Question

A box of mass 8 kg is placed on a rough inclined plane of inclined $\theta$. Its downward motion can be prevented by applying an upward pully 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. $\left(tan\theta \right)/3$
• B. $tan\theta$
• C. $\left(tan\theta \right)/2$
• D. $2tan\theta$

Answer 1

The correct option is A $\left(tan\theta \right)/3$

For downward direction

Force equation in perpendicular direction is given as

$N=mg\cos \theta$

$f^{{}^{\prime }}=\mu mg\cos \theta$

Force equation parallel to incline is

$f+f^{{}^{\prime }}=mg\sin \theta$

$f+\mu mg\cos \theta =mg\sin \theta$

$f=mg\sin \theta -\mu mg\cos \theta ..............\left(1\right)$

For upward direction

Force equation in perpendicular direction is given as

$N=mg\cos \theta$

$f^{{}^{\prime }}=\mu N$

$f^{\prime }=\mu mg\cos \theta$

Force equation parallel to incline is

$2f=mg\sin \theta +f^{\prime }$

$2f=mg\sin \theta +\mu mg\cos \theta ............\left(2\right)$

From equation (1),equation (2) becomes

$2mg\sin \theta -2\mu mg\cos \theta =mg\sin \theta +\mu mg\cos \theta$

$mg\sin \theta =3\mu mg\cos \theta$

$\tan \theta =3\mu$

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