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Apparent Weight and Elevator

  1. Oct 3, 2006 #1
    Can someone tell me what I'm doing wrong here?

    When you weigh yourself on good old terra firma (solid ground), your weight is 126 lb. In an elevator your apparent weight is 118 lb. What are the direction and magnitude of the elevator's acceleration?

    I have Wa = W-ma = mg-ma
    Wa=118lb
    m=126lb
    g=-9.8m/s

    So I have 118 = (126)(-9.8) - (126)a and solved for a but I'm not getting the right answer. I wasn't sure if the 9.8 should be negative or not but I tried both and neither is working. Any help would be appreciated
     
  2. jcsd
  3. Oct 3, 2006 #2
    Read the question carefully. lb is a unit of force.
     
    Last edited: Oct 3, 2006
  4. Oct 3, 2006 #3
    So the numbers I'm plugging in for m aren't the right numbers? How do I find them?
     
  5. Oct 3, 2006 #4
    Your weight is your mass times gravity. So if you want to find out your mass play around with Newton's second law.
     
  6. Oct 3, 2006 #5
    So since 126 = mg and g=9.8 then the mass is 12.857. And I did the same thing with the apparent weight and got 12.04. So I have the equation 12.04 = 12.857(-9.8) - (12.857)a and I'm solving for a and getting 10.736 which is getting marked wrong.
     
  7. Oct 3, 2006 #6

    SGT

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    You are mixing the units. Pound is a unit in the british system, while 9.8 m/s^2 is gravity acceleration in the International System. You should transform your pounds in Newtons, before calculating the mass.
     
  8. Oct 3, 2006 #7
    So 1 Newton = 0.224808943 lbs...so I multiply the 118 lbs by 0.224808943 and I get 26.526 N and then using weight=mg get mass? Because when I do that I'm getting 2.706 for Wa and that seems waaay off.
     
  9. Oct 3, 2006 #8
    You could also use g = 32ft/s and calculate mass in slugs, and then convert everything to the SI units in the final step.
    1N = 0.224808943 lbs. So while coverting from lbs->N, you don't use the same conversion factor.
     
  10. Oct 4, 2006 #9

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    You should divide 118 by 0.224808943 to obtain the weight in Newtons.
     
  11. Oct 4, 2006 #10
    So Wa=524.89N and W=560.476N. And using W=mg your apparent mass then is 53.56 and your actual mass is 57.19. Is that right?
     
  12. Oct 4, 2006 #11

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    There is no sense in apparent mass. The actual mass m is 57.19 kg. In the elevator the apparent weight is Wa = m(g-a), where a is the acceleration of the elevator.
     
  13. Oct 4, 2006 #12
    Well the apparent weight is given but do I use that number in Newtons or lbs when I'm trying to find acceleratin
     
  14. Oct 4, 2006 #13
    Whatever units you use, make sure you are using only one system. If you are asked to use a particular system, then use that.
     
  15. Oct 4, 2006 #14

    SGT

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    Since you already have the apparent weight in Newtons and the mass in kg, you can calculate a in m/s^2.
     
  16. Oct 4, 2006 #15
    So 524.89 = (57.19)(-9.8) - (57.19)a? Or is the 9.8 not supposed to be negative?
     
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