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I need a formula!

  1. Feb 14, 2008 #1
    Can anyone help me with a formula to calculate the force generated, when and object weighing "x", connected to a rope with 5% , free falls the length of the rope "l", or half the lenght of the rope? I am looking for fall factors with respect to climbing. This may be fairly simple?
     
  2. jcsd
  3. Feb 14, 2008 #2

    rbj

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    what's the "with 5%" about? the amount that the rope stretches per unit of force (like Hooke's law) is a needed parameter. then we can figger out what the time of decelleration is when the rope is pulled tight at the bottom of the fall. if the rope were steel cable that might not be expected to stretch measureably, the "object" hanging on it might break a few bones at the bottom of the fall.
     
  4. Feb 14, 2008 #3
    Assume that I have a 1m rope attached to the load at one end and, to a fixed anchor at the other. And the stretch in the rope will be 15% overall including knots.

    Does that help?
     
  5. Feb 14, 2008 #4
    Your problem's enunciate is unclear, please improve it.
    >connected to a rope with 5%
    What's that? Friction coefficient \mu=0.05?
    >free falls the length of the rope "l", or half the lenght of the rope?
    This doesn't make sense. For what you need the rope's length?
    Climbing: You need overcome your weight (assume zero your acceleration) and the friction coefficient between your hands and the rope must be sufficient for no-slipping. (and, of course, the rope must supports your weight!)
     
  6. Feb 14, 2008 #5
    Alright...... Rope tied to a 100kg load and to an anchor point. Drop the load from the same height as the anchor, thus the fall is 1m, the rope stretches 15%. Also drop the load from 1m above the anchor, thus the fall is 2 m.

    My question is How much force is generated?

    Load = 100kg
    Rope = 1m

    Drop distance = 1m
     
  7. Feb 14, 2008 #6

    stewartcs

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    Science Advisor

  8. Feb 14, 2008 #7
    Is it simply mass x gravity x length?
     
  9. Feb 14, 2008 #8

    stewartcs

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  10. Feb 14, 2008 #9
    Aah, okey! But why you didn't say that it is an elastic pendulum?
    So you have this pendulum starting from the rest in horizontal position (string tense) and because the gravity action it falls to the vertical position and strectching 15%? Please confirm.
     
  11. Feb 14, 2008 #10
    Because I'm not a scientist...... lol. just kidding.

    Not exactly. the load drops vertically, coming to a fairly abrupt stop at the rope. THe construction of the rope allows for 15% stretch.
     
  12. Feb 14, 2008 #11

    stewartcs

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    After thinking about the equation given in the link, it seems to be in error.

    Instead, using the principle of conservation of mechanical energy we should have,

    [tex] \Delta K + \Delta U_g + \Delta U_{rope} = 0 [/tex]

    [tex] \Delta K = 0 [/tex] since the object is stationary at the initial and final positions.

    Now let H be the distance from the anchor point to the object at the initial position, d be the distance the object falls, and L be the total length of the rope.

    [tex] \Delta U_g = mg \Delta y = mg(-2H - d) [/tex]

    and

    [tex] \Delta U_{rope} = \frac{1}{2} kd^2 [/tex]

    Now substituting these into the first equation we get,

    [tex] \frac{1}{2} kd^2 - mgd - 2mgH = 0[/tex]

    which, when solved using the quadradic equation, gives...

    [tex] d = \frac{mg + \sqrt{(mg)^2 + 4kmgH}}{k} [/tex]

    Note that only the principal root matters here.

    Now since the system will follow Hooke's Law we have,

    [tex] F_{max} = kd [/tex]

    Plugging in d to that gives,

    [tex] F_{max} = mg + \sqrt{(mg)^2 + 4kmgH} [/tex]

    Which is almost the same as what the link gave, except this is correct.

    The "springiness" factor can be described as [tex]k = \frac{ e_{rope}}{L} [/tex]

    Hence...

    [tex] F_{max} = mg + \sqrt{(mg)^2 + 2e_{rope}mg \frac{2H}{L}} [/tex]

    (2H is the distance the object falls before the rope begins to stretch)

    Hope that helps.

    CS
     
  13. Feb 14, 2008 #12

    stewartcs

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    BTW

    [tex] e_{rope} [/tex] is the elasticity of the rope. I looked up a book value for it and a typical one is ~20 kN.

    CS
     
  14. Feb 14, 2008 #13
    Finally I could understand your problem, redial.
    Let H<L be the distance between the fixed end point of the rope, of length L, and the weight tied by the another rope's end (so the rope is slack). The weight is released from the rest and falls freely a height L-H to the natural (not deformed) length L of the rope. The weight drops an additional height d during the deformation of the rope. Thus the total height that the weight drops is L-H+d. The corresponding potential energy is stored by the rope on the form of potential energy of deformation (assuming elastic this process), that is
    U_d=1/2.k.d^2,
    where d=0.15L is given. By conservation,
    U_w=U_d,
    or
    mg(L-H+d)=1/2.k.d^2.
    Notice that this equation is used to find k (the elastisc constant). Once you get k, the force due to deformation of the rope is
    F_d=k.d.
    (this is the force required to deform the rope 15% of its length; the additional potential energy due to height d, is also stored by the rope as energy of deformation).
     
  15. Feb 15, 2008 #14
    You can raise the doubt whether the force is k.d or mg+k.d. The latter one is simply FALSE! Let's make a free body diagram of the weight at an arbitrary rope's deformation x (this one is taken positive downwards from the rope's bottom end at the natural length L of the rope). So upwards is the force F_x exerted by the rope on the weight and downwards is the proper weight mg. So by the Newton's second law
    mg -F_x=m.d^2 x/dt^2.
    Therefore
    F_x=mg-m.d^2 x/dt^2.
    Observe that the acceleration is negative (the weight stops!) and F_x takes into account the weight action and the dynamics (inerce) one as well. But by the Hooke's law
    F_x=k.x
     
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