"Larry Jaques" wrote in message
...
| On Wed, 10 Sep 2008 18:04:02 -0400, with neither quill nor qualm,
| "Mark Dunning" quickly quoth:
|
|
| wrote in message
| ...
| When applying force perpendicular to a tight wire, the force
| multiplies by a large factor.
|
| I don't grok that, Ivan. A pull on the middle of the rope would be
| equal on all 3 ends, wouldn't it?
|
| No, this is not a simple straight line mechanics problem of pulleys
| acting as levers, it's a vector addition problem. When you pull the
| rope out of line at the center, the component of tension perpendicular
| at each end of the main rope has to be very high to match the small
| component needed to offset the sideways pull, and this approaches
| infinity the tighter the main rope gets.
|
| What's the math/law/formula, Mark?
|

The Resultant Force(Fr) produced along a straight line perpendicular to the
pull force(Fa) would equal to the Force Applied(Fa) X Cotangeant of the
Angle(A) between the straight line and the pulled rope angle. [Fr = Fa(Cot
A)]
The resulting tension(Ft) on the rope would be equal to Fa/Sin A. [Ft
=Fa/SinA]

It is Applied Trigonometry in Mechanics.