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4 masses and 3 pulleys

  1. Sep 18, 2016 #1
    1. The problem statement, all variables and given/known data
    Snap3.jpg

    m1 must stay in place. express m4 with the rest.
    What must be the maximum ratio ##\frac{m_1}{m_2}## in order m1 will stay in place.

    2. Relevant equations
    Mass-acceleration: F=ma

    3. The attempt at a solution
    2 cases, in the first ##m_1<m_2##, which is drawn above:
    $$T_1=m_1g,~~m_2g-T_1=m_2a_2~~\rightarrow~~a_2=\left( \frac{m_2-m_1}{m_2} \right)g$$
    $$\frac{1}{2}a_2=a_3,~~2T_1=T_3,~~T_3-T_4=m_3a_3~~\rightarrow~~T_4=2T_1-\frac{1}{2}m_3a_3$$
    $$T_4-m_4g=m_4a_3~~\rightarrow~~m_4=\frac{T_4}{a_3+g}=\frac{4m_1m_2-(m_2-m_1)m_3}{3m_2-m_1}$$
    Since here ##m_1<m_2## the denominator of m4 is positive, so the nominator must be positive too:
    $$4m_1m_2>(m_2-m_1)m_3,~~\frac{m_1}{m_2}\triangleq k~~\rightarrow~~k>\frac{m_3}{4m_2+m_3}$$
    In the second scenario ##m_1>m_2##:

    $$T_1-m_2g=m_2a_2~~\rightarrow~~a_2=\left( \frac{m_1-m_2}{m_2} \right)g$$
    $$T_4-T_3=m_3a_3~~\rightarrow~~T_4=2T_1+\frac{1}{2}m_3a_3$$
    $$m_4g-T_4=m_4a_3~~\rightarrow~~m_4=\frac{T_4}{g-a_3}=\frac{4m_1m_2+(m_1-m_2)m_3}{3m_2-m_1}$$
    Since here ##m_1<m_2## the nominator is positive so the also must be the denominator, thus:
    $$m_2-m_1~~\rightarrow~~\frac{m_1}{m_2}<3$$
    Which doesn't make sense since the ratio doesn't include m3 or m4 at least, as in the previous case.
     

    Attached Files:

  2. jcsd
  3. Sep 18, 2016 #2

    haruspex

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    Where did that 1/2 come from?

    Edit: I guess that was just a transcription error. Later lines look ok.

    In case 1 you found a lower bound for m1/m2. I see no contradiction. You were only asked for an upper bound.
     
    Last edited: Sep 18, 2016
  4. Sep 18, 2016 #3
    $$\left\{\begin{array}{l} T_3-T_4=m_3a_3 \\ 2T_1=T_3 \\ \frac{1}{2}a_2=a_3 \end{array}\right.~~\rightarrow~~T_4=2T_1-\frac{1}{2}m_3a_2$$
    But why doesn't ##\frac{m_1}{m_2}<3## include at least one of the rest? doesn't the second scenario depend on them also?
     
  5. Sep 18, 2016 #4

    haruspex

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    It depends how you read the question.
    If you read it as finding the max ratio such that a set of masses exists satisfying the condition then you get (anything up to but not including) 3.
    If you read it as the max ratio given the masses m2, m3, m4 then you get that it is less than (3m4+m3)/(4m2+m3+m4).
     
  6. Sep 19, 2016 #5
    $$m_4=\frac{4m_1m_2-(m_2-m_1)m_3}{3m_2-m_1}~~\rightarrow~~4m_1m_2+m_1m_3+m_1m_4=m_2m_3+3m_2m_4$$
    $$\frac{4m_1m_2+m_1m_3+m_1m_4}{m_2}=\frac{m_2m_3+3m_2m_4}{m_2}=4m_2\frac{m_1}{m_2}+m_3\frac{m_1}{m_2}+m_4\frac{m_1}{m_2}=m_3+3m_4$$
    $$(4m_2+m_3+m_4)\frac{m_1}{m_2}=m_3+3m_4~~\rightarrow~~\frac{m_1}{m_2}=\frac{m_3+3m_4}{4m_2+m_3+m_4}$$
    Why, specifically, ##\frac{m_1}{m_2}<...## and not ##\frac{m_1}{m_2}>...##?
    I didn't see that both scenarios yield the same equation ##m_4=\frac{4m_1m_2-(m_2-m_1)m_3}{3m_2-m_1}##. but from this equation i can deduce other things.
    First scenario:
    $$m_1<m_2~~\rightarrow~~3m_2-m_1>0~~\rightarrow~~4m_1m_2-(m_2-m_1)m_3>0~~\rightarrow~~\frac{m_1}{m_2}>\frac{m_3}{m_3+4m_2}$$
    Second scenario:
    $$m_1>m_2~~\rightarrow~~or\left\{\begin{array}{l} 3m_2-m_1>0~~\rightarrow~~4m_1m_2-(m_2-m_1)m_3>0~~\rm{exists~for~all~m_i} \\ 3m_2-m_1<0~~\rightarrow~~4m_1m_2-(m_2-m_1)m_3<0~~\rm{cannot~be} \end{array}\right.$$
    Does all this derivation apply to your:
    And also, if i just want that to happen, i have to: ##3m_2-m_1\neq 0##
     
    Last edited: Sep 19, 2016
  7. Sep 19, 2016 #6

    haruspex

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    Because there is a numerical upper bound (3) but no numerical lower bound greater than zero. If you make m3 and m4 arbitrarily small compared to m2 you can make m1/m2 arbitrarily close to 0.
    Yes, we are saying the same there.
     
  8. Sep 20, 2016 #7
    Why upper limit? in the scenario where ##m_1>m_2## i can make m3 and m4 big and they will pull up the pulley
     
  9. Sep 20, 2016 #8

    haruspex

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    The ratio m1/m2 can never exceed 3. Look at how the expression on the right varies as you change m2, m3 and m4. To maximise it you need a large m4 compared to m2 and m3. As m4 tends to infinity, what does the expression tend to?
     
  10. Sep 20, 2016 #9
    $$\frac{m_1}{m_2}=\frac{3\infty}{\infty}=3$$
    But now i don't understand what i have done in:
    $$m_1<m_2~~\rightarrow~~3m_2-m_1>0~~\rightarrow~~4m_1m_2-(m_2-m_1)m_3>0~~\rightarrow~~\frac{m_1}{m_2}>\frac{m_3}{m_3+4m_2}$$
    $$for:~m_3\rightarrow 0~~\Rightarrow~~\frac{m_1}{m_2}\rightarrow 0$$
    If m3 is small, and ##m_1<m_2##, m4 must be small but it doesn't appear.
     
  11. Sep 20, 2016 #10

    haruspex

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    As you wrote in post #5, the full equation is the same in both cases. In your analysis of m1<m2, you used that to get rid of m4 before letting m3 tend to zero. It only looks different because your analysis followed a different sequence.
     
  12. Sep 21, 2016 #11
    I thank you very much Haruspex
     
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