What are the forces at play in an elevator cab?

[rudransh verma]

Homework Statement

A passenger of mass $m=72.2 kg$ stands on a scale in an elevator cab. We are concerned with the scale reading when the cab is stationary and when it’s moving. ie $F_N=m(g+a)$.
When the cab is stationary or moving up with constant velocity what is its scale reading. ie $F_N=708 N$. This is the weight of the passenger.
What does the scale read if the cab accelerates with $3.20 m/s2$ upwards and downwards. ie $F_N=939$ and $477 N$.
For an upward acceleration the scale reading is greater than the passengers weight.
” That reading is a measurement of an apparent weight because it’s made in non inertial frame”.

I don’t understand this line.

During the upward acceleration what is the magnitude $F_{net}$ of the net force on the passenger and what is the magnitude $a_{p,cab}$ of his acceleration as measured in the frame of cab. Does $\vec F_{net}=m\vec a_{p,cab}$ ?

Relevant Equations

$\vec F=m\vec a$

I mentioned answers above in the question.

 

[jbriggs444]

As I understand it, the line that you do not understand is:

” That reading is a measurement of an apparent weight because it’s made in non inertial frame”.

This is a question about terminology. Not a question about physics. It is about what is meant by "apparent weight".

The idea is that if you have an object that is held at rest in an inertial frame supported by nothing but a scale, the force that you measure with the scale is the object's true weight. By Newton's second law, the measured support force it will be equal and opposite to the force of gravity on the object, the object's true weight. $\sum F = \text{(scale reading)} - mg = ma = 0$.

If you are using an non-inertial frame such as the rest frame of an accelerating elevator cab. Then you can still perform the same measurement procedure. You arrange for the object to remain at rest relative to your accelerating elevator cab. It is supported by the scale alone. You look at the scale reading and declare this as the object's weight.

If we look at this same scenario from the point of view of an inertial frame, we see that the "true weight" is not equal to the support force from the scale this time. $\sum F = \text{(scale reading)} - mg = ma$ but $ma$ is non-zero this time.

Still, from the point of view of an observer sitting in the elevator cab and not looking out through a window or acknowledging the flashing floor numbers, the scale reading seems like a perfectly good weight measurement in a higher (or lower) than normal gravitational field. It is an "apparent weight".

I am not a fan of this terminology. But it is just terminology. It does not change the physics.

 

[rudransh verma]

Not a question about physics. It is about what is meant by "apparent weight".

If I understand apparent weight correctly, a weight that is not measured at rest with respect to Earth such as the weight measured in moving cab.
An accelerating frame is non inertial frame but with respect to what? Earth?

 

[jbriggs444]

If I understand apparent weight correctly, a weight that is not measured at rest with respect to Earth such as the weight measured in moving cab.
An accelerating frame is non inertial frame but with respect to what? Earth?

We normally consider the rest frame of the Earth to be inertial, yes.

The question of exactly what constitutes an inertial frame can be picked apart and analyzed in fine detail, ending up with the mathematics of the curved space-time of general relativity. However, for the purposes of Newtonian physics:

A frame is "inertial" if objects obey Newton's laws when measured against it. Gravity counts as a real force.​

 

[rudransh verma]

A frame is "inertial" if objects obey Newton's laws when measured against it. Gravity is a real force.

Meaning the frame in which Newtons law hold is inertial. As in this case the inertial frame is Earth and non inertial is the elevator cab. But Newtons law will hold in an accelerating cab too. So it’s the accelerating frame that’s non inertial?

$\vec F_{net} = ma$ is measured from Earth but $\vec a_{p,cab}$ is measured from non inertial frame the cab and so F and a are not related to each other because they are both from different frames.

 

[jbriggs444]

But Newtons law will hold in an accelerating cab too.

Will they? All three? Where are the third law partner forces to the forces on the person in the accelerating cab?

 

[rudransh verma]

Will they? All three? Where are the third law partner forces to the forces on the person in the accelerating cab?

The mg+ma is not equal to mg. Ok. So third law doesn’t hold.
So a non accelerating frame where all the laws hold is inertial frame.

 

[jbriggs444]

The mg+ma is not equal to mg. Ok. So third law doesn’t hold.
So a non accelerating frame where all the laws hold is inertial frame.

