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B A simple pulley two mass system

  1. Apr 20, 2016 #1
    In a simple one pulley-two mass system I have two ways of solving for acceleration, but am not sure which one is the correct or proper way.
    61673?db=v4net.jpg

    #1)
    assume m2 has the greater mass of the two
    T-m1g = m1a
    m2g - T = m2a
    a = (m2g - m1g) / (m1+m2)

    #2)
    direction of up is positive, down is negative
    T-m1g = m1a
    T-m2g = m2a
    a = (m1g - m2g) / (m2-m1)

    The two accelerations are different. Which one is correct, and why?
     
  2. jcsd
  3. Apr 20, 2016 #2

    BvU

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    The rope has a fixed length. If m1 goes up, m2 goes down with the same speed. So ##v_1 = -v_2## and thereby ##a_1 = -a_2## . You only use one a in your second set. Apart from that, the second set is more clear (more consistent in signs of T and mg)
     
  4. Apr 20, 2016 #3
    Accelerations are equal to zero when summed up,
    m1 is moving up at a1
    m2 is moving up at -a2

    a1 + -a2 = 0
    a1 = a2
     
  5. Apr 21, 2016 #4

    BvU

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    Yes. A consequence of $$y_1+y_2={\rm \ constant} \Rightarrow v_1+v_2= 0 \Rightarrow a_1+a_2= 0 $$
    It is not wise to avoid negative numbers by redefining/assuming directions. Much better to keep one coordinate system definition and conclude directions from the signs of the calculated variables.
     
  6. Apr 21, 2016 #5
    I'm confused as to whether the second set is correct or not. Can you check the other part of post #3?
     
  7. Apr 21, 2016 #6
    #1 would work fine. If the acceleration came negative, then you know that your assumption that m2>m1 is wrong, and it would accelerate with -a in the direction opposite to what you assumed
    #2 is incorrect.
     
  8. Apr 22, 2016 #7
    this is correct!
    this is wrong
    specifically the equation
    is wrong because 'a' denotes mag of acceleration of m2 and mag of a vector is always positive
    since you already know that m2 has accln downwards , so it is known that m2g must be greater than T
    so since right side of eqn is +ve , so left side must also be +ve , so it must be (m2g-T) (since m2g > T)
    it is always better to use vectors
    T*j + mg*(-j)= m2a* (-j) j is unit vector in upward dir
     
    Last edited: Apr 22, 2016
  9. Apr 22, 2016 #8
    It all seems obvious now. These should be the equations paired with the second set.
    [itex]T\quad -\quad { m }_{ 1 }g\quad =\quad { m }_{ 1 }{ a }_{ 1 }\\ T\quad -\quad { m }_{ 2 }g\quad =\quad { m }_{ 2 }{ a }_{ 2 }\\ { a }_{ 2 }\quad =\quad { -a }_{ 1 }\\ \\ { a }_{ 1 }\quad =\quad \frac { { m }_{ 2 }g\quad -\quad { m }_{ 1 }g }{ { m }_{ 1 }\quad +\quad { m }_{ 2 } } [/itex]
    Which gives the correct answer! Thanks guys.
     
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