The problem had appeared in the Allahabad Untiversity examination, in the year 1972

Problem: A body resting on a rough horizontal plane, required a pull of 18 kg inclined at 30 degrees to the plane just to remove it. It was found that a push of 22 kg inclined at 30 degrees to the plane just removed the body. Determine and find the weight of the body and the coefficient of friction.

Solution:

Given - Angle of pull theta = 30 degree

Let W = weights of the body in kg.

mu = coefficient of friction,

F = Force of friction.

Case #1: Given pull, P =

18 kg,

Resolving the forces horzontally,

F = 18 cos 30 degree = 10 * 0.5 = 15.6 kg.

Now resolving the forces

vertically,

R = W - 18 sin 30 degree = W - 18 * 0.5 = W - 9 kg

Using the relation,

F = mu * R, with normal notations.

Therefore 15.6 = mu(W - 9)

Case #2: Given Push, P = 22 kg

Resolving the forces horizontally,

F= 22 cos 30 degree = 22 * 0.866 kg = 19.04 kg,

Now resolving the forces vertically,

R = W + 22 sin 30 degree = W + 22 * 0.5 = W + 11 kg,

Again using the standard notations,

F = mu * R,

19.04 = mu(W + 11)'

Solving equations in the first case and the second case, we get,

W = 99 kg and mu = 0.1732.

### Article Written By Swagatam

Swagatam is a blogger at Expertscolumn.com

Last updated on 25-07-2016 127