Actual Working of Weighing Machine (Normal vs gravitational force)

[Physicsnb]

Homework Statement

It's said that a weighing Machine( hereafter referred to as machine) measure the normal force applied by the body on the machine . But when a body is standing on machine 2 forces act simultaneously (namely the normal force and gravity) . At rest N=mg. So forces applied at the machine = 2mg (N+mg).
This is contradictory to the statement 1 . Explain in detai.

Relevant Equations

N= Normal force
Gravity = mg
N=mg (at rest)

I couldn't think of a possible explanation.

 

[MatinSAR]

So forces applied at the machine = 2mg (N+mg).

I think you forgot "=".
How ever, What is N here? Can you draw a free body diagram and show every force that acts on the machine?
Note that you should only draw forces that acts on machine.

See the picture. 3 forces act on machine. Can you identify them?

You can find forces if you think about following :
1. This machine has weight.
2. Someone is standing on it.
3. It's placed somewhere on a surface and pressing against it.

 

[hutchphd]

Homework Statement: It's said that a weighing Machine( hereafter referred to as machine) measure the normal force applied by the body on the machine . But when a body is standing on machine 2 forces act simultaneously (namely the normal force and gravity) . At rest N=mg. So forces applied at the machine = 2mg (N+mg).
This is contradictory to the statement 1 . Explain in detai.
Relevant Equations: N= Normal force
Gravity = mg
N=mg (at rest)

I couldn't think of a possible explanation.

Draw a free body diagram (I'll bet someone has told you that before !!!...... so where is it?). Your prof did not suggest this because she wants to teach you art.

 

[Steve4Physics]

But when a body is standing on machine 2 forces act simultaneously (namely the normal force and gravity)

Think more about the above statement. Remember, weight is gravitational attraction; it only acts on the attracting masses; gravitational attraction is not a contact force.

 

[Physicsnb]

I think you forgot "=".
How ever, What is N here? Can you draw a free body diagram and show every force that acts on the machine?
Note that you should only draw forces that acts on machine.View attachment 337306
See the picture. 3 forces act on machine. Can you identify them?

You can find forces if you think about following :
1. This machine has weight.
2. Someone is standing on it.
3. It's placed somewhere on a surface and pressing against it.

 

[Physicsnb]

These are the forces acting on it according to me

 

[Physicsnb]

Think more about the above statement. Remember, weight is gravitational attraction; it only acts on the attracting masses; gravitational attraction is not a contact force.

Won't the machine measure gravity too?
It makes some sense to me but I'm still puzzled as to why mg won't be taken into account when measuring weight by the machine.

 

[Physicsnb]

Draw a free body diagram (I'll bet someone has told you that before !!!...... so where is it?). Your prof did not suggest this because she wants to teach you art.

This is a rough FBD that I drew . Can you explain as to why mg isn't added to the final weight along with normal?

 

[MatinSAR]

View attachment 337314

If body pushes machine down with force $Mg$ then machine pushes the object up with force $Mg$.
So 3 forces act on body.

$N$ and $Mg$ upward.
$Mg$ downward.
As you see the net force cannot be 0 and the body should accelerate upward!

So that $Mg$ does not act on machine in my opinion. And please wait for @hutchphd and @Steve4Physics ideas. They will help more than me.

 

[Steve4Physics]

Won't the machine measure gravity too?
It makes some sense to me but I'm still puzzled as to why mg won't be taken into account when measuring weight by the machine.

Remembering Newton’s 3rd law:
- the body exerts a downwards normal reaction (contact force) of magnitude N on the machine;
- the machine exerts an upwards normal reaction (contact force) of magnitude N on the body.

If the body is in equilibrium, the downwards force on it (the pull of gravity, mg) balances the upward force on it (upwards reaction, N). So in equilibrium N=mg; these 2 forces have equal magnitudes.

The machine reads the magnitude of the downwards normal reaction, N, because this is the force actually pressing on it.

In equilibrium N has the same magnitude as the weight, so by measuring N, the machine is effectively measuring the weight.

Note this is only true in equilibrium. If you weigh yourself in an elevator while it accelerates, the reading depends on the acceleration.

