Question

The force required to just move a body up an 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. ${\tan }^{-1}\left(\mu /2\right)$
• C. ${\tan }^{-1}2\mu$
• D. ${\tan }^{-1}3\mu$

Answer 1

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

Where $F_1$,is the force required to just move the body up the inclineand $F_2$ is the force required for the body to prevent it from sliding.

Now,

$F_1=mg\sin \theta +\mu mg\cos \theta F_2=mg\sin \theta -\mu mg\cos \theta \Rightarrow \frac{F_1}{F_2}=2=\frac{mg\sin \theta +\mu mg\cos \theta }{mg\sin \theta -\mu mg\cos \theta }2(\sin \theta -\mu \cos \theta )=\sin \theta +\mu \cos \theta \sin \theta =3\mu \cos \theta \tan \theta =3\mu \theta ={\tan }^{-1}\left(3\mu \right)$