Question

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

The correct option is 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 free body diagram.
When
$\theta ={30}^{\circ },F-mg\sin {30}^{\circ }-\mu mg\cos {30}^{\circ }=0$
$F=mg\sin {30}^{\circ }+\mu mg\cos {30}^{\circ }$ ... (1)
When
$\theta ={60}^{\circ },F+\mu mg\cos {60}^{\circ }-mg\sin {60}^{\circ }=0$
$F=mg\sin {60}^{\circ }-\mu mg\cos {60}^{\circ }$ ... (2)
Now from equation (1) and (2)
$mg\sin {30}^{\circ }+\mu mg\cos {30}^{\circ }=mg\sin {60}^{\circ }-\mu mg\cos {60}^{\circ }$
$\Rightarrow mg\left(\frac{1}{2}\right)+\mu mg\left(\frac{\sqrt{3}}{2}\right)=mg\left(\frac{\sqrt{3}}{2}\right)-\mu mg\left(\frac{1}{2}\right)$
$\Rightarrow \mu [\frac{\sqrt{3}}{2}+\frac{1}{2}]=[\frac{\sqrt{3}}{2}-\frac{1}{2}]$
$\Rightarrow \mu (\sqrt{3}+1)=(\sqrt{3}-1)$
$\Rightarrow \mu =\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)$