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

AI Thread Summary
The discussion focuses on the stability of an Atwood machine when an elevator accelerates upwards at a constant velocity equal to gravitational acceleration, g. It is established that the relationship for mass M1 remains valid under acceleration, confirming stability due to balanced torques. The user questions the intuitive understanding of the system's motion in an accelerating frame and the validity of their initial equation regarding tensions. A correction is noted regarding a transcription error in the calculations, emphasizing the need for clarity in the equations used. Ultimately, the acceleration of the lift is likened to a change in gravitational force, which does not affect the stability conditions.
bananabandana
Messages
112
Reaction score
5

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!
 

Attachments

  • atwoodmachine.jpg
    atwoodmachine.jpg
    33.3 KB · Views: 782
Physics news on Phys.org
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.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

Similar threads

Back
Top