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A.P. French, Mechanics Problem

  1. Sep 13, 2005 #1
    I am using A.P. French, Mechanics, to supplement my Physics studies (i.e., on my own). I'm stumbling over problem 6-3 (b): A man of 80 kg jumps down from a window ledge 1.5 m above ground, bending his knees so that his center of gravity descends an additional distance h after his feet touch the ground, what must h be so that the average force exerted on him by the ground is only three times his normal weight?

    So, obviously, a=3 m/s^2, since we want F=(80 kg)(3 m/s^2). I use v^2 = 2g(1.5) to get the velocity at the point that the knees start to bend, so v^2 = 29.4 (m^2/s^2).

    Since I want a=3 m/s^2, I reason that all I need is to use 0 = v^2 - 2ah, with a=3 and v^2=29.4, to find the stopping distance at that acceleration, which should be h.

    This gives me 4.9 m, which seems excessive, and French gives h = 0.75 m, and I can't see how he ended up with that. It seems to me this would correspond to a final velocity of 2.12 m/s, which is way too low for a 1.5 m drop.

    Well, thank you for any suggestions or pointers.

    Dot
     
  2. jcsd
  3. Sep 13, 2005 #2

    Pengwuino

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    ... AP french? whats that have to do with physics
     
  4. Sep 13, 2005 #3

    Tom Mattson

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    A. P. French is the author of the book, ding-a-ling! :wink:

    You mean a=3g=29.4 m/s2, don't you?
     
  5. Sep 13, 2005 #4

    Pengwuino

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    :rofl: :rofl: :rofl: :rofl: :rofl:
     
  6. Sep 13, 2005 #5
    You mean a=3g=29.4 m/s2, don't you?[/QUOTE]

    Thanks Tom. What a gracious way to point out my foolish error.

    However, this would give h=0.5 m, not h = 0.75 m. So I am still puzzling...

    Thanks again,
    Dot
     
  7. Sep 13, 2005 #6

    Tom Mattson

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    I haven't worked out this problem yet, but here is one other thing:

    That would be true if the acceleration were constant, which it almost certainly is not.

    I'm not familiar with French, so let me ask you something: Do you at this point have energy methods (work-energy theorem, conservation of energy) at your disposal? How about impulse?
     
  8. Sep 13, 2005 #7
    The force exerted by the ground is 3mg and the downward gravity is mg
    Hence the deceleration is 3g-g=2g

    hence we have
    v^2=2(2g)(h)

    or
    h=v^2/(4g)
    and v^2=2g.1.5
    So
    h=2*1.5/4=3/4=0.75 metres or 75 centimetres
     
  9. Sep 13, 2005 #8
    Thank you. I find this simple problem very confusing, and I appreciate both of you helping me very much.

    Thanks again,
    Dot
     
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