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Newton's Third Law -- Normal vs. Tensile as Opposite of mg?

  1. Jun 28, 2015 #1
    Hello, I am returning to college after a ten year hiatus and am taking an online course on edx to try and refresh my knowledge a bit before the fall. I read a few other posts on Newton's Third law, but it seems I am falling short on this one concept.

    In the case of an object, m_1, which is at rest on some frictionless table. There is a rope connecting this object to another object, m_2, which is hanging over the edge. To be clear, the rope connects m_1 to m_2, m_1 is on the table and m_2 hangs in the air off the edge of the table. There could be a pulley if you like on the edge of the table assisting with the direction change of the rope.

    Assume I am drawing the free body diagrams for m_1 and m_2 separately, *not* as a single system. Also for simplifying explanation, assume a cartesian coordinate system with +y in the vertical at 90 and -y at 270, and +x at 0 and -x at 180.

    I place a downward arrow on each diagram for the F_mg. In the case of m_1, I have a opposite Normal force with an upward arrow and an arrow in the +x direction for the tension force on m_1 by the rope. It makes sense to me that the Normal force in this case is opposing the gravitational force.

    Why is it correct that m_2 only has the downward F_mg force (as mentioned) and an upward T force on m_2 from the rope? Why is the normal force not included here? As it stands, I am just "playing along" that the force just needs something to oppose it, so we have the Normal force and with m_2, the Tension force is sufficient, but I don't really understand.

    Is the Normal force only present in cases where there needs to be a "mathematical correction" and could any other provide the opposing force? What if there was a person pushing down on a box? There would be the arrow downward for F_mg, and would the applied contact force be sufficient as an opposing force where as there would be no normal force?

    Thanks in advance!
     
  2. jcsd
  3. Jun 28, 2015 #2

    Orodruin

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    The normal force is not the third law force pair with the gravitational force. Force pairs always act on different bodies and the third law partner of the gravitational force on one of the masses is the force with which that mass is acting on the Earth.

    Normal forces only appear when they are needed to satisfy physical constraints. For example, a force is necessary to stop m1 from falling through the table. The third law partner of this force is the force with which the mass m1 pushes the table.

    Bottom line is that you will never draw any third law pairs on the same free body diagram. If you draw two free body diagrams for two objects which affect each other with a force, the third law partner of the force on object 1 from object 2 will be an equal an opposite force on object 2 from object 1.
     
  4. Jun 28, 2015 #3
    This is exactly what I was confused about. Thank you very much!

    It sounds like I should read up a bit more on Third Pairs. Any other suggestions?
     
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