APPLIED MECHANICS – CVE 253

LESSON NOTES

Prof. A.A. Jimoh and Ibrahim, A.O.

Civil Engineering Department

University of Ilorin, Ilorin, Kwara State, Nigeria

I. INTRODUCTION TO MECHANICS

I.1 DEFINITION AND CONCEPT

Mechanics can be defined as the branch of physical science which studies the behaviour of a

body at rest or in motion under the action of forces.

It is basic to Engineering Sciences because of its use in the analysis and design of engineering

components and systems. A thorough knowledge of mechanics is a foundation tool for the

understanding of the physical phenomenon of engineering.

I.2 BRANCHES OF MECHANICS

Mechanics is a branch of physical sciences which is divided into three parts, namely: (i) Fluid

Mechanics; (ii) Mechanics of Deformable Bodies and (iii) Solid Mechanics.

Mechanics of Deformable bodies: This is the part of mechanics commonly referred to as

‘STRENGTH OF MATERIALS’. It studies the reaction of bodies to external forces i.e. it looks

into the resulting stress and/or deformation that arise as a result of application of external loads

(or forces).

Solid Mechanics: In this part of mechanics, the bodies are assumed to be rigid solids. A solid (or
rigid) body can therefore be defined as that in which the distance between any two points in the
body remains ‘appreciably’ unchanged.

Sub division of solid mechanics

This include: (i) Statics and (ii) Dynamics.

Statics is the study of the behaviour of bodies at rest under the action of balanced forces; while

dynamics is the study of the bahaviours of bodies in motion under the action of unbalanced

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forces. It is further divided into; (a) ‘Kinematics’ which deals with the motion of bodies without
any reference to the cause of motion.

(b) ‘Kinetics’ which deals with the relationship between forces and the resulting motion of
bodies on which they act.

The branch of science which deals with the study of different laws of mechanics as applied to
solution of engineering problems is called Applied Mechanics.

BASIC CONCEPTS

The basic concepts of mechanics are the concepts of time space mass and force.

Concept of time: This answers the question ‘when’. It gives the precise definition of when an

event takes place or will take place.

Concept of space: This answers the question ‘where. It gives the definition of the position. This

definition arises from the frame of reference that all point are defined from an imaginary origin

measured out in three direction called xyz axes.

Concept of mass: This answers the question ‘how much’ It gives the measure of quantity of a

body involved. e.g. 2kg of a body is placed on a tower.

Concept of force: This answers the question ‘what is the cause’ It defines what causes an action

to take place.

2.0 FORCE

Force is generally defined as ‘any effect or action’ that may cause a change in the state of rest or

motion of a body. It is a Vector quantity and could be direct or indirect (i.e. at a distance). It is
measure in Newton (N).

2.1 KINDS OF FORCES

1. Contact Forces: These are forces which are in direct contact with a body on which they

act e.g. force of push or pull and friction.

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2. Force Fields: these are forces that act on a body from a distance. It can be gravitational,
electrical or magnetic in nature. These forces have a space of action around them where any
body brought into such space feels the effect of the force creating the field.

2.2 PROPERTIES OF FORCES

For adequate description of a force, it must have three basic properties: (i) Magnitude: This is the

size of the force represented by a number with a unit e.g. 10N, 50N.

(ii) Direction: This is the line of action to which the force is acting (or the effect of the force can

be felt) e.g. 10N North 45o South. Diagrammatically it is shown by a line with an arrow head

pointing in the line of action.

(iii) Point of Action: This is the point on a body on which the force is acting. X10N

A force of 10N, acting on a body at point X in the direction of the east.

(iv) Sense or nature (push or pull).

2.2a REPRESENTATION OF FORCES:

Forces can be represented in two ways; Vector and Bow notation.

i. Bow notation ii. Vector

Figure 2.1: Force Representation

2.3: CLASSIFICATION OF FORCES

1. According to the effect produced by the force (external, internal, active and passive)

2. According to nature of the force (action-reaction, attraction-repulsion and tension-thrust)

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3. According to whether the force acts at a point or is distributed over a large area(concentrated
and distributed)

4. According to whether the force acts at a distance or by contact.

2.4 FORCE SYSTEMS

A force system is a collection of forces acting on a body in one or more planes.

According to the relative positions of the lines of action of the forces, the forces may be
classified

as follows :

Forces whose lines of action lie on the same plane are called coplanar forces. Spatial or space

forces have their lines of action lying in three dimensional space. Concurrent forces have their

lines of action passing through a common point, while for non concurrent forces, the lines of

action pass through different points. The following configurations are obtainable: (i) concurrent

coplanar (ii) Non concurrent coplanar and (iii) Spatial; concurrent and non concurrent.

