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Understanding how Newton's Third Law is applied?

  1. May 24, 2003 #1
    Hello!

    I've just begun attempting to teach myself physics, and I'm finding it difficult to understand Newton's third law. Would someone be so kind as to shed a little light on the situation?

    The idea of the equal and opposite reaction for every action makes perfect sense to me in some cases (a swimmer's motions push the water behind them, and in response, the water pushes them forward; I lean against a wall, the wall is pushed away from me and responds by exerting a force in the opposite direction that holds me up), however, I keep running into situations where it doesn't seem to make sense. For instance:

    1) I kick a pebble and a boulder with the exact same force. My foot experiences almost no noticeable response when I kick the pebble, but when I kick the boulder, I hurt my foot. How can these both be equal reactions?

    2) In outer space, a rocket is attached (by a cord, or by a beam, or by whatever would make this example logical without complicating things) to some object, let's say a crate. The rocket is trying to pull the crate. If the rocket fires, the action-reaction pair between the rocket's exhaust and the rocket subjects the rocket to some force x.
    Since the rocket is attached to the crate, the rocket also applies force x to the crate. Since the crate is attached to the rocket, it applies a force to slow the rocket down - which, by Newton's third law, is -x. The net force on the rocket is x - x = 0, and so, no matter how hard the rocket fires, it's never going to be able to escape equilibrium.
    Where's the error here?

    I would really appreciate help understanding where I'm going wrong with this law :wink: Thanks!
     
  2. jcsd
  3. May 24, 2003 #2

    drag

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  4. May 24, 2003 #3

    mathman

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    Without getting into a lot of detail, what you seem tohave missed in your examples is the mass of the object in question (Newton's law, force=mass times acceleration).
     
  5. May 24, 2003 #4
    Thanks for the great link!

    However, I am still unclear on where I am going wrong in the two examples I mentioned. I realize that the mass of the objects comes into play as part of the explanation, but I don't see where. Could someone offer some clarification here?
     
  6. May 24, 2003 #5
    Welcome to PF, zoro.

    1) Here, you're not actually kicking them with the same force -- the average force you put on the pebble is much smaller than that on the boulder. In general, force is a bad concept to use when dealing with collisions, because the actual collision forces are very dependent on how much the two colliding thingies deform. But here's what happens:

    Force = mass*acceleration. During the kick, the pebble accelerates quickly, but it is so light that the force needed is small. So the acceleration of your leg is negligible and you hardly feel like you hit anything. When you kick the much more massive boulder, the situation is reversed; the 3rd law means it accelerates much less than your leg. So your leg is stopped before the boulder moves very far at all.

    2) The rocket doesn't subject the crate to force X; it subjects it to some different force Y. Remember that if they are rigidly attached, that means they have to move at the same velocity always (otherwise one would be getting closer/further) hence their acceleration is the same. The total forces and accels are:

    F_rocket = X-Y
    F_crate = Y

    a_rocket = (X-Y)/m_rocket
    a_crate = Y/m_crate

    So if the two accels are equal,

    Y = X * m_crate/(m_rocket+m_crate)
    F_rocket = X * m_rocket/(m_rocket+m_crate)

    So the net force on the rocket -> 0 if the crate weighs a lot more, and -> X if the crate is really light.
     
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