# The hanging chain problem (catenary), numerically [closed]

I am supossed to solve the hanging chain in constant homogeneus gravity field: The chain of length $L_0$ is divided into $N$ parts which are homogenous with length $l$ and mass $m$ and connected by ideal joints. I am supposed to study the curve with different $N$ and $l$. It´s always true that $Nl=L_0=const.$ I´ve seen the analytical solution but I am supposed to do it numerically and I am supposed to get a set of inhomogenous equations a then solve them numerically. But I dont know how to get the equations. According to the task I am supposed to get a set of normal equations not differential equations. Can someone help me?

• Can you provide the analytical solution form start to end? You may do out HW in steps: 1. Define the problem, 2. Write appropriate equations. 3. Apply numerical methods. 4. Interpret the results. – Crowley Jan 11 '17 at 18:08
• This is a standard problem which is solved in many places on the internet. eg Googling your question I found this as #4. – sammy gerbil Jan 11 '17 at 18:29
• – PM 2Ring Dec 24 '18 at 19:43

The catenary shape equation is $$y(x) = y_C + a \left( \cosh \left( \frac{x-x_C}{a} \right) -1 \right)$$ where $$(x_C,y_C)$$ is the coordinate of the lowest point and $$a$$ is the so called catenary constant. At the supports the forces on the horizontal direction are $$H$$ and the unit weight of the cable is $$w = \rho g A$$ making the catenary constant $$a = \frac{H}{w}$$

Use the above to find the end-points of each segment. For a segment spanning between $$x_1$$ and $$x_2$$ in the horizontal direction with length $$\ell$$ and angle orientation $$\theta$$ the following is true

\begin{align} x_2 = x_1 + \ell \cos \theta \\ y_2 = y_1 + \ell \sin \theta \end{align}

Now for the balance of forces. The weight of each segment is $$W = w \ell$$ and the force equations

\begin{align} (T + \Delta T) \cos \left( \theta + \Delta \theta\right) - T \cos \left( \theta \right) & = 0 \\ (T + \Delta T) \sin \left( \theta + \Delta \theta\right) - T \sin \left( \theta \right) - w \ell & = 0 \end{align}

where $$T$$ and $$\theta$$ is the tension and angle on the left side of the segment and $$T+\Delta T$$ and $$\theta + \Delta \theta$$ the tension and angle on the right side.

This means that each segment recursively changes the tension and angle by

\begin{align} \Delta T & = \sqrt{T^2+w^2 \ell^2+2 T w \ell \sin \theta}-T \\ \Delta \theta & = \tan^{-1} \left( \frac{\cos \theta}{\sin\theta + \frac{T}{w \ell} } \right) \end{align}

You can check the results against the analytical form of the tension and angle \begin{align} T &= H \cosh\left( \frac{x-x_C}{a} \right) \\ \theta &= \sinh\left( \frac{x-x_C}{a} \right) \end{align}

Just first shot:

You may start with formula for steady state: $$\sum \mathbf F_i=0$$ where $\mathbf F_i$ are forces applied to the body under scope.

In our case there are forces between neighbours and gravity. Let's denote:

• $\mathbf F_{i+}$ the force applied to $i$-th "free" joint from the right neighbour,
• $\mathbf F_{i-}$ the force applied to $i$-th joint from the left neighbour,
• $\mathbf G_{i}$ the gravitational force applied to $i$-th joint.

For every joint equation $\mathbf F_{i+}+\mathbf F_{i-}+\mathbf G_{i}=\mathbf 0$ must be valid.

Decompose the vectors to their parts and solve for all $i$: $$F_{xi+}+F_{xi-}=0\\F_{yi+}+F_{yi-}+G_y=0$$

You may look for the angles of each massless rod and when you know their lengths and angles you can find positions of each joint.