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 force.
The force of friction is parallel to the surface and opposite to the direction of intended motion.
Types of Friction
R
(1) Static friction : The opposing force that comes into play when one body
P
tends to move over the surface of another, but the actual motion has yet not
F
started is called static friction.
(i) If applied force is P and the body remains at rest then static friction F = P.
mg
(ii) If a body is at rest and no pulling force is acting on it, force of friction on
it is zero.
(iii) Static friction is a self-adjusting force because it changes itself in
accordance with the applied force and is always equal to net external 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 upto 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.
Force of static friction Fs   s R . But force of maximum static friction is Fmax static 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 is defined as the ratio of force of limiting friction and
F
R
(b) Dimension : [ M 0 L0T 0 ]
normal reaction s 
(c) Unit : It has no unit.
(d) Value of  depends on material and nature of surfaces in contact that means whether dry or
wet ; rough or smooth polished or non-polished.
(e) Value of  does not depend upon apparent area of contact.
(f) Hence the force of limiting friction is a parameter which decides whether the object will slide over
or not. It is not always equal to force of friction.
(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 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) Kinetic friction does not depend upon the velocity of the body (provided velocity should not be
too large).
(v) 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 that 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
increases linearly with the applied force
(2) At point A the static friction is maximum. This represent
limiting friction ( Fl ) .
(3) Beyond A, the force of friction is seen to decrease slightly. The
portion BC of the curve represents the kinetic friction ( Fk ) .
Force of friction
 r is called coefficient of rolling friction. It would have the dimensions of length and would be
measured in metre.
* 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
centre of the wheel moves forward.
Graph Between Applied Force and Force of Friction
A
(1) Part OA of the curve represents static friction ( Fs ) . Its value
C
B
Fs
Fl
O
Applied force
(4) As the portion BC of the curve is parallel to x-axis therefore
kinetic friction does not change with the applied force, it remains
constant, whatever be the applied force.
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
tan  
R
S

F
P
Fl
R
tan  = s

Fk
mg
[As we know Fl   ]
s
R
or
  tan 1 (  L )
Hence coefficient of static 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.
By definition,  is called the angle of repose.
In limiting condition F  mg sin  and R  mg cos 
R
F
mg sin 
So F  tan 


mg
mg cos 
R
F
F
  s  tan   tan  [As we know
  s  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 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 :
(1) Minimum pulling force P at an angle  from the horizontal
F
By resolving P in horizontal and vertical direction (as shown in figure)
For the condition of equilibrium
F  P cos  and R  W  P sin 
P

R
P sin
P cos
W
By substituting these value in F  R
P cos    (W  P sin  )
 P cos   sin  (W  P sin  )
[As   tan  ]
cos 

P
W sin 
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)
For the condition of equilibrium
F  P cos  and R  W  P sin 
By substituting these value in F  R
 P cos   (W  P sin  )
 P cos   sin  (W  P sin  ) [As   tan  ]
cos 
W sin 

P
cos (   )
(3) Minimum pulling force P to move the body up on an inclined
plane
By Resolving P in the direction of the plane and perpendicular to the
plane (as shown in the figure)
P

R
P cos
F
P sin
W
R + P sin
P
P cos

F + W sin



W
W cos
For the condition of equilibrium
R  P sin   W cos 

R  W cos   P sin  and F  W sin   P cos 

F  P cos   W sin 
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 of inclined plane
R + P sin
F
P

P cos
+



W
W cos
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 

R  W cos   P sin  and
F  P cos   W sin 
By substituting these values in F  R and solving we get
W sin(   )
P
cos (   )
(5) Minimum force to avoid sliding of a body down on an inclined plane
P
R + P sin

W sin

F + P cos


W
W cos
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 
 R  W cos   P sin  and
P cos   F  W sin 
 F  W sin   P cos 
By substituting these values in F  R and solving we get
 sin (   ) 
P W 

 cos (   ) 
(6) Minimum force for motion along horizontal surface and its direction
P

R + P sin
Let the force P be applied at an angle  with the horizontal.
By resolving P in horizontal and vertical direction (as shown in figure)
For vertical equilibrium
R  P sin   mg

…(i)
R  mg  P sin 
and for horizontal motion
P cos  F
i.e. P cos   R
…(ii)
Substituting value of R from (i) in (ii)
P cos    (mg  P sin  )
 mg
…(iii)
P
cos    sin 
For the force P to be minimum (cos   sin  ) must be maximum i.e.
d
[cos    sin  ]  0
d
  sin    cos   0

F
P cos
mg
1
2


1
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

As tan    so from the figure, sin  
1  2
1
and cos  
1  2
By substituting these value in equation (iii)
 mg
 mg
 P  mg

