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JohnD66
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Join Date: Jun 2011
Posts: 1

Vertical speed component of a moving cable loop

11/03/2011 5:38 PM

Imagine a hanging loop of cable which would normally be in a catenary curve. If one end of the cable is being pulled in a horizontal direction at x metres per sec, what would be the equation to calculate the vertical component of the speed of the cable at any point along the suspended loop. I guess it would be a function of the tangential angle until the cable was taut, at which point that vertical component would be zero.
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Anonymous Poster #1
#1

Re: Vertical speed component of a moving cable loop

11/03/2011 6:35 PM

browse "cantenary" in wikipedia. it has all the calulations you should need.
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Mikerho
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#2

Re: Vertical speed component of a moving cable loop

11/03/2011 9:46 PM

Sounds suspiciously like a 2nd semester college calculus or a differential equations homework problem.

I'd help nudge you along but I would have to go "back to the books" for review, and that would take too much time out of kicking back and drinking my beer!

Seriously, don't wait for an answer. Tackle it with everything you've got (but take breaks; it helps to get away from it for a bit).

Sounds like you have some deriving to do. Just don't drink while doing so

Good luck

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nick name
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#3

Re: Vertical speed component of a moving cable loop

11/04/2011 9:02 AM

Consider the cable loop length as a constant and write the vertical coordinate as function of the distance between the ends. make the time derivative of it and you will have your answer.
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