Bart wrote:

A load on the line is 1000 lbs.

You have just lead a line through a block, around
an angle of 180 degrees. What is the load on the
sheave? [1 pt]

2000 lbs.

For an angle of 90 degrees, what would be the load on
the sheave? [1 pt]

2 x load x cos(angle/2) = 2000 cos(45) = 2000(.707) = 1,414 pounds

At what angle does the sheave load match the line load?
[1 pt]

Load = 2 x load x cos(angle/2)

cos(angle/2) = 1/2

angle = 2 inv_cos(1/2) = 2 * 60 = 120 degrees or pi/3 radians.


//Walt