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Newton's 3rd law and force equilibrium

  1. Nov 18, 2015 #1

    Assume a rock lying on the ground. Furthermore, let's assume both the rock and earth to be rigid bodies. I'm a bit confused about force equilibrium and Newton's 3rd law here. There's earth's gravity acting on the rock, and by Newton's 3rd law (or just by looking at the law of gravity) the rock's gravity acts on earth. But on the other hand, since the rock is at rest, the gravitational force pulling it towards the c.o.m. of earth needs to be balanced by a normal force of the ground acting on the rock (which is probably due to electrostatic interactions between the two objects).

    So what is the reactio to this normal force of the ground on the rock? Isn't it the rock pushing on the ground because of gravity? But then this force would have two opposing forces, first the attraction of the c.o.m. of the earth and second the normal force of the ground. I guess there's a flaw in my train of thought, but where?

  2. jcsd
  3. Nov 18, 2015 #2
    Yes. There are two forces on the rock: The gravitational attraction by the earth + the normal force by the ground. There are reactions to these two, the gravitational attraction by the rock on the earth + the normal force exerted (downwards) by the rock on the ground. The two gravitational forces are an action-reaction pair, and the two nrmal forces are an action-reaction pair.
  4. Nov 18, 2015 #3


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    It's not clear to me why you're getting confused. Gravity and the normal force are the equal and opposites that Newton's 3rd law demands.

    If there were no gravity, then the rock would sit on the Earth without exerting a force and the normal force would be zero.

    Note that bodies are not rigid at the molecular level, so gravity will have deformed the rock and the Earth to some degree, which is where the electrostatic repulsion comes from.
  5. Nov 18, 2015 #4
    Those two statements seem to contradict each other, don't they?
  6. Nov 18, 2015 #5
    No, gravity and normal force are not part of a 3rd law pair.
    The fact that gravity on the rock is balanced by the normal force on the rock does not make these two a pair in the sense of 3rd law.
    The fact that they act on the same body is a hint. The forces in a 3-rd law pair newer act on the same body.
  7. Nov 18, 2015 #6


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    Very true!
  8. Nov 18, 2015 #7
    I think I get it now, I forgot the normal force of the rock.

    A related question: Assume two mass points separated by a spring of some spring constant. How do I correctly calculate how much the spring is compressed?
  9. Nov 18, 2015 #8
    Why is the spring compressed?
  10. Nov 18, 2015 #9
    Because the two mass points attract each other.
  11. Nov 18, 2015 #10
    You need to know the spring constant. Then you can balance the gravitational attraction with the force exerted by the compressed spring. This gives the equilibrium configuration.
  12. Nov 18, 2015 #11
    You also need the length of the spring in the relaxed (uncompressed) state.
  13. Nov 19, 2015 #12
    Yes, suppose those quantities are given. What's unclear to me is if I need to consider the forces from both sides on the spring?
  14. Nov 19, 2015 #13


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    No, the F in the spring formula is the force at one side.
  15. Nov 19, 2015 #14
    Correct. In fact, there are always two forces, one on each side of the spring, otherwise, the spring itself would run away.
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