Right. In Einstein's 1905 paper introducing Special Relativity, he chose the phrasing "frame of reference where the equations of mechanics hold good".

I was specifically thinking that either:

1. You introduce a fictitious force equal to ma due to the acceleration of the cab. Now Newton's second law holds good. But this fictitious force has no third law partner. So Newton's laws do not all hold.

or

2. You do not introduce this fictitious force. But now, as you point out, an object subject to a downward force of mg and upward force of mg + ma has zero acceleration. So Newton's second law fails.

 

[rudransh verma]

Right. In Einstein's 1905 paper introducing Special Relativity, he chose the phrasing "frame of reference where the equations of mechanics hold good".

I was specifically thinking that either:

1. You introduce a fictitious force equal to ma due to the acceleration of the cab. Now Newton's second law holds good. But this fictitious force has no third law partner. So Newton's laws do not all hold.

or

2. You do not introduce this fictitious force. But now, as you point out, an object subject to a downward force of mg and upward force of mg + ma has zero acceleration. So Newton's second law fails.

Why $\vec F_{net}$ not equal to $m\vec a_{p,cab}$ ?
Is it because a is measured from Earth whereas $\vec a_{p,cab}$ is from cab ?

 

[jbriggs444]

Why $\vec F_{net}$ not equal to $m\vec a_{p,cab}$ ?

The net force is the vector sum of all of the forces on the object being weighed in the cab, right?

We can find that sum by identifying all of the forces and adding them up. We read a number from the scale. That number will turn out to be equal to $mg + ma$.

We calculate a number for gravity based on Newton's universal law of gravitation, $f=\frac{Gm_1m_2}{r^2}$. This number will turn out to be equal to $mg$.

So we have our sum. The net force on the object being weighed is $ma$.

Now let us look at $m\vec a_{p,cab}$ The acceleration of the object relative to the cab is zero. So this number is zero.

And we ask ourselves: is $\vec F_{net}$ (which has magnitude ma) equal to $\vec a_{p,cab}$ (which has magnitude zero)? Clearly not.

Is it because a is measured from Earth whereas $\vec a_{p,cab}$ is from cab ?

In the calculations above, $a$ is the acceleration of the reference frame. It is a measurable quantity. We just measured it with a scale. That is all an accelerometer is -- a spring scale and a reference mass.

 

[rudransh verma]

In the calculations above, a is the acceleration of the reference frame. It is a measurable quantity. We just measured it with a scale. That is all an accelerometer is -- a spring scale and a reference mass.

Can I say because $a$(and so Fnet) and $a_{p,cab}$ are measured from two different frames so Newtons law can not be valid for these inputs. $F_{net}=ma_{p,cab}$ doesn’t make sense.

 

[jbriggs444]

Can I say because $a$(and so Fnet) and $a_{p,cab}$ are measured from two different frames so Newtons law can not be valid for these inputs. $F_{net}=ma_{p,cab}$ doesn’t make sense.

I am not following what you are saying here.

You want to make a statement about Newton's second law. So write down Newton's second law. What object are you writing it down for? What is the acceleration of that object in your chosen reference frame? What forces is it subject to? What is the sum of those forces?

Does the law hold? Does $f_{net} = ma_{p,cab}$?

 

[Lnewqban]

For an upward acceleration the scale reading is greater than the passengers weight.
” That reading is a measurement of an apparent weight because it’s made in non inertial frame”.

I don’t understand this line.

When not moving, or moving up with constant velocity, the scale is pushing the feet of the passenger up with a force equal to $Fn=708~N$ (the natural weight of the passenger).

When the upwards movement starts, the scale must put additional force on the feet of the passenger in order to accelerate his body at the given rate of $3.20~m/s^2$.
That new acceleration is added to the omnipresent gravity acceleration, resulting on a total upwards acceleration of $9.81+3.20=13.01~m/s^2$.

 

[rudransh verma]

I am not following what you are saying here.

You want to make a statement about Newton's second law. So write down Newton's second law. What object are you writing it down for? What is the acceleration of that object in your chosen reference frame? What forces is it subject to? What is the sum of those forces?

