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Different tensions in a pulley system and acceleration

  1. Apr 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Here is the picture: http://www.webassign.net/giocp1/8-p-058.gif
    I am to find Tension 1 ( that between crate one and the pulley) and tension 2 (that between crate two and the pulley). The mass of the pulley is MP.

    2. Relevant equations
    See below




    3. The attempt at a solution
    I have found 4 simultaneous equations:
    T1=M1ax1
    Fn1-fg1=0
    T2-fg2=-m2ay2
    Torque:
    R(T1-T2)= I*angular acceleration
    I also found that with respect to the acceleration of m2, the angular acceleration of the pulley is (-ay2/R)
    Are these correct so far? It seems that after I do my algebra my answer for T1 does not make sense. For instance, it comes out non-dependent on mass 2.

    Thanks
     
  2. jcsd
  3. Apr 22, 2013 #2

    Simon Bridge

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    I don't think you have the angular acceleration of the pulley correct.
    How did you get it?
     
  4. Apr 22, 2013 #3
    linear acceleration= angular acceleration * R
     
  5. Apr 22, 2013 #4
    Instead, I first solved for the acceleration of the crates:
    I get, m2g/(m1+m2+.5mp)

    Then, I subbed it in for a in the equations for T1 and T2.

    I get,
    T1=m1*m2g/(m1+m2+.5mp)
    T2= (-m2*m2*g/(m1+m2+.5mp))+m2g
    Additionally, the expression for T2 looks "too negative."
    However, the next part of the question tells me to show that T1=T2 when mp =0. I can't seem to get this to work, though, with the expressions I derived.
     
    Last edited: Apr 22, 2013
  6. Apr 22, 2013 #5

    Simon Bridge

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    Oh I see what's happened -have you been careful about keeping the signs consistent?

    i.e. you appear to be making the +y and +x directions positive so m2 is going to pick up speed downwards, so it's acceleration is negative, while m1 is going to pick up speed to the right, so it's acceleration is positive. Thus ##a=a_1=-a_2##? But I cannot see any such reasoning in what you've told me.

    Why not pick your coordinate system to make the math easy?
    (It also makes it easier to mark ;) )

    i.e.
    for m1, pick +ve to the right so ##T_1=m_1a## ... see how we don't need the subscripts?
    for the pulley, pick +ve to be clockwise, to be consistent with m1.
    (You should express the entire thing in linear terms too - don't leave it as ##R(T_2-T_1)=I\alpha##)
    for m2, pick +ve downwards so that ##m_2g-T_2=m_2a##
    ... notice how I don't mess about with this ##F_g## stuff - just write out ##mg## right away?
    I label fbds like that too: having to keep track of notation just gives more opportunity to make mistakes.

    3 equations, 3 unknowns.
     
  7. Apr 23, 2013 #6

    So are my expressions for T1 and T2 correct, minus the sign problems?
     
  8. Apr 23, 2013 #7

    Simon Bridge

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    Can't tell without doing it for you - it is up to you to check.
    I was expecting to see an "R" in there from the moment of inertia of the pulley though.
     
  9. Apr 23, 2013 #8
    I did it the way you told me and I got the same answer for T2.
     
  10. Apr 23, 2013 #9

    Simon Bridge

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    You are getting:
    ... which is: $$T_2= -\frac{m_2^2g}{m_1+m_2+\frac{1}{2}m_p}+m_2g$$ ... well, looking closer, that should be a positive number. Perhaps the minus sign here is a red herring?

    I'm getting $$T_2 = \frac{m_2(m_p+2m_1)g}{m_p+2m_1 + 2m_2 }$$ ... which is the same.
    Looks like it you did track the minus signs correctly after all! (Though I could have made an algebra error myself ;) ).

    You can check my working:

    FBD for mass #1: ##T_1=m_1a## ...(1)
    FBD for mass #2: ##m_2g-T_2=m_2a## ...(2)
    FBD for pulley: ##R(T_2-T_1)=(\frac{1}{2}m_pR^2)(a/R) = \frac{1}{2}Rm_pa## ...(3)
    ... ahah: the R cancels out!

    ... from (3)
    ##\Rightarrow a = 2(T_2-T_1)/m_p## ...(4)

    ... sub into (1) and (2)
    ##T_1 = 2m_1(T_2-T_1)/m_p##
    ## \Rightarrow (1+2m_1/m_p)T_1 = (2m_1/m_p)T_2## ...(5)

    ##m_2g-T_2=2m_2 (T_2-T_1)/m_p ##
    ##\Rightarrow (1+2m_2/m_p)T_2 = m_2g+(2m_2/m_p)T_1## ...(6)

    To find ##T_2##:
    ... solve for ##T_1## in (5) and sub into (6) gives:

    $$\left ( 1+\frac{2m_2}{m_p}\right )T_2 = m_2g+\frac{2m_2}{m_p}\frac{2m_1/m_p}{1+2m_1/m_p}T_2\\
    \Rightarrow \left ( m_p+2m_2\right )(m_p+2m_1)T_2 = m_2m_p(m_p+2m_1)g+4m_1m_2T_2\\
    \Rightarrow T_2 = \frac{m_2(m_p+2m_1)g}{m_p+2m_1 + 2m_2 }$$
     
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