A catenary cable is one that is hung between two points not in the same vertical line. As depicted in figure 3, it is subject to no loads other than

its own weight. Thus, its weight (N/m) acts as a uniform load per unit length along the cable. A free-body diagram of a section AB is depicted in figure 3, where TĄ and TR are the tension forces at the end. Based on horizontal and vertical force balances, the following differential equation model of the cable can be derived asdx²1+ТА%3D Calculus can be employed to solve this equation for the height y of the cable as a function of distance x, y=\frac{T_{A}}{w} \cosh \left(\frac{w}{T_{A}} x\right)+y_{0}-\frac{T_{A}}{w} \text { where } \cosh x=\frac{1}{2}\left(e^{x}+e^{-x}\right) Determine the value of TA given the values for the parameters w = 12 and yo = 5, such that the cable height has a height of y = 15 at x = 40.