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Newtons Third Law

  1. Mar 18, 2008 #1
    Hey guys, was just wondering if you could explain the answer to this question. So basically we have a picture of a book resting on a table. On this book there is the weight force downwards and the reaction force upwards, both acting on the center of mass.
    The question:
    A class of students is asked to consider the relative magnitude of N and mg. The students are asked to give their reasoning.
    Frank states 'N and mg are equal in magnitude because N is the reaction force to mg inthe sense of Newtons Third Law"
    Emma states "N and mg are equal in magnitude because the net force on the book is zero. N is not the reaction force to mg in the seson of Newtons third law.'

    I instantly assumed that Frank would be correct however he isnt. I have read the explanation that was provided as to why he wasnt, but i just cant understand it. Can someone please help?

    P.S I would prefer not to post the explanation so i can get a response that doesnt make reference to that explanation, just so i will be able to understand it better. If needed, please request
     
  2. jcsd
  3. Mar 18, 2008 #2

    Doc Al

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    Staff: Mentor

    To find the "reaction" to any "action", express the interaction as body A exerting a force on body B. The "reaction" to that force is always: body B exerting an equal and opposite force on body A. Always identify the two bodies involved.

    For weight: What are the two bodies involved? What is the 3rd law pair of forces?

    For normal force: What are the two bodies involved? What is the 3rd law pair of forces?
     
  4. Mar 18, 2008 #3
    Okay so:

    Weight
    Body A: Earth
    Body B: Book

    Normal
    Body A: Table
    Body B: Book

    Okay so that clears that up, it obviously aint newtons third law as they are both being applied to the book and they arent an action-reaction. Just my last question would have to be what is this force that the book applies to the table? Where does it come from? If weight is the only force, neglicting air resistance, wouldn't it have to be weight?
     
  5. Mar 18, 2008 #4

    Doc Al

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    The book and table exert equal and opposite contact forces on each other. Since the book happens to be in equilibrium, that contact force will be equal to the book's weight. But that's not a consequence of Newton's 3rd law. (Weight is an interaction between book and earth, not a contact force with the table.)

    What if the book and table were in an accelerating elevator? The book's weight doesn't change, but the normal force (between book and table) does.
     
  6. Mar 18, 2008 #5
    Okay so what your suggesting is that we cant just assume that the normal force will be the same as the weight, unless it matches certain conditions. So for instance, with a ball bouncing. The normal force obviously doesn't equal to the weight force. Is this an example where the normal force is independent of the weight force?

    What i still dun seem to understand is the origin of this contact force. Back to the example with the ball. The netforce on the ball just before it hits the floor, assuming its mass is .1kg, is 1N downwards. Doesnt this mean that during the collision, the floor will experience 1N downwards
     
  7. Mar 18, 2008 #6

    Doc Al

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    That's right. The normal force will depend on the acceleration of the body.
    Sure. While the ball's in the air, the normal force is essentially zero. If it were just resting on the floor, the normal force would equal the weight. When it bounces against the floor, crushing the ball/floor molecules together, the normal force is much greater than the weight.

    Sure.
    No. Things change when they smash together. (The origin of the contact force is inter-molecular electromagnetic forces. Push your hands together--you are creating a contact force in the same manner.)
     
  8. Mar 18, 2008 #7
    Okay well lets consider motion in a horizontal plane. Lets say a car, accelerating, collides with a stationary truck. Will the acceleration affect the magnitude of the normal. Or will the normal force be the same as the force that provides the car with the acceleration.
     
  9. Mar 18, 2008 #8

    Doc Al

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    When I said that the normal force depends on the acceleration, I meant the acceleration during the collision or contact. (Not the acceleration of the car along the road prior to the collision.) When a moving car crashes into something, high contact forces are produced giving rise to rapid acceleration.
     
  10. Mar 18, 2008 #9
    And these high contact forces give rise to the normal reaction force? Please say im right...
     
  11. Mar 18, 2008 #10

    Doc Al

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    Staff: Mentor

    Those contact forces are the normal force.
     
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