A 72 kg passenger rides in an elevator that starts from rest on the ground floor

Question

A 72 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.

when t=2s, a = 2 m/s^2
when t=5s, a = 0 m/s^2
when t=8.5s, a = -3m/s^2

find:
(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

Explanation / Answer

N(a)= mg+ma = (72*9.8)+(72*2) = 849.6 upward N(b)= mg+ma = (72*9.8)+(72*(-3))= 489.6 upward N(c)= mg+ma = (72*9.8)+(72*2)= 849.6 downward

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