COMPUTATIONAL

ENGINEERING MECHANICS

FRICTION NOTES
UNIT-III COMPUTATIONAL ENGINEERING MECHANICS

UNIT – III
FRICTION
INTRODUCTION
It has been established since long that all surfaces of the bodies are never perfectly smooth.
It has been observed that whenever, even a very smooth surface is viewed under a microscope, it
is found to have some roughness and irregularities, which may not be detected by an ordinary
touch.
It will be interesting to know that if a block of one substance is placed over the level surface
of the same or different material, a certain degree of interlocking of the minutely projecting
particles takes place. This does not involve any force, so long as the block does not move or tends
to move. But whenever one of the blocks moves or tends to move tangentially with respect to the
surface, on which it rests, the interlocking property of the projecting particles opposes the
motion.
This opposing force, which acts in the opposite direction of the movement of the block, is called
force of friction or simply friction. Frictional force always acts tangentially at points of contact.

1.1. Classification of Friction

The friction may be classified in two main categories:
1. Statics Friction and 2. Kinetic Friction (Dynamic Friction)

1.1.1. Statics Friction

It is the friction experienced by a body when it is at rest. Or in other words, it is the friction
when the body tends to move.

1.1.2. Kinetic Friction (Dynamic Friction)

It is the friction experienced by a body when it is in motion. It is also called kinetic friction.
The dynamic friction is of the following two types:
i) Sliding friction: It is the friction, experienced by a body when it slides over another
body.
ii) Rolling friction: It is the friction, experienced by a body when it rolls over another
body.

1.2. Theory of Friction

To understand concept of theory of friction let us consider a block of weight W pulled
horizontally by a force P on a rough horizontal surface.
When a force P is increasing from zero, the frictional force F also goes on increasing to
maintain equilibrium.
At one stage, block is on the verge of sliding and frictional force F attains its maximum
value. (Limiting Friction)
Any further increase in applied force P, causes motion and value of F decreases rapidly to
a kinetic value.

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1.3. LIMITING FRICTION

It has been observed that when a body, lying over another body, is gently pushed, it does
not move because of the frictional force, which prevents the motion. It shows that the force of the
hand is being exactly balanced by the force of friction, acting in the opposite direction. If we again
push the body, a little harder, it is still found to be in equilibrium. It shows that the force of friction
has increased itself so as to become equal and opposite of the applied force. Thus the force of
friction has a remarkable property of adjusting its magnitude, so as to become exactly equal and
opposite to the applied force, which tends to produce motion.
There is, however, a limit beyond which the force of friction cannot increase. If the applied
force exceeds this limit, the force of friction cannot balance it and the body begins to move, in the
direction of the applied force. This maximum value of frictional force, which comes into play, when
a body just begins to slide over the surface of the other body, is known as limiting friction. It may be
noted that:

i) When the applied force 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.
Applied Force (P) < Frictional Force (Fr) ……….. (Stable Equilibrium)
ii) When the applied force is equal to the limiting friction, the body is on the verge of
motion (tends to move), and the friction is called limiting friction. Body is said to
have impending motion.
Applied Force (P) ≈ Frictional Force (Fr) ……….. (Limiting Equilibrium)
iii) When the applied force is greater than the limiting friction, the body starts moving,
and the friction is called kinetic friction (dynamic friction)
Applied Force (P) > Frictional Force (Fr) ……….. (Unstable Equilibrium)

1.4. Normal Reaction

It has been experienced that whenever a body, lying on a horizontal or an inclined surface,
is in equilibrium, its weight acts vertically downwards through its center of gravity. The surface,
in turn, exerts an upward reaction on the body. This reaction, which is taken to act perpendicular
to the plane, is called normal reaction and is, generally, denoted by R. It will be interesting to know
that the term ‘normal reaction’ is very important in the field of friction, as the force of friction is
directly proportional to it.

