Question

A body is placed on an inclined plane. The coefficient of friction between the body and the plane is $\mu$. The plane is gradually tilted up. If $\theta$ is the inclination of the plane, then frictional force on the body is

• A. constant upto $\theta ={\tan }^{-1}\left(\mu \right)$ and decrease after that
• B. increases upto $\theta ={\tan }^{-1}\left(\mu \right)$and decrease after that
• C. decreases upto $\theta ={\tan }^{-1}\left(\mu \right)$ and constant after that
• D. Increases upto $\theta ={\tan }^{-1}\left(\mu \right)$ and constant after that

Answer 1

The correct option is D Increases upto $\theta ={\tan }^{-1}\left(\mu \right)$ and constant after that

As $\theta$ increases, frictional force $(f=mg\sin \theta )$ will increase upto limiting value i.e. $\theta ={\tan }^{-1}\left(\mu \right)$
Here, from the FBD of block we can say that
$N=mg\cos \theta$, After limiting value, while $\theta$ will increase, the value of $\cos \theta$ will decrease which will result in decreasing the normal value and simultaneously the frictional force value.
Hence, the frictional force on the body will decrease after $\theta ={\tan }^{-1}\left(\mu \right)$