Block on Block Problem in Friction - JEE PDF Download

Clear all your JEE doubts with EduRev Join for Free

Join for Free

Case I: Bottom block is pulled and there is no friction between bottom block and the horizontal surface.

(i) When the bottom block is pulled upper block is accelerated by the force of friction acting upon it.
(ii) The maximum acceleration of the system of two blocks to move together without slipping is
α_max = μ_sg
(iii) The maximum applied force for which both the blocks move together is
F_max = μ_sg(m_u + m_B)
(iv) If α < m_sg blocks move together and applied force is
F = (m_B + m_u)a
In this case frictional force between the two block ,  f = m_u a
If a > μ _s g , blocks slip relative to each other and have different accelerations. The acceleration of the upper block is a_u = μ _kg and that for the bottom block is

Case II: Upper block pulled and there is no friction between bottom block and the horizontal surface.

(i) When the upper block is pulled, bottom block is accelerated by the force of friction acting on it.
(ii) The maximum acceleration of the system of two blocks to move together without slipping is

 where, μs = coefficient of static friction between the two blocks.
(iii) The maximum force for which both blocks move together is

(iv) If a < amax, blocks move together and frictional force between the two blocks is f = m_Ba
(v)  The applied force on the upper block is F =(m_B + m_u)a
(vi) If α > α_max blocks slide relative to each other and hence they have different accelerations.
(vii) The acceleration of the bottom block is
(viii) The acceleration of the upper block is

Solved Example

Example 1: In the situation shown in figure,
(a) What minimum force F will make any part or whole system move ?
(b) Find the acceleration of two blocks and value of friction at the two surface if F = 6 N.

Solution: Between 4 kg and ground:
f_(l1)= f(k₁)= μ1 (2+4)g = 6N
Between 2 kg and 4 kg:
f_(l2)=f(k₂)= μ₂ (2 g) = 4 N
(a) Limiting friction between 2 kg and 4 kg is less than that between 4 kg and ground. So 2 kg will slip over 4 kg if
(b) F > f(l₂)= 4N  ⇒ Fmin = 4N

f₁ > f₂ so it will not move
Its acceleration = 0

for 2 kg: a1 =(6-4)/2 = 1 ms^-2

Example 2: (a) Find the acceleration of the two blocks shown in figure.
(b) Find the friction force between all

Solution: Let us first calculate the limiting and kinetic friction between various surface. Between 3 kg and ground:
f(l₁) = f(k₁)= μ1 (2+3)g = 0.1×5g = 5N
Between 2 kg and 3 kg:
f(l₂) = f(k₂) = μ2 2g = 0.2×2g = 4N

Let us first assume that both blocks move together with common acceleration a.
a = (5+10 – f(k₁))/(2+3)
⇒ a = (15-5)/5 = 2 ms^-2
Now let us see how much friction force is required between 2 kg and 3 kg for common acceleration a.
5 – f₂ = 2 a
⇒ 5 – f₂ = 2 × 2
⇒ f₂ = 1 N

Since f₂ < f(l₂) ,
(1) Both blocks will move together with common acceleration a = 2 ms^-2
(2) Friction between 2 kg and 3 kg = f₂ = 1 N Friction between 3 kg and ground = f_(k1) = 5 N