Question

The force required just to move a body up an inclined plane is double the force required just to prevent the body sliding down. If the coefficient of friction is $0.25$ the angle of inclination of the plane is?

• A. $36.8$
• B. $45$
• C. $30$
• D. $53$

Answer 1

The correct option is A $36.8$

$Given:$The coefficient of friction $=0.25$

$Solution:$ The force required just to move a body up an inclined plane

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

The force required just to prevent the body sliding down

$F_2=mg(\sin \theta -\mu \cos \theta )$

Now,

$F_1=2F_2$

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

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

$\tan \theta =3\mu$

$\tan \theta =3\times 0.25$

$\tan \theta =0.75$

$\theta ={\tan }^{-1}0.75$

$\theta ={36.8}^0$

Hence, the angle of inclination of the plane is ${36.8}^0$