The Catenary

Suppose a length  
  of chain is hung from an uneven ceiling. The ends of the chain are fixed to two points some distance apart. What will be the shape of the chain?
Let the chain lie in the vertical  
  plane. An small length  
  of chain of wight per unit length  
  is subject to a vertical for  
\[T sin \theta =wds\]
 ; and a horizontal force  
\[T cos \theta =H\]
We must have  
\[\frac{Tsin \theta}{Tcos \theta}= \frac{wds}{H}= \frac{W}{H} ds= \frac{W}{H} \sqrt{(dx)^2+(dy)^2}\]
. Then  
\[\frac{Pd^2y}{dx^2} = \frac{w}{H} \sqrt{1+ (\frac{dy}{dx})^2}\]
  then we can write the equation as  
\[\frac{dp}{dx}= \frac{w}{H} \sqrt{1+p^2} \rightarrow \int \frac{1}{\sqrt{1+p^2}}dp = \frac{w}{H} \int dx\]


\[sinh^{-1} p = \frac{w}{H}x+c\]
We can set the  
  axis to go through the minimum of the curve, where  
\[\frac{dy}{dx}=sinh (\frac{w}{H} x)\]
. Integration now gives  
\[y=\frac{H}{w} cosh ( \frac{w}{H} x )+k\]
Choosing the  
  at a distance  
  below the lowest point of the chain, the 
  and the equation of the chain is  
\[y= \frac{H}{w} cosh ( \frac{w}{H} x)\]

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