P
min
2
2
1  2
1

1 

1  2
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
From the figure
ma  P  Fk
R
Fk
ma
P
mg
 a  P  Fk
m
(2) Acceleration of a block sliding down over 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.
R
F
ma
From the figure ma  mg sin   F

ma  mg sin   R

ma  mg sin    mg cos
 mg cos
mg sin
 Acceleration a  g [sin    cos  ]
 mg
Note : For frictionless inclined plane   0  a  g sin  .
ma
R
(3) Retardation of a block sliding up over a rough inclined plane
When angle of inclined plane is less than angle of repose, then for
the upward motion
ma  mg sin   F
 mg cos
mg sin + F
ma  mg sin    mg cos 
 mg
Retardation a  g [sin    cos ]
Note : For frictionless inclined plane   0  a  g sin 
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
(1) A force F is applied to the upper body, (2) A force F is applied to the lower body
m
M
F
A
B
L
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
(a) The body A will move on body B with acceleration (F/m).
aA  F / m
(b) The body B will remain at rest
aB  0
(c) If L is the length of B as shown in figure, A will fall from B after time t
2L
2mL
1 2


t

 As s  2 a t and a  F/m 
a
F


(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
(b) Combined system (m + M) will move together with common acceleration a A  a B  F
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 cause the
motion of B.
Free body diagram of A
F  Fk  m a A
i.e.
F  Fk
aA 
m
aA 
maA
(F   k mg )
m
A
F
Fk
Free body diagram of B
Fk  M a B
i.e.

aB 
Fk
M
MaB
FK
 mg
aB  k
M
B
Note : As both the bodies are moving in the same direction.
Acceleration of body A relative to B will be
MF   k mg (m  M )
a  a A  aB 
mM
So, A will fall from B after time
t
2L

a
2 m ML
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
FK
B
Fl
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 (= Fk) not Fl .
(2) A force F is applied to the lower body, then following four 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.

t
2L

a
A
2 ML
F
M
L
B
m
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)
(a) Both the body will move together with common acceleration a 
F
M m
(b) Pseudo force on the body A,
mF and
Fl   s mg
F   ma 
mM
(c) F   Fl  mF   mg  F   s ( m  M ) g
s
mM
So both bodies will move together with acceleration a A  a B 
F
if F   s [m  M ] g
mM
(iii) If friction is present between A and B only and F > Fl
(where Fl = s mg = limiting friction between body A and B)
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.
Free body diagram of A
ma A   k mg
i.e.
a A  k g
A
maA
Fk
Free body diagram of B
F  Fk  Ma B
i.e.
[F  k mg ]
aB 
M
MaB
FK
B
F
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 and F > Fl :
(where Fl = s(m+M)g = limiting friction between body B and surface)
The system will move only if F  Fl '' then replacing F by F  Fl  . The entire case (iii) will be valid.
However if F  F1  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.
O
r
Fl
R

y
A
mg sin
mg cos
h
mg
Let m = mass of the insect, r = radius of the bowl,  = coefficient of friction
for limiting condition at point A
......(i) and Fl  mg sin 
......(ii)
R  mg cos
Dividing (ii) by (i)
Fl

R
r2  y2

y
As Fl
tan  

or
y
 R 
r
1  2


1 
1 , 
h  r 1 

h  r  y  r 1 



1   2 
1   2 
Minimum Mass Hung from the String to Just Start the Motion
(1) When a mass m1 placed on a rough horizontal plane Another mass m2 hung from the string
connected by frictionless pulley, the tension (T) produced in string will try to start the motion of
R
mass m1 .
So
Fl
m1
T
T
m1g
m2
At limiting condition T  Fl
 m2 g  R  m2 g   m1 g
m2g
 m2  m1 this is the minimum value of m to start the motion.
2
Note: In the above condition Coefficient of friction   m2
m1
(2) When a mass m1 placed on a rough inclined plane Another mass m2 hung from the string
connected by frictionless pulley, the tension (T) produced in string will try to start the motion of
mass m1 .
At limiting condition
T
For m2 T  m2 g
…(i)
R
T
For m1 T  m1 g sin   F

T  m1 g sin   R
m1
m2
 T  m1 g sin   m1 g cos
…(ii)
m1g cos
m2g

m1g sin + F
From equation (i) and (ii) m2  m1 [sin    cos ]
this is the minimum value of m2 to start the motion
m1g
Note : In the above condition Coefficient of friction
 m2


 tan  
m
cos

 1

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 without sliding. Since the chain have uniform linear density therefore the ratio of
mass and ratio of length for any part of the chain will be equal.
m
mass hanging from the table
We know
 2 
m1
mass lying on the table
 For this case 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  l

l
l
by solving l  
(   1)
Coefficient of Friction Between a 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, S  u t  1 at 2
2
1
2
…(i)
S  ( g sin  ) t
2
[ As u  0 and a  g sin  ]
For rough wedge, S  u t  1 at 2
2
1
2
…(ii)
S  g (sin    cos  ) (nt )
2
[ As u  0 and a  g (sin    cos  )]
From equation (i) and (ii)
1
1
2
( g sin  ) t 2 = g (sin    cos ) ( nt )
2
2
 sin   (sin    cos  ) n 2
   tan  1  1 
 n2 


