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