Finding Acceleration of Masses in Elevator

In summary, the elevator is accelerating upwards at a rate of a0, with two masses hanging from a pulley. To find the acceleration of m1 as viewed from within the elevator, separate equations are needed for each mass. Using the constraint a1 = -a2, the correct equation can be written to find a1.
  • #1
yoni162
16
0

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?
 

Attachments

  • TwoMassesInAnElevator.png
    TwoMassesInAnElevator.png
    806 bytes · Views: 385
Physics news on Phys.org
  • #2
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.
 
  • #3
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!
 

1. What is acceleration in an elevator?

Acceleration in an elevator refers to the rate at which the elevator changes its velocity. It can be either positive (moving upward) or negative (moving downward).

2. How is acceleration calculated in an elevator?

The acceleration of an elevator can be calculated by dividing the change in velocity by the time it takes to make that change. This is represented by the equation a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. What factors affect the acceleration of an elevator?

The acceleration of an elevator is affected by the mass of the elevator and its contents, the force of the elevator motor, and the friction and air resistance within the elevator shaft.

4. How does the acceleration of an elevator change during its journey?

The acceleration of an elevator is not constant throughout its journey. It starts with a positive acceleration as it moves upward, reaches a constant velocity at the desired floor, and then decelerates to a stop. On the way down, the elevator will experience negative acceleration as it slows down and then reaches a constant velocity before coming to a stop again.

5. Can the acceleration of an elevator be felt by its passengers?

Yes, the acceleration of an elevator can be felt by its passengers in the form of a sensation of weightlessness or heaviness. This is due to the force of gravity and the fact that the elevator is accelerating or decelerating. However, in most modern elevators, the acceleration is smooth and gradual, so the sensation is minimal.

Similar threads

  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
578
Replies
25
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
2
Replies
42
Views
958
  • Introductory Physics Homework Help
Replies
3
Views
792
  • Introductory Physics Homework Help
Replies
33
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
885
Back
Top