FRICTION

Introduction.
If we slide or try to slide a body over a surface the motion is resisted by a bonding between the body and the
surface. This resistance is represented by a single force and is called friction.
The force of friction is parallel to the surface and opposite to the direction of intended motion.

Types of Friction.
(1) Static friction : The opposing force that comes into play when one body tends to move over the surface
of another, but the actual motion has yet not started is called static friction.
(i) If applied force is P and the body remains at rest then static friction R
F = P. P
F
(ii) If a body is at rest and no pulling force is acting on it, force of friction
on it is zero.
mg
(iii) Static friction is a self-adjusting force because it changes itself in
accordance with the applied force.
(2) Limiting friction: If the applied force is increased the force of static friction also increases. If the
applied force exceeds a certain (maximum) value, the body starts moving. This maximum value of static friction
up to which body does not move is called limiting friction.
(i) The magnitude of limiting friction between any two bodies in contact is directly proportional to the
normal reaction between them.
Fl  R or Fl   s R
(ii) Direction of the force of limiting friction is always opposite to the direction in which one body is at the
verge of moving over the other
(iii) Coefficient of static friction: (a)  s is called coefficient of static friction and defined as the ratio of force
F
of limiting friction and normal reaction  s 
R
(b) Dimension: [M 0 L0 T 0 ]
(c) Unit: It has no unit.
(d) Value of  s lies in between 0 and 1
(e) Value of  depends on material and nature of surfaces in contact that means whether dry or wet ; rough
or smooth polished or non-polished.
(f) Value of  does not depend upon apparent area of contact.
(3) Kinetic or dynamic friction: If the applied force is increased further and sets the body in motion,
the friction opposing the motion is called kinetic friction.
(i) Kinetic friction depends upon the normal reaction.
Fk  R or Fk   k R where  k is called the coefficient of kinetic friction
(ii) Value of  k depends upon the nature of surface in contact.
(iii) Kinetic friction is always lesser than limiting friction Fk  Fl  k   s
i.e. coefficient of kinetic friction is always less than coefficient of static friction. Thus we require more force
to start a motion than to maintain it against friction. This is because once the motion starts actually ; inertia of

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

rest has been overcome. Also when motion has actually started, irregularities of one surface have little time
to get locked again into the irregularities of the other surface.
(iv) Types of kinetic friction
(a) Sliding friction : The opposing force that comes into play when 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.
(b) Rolling friction : When objects such as a wheel (disc or ring), sphere or a cylinder rolls over a surface,
the force of friction comes into play is called rolling friction.
 Rolling friction is directly proportional to the normal reaction (R) and inversely proportional to the radius
(r) of the rolling cylinder or wheel.
R
Frolling   r
r
 r is called coefficient of rolling friction. It would have the dimensions of length and would be measured in
meter.
 Rolling friction is often quite small as compared to the sliding friction. That is why heavy loads are
transported by placing them on carts with wheels.
 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 center
of the wheel moves forward.
Graph Between Applied Force and Force of Friction.
(1) Part OA of the curve represents static friction (Fs ) . Its value increases linearly with the applied force
(2) At point A the static friction is maximum. This represent
A
limiting friction (Fl ) .
Force of friction

B C
(3) Beyond A, the force of friction is seen to decrease slightly. The
portion BC of the curve therefore represents the kinetic friction (Fk ) . Fl Fk

(4) As the portion BC of the curve is parallel to x-axis therefore

O Applied force
kinetic friction does not change with the applied force, it remains
constant, and whatever be the applied force.
Friction is a Cause of Motion.
It is a general misconception that friction always opposes the motion. No doubt friction opposes the motion
of a moving body but in many cases it is also the cause of motion. For example:
(1) In moving, a person or vehicle pushes the ground backwards (action)
and the rough surface of ground reacts and exerts a forward force due
Frictio
to friction which causes the motion. If there had been no friction there  n
Actio
will be slipping and no motion. n
(2) In cycling, the rear wheel moves by the force communicated to it by pedalling while front wheel moves
by itself. So, when pedalling a bicycle, the force exerted by rear wheel on ground makes force of friction act on it
in the forward direction (like walking). Front wheel moving by itself experience force of friction in backward
direction (like rolling of a ball). [However, if pedalling is stopped both wheels move by themselves and so
experience force of friction in backward direction.]

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

While pedalling Pedalling is stoped

(3) If a body is placed in a vehicle which is accelerating, the force of friction is the cause of motion of the
body along with the vehicle (i.e., the body will remain at rest in the accelerating vehicle until ma   s mg ). If there
had been no friction between body and vehicle the body will not move along with the vehicle.

a
smg ma

From these examples it is clear that without friction motion cannot be started, stopped or transferred from one
body to the other.
Advantages and Disadvantages of Friction.
(1) Advantages of friction
(i) Walking is possible due to friction.
(ii) Two body sticks together due to friction.
(iii) Brake works on the basis of friction.
(iv) Writing is not possible without friction.
(v) The transfer of motion from one part of a machine to other part through belts is possible by friction.
(2) Disadvantages of friction
(i) Friction always opposes the relative motion between any two bodies in contact. Therefore extra energy
has to be spent in overcoming friction. This reduces the efficiency of machine.
(ii) Friction causes wear and tear of the parts of machinery in contact. Thus their lifetime reduces.
(iii) Frictional force result in the production of heat, which causes damage to the machinery.
Methods of Changing Friction.
We can reduce friction
(1) By polishing.
(2) By lubrication.
(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 tires,
synthetic rubber is preferred because its coefficient of friction with the road is larger.

