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 $\overset{ \rightarrow }{P}$ is applied on the wedge as shown in figure, such that a block remains stationary with respect to wedge. The magnitude of force $\overset{ \rightarrow }{P}$ is

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

Answer 1

Answer: (A)

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

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