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# What the shape: A dangling rope.
Take a piece of rope or necklace, and lift the two ends up and let the rope dangle naturally by gravity. What is the shape made by this rope? The shape *looks parabolic*, but is it really a parabola? Call this shape the graph of some function $y = f(x)$. We would like to find out what $f(x)$ is.
![[---images/What the shape - dangling chain 2023-05-05 11.39.32.excalidraw.svg]]%%[[---images/What the shape - dangling chain 2023-05-05 11.39.32.excalidraw|🖋 Edit in Excalidraw]], and the [[Excalidraw/--- What the shape - hanging chain 2023-05-05 11.39.32.excalidraw.dark.svg|dark exported image]]%%
If you like to know the answer now, you can jump to the end. But let us see if we can derive this shape. Of course, we will need to make some **assumptions** and some simple physics, that everything is idealized in the following:
(1) The rope has **uniform density** $\rho$.
(2) Gravity force acts **downwards**, and is **proportional to the mass.**
(3) When things are **stationary, all forces cancel out**.
(4) The rope pulling creates a **tention force, tangent along the shape**.
(5) We assume this rope has a **nice** curve, given by some **nice** function $y=f(x)$. For us, it means differentiable as many times as we wish.
That's all the assumptions and physical knowledge we need! We don't even need to know any physical constants. But how should we proceed? What would the ancients do?
**Look at a tiny piece of course! An infinitestimal analysis.**
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Ok, with the shape of the rope being the graph of $y=f(x)$ for some $f(x)$, let us zoom in on a piece of the rope between $x$ and $x+\Delta x$. ![[---images/What the shape dangling rope 2023-05-05 12.15.10.excalidraw.svg]]%%[[---images/What the shape dangling rope 2023-05-05 12.15.10.excalidraw|🖋 Edit in Excalidraw]], and the [[Excalidraw/What the shape hanging rope 2023-05-05 12.15.10.excalidraw.dark.svg|dark exported image]]%%
Since the rope is not moving as it hangs in a static way, all the force on this piece of rope should cancel out to zero. So let us identify all the forces on it.
Since this tiny piece of rope is being pulled by the rope on its left, there is a tention force $\vec T_L$ to its left. Similarly this tiny piece is being pulled to the right, there is a tention force $\vec T_R$ to its right. And since gravity is acting on this piece, there is gravitational force $\vec F_G$ pulling it down. Let us put them all on this picture.
![[---images/What the shape dangling rope 2023-05-05 12.26.34.excalidraw.svg]]%%[[---images/What the shape dangling rope 2023-05-05 12.26.34.excalidraw|🖋 Edit in Excalidraw]], and the [[Excalidraw/What the shape hanging rope 2023-05-05 12.26.34.excalidraw.dark.svg|dark exported image]]%%
But what are these forces? Since gravity acts only downwards, the $x$-component of $\vec F_G$ is zero.
And as we assume that gravitational force is proportional to the mass of this piece of rope, and the rope is of uniform density, this means the gravitational force is just proportional to the **length** $\ell$ of this piece of rope. If this curve is differentiable, then this length is approximately $\ell = \sqrt{(\Delta x)^2 + (\Delta y)^2}$ . So we have $F \approx -k_0 \sqrt{(\Delta x)^2 + (\Delta y)^2}$, for some propotionality constant $k_0 > 0$. That is, $\vec F_G \approx \langle 0, -k_0 \sqrt{(\Delta x)^2 + (\Delta y)^2}\rangle$, where $\Delta y = f(x+\Delta x)-f(x)$.
What else do we know? We know the tension forces are tangent to the shape of the rope, so $\vec T_L$ is parallel to the line with slope $f'(x)$, while $\vec T_R$ is parallel to the line with slope $f'(x+\Delta x)$.
So this means $\vec T_L \| \langle 1, f'(x) \rangle$ and $\vec T_R \| \langle 1, f'(x+\Delta x) \rangle$, in other words $\vec T_L = \langle -k_1 , -k_1 f'(x)\rangle$ and $\vec T_R = \langle k_2, k_2 f'(x+\Delta x)\rangle$, for some constants $k_1, k_2 > 0$.
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Now, the $x$-components of the tension forces sum to zero, we have $k_1 = k_2$. Further, the $y$-components of these three forces sum to zero, so we have $$
-k_1f'(x)+k_1f'(x+\Delta x) - k_0 \sqrt{(\Delta x)^2 + (\Delta y)^2} \approx 0
$$But notice if we divide by $\Delta x$ then we have a difference quotient! Except it is of the second derivative:$$
k_1 \frac{f'(x+\Delta x) - f'(x) }{\Delta x}- k_0 \sqrt{(1 + (\frac{\Delta y}{\Delta x})^2} \approx 0
$$ If we let $\Delta x \to 0$, we get the following differential equation $k_1 y''- k_0 \sqrt{1+(y')^2} = 0$. By combining constants, we have a differential equation fo the form $$
y'' =C\sqrt{1+(y')^2}
$$for some constant $C$.
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Great! But how do we solve this second order differential equation, with derivative inside a square root? By staring at it we see that this is secretly a first order problem, if we let $u=y'$, then $u' = y''$ ! So recasting it in $u$, we get $u' =\frac{du}{dx}= -C \sqrt{1+u^2}$, which is a separable differential equation that we all love.
Separating the variable yields$$
\frac{du}{\sqrt{1+u^2}} = Cdx
$$ To integrate, we can use the hyperbolic identity $1+\sinh^2(t) = \cosh^2(t)$, setting $u = \sinh(t)$, $du = \cosh(t)dt$, to get$$
\int \frac{\cosh(t)dt}{\cosh(t)}=\int Cdx
$$Which gives$$
t=Cx+D
$$with some constant $D$. Now as $u=\sinh(t)$, we have $u=\sinh(Cx+D)$. And finally $y' = u$, so integrating once more gives $$
y(x)=\frac{1}{C}\cosh(Cx+D)+E
$$ In other words, the shape of this dangling rope is a **hyperbolic cosine, and not a parabola ! !** This shape is known as a **catenary curve**. The constants $C,D,E$ are dependent on the physical constants and initial conditions. of this rope, but we didn't need them to find this shape, using mostly **geometric** considerations !
#what-the-shape