Catenary Cable Problem: Timoshenko (2 solvers since last year only)

I posted this question last year on brilliant.org by modifying it just a little bit, however it is just the same problem as from the book you referred but I just added more calculations. I appreciate that you tried this problem.

Here, is my way of solution including the solution of the numerical method:

Consider right half of the section as we need to consider the vertical hanging length at the pulley end. As it is given $w$ is the weight per unit length, so let us assume that the total length of the cable is $L$ and the sagging length of cable be $2s$.

Weight of the half portion of the sagging cable : $W_s=ws$

Weight of the vertical hanging portion of the cable is given by : $W_h=w(L-2s)$

Let $H$ be the horizontal force of the tension in the cable at the bottom end of the cable. Thus, the tension at the just left of pulley is given by

$$T=\sqrt{(W_s)^2+H^2}=\sqrt{w^2s^2+H^2}$$

Furthermore, the vertical hanging length has also this same tension in the rope which is balanced by its weight. Thus,

$$T=w(L-2s)$$

As in the book, it is given the general formulas for the arc length and the vertical displacement of the cable by taking origin as the bottom most point in the cable

$$y(x)=\frac{H}{w}\left[\cosh\left(\frac{wx}{H}\right) - 1\right]$$

$$S(x)=\frac{H}{w}\sinh\left(\frac{wx}{H}\right)$$

Therefore, the length of the cable from bottom end to the pully end will be $$s=S\left(\frac{l}{2}\right)=\frac{H}{w}\sinh\left(\frac{wl}{2H}\right)$$

Equating the forces to attain the equilibrium $$\sqrt{w^2s^2+H^2}=w(L-2s)$$

Substituting $s$ and simplifying a little bit yields $$L=\frac{H}{w}\cosh\left(\frac{wl}{2H}\right)+\frac{2H}{w}\sinh\left(\frac{wl}{2H}\right) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;....(1)$$

We can notice that $l$ is constant; so $L=f(H/w)$ and let $$H/w=z$$

Thus, we for finding minimum $L_{min}$ we need to put $$\frac{\partial L}{\partial z} = 0$$

We have $$\frac{\partial L}{\partial z} = \left(1-\frac{l}{z}\right)\cosh\left(\frac{l}{2z}\right)+\left(2-\frac{l}{2z}\right)\sinh\left(\frac{l}{2z}\right) = 0$$ Let us use Bisection method for solving this equation as it is the easiest to use it in a software or online computing. Here is the link of the online computation done by me with 50 iterations: Bisection method with a=1 and b=2

In the above link I replace $l/z$ with $x$ to form an equation of the form $\phi(x)=0$ on which bisection method is to be applied with $\phi(1)=+ive$ and $\phi(2)=-ive$

Let me attach JPG of the result below:

So, transforming back in original variables we obtain $$\frac{l}{z}=1.7889033$$ or $$\frac{wl}{H}=1.7889033$$ Therefore, we obtained one of the results as asked in my modified problem in the link brilliant.org as $$\frac{H}{wl}=\boxed{0.55900}$$

Thus, we can say that $L$ will be minimum when $\frac{H}{wl}=0.55900$

Modify equation $(1)$ as below

$$\frac{L}{l}=\frac{H}{wl}\cosh\left(\frac{wl}{2H}\right)+\frac{2H}{wl}\sinh\left(\frac{wl}{2H}\right)$$

Hence, $$\frac{L_{min}}{l}=0.559\cosh(1.7889033/2)+2(0.559)\sinh(1.7889033/2) = \boxed{1.9367}$$

And lastly, we know that $$y(l/2)=f=\frac{H}{w}\left[\cosh\left(\frac{wl}{2H}\right)-1\right]$$ $$\frac{f}{l}=\frac{H}{wl}\left[\cosh\left(\frac{wl}{2H}\right)-1\right]=0.559(\cosh(1.7889033/2)-1) = \boxed{0.2389}$$

Therefore, the required answer will be $$a+b+c=0.559+1.9367+0.2389=\color{blue}{2.7346}$$

The general formula for the family of catenary curves between fixed points $(\pm \ell/2, 0)$ is $$y=a \cosh \frac{x}{a} - a \cosh \frac{\ell}{2a} $$ where higher values of the parameter $a$ (which has units of length) correspond to tauter curves.

