A Please Explain Elementary Physics Elevator Question

[Timbre]

TL;DR Summary

I was taught as everyone, the basic elevator changing velocity question. Yet nobody has ever explained this, it's held as obvious. Please explain.

Hello,

I'm joining this forum to ask two questions which have nagged me for some time. I am in no way trolling. They both are presumed obvious, yet don't make sense to me. Nobody will explain their positions, which is...uh...aka science. I also have a thread for the other question.


Yes, I'm questioning the most elementary physics question we're given in this world.

The classic elevator in motion question: A person is standing on a scale in an elevator that is in constant motion. The elevator begins to negatively accelerate to stop at the designated floor. What does the scale read when the elevator begins to decelerate (negatively accelerate), change velocity.

The answer is supposedly always that the scale reads less than it did when the person stood on it when the elevator was traveling at a constant rate.

I don't understand how physics uses acceleration of gravity yet they'll say that the rate of deceleration of the elevator does not factor.

Now, I'll say upfront that I understand this is a question used to illustrate a basic concept. And it presumes things like the distance from the center of the earth doesn't factor (less gravity further away), and factors like a spring in the scale (potentially exerting force to push the person and the elevator apart) also does not factor.

But I do factor the acceleration of gravity. Is the simple answer that this model is meant to be very simple and doesn't factor gravity as acceleration, it does so as a force, this is 101 physics, so that's why it doesn't make sense to me? I can accept that, and if that's the case, please tell me so and skip the rest of this.


The reason I don't understand this, is that gravity is moving things closer together, changing the rate that their distance differs. So why wouldn't that factor in the elevator example. Why can't the elevator decelerate at a rate that matches or is below that of the acceleration of gravity? And the scale would then read the person's weight while decelerating.


I could break it down if anyone would like: If we focus closer and closer in time, it appears paradoxical to me. So, the elevator is in X relation, distance from the person. The elevator begins to change velocity, slows. The person would continue in motion because no other force is acting on him, he's not tethered to the scale/elevator. But at that same moment he'd begin to move away from the elevator floor, gravity would also be moving him toward it.

Because gravity is a change in velocity, and thus the elevator's change in velocity could be canceled by gravity's, no? To a point. If the elevator decelerated more rapidly than 9.8 m/s/s, then it'd pull away from the person and the distance between person and elevator would increase, registered as "lighter" on the scale.

And exactly at 9.8 m/s/s? I'm unsure how to answer that, haven't thought it through. I assume the scale would read the same, as it'd match deceleration of the elevator/acceleration of gravity.

This also reflects a question of which physics equations to use, which I've never been clear on. Sometimes gravity is treated as a force and other times acceleration, and I've never understood how or why. I understand it's a force in a static equation, but acceleration when bodies are moving apart...but that's again to the crux of my elevator question, wouldn't we switch equations the moment the elevator began to decelerate? Why would we use the same equation with gravity as a force, in a static diagram, when the elevator begins to change velocity, and it's presumed in the supposed right answer that the bodies (elevator and person) are separating.

Thanks, All. :)

 

[kuruman]

Do you understand what a (bathroom) scale reads? If so, then please post it here in your own words. It is a good starting point and we'll take it from there. It is not at clear to me what question you are asking in post #1, but before attempting to answer your questions, we need first to ascertain what you don't understand.

 

[Timbre]

A bathroom scale reads weight, but we call that mass, force if using gravity as a constant static force, or in this situation it indicates the relative position of two objects, the person and the elevator (and assumes that the scale exerts no force to push the objects apart (such as an internal spring in the scale).

The question is why do we use an equation where gravity is a force, a constant, in a static equation, when answer to the question presumes that the bodies are moving apart, in a kinetic state, where gravity would be acceleration, not a constant force. (The "correct" answer is that the scale reads less when the elevator begins to decelerate than it did when at a constant velocity.)

 

[Lnewqban]

The classic elevator in motion question: A person is standing on a scale in an elevator that is in constant motion. The elevator begins to negatively accelerate to stop at the designated floor. What does the scale read when the elevator begins to decelerate (negatively accelerate), change velocity.

Without the interference of the elevator, that person would be freely falling.
That interference can be minimum if the cables break, or maximum when it starts moving upwards.

