Question

A block of mass 2kg rests on a rough inclined plane making an angle of 30 with the horizontal. The co-efficient of static friction between the block and the plane is 0.7. The frictional force on the block is

Answer 1

$\bold{F_{f} = Mu \times N}$

As, N = mg CosØ

$F_{f} = Mu \times mg CosØ$

$F_{f} = 0.7 \times 2 \times 10 \times 30$

$F_{f} = 0.7 \times 2 \times 10 \times \dfrac{\sqrt{3}}{2}$

$F_{f} = 12.12 N$

$\bold{Hence, Frictional Force = 12.12 N}$

Answer 2

Given,

Mass of the body, m=2kg

Angle of inclination between the block and the horizontal line is,ø=30°

coefficient of friction ,ų= 0.7

The force that tend to slide the body on the inclined plane is=mg sinø

=mg sin30°

=2×10×1/2(g=10m/s^2)

=10N

the limiting friction between the block and the hrozontal plane is

=ųmg cos30°

=0.7×2×10×0.865

=12.11N

So friction is equal to the applied force(•.•applied force <limiting friction)

So, frictional force is 10N