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}?

Falak Kinney 2021-02-11 Answered
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 ms2, (b) moving upward at a constant speed and (c) accelerating downward with an acceleration of 1.30 ms2?
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Expert Answer

Viktor Wiley
Answered 2021-02-12 Author has 84 answers

We have 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) Elevator is moving upward N = M( g + a) = 95( 9.8 + 1.8)=1102 kgm2/S


b) Elevator moving downward N = M( g - a) = 95( 9.8 -1.3)=807.5 kgm2/S

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