Angle of Friction.
Angle of friction may be defined as the angle which the resultant of limiting friction and normal reaction
makes with the normal reaction.
By definition angle  is called the angle of friction R
S
F 
tan   F P
R
mg

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

F
 tan  =  [As we know  ]
R
or   tan 1 ( )
Hence coefficient of limiting friction is equal to tangent of the angle of friction.
Resultant Force Exerted by Surface on Block.
In the above figure resultant force S  F 2  R 2
S  (mg ) 2  (mg ) 2

S  mg  2  1
when there is no friction (  0 ) S will be minimum i.e., S = mg
Hence the range of S can be given by, mg  S  mg  2  1
Angle of Repose.
Angle of repose is defined as the angle of the inclined plane with horizontal such that a body placed on it is
just begins to slide.
R
By definition  is called the angle of repose. F
In limiting condition F  mg sin 
mg sin 
and R  mg cos 

F mg cos 
So  tan   mg
R
F F
    tan   tan  [As we know    tan  ]
R R
Thus the coefficient of limiting friction is equal to the tangent of angle of repose.
As well as    i.e. angle of repose = angle of friction.
Calculation of Necessary Force in Different Conditions.
If W = weight of the body,  = angle of friction,   tan   coefficient of friction
then we can calculate necessary force for different condition in the following manner :
(1) Minimum pulling force P at an angle  from the horizontal
By resolving P in horizontal and vertical direction (as shown in figure)
For the condition of equilibrium R
F  P cos  and R  W  P sin  P P sin
By substituting these value in F  R F P cos

P cos    (W  P sin  )
sin  W
 P cos   (W  P sin  ) [As   tan  ]
cos 
W sin 
 P
cos (   )
(2) Minimum pushing force P at an angle  from the horizontal
By Resolving P in horizontal and vertical direction (as shown in the figure) R
For the condition of equilibrium
P F P cos
F  P cos  and R  W  P sin 

By substituting these value in F  R P sin
W

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

 P cos    (W  P sin  )
sin 
 P cos   (W  P sin  ) [As   tan  ]
cos 
W sin 
 P
cos (   )
(3) Minimum pulling force P to move the body up an inclined plane
By Resolving P in the direction of the plane and perpendicular to the plane (as shown in the figure)
For the condition of equilibrium
R  P sin   W cos  P R + P sin
 P cos
 R  W cos   P sin 
F + W sin
and F  W sin   P cos 

 W cos
 F  P cos   W sin  

By substituting these values in F  R and solving we get

W sin (  )
P
cos (   )
(4) Minimum force on body in downward direction along the surface of inclined plane to start its
motion
By Resolving P in the direction of the plane and perpendicular to the plane (as shown in the figure)
For the condition of equilibrium R + P sin
P F
R  P sin   W cos  P cos

+
 R  W cos   P sin  W sin 
 W cos
and F  P cos   W sin   W

By substituting these values in F  R and

solving we get
W sin(  )
P
cos (   )
(5) Minimum force to avoid sliding a body down an inclined plane
By Resolving P in the direction of the plane and perpendicular to the plane (as shown in the figure)
For the condition of equilibrium
P R + P sin F + P cos
R  P sin   W cos  

 R  W cos   P sin 
W sin 
and P cos   F  W sin    W
W cos

 F  W sin   P cos 
By substituting these values in F  R and solving we get

 sin (   ) 
P W 
 cos (   ) 

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

(6) Minimum force for motion and its direction

Let the force P be applied at an angle  with the horizontal.
By resolving P in horizontal and vertical direction (as shown in figure) P
For vertical equilibrium

R  P sin   mg
 R  mg  P sin  ….(i)
and for horizontal motion
P cos   F
i.e. P cos   R ….(ii) R + P sin

Substituting value of R from (i) in (ii)

F P cos
P cos    (mg  P sin  )
 mg
P ….(iii) mg
cos    sin 
For the force P to be minimum (cos    sin  ) must be maximum i.e.
d
[cos    sin  ]  0   sin    cos   0
d
 tan   
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
 1
As tan    so from the figure sin   and cos  
1 2
1 2
By substituting these value in equation (iii)
 mg  mg
P 
1  2
1 2 

1 2 1  2 
1
mg
 Pmin 
1 2
Acceleration of a Block against Friction.
(1) Acceleration of a block on horizontal surface
When body is moving under application of force P, then kinetic friction opposes its motion.
Let a is the net acceleration of the body
R
From the figure ma
Fk
ma  P  Fk P

P  Fk
 a
m mg

(2) Acceleration of a block down a rough inclined plane

When angle of inclined plane is more than angle of repose, the body placed on the inclined plane slides
down with an acceleration a.
From the figure ma  mg sin   F

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

R
 ma  mg sin   R ma
F

 ma  mg sin    mg cos 
 Acceleration a  g [sin    cos  ] mg sin  mg cos
 mg
Note :  For frictionless inclined plane   0  a  g sin  .
(3) Retardation of a block up a rough inclined plane
When angle of inclined plane is less than angle of repose, then for the upward motion
ma  mg sin   F ma
R
ma  mg sin    mg cos 
Retardation a  g [sin    cos  ]
Note :  For frictionless inclined plane   0  a  g sin  mg sin + F 
 mg
mg cos

Work Done Against Friction.

(1) Work done over a rough inclined surface
If a body of mass m is moved up on a rough inclined plane through distance s, then
Work done = force  distance
= ma  s ma
R
= mg [sin +  cos ]s
 mg s [sin    cos  ] s

mg sin +  mg
F  mg cos
(2) Work done over a horizontal surface
In the above expression if we put  = 0 then
Work done = force  distance
R
=Fs
=  mg s F P

It is clear that work done depends upon

(i) Weight of the body. mg
s
(ii) Material and nature of surface in contact.
(iii) Distance moved.

Motion of Two Bodies One Resting on the Other.

When a body A of mass m is resting on a body B of mass M then two
conditions are possible A m

(1) A force F is applied to the upper body, (2) A force F is applied to the L F
M B
lower body
We will discuss above two cases one by one in the following manner :
(1) A force F is applied to the upper body, then following four
situations are possible
(i) When there is no friction m A F
(a) The body A will move on body B with acceleration (F/m).
L
aA  F / m M B
(b) The body B will remain at rest
aB  0

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

(c) If L is the length of B as shown in figure A will fall from B after time t
2L 2mL  1 
t   As s  a t 2 and a  F/m 
a F  2 
(ii) If friction is present between A and B only and applied force is less than limiting friction (F < Fl)
(F = Applied force on the upper body, Fl = limiting friction between A and B, Fk = Kinetic friction between A and
B)
(a) The body A will not slide on body B till F  Fl i.e. F   s mg
F
(b) Combined system (m + M) will move together with common acceleration a A  a B 
M m
(iii) If friction is present between A and B only and applied force is greater than limiting friction (F > Fl )
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  k mg will oppose the motion of A while will cause the motion of B.
F  Fk  m a A Free body diagram of A Fk  M a B Free body diagram of B
maA MaB
F  Fk Fk
i.e. a A  i.e. aB  FK
m A F M
B
(F   k mg )  k mg
aA   aB 
m Fk M