1. Coplanar concurrent collinear force system. It is the simplest force system and

includes those forces whose vectors lie along the same straight line (refer Fig. 2.2)

Fig 2.2 Fig 2.3

2. Coplanar concurrent non-parallel force system. Forces whose lines of action pass

through a common point are called concurrent forces. In this system lines of action of all the

forces meet at a point but have different directions in the same plane as shown in Fig. 2.3.

3. Coplanar non-concurrent parallel force system. In this system, the lines of action of

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all the forces lie in the same plane and are parallel to each other but may not have same direction
as shown in Fig. 2.4.

4. Coplanar non-concurrent non-parallel force system. Such a system exists where the lines
of action of all forces lie in the same plane but do not pass through a common point. Fig. 2.5
shows such a force system.

Fig. 2.4 Fig. 2.5

5. Non-coplanar concurrent force system. This system is evident where the lines of action of all
forces do not lie in the same plane but do pass through a common point. An example of this force
system is the forces in the legs of tripod support for camera (Fig. 2.6).

Fig 2.6

6. Non-coplanar non-concurrent force system. Where the lines of action of all forces do

not lie in the same plane and do not pass through a common point, a non-coplanar non-
concurrent system is present.

2.5. FREE BODY DIAGRAMS

This is a diagram of isolated element or a portion of the body along with the net effects

of the system on it. It is useful in solving the forces and deformations of the system.

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 Fig. 2.7: Free Body Diagram

2.6. PARTICLE

A body whose dimensions are practically negligible is called a particle. Forces acting on the
particle are concurrent, the point through which they pass being the point representing the
particle.

2.7. RESULTANT FORCE

A resultant force is a single force which can replace two or more forces and produce the same

effect on the body as the forces.

2.8. COMPONENT OF A FORCE

Breaking up of a force into two parts is called the resolution of a force. The force which is
broken into two parts is called the resolved force and the parts are called component forces or
the resolutes. Generally, a force is resolved into the following two types of components :

1. Mutually perpendicular components

2. Non-perpendicular components.

1. Mutually perpendicular components. Let the force P to be resolved is represented in

magnitude and direction by oc in Fig. 2.8. Let Px is the component of force P in the direction oa

making an angle α with the direction oc of the force. Complete the rectangle oacb. Then the other

component Py at right angle to Px will be represented by ob which is also equal to ac.

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From the right-angled triangle oac

Fig 2.8 Fig 2.9

2. Non-perpendicular components. Refer Fig. 2.9. Let oc represents the given force P

in magnitude and direction to some scale. Draw oa and ob making angle α and β with oc.
Through c draw ca parallel to ob and cb parallel to oa to complete the parallelogram oacb. Then
the vectors oa and ob represent in magnitude and direction (to the same scale) the components P1

and P2 respectively.

2.9. PRINCIPLE OF RESOLVED PARTS

The principle of resolved parts states : “The sum of the resolved parts of two forces acting at a
point in any given direction is equal to the resolved parts of their resultant in that
direction”.

2.10. LAWS OF FORCES

The method of determination of the resultant of some forces acting simultaneously on a particle

is called composition of forces. The various laws used for the composition of forces are given as
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under :

1. Parallelogram law of forces

2. Triangle law of forces

3. Polygon law of forces.

1. Parallelogram law of forces. It states as under:

“If two forces, acting simultaneously on a particle, be represented in magnitude and

direction by the two adjacent sides of a parallelogram then their resultant may be
represented in magnitude and direction by the diagonal of the parallelogram which passes
through their point of intersection.”

Fig 2.10

2. Triangle law of forces. It states as under:

“If two forces acting simultaneously on a body are represented in magnitude and direction
by the two sides of triangle taken in order then their resultant may be represented in
magnitude and direction by the third side taken in opposite order.”

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3. Polygon law of forces. It states that:

“If a number of coplanar concurrent forces, acting simultaneously on a body are

represented in magnitude and direction by the sides of a polygon taken in order, then their
resultant may be represented in magnitude and direction by the closing side of a polygon,
taken in the opposite order”.

2.11. RESULTANT OF SEVERAL COPLANAR CONCURRENT FORCES

To determine the resultant of a number of coplanar concurrent forces any of the following two

methods may be used :

1. Graphical method (Polygon law of forces)

2. Analytical method (Principle of resolved parts).

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Example 2.1. Find the magnitude and direction of the resultant of two forces 40 N and 60 N

acting at a point with an included angle of 40° between them. The force of 60 N being horizontal.

