Q.
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 $\overrightarrow{\mathbf{P}}$ is applied on the wedge as shown in figure, such that a block remains stationary with respect to wedge. The magnitude of force $\overrightarrow{\mathbf{P}}$ is
A) $(M + m)g\ tan\ \alpha$
B) $g\ tan\ \alpha$
C) $mg\cos\alpha$
D) $(M + m)g{cosec}\alpha$
Explanation
(a) Since, $P = (M + m)a$
Now as in free body diagram of block,
[image:5_image156.jpeg]
$ma\cos{\alpha = mg\sin\alpha}\ $
$\therefore\ a = g\frac{\sin\alpha}{\cos\alpha} = g\tan\alpha$ or $P = (M + m)g\tan\alpha$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.