**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 f_{s} = F

**Case (iii):** If applied force F = (f_{s})_{max} block just ready to slide and frictional force (f_{s} )_{max} = f_{l} = μ_{s} N

F= μ_{s} 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 F_{ext}‘ > F_{ext} )

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.

N_{2} = mg ………..(1)

F – T – f = 0 …………..(2)

From Figure

N_{1} = N_{2}+ mg …………(3)

T – f = 0 ………………(4)

Again, f_{max} = μN_{2}

i.e. f_{max} = μmg ………..(5)

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

f ≤ f_{max} …………..(6)

Using equation (1) to (5) and the condition (6), F_{max} can be found out as,

F_{max} = 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 = m_{p} a_{p}

∴ T = F/2 …….(1) (because m_{p} ≅0)

From F.B.D. of the block

N = mg ……(2)

∴ T-f_{k} = ma

i.e T – μ_{k}N = 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/s^{2}

Solved Example: A block of mass ‘ m ’ is given horizontal velocity ‘ u ’. The coefficient of kinetic friction between block and surface is μ_{k} 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

f_{k} = μ_{k} N = μ_{k} mg

∴ m a = μ_{k} mg

⇒ a = μ_{k} g

(b) v² = u² + 2as

0 = u² – 2μ_{k} gs

⇒ $\large s = \frac{u^2}{2 \mu_k g}$

(c) v = u + at

0 = u – μ_{k} 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 μ_{k} = 0.1 between A and the wedge and the other surface is smooth, find the acceleration of the bodies.