Question

The greatest length of a certain chain which can be suspended from one end without breaking is I. If a catenary be formed with a length 2l/m
of this chain, show that the greatest distance possible between the points of support (supposed in the same horizontal line) will be

\[\frac{2l}{m}(m^2-1)^{1/2}\log\left(\frac{m+1}{m-1}\right)^{1/2}\]

Answer 1

Equation of a catenary:
\[f(x)=a\cosh\frac{x}{a}\quad(a\not=0)\]
The catenary has length \(\dfrac{2l}{m}\). Say we fix the support points at \((x_0,y_0)\) and \((-x_0,y_0)\) (here \(x_0\) is assumed to be positive; since catenaries are even functions, we can guarantee symmetry of these support points). Then,
\[\begin{align*}\frac{2l}{m}&=\int_CdS\\\\
&=2\int_0^{x_0}\sqrt{1+\sinh^2\frac{x}{a}}~dx\\\\
&=2\int_0^{x_0}\sqrt{\cosh^2\frac{x}{a}}~dx\\\\
&=2\int_0^{x_0}\cosh\frac{x}{a}~dx\\\\
&=2\bigg[\sinh\frac{x}{a}\bigg]_0^{x_0}\\\\
&=2\sinh\frac{x_0}{a}\end{align*}\]
The distance between the support points will be \(2x_0\), of course, so I would think that solving the above for \(2x_0\) and maximizing with respect to \(\dfrac{2l}{m}\) should do it. Is there something we can say about \(a\) with the given information? I'm not exactly up to date about the physical applications of catenaries.

Answer 2

Thank you for your answer. I have just reached that point, but I think the difficult step is to find the formula they give in the problem. There is an interesting link here,

http://www.sc.ehu.es/sbweb/fisica/solido/din_rotacion/catenaria/catenaria.htm

where the a parameter is linked with the physical quantities of the problem. No more data is given in the text.

Answer 3

Finally, I found the answer. Just using some formulas for the catenary,
\[y=c\cosh(x/c)\\
s=c\sinh(x/c)\\
y^2-s^2=c^2\] And,
\[l^2-(l/m)^2=c^2\] For the distace,
\[d=2c\log(\sec\theta+\tan\theta)=2c\log\left(\frac{l+l/m}{c}\right)\] So using the relation for c previously found, the relation asked holds.