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 move down the incline.
At equilibrium $\Rightarrow P+f=mg\sin \theta \Rightarrow f=mg\sin \theta -P$


Now as $P$ increases the value of $f$ decreases linearly with resepct to $P$.

$\Rightarrow$When $P=mg\sin \theta ,f=0$

$\Rightarrow$When $P$ is increased further, the block will have a tendency to move upwards along the inclined plane.
$\therefore$Therefore the frictional force acts downwards along the incline in this case.

at equilibrium
$P=f+mg\sin \theta$
$\therefore f=P-mg\sin \theta$
Now as $P$ increases, $f$ increases linearly w.r.t $P$.

$\therefore$Considering above mentioned points, if you consider up the plane +ve direction and down the plane as -ve direction, frictional force is represented by graph (a)