Sol. Ref. fig 2.10

Example 2.2. The angle between the two forces of magnitude 20 N and 15 N is 60° ; the 20 N

force being horizontal. Determine the resultant in magnitude and direction, if

(a) the forces are pulls ; and

(b) the 15 N force is a push and 20 N force is a pull.

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 Class
work.
The

resultant of two forces P and 30 N is 40 N inclined at 60° to the 30 N force.

Find the magnitude and direction of P.

Solution

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Assignment: 1. Find the magnitude of two forces such that, if they act at right angles, their

resultant is 5 N whilst when they act at an angle of 60°, their resultant is 37 N.

2. Two forces of magnitudes 3P, 2P respectively acting at a point have a resultant R.

If the first force is doubled, the magnitude of the resultant is doubled. Find the angle between the
forces.

Example 3. The following forces (all pull) act at a point :

(i) 25 N due North ;

(ii) 10 N North-East ;

(iii) 15 N due East ;

(iv) 20 N 30° East of South ;

(v) 30 N 60° South of West.

Find the resultant force. What angle does it make with East ?

Sol. The various forces acting at a point are shown in Fig. 2.25.

45°

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3.1 MOMENTS

The tendency of forces is not only to move the body but also to rotate the body. This rotational
tendency of a force is called moment. The force multiplied by the perpendicular distance from
the point to the line of action of the force is called moment about that point. Unit of moment

is equal to the force unit multiplied by the distance unit. It is measured in Nm.

Conventional sign of moment: Clockwise moment is taken to be negative and anticlockwise

moment is taken to be positive. For systems in equilibrium; Sum of clockwise moment must be
equal to sum of anticlockwise moment about a particular point of reference.

3.2. PRINCIPLE OF MOMENTS

The principle of moments may be stated as follows :

“When a body acted upon by several forces is in rotational equilibrium, the sum of the clockwise

moments of the forces about any point is equal to sum of the anti-clockwise moments of the
forces about the same point.”

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 Fig. 3.1

The moments of the forces P1, P2, P3 and P4 about O are given as under :

Moment of force P1 = P1× l1 clockwise (–)

Moment of force P2 = P2 × l2 anti-clockwise (+)

Moment of force P3 = P3 × l3 clockwise (–)

Moment of force P4 = P4 × l4 anti-clockwise (+)

The resultant of these moments will be equal to the algebraic sum of all the moments about O.

∴ The resultant moment = – P1l1 + P2l2 – P3l3 + P4l4

Since the body is in rotational equilibrium, according to the principle of moments the resultant

moment should be zero.

∴– P1l1 + P2l2 – P3l3 + P4l4 = 0orP1l1 + P3l3 = P2l2+ P4l4 or

Sum of anti-clockwise moments = sum of clockwise moments

This verifies the principle of moments.

3.4. EQUILIBRIUM CONDITIONS FOR BODIES UNDER COPLANAR

NONCONCURRENT FORCES

When a body is under the action of a coplanar non-concurrent force system it may rotate due

to resultant moment of the force system or it may set in a horizontal or vertical motion due
tohorizontal and vertical components of forces. The body, thus can only be in equilibrium if the
algebraic sum of all the external forces and their moments about any point in their plane is zero.

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Mathematically, the conditions of equilibrium may be expressed as follows :

1. ΣH = 0 (ΣH means sum of all the horizontal forces)

2. ΣV = 0 (ΣV means sum of all the vertical forces)

3. ΣM = 0 (ΣM means sum of all the moments)

3.5 METHOD OF DETERMINING THE MOMENT OF A FORCE

Depending on the nature of application of a force; there are three methods that can be used to

determine the moments of a force (1) by direct method: Using the definition that moment is force

multiplied by perpendicular distance.

(2) By Varignon’s Theorem: This theorem states that the moment of a force about any point is

equal to the sum of the moments produced by the rectangular components of the force.

For this theorem to be applied adequately; the forces concerned will first have to be resolved to

their rectangular components after which each moment due to individual component is

determined and then summed in accordance with the sign convention of moment.

(3) By the Principle of Transmissibility: This principle is applied by moving a force from its

point of application to any convenient point ( for ease of calculation) along its line of action.

WORKED EXAMPLE

Example 1: A man and a boy carry a weight of 300 N between them by means of uniform

pole 2 m long and weighing 100 N. Where must the weight be placed so that the man may bear
twice as much of the weight as that of the boy?

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 Figure 3.2

Sol. Refer to Fig. 3.2. The weight of the pole acts at 300 N

The weight of the pole acts at its centre G. Let boy bears weight W and the man bears 2W.