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.
S
v=0
u
Retarding force F  ma  R 
 a  g
ma   mg
From v 2  u 2  2aS  0  u 2  2 g S
[As v  0, a  g ]

S
u2
2g
or
S
P2
2m 2 g
[As momentum P = mu]
(ii) Time taken to come to rest
From equation v  u  a t  0  u   g t
[ As v  0, a   g ]
 t u
g
(2) On inclined road : When block starts with velocity u its kinetic energy
potential energy and some part of it goes against friction and after travelling
rest i.e. v = 0.
We know that retardation a  g [sin    cos ]
By substituting the value of v and a in the following equation
v 2  u 2  2a S
 0  u 2  2 g [sin    cos  ] S
will be converted into
distance S it comes to
v=0
S
u

u2
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]

S
m1
A
B
m1
m2
S2
S1
2
 FS  P
2m

m2
[Where P = momentum of block]
mg  S 
P2
2m
[As F =  mg]
P2
2m 2 g
In the given condition P and  are same for both the blocks.

S
 S1   m2 
 
So, S  1 ;
m2
2
S 2  m1 
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.
a
F
ma
m
R
F
M
m
CART
mg
Now block will remain static w.r.t. cart. If friction force R  mg
[As R  ma ]
 ma  mg
 a g

 a g
min

This is the minimum acceleration of the cart so that block does not fall.
and the minimum force to hold the block together
Fmin  ( M  m) a min
Fmin  ( M  m)
g

Understanding Concept:1. Determine the maximum acceleration of the train in which a box lying on its floor will remain
stationary, given that the co-efficient of static friction between the box and the train’s floor is 0.15.
2. See Fig. A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an
angle θ = 15° with the horizontal, the mass just begins to slide. What is the coefficient of static
friction between the block and the surface?
3. What is the acceleration of the block and trolley system shown in a Fig., if the coefficient of
kinetic friction between the trolley and the surface is 0.04? What is the tension in the string? (Take
g = 10 m s-2). Neglect the mass of the string.
4. Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against
a rigid wall (Fig.). The coefficient of friction between the bodies and the table is 0.15. A force of 200
N is applied horizontally to A. What are (a) the reaction of the partition (b) the action-reaction
forces between A and B ? What happens when the wall is removed? Does the answer to (b) change,
when the bodies are in motion? Ignore the difference between μs and μk.
5. A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the
block and the trolley is 0.18. The trolley accelerates from rest with 0.5 m s-2 for 20 s and then
moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer
on the ground, (b) an observer moving with the trolley.
6. The rear side of a truck is open and a box of 40 kg mass is
placed 5 m away from the open end as shown in Fig. The
coefficient of friction between the box and the surface below it is
0.15. On a straight road, the truck starts from rest and accelerates
with 2 m s-2. At what distance from the starting point does the box
fall off the truck? (Ignore the size of the box).
7. A brick slides on a horizontal surface. Which of the following will increase the magnitude of the
frictional force on it?
A. Putting a second brick on top
B. Decreasing the surface area of contact
C. Increasing the surface area of contact
D. Decreasing the mass of the brick
E. None of the above
8. The coefficient of kinetic friction:
A. is in the direction of the frictional force
B. is in the direction of the normal force
C. is the ratio of force to area
D. can have units of newtons
E. is none of the above
9. When the brakes of an automobile are applied, the road exerts the greatest retarding force:
A. while the wheels are sliding
B. just before the wheels start to slide
C. when the automobile is going fastest
D. when the acceleration is least
E. at the instant when the speed begins to change
