A catenary curve is the curve followed by a rope suspended at both ends in uniform gravity.

I thought I would try to solve it myself because it seemed like a good challenge, but almost immediately I got stuck on this question.

Given that nothing is moving, we can say

$$T_{left} + T_{right} + F_g = 0$$

where $T_{left}$ is the tension along the rope to the left and $T_{right}$ is the tension along the rope to the right. Fine and dandy.

Now given that the catenary curve is smooth between (but not including) its endpoints, $T_{left}$ should then always be colinear with $T_{right}$. So then the x-direction components cancel out and the net of the y-direction components cancels with gravity.

But what about at the very lowest point on the catenary curve? Intuitively, this should be the point where the tangent is perfectly orthogonal to gravity, and as such $\hat{T}_{right} = - \hat{T}_{left} = \hat{i}$.

Since both $T_{left}$ and $T_{right}$ are orthogonal to $F_g$...

what holds up the lowest point of a rope following a catenary curve?

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    $\begingroup$ In the limit, how small is this point? $\endgroup$
    – user184990
    Feb 15, 2018 at 19:22
  • 9
    $\begingroup$ What is the mass of this point? $0 \textrm{kg}$. So what is the force required to hold up this point? $0 \textrm{kg} * 9.8 \textrm{ms}^{-2} = 0 \textrm{N}$. $\endgroup$ Feb 15, 2018 at 23:59
  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$
    – Qmechanic
    Feb 16, 2018 at 13:26
  • $\begingroup$ Related: physics.stackexchange.com/q/421957/123208 $\endgroup$
    – PM 2Ring
    Dec 23, 2019 at 7:53

4 Answers 4


That lowest point is not exactly horizontally aligned with the neighboring points. It is slightly below.

If it wasn't, then, as you describe, there would be a net vertical force downwards. If that was the case, then the point would accelerate downwards until that vertical force was balanced. And then you would again reach the configuration I mentioned: the bottom point is not perfectly horizontally aligned with it's neighbors; it is slightly lower.

If you now would argue that we can reduce the size of that point towards being infinitesimal causing it's position to tend towards horizontal alignment, then you are forgetting that the mass also at the same time tends towards being infinitesimal; towards zero, basically. Having a mass of zero means no gravitational force downwards and thus no vertical net force anyways.

Such an ultimately lowest and infinitesimally small point does not need to be held up, because it isn't falling.

  • 6
    $\begingroup$ Nice, I hadn't thought of it that way but I guess as the mass of a given point approaches zero, the amount that it would need to sag below its neighbors also approaches zero. Calculus is so cool. :) $\endgroup$
    – Devsman
    Feb 15, 2018 at 19:57
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    $\begingroup$ Oh, so that's why no matter how hard you stretch the clothesline, it's never perfectly horizontal. $\endgroup$
    – Joker_vD
    Feb 16, 2018 at 12:36

The rope is continuous and of uniform density. The amount of mass any small segment has depends on how long this segment is.

At exactly the center you either have a segment of zero length and hence no mass and weight, or a segment of finite length, but with its ends curved up every so slightly providing just enough vertical force to support the weight.

That is what a catenary is by definition.

  • $\begingroup$ These are my sincere thank you for your recent answer. $\endgroup$
    – Sebastiano
    Feb 7, 2021 at 18:48

To solve such a problem you need to consider not a one point on a cable, but a short part of it and then move to the limit with length going to 0.

Then you can write equations just like you did and get:

$\vec{T}_{left} + \vec T_{right} = -m\vec g$

Let's consider only a symmetrical situation - part of cable at the bottom (still with non-zero length). fix angle $\alpha$ between $\vec{T}_{left}$ and horizontal line - for $\vec{T}_{right}$ we have the same angle.

Considering only vertical components of forces we get $\vec{T}_{left} \mathrm{sin}(\alpha) + \vec{T}_{right} \mathrm{sin}(\alpha) = mg$

Now we can take a limit with $m$ approaching 0. It's clear that $\alpha$ is going to 0 so that equation above holds. And that's what we're expecting. When you take shorter part of cable it gets flatter.

Short note on your approach: Take Newton's equation $m\vec a = \vec F = m\vec g$
You can't simply divide by m if it's equal to 0 and that's exactly your case.


The other answers give the right physical intuition, but I'll give the slightly more precise mathematical version. This question very nicely illustrates the importance of being consistent in the order to which you perform your Taylor expansions.

In this problem, you consider the tension as a vector ${\bf T}$ that varies continuously with the rope coordinate $s$, and think about the mathematical properties of the map ${\bf T}(s):\mathbb{R} \to \mathbb{R}^2$. In order to solve the problem, you need to consider the forces on very short segments of rope. More mathematically precisely, this means you Taylor expand the function ${\bf T}(s)$ to some finite order. If you imaging expanding to zeroth order and only consider the forces on individual points (as you are implicitly doing), then as the other answers point out, the two tensions cancel, gravity is zero, and the force-balance equation becomes trivial. So in order to get a useful equation, you need to expand everything to first order in $s$, and consider small but not "that" small stretches of rope. Once you do this, gravity becomes important, but also the two tensions at the two endpoints of the short stretch of rope are no longer collinear.

So you can legitimately tell two different stories: (a) gravity has no effect on really short segments of rope, and the tension is always collinear, or (b) gravity has an effect on short segments, but the tensions aren't collinear.

The first story is what you get at zeroth order in $ds$, and the second story is what you get at first order. Your problem is you you were (unknowingly) working at zeroth order in ${\bf T}(s)$ (by asserting that the tensions are collinear) but at first order in ${\bf W}(s) = \lambda\, ds\, {\bf g}$ (by asserting that the point at the bottom has weight). There are no paradoxes at either order, as long as you're consistent.

(This problem also illustrates another general physics principle, which is that the correct order to which to Taylor expand is often the lowest order that doesn't just give you "$0 = 0$".)


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