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Accelerating Atwood's Machine Problem

  1. Sep 22, 2004 #1
    I am having difficulty solving Problem 105 of Chapter 4 from Physics for Scientists and Engineers, 4th edition by Paul A. Tipler:

    The pulley in an Atwood's machine is given an upward acceleration [tex]\overrightarrow{a}[/tex]. Find the acceleration of each mass and the tension in the string that connects them.

    Though the problem doesn't say so explicity, we may assume that the pulley is massless and frictionless.

    For those who are unfamiliar with what an Atwood's machine looks like, try browsing the following link, which shows how to solve this problem in the absence of acceleration:


    Calling the two masses in the Atwood's machine [itex]m_1[/itex] and [itex]m_2[/itex], calling their respective accelerations (which are not identical) [itex]a_1[/itex] and [itex]a_2[/itex], and assuming that [itex]m_2 > m_1[/itex], I can easily obtain the following two equations for the tension, [itex]T[/itex]:

    [tex]F_{ym1} = m_1a_1 = T - m_1g[/tex]
    [tex]\Rightarrow T = m_1a_1 + m_1g[/tex]

    [tex]F_{ym2} = m_2a_2 = m_2g - T[/tex]
    [tex]\Rightarrow T = m_2g - m_2a_2[/tex]

    However, I haven't the slightest clue what the third equation is, though it obviously involves the acceleration [itex]a[/itex].

    Can someone please help?

    The answers for the two accelerations and the tension as given in the back of the book are:


    [tex]a_1 = \frac{m_2 - m_1}{m_1 + m_2}g + \frac{2m_2}{m_1 + m_2}a[/tex]
    [tex]a_2 = \frac{m_1 - m_2}{m_1 + m_2}g + \frac{2m_1}{m_1 + m_2}a[/tex]
    [tex]T = \frac{2m_1m_2}{m_1 + m_2}(g + a)[/tex]

    I should also note that when I take the answer for [itex]a_1[/itex], plug it into my first equation for [itex]T[/itex], and simplify, I do get the book's answer for the tension [itex]T[/itex]. Therefore, I am confident that the two equations I have derived are indeed correct.
     
  2. jcsd
  3. Sep 23, 2004 #2
    Hey...you do not need anymore equations...you wrote the ones you need already...

    remember how: m1a1 = T - m1g

    Well, plug the value of T you solved for, then simplify it (simplify and divide by m1) and you will have your a1 value. Repeat for a2.

    I did this real quick and got the right answer, so it does work.
     
    Last edited: Sep 23, 2004
  4. Sep 23, 2004 #3
    This is a 'simple' problem in using Newton's 2. law in an accelerating reference frame. You have to include a "ficticious force" somewhere to get the proper answers. I did a similar problem to this before (actually it was more complicated), but I still don't understand how to plug in this "ficticious force".
     
  5. Sep 23, 2004 #4
    What value of T?
     
  6. Sep 23, 2004 #5
    Oh.....sorry about that...I misread what NoPhysicsGenius wrote at the bottom...when he said he got the correct answer for T...I did not pay attention to the fact that he used an answer to get it...lol...nm my previous post then :)
     
  7. Sep 23, 2004 #6

    ehild

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    First, to avoid confusion, it is better to decide about positive direction. Let it be upward. In this case the gravitational force is negative and the tension of the rope means positive force for both masses.
    Think what would be the relation between the accelerations of both masses if the pulley were in rest with respect to the ground. The length of the string is constant. As much one mass raised up the same amount should the other one descend: the displacement of the second mass would be the opposite of the displacement of the first mass. The same is true for their velocity and for their acceleration:

    [tex]a_2 = -a_1 [/tex]


    If the pulley accelerates its accelerations adds up to that of the masses. Their relative accelerations with respect to the pulley should be [tex]a_r\mbox{ and } -a_r[/tex].

    [tex]a_1 = a +a _r[/tex]
    [tex]a_2 = a - a_r[/tex]
    [tex]F_{ym1} = m_1a_1 = T - m_1g[/tex]
    [tex]F_{ym2} = m_2a_2 = T - m_2g[/tex] .

    ehild
     
  8. Sep 23, 2004 #7
    If this is true, then the book's answer is wrong.

    According to the book, we have:

    [tex]a_1 = \frac{m_2 - m_1}{m_1 + m_2}g + \frac{2m_2}{m_1 + m_2}a[/tex]

    Applying [itex]a_2 = -a_1[/itex] would then give:

    [tex]a_2 = \frac{m_1 - m_2}{m_1 + m_2}g - \frac{2m_2}{m_1 + m_2}a[/tex]

    However, the book gives the following answer for [itex]a_2[/itex]:

    [tex]a_2 = \frac{m_1 - m_2}{m_1 + m_2}g + \frac{2m_1}{m_1 + m_2}a[/tex]

    Note that these two expressions for [itex]a_2[/itex] can only be equal if [itex]m_2 = -m_1[/itex], which is nonsensical.

    So there's something wrong here. And I believe it is that your assumption that the two accelerations will have equal magnitudes (but be opposite in direction) will be true if and only the masses [itex]m_1[/itex] and [itex]m_2[/itex] are equal. However, we are assuming [itex]m_2 > m_1[/itex], so that the two masses are not in a state of equilibrium, and the accelerations [itex]a_1[/itex] and [itex]a_2[/itex] will have unequal magnitudes.
     
