A 5.0 kilogram monkey hangs initially at rest from two vines, A and B. Each vine is 10 meters in length with negligible mass. Vine A is on the left with an angle of 30 degrees from the horizontal from the monkey and B is at 60 degrees from the horizontal of the angle next to the monkey. 1. Determine the tension of B while it is at rest 2. Determine the speed of the monkey. 3. Determine the tension of vine B at the lowest point through the swing. Here is what i got: 1. T=Wcos(theta) =(5)(9.81)(cos 60) =24.525N 2. PE(i)=KE(f) mgh=.5(m)(v^2) (5)(9.81)(10-10sin60)=(.5)(5)(v^2) v=5.127 3. T=mg+F(centripital) =(5)(9.81)+((5*5.127^2)/10m) =62.192 N Can anybody confirm this for me? I have been stuck on this problem for like 2 hours, Im confused if i calculated the tension in part 1 right or if i accidentlly did the tension for vine A instead. This is my worst part of physics, so any help would be appreciated!