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Teaching Force Concept Problem

  1. Feb 22, 2008 #1
    I am what is called an LA (Learning Assistant) for an introductory physics class and I'm a freshman. One problem in which I have in convincing the students is how forces are equal and opposite. For instance:
    A large truck breaks down out on the road and receives a push back into the town by a small compact car as shown in the figure below.

    While the car, still pushing the truck, is speeding up to get to cruising speed:
    A. the amount of force with which the car pushes on the truck is equal to that with which the truck pushes back on the car.
    B. the amount of force with which the car pushes on the truck is smaller than that with which the truck pushes back on the car.
    C. the amount of force with which the car pushes on the truck is greater than that with which the truck pushes back on the car.

    How do I explain the answer and concept to them? Every time I bring this up, they argue that Newton's Law of equal and opposite forces is violated because if they were to be equal and opposite, then the truck would not move.
     
  2. jcsd
  3. Feb 22, 2008 #2

    Doc Al

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    The equal and opposite force pairs in Newton's 3rd law act on different bodies. Thus they don't cancel out or produce equilibrium.
     
  4. Feb 22, 2008 #3
    yep :approve:
     
  5. Feb 26, 2008 #4
    Sorry for the late reply, I've been really busy, so if I were to explain this to a student, would I say that they are equal forces but produce different accelerations due to different masses?
     
  6. Feb 26, 2008 #5

    arildno

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    1. You should first make the student understand that the two forces act on DIFFERENT bodies.

    2. If he complains that the sum of the two forces is zero, and hence that "the acceleration is zero", question him on WHICH acceleration he's talking about.
    The acceleration of the SYSTEM comprised of the two bodies is, indeed, 0, but it does not at all follow from this that the bodies themselves have accelerations equal to zero.

    3. Then tell them that although the magnitude of each force is equal, the magnitudes of their accelerations will differ if they have different masses.
     
  7. Feb 26, 2008 #6

    kdv

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    Just a correction. If by system here we mean the car and truck, their acce;eration is not necessarily zero. They could be moving at constant velocity or they could have a nonzero acceleration. The net force on each object separately does not have to be zero.
    This is true in general but in this particular example, assuming that the car and truck always stay in contact, the two objects must have the same acceleration.
     
  8. Feb 26, 2008 #7

    kdv

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    The contact force between the two is not the only force acting on them so it's incorrect to say what you said. If the contact force was the only force, then you would be correct. To have such a case, you can imagine an astronaut very far from any massive object like a planet or a star, pushing against a second astronaut. In that case the only force betwene them is the contact force (if we neglect the force of gravity between them). They feel the same force but have different accelerations if they have different masses.

    Another example is an apple in free fall. It pulls on the Earth with the same force that the Earth pulls on the apple but the two objects have different accelerations.


    Going back to your problem, as the others said, the key point to emphasize in the third law is that the action-reaction forces act on different objects. The best way to explain the problem is to draw a free body diagram of the car an done for the truck and to show all the forces, pointing out the action-reaction pair. And then to emphasize that the acceleration of each object is given by the net force on that object divided by its mass.
    You should show two cases: one in which the car and truck move at constant velocity and one in which they have an acceleration. That should clarify things to them.
     
  9. Feb 26, 2008 #8

    Doc Al

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    Another thing to point out is that while it's certainly true (via Newton's 3rd law) that the force exerted by the car on the truck must be equal and opposite to the force that the truck exerts on the car, those forces are not the only forces acting on those bodies. Don't forget the force of the ground on the car!

    To figure out how a body accelerates one uses Newton's 2nd law, not the third law. And one must consider all the forces acting on the body.

    It will also be instructive to analyze forces on several systems:
    (a) the car
    (b) the truck
    (c) the "car + truck" considered as a single body

    Only in case (c) do those 3rd law forces "cancel out", since they are internal to the system and thus act on the same "body".
     
  10. Feb 26, 2008 #9

    arildno

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    You're right, I didn't bother to read the example at hand, and assumed it was a typical collision type issue. I readf his post 4, rather than the OP. Shame on me.
     
  11. Feb 26, 2008 #10
    You can do a simple demonstration to get them thinking. First, show them that a spring scale shows how much force is being exerted (set a 100N weight on the floor and have someone pull up slowly and watch the spring scale stretch and stretch and, just past 100N, voila!). Then set 2 students on something like furniture dollys, free to move with some friction. Give one of them two spring scales - one attached to the wall the other attached to a spring scale held by the other student. Make sure all 3 spring scales are basically colinear. Have the one with two spring scales try to pull the other guy to him and read all 3 scales. Then repeat, except omit the spring scale attached to the wall.

    This takes a little practice to get the friction right and to get the guy in the middle to stay still. Plus, you may have to do a little explaining about the friction. But, it might turn on some light bulbs.
    You can even take it a little farther by putting the guy in the middle on a board and the gut on the end on a piece of carpet so the frictions are dramatically different.
     
  12. Feb 26, 2008 #11
    Thanks guys.
     
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