If the elevator could accelerate upwards at three times the magnitude of the acceleration of gravity (3g), the scale would show three times the normal weight of the person.

If the elevator could be propelled down, accelerating downwards at three times the magnitude of the acceleration of gravity (3g), the scale would show no weight of the person, whose head would be pressing on the ceiling of the elevator with a force of 2mg.

 

[kuruman]

A bathroom scale reads weight, but we call that mass, force . . .

A bathroom scale uses a spring, or strain gauge or other such sensor to display the force pushing down on its surface divided by the acceleration of gravity $g = 9.8~\text{m/s}^2$. It does not read weight or mass. Weight is the force with which the Earth attracts a body and that force is the same regardless of whether the body is on a scale or not.

You can easily verify that by pushing down on a scale with your hand. The reading on the scale will depend on how hard you push down and is not your weight. Note that Newton's 3rd law says that if you push down on the scale with force $F$ the scale pushes up on your hand with force $F$.
So, if you are standing on a scale in an elevator the reading on the scale will be

1. equal to your weight divided by $g$ if the elevator is moving at constant velocity because the scale has to support only your weight;

2. greater than your weight divided by $g$ if the elevator is accelerating up because the scale has to support all your weight plus accelerate your body up against gravity;

3. less than your weight divided by $g$ if the elevator is accelerating down because the scale has to support only part of your weight while gravity does the rest;

4. zero if the elevator is in free fall because it doesn't matter whether you are standing on the scale or not since you cannot push down on it. Nevertheless, your weight is not zero because the Earth still attracts you with force $mg.$

 

[Herman Trivilino]

The reason I don't understand this, is that gravity is moving things closer together, changing the rate that their distance differs.

No it doesn't! Gravity tends to move things closer together, because the gravitational force is always attractive. Part of your confusion may be from mixing up the gravitational force with the gravitational acceleration. If the gravitational force is the only force acting on an object near Earth's surface, the object will accelerate at about 9.8 m/s², the so-called gravitational acceleration.

If another force also acts on that object, we can calculate its acceleration by looking at the net force. So, for example, if my mass is 100 kg, Earth exerts a downward gravitational force on me that has a magnitude of about 980 N. If I'm at rest standing on a bathroom scale the scale exerts an upward force on me of magnitude 980 N, we call this the normal force. Thus the net force acting on me is zero and my acceleration is zero.

A bathroom scale reads weight

No, it reads the magnitude of the normal force! So if I'm at rest (or equivalently moving with a constant velocity) it reads 980 N. But if I'm not at rest or not moving with a constant velocity it will not read 980 N. Let's say I'm accelerating and the scale reads 880 N. Therefore there's a downward gravitational force acting on me of magnitude 980 N and an upward normal force of magnitude 880 N. The net force is therefore downward and has a magnitude of 100 N. If I divide the net force by my mass, I get a downward acceleration that has a magnitude of 1.0 m/s². Thus I could be in an elevator moving upward and slowing down. (Or the elevator could be moving downward and speeding up).

 

[russ_watters]

The question is why do we use an equation where gravity is a force, a constant, in a static equation, when answer to the question presumes that the bodies are moving apart, in a kinetic state, where gravity would be acceleration, not a constant force. (The "correct" answer is that the scale reads less when the elevator begins to decelerate than it did when at a constant velocity.)

Sounds like you more or less understand it, but you're talking about it vaguely instead of picking specific examples and doing the math to figure out what happens. It's not complicated if you do that.

The scale reads force which arises in this situation from f=ma. Gravity (g=9.8 m/s^2)* is always there, and the elevator adds its own a: f=m(g+a). Let's say you weigh 50kg. With no additional acceleration the scale shows your weight as 490N. Pick a scenario and its acceleration, plug and chug.

For example, elevator decelerates at half a g. What does the scale read?

*Note: I guess technically the scale is reading the normal force applied by the scale, thus up is positive.

 

[Dale]

Sometimes gravity is treated as a force and other times acceleration, and I've never understood how or why.

Gravity is treated as a force classically.

In the special case that the object is in free fall, meaning that no other force acts on it, then by Newton’s second law $$ma=F=mg$$$$a=g$$ So in that case (free fall) and in that case only treating it as acceleration is equivalent to treating it as force.

In this case the object is not in free fall, so you treat gravity as a force.