 

[Orodruin]

View attachment 337314

This is not a free body diagram. A free body diagram isolates a part of the system and draws all of the forces acting on that part. This is not what you have drawn here.

Generally you seem to be confusing yourself. The gravitational force acts on the mass, not on the machine. The only force between body and machine is the contact (normal) force. The gravitational force acts on the body.

 

[Steve4Physics]

This is a rough FBD that I drew . Can you explain as to why mg isn't added to the final weight along with normal?

@Physicsnb, if you are still reading this thread, see if this helps.

We have:
- a body resting n a weighing machine (‘the machine’);
- the machine (say of negligible weight) resting on the earth.

Here are three free body diagrams – for the body, for the machine and for the earth. Forces always occur in (Newton 3rd law) pairs. Each pair is colour-coded in the diagrams. All the relevant forces are shown. I recommend studying the diagrams thoroughly!

 

[kuruman]

Here are three free body diagrams – for the body, for the machine and for the earth. Forces always occur in (Newton 3rd law) pairs. Each pair is colour-coded in the diagrams. All the relevant forces are shown. I recommend studying the diagrams thoroughly!

I disagree with the middle FBD. The machine itself has some mass and is attracted by the Earth, so there must be 3 forces acting on it. See post #2 by @MatinSAR.

 

[Steve4Physics]

I disagree with the middle FBD. The machine itself has some mass and is attracted by the Earth, so there must be 3 forces acting on it. See post #2 by @MatinSAR.

That's why I specifically said (in Post #12):

- the machine (say of negligible weight) resting on the earth.

I wanted to simplify the explanations/diagrams for the OP without losing the essential physics.

 

[Lnewqban]

...The machine itself has some mass and is attracted by the Earth, so there must be 3 forces acting on it....[/USER].

Blue arrows can be longer than the red arrows, representing the addition of the weights of machine and body, as the diagram is not to scale.

 

[gmax137]

Here are three free body diagrams – for the body, for the machine and for the earth. Forces always occur in (Newton 3rd law) pairs.

I hope the OP reads @Steve4Physics post. Much confusion arises when free body diagrams are done incorrectly, showing forces on more than one object.

 

[kuruman]

Much confusion arises when free body diagrams are done incorrectly, showing forces on more than one object.

And that happens when objects other than the system are drawn in the FBD as is the case in post #5. In this case there could be three choices of systems: person alone, machine alone and (person + machine). The sum of the forces shown in the first two should be what is shown in the composite two-component system. The figure below shows that addition and is drawn to scale. I considered a machine with mass because I don't think it complicates matters. Diagrams are drawn to scale and Newton's second law is shown below each FBD. In the last diagram I drew the force exerted by the floor acting at the CM for clarity.

 

[jbriggs444]

Homework Statement: It's said that a weighing Machine( hereafter referred to as machine) measure the normal force applied by the body on the machine . But when a body is standing on machine 2 forces act simultaneously (namely the normal force and gravity) .

Any object that remains at rest while under a downward force is necessarily subject to equal and opposite upward forces. Otherwise it would not remain at rest.

In a bathroom scale there are, to a reasonable approximation, only two forces acting on the scale. The downward force from the soles of the user's feet and the upward force from the floor under the scale.

Naturally, when subject to opposite forces like this, the scale is under stress. It squishes slightly. Springs are compressed. Load cells generate signals. Gears turn. Needles quiver. Readouts change.

You might well ask: "If there is 1000 N of downward force on the top of the scale and 1000 N of upward force on the bottom of the scale, is the scale measuring 1000 N of stress or 2000 N of stress?"

The answer is: "What does it matter?"

Consider calibration. We place a 100 kg standard weight on the scale and scribe a mark where the needle comes to rest. We remove the standard weight and place a human on the scale and see that the needle comes to rest in the same place. We have measured that person's weight at 100 kg. The numeric value of the stress on the scale never enters in.

But I suppose that you want to put numbers on the stress that the scale is under.

