Equilibrium Distance Between Carts with Spring and Chain

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Homework Statement


Two carts can slide along a horizontal rail without friction. The carts are connected:
(a) by an elastic spring of spring constant k and unstretched length l;
(b) by a chain of length l and linear density p.
The spring is going along the rail, the chain hangs in the vertical plane.
[PLAIN]http://img823.imageshack.us/img823/3803/project1f.jpg
Find the equation for the equilibrium distance, d, between the carts.


Homework Equations





The Attempt at a Solution


In the y direction:
F=0
lpg=2TsinA where A is the angle the chain makes with the spring
T=lpg/(2sinA)

In the x direction:
F=0
k(l-d)=lpg/(2tanA)
d=l-lpg/(2ktanA)

If I can find an equation that models the chain, then I can find the angle, but I am having trouble with the equation.
A catenary can be used to model it, which I believe is in the form (but I can be wrong):
f(x)=acosh(x/a), but I don't know how to find a value for of a
 
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You're right about the catenary,
f(x) = A \cosh\frac{x}{A}
Think about this: what do you know about the size of the catenary that might help you determine A?
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.

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