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# 18.2: The Intrinsic Equation to the Catenary

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

We consider the equilibrium of the portion AP of the chain, A being the lowest point of the chain (Figure XVIII.1). It is in equilibrium under the action of three forces: The horizontal tension $$T_{0}$$ at A; the tension $$T$$ at P, which makes an angle $$\psi$$ with the horizontal; and the weight of the portion AP. If the mass per unit length of the chain is $$\mu$$ and the length of the portion AP is $$s$$, the weight is $$\mu s\text{g}$$. It may be noted than these three forces act through a single point.

Clearly,

$T_{0}=T\cos\psi \label{18.2.1}$

and

$\mu s\text{g}=\ T\sin\psi, \label{18.2.2}$

from which

$(\mu s\text{g})^{2} + T_{0}^{2}=\ T^{2} \label{18.2.3}$

and

$\tan\psi=\frac{\mu\text{g}s}{T_{0}} \label{18.2.4}$

Introduce a constant $$a$$ having the dimensions of length defined by

$a=\frac{T_{0}}{\mu\text{g}}. \label{18.2.5}$

Then Equations $$\ref{18.2.3}$$ and $$\ref{18.2.4}$$ become

$T\ =\ \mu\text{g}\sqrt{s^{2}\ +\ a^{2}} \label{18.2.6}$

and

$s\ =\ a\tan\psi. \label{18.2.7}$

Equation $$\ref{18.2.7}$$ is the intrinsic equation of the catenary.