbs Fx0 R20000 A Ay A B C RAx0

R RAy2000 lbs Ax

3 ft 2 ft

R M A0 Ay

M A30002000(5)0 FBD

M A7,000 ft lbs

45

A simple free body diagram, shown above, of a block on a ramp illustrates this.

All external supports and structures have been replaced by the forces they generate. These include:

Mg: the product of the mass of the block and the constant of gravitation acceleration: its weight.

N: the normal force of the ramp. Ff: the friction force of the ramp.

The force vectors show direction and point of application and are labeled with their magnitude.

It contains a coordinate system that can be used when describing the vectors.

Example 3. Two loads 400N and 500N are suspended in a vertical plane by three springs as

shown in Figure. Find the tension in the strings

46

B A

30° 30°

C O

400N

500N

Fig for Example.1

Obviously tension in OB = TOB = 500N

Resolving the forces at O vertically 0 o

POA.sin30 + POB.sin30 = 400N Or, POA = 300N

Resolving the forces at O horizontally o O

POC + POA.cos30 = POB.cos30 o o

or, POC + 300N(cos30 ) = (500N)cos30 POC = 100√3 N Tension in OA = 300N Tension in OB = 500N Tension in OC = 100√3 N

Assignment

Problem. 1 An electric light fixture weighing 15 Newton hangs from a point C, by two strings AC

and BC. AC is inclined at 60° to the horizontal and BC at 45° to the vertical as shown in

Fig. /Prob.1.Using Lami’s theorem or otherwise determine the forces in the strings AC and BC.

47

O 60

A B

O 45

C

15N

Example2 Given: The loading car weight is 5500 lbs and its CG is at point G.

T

24" 30"

25o

G

25"

25"

Find: tension in cable and reactions at wheels.


1. Free body diagram can be applied only in a) Dynamic equilibrium problem b) Static equilibrium problem

c) Both static & dynamic equilibrium problems d) None of these

2. If the body is in equilibrium ,we may conclude that

a) No force acting in it b) Moment of all forces about any point is zero c) The resultant of all forces acting on it is zero

d) Both b&c

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3. The algebraic sum of the moments of two forces about any point in their plane is equal to the moment of their resultant about that point is known as a) Principle of moments b) Varignon’s theorem c) Lamis theorem d) None of these

4. Two coplanar couples having equal and opposite moments

a) Produce a couple and an unbalanced force

b) Are equivalent

c) Balance each other

d) None of these

5. A free body diagram of a body represents a) With its surroundings and external forces acting on it b) Isolated from its surroundings and all external forces acting on it c) Isolated from all external actions d) None of these

6.The force which meet at one point and their lines of action also lie on the same plane are

known as…………forces.

a) Coplanar concurrent b) Coplanar non-concurrent c) Non-coplanar non-concurrent d) None of these

49

Lecture 14 : Friction

Friction is the force distribution at the surface of contact between two bodies that prevents or

impedes sliding motion of one body relative to the other. This force distribution is tangent to the

contact surface and has, for the body under consideration, a direction at every point in the

contact surface that is in opposition to the possible or existing slipping motion of the body at that

point.

Types of friction

Dry friction resists relative lateral motion of two solid surfaces in contact. Dry friction is also subdivided into static friction between non-moving surfaces, and kinetic friction (sometimes called sliding friction or dynamic friction) between moving surfaces.

Lubricated friction or fluid friction resists relative lateral motion of two solid surfaces

separated by a layer of gas or liquid.

Fluid friction is also used to describe the friction between layers within a fluid that are

moving relative to each other.

Skin friction is a component of drag, the force resisting the motion of a solid body through a fluid.

Internal friction is the force resisting motion between the elements making up a solid

material while it undergoes deformation.

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Consider a solid block of mass m resting on a horizontal surface as shown. Assume that the

contacting surfaces are rough. As we gradually increase the load, the block remains static till the

load reaches a threshold value. Since the force in the x-direction has to be balanced, it is

apparent that as the magnitude of F increases from zero, the friction force also increases.

Friction force, is thus, self adjusting. However, the friction force cannot increase beyond a limit.

Thus there is a limiting value of friction. The maximum value of friction force, which comes into

play, when the motion is impending, is known as limiting friction. When the applied force is less

than the limiting friction, the body remains at rest and such frictional force is called static friction,

which may have any value between zero and the limiting friction. If the value of the applied force

exceeds the limiting friction, the body starts moving over the other body and the frictional resistance experienced by the body while moving is known as Dynamic friction. Dynamic

friction is found to be less than limiting friction. See the following animation to understand the

phenomenon of dry friction.

It is experimentally found that the magnitude of limiting friction bears a constant ratio to the

normal relation between the two surfaces and this ratio is called coefficient of Friction.

Coefficient of friction = = µ

Where F is limiting friction and N is the normal reaction between the contact surfaces.

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Columb Laws of friction:

(1) The force of friction always acts in a direction opposite to that in which the body tends to

move.

(2) Till the limiting value is reached, the magnitude of friction is exactly equal to the force which

tends to move the body.

(3) The magnitude of the limiting friction bears a constant ratio to the normal reaction between

the two surfaces.

(4) The force of friction depends upon the roughness/smoothness of the surfaces.

(5) The force of friction is independent of the area of contact between the two surfaces.

Angle of static friction:

Consider the block on the following surface.

The free body diagram is shown. The direction of resultant R measured from the direction of N is

specified by tan a=F/N. When the friction force reaches its limiting static value Fmax, the angle a

reaches a maximum.

52

Value of fs. Thus

tan fs = ms

The angle fs is called the angle of static friction.

Angle of kinetic friction:

When slippage is occurring, the angle a has a value fR corresponding to the kinetic friction force.

tan fR = mR

Cone of friction:

When a body is having impending motion in the direction of P the frictional force will be the

limiting friction and the resultant reaction R will make limiting friction angle a with the normal as

shown in the following figure. If the body is having impending motion in some other direction,

again the resultant reaction makes limiting frictional angle a with the normal in that direction.

Angle of Repose:

The maximum inclination of the plane on which a body, free from external forces, experiences

repose (sleep) is called Angle of Repose.

53

Now consider the equilibrium of the block shown above. Since the surface of contact is not

smooth, not only normal reaction, but frictional force also develops. Since the body tends to slide

downward, the frictional resistance will be up the plane.

∑forces normal to the plane =0, gives

N=Wcos θ …………1

∑forces normal to the plane =0, gives

F=Wsin θ ……………2

Dividing equ (2) by equ (1), we get

If N is the value of normal force when motion is impending, frictional force will be µN and hence

Hence, to avoid free sliding, the inclination angle should be less than the friction

angle.

54

Lecture 15 : Coefficient Of Friction

Coefficient of friction

The coefficient of friction (COF), also known as a frictional coefficient or friction coefficient, symbolized by the Greek letter µ, is a dimensionless scalar value which describes the ratio of the force of friction between two bodies and the force pressing them together. The coefficient of friction depends on the materials used; for example, ice on steel has a low coefficient of friction, while rubber on pavement has a high coefficient of friction. The coefficient of friction is an empirical measurement – it has to be measured experimentally, and cannot be found through calculations. Rougher surfaces tend to have higher effective values. Most dry materials in combination have friction coefficient values between 0.3 and 0.6. Values outside this range are rarer, but teflon, for example, can have a coefficient as low as 0.04. A value of zero would mean no friction at all, an elusive property – even magnetic levitation vehicles have drag. Rubber in contact with other surfaces can yield friction coefficients from 1 to 2. Occasionally it is maintained that µ is always < 1, but this is not true. While in most relevant applications µ < 1, a value above 1 merely implies that the force required to slide an object along the surface is greater than the normal force of the surface on the object. For example, silicone rubber or acrylic rubber-coated surfaces have a coefficient of friction that can be substantially larger than 1. Both static and kinetic coefficients of friction depend on the pair of surfaces in contact; their values are usually approximately determined experimentally. For a given pair of surfaces, the coefficient of static friction is usually larger than that of kinetic friction; in some sets the two coefficients are equal, such as teflon-on-teflon.

Kinetic friction

Kinetic (or dynamic) friction occurs when two objects are moving relative to each other and rub together (like a sled on the ground). The coefficient of kinetic friction is typically denoted as µk, and is usually less than the coefficient of static friction for the same materials. Examples of kinetic friction:

Kinetic friction is when two objects are rubbing against each other. Putting a book flat on a desk and moving it around is an example of kinetic friction.

55

Fluid friction is the interaction between a solid object and a fluid (liquid or gas), as the object moves through the fluid. The skin friction of air on an airplane or of water on a swimmer are two examples of fluid friction. This kind of friction is not only due to rubbing, which generates a force tangent to the surface of the object (such as sliding friction). It is also due to forces that are orthogonal to the surface of the object. These orthogonal forces significantly (and mainly, if relative velocity is high enough) contribute to fluid friction. Fluid friction is the classic name of this force. This name is no longer used in modern fluid dynamics. Since rubbing is not its only cause, in modern fluid dynamics the same force is typically referred to as drag or fluid resistance, while the force component due to rubbing is called skin friction. Notice that a fluid can in some cases exert, together with drag, a force orthogonal to the direction of the relative motion of the object (lift). The net force exerted by a fluid (drag + lift) is called fluid dynamic force (aerodynamic if the fluid is a gas, or hydrodynamic if the fluid is a liquid).

Application of Friction

1) Wedges

Wedges are small pieces of material with two of its opposite surfaces not parallel. They are used to lift heavy blocks, machinery, precast beam etc., slightly, required for final alignment or to make place for inserting lifting devices. The weight of the wedge is very small compared to the weight lifted. Hence, in all the problems, weight of wedges may be neglected. The following figure is showing a wedge:

Free body diagram of the weight to be lifted is shown below:

56

The body is acted upon by three forces: weight W, R1 (the resultant of normal wall reaction and

friction force) and R2 ( the resultant of the normal reaction of the wedge and the friction force). In

the free body diagram, the resolved components of R1and R2 are shown by thin lines. If the

friction angle is f, the R1and R2 will make angle with the respective normals to the surfaces.

Note that coefficient of friction is given by .

As the body is acted by three non-parallel forces, the forces must be coplanar and concurrent.

The relation between the forces can be found by Lami's theorem. The following figure shows the

three forces meeting at a common point:

57

By Lami's theorem:

Now, let us make the free body diagram of the wedge. Three forces acting on the wedge are

shown. R1and R2 are the resultants of normal and surface forces.

Hence three forces meet at a point as shown below.

58

The Lami's theorem gives us:

This analysis pertains to load being lifted by the wedge. If the load is lowered, the direction of

friction forces and P will reverse. The analysis is similar, except that f will be replaced by - .

For , P will be positive. That is some force will be needed to lower the load. In other

words, without applied P, the load W will not get lowered and the wedge is called self-locking.

59

Lecture 16 : Solved Examples

Example 1 A load of 150kN rests on a rough inclined plane (angle of inclination ). It can be

just moved up the plane if a 200kN force is applied horizontally or a force 125kN applied parallel

to the plane. Determine the inclination of the plane, and the coefficient of friction,.

Solution: See Fig.1 & 2

200kN

α

150kN

Fig.1 to Example

125kN

60

α

150kN

Fig.2 to Example

FBD when 200kN force is applied

R1

F1

200kN

150kN

P = W tan (+)

Or, 200kN = (150kN) tan (+)

Or, tan (+) = 200/150 = 1.333 0

Or, (+) = 54.31

R2 125kN

When 125kN force is applied

Sin (+) 6 1

F2

P = W

Cos

150kN

o Sin (53.1 )

Or, 125 (KN) = 150kN

Cos

o

Or, = 16.3 .

o o o

Therefore, = 53.1 - 16.3 = 36.8 (Ans).

Coefficient of Friction

o

= tan = tan 16.3 = 0.292 (Ans.)

Example2.

Two blocks with masses mA = 20 kg and mB = 80 kg are connected with a flexible cable that

passes over a frictionless pulley as shown in Fig. P9-4. The coefficient of friction between the

blocks is 0.25. If motion of the blocks is impending, determine the coefficient of friction between

block B and the inclined surface and the tension in the cable between the two blocks.

62

∑FY = An - 196.14 cos (35) = 0

An = 160.67 N

.

For impending motion of the block A: (Af = msAn)

+ ∑Fx = -T + 0.25(160.67) + 196.14 sin 35 = 0

T = 152.67 N

63

.

Block B weight: WB = mBg = 80 (9.807) = 784.56 N

= Bf/Bn= 257.17/803.34 = 0.320

∑Fy = Bn - 160.67 - 784.56 cos 35 = 0

Bn = 803.34 N

∑Fx = - Bf - 40.17 - 152.67 + 784.56 sin 35 = 0

Bf = 257.17 N

Assignment

1.Blocks A and B, of weight 50 N and 100 N, respectively rest on an inclined plane as shown in

the figure. The coefficient of friction between the two blocks is 0.3 and between block A and

inclined plane is 0.4. Find the value of for which either one or both the blocks start slipping. At

that instant, what is the friction force between B and A? Between A and inclined plane?

64

2. The block in figure is to be moved by applying a force T to a cable which slides over a fixed

pulley. Find the value of T which will cause sliding motion of the block.

3. In the figure shown, if all the contact surfaces are smooth, then the relation between P and W is

a. c. P=W tanα

b. d.

65


1. Static friction is always

a) Greater than dynamic friction

b) Less than dynamic friction

c) Equal to dynamic friction

d) None of the above

2. Coefficient of friction depends on

a) Area of contact

b) Shape of the surfaces

c) Strength of the surface

d) Nature of the surface

3. Frictional force required to move the body up the plane will rest, if it makes with the

inclined plane an angle

a) Equal to the angle of friction

b) Less than the angle of friction

c) Greater than the angle of friction


4.Frictional resistance depends upon

a) Speed of the body

b) Geometrical shape of the body

66

c) Nature of contacting surfaces

d) Weight of the body and nature of contacting surfaces

5. Angle of the friction is the

a) Angle between normal reaction and the resultant of normal reaction and the limiting

friction.

b) Ratio of limiting friction and normal reaction

c) Ratio of minimum friction force to friction force acting when the body is just about

to move.

d) Ration of minimum friction force to friction force acting when the body is in motion

6. Ratio of liming friction and normal reaction is known as

a) coefficient of friction

b) Angle of friction

c) Sliding friction


7. Kinetic friction is

a) The maximum force of friction when the body is about to move.

b) The force of friction when the body is in motion

c) The force of friction between two lubricated bodies

d) None of these.

67

8. Coulomb friction is the friction between

a) Bodies having relative motion

b) Two dry surfaces

c) Two lubricated surfaces

d) Solids and liquids

68

Lecture 17 : Distribution of Forces

Distribution of forces

We know that all forces are concentrated along their

lines of action and at their point of application.

Actually concentrated forces do not exist in the exact

since every external force applied mechanically to a

body is distributed over a finite contact area however

small.

Example

Force exerted by the road on automobile b

tyre is applied to the tyre over its entire

area of contact. It may be applicable if

the tyre is soft. When analyzing the forces

acting on the car as a whole, if the dimension

‘b’ of the contact area is negligible compared

With other pertinent dimension e.g. distance

between wheels, we may replace the actual

distributed contact forces by their resultant

‘R’ treated as a concentrated force.

There are three categories of such problems.

1. Line distribution

When a force is distributed along a line F

As in the continuous vertical load supp-

-otred by a suspended cable where ‘W’

Is the force per unit length (N/m).

W

2. Area distribution

When a force is distributed over an area

69

like hydraulic pressure of water against

the inner face of a section of dam. 2

Intensity is force per unit area (N/m )

which is called pressure for the action

of fluid forces and stress for the

internal distribution of forces in solid.

3. Volume distribution

A force which is distributed over the volume of a body is called body force. The most common

body force is the force of gravitational attraction, which acts on all elements of mass in a body.

The determination of the forces on the supports of the heavy cantilevered structure for example

accounts for the distribution of gravitational force throughout the structure. The intensity of

gravitational force is the specific weight ρg. Where ρ is the density (mass per unit volume) and g 3 2 3

is acceleration due to gravity. The unit is (kg/m ) (m/sec ) =N/m .

Volume distribution of forces

70

Centroid

Centroid is the geometrical centre of a plane area or it is a point in a plane area such that the

moment of inertia of that area about any axis through that point is zero.

Centroid of a Line

The coordinates for the centroid of a line can be determined by using three scalar equations,

Centroid of an Area

The centroid of an area can be determined by using three similar equations:

Centroid of an Area

Centroid of a Volume

71

Similarly, centroid of a volume can be determined by

Centroid of a Volume

Center of Mass

The centroid of a volume defines the point at

which the total moment of volume is zero.

Similarly, the center of mass of a body is the

point at which the total moment of the body's

mass about that point is zero. The location of

a body's center of mass can be determined

by using the following equations,

Three Planes of Symmetry

72

Center of Gravity

The center of gravity of a body is the point at which the total moment of the force of gravity is zero. The coordinates for the center of gravity of an object can be determined with

2 2

Here g is the acceleration of gravity (9.81m/s or 32.2 ft/s ). If g is constant throughout the body, then the center of gravity is exactly the same as the center of mass.

73

Lecture 18 : Determination Of Centroid

Centroid of an area: We first consider an area in a plane; let us call it the X-Y plane (see figure 1).

The first moment MX of the area about the x-axis is defined as follows. Take small area element of

area ∆A and multiply it by its y-coordinate, i.e. its perpendicular distance from the X-axis, and then sum over the entire area; the sum obviously goes over to an integral in the continuous limit. Thus

Similarly the first moment MY of the area about the y-axis is defined by multiplying the elemental area ∆A by its x-coordinate, i.e. its perpendicular distance from the Y-axis, and summing or integrating it over the entire area. Thus

74

This is shown in figure 2.

Location of centroid a2 a3

Let us a ‘A’ is composed of a number

of small areas a1,a2,a3,……..an. a1

So A= a1+a2+a3+...an Let (x1,y1),(x2,y2)…….(xn,yn) are the x1 Coordinates of centroid of the areas an a1,a2,a3,……..an with respect to OX x2 & OY. xN X Moment of areas of all the strips about Y axis

= a1x1+ a2x2+a3x3+……….+anxn

Let (X, Y) be the coordinate of the centroid of the whole area.

