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Elevator acceleration formula

  1. May 6, 2009 #1
    1. The problem statement, all variables and given/known data

    the question given is:

    What would the reading on a bathroom scale be if a 60kg pupil was accelerated upwards at a rate of 3m.s-2


    2. Relevant equations

    F=m(g-a)
    F=ma

    3. The attempt at a solution


    F=m(g-a)
    F=60kg(10m.s-2-3m.s-2)
    F=600-180
    F=420N which is equal to the force on the person
    The reading on the scale is therefore (42kg + 60kg) or 102kg but this seems too high

    Thanks for helping :wink:
     
  2. jcsd
  3. May 6, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    The resultant force,F is upwards. The forces acting are the normal reaction upwards,R, and the weight W. Can you make a simple relationship between F,W and R?
     
  4. May 6, 2009 #3
    Thanks for the help rock but i figured it out just before i saw your post.

    F=ma

    F=60x3
    F=180N
    =extra force added to weight

    therefore the total value showed on the scale is
    W= 180N + 600N (the mass converted to weight)
    W= 780N
    M= 78kg

    Thanks again for the help
     
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