1. The problem statement, all variables and given/known data Calculate the horizontal force P required to raise the 100-kg load. The coefficient of friction between the rope and fixed bars is 0.40. 2. Relevant equations T1/T2 = e^(βμ) Where β = the angle of the rope on the log, μ = 0.40, T1 = the bigger tension (P), and T2 = the smaller tension. 3. The attempt at a solution So upon first glance, I assumed that I needed to break this down into a FBD. I broke it into three different diagrams: 1.) the weight, 2.) the lower bar, and 3.) the upper bar. I then tried to solve the tensions from bottom to top. FBD of weight So this one was pretty simple, the block weighs (100)(9.81) = 981 N, so the tension in the cable has an upward force of 981 N. FBD of lower bar For this one, my drawing had two forces going forward (due to it moving), but I had the larger force at an angel going towards the upper bar. The smaller force was 981 N. To find the angle β, I used (3d)/2 and d/2 and used tan θ = ((3d)/2)/(d/2), which gave me θ = 71.565°. So I multiplied this times π/180 to get 0.39758 rads. So to find the larger tension I used the T1/T2 = e^(βμ) equation. T/981 = e^(.4*0.39758) and found T = 1616.771 N FBD of upper bar For the upper bar, my FBD had 1616.771 N coming from an angle from the lower bar, then going around the upper bar and having P horizontal like it's shown in the problem. For this one P is the bigger force and 1616.771 N is the smaller force. To find β, I added 71.565° + 90°, which equals 0.89758π rads. So I used the T1/T2 equation again: P/1616.771 = e^(.4*0.89758π), which gave me P = 4994 N, or 4.994 kN. This is pretty far off, since the book says the answer is 3.30 kN.