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Acting and reacting force and their work

  1. Oct 10, 2008 #1
    Is it true that when acting force apply work, the reacting force also apply one?
    Is it true that the acting and reacting force always apply work at the same time?
    Last edited: Oct 10, 2008
  2. jcsd
  3. Oct 11, 2008 #2


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    Science Advisor

    In Newton's theory, two masses M and m attract each other by gravity. We can call the force on M the "action" and the force on m the "reaction" (interchanging the names is fine too). Action and reaction are always equal in magnitude and opposite in direction. If M is very much bigger than m, in a given amount of time, M will barely move while m will pick up a lot of speed and move a big distance. So the gravitational field does less work on M is than it does on m.
  4. Oct 11, 2008 #3


    Staff: Mentor

    Remember W = f.d, and by Newton's 3rd law the f on two interacting bodies is equal, but there is nothing to constrain that the d must always be equal. Atyy gave a good example where the force is equal but not the d.
  5. Oct 11, 2008 #4
    well, i think you all misunderstood my question.

    my question is whether they apply work at the same time, not whether they apply the same(or equal) wok. In another words,if one of them apply work,is it true that the other one also apply one? OR if one don't apply,is it true that the other don't,either?

    Forgive my poor english expression ability. Thanks.
  6. Oct 11, 2008 #5


    Staff: Mentor

    The f is applied at the same time, but if d=0 then there is no work.
  7. Oct 13, 2008 #6
    in fact, this is the first step of my analysis.but i just coundn't find an example.
    well,i've got one now.
    For a central force system with 3 bodies(let's say 3 charge, or 3 celestial bodies),and with some proper conditions, we can keep the middle one stationary, so the d=0.
  8. Oct 14, 2008 #7
    Let's look at the simple case of an apple falling to earth (this is a better two-body problem... why put in a third if not needed for the concept?). Say the apple weighs 1 N and falls 1 m. The work done on the apple is 1 J. The weight is the gravitational force (mg). When the apple is falling, it also has a gravitational force on the Earth, with a value of 1 N (recall g=GM/r^2 (where g is the gravitational acceleration on the surface of the Earth, G is the gravitational constant, r is the radius of the earth plus the negligible height of the apple, and M is the mass of the earth) , so the weight is mg = GmM/r^2 . This action-reaction pair of forces is equal and opposite, but the force of the apple, however, deflects the Earth towards the apple in a negligible manner (the Earth has other forces on it, the mass is large, etc.). Therefore NO work is done on the earth.

    Always remember, work has to do with the process of exchanging energy. In this case, the Earth's gravitational field does work to change gravitational potential energy to kinetic energy. The energy of the earth doesn't change, so no work is done on it.
  9. Oct 14, 2008 #8
    but i can't tolerate such an approximation,negligible never equal to NO
    i agree with you but the last word. i am wondering how you deduce such a conclusion.
    if you can accept that the earth and the apple are equivalent,or there is an symmetry here, and form your point :"In this case, the Earth's gravitational field does work to change gravitational potential energy to kinetic energy.",we can also obtain the conclusion:the apple's gravitational field does work to change gravitational potential energy to kinetic energy of the earth.

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