I was trying to find how a uniform string of length $L$ fixed at a point (I assumed $(0,0)$) bends under gravity. I tried to minimise the potential energy within the constraint of the length of the string. The action I got was $$ S = \int^{x_f}_{0}{ \bigg(\mu g y \sqrt{1+ \bigg(\frac{dy}{dx}\bigg)^2} - \lambda \sqrt{1+ \bigg(\frac{dy}{dx}\bigg)^2}\bigg) dx}$$ where $\lambda$ is a Lagrange multiplier and $\mu$ is the linear mass density. The upper limit $x_f$ is not really known. I solved the Beltrami equation to find $y(x)$. I got $$ y(x)= \frac{1}{\mu g} \bigg(c_1 \cosh{\bigg(\frac{c_1x+c_1 c_2}{\mu g}\bigg)}- \lambda \bigg)$$ Where $c_1$ and $c_2$ are arbitrary constants. There must be three constraint relations to find $c_1 , c_2$ and $\lambda$. One of them is that $(0,0)$ is fixed. The other relation should be that the length is always constant. But to apply this one needs to know the coordinate of the tip of the string.

Since we do not know the coordinate of the tip, what other two constraint relations should this rod satisfy?

If there aren't any constraint relations, have I done something wrong or am I missing something very obvious?

  • $\begingroup$ Manvendra, It sounds like you are describing a cantilever supported beam. In other words, one end of the rod is, say, imbedded in a wall and the other end is free, and you are looking to determine how the rod deflects under its own weight. Is this the case? If not, you may need to include a diagram. $\endgroup$ – Bob D Apr 23 '19 at 22:24
  • $\begingroup$ Yes. I have got y(x). The problem I am facing is in determining the arbitrary constants. $\endgroup$ – Manvendra Somvanshi Apr 23 '19 at 22:25
  • $\begingroup$ So, you are trying to solve a problem using lagrangian that is normally solved via mechanics of materials? $\endgroup$ – Bob D Apr 23 '19 at 22:28
  • $\begingroup$ @BobD I was not aware that there already is a solution via mechanics. I just happened to be thinking about this problem. But now that you mention it, yes it seems I am. Any way is the solution correct? $\endgroup$ – Manvendra Somvanshi Apr 23 '19 at 22:34
  • $\begingroup$ I will post the mechanics solution. It doesn't answer your question (I know next to nothing about Lagrangians) but maybe you can make the necessary connections. $\endgroup$ – Bob D Apr 23 '19 at 22:55

If there aren't any constraint relations, have I done something wrong or am I missing something very obvious?

Yes I believe you have done something wrong, and something is obviously incorrect if I understand what your $y(x)$ function means.

I don’t know anything about the Lagrangian but if $y(x)$ is the downward deflection of the rod as a function of $x$, the distance from the fixed end of the rod, then the equation can’t be correct.

Refer to the diagram below showing the deflection of a uniformly loaded cantilever beam. For your case, the uniform load $w$ is the weight per unit length of the beam, or $\frac {mg}{L}$.

The second equation shows the deflection as a function of the distance from the fixed end of the beam (rod, etc.). If this analogous to your $y(x)$ and your $μg$ is analogous to my $mg/L$, then your equation says the deflection is inversely related to the load (weight) of the beam. This would be obviously incorrect.

In addition, there are two variables $E$, the modulus of elasticity (a property of the material,) and $I$, the moment of inertia (a function of the cross section area) that are essential in determining the deflection. So they would be constraints. I don’t see either in your equation, unless they are somehow related to your $C_1$ and $C_2$ or are buried $λ$.

Hope this helps.

enter image description here

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  • $\begingroup$ y(x) is not the deflection of the beam. The beam will be the curve y(x) when plotted. Or in other words in the chosen coordinate system the ordered pair (x,y) for all y=y(x), is a point in the rod. $\endgroup$ – Manvendra Somvanshi Apr 24 '19 at 20:02
  • $\begingroup$ As you and jacob1729 mentioned in his comment, one mistake that I have made is not to consider elastic potential energy. But that does not solve the problem of lack of constraints. $\endgroup$ – Manvendra Somvanshi Apr 24 '19 at 20:04
  • $\begingroup$ @ManvendraSomvanshi You said "I was trying to find how a uniform rod of length 𝐿 fixed at a point (I assumed (0,0)) bends under gravity". To me that is deflection. How can you determine deflection without including in your considerations the modulus of elasticity $E$ and the moment of inertia of the cross section? I guess I really don't understand what it is you are trying to determine. Anyway, good luck! $\endgroup$ – Bob D Apr 24 '19 at 22:27
  • $\begingroup$ Yes, you are right. I cannot find the 'correct' y(x). If I don't include $E$ and $I$ what I get is equation for a string fixed at a point. I must include $E$ for the correct result. But that is not the point of this question. Even if I include $E$ the number of constraint relations will be the same. $\endgroup$ – Manvendra Somvanshi Apr 24 '19 at 22:49
  • $\begingroup$ I should change my question to a string, because it is confusing. $\endgroup$ – Manvendra Somvanshi Apr 24 '19 at 22:55

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