  9. Sep 23, 2004 #8
    Solving for the tension, we would then obtain the following equations, only the first of which is consistent with what I derived for the tension:

    [tex]T = m_1a_1 + m_1g[/tex]
    [tex]T = m_2g + m_2a_2[/tex]

    In the link Atwood's machine (which I gave in my original post for the case of a nonaccelerating Atwood's machine), the instructor recommends the following, which is still applicable for the case of an accelerating Atwood's machine:

    Think about what the system will do. If the system is released from rest, the heavy mass will accelerate down and the lighter one will accelerate up.

    Align the coordinate systems with the acceleration. Each mass has its own coordinate system, but they must be consistent.

    Take +y down for mass [itex]m_2[/itex].
    Take +y up for mass [itex]m_1[/itex].

    Therefore, I believe that the original equations that I gave for the tension are the correct ones.

    Namely:

    [tex]T = m_1a_1 + m_1g[/tex]
    [tex]T = m_2g - m_2a_2[/tex]
     
  10. Sep 23, 2004 #9
    I don't even know what the expression for the fictitious force is, much less how to plug it in! :confused:

    Can anyone please lend some assistance?
     
  11. Sep 25, 2004 #10
    Oops! My bad! :blushing: I didn't read carefully enough ... You are saying (correctly) that [itex]a_2 = -a_1[/itex] in the case where the pulley is at rest relative to the ground. So please ignore my second post.

    Also, it turns out that my third post, which stresses making the coordinate systems of the two masses consistent, leads to difficulties when the pulley is accelerating. It is actually best to treat the upward direction as positive for both masses, as you recommended in your post. So please disregard my third post as well. :blushing:

    If we use your given equations, the correct answers can be achieved.

    First, solve both of the force equations for the tension [itex]T[/itex]:

    [tex]m_1a_1 = T - m_1g \Rightarrow T = m_1a_1 + m_1g[/tex]
    [tex]m_2a_2 = T - m_2g \Rightarrow T = m_2a_2 + m_2g[/tex]

    Now set the two equations for T equal to one another:

    [tex]m_1a_1 + m_1g = m_2a_2 + m_2g[/tex]

    Now substitute for [itex]a_1[/itex] and [itex]a_2[/itex]:

    [tex]m_1(a + a_r) + m_1g = m_2(a - a_r) + m_2g[/tex]
    [tex]\Rightarrow m_1a_r + m_2a_r = m_2g - m_1g + m_2a - m_1a[/tex]
    [tex]\Rightarrow a_r = \frac{m_2 - m_1}{m_1 + m_2}g + \frac{m_2 - m_1}{m_1 + m_2}a[/tex]

    Substituting [itex]a_r[/itex] into the expression for [itex]a_1[/itex] then gives:

    [tex]a_1 = a + a_r = \frac{m_1 + m_2}{m_1 + m_2}a + \frac{m_2 - m_1}{m_1 + m_2}g + \frac{m_2 - m_1}{m_1 + m_2}a[/tex]
    [tex]\Rightarrow a_1 = \frac{m_2 - m_1}{m_1 + m_2}g + \frac{2m_2}{m_1 + m_2}a[/tex]

    Substituting [itex]a_r[/itex] into the expression for [itex]a_2[/itex] then gives:

    [tex]a_2 = a - a_r = \frac{m_1 + m_2}{m_1 + m_2}a - [\frac{m_2 - m_1}{m_1 + m_2}g + \frac{m_2 - m_1}{m_1 + m_2}a][/tex]
    [tex]\Rightarrow a_2 = \frac{m_1 - m_2}{m_1 + m_2}g + \frac{2m_1}{m_1 + m_2}a[/tex]

    Finally, substituting [itex]a_1[/itex] into the first expression for [itex]T[/itex] then gives:

    [tex]T = m_1a_1 + m_1g = m_1[\frac{m_2 - m_1}{m_1 + m_2}g + \frac{2m_2}{m_1 + m_2}a] + \frac{m_1 + m_2}{m_1 + m_2}m_1g[/tex]
    [tex]\Rightarrow T = \frac{m_1m_2}{m_1 + m_2}g - \frac{{m_1}^2}{m_1 + m_2}g + \frac{2m_1m_2}{m_1 + m_2}a + \frac{{m_1}^2}{m_1 + m_2}g + \frac{m_1m_2}{m_1 + m_2}g[/tex]
    [tex]\Rightarrow T = \frac{2m_1m_2}{m_1 + m_2}g + \frac{2m_1m_2}{m_1 + m_2}a[/tex]
    [tex]\Rightarrow T = \frac{2m_1m_2}{m_1 + m_2}(g + a)[/tex]

    Thanks for your help, ehild. You were right all along.
     
  12. Sep 25, 2004 #11

    ehild

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    Nice to hear :smile:
    Fortunately, I found your answers only just now, so you worked out the problem by yourself.

    You can solve such Atwood's machine problems either that way as your instructor said
    but you can use the same coordinate system for the whole system as well. I prefer this latter method and in this case, when there was a third body, the pulley, accelerating upward, it was really better.

    ehild
     
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