FRICTION

1. FRICTION
When two bodies are kept in contact, electromagnetic forces act between the charged particles
(molecules) at the surfaces of the bodies. Thus, each body exerts a contact force on the other. The
magnitudes of the contact forces acting on the two bodies are equal but their directions are opposite
and therefore the contact forces obey Newton’s third law.

The direction of the contact force acting on a particular body is not necessarily perpendicular to the
contact surface. We can resolve this contact force into two components, one perpendicular to the
contact surface and the other parallel to it (figure. The perpendicular component is called the normal
contact force or normal force (generally written as N) and the parallel component is called friction
(generally written as f).
CONTACT FORCE= f 2  N2

2. REASONS FOR FRICTION

(i) nter-locking of extended parts of one object into the extended parts of the other object.
(ii) Bonding between the molecules of the two surfaces or objects in contact.

3. FRICTION FORCE IS OF TWO TYPES.

a. Kinetic b. Static
(a) Kinetic Friction Force
Kinetic friction exists between two contact surfaces only when there is relative motion between the
two contact surfaces. It stops acting when relative motion between two surfaces ceases.
DIRECTION OF KINECTIC FRICTION ON AN OBJECT 
t is opposite to the relative velocity of the object with respect to the other object in contact
considered.
Note that its direction is not opposite to the force applied it is opposite to the relative motion
of the body considered which is in contact with the other surface.

MAGNITUDE OF KINETIC FRICTION

The magnitude of the kinetic friction is proportional to the normal force acting between the two
bodies. We can write
fk = k N
where N is the normal force. The proportionality constant k is called the coefficient of kinetic
friction and its value depends on the nature of the two surfaces in contact.
Example 1. Find the direction of kinetic friction force

(a) on the block, exerted by the ground. (b) on the ground, exerted by the block.

Solution : (a)

(b)

where f 1 and f 2 are the friction forces on the block and ground respectively.

Example 2. In above example correct relation between magnitude of f 1 and f2 is

(A) f 1 > f 2 (B) f 2 > f 1
(C) f 1 = f 2 (D) not possible to decide due to insufficient data.
Solution : By Newton‘s third law the above friction forces are action-reaction pair and equal but opposite
to each other in direction. Hence (C).
Also note that the direction of kinetic friction has nothing to do with applied force F.
Example 3. All surfaces as shown in the figure are rough. Draw the friction force on A & B
A 10m/s

B 20m/s
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Solution :

Kinetic friction acts in such a way so as to reduce relative motion.

Example 4. Find out the distance travelled by the blocks shown in the figure before it stops.
10 m/s
10 kg
/////////////////////////////////////////
µk=0.5
Solution : N – 10 g = 0
N = 100 N N
f x = µkN
µ = µs = µk when not mentioned
f x = 0.5 × 100 = 50 N fk
Fx = ma
50 = 10 a  a = 5 10g
 v2 = u2 + 2as
02 = 102 + 2 (– 5) (S)  S = 10 m
Example 5. Find out the distance travelled by the block on incline before it stops. Initial velocity of the block
is 10 m/s and coefficient of friction between the block and incline is  = 0.5.
s
m/
10


37° fixed
µ
Solution : N = mg cos37°
 mg sin 37° + µN = ma
a = 10 m/s2 down the incline
Now v 2 = u2 + 2as
0 = 102 + 2(–10) S
S=5m

Example 6. Find the time taken in the above example by the block to reach the initial position.
Solution : a = g sin 37° – µg cos 37°
 a = 2 m/s2 down the incline
1 2 1 m
 S = ut + at S = × 2 × t2 5
2 2
 t = 5 sec.

