Civil Engg. Mechanics

1

ENGINEERING MECHANICS

[SUBJECT CODE-ME101]

FIRST YEAR FIRST SEMESTER

STUDY MATERIAL 2

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. 3

LECTURE CONTENTS

SL Lecture No. Lecture Title Page

1. Lecture 1

Introduction to mechanics, Statics,

Concept of Particle and Rigid body,

Force, Transmissibility of force

6-7

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

3. Lecture 3

Introduction to Vector Algebra,

Parallelogram Law, Addition and

Subtraction of vector

12-16

5. Lecture 4

Lami’s theorem, Free vector, Bound

vector, Representation of forces in term

of I,j & k

17-18

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

6. Lecture 6 Examples 21-22

7. Lecture 7

Two dimensional force system,

Resolution of forces, Moment,

Varignon’s theorem

23-24

8. Lecture 8 Couple, Example 25-29

9. Lecture 9

Resolution of a coplanar force by its

equivalent force-couple system,

Resolution of forces, Example 30-33

10. Lecture10 Example 34-38

11. Lecture11 Concept and equilibrium of forces, Free

body diagram

39-40

12. Lecture12 Free body diagram , example 41-42

13 Lecture13 Equation of equilibrium, example 43-49

14 Lecture14

Concept of friction, Columb law of

friction, Angle of repose 50-54 4

15 Lecture15 Coefficient of friction, Kinetic friction 55-59

16 Lecture16 Examples 60-68

17 Lecture17 Distributed force, Centroid & Centre of

Gravity

69-73

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

19 Lecture19 Centroid of an arc, Centroid of a

triangle, Example

79-82


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

22 Lecture22 Moment of inertia of plain figure,

Perpendicular axis theorem, Radius of

gyration

91-97


24 Lecture24 Concept of stress, different types of

stresses & strains, Hook’s law

102-110

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,

Rectilinear motion

127-129

29 Lecture29 Relative motion and example 130-138

30 Lecture30 Kinetics of particle, curvilinear motion,

example

139-141 5

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 6

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. 7

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.

8

Lecture 2 : Classification & Types Of Forces

1.1.1 CLASSIFICATION OF FORCE

The general classification of force is presented below:

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

■ Non-Concurrent

■ Parallel

Force

■ Concurrent

■ Non-Concurrent

■ Parallel

■ Non-Parallel 9

■ Non-Parallel

Table 1.2 TYPES & CHARACTERISTICS OF FORCE SYSTEM

Force System Line Diagram Characteristics

Coplanar forces

Y

x

Lines of action of all forces

lie on the same plane

Non-Coplanar forces y

F1

x

F fF

z F2

Lines of action of all forces

do not lie on the same

plane

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 10

Parallel Forces

F1

F2

F3

Lines of action all forces

are parallel to each other

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. 11

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 12

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 13

In OCD, OC

2

= OD

2

+DC

2

Or, R

2

= (OA+AD)

2

+DC

2

Or, R

2

= OA

2

+2. OA.AD + AD

2

+DC

2

Or, R

2

= P

2

+2PQcosα +Q

2

cos

2

α+ Q

2

sin

2

α

Or, R = (P

2

+ Q

2

+2PQcosα) ½

Direction of resultant vector R

Let Ø = angle made by resultant with OA

Then from OCD, tan Ø = CD/OD = Q sinα/ P+Qcosα

So, Ø = tan¯

1

[Q sinα/ 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. 14

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. 15

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:

R

2

= A

2

+ 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, 16

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

R

2

= (FX)

2

+ (FY)

2

(3)

and that the angle φ- the angle that R makes with the Xaxis may be determined from the equation

tanφ = Fy/Fx (4)

17

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). 18

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

Also F = √ (Fx

2

+ FY

2

+ FZ

2)

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

The three angles are related by, cos

2

α +cos

2

β +cos

2

γ = 1 or l

2

+m

2

+ n

2

= 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

Magnitude of unit vector F = l F l = √ (Fx

2

+ FY

2

+ FZ

2

)

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

19

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

(i) When θ = 0

0

, and vectors A & B are along same direction, A. B = AB cos0 = AB

(ii) When θ = 90

0

, 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

20

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

lFl = √ (20

2

+ 50

2

+ 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

OC

2

= OA

2

+ OB

2

B C

Or OB= √ (20

2-

10

2

) FB 10N

= 17.32N O A

Tanθ = 17.32/10 or θ = 60° 21

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°

Unit Vector = (40i+23.09j+18k)/ √ (40

2

+ 23.09

2

+18

2

) = 0.804i+.464j+.371k) (ans)

Assignment

1. Subtract the 20N force in example.1 from50N force. 400N

40° 500N 22

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


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

.

