## Help needed, converting weight to a falling force.

Hey all,
Just a quick overview of myself, I have recently started rock climbing (6 month) and I'm finding myself becoming more curious about the technical equipment involved.

I'm finding it hard to word the questions I'm trying to find the answers too. So I'm going to try and I'm hoping someone will understand.

I weigh 100kg's, How much of an impact force will I create If i was to free fall over different distances with the top distance being no more than 100m.

Can you convert the weight of a moving person into Kn ?(This being the impact force a rope can take, Most ropes can take an average of 9Kn.)

Below is a few things about the Climbing rope i'm currently using. hope this helps

Impact Force (kN): 8.90
Static Elongation (%): 9.10
Dynamic Elongation (%): 33.00

I will be honest although physics interests me I'v never really understood most of it and only done what was taught in school. So any direct answers would be appreciated, But answers with a description as to why would be even better.

Thanks all

Craig

P.s I'm hoping I'v worded this correctly, I doubt I have so any feed back or questions are more than welcome.

Thanks again

 PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus
 Recognitions: Homework Help As you fall, you gravitational potential energy is converted to kinetic energy - so you do $mgh=E_K$ neglecting air resistance. In general you can't so this will give you an over-estimate. You fall the length of your rope - the rope then stretches to take the strain - storing the kinetic energy ... then you bounce around a bit. It is very common to model the rope+climber as an undamped harmonic oscillator. For more details on how to use the rope impact http://en.wikipedia.org/wiki/Fall_factor There are usually lots of rules of thumb for this. The fastest way to find out if you'll break the rope is to tie a weight to the end of it and drop it off a bridge or something.
 When you fall you gain speed at roughly 10m/s^2 so after 1 sec your speed is 10m/s, after 2 sec your speed is 20m/s and so on. This means that you gain MOMENTUM ( mass x velocity) so after 1 sec of falling your momentum is 100 x 10 = 1000 Ns after 2 sec of falling your momentum is 2000 Ns and so on When you are brought to a halt (by rope, air bag, seat belt.... whatever) the force on you x the time for which it acts brings the momentum to zero (you stop!!) If the rope stops you in 1 sec then after falling for 1 sec the force to stop you is 1000N.... the same as your weight. You could call this '1g' After 2 sec of falling the force to stop you in 1 sec is 2000N.... 2g and so on You can see that it is important to extend the time to come to a stop for as long as possible to reduce the force you need to experience. This is the basic physics behind elastic ropes, seat belts, air bags, crumple zones etc..... to extend the time.

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## Help needed, converting weight to a falling force.

Does "Dynamic elongation" tell you the percentage stretch when 8.9kN is applied (at the end of your fall)?

 Interesting. Maybe I can use the above maths to work out my question on my thread ??? Thats if no one has answerd it yet. Wayne

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