10. A forward horizontal force of 12N is used to pull a 240-N crate at constant velocity across a
horizontal floor. The coefficient of friction is:
A. 0.5
B. 0.05
C. 2
D. 0.2
E. 20
11. The speed of a 4.0-N hockey puck, sliding across a level ice surface, decreases at the rate of
0.61m/s2. The coefficient of kinetic friction between the puck and ice is:
A. 0.062
B. 0.41
C. 0.62
D. 1.2
E. 9.8
12. A crate rests on a horizontal surface and a
woman pulls on it with a 10-N force. No matter what
the orientation of the force, the crate does not move.
Rank the situations shown below according to the
magnitude of the frictional force of the surface on the
crate, least to greatest.
A. 1, 2, 3
B. 2, 1, 3
C. 2, 3, 1
D. 1, 3, 2
E. 3, 2, 1
13. A crate with a weight of 50N rests on a horizontal surface. A person
pulls horizontally on it with a force of 10N and it does not move. To start it
moving, a second person pulls vertically upward on the crate. If the
coefficient of static friction is 0.4, what is the smallest vertical force for
which the crate moves?
A. 4N
B. 10N
C. 14N
D. 25N
E. 35N
14. A 40-N crate rests on a rough horizontal floor. A 12-N horizontal force is then applied to it. If
the coefficients of friction are s = 0.5 and k = 0.4, the magnitude of the frictional force on the
crate is:
A. 8N
B. 12N
C. 16N
D. 20N
E. 40N
15. A 24-N horizontal force is applied to a 40-N block initially at rest on a rough horizontal surface.
If the coefficients of friction are s = 0.5 and k = 0.4, the magnitude of the frictional force on the
block is:
A. 8N
B. 12N
C. 16N
D. 20N
E. 400N
16. A horizontal shove of at least 200N is required to start moving a 800-N crate initially at rest on
a horizontal floor. The coefficient of static friction is:
A. 0.25
B. 0.125
C. 0.50
D. 4.00
E. none of these
17. A force F (larger than the largest possible force of static friction) is applied to the left to an
object moving to the right on a horizontal surface. Then:
A. the object must be moving at constant speed
B. F and the friction force act in opposite directions
C. the object must be slowing down
D. the object must be speeding up
E. the object must come to rest and remain at rest
18. A bureau rests on a rough horizontal surface (s = 0.50, k = 0.40). A constant horizontal
force, just sufficient to start the bureau in motion, is then applied. The acceleration of the bureau
is:
A. 0
B. 0.98m/s2
C. 3.3m/s2
D. 4.5m/s2
E. 8.9m/s2
19. A car is traveling at 15m/s on a horizontal road. The brakes are applied and the car skids to a
stop in 4.0 s. The coefficient of kinetic friction between the tires and road is:
A. 0.38
B. 0.69
C. 0.76
D. 0.92
E. 1.11
20. A boy pulls a wooden box along a rough horizontal floor at
constant speed by means of a force P as shown. In the
diagram f is the magnitude of the force of friction, N is the
magnitude of the normal force, and Fg is the magnitude of the
force of gravity. Which of the following must be true?
A. P = f and N = Fg
B. P = f and N >Fg
C. P >f and N <Fg
D. P >f and N = Fg
E. none of these
21. A boy pulls a wooden box along a rough
horizontal floor at constant speed by means of a force
P as shown. In the diagram f is the magnitude of the
force of friction, N is the magnitude of the normal
force, and Fg is the magnitude of the force of gravity.
Which of the following must be true?
A. P = f and N = Fg
B. P = f and N >Fg
C. P >f and N <Fg
D. P >f and N = Fg
E. none of these
22. A 400-N block is dragged along a horizontal surface
by an applied force F as shown. The coefficient of kinetic
friction is k = 0.4 and the block moves at constant
velocity. The magnitude of F is:
A. 100N
B. 150N
C. 200N
D. 290N
E. 400 N
23. A block of mass m is pulled at constant velocity along a
rough horizontal floor by an applied force T as shown. The
magnitude of the frictional force is:
A. T cos 

