Question

In the figure, m_A = 2 kg and m_B = 4 kg. For what minimum value of F, A starts slipping over B : (g = 10 m/s²) –

$0$

• A. 24 N
• B. 36 N
• C. 12 N
• D. 20 N

Answer 1

Answer: (B)

36 N

Answer 2

Maximum frictional force between A and B could be

f₁ = µ₁ m_Ag = (0.2)(2)(10) N

f₁ = 4 N

Hence, maximum common acceleration till both the blocks move with same acceleration is

a =$\frac{f_{1}}{m_{A}}$=$\frac{4}{2}$= 2 m/s²

Now, taking (A + B) as the system:

(From weight = upthrust)

(f₂)_max = µ₂(m_A + m_B)g = 24 N

F – 24 = (m_A + m_B) a = 6 × 2 = 12

 F = 36 N