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

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Aicharya · Accepted Answer

The given situation is shown in the following figure,

The acceleration given to wedge to the left is $a$. The block has a pseudo acceleration to the right, pressing against the wedge because of which the block does not move. In this case, $\begin{array}{ll} \therefore & m g \sin \theta=m a \cos \theta \\ \Rightarrow & a=\frac{g \sin \theta}{\cos \theta} \quad \ldots (i) \end{array}$ Force exerted by the wedge on the block is equal to normal reaction on the block, $\begin{aligned} \text {i.e. } R & =m g \cos \theta+m a \sin \theta \\ & =m g \cos \theta+m \cdot g \frac{\sin \theta}{\cos \theta} \cdot \sin \theta \quad \text { [From Eq. (i)] } \\ & =\frac{m g}{\cos \theta}\left[\cos ^2 \theta+\sin ^2 \theta\right] \\ \Rightarrow R & =\frac{m g}{\cos \theta}=m g \sec \theta \end{aligned}$

A block of mass \(m\) is placed on a smooth wedge of inclination \(\theta\). The whole system is accelerated…

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