Question

A 75 kg person stands on a scale in an elevator. What does the scale read when
a) elevator is at rest
b) elevator goes up at a constant speed of 3 m/s
c) elevator falls at a constant speed of 5 m/s

Answer 1

i don't understand the work.
for b and c, i know that a=0 m/s^2

Answer 2

anyone? @agent0smith

Answer 3

Yeah I'll try to work it out for you :)

Answer 4

thank you.

Answer 5

This is one of those kinda trick question problems... as long as the elevator doesn't accelerate, the scale will always read the same.

Answer 6

If the elevator moves at constant speed, there is no change in the normal force from the floor acting on the person. It's the same as the elevator being at rest... does that make sense?

Answer 7

Compare it to when the elevator is accelerating - if the elevator is in freefall, what will the scale read?

Answer 8

The way they show the working is a bit confusing, though, but basically, in all three cases, the normal force N = mg, or N-mg = 0.

Answer 9

The net force on the person, F = ma, is equal to the sum of the normal force N acting upward, and the weight mg acting downward. So:

\[ma = N - mg\] so rearrange for N: \[N= ma+mg\] or \[N = m(a+g) \]

Answer 10

The scale basically calculates the weight of the person by using the normal force and dividing it by g to show the person's weight in kg (m-scale is the reading from the scale):
\[\frac{ N }{ g }= m _{scale}\]
so dividing the equation for N above by g:
\[m _{scale}= \frac{ N }{ g} = \frac{ m(a+g) }{ g }\]
hopefully this is helping, Jenny :P

Answer 11

so basically m(scale) and m are two different m's?
cuz if i look at it that way, it makes more sense to me.

Answer 12

Correct. m is the actual mass of the person, 75kg. Mass never changes, but weight depends on acceleration or gravity. The mass shown on the scale, is not *really* mass, unless the person is at rest.

For example, when the elevator accelerates upward at 9.8m/s, the mass on the scale would show as = 75(9.8+9.8)/9.8 = 150kg.

When the elevator is in freefall, or accelerating down at -9.8m/s, the mass on the scale would read = 75(-9.8+9.8)/9.8 = 0kg and the person feels weightless.

In this case though, a is 0 in all three cases.

Answer 13

ok thank you for clarifying :D

Answer 14

You're welcome :D

Scales measure weight, not mass. If you took that scale to the moon and stood on it at rest, it would not read 75kg, it'd read
\[m _{scale}=75(0+1.6)/9.8 = 12.2 kg\]
since acceleration due to gravity on the moon is 1.6m/s^2, but the scale is still dividing that force by 9.8m/s^2.

Answer 15

or on jupiter, the scale would read:
75(0+26.0)/9.8 = 199kg
since the value of g on jupiter is 26.0m/s^2, but again, the scale still divides everything by 9.8m/s^2 because it's designed for use on earth :P