# A uniform cable AB of line density σ hangs freely under gravity with its two ends A, B fixed at points on a horizontal line. The?

A uniform cable AB of line density σ hangs freely under gravity with its two ends A, B fixed at points on a horizontal line. The tension at the lowest point is T₀. Prove that in equilibrium the cable assumes the shape y = ccosh(x/c), where c = T₀/ σ (the x-axis being below AB and || to it and the y-axis passes through the lowest point.

[This problem can be solved by minimising the potential energy using the methods of the calculus of variations. This is NOT the preferred solution, which is one using entirely elementary methods].

### 1 Answer

- NickLv 65 years agoFavourite answer
I did an analysis of this problem about 3 Christmases ago (for fun I guess).

Anyway I will put the link to the pdf I created of my handwritten work.

https://drive.google.com/file/d/0BxfwtLhetXLqZDExY...

The notation isn't exactly the same as yours but you ought to be able to understand it. My x-axis includes the points A and B rather than touching the turning point as yours does, also your sigma is weight per unit length rather than my lambda which is mass per unit length.

Here's the Dropbox link:

https://www.dropbox.com/s/34zu0ovfsdy9ad4/catenary...

Also posted them on Flickr (the first two on the list but in reverse order):

https://www.flickr.com/photos/88769380@N06/?

That was a faff.