# Variation of Friction with applied Force

### How Friction varies with respect to applied Force ?

Consider a block of mass m place on a horizontal surface with normal reaction N.

Case (i): If applied force F = 0, the force of friction is zero.

Case (ii): If applied force F < (fs)max, the block does not move and the force of friction is fs = F

Case (iii): If applied force F = (fs)max block just ready to slide and frictional force (fs )max = fl = μs N

F= μs mg (since, N = mg); (at time t = 0)

Case (iv): If the above applied force continues to act with slightly greater magnitude (t ≠ 0) the body gets motion, static friction converts as kinetic friction and body possesses acceleration

$\large a = \frac{F_{ext}-f_k}{m} = \frac{f_l – f_k}{m}$

Case (v): If the applied force is greater than limiting friction the body starts moving and gets acceleration

$\large a = \frac{F_{ext}’-f_k}{m}$ (since , here Fext‘ > Fext )

Solved Example : A small block of mass ‘ m ‘ is placed on a Plank of mass ‘M’. The block is connected to plank with the help of a light string passing over a light smooth pulley. Shown in Figure.  The co-efficient of static friction between the block and plank is μ . The co-efficient of friction between the plank and the horizontal surface is zero. What maximum horizontal force F applied on the block of mass m can make the block and plank not to slide relatively ?

Solution:

The force diagram and the F.B.D. are shown in figure respectively taking the ground as reference frame.

N2 = mg     ………..(1)

F – T – f = 0    …………..(2)

From Figure
N1 = N2+ mg     …………(3)

T – f = 0    ………………(4)

Again, fmax = μN2

i.e. fmax = μmg    ………..(5)

For ‘ m ‘ not to slide relatively on ‘ M ‘

f ≤ fmax    …………..(6)

Using equation (1) to (5) and the condition (6), Fmax can be found out as,
Fmax = 2μmg

Solved Example : A block of mass m = 2 kg is accelerating by a force F = 20 N applied on a smooth light pulley as shown in the figure. If the coefficient of kinetic friction between the block and the surface is μ = 0.3, find its acceleration.

Solution:

From the F.B.D. of pulley shown in figure,

F – 2T = mp ap

∴ T = F/2    …….(1) (because mp ≅0)

From F.B.D. of the block

N = mg    ……(2)

∴ T-fk = ma

i.e T – μkN = ma    …………(3)

From (1), (2) and (3),

$\displaystyle \frac{F}{2} – \mu_k mg = ma$

$\displaystyle a = \frac{F}{2 m} – \mu_k g$

$\displaystyle a = \frac{20}{2 \times 2} – 0.3 \times 9.8$

a = 2.06 m/s2

Solved Example: A block of mass ‘ m ’ is given horizontal velocity ‘ u ’. The coefficient of kinetic friction between block and surface is μk then find

(a) retardation due to friction

(b) Distance covered by block before coming to rest

(c) Time taken by block to come to rest.

Solution:

(a) Friction force

fk = μk N = μk mg

∴ m a = μk mg

⇒ a = μk g

(b) v² = u² + 2as

0 = u² – 2μk gs

⇒ $\large s = \frac{u^2}{2 \mu_k g}$

(c) v = u + at

0 = u – μk gt

⇒ $\large t = \frac{u}{\mu_k g}$

Exercise : A block is moving with constant velocity when acted upon by a constant force F inclined at an angle θ to horizontal. Find the coefficient of kinetic friction.

Exercise : A block of mass m moves with constant speed down the inclined plane of inclination θ. Find the coefficient of kinetic friction.

Exercise : Two bodies A and B of masses 1 kg and 5 kg. Are connected by a light inextensible string passing over a smooth pulley. If μk = 0.1 between A and the wedge and the other surface is smooth, find the acceleration of the bodies.