View source for Chapter 4 Problem 70

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==Problem==
A 75.0-kg person stands on a scale in an elevator. What does the scale read (in N and in kg) when 

(a) the elevator is at rest, 

(b) the elevator is climbing at a constant speed of 3.0 m/s, 

(c) the elevator is descending at 3.0 m/s, 

(d) the elevator is accelerating upward at 3.0 m/s<math>^2</math> , 

(e) the elevator is accelerating downward at 3.0 m/s<math>^2</math>?

==Solution==
[[File:Chapter4Problem70s.png|200px|center|diagram]]

Newton’s second law for the person:

<math>\sum F =F_N-mg=ma\rightarrow F_N=m(g+a)</math>

The scale always take acceleration as <math>g</math>, thus

<math>\text{scale approximated mass}=\frac{F_N}{g}=m\frac{(g+a)}{g}</math>

===(a)===
<math>a=0</math>

<math>F_N=mg=\left(75.0 \textrm{kg}\right)\left(9.80 \textrm{m/s}^2\right)=7.35\times 10^2 \textrm{N}</math>

<math>\text{scale approximated mass}=m\frac{(g)}{g}=75.0 \textrm{kg}</math>

===(b)===
<math>a=0</math>;
<math>F_N=7.35\times 10^2 \textrm{N}</math>;
<math>\text{scale approximated mass}=75.0 \textrm{kg}</math>;

===(c)===
<math>a=0</math>;
<math>F_N=7.35\times 10^2 \textrm{N}</math>;
<math>\text{scale approximated mass}=75.0 \textrm{kg}</math>;

===(d)===

<math>F_N=m(g+a)=\left(75.0 \textrm{kg}\right)\left(12.80 \textrm{m/s}^2\right)=9.60\times 10^2 \textrm{N}</math>

<math>\text{scale approximated mass}=\frac{960\textrm{N}}{9.80 \textrm{m/s}^2}=98.0 \textrm{kg}</math>

===(e)===
<math>F_N=m(g+a)=\left(75.0 \textrm{kg}\right)\left(6.80 \textrm{m/s}^2\right)=5.1\times 10^2 \textrm{N}</math>

<math>\text{scale approximated mass}=\frac{510\textrm{N}}{9.80 \textrm{m/s}^2}=52.0 \textrm{kg}</math>