Question

A box of mass $8kg$ is placed on a rough inclined plane of inclination $\theta$. 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. $\frac{1}{3}\tan \theta$
• B. $3\tan \theta$
• C. $\frac{1}{2}\tan \theta$
• D. $2\tan \theta$

Answer 1

The correct option is A $\frac{1}{3}\tan \theta$

The minimum force required to just prevent downward motion

$F=mgsin\theta -\mu mgcos\theta$...(i)

the force rquired to start upward motion

$2F=mgsin\theta +\mu mgcos\theta$...(ii)

from (i) and (ii)

$\frac{1}{2}=\frac{tan\theta -\mu }{tan\theta +\mu }$

$tan\theta =3\mu$

$\mu =\frac{1}{3tan\theta }$