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1.5. A Block on Rough Surface

As we know that when a block is placed on a smooth surface, the normal reaction (N) or
(R) is perpendicular to surface but if it is placed on a rough surface, the reaction is not acting
normal to the surface. The reaction which is actually inclined to the surface can be resolved into
two components:
i) Along the surface ii) Normal to the surface

The component acting along the surface (tangential) is frictional force (Fr) and the
component normal to the surface is called normal reaction (N) or (R).

1.6. Coefficient of Friction (µ)

1.6.1. Coefficient of Static Friction (µs)

The magnitude of limiting static frictional force (Fs) is directly proportional to the
magnitude of normal reaction (R).
The relation can be expressed as,
Fs α R
∴ Fs = µ s × R
Where constant of proportionality is called coefficient of static friction and is denoted by “µs”
𝐅𝐬
∴ µs = ……. For Impending motion
𝐑
1.6.2. Coefficient of Kinetic Friction (µk)
When magnitude of P acting on a block is increased and if it becomes greater than
maximum friction force, the motion starts and magnitude of ‘F’ drops to a smaller value Fk known
as kinetic friction force.
The magnitude of this kinetic frictional force (Fk) is directly proportional to the magnitude
of normal reaction (R).
The relation can be expressed as,
Fk α R
∴ Fk = µk × R
Where constant of proportionality is called coefficient of kinetic friction and is denoted by
“µk”
𝐅𝐤
∴ µk = ……. For Actual motion
𝐑

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1.7. Angle of Friction (Φ)

1.7.1. Angle of Static Friction (Φs)

For impending motion the angle between normal reaction and resultant reaction is called
as angle of static friction.

By geometry, we have
Fs
tan Φs = ……. (refer fig.)
R
But We know,
Fs
= µs
R
So we can relate
µs = tan Φs

∴ Angle of Static Friction (Φs) = tan-1 (µs)

1.7.2. Angle of Kinetic Friction (Φk)

For actual motion when P = Fk, the angle between normal reaction and resultant reaction
is called as angle of kinetic friction.
∴ Angle of Kinetic Friction (Φk) = tan-1 (µk)

1.8. Angle of Repose (α)

It is defined as angle of inclined plane with horizontal at which body is just on verge of
sliding (impending motion) and so Fs = µs × R. Angle of Repose is denoted by ‘α’

1.9. Relation between Angle of Friction (Φ) and Angle of Repose (α)
To derive this relation consider F.B.D. of block of weight W on an inclined plane.

ΣFX = 0
µs . R - W sin α = 0

∴µs . R = W sin α ----eqn (i)

ΣFy = 0
R - W cos α = 0
∴ R = W cos α ----eqn (ii)
Put value of R in eqn (i)

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We get,
µs . W cos α = W sin α
∴µs = tan α
But we know that,
µs = tan Φs
∴tan Φs = tan α
∴ Φs = α
Therefore, Angle of Static Friction (Φs) and Angle of Repose (α) are numerically
equal.

1.10. Cone of Friction

It is an imaginary cone generated by revolving resultant reaction (R R) about the normal

reaction (R)

1.10.1. Properties of Cone of Friction

i) The radius of this cone is represented by frictional force (Fr)

ii) The semi-cone angle represents angle of friction (Φ)
iii) For co-planer forces, in order for motion not to occur the reaction R must be within
angle AOC
iv) For non-coplanar forces it must be within the cone of friction.

1.11. Laws of Friction (Coulomb’s Laws of Dry Friction)

Prof. Coulomb, after extensive experiments, gave some laws of friction, which may be
grouped under the following heads:
a) Laws of static friction, and
b) Laws of kinetic or dynamic friction.

a) Laws of Static Friction

Following are the laws of static friction:
i) The force of friction always acts in a direction, opposite to that in which the body tends to
move, if the force of friction would have been absent.

