Cable System Step Function Analysis

Cyrus
Messages
3,237
Reaction score
17
Cable System - Step Function

Hi, I want to do an experiment using two dissimilar ropes in my basement and see if I can produce some results using a rope analysis. Here’s the set up. I want to tie a light rope to a heavy rope, and be able to characterize the curvature the system takes on. I understand the basic equations involved that I must use to come up with a solution; however, they are based on a uniform weight distribution. Since I am tying a light rope to a very heavy rope, I have to use a step function, and I'm thinking of going the Laplace route to integrate. I would appreciate the help you could provide me with. I don't know if it will work or not, but I figure what the hell why not at least try. The laplace will make it very messy, Unfortunately (I think).
 
Last edited:
Physics news on Phys.org
The first function I know is that: dy/dx = \frac {1}{F_H} \int w(s)ds
Where, w(s), is the weight per unit length.
and I will state that:
w(s) = \varrho_1 for 0 \leq x \leq \alpha
and
w(s) = \varrho_2 for \alpha \leq x \leq \beta

where \alpha is where the two ropes are tied, and \beta is the end of the rope.
 
Last edited:
"I want to tie a light rope to a heavy rope, and be able to characterize the curvature the system takes on."

Why should there be any "curvature" at all? Just lay the ropes in a line on the floor!

If you are talking about wave motion of the two ropes then please say that.
 
No, this is not wave motion. Its a composite catenary system. Putting them on the floor would make no sense.
 
I don't think using laplace transform will do me a bit of good, because I don't have a differential equation I am working with. My integral would just be equivalent to having an expression for the total weight of the rope. I am thinking it would look like this: \int w(s)ds = w_1s + w_2u_{\alpha}(s) + C_1 I am just calling u-sub alpha the step function, or if you guys think that's bad notation, I can use < > and make it a singularity function, that turns on after you pass point beta along the rope. I am saying the two dissimilar ropes are tied together at alpha, and the weight of the rope suddenly 'jumps' much heavier (or lighter), after that point.

So now I have to integrate this mess:

x = \int \frac {ds} { [1+ \frac{1}{F^2_H} ( \int w(s)ds)^2 ]^{ \frac{1}{2}}}

Can someone please help me?

Idealy, what I REALLY want is one smooth function of s that I can write that has a built in step to it and gives me the same results, so the second integration is easier. Is that possible?
 
Last edited:
Catenary

You are (it seems) looking to describe the shape or curve that a flexible rope (cord, chain, telephone wire, whatever) would assume when both ends are suspended and the length of rope between them is acted on by gravity, (viz. catenary, and http://www-groups.dcs.st-and.ac.uk/~history/Curves/Catenary.html).

Good stuff.
 
Yeah, but those links are for a uniform weight distribution, not a step change, like two ropes tied together.
 

Similar threads

Back
Top