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Atwood's Machine with unknown masses

  1. Sep 28, 2009 #1
    1. The problem statement, all variables and given/known data

    A simple Atwood's machine uses two masses m1 and m2. Starting from rest, the speed of the two masses is 8.7 m/s at the end of 7.4 s. At that time, the kinetic energy of the system is 61 J and each mass has moved a distance of 32.19 m.

    Find the value of the heavier mass.

    Find the value of the lighter mass.

    2. Relevant equations

    a = (m2 - m1) / (m1 + m2)g
    Ke = 1/2mv2

    3. The attempt at a solution

    61 = 1/2m1v2 + 1/2m2v2
    61 = 1/2(8.7)2(m1+m2)
    61 = 37.845(m1+m2)
    m1+m2 = 1.612

    a = 8.7 / 7.4 = 1.176

    1.176 = (m2 - m1) / (m1 + m2)g
    10.976m1 = 8.624m2
    m1 = .786m2

    .786m2 + m2 = 1.612
    2m2 = 2.051
    m2 = 1.025

    m1 + 1.025 = 1.612
    m1 = .587

    I was sure that I solved it correctly, perhaps my algebra for solving the acceleration equation is incorrect.

    Thanks for any help.
     
  2. jcsd
  3. Sep 28, 2009 #2
    -------> Shouldn't this be 1.786m2=1.612?
     
  4. Sep 28, 2009 #3
    So you are saying my algebra for solving the acceleration equation is wrong? I got .786, not 1.786.
     
  5. Sep 28, 2009 #4
    It seems to me so.
    0.786m2 + m2 = 1.612
    (0.786+1)m2=1.612
    1.786m2=1.612
    Could you write down exact solution?
     
  6. Sep 28, 2009 #5
    I got .786 from this:

    1.176 = (m2 - m1) / (m1 + m2)g
    10.976m1 = 8.624m2
    m1 = .786m2
     
  7. Sep 28, 2009 #6
    the problem is: how do you get from

    [tex].786m_2 + m_2= 1.612 [/tex]

    to

    [tex] 2m_2 = 2.051 [/tex]
     
  8. Sep 28, 2009 #7
    [tex] 1.176 = \frac {m_2 - m_1} {m_1 + m_2} g [/tex]

    [tex] 10.976m_1 = 8.624m_2 [/tex]

    this step is also wrong
     
    Last edited: Sep 28, 2009
  9. Sep 28, 2009 #8
    I divided 1.612 by .786 then I thought I could combine like terms.

    I was kinda finicky on this step. Could you give me a hint on how to solve it algebraically?
     
  10. Sep 28, 2009 #9
    you have to divide ALL terms by .786. THat would give you [itex] m_2 + m_2/0.786 = 1.612/0.786 [/itex] I don't think that helps.

    [/QUOTE]

    multiply both sides by [itex] \frac {m_1 + m_2} {g} [/itex] (you already know what m_1+m_2 is)

    combine what you get with [itex] m_1 + m_2 = 1.612 [/itex]
     
  11. Sep 28, 2009 #10
    multiply both sides by [itex] \frac {m_1 + m_2} {g} [/itex] (you already know what m_1+m_2 is)

    combine what you get with [itex] m_1 + m_2 = 1.612 [/itex][/QUOTE]

    Okay, so I did this:

    1.176 = (m2 - m1) / (m1 + m2) g
    1.176 (m1 + m2) g = m2 - m1
    1.176 (15.7976) = m2 - m1
    18.578 = m2 - m1

    I don't know if this is correct because I multiplied m1 + m2 and g. I wasn't sure if I should divide by g or multiply since it is in the denominator.
     
  12. Sep 28, 2009 #11
    Okay, so I did this:

    1.176 = (m2 - m1) / (m1 + m2) g
    [/QUOTE]

    note that it's [tex] a = \frac {m_2 - m_1} {m_1 + m_2} g [/tex]

    not

    [tex] a = \frac {m_2 - m_1} {(m_1 + m_2) g} [/tex]

    so you have to multiply both sides by m_1 + m_2 and then divide by g
     
  13. Sep 28, 2009 #12
    Okay, here's what I came up with.

    1.176 = ((m2 - m1) / (m1 + m2))g

    (1.176 * 1.162) / g = .19344 = m2 - m1

    How does that look? I do not know what you mean when you say add it to m1 + m2 = 1.612.

    Thanks for the help.
     
  14. Sep 29, 2009 #13
    adding 2 equations:

    if you have A = B and C=D then A+C = B+D

    now you have .19344 = m2 - m1

    and m1 + m2 = 1.162
     
  15. Sep 29, 2009 #14
    A = .19344
    B = m2 - m1
    C = 1.162
    D = m1 + m2

    .19344 + 1.162 = (m2 - m1) + (m1 + m2)

    How does that set-up look?
     
  16. Sep 29, 2009 #15
    Looks OK. you can now add up the right and the left side of this equation.
     
  17. Sep 29, 2009 #16
    .19344 + 1.162 = (m2 - m1) + (m1 + m2)

    1.35544 = m2 + m1

    m1 = m2 - 1.35544


    Using the equation from Ke and Pe:

    m1 + m2 = 1.612

    m2 - 1.35544 + m2 = 1.612

    m2 + m2 = 2.96744

    2m2 = 2.96744

    m2 = 1.48372

    So m1:

    m1 + 1.48372 = 1.612

    m1 = .12828

    Is my math okay?
     
  18. Sep 29, 2009 #17
    the only reason for adding the 2 equations was that m1 would disappear.

    m1 - m2 + m1 + m2 = ?



    if you have 2 equations like 2x - y = 7 and 3x + y = 4 it's often easy to eliminate one
    of the variables by adding or subtracting them.
    if you add them you get 2x - y + 3x + y = 7 + 4 so you get 5x = 11 and x = 11/5

    then you substitute x = 11/5 in 2x - y = 7 to get y: 2 (11/5) - y = 7 => y = 22/5 -7 = 22/5 - 35/5 = - 13/5 = -2.6

    adding the equations works because y appears in one of them and -y in the other, so you
    know y will disappear.
     
  19. Sep 29, 2009 #18
    I understand now, thanks. I also went to my professor to discuss this question and he explained an easier way to do it, but similar.

    Thanks for all the help.
     
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