Gravitational Acceleration Problem

1. Apr 2, 2009

BandGeek13

1. The problem statement, all variables and given/known data
A 95 kg man jumps, therefore exerting a force of 1000 N on a bathroom scale. The scale is in a motionless elevator in Earth's gravity. What is the man's acceleration?

2. Relevant equations
Newton's 2nd Law: F=ma
Possibly Fg=mg ?

3. The attempt at a solution
I tried just using F=ma.
therefore, it'd be:
1000 N=(95kg)(a)
a=10 m/s^2

But this does not take into consideration the earth's gravity.

2. Apr 2, 2009

Redbelly98

Staff Emeritus
Welcome to PF.

"F=ma" uses the net force (vector sum of all forces acting on the man)

Your F uses the 1000N force that the scale exerts on the man. As you suspect, you need to add the force due to Earth's gravity. Just take the vector sum of the scale force and the force due to gravity.

3. Apr 2, 2009

BandGeek13

So would it be:

F(normal)-F(gravity)=ma
F(normal) - mg=ma
1000 N [up] - (95 kg)(9.8 N/kg [down])=(95 kg)(a)
1000 N [up] - 931 N [down] = (95 kg)(a)
69 N [up] = (95 kg)(a)
0.73 m/s^2 = a

This seems very low..

4. Apr 2, 2009

BandGeek13

wait!

69 N [up] = (95 kg)(a)
0.73 m/s^2 = a

is it:
1931 N [up] = (95 kg)(a)
20. m/s^2 = a

5. Apr 2, 2009

Redbelly98

Staff Emeritus
Looks good to me.

Note, the man's weight is about 930 N, so the scale would read 930 N when a=0.

The scale reading of 1000N is not much more than this, so expect a to be considerably less than g=9.8 m/s^2.

An acceleration of g=9.8 m/s^2 upward would require a scale reading of 2*930N, which would give Fnet = +930 N upward.

EDIT:
Nope. We have +1000N (upward) and -931N (downward), for Fnet=31N upward

6. Apr 2, 2009

Alright.
Thank you!!