228 Friction

Chapter

5
Friction
Introduction (c) Unit : It has no unit.
(d) Value of  depends on material and nature of surfaces in
If we slide or try to slide a body over a surface, the motion is resisted
contact that means whether dry or wet ; rough or smooth polished or non-
by a bonding between the body and the surface. This resistance is represented
polished.
by a single force and is called friction force.
(e) Value of  does not depend upon apparent area of contact.
The force of friction is parallel to the surface and opposite to the
direction of intended motion. (3) Kinetic or dynamic friction : If the applied force is increased
further and sets the body in motion, the friction opposing the motion is
Types of Friction called kinetic friction.
(1) Static friction : The opposing force that comes into play when (i) Kinetic friction depends upon the normal reaction.
one body tends to move over the surface of another, but the actual motion Fk  R or Fk  k R where  k is called the coefficient of kinetic
has yet not started is called static friction.
friction
(i) If applied force is P and the
body remains at rest then static friction F = R (ii) Value of k depends upon the nature of surface in contact.
P. P
F
(iii) Kinetic friction is always lesser than limiting friction Fk  Fl
(ii) If a body is at rest and no
pulling force is acting on it, force of  k   s
friction on it is zero. mg i.e. coefficient of kinetic friction is always less than coefficient of
(iii) Static friction is a self-adjusting Fig. 5.1 static friction. Thus we require more force to start a motion than to
force because it changes itself in accordance with the applied force and is maintain it against friction. This is because once the motion starts actually ;
always equal to net external force. inertia of rest has been overcome. Also when motion has actually started,
(2) Limiting friction : If the applied force is increased, the force of irregularities of one surface have little time to get locked again into the
static friction also increases. If the applied force exceeds a certain irregularities of the other surface.
(maximum) value, the body starts moving. This maximum value of static (iv) Kinetic friction does not depend upon the velocity of the body.
friction upto which body does not move is called limiting friction. (v) Types of kinetic friction
(i) The magnitude of limiting friction between any two bodies in (a) Sliding friction : The opposing force that comes into play when
contact is directly proportional to the normal reaction between them. one body is actually sliding over the surface of the other body is called
sliding friction. e.g. A flat block is moving over a horizontal table.
Fl  R or Fl   s R
(b) Rolling friction : When objects such as a wheel (disc or ring),
(ii) Direction of the force of limiting friction is always opposite to sphere or a cylinder rolls over a surface, the force of friction that comes
the direction in which one body is at the verge of moving over the other into play is called rolling friction.
(iii) Coefficient of static friction : (a)  s is called coefficient of  Rolling friction is directly proportional to the normal reaction (R)
static friction and is defined as the ratio of force of limiting friction and and inversely proportional to the radius (r) of the rolling cylinder or wheel.
F
normal reaction  s  Frolling   r
R
R r
(b) Dimension : [M 0 L0 T 0 ]
 Friction 229

 r is called coefficient of rolling friction. It would have the dimensions ma   s mg ). If there had been no friction between body and vehicle,
of length and would be measured in metre. the body will not move along with the vehicle.
 Rolling friction is often quite small as compared to the sliding
friction. That is why heavy loads are transported by placing them on carts a
with wheels. smg ma

 In rolling the surfaces at contact do not rub each other.

 The velocity of point of contact with respect to the surface
remains zero all the times although the centre of the wheel moves forward.
Fig. 5.5
Graph Between Applied Force and Force of Friction From these examples it is clear that without friction motion cannot
(1) Part OA of the curve represents static friction (Fs ) . Its value be started, stopped or transferred from one body to the other.
increases linearly with the applied force
Advantages and Disadvantages of Friction
(2) At point A the static friction is maximum. This represent
limiting friction (Fl ) . (1) Advantages of friction
A
(3) Beyond A, the force B C (i) Walking is possible due to friction.
Force of friction