Example 7. A block is given a velocity of 10 m/s and a force of

100 N in addition to friction force is also acting on the
block. Find the retardation of the block?

Solution : As there is relative motion 

 Kinetic friction will act to reduce this relative motion.
fk = µN = 0.1 × 10 × 10 = 10 N
100 + 10 = 10a
110
a= = 11 m/s2
10

Example 8. A body is projected up along a rough inclined plane from the bottom with some velocity. It
travels up the incline and then returns back. If the time of ascent is ta and time of descent is td,
then
(A) ta = td (B) ta > td (C) ta < td (D) data insufficient
Solution : Let velocity of projection be V and velocity of the block when it returns back = V'
then V > V' (since some K.E. is lost to friction)
Hence average velocity during ascent > average velocity during descent
 ta < td
 (b) STATIC FRICTION
 t exists between the two surfaces when there is tendency of relative motion but no relative motion
along the two contact surface.
For example consider a bed inside a room ; when we gently push the bed with a finger, the bed
does not move. This means that the bed has a tendency to move in the direction of applied force
but does not move as there exists static friction force acting in the opposite direction of the applied
force.

Example 10. What is value of static friction force on the block?

Solution : In horizontal direction as acceleration is zero.

Therefore  F = 0. 
 = 0

Direction of static friction force :

The static friction force on an object is opposite to its impending motion relative to the surface. .

(i)
Note : Here once again the static friction is involved when there is no relative motion between
two surfaces.

Example 11. In the following figure an object of mass M is kept on a rough

table as seen from above. Forces are applied on it as shown. Find
the direction of static friction if the object does not move.

Solution : In the above problem we first draw the free body diagram of find the resultant force.

As the object doe not move this is not a case of kinetic friction. The direction of static friction is
opposite to the direction of the resultant force FR as shown in figure by fs. Its magnitude is equal
to 25 N.
4. MAGNITUDE OF KINETIC AND STATIC FRICTION
Kinetic friction :

fk = k N
where N is the normal force. The proportionality constant k is called the coefficient of kinetic
friction and its value depends on the nature of the two surfaces in contact.

Static friction :
It is a variable and self adjusting force. However it has a maximum value called limiting friction.
fmax = sN
The actual force of static friction may be smaller than sN and its value depends on other forces acting
on the body. The magnitude of frictional force is equal to that required to keep the body at relative rest.

0  fs  f smax
s is called coefficient of static friction and k is called coefficient of kinetic friction.

s > k for a given pair of surfaces. If not mentioned then s = k can be taken. Value of  can be from
0 to .

Example 12. Find acceleration of block. Initially the block is at rest.

50 N

10 Kg

µ=0.5
Solution : zero
Example 13. Find out acceleration of the block. Initially the block is at rest.

Solution : N + 24 – 100 = 0 for vertical direction

 N = 76 N
Now 0  fs  µs N  0  fs  76 × 0.5
0  fs  38 N 32 < 38 Hence f = 32
 Acceleration of block is zero.

Example 14. Find out acceleration of the block for different ranges of F.
F
m

µ
Solution : 0  f  µSN  0  f  µSmg
a = 0 if F  µSmg
F  µMg
a= if F > µMg
M
Example 15. Find out acceleration of the block. Initially the block is at rest.
51N
10 kg

µS = 0.5
µk = 0.3
Solution : 0  fs  µSN
0  fs  50
Now 51 > 50  51
10
  Block will move but if the block starts moving then 30
kinetic friction is involved.
KF = µk N = 0.3 × 100 = 30 N
 51 – 30 = 10 a
 a = 2.1 m/s2
Example 16. Find out the minimum force that must be applied on the block vertically downwards so that the
block doesn’t move.
100N
10kg

µ=0.5
Solution : 100 – f s = 0
 fs = 100 .....(1) N F
F + 10 g = N  N = 100 + F .....(2)
10 kg
Now 0  f S  N 100N
100  0.5 N f
100  0.5 [100 + F] 10 g
200  100 + F ; F  100 N
 Minimum F = 100 N
Example 17. The angle of inclination is slowly increased. Find out the angle at which the block starts moving.

Solution : 0  f  µS N
mg sin > fsmax N f
mg sin  > µN 
 cos
mg sin > µ mg cos 
sin mg
 tan  > µ mg 
 = tan–1 µ
for tan  µ no sliding on inclined plane.
This method is used for finding out the value of µ
practically.
Example 18. Find out the acceleration of the block. If the block is initially at rest.