When the weight of 300 N acts at a distance x metres from the man.

ΣV = 0

W + 2W = 300 + 100 = 400

W = 133.3 N

Taking moments about A (boy)

2W × 2 = 300 (2 – x) + 100 × 1

4W = 600 – 300x + 100

4 × 133.3 = 700 – 300x

533.2 = 700 – 300x

300x = 166.8

∴ x = 0.556 m. (Ans.)

Example 2: The lever of a safety valve weighs 40 N and its c.g. lies at a distance 9 cm from

the centre of the valve which is at 8 cm from the fulcrum. If the diameter and weight of the valve
are respectively 7 cm and 60 N find the minimum weight that will be hung to the end of 80 cm
long lever to conserve steam in the boiler at a pressure of 100 N/cm2.

Fig. 3.3

Solution:

Area of the valve, A = πd2/4, where d is the diameter of the valve= π42× 7 = 38.48 cm2

Let W be the weight attached to the end A.

Taking moments about the fulcrum F, we get

W × 80 + 40 × (9 + 8) + 60 × 8 = p × A × 8

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(p = pressure of steam)

W × 80 + 680 + 480 = 100 × 38.48 × 8

∴ W = 370.3 N. (Ans.)

Class work:

A beam simply supported at both the ends carries load system as shown in Fig. 3.4. Find the
reactions at the two ends.

Solution:

Fig.3.4

Sol. Refer to Fig. 3.4. Since, all the loads acting on the beam are vertically downwards therefore,
the reactions at the ends shall be vertically upwards. Let RA and RB be the reactions at the ends
A and B respectively. Since, the beam AB is in equilibrium,

∴ ΣH = 0, ΣV = 0 and ΣM = 0.

There is no horizontal force.

There is no horizontal force.

∴ Considering vertical equilibrium of the beam (ΣV = 0)

RA+ RB= 2 × 4 + 6 + 2 or

RA + RB = 16

Taking moments about A(ΣM = 0), we get

RB × 8 = 2 × 4 × 42 + 6 × 4 + 2 × 6

= 16 + 24 + 12 = 52

∴ RB = 6.5 kN. (Ans.)

Substituting this value of RB

in eqn. (i), we get

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RA+ 6.5 = 16

or

RA = 9.5 kN. (Ans.)

3.6. COUPLE

A couple is pair of two equal and opposite forces acting on a body in a such a way that the lines
of action of the two forces are not in the same straight line. The effect of a couple acting on a rigid
body is to rotate it without moving it as a whole. The movement of the whole body is not possible
because the resultant force is zero in the case of forces forming a couple. The perpendicular
distance between the lines of action of two forces forming the couple is called the arm of couple.
Thus, in Fig. 3.2 two equal forces of magnitude P and acting at points A and B in the opposite
direction form a couple with AB as arm of the couple. The moment of a couple is known as torque
which is equal to one of the forces forming the couple multiplied by arm of the couple.

Fig 3.2

The following are the examples of couples in every day life.

1. Opening or closing a water tap. The two forces constitute a couple

2. Turning the cap of a pen.

3. Unscrewing the cap of an ink bottle.

4. Twisting a screw driver.

5. Steering a motor-car .

6. Winding a watch or clock with a key.

3.7. PROPERTIES OF A COUPLE

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1. The algebraic sum of the moments of the forces forming a couple about any point in their
plane is constant. Let two parallel and unlike forces be of magnitude P each forming a couple P ×
AB where points A and B are the points where forces P and P act. See Fig.3.7

Figure; 3.7

2. Any two couples of equal moments and sense, in the same plane are equivalent in their effect.

3. Two couples acting in one place upon a rigid body whose moments are equal but opposite in
sense, balance each other.

4. A force acting on a rigid body can be replaced by an equal like force acting at any other point
and a couple whose moment equals the moment of the force about the point where the equal like
force is acting.

5. Any number of coplanar couples are equivalent to a single couple whose moment is equal to
the algebraic sum of the moments of the individual couples.

3.8. ENGINEERING APPLICATIONS OF MOMENTS

Some of the important engineering applications of moments are discussed below :

1. The levers (simple curved, bent or cranked and compound levers).

2. The balance.

3. The common steel yard.

4. Lever safety valve.

3.8.1. The Lever. The lever is defined as a rigid bar, straight or curved which can turn

about a fixed point called the ‘fulcrum’. It works on the principle of moments i.e., when the lever
is in equilibrium the algebraic sum of the moments, about the fulcrum, of the forces acting on it

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is zero. The principle of lever was first developed by Archimedes. Some common examples of
the use of lever are :

(i) Crow bar ;

(ii) A pair of scissors ;

(iii) Fire tongs etc.

