Question

A 95.0kg person stands on a scale in an elevator. What is the apparent weight when the elevator is (a) accelerating upward with an acceleration of 1.80 m/s^{2}, (b) moving upward at a constant speed and (c) accelerating downward with an acceleration of 1.30 m/s^{2}?

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asked 2021-02-11
A 95.0kg person stands on a scale in an elevator. What is the apparent weight when the elevator is (a) accelerating upward with an acceleration of 1.80 \(\displaystyle\frac{{m}}{{s}^{{{2}}}}\), (b) moving upward at a constant speed and (c) accelerating downward with an acceleration of 1.30 \(\displaystyle\frac{{m}}{{s}^{{{2}}}}\)?

Answers (1)

2021-02-12

There are 2 forces acting on the person standing on thescale.
W = Mg the weight of the person due to gravity is actingdownward.
F(scale) = the reading of the scale is acting upward.
The resultant of these 2 forces provides the acceleration ofthe person on the scale.
So Mg - F(scale) = Ma
Solve for F(scale) from the values given. Obviously if a(acceleration of the elevator) = 0
then Ma = 0 and F(scale) = Mg.
a) When the elevator is moving upward then N = M( g + a)
= 95( 9.8 + 1.8)
\(= 1102\ kg-m^{2}/S\)
b) when the elevator moving downward then
N = M( g - a)
= 95( 9.8 -1.3)
\(= 807.5\ kg-m^{2}/S\)

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