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 $60^{\circ}$, then the coefficient of friction is

According to the question, angle of inclination $\theta=60^{\circ}$.
Now, force of friction, $f=\mu N=\mu m g \cos \theta$ and net retarding force, $\left(F_1\right)=m g \sin \theta+f$ $\therefore$ Net accelerating force down the inclined plane is
$\therefore$ External force needed (up the inclined plane) to maintain sliding motion is (net retarding force)
From Eqs. (i),(ii) and (iii) we get $m g \sin \theta+\mu m g \cos \theta=2(m g \sin \theta-\mu m g \cos \theta)$ or $3 \mu m g \cos \theta=m g \sin \theta$ or $$ \frac{\sin \theta}{\cos \theta}=3 \mu $$ or $$ \tan \theta=3 \mu \quad\left(\because \theta=60^{\circ}\right) $$ or $$ \tan 60^{\circ}=3 \mu $$ $$ \begin{aligned} \sqrt{3} & =3 \mu \\ \mu & =\frac{1}{\sqrt{3}} \end{aligned} $$ When inclination of plane is $60^{\circ}$ then the coefficient of friction, $\mu=\frac{1}{\sqrt{3}}$.