The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. If the angle of inclination of the plane is ${\text{60}}^0\text{,}$ then the coefficient of friction is ..

Upward motion : Friction force acts in downward direction

$\therefore {\mathrm{F}}_{\mathrm{u}}=\mathrm{mgsin}{60}^{\circ }+\mathrm{f}=\frac{\sqrt{3}}{2}\mathrm{mg}+\frac{\mathrm{\mu mg}}{2}=\mathrm{mg}\left[\frac{\sqrt{3}+\mathrm{\mu }}{2}\right]$

Downward motion : Friction force acts in upward direction

$\therefore {\mathrm{F}}_{\mathrm{d}}=\mathrm{mgsin}{60}^{\circ }-\mathrm{f}=\frac{\sqrt{3}}{2}\mathrm{mg}-\frac{\mathrm{\mu mg}}{2}=\mathrm{mg}\left[\frac{\sqrt{3}-\mathrm{\mu }}{2}\right]$

According to the question, ${\mathrm{F}}_{\mathrm{u}}=2{\mathrm{F}}_{\mathrm{d}}$

$\mathrm{mg}\left[\frac{\sqrt{3}+\mathrm{\mu }}{2}\right]=2\times \mathrm{mg}\left[\frac{\sqrt{3}-\mathrm{\mu }}{2}\right]$

$\text{ OR }\sqrt{3}+\mathrm{\mu }=2[\sqrt{3}-\mathrm{\mu }]\Rightarrow \mathrm{\mu }=\frac{1}{\sqrt{3}}$