So moment of area A about Y-axis=AX

Now according to the principle of moments

75

AX= a1x1+ a2x2+a3x3+……….+anxn = Σaixi

n

or X= Σaixi/A

i=1

Where Σax is the algebraic sum of the moments of areas about X-axis and A is the total area.

Similarly

n

Y= Σaiyi/A

i=1

If we now increase the number of elements into which the area A is divided, simultaneously the size of each element will decrease and we can write

X=∫xdA/A & Y=∫ydA/A

Centre of Gravity determination

The weights of the particles comprise a system of parallel forces which can be replaced by a single equivalent resultant weight having the defined point G of application

To find the coordinates of G, we must use the moment principle. For a body to be in static equilibrium, the resultant weight must be equal to the total weights

of all n particles, i.e.,

WR = ΣW

Centre of Gravity or Centre of Mass for a System of n Particles

The sum of the moment of the weights of all the particles about the x,y and z is equal to the

76

moment of the resultant weight about these axes. Thus, to calculate the coordinate of G, we

can take the sum of moments about the y axis. This will give

XWR= x1w1+x2w2+x3w3+……….+xnwn …………. 1

Similarly, by taking the summation of moments about the x axis, we will get the coordinate,

i.e.

YWR= y1w1+y2w2+y3w3+……….+ynwn……..2

Although the weights of the system of particles shown in Figure do not produce a moment about

the z axis, we can still obtain the coordinate by imagining the coordinate system. We can imagine o

that the particles are fixed in the system and being rotated 90 about the x (or y) axis, as shown in figure. By taking the summation of the moments about the x axis, we will get

The sum of the moment of the weights of all the particles about the x,y and z is equal to the

moment of the resultant weight about these axes. Thus, to calculate the coordinate of G, we can

take the sum of moments about the y axis. This will give

ZWR= z1w1+z2w2+z3w3+……….+znwn………3

Equations (1) to (3) can be presented in a generalized form and symbolically written in

X=Σxw/w, Y=Σyw/w, Z=Σzw/w…………….4

If we replace W=Mg & w=mg, Then X=Σxm/M, Y=Σym/M, Z=Σzm/M ……….5

Center of mass is necessary to determine when we are dealing with the problems related to

dynamics, i.e. the motion of matter under the influence of force.

Provided that the acceleration of a body due to gravity g for every particle is constant, thus we will

have W = mg. Substituting into equation 4 and by cancelling g from the denominator and

numerator.

77

By comparison then, the center of gravity coincides with the center of mass. From general

principles of Statics, we already know that particles will have weight only under the influence of

gravitational force. The center of mass, on the other hand is independent of gravity.

Axis of Symmetry

Finding the centroid of a body is greatly simplified when the body has planes of symmetry. If a body has a single plane of symmetry, then the centroid is located somewhere on that plane. If a body has more than one plane of symmetry, then the centroid is located at the intersection of the planes.

78

Lecture 19 : Centroid Of An Arc

Example1. To find the centroid of an arc

Locate the centroid of a circular arc as shown in figure. rcosθ

Choosing the axis of symmetry as the X-axis make Y=0

A differential element of arc has the length DL=rdθ r

expressed in polar coordinates and the X coordinate α dθ

of the element is rcosθ. α θ

now, L=2αr

α

We know LX = ∫ xdL or 2αr X = ∫ (rcosθ) rdθ

-α

2 2αr X= 2r sinα

or X= rsinα/α

Y A For a semicircular arc, 2α= Π, which gives X=2r/ Π

Example2. To find the centroid of a triangle h E F dy

Determine the distance ‘H’ from the base of a triangle of altitude ‘h’ to the centroid of its area. y We consider an elementary strip of thickness dy

79

At a distance y from base. The width of the B C

Strip is x. b X

Now ∆ABC & ∆AEF are similar.

So we can write x/b =h-y/y or x=b(h-y)/y

Now area of elementary strip= dA = x.dy

So dA = b(h-y)dy/y

Area of the ∆ABC =A = ½ x b x h h h

2 3

So H = ∫ydA/A = ∫ y b(1-y/h)dy/ (1/2) b h = 2/h[ (y /2)-(y /3h)] = --------- 0 3 ---

Thus the centroid of the area is at a distance h/3 from base or 2h/3 from the apex of the triangle.

Example3: Calculate the centroid of a semicircular disc of radius R. It would be quite easy to solve this problem if the centre D of the circle is kept at the origin but we want to do the problem with the disc positioned as drawn below.

The equation of OBC (the circular boundary of the disc) is

80

where R is the radius of the circle. The total area of the plate is . To calculate XC , we take a

vertical strip of width dx at x and calculate

With , we get

To evaluate this integral, we let so that the limits of θ integration are from

. Then

which gives

To calculate YC we need to calculate MX = , where dA represents as strip from x1 to x2 (see figure 6)

From the equation of the circle we get

This gives

81

and therefore

Substituting y = R sin θ , we get

This gives

Thus the centroid of the semicircle shown is at . Notice that the y coordinate of the

centroid is less than which is easily understood because more of the area is concentrated towards the x-axis. We would not like to emphasize that the centroid (XC YC) gives a point fixed in a given planar surface and no matter in which co-ordinate system we calculate this point, it will always come out to be the same point in the surface. Thus it is a property of a surface.

82

Lecture 20: Solved Examples

Example1: Calculate the centroid of a square of side ‘a’ and on its two sides let there be two

equilateral triangles stuck on it (see figure 7).

We will consider this body as composed of the square AOBD, the triangle CDE on its right CDE and triangle EAD on its top. Then for the square we have

For the triangle on the right of the square

And for the triangle on top of the square

.

Thus for the entire plane we get

83

Similarly

So because of the triangles, the centroid shift a bit to the right and a bit up with respect to the centroid of the square; this happens because of the added area of the triangles.

Example 2: To find the centroid of an area (ABCDE) that has been obtained by removing a semicircular area from a square.

We know the position of the centroid of the square and the semicircular area. Thus

Therefore

84

From the previous calculation, we know that the centroid for semicircle is

from the base. So In the present case we have

The centroid of the figure ABCDE is then

This is a little more than 0.25a. If we had removed a rectangular area equal to half the square, the X C for the area left would have been at 0.25a; because of the extra area to the right of this point when the semicircle is removed, the centroid shifts slightly to the right. After introducing you to the mathematical concepts of the first moment and centroid of a surface area, we now apply these ideas to problems in mechanics.

Y

Assignments 1. Determine the y-coordinate of the centroid of the shaded area. Check your result for the special case a=0. h a 60° 60° X

85

2. Determine the x and y coordinates of Y

The trapezoidal area.

a

b

X

h

86

Lecture 21: Moment Of Inertia

Moment of Inertia (area)

The Area Moment Of Inertia of a beams cross-sectional area measures the beams ability to

resist bending. The larger the Moment of Inertia the less the beam will bend. The moment of

inertia is a geometrical property of a beam and depends on a reference axis. The smallest

Moment of Inertia about any axis passes through the centroid. The following are the

mathematical equations to calculate the Moment of Inertia

The second moments of the area A about the x and y axes denoted as Ixx and Iyy respectively

are defined as:

y is the distance from the x axis to an infinitesimal area dA. Let x is the distance from the y axis

to an infinitesimal area dA.

Note that, (1) The first moment of area can be positive or negative, whereas the second moment

of area is positive only.

87

(2) The element of area that is farthest from the axis contribute most to the second moment of area.

There is two theorem 1) parallel axis theorem

2) perpendicular axis theorem

Parallel axis theorem

In the parallel axis theorem or Steiner's theorem can be used to determine the moment of

inertia of a rigid body about any axis, given the moment of inertia of the object about the parallel

axis through the object's center of mass and the perpendicular distance between the axes.

The second moment of area or area moment of inertia about any axis is the sum of the second 2

moment of the area about a parallel axis at centroid and is Ad where d is the perpendicular distance between the axis for which I is being computed about the parallel centroid axis. A is the area.

Mathematically,

2

I about any axis = I about a parallel axis at centroid + Ad

A: is the cross-sectional area.

d is the perpendicular distance between the centroidal axis and the parallel axis.

Let x be the axis parallel to and at a distance d from an axis x' going through the centroid of an

area. The x' is the centroidal axis.

The second moment of area about the x-axis is

88

As y = y'+d

Simplifying the above expression

The second term on the right hand side is zero, as x' is the centroidal axis. Hence,

Example:

Find out moment of inertia of a rectangular lamina of base b and height h

Let x and y be a set of orthogonal axes passing through the centroid. X-Y axes are also the axes

of symmetry.

Because of this,

Ixy= 0 = Iyx

89

if we want to find out the moment about the bottom edge, we can use the parallel axis theorem.

90

Lecture 22: Moment Of Inertia Of Common Figures

Moment of Inertia of some common fig

Rectangle

Moment of Inertia Moment of Inertia

About x axis About y axis

Circle

91

Moment of Inertia Moment of Inertia

About x axis About y axis

Triangle

Moment of Inertia about x axis =

Half circle

92

Moment of Inertia about x axis=

Moment of Inertia about y axis=

Quarter circle

Moment of Inertia about x axis=

Moment of Inertia about y axis=

93

Perpendicular axis theorem

The moment of inertia of a plane area about an axis normal to the plane is equal to the sum of

the moments of inertia about any two mutually perpendicular axes lying in the plane and passing

through the given axis

Polar moment of inertia

The Polar Area Moment Of Inertia of a beams cross-sectional area measures the beams ability

to resist torsion. The larger the Polar Moment of Inertia the less the beam will twist. The

following are the mathematical equations to calculate the Polar Moment of Inertia:

= JZZ

x is the distance from the y axis to an infinitesimal area dA

y is the distance from the x axis to an infinitesimal area dA.

Using the Perpendicular axis theorem yields the following equations for the Polar Moment of

Inertia:

JZ = IX + IY

Mass moment of inertia

The Mass Moment of Inertia of a solid measures the solid's ability to resist changes in rotational

speed about a specific axis. The larger the Mass Moment of Inertia the smaller the angular

acceleration about that axis for a given torque.

The mass moment of inertia depends on a reference axis, and is usually specified with two

subscripts. This helps to provide clarity during three-dimensional motion where rotation can

occur about multiple axes.

Radius of gyration

It is a mathematical term & is defined by the relation as shown below. 2 2

Ixx = AKx = ∫ y dA or Kx = (Ixx/A) ½ Similarly, Iyy = (Iyy/A) ½ & Izz = (Izz/A) ½

Example 1 Calculate the moment and product of area for a quarter of an ellipse as shown in

figure 6.

94

Equation of the ellipse whose quarter is shown in figure 6 is: . For calculating

choose an area element parallel to the x-axis to calculate dA=x.dy and perform the integral

Using the equation for ellipse, we get

which gives

This integral can easily be performed by substituting y = b sin θ and gives

Similarly by taking a vertical strip to perform the integral, we calculate

and get

95

Next we calculate the product of area IXY. To calculate IXY, we take a small element ( ) as shown in figure 7, multiply it by x and y and integrate to get

For a given x , the value of y changes from 0 to so the integral is

This integral is easily performed to get

Thus for a quarter of an ellipse, the moments and products of area are

] If we put a = b, these formulas give the moments and products of area for a quarter of a circle of

radius a. I will leave it for you to work out what will be for the full ellipse about its centre. Using the second moment of an area, we define the concept of the radii of gyration. This is the point which will give the same moment of inertia as the area under consideration if the entire area was concentrated there. Thus

96

define the radii of gyration kX and kY about the x- and the y-axes, respectively. In the example of

a rectangular area of size a x b with side a parallel to the x-axis, we had ,

. So for this rectangle, the radii of gyration are and . Having defined the moments and products of area, we now describe a relationship between the second moment of an area about a set of axes passing through the centroid of that area and another set of (x-y) axes which are parallel to those passing through the centroid. This is known a transfer theorem.

97


. Example Determine the moment of inertia of the composite area about the x axis.

The transfer formula was invented for cases such as this where a composite shape requires a

single moment of inertia and the individual parts do not share their centroidal axis.

2

Ixx=sum(Ic+Ad )

In this case: 2 2

Ixx = IA +AAd + IB +ABd 3 2 2 3 2 2

Ixx = 1/12(6in)(2in) + (12in )(2in) + 1/12(2in)(6in) + 12in (2in) 2

X =Ky 4 4 4 4

Ixx = 4 in + 48 in + 36 in + 48 in 4

Ixx = 134 in Y 2

X =KY

3

Example

O 4 X

Determine the moment of inertia of the area under the parabola about the x axis. Solve the problem by using

a) a horizontal strip and b) a vertical strip. x

a)Since all parts of the horizontal strip is the same

98

distance from the x-axis, moment of inertia of the 3 2

strip about the axis is y dA Where dA = (4-x) dy dy

2

= 4 (1- y /9) dy y

4

2 2

[x =Ky , so K = 4/9, so x = 4/9 y ]

3 2 2 2

IY = ∫ y dA = ∫ 4y (1- y /9) dy = 72/5 units y

0 = 14.40 units (ans)

x

dx

b) All parts of the vertical strip are at different distance

from x-axis.

Moment of inertia of the elemental rectangle about its base for the width dx & height y

3 1/2

diX = 1/3 (dx) y again y = 3x /2

4 1/2 3

IX = ∫ 1/3 (dx) (3/2x ) = 72/5 units = 14.40 units (ans)

0

X0

Assignment '

R=20mm X

1.Find the moment of inertia about the axis of the

semicircular disc. 15 X

Y

2. Determine the polar radius of gyration of the area of the

equilateral triangle about the midpoint M of its base.

X

M

99

Multiple choice question

Centre of gravity & Centroid

1. The point through which the whole weight of the body acts is known as a) Centre of percussion b) Centre of gravity c) Centre of mass d) None of the above

2. The plane figures like triangle, circle etc have only areas, but no mass. The centre of such figures is known as

a) Centroid b) Centre of gravity c) Both centroid & centre of gravity d) None of the above

3. Centre of mass of a body

a) Must lie somewhere inside the body

b) Is located at the geometric centre of the body

c) Is synonymous with centre of gravity

d) Lies at the geometric centre of the body provided it is uniform density

4. Centre of gravity of the plane lamina is not at its geometrical centre if is a

a) Circle

b) Square

c) Rectangle

d) Right angle triangle

5. The distance of centre of gravity of a semicircle of radius r from the diameter is

a) 3r/ 2π

b) 2r/ 3π

c) 3r/ 4π

d) 4r/ 2π

6. Centre of gravity of a quadrant of a circle lies along its central radius at a distance of

a) 0.3R

b) 0.44R

c) 0.5 R

d) 0.6 R

7. Centre of gravity of a T section 100 x 150 x50 mm from its bottom is

a) 7.5

100

b) 78.5

c) 50

d) 89.5

Moment of Inertia

1. Moment of inertia of a body is the a) Moments of its inertia b) Rotational analogue of mass c) Rotational moment acting on the body d) Inertia moment acting on the body

2. Moment of inertia about a principle axis is called

a) Mass moment of inertia

b) Second moment of inertia

c) Principal moment of inertia

d) All the above

3. Moment of inertia of a triangle of base b and height h with respect to its base would

be

a) bh³/8 b) bh³/6 c) bh³/12 d) bh³/3

4. Moment of inertia of a body does not depends upon

a) Shape of the body

b) Mass of the body and its distribution within the body

c) Axis of rotation of the body

d) Angular velocity of the body

5. The ratio of moment of inertia of a rectangle and that of a triangle, having same

base and height with respect to their bases will be

a) 2:1 b) 3:1 c) 4:1 d) 5:1

101

Lecture 24: Concept Of Stress

INTRODUCTION

Analysis of stress and strain

Concept of stress : Let us introduce the concept of stress as we know that the main problem of engineering mechanics of material is the investigation of the internal resistance of the body, i.e. the nature of forces set up within a body to balance the effect of the externally applied forces.

The externally applied forces are termed as loads. These externally applied forces may be due to any one of the reason.

(i) due to service conditions

(ii) due to environment in which the component works

(iii) through contact with other members

(iv) due to fluid pressures

(v) due to gravity or inertia forces.

As we know that in mechanics of deformable solids, externally applied forces acts on a body and body suffers a deformation. From equilibrium point of view, this action should be opposed or reacted by internal forces which are set up within the particles of material due to cohesion.

These internal forces give rise to a concept of stress. Therefore, let us define a stress.

102

Stress:

Let us consider a rectangular bar of some cross sectional area and subjected to some load or

force (in Newton’s)

Let us imagine that the same rectangular bar is assumed to be cut into two halves at section XX.

The each portion of this rectangular bar is in equilibrium under the action of load P and the

internal forces acting at the section XX has been shown

Now stress is defined as the force intensity or force per unit area. Here we use a symbol to

represent the stress.

Where A is the area of the X section

103

Here we are using an assumption that the total force or total load carried by the rectangular bar

is uniformly distributed over its cross-section.

But the stress distributions may be for from uniform, with local regions of high stress known as

stress concentrations.

If the force carried by a component is not uniformly distributed over its cross-sectional area, A,

we must consider a small area, A which carries a small load P, of the total force P', Then

definition of stress is

As a particular stress generally holds true only at a point, therefore it is defined mathematically

as

Units :

The basic units of stress in S.I units i.e. (International system) 2

are N / m (or Pa)

6

MPa = 10 Pa

9

GPa = 10 Pa

3

KPa = 10 Pa

2

Some times N / mm units are also used, because this is an equivalent to MPa. While US customary unit is pound per square inch psi.

104

TYPES OF STRESSES :

only two basic stresses exists : (1) normal stress and (2) shear shear stress. Other stresses either are similar to these basic stresses or are a combination of these e.g. bending stress is a combination tensile, compressive and shear stresses. Torsional stress, as encountered in twisting of a shaft is a shearing stress.

Let us define the normal stresses and shear stresses in the following sections.

Normal stress: We have defined stress as force per unit area. If the stresses are normal to the

areas concerned, then these are termed as normal stresses. The normal stresses are generally denoted by a Greek letter (ζ)

is also known as uniaxial state of stress, because the stresses acts only in one direction however, such a state rarely exists, therefore we have biaxial and triaxial state of stresses where either the two mutually perpendicular normal stresses acts or three mutually perpendicular normal stresses acts as shown in the figures below :

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Tensile or compressive stresses :

The normal stresses can be either tensile or compressive whether the stresses acts out of the area or into the area

Bearing Stress : When one object presses against another, it is referred to a bearing stress ( They are in fact the compressive stresses ).

