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(A) Let the angle of the inclined plane be $	heta = \sin^{-1}(1/l)$,which implies $\sin 	heta = 1/l$.To keep an object stationary relative to the in

Vedclass · Accepted Answer

$g/\sqrt{l^2 - 1}$

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(A) Let the angle of the inclined plane be $	heta = \sin^{-1}(1/l)$,which implies $\sin 	heta = 1/l$.To keep an object stationary relative to the inclined plane,the pseudo-force $ma$ acting horizontally must balance the component of gravity along the plane.The forces acting along the inclined plane are the component of gravity $mg \sin 	heta$ acting downwards and the component of the pseudo-force $ma \cos 	heta$ acting upwards.For the object to remain stationary,these forces must be equal: $mg \sin 	heta = ma \cos 	heta$.This simplifies to $a = g 	an 	heta$.Given $\sin 	heta = 1/l$,we can find $	an 	heta$ using the identity $	an 	heta = \frac{\sin 	heta}{\sqrt{1 - \sin^2 	heta}} = \frac{1/l}{\sqrt{1 - 1/l^2}} = \frac{1}{\sqrt{l^2 - 1}}$.Substituting this into the expression for acceleration,we get $a = \frac{g}{\sqrt{l^2 - 1}}$.

The horizontal acceleration that should be given to a smooth inclined plane of angle $\sin^{-1} (1/l)$ to keep an object stationary on the plane relative to the inclined plane is

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