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Sudden and static force

  1. Oct 29, 2013 #1
    I keep a book that his weight is 10kg(100 newton,lets say). My hand will feel a force of 100 newtons.
    Now lets say that the book is up of a desk,and with some force(lets say 60 newtons) I hit that.

    Why I will feel more pain the second time,while I feel less newtons?
    It is about the time?
     
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  3. Oct 29, 2013 #2

    arildno

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    Why do you feel less newtons??

    When you hit the book, you start out with a huge velocity difference between your hand and the book.
    That velocity difference must be reduced to 0 in a fraction of a second, i.e, a huge force couple is generated between your hand and the book.
     
  4. Oct 29, 2013 #3
    Ι feel less newtons,because I hit the book with less newtons,but I feel more pain from when I keep the book(that weight 10 kg)
     
  5. Oct 29, 2013 #4

    Drakkith

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    I'm having a very difficult time understanding you. Is the following correct?

    You hit a 10 kg book with your hand and feel some pain. I assume it's in the air or upright so it just flies off after you hit it.
    You then hit the same book with less force, but this time it's lying on a desk. You feel a greater amount of pain.

    Is that correct?
     
  6. Oct 29, 2013 #5
    Nope. See
    1)You just keep a book that weight 10 kg,so your hand feel 100 newtons.
    2)You leave that book somewhere,and you hit this book with force of 60 newtons.

    My hand will feel more newtons at 1,but more pain at 2.
     
  7. Oct 29, 2013 #6

    Drakkith

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    You don't hit the book with 60 newtons of force. You hit it with a certain velocity and the maximum force will depend on how quickly your hand decelerates. The force builds from zero to maximum before falling to zero again over the course of the deceleration time.
     
  8. Oct 29, 2013 #7

    sophiecentaur

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    Your thought experiment is not providing you with valid results, I'm afraid - presented as it is.

    It is not the 'Force' that counts in the context of collisions. People are always asking about the Force of an impact (vehicle collisions and injuries) but the damage done will depend upon the Kinetic Energy and Momentum involved rather than just the Force involved.
    When you apply 100N to keep the book suspended then the force is defined and you have an equilibrium situation. If the book is suspended in space and you 'hit it' with a force that you say is 60N, is this an instantaneous force? Is it applied for 1s, 1ms or 1μs? Whatever time it's applied for, if it's only 60N then there will be less deformation / damage / pain to your hand, simply because the force is less. If you have the book on Earth and can only give it an upwards force with your hand of 60N, then its weight will actually be pushing (accelerating) your hand towards the ground.

    If you hit a sharp corner of the book with your hand then the local pressure can give more pain than when you support the book on your palm - but that's another layer of complication.
     
  9. Oct 29, 2013 #8

    arildno

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    Hepic:
    Your hand starts out with velocity, ending with velocity 0 in time interval "dt".

    thus, the AVERAGE force experienced by it is mV/(dt), which can be very huge, if dt is tiny, V "ordinary".
     
    Last edited: Oct 29, 2013
  10. Oct 29, 2013 #9
    So to undestand. Does not matter only the force,but how quicly you hit that.
    Right?
     
  11. Oct 29, 2013 #10

    arildno

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    Not at all!
    The (average) force CANNOT be deduced as you believe it can.

    You can estimate it as the ratio between the momentum change and the time interval over which the momentum change happened.
     
  12. Oct 29, 2013 #11

    sophiecentaur

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    Hepic still appears to have his (her?) original problem of misplaced intuition. It would be possible to apply a force, as in the original model (say, with a small calibrated rocket jet, applied for some known time) but that is not the case with a hand hitting the book. You could even measure the force during a real impact but it would not be constant over the time of impact and would be only one of the factors involved in determining what would be happening.

    Vehicle insurance claims would be a very different thing if it were as simple as Hepic is suggesting.
     
  13. Oct 29, 2013 #12

    arildno

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    True enough.
    But understanding how the AVERAGE force can be found (and why it can be extremely huge) should dispel his idea that the magnitude of the force somehow could be calculated from the weight of the object.
     
  14. Oct 29, 2013 #13

    sophiecentaur

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    I think you mean that the peak force can be huge. An average force would presumably refer to the change of momentum divided by the duration of the contact. I can only hope that Hepic will follow this in a more formal and less intuitive way. It's the only way to get a proper understanding of this topic.
     
  15. Oct 29, 2013 #14

    arildno

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    The average force can be huge as well, not just the peak.
    Is mV/dt a huge number, the average force was huge (and the peak force even huger)
    And "change in momentum"/"time interval" IS the formal definition of average force over that time interval.

    :smile:
     
    Last edited: Oct 29, 2013
  16. Oct 29, 2013 #15

    ZapperZ

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    Please look up the concept of "Impulse".

    Zz.
     
  17. Oct 29, 2013 #16

    sophiecentaur

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    But aren't we starting off with a 60N scenario? That certainly limits the 'Average' force - and it could even be the limit for the Maximum force.
    And what is the time interval? Where do you actually say that the interaction of hand and book starts and finishes? What's the time profile of the force? My point is that the quantity Impulse is the relevant one, because it contains most of what's relevant and measurable in a situation like this.
    [Edit: You beat me to the term Impulse, Z]
     
  18. Oct 29, 2013 #17

    arildno

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    "My point is that the quantity Impulse is the relevant one"
    --
    Quibble.
    Since the impulse always equals the change in momentum.
    We have:
    [tex]\vec{I}=\int_{t_{0}}^{t_{1}}\vec{F}dt=\bigtriangleup\vec{p}[/tex]
    Average force equals:
    [tex]\vec{F}_{av}\equiv\frac{1}{\bigtriangleup{t}}\int_{t_{0}}^{t_{1}}\vec{F}dt=\frac{\bigtriangleup\vec{p}}{\bigtriangleup{t}}[/tex]

    And, this is by far the most practical way to find the average force, rather than demanding to use the instantaneous force distribution over the time interval.
    We don't know that distribution, nor do we need it, in order to compute the average force.
     
  19. Oct 29, 2013 #18

    arildno

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    OOPS! It seems that I accidentally in one post used "velocity" rather than "momentum".
    Damn!
     
  20. Oct 29, 2013 #19

    sophiecentaur

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    But does 'average force' matter? Is it of interest to anyone? It doesn't indicate the 'pain' or damage involved. It doesn't even represent anything about the Work done or energy input.
     
  21. Oct 29, 2013 #20

    arildno

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    It sure matters.
    Because OP seems hung up on measuring the forces involved in a particular situation, and does that in a totally skewed manner.

    Therefore, it DOES matter to point out to him WHAT sort of force he actually might calculate in a correct manner for a collision, and also to show how he is missing orders of magnitude of those forces by thinking they somehow are comparable to the object's weight.

    Beyond that, I certainly agree with you that they aren't particularly important.
     
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