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The Catenary

A ﬂexible inextensible string is suspended from two ﬁxed points, and takes up a position of equilibrium under the action of gravity. It is required to ﬁnd the equation of the curve in which it hangs.

Let $w$ be the weight of unit length, and $s$ the length of arc $AP$ measured from the lowest point $A$; then $ws$ is the weight of the portion $AP$. This is balanced by the terminal tensions, $T$ acting in the tangent line at $P$, and $H$ in the horizontal tangent. Resolving horizontally and vertically gives

$$Tcos\phi =H,\phantom{\rule{1em}{0ex}}Tsin\phi =ws,$$ |

in which $\phi $ is the inclination of the tangent at $P$; hence

$$tan\phi =\frac{ws}{H}=\frac{s}{c},$$ |

where $c$ is written for $\frac{H}{w}$, the length whose weight is the constant horizontal tension; therefore

$$\begin{array}{cc}\frac{dy}{dx}=\frac{s}{c},\phantom{\rule{1em}{0ex}}\frac{ds}{dx}=\sqrt{1+\frac{{s}^{2}}{{c}^{2}}},\phantom{\rule{1em}{0ex}}\frac{dx}{c}=\frac{ds}{\sqrt{{s}^{2}+{c}^{2}}},& \\ \frac{x}{c}={sinh}^{-1}\frac{s}{c},\phantom{\rule{1em}{0ex}}sinh\frac{x}{c}=\frac{s}{c}=\frac{dy}{dx},\phantom{\rule{1em}{0ex}}\frac{y}{c}=cosh\frac{x}{c},& \end{array}$$which is the required equation of the catenary, referred to an axis of $x$ drawn at a distance $c$ below $A$.

The following trigonometric method illustrates the use of the gudermanian: The “intrinsic equation,” $s=ctan\phi $, gives $ds=c{sec}^{2}\phi \phantom{\rule{0.3em}{0ex}}d\phi $; hence $dx=dscos\phi =csec\phi \phantom{\rule{0.3em}{0ex}}d\phi $; $dy=dssin\phi =csec\phi tan\phi \phantom{\rule{0.3em}{0ex}}d\phi $; thus $x=c{gd}^{-1}\phi ,y=csec\phi $; whence $\frac{y}{c}=sec\phi =secgd\frac{x}{c}=cosh\frac{x}{c}$; and $\frac{s}{c}=tangd\frac{x}{c}=sinh\frac{x}{c}$.

Numerical Exercise.—A chain whose length is 30 feet is suspended from two points 20 feet apart in the same horizontal; ﬁnd the parameter $c$, and the depth of the lowest point.

The equation $\frac{s}{c}=sinh\frac{x}{c}$ gives $\frac{15}{c}=sinh\frac{10}{c}$, which, by putting $\frac{10}{c}=z$, may be written $1.5z=sinhz$. By examining the intersection of the graphs of $y=sinhz$, $y=1.5z$, it appears that the root of this equation is $z=1.6$, nearly. To ﬁnd a closer approximation to the root, write the equation in the form $f(z)=sinhz-1.5z=0$, then, by the tables,

$$\begin{array}{llll}\hfill f(1.60)& =2.3756-2.4000=-.0244,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill f(1.62)& =2.4276-2.4300=-.0024,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill f(1.64)& =2.4806-2.4600=+.0206;\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$whence, by interpolation, it is found that $f(1.6221)=0$, and $z=1.6221$, $c=\frac{10}{z}=6.1649$. The ordinate of either of the ﬁxed points is given by the equation

$$\frac{y}{c}=cosh\frac{x}{c}=cosh\frac{10}{c}=cosh1.6221=2.6306,$$ |

from tables; hence $y=16.2174$, and
required depth of the vertex $=y-c=10.0525$
feet.^{14}

- Prob. 91.
- In the above numerical problem, ﬁnd the inclination of the terminal tangent to the horizon.
- Prob. 92.
- If a perpendicular $MN$ be drawn from the foot of the ordinate to the tangent at $P$, prove that $MN$ is equal to the constant $c$, and that $NP$ is equal to the arc $AP$. Hence show that the locus of $N$ is the involute of the catenary, and has the property that the length of the tangent, from the point of contact to the axis of $x$, is constant. (This is the characteristic property of the tractory).
- Prob. 93.
- The tension $T$ at any point is equal to the weight of a portion of the string whose length is equal to the ordinate $y$ of that point.
- Prob. 94.
- An arch in the form of an inverted catenary
^{15}is $30$ feet wide and $10$ feet high; show that the length of the arch can be obtained from the equations $coshz-\frac{2}{3}z=i$, $2s=\frac{30}{z}sinhz$.

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