Past the Breaking Point

The ideal cable will form a catenary $y(x) = a\cosh\left(\frac{x}{a}\right), \quad -\frac{D}{2} \leq x \leq \frac{D}{2}$ where $D$ is the distance between the two supports and $a$ is the parameter for the class of catenary curves.

By the arclength formula, we have $l = \int_{-D/2}^{D/2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx = \int_{-D/2}^{D/2} \sqrt{1 + \sinh^2\left(\frac{x}{a}\right)}\,dx = \int_{-D/2}^{D/2} \cosh\left(\frac{x}{a}\right)\,dx = a\sinh\left(\frac{x}{a}\right)\Big|_{-D/2}^{D/2} = 2a\sinh\left(\frac{D}{2a}\right)$

Let $\theta$ be the angle the cable makes with the horizontal where the cable meets a support. Then we have $\tan\theta = y'(D/2) = \sinh\left(\frac{D}{2a}\right)$ On the other hand, the greatest tension occurs at the supports, so we will assume the tension is $T$ at each support. Then the horizontal components of the tension cancel with one another while the sum of their vertical components must equal the magnitude of the cable's weight: $2T\sin\theta = mg$ It follows that $\tan\theta = \frac{mg}{\sqrt{4T^2 - m^2g^2}}$

This suffices to find a formula for $D$ in terms of $l,m,T$ . First we use all three equations to write $\frac{mg}{\sqrt{4T^2 - m^2g^2}} = \tan\theta = \sinh\left(\frac{D}{2a}\right) = \frac{l}{2a}$ so solving for $a$ yields $a = \frac{l\sqrt{4T^2 - m^2g^2}}{2mg}.$

Now solving for $D$ in $\frac{mg}{\sqrt{4T^2 - m^2g^2}} = \tan\theta = \sinh\left(\frac{D}{2a}\right)$ and then plugging in our value for $a$ yields $\begin{aligned} D &= \frac{l\sqrt{4T^2 - m^2g^2}}{mg}\cdot \sinh^{-1}\left(\frac{mg}{\sqrt{4T^2 - m^2g^2}}\right) \\ &= \frac{l\sqrt{4T^2 - m^2g^2}}{mg}\cdot \ln\sqrt{\frac{2T+mg}{2T-mg}} \\ &= \frac{l\sqrt{4T^2 - m^2g^2}}{2mg}\cdot \ln\frac{2T+mg}{2T-mg} \end{aligned}$