What is the shape of a hanging chain?

You may think that the graph of $\cosh{x}$ looks like a parabola, but it is slightly flatter. It is called a catenary, which is the shape formed by a hanging chain.

We can find many different “hanging strings” or “hanging chains” in our daily life. When a string has its two ends fixed, the gravity will bend it into a nice convex curve.

A hanging chain (Image Source: Wikipedia)
Hanging electric wires (Image Source: Wikipedia)

So here comes the question: what is the expression of this curve? Is it $y=x^2$ or $y=\sin(x)$ or something else?

It turns out the curve has an expression of $y=a\cosh\left(\frac{x}{a}\right)$, which is a hyperbolic function.

Hyperbolic functions are covered in A Level Further Mathematics Pure Core Student Book 2, Chapter 6 - Hyperbolic functions.

But how can we derive this result?

  1. Consider a catenary $y$, select a curve $AB$ on $y$ where point $A$ is the lowest point of $y$ and point $B$ is an arbitrary point on $y$. Show that $y’=\frac{l}{a}$ where $l$ is the length of $AB$ and $a$ is a constant.

  2. By considering a very short length of the chain $AB$ or otherwise, show that $l=\int_{0}^{x}\sqrt{1+y’^{2}}\mathrm{d}x$.

  3. Show that $y’’=\frac{1}{a}\sqrt{1+y’^{2}}$.

  4. Given that $y(0)=a$, solve the differential equation $y’’=\frac{1}{a}\sqrt{1+y’^{2}}$ and find $y$. (Hint: consider the subsitution $y’=p$)

Assume we have a thin and inextensible chain hangs stationary in the air. Let $A$ be the lowest point of the chain. We select another arbitrary point $B$ and let’s consider the chain $AB$. There are $3$ forces acting on the chain $AB$, a horizontal force $F$ to the left acting on point $A$, a force $T$ acting on point $B$ along its tangent and the gravity of the chain $AB$.

How to split a force into components in two perpendicular directions is covered in A Level Mathematics Student Book 2, Chapter 21 - Forces in context, Section 1 - Resolving forces

Since chain $AB$ is hanging stationary in the air, the resultant force is zero. Hence we have a pair of simultaneous equations.

$$ \begin{cases} F=T\cos(\theta) \\ G=T\sin(\theta) \end{cases} $$

Dividing these two equations gives us

$$ \begin{aligned} \tan(\theta)&=\frac{G}{F} \\ &=\frac{mg}{F} \\ &=\frac{\rho lg}{F} \\ &=\frac{l}{a} \left(\text{where a }=\frac{F}{\rho g}\right) \end{aligned} $$

$m$ is the mass of the chain, $\rho$ is the density of the chain, $l$ is the length of the chain, $g$ is the gravitational acceleration. We can measure the value of $F$, $\rho$ and $g$, so we can determine the value of $a$ and it is a constant.

Extension question
How can we measure the value of $F$?

Since $\tan(\theta)$ is the tangent to the curve, we can write it as


Now we have a differential equation.


But what is the length of the chain $AB$?

Consider a very short length $\mathrm{d}c$ of our curve $c$, as it is very short, we can approximate it to a striaght line.

By the Pythagoras’s Theorem, we have


Divide $\mathrm{d}x^2$ on both side.


Then we square root bot side.


Finally, we integrate both side to get our result.

$$\int_a^b1\ \mathrm{d}c=\int_a^b\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\mathrm{d}x$$

So the length of the curve between point $a$ and $b$ is $|c_{ab}|=\int_a^b\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\mathrm{d}x$.

We have $y’=\frac{l}{a}$ above, and now we know how to calculate the length of the chain.


And we get


Next, we diffierent both sides of the equation to get rid of the integral on the right.

$$ \begin{aligned} y’’=&\frac{1}{a}\left(\int_{0}^{x}\sqrt{1+y’^{2}}\mathrm{d}x\right)’ \\ =&\frac{1}{a}\sqrt{1+y’^{2}} \end{aligned} $$

Calculate the value of $\left(\int_{0}^{x}f(x)\mathrm{d}x\right)'$

What is the value of $\left(\int_{0}^{x}f(x)\mathrm{d}x\right)’$?

Let $g’(x)=f(x)$.

$$ \begin{aligned} \int_{0}^{x}f(x)\mathrm{d}x&=g(x)-g(0) \\ \left(\int_{0}^{x}f(x)\mathrm{d}x\right)’&=g’(x) - 0 \\ &=f(x) \end{aligned} $$

Differential equations are covered in A Level Mathematics Student Book 2, Chapter 13 - Differential equations.


Now we have a nice differential equation with some initial conditions: $y’(0)=0$; we might as well let $y(0)=a$ is a constant to keep our final result neat.

Let $y’=p$, the original equation turns into


We then seperate the variables, which gives us


Next we integrate both sides.


Which gives us


Prove $\int\frac{\mathrm{d}p}{\sqrt{1+p^2}}=\sinh^{-1}(p)+c$

Notice the hyperbolic identity $\cosh^{2}(x)-\sinh^{2}(x)=1$, it is obvious to make a subsitution $p=\sinh(u)$.

So we have


$$ \begin{aligned} \int\frac{\mathrm{d}p}{\sqrt{1+p^2}}&=\int\frac{\cosh(u)\mathrm{d}u}{\sqrt{1+\sinh^{2}(u)}} \\ &=\int\frac{\cosh(u)}{\cosh(u)}\mathrm{d}u \\ &=\int1\ \mathrm{d}u \\ &=u+c \end{aligned} $$

Where we have $u=\sinh^{-1}(p)$, so here we have proved this.


We have the initial condition that when $x=0$, $p=y’(0)=0$.


Which gives us the value of the constant $c_1=0$.

$$\sinh^{-1}(p)=\frac{x}{a}$$ $$p=\sinh\left(\frac{x}{a}\right)$$

Now we subsitute $p=y’(x)$ back into the equation and integrate both sides again.


$$\int1\ \mathrm{d}y=\int\sinh\left(\frac{x}{a}\right)\mathrm{d}x$$

Which gives us $$y=\frac{1}{a}\cosh\left(\frac{x}{a}\right)+c_2$$

Use our initial condition $y(0)=a$


So we get the value of $c_2=0$ as well.

The final expression for the curve of the hanging chain is


We made an assumption for our model: the chain is thin, inextensible with uniform density. What if the chain is extensible? Or nonuniform? How will these conditions change the result?