Puzzle for you

08162018, 09:45 PM
Post: #1




Puzzle for you
If you’d like, Calculate the distance between the poles 

08162018, 10:19 PM
Post: #2




RE: Puzzle for you
The rope is too short !


08162018, 11:29 PM
Post: #3




RE: Puzzle for you
Distance must be zero as the rope half lengh (3 feet) is the difference between the full height (5 feet) and the minimal height over ground (2 feet). The rope must hang straight down and up (sorry about my english)


08172018, 03:13 AM
Post: #4




RE: Puzzle for you
I assume it's a weightless rope hanging by the force of gravity.


08172018, 03:59 AM
(This post was last modified: 08172018 04:03 AM by Thomas Klemm.)
Post: #5




RE: Puzzle for you
We can model the rope as a catenary:
\(y=a\cosh(\frac{x}{a})\) where \(a\) is a scaling factor. Then with \(h\) as the difference of the heights and \(s\) as half the length of the rope we get: \(h=a(\cosh(\frac{x}{a})1)\) \(s=a\sinh(\frac{x}{a})\) With: \(u=\frac{x}{a}\) and \(v=\frac{h}{s}=\frac{a(\cosh(u)1)}{a\sinh(u)}=\tanh(\frac{u}{2})\) Thus \(u=2\tanh^{1}(v)=2\tanh^{1}(\frac{h}{s})\) This allows us to calculate: \(a=\frac{s}{\sinh(u)}=\frac{s}{\sinh(2\tanh^{1}(\frac{h}{s}))}=\frac{s^2h^2}{2h}\) Now we plug both \(a\) and \(u\) in to calculate: \(\begin{align*} x=a\cdot u &=\frac{s^2h^2}{2h}\cdot2\tanh^{1}(\frac{h}{s}) \\ &=\frac{s^2h^2}{h}\cdot\tanh^{1}(\frac{h}{s}) \\ &=s\cdot(\frac{s}{h}\frac{h}{s})\cdot\tanh^{1}(\frac{h}{s}) \\ &=s\cdot(\frac{1}{v}v)\cdot\tanh^{1}(v) \end{align*}\) Example: \(h=3\) \(s=4\) \(x=4\cdot(\frac{4}{3}\frac{3}{4})\cdot\tanh^{1}(\frac{3}{4})\approx 2.27022850723\) Limit \(h\to s\) For \(v\to1\) the value \(\tanh^{1}(v)\to\infty\) but at the same time \(\frac{1}{v}v\to0\). Thus we can apply L'Hôpital's rule to find: \(\begin{align*} \lim_{v\to1}(\frac{1}{v}v)\cdot\tanh^{1}(v) &= \lim_{v\to1}\frac{1v^2}{v}\cdot\tanh^{1}(v) \\ &= \lim_{v\to1}\frac{\tanh^{1}(v)}{\frac{v}{1v^2}} \\ &= \lim_{v\to1}\frac{\frac{1}{1v^2}}{\frac{1+v^2}{(1v^2)^2}} \\ &= \lim_{v\to1}\frac{1v^2}{1+v^2}=\frac{11}{1+1}=0 \end{align*}\) And thus: \(x=s\cdot0=0\) Which we already knew. Cheers Thomas 

08172018, 12:02 PM
(This post was last modified: 08172018 12:37 PM by Albert Chan.)
Post: #6




RE: Puzzle for you
(08172018 03:59 AM)Thomas Klemm Wrote: We can model the rope as a catenary: Out of curiosity, had the curve a parabola, with same h and x, s ~ 3.409 (less rope) With a fixed h and x, can I assume catenary always curvier than a parabola ? Regarding this puzzle, my original guess were also 0 distance between poles. But, the rope cannot even fit between the poles ... Edit: catenary is indeed curvier (flatter bottom, steeper rise): http://mathforum.org/library/drmath/view/65729.html 

08192018, 06:57 PM
Post: #7




RE: Puzzle for you
(08162018 09:45 PM)Zaphod Wrote: a.) In engineering practice the headroom at least 4.5 meter (approx 15 feet). b.) 2 feet is less than 15 feet. c.) Therefore this is not a real life problem. d.) Engineers do not waste their time to a non reallife fictious problems. e.) I am an engineer. f.) Therefore in my sight this problem is useless and only playing with numbers  I do not want to do anything with it. Maybe if you have data of distance of poles vs headroom below the rope, in this case this will be more interesting. Csaba 

08192018, 07:53 PM
Post: #8




RE: Puzzle for you  
08242018, 08:02 PM
Post: #9




RE: Puzzle for you
(08162018 11:29 PM)HPCollection Wrote: Distance must be zero as the rope half lengh (3 feet) is the difference between the full height (5 feet) and the minimal height over ground (2 feet). The rope must hang straight down and up (sorry about my english) (08172018 03:59 AM)Thomas Klemm Wrote: We can model the rope as a catenary: Both these 

12222018, 09:22 PM
Post: #10




RE: Puzzle for you
(08172018 03:59 AM)Thomas Klemm Wrote: Example: Just stumbled upon this video where the same problem (although scaled by 10) is solved in a slightly different way: Can You Solve Amazon's Hanging Cable Interview Question? Cheers Thomas 

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