Power arm. The perpendicular distance between the fulcrum and the line of action of forces is
known as power arm.Weight arm. The perpendicular distance between the fulcrum and the
point where the load/weight acts is called weight (or load) arm.

Fig. 3.8

The principle of moments is applicable when the lever

Mechanical advantage of the lever= W/P=a/b=power arm/weight arm

Power × power arm = weight × weight arm.

The levers may be of the following types :

(a) Simple levers

(b) Curved bent or cranked levers

(c) Compound levers.

(a) Simple levers. A simple lever is one which has only one fulcrum. These levers consist of

only one power arm and one weight arm.

Depending upon the position of their fulcrum, load and effort, simple levers are further divided

into the following three types :

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Type 1. Fulcrum between W and P

Reaction R on the fulcrum = P + W

Type 2. Weight between F and P

Reaction R on the fulcrum = W – P

Type 3. Power between W and F

Reaction R on the fulcrum = P – W

Fig.3.9

WORKED EXAMPLE

Fig. 3.10 shows a bell crank lever. Determine the magnitude of the load lifted perpendicular to
the weight arm FB, 5 cm long when a force of 100 N is applied at A perpendicular to power arm
FA, 20 cm long.

Fig. 3.10

Solution:

Sol. Taking moments about F,

100 × 20 = W × 5

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∴W = 400 N. (Ans.)

Ex 2:Four forces equal to 10 N, 20 N, 30 N and 40 N are respectively acting along the four sides
(1 m each) of a square ABCD, taken in order. Find the magnitude, direction and position of the
resultant force. See fig. 3.11

Fig. 3.11

Solution:

Magnitude of the resultant force, R = ?

Resolving all the forces horizontally,

ΣH = 10 – 30 = – 20 N

and resolving all the forces vertically,

ΣV = 20 – 40 = – 20 N

Now, resultant force R = 28.28 N

i.e., R = 28.28 N. (Ans.)

Direction of the resultant force ?

Let θ = angle which the resultant makes with the horizontal

∴ tan θ = ΣV/ΣH=-20/-20 = 1

Or θ = 45°

Since ΣH as well ΣV is –ve, therefore θ lies between 180° and 270°

∴ Actual θ =180 + 45° = 225°. (Ans.)

Position of the resultant force = ?

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Let x = perpendicular distance between A and the line of the resultant force.

Now taking moments about A, we get

ΣMA: 28.28 × x = 10 × 0 + 20 × 1 + 30 × 1 + 40 × 0 = 50

x = 50/ 28.28

= 1.768 m. (Ans.)

4.0 FRICTION

4.1. CONCEPT OF FRICTION

Force of friction or frictional force may be defined as the opposing force which is called

into play in between the surfaces of contact of two bodies, when one body moves over the
surface of another body. (Fig. 4.1).

Fig. 4.1

There are appliances and devices known as friction devices such as belts and ropes, friction

clutches, jib and cotter joints, brakes, nuts and bolts, in which friction is desirable and efforts are

made to maximise it.

4.2. CHARACTERISTICS OF FRICTIONAL FORCE

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The force of friction or frictional force entails the following characteristics :

(i) It is self-adjusting. As tractive force P increases, the frictional force F also increases, and

at any instant only as much frictional force comes into play as is necessary to prevent the motion.

(ii) It always acts in a direction opposite to the motion (i.e., always opposes the tractive force).

(iii) It is a passive force (since it exists only if the tractive force P exists).

6.3. TYPES OF FRICTION

Friction may be classified as follows :

1. Friction in unlubricated surfaces

2. Friction in lubricated surfaces.

Friction in unlubricated surfaces. The friction that exists between two unlubricated surfaces

is called solid friction or dry friction.

It may be of the following two types :

(i) Sliding friction

(ii) Rolling friction.

The friction that exists when one surface slides over the other is called sliding friction.

The friction that exists between two surfaces separated by balls or rollers, is called the rolling

friction. It may be remembered that rolling friction is always less than the sliding friction.

Friction in lubricated surfaces. It may further be subdivided as follows :

(i) Boundry (or greasy or non-viscous) friction

(ii) Viscous friction.

If in between two rubbing surface there exists a thin film or layer of an oil or a lubricant, the

oil gets absorbed in the surfaces and such a film is known as absorbed film. Instead of metal to
metal contact of the surfaces, there is a contact between thin layer of the oil and obviously the
frictional force is reduced. In such a case the frictional force is known as boundary friction. In
this chapter we shall only deal with friction between unlubricated parts.

