Question

A wooden wedge of mass $M$ and inclination angle $\alpha$ 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 thrust on the pulleyis nearly block remains stationary with respect to wedge. The magnitude of force $P$ is

• A. $(M+m) g \tan \alpha$
• B. $g \tan \alpha$
• C. $m g \cos \alpha$
• D. $(M+m) g \,\text{cosec} \,\alpha$

Answer 1

Answer: (A)

Since, $P=(M+m) a$
Now as in free body diagram of block,

$\therefore a=g \frac{\sin \alpha}{\cos \alpha}=g \tan \alpha $
or $ P=(M+m) g \tan \alpha$