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(A) When the box is on the verge of toppling,the normal reaction $N$ acts through the edge $O$ about which it tends to rotate. The torque due to the a

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$\frac{a}{2 H}$

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(A) When the box is on the verge of toppling,the normal reaction $N$ acts through the edge $O$ about which it tends to rotate. The torque due to the applied force $F$ about $O$ must balance the torque due to the weight $mg$ about $O$.For toppling to occur:$F 	imes H = mg 	imes \frac{a}{2} \quad \dots(1)$For sliding to occur,the applied force must overcome the maximum static friction:$F = \mu mg \quad \dots(2)$At the critical height $H$,the box is on the verge of both toppling and sliding simultaneously. Substituting $F = \mu mg$ from equation $(2)$ into equation $(1)$:$(\mu mg) 	imes H = mg 	imes \frac{a}{2}$Dividing both sides by $mg$:$\mu H = \frac{a}{2}$Therefore,the coefficient of friction is:$\mu = \frac{a}{2 H}$

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