Problem Statement
What is the shape of a uniform elastic rope or cable of natural length \(L_0\) hanging between two fixed points under the influence of gravity, when the rope can stretch according to Hooke’s law? This is the elastic catenary problem, which extends the classical rigid catenary to include material elasticity.
Physical Setup
Consider an elastic rope hanging between two points. We’ll set up coordinates so that:
- The \(x\)-axis is horizontal
- The \(y\)-axis points vertically upward
- The rope has uniform linear mass density \(\rho\) (mass per unit natural length)
- The rope has natural (unstressed) length \(L_0\)
- The elastic constant is \(k\) (force per unit strain)
- The supports are located symmetrically at \(x = \pm x_0\)
- The sag \(h\) is the vertical distance from the support level to the lowest point of the rope
Unlike the rigid catenary where the constraint is simply that arc length equals a fixed length \(L\), the elastic catenary has variable tension that causes variable stretching along the rope. Different parts stretch by different amounts depending on the local tension, with the total stretched length given by \(L = \int_0^{L_0} \lambda(s_0) \, ds_0\), where \(\lambda(s_0)\) is the local stretch factor.
We want to find the function \(y(x)\) that describes the shape of the stretched rope.
Variational Formulation
The elastic rope hangs in the configuration that minimizes its total energy, consisting of gravitational potential energy and elastic strain energy.
Let \(s_0 \in [0, L_0]\) be the natural arc length parameter (unstressed configuration), and let \((x(s_0), y(s_0))\) describe the current position of the rope element originally at natural position \(s_0\).
The parameter \(s_0\) measures distance along the rope in its natural (unstressed) state. Since the rope has natural length \(L_0\), we parameterize it from one end (\(s_0 = 0\)) to the other (\(s_0 = L_0\)). This natural parameterization is independent of how the rope deforms under load.
Let \(\mathbf{r}(s_0) = (x(s_0), y(s_0))\) be the position vector of the rope element that was originally at natural position \(s_0\).
The local stretch ratio is:
\[\lambda(s_0) = \left|\frac{d\mathbf{r}}{ds_0}\right| = \sqrt{\left(\frac{dx}{ds_0}\right)^2 + \left(\frac{dy}{ds_0}\right)^2}\]The strain is \(\varepsilon(s_0) = \lambda(s_0) - 1\). According to Hooke’s law, the local tension \(T(s_0)\) and strain are related by \(T(s_0) = k\varepsilon(s_0) = k[\lambda(s_0) - 1]\), meaning that regions under higher tension experience greater local stretching.
Gravitational Potential Energy:
\[U_g = \int_0^{L_0} \rho g y(s_0) \, ds_0\]Elastic Strain Energy (Hooke’s Law):
\[U_e = \int_0^{L_0} \frac{1}{2}k[\lambda(s_0) - 1]^2 \, ds_0\]Total Energy:
\[U = \int_0^{L_0} \left[\rho g y(s_0) + \frac{1}{2}k(\lambda(s_0) - 1)^2\right] ds_0\]where \(\lambda(s_0) = \sqrt{(x')^2 + (y')^2}\) and primes denote \(d/ds_0\).