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4.4. STATIC AND DYNAMIC FRICTION

Static friction. The static friction is the friction offered by the surfaces subjected to external

forces until there is no motion between them. Dynamic friction. The dynamic friction is the
friction experienced by a body when it is in motion. It is also known as kinetic friction and is
always less than static friction (the kinetic friction is about 40 to 75 per cent of the limiting static
friction).

4.5. LIMITING FRICTION

Hence, limiting force of friction may be defined as the maximum value of friction force which

exists when a body just begins to slide over the surface of the other body. When the applied force
or tractive force P is less than the limiting friction, the body remains at rest, and the friction is
called static friction, which may have any value between zero and limiting friction.

4.6. LAWS OF FRICTION

4.6.1. Law of static friction

The laws of static friction are as follows :

1. The frictional force always acts in a direction opposite to that in which the body tends to

move.

2. The frictional force is directly proportional to the normal reaction between the surfaces.

3. The frictional force depends upon the nature of surfaces in contact.

4. The frictional force is independent of the area and shape of the contacting surfaces.

4.6.2. Laws of dynamic or kinetic friction

1. The frictional force always acts in a direction opposite to that in which the body moves.

2. The frictional force is directly proportional to the normal reaction between the two

contacting surfaces.

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3. The magnitude of force of dynamic friction bears a constant ratio to the normal reaction

between two surfaces but the ratio is slightly less than that in case of limiting friction.

4. The frictional force remains constant for moderate speeds but it decreases slightly with the

increase of speed.

4.7. ANGLE OF FRICTION

Refer to Fig. 4.2. It is defined as the angle which the resultant (R) of normal reaction (N) and

limiting force of friction (F) makes with the normal (N). It is denoted by φ.

4.8. CO-EFFICIENT OF FRICTION

It is defined as the ratio of limiting force of friction to the normal reaction between the two

bodies. It is denoted by µ.

4.9. ANGLE OF REPOSE

Refer to Fig. 4.9. Consider a body of weight W acting on a rough horizontal plane inclined at
angle α. The body is in equilibrium under the action of the following forces:

Fig. 4.9

(i) Weight, W (which may be resolved into

two components

W sin α and W cos α as shown in Fig. 4.9)

(ii) Normal reaction, N and

(iii) Frictional force, F (= µN).

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In the limiting condition when the block is about to slide down the inclined plane, the frictional
force must act up the plane and for equilibrium ; considering the forces along and perpendicular
to the plane.

F = W sin α

N = W cos α

F/N= Wsin α/Wcosα = tan α

F/N=µ = tan φ

is the angle of friction.

The angle α is called angle of repose and is equal to the angle of friction when the body is in the

condition of limiting equilibrium on an inclined plane.

4.10. CONE OF FRICTION

If the line OA of Fig. 4.10 making the maximum angle of friction φ with the normal is revolved
about OB as an axis, the cone generated is called the cone of friction.

Fig. 4.10

4.11. MOTION OF BODY ON HORIZONTAL PLANE

Fig. 4.11 shows a body lying on a horizontal plane under the influence of force P which is
inclined at angle θ to the surface (or horizontal plane). The value of force P can be determined by
considering the limiting equilibrium.

Resolving the forces parallel to the plane (i.e.,

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 Fig. 4.11

horizontally), we get

F = P cos θ or F =µN = P cos θ...(i)

Resolving the forces perpendicular to the plane (i.e.,

vertically), we get

N + P sin θ = W

N = W – P sin θ

P= W sinφ /cos (θ- φ)

4.12. MOTION UP AN INCLINED PLANE

Fig. 4.12 shows a body lying on an inclined plane under the influence of a force P. Let W be
weight of the body, α be the inclination of the plane to the horizontal and µ be the co-efficient of
friction. The value of P can be determined by considering the limiting equilibrium as follows :

Case I. When angle of inclination of the force to the plane is θ :

Case II. When the force is parallel to the plane :

θ=0

Case III. When there is no force of friction :

µ=0

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∴ tan φ = 0

φ=0

Example 4.1. A pull of 25 N at 30° to the horizontal is necessary

to move a block of wood on a

horizontal table. If the co-efficient of friction between the bodies in contact is 0.2, what is the
weight of the block ?

