loginatnine
Mar 15, 2012, 2:16 AM
Views: 6849
Registered: Jun 16, 2010
Posts: 4
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ptlong2 wrote: It's easy enough to calculate. If the climber falls a distance x the potential energy is mgx. This has to equal the potential energy of the rope which is 1/2*k*x^2. Solve for x and you have x=2mg/k. So, F=kx=2mg.
Your analysis here is valid for a FF=0 where the rope stretch will be equal to the fall height. That's x in your equation.
In a real fall, it will be something like mg(H+x) = 1/2*k*x^2 where x is the rope stretch and H is the length of the fall before the climber is catch by the rope.
(This post was edited by loginatnine on Mar 15, 2012, 2:18 AM)
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