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$A$ horizontal force $F = mg/3$ 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 $\mu_S = 1/2$. The distance between lines of action of $‘mg’$ and normal reaction $‘N’$ is :
$a/2$
$a/3$
$a/4$
None
Solution
In this case, torque of $\mathrm{mg}$ is balanced by torque due to normal reaction and applied force $F.$
$\therefore \mathrm{N}\left(\frac{\mathrm{a}}{2}-\mathrm{x}\right)+\mathrm{Fa}=\mathrm{mg} \frac{\mathrm{a}}{2}$
or $\mathrm{N}\left(\frac{\mathrm{a}}{2}-\mathrm{x}\right)+\frac{\mathrm{mg} \mathrm{a}}{3}=\frac{\mathrm{mga}}{2}$
or $\mathrm{N}\left(\frac{\mathrm{a}}{2}-\mathrm{x}\right)=\frac{\mathrm{mga}}{2}-\frac{\mathrm{mga}}{3}=\frac{\mathrm{mga}}{6}$
But $\mathrm{N}=\mathrm{mg}$
$\therefore \quad \mathrm{mg} \frac{\mathrm{a}}{2}-\mathrm{mgx}=\frac{\mathrm{mga}}{6}$
or $\quad x=\frac{a}{2}-\frac{a}{6}=\frac{3 a-a}{6}=\frac{a}{3}$
