Question

A force of 250 N pulls a body of weight 500 N up an inclined plane, the force being applied parallel to the plane. If the inclination of the plane to the horizontal is 15°, find the coefficient of friction.

Answer 1

Answer:0.25N

Explanation:

Answer 2

Answer:

The co-efficient of friction (  $\mu$) = 0.25 (Approx)  

Explanation:

Provided that:

Force being applied (F) = 250 N

The inclination of the plane to the horizontal = 15°

Weight of the body (N = mg) =500 N

Let, the co-efficient of friction on the inclined surface = $\mu$

As shown in the figure,If the force is pulling the body of weight 500 N up the inclined plane with uniform velocity, then from the equilibrium condition we can say;

$F=\mu \:mg\cos\theta+mg\sin\theta\\or, F-mg\sin\theta =\mu\:mg\cos\theta\\\\or, \frac{F-mg\sin\theta}{mg\cos\theta} =\mu\\\\So, \mu=\frac{(250-500\sin\(15 \degree)}{(mg\cos\(15 \degree)} \\\\\mu=0.249688897$

Hence, the co-efficient of friction (  $\mu$) = 0.25 (Approx.)  

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