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 $P_1=mg(\sin \theta -\mu \cos \theta )$ to $P_2=mg(\sin \theta +\mu \cos \theta )$, the frictional force $f$ versus $P$ graph will look like

• A.
• B.
• C.
• D.

Answer 1

The correct option is A


As $\tan \theta >\mu$, the block has a tendency to slip down the incline. Therefore, friction will act up the inclined plane.
$P+f=mg\sin \theta$
$\Rightarrow f=mg\sin \theta -P.$

At $P=P_1$,
$f=mg\sin \theta -mg\sin \theta +\mu mg\cos \theta$
$=\mu mg\cos \theta =f_{max}$
Now as $P$ increases, $f$ decreases linearly with respect to $P$.
When $P=mg\sin \theta ,f=0.$

When $P$ is increased further, the block has a tendency to move upwards along the incline. Therefore the frictional force acts downward along the incline.


i.e $P=f+mg\sin \theta$
$\Rightarrow f=P-mg\sin \theta$
Now as $P$ increases, $f$ increases linearly w.r.t. $P$.
At $P=P_2$,
$f=mg\sin \theta +\mu mg\cos \theta -mg\sin \theta$
$=\mu mg\cos \theta =f_{max}$ in the opposite direction.
This is repesented by graph (a).