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A block of mass m lies on wedge of mass M, which lies on fixed horizontal surface. The wedge is free to move on the horizontal surface. A horizontal force of magnitude F is applied on block as shown, neglecting friction at all surfaces, the value of force F such that block has no relative motion w.r.t. wedge will be: (where g is acceleration due to gravity)

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$\frac{m}{M} (M + m) g \tan θ$

$(M + m) g \tan θ$

$(M + m) g \cot θ$

$\frac{m}{M} (M + m) g \cot θ$

answer is C.

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Detailed Solution

For no relative motion between wedge and block, let the acceleration of both block and wedge be 'a' towards left.

From FBD of block; $N \cos θ = m g$ ...(i)
and $F - N \sin θ = m a$ ...(ii)
From FBD of wedge; $N \sin θ = M a$ ...(iii)
from equations (i), (ii) and (iii)

solving we get $F = \frac{m}{M} (M + m) g \tan θ$

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