23

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

FX = Fcosθ & FY = Fsinθ F = √ (FX

2

+ Fy

2

), θ = tan¯

1

(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. 24

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. 25

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. 26

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

27

(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. 28

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 xdirection 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

29

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 1 Y/ Rx = tan 1 ∑FY / ∑FX 30 R̄� ̄�

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

31

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

Exmpl.5)

Applying Lami’s theorem:

= sin =

Or, T1 = 15kN (sin150

o

) / sin75

o

= 7.76kN Ans.

& T2 = 15kN (sin 135

o

) / sin 75

o

= 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

100kN –100kN (cos45

o

) – P = 0

Or, P = 29.3kN ns.

Resolving the forces vertically and computing Σ V = 0

200kN –100kN (sin45

o

) –Q = 0

Or, Q = 129.3kN Ans.

15kN

sin75

o

T1

sin150

o

T2

sin135

o32

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 33

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.

34


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

direction of the resultant force.

10°

Ø

Parallelogram Law. The parallelogram law of addition is shown in

Figure Two unknowns are the magnitude of FR and the angle θ(theta).

FR = √ (100 N)

2

+ (150 N)

2

– 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

F2=150N

F1= 100N

15°

15°

10°

θ

90°-25°=65°

125° 65°

150N

100N

FR 35

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.

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 .

A

B

C

500N

θ

30°

FAC=400N

FAB

500N

θ

60°

30°

FAC=400

500N

FAB

θ 60°

Ø

FIG A FIG B36

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.

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

40N 50N

T1

T2

T3

35° θ37

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

Assignment

1A. The vertical force F acts downward at A on the two-membered

frame. Determine the magnitudes of the two

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.

2. For the stepladder shown in Figure, sides AC and CE are

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 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.


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

A

B

C

C

D38

2. Two non-collinear parallel equal forces acting in opposite direction

a) Balance each other

b) Constitute a couple

c) Constitute a moment


3. The force polygon for coplanar concurrent forces

a) Must Close

b) Must not close

c) Anyone 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


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

39

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

and the total force on each particle is permanently zero.

[1]

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

interchangeably with "mechanical equilibrium", as defined next.

[2]

A standard definition of mechanical equilibrium for a particle is:

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

that the net force acting upon the particle is zero.

[3]

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

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

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

[3]

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. 40

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. 41

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. 42

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 threeforce members.

43

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

A B

C D

FBD 44

P Q S

A

x

C D

W

A

y

B

y

y y

x x

A y

F A

F A

M B

Σ = =>

Σ = =>

Σ = =>

0

0

0

Additional equations could be written.

Σ = 0 MB

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

We can use Σ = 0 MB

to replace one of the above 3.

Example 1.Given

Find: Reactions

2 m

2 kN

2 kN

1.5 m

1.5 m

A B

4 kN

2 2 0

0

= −

+ + =

Σ =

Ax

Ax

x

R

R

F

3 0

0

− + + =

Σ =

Ay B

y

R R

F45

FBD

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

Example2.Given

Find: Reactions at A

FBD

2 m

2 kN

2 kN

1.5 m

1.5 m

A B

R

B

R

Ay

R

Ax

3 ft 2 ft

2000 lbs

A C B

3000 ft lbs

3 ft 2 ft

2000 lbs

A B C

3000 ft lbs

R

Ay

R

Ax

MA

0

0

=

Σ =

Ax

x

R

F

2000 lbs

2000 0

0

=

− =

Σ =

Ay

Ay

y

R

R

F

,7 000 ft lbs

3000 2000 )5( 0

0

=

+ − =

Σ =

A

A

A

M

M

M46

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 47

Fig for Example.1

Obviously tension in OB = TOB = 500N

Resolving the forces at O vertically

POA.sin30

0

+ POB.sin30

o

= 400N

Or, POA = 300N

Resolving the forces at O horizontally

POC + POA.cos30

o

= POB.cos30

O

or, POC + 300N(cos30

o

) = (500N)cos30

o

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.

