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Apparent weight loss 2nd Newtons Law elevator problem

  1. Mar 4, 2016 #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


    3. The attempt at a solution
    I understand that mg - Fn = m(0.20g) happens and then solving for Fn = 0.80mg.

    I also understand that F'n is -0.80mg.

    What I don't understand is why the "scale only needs to exert a force of only 0.80mg and that it will only give a reading of 0.80m"; why did the g get eliminated?
     
  2. jcsd
  3. Mar 4, 2016 #2
    Because you want the net force to be downward so it can actually accelerate downward and n is just the force that the scale will push you up with. It also equal to the apparent weight because the springs in the scale gets compressed for example. Draw a FBD and try to figure what I am saying
     
  4. Mar 4, 2016 #3
    I still don't understand. My book says the F'n=0.80mg downward. Her weight is still mg=(65kg)(9.8m/s^2)=640N. But the scale, needing to exert a force of only 0.80mg and that it will only give a reading of 0.80m=52kg". I don't understand why it's only giving a reading of 0.80m, instead of 0.80mg. WHy was g dropped?
     
  5. Mar 4, 2016 #4
    I still don't understand. My book says the F'n=0.80mg downward. Her weight is still mg=(65kg)(9.8m/s^2)=640N. But the scale, needing to exert a force of only 0.80mg and that it will only give a reading of 0.80m=52kg". I don't understand why it's only giving a reading of 0.80m, instead of 0.80mg. WHy was g dropped?
     
  6. Mar 4, 2016 #5

    haruspex

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    The scale reads in kg, not Newtons. But the spring inside it is sensitive to force, not mass. So the scale interprets a force of mg as a mass m, and displays it as such.
     
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