[a = g tan \theta] A block of mass m is placed on a smooth inclined wedge ABC of inclination \theta as shown in the f

For Competitive Exams

A block of mass m is placed on a smooth inclined wedge ABC of inclination $\theta$ as shown in the figure. The wedge is given an acceleration a towards the right. The relation between a and $\theta$ for the block to remain stationary on the wedge is

a = g cos $\theta$

a = $\dfrac{g}{sin \theta}$

a = $\dfrac{g}{cosec \theta}$

a = $g tan \theta $

Explanation:

In non-inertial frame,
$ N sin \theta = ma \cdot\cdot\cdot(i)$
$N \cos \theta = mg \cdot\cdot\cdot(ii)$
$\tan \theta = \dfrac{a}{g}$
$ a = g \tan \theta$

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