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Elevator and scale example

  1. Jun 2, 2013 #1
    1. The problem statement, all variables and given/known data

    A 65-kg woman descends in an elevator that briefly acclerates at 0.20g downward when leaving a floor. She stands on a scale that reads in kg. (a) During this acceleration, what is her weight and what does the scale read?

    2. Relevant equations

    ƩF = ma
    mg - Fn = m(0.20g)

    3. The attempt at a solution

    I have an answer that states that the scale needs to exert a force of 0.80mg which will give a reading of 0.80m = 52kg, but I do not understand why this makes sense.
     
  2. jcsd
  3. Jun 2, 2013 #2
    What doesn't make sense about it?
     
  4. Jun 2, 2013 #3
    why do we just eliminate 'g' to find the scale reading?
     
  5. Jun 2, 2013 #4

    PhanthomJay

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    That is because the scale is calibrated in kilograms which is a mass unit and not a force unit. 52 kg of mass is 52g or 520 Newtons of force (weight) using g = 10 m/sec/sec.
     
  6. Jun 3, 2013 #5
    Ok. Can you also explain why the formula is mg - Fn = m(0.20g) and NOT Fn - mg = m(0.20g) ??
     
  7. Jun 3, 2013 #6

    CAF123

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    The acceleration of the elevator is downward so it is convenient to set up the 1D coordinate system such that downwards is the positive direction. This leads to the eqn mg - Fn = 0.20mg. Alternatively, if you take downwards as negative, then you have Fn - mg = -0.20mg which is equivalent.
     
    Last edited: Jun 3, 2013
  8. Jun 3, 2013 #7
    Ok so if we use mg - Fn = m(0.20g) and we solve for Fn we will get:

    Fn = 0.80mg

    Since we took downwards to be the positive direction then Fn = 0.80mg will be downwards since it's positive. I know I'm incorrect here, but mathematically I cannot figure out why.
     
  9. Jun 3, 2013 #8

    CAF123

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    Yes

    You expect a positive number here. By setting up the eqn in the first place, you knew that the normal force was directed in the negative direction in this case. i.e when doing a free body diagram you identify all the forces then use NII.

    The vector equation is ##m \underline{a} = mg \hat{y} - F_n \hat{y}## from which you extract the scalar equation ##ma_y = mg - F_n##. ##F_n## is the magnitude of the vector ##F_n \hat{y}## and, as such, is non negative.
     
  10. Jun 3, 2013 #9
    I'm just confused cause when we used the equations for mechanics (distance, speed, time) our final result would be either positive or negative which would indicate the direction, but now that we are using Newton's equation it seems to not be the case.

    Is this because with Newton's equations we are, in a way, creating the equation before we calculate? as opposed to the mechanic equations which, in a way, were always static.
     
  11. Jun 3, 2013 #10

    tiny-tim

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    hi e-zero! :smile:

    (try using the X2 button just above the Reply box :wink:)
    nope

    when you wrote mg - Fn, you were already assuming that Fn was upward

    if your Fn was downward, you would have written mg + Fn = m(0.20g),

    giving you Fn = -m(0.80g),

    proving that your normal force was -m(0.80g) downward, which is m(0.80g) upward!! :wink:
     
  12. Jun 3, 2013 #11
    Oh I see. I totally missed that. Fn is subtracted because its upward and in this case upward is negative.
     
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