First slide

Connected bodies

Question

In the diagram shown the table top is rough and coefficient of friction between any of the blocks and table top is $μ$ . The blocks A and B are on the verge of slipping. If position of the blocks A and B are interchanged and the system is released from rest, then acceleration of the blocks is

Moderate

A

$\frac{g (1 - μ)}{(1 + μ)}$
B

$μg$
C

$\frac{g}{(1 + μ)}$
D

$g (1 - μ)$

Solution

In the first case, $mg = μ \times Mg \Rightarrow m = μM$
In the second case, resultant driving force on the blocks $= Mg - μmg = Mg (1 - μ^{2})$
$∴$ Acceleration, $a = \frac{Mg (1 - μ^{2})}{(M + m)} = \frac{Mg (1 - μ^{2})}{M (1 + μ)} = g (1 - μ)$

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