1. The problem statement, all variables and given/known data The uniform I-beam has a mass of 74 kg per meter of length and is supported by the rope over the fixed 300-mm drum. If the coefficient of friction between the rope and drum is 0.50, calculate the least value of the force P which will cause the beam to tip from its horizontal position. 2. Relevant equations T1/T2 = e^(βμ) Where β = the angle of the rope on the drum, μ = 0.50, T1 = the bigger tension, and T2 = the smaller tension 3. The attempt at a solution So to start off, I added the lengths up to see that the beam is 1.8 m long, I then used that to determine the weight = 1.8*74*9.81 = 1306.7 N. Also the rope on the drum suggests the angle β = π rads and μ is known to be 0.50. I began by making my FBD, which I believe is probably where I've made my mistake. I broke it into two separate FBD, one with just the beam and one with just the drum. FBD of beam The FBD of the beam I saw 4 forces acting on it. the weight (1306.7 N) acting downward in the middle of the beam, the force P acting downward (as indicated in the problem), and the two ropes which I called tension 1 (left rope) and tension 2 (right rope) both with forces acting upward due to them being tension ropes (therefore they cannot be downward, I believe). I used the moment equation ∑M = 0 at T1 and T2. ∑M =0 at T2 gives me: (-.75)(-P) + (.15)(-1306.7) + (.3)(T1) ⇒⇒ T1 = (1.05P + 196)/0.3 ∑M =0 at T1 gives me: (-1.05)(-P) + (-.15)(-1306.7) + (-.3)(T1) ⇒⇒ T2 = (-.75P + 196)/0.3 FBD of drum For this FBD I just had T1 going downward (due to the tipping) and T2 going upward. I used the T1/T2 equation to have (1.05P+196)/(-.75+196) = e^(.5π), which gave me P = 244.5 N. This is incorrect, as the book says the answer should be P = 160.3 N.