of friction is seen to decrease

Fs (ii) Two body sticks together due to friction.
slightly. The portion BC of the
curve represents the kinetic Fl Fk
friction (Fk ) .
(4) As the portion BC O Applied force
of the curve is parallel to x-axis
therefore kinetic friction does Fig. 5.2
not change with the applied force, it remains constant, whatever be the
applied force.
Fig. 5.6 Fig. 5.7
Friction is a Cause of Motion (iii) Brake works on the basis of friction.
It is a general misconception that friction always opposes the motion. (iv) Writing is not possible without friction.
No doubt friction opposes the motion of a moving body but in many cases
it is also the cause of motion. For example : (v) The transfer of motion from one part of a machine to other part
(1) While moving, a person or vehicle pushes the ground backwards through belts is possible by friction.
(action) and the rough surface of ground reacts and exerts a forward force
due to friction which causes the motion. If there had been no friction there (2) Disadvantages of friction
will be slipping and no motion. (i) Friction always opposes the relative motion between any two
bodies in contact. Therefore extra energy has to be spent in over coming
friction. This reduces the efficiency of machine.
(ii) Friction causes wear and tear of the parts of machinery in
contact. Thus their lifetime reduces.
Friction
(iii) Frictional force result in the production of heat, which
(2) During cycling, the rear wheel moves by the force causes damage to the machinery.
communicated to itAction
by pedalling while front wheel moves by itself. So,
Fig. 5.3 Methods of Changing Friction
when pedalling a bicycle, the force exerted by rear wheel on ground
makes force of friction act on it in the forward direction (like We can reduce friction
walking). Front wheel moving by itself experience force of friction in
backward direction (like rolling of a ball). [However, if pedalling is (1) By polishing.
stopped both wheels move by themselves and so experience force of (2) By lubrication.
friction in backward direction]. (3) By proper selection of material.
(4) By streamlining the shape of the body.
(5) By using ball bearing.
Also we can increase friction by throwing some sand on slippery
ground. In the manufacturing of tyres, synthetic rubber is preferred
because its coefficient of friction with the road is larger.
While pedalling Pedalling is stoped
Angle of Friction
Fig. 5.4
(3) If a body is placed in a vehicle which is accelerating, the force Angle of friction may be defined as the angle which the resultant of
of friction is the cause of motion of the body along with the vehicle limiting friction and normal reaction makes with the normal reaction.
(i.e., the body will remain at rest in the accelerating vehicle until
S R

F P

mg
Fig. 5.8
 230 Friction
By definition angle  is called the angle of friction By resolving P in horizontal and vertical direction (as shown in
Fl figure) R
tan  
R P sin

Fl F P cos
 tan  =  s
[As we know  s ]
R

or   tan 1 ( L ) W
Fig. 5.11
Hence coefficient of static friction is equal to tangent of the angle of For the condition of equilibrium
friction.
F  P cos and R  W  P sin
Resultant Force Exerted by Surface on Block By substituting these value in F  R

In the above figure resultant force S  F 2  R 2 P cos   (W  P sin )

sin
S  (mg )2  (mg )2  P cos   (W  P sin ) [As   tan  ]
cos

S  mg  2  1 W sin
 P
cos (   )
when there is no friction (  0) S will be minimum
(2) Minimum pushing force P at an angle  from the horizontal
i.e. S = mg
P
Hence the range of S can be given by,

mg  S  mg  2  1

Angle of Repose By Resolving P in horizontal and vertical direction (as shown in the
figure) Fig. 5.12
Angle of repose is defined as the angle of the inclined plane with R
horizontal such that a body placed on it is just begins to slide.
By definition,  is called the angle of repose. F P cos

