A block of mass 2kg rests on a rough inclined plane making an angle 30 degree on horizantal static friction is 0.6 find frictional force between them
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the force applied on the body that is on the inclined plane is given,as
F=mg sin θ
F=2×9.8×sin 30
=9.8
the limiting friction force between the block and the inclime plane is given as,
f=mumg cos θ
f=0.6×2×9.8 cos 30
=10.18 N
since the limiting friction force is greater than the force that tends to slide the body
Thus,the body will be at rest and the force of friction on the block is 9.8 N
since
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