# A uniform cube of side a and mass m rests on a rough horizontal table

Problem: A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force F is applied normal to one of the faces at a point that is directly above the centre of the face, at a height 3a/4 above the base. The minimum value of F for which the cube begins to tip about the edge is________ (Assume that the cube does not slide.)  (IIT JEE 1984)

Solution: Let the cube topples about the point O.

The forces acting on the cube are applied force $F$, weight $mg$, frictional force $f$ and normal reaction $N$. The centre of mass of the cube lies on its geometrical centre because the cube is uniform. Thus, perpendicular distance of the weight $mg$ from the point O is $a/2$. The applied force $F$ is minimum when the cube is about to topple. At this instant, only point O of the cube is in contact with the ground. Thus, normal reaction $N$ passes through the point O. The torques of $N$ and $f$ about the point O are zero because these forces pass through this point. When the cube is about to topple, net torque on the cube about the point O is zero i.e., \begin{align} F(3a/4)-mg(a/2)=0, \end{align} which gives, \begin{align} F=2mg/3.\nonumber \end{align}