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Homework Help: Weight on an elevator question torturing me

  1. Oct 9, 2009 #1
    A 88.0 kg person stands on a scale in an elevator. What is the apparent weight in each of the following situations?
    (a) when the elevator is accelerating upward with an acceleration of 1.40 m/s2

    (b) when the elevator is moving upward at a constant speed

    (c) when the elevator is accelerating downward with an acceleration of 1.70 m/s2

    in my text it talks about apparent weight on an elevator and gives the equation:

    Force Normal = mg + ma

    then it doesn't say a darn thing afterwords. no example or anything.

    my attempted solutions:

    a) a= 9.8 + 1.4 = 11.2 m/s^2

    F = (88kg x 9.8m/s^2)+(88kg x 11.2m/s^2)= 1795.2 N

    b) the elevator isn't moving so the apparent weight would be the same as the normal weight
    w = mg = 88 x 9.8 = 862.4N

    c) a = 9.8 - 1.7 = 8.1m/s^2

    F = (88 x 9.8)+(88 x 8.1)= 1575.2n

    i know a and c can't make sense because they are worked the same way and c is greater b. c should be less than b

    am i using the formula wrong?
  2. jcsd
  3. Oct 9, 2009 #2


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    Science Advisor

    No, a= 1.4. g is 9.8

    This should be either 88kg x 9.8 m/s^2+ 88kg*1.4m/s^2 or 88kg(9.2+ 1.4)= 88kg(11.2m/s)

    Yes, that is correct.

    Again you are using g twice. a= -1.7 so either 88*9.8- 88*1.7 or 88(9.8- 1.7).

    "a" is the acceleration of the elevator as given. You do not add 9.8 to that to get a.
  4. Oct 9, 2009 #3
    thank you my friend.
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