1. A 700 N person stands on a bathroom scale in an elevator
A 700 N person stands on a bathroom scale in an elevator. Draw in the two forces that are acting on the passenger. Fg
FNorm

2. Fg = FNorm
While that passenger waits on the elevator to begin his ascent upward, what is the reading on the scale?
700 N Fg
FNorm

3. Fg = FNorm
While that passenger is moving at a constant velocity, what is the reading on the scale?
700 N Fg
FNorm

4. Consider the instant when the elevator initially begins its upward ascent. If its acceleration is 3 m/s2, what is the passenger’s “apparent weight”?
(Solution on next slide) a Fg
FNorm

5. a F = Fnorm – Fg And F = ma So Fnorm – Fg = ma
Side note: m = Fg /g
= 700N = m/s2
71.4 kg
F = Fnorm – Fg a
And F = ma
So Fnorm – Fg = ma
Apparent Weight means solve for Fnorm …
Fnorm = Fg + ma
Fnorm = 700N + (71.4 kg)(3m/s2) Fg
FNorm
Fnorm = N

6. Consider the time interval when the elevator begins to slow down (while going upward). If its acceleration is 3 m/s2, what is the passenger’s “apparent weight”? (first you must determine the direction of acceleration) a Fg
FNorm

7. a F = Fnorm – Fg And F = ma (downward) So Fnorm – Fg = - ma
Side note: m = Fg /g
= 700N = m/s2
71.4 kg
F = Fnorm – Fg a
And F = ma (downward)
So Fnorm – Fg = - ma
Apparent Weight means solve for Fnorm …
Fnorm = Fg - ma
Fnorm = 700N - (71.4 kg)(3m/s2) Fg
FNorm
Fnorm = N

8. Consider the instant when the elevator initially begins its downward descent. If its acceleration is 3 m/s2, what is the passenger’s “apparent weight”? a Fg
FNorm

9. Consider the time interval when the elevator begins to slow down (while going downward). If its acceleration is 3 m/s2, what is the passenger’s “apparent weight”? (first you must determine the direction of acceleration) a Fg
FNorm