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Homework Help: Maximum elastic energy in rope

  1. Nov 7, 2012 #1

    If a climber attached by rope happened to fall what would be the maximum elastic energy stored in the rope?
    I believe all the potential energy due to altitude would be converted into elastic in the rope (once the elastic energy is maximum and the kinetic energy equals zero), would it not? Or is the total energy stored in the rope be equal to the sum of the potential and elastic?
  2. jcsd
  3. Nov 7, 2012 #2
    May someone please help me sort this out?

    I am trying to find out whether a person weighing 784.8 N could survive the fall, given that the initial length of the cord (L0) is 2 m, the person falls 4 m down, the diamater of the cord is 10.5 mm, and pi*diameter^2*Y (=Young modulus) is known to be 750 Pounds.

    I wrote down the energy balance thus:

    Mg(L0 + delta L) = [0.5*A (=surface area of cord)*Y]/L0

    Which could be written as:
    Mg(L0 + delta L) = [pi*diameter^2*Y*(delta L)^2]/8L0

    Now, could I deduce whether the person will survive the fall from this expression? Or, ought I to write down a force balance, thus:

    Mg = [A*Y*(delta L)]/L0

    Whence I could find the maximum M the cord could possibly bear?

    I am also given that the maximum force possible to be exerted without the person getting hurt is 2650 Pounds.

    Please advise on this.
  4. Nov 7, 2012 #3
    I made a mistake. The initial formulation of the energy balance ought to be:
    Mg(L0 + delta L) = [0.5*A (=surface area of cord)*Y]*(delta L)^2/L0
  5. Nov 7, 2012 #4


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    So are you ok now or do you still need help?
  6. Nov 7, 2012 #5
    Thank you for replying!
    I have actually changed the equations since posted, to:

    (1) Mg(h + Xmax) = [pi*diameter^2*Xmax^2*Y]/8L0
    (2) Fmax = [pi*diameter^2*Y*Xmax]/4L0

    I am quite positive they are correct, yet by all means do please confirm.
    Another thing is that (pi*diameter^2*Y) is given as 7500 Pounds (and not as 750 Pounds, as I initially stated). Pardon me for that.
    In case a 784.8 N person were to fall from that height, I got that the maximum force would be equal to 1340 Pounds (Y = 96.31*10^6 Pa, maximum elongation = 1.43 m), which is less than 2650 (=the maximum force possible without any harm, thus the question states). Furthermore, a 981 N person would also not be harmed, as the maximum force then would be equal to 1525 Pounds.
    Do these results seem reasonable to you, or have I made some mistakes? May you kindly confirm?
  7. Nov 7, 2012 #6
    A few further remarks, if I may: I used 1 Pound force = 4.48 Newtons, thus I got to Y = 96.31*10^6. I had to solve the quadratic equation for the energy in order to determine the maximum elongation.
  8. Nov 7, 2012 #7


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    I'm finding what you've written rather confusing. It would help me if you could just use algebra and refrain from putting in the values of the constants until the end.
    I was confused about the 4m, but I think I understand it: the climber falls when 2m above the point of attachment of rope to rock, so the climber falls 4m before the rope becomes taut. Is that it?
  9. Nov 7, 2012 #8
    No values whatsoever have yet been substituted into the equations! The 4 you are referring to has nothing to do with the height. It is merely pi*radius^2=pi*(diameter^2)/4.
  10. Nov 8, 2012 #9
    The dynamic kernmantle rope used for fall protection by climbers is specifically designed to absorb quite a lot of energy. To complicate things further, the climber has a whole lot of choices as to which rope to buy, each with different energy absorption characteristics. That is why the rope suppliers will provide this engineering test data to the interested customer.

    Once you have this information, you can determine how much energy is absorbed by the rope, and how much of it is returned to the climber as the elastic energy that results in the rebound.

    Energy absorbed will heat the rope. If the fall is too high, the amount of energy transmitted to the rope will be high enough to at least partly melt the rope fibers. A friend survived a 160 ft fall with nearly no injury because he had designed sufficient energy absorption into his support system. A large part of his energy absorbing system was the rope itself, but it ruined the rope. It partly melted and then became stiff as a rod upon cooling. It also lost about 30% if the rope's original diameter.
    Last edited: Nov 8, 2012
  11. Nov 8, 2012 #10


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    Sorry, I should have made that clearer. The equations you posted are ok, but you didn't develop them into an equation for the answer - just quoted a numerical answer.
    No I mean this 4: "the person falls 4 m down". The only way I can make sense of this is that the climber slips when 2m above the point at which the rope is anchored, i.e. it is taut (just) in what one might call the 12 o'clock position. The climber therefore falls 4m before the rope becomes taut again, now in the 6 o'clock position. The total PE lost by climber is therefore Mg(h+x) where h = 4. Perhaps you thought that was obvious. Perhaps it should have been.
    I don't understand why you need to know the rope's diameter if you already know the value of k = A*Y.
    Fwiw, I get Fmax = Mg(1+√(1+2hk/MgL)) where k = 7500 lb wt.
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