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Atwood Machine in Elevator

  1. Apr 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Please see attached for diagram. We know that the elevator arm is horizontal when the lift is stationary, with ## M_{1}=\frac{4M_{2}M_{3}}{(M_{2}+M_{3})}## It wants us to find out if this is still the case when the lift is accelerated upwards at a constant velocity ##g##

    2. Relevant equations


    3. The attempt at a solution
    Let the tension in the ##M_{1}## rope be ##T_{1}##, that in the rope between the lift and the support for ##M_{2}## and ##M_{3}## be ##T_{A}## and that in the rope between masses ##M_{2}## and ##M_{3}## be ##T_{B}## Take up as positive (i.e lift is moving up at ## +g##)

    Look at ##M_{1}## first:
    $$ T_{1}-M_{1}g=M_{1}g \implies T_{1}=2M_{1}g $$

    Then look at the ##M_{2},M_{3}## system to work out ##T_{B}##:
    $$ T_{B}-M_{2}g=M_{2}(g+a) $$
    $$ T_{B}-M_{3}g=M_{3}(g-a) $$
    Substitute for ## T_{B}## and rearrange for ## a##:
    $$ a = \frac{M_{3}-M_{2}}{M_{3}+M_{2}}g $$
    $$ T_{B} = 2M_{2}g+M_{2}a = \frac{4M_{3}M_{2}g}{(M_{3}+M_{2}} $$
    Therefore:
    $$ T_{A} = 2\times T_{B} = \frac{8M_{3}M_{2}g}{(M_{3}+M_{2})} $$
    So the thing is stable when accelerating cosntantly as the torque to ##T_{1}## is balanced by that due to ##T_{A}##, from our relationship for ##M_{1}##.
    But I have two questions - one is there not some easier (intuitive way) to think about the motion of the system in the accelerating frame that I wasn't smart enough to find? - and two - when I first did the problem I wrote down:
    $$ T_{A}-2T_{B} = (M_{2}+M_{3})g $$ - I figure this is false because there is no mass in the rope, so that even if it is accelerating with speed ##g## then we don't need to worry about ##M_{2}## and ##M_{3}## as they are already taken care of by the tension?

    Thanks!
     

    Attached Files:

  2. jcsd
  3. Apr 28, 2015 #2

    haruspex

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    You made a transcription error when typing this out. a should be twice what you wrote, but you get the right TB in the next line.
    Seems to me that the acceleration of the lift is equivalent to a change in the value of g. Since g does not feature in the balance equation for the stationary case, changing it cannot change the equation.
    For the second question, you'll need to explain your rationale for writing that equation.
     
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