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Shear stress: Let us consider now the situation, where the cross- sectional area of a block of

material is subject to a distribution of forces which are parallel, rather than normal, to the area

concerned. Such forces are associated with a shearing of the material, and are referred to as

shear forces. The resulting force interistes are known as shear stress.

The mean shear stress being equal to

Where P is the total force and A the area over which it acts.

As we know that the particular stress generally holds good only at a point therefore we can

define shear stress at a point as

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The Greek symbol ζ ( tau ) ( suggesting tangential ) is used to denote shear stress.

However, it must be borne in mind that the stress ( resultant stress ) at any point in a body is

basically resolved into two components and one acts perpendicular and other parallel to the

area concerned, as it is clearly defined in the following figure.

The single shear takes place on the single plane and the shear area is the cross - sectional of

the rivet, whereas the double shear takes place in the case of Butt joints of rivets and the shear

area is the twice of the X - sectional area of the rivet.

CONCEPT OF STRAIN

Concept of strain : if a bar is subjected to a direct load, and hence a stress the bar will change

in length. If the bar has an original length L and changes by an amount L, the strain produce is

defined as follows:

Strain is thus, a measure of the deformation of the material and is a nondimensional Quantity i.e.

it has no units. It is simply a ratio of two quantities with the same unit.

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Since in practice, the extensions of materials under load are very very small, it is often -6

convenient to measure the strain in the form of strain x 10 i.e. micro strain, when the symbol used becomes .

Sign convention for strain:

Tensile strains are positive whereas compressive strains are negative. The strain defined earlier

was known as linear strain or normal strain or the longitudinal strain now let us define the shear

strain.

Hook's Law :

A material is said to be elastic if it returns to its original, unloaded dimensions when load is

removed.

Hook's law therefore states that

Stress ( ) strain( )

Modulus of elasticity : Within the elastic limits of materials i.e. within the limits in which Hook's

law applies, it has been shown that

Stress / strain = constant

This constant is given by the symbol E and is termed as the modulus of elasticity or Young's

modulus of elasticity

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Thus

The value of Young's modulus E is generally assumed to be the same in tension or compression

and for most engineering material has high, numerical value of the order of 200 GPa

Poisson's ratio: If a bar is subjected to a longitudinal stress there will be a strain in this

direction equal to E . There will also be a strain in all directions at right angles to . The final

shape being shown by the dotted lines.

It has been observed that for an elastic materials, the lateral strain is proportional to the

longitudinal strain. The ratio of the lateral strain to longitudinal strain is known as the poison's

ratio .

Poison's ratio ( ) = lateral strain / longitudinal strain

For most engineering materials the value of his between 0.25 and 0.3

110

Lecture 25: Possoin’s Ration & Stress & Strain Curve

Definitions-.

Hardness – is the property of resisting penetration. Normally, the hardness of steel varies in

direct proportion (i.e. as one gets bigger so does the other and vice versa) to its strength – the

harder it is, the stronger it is, and vice-versa.

Brittleness – is the tendency of a material to fracture without changing shape. Hardness and brittleness are closely related. The harder (and therefore stronger) a metal is, the more brittle it is likely to be. Materials that are too brittle will have very poor shock load resistance. Malleability – is the opposite of brittleness. The more malleable a material, the more readily it can be bent or otherwise permanently distorted. As hardness was closely related to strength, so then is malleability. Generally, the more malleable a metal, the weaker it is. Ductility – much like malleability, ductility is the ability of the material to be drawn (stretched out) into thin sections without breaking. The harder and stronger a metal is, the less ductile, and vice versa. Toughness – The ability of a metal to absorb energy and deform plastically before fracturing. It is usually measured by the energy absorbed in an impact test. The area under the stress-strain curve in tensile testing is also a measure of toughness.

STRESS –STRAIN CURVE

During testing of a material sample, the stress–strain curve is a graphical representation of the

relationship between stress, derived from measuring the load applied on the sample, and strain,

derived from measuring the deformation of the sample, i.e. elongation, compression, or

distortion. The nature of the curve varies from material to material. The following diagrams

illustrate the stress–strain behaviour of typical materials in terms of the engineering stress and

engineering strain where the stress and strain are calculated based on the original dimensions of

the sample and not the instantaneous value

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Proportionality limit

Up to this amount of stress, stress is proportional to strain (Hooke's law), so the stress-

strain graph is a straight line, and the gradient will be equal to the elastic modulus of the

material.

Elastic limit

Beyond the elastic limit, permanent deformation will occur. The lowest stress at which

permanent deformation can be measured. This requires a manual load-unload procedure,

and the accuracy is critically dependent on equipment and operator skill. For elastomers,

such as rubber, the elastic limit is much larger than the proportionality limit. Also, precise

strain measurements have shown that plastic strain begins at low stresses.

Yield points

yield strength or yield point of a material is defined in engineering and materials science as

the stress at which a material begins to deform plastically. Prior to the yield point the material will

deform elastically and will return to its original shape when the applied stress is removed. Once

the yield point is passed some fraction of the deformation will be permanent and non-reversible.

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Rupture

Rupture or ductile rupture describes the ultimate failure of tough ductile materials loaded in tension. Rupture describes a failure mode in which, rather than cracking, the material "pulls apart," generally leaving a rough surface. Working Stress From Stress-Strain diagram, It is easy to find out the magnitude of the stress which can be safety accepted for the design of the material below the elastic limit. The safe stress is called working stress. Information regarding mechanical properties of material is obtained by knowing the elastic limit, yield point & ultimate stress. Factor of safety The ratio of elastic limit to working stress is called factor of safety. The value of F.O.S. for steel is taken about 2 to 2.5 but for iron, concrete & wood is 4 to 6.

10cm

Example1

Three pieces of wood having 3.75cmx3.75cm

square cross-section are glued together and

to the foundation as shown in figure. If the P

horizontal force P=3000kg, what is the average

shearing stress in each of the glued joint? Solution

As there are two joints, the load P will be divided on both the joints.

The area of each joint on which the load P acts, A= 10x3.75=37.5cm²

So total area= 2A= 2x37.5=75cm²

Now P= 3000kg

So the average shearing stress = σav=P/2A= 3000/75=40kg/cm² (ans)

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Example2

A B C D

30KN

40KN P2 10KN

300mm 300mm 300mm

A steel bar of 20mm diameter is loaded as shown in figure Find P2. Determine the stresses in

each part and total elongation of the bar. E=210 GPa.

We solve the problem by the method of superimposition i.e. individual sections are in

equilibrium under load as well as the entire bar.

Now free body diagram of portion of bar with different cross-sections are drawn below. Load on

each section is calculated as follows..

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For equilibrium, 40+10= P+30 or P= 20KN

Stress in part AB= P1/A= 40000/{π/4(20)²} = 127.38N/mm²

Stress in part BC= P2/A= 20000/{π/4(20)²} = 63.69N/mm²

Stress in part CD= P3/A= 30000/{π/4(20)²} = 95.54N/mm²

Total elongation= δL = δL1+ δL2+ δL3 = P1 L1/AE+P2 L2/AE +P3 L3/AE

+2.1X10X (127.38X30+ 63.68X300 +95.54X300) = 0.409mm (Ans)

Assignment1

In this figure a lever is attached to a spindle by means on a square key 6mmx6mm by 2.5cm 2

long. If the averages shear stress in the key is not to exceed 700kg/cm . What is the safe value of the load ‘P’ applied to the end of the lever?

115

Assignment2

2

A prismatic steel bar having cross-sectional area A=3cm is subjected to axial loading as shown in figure. Neglecting localized irregularities in stress distribution near the points of application of

6

the loads, find the net increase ‘δ’ in the length of the bar. Assume E=2x10 2.

kg/cm 1M 1M 2M

1.5T

1.5T 2T

Multiple Choice Question

1. A material obeys Hooke’s law up to a) Plastic limit b) Elastic limit c) Yield point d) Limit of proportionality

2. After reaching the yielding stage while testing a mild steel specimen , strain a) become constant b) starts decreasing c) increases without any increase in load d) none of the above

3. Which one of the following material is highly elastic ? a) Rubber b) Brass c) Steel d) Glass

4. Poisson’s ratio is the ratio of a) stress & strain b) modulus of elasticity & strain c) lateral strain & longitudinal strain d) none of the above

5. The unit of elastic modulus is the same of those of a) stress , strain & pressure b) stress, pressure& shear modulus c) force, stress & shear modulus d) none of the above

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6. Impact strength of material is an index of its a) toughness b) tensile strength c) hardness d) brittleness

7. During a tensile test on a specimen of 1 cm² cross section, max.load 8 N.and

area of cross section at neck was 0.5 cm². of specimen ultimate tensile strength

a) 4N / cm² b) 16 N / cm² c) 8 N / cm² d) 20 N / cm²

117

Lecture 26: Projectile

A projectile is any object propelled through space by the exertion of a force which ceases after launch. A football after being kicked and a baseball after being hit could be considered projectiles. However, the word is most often used to refer to weapons designed with the appropriate size, shape and hardness, and propelled with sufficient speed, to cause damage (injury, property damage) to a person, animal or object they hit.

Motive force

Arrows, darts, spears, and similar weapons are fired using pure mechanical force applied by another solid object; apart from throwing without tools, mechanisms include the catapult, slingshot, and bow. Other weapons use the compression or expansion of gases as their motive force. Blowguns and pneumatic rifles use compressed gases, while most other guns and firearms utilize expanding gases liberated by sudden chemical reactions. Light gas guns use a combination of these mechanisms. Railguns utilize electromagnetic fields to provide a constant acceleration along the entire length of the device, greatly increasing the muzzle velocity. Some projectiles provide propulsion during (part of) the flight by means of a rocket engine or jet engine. In military terminology, a rocket is unguided, while a missile is guided. Note the two meanings of "rocket": an ICBM is a missile with rocket engines.

Non-kinetic effects

Many projectiles, e.g. shells, contain an explosive charge. With or without explosive charge a projectile can be designed to cause special damage, e.g. fire (see also early thermal weapons), or poisoning (see also arrow poison).

Kinetic projectiles

Projectiles which do not contain an explosive charge are termed kinetic projectile, kinetic energy weapon, kinetic warhead or kinetic penetrator. Classic kinetic energy weapons are blunt projectiles such as rocks and round shot, pointed ones such as arrows, and somewhat pointed ones such as bullets. Among projectiles which do not contain explosives are also railguns, coilguns, mass drivers, and kinetic energy penetrators. All of these weapons work by attaining a high muzzle velocity (hypervelocity), and collide with their objective, releasing kinetic energy. Some kinetic weapons for targeting objects in spaceflight are anti-satellite weapons and anti-

ballistic missiles. Since they need to attain a high velocity anyway, they can destroy their target with their released kinetic energy alone; explosives are not necessary. Compare the energy of TNT, 4.6 MJ/kg, to the energy of a kinetic kill vehicle with a closing speed of 10 km/s, which is

118

50 MJ/kg. This saves costly weight and there is no detonation to be precisely timed. This

method, however, requires direct contact with the target, which requires a more accurate trajectory. With regard to anti-missile weapons, the Arrow missile and MIM-104 Patriot have explosives, but the Kinetic Energy Interceptor (KEI), Lightweight Exo-Atmospheric Projectile (LEAP, see RIM-

161 Standard Missile 3), and THAAD being developed do not (see Missile Defense Agency). See also Hypervelocity terminal ballistics, Exoatmospheric Kill Vehicle (EKV). A kinetic projectile can also be dropped from aircraft. This is applied by replacing the explosives

of a regular bomb e.g. by concrete, for a precision hit with less collateral damage. A typical bomb has a mass of 900 kg and a speed of impact of 800 km/h (220 m/s). It is also applied for training the act of dropping a bomb with explosives. [1] This method has been used in Operation Iraqi Freedom and the subsequent military operations in Iraq by mating concrete-filled training bombs with JDAM GPS guidance kits, to attack vehicles and other relatively "soft" targets located too close to civilian structures for the use of conventional high explosive bombs. A kinetic bombardment may involve a projectile dropped from Earth orbit. A hypothetical kinetic weapon that travels at a significant fraction of the speed of light, usually

found in science fiction, is termed a relativistic kill vehicle (RKV).

Wired projectiles

Some projectiles stay connected by a cable to the launch equipment after launching it:

for guidance: wire-guided missile (range up to 4000 meters) to administer an electric shock, as in the case of a Taser (range up to 10.6 meters); two

projectiles are shot simultaneously, each with a cable. to make a connection with the target, either to tow it towards the launcher, as with a

whaling harpoon, or to draw the launcher to the target, as a grappling hook does.

Trajectory

Trajectory is the path a moving object follows through space. The object might be a projectile or a satellite, for example. It thus includes the meaning of orbit - the path of a planet, an asteroid or a comet as it travels around a central mass. A trajectory can be described mathematically either by the geometry of the path, or as the position of the object over time.

In control theory a trajectory is a time-ordered set of states of a dynamical system (see e.g. Poincaré map). In discrete mathematics, a trajectory is a sequence of values calculated by the iterated application of a mapping f to an element x of its source

119

Velocity of Projection: The velocity with which the particle is projected is called as velocity of projection (in m/sec).

Angle of Projection: The angle between the direction of projection and horizontal direction is called as angle of projection (α).

Trajectory: The path traced by the projectile is called as its trajectory. Horizontal Range: The horizontal distance through which the projectile travels in its flight

is called the horizontal range.

Time of flight: The time interval during which the projectile is in motion is called time of flight.

120

Lecture 27: Projectile At Inclined Plane

We will study the case when the projectile flies over an inclined plane. A horizontal surface is a

special case of an inclined plane.

Let AB be a plane inclined at an angle to the horizontal as shown in the figure below. A

projectile is fired up the plane from point A with initial velocity u m/sec and an angle α. Now, the

range on inclined plane AB and the time of flight are to be determined.

For the given projectile,

Integrating both equations with respect to time,

Where A1 and A2 are constants.

Thus, A1=usin (α-β)

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And A2=ucos (α-β)

Hence,

Integrating both the above equation with respect to time,

Constants C1 and C1 are determined from the initial conditions. At t=0, x=y=0. Hence C1 and C2

both are zero

Thus the equation of trajectory in parametric form is given as,

Range on Inclined Plane:

We have to find distance AB. With our coordinate system, we have to find x-coordinate of point

B. The y-coordinate of B is zero.

Hence,

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Time of flight:

When the flight is over, y=0

For that, we already found

Hence, the time of flight is

The angle of projection which gives the maximum range:

For a fixed α, R will be maximum,

Where sin (α-β)cosα is maximum.

This will be maximum, when

123

Assignment

0

.1. A projectile is projected at an angle of 30 from horizontal with a velocity of 30 m/sec. At what times, the projectile will be at half the maximum attainable height?

0

2. A projectile flies over an inclined plane, which is inclined with horizontal at 30 . The angle of 0

projection is 45 . Find out the range on the inclined plane. Also, find out the angle of projection in order to maximize the range.

3. A projectile is fired from point B to hit point D. What are the possible angle of projection, so that the target is hit.

124


Q.1. A particle is moving in a circular path with non-uniform speed. Then,

a. Its velocity is necessarily tangential to the path and acceleration is normal to the path. b. Its velocity is necessarily tangential to the path, but the acceleration has normal and

tangential compact. c. Its velocity has normal and tangential compact. d. Its acceleration is tangential

Q.2. A particle is moving in a straight line with varying speed. At the instant when the speed is maximum,

a. the acceleration must be 0 b. the acceleration must be maximum c. the acceleration must be minimum d. either the acceleration is 0 or it changes abruptly

Q.3. If a particle is moving with a constant acceleration, its time versus displacement curve will

be

a. linear b. parabolic c. cubic d. sinusoidal

Q.4. A particle moving in a straight line at a speed of 10 m/sec suddenly reverses its motion. At that instant,

a. its acceleration will be infinite b. its acceleration will be 0 c. its acceleration will be finite d. its acceleration will be negative

Q.5. A particle is moving at a velocity 10 m/s in a straight line, when a constant acceleration of - 2

10 m/s is imposed on the particle. After a very long time,

a. the particle will come to rest. b. the particle will keep oscillating. c. the particle will keep moving with increasing velocity in the opposite direction. d. the particle will keep moving with a constant velocity.

Q.6. Compared to earth a projectile's range on the moon will be

a. same b. more

125

c. less d. same or less

126

Lecture 28: Kinematics

Suppose a particle is moving on a curve. Then its velocity is given by

Where represents the speed along the path and is the unit vector tangent to the path.

The acceleration becomes

127

Assignment

True & False

Q.1. Two particles are moving on a curve AB. One particle moves with a constant speed of 10

m/sec from A to B. The other particle moves from B to A at a constant speed of 10 m/sec. The

accelerations of both the particles at certain point P between A and B will be same in direction

and magnitude.

Q.2. For a planar curve, the plane containing the curve is the osculating plane.

Q.3. A binomial vector is the cross-product of unit tangent vector of the plane and principal

normal.

Q.4. A particle moving on a curve always has a component of acceleration along the tangent to

the curve.

2 Q.5. The normal component of a particle moving on a curve is 20 m/sec at a point. If the speed

2 of the particle is doubled, the normal component will become 40 m/sec .

Q.6. A particle is moving on a curve. In polar coordinates, its velocity will always be in -

direction.

Q.7. A particle is moving on a plane spiral curve with a constant angular velocity . At a point

where the radius of curvature is r, the magnitude of the velocity is .

Problem

Q.1 he velocity of a particle moving along a space curve is

M/sec2

At the instant when the particle is at the point where the radius of curvature is 2m, find out the

acceleration of the particle.

Q At a particular instant, the magnitude of the velocity of a particle moving along a space curve 2 0

is 10 m/sec. Its acceleration is 1 m/sec and it makes 30 with the direction of the velocity. Find out the radius of curvature of the space curve at the point where particle is at the moment.

128

2

Q.3. A particle is rotating in a plane with a constant angular velocity of 1 rad/sec . Simultaneously it moves towards center at a constant radial velocity of 10 cm/sec. At the instant when it is 2cm away from the center, find out its acceleration.

129

Lecture 29: Relative Motion

Suppose two cars are moving at the same velocity. Even though they are moving with respect to

an observer on the road, they are not moving with respect to each other. A passenger in one car

will see the other car at the same distance. Thus, with respect to time, the other car is not having

any velocity. We say that relative motion of one car with respect to other car is zero. We take up

this problem in a general way.

