solved thank you : ) 1. The problem statement, all variables and given/known data A bucket of water of mass 14.2 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.350 m with mass 12.8 kg . The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 10.7 m to the water. You can ignore the weight of the rope. no friction m =14.2 kg r = 1/2 0.350 m M = 12.8 kg Δy = 10.7 m g = 9.8 m/s2 2. Relevant equations t = Iα F = ma I = 1/2 mr2 a = rα 3. The attempt at a solution What is the tension in the rope while the bucket is falling? F = ma the sum of the forces in the y direction for the bucket is (T - mg) so T - mg = ma T = ma + mg a is unknown t = F x r = Iα since the string is pulling on the windlass and its mass is negligible (F = T) so t = T x r = Iα I = 1/2 Mr2 substitute this into the equation and simplify (did this for a previous question without values to find α) T = 1/2 Mrα set equations 1 and 2 equal to each other and solve for a ma + mg = 1/2 Mrα ma = 1/2 Mrα - mg α is unknown a = -rα α = -a/r substitute into equation 3 ma = 1/2 Mr(-a/r) - mg ma = -1/2 Ma - mg ma + 1/2 Ma = - mg a(m + 1/2 M) = - mg a = - (mg)/(m + 1/2 M) a = - (14.2 kg*9.8 m/s2)/(14.2 kg + 1/2*12.8 kg) a = - 6.76 m/s2 substitute back into equation 1 T = ma + mg T = 14.2 kg* -6.76 m/s2 + 14.2 kg*9.8 m/s2 T = 43.17 N This is the correct answer : ) thank you for your help conclusion: reason for mistake, 1. sign mistake in ( T = ma - mg ) fixed to ( T = ma + mg ) 2. a = -rα ; the sign conventions for this equation were wrong. I didn't really understand this equation, so I will study it more so that I don't make this mistake again.