The cord passing over the two small pegs A and B of the square board is subjected to a tension of 100 N. Determine the required tension P acting on the cord that passes over pegs C and D so that the resultant couple produced by the two couples is 15 N•m acting clockwise. Take θ = 15°.

#### Solution:

Let us assume clockwise moments are positive and write an equation as follows:

\circlearrowright^+ 15=100\cos30^0(0.3)+100\sin30^0(0.3)-p\cos15^0(0.3)-p\sin15^0(0.3)

(solve for p)

p=70.7 N

When solving this problem, remember that each force was broken into x and y components. These components were then used to calculate the moment.