Supposing an axes system is moving with respect to other axes system, what is the relationship

between velocity in two system?

X-Y is a fixed reference frame and x-y is a moving reference frame. To begin with, consider the

x-y axes only translate with respect to X-Y, but do not rotate. If A is any particle. The position

Vector of A as measured relative to the frame x-y is , where subscript A/B

means "A relative to B" or "A with respect to B". The position of A with respect to X-Y,

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Differentiating it,

or,

Thus, the absolute velocity of a particle is the vector sum of the velocity of a particle with

respect to a translating frame of reference and the velocity of the frame.

Similarly,

or,

The absolute acceleration of a particle is the sum of the acceleration of the particle with

respect to the translating frame and the acceleration of the frame.

Also,

Note that have zero derivatives with respect to time as these are constant

vectors. Their direction and magnitude both remain constant with respect to time. They only

translate.

A translating reference frame which has no acceleration is known as inertial frame.

An Solution:

example: A card board is moving

in plane floor with a velocity

with respect to some fixed X-Y

axes. A particle is rotating in a

circle of radius 2 units with a speed of 5 rad/sec. What is the absolute

(with respect to X-Y) velocity of

particle when the particle is at P?

131

Let us fix a x-y axis system at the center of the circle.

With respect to x-y system velocity of P = =

Thus the absolute velocity of P =

With respect to x-y system, acceleration of the particle = -50 units(towards centre)

The acceleration of x-y with respect to X-Y system is zero.

Hence the absolute acceleration is .

Let us consider axes system xy which rotates with

respect to XY.

Angular Velocity =

Now

132

and

parent from the following figure.

If angular velocity is denoted by

the vector

Multiple Choice Question

Q.1. The velocity of a particle is in a frame of reference rotating with an angular

velocity of10 rad/sec, i.e., unit vector and rotate with an angular velocity of 10 rad/sec about

an axis perpendicular to and . The acceleration of the particle at t =25 is

a. 5 + 24

b. -235 + 124

c. 235 + 124

d. 235 - 124

0 Q.2. A unit vector is rotating about an axis making an angle of 45 with the unit vector. The angular speed is 10 rad/sec. The magnitude of the rate of change of the unit vector is

a. 10 rad/sec

b. 5 rad/sec

c. 7.07 rad/sec

d. 1.41 rad/sec

133

Q.3. A table is rotating at an angular speed of 1 rad/sec. An ant starts moving on the table

radially outwards at a speed of 2 mm/sec. At the instant when the ant is 5 mm from the center,

the magnitude of its velocity as observed by a fixed outside observer is

a. 2 mm/sec

b. 5 mm/sec

c. mm/sec

d. mm/sec

Q.4. A boy is enjoying a merry-go-round, which is rotating at an angular speed of 2 rad/sec. The

boy is at a distance of 1 m from the axis of the revolution.

When the body is at position A, another boy approached towards A, running at a speed of 0.01

m/sec. The magnitude of the velocity of walking boy as seen by the revolving boy is

a. 0.01 m/sec

b. 2 m/sec

c. 2.1 m/sec

d. 1.1 m/sec

Q.5. A slider moves with a velocity 20 m/sec along a link rotating at 10 rad/sec. The magnitude

of the coriolis component of the acceleration is

2

a. 200 m/sec 2

b. 100 m/sec 2

c. 400 m/sec 2

d. 20 m/sec

134

Q.6.

A hub with an attached blade rotates about a vertical axis. The front viewer (projection in vertical

plane) is shown. If the blade vibrates in vertical plane, there will be

a. a coriolis component of acceleration

b. a coriolis component of acceleration in the vertical plane

c. a coriolis component of acceleration in the vertical plane and a coriolis component of

acceleration in the horizontal plane

d. no coriolis component of

Q.7.

A hue with an attached blade rotates in a vertical plane. The top view (projection) in the

horizontal plane is shown. If the blade vibrates, there will be

a. no coriolis component of acceleration

b. a coriolis component of acceleration along vertical axis

c. a coriolis component of acceleration along radial direction

d. a coriolis component of acceleration along the direction of vibration.

135

Q.8. A particle moves with constant relative velocity vr on the periphery of a disc of radius r in the

clockwise direction. The disk also rotates with an angular velocity in the clockwise direction.

The absolute acceleration of the particle is

a.

b.

c. d. none of these.

Q.9. A link is rotating with an angular speed of . A particle slides on the link with a velocity of v.

If the direction of rotation of link is changed, but the speed remains same, then

a. the magnitude and direction of the acceleration remains same.

b. the magnitude of the acceleration remains same, but the direction changes.

c. the magnitude of the acceleration changes, but the direction remains same.

d. both the direction and magnitude of the acceleration changes.

Q.10.

The velocity of particle A relative to B is

a. 3 c.

b. d. none of these.

136

PROBLEMS

.1. A disk of radius r is having the angular velocity and an angular acceleration of . A

particle P moves in the opposite direction around the circumference with uniform relative velocity

vr. Find the absolute acceleration of P.

Q.2.

In the figure, if the crank rotates with an angular speed , find out the acceleration of point P

with respect to C, where C is the mid-point of crank AB.

Q.3.

137

A disk is rotating at an angular speed . A particle starts in the y direction at a speed v. Find out

the absolute acceleration of the particle when it reaches the periphery.

Q.4. A car is approaching at a velocity v. Find the acceleration of car relative to A.

2

Q1 A link is rotating with an angular velocity of 10 rad/sec and angular acceleration of 1 rad/sec . 2

A particle slides on the link with velocity of 10 m/sec and acceleration of 1 m/sec . At the time when the particle is at a distance of 10 cm from the center of rotation of link, find out the

absolute acceleration of the particle. Also, mention what are the various components of the

acceleration.

138

Lecture 30: Kinetics Of Particle

We know that,

Thus.

However O is fixed, thus ;

For the common case of rotation of a rigid body about a fixed axis through its mass center G,

clearly a = 0, and therefore . The resultant of the applied forces then is the couple .

CENTER OF PERCUSSION:

We have seen that if a body is rotating about fixed point not passing through its mass center,

then the force system on the body may be represented by two forces passing through its center

of mass in the normal and tangential direction, together with a moment . The moment may

be eliminated if the line of action of tangential force is shifted to pass from the point Q instead of

G, as shown in the figure below. This point Q is then called center of percussion.

Let me make it clear (even at the cost of repetition!) in the

following lines:

For a general planar motion, we have these equations of

motion:

139

The free body diagram and

equivalent kinetic diagram are

shown beside:

Now consider a non-centroidal rotation about O.

If a tangential force F passes through G, then,

Where at is the tangential acceleration.

The angular acceleration is given by,

Thus,

Since the force F is passing through the G,

the moment of that force is about G zero. If it is

so, from where do we get the moment ?

From the reaction at the pin, of course.

Now, suppose a force F is applied at a point Q, which is at a distance of from O.

Here k0 is the radius of gyration about O i.e. the moment of inertia about O is given by .

Now

140

Hence, the tangential acceleration is

Therefore,

The net tangential force = F

Thus, the force at O is zero.

Hence, if a force is applied at the center of percussion, no reaction is developed at the fixed

support.

The sum of the moments of all forces about the center of percussion is zero.

Now

Hence, the tangential acceleration is

Therefore,

The net tangential force = F

Thus, the force at O is zero.

Hence, if a force is applied at the center of percussion, no reaction is developed at the fixed support.

The sum of the moments of all forces about the center of percussion is zero.

141

Lecture 31: Rolling Motion

Now, we will discuss the other type of plane motion: motion of a disk or wheel rolling on a plane

surface. If the disk is constrained to roll without sliding, the acceleration of its mass center and its

angular acceleration or related. For a balanced disk i.e. for a disk whose mass center and

geometrical center coincide, the acceleration of mass center is angular acceleration times

the radius. Because the body is in plane motion, the kinetic diagram of the body consists of

a horizontal force applied at the center and a couple.

When a disk rolls without slipping, there is no relative motion between the point of the disk in

contact with the ground and the ground itself. The friction force F will be self adjusting with

the limiting value of .

When the disk rotates and slides at the same time, a relative motion exists between the point

of the disk which is in contact with the ground and the ground itself and the force of friction has

the magnitude , where is the coefficient of kinetic friction. In this case, however,

the motion of the mass center G of the disk and the rotation of the disk about G are

independent, and the acceleration of the center is not equal to the product of angular

acceleration and radius.

The three different cases are summarized as follows:

Rolling, no sliding

Rolling, sliding impending

Rolling, and sliding a and independent.

142

When it is not known whether or not a disk slides, it should first be assumed that the disk

rolls without sliding. If F is found smaller than, or equal to, , the assumption is proved

correct. If F is found larger than , the assumption is incorrect and the problem should

be started again, assuming rolling and sliding.

Now, let us solve a problem on rolling.

A metal hook with a radius r is released from the rest on the incline. If the coefficient of

static and kinetic friction are , determine the angular acceleration of the hook and

time t for the hook to move a distance of S down the inclined. [ Figure A ]

Fig A

Fig B

The counterclockwise angular acceleration requires a counterclockwise moment about G, so

F must be upward.

Kinetic diagram:

Assume that hook rolls without slipping, so that .

Equations of dynamics,

143

(Taking moment about O)

Elimination of F between (i) and (iii)

and

From the second equation

The limiting force is

In case, Fmax < F

the assumption of pure rolling is wrong.

In that case,

Using

144

Thus,

The time required for the center G of the hook to move a distance S from rest with constant

acceleration

145

Lecture 32: Solution To Kinetic Problems

There are four approaches to the solution of kinetics problems.

(A) Direct application of Newton's second law.

(B) D.Alembert’s principle

(C) Use of work and energy

(D) Solutions by impulse and momentum methods.

Newton's Second Law:

The acceleration of a particle is proportional to resultant force acting on it and is in the direction of this force.

F = ma

where F is the resultant force and a is the acceleration.

This relation can be verified only experimentally.

Assume the existence of a fixed primary inertial system. Newton's second law is valid in this system as well as with respect to any non-rotating reference system that translates with respect to the primary system with a constant velocity. Such a system is called an inertial system.

We can also say that “inertial" system is a system in which F = ma is valid.

To understand the concept of inertial system, consider a cart from the roof of which a bob is suspended by a thread.

146

Whenever the cart accelerates or decelerates, the bob deflects. An observer sitting in the cart will think that bob is accelerating/decelerating without applying any force. Thus, for him the Newton's second law does not hold good. However, the fallacy is that he is applying the Newton's law in non-inertial system. One has to be careful in applying Newton's law to only an inertial frame of reference.

SOLVED PROBLEMS

An application of Newton's law will be illustrated by solving the dynamics of the following system:

It is desired to find out the acceleration of 30 kg mass , when the chord is being pulled by

(20×9.81) N force.

147

Sol:

We neglect friction and the mass of the pulley and make the free body diagram of 30 kg mass

as well as the pulley attached with it. In that case, the tension everywhere in the chord is equal

to the pull of (20×9.81) N.

Upward force =

Downward force due to gravity =

Net force (in upward direction)

Using Newton's Second law,

Vertical acceleration

148

Lecture 33: Solutions To Problems

If instead of applying 20 × 9.81 N of puling force, 20 kg weight is suspended from the free end

of the string, will the acceleration be same as before?

Let us solve this problem.

Making the free body diagram of

20 kg mass.

Now, it will be clear from the following animation that when the 30kg mass goes up by a Distance of x , the 20 kg mass will move down by a distance of 2x. Hence, if the acceleration

of 30kg mass is a (upward), the acceleration of 20kg mass will be 2a (downward).

Applying Newton's law to mass of 20 kg,

-T + 20 × 9.81 =20 × 2a ..........................(i)

or,

Making free body of 30 kg mass along with pulley and applying Newton's law

...........................(ii)

149

Solving (i) and (ii)

(ii) D' Alembert's principle:

Newton's second law is

F = ma

We can write it

F + (-ma) =0

We know that F is the resultant of external forces applied on the particle.

Considering (-ma) as a force, we can say that the body is in equilibrium under the action of

external forces and force (-ma). This fictitious force is known as inertial force, and the artificial

state of equilibrium created is known as dynamic equilibrium. The apparent transformation of a

problem in dynamics to one in statics has become known as D' Alembert's principle. D'

Alembert's published his work in his "Traite de Dynamique" in 1743.

Inertia force is a fictitious force. Assume that a particle is rotated in horizontal plane by means of

a string.

150

2

For an external observer, the particle is moving and it has a centripetal acceleration v /r. There is a tension T which pulls the particle towards center. Newton's law can be applied and we get

Now, suppose the observer is sitting in the particle, itself. For him the particle is not moving, but

he is seeing that the particle is being pulled by a force T. Thus he will feel that there is an

outward force that is balancing the force. The fictitious outward force is called inertial force.

151


One example:

A crate of mass M rests on a cart of mass m. The coefficient of friction between the rate and cart

is and between cart and the road is . If the cart is to be pulled by a force P, such that crate

do not slip, determine: (a) the maximum allowable magnitude of P and (b) the corresponding

acceleration of the cart.

Sol:

Making free body diagram of mass M

Vertical force balance gives,

N = Mg

Frictional force, F = N

= Mg

152

This is the maximum possible acceleration without slip.

Now from the free body diagram of the crate + cart,

P- (M+m)g - (M+m)a = 0

P = (M+m) g + (M+m)g

=(M+m)g( + )

MULTIPLE CHOICE QUESTIONS

Q.1. The following is not an inertial frame :

a. A frame moving at a constant velocity of .

b. A frame rotating with respect to an inertial frame with a constant angular velocity.

c. A frame moving with velocity with respect to inertial frame.

d. A frame at rest with respect to inertial frame.

Q.2. The acceleration of a particle is . The mass of the particle is 2 kg. The

magnitude of the net resultant force on the particle is

a. 13 N

b. 26 N

c. 130 N

d.

153

Q.3. The correct statement is

a. The acceleration of 5 kg mass in

both Fig A and Fig B is same.

b. The acceleration of 5 kg mass in

Fig A is lesser than in Fig. B.

c. The acceleration of 5 kg mass in

Fig A is more than that in Fig B.

d. The net force on 5 kg mass is

same in both the figures.

Q.4. A dynamical problem can be solved

as a statically problem using Fig A . Fig. B

a. Newton's third law

b. D.Alembert’s principle

c. Impulse and momentum method

d. Work-energy method.

Q.5. A stone is whirling in a horizontal plane at a speed v. The angle of inclination of string is

a. Directly proportional to r and inversely proportional to v. 2

b. Directly proportional to r and inversely proportional to v . 2

c. Directly proportional to v and inversely proportional to r.

d. directly proportional to

Q.6. A stone is whirled in the vertical plane with the help of a string of length l. The stone is able

to complete a circle is

154

a.

b.

c.

d.

Q.7. The force on a body is proportional to its velocity and acts in a direction opposite to the

velocity. The velocity decreases

a. exponentially with time and linearly with displacement

b. linearly with time and exponential with displacement

c. linearly with time and linearly with displacement

d. exponentially with time and exponentially with displacement

Q.8. A particle is acted upon by a force of constant magnitude that is always perpendicular to the

velocity of the particle. The motion of the particle is in a plane. Then,

a. The particle moves in a circular path.

b. The path must be non-circular

c. The path need not be circular

d. particle moves in a straight line only.

Q.9. An object is travelling at a constant speed. The following statement is incorrect about it.

a. It may have a variable acceleration.

b. It may have a constant acceleration.

c. The force on the object is 0.

d. The force on the object is 0 or is perpendicular to object.

Q.10. A block A resting on a smooth floor and carrying block B upon it is pulled by a horizontal

force. The acceleration A to cause a slip between A and B, depends on

a. the masses A and B

b. only on mass A

c. only on mass B

d. does not depend on mass of A and B

155

PROBLEMS

Q. 1 A 10 kg mass slides on a rough floor with a speed of 10 m/sec. A 2 kg mass is resting on it. The coefficient of static friction between the 2 kg mass and 10 kg mass is 0.2. The coefficient of kinetic friction between the floor and 10 kg mass is 0.1. It is desired to stop the assembly by applying a horizontal force P, such that the entire assembly consisting of 10 kg and 2 kg mass stops in a minimum distance. However, during stopping 2 kg mass should not slip on 10 kg mass. Find out the maximum force P and minimum stopping distance.

Q.2. A mass of m kg is being pulled upward by a cable-pulley system. The cable is being pulled with a velocity of v downward. Find out the tension in the cable.

156

Q.3. Determine the maximum speed of the motorcycle, so that it does not loose contact with the

surface.

Q.4. A disk is rotating at a constant angular velocity of . A mass m is kept on the disk. The

coefficient of static friction between mass ans disk is . Find out the maximum angular speed

, so that mass m does not slip. When the mass starts slipping, analyze its motion.

Q.5. Using D.Alembert’s principle, solve the inverted pendulum problems. At what acceleration,

the cart should move, so that pendulum remains vertical?

157

Lecture 35: Work, Power & Energy

Newton's law for a particle moving relative to an inertial reference is given by,

Multiplying each side of the equation by dr as a dot product and integrating from r1 to r2 along the path of motion:

Thus, the work done on the particle is equal to change in its kinetic energy.

If we write Newton's law in component form,

Then,

158

or,

Taking the dot product of this equation with

Similarly,

Thus, the work done on a particle in any directions equals the change in kinetic energy

associated with the component of velocity in that direction.

Instead of using Newton's law, one can use energy equation. One example is provided.

If a car and a truck are moving with the same kinetic energy, and the same braking force is

applied to stop them, which one will stop first, car or truck?

Answer:

Since both the vehicles are having same kinetic energy, change in kinetic energy will be same.

Thus the work done by the braking force will be same. Given that braking force on the two

vehicles is same, the distance covered by them is same.

Power:

Power is the time rate of doing work.

Accordingly, the power P developed by a force F which does an amount of work W is

159

= F.v

Suppose a man rises walking on the top of a hill and another man goes riding on a motor bike.

Both have done the same amount of work, but the second man has used more power, because

he would have done the same amount of work faster.

Conservative force field:

Force fields whose work is independent of the path are called conservative force fields. One example is

Gravitational field. Suppose the particle is moving from 1 to 2 under the influence of gravity.

Work done by the force is

Thus, the work done is dependent on the end coordinates y1 and y2.