Solution : (FBD of the block excluding friction)

N
75
kg
10
°
37
° c os
37 Mg
sin
Mg
N = 10 g cos 37° = 80 N
Now 0  fs  µN ; 0  fs  0.5 × 80  fs  40 N
N
75

60 80
We will put value of f in the last i.e. in the direction opposite to resultant of other forces. f acts
down the incline and its value is of = 75 – 60 = 15 N
So acceleration is zero
Example 19. In the above problem how much force should be added to 75 N force so that block starts to
move up the incline.
Solution :  60 + 40 = 75 + f extra  fs = 25 N
Example 20. In the above problem what is the minimum force by which 75 N force should be replaced with
so that the block does not move.
Solution : In this case the block has a tendency to move downwards.
Hence friction acts upwards.
F
40

60
 F + 40 = 60  F = 20 N
Example 21. Top view of a block on a table is shown (g = 10 m/s2).

Find out the acceleration of the block.

Solution :
40N

40 g FR = 40 2  56N
Now f s  µN  f s  50 ; FR > f smax

Hence the block will move. a =

40 2  50
10

= 4 2  5 m/s2 
Example 22. Find minimum µ so that the blocks remain stationary.

50kg

100kg
Solution : T = 100 g = 1000N
 f = 1000 to keep the block stationary
Now f max = 1000N 1000N
µN = 1000
f
µ=2
Can µ be greater than 1 ?
Yes 0 < µ 
Example 23. Find out minimum acceleration of block A so that the 10 kg block doesn’t fall.

Solution : Applying NL in horizontal direction

N = 10 a ......(1) µN
Applying NL in vertical direction
a
10 g = µ N ......(2) N 10
10 g = µ 10 a from (1) & (2)
g 10 g
 a = = 20 m/s2
µ

Example 24. Find the tension in the string in situation as shown in the figure below. Forces 120 N and 100 N
start acting when the system is at rest and the maximum value of static friction on 10 kg is 90 N
and that on 20 kg is 60N?
 Solution : (i) Let us assume that system moves towards left then as it is clear from FBD, net force in
horizontal direction is towards right. Therefore the assumption is not valid.

Above assumption is not possible as net force on system comes towards right. Hence
system is not moving towards left.
(ii) Similarly let us assume that system moves towards right.

Above assumption is also not possible as net force on the system is towards left in this
situation.
Hence assumption is again not valid.
Therefore it can be concluded that the system is stationary.

Assuming that the 10 kg block reaches limiting friction first then using FBD’s.

120 = T + 90  T = 30 N
Also T + f = 100
 30 + f = 100
f = 70 N which is not possible as the limiting value is 60 N for this surface of block.
   Our assumption is wrong and now taking the 20 kg surface to be limiting we have

T + 60 = 100 N  T = 40 N
Also f + T = 120 N  f = 80 N
This is acceptable as static friction at this surface should be less than 90 N.
Hence the tension in the string is T = 40 N.

22.2 Contact force

Let f be the force of friction and N the normal reaction, then the net contact force by the surface on the
object is Fsurface  N2  f 2 . Its minimum value (when f = 0) is N and maximum value (when f = N) is

N 1  2

Therefore N  Fsurface  N 1  2
Fsurface
N

Applied force
f
22.7 Pulley with friction between block and surface
N
a
Case-I : For mass m1 : T – m1g = m1a
m1g m1 T
For mass m2 : m2g – T= m2a
m1g T
on solving,
m2 a
(m2  m1 )g m1m2 (1  )g
Acceleration a  r T
(m1  m2 ) (m1  m2 )
m2g

Case-II : For mass m1 : m1g – T = m1a and N = m2g cos  N = m2 g cos 

For mass m2 : T–m2g cos – m2g sin = m2a

on solving,

 m  m2 (sin    cos )  T
Acceleration a   1 g
 (m1  m2 )  m1 a

m1m2 (1  sin    cos )g 

Tension T  m2g m1g
(m1  m2 )

