Solution of a Simple Pulley Problem

A sheave assembly is shown at the right. The two lines EP and EQ are both inclined at an angle of 15 degrees from the vertical. The weight W weighs 5,000 pounds and the sheave weighs 60 pounds. Determine the force in the cable section CP and the cable section DQ. Assume the pulley is frictionless.

We need to start by noting the following:

• The forces in the lines are equal because they are the same cord wrapped around the frictionless pulley. We can thus say F_EP = F_EQ.
• We will refer to the angle as α, which again are equal to each other.

From these, we can designate the free body diagram as the triangle PQE, and the sum of the forces in the y-direction are

F_EP cosα + F_EQ cosα – W – W_sheave = 0

Noting the symmetries,

2F_EP cosα = W + W_sheave

Solving,

F_EP = (W + W_sheave)/(2 cosα) = (5000 + 60)/(2 cos 15) = 2619 lbs.

The solution from the source (Smith’s Mechanic) is shown above.