Chapter 4 Problem 70

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$^{2}$ ,

(e) the elevator is accelerating downward at 3.0 m/s$^{2}$?

Solution

Newton’s second law for the person:

$\sum F=F_{N}-mg=ma\rightarrow F_{N}=m(g+a)$

The scale always take acceleration as $g$, thus

${\text{scale approximated mass}}={\frac {F_{N}}{g}}=m{\frac {(g+a)}{g}}$

(a)

$a=0$

$F_{N}=mg=\left(75.0{\textrm {kg}}\right)\left(9.80{\textrm {m/s}}^{2}\right)=7.35\times 10^{2}{\textrm {N}}$

${\text{scale approximated mass}}=m{\frac {(g)}{g}}=75.0{\textrm {kg}}$

(b)

$a=0$; $F_{N}=7.35\times 10^{2}{\textrm {N}}$; ${\text{scale approximated mass}}=75.0{\textrm {kg}}$;

(c)

$a=0$; $F_{N}=7.35\times 10^{2}{\textrm {N}}$; ${\text{scale approximated mass}}=75.0{\textrm {kg}}$;

(d)

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

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

(e)

$F_{N}=m(g+a)=\left(75.0{\textrm {kg}}\right)\left(6.80{\textrm {m/s}}^{2}\right)=5.1\times 10^{2}{\textrm {N}}$

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