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Inclined Pulley system, little information given.

  1. Oct 23, 2009 #1
    1. The problem statement, all variables and given/known data
    I have dutifully reproduced the problem in OODraw please forgive the poor aesthetic quality but I believe all the important information is visible. fnYb6.jpg


    2. Relevant equations
    [tex]mgL-Lsin(\theta)m'g = \frac{mv^2}{2} + \frac{m'v'^2}{2} [/tex]
    [tex] T=T'[/tex]
    [tex] v=v'[/tex]
    [tex] \therefore v^2 = \frac{2gl(m-m'sin(\theta)}{m+m'}[/tex]
    [tex] v = \sqrt{\frac{2gl(m-m'sin(\theta)}{m+m'}}[/tex]

    Also:

    [tex]m'a = T-m'gsin(\theta)[/tex]
    [tex]ma = mg - T[/tex]
    [tex]\therefore m(g-a)=m'(a+gsin(\theta))[/tex]


    3. The attempt at a solution

    I assumed that T must be equal to T' as it is the same rope and hence can only have one tension? It just seemed to make more sense. The Lsin(theta) bit comes from the assumption that the rope length remains constant and hence m' is dragged parallel to the slope the same distance that m falls. I also assumed that v' must equal v as otherwise the rope would break, I just don't see how it could make physical sense if they weren't equal.

    I need an expression for V in terms of L and theta and possibly g. That makes the second part of the question solvable. But I cannot see how to cancel the expressions for the masses out of my equations.

    Any help?
     
    Last edited: Oct 23, 2009
  2. jcsd
  3. Oct 23, 2009 #2
    You're right, there is no way to cancel m or m', because the speed depends on them.

    The best you can do is set [tex] M = m/m' [/tex] to get

    [tex] v = \sqrt{\frac {2 g l (M - sin(\theta))} {M+1} } [/tex]
     
  4. Oct 23, 2009 #3
    Hmmm.. that doesn't help with the second part of the question though where i have to show that it is equal to 1.5ms^-1 for certain parameters, because I am not given any info about the masses or their ratios.

    I wonder if we are meant to assume m=m' and the question has just been given poorly. I will see what I can get with that assumption.
     
  5. Oct 23, 2009 #4
    Putting m=m' we get that:
    [tex] v= \sqrt{gL(1-sin(\theta))} [/tex]

    However for L=1, theta = 20 degrees, this yields a value of 2.54ms^-1 not 1.5ms^-1 as specified in the question.

    Therefore m cannot equal m'.

    Unfortunately given the fact that I have no information about the masses I cannot see a solution to this problem!?
     
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