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

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(B) To find the minimum force $F$ required to topple the cube,we consider the torque about the edge of the cube about which it will topple (point $D$)

Vedclass · Accepted Answer

$\frac{2}{3} mg$

Vedclass · Answer

(B) To find the minimum force $F$ required to topple the cube,we consider the torque about the edge of the cube about which it will topple (point $D$).The torque due to the applied force $F$ about point $D$ is given by $	au_F = F 	imes \frac{3a}{4}$.The torque due to the gravitational force $mg$ (acting at the center of mass of the cube,which is at a horizontal distance of $\frac{a}{2}$ from the edge $D$) about point $D$ is $	au_g = mg 	imes \frac{a}{2}$.For the cube to begin to topple,the torque due to the applied force must be equal to the torque due to gravity:$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$Thus,the minimum force required is $\frac{2}{3} mg$.

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