Question

The force required to just move a body up the inclined plane is double the force required to just prevent the body from sliding down the plane. The coefficient of friction is $\mu$. The inclination $\theta$ of the plane is

• A. ${\tan }^{-1}\mu$
• B. ${\tan }^{-1}\left(\frac{\mu }{2}\right)$
• C. ${\tan }^{-1}2\mu$
• D. ${\tan }^{-1}3\mu$

Answer 1

The correct option is D ${\tan }^{-1}3\mu$

$N=mg\cos \theta$

$fs=N\mu =\mu mg\cos \theta$

$mg\sin \theta =fs+f$

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

$f=mg(\sin \theta -\mu \cos \theta )$ ...(1) [ref. image 1]

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

$f^{\prime }=\mu mg\cos \theta +mg\sin \theta$ ...(2) [ref. image 2]

$f^{\prime }=2f$ ...(3)

$\therefore 2mg(\sin \theta -u\cos \theta )=mg(\sin \theta +\mu \cos \theta )$

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

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

$\tan \theta =3\mu$

$\theta ={\tan }^{-1}\left(3\mu \right)$.