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Vector Component

A 4 kg block ...

Question

A 4 kg block A is placed on the top of 8 kg block B which rests on a smooth table,.

If block A just slips on block B when a force of 12 N is applied on A, then the maximum horizontal force F that can be applied on B so that both blocks move together is

A

12N

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B

24N

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C

36N

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D

48N

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Solution

The correct option is B 24N
When Force is applied on A.
$\frac{F_{A}}{m_{A} + m_{B}} = \frac{μ m_{A} g}{m_{B}}$ ........ (i)
When Force is applied on B.
$\frac{F_{B}}{m_{A} + m_{B}} = \frac{μ m_{A} g}{m_{A}}$ ......... (ii)
eqn (i)/(ii) $\frac{F_{A}}{F_{B}} = \frac{m_{A}}{m_{B}} \Rightarrow F_{B} = \frac{8}{4} \times 12 = 24 N$

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