Laws of Motion - Result Question 30
30. 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
[2004]
(a) $mg / \cos \theta$
(b) $mg \cos \theta$
(c) $mg \sin \theta$
(d) $mg$
Show Answer
Answer:
Correct Answer: 30. (a)
Solution:
- (a) According to the condition,
$N=m a \sin \theta+m g \cos \theta$
Also, $m g \sin \theta=m a \cos \theta$
From (1) & (2), $a=g \tan \theta$
$\therefore N=m g \frac{\sin ^{2} \theta}{\cos \theta}+m g \cos \theta$.
$=\frac{m g}{\cos \theta}(\sin ^{2} \theta+\cos ^{2} \theta)=\frac{m g}{\cos \theta}$ or, $N=\frac{m g}{\cos \theta}$
The condition for the body to be at rest relative to the inclined plane, $a=g \sin \theta-b con \theta=0$ Horizontal acceleration, $b=g \tan \theta$.
