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Newton's third law and free body diagrams

  1. May 27, 2009 #1
    1. The problem statement, all variables and given/known data
    "Three blocks (m1, m2 and m3) are in contact with each other on a frictionless, horizontal surface. A horizontal force is applied to m1.
    http://i4.photobucket.com/albums/y111/kathy_felldown/sb-pic0556.png
    What is the net force on block 1? "

    When drawing free body diagrams for this system, I know that you are only supposed to include forces acting upon THAT object, but I wasn't sure if I always needed to take Newton's third law into account. For example, would the FBD for object m1 have these forces:
    - weight pointing down
    - normal pointing up
    - applied force pointing right
    - force exerted by m2 ON m1... but I have no idea how to calculate this value!

    What confuses me about Newton's third law is that if there is always an equal and opposite force, wouldn't that mean that the two objects that are in contact, wouldn't move at all even if a force is applied in one direction? Because isn't the force that's applied on m2 by m1 equal and opposite to the applied force, since the applied force is what is "forcing" m1 to move m2? Or is the force exerted on m2 by m1 NOT EQUIVALENT to the applied force?


    2. Relevant equations
    Fnet = ma


    3. The attempt at a solution
    I found acceleration by calculating the Fnet = m(total)*a
    a = 17.9 / (1.61 + 3.39 + 4.16) = 1.95m/s/s

    Just for curiosity, I tried the Fnet = (1.61kg)(1.95m/s/s) = 3.14N and it was correct.
    I have no idea why the applied force, 17.9N is not considered in this... OR the force of m2 on m1 (according to Newton's 3rd law)
     
    Last edited: May 27, 2009
  2. jcsd
  3. May 27, 2009 #2
    You seem to have identified all the forces exerted ON m1.
    These are what you need to add when applying Newton's 2nd
    Law to m1.
    The force exerted by m1 on m2 is NOT a force exerted on m1!
     
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