Question

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. The coefficient of friction when the angle of inclination of the plane is ${60}^o$, will be:

• A. $\frac{1}{3}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

$\frac{1}{\sqrt{3}}$

Let the coefficient of friction be $\mu$

Thus friction force acting $f=\mu mg$ $cos{60}^o=\frac{\mu mg}{2}$

Upward motion : Friction force acts in downward direction

$\therefore$ $F_u=mgsin{60}^o+f=\frac{\sqrt{3}}{2}mg+\frac{\mu mg}{2}=mg[\frac{\sqrt{3}+\mu }{2}]$

Downward motion : Friction force acts in upward direction

$\therefore$ $F_d=mgsin{60}^o-f=\frac{\sqrt{3}}{2}mg-\frac{\mu mg}{2}=mg[\frac{\sqrt{3}-\mu }{2}]$

According to the question, $F_u=2F_d$

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

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