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Newton's Laws of Motion

  1. Jul 3, 2006 #1
    Two bodies of masses m1 and m2, are released from the position shown in the figure. If the mass of the smooth-topped table is m3, find the reaction of the floor on the table while the two bodies are in motion. Assume the table does not move.

    I attempt to find out all the forces acting. There's tension of the String, w1 and w2, w3, normal reaction force and maybe friction. But how does movement of m2 has anything to do with the normal reaction force?

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    Last edited: Jul 3, 2006
  2. jcsd
  3. Jul 3, 2006 #2
    did they which of m1 or m2 is heavier?
  4. Jul 3, 2006 #3
    is there any friction present at all?
  5. Jul 3, 2006 #4
    That should be irrelevant.

    Since the problem says that the table is "smooth-topped" we should assume that there is no friction.
  6. Jul 3, 2006 #5
    it would be relevant if friction wus present.
    anyway, i think that if the table stay at one place, then the reactive force of the floor would be g(m1+m2+m3). that wut i think but i am not very sure at alll!
  7. Jul 3, 2006 #6


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    As d_leet states this is irrelavent.
    Last edited: Jul 3, 2006
  8. Jul 3, 2006 #7
    yea, you can the ans to my ques from the problem...sorry....my mistake
  9. Jul 4, 2006 #8
    I am not sure whether friction is present or not. The answer given is ((m1m2/m1+m2)+m2+m3)g. And friction is include in the answer. The friction is m1m2g/m1+m2

    Why is friction involved since the table is smooth? Is there something wrong with the question?

    And yes, it is not stated whether m1 is greater than m2 or not.
  10. Jul 4, 2006 #9


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    The quantity:

    [tex] \frac{m_1 m_2} {m_1+m_2} [/tex]

    is in fact the reduced mass of the masses m1 and m2.

    Hopefully this hint will help you answer the question.
    Last edited: Jul 4, 2006
  11. Jul 4, 2006 #10


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    Friction is needed between the table and the floor, in order to keep the table stationary with respect to the floor.
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