Hanging Wire (Catenary)

Geometry Level 4

A wire is to be hung between two poles that are separated by a horizontal distance of $10$ m. The two hanging points have the same elevation above the ground. What length of wire will result in the point half way between the two poles to be $1$ m below the hanging points ?

The answer is 10.2619.

2 solutions

Steven Chase
Aug 6, 2019

Here is an outline of my solution:

Catenary equation:

$y = \frac{a}{2} \Big( e^{x/a} + e^{-x/a} \Big)$

$y$ position at $x = 0$ :

$y(0) = \frac{a}{2} (1 + 1) = a$

$y$ position at $x = 5$ :

$y(5) = a + 1 = \frac{a}{2} \Big( e^{5/a} + e^{-5/a} \Big)$

Solving numerically for $a$ results in $a \approx 12.663$ . Then integrating the arc length over the range $(-5,5)$ results in a total length $L \approx 10.262$

There is no need to integrate numerically for the arc-length. If we have $y \; = \; a\cosh(\tfrac{x}{a}) \hspace{2cm} |x| \le b$ then the arclength is $L \; = \; \int_{-b}^b \sqrt{1 + (y')^2}\,dx \; = \; 2a\sinh \tfrac{b}{a}$ With $b=5$ and $y(b)-y(0)=1$ we obtain $a = 12.6632$ , as you say. The arclength is then just $2a\sinh\tfrac{5}{a} = 10.2619$ .

Mark Hennings - 1 year, 10 months ago

A Former Brilliant Member
Mar 5, 2020

Hanging Wire (Catenary)

The answer is 10.2619.

2 solutions

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