One catenary problem - continued (again!)

Forum|Forum|12 years ago
November 6, 2013
3 replies
25281 views

For some reason when I try to post a reply to Fred here: http://communities.ptc.com/message/225402#225402 it seems to wipe out all the other posts in that thread. Hence I’ve started this continuation.

Fred said:

I agree. It would only be correct if Fx were the tension in the cable, but then some other equations would be wrong (though your equation above still has mismatched units Fred!).

Alan

Other

A

AlanStevens

Author

19-Tanzanite

I should have added that the "this equation" to which Fred is referring is:

Alan

V

ValeryOchkov

24-Ruby IV

Sorry

g*G

And second!

Where ara the sheet and the Animation

A

AlanStevens

Author

19-Tanzanite

Valery Ochkov wrote:
Where ara the sheet and the Animation

See attached.

Alan

1_chainHHb.avi

2_Chain-Load-H-Hb-xmcd.zip

F

Fred_Kohlhepp

23-Emerald I

Not tension, only the horizontal component:

Sorry, my image didn't come thru well. alpha is the angle of the tangent to the chain at the weight,

tangent of aplha is equal to the slope (dy/dx) AND the ratio of vertical tension to horizontal tension. If the weight G is shared equally (only at the midpoint) then the math follows.