Chapter 8, Problem 17P

A catenary cable is one that is hung between two points not in the same vertical line. As depicted in Fig. P8. 17a, it is subject to no loads other than its own weight. Thus, its weight $\left(\text{N}/\text{m}\right)$ acts as a uniform loads per unit length along the cable. A free-body diagram of a section AB is depicted in Fig. P8.17b where $T_A\text{ and }{T}_B$ 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:

$\frac{d^{2}y}{d{x}^2}=\frac{w}{T_A}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}$

Calculus can be employed to solve this equation for the height y of the cable as a function of distance x,

$y=\frac{T_A}{w}\text{cosh }\left(\frac{w}{T_A}x\right)+y_0-\frac{T_A}{w}$

where the hyperbolic cosine can be computed by

$\text{cosh }x=\frac{1}{2}\left(e^x+e^{-x}\right)$

Use a numerical method to calculate a value for the parameter $T_A$ given values for the parameters $w=10\text{ and }{y}_0=5$, such that the cable has a height of $y=15\text{ at }x=50$.

FIGURE P8.17: (a) Forces acting on a section AB of a flexible hanging cable. The load is uniform along the cable (but not uniform per the horizontal distance x), (b) A free body diagram of section AB.