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The Hanging Cable

Solution

First question: the center of the cable is 10m above the ground

Using just half of the picture for simplicity, you can easily notice that:

If the cable is 80 m, then half of it is 40 m. But notice 40 m from the top of a 50 m pole is already 10 m above the ground. The cable therefore is hanging directly downward! The cable has to be doubled back upon itself, and the two poles must be coincident and 0 m apart!

This part is actually a trick question: the two poles are a distance of 0 apart. No physics was required to solve this one, just logical thinking!

Second question: the center of the cable is 20m above the ground

The equation for a catenary tangent/touching the ground/x-axis is: y = a cosh(x/a) – a

The arc length can be written as: a sinh(x/a)

The parameter a is unknown, and we want to solve for the value of x of the pole.

Note the top of the pole is (x, 30), so we get one equation:

a cosh(x/a) = 30 + a

We also know that half of the cable length is 40 m, so using the equation for arc length of a catenary, we get another equation:

a sinh(x/a) = 40

We divide both equations by a to get:

cosh(x/a) = (30 + a)/a
sinh(x/a) = 40/a

We then use the hyperbolic identity:

cosh2 t – sinh2 t = 1

Substituting, we get:

(30 + a)2/a2 – (40/a)2 = 1

We can solve this to get:

a = 35/3

Then we use the equation:

(35/3) sinh (x/(35/3)) = 40

x = (35/3) arcsinh(120/35)

x = (35/3) ln(120/35 + √((120/35)2 + 1)) ≈ 22.7

As this is half the hanging chain, the distance between the two poles is then double this value:

2x = (70/3) ln(7) ≈ 45.4 m

Did you get the right answer?