Catenary Cable Problem: Timoshenko (2 solvers since last year only) 
I was doing this amazing problem Chapter 4, Problem 10 from book Engg Mechanics Revised 4E by Timoshenko,  and here is the link having the modified problem which resembles a lot from book.
  Timoshenko modified problem
  Let me attach the image of the question in the above link also for better visual appearance.


Last year I couldn't solve this problem, but today i see this question again. so i really wanted to solve it and wanna know how do we get to the solution! 
I have the final answer but dont know how to get these.
I got some complex term involving $\coth$ function dont remember exactly.. as i solved this one last year. This is an example of catenary cable. So catenary equations are useful, but needs more mathematics than that.

It's been more than a year, only 2 persons were able to solve this

Let me attach the original problem from the book:
which is almost the same as in the link i shared, solve this please. I feel this is one of the most hard problem in the book.
Bounty started : 19/06/2019

 A: 
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
$$a+b+c=0.559+1.9367+0.2389=\color{blue}{2.7346}$$
A: 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 $$(1 - \xi^2) L_\text{min}^2 = \frac{3}{4}.$$
A: Too long for comments.
Starting from Aman Rajput's answer, we need to find the zero of
$$\frac{\partial L}{\partial z} = \left(1-\frac{l}{z}\right)\cosh\left(\frac{l}{2z}\right)+\left(2-\frac{l}{2z}\right)\sinh\left(\frac{l}{2z}\right)$$ Let $z=\frac{l}{2 x}$ and consider
$$f(x)=(2-x) \sinh (x)+(1-2 x) \cosh (x)$$ Using Taylor expansions, we have
$$f(x)=1-\frac{x^2}{2}+O\left(x^3\right)$$ giving as a first guess $x=\sqrt 2$ which is too large; but, for this value $f(x)<0$ and $f''(x)<0$ too; so, by Darboux theorem, using Newton method, we shall converge without any overshoot of the solution.
But we also can notice that $f(1)=-\frac 1e$ and that $f''(1)=-3e$. So $1$ is a better starting point. We could even improve the guess expanding $f(x)$ as a Taylor series around $x=1$ and get
$$f(x)=(\sinh (1)-\cosh (1))+(x-1) (-2 \sinh (1)-\cosh (1))+O\left((x-1)^2\right)$$ giving as a better estimate
$$x_0=1-\frac{2}{3 e^2-1}$$
Below are reproduced the successive iterates
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 0.9055140503 \\
 1 & 0.8945866887 \\
 2 & 0.8944516731 \\
 3 & 0.8944516527
\end{array}
\right)$$
Just for the fun of it, we could have an almost exact solution building the $[1,6]$ Padé approximant of $f(x)$ around $x=1$ and get
$$x=1-\frac{10 \left(1+9 e^2 \left(-14+78 e^2-90 e^4+27 e^6\right)\right)}{1+45 e^2
   \left(17-200 e^2+468 e^4-351 e^6+81 e^8\right)}\approx 0.8944516695$$
A: Let points A and B have $x_A = -l/2$ and B the symmetrical of A to origin. Not caring about a vertical displacement, the wire between A and B has the equation $y(x)=a \cosh (x/a)$. It is the parameter $a$ we seek to find for minimum length of the wire.
Some pre-calculations: 
(1) the length $l_1$ of the wire between A and B comes from the usual calculation $l_1 = 2 a \sinh (l/2a)$;
(2) the equilibrium at point B means the tension in the wire equals the weight of the vertical wire (which has, say, length $l_2$). If the slope of the chain on the left has angle $\tan \alpha = y'(l/2)$, then the vertical component of the tension must balance the weight of half wire: $$T \sin \alpha = \rho g \frac{l_1}{2},$$ where $\rho$ is the mass per unit length of the wire (not needed).
Replacing all the above, one gets: $$l_2 = \frac{1}{2}l_1 \coth\frac{l}{2a}.$$