But in a free fall case you can still choose to treat gravity as a force. They are equivalent, so if you find switching to be confusing then just always treat it as a force.

 

[A.T.]

I was taught as everyone, the basic elevator changing velocity question. Yet nobody has ever explained this, it's held as obvious.

If it's not obvious, do the math, systematically without any intuitive reasoning. Make sure you use a consistent sign convention. All your confusion seems to stem from an attempt to describe the quantitative relationships in natural a language, which is not precise enough.

Sometimes gravity is treated as a force and other times acceleration,

Newtonian gravity is a force, that is proportional to the mass of the body it's acting at. Thus the gravitational field strength (normalized force) is defined as force / mass, which per Newton's 2nd law of motion represents the acceleration of a free falling object, with no forces other than gravity acting on it.

 

[Herman Trivilino]

Gravitational acceleration (g=9.8 m/s^2)* is always there,

I know what you mean. $g$ always has a magnitude of 9.8 m/s² but objects don't always accelerate at this rate, they have to be in free fall.

 

[Herman Trivilino]

Sometimes gravity is treated as a force and other times acceleration, and I've never understood how or why.

This lies at the root of your misunderstanding. Gravity is the name we give to the phenomenon.

There's gravitational force and gravitational acceleration. Two separate things.

When you're in an introductory physics course the students and often the instructor will refer to the quantity $g$ as "gravity", which is a mistake that has likely led to your misunderstanding. Just because the quantity $g$ appears in an equation does not mean an object is actually accelerating at that rate.

For this and other reasons I always refer to $\vec{g}$ as the free fall acceleration, and write its magnitude as $g$ which always has a value of +9.8 m/s².

 

[russ_watters]

I know what you mean. $g$ always has a magnitude of 9.8 m/s² but objects don't always accelerate at this rate, they have to be in free fall.

I've made a slight edit but I'm open to suggestions on the wording. Your weight barring additional accelerations (force applied by the scale) is the force required to cancel/oppose gravitational acceleration.

...though ilI don't want this to become more about semantics when the math is what matters. As long as the wording gets to the right math, I think it's ok. Similarly with sign conventions.

 

[Timbre]

I know what you mean. $g$ always has a magnitude of 9.8 m/s² but objects don't always accelerate at this rate, they have to be in free fall.

Thank you. I think this and your next post answers my question.

To repeat it back to illustrate that I understand it:

The gravity as a force and acceleration are two different things, but they're combined in elementary physics shorthand. An object must be in freefall for it to accelerate and for that 9.8 m/s/s change in velocity, aka acceleration, to come into play.

"Acceleration" is also the term used for the elevator leaving a floor and approaching steady-state cruising speed (if you will), or decelerating (negative acceleration) to stop at the desired floor. This form of "acceleration" IS measured on the scale.

But the 9.8 m/s/s gravity "acceleration", while using the same term, is not apparent on the scale because the two objects (person and elevator/scale) are touching, are not in freefall, which is required to factor in the gravity acceleration.

So gravity only as a force, the appropriate equations of using it as a force, apply to the elevator scenario. Because the objects, person/scale/elevator are always touching.

Is that a (pedantically lol) accurate summary?

 

[kuruman]

. . . but they're combined in elementary physics shorthand.

If by "combined in elementary physics shorthand" you mean Newton's second law $\mathbf F_{\text{net}}=m\mathbf a$, then yes.

An object must be in freefall for it to accelerate . . .

You get in your car, start the engine and push on the gas pedal to get the car to accelerate from rest without being in free fall (non-pedantically lol). Do not generalize unnecessarily because imprecise language leads to faulty thinking and hence confusion.

 

[A.T.]

.... to come into play ... to factor in ....

Instead of using those vague terms you should learn the simple formulas, and apply them.

 

[Herman Trivilino]

The gravity as a force and acceleration are two different things, but they're combined in elementary physics shorthand.

No, they are not combined. Force and acceleration are two distinct things and they are very much kept separate.

The terminology is what's confusing you. As I said before, gravity is a phenomenon. We can describe it using many tools. Gravitational force and gravitational acceleration are just two of those tools.

An object must be in freefall for it to accelerate and for that 9.8 m/s/s change in velocity, aka acceleration, to come into play.