400N

500N

B

C

O

30° 30°

A48

15N

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

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

B

A

45

O

60

O

C

30"

25

o

G

25"

24"

25"

T49

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

50

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. 51

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. 52

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. 53

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. 54

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

N and hence

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

angle.

55

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

, 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 k,

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. 56

• 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: 57

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: 58

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. 59

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.

60


Example 1 A load of 150kN rests on a rough inclined plane (angle of inclination α ). 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, α and the coefficient of friction,µ.

Solution: See Fig.1 & 2

Fig.1 to Example

200kN

α

150kN

125kN

α

150kN61

Fig.2 to Example

FBD when 200kN force is applied

150kN

P = W tan (α+Φ)

Or, 200kN = (150kN) tan (α+Φ)

Or, tan (α+Φ) = 200/150 = 1.333

Or, (α+ Φ) = 54.31

0

When 125kN force is applied

Sin (α+Φ)

R1

200kN

F1

F2

R2

125kN62

P = W

CosΦ

Or, 125 (KN) = 150kN

Or, Φ = 16.3

o

.

Therefore, α = 53.1

o

- 16.3

o

= 36.8

o

(Ans).

µ = tanΦ = tan 16.3

o

= 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.

Sin (53.1

o

)

CosΦ

Coefficient of Friction

150kN63

∑FY = An - 196.14 cos (35°) = 0

An = 160.67 N

For impending motion of the block A: (Af = ms

.

An)

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

T = 152.67 N 64

Block B weight: WB = mB

.

g = 80 (9.807) = 784.56 N

µ = Bf/Bn= 257.17/803.34 = 0.320

Free body diagram for block B:

∑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? 65

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.

b. d. 66


1. Static friction is always

a) Greater than dynamic friction

b) Less than dynamic friction

c) Equal to dynamic friction


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 67

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. 68

8. Coulomb friction is the friction between

a) Bodies having relative motion

b) Two dry surfaces

c) Two lubricated surfaces

d) Solids and liquids

69

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

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

As in the continuous vertical load supp-

-otred by a suspended cable where ‘W’

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

2. Area distribution

When a force is distributed over an area

b

W70

like hydraulic pressure of water against

the inner face of a section of dam.

Intensity is force per unit area (N/m

2

)

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

is acceleration due to gravity. The unit is (kg/m

3

) (m/sec

2

) =N/m

3

.

Volume distribution of forces71

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

Centroid of an Area

The centroid of an area can be

determined by using three similar

equations:

Centroid of a Volume72

Centroid of a Volume

Similarly, centroid of a volume can be

determined by

Center of Mass

Three Planes of Symmetry

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,

73

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

Here g is the acceleration of gravity (9.81m/s

2

or 32.2 ft/s

2

). If g is constant throughout the body,

then the center of gravity is exactly the same as the center of mass. 74

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 75

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 76

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 77

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

that the particles are fixed in the system and being rotated 90

o

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. 78

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.

79

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

2αr X= 2r

2

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 80

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

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

2

/2)-(y

3

/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 81

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 82

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.

83

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 84

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 85

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 86

2. Determine the x and y coordinates of Y

The trapezoidal area.

a

b

X

h 87

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. 88

(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

moment of the area about a parallel axis at centroid and is Ad

2

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,

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

2

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 89

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 = Iyx90

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

91

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 92

Moment of Inertia Moment of Inertia

About x axis About y axis

Triangle

Moment of Inertia about x axis =

Half circle93

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= 94

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.

Ixx = AKx

2

= ∫ y

2

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. 95

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 96

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 97

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. 98


.

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.

