Question

The force required to just move body up the plane is double the force required to just prevent the body from sliding down the plane. The coefficient of friction is $\mu$ and the force is parallel to the plane. The inclination of the plane is $\theta$, then $\tan \theta =$

• A. $\mu$
• B. $\frac{\mu }{2}$
• C. $2\mu$
• D. $3\mu$

Answer 1

The correct option is D $3\mu$

Force required to prevent the body from sliding

$F=mg\sin \theta -\mu mg\cos \theta$

Force required to just move the body up

$F_1=mg\sin \theta +\mu mg\cos \theta$

It is given

$F_1=2f$

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

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