An object in free fall near Earth's surface has an acceleration of 9.8 m/s², but there are many other situations having nothing to do with gravity where an object can have an acceleration of 9.8 m/s². In all cases it means the velocity is changing at a rate of 9.8 m/s per second. In the case of constant acceleration it means the velocity changes by 9.8 m/s every second.

"Acceleration" is also the term used for the elevator leaving a floor and approaching steady-state cruising speed (if you will), or decelerating (negative acceleration) to stop at the desired floor. This form of "acceleration" IS measured on the scale.

The term is not "also" used for anything else. It always refers to the rate of change of velocity.

But the 9.8 m/s/s gravity "acceleration", while using the same term, is not apparent on the scale because the two objects (person and elevator/scale) are touching, are not in freefall, which is required to factor in the gravity acceleration.

In Post #6 I mentioned that if I have a mass of 100 kg then I have a weight of 980 N, this is calculated using $w=mg$. It's true regardless of my state of motion. It's not "required" that I have an acceleration of 9.8 m/s².

So gravity only as a force, the appropriate equations of using it as a force, apply to the elevator scenario. Because the objects, person/scale/elevator are always touching.

Is that a (pedantically lol) accurate summary?

No. The gravitational force applies to any situation where gravity is present, you don't have to be in an elevator and you don't have to be touching anything. But you can.

I suggest that instead of trying to write generalizations of what I wrote, you ask specific questions about what I wrote. You have to understand something before you can make generalizations about it.

 

[russ_watters]

You can rewrite the force equation to f=w+ma to mostly avoid dealing with the concept of gravitational acceleration if it trips one up. Whatever gets one to math that provides correct answers about what the scale reads is ok. I just prefer not sweeping under the rug where the "w" comes from.

 

[Timbre]

No, they are not combined. Force and acceleration are two distinct things and they are very much kept separate.

The terminology is what's confusing you. As I said before, gravity is a phenomenon. We can describe it using many tools. Gravitational force and gravitational acceleration are just two of those tools.


An object in free fall near Earth's surface has an acceleration of 9.8 m/s², but there are many other situations having nothing to do with gravity where an object can have an acceleration of 9.8 m/s². In all cases it means the velocity is changing at a rate of 9.8 m/s per second. In the case of constant acceleration it means the velocity changes by 9.8 m/s every second.


The term is not "also" used for anything else. It always refers to the rate of change of velocity.


In Post #6 I mentioned that if I have a mass of 100 kg then I have a weight of 980 N, this is calculated using $w=mg$. It's true regardless of my state of motion. It's not "required" that I have an acceleration of 9.8 m/s².


No. The gravitational force applies to any situation where gravity is present, you don't have to be in an elevator and you don't have to be touching anything. But you can.

I suggest that instead of trying to write generalizations of what I wrote, you ask specific questions about what I wrote. You have to understand something before you can make generalizations about it.

I realized after I wrote that that I could've stated this better: That gravity is a force. And it's effects are different depending on whether an object is touching something else or in freefall. Sound good? I think I understand this, but maybe you'll correct that. Avoiding semantics.

If touching, it's a force holding an object.

It's the same force, but in lieu of anything opposing it (like an elevator floor or even wind resistance) it will accelerate the object, because it's always applying the force. It doesn't "get the object up to speed" then stop applying that force, it's always present thus the object will always accelerate. Until that object is against another equal and opposite force, such as a thick fluid like air at 120mph or so, when something may reach terminal velocity. Or a floor of an elevator.

Then the force is apparent as a static force, holding something. But gravity is always a force, but it acts differently depending on the object's contact with another opposing object?

(A skydiver with a scale under his feet at terminal velocity and the same person standing on that scale in an elevator, the scale will read the same.)

So that skydiver, if he exited a stationary hot air balloon, would accelerate initially at 9.8 m/s/s. Because no wind resistance to speak of. But as he was at, say 100mph, he'd be accelerating less rapidly, and the reading on the scale would be increasing, until he reached equilibrium called terminal velocity. The scale reading part of his weight while he's still accelerating before terminal velocity, indicates a split of the force of gravity, part of it is a constant force on the scale, part of it is still accelerating. So at initial jump out of the balloon, it's pure acceleration, and at terminal it's pure static force, but in principle/effect it's always the same force.