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B Newton's 3rd law and work

  1. Aug 3, 2016 #1
    Newton`s 3rd law says, every action has an equal but opposite reaction. So when we push something and it moves (towards right, left, wherever). Doesn't it violates Newton`s 3rd law because the object being pushed should push us back at equal force but opposite in direction.
     
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  3. Aug 3, 2016 #2

    BvU

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    It does ! You feel the resistance, don't you ?
     
  4. Aug 3, 2016 #3
    Yes, but then it should remain there all the time, no matter how much energy (force) I apply. It should always cancel it out and do not move?
     
  5. Aug 3, 2016 #4

    BvU

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    I was referring to the situation where it moves. If it moves at a constant velocity, the only force you need to exercise is to overcome the friction with the floor (Newton first law).
     
  6. Aug 3, 2016 #5

    A.T.

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  7. Aug 5, 2016 #6

    VMP

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    Suppose that Newton's 3rd law isn't true i.e.
    [tex]\vec{F}_{1}\neq -\vec{F}_{2}[/tex]
    where
    [tex]\vec{F}_{1}[/tex]
    is the force that first box exerts on second and
    [tex]\vec{F}_{2}[/tex]
    is the force that second box exerts on the first box.
    We can mark two points one representing the box 1 of mass [tex]m_{1}[/tex] and the other representing the box 2 of mass [tex]m_{2}.[/tex]
    If we apply force* [tex]\vec{F}[/tex] then acceleration of system [tex]m_{1}+m_{2}[/tex] will be: [tex]\vec{a}=\frac{\vec{F}}{m_{1}+m_{2}}.[/tex]

    Box 1 is accelerating with acceleration [tex]\vec{a}_{1}=\frac{\vec{F}+\vec{F}_{2}}{m_{1}}[/tex]

    and box 2 is accelerating with acceleration [tex]\vec{a}_{2}=\frac{\vec{F}_{1}}{m_{2}}.[/tex]

    Based on daily observations we know that [tex]\vec{a}=\vec{a}_{1}=\vec{a}_{2}\; (1)[/tex]
    I haven't recorded any physical box going through another one without having any effect.

    From equation (1) we can easily see that:

    [tex]\vec{a}=\vec{a}_{1}[/tex]
    [tex]\vec{a}=\vec{a}_{2}.[/tex]

    Giving us a system of equations:

    [tex]\frac{\vec{F}}{m_{1}+m_{2}}=\frac{\vec{F}+\vec{F}_{2}}{m_{1}}[/tex]
    [tex]\frac{\vec{F}}{m_{1}+m_{2}}=\frac{\vec{F}_{1}}{m_{2}}[/tex]

    With a little bit of algebra it can be shown that [tex]\vec{F}_{1}= -\vec{F}_{2}.[/tex]
    This is a contradiction with our initial supposition,
    our starting premise was wrong.

    Okay, now that the "formal" part is out of the way, I hope you can see that Newton's 3rd law just says "objects don't pass through each other".

    *If we apply force of magnitude F the system will react with force equal in magnitude and opposite in direction ( no worries :) ).

    Cheers
     
  8. Aug 5, 2016 #7

    sophiecentaur

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    It isn't Newton's Law that's wrong. It's how you are trying to apply it. N3 simply states that you can't push against 'nothing'; if you are pushing against something then the reaction force is the same as the force you are applying. This is true if there is no resulting relative movement or if the object is totally free to move in space. Matters of Equilibrium, where forces are equal and opposite have nothing to do with N3.
    Airing of this problem with N3 is a regular occurrence on PF. It seems to be one of those things that we all have to struggle with and then we see the light. :smile:
     
  9. Aug 5, 2016 #8
    If you push a cart forward, and you are standing on rollerskates, you will move backward. If you stand firmly on the ground and you push the cart forward, then you will push the ground backward (a tiny little bit).
     
  10. Aug 5, 2016 #9
    This is a common misconception. It can be solved simply by realizing that the two opposing forces act on different objects. The reaction force from the object acts on you, but your force acts on the object, so they don't cancel out, because they're not acting on the same thing.
     
  11. Aug 5, 2016 #10

    sophiecentaur

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    Forget the roller skates. How do you know you are pushing the cart? You can Feel the Force against your hands. That's the reaction force, whether you slide back on skates or stay with your feet stuck on the floor. Of course, the actual values of those forces may well different. You couldn't produce as much force if you were on the skates and it wouldn't last for long, either.
     
    Last edited: Aug 6, 2016
  12. Aug 23, 2016 #11
    I think the main problem is that there is a fundamental confusion over "net force". you push on a box and it pushes back in space with equal force (jets on the back) there is no movement. but, if there are no jets on the back, the block still pushes back with a force equal to that of what it takes to accelerate it based on the applied force. the mass x the acceleration will equal that force applied and the box will move and accelerate.
    is that a fair explanation?
    edit: or if there is no acceleration and it is moving, then the force will equal what the friction on the box is. OR, there is no more force applied and it is still moving, goes back to N1, where something in motion continues in motion until ...................
     
  13. Aug 23, 2016 #12

    jbriggs444

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    "Net force" is a concept that arises in Newton's second law. F=ma where F is the net force on a particular object, m is the mass of that object and a is the acceleration of that object.

    Newton's third law says that forces come in pairs: a on b and b on a. But only one member of the pair acts on a particular second-law object. For the purposes of Newton's second law, the "net force" only includes one member of the pair, not both. (*)

    Any consideration of an object pushing back because of "jets on the back" misses the point. Forces come in pairs regardless of whether there are jets on the back of one or both.

    (*) If you try to lift yourself by your own bootstraps, both members of the third law pair act on the same object and do indeed result in a "net force" of zero. This suggests why it is futile to try to lift yourself in such a manner.
     
  14. Aug 23, 2016 #13

    sophiecentaur

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    The confusion is in failing to realise the N2 and N3 are two different laws. Action and reaction are always equal. That has nothing at all to do with whether or not two Non N3 forces are in equilibrium and happen to be equal.
    Anyone who finds an apparent paradox or self contradiction here would be advised to look at their own understanding rather than looking for flaws that no one has yet managed to spot. You have to read the formal wording of the laws and seeing what good text books tell you, rather than trying to use the random sorts of interpretations you can find on the Internet.
     
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