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.

enter image description here

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

enter image description here

  • $\begingroup$ Can you write the full reference to the book in question (title, etc.)? $\endgroup$
    – Chip
    Jun 18, 2019 at 2:41
  • $\begingroup$ Related/partial solution here: physics.stackexchange.com/questions/481379/… $\endgroup$ Jun 18, 2019 at 2:55
  • $\begingroup$ @EricTowers problem you attached is when either $a$ or $L$ is unknown, here both are unknowns , now try to solve. $\endgroup$ Jun 18, 2019 at 9:40
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    $\begingroup$ @learningstudent thanks for adding the book reference. $\endgroup$
    – Chip
    Jun 19, 2019 at 1:56

5 Answers 5


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


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


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

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


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: enter image description here

enter image description here

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


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 $$a+b+c=0.559+1.9367+0.2389=\color{blue}{2.7346}$$

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    $\begingroup$ wow! amazing solution neat and clean , thanks for your effort :) Didn't think the one who posted there will post a solution here $\endgroup$ Jun 21, 2019 at 7:55

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 $$(1 - \xi^2) L_\text{min}^2 = \frac{3}{4}.$$

  • $\begingroup$ The answer of $L$ is given in the book in terms of $l$. which means you can easily find $L/l$ without finding it numerically. and hence we can find other term $H/wl$ and moreover the answer of $f/l$ is also given in the book directly. So please show your calculation how you find $$L$$ $\endgroup$ Jun 20, 2019 at 5:19
  • $\begingroup$ What solution does the book give? The best I can do is an answer in terms of inverses of transcendental functions. $\endgroup$ Jun 20, 2019 at 13:53
  • $\begingroup$ no issue now, everything is fine, $\endgroup$ Jun 20, 2019 at 14:44
  • $\begingroup$ i checked your answer equations give -ive values for $f/l$ and moreover your differentiation seems not correct. $\endgroup$ Jun 21, 2019 at 7:54
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    $\begingroup$ I said the sag-to-span ratio is $-y(0)/\ell$. Did you include the negative sign? $\endgroup$ Jun 21, 2019 at 15:15

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:

fixed-point iteration convergence.

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.


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 $$x=1-\frac{10 \left(1+9 e^2 \left(-14+78 e^2-90 e^4+27 e^6\right)\right)}{1+45 e^2 \left(17-200 e^2+468 e^4-351 e^6+81 e^8\right)}\approx 0.8944516695$$


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
  • $\begingroup$ Interesting! However: (1) For clarity, the term 'lagrangian' you use can be misleading (this is a mechanics problem, so Lagrangian could easily be understood as the one used in the analytical mechanics) (2) Would you please explain how did you arrive at your expression for the said 'lagrangian'? (and why does the constraint look like you write it). Thank you! (PS: MATHEMATICA has 'FindRoot' or similar to directly compute the numerical solutions). $\endgroup$
    – Chip
    Jun 23, 2019 at 8:31
  • $\begingroup$ @Chip I have before a deleted post in which the development of the necessary formulation was done. It was down-voted alleging that the presented work was already into the text books. So I presented a short version assuming the results. Regarding the 'lagrangian' this is a mainly mathematics driven site, so no ambiguity. Regarding the MATHEMATICA indication of Newtonś method, the purpose is to show transparency as to the numerical methods. Thanks for asking. $\endgroup$
    – Cesareo
    Jun 23, 2019 at 9:37

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