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The height of a string in a gravitational field in 2-dimensions is bounded by $h(x_0)=h(x_l)=0$ (nails in the wall) and also $\int_0^l ds= l$. ($h(0)=h(l)=0$, if you take $h$ as a function of arc length) .

What shape does it take?

My try so far: minimise potential energy of the whole string, $$J(x,h, \dot{h})=\int_0^l gh(x) \rho \frac{ds}{l}=\frac{g \rho }{l}\int_0^l h(x) \sqrt{1+\dot{h}^2} dx$$

With the constraint $$\int_0^l \sqrt{1+\dot{h}^2} dx- l=0$$ If it helps, it's evident that $\dot{h}(\frac{l}{2})=0$.

Generally, this kind of equation is a case of a constrained variational problem, meaning that the integrand in $$\int_0^l \frac{g \rho }{l}h(x) \sqrt{1+\dot{h}^2} +\lambda(\int_0^l \sqrt{1+\dot{h}^2} dx- l)dx$$

Must satisfy the Euler Lagrange equation. The constraint must also be satisfied.

But, in truth, by this point I am clueless. $\lambda$ is worked through $\nabla J = \lambda \nabla(\int_0^l \sqrt{1+\dot{h}^2} dx- l)$. I have tried this , but get nonsensical answers.

Is this method the best? If so, in what ways am I going about it wrongly thusfar?

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    $\begingroup$ The shape is a catenary. There must be a thousand derivations out there in Googleland. Just Google for "catenary". $\endgroup$ – John Rennie Jan 17 '13 at 18:35
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    $\begingroup$ It seems like you are on the right track, just apply the Euler Lagrange equations. where are you getting stuck at? Also note that the way you have written the functional, you are considering the string in a region of space where the gravitational field strength is constant (is that what you want?). So when you say 2-dimensions it doesn't really matter, and qualitatively the behavior is the same as in 3-d, and we know what the shape is: a catenary. $\endgroup$ – nervxxx Jan 17 '13 at 18:37
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    $\begingroup$ See also Wikipedia. $\endgroup$ – Qmechanic Jan 17 '13 at 19:52
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    $\begingroup$ You don't really need to know what $\lambda$ is, and also you shouldn't be using the arc-length parameterization (you don't know what it is apriori). but the action you have is still ok as long as the integral runs from $x = 0$ to $x = x_l$, not $0$ to $l$. Applying the EL eqns on your action should still give you the equation of the catenary. Here's how it's done: planetmath.org/encyclopedia/… $\endgroup$ – nervxxx Jan 22 '13 at 18:03
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    $\begingroup$ I think I understand now- I made a mistake having the integral within the integral. $\endgroup$ – Meow Jan 22 '13 at 18:56
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The 2D problem of a cable hanging over a span $S$ with end-point height difference of $h$ is solved with the following shape:

$$ y(x) = y_c + a \left( \cosh \left( \frac{x-x_c}{a} \right) - 1 \right) $$

Where $(x_c,y_c)$ is the lowest point on the catenary and $a$ the catenary constant ($a = \frac{H}{\lambda\,g}$, $H$ horizontal tension, $\lambda$ is mass per unit length and $g$ is gravity)

With the end-points $y(0)=0$ and $y(S)=h$ the solution for the lowest point on the catenary is

$$x_c = \frac{S}{2} + a \sinh^{-1} \left( \frac{h\; {\rm e}^\frac{S}{2 a}}{a\, (1-{\rm e}^\frac{S}{a})} \right) $$

$$ y_c = -a \left( \cosh \left( \frac{x_c}{a} \right) - 1 \right) $$

  • an example solution is shown below: enter image description here

The 3D problem is solved by a change of coordinates in order to make gravity vertical and the two end points in-plane. The apply the 2D solution from above.

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Your approach so far is correct. Now, the first thing to do is to create a coordinate axes. In this case, I will say that the “poles” are at x=-a and x=a, and y=0 is the point where the rope is attached to the “poles”. Now, we can assume constant linear density μ. This is not strictly necessary, but it makes the calculations quite easier. Now, we can set up our constraints in this problem, which would be the length in this case.

$$ J \equiv L = \int_{string} dS = \int_{-a}^{a} \sqrt{1 + (\frac{dy}{dx})^2} dx$$

From now on, in order to maintain brevity, I will write $\frac{dy}{dx}$ as y’. Now, we need to find the quantity that needs to be minimized in this problem. In this case, we want to minimize the potential gravitational energy, $U_g$. So now we need to find the value of our differential, $dU_g$. Because $U_g = mgy$, it is easy to see that $dU_g = (μ dS) \cdot gy$. Putting this into an integral:

$$ U_g = \int_{-a}^{a} μgy \sqrt{1 + (y’)^{2}} dx$$

Now we need to implement our length constraint:

$$ K \equiv U_g + λJ = \int_{-a}^{a} [μgy \sqrt{1 + (y’)^{2}} + λ \sqrt{1 + (y’)^2}] dx$$

Now upon inspection, we can consider the function $$F(x, y, y') \equiv μgy \sqrt{1 + (y’)^{2}} + λ \sqrt{1 + (y’)^2}$$

Now notice that $F$ does not depend explicitly on $x$, so we can use the Beltrami Identity:

$$F - y' \cdot \frac{\partial F}{\partial y'} = C$$

Applying this identity,

$$μgy(1+(y')^2) + λ(1+(y')^2) - μgy(y')^2 - λ(y')^2 = C\sqrt{1+(y')^2}$$ $$μgy + λ = C\sqrt{1+(y')^2}$$ $$(μgy + λ)^2 = C^2 + C^2(y')^2$$ $$y' = \sqrt{\frac{(μgy + λ)^2}{C^2} - 1} \Rightarrow dx = \frac{1}{\sqrt{\frac{(μgy + λ)^2}{C^2} - 1}} dy$$

Now, although this integral might look challenging, it can be made quite easier with a simple substitution:

$$let\ \cosh u = \frac{μgy + λ}{C} \Rightarrow \sinh u\ du = \frac{μg}{C} dy$$ $$x+K_1 = \frac{Cu}{μg} \Rightarrow x+K_1 = \frac{C}{μg} \cosh ^{-1}(\frac{μgy + λ}{C})$$ $$\cosh (\frac{μg}{C} (x+K_1)) = \frac{μgy + λ}{C}$$ $$y = \frac{C}{μg} \cosh (\frac{μg}{C} (x+K_1)) - \frac{λ}{μg}$$

Now in order to solve for some of these constants, we can apply our boundary condtions.

When x = $\pm a$, $y = 0$:

$$0 = \frac{C}{μg} \cosh (\frac{μg}{C} (a+K_1)) - \frac{λ}{μg} = \frac{C}{μg} \cosh (\frac{μg}{C} (-a+K_1)) - \frac{λ}{μg}$$ $$0 = \cosh (\frac{μg}{C} (a+K_1)) = \cosh (\frac{μg}{C} (-a+K_1))$$ $$K_1 = 0$$ $$ \frac{C}{μg} \cosh (\frac{μga}{C}) - \frac{λ}{μg} \Rightarrow λ = C \cosh (\frac{μga}{C})$$ $$\boxed{\therefore\ y = \frac{C}{μg}( \cosh (\frac{μgx}{C}) - \cosh (\frac{μga}{C}))}$$

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