Question

A box of mass $\text{8 kg}$ is placed on a rough inclined plane of angle $\theta$ with horizontal. Its downward motion can be prevented by applying an upward pull $F$ and it can be made to slide upwards 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. $\frac{\tan \theta }{2}$
• D. $2\tan \theta$

Answer 1

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

FBD for both the cases are shown below





Case(i): when the block is prevented from downward motion by applying force $F$
$\Rightarrow F+f=mg\sin \theta$
Case (ii):when the block is made to slide upward by applying force $2F$
$\Rightarrow 2F=mg\sin \theta +f$
On solving both equations we get,
$3f=mg\sin \theta$
Since $f=\mu N=\mu mg\cos \theta$
$\Rightarrow 3f=3\mu mg\cos \theta =mg\sin \theta$
$\Rightarrow \mu =\frac{\tan \theta }{3}$