An equilateral prism of mass m rests on a rou | vedclass

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(A) To topple the prism, it must rotate about the edge $C$.The force $F$ is applied at the top vertex $A$ at a height $h = a\frac{\sqrt{3}}{2}$ from t

Vedclass · Accepted Answer

$\frac{mg}{\sqrt{3}}$

Vedclass · Answer

(A) To topple the prism, it must rotate about the edge $C$.The force $F$ is applied at the top vertex $A$ at a height $h = a\frac{\sqrt{3}}{2}$ from the base.The weight $mg$ acts at the center of mass, which is at a horizontal distance of $a/2$ from the edge $C$.For the prism to be on the verge of toppling, the torque due to force $F$ about point $C$ must balance the torque due to gravity $mg$ about point $C$.Torque due to $F$ about $C$ = $F 	imes h = F 	imes a\frac{\sqrt{3}}{2}$.Torque due to $mg$ about $C$ = $mg 	imes \frac{a}{2}$.Equating the torques: $F 	imes a\frac{\sqrt{3}}{2} = mg 	imes \frac{a}{2}$.Solving for $F$, we get $F = \frac{mg}{\sqrt{3}}$.

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