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ii) The magnitude of the force of friction is exactly equal to the force, which tends to move
the body.
iii) The magnitude of the limiting friction bears a constant ratio to the normal reaction
between the two surfaces. Mathematically:
Fmax
= constant (i.e. µs)
R
Where, Fmax = Limiting friction, and
R = Normal reaction.
iv) The force of friction is independent of the area of contact between the two surfaces.
v) The force of friction depends upon the roughness of the surfaces.

b) Laws of Kinetic Friction or Dynamic Friction

Following are the laws of kinetic or dynamic friction:
i) The force of friction always acts in a direction, opposite to that in which the body is
moving.
ii) The magnitude of kinetic friction bears a constant ratio to the normal reaction
between the two surfaces. But this ratio is slightly less than that in case of limiting
friction.
Fk
= constant (i.e. µk)
R
*** but the value of µk is always less than value of µs.
∴ µk < µs

iii) For moderate speeds, the force of friction remains constant. But it decreases slightly
with the increase of speed.

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BELT FRICTION (BELT DRIVES)

INTRODUCTION
The belts are used to transmit power from one shaft to another by means of pulleys which
rotate at the same speed or at different speeds. The amount of power transmitted depends upon
the following factors:
1. The velocity of the belt.
2. The tension under which the belt is placed on the pulleys.
3. The arc of contact between the belt and the smaller pulley.
4. The conditions under which the belt is used.
It may be noted that:
a) The shafts should be properly in line to insure uniform tension across the belt section.
b) The pulleys should not be too close together, in order that the arc of contact on the smaller
pulley may be as large as possible.
c) The pulleys should not be so far apart as to cause the belt to weigh heavily on the shafts,
thus increasing the friction load on the bearings.
d) A long belt tends to swing from side to side, causing the belt to run out of the pulleys,
which in turn develops crooked spots in the belt.
e) The tight side of the belt should be at the bottom, so that whatever sag is present on the
loose side will increase the arc of contact at the pulleys.
f) In order to obtain good results with flat belts, the maximum distance between the shafts
should not exceed 10 meters and the minimum should not be less than 3.5 times the
diameter of the larger pulley.

1.1. Types of Belts

Though there are many types of belts used these days, yet the following are important
from the subject point of view:
1. Flat Belt: The flat belt as shown in Fig. (a), is mostly used in the factories and workshops,
where a moderate amount of power is to be transmitted, from one pulley to another when
the two pulleys are not more than 8 meters apart.
2. V- Belt: The V-belt as shown in Fig. (b), is mostly used in the factories and workshops,
where a great amount of power is to be transmitted, from one pulley to another, when the
two pulleys are very near to each other.
3. Circular Belt or Rope: The circular belt or rope as shown in Fig. (c) is mostly used in the
factories and workshops, where a great amount of power is to be transmitted, from one
pulley to another, when the two pulleys are more than 8 meters apart.

If a huge amount of power is to be transmitted, then a single belt may not be sufficient. In
such a case, wide pulleys (for V-belts or circular belts) with a number of grooves are used. Then

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a belt in each groove is provided to transmit the required amount of power from one pulley to
another.

1.2. Types of Flat Belt Drives

The power from one pulley to another may be transmitted by any of the following types
of belt drives.
1. Open Belt Drive: The open belt drive, as shown in Fig., is used with shafts arranged
parallel and rotating in the same direction. In this case, the driver A pulls the belt from
one side (i.e. lower side RQ) and delivers it to the other side (i.e. upper side LM). Thus the
tension in the lower side belt will be more than that in the upper side belt. The lower side
belt (because of more tension) is known as tight side whereas the upper side belt (because
of less tension) is known as slack side, as shown in Fig.

2. Crossed Belt Drive: Crossed or twist belt drive. The crossed or twist belt drive, as shown
in Fig., is used with shafts arranged parallel and rotating in the opposite directions. In this
case, the driver pulls the belt from one side (i.e. RQ) and delivers it to the other side (i.e.
LM). Thus, the tension in the belt RQ will be more than that in the belt LM. The belt RQ
(because of more tension) is known as tight side, whereas the belt LM (because of less
tension) is known as slack side, as shown in Fig.

