Catenary supporting a beam problem

[jayman16]

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.

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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?

 

[Valerie Park]

Using the catenary equations, you can solve for the sag of the cable at point C, which is half the length of the cable.The sag of the cable at C is given by:s = csinh(x/c) where x is the distance from A to C.Since we know the tension at A and the weight of the cable per unit length, we can calculate the constant c as follows:c = Ta/ (w*g) where w is the weight of the cable per unit length and g is the acceleration due to gravity.Substituting this value into the equation above, we get:s = (Ta/ (w*g)) * sinh(2.25/ (Ta/ (w*g)) The length of the cable AB is then given by:AB = 2 * s = 2 * (Ta/ (w*g)) * sinh(2.25/ (Ta/ (w*g)) In this case, w = 10kg/m and g = 9.81m/s2, so we get:AB = 2 * (2452.5 / (10*9.81)) * sinh(2.25 / (2452.5 / (10*9.81)) AB = 13.3m