What is the Equilibrium Shape of a String Suspended Between Two Points?

Since you didn't want to bother with the actual value of ##c## it is OK to set ##x_0 = 0## and ##\bar c = 1##, for example. But it would be better to use ##x_0 = x_{min}## and ##\bar c = y_{min}##, of course.
  • #1
CAF123
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Homework Statement


An inextensible string of length ##\ell##, whose mass per unit length is ##\rho## is hung between two points at the same height a distance ##d## apart (##d < \ell)##. Using the Euler-Lagrange equations and the first integrals, find the shape of the string when it is in equilibrium.

Relate the constants of integration to ##\ell## and ##d##.

Homework Equations


Need to find a function##F## such that ##I[y] = \int_{x_1}^{x_2} F(y,y',x)\,dx## and ##I[y]## is minimized. For ##F \neq F(x), ## $$y' \frac{\partial F}{\partial y'} - F = \text{const}$$

The Attempt at a Solution


Since l>d, the string will be elongated or distorted between the two ends. Suppose the shape is described by ##y(x)##. In equilibrium, the net force on each segment of the string will vanish. Equivalently, the potential energy of each segment of the string is at a minimum and this is easier to deal with.

Take the zero of potential to be at the highest point of the string between the two supports. Then a small segment d##\ell## of the string will be of length ##\sqrt{1+y'^2}##. So ##dm = \rho d\ell = \rho \sqrt{1+y'^2}##. Since zero of potential was assigned at the highest point, all points on the string are below this, so the potential energy of a segment is then ##U = - \rho \sqrt{1+y'^2}gy##, with the y-axis vertical. Is this correct?

If I then use Euler-Lagrange, since ##U \neq U(x)##, I use the formula in the relevant equations to get that the equilibrium length satisfies ##y/\sqrt{1+y'^2} = \text{const}##, which when integrated gives ##y ## ~ ##\cosh x##, which seems reasonable given the form of this graph.
 

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  • #2
Looks good to me. Cosh is also called Catenary
 
  • #3
My integral is $$\int_{y_o}^{y} \frac{\text{d}y'}{\sqrt{\left(\frac{y'}{C}\right)^2 - 1}} = \int_{x_o}^x \text{d}x',$$ where ##C## is the constant on the RHS of the Euler-Lagrange first integral.

I set ##y_o = x_o = 0## arbritarily in my set up. Carrying the integration through, I obtain ##C(\theta(y) - \theta(y_o)) = x - x_o## using the substitution ##y = C \cosh \theta##. My problem is there does not exist a ##\theta## satisfying ##y_o = 0 = C \cosh \theta##, so I cannot change the integration limits.
 
  • #4
In terms of relating the constants of integration to l and d (the second part of the question), I can see one boundary condition that is y(x = 0, d) = 0. (it was given both ends were at the same height at both ends, and in my set up I chose these points to coincide with the x axis). I only had one integration constant, so this should be enough but I did not get any relation involving l.
 
  • #5
Is there something wrong with setting ##x_o = y_o = 0##? Physically I think no, but if I am to use the substitution I did then maybe I need to change this.
 
  • #6
CAF123 said:
My integral is $$\int_{y_o}^{y} \frac{\text{d}y'}{\sqrt{\left(\frac{y'}{C}\right)^2 - 1}} = \int_{x_o}^x \text{d}x',$$ where ##C## is the constant on the RHS of the Euler-Lagrange first integral.
I set ##y_o = x_o = 0## arbritarily in my set up. Carrying the integration through, I obtain ##C(\theta(y) - \theta(y_o)) = x - x_o## using the substitution ##y = C \cosh \theta##. My problem is there does not exist a ##\theta## satisfying ##y_o = 0 = C \cosh \theta##, so I cannot change the integration limits.
A few questions to get you unstuck again:
Whatever happened to ##\ell## itself ? Can't just disappear !
Why the integral from ##y_0## to ##y## (meaning from 0 to 0 ?) and not from ##y'_0## to ##y'_{x_2}## ?
And (from post 4): "y(x = 0, d) = 0" is really two boundary conditions, rght ?
Then: you introduce ##\theta## with ##y = C \cosh \theta##; what is ##\theta## ?
Previously you had ##y ## ~ ##\cosh x##; would ##y ## ~ ##\cosh (x-d/2) ## be OK too ?

From post 1: ##dm = \rho d\ell = \rho \sqrt{1+y'^2}##, but you mean ## \rho \sqrt{1+y'^2}\, dx##, right ? You can't have a differential on the left and not on the right hand side !
Similarly: ##U = - \rho \sqrt{1+y'^2}gy##, is a value on the left and function on the right. You mean ## U = \int_0^d F dx \ ## with ##F = - \rho \sqrt{1+y'^2}gy\ ##, right ?
 
