Question

A block placed on a rough inclined plane of inclination $(\theta ={30}^0)$ can just be pushed upwards by applying a force "F" as shown. If the angle of inclination of the inclined plane is increased to $(\theta ={60}^0)$, the same block can just be prevented from sliding down by application of a force of same magnitude. The coefficient of friction between the block and the inclined plane is

• A. $\frac{\sqrt{3}+1}{\sqrt{3}-1}$
• B. $\frac{2\sqrt{3}-1}{\sqrt{3}+1}$
• C. $\frac{\sqrt{3}-1}{\sqrt{3}+1}$
• D. None of these

Answer 1

Answer: (C)

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

The friction force in the first case is in the downward direction and in the second case it is in the upward direction. From FBD,
when $\theta ={30}^o,F-mg\sin {30}^o-\mu mg\cos {30}^o=0......\left(1\right)$
when $\theta ={60}^o,F+\mu mg\cos {60}^o-mg\sin {60}^o=0......\left(2\right)$
using (1), (2) becomes, $mg\sin {30}^o+\mu mg\cos {30}^o=-\mu mg\cos {60}^o+mg\sin {60}^o$
or
$\mu (\cos {60}^o+\cos {30}^o)=\sin {60}^o-sin{30}^o$
or
$\mu (\frac{1}{2}+\frac{\sqrt{3}}{2})=\frac{\sqrt{3}}{2}-\frac{1}{2}$
$\therefore \mu =\frac{\sqrt{3}-1}{\sqrt{3}+1}$