If, for some screwball reason, we wanted to express the Cauchy stress tensor in terms of total force instead of one sided force, the net effect would be to multiply each component of the stress tensor matrix by two. Then when we went to extract a one-sided force by taking the matrix product of the stress tensor with a directed area element, we would need to divide by two. The net effect would be a wash.

We conventionally use one-sided forces to populate the components of the stress tensor.

 

[kuruman]

OK, guys, let's take apart the following passage

But when a body is standing on machine 2 forces act simultaneously (namely the normal force and gravity) . At rest N=mg. So forces applied at the machine = 2mg (N+mg).

I will rewrite it using subscripts to indicate what I am talking about because I am not convinced that ignoring the mass of the machine is a simplification or a reasonable approximation here. My modifications are shown in red.

But when a body is standing on the machine 2 forces act simultaneously (namely the normal force and gravity). At rest $N=m_{\color{red}{\text{person}}}\color{black}~g.$ So the sum of the downward forces applied at the machine is $$\color{red}{F_{\text{down}}}=m_{\color{red}{\text{person}}}\color{black}~g +m_{\color{red}{\text{machine}}}\color{black}~g.$$This last equation exposes the dastardly manner in which the author tries to pull a fast one on the reader. The author (a) uses the same symbol $m$ for the mass of both the machine and the person conveniently writing $F_{\text{down}}=2mg$ as if the masses were the same; (b) shoves under the rug the fact that of the two downward forces only the normal force results in a machine reading.

 

[Orodruin]

You might well ask: "If there is 1000 N of downward force on the top of the scale and 1000 N of upward force on the bottom of the scale, is the scale measuring 1000 N of stress or 2000 N of stress?"

A correctly calibrated scale would show 1000 N.

If, for some screwball reason, we wanted to express the Cauchy stress tensor in terms of total force instead of one sided force

It is unclear to me what you even mean by this. The stress tensor is a well defined object that gives you the force per area across a given in an object. There is no sidedness apart from the directionality of the surface.

 

[Orodruin]

OK, guys, let's take apart the following passage

I will rewrite it using subscripts to indicate what I am talking about because I am not convinced that ignoring the mass of the machine is a simplification or a reasonable approximation here. My modifications are shown in red.

But when a body is standing on the machine 2 forces act simultaneously (namely the normal force and gravity). At rest $N=m_{\color{red}{\text{person}}}\color{black}~g.$ So the sum of the downward forces applied at the machine is $$\color{red}{F_{\text{down}}}=m_{\color{red}{\text{person}}}\color{black}~g +m_{\color{red}{\text{machine}}}\color{black}~g.$$This last equation exposes the dastardly manner in which the author tries to pull a fast one on the reader. The author (a) uses the same symbol $m$ for the mass of both the machine and the person conveniently writing $F_{\text{down}}=2mg$ as if the masses were the same; (b) shoves under the rug the fact that of the two downward forces only the normal force results in a machine reading.

I think you got the wrong idea of what the OP is saying. The OP is saying that both the gravity of the person and the normal force of the person is acting on the scale (which is obviously wrong - the gravity on the person acts on the person, not on the scale). This interpretation is corroborated by the OP’s (attempt at a) free body diagram in #5. The mass of the scale was never an issue.

 

[MatinSAR]

OK, guys, let's take apart the following passage

Hello. At first I apologize for replying to this post . I don't want to add anything to your post but I want to explain my idea in post #2 wich you have mentioned. I guess OP is saying that $N_1=2 M_{person} g$ because he thinks that person's weight is pushing the scale down so $M_{person} g$ is also acting on the scale.

According to your FBD in this case with OP's misunderstanding $F_{down}$ should be $2 M_{person} g+ M_{Machine} g$ so $N_1$ which we're looking for should be $2M_{person} g$.

And I think it doesn't matter if we ignore the mass of the scale or not, because :
1. It's effect is neutralized by the normal force of the elevator's floor ($N_2$in your FBD).
2. We want to find $N_1$ and $N_2$ is not important here.

 

[kuruman]

I think you got the wrong idea of what the OP is saying.