B. T sin 
C. zero
D. mg
E. mg cos 
24. A block of mass m is pulled along a rough horizontal floor
by an applied force T as shown. The vertical component of the
force exerted on the block by the floor is:
A. mg
B. mg - T cos
C. mg + T cos 
D. mg - T sin 
E. mg + T sin 
25. A 12-kg crate rests on a horizontal surface and a boy pulls
on it with a force that is 300 below the horizontal. If the
coefficient of static friction is 0.40, the minimum magnitude force he needs to start the crate
moving is:
A. 44N
B. 47N
C. 54N
D. 56N
E. 71N
26. A crate resting on a rough horizontal floor is to be moved horizontally. The coefficient of static
friction is 0.40. To start the crate moving with the weakest possible applied force, in what direction
should the force be applied?
A. Horizontal
B. 240below the horizontal
C. 220 above the horizontal
D. 240 above the horizontal
E. 660 below the horizontal
27. A 50-N force is applied to a crate on a horizontal rough floor, causing it to move horizontally. If
the coefficient of kinetic friction is 0.50, in what direction should the force be applied to obtain the
greatest acceleration?
A. Horizontal
B. 600above the horizontal
C. 300 above the horizontal
0
0
D. 27 above the horizontal
E. 30 below the horizontal
28. A professor holds an eraser against a vertical chalkboard by pushing horizontally on it. He
pushes with a force that is much greater than is required to hold the eraser. The force of friction
exerted by the board on the eraser increases if he:
A. pushes with slightly greater force
B. pushes with slightly less force
C. stops pushing
D. pushes so his force is slightly downward but has the same magnitude
E. pushes so his force is slightly upward but has the same magnitude
29. A horizontal force of 12N pushes a 0.5-kg book against a vertical wall. The book is initially at
rest. If the coefficients of friction are s = 0.6 and k = 0.8 which of the following is true?
A. The magnitude of the frictional force is 4.9N
B. The magnitude of the frictional force is 7.2N
C. The normal force is 4.9N
D. The book will start moving and accelerate
E. If started moving downward, the book will decelerate
30. A horizontal force of 5.0N pushes a 0.50-kg book against a vertical wall. The book is initially at
rest. If the coefficients of friction are s = 0.6 and k = 0.80, the magnitude of the frictional force is:
A. 0
B. 4.9N
C. 3.0N
D. 5.0N
E. 4.0N
31. A horizontal force of 12N pushes a 0.50-kg book against a vertical wall. The book is initially at
rest. Ifs = 0.6 and k = 0.80, the acceleration of the book in m/s2 is:
A. 0
B. 9.4m/s2
C. 9.8m/s2
D. 14.4m/s2 E. 19.2m/s2
32. A horizontal force of 5.0N pushes a 0.50-kg block against a vertical wall. The block is initially
at rest. If s = 0.60 and k = 0.80, the acceleration of the block in m/s2 is:
A. 0
B. 1.8
C. 6.0
D. 8.0
E. 9.8
33. A heavy wooden block is dragged by a force F along a
rough steel plate, as shown below for two possible situations.
The magnitude of F is the same for the two situations. The
magnitude of the frictional force in (ii), as compared with that
in (i) is:
A. the same
B. greater
C. less
D. less for some angles and greater for others
E. can be less or greater, depending on the magnitude of the applied force.
34. A block is first placed on its long side and then on
its short side on the same inclined plane, as shown.
The block slides down the plane on its short side but
remains at rest on its long side. A possible explanation
is:
A. the short side is smoother
B. the frictional force is less because the contact area is
less
C. the center of gravity is higher in the second case
D. the normal force is less in the second case
E. the force of gravity is more nearly down the plane in the second case
35. A box rests on a rough board 10 meters long. When one end of the board is slowly raised to a
height of 6 meters above the other end, the box begins to slide. The coefficient of static friction is:
A. 0.8
B. 0.25
C. 0.4
D. 0.6
E. 0.75
36. A block is placed on a rough wooden plane. It is found that when the plane is tilted 30 to the
horizontal, the block will slide down at constant speed. The coefficient of kinetic friction of the
block with the plane is:
A. 0.500
B. 0.577
C. 1.73
D. 0.866
E. 4.90
37. A crate is sliding down an incline that is 350 above the horizontal. If the coefficient of kinetic
friction is 0.40, the acceleration of the crate is:
A. 0
B. 2.4m/s2
C. 5.8m/s2
D. 8.8m/s2
E.10.3m/s2
38. A 5.0-kg crate is resting on a horizontal plank. The coefficient of static friction is 0.50 and the
coefficient of kinetic friction is 0.40. After one end of the plank is raised so the plank makes an
angle of 250 with the horizontal, the force of friction is:
A. 0
B. 18N
C. 21N
D. 22N
E. 44N
39. A 5.0-kg crate is resting on a horizontal plank. The coefficient of static friction is 0.50 and the
coefficient of kinetic friction is 0.40. After one end of the plank is raised so the plank makes an
angle of 300 with the horizontal, the force of friction is:
A. 0
B. 18N
C. 21N
D. 22N
E. 44N
40. A 5.0-kg crate is on an incline that makes an angle of 300 with the horizontal. If the coefficient
of static friction is 0.50, the minimum force that can be applied parallel to the plane to hold the
crate at rest is:
A. 0
B. 3.3N
C. 30N
D. 46N
E. 55N
41. A 5.0-kg crate is on an incline that makes an angle of 300 with the horizontal. If the coefficient
of static friction is 0.5, the maximum force that can be applied parallel to the plane without moving
the crate is:
A. 0
B. 3.3N
C. 30N
D. 46N
E. 55N
42. Block A, with mass mA, is initially at rest on a horizontal floor. Block B, with mass mB, is
initially at rest on the horizontal top surface of A. The coefficient of static friction between the two
