Question

determine the force that a 65.0 kg man exerts on the bottom of an elevator that is
a. at rest.
b. accelerating up at 4.90 m/s².
c. moving at constant velocity down.
d. moving down at 1.23 m/s².

Explanation:

Step1: Recall Newton's second law

Newton's second law is \(F = ma\), where \(F\) is the net force, \(m\) is the mass, and \(a\) is the acceleration. The force exerted by the man on the elevator floor \(F_N\) (normal force) and the force of gravity \(F_g=mg\) act on the man. The net force \(F_{net}=F_N - F_g\) (taking up as positive).

Step2: Solve part (a)

When the elevator is at rest, \(a = 0\).
By Newton's second law \(F_{net}=ma\), so \(F_N -mg=0\).
Given \(m = 65.0\space kg\) and \(g = 9.8\space m/s^{2}\), then \(F_N=mg\).
\(F_N=(65.0\space kg)\times(9.8\space m/s^{2}) = 637\space N\)

Step3: Solve part (b)

When the elevator is accelerating up \(a = 4.90\space m/s^{2}\).
By Newton's second law \(F_N -mg=ma\).
\(F_N=m(g + a)\)
Substitute \(m = 65.0\space kg\), \(g = 9.8\space m/s^{2}\), \(a=4.90\space m/s^{2}\)
\(F_N=(65.0\space kg)\times(9.8 + 4.90)\space m/s^{2}=(65.0\space kg)\times(14.7\space m/s^{2})=955.5\space N\)

Step4: Solve part (c)

When the elevator is moving at constant velocity \(a = 0\).
By Newton's second law \(F_N -mg=0\), so \(F_N=mg\)
\(F_N=(65.0\space kg)\times(9.8\space m/s^{2}) = 637\space N\)

Step5: Solve part (d)

When the elevator is moving down with \(a = 1.23\space m/s^{2}\), taking up as positive \(a=- 1.23\space m/s^{2}\)
By Newton's second law \(F_N -mg=ma\)
\(F_N=m(g + a)\)
Substitute \(m = 65.0\space kg\), \(g = 9.8\space m/s^{2}\), \(a=-1.23\space m/s^{2}\)
\(F_N=(65.0\space kg)\times(9.8-1.23)\space m/s^{2}=(65.0\space kg)\times(8.57\space m/s^{2}) = 557.05\space N\)

Answer:

a. \(637\space N\)
b. \(955.5\space N\)
c. \(637\space N\)
d. \(557.05\space N\)