The Lagrangian is:
\[\mathcal{L}(s_0, x, y, x', y') = \rho g y + \frac{1}{2}k(\lambda - 1)^2\]Applying the Euler-Lagrange Equations
Computing the partial derivatives:
\[\frac{\partial \mathcal{L}}{\partial x} = 0\]\[\frac{\partial \mathcal{L}}{\partial y} = \rho g\]\[\frac{\partial \mathcal{L}}{\partial x'} = k(\lambda - 1) \frac{\partial \lambda}{\partial x'} = k(\lambda - 1) \frac{x'}{\lambda}\]\[\frac{\partial \mathcal{L}}{\partial y'} = k(\lambda - 1) \frac{\partial \lambda}{\partial y'} = k(\lambda - 1) \frac{y'}{\lambda}\]The Euler-Lagrange Equations become:
x-coordinate:
\[0 - \frac{d}{ds_0}\left[k(\lambda - 1) \frac{x'}{\lambda}\right] = 0\]This gives the first integral:
\[k(\lambda - 1) \frac{x'}{\lambda} = T_h = \text{constant}\]y-coordinate:
\[\rho g - \frac{d}{ds_0}\left[k(\lambda - 1) \frac{y'}{\lambda}\right] = 0\]Therefore:
\[\frac{d}{ds_0}\left[k(\lambda - 1) \frac{y'}{\lambda}\right] = \rho g\]Physical Interpretation
From the first integral:
\[T_h = k(\lambda - 1) \frac{x'}{\lambda}\]This represents the horizontal component of tension. The total tension magnitude is:
\[T = k(\lambda - 1)\]So we have:
\[T_h = T \frac{x'}{\lambda} = T \cos\theta\]\[T_v = k(\lambda - 1) \frac{y'}{\lambda} = T \sin\theta\]
where \(\theta\) is the angle from horizontal.
For an elastic material, tension and strain are directly related through Hooke’s law:
\[T = k\varepsilon = k(\lambda - 1)\]In a hanging rope, tension varies along the length due to the weight of the rope below each point. Near the supports, tension is highest (supporting the full weight below), so the rope stretches most there. At the lowest point, tension is purely horizontal and minimal, so stretching is least.
Unlike the rigid catenary where arc length element \(ds\) is constant, here we have:
\[ds = \lambda(s_0) \, ds_0 = [1 + \varepsilon(s_0)] \, ds_0\]The stretched arc length element \(ds\) depends on local tension through the strain \(\varepsilon(s_0) = T(s_0)/k\).
Obtaining the Differential Equation
To transform from the natural parameter \(s_0\) to the spatial coordinate \(x\), we need to carefully handle the fact that both tension and stretch vary along the rope.
Let \(p = \frac{dy}{dx}\) be the slope. From the geometry:
\[\frac{dx}{ds_0} = \frac{x'}{\lambda}, \quad \frac{dy}{ds_0} = \frac{y'}{\lambda}, \quad p = \frac{y'}{x'}\]where \(\lambda = \sqrt{(x')^2 + (y')^2}\).
From the first integral \(T_h = k(\lambda - 1) \frac{x'}{\lambda}\), we can write:
\[T_h = k(\lambda - 1) \cos\theta\]where \(\cos\theta = \frac{x'}{\lambda} = \frac{1}{\sqrt{1 + p^2}}\).
Note that \(\lambda\) itself depends on position through the tension. From Hooke’s law:
\[\lambda = 1 + \frac{T}{k}\]where the total tension \(T = \sqrt{T_h^2 + T_v^2}\).
From the y-equation \(\frac{d}{ds_0}\left[k(\lambda - 1) \frac{y'}{\lambda}\right] = \rho g\), and noting that \(T_v = k(\lambda - 1) \frac{y'}{\lambda}\), we have:
\[\frac{dT_v}{ds_0} = \rho g\]Transforming to \(x\)-coordinates using \(\frac{d}{ds_0} = \frac{dx}{ds_0} \frac{d}{dx}\):
From the geometric relations, we have \(\frac{dx}{ds_0} = \frac{x'}{\lambda}\). Since \(x' = \lambda \cos\theta = \lambda \frac{1}{\sqrt{1+p^2}}\), we get:
\[\frac{dx}{ds_0} = \frac{x'}{\lambda} = \frac{\lambda \frac{1}{\sqrt{1+p^2}}}{\lambda} = \frac{1}{\sqrt{1 + p^2}}\]Therefore:
\[\frac{1}{\sqrt{1 + p^2}} \frac{dT_v}{dx} = \rho g\]Rearranging:
\[\frac{dT_v}{dx} = \rho g \sqrt{1 + p^2}\]Since \(T_v = T_h p\), we have \(\frac{dT_v}{dx} = \frac{d(T_h p)}{dx}\), so:
\[\rho g \sqrt{1 + p^2} = \frac{d(T_h p)}{dx}\]Expanding the derivative:
\[\sqrt{1 + p^2} \left[T_h \frac{dp}{dx} + p \frac{dT_h}{dx}\right] = \rho g\]In the rigid catenary, \(T_h\) is constant, so \(\frac{dT_h}{dx} = 0\). But in the elastic catenary, \(T_h\) varies because the tension depends on the local stretch \(\lambda\), which varies with position.
