1. The problem statement, all variables and given/known data The 180-lb tree surgeon lowers himself with the rope over a horizontal limb of the tree. If the coefficient of friction between the rope and the limb is 0.60, compute the force (P) which the man must exert on the rope to let himself down slowly. 2. Relevant equations T1/T2 = e^(βμ) Where β = the angle of the rope on the log, μ = 0.60, T1 = the bigger tension (180 lbs), and T2 = the smaller tension (P) 3. The attempt at a solution So I looked at the picture and I believe β = 180° which = π rads. Then I plugged those into my equation and got: 180/P = e^(.6π), this gave me P = 27.33 lbs, which is incorrect. The correct answer should be 23.7 lbs, but I have no idea what I'm doing wrong. I'm guessing my β is the wrong number, but I'm not sure how to find the correct number if that's the case.