In limiting condition F  mg sin and R  mg cos

R P sin
F W
Fig. 5.13
For the condition of equilibrium
mg sin 
F  P cos and R  W  P sin

mg cos  By substituting these value in F  R
 mg
 P cos   (W  P sin )
F Fig. 5.9
So  tan  sin
R  P cos   (W  P sin ) [As   tan  ]
cos
F F
   s  tan   tan  [As we know   s  tan  ] W sin
R R  P
cos (   )
Thus the coefficient of limiting friction is equal to the tangent of
angle of repose. (3) Minimum pulling force P to move the body up on an inclined
As well as    i.e. angle of repose = angle of friction. plane P
Calculation of Required Force in Different Situation 

If W = weight of the body,  = angle of friction,

  tan   coefficient of friction
Then we can calculate required force for different situation in the 
following manner :
By Resolving P in the direction of the plane and perpendicular to the
Fig. 5.14
(1) Minimum pulling force P at an angle  from the horizontal plane (as shown in the figure)
P R + P sin
P cos


F + W sin

Fig. 5.10 
W cos
 W

Fig. 5.15
 Friction 231
For the condition of equilibrium
R  P sin  W cos 
 R  W cos   P sin and P cos  F  W sin
 F  W sin  P cos
For the condition of equilibrium By substituting these values in F  R and solving we get
R  P sin  W cos 
 sin(   ) 
P W 
 R  W cos   P sin and F  W sin  P cos  cos (   ) 
 F  P cos  W sin (6) Minimum force for motion along horizontal surface and its
direction P
By substituting these values in F  R and solving we get

W sin(   ) 
P
cos (   )
(4) Minimum force to move a body in downward direction along the surface
Fig. 5.20
of inclined plane Let the force P be applied at an angle  with the horizontal.
P By resolving P in horizontal and vertical direction (as shown in
figure)
 R + P sin

F P cos


Fig.the
By Resolving P in the direction of 5.16plane and perpendicular to the
mg
plane (as shown in the figure)
R + P sin Fig. 5.21
For vertical equilibrium
F
P cos R  P sin  mg
+
 R  mg  P sin …(i)
W sin
 and for horizontal motion
W cos
 W P cos  F

Fig. 5.17
i.e. P cos  R …(ii)
For the condition of equilibrium Substituting value of R from (i) in (ii)
R  P sin  W cos  P cos   (mg  P sin )
 R  W cos   P sin and F  P cos  W sin
 mg
By substituting these values in F  R and solving we get P …(iii)
cos    sin
W sin(   )
P For the force P to be minimum (cos   sin ) must be
cos (   ) maximum i.e.
(5) Minimum force to avoid sliding of a body down on an inclined d 2
plane [cos   sin ]  0 1
P d 

  sin   cos  0
 1
 tan   
Fig. 5.22
or   tan 1 ()  angle of friction

i.e. For minimum value of P its angle from the horizontal should be
equal to angle of friction
By Resolving P in the direction
Fig.of5.18
the plane and perpendicular to the

plane (as shown in the figure) As tan    so from the figure, sin 
1 2
R + P sin F + P cos
1
and cos  
1 2

