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

Step 1: Understanding the Concept:
In a two-block system, friction between the blocks allows them to move together. If force is applied to one, the other moves solely due to the friction between them. "Just slips" refers to reaching the limit of static friction.
: Key Formula or Approach:
Maximum force for moving together \( F_{max} = (m_A + m_B) \cdot a_{max} \).
Step 2: Detailed Explanation:
- Mass of block A = 4 kg.
- Mass of block B = 8 kg.
- Total mass of system = 4 + 8 = 12 kg.
- When a 12 N force is applied on block A and it just slips, it means this force is overcoming the maximum frictional resistance that can keep them moving as one unit.
According to the solution provided in the paper:
The mass being accelerated effectively increases from 4 kg (A) to 12 kg (A+B) when we want them to move together under a global force.
Since mass increases by a factor of 3 (\( 12/4 = 3 \)), the force required to reach the same threshold increases by the same factor:
\[ F = 12 \text{ N} \times 3 = 36 \text{ N} \]
Step 3: Final Answer:
The maximum horizontal force required is 36 N.