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

MF   k mg (m  M )
Acceleration of body A relative to B will be a  a A  a B 
mM
2L 2 m ML
So, A will fall from B after time t  
a MF   k mg (m  M )
(iv) If there is friction between B and floor
(where Fl    (M  m ) g = limiting friction between B and floor, Fk = kinetic friction between A and B)
B will move only if Fk  Fl and then Fk  Fl  M a B
MaB
However if B does not move then static friction will work (not limiting friction) FK
between body B and the floor i.e. friction force = applied force (= Fk) not Fl . B
(2) A force F is applied to the lower body, then following four Fl
situations are possible
(i) When there is no friction
(a) B will move with acceleration (F/M) while A will remain at rest (relative to ground) as there is no pulling
force on A.
F
a B    and a A  0
M
(b) As relative to B, A will move backwards with acceleration (F/M) and so will fall from it in time t.
2L 2 ML
 t 
a F
(ii) If friction is present between A and B only and F < Fl
(where F = Pseudo force on body A and Fl = limiting friction between body A and B)
F
(a) Both the body will move together with common acceleration a 
M m

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

mF
(b) Pseudo force on the body A, F   ma  and Fl   s mg
mM
mF
(c) F   Fl    s mg  F   s (m  M ) g
mM
F
So both bodies will move together with acceleration a A  a B  if F   s [m  M ] g
mM
(iii) If friction is present between A and B only and F > Fl
(where Fl = s (m + M)g = limiting friction between body B and surface)
Both the body will move with different acceleration. Here force of kinetic friction  k mg will oppose the
motion of B while will cause the motion of A.
ma A  k mg Free body diagram of A F  Fk  Ma B Free body diagram of B
MaB
i.e. a A   k g [F   k mg ]
maA i.e. a B  FK
A M F
Fk B

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

Acceleration of body A relative to B will be
 F   k g(m  M ) 
a  a A  a B   
 M 
Negative sign implies that relative to B, A will move backwards and will fall it after time
2L 2 ML
t 
a F   k g(m  M )

(iv) If there is friction between B and floor: The system will move only if F  Fl' then replacing F by
F  Fl . The entire case (iii) will be valid.
However if F  Fl the system will not move and friction between B and floor will be F while between A and
B is zero.
Motion of an Insect in the Rough Bowl.
The insect crawl up the bowl up to a certain height h only till the component of its weight along the bowl is
balanced by limiting frictional force.
Let m = mass of the insect, r = radius of the bowl,  = coefficient of friction
for limiting condition at point A
O
R  mg cos  ......(i) and Fl  mg sin  ......(ii) r

Fl R
Dividing (ii) by (i) y

As Fl  R 
Fl A
tan    mg sin h
R
mg cos
mg
r2  y2 r
  or y 
y 1 2
 1   1 
So h  r  y  r 1   ,  h  r 1  
 1 2   1 2 
   

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

Minimum Mass Hung From the String to Just Start the Motion.
(1) When a mass m1 placed on a rough horizontal plane : Another mass m 2 hung from the string
connected by pulley, the tension (T) produced in string will try to start the motion of mass m 1 .
At limiting condition
T  Fl R
T
 m 2 g  R F1 m1

 m 2 g   m1 g m1g T

 m 2  m 1 this is the minimum value of m 2 to start the motion. m2

m2
Note :  In the above condition Coefficient of friction  
m1
(2) When a mass m1 placed on a rough inclined plane: Another mass m 2 hung from the string
connected by pulley, the tension (T) produced in string will try to start the motion of mass m 1 .
At limiting condition
For m 2 T  m2g ...... (i)
T
R
For m 1 T  m 1 g sin   F  T  m 1 g sin   R T
m1
 T  m 1 g sin   m 1 g cos  ......(ii) m2

From equation (i) and (ii) m 2  m 1 [sin    cos  ] m1g sin + F  m1g m2g
cos
this is the minimum value of m 2 to start the motion m1g

Note :  In the above condition Coefficient of friction

 m2 
  tan  
 m 1 cos  
Maximum Length of Hung Chain .
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 with out sliding. Since the chain have uniform linear density therefore the ratio of mass or ratio of length
for any part of the chain will be equal.
m mass hanging from the table ( l – l
We know   2  )
m1 mass lying on the table
l

 For this expression we can rewrite above expression in the following manner
length hanging from the table
 [As chain have uniform linear density]
length lying on the table
l

l  l
l
by solving l  
(  1)

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

Coefficient of Friction Between Body and Wedge .

A body slides on a smooth wedge of angle 
and its time of descent is t.
S S
Smooth wedge Rough wedge
 

If the same wedge made rough then time taken by it to come down becomes n times more (i.e. nt)
The length of path in both the cases are same.

For smooth wedge For rough wedge

1 1
S  u t  at 2 S  u t  at 2
2 2
1 1
S  (g sin  ) t 2 .....(i) S  g (sin    cos  ) (nt ) 2 .....(ii)
2 2
[As u  0 and a  g sin  ] [As u  0 and a  g (sin    cos  )]
1 1
From equation (i) and (ii) (g sin  ) t 2 = g (sin    cos  ) (nt ) 2
2 2
 sin   (sin    cos  ) n 2
 1 
   tan  1 
 n 2 
Stopping of Block Due to Friction.
(1) On horizontal road
(i) Distance travelled before coming to rest: A block of mass m is moving initially with velocity u on
a rough surface and due to friction it comes to rest after covering a distance S.
Retarding force F  ma  R
 ma   mg
 a  g . S
v=0
u
From v 2  u 2  2aS  0  u 2  2  g S [As v  0, a  g]
u2
 S 
2 g
P2
or S  [As momentum P = mu]
2 m 2 g
(ii) Time taken to come to rest
From equation v  u  a t  0  u   g t [As v  0, a   g]
u
t 
g
(iii) Force of friction acting on the body
We know, F = ma
(v  u)
So, Fm
t

