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"You can't push a rope" is a wisdom saying that some engineering teachers pass along to their students. Since I'm not an engineer, I can only guess at what they mean, but it sounds to me like code for a mathematical principle (but from control theory? numerical analysis? ergodic theory? dynamical systems? statistics?)

1) I would appreciate having any mathematicians who work with engineers hazard a general framework for a rigorous formulation of what the engineers mean by this slogan.

2) I would like to know of any deep theorems that (of the no-go variety) that naturally fall under this rubric (whether engineers know about these theorems or not).

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    $\begingroup$ Perhaps you can post it as an answer to mathoverflow.net/questions/3559/… and wait for comments ;) $\endgroup$
    – Johan Wästlund
    Commented Jul 10, 2012 at 7:08
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    $\begingroup$ When I was an engineering student, I had a professor who said it, and meant it literally. I remember drawing structure diagrams with arrows representing forces at all the nodes. A solid beam could have compression and tension forces at either end, but ropes or cables could only have tension. $\endgroup$
    – Amit Kumar Gupta
    Commented Jul 10, 2012 at 7:09
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    $\begingroup$ I remember once pushing a rope when it was on the ground... perhaps the saying has a second part "unless it is on the ground". :-) $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 10, 2012 at 8:46
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    $\begingroup$ @Alexander: Веревку невозможно толкать. $\endgroup$
    – Boris Bukh
    Commented Jul 10, 2012 at 12:55
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    $\begingroup$ Actually, it would be interesting to solve it as a control theory problem. Model a rope as a large number of points connected by short incompressible rods, and suppose that initially they are perfectly aligned and horizontal. Then you can "push the rope" by applying a horizontal force which will be propagated through the rods. But this is an unstable equilibrium; is it possible to determine vertical "control" forces that keep the rope in a neighbourhood of the unstable equilibrium point, while pushing the rope horizontally at the same time? $\endgroup$
    – Federico Poloni
    Commented Jul 10, 2012 at 14:07

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The rigorous mathematical context is stability. A straight rope in either tension or compression is a valid solution of the underlying PDE, but in compression this solution is unstable, so it cannot be realized in practice.

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See http://engsci.unavoidable.ca/civ102/CIV102-Notebook.pdf pages 26 and 27 for a mathematical discussion of why you can't push a rope (from a civil engineering course).

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    $\begingroup$ For those like me on somewhat slow connections: that pdf is large (ca 70 Mb). $\endgroup$
    – user9072
    Commented Jul 17, 2012 at 9:47
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It's not as mystical as you seem to think. Simply, for a rope to be an efficacious component of a system that is performing work, the rope must be in tension (pulled on), not pushed on. There is a similar saying among engineers: sh-- doesn't flow uphill.

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  • $\begingroup$ It doesn't matter whether it's doing work or not. The idea is simply that a rope can sustain tension but not compression. $\endgroup$
    – user21349
    Commented Dec 11, 2015 at 20:38
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I've been an engineer, and now I am an attorney. I heard an engineer say this recently, and I thought about from both prospectives. As an attorney, I must disagree because there is an assumption that is left out of the saying, if inserted changes the conclusion:

You can't push with the LENGTH of a rope, only its width; and, You can't pull with the width of a rope, only its LENGTH.

Think about placing a coil of rope between a stuck vehicle and another vehicle, as a cushion between them.

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    $\begingroup$ You are right, but this is not an answer to the question. $\endgroup$
    – Helene Sigloch
    Commented Feb 14, 2016 at 16:11

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