Calculating your weight in an elevator

[Chenkel]

TL;DR Summary

This is a less common question to a classical physics problem, involving measurement of the weight of an object on a moving platform.

I was watching one of Walter Lewin's lectures, he gave an example of a scale placed at your feet in a moving platform, apparently your weight changes when the moving platform accelerates downward, my question is if my reasoning is correct. I'm wondering why your feet stay in contact with the scale when there's a down force applied to the elevator, but not to you. My reasoning goes like this, unless the platform is accelerating down faster than gravity, you will be accelerating faster than the platform, so the moment the platform moves a vertical distance downward from your feet, your feet will move through that distance to come back in contact with the platform. If your body and feet are in contact with the scale, then you can write $mg - F_s = ma$ where $F_s$ is the reaction force of the scale on your body, then you can find $F_s = m(g - a)$ this has the same magnitude as the force of the body onto the scale, according to Newton's third law, so we will read a 'reduced' weight when the platform is accelerating downwards.

Let me know what you guys think, thanks!

 

[Delta2]

Your view is correct I believe, if you want an alternative view:
Because our feet cannot penetrate into the scale because feet and scale are solid objects, they both move down with the same acceleration. But if the only force to our feet was that of gravity then they would move downwards with g which is greater than the acceleration a. Hence there must be a force (which is a contact normal force ) $F_s$ from the scale to our feet such that $mg-F_s=ma$. If $a>g$ then $F_s$ must be negative (so that that equation still holds), that is it must have direction same as gravity, but contact forces can be only repulsive and not attractive. Hence in this case it becomes $F_s=0$ and the equation doesn't hold but we don't need it to hold as long as our feet and the scale don't move with the same acceleration. And yes this is what it will happen, we will fall with acceleration g, while the elevator falls with acceleration a>g (and eventually we ll hit the ceiling of the elevator lol).

 

[Mister T]

so we will read a 'reduced' weight when the platform is accelerating downwards.

$F_s$ is reduced, but $mg$ stays the same.

So, it's a matter of semantics as to whether the thing you call weight is $mg$ or $F_s$.

 

[kuruman]

$F_s$ is reduced, but $mg$ stays the same.

So, it's a matter of semantics as to whether the thing you call weight is $mg$ or $F_s$.

I disagree. There are two distinct forces here. The weight is the force with which the Earth attracts the mass. Near the surface of the Earth it is constant and equal to $mg$. $F_s$ is the normal force exerted by whatever on the surface of the scale. The scale reports the magnitude of that force. It just happens that the reading on the scale is equal to one's weight when one is just standing on the scale and the scale is not accelerating. It is not one's weight when one is pushing down on the scale with one's hand.

 

[hmmm27]

I'm wondering why your feet stay in contact with the scale when there's a down force applied to the elevator

Elevators - as far as I know - have a rope attached to the top, not the bottom. The mechanism does not apply a downwards force to the cab, it allows the cab to succumb to the acceleration of gravity by a limited amount.

Therefore, the most downwards acceleration an elevator will have = g, and there's no reason for your feet to leave the scale.

 

[Delta2]

Elevators - as far as I know - have a rope attached to the top, not the bottom. The mechanism does not apply a downwards force to the cab, it allows the cab to succumb to the acceleration of gravity by a limited amount.

Therefore, the most downwards acceleration an elevator will have = g, and there's no reason for your feet to leave the scale.

You can make elevators that have ropes both up and down, so you can move the elevator with acceleration a>g as long as the down rope is pulling.

 

[hmmm27]

You can make elevators that have ropes both up and down, so you can move the elevator with acceleration a>g as long as the down rope is pulling.

I'm not sure that would be a popular mode of transportation. In a previous post, the OP evinced the opinion that if a force holding 2 objects together is lessened (but still >= 0), the objects will separate.

 

[Delta2]

I'm not sure that would be a popular mode of transportation.

Popular or not, just saying you can make it.

In a previous post, the OP evinced the opinion that if a force holding 2 objects together is lessened (but still >= 0), the objects will separate.

I don't think this is always the case, but which thread is that so I can read the whole context...

 

[hmmm27]

Popular or not, just saying you can make it.