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

mu
F [As v = 0]
t
 u 
F   mg  As t 
 g 
(2) On inclined road : When block starts with velocity u its kinetic energy will be converted into potential
energy and some part of it goes against friction and after travelling distance S it comes to rest i.e. v = 0.
And we know that retardation a  g [sin    cos  ]
By substituting the value of v and a in the following equation v=0

v  u  2a S
2 2
S
u
 0  u  2 g [sin    cos  ] S
2


u2
S 
2 g (sin    cos  )
Stopping of Two Blocks Due to Friction.
When two masses compressed towards each other and suddenly released then energy acquired by each
block will be dissipated against friction and finally block comes to rest
i.e., F × S = E [Where F = Friction, S = Distance covered by block, E = Initial kinetic energy of the
block]
P2
 FS  [Where P = momentum of block] A B
2m
m1 m1 m2 m2
P2
 mg  S  [As F =  mg] S1 S2
2m
P2
 S 
2 m 2 g
In a given condition P and  are same for both the blocks.
2
1 S m 
So S  2  1   2 
m S 2 m1 
Velocity at the Bottom of Rough Wedge. :
A body of mass m which is placed at the top of the wedge (of height h) starts moving downward on a rough
inclined plane.
Loss of energy due to friction = FL (Work against friction)
PE at point A = mgh u=0
1 m
KE at point B = mv 2 A
2 L

By the law of conservation of energy h

1 m B
i.e. mv 2  mgh  FL v
2
2
v (mgh  FL)
m

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

Sticking of a Block With Accelerated Cart.

When a cart moves with some acceleration toward right then a pseudo force (ma) acts on block toward left.
This force (ma) is action force by a block on cart.
Now block will remain static w.r.t. block. If friction force R  mg a
 ma  mg [As R  ma ] F
F M
g m
 a ma m R CART

g mg
 a min 

This is the minimum acceleration of the cart so that block does not fall.
and the minimum force to hold the block together
g
Fmin  (M  m ) a min OR Fmin  (M  m )

Sticking of a Person With the Wall of Rotor.
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 w, 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)
F
 R = mg
R
  Fc  mg [Here, Fc= centrifugal force] FC
mg
 m  min
2
r  mg
g
  min 
r

Sample problems based on fundamentals of friction

Problem 1. If a ladder weighing 250N is placed against a smooth vertical wall having coefficient of friction between it
and floor is 0.3, then what is the maximum force of friction available at the point of contact between the
ladder and the floor
(a) 75 N (b) 50 N (c) 35 N (d) 25 N
Solution : (a) Maximum force of friction Fl  s R  0.3  250  75 N
Problem 2. On the horizontal surface of a truck (  = 0.6), a block of mass 1 kg is placed. If the truck is accelerating at
the rate of 5m/sec2 then frictional force on the block will
be
(a) 5 N (b) 6 N (c) 5.88 N (d) 8 N
Solution : (a) Limiting friction  s R  smg  0.6  1  9.8  5.88 N
When truck accelerates in forward direction at the rate of 5m / s 2 a pseudo force (ma ) of 5N works on block
in back ward direction. Here the magnitude of pseudo force is less than limiting friction So, static friction
works in between the block and the surface of the truck and as we know, static friction = Applied force =
5N.

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

Problem 3. A block of mass 2 kg is kept on the floor. The coefficient of static friction is 0.4. If a force F of 2.5 N is
applied on the block as shown in the figure, the frictional force between the block and the floor will be [MP
PET 2000]
(a) 2.5 N
F
(b) 5 N
(c) 7.84 N
(d) 10 N
Solution : (a) Applied force = 2.5 N and limiting friction = mg = 0.4 × 2 × 9.8 = 7.84 N

As applied force is less than limiting friction. So, for the given condition static friction will work.
Static friction on a body = Applied force = 2.5 N.
Problem 4. A block A with mass 100 kg is resting on another block B of mass 200 kg. As shown in figure a horizontal
rope tied to a wall holds it. The coefficient of friction between A and B is 0.2 while coefficient of friction
between B and the ground is 0.3. The minimum required force F to start moving B will be

(a) 900 N
(b) 100 N
A
(c) 1100 N A fAB
(d) 1200 N B F
Solution : (c) Two frictional force will work on block B. B F
fBG Groun
F  f AB  fBG   AB m a g  BG (m A  m B )g d
= 0.2 × 100 × 10 + 0.3 (300) × 10
= 200 + 900 = 1100N. (This is the required minimum force)
Problem 5. A 20 kg block is initially at rest on a rough horizontal surface. A horizontal force of 75 N is required to set
the block in motion. After it is in motion, a horizontal force of 60 N is required to keep the block moving
with constant speed. The coefficient of static friction is

(a) 0.38 (b) 0.44 (c) 0.52 (d) 0.60

Fl 75
Solution : (a) Coefficient of static friction S    0 .38 .
R 20  9 .8
Problem 6. A block of mass M is placed on a rough floor of a lift. The coefficient of friction between the block and the
floor is . When the lift falls freely, the block is pulled horizontally on the floor. What will be the force of
friction
(a)  Mg (b)  Mg/2 (c) 2 Mg (d) None of these
Solution : (d) When the lift moves down ward with acceleration 'a' then effective acceleration due to gravity
g' = g – a
g'  g  g  0 [As the lift falls freely, so a = g]
So force of friction  mg '  0
Sample problems based on angle of friction and angle of repose

Problem 7. A body of 5 kg weight kept on a rough inclined plane of angle 30o starts sliding with a constant velocity. Then
the coefficient of friction is (assume g = 10 m/s2)
(a) 1 / 3 (b) 2 / 3 (c) 3 (d) 2 3
Solution : (a) Here the given angle is called the angle of repose
1
So,   tan 30 o 
3

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

Problem 8. The upper half of an inclined plane of inclination  is perfectly smooth while the lower half is rough. A body
starting from the rest at top comes back to rest at the bottom if the coefficient of friction for the lower half
is given
(a)  = sin  (b)  = cot  (c)  = 2 cos  (d)  = 2 tan 
Solution : (d) For upper half by the equation of motion v 2  u 2  2as
v 2  0 2  2(g sin  )l / 2  gl sin  [As u  0, s  l / 2, a  g sin  ]
For lower half Smooth

