Friction 1
FRICTION
Friction is defined as the contact resistance exerted by
one body upon another body when one body moves or
tends to move past another body. This force which opposes
the movement or tendency of movement is known as
frictional resistance or friction. Friction is due to the
resistance offered by minute projections at the contact
surfaces. Hence friction is the retarding force, always
opposite to the direction of motion. Friction has both
advantages & disadvantages.
Disadvantages ---- Power loss, wear and tear etc.
Advantages ---- Brakes, traction for vehicles etc.
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Friction 2
W
P
F (Friction)
N
Hills & Vales
Magnified Surface
Frictional resistance is dependent on the amount of wedging
action between the hills and vales of contact surfaces. The
wedging action is dependent on the normal reaction N.
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Friction 3
Frictional resistance has the remarkable property of
adjusting itself in magnitude of force producing or tending
to produce the motion so that the motion is prevented.
When P = 0, F = 0  block under equilibrium
When P increases, F also increases proportionately to
maintain equilibrium. However there is a limit beyond
which the magnitude of this friction cannot increase.
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Friction 4
When the block is on the verge of motion(motion of the
block is impending) F attains maximum possible value,
which is termed as Limiting Friction. When the applied
force is less than the limiting friction, the body remains at
rest and such frictional resistance is called the static friction.
Further if P is increased, the value of F decreases rapidly
and then remains fairly a constant thereafter. However at
high speeds it tends to decrease. This frictional resistance
experienced by the body while in motion is known as
Dynamic friction OR Kinetic Friction.
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Friction 5
Dynamic Friction
Sliding friction friction experienced
when a body slides over another
surface.
Rolling friction  friction experienced
by a body when it rolls over a surface.
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Friction 6
FN
W
Fmax = N
P
Where Fmax = Limiting Friction
Fmax

N
(limiting friction)
N= Normal Reaction between the
contact surfaces
 =Coefficient of friction
R
  = Fmax
N
Note : Static friction varies from zero to a maximum value. Dynamic
friction is fairly
a constant.
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Friction 7
Angle of Friction
The angle between N & R
depends on the value of F. This
angle θ, between the resultant
R and the normal reaction N is
termed as angle of friction. As
F increases, θ also increases
and will reach to a maximum
value of  when F is Fmax
(limiting friction)
W
P
Fmax

(limiting friction)
N
R
i.e. tan = (Fmax )/N = 
Angle  is known as Angle of limiting Friction.
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Friction 8
Angle of limiting friction is defined as the angle between the
resultant reaction (of limiting friction and normal reaction)
and the normal to the plane on which the motion of the body
is impending.
Angle of repose
When granular material is heaped, there exists a limit for the
inclination of the surface. Beyond that angle, the grains start
rolling down. This limiting angle upto which the grains
repose (sleep) is called the angle of repose of the granular
material.
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Friction 9
Significance of Angle of repose:
The angle that an inclined plane makes with
the horizontal, when the body supported on the
plane is on the verge of motion due to its self weight is equal to the angle of repose.
Angle of repose is numerically equal to
Angle of limiting friction
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Friction 10
Laws of dry friction
1. The magnitude of limiting friction bears a constant ratio
to the normal reaction between the two surfaces.
(Experimentally proved)
2. The force of friction is independent of the area of
contact
between the two surfaces.
3. For low velocities the total amount of friction that can
be developed is practically independent of velocity.
It is less than the frictional force corresponding
to impending motion.
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Friction 11
FRICTION IN BELT/ ROPE DRIVES
The transmission of power by means of belts or rope drives
is possible only because of friction between the wheels and
the belt. Tension in the belt is more on the side it is pulled
and less on the other side. Accordingly they are called as
tight side and slack side.

T2 (Tight side)
Pull
W
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T1 (Slack side)
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RELATIONSHIP BETWEEN TIGHTSIDE AND
SLACKSIDE FORCES IN A ROPE
Friction 12
A load W is being pulled by a force P over a fixed drum. Let the force
on tight side be T2 and on slack side be T1. (T2>T1 because of
frictional force between drum and the rope). Let  be the angle of
contact in radians between rope and the drum. Consider an elemental
length of rope as shown. Let T be the force on slack side and T+dT
on tight side. There will be normal reaction N on the rope in the radial
direction and frictional force F= N in the tangential direction.