I am aware of what OP is saying in post #5. OP is in agreement with the apparent paradox that the author of the problem, a person separate from the OP, formulated. In the author's statement of the problem, as reported by the OP in post #1, we are provided with the following reasoning

1. The normal force on the person standing up on the scale is $N=mg$.
2. There are two forces acting simultaneously on the machine: the normal force and gravity
3. Therefore, "the forces applied at the machine" are $2mg.$

The task assigned by the author is to explain in detail the contradiction between statements 3 and 1. OP's initial response was

I couldn't think of a possible explanation.

Here is my analysis.

Statement 1 is true and correct. Here $m$ is the mass of the person standing on the machine.

In statement 2 we have the assertion that there are two forces acting on the machine, gravity and the normal force. Actually there are three forces. The floor exerts a third force that is sufficient to support the weight of the machine and the person. However this third force can be ignored because it is not essential to the author's argument which in concerned with downward forces only. Nevertheless, statement 2 sets up the contradiction by the reference to gravity acting on the machine. Everybody knows that the force of gravity is $mg$, but what $m$ is this? The author does not elaborate on purpose. However, being smart, we know that this $m$ should be the mass of the machine which is different from $m$ in statement 1.

In statement 3 the author springs the trap that was set up in statement 2. The author adds the two different masses to get $2mg$ as the downward force on the machine. This is wrong because the masses are not the same as we have seen. Having erroneously obtained $2mg$, the author then claims that a contradiction exists with statement 1. In so doing, the author implies that this $2mg$ is the reading on the machine. It is not. The force of gravity acting on the machine does not produce a reading because it does not act at the platform where the person stands, an inference that the author wants us to make.

If I were a machine-mass denier, my argument might appear to be that the force of gravity acting on the machine does not really count, therefore the only downward force on the machine is the normal force $N=mg$ and not $2mg.$ That would appear like an attempt to say that the contradiction does not exist because $mg+mg$ is not really $2mg$ since one $mg$ can be ignored. Denial of the contradiction is not a convincing argument to someone who is grappling with these ideas. The way to do it convincingly is, as posted in #21, to show explicitly that the two masses are not additive to get $2mg.$

 

[Orodruin]

OP is in agreement with the apparent paradox that the author of the problem, a person separate from the OP, formulated.

I do not believe that this is the case. This thread seems to me to be a typical example of someone making up their own problem and posting it.

1. The normal force on the person standing up on the scale is N=mg.
2. There are two forces acting simultaneously on the machine: the normal force and gravity
3. Therefore, "the forces applied at the machine" are 2mg.

The task assigned by the author is to explain in detail the contradiction between statements 3 and 1. OP's initial response was

Again, this is your reading of the OP resting on the assumption that the problem is not one posed by thd OP to themselves. I believe that it is not the case. It is quite common for posters to make ip their own problems and many times they are badly formulated. I disagree with your reading of the OP as the most likely one. To me (and seemingly to most posters in this thread), the most likely thing is that the OP has made up a problem based on a misconception of theirs (that both the person’s gravity and their normal force acts on the scale - supported by the attempt at a fbd in post #5) and is trying to clear that up.

Your analysis is of course fine, but not at the heart of the issue as most people are reading the OP.

 

[Physicsnb]

OK, guys, let's take apart the following passage

I will rewrite it using subscripts to indicate what I am talking about because I am not convinced that ignoring the mass of the machine is a simplification or a reasonable approximation here. My modifications are shown in red.

But when a body is standing on the machine 2 forces act simultaneously (namely the normal force and gravity). At rest $N=m_{\color{red}{\text{person}}}\color{black}~g.$ So the sum of the downward forces applied at the machine is $$\color{red}{F_{\text{down}}}=m_{\color{red}{\text{person}}}\color{black}~g +m_{\color{red}{\text{machine}}}\color{black}~g.$$This last equation exposes the dastardly manner in which the author tries to pull a fast one on the reader. The author (a) uses the same symbol $m$ for the mass of both the machine and the person conveniently writing $F_{\text{down}}=2mg$ as if the masses were the same; (b) shoves under the rug the fact that of the two downward forces only the normal force results in a machine reading.

You seem to have misinterpreted my question.
When a person is standing on it Normal force equal to m(person)*g acts downwards along with m(person)g .
So , f(downwards)= N+m(person)g= 2m(person)g.

