Re: [Phys-l] nifty question (solution)

[brian whatcott <betwys1@sbcglobal.net>]

 Some thoughts, for what they're worth:
a catenary lifted at the center to support height provides two similar 
catenaries of half width and half depth of the original. Two part CoGs 
at half the original vertical distance from the support level
each carrying half the original hung weight implies increased potential 
energy.(obviously)
A catenary lifted to support height, where one half  has a higher CoG 
than the other exerts unequal reaction on the center support. An extreme 
division of cable lengths in each half span reduces the CoG height 
nearer the original position, and so is energetically favored, for 
sliding friction low enough at the center support. I suggest the center 
lifted catenary is always energetically unfavorable for low enough 
sliding friction at that support.

Brian W

On 9/24/2010 7:26 AM, curtis osterhoudt wrote:
> The reason I'd asked "what happens if you pulled up, instead of down?", is that,
> for at least some range of up-pulls, I suspect the situation becomes unstable in
> equilibrium, and the cable will slip to one side or the other given something to
> break the left-right symmetry. I wondered if anyone might venture to show that,
> or disagrees with my guess.
>
>
>   /**************************************
> "The four points of the compass be logic, knowledge, wisdom and the unknown.
> Some do bow in that final direction. Others advance upon it. To bow before the
> one is to lose sight of the three. I may submit to the unknown, but never to the
> unknowable." ~~Roger Zelazny, in "Lord of Light"
> ***************************************/
>
>
>
>
> ________________________________
> From: Carl Mungan<mungan@usna.edu>
> To: phys-l@carnot.physics.buffalo.edu
> Sent: Fri, September 24, 2010 6:00:35 AM
> Subject: Re: [Phys-l] nifty question (solution)
>
> Okay, looks like most people got it, and several people even
> anticipated the second question below. Reference for these questions
> is given in the last line below.
>
> > Consider an ideal (ie. cannot stretch, and can bend freely) cable
> > suspended from two points of equal height (that are closer together
> > than the cable's length). The cable adopts the shape of a catenary
> > (hyperbolic cosine).
> >
> > Now suppose you hang a weight from the center of the cable. Does the
> > cable's center of mass move up, down, or remain at the same height?
> The center of mass goes up. Here's a simple explanation:
>
> Suppose instead of suddenly attaching the hanging mass m to the
> cable, I instead apply a downward force F to the center of the cable
> that slowly increases from 0 up to mg. (This guarantees the process
> is quasistatic, just like what we do in thermo when we compress a
> piston in a gas-filled cylinder.) Clearly I do positive work on the
> cable. Thus the energy of the cable increases. Since the process was
> quasistatic, there is never any KE of the cable. Further, the cable
> is ideal and thus doesn't stretch, so it cannot gain elastic PE. All
> that's left is for the cable to gain gravitational PE, and thus its
> COM must rise.
>
> > From the same readings, here's another (much easier) nifty question.
> Suppose instead of pulling downward on the center of the cable, I
> pull upward, until the center of the cable is at the same height as
> the two ends of the cable. What upward force F am I applying at that
> instant?
>
> Reference: AJP 58:1110 (1990). -Carl