A 4 kg block A is placed on the top of a 8 kg block B 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 on B to make both A and B move together, is

The correct answer is 36 N.

Let's carefully set up the problem before solving it step by step.

Given information:
- Mass of block A, $m_A$ = 4 kg (placed on top of B)
- Mass of block B, $m_B$ = 8 kg (rests on smooth table)
- A just slips on B when a horizontal force of 12 N is applied on A
- The table is smooth (no friction between B and table)

Step 1: Find the Maximum Friction Force Between A and B

When we say "A just slips on B" under a 12 N applied force, it means the applied force has reached the exact threshold at which the static friction between A and B can no longer hold A in place relative to B. At this moment, the friction between A and B is at its maximum (limiting) value.

In this scenario, the limiting friction force between A and B equals the applied force that causes slipping:

$f_{\max }=12\text{N}$

This is the maximum force that friction between the two blocks can ever exert on either block, regardless of the direction of the applied force.

Step 2: Find the Maximum Possible Acceleration (without slipping)

Now, the question asks: what is the maximum force applied on B such that both A and B move together (i.e., A does not slip on B)?

When force is applied on B, the only horizontal force acting on block A is the friction exerted on it by block B. For the two blocks to move together without slipping, this friction must be sufficient to accelerate A.

Using Newton's Second Law for block A alone:

$f_{\max }=m_A\times a_{\max }$

$12=4\times a_{\max }$

$a_{\max }=\frac{12}{4}=3\text{m/s²}$

This is the maximum acceleration the system (A + B together) can have before A starts to slip on B.

Step 3: Find the Maximum Force on B

If the force F is applied on B and both blocks move together as a single system, we apply Newton's Second Law to the entire combined system. The table is smooth, so there is no friction from the table on B.

$F=(m_A+m_B)\times a_{\max }$

$F=(4+8)\times 3$

$F=12\times 3$

$F=36\text{N}$

Summary of Logic:

The key insight is that the maximum friction between A and B (12 N) is a fixed physical property of the surface contact. Whether the force is applied on A or on B, this limiting friction value does not change. When force is applied on B, friction is the only agent that can "drag" A along — and it can exert a maximum of 12 N on A, producing a maximum system acceleration of 3 m/s². For the whole 12 kg system to achieve this acceleration, the applied force on B must be F = 36 N.