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 `tanthetagtmu`. 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(sintheta-mucostheta)` to `P_2=mg(sintheta+mucostheta)`, the frictional force f versus P graph will look like
A.
B.
C.
D.

Answer 1

Correct Answer - A
As `tan thetagtmu`, the block has a tendensy to move down the incline. Therefore a force P is applied upwards along the incline. Here, at equilibrium `P+f=mg sin theta implies f= mg sin theta-P`

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 downwards along the incline.
Here, at euilibrium `P=f+mg sin theta`
`:.` `f=P-mg sin theta`
Now as P increases, f increases linearly w.r.t P.
This is respresented by graph(a).