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Shape formed by a string lighter than air
By Brad Rodgers (P1930) on Monday, November 13, 2000 - 08:39 pm:

What shape would a string of length l, that is lighter than air, form, if its ends were tied down and separated by a distance of s. I'm sure it's a parabola, and by the looks of what I'm envisioning it to be, it would be something along the lines of

y=(s-l)x2+ a variable dependent upon s.

I can't seem to be able to find out why though.

What is the equation, and why?

Thanks,

Brad

By Anonymous on Monday, November 13, 2000 - 09:03 pm:

Catenary.
Like the shape a loose chain takes.
Similar to a cosh function

By Dan Goodman (Dfmg2) on Tuesday, November 14, 2000 - 02:26 am:

A hanging chain forms a catenary. I'm not sure exactly what shape something that is lighter than air would take up, you'd have to set up a physical model, however if it turns out that it is just like gravity acting in a different direction then it would be an upside down catenary.

By Brad Rodgers (P1930) on Tuesday, November 14, 2000 - 04:05 am:

Thanks. I was able to find a derivation for a hanging chain. It comes up with the general result of

y=a cosh(x/a)

where a is a constant. I think we might be able to apply this to the above example (assuming that the two ends were at equal altitudes, y=0)

Then,

0=a cosh(s/2a)

And

dl/dx=(1+a sinh(x/a))

This should end up solving for an equation to match the string, although you'll have to change a to -a in the final equation.

I do have a few questions regarding the text I read, where does the idea that the horizontal force is

Tcos(o)

where T is the tension at a point, and o is the angle that the string's tangent would meet with the x axis at the same point come from?

Thanks,

Brad

By Dan Goodman (Dfmg2) on Tuesday, November 14, 2000 - 07:55 pm:

Right, the tension bit is basically mechanics stuff which I don't understand particularly well. However, I'll do my best; in a string, the tension (which is constant along the string =T) acts in the direction of the tangent of the string (I think). So the horizontal force acting on the string can be found by simply resolving the forces in the x and y direction.

You're right, the catenary proof isn't in the notes, but you can set up a calculus of variations problem to do it, you want to minimise potential energy
(ò0 a-mgy dx, where m is the density of the string)
subject to the constraint that y(0)=y(a)=0 and ò0 asqrt(1+(dy/dx)2)dx=L the length of the string. If you can't do this (and it is hard) post again and I'll see if I can help.

By Michael Doré (Md285) on Wednesday, November 15, 2000 - 12:21 am:

The catenary proof is pretty easy to prove using instrinsic co-ordinates, and by resolving forces. It's a bit late so I'll just sketch the proof:

With intrinsic co-ordinates, instead of representing a curve as y = f(x) in Cartesian co-ordinates, we specify the curve by s = g(t), so that an arbitrary point on the curve with intrinsic co-ordinates (s,t) satisfies the following:

1) At this point the curve's tangent makes an angle t with the x-axis.

2) The length of the curve from its lowest point to this point is s.

Now it is clear in this case with a hanging chain that no two points on the curve will have equal gradient (in sign and magnitude). So we can uniquely specify our curve by the formula s = g(t).

Let the tension in the chain at the point (s,t) be T(t).

Now consider the piece of curve from the lowest point of the chain to point with intrinsic co-ordinates (s,t). The mass of this piece is ps (where p = length density of chain). The weight of this section is therefore psg and must be balanced by the vertical component of the tension at the point (s,t) (note there is no vertical tension provided at the lowest point on the curve, as its gradient is zero):

T(t) sin t = psg

We can also resolve horizontally on this piece of curve:

T(0) cos 0 = T(t) cos t

(i.e. the two horizontal tensions must cancel for equilibrium)

So T(t) = T(0)/cos t

So our equation is:

psg = sin t T(0)/cos t

Which is of the form:

s = A tan t

where A is a constant.

This is (I hope) the equation for a catenary in intrinsic co-ordinates.

To get this back into Cartesian co-ordinates you can use the relation:

s = ò[sqrt(1 + (dy/dx)2)dx] = A tan t = A dy/dx

So differentiating and squaring:

1 + (dy/dx)2 = A2 (d2y/dx2)2

which by way of z = dy/dx becomes:

1 + z2 = A2(dz/dx)2

which can be solve by taking the square root of each side. The final answer is z = k sinh x so y = k cosh x for some constant k (we needn't bother with integration constants), which is the catenary.

Sorry if that was poorly explained,

Michael