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Force of a falling object

  1. Apr 22, 2012 #1
    F=ma
    eg, a static bag of a 1kg weight has a force on the table of .98N.
    if it were to fall, would its force hitting the ground be the sum of the standard g of 9.8m/s+ 9.8t?
    or do you scrap that idea and use K.E. instead?
    and also, if it reaches terminal velocity and has an acceleration of 0, surely F=ma isnt used as the result would be 0 force acting on the surface it lands

    been a while since i was taught this in college
     
  2. jcsd
  3. Apr 22, 2012 #2
    it's normal component of velocity will become zero after hitting.use impulse momentum equation to find out the force exerted by ground
     
  4. Apr 22, 2012 #3
    No, it would be F=mg in the direction of falling, and some air resistance force in the opposite direction (if you taking air friction into account).

    Whatever you use, you have to have the same results.

    No. When it reaches terminal velocity, it have constant speed due the equality of F=mg and the air ressistance force (which depend od velocity). But when it hits the ground, it's velocity change from given velocity to the zero, and force acting in that case is change on momentum over time.
     
  5. Apr 22, 2012 #4
    that maK.E.s total sense. thanks for that, of course its the same results, i wasnt thinking straight obviously
     
  6. Apr 24, 2012 #5

    haruspex

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    Don't confuse force with impulse.
    If a mass M hits the ground at speed V the impulse (change in momentum) is M.V.
    This is not a force; it is the integral of the force over the duration of the impact.
    The force, as a function of time, will depend strongly on the nature of the bodies. If either is soft then the impulse is spread out, giving moderate forces over an extended period. If both hard, you'll get a shorter, sharper peak.
     
  7. Apr 24, 2012 #6
    well I think it is better to ask for impulse in this case because it's normal velocity will become zero just after the impact may be very less time and after that it will exert only it's weight.
     
  8. Apr 24, 2012 #7

    haruspex

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    It is still valid to ask about force. You might want to know whether an object will break on impact, and that's a question of the peak force generated. But in general it's much harder to figure out because it depends on the detailed characteristics of the objects in collision.
     
  9. Apr 26, 2012 #8
    ok,then let me know what will be the force on a body if it's momentum changes by a finite amount in an infinitesimal time.well,that is the place to talk about impulse .
     
  10. Apr 26, 2012 #9

    haruspex

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    momentum change = ∫force.dt (as vectors)
    If the time delta is zero then the momentum change must be too.
    In practice, therefore, the time delta is never 0, but it might be very short.
     
  11. Apr 28, 2012 #10
    that is what I am saying if time delta is very short then for a finite momentum change the force should be some sort of delta function and when object hits the ground it will be very large as is the case with delta function when it's argument becomes zero but neverthless the integral is finite so it is better to work with impulse rather than force.
     
  12. Apr 29, 2012 #11

    haruspex

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    Delta functions are theoretical devices. They can give correct mathematical answers because they represent a limit which will be approximated by reality, but as with all limit processes you have to treat them carefully.
    A nonzero impulse imparted in zero time (a Green's function) would imply an infinite force (a delta function). Clearly that does not happen in the real world. If you want to know the change in momentum of the target, the duration of the impact isn't critical; you can treat it as a delta function if you like. But if you want to know the maximum force (e.g. for determining whether something will break) then you can't treat it as a delta function; doing so would imply that tapping on a concrete wall would shatter it.
    As far as I am aware, it is possible that neither infinities nor infinitesimals occur in the real world.
     
  13. Apr 30, 2012 #12

    K^2

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    In a linear elastic collision, the force profile follows a sin² shape. In practice, most collisions do fit that model pretty well, even the inelastic collisions for some reason. So you can estimate the peak force as 2m(Δv)/(Δt). To estimate Δt, you need to know something about elastic properties of materials involved.
     
  14. Apr 30, 2012 #13
    all right, you live in real world.moreover sin^2 of what
     
  15. Apr 30, 2012 #14

    haruspex

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    Do you have a reference for that? I would have expected just a sin function, i.e. like a normal spring. sin² seems unlikely.
    On the basis of sin(w.t), the peak force would be M.π/(2T), where M is the absorbed momentum and T is the time to absorb it.
     
  16. May 1, 2012 #15

    K^2

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    Erm. You are absolutely right. It should have been Sin, not Sin². Not even sure where I got the square from. Maybe I was thinking about energy.

    The real collision's force profile does level off at zero smoothly, unlike the sin, due to the fact that contact surface changes a bit as object deforms. So T has to exclude these areas, but otherwise, it's not a bad fit.
     
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