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**standardflop****Member**- Registered: 2005-11-11
- Posts: 2

Hello

A cable is attached at two (nonemoving)fixpoints. Now the cables midpoint is pulled downward, thereby destroying the catenary curve.

Question: How will the cables center of mass (CM) move: op, down or stay fixed?

My thoughts:

I find it intuituitive that the CM will assume the lowest possible position in the catenary (cosh y) curve, but how can i "proove" this? and is it possible to describe the hight of CM as function of the downward midpoint-displacement?

I also wonder if the catenary curve really is "destroyed" - can one not argue, that catenary curve is just altered to fit the new lowered midpoint?

Thank You

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,619

Off the cuff - you would need to solve for the new curve created. There would be a "discontinuity" at the middle.

A "free body" diagram of half the curve would probably solve it all. With forces applied at the end: the new downwards force and a "mirrored" opposite pulling force exerted by the other half.

(That free body curve would also be a catenary, just with special end forces)

I know it can be done, but am not an expert at catenary mathematics.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**standardflop****Member**- Registered: 2005-11-11
- Posts: 2

MathsIsFun: that sounds really interesting...if you get it working id really like to hear about it..

my approach it a bit more primitive, i came to think the following -

suppose the cables center of mass is located in y1. applying a downward force F to the midpoint i equivalent to placing a weight at that point of the chain. the new systems (cable + weight ) center of mass will be located somewhere under y1, if i.e. the weight weighs w [N] and the m_cable=20kg the new center of mass will lie at

y2 = ( y1*m_cable + (y1-d)w/g ) / ( 2 ), where d is the distance between y1 and the weight.

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