blocks is s. Block A is pulled with a horizontal force. It begins to slide out from under B if the
force is greater than:
A. mAg
B. mBg
C. s mA g
D. s mB g
E. s (mA +mB) g
43. The system shown remains at rest. Each block weighs 20 N.
The force of friction on the upper block is:
A. 4N
B. 8N
C. 12N
D. 16N
E. 20N
44. Block A, with a mass of 50 kg, rests on a horizontal table top. The
coefficient of static friction is 0.40. A horizontal string is attached to A
and passes over a massless frictionless pulley as shown. The smallest
mass mB of block B, attached to the dangling end, that will start A
moving when it is attached to the other end of the string is:
A. 20 kg
B. 30 kg
C. 40 kg
D. 50 kg
E. 70 kg
45. Block A, with a mass of 10 kg, rests on a 350 incline. The coefficient
of static friction is 0.40. An attached string is parallel to the incline and
passes over a massless, frictionless pulley at the top. The largest mass
mB of block B, attached to the dangling end, for which A begins to
slide down the incline is:
A. 2.5kg
B. 3.5kg
C. 5.9kg
D. 9.0kg
E. 10.5kg
46. Block A, with a mass of 10 kg, rests on a 350 incline. The coefficient
of static friction is 0.40. An attached string is parallel to the incline and
passes over a massless, frictionless pulley at the top. The largest mass
mB, attached to the dangling end, for which A remains at rest is:
A. 2.5kg
B. 3.5kg
C. 5.9kg
D. 9.0kg
E. 10.5kg
47. Block A, with a mass of 10 kg, rests on a 300 incline. The coefficient of
kinetic friction is 0.20. The attached string is parallel to the incline and
passes over a massless, frictionless pulley at the top. Block B, with a mass
of 8.0 kg, is attached to the dangling end of the string. The acceleration of B
is:
A. 0.69m/s2, up the plane
B. 0.69m/s2, down the plane
C. 2.6m/s2, up the plane
D. 2.6m/s2, down the plane
E. 0
48. Block A, with a mass of 10 kg, rests on a 300 incline. The coefficient
of kinetic friction is 0.20. The attached string is parallel to the incline
and passes over a massless, frictionless pulley at the top. Block B, with
a mass of 3.0 kg, is attached to the dangling end of the string. The
acceleration of B is:
A. 0.20m/s2, up
B. 0.20m/s2, down
C. 2.8m/s2, up
D. 2.8m/s2, down
E. 0
49. A 1000-kg airplane moves in straight flight at constant speed. The force of air friction is 1800
N. The net force on the plane is:
A. zero
B. 11800N
C. 1800N
D. 9800N
E. none of these
50. Why do raindrops fall with constant speed during the later stages of their descent?
A. The gravitational force is the same for all drops
B. Air resistance just balances the force of gravity
C. The drops all fall from the same height
D. The force of gravity is negligible for objects as small as raindrops
E. Gravity cannot increase the speed of a falling object to more than 9.8m/s
51. A ball is thrown downward from the edge of a cliff with an initial speed that is three times the
terminal speed. Initially its acceleration is
A. upward and greater than g
B. upward and less than g
C. downward and greater than g
D. downward and less than g
E. downward and equal to g
52. A ball is thrown upward into the air with a speed that is greater than terminal speed. On the
way up it slows down and, after its speed equals the terminal speed but before it gets to the top of
its trajectory:
A. its speed is constant
B. it continues to slow down
C. it speeds up
D. its motion becomes jerky
E. none of the above
53. A ball is thrown upward into the air with a speed that is greater than terminal speed. It lands
at the place where it was thrown. During its flight the force of air resistance is the greatest:
A. just after it is thrown
B. halfway up
C. at the top of its trajectory
D. halfway down
E. just before it lands.
54. In a situation the contact force by a rough horizontal surface on a body placed on it has
constant magnitude. If the angle between this force and the vertical is decreased , the force and the
vertical is decreased, the friction force between the surface and the body will
(a) increse
(b*) decrease
(c) remain the same (d) may increase or decrease
55. While walking on ice, one should take small steps to avoid slipping . This is because smaller
steps to avoid slipping. This is because smaller steps ensure
(a) larger friction
(b*) smaller friction (c) larger normal force (d) smaller normal force
56. A body of mass M is kept on a rough horizontal surface (friction coefficient = ). A person is
trying to pull the body by applying a horizontal force but the body is not moving. The force by the
surface on A is F where
(a) F = Mg
(b) F = Mg
(c*) Mg  F  Mg 1   2
(d) Mg  F  Mg 1   2
57. A scooter starting from rest moves with a constant acceleration for a time Δt1, then with a
constant deceleration for the next Δt2 and finally with a constant deceleration for the next Δt3 to
come to rest. A 500N man sitting on the scooter behind the driver manges to stay at rest with
respect to the scooter without touching any other part. The force exerted by the seat on the man is
(a) 500 N throughout the journey
(b*) less than 500N throughout the journey
(c) more than 500N throughout the journey (d) > 500 N for time Δt1 and Δt3 and 500 N for Δt2.
58. Consider the situation shown in figure. The wall is smooth but the surface of
A and B in contact are rough. The friction on B due to A in equilibrium
(a) is upward
(b) is downward
(c) is zero
(d*) the system cannot remain in equilibrium.
59. Suppose all the surfaces in the previous problem are rough. The direction of friction on B due
to A
(a*) is upward
(b) is downward
(c) is zero
(d) depends on the masses of A and B.
60. Two cars of unequal masses use similar tyres. If they are moving at the same initial speed, the
