A uniform cube of side a and mass m rests on | vedclass

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(B) For the cube to begin to tilt about the edge $O$,the normal reaction from the table must act at the edge $O$ itself.Taking the torque (moment) abo

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$\frac{2mg}{3}$

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(B) For the cube to begin to tilt about the edge $O$,the normal reaction from the table must act at the edge $O$ itself.Taking the torque (moment) about the edge $O$:The torque due to the applied force $F$ is $	au_F = F 	imes \frac{3a}{4}$ (tending to rotate clockwise).The torque due to the weight $mg$ of the cube is $	au_g = mg 	imes \frac{a}{2}$ (tending to rotate counter-clockwise,as the weight acts at the centre of mass).For the cube to be on the verge of tilting,the net torque about $O$ must be zero:$F 	imes \frac{3a}{4} = mg 	imes \frac{a}{2}$Solving for $F$:$F = mg 	imes \frac{a}{2} 	imes \frac{4}{3a}$$F = \frac{2}{3} mg$

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