A little consideration will show that at a point where the belt crosses, it rubs against each
other and there will be excessive wear and tear. In order to avoid this, the shafts should be placed
at a maximum distance of 20 b, where b is the width of belt and the speed of the belt should be
less than 15 m/s.

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1.3. Velocity Ratio of a Belt Drive

It is the ratio between the velocities of the driver and the follower or driven. It may be
expressed, mathematically, as discussed below:
Let, d1 = Diameter of the driver,
d2 = Diameter of the follower,
N1 = Speed of the driver in r.p.m.,
N2 = Speed of the follower in r.p.m.,
∴ Length of the belt that passes over the driver, in one minute
= π d1 N1
Similarly, length of the belt that passes over the follower, in one minute
= π d2 N2
Since the length of belt that passes over the driver in one minute is equal to the length of
belt that passes over the follower in one minute, therefore
∵ π d1 N1 = π d 2 N2
𝐍𝟐 𝐝𝟏
As velocity ration is =
𝐍𝟏 𝐝𝟐
The above relation shows that speed of the pulley is inversely proportional to the
diameter of the pulley.
Also, When thickness of the belt (t) is considered, then velocity ratio,
𝐍𝟐 𝐝𝟏 +𝐭
=
𝐍𝟏 𝐝𝟐 +𝐭
Notes: 1. The velocity ratio of a belt drive may also be obtained as discussed below:
We know that the peripheral velocity of the belt on the driving pulley,
πd1 N1
𝑣1 = m/s
60
and peripheral velocity of the belt on the driven pulley,
πd2 N2
𝑣2 = m/s
60
When there is no slip, then ν1 = ν2.
πd1 N1 πd2 N2 N2 d1
∴ =  =
60 60 N1 d2
2. In case of a compound belt drive as shown in Fig., the velocity ratio is given by
Speed of last driven Product of diameters of drivers 𝐍𝟒 𝐝𝟏 𝐱 𝐝𝟑
= 
𝐍𝟏
= 𝐝𝟐 𝐱 𝐝𝟒
Speed of first driver Product of diameters of drivens

1.4. Length of the Belt Drives

Though there are many types of belts used these days, yet the following are important
from the subject point of view:

i) Length of Open-Belt Drive: Total length of the belt is the sum of three parts,
1. The length of belt in contact with the driver (larger) pulley (APB),
2. The length of belt in contact with the driven (smaller) pulley (CQD), and
3. The length of belt NOT in contact with the driver and driven pulley (BC + AD)

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ii) Length of Cross-Belt Drives: Total length of the belt is sum of three parts :
1. The length of belt in contact with the driver (larger) pulley (APB),
2. The length of belt in contact with the driven (smaller) pulley (CQD) and
3. The length of belt NOT in contact with the driver and driven pulley (BD + AC).

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1.5. Derivation for relation between the Tight side, Slack Side, Lap Angle and Coefficient of
Friction for a Flat Belt (Ratio of Driving Tensions for Flat Belt Drive)
Consider a flat belt passing over a fixed drum and belt is just about to slide towards right
(impending motion).
Let, T1 = Tension on Tight Side (Greater
Tension)
T2 = Tension on Slack Side (Smaller Tension)
θ= Lap Angle OR Angle of contact in Radians
𝜇𝑠 = Co-efficient of Static Friction.
Relation between T1 and T2 can be given as:
T1
= 𝑒 μs .θ
T2

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***Derivation:
Consider a belt wrapped around a fixed pulley with friction. The slack side tension T2
increases along the surface of contact to T1 tight side tension because of friction.
Consider the FBD of an element of belt subtending an angle dθ at the centre. Let T be the
tension on one face of element which increases to (T + dT) on the other face.