minimum stopping distance
(a) is smaller for the heavier car
(b) is smaller for the lighter car
(c*) is same for both cars
(d) depends on the volume of the car.
61. In order to stop a car in shortest distance on a horizontal road, one should
(a) apply the brakes very hard so that the wheels stop rotating
(b*) apply the brakes hard enough to just prevent slipping
(c) pump the brakes (press and release)
(d) shut the engine off and not apply brakes .
62. A block A kept on an inclined surface just begins to slide if the inclination is 30º. The block is
replaced by another block B and it is found that it just begins to slide if the inclination is 40º.
(a) mass of A > mass of B
(b) mass of A < mass of B
(c) mass of A = mass of B
(d*) all the three are possible.
63. A boy of mass M is applying a horizontal force to slide a box of mass M’ on a rough horizontal
surface . The coefficient of friction between the shoes of the boy and the floor is  and that between
the box and the floor is ’. In which of the following cases it is certainly not possible to slide the box
?
(a*)  <’, M < M’
(b)  >’, M < M’
(c)  <’, M > M’
(d)  >’, M > M
64. Let F, FN and f denote the magnitudes of the contact force , normal force and the friction
exerted by one surface on the other kept in contact. If none of these is zero,
(a*) F > FN
(b*) F > f
(c) FN > f
(d*) FN – f < F < FN + f
65. The contact force exerted by a body A on another body B is equal to the normal force between
the bodies. We conclude that
(a) the surface must be frictionless
(b*) the force of friction between the bodies is zero
(c) the magnitude of normal force equals that of friction
(d*) the bodies may be rough but they don’t slip on each other.
66. Mark the correct statements about the friction between two bodies.
(a) Static friction is always greater than the kinetic friction.
(b*) Coefficient of static friction is always greater than the coefficient of kinetic friction.
(c*) Limiting friction is always greater than the kinetic friction.
(d*) Limiting friction is never less than static friction.
67. A block is placed on a rough floor and a horizontal force F is applied on it. The force of friction f
by the floor on the block is measured for different values of F and a graph is plotted between them.
(a) The graph is a straight line of slope 45º
(b) The graph is straight line parallel to the F-axis.
(c*) The graph is a straight line of slope 45º for small F and a straight line parallel to the F-axis for
large F.
(d*) There is a small kink on the graph.
68. Consider a vehicle going on a horizontal road towards east. Neglect any force by the air. The
frictional forces on the vehicle by the road.
(a*) is towards east if the vehicle is accelerating
(b*) is zero if the vehicle is moving with a uniform velocity
(c) must be towards east.
(d) must be towards east.
FRICTION
{H Series}
1. A body of mass 400 g slides on a rough horizontal surface. If the frictional force is 3.0 N, find (a)
the angle made by the contact force on the body with the vertical and (b) the magnitude of the
contact force. Take g = 10m/s2.
2. The coefficient of static friction between a block of mass m and an incline is s = 0.3. (a) What
can be the maximum angle  of the incline with the horizontal so that the block does not slip on
the plane? (b) If the incline makes an angle /2 with the horizontal, find the frictional force on the
block.
3. A horizontal force of 20 N is applied to a block of mass 4 kg resting on a rough horizontal table.
If the block does not moue on the table, how much frictional force the table is applying on the
block? What can be said about the coefficient of static friction between the block and the table?
Take g = 10m/s2.
4. The coefficient of static friction between the block of 2 kg and the table shown in figure1 is s =
0.2. What should be the maximum value of m so that the blocks do not move? Take g = 10m/s2.
The string and the pulley are light and smooth.
Fig.1
5. The coefficient of static friction between the two blocks shown in figure 2 is µ and the table is
smooth. What maximum horizontal force F can be applied to the block of mass M so that the
blocks moue together?
Fig.2
6. A block slides down an incline of angle 30° with acceleration g/4. Find the kinetic friction
coefficient.
7. A block of mass 2.5 kg is kept on a rough horizontal surface. It is found that the block does not
slide if a horizontal force less than 15 N is applied to it. Also it is found that it takes 5 seconds to
slide through the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed to
start the motion. Taking g =10 m/s2, calculate the coefficients of static and kinetic friction between
the block and the surface.
8. A block placed on a horizontal surface is being pushed by a force F making an angle  with the
vertical. If the friction coefficient is µ, how much force is needed to get the block just started.
Discuss the situation when tan<.
9. Find the maximum value of M/m in the situation shown in figure 3 so that the system remains
at rest. Friction coefficient at both the contacts is Discuss the situation when tan < µ.
Fig.3
10. Consider the situation shown in figure 4. The horizontal surface below the bigger block is
smooth. The coefficient of friction between the blocks is µ. Find the minimum and the maximum
force F that can be applied in order to keep the smaller blocks at rest with respect to the bigger
block.
Fig.4
11. Figure 5 shows two blocks connected by a light string placed on the two inclined parts of a
triangular structure. The coefficients of static and kinetic friction are 0.28 and 0.25 respectively at
each of the surfaces. (a) Find the minimum and maximum values of m for which the system
