Question

1. solve for the unknown

solve for f scale for the elevator at rest: f net = ma = 0 n.
(84.0 kg)(9.8 n/kg)
=823.2 n
solve for f scale for the elevator accelerating upward: f net = ma.
(84.0 kg)(2.25 m/s²)+(84.0 kg)(9.8 n/kg)
=1,012.2 n

Explanation:

Step1: Analyze when elevator is at rest

When the elevator is at rest, \(F_{net}=ma = 0N\). According to Newton's second - law \(F_{scale}-mg=ma\). Since \(a = 0\), then \(F_{scale}=mg\). Given \(m = 84.0kg\) and \(g=9.8N/kg\), we have \(F_{scale}=(84.0kg)\times(9.8N/kg)=823.2N\).

Step2: Analyze when elevator is accelerating upward

When the elevator is accelerating upward, \(F_{net}=ma\). Using Newton's second - law \(F_{scale}-mg = ma\), so \(F_{scale}=m(a + g)\). Given \(m = 84.0kg\), \(a=2.25m/s^{2}\), and \(g = 9.8m/s^{2}\), then \(F_{scale}=84.0kg\times(2.25m/s^{2}+9.8m/s^{2})=84.0\times12.05N = 1012.2N\).

Answer:

When the elevator is at rest, the scale reading \(F_{scale}=823.2N\). When the elevator is accelerating upward, the scale reading \(F_{scale}=1012.2N\).