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#1 2005-11-12 03:35:36
- standardflop
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cable and center of mass
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
#2 2005-11-12 06:58:52
- MathsIsFun
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Re: cable and center of mass
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
#3 2005-11-12 12:57:29
- standardflop
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Re: cable and center of mass
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.