Finding Acceleration of Masses in Elevator

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SUMMARY

The discussion focuses on calculating the acceleration of mass m1 in an upward-accelerating elevator, where two masses are connected by a pulley. The user initially sets up the equations of motion for both masses, m1 and m2, but struggles with the sign conventions and the relationship between their accelerations. The correct approach involves using the equations T - m2g - m2a0 = m2a2 and a1 = -a2, which accurately reflects the system's dynamics. The user acknowledges the correction and thanks the contributor for the clarification.

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  • Understanding of Newton's second law of motion
  • Familiarity with forces acting on objects in a pulley system
  • Knowledge of acceleration and its relation to mass and force
  • Basic grasp of sign conventions in physics equations
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  • Study Newton's laws of motion in detail
  • Learn about tension in pulley systems and its effects on connected masses
  • Explore advanced dynamics problems involving non-inertial reference frames
  • Investigate the implications of acceleration on forces in different contexts
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Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in understanding dynamics in non-inertial reference frames.

yoni162
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Homework Statement


An elevator is accelerating upwards with acceleration a0. Two masses are hanging down from a pulley, connected to each other. I need to find the acceleration of m1, as viewed from within the elevator.

Homework Equations


The Attempt at a Solution


If I look at each mass separately, each one is subjected to 3 forces: mg, T, and ma0, when T is obviously pointed upwards, and mg & ma0 are pointed downwards..
Now, if I want to find a1 --> m1's acceleration as viewed from within the elevator, I write:

m1
T-m1g-m1a0=m1a1

When I'm assuming a1 is in the positive direction of the Y axis. Now to m2:

m2
m2g+m2a0-T=m2a2

When I'm assuming that a2 will be in the opposite direction of the Y axis (since m1 will go up, m2 will go down). That's why I write forces the are in the positive direction of Y with a (-) before them, and forces which are directed in the negative direction of Y with a + before them. I know that's only a guess, but if I'm wrong and m1 will go down, a1 will turn out to be negative, no? I've used this method when solving 'inertial' problems and it works..
Also, I'm assuming the a1=-a2, again, like in 'inertial' situations..only here I'm not so sure..is this assumption true? Because when I solve the equations like they are now, I get that:

(m2-m1)(a0+g)=a1(m1-m2)

a1=-a0-g

Which isn't true according tow the answers..where am I wrong?
 

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yoni162 said:
m2
m2g+m2a0-T=m2a2

When I'm assuming that a2 will be in the opposite direction of the Y axis (since m1 will go up, m2 will go down). That's why I write forces the are in the positive direction of Y with a (-) before them, and forces which are directed in the negative direction of Y with a + before them. I know that's only a guess, but if I'm wrong and m1 will go down, a1 will turn out to be negative, no? I've used this method when solving 'inertial' problems and it works..
OK, but...
Also, I'm assuming the a1=-a2, again, like in 'inertial' situations..only here I'm not so sure..is this assumption true?
Now you've reversed the opposite direction you just gave things above. So you've outsmarted yourself!

Write your 2nd equation (using the usual sign conventions) as:
T -m2g -m2a0 = m2a2

And then use a1 = -a2 as your constraint. That already incorporates the fact that if one goes up, the other goes down.
 
Doc Al said:
OK, but...

Now you've reversed the opposite direction you just gave things above. So you've outsmarted yourself!

Write your 2nd equation (using the usual sign conventions) as:
T -m2g -m2a0 = m2a2

And then use a1 = -a2 as your constraint. That already incorporates the fact that if one goes up, the other goes down.

Hah! you're right..thanks a lot!
 

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