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Newton's Third Law Question

  1. Feb 7, 2008 #1
    Ok, i'm getting a lot of mixed messages about the reactionary forces involved with the 3rd law. if i push a box at a constant speed with 100N of force, is the friction that resists the motion the opposite but equal (100N) reaction force? isn't the friction force LESS than the push force because the box is moving? is the reaction force something different (i.e. the box itself pushing back)?
     
  2. jcsd
  3. Feb 7, 2008 #2

    G01

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    Yes, the reaction force is NOT the friction force. As you said, the reaction force of you pushing on the block is the force of the block pushing back on you.

    Remember that the action force and the reaction force act on different objects. The action force acts on the block, while the reaction force acts on the person, the source of the action force. Consider this, and tell me why the force of friction can't be the reaction force for the pushing force.
     
    Last edited: Feb 7, 2008
  4. Feb 7, 2008 #3
    No. I would draw a free body diagram, and review your class notes one more time and think about it, and post any new thoughts you might have.

    Hint: Think about things one at a time. You are confusing yourself because you are trying to think of many senarios all at once. Think methodically.
     
    Last edited: Feb 7, 2008
  5. Feb 7, 2008 #4
    i guess my main hangup is that i don't see how any object moves at all if an equal force opposes the push force, plus friction and inertia
     
  6. Feb 7, 2008 #5
    scratch that, i understand when the object pushing is greater in mass, but otherwise, i'm still confused
     
  7. Feb 7, 2008 #6
    is the motion of the object simply a byproduct when the force of the push becomes greater than the force of friction, and regardless of the motion, the box will always push back at the same force as the push action force?
     
  8. Feb 7, 2008 #7
    I know for a fact your physics book does not make this statement. Crack it open.

    Slow down speed racer. One thing at a time. Look for the answers to your questions in your book one at a time. Not all at once.
     
  9. Feb 7, 2008 #8
    i have been reading, i've read every chapter up to this question and the only reason i've come here is because i've exhausted myself pouring over the pages and not finding answers pertaining to this example and i need a human source that is more flexible and able than text.
     
  10. Feb 7, 2008 #9
    Tackle your questions one at a time. Type what your book says about action/reaction forces. Look at the picture next to the description very carefully.
     
  11. Feb 7, 2008 #10
  12. Feb 7, 2008 #11
    i think this excerpt relates the best to the concept "the tires of a car push against the road while the road pushes back on the tires...the reaction force is what accounts for motion in this example. This force depends on friction; a person or car on ice, for example, may be unable to exert the action force to produce the needed reaction force. Neither force exists without the other." there is no picture.
     
  13. Feb 7, 2008 #12
    so obviously the action-reaction is where two objects (me and the box) touch, with our forces pushing at each other. But my problem is with the sustained motion and the force of friction. The friction is needed in order for the action-reaction to take place. Maybe the friction merely reduces the force involved in the action-reaction, but when the force is no longer canceled out, it moves....
     
  14. Feb 7, 2008 #13
    You should review newtons FIRST law.
     
  15. Feb 7, 2008 #14
    yeah, isn't what i just said basically about inertia? where the box has moving equilibrium once the pushing force equals the friction force. It's in equilibrium at that point because no net force is acting on the box because friction and push cancel each other.
     
  16. Feb 7, 2008 #15
    ok, so the box is in dynamic equilibrium...so what's the difference between the force of the box pushing back and the force of friction?
     
  17. Feb 7, 2008 #16

    russ_watters

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    You just asnwered that question. It's in dynamic equilibrium.
     
  18. Feb 7, 2008 #17
    What do you mean?
     
  19. Feb 8, 2008 #18
    both the force of friction and the reaction force equal and oppose (essentially cancel out) the push force, how can there be two opposing forces (one acting on the box and one acting on the pusher)?
     
  20. Feb 8, 2008 #19
    This statement is not true. What is the push force?

    I know you saw a picture of that football being punted. I expect you to answer this on your own.
     
  21. Feb 8, 2008 #20
    the push force is the force of the interaction between the box and the pusher going both ways. the friction force is the force of the interaction between the box and the ground.
     
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