𝑻𝒊𝒈𝒉𝒕 𝑺𝒊𝒅𝒆
∴ In general, = 𝒆𝛍 𝐬 . 𝛉
𝑺𝒍𝒂𝒄𝒌 𝑺𝒊𝒅𝒆

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The above expression gives the relation between the tight side and slack side tensions, in
terms of coefficient of friction and the angle of contact.

*** Points to be Remembered:

1. The above relation is used only when the belt is about to slip.
𝑻𝒊𝒈𝒉𝒕 𝑺𝒊𝒅𝒆
2. If belt is actually slipping, then use = 𝒆 𝛍𝐤 . 𝛉
𝑺𝒍𝒂𝒄𝒌 𝑺𝒊𝒅𝒆
3. The lap angle or the angle of contact 𝛽 must be in radians.
4. The tight side represents the larger tension in part of belt which pulls and slack side
represents the smaller tension in part which resists.

1.6. Centrifugal Tension

Since the belt continuously runs over the pulleys, therefore, some centrifugal force is
caused, whose effect is to increase the tension on both the tight as well as the slack sides.
The tension caused by centrifugal force is called centrifugal tension. At lower belt speeds
(less than 10 m/s), the centrifugal tension is very small, but at higher belt speeds (more than 10
m/s), its effect is considerable and thus should be taken into account.
We know if a particle of mass M is rotated in a circular path of radius r at a uniform
velocity v, a centrifugal force acts radially outwards and its magnitude is equal to
𝑴𝒗𝟐
FC =
𝒓
In pulley belt arrangement also as the belt material is moving with certain velocity
on a pulley, it develops force which acts outwards from the centre of the pulley. This force
is called the centrifugal force (Fc).
Let us consider an elemental length of belt which subtends an elemental angle dθ
at the centre of the pulley as shown in Fig.
Let v = Velocity of the belt in m/s
r = Radius of pulley over which the belt runs
m = Mass of the belt per metre length
Tc = Centrifugal tension acting on belt
Fc = Centrifugal force acting radially outward

Elemental length of the belt AB = r dθ

Mass of the belt AB = m . r . dθ
Mv2 v2
Centrifugal force FC = = m . r . dθ
r r

Resolving the forces horizontally and applying

limiting equilibrium condition, we have
dθ
FC = 2 𝑇𝑐 𝑠𝑖𝑛
2
Equating both equations

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1.7. Initial Tension in the Belt

Initially the belt is mounted such that a firm grip develops between the belt and pulley.
Therefore, in stationary condition some amount of tension is developed in the belt. This tension
is known as initial tension. This initial tension helps to prevent slipping of the belt on the pulley,
which would result in loss of power and excessive wear. The initial tension is equal in both parts
of the belt.
Under working condition, tension in the two side of the belt will change. The tight side of
the belt stretches until the pull is increased from T0 to T1 and slack side shortens until the pull is
decreased from T0 to T2.

1.8. Power Transmitted and Condition for Maximum Power Transmitted

Let T1 = Tension on tight side
T2 = Tension on slack side
v = Linear velocity of belt
Power = Force ´ Velocity
∴ Power transmitted by the belt is given by
P = (T1 – T2) v

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1.9. Slip of Belt

We have assumed the motion of belts and pulleys has a firm frictional grip between the
belts and the pulleys. But sometimes, the frictional grip becomes insufficient. This may cause
some forward motion of the driver without carrying the belt with it. This is called slip of the belt
and is generally expressed as a percentage.
The result of the belt slipping is to reduce the velocity ratio of the system. As the slipping
of the belt is a common phenomenon, thus the belt should never be used where a definite velocity
ratio is of importance (as in the case of hour, minute and second arms in a watch).
Let s1 % = Slip between the driver and the belt, and
s2 % = Slip between the belt and follower,

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TYPE-I: Problems on General Type of Friction

Que 3.1.1. Determine whether the block shown in Fig. is in equilibrium and find
magnitude and direction of friction force. Take 𝜇𝑠 = 0.3 and 𝜇𝑘 = 0.2

Solution:

Que 3.1.2. Knowing that co-efficient of static friction is 0.25, determine (a) The minimum
value of P required to start the block moving up the incline (b) The corresponding value
of 𝛽. Refer Fig.
Solution:
Draw F.B.D. of block assuming x-axis along the plane and y-axis as perpendicular
to plane. Apply conditions of equilibrium for upward impending motion.

