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Elevator at rest compared to when its accelerating

  1. May 12, 2012 #1
    1. The problem statement, all variables and given/known data

    When the elevator is at rest, the scale reads W.

    Suppose the elevator now accelerates down-ward at a constant rate of 0.4 g.
    What is the ratio of the new scale reading to the value W of the scale reading when the
    elevator is at rest?



    2. Relevant equations

    Force equations:

    F = ma
    Fnet = 0

    3. The attempt at a solution

    What I did is to draw a force diagram that resulted:

    At rest:
    Fn - mg = 0
    mg = Fn

    Accelerating:
    Fn - mg = -ma, where a = .4g
    mg = Fn + .4mg

    I set Fn = mg = 1, so ratio becomes to 1.4 -> (1 + .4)/(1)

    The real answer is .6, but I'm not really sure how to get that. I know if I do:

    Fn - mg = ma
    mg = Fn - .4mg, the ratio then becomes to .6

    Why is it "ma" instead of "-ma" when the elevator is accelerating downwards?
     
  2. jcsd
  3. May 12, 2012 #2
    Call the normal reaction for the first case to be Fn1 and second Fn2.

    Also, for the second equation Fn2 is not equal to mg, so you cannot let it be 1.

    Bear in mind, the weight read by the scale is not mg. So, which force does the weighing scale actually show? (what effects it? think of Newton's third law! :smile:)
     
    Last edited: May 12, 2012
  4. May 12, 2012 #3
    I see. It just hit me that a weighing scale shows the normal force and not the gravitational force.

    So I did my work right, just not the final answer!

    mg = Fn + .4mg
    Fn = mg - .4mg, which then becomes .6.

    Got it. Thanks again!
     
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