Ixx=sum(Ic+Ad

2

)

In this case:

Ixx = IA +AAd

2

+ IB +ABd

2

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

3

+ (12in

2

)(2in)

2

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

3

+ 12in

2

(2in)

2

Ixx = 4 in

4

+ 48 in

4

+ 36 in

4

+ 48 in

4

Ixx = 134 in

4

Y

X =KY

2

3

Example

Determine the moment of inertia of the area under the

parabola about the x axis. Solve the problem by using O 4 X

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

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

X =Ky

299

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

strip about the axis is y

2

dA

Where dA = (4-x) dy dy

= 4 (1- y

2

/9) dy y

4

[x =Ky

2

, so K = 4/9, so x = 4/9 y

2

]

3

IY = ∫ y

2

dA = ∫ 4y

2

(1- y

2

/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

diX = 1/3 (dx) y

3

again y = 3x

1/2

/2

4

IX = ∫ 1/3 (dx) (3/2x

1/2

)

3

= 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

100

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


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


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) 7.5 101

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

102

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. 103

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 104

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)

are N / m

2

(or Pa)

MPa = 10

6

Pa

GPa = 10

9

Pa

KPa = 10

3

Pa

Some times N / mm

2

units are also used, because this is an equivalent to MPa. While US

customary unit is pound per square inch psi. 105

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 : 106

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 ). 107

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 108

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. 109

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

convenient to measure the strain in the form of strain x 10

-6

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 110

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

111

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 value112

Proportionality limit

Up to this amount of stress, stress is proportional to strain (Hooke's law), so the stressstrain 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.113

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.

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

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)

P

10cm 114

Example2

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..

A B C D

P2 10KN

300mm 300mm 300mm

30KN

40KN 115

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

long. If the averages shear stress in the key is not to exceed 700kg/cm

2

. What is the safe value

of the load ‘P’ applied to the end of the lever? 116

Assignment2

A prismatic steel bar having cross-sectional area A=3cm

2

is subjected to axial loading as shown

in figure. Neglecting localized irregularities in stress distribution near the points of application of

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

6

kg/cm

2.

1M 1M 2M

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


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


1.5T

1.5T

2T117

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 118

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 antiballistic 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 119

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 120

•

• 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.

121

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) 122

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, 123

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 124

Assignment

.1. A projectile is projected at an angle of 30

0

from horizontal with a velocity of 30 m/sec. At

what times, the projectile will be at half the maximum attainable height?

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

0

. The angle of

projection is 45

0

. 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.

125


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 -

10 m/s

2

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 126

c. less

d. same or less 127

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 128

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.

Q.5. The normal component of a particle moving on a curve is 20 m/sec

2

at a point. If the speed

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

2

.

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.1The velocity of a particle moving along a space cur ve 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.

Q2 At a particular instant, the magnitude of the velo city of a particle moving along a space curve

is 10 m/sec. Its acceleration is 1 m/sec

2

and it makes 30

0

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. 129

Q.3. A particle is rotating in a plane with a constant angular velocity of 1 rad/sec

2

.

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.

130

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, 131

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

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?

Solution:132

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 133

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

Q.2. A unit vector is rotating about an axis making an angle of 45

0

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 134

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

a. 200 m/sec

2

b. 100 m/sec

2

c. 400 m/sec

2

d. 20 m/sec

2

135

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. 136

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. 137

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. 138

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.

QA 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

2

. 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. 139

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: 140

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 141

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. 142

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. 143

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, 144

(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 145

Thus,

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

acceleration 146

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. 147

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. 148

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 149

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) 150

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. 151

For an external observer, the particle is moving and it has a centripetal acceleration v

2

/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. 152


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

= Mg153

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. 154

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

a. Newton's third law

b. D.Alembert’s principle

c. Impulse and momentum method

d. Work-energy method.

Fig A . Fig. B

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.

b. Directly proportional to r and inversely proportional to v

2

.

c. Directly proportional to v

2

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 155

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 156

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. 157

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? 158

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, 159

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 160

= 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 161

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. 162

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 163

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. 164

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. 165

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 166

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,

(KE)1 = 1/2 mV

2

After impact

(KE)2 = 1/2 (M+m) Vf

2

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 167

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. 168

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 169

Therefore, the angular velocity of the particle at the radius of r/2 is 4

The resultant speed of the particle is

Civil Engg. Mechanics

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