A uniform cube of mass m and side a is placed on a frictionless...

Consider the below diagram Moment of the force $F$ about point (A, \tau_1=F \times a) (anti-clockwise) Moment of weight $m g$ of the cube about point $A$, (\tau_2=m g \times \frac{a}{2} \text{ (clockwise) }) Cube will not exhibit motion, if (\tau_1=\tau_2) ((\because) In this case, both the torque will cancel the effect of each other] ($$\begin{array}{ll} \therefore & F \times a=m g \times \frac{a}{2} \\ \Rightarrow & F=\frac{m g}{2} \end{array}$$) Cube will rotate only when, (\tau_1>\tau_2) (\Rightarrow \quad F \times a>m g \times \frac{a}{2} \Rightarrow F>\frac{m g}{2})