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.
Friction 2
W
P
F (Friction)
N
Hills &
Valves
Magnified Surface
Frictional resistance is dependent on the amount of wedging
action between the hills and valves of contact surfaces. The
wedging action is dependent on the normal reaction N.
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.
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.
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.
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.
Friction 7
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)
Angle of Friction
W
P
Fmax

(limiting friction)
N
i.e. tan = (Fmax )/N = 
Angle  is known as Angle of limiting Friction.
R
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.
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
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.
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
T1 (Slack side)
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
T1
T
N F
d
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
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
FRICTION-Numerical Problems
Friction 15
(1) If coefficient of friction is 0.20 between the contact surfaces
a) Find the force P just to cause motion to impend up the plane
b) Find the force P just to prevent motion down the plane
c) Determine the magnitude and direction of the friction
if P = 80N.
200N
P
30°
Friction 16
a)
Imp. motion
30°
Y +ve
X + ve
200N
60°
F1 = N1 = 0.20N1
P
P
N1
Fy = 0
N1 – 200 Sin60 = 0  N1= 173.2 N
F1 = 0.20N1 = 0.20 173.20 = 34.64N
Fx = 0
P – 200Cos60 – F1 = 0  P = 134.64 N
30°
Friction 17
b)
Y +ve
30°
60°
P
F1 = 0.20N1
Fx = 0
P – 200Cos60 + 0.20 173.2 = 0
N1=173.2
 P = 65.36 N
X + ve
Friction 18
C) Block will be under rest for the value of P between
134.64 & 65.36N..
Y +ve
Given, P = 80N
60°
Imp. motion
P=80
F1
Fx = 0
80 – 200Cos60 + F1 = 0
N1
 F1 = 20 N
X + ve
Friction 19
2) Compute the magnitude of P that will cause the motion
to impend up the plane. Coefficient of friction, μ = 0.20
200N
200N
30°
P
P
30°
F1 = 0.20N1

N1
tan  =  = 0.20
  = 11.3°
R
Friction 20
200N
41.3°
P
48.70°
R
200
30 +  = 41.30°
R
200/Sin48.70 = P/ Sin41.3°  P = 176N
48.7°
P
Friction21
3) Block A weighing 1000N rests over block B of weight
2000N as shown in fig. Block A is tied to the wall with a
horizontal string. If coefficient of friction between A & B is
0.25 and between B and the floor is 0.33, what should be the
value of P just to move the block B ?
A
B
P
Friction22
RELATIVE
MOTION
FBD of Block A
Y +ve
1000N
FBD of Block B
Imp. motion
N1=1000
F1
T
A
X +ve
F1
Block A: Fy = 0
B
N1
 N1 - 1000 = 0
N1 = 1000 N
F1 = 1 N1
= 0.25 1000
= 250 N
Fx = 0
F – T = 0  T= 250 N
2000
P
F2
N2
Friction23
FBD of Block B
Imp. motion
Y +ve
N1=1000
F1
B
2000
P
X +ve
F2
N2
Block A: Fy = 0
Fx = 0
N2 - N1 - 2000 = 0
N2 - 1000 -2000 = 0
N2 = 3000 N
P - F1 -F2 = 0
P- 250 - 0.33 N2 = 0
P - 250 -0.33  3000 = 0
 P = 1250 N
Friction24
4) The bodies shown in the following figure are separated by
an uniform strut weighing 100N which is attached to the
bodies with frictionless pins. Coefficient of friction under
each body is 0.30. Determine the force P that will just start
the system rightward. Weight of block A= 400 N, B= 200N
FBD of the Strut
tan  = 0.30
  = 16.7°
P
A
B
30°
100 N
T
50 N
45°
T
100/2 = 50 N
Friction25
FBD of B
200+50
250
R
250
30
T
30°
F2
16.70
45°
N2
61.7° R
60°
T
45+16.70
R
250/Sin58.3° = T/Sin61.70°
 T = 258.72N
61.7°
58.3°
60°
T
Friction26
FBD of A
Y +ve
400+50
T = 258.72
30°
P
X +ve
F1 = 0.30N1
N1
Fy = 0
Fx = 0
N1 - 450 - 258.72Sin30 = 0
N1 = 579.36N
P - F1- 258.72Cos30 = 0
P-0.30 579.36-258.72cos30=0
P- 173.81- 224.06=0
 P = 397.87 N
Friction27
5)What horizontal force P is required on the wedges B and
C just to raise the weight 1000N resting on A. Angle of limiting
friction between all contact surfaces is 10o.
A
P
B
15°
15°
C
P
Friction28
FBD of A
1000N
1000N
25°
1000N
F1
R1
10°
N1 15°
F2
R2
10°
15°
N2
R2
130°
25°
10+15
R1 =25°
R1
25°
R2
1000/Sin130 = R1/ Sin25 = R2/ Sin25
 R1 = R2 = 551.69N
FBD of B
Friction29
N1
15°
10° R = 551.69
1
P
R1=551.69
F1
25°
P
10°
N3
R3
65°
10°
80°
65°
80°
F3
R3
35°
P
R3
P/Sin35 = 551.69/ Sin80
 P = 321.32N
Note: FBD of Block (C) can also be considered. No need to
consider the FBD of both the blocks (B) & (C).
Friction30
6) Determine the force P required just to start the wedge
A shown in the figure. Angle of limiting friction between all
contact surfaces is 15°.
P
 = 15°
2000N