Example 31. Initially the system is at rest. find out minimum value of F for which sliding starts between the
two blocks.
µ=0.5
10
Smooth
20 F
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Solution : At just sliding condition limiting friction is acting.
10
F – 50 = 20 a .......(1) f = 50
f = 10 a .......(2)
50 = 10 a f = 50
 a = 5 m/s2 20 F
hence F = 50 + 20 × 5 = 150 N
 Fmin = 150 N

22.3 Angle of friction () N

The angle which the resultant of the force of limiting friction fL and normal
reaction N makes with the direction of normal reaction N.
f
tan   L   s    tan 1 s  Applied
N force
fL
22.4 Angle of Repose or Angle of Sliding
It is defined as the minimum angle of inclination of a plane with the
horizontal at which a body placed on it just begins to slide down or W
equivalently the maximum angle of inclination of plane with the
horizontal at which a body placed on it does not slide. N fL
fL = mgsin ...(i) and N = mgcos ...(ii)
Dividing (i) by (ii)
f mg sin 
so s  L   tan  r = tan–1s
N mg cos  mg cos

This fact is used for finding the coefficient of static friction in the laboratory.  mg
 TWO BLOCK PROBLEMS

Example 27. Find the acceleration of the two blocks. The system is initially at rest and the friction coefficient
are as shown in the figure?
µ=0.5 10 A F = 50 N
Smooth
10 B
//////////////////////////////////////////////////
Solution : Method of solving
Step 1 : Make force diagram.
Step 2 : Show static friction force by f because value of friction is not known.
Step 3 : Calculate separately for two cases.
Case 1 : Move together
Step 4 : Calculate acceleration.
Step 5 : Check value of friction for above case.
Step 6 : If required friction is less than available it means they will move together else move
separately.
Step 7: (a) above acceleration will be common acceleration for both
Case 2 : Move separately
Step 7(b) If they move separately then kinetic friction is involved. Whose value is µN.
Step 8 : Calculate acceleration for above case.
f
A 50
B
f
f max = µN  f  50 N (available friction)

Move together Move separately

50
(i) a = = 2.5 m/s2 No need to calculate
10  10
(ii) Check friction for B : f = 10 × 2.5 = 25
25 N is required which is less than available friction hence they will move together.
and aA = aB = 2.5 m/s2
Example 28. Find the acceleration of the two blocks. The system is initially at rest and the friction coefficient
are as shown in the figure?
µ=0.5 10 A 101 N
smooth
10 B
//////////////////////////////////////////////////
Solution : f max = 50 N  f  50 N
f
A 101
B
f
101
(i) If they move together a = = 5.05 m/s2
20
(ii) Check friction on B
f
10
f = 10 × 5.05 = 50.5 (required)
50.5 > 50 (therefore required > available). Hence they will not move together.
(iii) Hence they move separately so kinetic friction is involved.
fk = µN=50
A 101
B
fk=50
101 50 50
 for aA = = 5.1 m/s2 aB = = 5 m/s2
10 10
Also aA > aB as force is applied on A.
 Example 29. Find the acceleration of the two blocks. The system is initially at rest and the friction coefficient
are as shown in the figure?
µ=0.5 F = 60 N
10
Smooth
20
///////////////////////////////////////////
Solution : Move Together Move Separately
60
a= = 2 m/s2 No need to calculate.
30
Check friction on 20 kg.
f = 20 x 2
f = 40 (which is required)
40 < 50 (therefore required < available)
 will move together.

22.6 Acceleration of a Body Down a Rough Inclined Plane

If angle of inclination is greater than the angle of repose, then the body accelerates down the incline.

Net force on the body down the inclined plane


Fnet = mg sin – fr

applying Newton's second laws of motion 

ma = mg sin – kN = mg sin – kmg cos

ma = mg [sin – kcos] 
mgcos
 mg
a = g [ sin – kcos] hence a < g 