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Physics teacher demonstrated Newton's 3rd Law

  1. Jan 8, 2007 #1
    Today in class, my physics teacher demonstrated Newton's 3rd Law by nailing a hammer in a piece of wood.

    She drew a free body diagram and there were three force pairs: force pair between hammer and nail, force pair between nail and wood from gravity/normal force, but however, she also put a 3rd force pair that she said was the force of the nail on the wood.

    I am confused by this 3rd force pair as she stated the force exerted by the hammer was 10 N, but the force of the nail on the wood was only 5 N. How is this logical? Isn't this 3rd force pair just from the nail pushing against the wood? I thought this force just equals the 10 N?

    What is really confusing me is, if forces come in pairs, shouldn't the nail exert a force of 10 N from the 10 N exerted by the hammer? So the nail is exerting 10 N on the wood block and the wood block is exerting 10 N back on the nail. How can hammering a nail into wood be possible?
     
    Last edited: Jan 8, 2007
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  3. Jan 8, 2007 #2

    Doc Al

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    If the hammer hits the nail with a force of 10 N, then the nail hits the hammer with a force of 10 N. These are 3rd-law pairs.

    But the nail also exerts a force on the wood, and thus the wood exerts an equal and opposite force on the nail. These are also 3rd-law pairs.

    But these are completely different force pairs. There's no reason to think that the wood and the hammer exert equal force on the nail.

    The nail interacts with three different objects: (1) the earth, through its weight; (2) the wood; (3) the hammer. These are separate interactions, each involving a different 3rd-law pair of forces.
     
  4. Jan 8, 2007 #3
    Shouldn't the nail have a force of 10 N as it is moving down? As it is moving down on the wood, isn't it exerting 10 N on the wood? So then this force pair must cancel out the acceleration of the nail.
     
  5. Jan 8, 2007 #4
    If the hammer hits the nail with 10 N force and the nail hits the hammer with 10 N force then no work is done, right? Meaning the nail does not move.

    If the 10 N force moves from the hammer through the nail into the wood and the wood resists with less then 10 N force (or hits back) allowing the nail to force its way into the wood then work is done. Would there be a less then 10 N force measured back up to the hammer? Driving a nail into wood creates heat and there is movement from the wood as the nail wedges its way in.

    I Have never measured the actual force in newtons of a hammer strike but I have driven many nails and I can tell you there is a large difference in the feel from the hammer when the nail moves as it is driven partialy into the wood as compaired to when the nail does not move at all, like when it hits concrete. A good carpenter can feel the difference between driving a nail into plywood and hitting or missing the joist below by the resistance.
     
  6. Jan 8, 2007 #5

    russ_watters

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    Not correct. The inertia of the nail is what provides the 10N of force (resistance) to the hammer while it is accelerating via f=ma.

    Force at a contact point is always the same for both objects.
    That's what the problem says and the problem is correctly constructed (if oversimplified...).
     
    Last edited: Jan 8, 2007
  7. Jan 8, 2007 #6

    russ_watters

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    The force it exerts on the wood will counter its acceleration. But force isn't something an object carries with it, so the force at one end need not equal the force at the other. The problem included the force from the hammer and the force from the wood - subtract them and you get the force left over to accelerate the nail.
     
  8. Jan 8, 2007 #7

    Einstein said for every action there is an opposite and equal reaction.
    If you apply a downward force of 5 n, the resistant force of the wood also equals 5n. And 5n plus 5n equals 10N.
    But then, for the nail to penetrate the wood, it woud need a force potentail greater than the resistance of the wood.
    Basically, if wood had a resistance fator of 5N and a nail was hit with the force of 5N, it would not penetrate the wood. The force would need to be greater and thus it would be, FxR=DT. Force times distance travelled.
    You do have your teacher on this one.
    The actual example used was if you hit a book with your fist, the book would not move because it's resistance equals the force of your hand.
    Set a book on a sturdy table and slam your fist onto it. See what hurts.
    Let me know, okay ?
     
  9. Jan 9, 2007 #8

    disregardthat

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    Resistance in a fctor of 5N? The resistance the wood have more than just the force added to it to do. The shape of the nail, or the area that are pushing has a lot to say. Like it will not be easy to hit the nail down into the wood upside down for example.

    The force that is added to the nail is making it move at a certain acceleration, but the wood are resisting the nail, and the nail slows down, the deceleration of mass(nail) here equals the force that is 'sent' back to the hammer. It is of course less than 10N(assuming the hammer hit the nail by 10N). If not, the nail would not move.

    Can someone explain the F=ma?
    If you hit the nail in 10N, and let's say the nail has a mass of 0.04 kg the acceleration would be:

    10N=0.04kg * (x)m/s^2

    10N/0.04kg=(x)m/s^2

    250N/kg=250m/s^2

    The acceleration is 250m/s^2 ?
    For how long, and what's the velocity if this nail was hit without resistance?
     
    Last edited: Jan 9, 2007
  10. Jan 9, 2007 #9

    russ_watters

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    This is where the problem breaks down. In reality, for a hammer and nail, the force is applied for an extremely short time and the force is extremely high - thousands of newtons for miliseconds.
     
  11. Jan 9, 2007 #10
    So a force doesn't cause permanent acceleration right? The amount a force accelerates an object by is dependent on the amount of time the force is applied?

    In other words, for 1 kg object, 1 N can be thought of as the presence of acceleration that is equal to 1 m/s^2? And that force must remain acting on the object for the object to continue accelerating at 1 m/s^2 right?
     
  12. Jan 9, 2007 #11
    When considering the work done on an object by a force, or multiple forces for that matter, there is no need to look at forces exerted by the object on other objects, only forces acting on the object.
     
  13. Jan 10, 2007 #12

    Hootenanny

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    Correct. Look up the term Impulse
     
  14. Jan 10, 2007 #13

    disregardthat

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    I thought time was a factor here, so the information the teacher gave is not complete, she must say for how long the 10N at average was affecting the nail. If it was 10N at average for a quarter of a second, I guess the velocity would be 62.5m\s(if nothing affected it). And if the wood gives a continous force back to the nail at 5N, it will slow down with 125m\s^2.

    So, the nail would stop when it's velocity reaches zero. It has a velocity of 62.5m/s after the presumable 0.25 second it took to accelerate it at 62.5m/s^2. The deceleration of 125m\s would take this much time:

    125 = 62.5/s^2
    125s^2 = 65.5
    s^2=62.5/125
    s=quadr(0.5)
    s=0.7
    It will theoretically take 0.7 seconds for the nail to stop, if it had no head on it, the head will make the resistance much greater when it hits the wood.

    Would this be correct, or did I miss an important part here?
     
  15. Jan 10, 2007 #14

    Doc Al

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    I don't quite see the point of your calculation. For one, you assume that the hammer and wood exert their forces on the nail at different times. And, as Russ said, the forces given in this example are wildly off the mark. If you knew the net force on the nail at any given instant you could calculate its acceleration. But that's not the intent of the illustration, which is just meant to point out the correct 3rd-law force pairs. Sure, if the center of mass of the nail is going to move, there must be some net force on it for some small amount of time. But you can't take the data given very seriously. (She could have been a bit more careful.)
     
  16. Jan 11, 2007 #15

    disregardthat

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    Well, how much data would you need to have to calculate it, and how could you do it?
     
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