A cable weighing 10kg/m supports a 250kg beam(AC) hinged at the wall(BC). The beam is 4.5m long. Find the length of cable AB. BC is the wall. AC is the beam. Distance AC is 4.5m. Cable runs from A to B and sags forming a catenary. The angle of the cable at A is 30 degrees from the horizontal. |B | | | | | | | | | ______________________________|A C Let Ta = Tension of cable at A Taking moments about C( since beam is in equilibrium moments about any point = 0) Tasin30(4.5) - 250(9.81)(2.25) = 0 Ta = 2452.5 that's all i have done so far. To find the length of cable AB i am supposed to use equations related to the catenary like sinh and cosh functions, s = csinh(x/c) and y = ccosh(x/c) but i do not know how to proceed after finding the tension at A. Can anyone help me?