By some standard results on catenaries, if we denote the gravitational constant by $g$, the mass per unit length of the cable by $\mu$, and half the length of the catenary by $s$, then the tension at the bottom of the catenary is $\mu g a$ and the force that the catenary exerts on the pulley is $F_\text{left} = \mu g \sqrt{s^2 + a^2}$, which must balance the force $F_\text{right} = \mu g (L - 2s)$ exerted by the hanging portion of the cable. (Note that $w$ in the statement of the Brilliant problem, the weight force per unit length, equals $\mu g$.) We can calculate $s$ in terms of the other parameters: $$s = \int_0^{\ell/2} \sqrt{1 + y'^2 }\, dx = \int_0^{\ell/2} \cosh \frac{x}{a} \,dx = a \sinh \frac{\ell}{2a}.$$ This lets us rewrite the force-balancing equation: \begin{align*} F_\text{left} &= F_\text{right} \\ \sqrt{s^2 + a^2} &= L - 2s \\ a \sqrt{1 + \sinh^2 \frac{\ell}{2a}} &= L - 2a \sinh \frac{\ell}{2a} \\ a \cosh \frac{\ell}{2a} + 2a \sinh \frac{\ell}{2a} &= L \tag{*} \end{align*} and the smallest value of $L$ for which an equilibrium exists is thus the minimum of the LHS in the region $a > 0$. I think this has to be done numerically: $\frac{d}{da}$ of the LHS is $$\left( 2 - \frac{\ell}{2a}\right) \sinh \frac{\ell}{2a} + \left( 1 - \frac{\ell}{2a} \right) \cosh \frac{\ell}{2a},$$ and setting this equal to $0$ gives a transcendental equation. I've searched in vain for a way to use this relationship to get a minimum value of $L$ even if we can't find $a$. Nevertheless, numerical solutions give $$\frac{L}{\ell} = 1.9367$$ and $$\frac{a}{\ell} = 0.559002$$ from which the sag-to-span ratio $-y(0)/\ell$ can be given as $0.238924$, which would give $2.735$ as the answer to the Brilliant problem. Brilliant gives the answer as $2.742$, which is within numerical error.

Update: Just a note that it's possible to derive a simple, non-transcendental relationship between the minimum value of $L$ and the minimizing value of $a$, which makes it very unlikely that the minimum value of $L$ has a convenient closed form.

Set $\ell = 1$ for convenience (one may show easily that the minimizing value of $a$ and the minimum value of $L$ scale in direct proportion with $\ell$), and substitute $x = \ell/2a$. Thus, we must minimize $$\frac{2 \sinh x + \cosh x}{2x} = \frac{3 e^x - e^{-x}}{4x}$$ and we know that the minimum $L_\text{min}$ occurs at $x = \xi$ where $$\left. \frac{d}{dx}\right|_{x=\xi} \frac{3 e^x - e^{-x}}{4x} = \frac{(3\xi-3) e^{\xi} + (\xi+1) e^{-\xi}}{4\xi^2} = 0.$$

From this, we have $$3 e^\xi - e^{-\xi} = 3\xi e^\xi + \xi e^{-\xi}$$ and thus $$L_\text{min} = \frac{3 e^\xi - e^{-\xi}}{4\xi} = \frac{3 e^\xi + e^{-\xi}}{4} = \cosh \xi + \frac{1}{2} \sinh \xi$$ But we know from (*) that $$\xi L_\text{min} = \sinh \xi + \frac{1}{2} \cosh \xi$$ By adding or subtracting these equations, we get the system: \begin{align*} (1 + \xi) L_\text{min} &= \frac{3}{2} e^{\xi} \\ (1 - \xi) L_\text{min} &= \frac{1}{2} e^{-\xi} \\ \end{align*} Multiplying these equations gives $$(1 - \xi^2) L_\text{min}^2 = \frac{3}{4}.$$