# HW Help, Due 11:55 (EST) [4:55 GMT]

1. Jan 29, 2004

### um_alim

I've been trying this question over and over, getting the same exact answer (0.270 m/s^2), however, it doesn't seem to be the right answer.

A man stands on a scale in an elevator that is accelerating upward. The scale reads 795.5 N. When he picks up a 27.0 kg box, the scale reads 1067.4 N. The man's mass is 79.0 kg. What is the acceleration of the elevator?

2. Jan 29, 2004

### deltabourne

Net_F=Fn-m*g=m*a
we know the scale reads 795.5, so that is the normal force.. and his weight is 79.0*9.81 = 774.99 kg*m/s^2

so we get 795.5-774.99 = 79.0kg*a
or a = 0.2596202531645569620253164556962 = 0.260 m/s^2

if you do it with the box, you end up with the same answer.. maybe you had a math error?

3. Jan 29, 2004

### um_alim

Oh, also, here's my work for the problem:

We know that the man, man and the box, situations are all accelerating at the same acceleration. So, we only need to find the acceleration of one situation. (I did the one with the box, but both give relatively the same answer +/- 0.001 m/s^2)

Apparent weight = mass x gravity + mass x acceleration

- solve for acceleration

795.5 N = 79.0 kg x 9.8 m/s^2 + 79.0 kg x (acceleration)
795.5 N = 774.2 N + 79.0 kg x (acceleration)
795.5 N - 774.2 N = 79.0 kg x (acceleration)
21.3 N = 79.0 kg x (acceleration)

[divide by 79.0 kg]

acceleration = 0.2696202532

Which should be the answer, right? But the computer doesn't accept that answer.

4. Jan 29, 2004

### um_alim

That's right! 0.260! Thanks!

5. Jan 29, 2004

### deltabourne

edit: :)

it seems as if it wants you to use 9.81, not 9.8

6. Jan 29, 2004

### um_alim

Yeah, I can't believe it! The textbook itself uses 9.8 m/s^2, yet it won't except answers using those numbers.