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Que 3.1.3. Two blocks are separated by a

uniform strut attached to each block.
Block A weighs 400 N, block B weighs 300
N and weight of strut AB is 200 N. If µB =
0.25 find minimum co-efficient of friction
under A to prevent motion. Refer Fig.

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Solution:
*** Note: Since the strut is multi-force member, the directions of the pin reaction at A and
B are unknown. Instead of selecting the components as horizontal and vertical as we
usually do, here we take them as vertical and along the strut.

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Que 3.1.4. Two blocks are connected by a

horizontal link AB are supported on two rough
planes. The co-efficient of friction for block A is
0.4 and angle of friction for block B is 150 What
is the smallest weight W of block A for which
equilibrium of the system can exist? Refer
Fig.

Solution:

Que 3.1.5. : Two blocks A and B each of weight 100 N are connected by a slender bar of
negligible weight. If 𝜇𝑠 = 0.3 at all surfaces of contact. Determine the largest value of P to
maintain equilibrium. Refer Fig.

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Solution:

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Que 3.1.6. Determine the force P to cause

motion to impend. Take masses of blocks
A and B as 9 kg and 4 kg respectively and
the coefficient of sliding friction as 0.25.
The force P and the rope are parallel to the
inclined plane as shown in Fig. Assume
pulley to be frictionless.

Solution:

Que 3.1.7. Find force P required to pull

block B (shown in Fig.). Coefficient of
friction between A and B is 0.3 and between
B and floor is 0.25. Weights of A = 20 kg and
B = 30 kg.

Solution:

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Que 3.1.8. Two identical blocks A and B are

connected by a rod and rest against vertical and
horizontal planes, respectively, as shown in Fig. 8.8.
If sliding impends when θ = 450, determine the
coefficient of friction 𝜇, assuming it to be the same
at both floor and wall.

Solution:

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Que 3.1.9. Two blocks A and B of weight 500 N

and 750 N, respectively are connected by a cord
that passes over a frictionless pulley, as shown
in Fig. The coefficient of friction between the
block A and the inclined plane is 0.4 and that
between the block B and the inclined plane is
0.3. Determine the force P to be applied to block
B to produce the impending motion of block B
down the plane.
Solution:

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Que 3.1.10. Find the value of q if the

blocks A and B shown in Fig. have
impending motion. Given block A = 20 kg,
block B = 20 kg, 𝜇𝐴 = 𝜇𝐵 = 0.25.

Solution:

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Que 3.1.11. Two blocks A = 100 N and B

= 150 N are resting on ground as shown
in Fig. Coefficient of friction between
ground and block B is 0.10 and that
between block B and A is 0.30. Find the
minimum value of weight P in the pan so
that motion starts. Find whether B is
stationary w.r.t. ground and A moves or
B is stationary w.r.t. A.
Solution:

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Que 3.1.12. Two loads W1 and W2 resting on

two inclined planes are connected by a
horizontal link AB as shown in Fig. If W1 =
981 N find minimum value of W2 for which
equilibrium can exist. The angle of limiting
friction is 200 at all surfaces of contact.

Solution:

Que 3.1.13. The blocks are placed on the surface

one above the other as shown in Fig. The static
coefficient of friction between the blocks and
block C and surface is also shown. Determine the
maximum value of P that can be applied before
any slipping takes place.

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Solution:

Que 3.1.14. Find the value of M which may be applied to

the cylinder if it is not to spin. Weight of cylinder is W and
its radius is r. The co-efficient of friction µ is the same at
A and B. Refer Fig.

