Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Acceleration of an Elevator

  1. Feb 4, 2010 #1
    1. The problem statement, all variables and given/known data

    A person stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.76 of the person's regular weight.

    Calculate the magnitude of acceleration of the elevator.

    2. Relevant equations

    F=ma
    N=ma-mg

    3. The attempt at a solution

    Ok, so the scale in the elevator shows weight (mg), the elevator accelerates downwards, and the scale shows .76 of the persons weight.

    [tex]\Sigma[/tex]F=ma
    mg-N=ma
    a=(mg-N)/m
    a=(m(9.8)-.76)/m

    So now I have 2 unknowns in one equation. I need to find A, but I do not know the mass of the person.

    I figured that if the weight shown by the person is .76 of his normal, I should be able to figure this out if i just assign a random value to "m"... correct?

    so plugging a number in for m i get...
    a=((10)(9.8)-.76(10))/(10)
    a=9.04

    This does not possibly seem correct. if it were true, the elevator would nearly be in freefall, and the scale would be reading something around .2 of the persons weight, right?

    What am i doing wrong?
     
  2. jcsd
  3. Feb 4, 2010 #2
    0.76 is the ratio of the persons weight in the accelerating elevator to his normal weight
    m(g-a) / mg
     
  4. Feb 4, 2010 #3
    Ok, I did not realize that it was a ratio.
    so now I solve for a

    m(g-a)/mg=.76
    m(g-a)=.76mg
    g-a=.76g
    a=g-.76g
    a=9.8-.76(9.8)
    a=2.5

    Thank you very much, this was the correct answer.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook