A cubical block of side L rests on a rough ho | vedclass

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(C) To find the minimum force required to topple the block,we consider the torque about the edge of the block where it is about to rotate (the pivot p

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$mg/2$

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(C) To find the minimum force required to topple the block,we consider the torque about the edge of the block where it is about to rotate (the pivot point $P$).At the critical condition for toppling,the normal reaction $N$ acts through the pivot point $P$.The torque due to the gravitational force $mg$ about point $P$ is $	au_{mg} = mg 	imes (L/2)$.The torque due to the applied horizontal force $F$ about point $P$ is $	au_{F} = F 	imes L$.For the block to topple,the torque due to the applied force must be greater than the torque due to gravity:$	au_{F} > 	au_{mg}$$F 	imes L > mg 	imes (L/2)$$F > mg/2$Thus,the minimum force required to topple the block is $F = mg/2$.

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