remains at rest. (b) Find the acceleration of either block if m is given the minimum value calculated
in the first part and is gently pushed up the incline for a short while.
Fig.5
12. If the tension in the string in figure 6 is 16 N and the acceleration of each block is 0.5 m/s2,
find the friction coefficients at the two contacts with the blocks.
Fig.6
13. The friction coefficient between the table and the block shown in figure 7 is 0.2. Find the
tensions in the two strings.
Fig.7
14. The friction coefficient between a road and the tyre of a vehicle is 4/3. Find the maximum
incline the road may have so that once hard brakes are applied and the wheel starts skidding, the
vehicle going down at a speed of 36 km/hr is stopped within 5 m.
15. Figure 8 shows two blocks in contact sliding down an inclined surface of inclination 30°. The
friction coefficient between the block of mass 2.0 kg and the incline is 1, and that between the
block of mass 4.0 kg and the incline is 2. Calculate the acceleration of the 2.0 kg block if (a) 1 =
0.20 and 2 = 0.30, (b) 1 = 0.30 and 1= 0.20. Take g = 10 m/s2.
Fig.8
16. Two masses Ml and M2 are connected by a light rod and the system is slipping down a rough
incline of angle  with the horizontal. The friction coefficient at both the contacts is . Find the
acceleration of the system and the force by the rod on one of the blocks.
17. A block of mass M is kept on a rough horizontal surface. The coefficient of static friction
between the block and the surface is . The block is to be pulled by applying a force to it. What
minimum force is needed to slide the block ? In which direction should this force act?
18. The friction coefficient between the board and the floor shown in figure 9 is . Find the
maximum force that the man can exert on the rope so that the board does not slip on the floor.
Fig.9
19. A 2 kg block is placed over a 4 kg block and both are placed on a smooth horizontal surface.
The coefficient of friction between the blocks is 0.20. Find the acceleration of the two blocks if a
horizontal force of 12 N is applied to (a) the upper block, (b) the lower block. Take g=10m/s2.
20. Find the accelerations a1, a2, a3 of the three block; shown in figure 10 if a horizontal force of 10
N ii applied on (a) 2 kg block, (b) 3 kg block, (c) 7 kg block. Take g=10m/s2.
Fig.10
21. The friction coefficient between the two blocks shown it y figure 11 is  but the floor is smooth.
(a) What maximum horizontal force F can be applied without disturbing the equilibrium of the
system? (b) Suppose the horizontal force applied is double of that found in part (a). Find the
accelerations of the two masses.
Fig.11
22. Suppose the entire system of the previous question is kept inside an elevator which is coming
down with an acceleration a <g. Repeat parts (a) and (b).
23. Consider the situation shown in figure 11. Suppose small electric field E exists in the space in
the vertically upward direction and the upper block carries a positive charge Q on its top surface.
The friction coefficient between the two blocks is N but the floor is smooth. What maximum
horizontal force F can be applied without disturbing the equilibrium?
24. A block of mass m slips on a rough horizontal table under the action of a horizontal force
applied to it. The coefficient of friction between the block and the table is . The table does not
move on the floor. Find the total frictional force applied by the floor on the legs of the table: Do you
need the friction coefficient between the table and the floor or the mass of the table?
25. Find the acceleration of the block of mass M in the 1ituation of figure 12. The coefficient of
friction between the two blocks is  and that between the bigger block and the ground is µ2.
Fig.12
26. A block of mass 2 kg is pushed against a rough vertical wall with a force of 40 N, coefficient of
static friction being 0.5. Another horizontal force of 15 N, is applied on the block in a direction
parallel to the wall. Will the block move? If yes, in which direction? If no, find the frictional force
exerted by the wall on the block.
27. A person (40 kg) is managing to be at rest between two vertical walls by pressing one wall A by
his hands and feet and the other wall B by his back 13. Assume that the friction coefficient
between his body and the walls is 0.8 and that limiting friction acts at all the contacts. (a) Show
that the person pushes the two walls with equal force. (b) Find the normal force exerted by either
wall on the person. Take g=10m/s2.
Fig.13
28. Figure 14 shows a small block of mass m kept at the left end of a larger block of mass M and
length L. The system can slide on a horizontal road. The system is started towards right with an
initial velocity v. The friction coefficient between the road and the bigger block is µ and that
between the blocks is 2. Find the time elapsed before the smaller block separates from the bigger
block.
Fig.14
ANSWERS
1.
2.
3.
4.
5.
6.
7.
8.
9.
(a) 370 (b) 5.0N
(a) tan-1(0.3) (b) mg sin(/2)
s≥0.5
0.4 kg
g(M+m)
1/2 3
s = 0.60,k = 0.52
mg/ (sin cos)
/( sin cos)
10. 1   M  2m g and 1   (M  2m) g
1 
11.
12.
13.
14.
15.
16.
17.
1- 
(a) 9/8 kg, 32/9 kg (b) 0.31m/s2
1 = 0.75,2 = 0.06
96 N in the left string and 68 N in the right string
160
2.7 m/s2, 2.4 m/s2.
a = g(sin cos), zero
mg/( 1   2 ) at an angle tan-1 with the horizontal.
18. m) g/ (1+ )
19. (a) upper block 4 m/s2, lower block 1 m/s2.
(b) both blocks 2 m/s2.
20. (a) a1 = 3 m/s2, a2 = a3 = 0.4 m/s2
(b) a1 = a2 = a3 = 5/6 m/s2 (c) same as (b)
21. (a) 2mg (b) 2mg/(M + m) in opposite direction.
22. (a) 2mg (b) 2m(g-a) /(M + m)
23. 2mg-QE)
24. mg
25.
[ 2m   2 M  m ]g
M  m[5  2(1   2 )]
26. it will move at an angle of 530 with the 15 N force.
27. (b) 250 N
28.
4Ml
M  m g