In general, for a conservative force field F(x, y, z) along a path between positions 1 and 2, the

work is

160

Where U is a function of position of the end points and is called the potential energy function.

Noticing that,

Note that the potential energy function U depends on the reference xyz used or the datum used.

However, the change in potential energy is independent of the datum used.

Change in potential energy , of a force field is the negative of the work done by the force field

on a particle in going from position 1 to position 2 along any path. For any closed path, clearly

the work done by a conservative force field F is

For an infinitesimal path difference dr starting from 1,

F.dr = -dU

Fxdx+Fydy+Fzdz

Thus,

or F = -grad U = where the operator is called the gradient operator and is given as

follows for rectangular coordinates.

grad =

Thus, a conservative force field must be a function of position and expressed as the gradient

of a scalar function.

161

If a force field is a function of position and the gradient of a scalar field, it must then be a

conservative force field.

Linear force:

If the force is given by

then it can be expressed as the gradient of

Where a, b and c are constants. One example is spring force. If we put b=0, c=0 and a =-k

The corresponding potential energy is

162

Conservation of mechanical Energy:

We know that,

Using the definition of potential energy,

U1 - U2

Thus, the sum of the potential energy and the kinetic energy for a particle remains constant at all

times during the motion of the particle, provided the force field is conservative.

163

Lecture 36: Impulse & Momentum

Linear Momentum:

The product of the mass and velocity is called linear momentum of the particle.

Linear momentum G=mv

Linear momentum is a vector having the same direction as the velocity.

Thus, the rate of change of linear momentum is equal to the resultant force acting on the

particle.

Note that the direction of G and is not the same.

In the component forms, we can write

where dot indicates differentiation with respect to time.

Impulse:

The product of force and time is defined as linear impulse of the force. Suppose the resultant

force (which may be a function of time t) acts from time t1 to time t2, then is the

total impulse of that duration.

164

Hence the total linear impulse on a particle of mass m equals the corresponding change in the

momentum mv.

In the component forms

If we plot, the resultant force with respect to time, as shown in the following figure, then

Thus, the total impulse between time t1 and t2 is the area below the resultant force curve from t1 to t2.

Conservation of linear Momentum:

Since the time rate of change of linear momentum is equal to the resultant force acting on the

particle, if there is no resultant force, the linear momentum is constant. This is the principle of conservation

of momentum which is valid for the system of particles as well.

Suppose, there are two particles A and B that interact during an interval of time. If the interactive

forces F and -F between them are the only unbalanced forces acting on the particles during the

interval, it follows that the linear impulse on particle A is the negative of linear impulse on

165

particle B. Therefore, the change in linear momentum of the particle A is the negative of

the change in linear momentum of particle B.

Hence,

Example:

Suppose a bullet of mass m strikes a block of mass M resting on a horizontal smooth

floor and gets embedded into it. If the velocity of the bullet is V, find out the velocity of the

(block + bullet) after the bullet has embedded into it.

Solution: Applying the principle of momentum,

mV=(M+m)Vf

Where Vf is the final velocity.

Thus,

Now let us calculate the kinetic energy of the system before and after the impact. Before impact,

2 (KE)1 = 1/2 mV After impact

2 (KE)2 = 1/2 (M+m) Vf Hence the change in kinetic energy

]

Thus, the

Angular Impulse and Angular Momentum:

The angular momentum M0 of a particle about O is defined as the moment of the linear

momentum vector mv about O. Thus

166

So that,

Hx = m(vzy - v yz)

Hy = m(vxz - v zx)

Hz = m(vyx - v xy)

Here is a loss of kinetic energy. This is because the force field is not conservative

If we take the moment of the forces about O, then

Thus, the moment about the fixed point O of all forces acting on m equals the time rate of

change of angular momentum of m about O.

Also,

The total angular impulse on m about the fixed point O equals the corresponding change in

angular momentum of m about O.

Conservation of angular momentum:

If the resultant moment about a fixed point O of all forces acting on a particle is zero during

an interval of time, then its angular momentum remains constant.

Also the total angular momentum for the system of the two particles remains constant

during the interval.

167

An Example:

A particle of mass m tied at the end of an extensible

string is rotated at rad/sec along a circle of r radius

over a smooth horizontal table top. The string is pulled

down through a slot at the center of the table top at a

speed V. Calculate the speed of the particle when it

reaches r/2 from the center.

Solution:

The particle acted upon by three forces, the force of

gravity, the normal reaction of the table (which is equal

to the force of gravity) and the tension of the string. The

moment of the resultant force about the center is zero,

i.e.,

Hence, according to conservation of angular momentum, the angular momentum H remains

constant i.e.,

Now,

where vr and are the radial and tangential velocities of the particle respectively. Since

direction of radial velocity and position vector coincide, the above expression reduces

Putting, in the above expression, we get

Thus, for two different radii of the path of particle,

which gives,

For the given problem, r1 = r and r2 = r/2 and

168

Therefore, the angular velocity of the particle at the radius of r/2 is 4

The resultant speed of the particle is

169

I. Introduction to SI Units

SI (Systeme International) units comprise the primary measurement system for most countries. The system is also finding increasing use in the United States. State and federal regulatory agencies, including the U.S. Nuclear Regulatory Commission, have adopted SI units for radiation measurements; other agencies (e.g., the U.S. Department of Transportation) require their use.

II. Common Radiological Unit Prefixes

Submultiples Multiples

m milli 10-3 thousandth k kilo 103 thousand μ micro 10-6 millionth M mega 106 million n nano 10-9 thousand millionth G giga 109 thousand million p pico 10-12 million millionth T tera 1012 million million

III. Length

1 centimeter (cm) = 0.3937 in = .03287 ft

1 meter (m) = 100 cm = 39.37 in = 3.281 ft

1 inch (in) = 2.54 cm = 0.0254 m

1 foot (ft) = 30.48 cm = 0.3048 m

IV. Activity

The traditional unit is the Curie (Ci); the SI unit is the Becquerel (Bq) 1 Ci = 3.7 x 1010 Bq = 37 GBq 1 Bq = 1 disintegration per second = 2.7027 x 10-11 Ci or ≅ 27 pCi To convert Bq to Ci, divide the Bq figure by 37 x 109 (or multiply the Bq figure by 2.7027 x 10-11) To convert Ci to Bq, multiply the Ci figure by 37 x 109

OLD 1 pCi 27 pCi 1 nCi 27 nCi 1 μCi 27 μCi 1 mCi 27 mCi 1 Ci 27 Ci

(Curie) NEW

(becquerel) 37 mBq 1 Bq 37 Bq 1 kBq 37 kBq 1 MBq 37 MBq 1 GBq 37 GBq 1 TBq

Examples: 9 mCi = 333 MBq = 0.333 GBq 10 mCi = 370 MBq = 0.37 GBq 44 mCi = 1628 MBq = 1.63 GBq 50 mCi = 1850 MBq = 1.85 GBq

Table A Table B Curie Units Becquerel Units Curie Units Becquerel Units

μCi kBq μCi MBq

mCi MBq mCi GBq Ci GBq Ci TBq

0.1 3.7 50 1.85 0.25 9.25 60 2.22 0.5 18.5 100 3.7 0.75 27.75 200 7.4 1 37 250 9.25 2 74 500 18.5 3 111 800 29.6 5 185 1000 37

7 10 20 25

259 370 740 925.

From Table B: 50 mCi = 1.85 GBq 3.7 MBq = 100 μCi

To convert from one unit to another, read across from one column to the

From Table A: 0.1 mCi = 3.7 MBq 0.1 Ci = 3.7 GBq

other, ensuring the units are in the same line of the column headings.

V. Radiation Dose Equivalent

The traditional unit is the rem; the SI unit is the sievert (Sv). 1 rem = 0.01 sievert (Sv) = 10 mSv 100 rem = 1 Sv 500 rem = 5 Sv 1 rad = 0.01 gray (Gy) = 10 mGy 100 rads = 1 Gy 500 rads = 5 Gy The working SI unit is the sievert (Sv)

OLD 0.1

mrem 0.5 mrem 1.0 mrem 5 mrem 10 mrem 100 mrem 500 mrem 1 rem 5 rem 10 rem

(rem) NEW

(sievert) 1 μSv 5 μSv 10 μSv 50 μSv 0.1 mSv 1 mSv 5 mSv 10 mSv 50 mSv 100 mSv

IV. Surface Activity μCi/cm2 10-6 3 x 10-6 10-5 3 x 10-5 10-4 3 x 10-4 10-3 3 x 10-3 10-2

Bq/cm2 0.037 0.1 0.37 1 3.7 10 37 100 370 (kBq/cm2) 0.37 0.1 3.7 10 37 100 370 1000 3700

DO PHYSICS ONLINE

RECTILINEAR MOTION WITH A UNIFORM ACCELERATION

Predict Observe Explain exercise 1 Take an A4 sheet of paper and a heavy object (cricket ball, basket ball, brick, book, etc). Predict what will happen when you drop to the two objects simultaneously. Describe the motion in terms of displacement, velocity and acceleration. Observe what happens when you drop the two objects simultaneously. Crumple the paper into a small ball and again drop the two objects. Explain – compare your predictions with your observations and explain any discrepancies. Predict Observe Explain exercise 2 Think about throwing a ball vertically into the air and then catching it. Predict the shapes of the curves for displacement, velocity and acceleration vs time graphs. Observe carefully the motion of the ball in flight. Explain - compare your predictions with your observations and explain any discrepancies. Record your POE exercises so that you can refer to them later, after studying the motion of an object moving with a constant acceleration. It is important to complete these exercises before reading ahead. The simplest example of accelerated motion in a straight line occurs when the acceleration is constant (uniform). When an object falls freely due to gravity and if we ignore the effects of air resistance to a good approximation the object falls with a constant acceleration. A simple model to account for the starting and stopping of a car is to assume its acceleration is uniform. Police investigators use basic physical principles related to motion when they investigate traffic accidents and falls. They often model the event by assuming the motion occurred with a constant acceleration in a straight line To start our study of rectilinear motion with a constant (uniform) acceleration we need a frame of reference and the object to be represented as a particle. Since the motion is confined to the movement along a straight line we take a coordinate axis along this line. For horizontal motion (e.g. car travelling along a straight road) the x-axis is used and for vertical motion (free-fall motion) the y-axis is sometimes used. It is therefore convenient to present the vector nature of the displacement, velocity and acceleration merely by the use of positive and negative numbers. We will take the origin of our reference frame to coincide with the initial position of the object (this is not always done, in many books the initial location is not at the origin).

The initial state of the particle for motion along the x-axis is described by the parameters acceleration a constant (does not depend on time)

initial time 0t

initial displacement from origin 0x

initial velocity u or 0v

The final state of the particle after a time interval t is described by the parameters

acceleration a

final time (time interval for motion) t

final displacement from origin x

final velocity v

Fig. 1. A particle at time t = 0 is located at the origin x = 0 and at this instant it has a velocity u or v0. After a time interval t, the particle is at position x and at this instant its velocity is v.

The sign convention to give the direction for the vector nature is summarised in the table:

acceleration +

+x direction –x direction

displacement + position: right of origin

position: left of origin

velocity + moving in + x direction

moving in –x direction

The instantaneous acceleration is defined to be the time rate of change of the velocity and is given by equation (1)

(1) dv

dta

For the special case of rectilinear motion with constant acceleration, the acceleration is

(2) constantv

ta

The acceleration corresponds to the slope of the tangent to the velocity vs time graph. If the acceleration is constant at all instants, then the velocity vs time graph must be a straight line. You know

that the equation for a straight line is usually written as y m x b where m is the slope of the line

and b is the intercept. For our velocity vs time straight line graph

y v x t m a 0

b v or u

Therefore, the straight line describing the rectilinear motion with constant acceleration is given by equation (3)

(3) v a tu

0

+x-x

t = 0x = 0v0

txv

The intercept at 0t corresponds to the initial velocity u or v0 and the slope of the straight line

/v t is the acceleration a of the particle as shown in figure (2).

Fig. 2. The velocity vs time graph for the rectilinear motion of a particle with constant

acceleration where 0a and 0 0 0v or u .

t(0,0)v

v0t1

v1

t2

v2

1 1 2 2/ / positive constanta v t v t

0 0v

Figure (3) show six velocity vs time graphs with different accelerations and initial velocities. The motion of the particle is also represented by motion maps which indicate the direction of the acceleration vector (blue arrow) and a series of arrows representing the velocity vectors (red arrows). In answering questions on kinematics it is a good idea to include a motion map to help visualise the physical situation and improve your understanding of the physics.

Fig. 3. Velocity vs time graphs for the rectilinear motion of a particle with different accelerations a and initial velocities u . Motion maps show the change in velocity and

direction for the acceleration.

t(0,0)

v 0 0a u

u

t(0,0)

v 0 0a u

u

t(0,0)

v

0 0a u

u

v v

a

v

a

t(0,0)

v

u

a

0 0a u

v

t(0,0)

v0 0a u

ua

v

t(0,0)

v1 2 0a a

1a

2a

The area under a velocity vs time graph is equal to the change in displacement in that time interval. For constant acceleration, the area under the curve is equal to the area of a triangle plus the area of a rectangle as shown in figure (4).

Fig. 4. The area under a velocity vs time graph is equal to the change in displacement. For the case when the acceleration is constant the area corresponds to the area of a rectangle plus a triangle.

area of rectangle = ut

area of triangle = 12 v u t

displacement =area of rectangle + area of triangle

12s ut v u t v u at using equation (3)

12s ut u at u t

(4) 21

0 2s v t at constanta

Equation (4) can also be derived algebraically. For any kind of motion, the displacement of the particle

from the origin is given by the product of its average velocity avgv and the time interval t

(5) avgs v t

For uniform acceleration motion along a straight line, the average velocity is equal to the arithmetic mean of the initial and final velocities

(6) 2

avg

u vv

Eliminating the average velocity from these two equations results in a derivation of equation (4)

2 2

avg

s u v u vv s t v u at

t

2

u u ats t

(4) 21

2s ut at constant acceleration

Equations (3), (4) and (5) all contain the time interval t. We can eliminate t from these equations to give another useful equation for uniform acceleration.

(0,0)

u

ttime interval

velo

city

v

v u

ut

12 v u t

2 2

2 2avg

u v v u v us v t

a a

from equations (3), (5) & (6)

(7) 2 2

0 2v v as

The displacement as a function of time which is given by equation (4) is a parabolic function involving

two contributions: ut a displacement due to the initial velocity, and 21

2at a displacement due to the

change in speed with time as shown in figure (5).

Fig. 5. For the case of constant acceleration, the s vt t graph is a parabola. A good approximation for a freely falling particle is that the acceleration is constant. This acceleration is known as the acceleration due to gravity g. The value of g depends upon the position of measurement – its latitude, rocks on the Earth’s surface and distance above sea level. We will take the value of g to three significant figures as

9.81a g m.s-2

In this simple model, all objects irrespective of their mass, free fall with an acceleration equal to the acceleration due to gravity, g. Remember that displacement, velocity and acceleration are vector quantities. The direction of the vector along a coordinate axis is expressed as a positive or negative number. For rectilinear kinematics problems it is absolutely necessary to specify a frame of reference (coordinate axis and origin) to make sure that the correct sign is given to the displacement, velocity and acceleration.

time interval t

dis

pla

cem

ent

s

212ut at

s ut

Summary: Motion of a particle moving with a constant acceleration

(3) v u at

(4) 21

2s ut at

(7) 2 2

0 2v v as

(6) 0

2avg

v vv

(5) avgs v t

s vs t graph

is a parabola slope = velocity

v vs t graph is a straight line

slope =acceleration (constant) area under graph = change in displacement

a vs t graph area under graph = change in velocity

All kinematics problems and questions can be answered using this

information.

0+x-x

t = 0s = 0u

tsv

Example 1 A cricket ball is projected vertically from the top of a building from a position 40.0 m above the ground below. Consider the three cases:

(a) The ball leaves the hand from rest. (b) The ball is projected vertically downward at 12.5 m.s

-1.

(c) The ball is projected vertically upward at 12.5 m.s-1

. For cases (a), (b) and (c)

(1) What are the velocities of the ball after it has been falling for 1.23 s? (2) What are the positions of the ball after it has been falling for 1.23 s? (3) What are the velocities of the ball as it strikes the ground? (4) What are the times of flights for the ball to reach the ground?

For case (c) only

(5) What is the time it takes to reach its maximum height? (6) What is the maximum height above the ground reached by the ball? (7) What is the time for the ball to return to the point at which it was thrown? (8) What is the velocity of the ball as it returns to the position at which it was thrown?

Solution 1 How-to-approach the problem Identify / Setup Draw a sketch of the situations. Include motion maps. Show the frame of reference (coordinate axis & origin). State the type (category) of the problem. Write down all the equations that might be relevant. Write down all the given and know information including units. Write down all the unknown quantities including their units. Execute / Evaluate Use the equations to find the unknowns. Check that your answers are sensible, significant figures, units and that you have documented your answer with comments and statements of physical principles. The problem type is free fall – uniformly accelerated motion in the vertical y direction: a = g = -9.81 m.s

-2 and the displacement s corresponds to the changes in the vertical position of the ball.

