Let’s say we have a situation like this:
We need to find the distance between the 2 poles. Ignore the heading of the image – ‘Amazon Interview Question’ since this image is taken from the internet. We assume that the cable is of uniform density, and that the height of its hanging is the same at both the poles. Let us put this into a coordinate geometry perspective, we choose the origin at exactly the midpoint of the wire.
In general, the shape that a wire/cable/string takes under its own weight when it’s losely hung from 2 points on 2 poles at equal heights is called a Catenary Curve. In case you’re interested to derive the equation of the same, here’s a popular video that illustrates this.
We’d need to make use of Hyperbolic functions for this problem. To simplify this case, and to give you a gist of the basic equation of a catenary curve, which is defined by
The arc length is \(a\sinh \left ( \frac{x}{a} \right )\). We can assume \(a\) to be an unknown real number.
Let’s assume that the vertical height of the midpoint of the cable is \(20m\) instead of \(10m\) from the ground.
We have 2 things to note:
a) The top of the pole is at point \((x,30)\), thus we have our first equation:
b) We know the length of the arc to be \(40m\). So we have
We thus have 2 equations to derive \(x\) & \(a\).
Let’s make use of an hyperbolic identity to solve this : \((\cosh t)^2-(\sinh t)^2 =1\). Thus dividing both the equations by \(a\) and applying this identity we have
\(\left ( \frac{30+a}{a} \right )^2- \left ( \frac{40}{a} \right )^2=1\). Solving this equation we derive \(a=\frac{35}{3}\). Plugging this into our equation we have \(x \approx 22.7\) (We may need calculators for this).
Test your concepts:
Can you now solve the original problem where the vertical height of the midpoint of the cable is at a distance of 10m from the ground? Do we really need to use the Catenary equation to solve this problem?
<<In case there is any confusion over any example (solved or not), problem or concept, please feel free to create a thread in the appropriate forum. Our memebrs/experts will try their best to help solve the problem for you>>