Question

A horizontal force of value $\frac{mg}{6}$ is applied on the upper surface of a uniform cube of mass $m$ and side $a$ which is resting on a rough horizontal surface having $\mu =\frac{1}{2}.$ The distance between lines of action of mg and normal reaction is:

• A. $\frac{a}{2}$
• B. $\frac{a}{3}$
• C. $\frac{a}{4}$
• D. None of these

Answer 1

The correct option is D None of these

The situation can be pictured as shown in the figure
Given $F=\frac{mg}{6}$
For horizontal equilibrium, force in horizontal direction should be zero.
$\therefore F=f$
$\therefore f=\frac{mg}{6}$
For vertical equilibrium, force in vertical direction should be zero.
$\therefore N=mg$
For rotational equilibrium about c, torque about c should be zero.
$\therefore \frac{mg}{6}\times \frac{a}{2}\times 2=N\times d$
$\therefore \frac{mg}{6}\times \frac{a}{2}\times 2=mg\times d$
$\therefore d=\frac{a}{6}$
Therefore, the distance between the line of action of the normal and the gravitational force is $\frac{a}{6}$