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°.
Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.
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.