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Assume I have a chain of uniform mass, $M_c$, suspended between two points such that it will follow the trajectory of a catenary curve when only supporting its own weight. Let's also say that this cable is actually supporting a suspension bridge of uniform mass, $M_b$, running parallel to the cable along a straight line. In the limit where the mass of the cable is negligible compared to the mass of the suspension bridge, i.e. where $M_c << M_b$, the cable should follow a parabolic trajectory.

However, I'm curious what happens when the weight of the cable and the weight of the bridge are within an order of magnitude or so of one another, i.e. where $M_c$ ~ $M_b$. How would I derive a general equation for the curvature of the cable as a function of $\frac{M_c}{M_b}$?

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Let $y(x)$ be the curve describing the shape of the cable. Let $T(x)$ be the tension in the cable. Consider a small segment extending from $x$ to $x+dx$. The horizontal component of the tension at the two ends of this segment must cancel out, so $T_x$ must be constant. If $\theta$ is the angle the cable makes with the vertical, so that $y'=\tan\theta$, then $T\cos\theta$ is constant. In other words, $$ {T\over \sqrt{1+y'^2}}=\alpha $$ where $\alpha$ is some constant.

Now consider vertical forces. The tension forces on the two ends of our small length element are $$ T_y(x+dx)-T_y(x)=T_y'(x)\,dx=(T\sin\theta)'\,dx=\left(Ty'\over \sqrt{1+y'^2}\right)'dx= \alpha y''\,dx. $$ This force must balance the weight of the cable and of the hanging bridge below it: $$ \alpha y''\,dx=\beta\sqrt{1+y'^2}\,dx+\gamma\, dx. $$ Here $\beta,\gamma$ are the linear mass densities of the cable and bridge respectively.

So the equation we need to solve is $$ y''=b\sqrt{1+y'^2}+c, $$ where $b=\beta/\alpha,c=\gamma/\alpha$.

Let $m(x)=y'(x)$ be the slope. Then this is a first-order separable equation for $m$, with solution $$ x=\int{dm\over b\sqrt{1+m^2}+c}. $$ Mathematica cheerfully does this integral: $$ x=\frac{\frac{c \tan ^{-1}\left(\frac{c m}{\sqrt{m^2+1} \sqrt{b^2-c^2}}\right)}{\sqrt{b^2-c^2}}-\frac{c \tan ^{-1}\left(\frac{b m}{\sqrt{b^2-c^2}}\right)}{\sqrt{b^2-c^2}}+\sinh ^{-1}(m)}{b}. $$ Now "all" you have to do is invert this to get $m(x)$, and integrate $y(x)=\int m(x)dx$. There's no nice closed-form solution for this, unfortunately. But Mathematica does verify that the solution goes to a catenary as $c\to 0$ and to a parabola as $c\to\infty$, so I think it's right.

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