Question

A weight 500N just starts moving down a rough inclined plane supported by force 200N acting parallel to the plane and it is at the point of moving up the plane when pulled by a force of 300N parallel to the plane. Find the inclination of the plane and the coefficient of friction between the inclined plane and the weight.

Answer 1

In first case body is moving down the plane, so frictional force is acting up the plane Let θ be the angle of inclination and µ be the coefficient of friction.

Sum of forces parallel to plane

= 0 200 + µR = 500 sinθ ...(i)

Sum of forces perpendicular to plane

= 0 R = 500 cosθ ...(ii)

Putting the value of (ii) in equation (i) 200 + 500µcos?

= 500 sinθ ...(iii)

Now 300N is the force when applied to block, it move in upward direction.

Hence in this case frictional force acts downward.

Sum of forces perpendicular to plane = 0

R = 500 cosθ ...(vi)

Sum of forces parallel to plane

= 0 300 = µR + 500 sinθ ...(v)

Putting the value of (iv) in equation (v) 300

= 500µcosθ + 500 sinθ ...(vi)

Adding equation (iii) and (vi),

We get Sinθ = 1/2; or θ = 30°

Putting the value in any equation we get

µ = 0.115.