Does the law hold? Does $f_{net} = ma_{p,cab}$?

Ok! So choosing Earth as the reference frame we calculated $\vec F_{net}$ which is equal to $ma$. All the measurements are relative to earth. Whereas in $m\vec a_{p,cab}$ $a_{p,cab}$ is measured relative to cab which is zero.
So because one part of N second law is from different frame as compared to other we cannot equate it or as you said it they are not equal. Right?

 

[rudransh verma]

When not moving, or moving up with constant velocity, the scale is pushing the feet of the passenger up with a force equal to $Fn=708~N$ (the natural weight of the passenger).

When the upwards movement starts, the scale must put additional force on the feet of the passenger in order to accelerate his body at the given rate of $3.20~m/s^2$.
That new acceleration is added to the omnipresent gravity acceleration, resulting on a total upwards acceleration of $9.81+3.20=13.01~m/s^2$.

So when it is at rest it’s $F_N=mg$ , the true weight but when it starts accelerating $F_N= mg+ma$, the apparent weight. The scale is pushing us up + the force of the lift being accelerated upwards.

 

[jbriggs444]

Ok! So choosing Earth as the reference frame we calculated $\vec F_{net}$ which is equal to $ma$

You wrote down the reading on a bathroom scale without any need to choose a reference frame. You calculated the force of gravity without any need to choose a reference frame.

Under the model of Newtonian mechanics, forces are invariant. They do not depend on the choice of reference frame.

Whereas in $m\vec a_{p,cab}$ $a_{p,cab}$ is measured relative to cab which is zero.

Yes. Coordinate acceleration is frame-relative. This was measured against the cab standard of rest where it was found to be zero.

So because one part of N second law is from different frame as compared to other we cannot equate it or as you said it they are not equal. Right?

All measured quantities here are either invariant or are relative to our chosen frame. We are not mixing frames in our measurements or calculations.

 

[rudransh verma]

You wrote down the reading on a bathroom scale without any need to choose a reference frame. You calculated the force of gravity without any need to choose a reference frame.

I am just saying its all done relative to earth.

forces are invariant. They do not depend on the choice of reference frame.

If acceleration is frame relative and force depends on acceleration how is force not frame relative?

All measured quantities here are either invariant or are relative to our chosen frame. We are not mixing frames in our measurements or calculations.

So $F_{net}=ma$ is same force irrespective of cab and earth. It doesn't make sense to think of force with respect to any frame?

 

[jbriggs444]

I am just saying its all done relative to earth.

I am saying that it is not. In Newtonian mechanics, force is an invariant. It is not relative to anything.

If acceleration is frame relative and force depends on acceleration how is force not frame relative?

Force is defined in terms of acceleration with respect to any inertial frame. It does not matter which inertial frame you pick. That makes it not frame relative.

So $F_{net}=ma$ is same force irrespective of cab and earth. It doesn't make sense to think of force with respect to any frame?

Right. It does not make sense to think of force with respect to a frame. You already saw this when you noticed that the force of gravity and the reading on the scale were the same when the elevator was at rest, when it was ascending at a constant rate and when it was descending at a constant rate.

Three different frames. But the forces were the same in all three.

Even in a non-inertial frame, the force is still invariant. By that I mean that if the spring scale says 100 Newtons then the force from the scale is 100 Newtons, regardless of what frame of reference we are using.

 

[rudransh verma]

You already saw this when you noticed that the force of gravity and the reading on the scale were the same when the elevator was at rest, when it was ascending at a constant rate and when it was descending at a constant rate.

But reading on scale increased and decreased with accelerating or de accelerating cab.
So you are saying addition/subtraction of factor ma to mg doesn’t mean force is relative to any frame?
That is increased force from the cab to the passenger.

 

[Lnewqban]

Inertial frame of reference: An observer standing on the ground can see the elevator moving (if exterior to a building) and any acceleration and associate force (as mass remains the same) makes perfect sense.

Non-inertial frame of reference: Not so much to a person trapped inside the box, feeling misterious forces but seeing no movement relative to ground (real experience in most elevators).

 

[jbriggs444]

But reading on scale increased and decreased with accelerating or de accelerating cab.

