Question

A block of mass $m$ rests on rough inclined plane with inclination at ${40}^{\circ }$ to the horizontal and is just about to slip. What is the coefficient of friction between the plane and the block?

• A. $\frac{1}{\tan {40}^{\circ }}$
• B. $\tan {20}^{\circ }$
• C. $\tan {40}^{\circ }$
• D. $\frac{1}{\tan {20}^{\circ }}$

Answer 1

The correct option is C $\tan {40}^{\circ }$


Let $mg\sin \theta$ and $mg\cos \theta$ is componenet of weight of block in normal and parallel to the inclined plane respectively. Block is at rest
Net force normal to inclined plane is
$N=mg\cos \theta$
Net force in the plane of inclined plane is
$f=mg\sin \theta ...\left(i\right)$
$\because$ Block is about to slip
$\therefore$ $f=\text{limiting friction}=\mu N$
$\Rightarrow f=\mu mg\cos \theta ...\left(ii\right)$
from eq $\left(i\right)$ and $\left(ii\right)$
$\mu mg\cos \theta =mg\sin \theta$
$\mu =\tan \theta =\tan {40}^{\circ }$