0  u  2 g(sin    cos  ) l /2 [As v  0, s  l / 2, a  g (sin    cos  ) ]

2
Rough

 0  gl sin   gl(sin    cos  ) [As final velocity of upper half will be equal to the initial velocity of lower half]
 2 sin    cos     2 tan 

Sample problems based on force against friction

Problem 9. What is the maximum value of the force F such that the block shown in the arrangement, does not move (
 1/2 3 )
R
(a) 20 N F m = 3kg
f F cos 60°
(b) 10 N 60o
(c) 12 N
(d) 15 N
W+F sin 60°
Solution : (a) Frictional force f  R
 F cos 60  (W  F sin 60 )

 F cos 60 
2 3
1
 3 g  F sin 60 
 F  20 N .
Problem 10. A block of mass m rests on a rough horizontal surface as shown in the figure. Coefficient of friction between
the block and the surface is . A force F = mg acting at angle  with the vertical side of the block pulls it. In
which of the following cases the block can be pulled along the surface
(a) tan    R+mg
(b) cot    mg = F cos
 f mg sin  =p
(c) tan  / 2   m
(d) cot  / 2  
Solution : (d) For pulling of block P  f
mg
 mg sin   R  mg sin    (mg  mg cos  )
 sin    (1  cos  )
     
 2 sin cos    2 sin 2   cot   
2 2  2 2
Sample problems based on acceleration against friction
Problem 11. A body of mass 10 kg is lying on a rough plane inclined at an angle of 30o to the horizontal and the coefficient
of friction is 0.5. The minimum force required to pull the body up the plane is
(a) 914 N (b) 91.4 N (c) 9.14 N (d) 0.914 N
Solution : (b) F  mg (sin    cos  )  10  9.8 (sin 30  0.5 cos 30 )  91 .4 N
Problem 12. A block of mass 10 kg is placed on a rough horizontal surface having coefficient of friction  = 0.5. If a
horizontal force of 100 N is acting on it, then acceleration of the block will be

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

(a) 0.5 m/s2 (b) 5 m/s2 (c) 10 m/s2 (d) 15 m/s2

Applied force – kinetic friction 100  0 .5  10  10
Solution : (b) a  = 5m/s2.
mass 10
Problem 13. A body of weight 64 N is pushed with just enough force to start it moving across a horizontal floor and the
same force continues to act afterwards. If the coefficients of static and dynamic friction are 0.6 and 0.4
respectively, the acceleration of the body will be (Acceleration due to gravity = g)
g g
(a) (b) 0.64 g (c) (d) 0.2 g
6 .4 32
64
Solution : (d) Limiting friction = Fl   s R  64 = 0.6 m g  m = .
0 .6 g
64
64  0 . 4 
Applied force – Kinetic friction 64   K mg 0 . 6 = 0.2g
Acceleration = = =
Mass of the body m 64
0 .6 g
Problem 14. If a block moving up at   30 o with a velocity 5 m/s, stops after 0.5 sec, then what is

(a) 0.5 (b) 1.25 (c) 0.6 (d) None of these
u
Solution : (c) From v  u  at  0  u  at  t 
a
u
for upward motion on an inclined plane a  g(sin    cos  ) t 
g(sin    cos  )
Substituting the value of   30 o , t  0.5 sec and u  5 m / s , we get   0.6
Sample problems based on work done against friction
Problem 15. A body of mass 5kg rests on a rough horizontal surface of coefficient of friction 0.2. The body is pulled
through a distance of 10m by a horizontal force of 25 N. The kinetic energy acquired by it is (g = 10 ms2)

(a) 330 J (b) 150 J (c) 100 J (d) 50 J

Solution : (b) Kinetic energy acquired by body = Total work done on the body – Work done against friction
= F  S  mgS = 25 × 10 – 0.2 × 5 × 10 ×10 = 250 – 100 = 150 J.
Problem 16. 300 Joule of work is done in sliding a 2 kg. block up an inclined plane to a height of 10 meters. Taking value
of acceleration due to gravity ‘g’ to be 10 m/s2, work done against friction is
(a) 100 J (b) 200 J (c) 300 J (d) Zero
Solution : (a) Work done against gravity = mgh = 2 × 10 × 10 = 200J
Work done against friction = Total work done – Work done against gravity = 300 – 200 = 100J.
Problem 17. A block of mass 1 kg slides down on a rough inclined plane of inclination 60 o starting from its top. If
the coefficient of kinetic friction is 0.5 and length of the plane is 1 m, then work done against friction
is (Take g = 9.8 m/s 2)
(a) 9.82 J (b) 4.94 J (c) 2.45J (d) 1.96 J
1
Solution : (c) W  mg cos  .S  0.5  1  9.8  = 2.45 J.
2
Problem 18. A block of mass 50 kg slides over a horizontal distance of 1 m. If the coefficient of friction between their
surfaces is 0.2, then work done against friction is
(a) 98 J (b) 72J (c) 56 J (d) 34 J
Solution : (a) W  mgS  0.2  50  9.8  1  98 J .

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

Sample problems based on body resting on another

Problem 19. A 4 kg block A is placed on the top of a 8 kg block B which rests on a smooth table. A just slips on B when a
force of 12 N is applied on A. Then the maximum horizontal force on B to make both A and B move together,
is
(a) 12 N (b) 24 N (c) 36 N (d) 48 N
Solution : (c) Maximum friction i.e. limiting friction between A and B, Fl = 12 N.
If F is the maximum value of force applied on lower body such that both body move together
It means Pseudo force on upper body is just equal to limiting friction
 F   4 
F'  Fl  m    F  12  F  36 N .
m  M  4 8 
Problem 20. A body A of mass 1 kg rests on a smooth surface. Another body B of mass 0.2 kg is placed over A as shown.
The coefficient of static friction between A and B is 0.15. B will begin to slide on A if A is pulled with a force
greater than
(a) 1.764 N (b) 0.1764 N B
(c) 0.3 N (d) It will not slide for any F
A
Solution : (a) B will begin to slide on A if Pseudo force is more than limiting friction
 F   F 
F'  Fl  m    s R  m    0 .15 mg F  1.764 N
m  M  m  M 
Problem 21. A block A of mass 2 kg rests on another block B of mass 8 kg which rests on a horizontal floor. The coefficient
of friction between A and B is 0.2, while that between B and floor is 0.5. When a horizontal force of 25 N is
applied on the block B, the force of friction between A and B is [IIT-JEE 1993]
(a) Zero (b) 3.9 N (c) 5.0 N (d) 49 N
Solution : (a) Limiting friction between the block B and the surface
FBS  BS .R  0.5 m  M  g  0.5(2  8)10  50 N
A 2kg
25 N
but the applied force is 25 N so the lower block will not move i.e. B 8kg
there is no pseudo force on upper block A. Hence there will be no
Surface
force of friction between A and B.
Problem 22. An insect crawls up a hemispherical surface very slowly (see the figure). The coefficient of friction between
the insect and the surface is 1/3. If the line joining the centre of the hemispherical surface to the insect
makes an angle  with the vertical, the maximum possible value of  is given by [IIT-JEE (Screening) 2001]
(a) cot   3
(b) tan   3

