A block of mass m is placed on a smooth incli | vedclass

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(C) To keep the block stationary on the wedge,we analyze the forces in the non-inertial frame of the wedge.$1$. The forces acting on the block are: gr

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$a = g 	an 	heta$

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(C) To keep the block stationary on the wedge,we analyze the forces in the non-inertial frame of the wedge.$1$. The forces acting on the block are: gravitational force $mg$ (downwards),normal reaction $N$ (perpendicular to the surface),and pseudo force $ma$ (towards the left).$2$. Resolving the forces into components parallel and perpendicular to the inclined surface:- Perpendicular to the surface: $N = mg \cos 	heta + ma \sin 	heta$. However,for the block to be stationary relative to the wedge,the net force along the incline must be zero.- Parallel to the incline: $ma \cos 	heta = mg \sin 	heta$.$3$. From the parallel component equation:$ma \cos 	heta = mg \sin 	heta$$a \cos 	heta = g \sin 	heta$$a = g \frac{\sin 	heta}{\cos 	heta}$$a = g 	an 	heta$

$A$ 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

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