Equilibrium Distance for Spring & Chain Homework

[CornMuffin]

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

here is what I tried, but my final equation for the equalibrium distance d is not right

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
A catenary can be used to model it, which I believe is in the form:
f(x)=acosh(x/a), -d/2<x<d/2 so I have to find the value of a

I'll set the y-axis through the center of the spring, and the x-axis length l above the spring, because I thought that would make the calculations easier.

The arclength is l so,
$$l = \int^{-d/2}_{d/2} \sqrt{1+(f'(x))^2}\,dx$$
$$l = \int^{-d/2}_{d/2} \sqrt{1+(sinh(x/a))^2}\,dx$$
$$l = \int^{-d/2}_{d/2} cosh(x/a)\,dx$$
$$l = asinh(\frac{d}{2a})-asinh(-\frac{d}{2a})$$
$$l = 2asinh(\frac{d}{2a})$$


$$-l=f(d/2)$$
$$-l=acosh(\frac{d}{2a})$$
$$a=-\frac{l}{cosh(\frac{d}{2a})}$$


$$l = 2asinh(\frac{d}{2a})$$
$$\implies l = -\frac{2l}{cosh(\frac{d}{2a})}sinh(\frac{d}{2a})$$
$$1/2 = tanh(\frac{d}{2a})$$
$$\implies \frac{d}{2a} = (1/2)ln(\frac{1+1/2}{1-1/2}) = (1/2)ln3$$
$$a=d/ln3$$


So,
$$f(x)=\frac{d}{ln3} cosh(\frac{xln3}{d})$$
$$f'(x)=sinh(\frac{xln3}{d})$$
$$f'(d/2)=sinh(\frac{ln3}{2})=\frac{\sqrt{3}}{3}$$

so,
$$A=tanh^{-1}(\frac{\sqrt{3}}{3})$$

From the equation on top:
$$d=l-\frac{lpg}{2ktanhA}$$
$$d=l-\frac{lpg}{2ktanh(tanh^{-1}(\frac{\sqrt{3}}{3}))}$$
$$d=l-\frac{lpg\sqrt{3}}{k}$$


but this final result can't be right, when pg/k is large, d should be small, but according to the equation, d is negative

 

[mark726]

, which doesn't make sense. Can anyone help me out?

Thank you for sharing your attempt at solving this problem. It seems like you have made a good start, but there are a few issues with your approach that may be leading to the incorrect final equation for d.

Firstly, when considering the forces in the y-direction, you have assumed that the chain is at an angle A, but this may not necessarily be the case. The chain may have a different angle depending on the equilibrium position of the carts. Therefore, it may be more accurate to say that the force in the y-direction is equal to the weight of the chain (lpd) multiplied by the sine of the angle between the chain and the x-axis, rather than assuming a set angle A.

Secondly, in your calculation for the length of the chain, you have assumed that the chain is a catenary, which is a reasonable approximation for a chain hanging under its own weight. However, in this problem, the chain is also connected to the spring, which will affect the shape of the chain. In this case, it may be more accurate to use a differential equation to model the shape of the chain and then solve for the angle A.

Lastly, in your final equation for d, you have assumed that the chain is at its maximum tension when the carts are at their equilibrium position. However, this may not be the case as the tension in the chain may vary depending on the position of the carts. Again, it may be more accurate to use a differential equation to model the tension in the chain and then solve for the equilibrium position of the carts.

Overall, your approach shows good thinking and problem-solving skills, but it may need some adjustments to account for the specific conditions of this problem. I hope this helps and good luck with your further attempts at solving this problem.