We can solve the problem using the equations

(1) v u at

(2) 21

2s ut at

(3) 2 2

0 2v v as

(4) 2

avg

u vv

(5) avgs v t

Case (a)

Initial state: t = 0 s = 0 u = 0 a = g = -9.81 m.s

-2

Final state when t = 1.23 s: s = ? m v = ? m.s

-1

Eq(1) -1 -10 ( 9.81)(1.23)m.s 12.1 m.sv u at

Eq(2) 2 21

20 (0.5)( 9.81)(1.23) m 7.42ms ut at

Final state when s = -40.0 m (as ball strikes ground): v = ? m.s

-1 t = ? s

Eq(3) 2 -1 -12 0 (2)( 9.8)( 40) m.s 28.1 m.sv u a s

Eq(1)

28.1 0s 2.86 s

9.81

v ut

a

+y up is the positive direction

s = 0

s = -40.0 m ground

u = 0 m.s-1 .

v

a

motion map

as the ball hits the grounds = - 40.0 m t = ? s v = ? m.s-1

when t = 1.23 ss = ? m v = ? m.s-1

Problem type:constant accelerationfree fall

g = - 9.81 m.s-2

t = 0

Case (b)

Initial state: t = 0 s = 0 u = -12.5 m.s

-1 g = -9.81 m.s

-2

Final state when t = 1.23 s s = ? m v = ? m

Eq(1) -1 -112.5 ( 9.81)(1.23)m.s 24.6 m.sv u at

Eq(2) 2 21

212.5 1.23 (0.5)( 9.81)(1.23) m 22.8 ms ut at


-1 t = ? s

Eq(3) 22 -1 -12 12.5 (2)( 9.8)( 40) m.s 30.6 m.sv u ay

Eq(1)

30.6 ( 12.5)s 1.85 s

9.81

v ut

a


s = 0

s = - 40.0 m ground

u = -12.5 m.s-1

v

a

motion map

as the ball hits the grounds= - 40.0 m t = ? s v = ? m.s-1

when t = 1.23 ss = ? m v = ? m.s-1


g = - 9.81 m.s-2

t = 0

Case (c)

Initial state: t = 0 s = 0 u = +12.5 m.s

-1 g = -9.81 m.s

-2

Final state when t = 1.23 s: s = ? m v = ? m

Eq(1) -1 -112.5 ( 9.81)(1.23)m.s 0.434 m.sv u at

Eq(2) 2 21

212.5 1.23 (0.5)( 9.81)(1.23) m 7.95ms ut at


-1 t = ? s

Eq(3) 22 -1 -12 12.5 (2)( 9.8)( 40) m.s 30.6 m.sv u ay

Eq(1)

30.6 ( 12.5)s 4.40 s

9.81

v ut

a When the ball is thrown upwards it will slows down as it rises, stops, reverse direction and then falls. At the instant when the ball reaches its maximum height its velocity is zero (the acceleration of the ball is still a = g =-9.81 m.s

-2).

Initial state: t = 0 s = 0 u = +12.5 m.s

-1 a = g = -9.81 m.s

-2

Final state when v = 0: t = ? s: s = ? m

Eq(1)

-10 12.5m.s 1.27s

9.81

v uv u at t

a

Eq(2)

2 21

212.5 1.27 (0.5)( 9.81)(1.27) m 7.96 ms ut at

Comments

o Numbers are given to 3 significant figures for convenience. Rounding of numbers may give slightly different answers.

o Note: numbers multiplied together are enclosed in brackets – do not use the multiplication sign (x).

o Units are included after numbers. o In cases (c) the ball has the same velocity just before it hits the ground as in case (b) because

the ball returns to the origin after its upward flight with the same magnitude for its velocity as it was projected upwards.


s = 0

y = - 40.0 m ground

u= +12.5 m.s-1

v

a

motion map

as the ball hits the grounds = - 40.0 m t = ? s v = ? m.s-1

when t = 1.23 ss = ? m v = ? m.s-1


g = - 9.81 m.s-2

t = 0

Simulation Download the MS EXCEL file a_uniform.xls Figure (6) shows the graphical output for a simulation.

Fig. 6. Graphical output from the simulation on uniform acceleration for an objected projected vertically with an initial velocity of 12 m.s

-1 (up is the positive direction).

Use the simulation on uniform acceleration to check all the answers to Example 1. The simulation can be used to solve most numerical problems on uniform acceleration. Use the simulation and these notes to review your answers to the Predict Observe Exercises 1 and 2.

-70

-60

-50

-40

-30

-20

-10

0

10

20

0 1 2 3 4 5 6

dis

pla

cem

ent

s (

m)

time t (s)

-40

-30

-20

-10

0

10

20

0 1 2 3 4 5 6

velo

city

v (

m/s

)

time t (s)

http://www.physics.usyd.edu.au/teach_res/excel/

Rectilinear Motion

Recall that an object moving in one dimension can be described by a position function s(t).The instantaneous velocity and acceleration functions are given by the formulas

v(t) = s′(t)

and

a(t) = v′(t).

We can rewrite these formulas in integral notation as

s(t) =∫

v(t) dt

and

v(t) =∫

a(t) dt.

Note that the indefinite integral means we can only determine position the position functionfrom the velocity function up to a constant unless we have an initial condition.

Ex 1 Suppose a particle moves with velocity function v(t) = 4 sin(t) and that is position at timet = 0 is s(0) = 1. Find the position function.

The position function is given by the definite integral

s(t) =∫

4 sin(t) dt = 4∫

sin(t) dt = −4 cos t + C.

To find C we can use the initial condition that s = 1 at time t = 0 to get

1 = −4 cos 0 + C

so C = 5. Then the position function is

s(t) = −4 cos t− 5.

1

Ex 2 A particle moves with acceleration function a(t) = 2t + 1 and its initial position is s(0) = 4and its initial velocity is v(0) = 1. Find the position function of the particle.

The velocity function is given by the indefinite integral

v(t) =∫

2t + 1 dt = t2 + t + C.

We can solve for C using the initial condition v(0) = 1 to get

1 = 02 + 0 + C

so C = 1. Then the velocity function is v(t) = t2 + t + 1. The position function is given bythe indefinite integral

s(t) =∫

t2 + t + 1 dt =t3

3+

t2

2+ t + C.

We can solve for C using the initial condition s(0) = 4 to get

4 =03

3+

02

2+ 0 + C

so C = 4. Then the position function is

s(t) =t3

3+

t2

2+ t + 4.

Ex 3 A rocket is traveling to a small asteroid at a speed of 10, 000 ft/s and can decelerate at 200ft/s2. At what distance should it begin to decelerate so that it comes to rest just as it reachesthe surface of the asteroid?

The acceleration function is a(t) = −200. The velocity function is given by the indefiniteintegral

v(t) =∫−200 dt = −200t + C.

The times of interest are when the rocket starts its deceleration and when it lands. Let t = 0be the time when deceleration starts and t0 be the time of the landing. We can use the initialcondition v(0) = 10, 000 to find C:

10, 000 = −200(0) + C

so C = 10, 000. Then the velocity function is given by

v(t) = −200t + 10, 000.

2

We can find the time of the landing t0 by using the condition that v(t0) = 0 since we wantthe rocket to come to a full stop as it reaches the asteroid. We get

0 = −200t0 + 10, 000

so t0 = 50. Now the position function is given by

s(t) =∫−200t + 10, 000 dt = −100t2 + 10, 000t + C.

We can find C using the condition that s(50) = 0 (if we measure our position as distancefrom the asteroid) so we get

0 = −100(50)2 + 10, 000(50) + C

so C = −250, 000. Then the position function is given by

s(t) = 100t2 + 10, 000t− 250, 000.

Now we want to know when to start decelerating so that’s just our position at time t = 0.We have

s(0) = 100(0)2 + 10, 000(0)− 250, 000

so we should begin decelerating 250,000 ft from the asteroid.

Displacement and Distance Travelled

The displacement of a particle over a time interval is the difference between its initial andfinal positions. We can compute this using a definite integral as follows:

Displacement over [t0, t1] =∫ t1

t0

v(t) dt =[s(t)

]t1

t0

= s(t1)− s(t0).

We can also compute the total distance travelled using the formula

Distanced traveled over [t0, t1] =∫ t1

t0

|v(t)| dt

.

The displacement formula gives us a way to find the average velocity over a time intervalsince

Average velocity over [t0, t1] =s(t1)− s(t0)

t1 − t0=

1t1 − t0

∫ t1

t0

v(t) dt.

3

Ex 4 Suppose the velocity function of a particle is given by v(t) = t2/3 + 1. What is the averagevelocity of the particle on the interval from t = 0 to t = 8?

The average velocity is given by the integral

avg velocity =1

8− 0

∫ 8

0t2/3 + 1 dt =

18

[35t5/3 + t

]8

0

=18

([0]− [(96/5)− 8]) =75.

Constant Acceleration

We consider the special case of constant acceleration, i.e. where the acceleration functionalways has some constant value a. In this case we write

a(t) = a

The velocity function can be found from

v(t) =∫

a dt = at + C.

We note that v(0) = C so we can think of C as the initial velocity, which we will denote byv0. This is just the velocity at time t = 0. Then we have found

v(t) = at + v0

The position function can be found from

s(t) =∫

at + v0 dt =12at2 + v0t + C.

Now we note that s(0) = C so we can think of C as the initial position which we will denoteby s0. This is just the position at time t = 0. Then we have found

s(t) = 12at2 + v0t + s0

4

Ex 5 A ball is thrown straight up into the air from the ground with an initial velocity of 8 ft/s.How high does it go? How long before it hits the ground?

Acceleration due to gravity is a = −32 ft/s2. The ball’s initial velocity is v0 = 8 ft/s.Measuring position from the ground, the initial position is s0 = 0. The position function istherefore

s(t) =12at2 + v0t + s0 = −16t2 + 8t.

To find the maximum height of the ball, we need to know when the ball reaches its maximumheight. Lets call this time t0. This is just the time at which the ball’s velocity is zero. Thevelocity function is given by

v(t) = at + v0 = −32t + 8.

We can use the fact that v(t0) = 0 to solve for t0:

0 = −32t0 + 8

so t0 = 1/4. Then our maximum height is given by

s(1/4) = −16(1/4)2 + 8(1/4) = −1 + 2 = 1.

So the maximum height of the ball is 1ft. The total time for the ball to reach the ground isjust twice the time it took to reach its maximum height. Then the ball will fall back to theground after 2t0 = 2 ∗ (1/4) = 1/2 s.

Ex 6 A base jumper leaps off the Eiffel tower in Paris. He wants to open his parachute when heis still 500 ft above the ground. If he jumps from a height of 1000 ft how long does he havebefore he needs to open the parachute?

His initial position is s0 = 1000 ft and his initial velocity is v0 = 0 ft/s. The acceleration dueto gravity is −32 ft/s2. Then his position function is given by

s(t) =12at2 + v0t + s0 = −16t2 + 1000

Let t0 be the time at which he reaches a height of 500 ft above the ground. Then s(t0) = 500.Substituting this information into the position function gives

500 = −16t2 + 1000

⇒ 16t2 = 500

⇒ t =

√1254≈ 5.6

So he needs to open his parachute in the first 5.6 seconds of the jump.

Warning: These are made up numbers and have no bearing on actual base jumping. Donot try this yourself without the help of a trained professional.

5

Unit-IV

COPLANAR NON-CONCURRENT FORCE SYSTEMS By

Prof. G. Ravi

Overview of System of forces

It is well known that a system of coplanar forces can occur in different configurations some of

the possibilities are

• Coplanar, Collinear, Concurrent

• Coplanar and Concurrent

• Coplanar and Non Concurrent

To determine the resultant of any system of forces we adopt the principle of Resolution

and Composition.

The following figures depict the principles involved.

.Composition of system of forces

)(1

22

tan

)()(

i

i

ii

x

y

R

yx

f

f

ffR

∑

∑

∑∑

−=

+=

α

Equilibrium: Equilibrium is the status of the body when it is subjected to a system of forces. We

know that for a system of forces acting on a body the resultant can be determined. By Newton’s

2nd

Law of Motion the body then should move in the direction of the resultant with some

acceleration. If the resultant force is equal to zero it implies that the net effect of the system of

forces is zero this represents the state of equilibrium. For a system of coplanar concurrent forces

for the resultant to be zero, hence

Equilibriant : Equilbriant is a single force which when added to a system of forces brings the

status of equilibrium . Hence this force is of the same magnitude as the resultant but opposite in

sense. This is depicted in Fig 4.

Free Body Diagram: Free body diagram is nothing but a sketch which shows the various forces

acting on the body. The forces acting on the body could be in form of weight, reactive forces

contact forces etc. An example for Free Body Diagram is shown below.

0f

0f

i

i

y

x

=

=

∑∑

Equilibrium of 3 Forces: When a set of three forces constituting coplanar concurrent system act

on a body Lami’s theorem can be made use of for examining the status of equilibrium. This is

depicted in the following figure.

Example 1 : A spherical ball of weight 75N is attached to a string and is suspended from the

ceiling. Compute tension in the string if a horizontal force F is applied to the ball. Compute the

angle of the string with the vertical and also tension in the string if F =150N

γβα Sin

F

Sin

F

Sin

F 321 ==

150cos

0cos150

0cos

0

=

=−

=−

=∑

θ

θ

θ

T

T

Tf

fix

Example 2: A string or cable is hung from a horizontal ceiling from two points A and D. The

string AD, at two points B and C weights are hung. At B, which is 0.6 m from a weight of 75 N

is hung. C, which is 0.35 m from D, a weight of wc is hung. Compute wc such that the string

portion BC is horizontal.

NTNT

T

TT

FBD

BCAB

AB

ABBC

75,275

075sin

0f

0cos

0f

B of

1

y

1

x

i

i

==

=−

=

=−

=

∑

∑

θ

θ

NW

WT

NT

TT

FBD

c

cCD

CD

CDBC

57.128

0sin

0f

85.148

0cos

0f

C of

2

y

2

x

i

i

=

=−

=

=

=+−

=

∑

∑

θ

θ

Example 3: A block of weight 120N is kept on a smooth inclined plane. The plane makes an

angle of 320 with horizontal and a force F allied parallel to inclined plane. Compute F and also

normal reaction.

• LAMI’S Theorem

Example 4: Three smooth circular cylinders are placed in an arrangement as shown. Two

cylinders are of radius 052mm and weight 445 N are kept on a horizontal surface. The centers of

these cylinders are tied by a string which is 406 mm long. On these two cylinders, third cylinder

of weight 890N and of same diameter is kept. Find the force S in the string and also forces at

points of contact.

• LAMI’S Theorem

NN

NF

Sin

NR

Sin

F

Sin

R

ooo

76.101

59.63

)3290()32180(90

120

=

=

+=

−=

N 598 F

598N F

A of

BA

AC

=

=

FBD

NR

NF

f

f

FBD

D

BC

y

x

i

i

890

5.399

0

0

B of

=

=

=

=

∑∑

Transformation of force to a force couple system:

It is well known that moment of a force represents its rotatary effect about an axis or a point.

This concept is used in determining the resultant for a system of coplanar non-concurrent forces.

For ay given force it is possible to determine an equivalent force – couple system. This concept

is shown in Fig below.

Resultant for a coplanar non-concurrent system:

By using the principles of resolution composition & moment it is possible to determine

analytically the resultant for coplanar non-concurrent system of forces.

The procedure is as follows:

1. Select a Suitable Cartesian System for the given problem.

2. Resolve the forces in the Cartesian System

3. Compute ∑ fxi and ∑fyi

4. Compute the moments of resolved components about any point taken as the moment

centre O. Hence find ∑ M0

22

+

= ∑∑

iyf

ixfR

=

∑

∑

ixf

iyf

R tan 1-α

5. Compute moment arm

6. Also compute x- intercept as

7. And Y intercept as

Example 1: Compute the resultant for the system of forces shown in Fig 2 and hence compute

the Equilibriant.

R

Md

o

R

∑=

∑∑

=ix

o

Rf

MX

∑∑

=ix

o

Rf

My

KN 28.8

60 cos 32 - 44.8 o

=

=∑ ixf

KNM

M

f

oo

o

yi

34.62

)3(60sin32)4(60cos32)3(4.14

49.83

KN 44.6 R

KN 34.11 -

60sin 32- 14.4 - 8

o

R

o

−=

−+−=+

=

=

=

=

∑

∑

ς

α

m 164.28.28

34.62y

m 827.1 11.34

34.62 x

m 396.1 64.44

34.62d

R

R

R

==

==

==

Example 2: Find the Equilibriant for the rigid bar shown in Fig 3 when it is subjected to forces.

• Resultant and Equilibriant

Equilibrium: The concept of equilibrium is the same as explained earlier. For a system of

Coplanar Non concurrent forces for the status of equilibrium the equations to be satisfied are

The above principles are used in solving the following examples.

;90

516

0

o

R

y

x

KNf

f

i

i

=

−=

=

∑∑

α

KNM 1462-

)4(344)2(172)1(430

=

−+−=+∑ AMς

;0 ;0 ;0 === ∑∑∑ oyx Mffii

Example 3: A bar AB of length 3.6 m and of negligible weight is acted upon by a vertical force

F1 = 336kN and a horizontal force F2 = 168kN shown in Fig 4. The ends of the bar are in

contact with a smooth vertical wall and smooth incline. Find the equilibrium position of the bar

by computing the angle θ.

• Eq. 1 gives HA=420 KN

• Beams – Laterally loaded bending

• Supports – Hinge, Roller, Fixed

• Equilibrium Concept for support reactions

• Equations are

o87.36

2.19.0tan

=

=

α

α

;420

013.53sin

0

)1..(..........013.53cos

0

1

2

KNR

FR

f

RFH

f

B

o

B

y

o

BA

x

i

i

=

=−

=

=−−

=

∑

∑

o

A

B

H

M

3.28

0.538 tan

0 cos 705.6 sin 1310.4-

0)sin (1.2 168 -)cos1.2(336)sin6.3(

;0

=

=

=+

=+−

=+∑

θ

θ

θθ

θθθ

ς

;0 ;0 ;0 === ∑∑∑ oyx Mffii

SUPPORT REACTIONS IN BEAMS: Beams are structural members which are generally

horizontal. They are subjected to lateral forces which act orthogonal to the length of the member.

There are various types of mechanisms used for supporting the beams. At these supports the

reactive forces are developed which are determined by using the concept of equilibrium. The

different types of supports are depicted in the table below.

SUPPORT REACTION NO.OF REACTIONS

ROLLER

(1)

HINGE

(2)

FIXED

(3)

VA

TYPES OF LOADS ACTING ON BEAMS: There are various types of forces or loads which

act on beams. They are (a) Concentrated or point load (b) Uniformly distributed load (UDL) (c)

Uniformly varying load (UVL) (d) Arbitrary distributed load. The methodology of converting

UDL, UVL to equivalent point load is shown in the Fig below.

Some example problems of determining support reactions in beams are illustrated next.

Example 4: Determine the support reactions for the beam shown in Fig 7 at A and B.


;0

;0

;0

=

=

=

∑∑∑

o

y

x

M

f

f

i

i

KNV

KNV

V

M

KNVV

VV

A

B

B

A

BA

BA

7.23

3.43

0)10()9(32)5(25)2(10

0

;67

0322510

=

=

=+−−−

=+

=+

=+−−−

∑ς

;35V

45V

0)8(V)7(40)2(400

80VV

0V40-40-;0

0H ;0

A

B

B

BA

B

A

KN

KN

M

Vf

f

A

Ay

x

i

i

=

=

=+−−=

=+

=+=

==

∑∑

ς



Review

• Coplanar system of Forces.

