Does Acceleration Affect the Stability of an Atwood Machine in an Elevator?

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SUMMARY

The discussion centers on the stability of an Atwood machine when the elevator is accelerated upwards at a constant velocity equal to gravitational acceleration (g). The analysis reveals that the tension in the ropes, specifically T1 and TA, remains balanced, indicating that the system is stable under these conditions. The derived equations show that the relationship between masses M2 and M3, along with the tension calculations, confirm that the system's stability is unaffected by the elevator's acceleration. Additionally, the participants address misconceptions regarding the role of mass in the tension equations.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with tension in ropes and pulleys
  • Knowledge of gravitational acceleration (g)
  • Basic algebra for manipulating equations
NEXT STEPS
  • Explore the concept of non-inertial reference frames in physics
  • Study the dynamics of systems with multiple masses and pulleys
  • Learn about the effects of acceleration on tension in mechanical systems
  • Investigate the implications of changing gravitational forces in physics problems
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Students studying classical mechanics, physics educators, and anyone interested in understanding the dynamics of accelerated systems and tension in Atwood machines.

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Homework Statement


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##

Homework Equations

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##) [/B]
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!
 

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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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