(c) sec   3
(d) cosec   3

Solution : (a) From the above expression, for the equilibrium R  mg cos  and F  mg sin  .
1
Substituting these value in F  R we get tan    or cot   3.


ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

Sample problems based on hung mass

Problem 23. Two blocks of mass M1 and M 2 are connected with a string passing over a pulley as shown in the figure.
The block M1 lies on a horizontal surface. The coefficient of friction between the block M1 and horizontal
surface is . The system accelerates. What additional mass m should be placed on the block M1 so that the
system does not accelerate
M 2  M1 m
(a)

M1
M2
(b)  M1

M1 M2
(c) M 2 

(d) (M 2  M1 )
Solution : (b) By comparing the given condition with general expression
M2 M M
  m  M 1  2  m  2  M1
m  M1  
Problem 24. The coefficient of kinetic friction is 0.03 in the diagram where mass m 2  20 kg and m1  4 kg . The
acceleration of the block shall be (g  10 ms 2 )
T
20 kg m2
(a) 1.8 ms 2
T

(b) 0.8 ms 2
m1 4 kg

(c) 1.4 ms 2

(d) 0.4 ms 2 m2a

F T
Solution : (c) Let the acceleration of the system is a m2

From the F.B.D. of m 2

T  F  m 2 a  T  m 2 g  m 2a
T
 T  0.03  20  10  20 a  T  6  20 a .....(i)
From the FBD of m1
m1 m1 a
m1 g  T  m 1 a
 4  10  T  4 a  40  T  4 a ....(ii) F
m1g
Solving (i) and (ii) a  1.4 m / s 2 .

Problem 25. A heavy uniform chain lies on a horizontal table top. If the coefficient of friction between the chain and the
table surface is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of
the table is
(a) 20% (b) 25% (c) 35% (d) 15%
    0 . 25 
Solution : (a) From the expression l'    l   l [As  = 0.25]
   1   0 . 25  1 
0 . 25 l
 l'  l  = 20% of the length of the chain.
1 .25 5

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

Problem 26. A body takes just twice the time as long to slide down a plane inclined at 30o to the horizontal as if the plane
were frictionless. The coefficient of friction between the body and the plane is
3 4 3
(a) (b) 3 (c) (d)
4 3 4
 1   1  3
Solution : (a)   tan   1    tan 30  1  2   .
 n   2 
2
4

Sample problems based on motion of body on rough surface

Problem 27. A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10s. Then
the coefficient of friction is

(a) 0.01 (b) 0.02 (c) 0.03 (d) 0.06

u 6
Solution : (d) v  u  at  u  g t  0     0 . 06 .
gt 10  10
Problem 28. A 2 kg mass starts from rest on an inclined smooth surface with inclination 30o and length 2 m. How much
will it travel before coming to rest on a surface with coefficient of friction 0.25 [UPSEAT 2003]
(a) 4 m (b) 6 m (c) 8 m (d) 2 m
Solution : (a) v  u  2aS  0  2  g sin 30  2
2 2
v 20
Let it travel distance ‘S’ before coming to rest
v2 20
S   4 m.
2 g 2  0 .25  10
Sample problems (Miscellaneous)
Problem 29. A motorcycle is travelling on a curved track of radius 500m if the coefficient of friction between road and tyres
is 0.5. The speed avoiding skidding will be [MH CET (Med.)
2001]
(a) 50 m/s (b) 75 m/s (c) 25 m/s (d) 35 m/s
Solution : (a) v  rg  0.5  500  10  50 m/s.
Problem 30. A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction
between the block and the wall is 0.2. The weight of the block is
(a) 2 N F
(b) 20 N
10 N
(c) 50 N R 10 N
(d) 100 N
Solution : (a) For equilibrium W

Weight (W) = Force of friction (F)

W  R  0.2  10 = 2N

Problem 31. A body of mass 2 kg is kept by pressing to a vertical wall by a force of 100 N. The friction between wall and
body is 0.3. Then the frictional force is equal to
(a) 6 N (b) 20 N (c) 600 N (d) 700 N
Solution : (b) For the given condition Static friction = Applied force = Weight of body = 2 × 10 = 20 N.

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

Problem 32. A fireman of mass 60kg slides down a pole. He is pressing the pole with a force of 600 N. The
coefficient of friction between the hands and the pole is 0.5, with what acce leration will the fireman
slide down (g = 10 m/s 2 )
(a) 1 m/s2 (b) 2.5 m/s2 (c) 10 m/s2 (d) 5 m/s2
Solution : (d) Friction = R = 0.5 × 600 = 300 N, Weight = 600 N F
W  F 600  300
ma = W – F  a   R 600 N
m 60
W
 a  5 m/s 2
Problem 33. The system shown in the figure is in equilibrium. The maximum value of W, so that the maximum value of
static frictional force on 100 kg. body is 450 N, will be
(a) 100 N 100 kg 45o

