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Elevator physics

  1. May 10, 2012 #1
    1. The problem statement, all variables and given/known data

    an elevator descends with a = 2m/s^2. there is a wedge on it. the wedge has an angle of 30 degrees. at the top of the incline of the wedge a small block is released with zero initial velocity from the elevator's frame of ref. after 1 sec, how much does the block travel along the incline of the wedge with respect to the wedge , given co-efficient of friction is 0.1 ?

    2. Relevant equations

    N = mgcos30, friction = co-efficient of friction x N

    3. The attempt at a solution

    the main problem I have is with the wording...does it mean that the elevator is going at g+2 m/s^2 or 2 m/s^2 ? I assumed the latter as the former would result in the wedge colliding with the top of the elevator.

    what would happen in case A and case B ? would the results be different ?
     
  2. jcsd
  3. May 10, 2012 #2
    First you have to find Normal force.
    If you weigh yourself in the elevator descending at 2m/s2, what's the reading.
    Now take it as if you're on this earth with different g.
     
  4. May 10, 2012 #3
    An elevator normally has a counterweight, that travels in the opposite direction as the car. The counterweight in all the elevators that I've worked on is heavier than an empty car, but a loaded car might be different. So the car would have an effective weight of whatever the difference between the car and the counterweight is. So in a real elevator, would the car stop much more quickly?

    I'm guessing the counterweight is the reason that the acceleration of the car is so much smaller than g, but the fact that the acceleration is as large as it is would indicate that it is probably overloaded.
     
  5. May 11, 2012 #4
    well, taking g = 10 to simplify, we get an effective within-elevator g of 8, and since acc. down the slope = gsin30 or 4, minus the acceleration loss due to friction, you get a net acceleration of approximately 3.3 m/s^2 down the slope, resulting in about 1.65 meter moved in 1 sec.

    this assumes that acceleration of elevator is 2, not g+2...it's more of a wording issue really. should I assume it to be g+2 or 2 ?
     
  6. May 11, 2012 #5
    Why would you assume that the acceleration is anything other than the number they give you? (Not to mention that an elevator descending at 12 m/s would be quite a shock to the riders!)
     
  7. May 11, 2012 #6
    Quite clearly, acceleration of the elevator is 2 m/s2 downwards and not (10+2)m/s2. :smile:
     
  8. May 11, 2012 #7
    Taking downward as positive
    a=acceleration of elevator

    Mg-N=Ma
    N=M(g-a)
    Mg'=M(10-2)
    g'=8

    Evaluation:
    If the acceleration downward is equal to g, the apparent weight will be zero(g'=0), free fall.
    If a>g then everything inside the elevator is flying up(negative =motion-up).
     
    Last edited: May 11, 2012
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