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From Wikipedia:

To answer this question: "What is the shape of a chain suspended at both ends?" we can use the variational principle that the shape must minimize the gravitational potential energy.

The gravitational potential energy is highest at the ends. Then why isn't the chain shaped like a V?

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    $\begingroup$ The gravitational potential energy is highest at the ends. Why does this observation suggest that the shape should be a "V"? $\endgroup$
    – sammy gerbil
    Aug 9, 2018 at 20:07
  • $\begingroup$ @sammygerbil: I imagine the reasoning might have been something like: $U = mgh$, and $m$ and $g$ are uniform, so if $U$ is maximized/minimized, then so should be $h$. $\endgroup$
    – user541686
    Aug 10, 2018 at 1:16
  • $\begingroup$ Are the other comments, including the one I added today, available to see somewhere else on the site? $\endgroup$
    – Philip Wood
    Aug 14, 2018 at 19:47
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    $\begingroup$ I notice the bounty was added because of the diagram in PhilipWood's answer, but the edit history of that post suggests that the image was actually added by ace. $\endgroup$
    – user191954
    Aug 15, 2018 at 13:01
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    $\begingroup$ The chain is shaped like a “V” (for a two-link chain)! $\endgroup$
    – Gilbert
    Dec 8, 2020 at 17:21

4 Answers 4

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The V shape makes sense for an ideal chain with all of its mass concentrated at the midpoint (with the rest of the chain being massless).

But now consider a chain with its mass distributed over 3 points, e.g. equal masses at the quarter-way points and the midpoint. The quarter point masses will pull the V out of shape, introducing their own bends into the chain. These new bends must change the position of the midpoint mass, and by symmetry the movement must be vertical. Before we added the new masses the midpoint mass was at the lowest possible position, so adding the quarter point masses must lift the midpoint mass.

This is an easy experiment to perform, using some light thread and a few nuts (the metal ones, not the edible kind :) ).

Now subdivide the chain into 8 sections, add nuts to the midpoints of the new sections, and we'll get new bends. Repeat the process indefinitely, and we get the catenary.

Here are the first few steps of this process, calculated using Python. All of the chains in this diagram have a length of 400 units, the background squares are 10x10 units.

Some discrete catenaries

You can see the original SVG version of this diagram on GitHub.

The diagram below shows the forces acting on each of the masses. It also shows the vector sums for the points on the right side of the chain. The tension forces on each segment of the chain are directed along the chain. They must be equal and opposite, and the forces at each mass must sum to zero. By symmetry, the forces on the left side of the chain must mirror those on the right side. Note that the shape of the chain is unchanged if we scale all the forces by the same amount.

Discrete catenary, with vectors

The SVG version of this diagram is here.

For $i \ge 0$, let $(x_i, y_i)$ be the force vector pointing upwards to the right from mass $i$. Let the weight of each mass be 2 units, so the weight vector at each point is (0, -2).

At mass #$0$ we have $$(-x_0, y_0) + (x_0, y_0) = (0, 2)$$ So $y_0 = 1$

At mass #$1$ we have $$(-x_0, -y_0) + (x_1, y_1) = (0, 2)$$ So $x_1 = x_0$ and $y_1 = 2 + y_0 = 3$

For $i \gt 0$, at mass #$i$ we have $$(-x_{i-1}, -y_{i-1}) + (x_i, y_i) = (0, 2)$$ So $x_i = x_{i-1}$ and $y_i = 2 + y_{i-1}$

Thus the $x$ components of each vector $(x_i, y_i)$ on the right side of the chain are identical, and the $y$ components form the arithmetic progression 1, 3, 5, 7...

Let $r_i = \sqrt{x_i^2 + y_i^2}$, and let $s$ be the length of a chain segment.