• Concurrent, Non Concurrent.

• Resultant, Equilibrium.

• Concept of Equilibrium.

• Examples.

• Analysis of Trusses

KNH

H

f

A

A

xi

32.17

032.17

;0

=

=−

=∑

KN 10 KN; 45

0)11(10)9(15)8()6(2025210

0

55

010152010

0

==

=−−+−+×−

=+

=+

=+−−−−

=

∑

∑

AB

B

A

BA

BA

y

VV

V

M

VV

VV

fi

ς

;62.4

;24.9

;12

0)6(20)10(

;0

20866.0

030cos20;0

5.0

030sin

;0

0

KNH

KNR

KNV

V

M

RV

RVf

RH

RH

f

A

B

A

A

B

BA

o

BAy

BA

BA

x

i

i

=

=

=−

=+−

=+

=+

=+−=

=

=−

=

∑

∑

∑

ς

ANALYSIS OF PLANE TRUSSES: Trusses are special structures which are formed by joining

different members. Trusses are used as part of roofing systems in industrial buildings, factories

workshops etc. Prominent features of trusses are

• Trusses are articulated Structures.

• The basic Geometry used in a truss is a triangle.

• Every member is pin connected at ends.

• Trusses carry loads only at joints. Joints are junctions where members meet.

• Self weight is neglected.

• The forces in various members of the truss are axial in nature.

A typical figure of a plane truss and the scheme by which truss configuration is arrived at is

shown by the following figures.

Plane Trusses

Truss configuration

• • A truss is said to be perfect if m= 2 j – 3 where m � Members; j � Joints

•

Analysis of Trusses: Analysis of trusses would imply determining forces in various members.

These forces will be in the form of Axial Tension (or) Compression. The

Equilibrium concept is made use of for analyzing the trusses. The two methods of analysis are

1. Method of Joints.

2. Method of Sections.

These two methods of analysis are illustrated by the following examples

Example 1: • Analyse the truss shown in Figure and hence

compute member forces

• Step 1: Draw FBD

• Step 2: Compute support Reactions (HA, VA,

VB).

• Draw FBD’s of Joints to compute member

forces.

• ∑fxi=0

• ∑fyi=0

• HA= - 10 KN

• VA+VB =27.32

• ζ + ∑MA = 0

• -17.32(3) - 10(3) - 10(2.25) + 6VB=0

• VB = 17.41 KN; VA= 9.91 KN

• FBD of joint A

• ∑fxi=0

• -10+PAC cos θ + PAD = 0

• ∑fyi=0; VA + PACsin θ =0

• PAC =-16.52 KN

• PAD=23.21 KN

• ∑fxi=0

• -PAD + PDB = 0

;87.36

3

25.2 tan

o

AD

CD

=

==

θ

θ

• PDB = 23.21 KN

• ∑fyi=0

• -10+PCD = 0

• PCD = 10 KN

• ∑fxi=0

• -PBD – PBC cos θ =0

• PBC = -29.02 KN

• ∑fyi=0

• VB +PBC sin θ = 0

• 17.41 – 29.02 sin θ = 0

•

Sl.No Member Force Nature

1 AC 16.52 C

2 AD 23.21 T

3 CB 29.02 C

4 CD 10 T

5 DB 23.21 T

Example 2 : Analyse the truss shown in figure and hence compute member forces.

• ∑fxi=0

• HA-10+10=0; HA = 0

• ∑fyi=0

• VA+ VB – 20= 0

• VA+ VB= 20

• ζ + ∑MA = 0

• 10(4)-20(3)+10(4)+VE(6)=0

• VE = 10 KN;

• VA =10 KN;

• Symmetrical

o Geometry ;

o Loads

• ∑fxi = 0

• PAC=0

• ∑fyi = 0

• PAB + 10 =0

• PAB = - 10KN

• tan θ = 4/3

• θ=53.13o

• ∑fxi = 0

• -10 + PBD+PBC cos θ =0

• PBD +0.6PBC =10

• ∑fyi = 0

• -PBA− PBC sin θ =0

• -(-10)-0.8 PBC = 0

• PBC= 12.5 KN

• PBD =2.5 KN

• ∑fxi = 0

• -PDF – PDB = 0

• PDF = -2.5 KN

• ∑fyi = 0

• PDC=0

• Symmetrical

Sl.No Member Force Nature

1 AB, EF 10 KN C

2 AC, CE 0 -

3 BC, FC 12.5 KN T

4 BD, FD 2.5 KN T

5 DC 0 -

Example 3: Analyse the truss shown in figure and hence compute member forces.

• Isosceles triangle;

• CD = DB = a

• ∑fxi = 0 HA = 0

• ∑fyi = 0

• VA+VB = 5

• + ∑MA=0

• -5(2a)+VB(3a) = 0

• VB = 3.33 KN; VA = 1.67 KN

ac

dCaooo

2

;60sin90sin30sin

=

==

• ∑fxi = 0

• PAC cos 300 + PAD = 0

• ∑fyi = 0

• 1.67+PAC sin 300 = 0

• PAC = -3.34 KN

• PAD = 2.89 KN

• ∑fxi = 0

• -PDC cos 600 -2.89 +PDB = 0

• ∑fyi = 0

• PDC sin 600 – 5 = 0

• PDC = 5.77 KN

• PDB = 5.77 KN

• ∑fxi = 0 – PBC cos 300 –5.77 = 0

• PBC = -6.66 KN

•

Sl. No Member Force Nature

1 AC -3.34 KN C

2 AD 2.89 KN T

3 BC 6.66 KN C

4 BD 5.77 KN T

5 CD 5.77 KN T

• Method of Sections: Another method of analysis of trusses is method of sections wherein

which the concept of equilibrium of a system of coplanar non concurrent forces is made use

of. The concept of free body diagram is an important part in this method. This method will be

very useful when only few member forces are required. The equation of moment equilibrium

becomes an important tool in this method. The method is illustrated in following figure.

PROCEDURE FOR METHOD OF SECTIONS

• Step 1: Compute support reactions (if need be).

• Step 2: Place the section to cut not more than three members.

• Step 3: Write FBD, unknown forces away from section(T).

• Step 4: Use equilibrium concept to get member forces

This procedure is used for analyzing some examples as shown below.

Example 4 : Compute the forces in members EC, FC and FD of the truss shown in figure.

• tan θ = ¾; sin θ = 0.6; cos θ = 0.8

• ∑MF = 0

• - 20(3)+PEC(4) = 0

• PEC = 15 KN (T)

• ∑fxi = 0; 20-PFC cos θ = 0

• PFC = 25 KN (T);

• ∑fyi = 0; -PEC - PFC sin θ – PFD = 0;

• PFD = - 30 KN

• = 30 KN (C)

Example 5 : Compute the forces in members BE, BD and CD of truss shown in Figure.

•

• ζ + ∑MB =0

• -20(3)-PCD(BC) = 0

• PCD = -34.64 KN = 34.64 KN (C)

• ∑fxi=0

• - PCD –PBD cos300 - PBE cos300 =0

• PBD+PBE=40

• ∑fyi=0

• PBE − PBD=80

• Solve to get PBE = 60 KN; PBD = -20=20 KN (C)

Example 6: Compute the forces in members BD, CD and CE of the truss shown in figure.

• Support reactions

• ∑fxi=0; HA + 24 =0

• HA = -24 KN

• ∑fyi=0; VA + VB = 40+31+40=111 KN

• ζ + ∑MA=0

• -40(3.6)-31(2)(3.6)-40(3)3.6-24(2.7)+4(3.6)VB = 0

• VB = 60 KN; VA = 51 KN

• ζ + ∑MC=0

• -VA (3.6)- PBD(2.7) = 0

• PBD= - 68 KN;

• = 68 KN (c)

• ζ + ∑MD=0

• - VA(2)(3.6)+2.7HA+40(3.6)+PCE(2.7)=0

• PCE=106.67 KN (T)

• ∑fyi=0; 51 -40 + PCD sin θ =0; PCD = -

18.33 KN=18.33 KN(C).

Lecture 5

Equilibrium Static deals primarily with the description of the force conditions necessary and sufficient to

maintain the equilibrium of engineering structures.

When body is equilibrium, the resultant of all forces acting on it is zero. Thus, the resultant force

R and the resultant couple m are both zero, and we have the equilibrium equations

These requirements are both necessary and sufficient conditions for equilibrium.

All physical bodies are three-dimensional, but we can treat many of them as two-dimensional

when the forces to which they are subjected act in a single plane or can be projected onto a single plane.

When this simplification is not possible, the problem must be treated as three

System Isolation And The Fee- body Diagram

Before we apply Eqs.3/1, we must define unambiguously the particular body or mechanical

system to be analyzed and represent clearly and completely all forces actins oz the body. Omission of a

force which acts on the body in question, or inclusion of a force which does not act on the body, will give

erroneous results. A mechanical system is defined as a body or group of bodies which can be

conceptually isolated from all other bodies. A system may be a single body or a combination of connected

bodies. The bodies may be rigid or non rigid. The system may also be an identifiable fluid mass, either

liquid or gas, or a combination of fluids and solids. In statics we study primarily forces which act on rigid

bodies at rest, although we also study forces actins on fluids in equilibrium. Once we decide which body

or combination of bodies to analyze, we then treat this body or combination as a single body isolated

from all surrounding bodies. This isolation is accomplished by means of the free body diagram, which is

a diagrammatic representation of the isolated system treated as a single body. The diagram shows all

forces applied to the system by mechanical contact with other bodies, which are imagined to be removed

If appreciable body forces are present. Such as gravitational or magnetic attraction, then these forces

must also be shown on the free-body diagram of the isolated system.

Only after such a diagram has been carefully drawn should the equilibrium equations be written.

Because of its critical importance, we emphasize here that

Before attempting to &aw a free-body diagram, we must recall the basic characteristics of force. These

chrematistics were described in Art. 2/2, with primary attention focused on the vector properties of force.

Forces can be applied either by direct physical contact or by remote action. Forces car be either internal

or external to the system under consideration. Application of force is accompanied by reactive force, and

both applied and reactive forces may be either concentrated or distributed. The principle of

34

transmissibility permits the treatment of force as a sliding vector as far as its external effects on a rigid

body are concerned.

We will now use these force characteristics to develop conceptual models of isolated mechanical systems.

These models enable us to write the appropriate equations of equilibrium, which can then be analyzed.

Modeling the Action of Forces Figure 1 shows the common types of force application on mechanical systems for analysis in two

dimensions. Each example shows the force exerted on the body to be isolated, by the body to be removed.

Newton's third law, which notes the existence of an equal and opposite reaction to every action, must be

carefully observed. The force exerted on the body in question by a contacting or supporting member is

always in the sense to oppose the movement of the isolated body which would occur if the contacting or

supporting body were removed.

Figure1

35

Typical examples of actual supports that are referenced to Fig.1 are shown in the following sequence of

photo

In Fig. 1, Example 1 depicts the action of a flexible cable, belt, rope, or chain on the body to

which it is attached. Because of its flexibility, a rope or cable is unable to offer any resistance to bending,

shear, or compression and therefore exerts only a tension force in a direction tangent to the cable at its

point of attachment. The force exerted by the cable on the body to which it is attached is always away

from the body. When the tension T is large compared with the weight of the cable, we may assume that

the cable forms a straight line. When the cable weight is not negligible compared with its tension, the sag

of the cable becomes, important, and the tension in the cable changes direction and magnitude along its

length.

When the smooth surfaces of two bodies are in contact. as in Example2 The force exerted by one on the

other is normal to the tangent

to the surfaces and is compressive, Although no actual surfaces are perfectly smooth, we can assume this

to be so for practical purposes in many instances. 36

37

When mating surfaces of contacting bodies are rough, as in Example3 , the force of contact is not

necessarily normal to the tangent to the surfaces, but may be resolved into a tangential or frictional

component F and a normal component N.

Example 4 illustrates a number of forms of mechanical support which effectively eliminate tangential

friction forces. ln these cases the net reaction is normal to the supporting surface Example 5 shows the

action of a smooth guide on the body it supports. There cannot be any resistance parallel to the guide

Example 6 illustrates the action of a pin connection. Such a connection can support force in any

direction normal to the axis of the pin We usually represent this action in terms of two rectangular

components. The correct sense of these components in a specific problem depends on how the member is

loaded. when not otherwise initially known, the sense is arbitrarily assigned and the equilibrium equation

are then written. If the solution of these equations yields a positive algebraic sign for the force

component, the assigned sense is correct. A negative sign indicates the sense is opposite to that initially

assigned.

If the joint is free to turn about the pin, the connection can support only the force R. If the joint is

not free to turn, the connection can also support a resisting couple M. The sense of M is arbitrarily shown

here, but the true sense depends on how the member is loaded.

Example 7 shows the resultants of the rather complex distribution of force over the cross section

of a slender bar or beam at a built-in or fixed support. The sense of the reactions F and V and the bending

couple M in a given problem depends of course, o how the member is loaded.

One of the most common forces is that due to gravitational attraction, Example 8. This force

affects all elements of mass in a body and is, therefore. distributed throughout it. The resultant of the

gravitational forces on all elements is the weight W = mg of the body, which passes through the center of

mass G and is directed toward the center of the earth for earthbound structures The location of G is

frequently obvious from the geometry of the body, particularly where there is symmetry. When the

location is not readily apparent, it must be determined by experiment or calculations.

Similar remarks apply to the remote action of magnetic and electric forces. These forces of remote

action have the same overall effect on a rigid body as forces of equal magnitude and direction applied by

direct.

Example 9 illustrates the action of a linear elastic spring and of a nonlinear spring with either

hardening or softening characteristics. The force exerted by a linear spring, in tension or compression, is

given by F = kx, where k is the stiffness of the spring and x is its deformation measured from the neutral

or unreformed position.

The representations in Fig. 1 are not free-body diagrams, but are merely elements used to

construct free body diagrams. Study these nine conditions and identify them in the problem work so that

you can draw the correct free-body diagrams.

38

Construction of Free-Body Diagrams The full procedure for drawings a free-body diagram which isolates a body or system consists of the

following steps

Step 1. Decide which system to isolate The system chosen should usually involve one or more of the

desired unknown quantities.

Step 2. Next isolate the chosen system by drawing a diagram which represent its complete external

boundary. This boundary defines the isolation of the system from all other attracting or contacting bodies,

which are considered removed This step is often the most crucial of all. Make certain that you have

completely isolated the system before proceeding with the next step.

Step 3. Identify{y all forces which act oz the isolated system as applied by the removed contacting and

attracting bodies, and represent them in their proper positions on the diagram of the isolated system Make

a systematic traverse of the entire boundary to identify all contact forces. Include body forces such as

weights, where appreciable. Represent . all known forces by vector arrow, each with its! Proper

magnitude, direction, and sense indicated. Each unknown force should be represented by a vector arrow

with the unknown magnitude or direction indicated by symbol. if the sense of the vector is also unknown,

you must arbitrarily assign a sense. The subsequent calculations with the equilibrium equations will yield

a positive quantity if the incorrect sense was assumed and a negative quantity if the incorrect sense was

assumed. it is necessary to be consistent with the assigned characteristics of unknown forces throughout

all of the calculations. If you are consistent, the solution of the equilibrium equations will reveal the

correct senses.

Step 4. Show the choice of coordinate axes directly on the diagram Pertinent dimensions may also be

represented for convenience. Note, however., that the free-body diagram serves the purpose of focusing

attention on the action of the external forces, and therefore the diagram should not be cluttered with

excessive extraneous information. Clearly distinguish force arrows from arrows representing quantities

other than forces. for this purpose a colored pencil may be used.

Completion of the foregoing four steps will produce a correct free-body diagram to use in applying the

governing equations, both in statics and in dynamics. Be careful not to omit from the free-body diagram

certain forces which may not appear at first glance to be needed in the calculations. lt is only through

complete isolation and a systematic representation of all eternal forces that a reliable accounting of the

effects of all applied and reactive forces can be made. very often a force which at first glance may not

appear to influence a desired result does indeed have an influence. Thus. the only safe procedure is to

include on the free-body diagram all forces whose magnitudes are not obviously negligible. The free-

body method is extremely important in mechanics because it ensures an accurate definition of a

mechanical system and focuses

attention on the exact meaning and application of the force laws of statics and dynamics. Review the

foregoing four steps for constructing a free-body diagram while studying the sample free-body diagrams

shown in Fig. 2 .

Examples of Free-Body Diagrams Figure 2 gives four examples of mechanisms and structures together with their correct free-body

diagrams. Dimensions and magnitudes are omitted for clarity. In each case we treat the entire system as a

single body, so that the internal forces are not shown The characteristics of the various types of contact

forces illustrated in Fig 1 are used in the four examples as they apply

Figure 2

39

Examples Example 1 Determine the magnitudes of the forces C and T, which, along with the other Forces shown, act on the

bridge-truss joint.

40

Solution

The given sketch constitutes the free-body diagram of the isolated

section of the joint in question and shows the five forces which ere

in equilibrium

Solution 1 (scalar algebra): for the x-y axes as shown we have

Example 2 Calculate the tension t in the cable which supports the 500-kg mass

with the pulley arrangement shown. Each pulley is free to rotate about

its bearing, and the weights of all parts are small compared with the

load. Find the magnitude of the totl force on the bearing of pulley C.

41

Example 3 Determine the magnitude T of the tension in supporting cable and the magnitude of the force on pin at A for the jip crane shown. The beam AB is a standarad 0.5-m I-beam with a mass of 95 kg per meter of length.

42

Example 4 The link shown in Fig. a is pin-connected at A and rests against a smooth support at B. Compute the

horizontal and vertical components of reaction at pin A.

Solution

43

Problems

44

S.N _RECTILINEAR MOTION

We should more properly call vet)the instantaneous velocity functionto distinguish instantaneous velocity.from average velocity. However, wewill follow the standard practice of re-ferring to it as the "velocity function,"leaving it understood that it describesinstantaneous velocity.

In this section we will continue the study of rectilinear motion that we began in Section3.1. We will define the notion of "acceleration" mathematically, and we will show howthe tools of calculus developed earlier in this chapter can be used to analyze rectilinearmotion in more depth.