(b) 250 N
(c) 450 N W
(d) 1000 N
W
Solution : (c) For vertical equilibrium T1 sin 45 o  W  T1 
sin 45 o T1 sin 45o T1
W F T2 45o
For horizontal equilibrium T2  T1 cos 45 o  cos 45 o = W
sin 45 o T1 cos 45o

and for the critical condition T2 = F  W = T2 = F = 450 N W

Practice Problems
 Basic level
1. When a body is moving on a surface, the force of friction is called [MP PET 2002]
(a) Static friction (b) Dynamic friction (c) Limiting friction (d) Rolling friction
2. Which one of the following is not used to reduce friction [Kerala (Engg.) 2001]
(a) Oil (b) Ball bearings (c) Sand (d) Graphite
3. A block of mass 10 kg is placed on an inclined plane. When the angle of inclination is 30o, the block just begins to slide down the
plane. The force of static friction is
[Kerala (Engg.) 2001]
(a) 10 kg wt (b) 89 kg wt (c) 49 kg wt (d) 5 kg wt
4. A vehicle of mass m is moving on a rough horizontal road with momentum P. If the coefficient of friction between the tyres and
the road be , then the stopping distance is [CBSE PMT 2001]

P P2 P P2
(a) (b) (c) (d)
2 m g 2 m g 2 m 2 g 2 m 2g

5. A box is lying on an inclined plane what is the coefficient of static friction if the box starts sliding when an angle of inclination is
60o
[KCET (Engg./Med.) 2000]
(a) 1.173 (b) 1.732 (c) 2.732 (d) 1.677
6. A brick of mass 2 kg begins to slide down on a plane inclined at an angle of 45o with the horizontal. The force of friction will be
[CPMT 2000]
(a) 19.6 sin 45o (b) 19.6 cos 45o (c) 9.8 sin 45o (d) 9.8 cos 45o
7. To avoid slipping while walking on ice, one should take smaller steps because of the [BHU 1999]
(a) Friction of ice is large (b) Larger normal reaction
(c) Friction of ice is small (d) Smaller normal reaction

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

8. Two bodies having the same mass, 2 kg each have different surface areas 50 m 2 and 100 m 2 in contact with a horizontal plane. If
the coefficient of friction is 0.2, the forces of friction that come into play when they are in motion will be in the ratio
[EAMCET (Med.) 1999]
(a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 1 : 4
9. Starting from rest, a body slides down a 45 o inclined plane in twice the time it takes to slide down the same distance in the absence
of friction. The coefficient of friction between the body and the inclined plane is [CBSE PMT 1990]
(a) 0.33 (b) 0.25 (c) 0.75 (d) 0.80
10. Brakes of very small contact area are not used although friction is independent of area, because friction
(a) Resists motion (b) Causes wear and tear
(c) Depends upon the nature of materials (d) Operating in this case is sliding friction
11. The angle between frictional force and the instantaneous velocity of the body moving over a rough surface is
(a) Zero (b) /2
(c)  (d) Equal to the angle of friction
12. What happens to the coefficient of friction, when the normal reaction is halved
(a) Halved (b) Doubled
(c) No change (d) Depends on the nature of the surface
13. What can be inferred regarding the limiting frictional force in the following four figures
R R R
R

A C
B D

mg mg
(a) FA  FB  Fmg
C  FD (b) FA  FB  FC  FD (c) FA  FB  FC  Fmg
D (d) FA  FB  FC  FD

14. A force of 98 Newton is required to drag a body of mass 100 kg on ice. The coefficient of friction will be
(a) 0.98 (b) 0.89 (c) 0.49 (d) 0.1
15. A 60 kg body is pushed with just enough force to start it moving across a floor and the same force continues to act afterwards. The
coefficients of static and sliding friction are 0.5 and 0.4 respectively. The acceleration of the body is
(a) 6 m / sec2 (b) 4.9 m / sec2 (c) 3.92 m / sec2 (d) 1 m / sec2

16. A particle is projected along a line of greatest slope up a rough plane inclined at an angle of 45 o with the horizontal. If the
1
coefficient of friction is , then the retardation is
2
g g g  1 g  1
(a) (b) (c) 1  2  (d) 1  2 
2 2 2 2   2  
17. A block moves down a smooth inclined plane of inclination . Its velocity on reaching the bottom is v. If it slides down a rough
inclined plane of same inclination its velocity on reaching the bottom is v/n, where n is a number greater than 0. The coefficient of
friction  is given by
1 1
 1   1   1 2  1 2
18. (a)   tan  1  2  (b)   cot  1  2  (c)   tan  1  2  (d)   cot  1 
 n   n   n   n 2 

19. Consider a car moving along a straight horizontal road with a speed of 72 km/hr. If the coefficient of static friction between the
tyres and the road is 0.5, the shortest distance in which the car can be stopped is (g  10 m / s )
2

(a) 30 m (b) 40 m (c) 72 m (d) 20 m

20. All the surfaces shown in the figure are rough. The direction of friction on B due to A is
(a) Zero F
B
(b) To the left
A

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

(c) Upwards
(d) Downwards
21. A body of mass M just starts sliding down an inclined plane (rough) with inclination , such that tan = 1/3. The force acting on
the body down the plane in this position is
Mg 2 Mg
(a) Mg (b) (c) Mg (d)
3 3 10
 Advance level
22. Consider the following statements
Assertion (A) : It is difficult to move a cycle along the road with its brakes on.
Reason (R) : Sliding friction is greater than rolling friction.
Of these statements [AIIMS 2002]
(a) Both A and R are true and the R is a correct explanation of the A
(b) Both A and R are true but the R is not a correct explanation of the A
(c) A is true but the R is false
(d) Both A and R are false
(e) A is false but the R is true
23. A body is sliding down an inclined plane having coefficient of friction 0.5. If the normal reaction is twice that of the resultant
downward force along the incline, the angle between the inclined plane and the horizontal is [EAMCET (Engg.) 2000]
(a) 15o (b) 30o (c) 45o (d) 60o
24. A block of mass 2 kg rests on a rough inclined plane making an angle of 30 o with the horizontal. The coefficient of static friction
between the block and the plane is 0.7. The frictional force on the block is [IIT-JEE 1980]