You changed the physical situation. An elevator accelerating upward is a different physical situation from an elevator accelerating downward. That has nothing to do with changing frames and everything to do with measuring two different things.

You change frames of reference by writing different things down with pencil on paper. Not by changing anything physical.

 

[Steve4Physics]

Newton’s 2nd law works in an accelerating frame if we include the fictitious (pseudo) force caused by the acceleration.

If the cab has acceleration $\vec a_{cab}$ then, in the frame of the cab, the passenger of mass m experiences a fictitious (pseudo) force $\vec P = -m\vec a_{cab}$. Note the minus sign: if the cab accelerates upwards, the pseudo force acts downwards and vice versa.

The 2 real forces on the passenger are their weight ($\vec W$) and the normal reaction of the scales ($\vec R$).

In the accelerating cab’s frame, we note that the passenger's acceleration is zero (they are stationary with respect to the cab). The 3 forces (2 real and 1 fictitious) acting on the passenger are therefore balanced. So N2L gives:
$\vec R + \vec W + \vec P = 0$

In 1-dimension, taking g = -10m/s² we can write
$R + mg +(– ma_{cab}) = 0$
$R = m(a_{cab} -g)$ (remember we are taking g as negative, so -g is positive)

On the ground, An observer sees the passenger’s acceleration to be (the same as the cab's) $a_{cab}$ and only 2 forces ($\vec W$ and $\vec R$) acting, so N2L gives:
$R + mg = ma_{cab}$
$R = m(a_{cab} -g)$
which is exactly the same result as when calculated in the non-inertial frame.

 

[rudransh verma]

When the upwards movement starts, the scale must put additional force on the feet of the passenger in order to accelerate his body at the given rate of $3.20~m/s^2$.
That new acceleration is added to the omnipresent gravity acceleration, resulting on a total upwards acceleration of $9.81+3.20=13.01~m/s^2$.

I really thought about it and I don’t understand how an upward acceleration or upward force increases weight of the body?
Suppose there is a traditional weighting scale with springs like this .
I don’t see how an upward acceleration will stretch the spring even more?

 

[haruspex]

I don’t see how an upward acceleration will stretch the spring even more?

An upward acceleration means the vertical force exerted by the scales exceeds mg. Therefore the force the mass exerts on the scales exceeds mg.

 

[rudransh verma]

An upward acceleration means the vertical force exerted by the scales exceeds mg.

How can you say that?

Therefore the force the mass exerts on the scales exceeds mg.

First the mass exerts the force and in return the scale exerts force. Aren't you saying just opposite?

 

[jbriggs444]

How can you say that?

First the mass exerts the force and in return the scale exerts force. Aren't you saying just opposite?

The force that the mass exerts on the scale is equal (and opposite) to the force that the scale exerts on the mass. That is Newton's third law.

Neither one causes the other. They are part and parcel of a single interaction.

 

[Lnewqban]

I really thought about it and I don’t understand how an upward acceleration or upward force increases weight of the body?

The upward force on the scale, which has a mass, is the first thing that happens.
Then, the momentum of the scale increases and upwards acceleration begins.
The inertia of the passenger’s body resists that.
The spring is caught in between the accelerating scale and the body that resists that acceleration.

Please, see:
https://en.m.wikipedia.org/wiki/Newton's_laws_of_motion#Second_law

 

[rudransh verma]

The spring is caught in between the accelerating scale and the body that resists that acceleration.

I don’t want to say thanks because that will end this topic. But Thanks.
So the spring is pulled from both sides. So magical!
In the downward acceleration the spring is compressed from both sides.

 

[jbriggs444]

So the spring is pulled from both sides.

Of course, if we approximate springs as being massless then there is no such thing as a spring that is pushed or pulled from one side only. It will always be pulled from both sides, pushed from both sides or just be sitting there doing nothing.

Newton's second law.

 

[rudransh verma]

Right. In Einstein's 1905 paper introducing Special Relativity, he chose the phrasing "frame of reference where the equations of mechanics hold good".

