A 48 kg passenger rides in an elevator that starts from rest on the ground floor of a building at t = 0 and rises to the top floor during a 10 s interval. The acceleration of the elevator as a function of the time is shown in Fig. 3-32, where positive values of the acceleration mean that it is directed upward. Give the magnitude and direction of the following forces.
Fig. 3-32 (in words)
-max positive acceleration = 2 m/s^2 at t = 2 s
-max negative acceleration = -3 m/s^2 at t = 8.5 s
-0 acceleration from 4 s to 7 s
(a) the maximum force on the passenger from the floor
(b) the minimum force on the passenger from the floor
(c) the maximum force on the floor from the passenger
F = ma
The Attempt at a Solution
For part a), I thought it would just be F =(48kg)(2), but that was not correct. I know it's in an upward direction, but I can't seem to find the magnitude.
Part b) I figured that the minimum force should be when the acceleration = 0, so I tried F=(48)(0) = 0. But it just doesn't seem logical to me...
Part c) I think the maximum force should be when a = -3, but I plugged it into the formula and it was not correct.
Any help would be appreciated, thanks!