Weight on an elevator question torturing me

1. Oct 9, 2009

mr.coon

A 88.0 kg person stands on a scale in an elevator. What is the apparent weight in each of the following situations?
(a) when the elevator is accelerating upward with an acceleration of 1.40 m/s2

(b) when the elevator is moving upward at a constant speed

(c) when the elevator is accelerating downward with an acceleration of 1.70 m/s2

in my text it talks about apparent weight on an elevator and gives the equation:

Force Normal = mg + ma

then it doesn't say a darn thing afterwords. no example or anything.

my attempted solutions:

a) a= 9.8 + 1.4 = 11.2 m/s^2

F = (88kg x 9.8m/s^2)+(88kg x 11.2m/s^2)= 1795.2 N

b) the elevator isn't moving so the apparent weight would be the same as the normal weight
w = mg = 88 x 9.8 = 862.4N

c) a = 9.8 - 1.7 = 8.1m/s^2

F = (88 x 9.8)+(88 x 8.1)= 1575.2n

i know a and c can't make sense because they are worked the same way and c is greater b. c should be less than b

am i using the formula wrong?

2. Oct 9, 2009

HallsofIvy

Staff Emeritus
No, a= 1.4. g is 9.8

This should be either 88kg x 9.8 m/s^2+ 88kg*1.4m/s^2 or 88kg(9.2+ 1.4)= 88kg(11.2m/s)

Yes, that is correct.

Again you are using g twice. a= -1.7 so either 88*9.8- 88*1.7 or 88(9.8- 1.7).

"a" is the acceleration of the elevator as given. You do not add 9.8 to that to get a.

3. Oct 9, 2009

mr.coon

thank you my friend.