Q. A body is sliding down an inclined plane having coefficient of friction 0.5. If the normal reaction is twice that of the resultant downward force along the incline, the angle between the inclined plane and the horizontal is
A) $15{^\circ}$
B) $30{^\circ}$
C) $45{^\circ}$
D) $60{^\circ}$
Explanation
(c) Resultant downward force along the incline
$= mg\ (\sin\theta - \mu\cos\theta)$
Normal reaction $= mg\cos\theta$
Given : $mg\cos\theta = 2\ mg(\sin\theta - \mu\cos\theta)$
By solving $\theta = 45{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.