The horizontal acceleration that should be gi | vedclass

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(A) Let the horizontal acceleration given to the inclined plane be $a$. To keep the object stationary relative to the plane,the pseudo force $ma$ acti

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the horizontal acceleration given to the inclined plane be $a$. To keep the object stationary relative to the plane,the pseudo force $ma$ acting on the object in the horizontal direction must balance the component of gravity along the plane.Resolving the pseudo force $ma$ into components parallel and perpendicular to the inclined plane,the component parallel to the plane is $ma \cos 	heta$.The component of gravity acting down the plane is $mg \sin 	heta$.For the object to remain stationary relative to the plane,these two forces must be equal:$ma \cos 	heta = mg \sin 	heta$$a = g 	an 	heta$Given $\sin 	heta = \frac{1}{l}$,we can construct a right-angled triangle where the opposite side is $1$ and the hypotenuse is $l$. The adjacent side is $\sqrt{l^2 - 1^2} = \sqrt{l^2 - 1}$.Therefore,$	an 	heta = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{1}{\sqrt{l^2 - 1}}$.Substituting this into the expression for $a$:$a = g \left(\frac{1}{\sqrt{l^2 - 1}}ight) = \frac{g}{\sqrt{l^2 - 1}}$.

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

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