 

[Physicsnb]

I am aware of what OP is saying in post #5. OP is in agreement with the apparent paradox that the author of the problem, a person separate from the OP, formulated. In the author's statement of the problem, as reported by the OP in post #1, we are provided with the following reasoning

1. The normal force on the person standing up on the scale is $N=mg$.
2. There are two forces acting simultaneously on the machine: the normal force and gravity
3. Therefore, "the forces applied at the machine" are $2mg.$

The task assigned by the author is to explain in detail the contradiction between statements 3 and 1. OP's initial response was

Here is my analysis.

Statement 1 is true and correct. Here $m$ is the mass of the person standing on the machine.

In statement 2 we have the assertion that there are two forces acting on the machine, gravity and the normal force. Actually there are three forces. The floor exerts a third force that is sufficient to support the weight of the machine and the person. However this third force can be ignored because it is not essential to the author's argument which in concerned with downward forces only. Nevertheless, statement 2 sets up the contradiction by the reference to gravity acting on the machine. Everybody knows that the force of gravity is $mg$, but what $m$ is this? The author does not elaborate on purpose. However, being smart, we know that this $m$ should be the mass of the machine which is different from $m$ in statement 1.

In statement 3 the author springs the trap that was set up in statement 2. The author adds the two different masses to get $2mg$ as the downward force on the machine. This is wrong because the masses are not the same as we have seen. Having erroneously obtained $2mg$, the author then claims that a contradiction exists with statement 1. In so doing, the author implies that this $2mg$ is the reading on the machine. It is not. The force of gravity acting on the machine does not produce a reading because it does not act at the platform where the person stands, an inference that the author wants us to make.

If I were a machine-mass denier, my argument might appear to be that the force of gravity acting on the machine does not really count, therefore the only downward force on the machine is the normal force $N=mg$ and not $2mg.$ That would appear like an attempt to say that the contradiction does not exist because $mg+mg$ is not really $2mg$ since one $mg$ can be ignored. Denial of the contradiction is not a convincing argument to someone who is grappling with these ideas. The way to do it convincingly is, as posted in #21, to show explicitly that the two masses are not additive to get $2mg.$

You seem to have misinterpreted my question.
When a person is standing on it Normal force equal to m(person)*g acts downwards along with m(person)g .
So , f(downwards)= N+m(person)g= 2m(person)g.

 

[PeroK]

You seem to have misinterpreted my question.
When a person is standing on it Normal force equal to m(person)*g acts downwards along with m(person)g .
So , f(downwards)= N+m(person)g= 2m(person)g.

We need to get back to helping the OP on these questions!

 

[Orodruin]

When a person is standing on it Normal force equal to m(person)*g acts downwards along with m(person)g .
So , f(downwards)= N+m(person)g= 2m(person)g.

The answer to which has been given several times in this thread. The only downwards force acting on the scale is the normal force from the person standing on it (neglecting the scale mass for the time being). The gravitational force on the person does not act on the scale, it acts on the person. The force balance for the person tells us that the normal force from the scale on the person is equal to mg (assuming equilibrium). Newton's third law tells us that the normal force from the person on the scale is equal in magnitude, but opposite in direction, to the normal force from the scale on the person since they form a third law pair.

 

[PeroK]

You seem to have misinterpreted my question.
When a person is standing on it Normal force equal to m(person)*g acts downwards along with m(person)g .
So , f(downwards)= N+m(person)g= 2m(person)g.

Forget the scale. If you are standing on the ground, how many forces are acting on you? Please specify them and give their magnitude and direction.

Hint: as you are at rest, Newton's second law says that the total net force on you is zero.

 

[kuruman]

You seem to have misinterpreted my question.

I misinterpreted who asked this question, initially. I thought it was someone other than you and I now stand corrected.

Now to address this

When a person is standing on it Normal force equal to m(person)*g acts downwards along with m(person)g .
So , f(downwards)= N+m(person)g= 2m(person)g.

