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Pulley -- Solving for an unknown mass

  1. Jul 8, 2015 #1
    1. The problem statement, all variables and given/known data


    If I have a mass attached to each side of an atwood's machine and then add some mass, x, to one side, how could you determine x, the mass added to the atwood's machine?



    2. Relevant equations




    F = ma I would guess.



    3. The attempt at a solution



    Attempt 1:

    [itex] F_{2net}= m_{2}g - T_{2} =[/itex]

    [itex] (m_{2} + x)a_{net} = m_{2}g - T_{2} [/itex]

    [itex] = (m_{1})(a_{net}) + x(a_{net}) = m_{2}g - (T_{2}) [/itex]

    [itex] = (m_{1})(a_{net}) + x(a_{net}) = m_{2}g - (T_{2}) [/itex]

    [itex] = (m_{1})(a_{net}) + x(a_{net}) = m_{2}g - (m_{2}g - m_{2}a_{net}) [/itex]

    [itex] = w_{2} = 0 [/itex]

    Everything cancels



    Attempt 2:


    [itex] F_{2net}= m_{2}g - T_{2} [/itex]

    [itex] (m_{2} + x)a_{net} = (m_{2} +x)g - ((m_{2}+x)g - (m_{2} + x)a_{net}) [/itex]

    [itex] (m_{2}a_{net} + xa_{net}) = (m_{2}g +xg) - ((m_{2}g+xg) - (m_{2}a_{net} + xa_{net})) [/itex]


    Which cancels out to 0 = 0
     
  2. jcsd
  3. Jul 8, 2015 #2

    Doc Al

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    Staff: Mentor

    Not quite clear what you are doing. First question: Do you start with equal masses on each side?
     
  4. Jul 8, 2015 #3
    Suppose I have 5kg, m1, on the left side and 10kg, m2, on the right side. Then, a mass of unknown value, m?, is added to the 10kg side. I guess I could observe the change in net acceleration and somehow determine how much mass was added to the machine.
     
  5. Jul 8, 2015 #4

    Doc Al

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    Sounds reasonable. To calculate the acceleration in each case, apply Newton's 2nd law to each mass separately.
     
  6. Jul 8, 2015 #5
    [itex]a_{net1} = \frac{m_{1}-m_{2}}{m_{1}+m_{2}}g[/itex]

    [itex]a_{net2} = \frac{m_{1}-(m_{2}+x)}{m_{1}+(m_{2}+x)}g[/itex]

    Is it only possible to find x if we experimentally observe a_{net1} and a_{net2} or is knowing just a_{net1} enough to solve for x?
     
  7. Jul 8, 2015 #6

    Doc Al

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    Look at your equations. The first one (for ##a_1##) doesn't not even involve x.
     
  8. Jul 8, 2015 #7
    Makes sense. When I solve for x (in the second equation involving x), the result is:

    [itex] x = \frac {a_{2}(m_{1}) + a_{2}(m_{2}) - m_{1}g + m_{2}g}{g-a_{2}}[/itex]

    Does this look correct?
     
  9. Jul 8, 2015 #8

    Doc Al

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    Just to keep from going nuts, realize that this equation:
    Assumes that ##m_1## is the heavier one. (Otherwise the acceleration would be negative.)

    Since you want to add x to the heavier side, rewrite your second equation accordingly. But I think you've got the right idea.
     
  10. Jul 8, 2015 #9
    Ah yes, appreciate the note on the arrangement.


    [itex] x = \frac {a_{2}(m_{2}) + a_{2}(m_{1}) - m_{2}g + m_{1}g}{g-a_{2}}[/itex]

    Is this more like it?
     
  11. Jul 8, 2015 #10
    You know what, I just realized that post #9 is not correct either. The idea is right, as you said, but not quite the right arrangement. I have to start with ((m2 +x) - m1)/ ((m2 + x) + m2) = a_net.

    Thank you.
     
  12. Jul 9, 2015 #11

    Doc Al

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    You have a typo in the denominator.

    Here, m2 is the heavier mass and you are adding x to it. Good! Now you should be able to solve for x.
     
  13. Jul 10, 2015 #12
    Thank you, Doc Al. Your guidance on this topic was very helpful.
     
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