Not saying you couldn't make it ; just saying it's not the sort of thing you'd want to forget about early Monday morning.

I don't think this is always the case, but which thread is that so I can read the whole context...

Sry, my bad : Newton's third law (equal and opposite force)

 

[kuruman]

Elevators - as far as I know - have a rope attached to the top, not the bottom. The mechanism does not apply a downwards force to the cab, it allows the cab to succumb to the acceleration of gravity by a limited amount.

Therefore, the most downwards acceleration an elevator will have = g, and there's no reason for your feet to leave the scale.

The building (now demolished) where I worked for a number of years was opened in 1970 and had a single elevator that functioned as a freight and people elevator. Its construction was unconventional in that the cab was pushed by a telescoping piston from below. The building was 5 stories high with the physics department on the top 2 floors. We could move heavy equipment without worrying about exceeding weight limits. I got stuck in it a few times and had no worry of dropping to a premature death before help arrived.

If you think about it for a moment, the free body diagram of the contents in an accelerating elevator is the same regardless whether the upward vertical force (in addition to gravity) acting on the elevator is pushing from below or pulling from above. The contents will "succumb to gravity" identically and wouldn't know the difference. I can attest to that.

 

[PeroK]

The weight is the force with which the Earth attracts the mass.

I don't necessarily disagree with this definition, but it is at odds with the concept of weightlessness when in freefall.

IMO, the whole concept of weight should be expunged from physics textbooks. There is mass and there is force and that's that.

 

[Delta2]

I don't necessarily disagree with this definition, but it is at odds with the concept of weightlessness when in freefall.

IMO, the whole concept of weight should be expunged from physics textbooks. There is mass and there is force and that's that.

Well I remember my secondary high school book on this subject. It was saying that:
"Weight is the force with which the Earth attracts a body or our body, however the sense we have for our weight comes from the normal reaction force of the surface we stand upon"

 

[PeroK]

Well I remember my secondary high school book on this subject. It was saying that:
"Weight is the force with which the Earth attracts a body or our body, however the sense we have for our weight comes from the normal reaction force of the surface we stand upon"

It's either one or the other. Whoever wrote that book can't have his baklava and eat it!

 

[Delta2]

It's either one or the other

I disagree, both statements are very true for me. If our weight is the only force on us then we do free fall and we don't have any sense of our weight because to have some sense of a force on us there must be a contact normal reaction force on us.

 

[kuruman]

I don't necessarily disagree with this definition, but it is at odds with the concept of weightlessness when in freefall.

IMO, the whole concept of weight should be expunged from physics textbooks. There is mass and there is force and that's that.

IMO the concept of weightlessness when in free fall is fraught. That's because the Earth does not stop attracting an object when the Earth force is the only force on the object and neither does the object stop attracting the Earth. The idea of weightlessness is misleading because it's about the removal of the normal force that prevents the object from moving closer to the center of the Earth not the removal of the Earth that exerts that force.

Saying that an astronaut in orbit around the Earth is "weightless" misleads the innocent into thinking that the astronaut is too far from the center of the Earth to experience an appreciable attraction.

I agree with you, the concept of weight should be expunged from physics textbooks and we should stick with mass and forces. However, I hope that you will agree with me that the concept of weightlessness should also go with it otherwise someone wants to have his crumpet and eat it too.

 

[nasu]

It's either one or the other. Whoever wrote that book can't have his baklava and eat it!

But you can eat it and "sense" it in your stomach.

 

[erobz]

If your body and feet are in contact with the scale, then you can write $mg - F_s = ma$ where $F_s$ is the reaction force of the scale on your body, then you can find $F_s = m(g - a)$ this has the same magnitude as the force of the body onto the scale, according to Newton's third law, so we will read a 'reduced' weight when the platform is accelerating downwards.

Let me know what you guys think, thanks!

I think technically for the frame of the elevator you should first write:

$$ m\vec{g} + \vec{F_s} -m\vec{A} = m\vec{a} $$

Where

$\vec{A}$ is the acceleration of the frame relative to an inertial frame.

$\vec{a}$ is the acceleration of the object relative to the frame

For the case of the persons standing still on the scale in the elevator $\vec{a}=0$, and $\vec{A}$ is the acceleration of the elevator so that:

$$\vec{F_s} = m( \vec{A} - \vec{g})$$

This may seem trivial, but it is important when you start to examine “motion” inside the accelerating frame.