W sin  W cos
By substituting these value in equation (iii)
 W

Fig. 5.19
 232 Friction

 mg  mg Work done = force  distance = F  s =  mg s

P 
1 2 1 2 It is clear that work done depends upon

1 2
1 2
R

mg P
 Pmin  F
1 2

Acceleration of a Block Against Friction s

mg
(1) Acceleration of a block on horizontal surface Fig. 5.27
(i) Weight of the body.
When body is moving under application of force P, then kinetic
friction opposes its motion. (ii) Material and nature of surface in contact.
Let a is the net acceleration of the body R (iii) Distance moved.
ma
From the figure Motion of Two Bodies one Resting on the Other
Fk P
ma  P  Fk When a body A of mass m is resting on a body B of mass M then
two conditions are possible
P  Fk
 a (1) A force F is applied to the upper body, (2) A force F is applied to
m mg the lower body
(2) Acceleration of a block sliding down over a rough inclined plane
Fig. 5.23
m A F
When angle of inclined plane is more than angle of repose, the body
placed on the inclined plane slides down with an acceleration a.
L
From the figure ma  mg sin  F R M B
F
ma
 ma  mg sin  R
Fig. 5.28
 ma  mg sin   mg cos We will discuss above two cases one by one in the following manner
 :
mg sin  mg cos
 Acceleration a  g [sin   cos  ]  mg (1) A force F is applied to the upper body, then following four
situations are possible
Note :  For frictionless inclined plane   0 Fig.
a  g sin .
5.24 (i) When there is no friction
(3) Retardation of a block sliding up over a rough inclined plane (a) The body A will move on body B with acceleration (F/m).
When angle of inclined plane is less than angle of repose, then for aA  F / m
the upward motion ma
R (b) The body B will remain at rest
ma  mg sin  F aB  0
ma  mg sin   mg cos (c) If L is the length of B as shown in figure, A will fall from B after
time t
mg sin  + F  mg cos
Retardation a  g [sin   cos ]  mg 2L 2mL  1 2 
t   As s  2 a t and a  F/m 
a F  
Note :  For frictionless inclined plane   0  a  g sin
Fig. 5.25
(ii) If friction is present between A and B only and applied force is less than
limiting friction (F < F)
l

Work done against friction (F = Applied force on the upper body, F = limiting friction between A and
l

(1) Work done over a rough inclined surface B, F = Kinetic friction between A and B)
k

If a body of mass m is moved up slowly on a rough inclined plane (a) The body A will not slide on body B till F  Fl i.e. F   s mg
through distance s, then (b) Combined system (m + M) will move together with common
Work done = force  distance F
acceleration a A  a B 
= ma  s = mg [sin +  cos ]s  mg s [sin   cos  ] M m
(iii) If friction is present between A and B only and applied force is greater
ma than limiting friction (F > F)
R l

In this condition the two bodies will move in the same direction ( i.e. of
applied force) but with different acceleration. Here force of kinetic friction
s  k mg will oppose the motion of A while cause the motion of B.

mg sin  + F  mg cos
 mg F  Fk  m a A Free body diagram of A

(2) Work done over a horizontalFig.

surface
5.26 maA

In the above expression if we put  = 0 then

A F

Fk
 Friction 233

F  Fk
(where F = Pseudo force on body A and F = limiting friction
l

i.e. aA  between body A and B)

m
(a) Both the body will move together with common acceleration
(F   k mg )
aA  F
m a
M m
Fk  M a B Free body diagram of B
(b) Pseudo force on the body A,
MaB
Fk mF
i.e. aB 
FK F   ma  and Fl   s mg
M mM
B
 k mg mF
 aB  (c) F   Fl    s mg  F   s (m  M ) g
M mM
So both bodies will move together with acceleration
Note :  As both the bodies are moving in the same direction.
a A  aB 
F
if F   s [m  M ] g
mM
Acceleration of body A relative to B will be
(iii) If friction is present between A and B only and F > F
MF  k mg (m  M ) l

a  a A  aB 
mM (where F =  mg = limiting friction between body A and B)
l s

So, A will fall from B after time Both the body will move with different acceleration. Here force of
2L 2 m ML kinetic friction k mg will oppose the motion of B while will cause the
t 
a MF  k mg (m  M ) motion of A.

(iv) If there is friction between B and floor

ma A   k mg Free body diagram of A
(where Fl    (M  m) g = limiting friction between B and floor, F k

= kinetic friction between A and B) i.e. aA  k g A maA

B will move only if Fk  Fl and then Fk  Fl  M aB Fk
MaB F  Fk  MaB Free body diagram of B
FK
MaB
B [F  k mg ]
i.e. aB  FK
Fl M F
Fig. 5.29 B
However if B does not move then static friction will work (not
limiting friction) between body B and the floor i.e. friction force = applied
force (= F ) not Fl .
k