F
d/2
T+dT
P
w
T2
T
N F
d
T1
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Friction 13
 Forces in radial direction = 0
N-T Sin d/2 – (T+dT)Sin d/2 = 0 { Sin d/2 = d/2 as d is small }
 N-T d/2- (T+dT) d/2 = 0
i.e. N = ( T+dT/2) d ------(1)
We know that F = N
 F =  ( T+dT/2) d-----(2)
 Forces in tangential direction = 0
(T+dT) Cos d/2 = F + T Cos d/2 { Cos d/2 = 1 as d is small }
 T + dT = F + T
i.e.
dT = F------(3)
From (2) & (3)
dT =  ( T+dT/2) d
Neglecting small quantity of higher order, dT = T d
dT/T =  d
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Friction 14
Integrating both sides,
T2

T
1
0
 dT/T =   d

T2
(log T) = ()
T
1
0
Log (T2/T1) = 

T2/ T1 = e
where T = Force on tightside
2
T = Force on slackside
1
 = Angle of contact in radians
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Friction15
EXERCISE PROBLEMS
1 ) For the block shown in fig., determine the smallest
force P required
a) to start the block up the plane
b) to prevent the block moving down the plane.
Take μ = 0.20
[Ans.: (a) Pmin = 59.2N
(b) Pmin = 23.7N θ = 11.3o]
P

100N
25
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Friction 16
2) A block of weight 2000 N is attached to a cord
passing over a frictionless pulley and supporting a
weight of 800N as shown in fig. If μ between the block
and the plane is 0.35, determine the unknown force P
for impending motion
(a) to the right
(b) to the left
[Ans.: (a) P = 132.8N (b) P = 1252N]
2000N
800N
30
P
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Friction 17
3) Determine value of angle θ to cause the motion of
500N block to impend down the plane, if μ for all
contact surfaces is 0.30.
[Ans.: θ = 28.4°]
200N
500N
=?
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Friction 18
4) In Figure, μ between rope and the fixed drum and
between all contact surfaces is 0.20. Determine the
minimum weight W to prevent the downward motion
of the 1000N body.
[Ans. : T1 = 0.76W, T2 = 1.424W, W = 253N]
W
1000N
3
4
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Friction 19
5) A horizontal bar 10m long and of negligible weight
rests on rough inclines as shown in fig. If angle of
friction is 15o, how close to B may the 200N force be
applied before the motion impends.
[Ans.: x = 3.5m]
100N
2m
A
200N
X=?
B
30
60
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Friction 20
6) Determine the vertical force P required to drive the
wedge B downwards in the arrangements shown in fig.
Angle of friction for all contact surfaces is 12o.
[Ans.: P = 328.42N]
P
1600N
A
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B
20
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Friction 21
7) Determine the force P which is necessary to start the
wedge to raise the block A weighing 1000N. Self
weight of the wedge may be ignored. Take angle of
friction,  = 15o for all contact surfaces.
[Ans.: P = 1192N]
A
20
P
wedge
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Friction 22
8) A ladder of weight 200N, 6m long is supported as shown
in fig. If μ between the floor and the ladder is 0.5 &
between the wall and the ladder is 0.25 and it supports a
vertical load of 1000N, determine
a) the least value of α at which the ladder may be placed
without slipping
b) the reactions at A & B
[Ans.: (a) α = 56.3o (b) RA = 1193 N, RB = 550N]
1000N
B
5m

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A
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Friction 23
9) An uniform ladder of weight 250N is placed against a
smooth vertical wall with its lower end 5m from the wall. μ
between the ladder and the floor is 0.3. Show that the ladder
remains in equilibrium in this position. What is the frictional
resistance on the ladder at the point of contact between the
ladder and the floor?
Smooth wall
[Ans.: FA = 52N]
B
12m
A
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5m
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Friction 24
10)A ladder of length 5m weighing 500N is placed at 45o
against a vertical wall. μ between the ladder and the
wall is 0.20 & between ladder and ground is 0.50. If a
man weighing 600N ascends the ladder, how high will
he be when the ladder just slips. If a boy now stands on
the bottom rung of the ladder, what must be his least
weight so that the man can go to the top of the ladder.
[Ans.: (a) x = 2.92m (b) Wboy = 458N]
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