To calculate the coordinates of each mass first choose a location $P_0$ for mass #$0$. Then $P_{i+1} = P_i + (x_i, y_i)s/r_i$

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Aug 14, 2018 at 19:28
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Sorry to put in my contribution so late in the proceedings, but I don't think anyone has yet presented this simple argument based on forces…

Suppose the chain did hang in a V shape. Consider a small portion P of the chain. It is acted upon by a downward force, W, of gravity and by forces tangential to the chain from neighbouring parts of the chain. Now draw a line AB through P at right angles to the chain at P, and in the plane of the chain as a whole. You'll see immediately that W has an unbalanced component along AB, so P can't be in equilibrium.

Diagram of forces

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Aug 14, 2018 at 19:30
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The V has a higher gravitational potential energy than the catenary does.

To see this, consider pulling down at the center of the catenary very hard. As you pull down, the chain deforms into a V. But you have to do work to pull it down, which means energy is going into the system and its center of mass is moving up. Hence the catenary has a lower center of mass.

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    $\begingroup$ @knzhou sorry, but you seem to be assuming what you want to prove. The fact that "you have to do work to pull it down" is equivalent to the fact that the catenary has less potential energy. You just restated the problem with different words. Or did I misunderstand your argument? $\endgroup$
    – AccidentalFourierTransform
    Aug 9, 2018 at 20:34
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    $\begingroup$ @AccidentalFourierTransform I think this is meant to be an experimental demonstration. That is, it's not assuming the catenary has lower PE, it's saying that you can prove it has lower PE by building one and pulling on it. $\endgroup$
    – David Z
    Aug 9, 2018 at 20:38
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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind ♦
    Aug 10, 2018 at 13:44
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That would require the concavity of the rope to be infinite at the vertex.

But the macroscopic world is smooth and local. Infinities don't come out of nowhere. This should be consistent with both your general experience and your knowledge of physical laws. $F = ma$ doesn't allow $a$ to be infinite unless $F$ is infinite at the same point.

So, for you to have infinite concavity at the vertex, some phenomenon with infinite magnitude needs to be causing that to happen at the vertex.

So the question is, is there anything of infinite magnitude going on at the vertex?
I think you'll agree that the answer is no. Hence, you can't expect infinite concavity there.

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    $\begingroup$ This question is about chains, which can easily be "bent" by 90°. $\endgroup$
    – Todd Sewell
    Aug 10, 2018 at 8:00
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    $\begingroup$ Given that we're considering a static equilibrium, v=0 and a=0. Zero is quite finite. $\endgroup$
    – MSalters
    Aug 10, 2018 at 9:50
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    $\begingroup$ @MSalters: You guys really need to stop trying to shoot down answers just for the sake of it. This was not testimony under penalty of perjury that nobody could come up with a quantity that's infinite in the macroscopic world. You know perfectly well that the OP was clearly not going to cite macroscopic quantum effects as a counterargument for me to try to preempt pedantry on that front. I just wrote "the macroscopic world is smooth" to mean "yeah just humor the asterisk for a sec and ignore exotic quantum stuff, I think you'll agree this'll give you the right intuition here". $\endgroup$
    – user541686
    Aug 10, 2018 at 12:48
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    $\begingroup$ @ctrl-alt-delor: We're talking about ideal chains here, with zero thickness etcetera. You can bend those in a 90 degree angle, no problem at all. Just take a chain of length L, and hang it from one end above a flat frictionless surface, suspended at a height h<L. There will be a length L-h of chain lying on the flat surface, at a 90 degree angle to the suspended part of the chain. $\endgroup$
    – MSalters
    Aug 10, 2018 at 12:49
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    $\begingroup$ Even if one didn't allow infinitely sharp bends, it's clear that the bending in the shape a rope/chain forms is being determined by something other than the limits on bending radius you're describing here. It is neither an exact V, nor a mostly-exact V with a sharp bend in the middle and zero curvature elsewhere. The nonzero curvature in areas that could have been straight is what the OP was asking about. $\endgroup$
    – Sneftel
    Aug 13, 2018 at 11:37

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