Question

$A$ block of mass $m$ is on an inclined plane of angle $\theta$. The coefficient of friction between the block and the plane is $\mu$ and tan $\theta >\mu .$ The block is held stationary by applying a force $P$ parallel to the plane. The direction of force pointing up the plane is taken to be positive. As $P$ is varied from $P_1=mg$ $\left(\sin \theta -\mu \cos \theta \right)$ to $P_2=mg\left(\sin \theta +\mu \cos \theta \right)$, the frictional force $f$ versus $P$ graph will look like

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

when
$P=mg(\sin \theta -\mu \cos \theta )$
$f=\mu mg\cos \theta$ (upwards)
when $P=mg\sin \theta ;f=0$
and when $P=mg(\sin \theta +\mu \cos \theta )$
$f=\mu mg\cos \theta$ (downwards)
Hence friction is first positive, then zero and then negative.