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Catenary: shape of hanging chain
By Tania Jacob on January 6, 1999:

The catenary, y=a cosh(x/a) is the shape of a perfectly flexible chain suspended by its ends and acted on by gravity (not a parabola)...what is so special about its equation that makes it exactly right for our gravity?? I know cosh(x)=(ex+e-x)/2, so does e have some property that relates it to earth's gravity and does that mean that all the equations involving e are somehow 'special' to earth? (it would be terrible if that was true!)
Tania

By Simon Munday on January 10, 1999:

Hello Tania,

The only difference between gravity on the Earth and gravity on other planets is the strength. You know for example that gravity will be much weaker on the moon, but otherwise it is exactly the same. You seem to be concerned that this will alter the shape of the chain, specifically that is will not be a catenary any more.

To get an idea of what is going on and to simplify the problem, consider a perfectly flexable chain suspended in a uniform gravitational field. This approximates the situation on Earth if the strength of the field is set to imitate the strength of gravity, and by changing the value of the strength of the field you can see what will happen on different planets.

You would use this idealized system to set up a differential equation (which I guess you haven't done yet) and solve this equation to get a catenary. Now altering the constant that determines the strength of the field only alters a term of the differential equation by a constant factor, and it turns out that this always leads to y=a×cosh(x/b), just with different values of a and b for different values of the gravitational field constant.

So the catenary is not unique to Earth and there is no relation between the strength of Earth's gravity and the value e. You can be sure that whenever you get a mathematical constant (e.g. e) directly relating to a physical constant (e.g.) g you are doing something wrong.

I hope this helps

Simon & Richard.

By Tania Jacob on January 12, 1999:

Thanks, that does help but I guess it will be just one of those things I will have to think about for a long time. Just another question, I would have thought that the equation defining how a chain hangs due to gravity would be just a slight variation of the equation defining how an object falls due to gravity but it turns out this is not the case right? I think I read somewhere that the equation defining the path of an object due to gravity is parabolic. So is it just that I'm thinking the wrong way about these equations and gravity??

By Richard Dwight on January 13, 1999:

Tania,

The path of a free body under gravity is different to the shape of a hanging chain because there
are extra forces acting on each link of the chain, specifically the tension in the chain. This
makes it into a completely different problem with a different solution. You are right that the path
of an object due to gravity is a parabola.

You should think of gravity simply as a force acting on masses, and not try to consider real-world
intuition at the same time. Treat it mathematically and use physical interpretation when the
maths is done. If you can't do the maths yet then you should learn it before you think about these
problems, it helps a lot.

Note that the problem of finding the shape of a hanging chain is quite a difficult problem, while
the path of a falling body is relatively easy, so you should start there.

Hope this helps,
R

By Tania Jacob on January 15, 1999:

Ok, thanks. I guess I will stop trying to mix maths with physics ahead of what I've learned. Maybe someday it will all make sense! :)