 

[kuruman]

I think technically for the frame of the elevator you should first write:$$ mg- F_s -m\vec{A} = m\vec{a} $$

Technically, one should not first write an equation that adds scalars and vectors together.

 

[erobz]

Technically, one should not first write an equation that adds scalars and vectors together.

Yeah, the thing is I assumed a direction for $g$ and $F_s$. The direction that the elevator travels, or one might move in it remains variable, hence the vector notation. Perhaps there is a more appropriate way to handle it, I’ve just changed all to vectors. Sorry if my abuse of notation caused any confusion.

 

[kuruman]

Yeah, the thing is I assumed a direction for $g$ and $F_s$. The direction that the elevator travels, or one might move in it remains variable, hence the vector notation. Perhaps there is a more appropriate way to handle it, I’ve just changed all to vectors. Sorry if my abuse of notation caused any confusion.

I think a more appropriate way to handle it is not to worry about acceleration in a non-inertial frame unless you do have to worry about it. In other words set it equal to zero when it is zero.

 

[erobz]

I think a more appropriate way to handle it is not to worry about acceleration in a non-inertial frame unless you do have to worry about it. In other words set it equal to zero when it is zero.

We’ll, I was just recalling when I was confused about it. You get used to seeing $F = ma$ in almost all practical engineering problems, so “non-inertial frames” is this thing that gets buried in the pile of trivia type knowledge, and it truly doesn’t belong there. It’s not until you try an “elevator problem” that you realize when you set $a=0$ that delicate illusion of Newton’s Second has you questioning everything you thought you knew! I’m sure it’s not that way for physicists (and I can’t speak for all undergraduate engineering programs), but it was that way for me.

 

[kuruman]

A physicist will tell you that if you understand Newton's second law in an inertial frame it should be easy to understand it in a non-inertial frame. In the inertial frame, the sum of all the forces acting on the system is $\mathbf{F}_{\text{net}}=m\mathbf{a}$ where the left hand side is the sum of all the external forces acting on the system. The right hand side is the mass times the acceleration of the system relative to the inertial frame.

Now imagine being the accelerating system, i.e. at rest relative to the system. You still experience the same external forces however you are at rest. Clearly the net force on you is zero. The only that can be consistent with Newton's second law is to have an additional force and write $\mathbf{F}_{\text{net}} +(- m\mathbf{a})=0.$ In other words, in the non-inertial frame there must be an additional force, called an inertial or sometimes fictitious force, that you experience only because you are seeing yourself at rest in that frame.

How does that apply to you standing on a bathroom scale accelerating down? The external force acting on you is gravity $mg$ to which you add the inertial force $(-ma)$ to get the reading on the scale $m(g-a).$

 

[erobz]

A physicist will tell you that if you understand Newton's second law in an inertial frame it should be easy to understand it in a non-inertial frame.

Yeah, they tell you that…then they turn that little $ma$ into a multi page derivation for all the forces experienced let’s say in a rotating sphere using polar coordinates!

 

[kuruman]

Yes, a rotating non-inertial frame is a bit more complicated but not that bad. The multipage derivation is there to show you that it was not pulled out of a hat and that it is correct. Physicists are averse to writing things down without proper justification.

 

[jbriggs444]

Yes, a rotating non-inertial frame is a bit more complicated but not that bad. The multipage derivation is there to show you that it was not pulled out of a hat and that it is correct. Physicists are averse to writing things down without proper justification.

The definition of "weight" that I've always preferred is the net inertial force that is present on an object due to one's choice of adopting a non-inertial frame of reference. For example, when we adopt a lab frame on the surface of the earth, we get a vector-valued "weight" given by $m$ multiplied by local $g$ with a direction given by the local "vertical". No calculation is required. Most conveniently, one can simply measure the magnitude of "weight" with a properly calibrated spring scale and its direction with a spirit level or plumb bob.

Under this definition, an astronaut is "weightless" when one adopts the spacecraft 's free fall rest frame. By contrast, one can blink and find that the astronaut is no longer "weightless" when one adopts the ECI frame.