Question

5
S
43. A block of mass m is on an inclined plane of angle
0. The coefficient of friction between the block and
the plane is u and tano > u. 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 mg(sine – u cose) to P2 = mg(sino + р
coso), the frictional force f versus P graph will
looks like
[IIT-JEE 2010]
​

Answer 1

Answer

Correct option is

A

mgsinθ+μmgcosθ

Body tends to move upward i.e., in the direction of f

1

hence frictional for f

r

is in the opposible direction of f

2

FBD of block

as for equillibrium

N

2

mgcosθ=f

r

= μmgcosθ

minimum force f

1

such that block start moving up ward

F

1

=mgsinθ+f

r

F

1

=mgsinθ+μmgcosθ.