August 2020

Question: What is the shape of a flexible rope hanging from nails at each end and sagging under the gravity?

Answer:

First, observe that no matter how the rope hangs, it will have a lowest point $A$ (see Fig. 1)

Fig. 1

It follows that the hanging rope can be placed in a coordinate system whose origin coincides with the lowest point $A$ and the tangent to the rope at $A$ is horizontal:

Fig. 2

At $A$ , the rope to its left exerts a horizontal force. This force (or tension), denoted by $T_0$ , is a constant:

Fig. 3

Shown in Fig. 3 also is an arbitrary point $B$ with coordinates $(x, y)$ on the rope. The tension at $B$ , denoted by $T_1$ , is along the tangent to the rope curve. $\theta$ is the angle $T_1$ makes with the horizontal.

Since the section of the rope from $A$ to $B$ is stationary, the net force acting on it must be zero. Namely, the sum of the horizontal force, and the sum of the vertical force, must each be zero:

$\begin{cases}T_1cos(\theta)=T_0\quad\quad\quad(1)\\ T_1\sin(\theta) = \rho gs\;\;\quad\quad(2)\end{cases}$

where $\rho$ is the hanging rope’s mass density and $s$ its length from $A$ to $B$ .

Dividing (2) by (1), we have

$\frac{T\sin(\theta)}{T\cos(\theta)} = \tan(\theta) = \frac{\rho g}{T_0}s\overset{k=\frac{\rho g}{T_0}}{\implies} \tan(\theta) = ks.\quad\quad\quad(3)$