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 parall...

Assessment: WAEC SSCE - Further Mathematics - 2024 (Essay) Subject: Further Mathematics

Question 1 Report

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\(^{-1}\)

Answer Details

From the diagram above;

fr + mg sin \(\theta\) = 420

fr = 420 - 40 x 10 sin 30º

fr = 420 - 200 = 220N

(b) coefficient of friction(μ)  = \(\frac{\text{frictional force}}{\text{normal reaction}}\) = \(\frac{fr}{R}\)

μ =  \(\frac{fr}{mg cos \theta}\) =  \(\frac{220}{400 cos 30}\) = 0.63508

Therefore, μ = 0.635.

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