Physics - Rotation Question with Solution | TestHub

For the cube to just topple, it will pivot about the edge furthest from the applied force. Let's call this point P. The torque due to the applied force $F$ about P is $${\tau }_F=F\times a$$. The torque due to gravity (weight $Mg$) about P is $${\tau }_{Mg}=Mg\times \frac{a}{2}$$. (The center of mass is at $a/2$ from the base, and $a/2$ from the edge P horizontally).

For toppling to occur, the torque due to $F$ must be greater than or equal to the torque due to $Mg$. For minimum force to just topple, $${\tau }_F={\tau }_{Mg}$$, so $$F\times a=Mg\times \frac{a}{2}$$. This gives $$F=\frac{Mg}{2}$$. We assume friction is sufficient to prevent sliding before toppling.