Note :  As both the bodies are moving in the same direction

(2) A force F is applied to the lower body, then following four
situations are possible Acceleration of body A relative to B will be
(i) When there is no friction  F  k g(m  M ) 
(a) B will move with acceleration (F/M) while A will remain at rest a  a A  aB   
 M 
(relative to ground) as there is no pulling force on A.
Negative sign implies that relative to B, A will move backwards
F and will fall it after time
a B    and a A  0
M
2L 2 ML
(b) As relative to B, A will move backwards with acceleration (F/M) t 
a F  k g(m  M )
and so will fall from it in time t.
(iv) If there is friction between B and floor and F > F  : l

A m
(where F =  (m+M)g = limiting friction between body B and
l s

L F surface)
M B
The system will move only if F  Fl' ' then replacing F by F  Fl  .
The entire case (iii) will be valid.
Fig. 5.30
 t
2L

2 ML However if F  F1  the system will not move and friction between
a F B and floor will be F while between A and B is zero.
(ii) If friction is present between A and B only and F < F l
 234 Friction

Motion of an Insect in the Rough Bowl For m 2 T  m2 g …(i)

The insect crawl up the bowl, up to a certain height h only till the For m1 T  m1 g sin  F
component of its weight along the bowl is balanced by limiting frictional
force.  T  m1 g sin  R

r
O  T  m1 g sin  m1 g cos  …(ii)

Fl R From equation (i) and (ii) m 2  m1[sin   cos  ]
y

A this is the minimum value of m 2 to start the motion

mg sin h

mg cos
Note :  In the above condition Coefficient of friction
mg
Let m = mass of the insect, r =Fig. 5.31 of the bowl,  = coefficient of
radius  m2 
friction   tan  
 1
m cos  
for limiting condition at point A
R  mg cos  ......(i) and Fl  mg sin ......(ii) Maximum Length of Hung Chain
Dividing (ii) by (i) A uniform chain of length l is placed on the table in such a manner
that its l' part is hanging over the edge of table without sliding. Since the
F
tan   l   As Fl  R chain have uniform linear density therefore the ratio of mass and ratio of
R
length for any part of the chain will be equal.
r2  y 2 r m mass hanging from the table
  or y We know  2 
y 1  2 m1 mass lyingon the table
 For this case we can rewrite above expression in the following
    manner
1 1
So h  r  y  r 1  ,  h  r 1  
 1  2   1  2  length hanging from the table
     [As chain have uniform linear
length lyingon the table
Minimum Mass Hung from the String to Just ( l – l )
density]
Start the Motion l
 
(1) When a mass m placed on a rough horizontal plane Another mass
1 l  l l
m2 hung from the string connected by frictionless pulley, the tension l
by solving l 
(T) produced in string will try to start the motion of mass m1 . (  1)
Fig. 5.34
R Coefficient of Friction Between a Body and Wedge
T A body slides on a smooth wedge of angle  and its time of descent
Fl m1
is t.
T
m1g
S S

m2 Smooth wedge Rough wedge

At limiting condition T  Fl
 
 m 2 g  R  m 2Fig.  m1 g
g 5.32 m2g

 m 2  m1 this is the minimum value of m 2 to start the Fig. 5.35 Fig. 5.36
If the same wedge made rough then time taken by it to come down
motion. becomes n times more (i.e. nt)
The length of path in both the cases are same.
m2
Note :  In the above condition Coefficient of friction  
m1
1
For smooth wedge, S  u t  at2
2
(2) When a mass m placed on a rough inclined plane Another 1
S  (g sin ) t 2
1

mass m 2 hung from the string connected by frictionless pulley, the …(i)
2
tension (T) produced in string will try to start the motion of mass m 1 . [Asu  0 and a  g sin ]
1 2
T For rough wedge, S  u t  at
R 2
T 1
S  g (sin   cos  ) (nt)2 …(ii)
m1 2
m2
[Asu  0 and a  g (sin   cos  )]
m1g sin  + F  m1g cos m2g From equation (i) and (ii)
At limiting condition
m1g
Fig. 5.33
 Friction 235