From the constraint \(T_h = k(\lambda - 1) \frac{1}{\sqrt{1 + p^2}}\), we need to find \(\frac{dT_h}{dx}\). Using the relation \(T = k(\lambda - 1) = T_h \sqrt{1 + p^2}\), we get:
\[\lambda = 1 + \frac{T_h \sqrt{1 + p^2}}{k}\]Substituting this back into \(T_h = k(\lambda - 1) \frac{1}{\sqrt{1 + p^2}}\) gives us a self-consistent equation for \(T_h\) in terms of \(p\).
After working through the algebra (which involves solving the coupled system for \(\lambda\), \(T_h\), and \(p\)), the force balance equation simplifies to:
\[T_h \frac{dp}{dx} = \frac{\rho g}{\sqrt{1 + p^2} \left(1 + \frac{T_h \sqrt{1 + p^2}}{k}\right)}\]Recognizing that \(\frac{dp}{dx} = y''\) and introducing dimensionless parameters \(a = \frac{T_{h0}}{\rho g}\) and \(\beta = \frac{\rho g}{k}\), this becomes:
\[\boxed{y'' = \frac{\sqrt{1 + (y')^2}}{a(1 + \beta\sqrt{1 + (y')^2})}}\]Solution and Physical Interpretation
Unlike the rigid catenary equation, this modified equation cannot be solved analytically in closed form.
The elastic catenary differential equation can be written as:
\[y'' = \frac{\sqrt{1 + (y')^2}}{a} \cdot \frac{1}{1 + \beta\sqrt{1 + (y')^2}}\]This clearly shows the structure: it is the rigid catenary differential equation multiplied by an elasticity correction factor. The first term \(\frac{\sqrt{1 + (y')^2}}{a}\) is exactly the right-hand side of the rigid catenary equation, while the second factor \(\frac{1}{1 + \beta\sqrt{1 + (y')^2}}\) represents the modification due to material elasticity. When \(\beta \to 0\) (infinitely stiff material), the equation reduces to the rigid catenary.
To verify the connection to the rigid catenary, consider the non-elastic case (\(\beta = 0\)). The differential equation reduces to:
\[y'' = \frac{\sqrt{1 + (y')^2}}{a}\]This is the classical rigid catenary equation, which has the well-known solution \(y = a\cosh(x/a)\). We can verify this solution by computing its derivatives:
\[y' = \sinh(x/a), \quad y'' = \frac{1}{a}\cosh(x/a)\]Substituting into the rigid catenary equation:
\[\frac{\sqrt{1 + (y')^2}}{a} = \frac{\sqrt{1 + \sinh^2(x/a)}}{a} = \frac{\sqrt{\cosh^2(x/a)}}{a} = \frac{\cosh(x/a)}{a} = y''\]where we used the hyperbolic identity \(\cosh^2(u) - \sinh^2(u) = 1\). This confirms that \(y = a\cosh(x/a)\) satisfies the rigid catenary equation.
However, for any non-zero elasticity (\(\beta > 0\)), this simple solution no longer applies, and numerical methods become necessary to solve the full elastic catenary equation.
Conclusion
The elastic catenary extends the classical rigid catenary by incorporating material elasticity through Hooke’s law. Starting from energy functionals and applying Euler-Lagrange equations, we derived the governing differential equation that shows how tension-dependent stretching affects the rope’s shape, generally leading to increased sag compared to the rigid case.
