What is the shape of a hanging rope?

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)

Since

\tan(\theta) = \frac{dy}{dx}, the slope of the curve at B,

and

s = \int\limits_{0}^{x}\sqrt{1+(\frac{dy}{dx})^2},

we rewrite (3) as

\frac{dy}{dx} = k \int\limits_{0}^{x}\sqrt{1+(\frac{dy}{dx})^2}\;dx

and so,

\frac{d^2y}{dx^2}=k\cdot \frac{d}{dx}(\int\limits_{0}^{x}\sqrt{1+(\frac{dy}{dx})^2}\;dx)=k\sqrt{1+(\frac{dy}{dx})^2}

i.e.,

\frac{d^2y}{dx^2}=k\sqrt{1+(\frac{dy}{dx})^2}.\quad\quad\quad(4)

To solve (4), let

p = \frac{dy}{dx}.

We have

\frac{dp}{dx} = k\sqrt{1+p^2} \implies \frac{1}{\sqrt{1+p^2}}\frac{dp}{dx}=k.\quad\quad\quad(5)

Integrate (5) with respect to x gives

\log(p+\sqrt{1+p^2}) = kx + C_1\overset{p(0)=y'(0)=0}{\implies} C_1 = 0.

i.e.,

\log(p+\sqrt{1+p^2}) = kx.\quad\quad\quad(6)

Solving (6) for p yields

p = \frac{dy}{dx} =\sinh(kx).\quad\quad\quad(7)

Integrate (7) with respect to x,

y = \frac{1}{k} \cosh(kx) + C_2\overset{y(0)=0,\cosh(0)=1}{\implies}C_2=-\frac{1}{k}.

Hence,

y = \frac{1}{k}\cosh(kx)-\frac{1}{k}.

Essentially, it is the hyperbolic cosine function that describes the shape of a hanging rope.


Exercise-1 Show that \int \frac{1}{\sqrt{1+p^2}} dp = \log(p + \sqrt{1+p^2}).

Exercise-2 Solve \log(p+\sqrt{1+p^2}) = kx for p.

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