- #1

mikeym

- 16

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I'm a climber and I'm trying to understand the physics involved in climbing and falling. In particular I'm trying to understand the factors that are used within the climbing community to describe it and to unite this with my (high school) physics understanding. I also would like to be able to calculate the forces involved in a couple of simple examples.

The first and foremost term used in climbing to determine the severity of a fall is the fall factor which can be seen from this diagam:

www[dot]getoutofglasgow .com/mikey/fallfactor.png

where the

*fall factor = length of fall / length of rope*

It is understood that the fall factor is a number <2 where the closer to 2 it is the more force is applied to the protection as a result of the fall. This happens where the climber falls from above the belayer before putting the rope through a piece of protection.

Now From what I remember of high school physics

The force applied by the rope is

*F = kx*where k is the spring constant of the rope (I'm not sure how to get this) and x is the stretch on the rope.

No this has to absorb the kinetic energy of the falling climber who's KE can be calculated from.

v^2 = v0^2 + 2ad

where v0 = 0 for a slipping climber and a = g or 9.81 m/s^2 for a earthly climber

so v^2 = 2gd

and KE = 1/2 * m v^2

so KE = m g d

Now this is where things get confused for me.

Can I say

*KE = Fd*

where d is the stretch distance x and

*F = kx*to give

*KE = k x^2*and therefore

*k = mgd / x^2*

and for force

*F = mgd / x*

?

Other questions I have are about the friction on the protection and how this can be accounted for. The force applied on the top protection (which acts as a pulley) and on the belayer. And how to go about calculating k for ropes from manufacturers information - and how k varies as the length of the rope extends as there is more rope to stretch.