Laws Of Motion Ques 4
4. 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 ( $\mathrm{g}$ is acceleration due to gravity) will be
[2004]
(a) $\mathrm{mg} / \cos \theta$
(b) $\mathrm{mg} \cos \theta$
(c) $m g \sin \theta$
(d) $\mathrm{mg}$
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Answer:
Correct Answer: 4.(a)
Solution: (a) According to the condition,
$\mathrm{N}=m a \sin \theta+m g \cos \theta$ $\quad$ ……..(1)
Also, $m g \sin \theta=m a \cos \theta$ $\quad$ ……..(2)
From (1) & (2), $a=g \tan \theta$
$ \begin{aligned} \therefore N & =m g \frac{\sin ^2 \theta}{\cos \theta}+m g \cos \theta . \\ & =\frac{m g}{\cos \theta}\left(\sin ^2 \theta+\cos ^2 \theta\right)=\frac{m g}{\cos \theta} \end{aligned} $
or, $N=\frac{m g}{\cos \theta}$
The condition for the body to be at rest relative to the inclined plane, $a=g \sin \theta-b \operatorname{con} \theta=0$
Horizontal acceleration, $b=g \tan \theta$.
BETA