  • #7
BvU said:
From post 1: ##dm = \rho d\ell = \rho \sqrt{1+y'^2}##, but you mean ## \rho \sqrt{1+y'^2}\, dx##, right ?
Indeed, so a small segment of string of length dl contributes ##dU = -\rho \sqrt{1+y'^2} gy \,dx## to the total potential energy of the system. To obtain the total potential energy of the system, integrate over all x. I.e ##U = \int_0^d -\rho \sqrt{1+y'^2} gy \,dx##. Now, since the integrand is not a function of x, I can still apply the relevant Euler-Lagrange first integral.

A few questions to get you unstuck again:
Whatever happened to ##\ell## itself ? Can't just disappear !
Yes, I think I need to incorporate it into the boundary conditions.
Why the integral from ##y_0## to ##y## (meaning from 0 to 0 ?) and not from ##y'_0## to ##y'_{x_2}## ?
##y_o=0## and ##y## stood for some arbritary y in the range from 0 to the y value corresponding to the lowest point of the string.
And (from post 4): "y(x = 0, d) = 0" is really two boundary conditions, rght ?
Yes.
Then: you introduce ##\theta## with ##y = C \cosh \theta##; what is ##\theta## ?
I did not really think of a physical realization of ##\theta##, I was using the substitution in order to solve the integral.
 
  • #8
Ah, I'm sorry, you are doing the right thing. I got confused by your using ##y'## for ## dy/dx ## on one occasion and for the name of an integration variable on another.

Indeed the "equilibrium length" (i.e. the functional that gives an extremum for ##U##) satisfies the differential equation $${\rho g \, y \over \sqrt{1+y'^2} } = c, $$ which solves to $${dx\over dy} = { \bar c \over\sqrt{y^2-\bar c^2}}$$with ##\bar c ={c\over \rho g}##.

Then you set ##y = \bar c \,\cosh \theta##, hence ## dy = \bar c \, \sinh \theta d\theta ## to get an integral over ##\bar \theta## :
$$x = \bar c \int_{\theta_0}^\theta {\sinh \bar\theta\over \sinh \bar\theta}\, d\bar\theta = \bar c \, \theta + b $$

But now I've painted myself in the same corner as you. There must be something wrong with our y: The minimum surface of revolution has the same differential equation and the same cosh answer, but there is of course a condition y > 0 (see e.g. page 6 of http://math.hunter.cuny.edu/mbenders/cofv.pdf) that we are lacking. We can shift the x such that e.g. b = 0, but I don't know about y.

And I also need ##\ell## back in, probably via the ##\int ds ##...
 
  • #9
Found a solution that indeed makes use of the constraint, but I don't know if it is going to make you happy.

Credit to Frank Porter at CalTech for this link that combines Euler-Lagrange with Lagrange multipliers method for optimization with the ##\int d\bar \ell = \ell## constraint.

##\bar c## and b as integration constants now get company from a third unknown (the Lagrange multiplier ##\lambda##), so that y0=0, y2=0 and ##\ell## can be accommodated comfortably.
 
  • #10
I think I can use the boundary conditions to determine ##\bar{c}## and ##b = -\bar{c}\theta_0##, but the problem is I cannot do this in practice because there is no theta satisfying coshθ=0.

Also, I am yet to find a boundary condition involving l. (as per the question requirements).

Edit: we typed past each other - thanks, but the solution uses Lagrange multipliers and I didn't cover this in my course. I am speaking to my professor on Thursday and I will ask him then unless someone else chips in. Then I will let you know.
 

What is the equilibrium shape of a string?

The equilibrium shape of a string is the shape it takes when all forces acting upon it are balanced and it is not moving. This is also known as the resting shape or natural shape of the string.

What factors affect the equilibrium shape of a string?

The equilibrium shape of a string is affected by factors such as tension, weight, and elasticity of the string. Tension is the force applied to the string, weight is the force of gravity acting on the string, and elasticity is the string's ability to return to its original shape after being stretched.

How is the equilibrium shape of a string calculated?

The equilibrium shape of a string can be calculated using the principle of virtual work, which takes into account the forces acting on the string and the displacement of the string from its resting position. This calculation can be complex and may require advanced mathematical techniques.

Can the equilibrium shape of a string change?

Yes, the equilibrium shape of a string can change if the forces acting upon it change. For example, if the tension or weight of the string is altered, the equilibrium shape will also change. Additionally, if the string is stretched beyond its elastic limit, it may permanently change its equilibrium shape.

What is the practical application of understanding the equilibrium shape of a string?

Understanding the equilibrium shape of a string is important in fields such as engineering and physics, where strings are used in various applications. It can help engineers design structures that can withstand certain forces and physicists can use it to study the behavior of strings in different situations.

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