REVIEW OF TERMINOLOGYRecall from Section 3.1 that a particle that can move in either direction along a coordinateline is said to be in rectilinear motion. The line might be an x-axis, a y-axis, or a coordinateline inclined at some angle. In general discussions we will designate the coordinate line asthe s-axis. We will assume that units are chosen for measuring distance and time and thatwe begin observing the motion of the particle at time t = O. As the particle moves alongthe s-axis, its coordinate s will be some function of time, say s = set). We call set) the

position function of the particle: and we call the graph of s versus t the position versustime curve. If the coordinate of a particle at time t] is S(tl) and the coordinate at a latertime t2 is S(t2), then S(t2) - S(tl) is called the displacement of the particle over the timeinterval [tl, t2]. The displacement describes the change in position of the particle.

Figure 5.8.1 shows a typical position versus time curve for a particle in rectilinear motion.We can tell from that graph that the coordinate of the particle at time t = 0 is so, and wecan tell from the sign of s when the particle is on the negative or the positive side of theorigin as it moves along the coordinate line.

~ Example 1 Figure 5.8.2a shows the position versus time curve for a particle movingalong an s-axis. In words, describe how the position of the particle changes with time.

Solution. The particle is at s = -3 at time t = O. It moves in the positive direction untiltime t = 4, since s is increasing. At time t = 4 the particle is at position s = 3. At that timeit turns around and travels in the negative direction until time t = 7, since s is decreasing.At time t = 7 the particle is at position s = -1, and it remains stationary thereafter, sinces is constant for t > 7. This is illustrated schematically in Figure 5.8.2b ...•

~sI i I I

1/ '\

1-- ,__ ,J i\ }--- - -i/ t

1 1\., I)

L.••••

I-

54321o

-1-2-3-4-5

o 1 2 3 4 5 6 7 8 9 10

(a)

t ? 7: ~:

• VELOCITY AND SPEEDRecall from Formulas (6) and (7) of Section 3.1 and Formula (4) of Section 3.2 that theinstantaneous velocity of a particle in rectilinear motion is the derivative of the positionfunction and the instantaneous speed is the absolute value of the instantaneous velocity.Thus, if a particle in rectilinear motion has position function s (t), then we define its velocityfunction vet) to be

I dsvet) = s (t) = -

dt

Iv(t)1 = Is'(t)1 = )~; IThe sign of the velocity tells which way the particle is moving-a positive value for v (t)

means that s is increasing with time, so the particle is moving in the positive direction, and

a negative value for vet) means that s is decreasing with time, so the particle is movingthe negative direction. If vet) = 0, then the particle has momentarily stopped. The spefunction, which is always nonnegative, tells us how fast the particle is moving but notdirection of motion.

~ Example 2 Let s (t) = t3 - 6t2 be the position function of a particle moving alean s-axis, where s is in meters and t is in seconds. Find the velocity and speed functioand show the graphs of position, velocity, and speed versus time.

Solution. From (1) and (2), the velocity and speed functions are given by

dsvet) = - = 3t2 - 12t and Iv(t)1 = 13t2 - 12tl

dt

The graphs of position, velocity, and speed versus time are shown in Figure 5.8.3. Obsethat velocity and speed both have units of meters per second (m/ s), since s is in meters (and time is in seconds (s) .•••

The graphs in Figure 5.8.3 provide a wealth of visual information about the motiorthe particle. For example, the position versus time curve tells us that the particle is onnegative side of the origin for 0 < t < 6, is on the positive side of the origin for t > 6, ;is at the origin at times t = 0 and t = 6. The velocity versus time curve tells us thatparticle is moving in the negative direction if 0 < t < 4, is moving in the positive directif t > 4, and is momentarily stopped at times t = 0 and t = 4 (the velocity is zero at thtimes). The speed versus time curve tells us that the speed of the particle is increasingo < t < 2, decreasing for 2 < t < 4, and increasing again for t > 4.

II ACCELERATIONIn rectilinear motion, the rate at which the instantaneous velocity of a particle changes \i

time is called its instantaneous acceleration. Thus, if a particle in rectilinear motionvelocity function vet), then we define its acceleration function to be

I dvaCt) = v (t) = -

dt

Alternatively, we can use the fact that vet) = S'(t) to express the acceleration functioterms of the position function as

d2saCt) = S"(t) = -2

dt

~ Example 3 Let set) = t3 - 6t2 be the position function of a particle moving alan s-axis, where s is in meters and t is in seconds. Find the acceleration function aCt),show the graph of acceleration versus time.

Solution. From Example 2, the velocity function of the particle is vet) = 3t2 - 121the acceleration function is dv

aCt) = - = 6t - 12dt

and the acceleration versus time curve is the line shown in Figure 5.8.4. Note that inexample the acceleration has units of m/ S2, since v is in meters per second (m/ s) and 1

is in seconds (s). •••

Ifa (t) = 0 over a certain time interval,what does this tell you about the mo-tion of the particle during that time?

ill SPEEDING UP AND SLOWING DOWNWe will say that a particle in rectilinear motion is speeding up when its speed is increasingand is slowing down when its speed is decreasing. In everyday language an object that isspeeding up is said to be "accelerating" and an object that is slowing down is said to be"decelerating"; thus, one might expect that a particle in rectilinear motion will be speedingup when its acceleration is positive and slowing down when it is negative. Although this istrue for a particle moving in the positive direction, it is not true for a particle moving in thenegative direction-a particle with negative velocity is speeding up when its accelerationis negative and slowing down when its acceleration is positive. This is because a positiveacceleration implies an increasing velocity, and increasing a negative velocity decreasesits absolute value; similarly, a negative acceleration implies a decreasing velocity, anddecreasing a negative velocity increases its absolute value.

The preceding informal discussion can be summarized as follows (Exercise 37):

THE SIGN OF ACCELERATION. A particle in rectilinear motion iswhen its velocity and acceleration have the same sign and slowing down

opposite signs.

~ Example 4 In Examples 2 and 3 we found the velocity versus time curve and theacceleration versus time curve for a particle with position function s (t) = t3 - 6t2• Usethose curves to determine when the particle is speeding up and slowing down, and confirmthat your results are consistent with the speed versus time curve obtained in Example 2.

Solution. Over the time interval 0 < t < 2 the velocity and acceleration are negative, sothe particle is speeding up. This is consistent with the speed versus time curve, since thespeed is increasing over this time interval. Over the time interval 2 < t < 4 the velocityis negative and the acceleration is positive, so the particle is slowing down. This is alsoconsistent with the speed versus time curve, since the speed is decreasing over this timeinterval. Finally, on the time interval t > 4 the velocity and acceleration are positive, sothe particle is speeding up, which again is consistent with the speed versus time curve. .••

flI ANALYZING THE POSITION VERSUS TIME CURVEThe position versus time curve contains all of the significant information about the positionand velocity of a particle in rectilinear motion:

• If s (t) > 0, the particle is on the positive side of the s-axis.

• If set) < 0, the particle is on the negative side of the s-axis.• The slope of the curve at any time is equal to the instantaneous velocity at that time.

• Where the curve has positive slope, the velocity is positive and the particle is moving inthe positive direction.

• Where the curve has negative slope, the velocity is negative and the particle is movingin the negative direction.

• Where the slope of the curve is zero, the velocity is zero, and the particleis momentarilystopped.

Information about the acceleration of a particle in rectilinear motion can also be deducedfrom the position versus time curve by examining its concavity. For example, we knowthat the position versus time curve will be concave up on intervals where Sll (t) > 0 andwill be concave down on intervals where Sll(t) < O. But we know from (4) that Sll(t) is

the acceleration, so that on intervals where the position versus time curve is concave up the i

particle has a positive acceleration, and on intervals where it is concave down the particlehas a negative acceleration.

Table 5.8.1 summarizes our observations about the position versus time curve.

POSITION VERSUSTIME CURVE

BEHAVIOR OF THE PARTICLE

AT TIME t= toCHARACTERISTICS OF THE

CURVE AT t = to

• S(tO) > 0• Curve has positive slope.• Curve is concave down.

• s(tO) > 0• Curve has negative slope.• Curve is concave down.

• s(tO) < 0• Curve has negative slope.• Curve is concave up.

• Particle is on the positive side of the• Particle is moving in the positive directlLon.• Velocity is decreasing.• Particle is slowing down.

• Particle is on the positive side of the• Particle is moving in the negative direction.• Velocity is decreasing.• Particle is speeding up.

• Particle is on the negative side of the• Particle is moving in the negative direction.• Velocity is increasing.• Particle is slowing down.

• s(to) > 0• Curve has zero slope.• Curve is concave down.

• Particle is on the positive side of the origin.• Particle is momentarily stopped.• Velocity is decreasing.

~ Example 5 Use the position versus time curve in Figure 5.8.2 to determine when theparticle in Example 1 is speeding up and slowing down.

Solution. From t = 0 to t = 2, the acceleration and velocity are positive, so the particle isspeeding up. From t = 2 to t = 4, the acceleration is negative and the velocity is positive, sothe particle is slowing down. At t = 4, the velocity is zero, so the particle has momentarilystopped. From t = 4 to t = 6, the acceleration is negative and the velocity is negative, sothe particle is speeding up. From t = 6 to t = 7, the acceleration is positive and the velocityis negative, so the particle is slowing down. Thereafter, the velocity is zero, so the particlehas stopped. ••

~ Example 6 Suppose that the position function of a particle moving on a coordinateline is given by s(t) = 2t3 - 2lt2 + 60t + 3. Analyze the motion of the particle for t ::::O.

v(t) = Sl(t) = 6P - 42t + 60 = 6(t - 2)(t - 5)a(t) = v/(t) = 12t - 42 = 12 (t - D

• Direction of motion: The sign analysis of the velocity function in Figure 5.8.5 showsthat the particle is moving in the positive direction over the time interval 0 :s t < 2,stops momentarily at time t = 2, moves in the negative direction over the time interval2 < t < 5, stops momentarily at time t = 5, and then moves in the positive directionthereafter.

t)

++++++++0------------0++++++++Positive

direction

Sign of vCt) = 6(t - 2)(t - 5)Direction of motionPositive

direction

Figure 5.8.5

Negativedirection

• Change in speed: A comparison of the signs of the velocity and acceleration functions isshown in Figure 5.8.6. Since the particle is speeding up when the signs are the same andis slowing down when they are opposite, we see that the particle is slowing down overthe time interval 0 :s t < 2 and stops momentarily at time t = 2. It is then speedingup over the time interval 2 < t < ~. At time t = ~the instantaneous acceleration iszero, so the particle is neither speeding up nor slowing down. It is then slowing downover the time interval ~ < t < 5 and stops momentarily at time t = 5. Thereafter, it isspeeding up.

7251! f )0

++++++++0------------0++++++++ Sign of v(l) = 6(1- 2)(1- 5)

Sign of a(l) = 12(1 - D-------------0+++++++++++++7o 2 2 5

I I ! I

Slowingdown

Figure 5.8.6

Speedingup

Slowingdown

Speedingup

Conclusions: The diagram in Figure 5.8.7 summarizes the above information schematically.The curved line is descriptive only; the actual path is back and forth on the coordinate line.The coordinates of the particle at times t = 0, t = 2, t = ~,and t = 5 were computed fromset). Segments in red indicate that the particle is speeding up and segments in blue indicatethat it is slowing down. .•• .

t=5( 1= 2~

•• 2

: • )1= 2t = 0 • ~ s••

Figure 5.8.7 0 3 28 41.5 55

<D For a particle in rectilinear motion, the velocity and positionfunctions v (t) and s (t) are related by the equation .and the acceleration and velocity functions aCt) and v(t) arerelated by the equation .

2. Suppose that a particle moving along the s-axis has posi-tion function set) = 7t - 2t2• At time t = 3, the particle'sposition is ', its velocity is , its speed is____ .,and its acceleration is _

3. A particle in rectilinear motion is speeding up if.t~e sign~ ofits velocity and acceleration are , and It IS slowmgdown if these signs are ---

4. Suppose that a particle moving along the s-axis has positiofunction s(t) = t4 - 24t2 over the time interval t ~ O. Thparticle slows down over the time interval(s) -~-~--_.

~--'·_--~----'--------II .I 1. The graphs of three position functions are shown in the

accompanying figure. In each case determine the signs ofthe velocity and acceleration, and then determine whetherthe particle is speeding up or slowing down.

S. Sketch a reasonable graph of s versus t for a mouse thatis trapped in a narrow corridor (an s-axis with the pos-itive direction to the right) and scurries back and forthas follows. It runs right with a constant speed of 1.2mls for awhile, then gradually slows down to 0.6 mis,then quickly speeds up to 2.0 mis, then gradually slowsto a stop but immediately reverses direction and quicklyspeeds up to 1.2 ml s.

(a)Figure Ex-}

6. The accompanying figure shows the graph of s versus tfor an ant that moves along a narrow vertical pipe (ans-axis with the positive direction up).(a) When, if ever, is the ant above the origin?(b) When, if ever, does the ant have velocity zero?(c) When, if ever, is the ant moving down the pipe?

2. The graphs of three velocity functions are shown in theaccompanying figure. In each case determine the sign ofthe acceleration, and then determine whether the particleis speeding up or slowing down.

(a)Figure Ex-2

3. The position function of a particle moving on a horizontalx-axis is shown in the accompanying figure.(a) Is the particle moving left or right at time to?(b) Is the acceleration positive or negative at time to?(c) Is the particle speeding up or slowing down at

time to?(d) Is the particle speeding up or slowing down at

time tl?

7. The accompanying figure shows the graph of velocityversus time for a particle moving along a coordinate line.Make a rough sketch of the graphs of speed versus timeand acceleration versus time.

-U\TIIT:5)

15

I~t

2 3 4 5 6 Figure Ex-7

8. The accompanying figure shows the position versus timegraph for an elevator that ascends 40 m from one stop tothe next.(a) Estimate the velocity when the elevator is halfway

up to the top. --(b) Sketch rough graphs of the velocity versus time curve

and the acceleration versus time curve.

4. For the graphs in the accompanying figure, match theposition functions with their corresponding velocityfunctions.

J~lEo 5 10 15 20 25

Time t (5) Figure Ex-8

(I)Figure Ex-4

9. The accompanying figure shows the velocity versus timegraph for a test run on a Pontiac Grand Prix GTP. Using thisgraph, estimate(a) the acceleration at 60 mi/h (in ft/sz)(b) the time at which the maximum acceleration occurs.

120~ 100.5 80;:;, 60z;-'u 400

~ 20

-l- --Ifr

-j -- --- -t- -iI

/-- --o 5 10 15 20 25 30

Time t (5) Figure Ex·9

11-12 The function s (t) describes the position of a particlemoving along a coordinate line, where s is in meters and t isin seconds.(a) Make a table showing the position, velocity, and accel-

eration to two decimal places at times t = 1,2,3,4,5.(b) At each of the times in part (a), determine whether the

particle is stopped; if it is not, state its direction ofmotion.

(c) At each of the times in part (a), determine whether theparticle is speeding up, slowing down, or neither.

nt11. s(t) = sin "4

13-18 The function s (t) describes the position of a particlemoving along a coordinate line, where s is in feet and t is inseconds.(a) Find the velocity and acceleration functions.(b) Find the position, velocity, speed, and acceleration at

time t = 1.(c) At what times is the particle stopped?(d) When is the particle speeding up? Slowing down?(e) Find the total distance traveled by the particle from time

t = 0 to time t = 5.

13. s(t) = t3 - 3t2, t 2: 0

14. s(t) = t4 - 4t2 + 4, t 2: 0

15. set) = 9 - 9cos(nt/3), O::s t ::S5t

16. set) = t2 +4' t 2: 0

E3 19. Let s(t) = t / (t2 + 5) be the position function of a particlemoving along a coordinate line, where s is in meters and tis in seconds. Use a graphing utility to generate the graphsof set), vet), and aCt) for t 2: 0, and use those graphs whereneeded.(a) Use the appropriate graph to make a rough estimate of

the time at which the particle first reverses the directionof its motion; and then find the time exactly.

(b) Find the exact position of the particle when it first re-verses the direction of its motion.

(c) Use the appropriate graphs to make a rough estimate ofthe time intervals on which the particle is speeding upand on which it is slowing down; and then find thosetime intervals exactly.

21- 28 Aposition function of a particle moving along a coor-dinate line is given. Use the method of Example 6 to analyzethe motion of the particle for t 2: 0, and give a schematicpicture of the motion (as in Figure 5.8.7).

21. s = -4t + 3

23. s = t3 - 9t2 + 24t

25. s = 16te-(t2/8)

22. s = 5t2 - 20t

24. s = t3 - 6tZ + 9t + I25

26. s = t +--t+2

S= (cos t, O::s t < 2n

27.I, t2:2n

{2t(t - 2f,

28. s =13-7(t-4)z, t2:3

30. Let s = 1001 (t2 + 12) be the position function of a particlemoving along a coordinate line, where s is in feet and t is inseconds. Find the maximum speed of the particle for t 2: 0,and find the direction of motion of the particle when it hasits maximum speed.

31-:52 A position function of a particle moving along aordinate line is provided. (a) Evaluate s and v when a(b) Evaluate s and a when v = O.

31. s = In(3t2 - 12t + 13) 32. s = t3 - 6t2 + 1E3 33. Let s = J2t2 + 1be the position function of a particle

ing along a coordinate line.(a) Use a graphing utility to generate the graph of v

t, and make a conjecture about the velocity of theticle as t -+ +00.

(b) Check your conjecture by finding lim v.t-+ +00

34. (a) Use the chain rule to show that for a particle inear motion a = v(dvlds).

(b) Let s = J3t + 7, t 2: O. Find a formula for v inof s and use the equation in part (a) to find theation when s = 5.

35. Suppose that the position functions of two particles, PIP2, in motion along the same line are

Sl = ~tZ - t + 3 and S2 = _~tZ + t + 1respectively, for t 2: O.(a) Prove that Pj and P2 do not collide.(b) How close do PI and Pz get to one another?(c) During what intervals of time are they moving in

site directions?

36. Let SA = 15t2 + lOt + 20 and SB = 5t2 + 40t, t 2: 0, bethe position functions of cars A and B that are moving alongparallel straight lanes of a highway.(a) How far is car A ahead of car B when t = O?(b) At what instants of time are the cars next to one another?(c) At what instant of time do they have the same velocity?

Which car is ahead at this instant?37. Prove that a particle is speeding up if the velocity and accel-

eration have the same sign, and slowing down if they haveopposite signs. [Hint: Let r(t) = Iv(t)1 and find r' (t) usingthe chain rule.]

Ce 202 Engineering Mechanics

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