I was specifically thinking that either:

1. You introduce a fictitious force equal to ma due to the acceleration of the cab. Now Newton's second law holds good. But this fictitious force has no third law partner. So Newton's laws do not all hold.

or

2. You do not introduce this fictitious force. But now, as you point out, an object subject to a downward force of mg and upward force of mg + ma has zero acceleration. So Newton's second law fails.

Why it is necessary to have all three laws hold to be a inertial frame. What is the meaning of inertial frame.
Also what is the relation between frames and it’s speed/velocity ?(You said rest frame , frame moving up with constant v, moving down with constant v are three different frames). Why are they different?

 

[jbriggs444]

What is the meaning of inertial frame.

Other threads in these forums have gone on deeply and at length (some with more than a little heat) about what is meant by the term "frame of reference". I will try to keep thing simpler than that.

One can get through high school (U.S. years 10, 11 and 12) with the notion of "frame of reference" as another name for a cartesian coordinate system. Like a piece of graph paper. The graph paper is free to move, but one is expected to pretend that it is at rest. You can write down the positions of all of your objects of interest in terms of the x and y coordinates found on the graph paper at events of interest. Of course, one is expected to record time stamps as well.

A piece of graph paper can move. It can move smoothly. It can accelerate. It can rotate. A piece of graph paper that translates at constant velocity without rotation can be thought of as defining an inertial frame.

This intuition was enough to get me through first year physics. But it left me somewhat uncomfortable. A frame of reference is not really a full blown coordinate system. Instead, it is more like a standard of rest.

One is not limited to cartesian coordinates. One can still have a standard of rest with polar coordinates. Nor is one limited to a particular scale -- a coordinate system implies a choice of units. But a frame of reference does not need a unit choice. What we need to have a frame of reference is a bit less than what we need to have for a coordinate system.

The picture I like these days is that a frame of reference is a standard of rest that applies at every point (event) in the space-time being considered. It gives you a standard for what "at rest" means at that event. It also gives you a standard for direction and speed. So if you have any object of interest at any point and at any time, the "frame of reference" is the standard that let's you say "that object was moving in such and such direction at such and such speed".

For example, if you have a frozen lake with bunch of stakes pounded in and on each stake is a speed detector and a compass, you have yourself the realization of a frame of reference.

An inertial frame of reference is one where objects free of external forces would have the same velocity vector each time they pass near a stake on the ice. That's Newton's first law.

Newton's second and third laws clarify what we mean by the term "force".

 

[rudransh verma]

@jbriggs444 I have posted a thread named “
'Inertial vs non inertial frames'
https://www.physicsforums.com/threads/inertial-vs-non-inertial-frames.1010363/
” please see it. It says something more about frames. What you wrote is very elementary but I am relieved to know that I don’t have to know much to solve 10+2 problems. I was very worried I won’t be able to solve problems without a firm grasp on frames.
This is the video by the way of that thread.

 

[rudransh verma]

All measured quantities here are either invariant or are relative to our chosen frame. We are not mixing frames in our measurements or calculations.

How is it possible? The acceleration a is relative to Earth whereas the $a_{cab}$ is relative to cab.

 

[Steve4Physics]

How is it possible? The acceleration a is relative to Earth whereas the $a_{cab}$ is relative to cab.

Hi @rudransh verma. Here’s an exercise for you.

A person, P, in an elevator cab’, stands on weighing-scales calibrated in Newtons.
Take:
g = 10m/s² (downwards relative to the ground)
Mass of P= 50kg
Reading on scales = 600N

1. For an observer in the ground frame of reference:
a. What is the gravitational force on P?
b. What force is exerted by the scales on P?
c. Are there any other (real or fictitious) forces on P in this frame? If so, describe.
d. What is the effective resultant force on P measured in this frame?
e. What is P’s acceleration in this frame?

2. For an observer in the cab’ frame of reference:
a. What is the gravitational force on P?
b. What force is exerted by the scales on P?
c. Are there any other (real or fictitious) forces on P in this frame? If so, describe.
d. What is the effective resultant force on P measured in this frame?
e. What is P’s acceleration in this frame?

Edit - minor changes.

 

[jbriggs444]

How is it possible? The acceleration a is relative to Earth whereas the $a_{cab}$ is relative to cab.

I am not inclined to go back and unravel the chain of references to figure out exactly what misunderstanding we were working on.