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Help needed, converting weight to a falling force.

  1. Feb 11, 2012 #1
    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


    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
  2. jcsd
  3. Feb 12, 2012 #2

    Simon Bridge

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    As you fall, you gravitational potential energy is converted to kinetic energy - so you do [itex]mgh=E_K[/itex] 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

    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.
  4. Feb 12, 2012 #3
    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.
  5. Feb 12, 2012 #4


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    Does "Dynamic elongation" tell you the percentage stretch when 8.9kN is applied (at the end of your fall)?
  6. Feb 12, 2012 #5

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

  7. Feb 12, 2012 #6


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    Unfortunately, "falling force" doesn't really mean anything, as a term in Physics and I was trying to ascertain exactly what the terms in the OP really mean (hence my question).
    In any case, no muscles are involved in this question. :smile: I have a feeling that we won't be dealing with acres and acres of non-Physics reasoning, once we get to the bottom of the question.
  8. Feb 12, 2012 #7


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    I've looked up the terms and done a rough calculation based on the gravitational potential energy lost on the journey to the bottom of your fall being equal to the elastic energy stored on the stretched rope. For your rope, if you (100kg) fall 100m to the end of the rope, it would appear to stretch by about 31m (!!!! boinnnng!!!) and the force on you, at the very bottom, will be about 8.4kN. This means you will be experiencing a pull of about 8g. It will come on gradually, of course, as the rope tightens. If your harness doesn't fit properly (or even if it does) that's a pretty horrific force it will be exerting on you.
    I will check this tomorrow and try to give you a way of working out the forces for any drop height.
  9. Feb 13, 2012 #8


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    I see it now. All other things being equal, it makes no difference how far you fall to the max force on your harness. That's why the figure they publish is so useful.
    Hard to believe? Read on. . . . .
    I went to the trouble of calculating for a 10m rope length and got the same sort of answer as for a 100m rope and then started thinking and calculating. Here's the arm-waving argument that may explain it. The Kinetic Energy gained as you fall is proportional to the height fallen. This energy gets stored in the stretched rope. (Thankfully, there is significant internal friction so you don't just bounce up and down for ever!). A longer rope will stretch by the same percentage of its length as a similar short rope (the short rope will be 'stiffer', with a higher 'spring constant') so you will take a proportionally longer distance to come to a halt. The force at the end of the ride will be the same in all cases.

    But this isn't the whole story. As far as the experience goes, you will suffer a lot less from a short fall on a short rope, because the braking forces will be acting on you for a much shorter time so a long fall is not to be recommended. Of course, a short fall on an already long rope, will be much more gentle - which is what we would expect and is why your rope is always pulled in as much as poss.
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