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Elevator and acceleration problem

  1. Apr 15, 2016 #1
    1. The problem statement, all variables and given/known data
    A 60.0 kg student is riding the elevator at the CN Tower in Toronto. Calculate her apparent weight if she is accelerating upwards at a rate of 5.20 m/s2. Calculate her apparent weight if she is riding down the elevator. Include a solution with the derived equation for each.

    2. Relevant equations

    3. The attempt at a solution
    I have drawn the free body diagram for the first part of the question and I think that the equation that I should use is Fn-Fg = ma. However I'm not sure how to get the normal force, do I substitute the numbers in and solve for it, or is there another way I get it? Any help would be much appreciated, I go to a self-directed school and there are no physics teachers here today for me to ask!
  2. jcsd
  3. Apr 15, 2016 #2
    Looking at it again, I think I understand. Am I trying to find the normal force?
  4. Apr 15, 2016 #3


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    Apparent weight is the normal force, yes.
  5. Apr 15, 2016 #4
    Ok! So now I have:
    Fn = ma +fg
    Fn = (60kg)(5.20m/s2) + (60kg)(9.81m/s2)
    Fn = 312N + 588.6N
    Fn = 900.6N
    Then do I divide that by 9.81m/s2 to get it into kg? The answer would be 91.8 kg.
  6. Apr 15, 2016 #5


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    That depends. Technically kg is the unit of mass and weight is measured in N. You'll have to ask whoever set the question what they meant. I would play safe and leave it in N.
  7. Apr 15, 2016 #6
    I see. I'll ask someone on Monday. Thank you so much for your help!
  8. Apr 15, 2016 #7
    Some people use kgf as a measure of force. 1 kgf is the force that gravity would impose on a mass of 1 kgm. To convert N to kgf, divide by 9.81.
  9. Apr 16, 2016 #8
    When we want to find the weight of a body, we can sort of imagine a weighing machine kept under the body. The weighing machine shows the Normal force exterted on it. So yes, we essentially need to find the normal force between the body and the contact. That would be given by the sum of the gravitational force [acts downward,obviously] and the pseudo force exerted due to the upward acc of the elevator [acts downward]
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