Q. A block of mass $m$ is placed on a smooth wedge of inclination $\theta$. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block ($\ g$ is acceleration due to gravity ) will be
A) $mg\cos\theta$
B) $mg\sin\theta$
C) $mg$
D) $mg/\cos\theta$
Explanation
(d) When the whole system is accelerated towards left then pseudo force $(ma)$ works on a block towards right
For the condition of equilibrium
$mg\sin\theta = ma\cos\theta \Rightarrow a = \frac{g\sin\theta}{\cos\theta}$
$\therefore$ Force exerted by the wedge on the block
$R = mg\cos\theta + ma\sin\theta$
$R = mg\cos{\theta + m\ \left( \frac{g\sin\theta}{\cos\theta} \right)}\sin\theta = \frac{mg(\cos^{2}{\theta + \sin^{2}{\theta)}}}{\cos\theta}$
$R = \frac{mg}{\cos\theta}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.