Sol. Let W = weight of the body

P = effort applied (= 25 N)

N = normal reaction

µ = co-efficient of friction (= 0.2)

Refer to Fig. 4.12

Fig. 4.12

Resolving forces parallel to the plane,

F = P cos 30°or µN = P cos 30° ...(i) (∵ F = µN)

and resolving forces perpendicular to the plane,

N + P sin 30° = W

Or N = W – P sin 30°...(ii)

Substituting the value of N in eqn. (i), we get

µ (W – P sin 30°) = P cos 30°or

0.2 (W – 25 × 0.5) = 25 × 0.866 or 0.2 (W – 12.5) = 21.65

or W = 21.65/05 + 12.5

W = 120.75 N (newtons). (Ans.)

29
Class work: A body resting on a rough horizontal plane required a pull of 18 N inclined at 30° to
the plane just to move it. It was found that a push of 20 N inclined at 30° to the plane just moved
the body. Determine the weight of the body and co-efficient of friction.

Example 4.2: An object of weight 100 N is kept in position on a plane inclined at 30° to the

horizontal by the applied force P. The co-efficient of friction of the surface of the inclined plane
in 0.25. Determine the minimum magnitude of the force P.

Sol. Fig. 4.13 shows the various force acting on the object.

Fig. 4.13

Presuming that P is sufficient to hold the block W on the inclined plane, we have

Resolving all the forces parallel and perpendicular to the plane,

µN + P cos 30° = W sin α 0.25 N + P cos 30° = W sin 30°...(i)

and N = W cos α + P sin 30° or N = W cos 30° + P sin 30°...(ii)

From (i) and (ii), we get

0.25 (W cos 30° + P sin 30°) + P cos 30° = W sin 30°

0.25 (100 cos 30° + P sin 30°) + P cos 30° = 100 sin 30°

0.25 (86.67 + 0.5 P) + 0.867 P = 50

21.67 + 0.125 P + 0.867 P = 50

0.992 P = 28.33 N (newton)

P = 28.56 N (newtons). (Ans)

4.13. SCREW FRICTION

The screw friction is the friction experienced by screw threads made by cutting a continuous
helical groove on a cylindrical surfaces such as screws, bolts, nuts, studs etc. These parts are
widely used in various machines and structures for fastening. The screw threads may be of

30
 (i) V shape and (ii) square shape.

The V threads are stronger and offer more frictional force to motion than square threads. V-
threads are used for lighter load and square threads are used for heavier loads.

Square threads are used in screw jacks, vice screws etc. Screw jack is a device used for raising
/lifting heavy loads by applying a small effort at its handle. It works on the principle of an
inclined plane. See Figure 4.13

Fig. 4.13

(i) Effort required to lift the load

Let P = effort required to lift the load when applied at mean radius ;

W = weight of the body to the lifted ; and

µ = co-efficient of friction between the screw and the nut.

Resolving the forces parallel to the plane,

P cos α = W sin α + F or P cos α = W sin α + µN...(1)

and resolving the forces perpendicular to the plane,

N = P sin α + W cos ...(2)

tan α =P/(πdm)

(ii) Effort required to lower the load.

31
 Resolving perpendicular and parallel to the plane,

ii. Efficiency of a screw jack. We know that effort P required at the mean radius of a
screw jack to lift the load W is given by, P = W tan (α + φ). If there is no friction,
φ=0

Example 4.13. A screw jack carries a load of 4 kN. It has a square threaded single start screw of
20 mm pitch and 50 mm mean diameter. The co-efficient of friction between the screw and its
nut is 0.29. Calculate the torque required to raise the load and efficiency of the screw. What is
the torque required to lower the load ?

Sol. Load,

W = 4 kN

Mean diameter,dm = 50 mm = 5 cm

Pitch,p = 20 mm = 2 cm

Co-efficient of friction, µ = 0.29

Torque required to raise the load,

32
5.0

CENTROID

Centre of gravity of the body may be defined as the point through which the whole weight of a
body may be assumed to act. The centre of gravity of a body or an object is usually denoted by
c.g. or simply by G. The position of c.g. depends upon shape of the body and this may or may
not necessarily be within the boundary of the body.

The centroid or centre of area is defined as the point where the whole area of the figure is
assumed to be concentrated. Thus, centroid can be taken as quite analogous to centre of gravity
when bodies have area only and not weight.

5.1. POSITIONS OF CENTROIDS OF PLANE GEOMETRICAL FIGURES

Table 5.1 gives the positions of centroids of some plane geometrical figures.