A
B
500N
Friction 31
FBD of B
F2 = 0.27N2
2000N
500 N
N2
F1 = 0.27N1
N1
Y +ve
= 15°
tan  = 
  = 0.27
X +ve
Fx = 0
N2 - 500- 0.27N1 = 0
N2 = 500 + 0.27N1 ---------(1)
Fy = 0
N1- 2000 - 0.27N2 = 0
0.27N2 = N1 -2000
N2= 3.70N1 -7407.41 ------(2)
From (1) & (2)
500 + 0.27N1 = 3.70N1 – 7407.41
3.43N1 = 7907.41
 N = 2305.37 N
N2=1122.45N
Friction 32
FBD of Wedge A
P
R2=  1122.452 + 303.062
= 1162.64
R2=1162.64
P
N2=1122.45
=15°
15°
N3
15°
R3
75°
R2=1162.64
F3
F2=0.27N2
P
30°
15°
=303.06
P/Sin45
= 1162.64/ Sin60
 P = 949.29N
45°
R3
60°
75°
60°
R2=1162.64
R3
Friction 33
7) Determine the minimum value of P to prevent the blocks
from slipping. Neglect the weights of the link rods. Coefficient of friction for all contact surfaces is 0.25.Find the
frictional force under the block B and comment on the result.
WA= WB=2000N
A
Pin Joint
30
60
C
B
P
Pin Joints
Friction 34
Fx = 0
N1 - T1 Cos30= 0
N1 = T1 Cos30 = 0.866T1-------(1)
FBD of A
2000
N1
30°
T1
F1=0.25 N1
Fy = 0
- 2000 + F1 + T1Sin30 = 0
-2000 +0.25N1 + 0.5T1= 0------(2)
Y +ve
X +ve
From (1) & (2)
-2000 + 0.25(0.866T1 )+ 0.5T1 = 0
 T1 = 2791.32 N
Friction 35
Joint (C)
T1 = 2791.32
30°
T2
P
90°
30°
60°
60°
T2
T1
P/Sin90 = 2791.32/Sin60
= T2/Sin30
P=3223.14 N
T2 = 1611.57N
P
Friction 36
FBD of B
2000
F2
N2
Fy = 0
T2=1611.57 N - 2000 – 1611.57Sin60 = 0
2
Y +ve

