Wooden wedge of mass M and inclination angle $α$ rests on a smooth floor. A block of mass m is kept on wedge. A force P is applied on the wedge as shown in figure, such that a block remains stationary with respect to wedge. The magnitude of force P is

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$(M + m) g \cos e c α$

$g \tan α$

$m g \cos α$

$(M + m) g \tan α$

answer is A.

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Detailed Solution

Since, P = (M+m)a

Considering the observer on the block, the block will be at rest w.r.t that observer

Now as in free body diagram of block,

$m a \cos α = m g s i n α$

$∴ a = g \frac{\sin α}{\cos α} = g \tan α$

$\Rightarrow P = (M + m) g \tan α$

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