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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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Verified Answer
The correct answer is:
\(m g \sec \theta\)
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,
\(\begin{aligned}
\text {i.e. } R & =m g \cos \theta+m a \sin \theta \\
& =m g \cos \theta+m \cdot g \frac{\sin \theta}{\cos \theta} \cdot \sin \theta \quad \text { [From Eq. (i)] } \\
& =\frac{m g}{\cos \theta}\left[\cos ^2 \theta+\sin ^2 \theta\right] \\
\Rightarrow R & =\frac{m g}{\cos \theta}=m g \sec \theta
\end{aligned}\)
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,
\(\begin{aligned}
\text {i.e. } R & =m g \cos \theta+m a \sin \theta \\
& =m g \cos \theta+m \cdot g \frac{\sin \theta}{\cos \theta} \cdot \sin \theta \quad \text { [From Eq. (i)] } \\
& =\frac{m g}{\cos \theta}\left[\cos ^2 \theta+\sin ^2 \theta\right] \\
\Rightarrow R & =\frac{m g}{\cos \theta}=m g \sec \theta
\end{aligned}\)
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