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Homework Help: Newton Third Law general equation

  1. Oct 18, 2006 #1
    I have a picture provided. You can assume that the table is frictionless. I need an expression for the acceleration of m1.
     

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    Last edited: Oct 18, 2006
  2. jcsd
  3. Oct 18, 2006 #2

    HallsofIvy

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    We don't do your homework for you. What are the forces on each mass?
     
  4. Oct 19, 2006 #3
    here is what I have so far:

    Fnetx for m1 is T=m1a
    Fnety for m1 is N-m1g=0

    I guess my real dilemma is i don't really know what the forces on m2 are. I believe there is no acceleration constraint since m2 is on the second pulley. Does that mean that the tension and weight are equal to m2a, with m2 having a different acceleration than m1?
     
  5. Oct 19, 2006 #4

    Doc Al

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    OK. (All you need here is Fnetx.)

    Treat m2 and its pulley as one object. I see 3 forces acting on it.
    Of course there is an acceleration constraint, which you need to figure out. When m2 drops X distance, how far must m1 move?

    You will apply Newton's 2nd law to m2, just like to m1. Yes, a2 is different from a1, but there is a simple relationship.
     
  6. Oct 19, 2006 #5
    so a2 =-a1? I didn't understand if the mass on the pulley made it different from other problems. So there is tension T from the rope attached to the wall, the weight force m2g, and the tension from m1 which is m1a1?
     
  7. Oct 19, 2006 #6

    Doc Al

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    No. (But the minus sign is good.) Answer my question earlier: When m2 drops X distance, how far must m1 move? (Get a piece of string and figure it out.)

    The tension will be the same throughout the rope. The only things exerting a force on m2 are: (1) the ropes; (2) gravity.

    Set up an equation for each mass. (After you've found the acceleration constraint.) Then solve them together to find the accelerations.
     
  8. Oct 19, 2006 #7
    ok, so I think I got it now.

    a2 = (-1/2)a1

    so u have for m2 in y direction:
    2T-m2g=m2a2 ; since there are two ropes pulling on m2

    i already know that T =m1a1 and the acceleration constraint, so plugging this in i get:
    2m1a1-m2g=-m2a1/2

    4m1a1-2mg=-m2a1

    a1(4m1 + m2)= 2mg

    a1=2mg/(4m1+m2)
    does this look ok?
     
  9. Oct 19, 2006 #8

    Doc Al

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    Excellent!
     
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