1. The problem statement, all variables and given/known data Blocks A and B are connected by a cable that passes over support C. Friction between the blocks and the inclined surfaces may be neglected. Knowing that motion of block B up the incline is impending when mB = 4.5 kg, determine: (a) the coefficient of static friction between the rope and the support (b) the largest value of mB for which equilibrium is maintained 2. Relevant equations T2 = T1e^(μsβ) (1) β is in angle in radians W = mg 3. The attempt at a solution My problem is that for part (b), my answer is wrong. I would like to know where I went wrong in my calculations. Thanks for the help in advance. I found the forces working on Block A ƩFx = 0 → TA - WAsin(16) = 0 TA = 24.336 N Then, Block B ƩFx = 0 → WBsin(16) - TB = 0 TB = 12.168 N Then, I looked at support C where I found the angle to be 32°, and I used (1) to find the static friction. β = (32)(∏/180) ≈ 0.559 TA = TBe^(μsβ) 24.336 N = (12.168 N)e^(μsβ) e^(μsβ) = (24.336 N)/(12.168 N) μsβ = ln [(24.336 N)/(12.168 N)] μs = (1/0.559)*ln [(24.336 N)/(12.168 N)] μs ≈ 1.240 Then, since I know TB has to be greater than itself to upset the equilibrium, I did TB = TAe^(μsβ) TB = (24.336 N)e^[(1.240)(0.559)] TB ≈ 48.673 N Then, to get the mass I divided the TB by g = 9.81 mB = (48.673)/(9.81) ≈ 4.962 kg That is wrong.