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. If the coefficient of friction is $\mu$, then 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 ${\tan }^{-1}3\mu$

For both the cases, friction force, $f=\mu N=\mu mg\cos \theta$

Force required to pull upwards,
$F_1=(mg\sin \theta +\mu mg\cos \theta )$

Force required to stop sliding,
$F_2=(mg\sin \theta -\mu mg\cos \theta )$
Since $F_1=2F_2$
$(mg\sin \theta +\mu mg\cos \theta )=2(mg\sin \theta -\mu mg\cos \theta )$
$\Rightarrow \tan \theta =3\mu$
$\theta ={\tan }^{-1}\left(3\mu \right)$