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Solution:

Que 3.1.15. Block A of mass 30 kg rests on block

B of mass 40 kg, as shown in Fig. Block A is
restrained form moving by a horizontal rope tied
at point C. What force P applied parallel to the
plane inclined at 300 with horizontal is necessary
to start block B sliding down the plane. Take
coefficient of friction for all surfaces as 0.35.

Solution:

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TYPE-II: Problems on Belt Friction

Que 3.2.1. For given arrangement, find the angle α for which load begins to slip. Take m
= 0.3 between fixed drum and belt.
Solution:

Que 3.2.2. A lever CD is connected to

cylindrical drum A through a belt, as
shown in Fig. such that the drum does
not rotate. The coefficient of friction
between the belt and the drum is 0.3.
A boy exerts a 100 N upward push on
the lever at C.
Determine:
(i) The maximum weight W
that the boy can lift, and
(ii) The maximum weight W
that the boy can hold.
Solution:

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UNIT-III COMPUTATIONAL ENGINEERING MECHANICS

Que 3.2.3. A belt having cross section of dimension 50 mm x 6 mm has an angle of lap of
1900 on a pulley with diameter 800 mm. The pulley rotates at 240 rpm. If the material of
the belt has a density of 1120 kg/m3 and withstand a maximum tensile stress of 1500
kN/m2 . Find (i) centrifugal tension and (ii) power transmitted. Take 𝜇= 0.3 between the
belt and pulley.
Solution:

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UNIT-III COMPUTATIONAL ENGINEERING MECHANICS

Que 3.2.4. A leather belt transmits 30 kW from a pulley 75 cm diameter which runs at
500 rpm. The belt is in contact with the pulley over an arc of 1500 and coefficient of
friction is 0.3. If the mass of belt is 1 kg per metre and if the safe stress is not to exceed
245 N/cm2. Find the minimum area of the belt.
Solution:

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UNIT-III COMPUTATIONAL ENGINEERING MECHANICS

Que 3.2.5. In an open-belt drive, the angle of lap on the smaller pulley is 1720. The smaller
pulley has a diameter of 800 mm and rotates at 600 rpm. Consider m between the pulley
and belt as 0.3. Neglecting centrifugal tension, find the power that can be transmitted if
the initial tension in the belt is 1200 N.
Solution:

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UNIT-III COMPUTATIONAL ENGINEERING MECHANICS

Que 3.2.6. Two parallel shaft, which are 6 m apart, are to be connected by a belt running
over pulleys of diameter 46 cm and 30 cm respectively. Determine the length of the belt
considering both open and crossed belt drives.
Solution:

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UNIT-III COMPUTATIONAL ENGINEERING MECHANICS

Que 3.2.7. A belt 100 mm wide and 8 mm thick is transmitting power at a belt speed of
1600 m/minute. The angle of lap of the smaller pulley is 165° and coefficient of friction is
0.3. The maximum permissible stress in the belt is 2 N/mm2 and the mass of the belt is
0.9 kg/m. Find the power transmitted and the initial tension in the belt. Also find the
maximum power that can be transmitted and the corresponding belt speed.
Solution:

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UNIT-III COMPUTATIONAL ENGINEERING MECHANICS

Que 3.2.8. A cross belt drive has the following details:

Thickness of belt = 8 mm
Maximum permissible stress in the belt = 2 MPascal
Diameter of driver pulley = 400 mm
Diameter of driven pulley = 800 mm
Density of belt material = 1500 kg/m3
Coefficient of friction between belt and pulley = 0.25
If the angular velocity of driver pulley is 1200 rpm, find the power transmitted in
watt per unit mm width of the belt. If it is desired to transmit power at maximum level,
which is 16 kW, find the width of the belt and angular velocity in rpm of the driver pulley.
Solution:

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UNIT-III COMPUTATIONAL ENGINEERING MECHANICS

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