Question

The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefficient of friction between the body and the inclined plane is $\frac{1}{2 \sqrt{3}}$ , then the angle of the inclined plane is

• A. $60^\circ $
• B. $45^\circ $
• C. $30^\circ $
• D. $15^\circ $

Answer 1

Answer: (C)

Minimum force required to move a body up a rough inclined plane,
$F_{1}=mg\left(\right.sin \theta +μcos⁡\theta \left.\right)$
Minimum force required to prevent the body from sliding down the rough inclined plane,
$F_{2}=mg\left(sin \theta - μcos ⁡ \theta \right)$
According to the question,
$F_{1}=3F_{2}$
$mg\left(sin \theta + μcos ⁡ \theta \right)=3\left(mgsin ⁡ \theta - μmgcos ⁡ \theta \right)$
$4\mu cos \theta =2sin⁡\theta $
$sin \theta = 2 \mu cos ⁡ \theta $
$tan \theta = 2 \mu = \frac{1}{\sqrt{3}}$
$\theta =30^\circ $