Now, we seek to minimize $$l_1+l_2 = f(a) = a \left(2 \sinh\frac{l}{2a} + \cosh\frac{l}{2a} \right).$$
From the usual $f'(a)=0$ one gets the nonlinear equation for $x=l/2a$: $$3 e^{2x}=\frac{1+x}{1-x}.$$
An easy way to solve it numerically would be to rewrite it as: 
$$x = \frac{3e^{2x}-1}{3e^{2x}+1} := G(x),$$ and use a fixed point iteration. Namely, start with some value of x (which, from the other posts, should be close to 1, say we start with $x_0 = 0.5$) and do $x_{i+1} = G(x_i)$ until desired convergence. The iteration results, given as pairs $(i, \; x_i)$, read: 
.
So, for minimum length, one has $x = \frac{l}{2a} \approx 0.8944$. 
The sag-span ratio reads $ \frac{f}{l} = \frac{y(l/2)-y(0)}{l} = (\cosh\frac{l}{2a}-1)\frac{a}{l}$, and can be now easily computed ($ \frac{f}{l} \approx 0.4274 \frac{a}{l}$) or see simplified expressions in the other posts, based on the minimum condition $f'(a)=0$.
More sophisticated approaches (not assuming known the shape of the hanging wire), would probably require to compute the vertical position of the center of gravity of the whole wire as a functional of the shape of the left hand-side wire, $y(x)$, and minimize it, under the equilibrium condition and then seeking to minimize the obtained possible wire lengths? That would require the use of calculus of variations with constraints.
A: Given the variables
$$
\begin{array}{rcl}
H & = & \text{Horizontal equilibrium force}\\
T & = & \text{Cable tension at  the pulley}\\
Q & = & \text{Total weight at half the sagging}\\
s_0 & = & \text{Cable length hanging at the pulley right side}\\
s & = & \text{Cable length for half the symmetrical sagging}\\
l & = & \text{Half the distance between the sagging extrema}\\
\alpha & = & \frac{H}{w}\\
s & = & \alpha\sinh\left(\frac{l}{\alpha}\right)\\
L & = & \text{Total cable length}\\
L & = & 2s+s_0 \text{}
\end{array}
$$
the minimization problem can be stated as
$$
\min_{s_0,\alpha}L\ \ \ \text{s. t.}\ \ \ T^2= Q^2+H^2
$$
The lagrangian reads
$$
L(s_0,\alpha,\lambda)=2s+s_0+\lambda w^2\left(s_0^2-\alpha^2-\alpha^2\sinh^2\left(\frac{l}{\alpha}\right)\right)
$$
or calling $\mu = \lambda w^2$
$$
L(s_0,\alpha,\mu)=2s+s_0+\mu\left(s_0^2-\alpha^2\cosh^2\left(\frac{l}{\alpha}\right)\right)
$$
so the stationary points are the solutions for
$$
\nabla L = 0 = \left\{
\begin{array}{l}
 \alpha ^2 \mu  \cosh ^2\left(\frac{l}{\alpha }\right)+l \cosh \left(\frac{l}{\alpha }\right)-\alpha 
   l \mu  \sinh \left(\frac{l}{\alpha }\right) \cosh \left(\frac{l}{\alpha }\right)-\alpha  \sinh
   \left(\frac{l}{\alpha }\right) \\
 2 \mu  s_0+1 \\
 s_0^2-\alpha ^2 \cosh ^2\left(\frac{l}{\alpha }\right) \\
\end{array}
\right.
$$
so substituting into the first equation.
$$
\mu = -\frac{1}{2s_0}\\
s_0 = \alpha\cosh\left(\frac{l}{\alpha}\right)
$$
we obtain
$$
(2 \alpha -l) \sinh \left(\frac{l}{\alpha }\right)+(\alpha -2 l) \cosh \left(\frac{l}{\alpha }\right)=0
$$
Calling now $\beta = \frac{\alpha}{l}$ we obtain finally
$$
(2\beta-1)\sinh \left(\frac{1}{\beta }\right)+(\beta -2) \cosh \left(\frac{1}{\beta }\right)=0
$$
Follows a MATHEMATICA script which uses the Newton algorithm to obtain the solution $\beta^* = 1.1180034124$.
Clear[F, dF, X, X0, X1, F0, dF0]
F = (beta - 2)  Cosh[1/beta] + (2 beta - 1)  Sinh[1/beta]
dF = D[F, beta];
n = 10;
beta0 = 1.;
error = 10^-10;
For[i = 1, i <= n, i++,
 dF0 = dF /. {beta -> beta0};
 F0 = F /. {beta -> beta0};
 beta1 = beta0 - F0/dF0;
 If[Abs[beta1 - beta0] < error, Print["i = ", i, " beta = ",beta1]; Break[]];
 beta0 = beta1
]

