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

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

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

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

a F = Fnorm – Fg And F = ma So Fnorm – Fg = ma Side note: m = Fg /g = 700N = 9.8 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 = 914.3 N

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

a F = Fnorm – Fg And F = ma (downward) So Fnorm – Fg = - ma Side note: m = Fg /g = 700N = 9.8 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 = 485.7 N

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

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