Question

A body of mass $m$ is kept on a rough fixed inclined plane of angle of inclination $\theta$. It remains stationary. Then magnitude of force acting on the body by the inclined plane is equal to

• A. $\mu\,mg\,cos\,\theta $
• B. $mg\,cos\,\theta $
• C. $mg\,sin\,\theta $
• D. $mg $

Answer 1

Answer: (D)

Friction force by the inclined plane is $=\mu mg \cos \theta= mg \sin \theta$ up the inclined plane since the block does not slide.
Normal force perpendicular to inclined plane is $m g \cos \theta$.
So, Total force is vector sum of two forces is equal to $mg$.
$
\begin{array}{l}
N =m g \cos \theta, f _{ s }=m g \sin \theta \\
R ^{2}= N ^{2}+ f _{ s }^{2} \Rightarrow R = mg
\end{array}
$