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Acceleration of an Elevator

  • Thread starter Cllzzrd
  • Start date
  • #1
7
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Homework Statement



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.

Homework Equations



F=ma
N=ma-mg

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?
 

Answers and Replies

  • #2
48
0
0.76 is the ratio of the persons weight in the accelerating elevator to his normal weight
m(g-a) / mg
 
  • #3
7
0
0.76 is the ratio of the persons weight in the accelerating elevator to his normal weight
m(g-a) / mg
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.
 

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