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Modified Atwood Confusion

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data
    A Modified Atwood's Machine has a 10 N cart on a frictionless, horizontal track with a 10 N hanging weight attached to a string connecting the two weights. A second track is set up with the hanging weight replaced by a person who can maintain a 10 N pull on the string (as measured with a force probe). Which set-up has the greater acceleration?


    2. Relevant equations
    N/A


    3. The attempt at a solution
    Wouldn't they have equal acceleration since the force pulling them is the same?
     
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  3. Oct 22, 2008 #2

    PhanthomJay

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    Same externally applied force, but one system has twice the mass of the other. Draw a free body diagram of each mass for the first case and identify forces and solve for acceleration using newton 2. Then look at case 2. Are the accelerations equal?
     
  4. Oct 22, 2008 #3
    [tex] T = ma [/tex]
    and for the hanging block
    [tex] -T + mg = ma [/tex]
    [tex] T = mg - ma[/tex]

    plugging it in to the first one:

    [tex] mg - ma = ma[/tex]
    [tex] a = g/2[/tex]

    For case 2 we would have:

    [tex] T = ma [/tex]

    [tex] -T + F_{applied} = ma[/tex]
    [tex]T = F_{applied} - ma[/tex]

    plugging in for the first 1

    [tex]F_{applied} - ma = ma[/tex]
    [tex]a = \frac{10}{2*m}[/tex]

    Both those equations yield the same acceleration. So they would be equal. Is that correct?
     
  5. Oct 22, 2008 #4

    PhanthomJay

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    This is correct
    correct, for the cart
    but for this analysis of the force on the hanging rope, what is the value of 'm' to use? Is there any mass involved here?
     
  6. Oct 22, 2008 #5
    I see, I forgot to differentiate which mass was which. There is an applied force of 10N, but no mass is used, how does that work?
     
  7. Oct 22, 2008 #6

    PhanthomJay

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    well, you can use your equation [tex]T = F_{applied} - ma[/tex] if you want, and set m=0 to solve for T. Then solve for the acceleration by plugging T into your equation for the cart. That's one way of doing it.
     
  8. Oct 22, 2008 #7
    I think I got it. There would be more acceleration for the person pulling. Right?
     
  9. Oct 22, 2008 #8

    PhanthomJay

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    Yes. A simpler way is to realize that the applied force at the hanging end (10N) is just the tension force in the string (10N), which is the same tension force accelerating the cart (thus, a=g for this case).
     
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