1. The problem statement, all variables and given/known data "A derrick boom is guyed by cables AC and AD. A worker is lifting a 20kg block by pulling on a rope that passes through the pulley at A. Knowing that the boom AB exerts a force A that is directed from B to A, determine this force and the force in each of the two cables." http://img213.imageshack.us/img213/4463/2673dsc0036ep9.jpg [Broken] ANSWER: (From back of Textbook) F(AB) = 1742N T(AC) = 1517N T(AD) = 403N 2. Relevant equations F(net) in every direction = 0 (Hence statics) F(AB) = (TensionAB)(Unit Vector Lambda) Lamda = AB(position vector from A to B) ---- |AB|(Magnitude of AB) 3. The attempt at a solution Coordinates: A = 0x + 6y + 0z B = -6x + 0y -3z C = -10.5x + 0y -8z D = -6x + 0y -7z E = +6x +1.5y +0z All forces of all the ropes and the weight force must be equal to the one supporting object: the boom(Tab). I drew a triangle with AE as the hypotenuse. Tan(theta) = (opp/adj) = [(6-1.5)/6] (theta) = arctan(4.5/6) (theta) = 36.87degrees [(4.5^2) + (6^2)]^(1/2) = Hyp Hyp = 7.5 W = (m)(g) = 196N = Tae .: (196N/7.5m) = (Taex/6m) = (Taey/4.5m) Taex = (6m)(196/7.5m) = 156.8N Taey = (4.5m)(196/7.5m) = 117.60N Sum Fx = 0 = Tab(6/9) - Tac(10.5/14.5) - Tad(6/11) - 156.8N Sum Fy = 0 = Tab(6/9) - Tac(6/14.5) - Tad(6/11) -(9.8*20kg) - 117.60N Sum Fz = 0 = Tab(3/9) - Tac(8/14.5) - Tad(7/11) From here I used simple substitution and elimination to solve for the variable Tab, Tac and Tad; but I keep getting numbers that are off by around 150N for each value. I have tried this problem so many times, can somebody check to see if my equations above are correct? I am assuming my problem lies there and not in my simultaneous equation solving skills.