A body of mass 40 kg is placed on a rough inclined plane which makes an angle of 30\(^0\) with the horizontal. If a force of 420 N is applied upwards parallel to the plane. find the:
(a) maximum friction force that will keep the body in equilibrium;
(b) coefficient of friction.[Take g = 10ms\(^{-1}\)
Explanation
From the diagram above;
fr + mg sin \(\theta\) = 420
fr = 420 - 40 x 10 sin 30ยบ
fr = 420 - 200 = 220N
(b) coefficient of friction(ฮผ) = \(\frac{\text{frictional force}}{\text{normal reaction}}\) = \(\frac{fr}{R}\)
ฮผ = \(\frac{fr}{mg cos \theta}\) = \(\frac{220}{400 cos 30}\) = 0.63508
Therefore, ฮผ = 0.635.