This came from a physics wave book. I have a static, massive wire on the x-axis, with a y-displacement due to a force per unit length $F_y$. I start with the equation $F_y = T_0\frac{\partial^2(\eta)}{\partial(x)^2}$, where $\eta$ is a small displacement in the y direction and $T_0$, with units of force, is the tension supporting the wire. I need to find how twire is shaped, so I need to integrate both sides twice with respect to x. $$F_y\iint{{dx}^2} = T_0\iint{\frac{\partial^2(\eta)}{\partial(x)^2}{dx^2}}$$

Since these are both indefinite integrals this is rather easy and I get $F_y x^2 = T_o\eta +Cx+D$. But if I were to attempt definite integrals, how exactly would I proceed? If both are definite... $$F_y\int_{x_1}^{x_2}\int_{x_1}^{x_2}{{dx}^2} = T_0\int_{x_1}^{x_2}\int_{x_1}^{x_2}\frac{\partial^2(\eta)}{\partial(x)^2}{dx^2}$$ Then this is how I think the fundamental theorem of calculus would apply $$F_y(x_2-x_1)\int_{x_1}^{x_2}{dx} = T_0\int_{x_1}^{x_2}\left(\frac{\partial(\eta(x_2))}{\partial(x)}-\frac{\partial(\eta(x_1))}{\partial(x)}\right){dx}$$ But this seems very wrong. Having $\eta$ as a function of only the endpoints seems odd. Intuitively I would indefinite integrate this once, then definite integrate the second time to limit it. How do you apply the F.T.C here?

Also I've seen from other questions that the curve is not in fact parabolic as my first equation suggests, what assumption am I making that threw this off?

Any help is appreciated, I feel this hit's on a key concept I missed in calculus.


1 Answer 1


The assumption that the tension $T$ is constant is what's violated here. Because the cable has a weight, that isn't true.

The shape of the cable is not a parabola but a catenary, i.e a shape descrbed in terms of the hyperbolic cosine $\cosh$. The way to solve for the shape of the cable is to find and solve a differential equation that describes the shape of the cable. In order to do that , let $w_l$ be the weight per unit length of the cable and draw a free-body diagram of a short length of the cable at a distance $l$ from the left end, near the lowest point in the cable:

enter image description here

where $T_0$ is the tension at the lowest point of the cable , $T_l$ is the tension at the point a distance $l$ along the cable, and $\theta$ is the angle that $T_l $ makes with the horizontal. Now from the figure we see that

$\cos\theta = \dfrac{T_0}{T_l}$

and also that $T_l\sin\theta = W(l)$, the weight of the cable up to length $l$.

so after a little algebra we get

$T_l = \sqrt{{T_0}^2 +{{{w_l}^2}{l^2}}}$

Now consider a small length $dl$ of the cable. We know $dx =dl\cos\theta$ so

$dx = {\dfrac{{T_0}}{\sqrt{{T_0}^2 +{{{w_l}^2}{l^2}}}}dl}$

Doing the line integral of this along the cable from $0$ to $l$ gives us an expression for $x$ in terms of $l$:

$x(l) = {\dfrac{{T_0}}{{w_l}}}\sinh^{-1}(\dfrac{w_ll}{T_0})$

Using that equation, and the fact that $dy=dx\tan\theta = {\dfrac{W(l)}{T_0}}$ and also integrating, we arrive after a little algebra for an equation for $y(l)$ in terms of $x(l)$:

$y(l) = {\dfrac{{T_0}}{{w_l}}}\cosh(\dfrac{w_lx}{T_0})$

So the shape of the cable isn't a parabola, but a catenary curve which is described as a function of hyperbolic cosine.

  • $\begingroup$ Thank you for the response. I think the dx formula should have a T_0 not a T_l. Also I get confused starting at the dy=dltan(theta) line, that just does not seem to be true and I cannot find out how you got to the final y(l) equation. $\endgroup$ Commented Jun 16, 2014 at 4:56
  • $\begingroup$ Good catch on the typos (dy should be dx tan(theta), not dl), fixed them. $\endgroup$
    – paisanco
    Commented Jun 16, 2014 at 12:04

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