(a) 9.8 N (b) 0.7  9.8  3 N (c) 9.8  3 N (d) 0.7  9.8 N
25. A body of weight W is lying at rest on a rough horizontal surface. If the angle of friction is , then the minimum force required to
move the body along the surface will be
(a) W tan (b) W cos (c) W sin (d) W cos
26. A block of mass M is placed on a rough horizontal surface as shown in the figure. A force F = Mg acts on the block. It is inclined to
the vertical at an angle . The coefficient of friction is . The block can be pushed along the surface only when
(a) tan   
F = Mg
(b) cot    
M
(c) tan  / 2  
(d) cot  / 2  
27. A plane is inclined at an angle  with the horizontal. A body of mass m rests on it. If the coefficient of friction is , then the
minimum force that has to be applied parallel to the inclined plane to make the body just move up the inclined plane is
(a) mg sin  (b)  mg cos
(c)  mg cos – mg sin (d)  mg cos + mg sin
28. A block of mass m is placed on another block of mass M which itself is lying on a horizontal surface. The coefficient of friction
between the two blocks is 1 and that between the block of mass M and horizontal surface is  2 . What maximum horizontal force
can be applied to the lower block so that the two blocks move without separation
(a) (M + m) (2  1 )g
m
(b) (M – m) (2  1 )g
M
(c) (M – m) (2  1 )g
(d) (M + m) (2  1 )g
29. A block of mass M1 is placed on a slab of mass M2. The slab lies on a frictionless horizontal surface. The coefficient of static friction
between the block and slab is 1 and that of dynamic friction is 2. A force F acts on the block M1. Take g = 10 ms–2. If M1 = 10 kg,
M2 = 30 kg, 1 = 0.5, 2 = 0.15 and F = 40 N, what will be the acceleration with which the slab will move

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

(a) 5 ms 2
(b) 2 ms 2
(c) 1 ms 2
(d) Zero
30. In the above problem if F = 100 N, what will be the acceleration with which the slab will move
(a) 5 ms 2 (b) 2 ms 2 (c) 1 ms 2 (d) None of these
31. A block X of mass 4 kg is lying on another block Y of mass 8 kg. As shown in the figure. When the force acting on X is 12N, block
X is on the verge of slipping on Y. The force F in Newton necessary to make both X and Y move simultaneously will be
(a) 36
(b) 3.6
(c) 0.36
(d) 3.6
32. Two masses 10 kg and 5 kg are connected by a string passing over a pulley as shown. If the coefficient of friction be 0.15, then the
minimum weight that may be placed on 10 kg to stop motion is

(a) 18.7 kg
(b) 23.3 kg
(c) 32.5 kg
(d) 44.3 kg
33. Two blocks of mass M1 and M2 are connected with a string which passes over a smooth pulley. The mass M1
is placed on a rough inclined plane as shown in the figure. The coefficient of friction between the block and
the inclined plane is . What should be the maximum mass M2 so that block M1 slides downwards
(a) M 2  M1 (sin    cos  )
(b) M 2  M 1 (sin    cos  )
(c) M 2  M1 /(sin    cos  )
(d) M 2  M1 /(sin    cos  )
34. A car starts from rest to cover a distance s. the coefficient of friction between the road and the tyres is . The minimum time in
which the car can cover the distance is proportional to
1 1
(a)  (b)  (c) (d)
 
35. An engine of mass 50,000 kg pulls a coach of mass 40,000 kg. If there is a resistance of 1 N per 100 kg acting on both the engine
and the coach, and if the driving force of the engine be 4,500 N, then the acceleration of the engine is
(a) 0.08 m / s 2 (b) Zero (c) 0.04 m / s 2 (d) None of these
36. In the above question, then tension in the coupling is
(a) 2,000 N (b) 1,500 N (c) 500 N (d) 1000 N
37. An aeroplane requires for take off a speed of 72 km/h. The run of the ground is 100m.The mass of the plane is 10 4 kg and the
coefficient of friction between the plane and the ground is 0.2. The plane accelerates uniformly during take off. What is the
acceleration of the plane
(a) 1 m/s2 (b) 2 m/s2 (c) 3 m/s2 (d) 4 m/s2
38. The force required to just move a body up an inclined plane is double the force required to just prevent it from sliding down. If 
is angle of friction and  is the angle which incline makes with the horizontal then
(a) tan  = tan  (b) tan  = 2 tan  (c) tan  = 3 tan  (d) tan  = 3 tan 
39. A body is on a rough horizontal plane. A force is applied to the body direct towards the plane at an angle  with the vertical. If  is
the angle of friction then for the body to move along the plane
(a)    (b)    (c)    (d)  can take up any value

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

 FRICTION

40. In the arrangement shown W1  200 N, W2  100 N,   0 .25 for all surfaces in contact. The block W1 just slides under the block
W2
(a) A pull of 50 N is to be applied on W1
(b) A pull of 90 N is to be applied on W1

(c) Tension in the string AB is 10 2 N

(d) Tension in the string AB is 20 2 N

41. A board of mass m is placed on the floor and a man of mass M is standing on the board as shown. The coefficient of friction between
the board and the floor is . The maximum force that the can exert on he rope so that the board does not slip on the floor is

(a) F  (M  m )g
(b) F  mg
Mg
(c) F
 1
(M  m )g
(d) F 
 1
42. A body slides over an inclined plane forming an angle of 45° with the horizontal. The distance x travelled by the body in time t is
described by the equation x  kt 2 , where k  1.732 . The coefficient of friction between the body and the plane has a value
(a)   0 .5 (b)   1 (c)   0 .25 (d)   0 .75
43. Two blocks A and B of masses m and M respectively are placed on each other and their combination rests on a fixed horizontal
surface C. A light string passing over the smooth light pulley is used to connect A and B as shown. The coefficient of sliding friction
between all surfaces in contact is . If A is dragged with a force F then for both A and B to move with a uniform speed we have
(a) F  (M  m )g
(b) F  mg
(c) F  (3 M  m )g
(d) F  (3m  M )g
44. A force of 100 N is applied on a block of mass 3 kg as shown in figure. The coefficient of friction between the surface of the block
is 1/4. The friction force acting on the block is

(a) 15 N downwards (b) 25 N upwards (c)20 N downwards (d)20 N upwards

F = 100 N

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil

FRICTION

Answer Sheet (Practice problems)

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

b c d d b a c a c b

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

c c a d d c a b c d

21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

a c a a c d d c d a

31. 32. 33. 34. 35. 36. 37. 38. 39. 40.

b b d c a b c a b, d d

41. 42. 43. 44

a d c b

ADITYA VIDYASHRAM RESIDENTIAL SCHOOL P.V.V.SATYANARAYANA M.Sc, M.Phil