What Does the Scale Read in Different Elevator Scenarios?

In summary: Whereas, when you write W = mg and F = ma, the W always has the greater magnitude and therefore mg = ma.
  • #1
nofinga911
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Homework Statement


A 192 lb person stands on a scale in an elevator. What does the scale read
a. When the elevator is at rest.
b. when the elevator is accelerating up at 4ft/sec^2
c. when elevator is accelerating down at 3ft/sec^2


Homework Equations





The Attempt at a Solution



Im not really sure how to do this, can someone help me and show the formulas as well so I can learn it? Thanks
 
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  • #2
When the elevator is at rest, the scale will read 192 pounds.
When it is accelerating upward, it will read higher because you not only have the usual mg force pressing the person down but also the ma force necessary to accelerate him upward.

This problem has some messy units - be careful with them. It will be quite difficult to figure out the mass. I'm just old enough to remember that when we had pounds the unit of mass was the "slug". Not joking! Maybe convert everything into metric units?
 
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  • #3
The unit of mass in the US system of measure, is still the slug. 1 slug weighs about 32 pounds : W=mg, where m is in slugs, g is 32ft/sec^2, and W is in pounds.

Nofinga911: What have you tried so far? The formulas will not mean much without your initial attempt.
 
  • #4
PhanthomJay said:
The unit of mass in the US system of measure, is still the slug. 1 slug weighs about 32 pounds : W=mg, where m is in slugs, g is 32ft/sec^2, and W is in pounds.

Nofinga911: What have you tried so far? The formulas will not mean much without your initial attempt.
Thats just it, I don't know the formulas...It would help me if I saw it done by someone else with the formulas this way I have some time to studyit before my test tomorrow..it was one of two out of the 20 questions I was unable to get. I would be able to break the formula down if I saw it done.
 
  • #5
The formulas you need are F = mg and F = ma.
Yes, let us see you put the numbers in and do some multiplying!
Do you have the number for g in ft/s^2 ?
 
  • #6
The problem statement said that this was a 192 lb person, which means that we can assume that at rest on the surface of the earth, this person weight 192 lb.

Delphi51, how do you figure that for the elevator at rest, the scale will read 10 lb? Why not 192 lb?

By the way, Delphi51, I think you probably meant to say W=mg and F=ma, did you not?
 
  • #7
Terribly sorry - I meant to type 192! That must have been very confusing.

What is the difference between F = mg and W = mg?
Weight is the Force of gravity so either letter should be okay.
The two forces must be added to get the answers.
 
  • #8
What is the difference between F = mg and W = mg?

The difference, as I see it, is that when you write F = mg and F = ma, there is a natural equivalence implied that therefore mg = ma and therefore g = a.
 

FAQ: What Does the Scale Read in Different Elevator Scenarios?

What is the "Elevator/ Scale problem"?

The "Elevator/ Scale problem" is a classic thought experiment in physics that explores the concept of apparent weight. It involves a person standing on a scale inside an elevator that is either accelerating upwards or freefalling downwards.

How does the apparent weight change when the elevator is accelerating upwards?

When the elevator is accelerating upwards, the apparent weight of the person on the scale increases. This is because the normal force from the scale must increase to match the person's weight and the force of acceleration.

How does the apparent weight change when the elevator is freefalling downwards?

When the elevator is freefalling downwards, the apparent weight of the person on the scale decreases. This is because the normal force from the scale decreases as the person and the scale are in a state of weightlessness due to the lack of contact with the floor.

What happens to the apparent weight when the elevator is at a constant velocity?

When the elevator is at a constant velocity, the apparent weight of the person on the scale remains the same as their actual weight. This is because there is no acceleration or deceleration occurring, so the normal force from the scale is equal to the person's weight.

How does the mass of the person affect the apparent weight in the "Elevator/ Scale problem"?

The mass of the person does not affect the apparent weight in the "Elevator/ Scale problem." The apparent weight is solely dependent on the acceleration or deceleration of the elevator and is independent of the person's mass.

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