| Posted on Friday, 06 June, 2003 - 05:08 pm: |
How can we show that the curve a hanging chain assumes when held at its edges is of the form:
y(x) = a cosh(x/a)
Marcos
P.S. Ideally, all I'd like are some hints to set me in the right direction
| Posted on Friday, 06 June, 2003 - 05:18 pm: |
Marcos, have you heard of the Euler-Lagrange equation?
| Posted on Friday, 06 June, 2003 - 05:33 pm: |
If not, then don't despair: finding the shape of the catenary can be done using ordinary statics. It's easiest to work first in terms of intrinsic co-ordinates (arc length s and tangent angle y). You should be able to work out a differential equation for these two by resolving forces on an 'infinitesimal' element of chain. Intrinsic co-ordinates are particularly useful here as the mass of the element is related directy to ds. Matthew.
| Posted on Friday, 06 June, 2003 - 06:44 pm: |
Matthew,
No problem...
I don't really feel comfortable with the Euler-Lagrange equation (yet, at least) so, I guess it's intrinsic coords! (Boy, have I been waiting for a chance to use these in a real problem...)
My mechanics skill is pretty close to zero but I get something like,
s = ds[cot(y)cot(dy) - 1]
I got this by considering the component of the weight of the chain up to the element causing the tension on one side, the weight of the actual element and I resolved perpendincular to the other tension acting where the tangential angle is (y + dy)
I somehow doubt this is correct though (I don't want to cheat and work out the differential equation from the equation of a catenary)...
Thanks for the reply,
Marcos
| Posted on Friday, 06 June, 2003 - 08:14 pm: |
Sorry to cut into the discussion Marcos,
But i am simply not able to see how you got that equation of yours.Could you add some steps and clarify that please?
love arun
| Posted on Friday, 06 June, 2003 - 10:27 pm: |
It must be seriously flawed I'm guessing since (I cheated) the differential equation should lead to ds/dy = sec2y...
Here's a diagram of what I did before:
Firstly, let's assume that the curve has uniform density r (i.e. mass =rs). Now, M is due to the weight of the chain (which is rgs) but only the resolved part along the tangent (on the left of the element) is taken into account. So, |M|=rgssin(y) Also, |W=rg(ds) Resolving perpendicularly to T, |M|sin(dy)=|W|cos (y+dy) This leads to the result in bold I posted before... Marcos P.S. I'm pretty sure that the whole thing is wrong but I wanted to post it so people can see where I'm going wrong...
| Posted on Friday, 06 June, 2003 - 10:56 pm: |
I think this was solved already in this forum...
Try searching Asked NRich
Yatir
| Posted on Saturday, 07 June, 2003 - 05:56 am: |
Your working seems to be right. (I haven't given much thought to it though.) However, your final result actually gives, ds/dy = sec(y)
(but this is not the required result as such)
love arun
| Posted on Saturday, 07 June, 2003 - 07:19 am: |
Arun, I'm probably being quite stupid but I don't see how you got that... I get: ssin(y)sin(dy)=ds[cos(y) cos(dy)-sin(y)(dy)] Dividing by sin(y)sin(dy) gives the thing in my second post. Am I meant to assume that sin(dy)=0, cos(dy)=1? If so, the closest thing I get to what you have is ds=ssec(y)...
(Hi and) thanks Yatir, I'll have a look
,Marcos
| Posted on Saturday, 07 June, 2003 - 07:34 am: |
Since dy is very small, sin(dy)=dy (not 0) and cos(dy)=1 Also since dy and ds are very small, the term containing dy* ds can be neglected.
love arun
| Posted on Saturday, 07 June, 2003 - 07:46 am: |
D'oh! (I'm not with it these days) But still, Don't you get ds/dy = stan(y) (I bet I've done another stupid mistake...)
Marcos
| Posted on Saturday, 07 June, 2003 - 11:02 am: |
ds/dy = stan(y) implies the result that Arun quoted, that ds/dy = sec(y). The mistake, though, is further back. The weight of the rest of the chain doesn't act vertically on the point of interest, so we can't resolve it on to M. Instead, it is the resolved component of M (or rather, the equal and opposite force to M) which must balance the weight of the rest of the chain. This means that you are multiplying by sin(y) when you should be dividing by it. I think that making this change allows you to derive the correct equation. However, I've been suggesting a more difficult method than is necessary. I'd imagined that you would get two differential equations relating s, y and the tension in the chain, by resolving at the point of interest in two perpendicular directions. This makes for very hard work. You in fact took a short cut, obtaining the tension as a function of s and y by considering the weight of all of the chain underneath the element (but resolving the wrong way round). This method can be taken further to get the standard method of doing the problem, by considering the total forces on the section of chain from the midpoint, at which it is horizontal (s=0) to the point of interest. For the three forces are the weight of the whole section, the tension at s=0 (which is a constant) and the tension at the point of interest (which is unknown). Resolving perpendicular to the latter gives us the intrinsic equation directly. Matthew.
| Posted on Saturday, 07 June, 2003 - 11:27 am: |
Yeah, thanks Matthew. I get it now. I get the catenary equation too... One last thing: I get ds/dy = s/sin(y)cos(y). I know this is equivalent to ds/dy = sec2y (by working backwards) but how do I derive the latter from the former? [likewise ds/dy = stan(y) becomes ds/dy = sec(y)]
Marcos
| Posted on Saturday, 07 June, 2003 - 06:03 pm: |
Think of them as differential equations, and use separation of variables. ds/dy = stan(y) is easier to solve, but for the other one you just need a standard identity, and a slightly obscure trigonometric integral. Matthew.