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 $ma$ act

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 acting down the plane.Resolving the forces parallel to the inclined plane:The component of the pseudo force $ma$ along the plane is $ma \cos 	heta$.The component of gravity $mg$ along the plane is $mg \sin 	heta$.For the object to remain stationary relative to the plane, these forces must be equal: $ma \cos 	heta = mg \sin 	heta$.Thus, $a = g 	an 	heta$.Given $\sin 	heta = \frac{1}{l}$, we can find $	an 	heta$. In a right-angled triangle with opposite side $1$ and hypotenuse $l$, the adjacent side is $\sqrt{l^2 - 1}$.Therefore, $	an 	heta = \frac{1}{\sqrt{l^2 - 1}}$.Substituting this into the expression for $a$, we get $a = \frac{g}{\sqrt{l^2 - 1}}$.

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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