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Help calculating maximum loads

  1. Jun 21, 2017 #1
    So I am trying to solve a problem, I’ve done a simple solution but I want a more comprehensive answer and don’t know how to do the math.

    The problem involves climbing anchors and the difference between max loads of an equalized vs non equalized anchor.

    The two different situations, I’ll try to keep as simple as I can and still represent likely real life scenarios:

    1. An anchor with 2 points perfectly equalized.
    2. An anchor with 2 points but .5m of fall between anchor points, meaning that the load of a fall comes down on one anchor point until that point breaks, then there will be a .5 meter fall before the load comes down on the second anchor point.

    What I’m trying to figure out is how much more total force can be expected from #2 vs #1 in a fall that generates 15kn of maximum load in #1

    I did the basic calculations (f=ma) and the additional force is negligible:

    I assumed a 100kg weight falling an arbitrary distance to generate a force of 15kn. Each anchor point breaks at 10kn.

    In #1 the anchor doesn’t break.
    In #2 the first anchor point breaks at 10kn (dissipating the force as heat in the components) and the second anchor point experienced a load of 5kn plus the force of 100kg falling .5 meters (~100n)
    For 5.1kn force, and the second anchor point does not break.

    Where I get lost is the fact that the rope stretches (and recoiles) and the disparity that that causes, which I do not understand.

    According to an online force calculator for climbing rope falls (That presumably factors in rope stretch)

    The force generated by a .5m fall is .5.6kn to 11.25kn depending on the length of rope Resulting in a force of at least 10.6kn and the second anchor point breaks. This doesn’t apply to the problem precisely because, I assume, it is calculating an acceleration from 0m/s to 1m/s with the rope unstretched.

    I would expect that there will be no significant additional load but I don’t know how to work it out, and the online calculator seems to suggest otherwise, I’m stuck here. I don’t understand where all the extra force can come from.
  2. jcsd
  3. Jun 22, 2017 #2
    Any rope you use is going to have some elasticity to it. How much it stretches and rebounds depends on the material you use.

    You need to consider that when the rope is fully extended and begins to stretch, this causes a massive spike in the force on that rope. A less elastic rope will mean that the force spike is shorter, but also larger. The total energy absorbed by the system is always the same, no matter how elastic your rope is. I'm kind of curious as to how you came to the results you got.

    A good way to think about it is that a rope that is undergoing such a stress is essentially a spring (not exactly, but a close enough analogy for this purpose). A very 'stiff' rope can be thought of as a spring with a very large spring constant, whereas a very elastic rope can be thought of as a spring with a relatively small spring constant.

    Say, for example, we're dropping the 100 kg weight from a height of 10 m, and we have it hanging off of a single anchor point that snaps at 20 kN. After the drop, the weight will be moving downwards at a speed of 14 m/s, and has a kinetic energy of 9800 J.
    For simplicity's sake, let's also assume that gravity stops acting on the weight as soon as the weight fell the 10 meters.

    As a first example, let's say our rope has a 'spring constant' of 10 kN/cm. After absorbing the kinetic energy of the weight, the rope will be stretched by 14 cm and exert a force of 140 kN (!!) on both the weight and the anchor, and the anchor snaps. This is how most ropes will behave, especially materials such as steel cable.

    Now, let's assume your rope is basically an elastic band and has a spring constant of 10 N/cm. Again, absorbing the kinetic energy, it will stretch by 4.42 meters and exert a force of 4.42 kN and thus your anchor holds.

    Basically, the stiffer your rope is, the larger the maximum force will be.

    I calculated the stretch distance as x = √2E/k , where x is the distance stretched in meters, E is the energy the spring absorbs and k is the spring constant.
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