Question

A 4 kg block A is placed on the top of a block B of mass 8 kg, which rests on a smooth table. A just slips on B when a force of 12 N is applied on A. Then, the maximum horizontal force required to make both A and B move together is:

• A. 12 N
• B. 24 N
• C. 36 N
• D. 48 N

Hint:

For two-block systems, use limiting friction condition and relate acceleration of both blocks carefully.

Answer 1

The Correct Option is C

Concept: Friction provides acceleration to the lower block. At limiting condition: \[ f_{max} = \mu m g \] Also, acceleration: \[ a = \frac{F}{m} \]

Step 1: Find coefficient of friction.
When 12 N is applied on A (mass 4 kg), it just slips: \[ a = \frac{12}{4} = 3 \, \text{m/s}^2 \] Friction force: \[ f = m a = 4 \times 3 = 12 \, \text{N} \] \[ f_{max} = \mu (4g) = 12 \Rightarrow \mu = \frac{12}{4g} = \frac{3}{g} \]

Step 2: Maximum force for no slipping.
Total mass: \[ M = 4 + 8 = 12 \, \text{kg} \] Acceleration: \[ a = \frac{F}{12} \] Friction needed to move A: \[ f = m_A a = 4 \cdot \frac{F}{12} = \frac{F}{3} \] For no slipping: \[ \frac{F}{3} \leq f_{max} = \mu (4g) = 12 \]

Step 3: Solve for F.
\[ \frac{F}{3} = 12 \Rightarrow F = 36 \, \text{N} \]