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. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${37}^{\circ }$
• D. ${53}^{\circ }$

Answer 1

The correct option is C ${37}^{\circ }$

Given: $F_1=2F_2$
Case $\left(i\right)$
When body moves in upward direction then friction acts along the plane in downward direction.
Given: $F_1=2F_2$
Case $\left(ii\right)$
To avoid slipping down of the body case, friction acts along the plane in upward direction
In case $\left(i\right)$
$F_1=mg\sin \theta +\mu mg\cos \theta$
In case $\left(ii\right)$
$F_2=mg\sin \theta -\mu mg\cos \theta$
$F_1=2F_2$
$mg\sin \theta +\mu mg\cos \theta =2mg\sin \theta -2\mu mg\cos \theta$
$3\mu \cos \theta =\sin \theta$
$\therefore \theta ={\tan }^{-1}3\mu$
$\theta ={\tan }^{-1}3\left(0.25\right)=ta{n}^{-1}=0.75={37}^{\circ }$