N
=3395.60
2
60°
Fx = 0
X +ve
F2 – 1611.57Cos60 = 0
F2 = 805.79 N (Friction Developed
under block B)
Limiting friction = N2 = 0.25 x 3395.60 = 848.92N
Limiting friction is greater than Friction developed.
Hence the block B is at rest.
Friction 37
8)An uniform ladder of length 7m rests against a vertical
wall with which it makes an angle of 45o. Coefficient of
friction between the ladder and the wall is 1/3 and between
ladder and the floor is 1/2. If a person whose weight is half
that of the ladder ascends it, how high will he be when the
Y +ve
ladder just slips?
 Fx=0
FA-NB=0
Fy=0
NB
B
0.5NA-NB=0
7m
NB=0.5NA--------(1)
FB
0.5W
NA-W-0.5W+FB=0
NA+0.33NB=1.5W------(2)
a
X +ve
FA
3.5m W
45
A
7cos45
NA
7sin45
Friction 38
From (1) & (2)
NA+0.33(0.5NA)=1.5W
NA=1.29W
NB=0.64W
+ ve moment
MB=0
(FA 7sin45)-(NA  7cos45)+(W 3.5cos45)+(0.5W
acos45)=0
a = 2m from the top
Friction 39
9) An uniform ladder 3m in length and weighing 180N is
placed against a wall with its end A at the floor and the
other end B on the wall, ladder AB making 60 with the
floor. Coefficient of friction between the wall and ladder
is 0.25 and between floor and ladder is 0.35.In addition to
the self weight,the ladder has to support a person
weighing 900N at its top B. To prevent slipping, a force P
is applied horizontally at A at the level of the floor. Find
the minimum force P required for this condition. Find
also the minimum angle α at which the above ladder with
the person at the top should be placed to prevent slipping
without the horizontal force P.
Friction 40
a) When  = 60°
Y +ve
Fx = 0, FA+P-NB=0
0.35NA+P-NB=0
NB=P+0.35NA---------(1)
Fy=0,
X +ve
NA-180-900+FB=0
NA+0.25NB=1080----(2)
900N
B N
B
MB=0
0.35NA  3sin60+P 3sin60+180 x
1.5cos60 -NA 3cos60=0--------(3)
3m
From(1),(2)&(3)
1.5m
NB=499.16N
NA=955.21N
P=164.80N
A
FA
FB
 180 N
3cos
N
3sin
Friction 41
(b) Force P is removed, =?
Y +ve
Fx=0
X +ve
FA-NB=0
0.35NA-NB=0 0.35NA=NB---(1)
Fy=0
NA-180-900+FB=0 NA+0.25NB=1080-----(2)
MB=0
0.35NA 3sin+180 1.5cos-NA 3cos=0----(3)
From(1), (2) &(3)
=68.95
Friction 42
10) A flexible cable which supports a load of 981 N is passed over
a fixed circular drum and subjected to a force P to maintain
equilibrium. Coefficient of friction between the cable and the
drum is 0.30.
(a)For  = 0, determine the maximum and minimum value of
P which may have to be applied in order not to raise or
lower the load.
(b)For P = 500N, determine the minimum value which the
angle  may have such that the load begins to slip.

P
981N
Friction 43
(a) T2/T1 = e

 = 0, angle of contact  = 90 = /2
radians.
For impending motion (upward)
T2 = Pmax
Pmax/981= e 0.3/2
T1 = 981 N
Pmax=1572 N
For impending motion (down ward)
T2 = 981N
T1 = Pmin
981/Pmin= e0.3/2
Pmin=612N
(b) P = 500 N  P = T1
T2 = 981 N
981/500=e0..3
=2.247radians=128.7
  = -90= 38.7
P
981N
Friction 44
11) A cord is wrapped thrice round a capstan A and twice
round a capstan B. Finally the cord goes over a barrel and
supports a weight of 40N. What force P is required to
maintain this load? Take coefficient of friction = 0.1 for all
contact surfaces.
A
B
Total angle = 5xx360 + 90°
= 10.5  radians
40 > P  T2 = 40, T1 = P
T2/T1= e
40/P = e 0.1(10.5)
P =1.477N
P
40N
Friction 45
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
Friction 46
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
Friction 47
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
=?
Friction 48
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
Friction 49
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
Friction 50
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
B
20
Friction 51
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
Friction 52
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
5m

A
B
Friction 53
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
5m
Friction 54
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]