Table 5.1 Centroids of Plane Geometrical Figures

33
Table 5.2 gives the positions of centre of gravity of regular solids.

34
5.3. (a) CENTROIDS OF COMPOSITE AREAS

The location of the centroid of a plane figure can be thought of as the average distance of the
area to an axis. Usually the axes involved will be the X and Y-axes. In determining the location
of the centroid it is found advantageous to place the X-axis through the lowest point and the Y-
axis through the left edge of the figure. This places the plane area entirely within the first
quadrant where x and y distances are positive (Fig. 5.2). Then divide the area into simple areas
such as rectangles, triangles, etc. (Fig. 5.3).

35
Fig. 5.2; Centroid of a composite area. Fig 5.3:Composite area divided into simple areas.

Take the moment of each single area about the Y-axis. Sum up the moments about the Y-axis.
Since the centroid of the composite figure is the point at which the entire area about the Y-axis
must be equal to the moments of its component parts about the Y-axis, therefore

(a + a + .... + a) x = ax + ax + ...... + ax

Note. If a hole exists in the plane figure, treat it as a negative area. The moment of a negative
area will be negative provided that the entire figure lies in the first quadrant.

5.3. (b) CENTRE OF GRAVITY OF SIMPLE SOLIDS

The weight of the body is a force acting at its own centre of gravity and directed towards the

centre of the earth. The position of the centres of bodies weighing W1, W2, W3 etc. is found in
the same manner as the resultant of parallel forces.

36
Example 5.1. Find out the position of the centroid of L section as shown in Fig. 5.40.

Fig. 5.40

Sol. Refer to Fig. 5.40.

Divide the composite figure into two simple areas :

(i) Rectangle (16 cm × 4 cm)

(ii) Rectangle (8 cm × 4 cm)

To determine the location of the centroid of the plane figure we have the following table :

x bar = Σax/Σa

=384/96

= 4 cm. (Ans.)

y bar = Σay/Σa

=576/96

= 6 cm. (Ans.)

37
Example 5.2. Determine the location of the centroid of the plane figure shown in Fig. 5.5.

Solution

Divide the composite figure into three simple areas :

(i) a rectangle (7 cm × 5 cm)...(1)

(ii) a quadrant (2 cm radius)...(2)

(iii) a circle (3 cm dia.)...(3)

Fig. 5.5

The rectangle is a positive area. The quadrant and hole are treated as negative areas.

To determine the location of the centroid of the plane figure, we have the following table:

6.0

Moment of Inertia (First)

The cumulative product of area and square of its distance from an axis is called the moment of

inertia of a section about that axis. It can be expressed as

38
where Ix = moment of inertia (M.O.I.) of the section about the x-axis, and

y = the distance of infinitesimal area da from the x-axis as shown in Fig. 6.1

Fig. 6.1

Similarly, the moment of inertia of a section about the y-axis is given by

The cumulative product of area and square of its distance from a point is known as the polar

moment of inertia. It is given by

39
 Table 6.1. Moments of Inertia for Simple Areas

6.1 : SECOND MOMENT OF INERTIA

6.2 THEOREM OF PARALLEL AXES

The theorem of parallel axes states as follows :

“The moment of inertia of a lamina about any axis in the plane of the lamina equals the sum

of moment of inertia about a parallel centroidal axis in the plane of lamina and the product of the

area of the lamina and square of the distance between the two axes.”

40
 Ilm = Ixx+ Ah2

6.3. THEOREM OF PERPENDICULAR AXES

The theorem of perpendicular axes states as follows :

“If Iox and Ioy be the moments of inertia of a lamina about mutually perpendicular axes OX

and OY in the plane (see fig 6.2) of the lamina and Ioz be the moment of inertia of the lamina
about an axis (oz) normal to the lamina and passing through the point of intersection of the axes
OX and OY,

then Ioz = Iox + Ioy

Fig 6.2

6.4. RADIUS OF GYRATION OF THE SECTION

One of the properties of cross-section which influence the structural behaviour of the members

is radius of gyration.

where Ii= moment of inertia about ith axis ; and

ki= radius of gyration of area about ith axis.

Members, when subjected to axial forces tend to buckle. The load at which members will

buckle is proportional to the square of the radius of the gyration. The radius of gyration is usually

referred to with respect to centroidal axes system of the reaction.

Example 6.1:Find the moment of inertia about the centroidal axes XX and YY of the section

shown in Fig. 6.2

41
 Fig. 6.2

Solution

42
Second Moment of Area

Example 6.2. For the shaded area shown in Fig. 6.3, find the following :

(i) The position of the centroid

(ii) The second moment of area about the base

(iii) The radius of gyration about the base.

43
 Fig 6.3

Sol.

(i) The position of the centroid :

To determine the location of the centroid of the shaded area we have the table given below :

(ii) The second moment of area (M.O.I.) about the base, ILL

44
45