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Delamber force

  1. Nov 27, 2009 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations

    3. 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:


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


    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:



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

    Attached Files:

  2. jcsd
  3. Nov 27, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    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.
  4. Nov 27, 2009 #3
    Hah! you're right..thanks a lot!
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