You are considering the downward forces acting on the machine. Forces act at specific places and have their origins at specific entities outside the system that you are looking at. I will ask a few questions and provide the answers that you have so far with evaluations to each.

Question 1: What entities outside the machine are acting on it?
Answer 1: The person and the Earth.
Correct.

Question 2: What is the force exerted by the person on the machine?
Answer 2: It is the normal force $N$ and I know that $N=mg.$
Correct.

Question 3: What is the force exerted by gravity on the machine?
Answer 3: The same as the weight of the person $mg.$
Incorrect.

To see why is incorrect, consider the following scenario. The person gets off the machine, lifts it off the ground and lets go.

Question 4: While the machine is in the ir, is the force exerted by gravity on the machine different from what it was when the person was standing on it? Why or why not?
Answer 4: ___________________________ .

We need to get back to helping the OP on these questions!

Right on. See above. This is my last post here on the "too many cooks spoil the bouillabaisse" principle.

 

[Steve4Physics]

When a person is standing on it Normal force equal to m(person)*g acts downwards along with m(person)g .
So , f(downwards)= N+m(person)g= 2m(person)g.

At the risk of being one the excessive number of cooks...

Gravitational attraction acts between 2 masses, A and B. Each attracts the other in accordance with Newton's 3rd law.

If you put a 3rd object, C, in-between A and B, the gravitational force between A and B simply 'passes through' C without affecting C. So C is 'transparent' to this force. There is no gravitational force on C due to the gravitational attraction between A and B.

(There will be additional forces of gravitational attraction between A and C and between B and C, but that's a different issue.)

If A is a person, B is the earth and C is a weighing machine, then the weighing machine is 'transparent' to the forces between A and B. The gravitational attraction force of B on A 'passes through' C; it does not exert a force on C.

 

[jbriggs444]

You seem to have misinterpreted my question.
When a person is standing on it Normal force equal to m(person)*g acts downwards along with m(person)g .
So , f(downwards)= N+m(person)g= 2m(person)g.

A rope is being tugged by two teams with force $\vec{F_1}$ rightward by the one team and a equal but opposite force $\vec{F_2}$ to the left by the other team. The magnitude of both forces is $F$.

What is the the tension in the rope?

a) $2F$
b) $F$
c) $0$
d) Something else
e) Cannot be determined from the given information

The same rope is being tugged by team left with the same force $F$ as before. Team right ties their end to a handy wall and goes away to eat lunch.

Now what is the tension in the rope?

a) $2F$
b) $F$
c) $0$
d) Something else
e) Cannot be determined from the given information

What is the net force on the rope in the two scenarios above? Is this different from the tension in the rope?

 

[Orodruin]

Answer 3: The same as the weight of the person

Again, I do not think this is an accurate interpretation of the OP given the attempted free body diagram in #5. The OP (wrongly) believes the gravity on the person acts on the scale. The gravity on the scale was never considered by the OP.

 

[kuruman]

The gravity on the scale was never considered by the OP.

That is not how I would interpret this OP statement in post #1 (color emphasis mine)

But when a body is standing on machine 2 forces act simultaneously (namely the normal force and gravity[/color]) .

(Even though I said I would not post any more here, I had to reply.)

 

[Lnewqban]

Was that person inside the original square?

 

[Orodruin]

That is not how I would interpret this OP statement in post #1 (color emphasis mine)

(Even though I said I would not post any more here, I had to reply.)

That is an interpretation, but the OP also says “when a person stands on”. If you look at the attempted FBD in #5, it certainly seems like the mg force is related to the person rather than the scale.

I am not saying I cannot see your interpretation. I am saying it was a bit over the top to call out everyone ignoring the mass of the scale when their interpretation of the OP was equally valid (and the one OP intended).

 

[kuruman]

I am saying it was a bit over the top to call out everyone ignoring the mass of the scale when their interpretation of the OP was equally valid (and the one OP intended).

It certainly was not my intention to "call out everyone" and, if my post seemed to indicate that, I apologize to all. My point of view is that it makes no sense to draw a free body diagram for a system with zero (or negligible) mass. FBDs serve as guides for writing Newton's second law which requires a non-zero mass to make sense. I will say no more.