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Homework Help: Weight on an Elastic thread question.

  1. Jul 26, 2010 #1
    From http://star.tau.ac.il/QUIZ/ [Broken] website:

    This problem can also be found on page 32 http://kvant.mirror1.mccme.ru/1990/05/resheniya_zadach_f1206_-_f1212.htm [Broken], but it's in Russian so might be hard to read...

    Can someone explain this solution? Is it always true that F=Fo/2? Don't we need to know some details about the thread?

    I would think this is just a nonhomogeneous differential equation with a force "F".

    What do you guys think about the solution? Does it always hold true in general?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 26, 2010 #2

    K^2

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    The wording is a little bit weird, but yes, it's exactly F0/2.

    Simplest way to see this. If you gradually apply a force and increase it to F, you displace the mass by F/k. This new point is the new equilibrium point.

    If you instantly add point F, the equilibrium is still shifted by F/k, but mass has not moved yet. It will accelerate towards equilibrium point, overshoot it, and pass it by another F/k. So now the mass is 2F/k from point where it was before you applied force, and that means total force on the thread is 2F.

    So by applying a force instantly, you cause maximum force to be twice the applied force, and that means you can break the thread by applying only 1/2 of the force you need to break it.

    Edit: It's not clear from wording of the problem, but the Quant's solution starts out by stating that they treat the thread as obeying Hook's Law, which is fair for a tight thread with low elasticity.
     
  4. Jul 26, 2010 #3
    K^2....

    Thanks for the reply....

    Okay....

    Makes sense...

    How do we know it will overshoot by exactly F/k?

    That makes sense assuming the overshoot is exactly F/k.....

    That seems reasonable....
     
  5. Jul 26, 2010 #4

    K^2

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    Start with equation of motion.
    [tex]m\ddot{x} + kx - F - mg = 0[/tex]

    Suppose the following is a solution.
    [tex]x = A cos(\omega t) + D[/tex]

    Substituting...
    [tex]- \omega^2 m A cos(\omega t) + k A cos(\omega t) + kD - F - mg = 0[/tex]

    Since it must hold for all t, there are actually two equations there.
    [tex]k = \omega^2 m[/tex]
    and
    [tex]kD = F + mg[/tex]

    Amplitude A is determined by initial condition. The oscillations have frequency ω=sqrt(k/m), no surprise. And the oscillations take the mass to ±A from point x=D=(mg+F)/k, which is the equilibrium point. Since the object starts at old equilibrium point (mg/k) and new equilibrium point is (mg+F)/k, amplitude A is trivially given by A=(mg+F)/k-mg/k = F/k.

    So the maximum is achieved at x=D+A=mg/k+2F/k, the total force there is mg+2F. If mg+2F = mg+F0, the thread breaks. So F=F0/2.
     
  6. Jul 27, 2010 #5
    Got it....makes perfect sense.

    Thanks.
     
  7. Jul 27, 2010 #6
    Hi :)

    I have no issue with the solution Fo/2 as the theoretical limiting case.

    However, the example is simplified and omits many real-world factors such as damping ratio of the thread, energy dissipation of any kind (such as generation of heat), external resistance such as conducting the experiment in treacle rather than a vaccuum (drag coefficient), etc etc...

    Are all here agreed that the solution Fo/2 is the theoretical limiting case, and that in any real-world scenario, solutions can tend to Fo/2, but cannot reach it exactly ?
     
  8. Jul 27, 2010 #7

    K^2

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    For a tight thread, all of these coefficients are not going to play a big role in a single oscillation. I suppose, you can "break" the solution by introducing a material that has a lot of viscous give before it snaps.

    So yeah, it's an approximation, but it's an extremely good one.
     
  9. Jul 27, 2010 #8
    Thanks.

    An example I've used in other venues is using natural rubber as the material for the elastic thread.

    Natural rubber has a damping ratio range of 0.01 to 0.08.

    Incorporating damping ratio for xmax results in the following in other texts I've looked at...

    xmax = [tex]\stackrel{F}{k}[/tex](1+e-[tex]\zeta[/tex][tex]\pi[/tex])

    Not used to the formatting here I'm afraid - xmax=F/k*(1+e^-zeta*pi)

    Where [tex]\zeta[/tex] (zeta) is the damping ratio.

    Applying 0.08 to (1+e-[tex]\zeta[/tex][tex]\pi[/tex]) = 1.78

    Using damping ratio of zero results in 2 as expected.


    I agree that Fo/2 is a reasonable approximation for general use, but do think that 1.78 is a quite significant difference to 2.

    Any issues there ?

    ETA: It would also be useful to have opinion on Fo/2 being an unattainable limiting case in any real-world scenario. I cannot think of any real-world situation which would allow for exactly Fo/2 to be reached. Splitting hairs of course, but it would be useful to have additional confirmation. Would imply a material with a damping ratio of exactly zero.
     
    Last edited: Jul 27, 2010
  10. Jul 27, 2010 #9
    Can anyone confirm or deny what femr2 is saying here?
     
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