Question

A box of mass $\text{8 kg}$ is placed on a rough inclined plane of angle $\theta$. Its downward motion can be prevented by applying an upward pull $F$ and it can be made to slide upwards(constant speed) by applying a force $2F$. The coefficient of friction between the box and the inclined plane is

• A. $\frac{\tan \theta }{3}$
• B. $3\tan \theta$
• C. $\tan \frac{\theta }{3}$
• D. $2\tan \theta$

Answer 1

The correct option is A $\frac{\tan \theta }{3}$


Case(i) When block is about to slip down the inclined plane, friction acts up the plane
$F+f=mg\sin \theta --\left(i\right)$


Case(ii) When block is sliding up the inclined plane, friction acts down the plane.

$2F=mg\sin \theta +f--\left(ii\right)$

Taking $2\times \left(i\right)-\left(ii\right)$ gives
$3f=mg\sin \theta$
$\Rightarrow f=\frac{mg\sin \theta }{3}=\mu mg\cos \theta$ $(\because f=\mu N=\mu mg\cos \theta )\text{in both cases}$
$\Rightarrow \mu =\frac{\tan \theta }{3}$