This question appeared in the Bombay University exams in the year 1973.
Problem: A rectangular prism block, weighing 15 kg, is resting on a rough inc lined plane. The block is tied up by a horizontal string, which has a tension of 5 kg. Find, the frictional force on the block, the normal reaction of the inclined plane, the coefficient of friction between the surfaces of contact.
Solution:
Given: Weight of the prism, W = 15 kg
Tension in string = 5 kg,
Frictional force on the block
Let F = Frictional force,
R = Normal reaction of the inclined

plane,
Mu = Coefficient of friction,
A little consideration shows that since the string is in tension means that the block is trying to slide downward and the frictional

force is trying to oppose this toward the upward direction.
The block is in equilibrium because of the following forces.
Its own weight,
Horizontal force of 5 kg as a result of the tension over the string,
Normal reaction R over the plane,
The force of friction F acting upward.
Resolving the forces along the plane of inclination, we get:
F + 5 cos 45 degree = 15 sin 45 degree
F = 15 sin 45 degree – 5 cos 45 degree = 7.07 kg.
Normal reaction of the inclined plane
Resolving the forces perpendicular to the plane gives:
R = 5 sin 45 degree + 15 cos 45 degree = 14.14 kg
Using the standard relation we get,
F = mu * R
Therefore mu = F / R = 7.07 / 14.14 = 0.5 is the answer.

### Article Written By Swagatam

Swagatam is a blogger at Expertscolumn.com

Last updated on 25-07-2016 563

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