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Potential in caternary problem 
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#1
Apr2312, 07:31 AM

P: 5

a flexible but inextensible chain having uniform mass density is suspended between two points (of course not vertically aligned). Find the shape of the equilibrium of the chain.
The chain will settle down to a position of minimal potential energy. Let the suspending points be (a,y(a)) and (b,y(b)) where (without loss of generality) b>a and y(b)>y(a). The potential energy delta p (relative to y(a)) of a portion of chain corresponding to small delta x is gievn by delta p = y(1+y'^2)^(1/2) delta x. I dont understand where this potential comes from 


#2
Apr2312, 10:22 AM

P: 84

I don't recognise it either apart from the square root but its often difficult to work backwards from part of an answer several good derivations can be found at
http://en.wikipedia.org/wiki/Catenary Regards Sam 


#3
Apr2412, 07:53 AM

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P: 9,657

y' is the slope, tan(θ). Squaring, adding 1 and sqrting gives sec(θ).
Multiplying that by dx gives ds, the length of the section of chain (hypotenuse). Taking the density to be 1, that's also its mass. The final factor, y, seems intended to be its height above the reference point, y(a), but that seems a very confusing notation. 


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