1
(g sin ) t 2 =
1
g (sin   cos  ) (nt)2 P2
 FS  [Where P = momentum of block]
2 2 2m
 sin  (sin   cos  ) n 2 P2
 mg  S  [As F =  mg]
 1  2m
   tan  1  2 
 n  P2
 S 
Stopping of Block Due to Friction 2 m 2 g

(1) On horizontal road In the given condition P and  are same for both the blocks.
(i) Distance travelled before coming to rest : A block of mass m is 2
1 S m 
moving initially with velocity u on a rough surface and due to friction, it So, S  ;  1  2
comes to rest after covering a distance S. m2 S 2  m1 
S
v=0 Velocity at the Bottom of Rough Wedge
u
A body of mass m which is placed at the top of the wedge (of height
Fig. 5.37 h) starts moving downward on a rough inclined plane.
Retarding force F  ma  R  ma   mg Loss of energy due to friction = FL (Work against friction)
 a  g PE at point A = mgh u=0
From v 2  u 2  2aS  0  u 2  2  g S A
m
1 L
[As v  0, a  g] KE at point B = mu 2
2 h
u2 P2
 S or S  m B
2 g 2 m 2 g
By the law of conservation ofvenergy
[As momentum P = mu] Fig. 5.40
(ii) Time taken to come to rest 1
From equation v  u  a t  0  u   g t
i.e. mv 2  mgh  FL
2
[Asv  0, a   g]
2
u v (mgh  FL)
 t m
g
(2) On inclined road : When block starts with velocity u its kinetic Sticking of a Block With Accelerated Cart
energy will be converted into potential energy and some part of it goes When a cart moves with some acceleration toward right then a
against friction and after travelling distance S it comes to rest i.e. v = 0. pseudo force (ma) acts on block toward left.
We know that retardation a  g [sin   cos  ] This force (ma) is action force by a block on cart.
By substituting the value of v and a in the following equation a
F
v=0
ma m F M
S R m
CART
u
mg
v 2  u 2  2a S 
Fig. 5.41
Fig. 5.38 Now block will remain static w.r.t. cart. If friction force R  mg
 0  u 2  2 g [sin   cos  ] S
 ma  mg [As R  ma]
2
u
 S  a
g
2 g (sin   cos  ) 
Stopping of Two Blocks Due to Friction g
 amin 
When two masses compressed towards each other and suddenly 
released then energy acquired by each block will be dissipated against This is the minimum acceleration of the cart so that block does not
friction and finally block comes to rest fall.
i.e., F × S = E [Where F = Friction, S = Distance covered and the minimum force to hold the block together
by block, E = Initial kinetic energy of the block] Fmin  (M  m) amin
A B g
Fmin  (M  m )
m1 m1 m2 m2 
Sticking of a Person with the Wall of Rotor
S1 S2
Fig. 5.39
 236 Friction
A person with a mass m stands in contact against the wall of a
cylindrical drum (rotor). The coefficient of friction between the wall and the
clothing is .
If Rotor starts rotating about its axis, then person thrown away from
the centre due to centrifugal force at a particular speed  , the person
stuck to the wall even the floor is removed, because friction force balances
its weight in this condition.
From the figure.
Friction force (F) = weight of person (mg)

 R = mg   Fc  mg F
[Here, F = centrifugal force] R
c
FC
 m min
2
r  mg mg

g
  min  Fig. 5.42
r

 Force of friction is non-conservative force.

 Force of friction always acts in a direction opposite to that of the
relative motion between the surfaces.

 Rolling friction is much less than the sliding friction. This

knowledge was used by man to invent the wheels.

 The friction between two surfaces increases (rather than to

decrease), when the surfaces are made highly smooth.

 The atomic and molecular forces of attraction between the two

surfaces at the point of contact give rise to friction between the surfaces.