What is the Correct Reading on the Scale in This Mass/Scale Puzzle?

[DaveC426913]

Here are two free-body diagrams of the scale. One for your scenario (left) and one for the scale suspended from the ceiling (right). As you said, the scale hanging from the ceiling will read 10 kg or 100 N.

Oh. I see.

And now I see why it has to be in Newtons, rather than in kg. The right diagram wouldn't make sense in kg.

 

[jbriggs444]

3. What I have been asking for is - not any more explanations - but a labeled FBD (or merely a mathematical solution) that results in 10kg.

You are looking for the wrong thing. The problem is with your intuition. The way to fix it is to fix the intuition.

We already have drawings. More drawings or labelled free body diagrams will not help. Nor will mathematics help without a definition for tension first.

We need to attack the intuition that thinks that two opposing momentum flows add to produce tension. They don't.

 

[Nugatory]

We need to attack the intuition that thinks that two opposing momentum flows add to produce tension.

I agree, but I find that the free body diagram of the two components of the scale (hook attached to the pointer, case) is how to do that.

 

[Charles Link]

I'm joining this a little late, but I see one thing that may help explain things that may not have been mentioned previously: These problems with tension T in a rope treat the rope a little as having some kind of magic, when the rope is very much like a spring at the molecular level, where it stretches ever so slightly (when pulled on) and thereby generates a restoring force. That I think is what is sometimes being demonstrated when a spring balance is inserted in the chain, but the details of the rope itself is often omitted from the explanation.

 

[FactChecker]

3. What I have been asking for is - not any more explanations - but a labeled FBD (or merely a mathematical solution) that results in 10kg.

Your logic was fine from the beginning, so the question is how to make it more formal.
Since there is no net torque on the pulleys, couldn't you just make an FBD diagram as your first post described, freezing one pulley and ignoring the weight on that side?
PS. I really don't see anything too informal in your original post, I am not trained in these things.

 

[PAllen]

You are looking for the wrong thing. The problem is with your intuition. The way to fix it is to fix the intuition.

We already have drawings. More drawings or labelled free body diagrams will not help. Nor will mathematics help without a definition for tension first.

We need to attack the intuition that thinks that two opposing momentum flows add to produce tension. They don't.

Going with this idea, consider two rockets with equal thrust pulling on both end of a string. This presents the same ‘seeming’ conundrum. Replace one rocket with the string attached to a post a you readily see that the tension is equal to the force by one rocket. Replace the tied end with a rocket and somehow you want to add forces. This does seem to be the key false intuition.

 

[DaveC426913]

Going with this idea, consider two rockets with equal thrust pulling on both end of a string. This presents the same ‘seeming’ conundrum. Replace one rocket with the string attached to a post a you readily see that the tension is equal to the force by one rocket. Replace the tied end with a rocket and somehow you want to add forces. This does seem to be the key false intuition.

Yes.

Now, can you demonstrate that in one diagram, without resorting to altering the scenario (the string) and then explaining that one?

 

[PAllen]

The tension in the string is the result of one end fixed and the other end under force. The nature of the fixing of one end is irrelevant - whether it is a rocket, glue, or a knot.
All three simply apply the force needed to prevent the string from moving.

 

[Herman Trivilino]

The tension in the string is the result of one end fixed and the other end under force. The nature of the fixing of one end is irrelevant - whether it is a rocket, glue, or a knot.
All three simply apply the force needed to prevent the string from moving.

That is part of the pedagogical point of the original scenario, the other part requires a device for measuring that force. I've seen students respond, after having all this explained to them, that they didn't understand that that's how force-measuring devices work.

But that is simply not true. Anyone who has ever used a scale in the produce section has seen how force-measuring devices work.

So the genuine pedagogy here is understanding something about how force-measuring devices work. Anyone who thinks that two forces, each of magnitude $F$ applied in opposite directions to a force-measuring device, cause the device to read $2F$ has a fundamental misunderstanding of how these devices work.

Stand on your bathroom scale and see that it reads 200 lb. You are pushing down on the scale with a force of 200 lb. The bathroom floor is pushing up on the scale with a force of 200 lb. Yet the scale doesn't read 400 lb! Why not? Because that's not the way force-measuring devices work!

 

[PAllen]

That is part of the pedagogical point of the original scenario, the other part requires a device for measuring that force. I've seen students respond, after having all this explained to them, that they didn't understand that that's how force-measuring devices work.

But that is simply not true. Anyone who has ever used a scale in the produce section has seen how force-measuring devices work.

So the genuine pedagogy here is understanding something about how force-measuring devices work. Anyone who thinks that two forces, each of magnitude $F$ applied in opposite directions to a force-measuring device, cause the device to read $2F$ has a fundamental misunderstanding of how these devices work.

Just replace string with spring in the argument. If the spring moves freely, it measures 0. If held stationary, it measures the force - on either end - they must be equal.

 

[Demystifier]

I can intuit the answer but I can't formalize it.

So far nobody formalized it, so let it be me. Let me denote the mass on the left as $m_1$ and the mass on the right as $m_2$. The corresponding gravitational potential energies are
$$V_1=m_1gz_1, \;\;\; V_2=m_2gz_2$$
where $z$ is the vertical position coordinate. The potential energy of the elastic spring is
$$V_{\rm spring}=\frac{k}{2}(L-L_0)^2$$
where $L$ is the length of the spring and $L_0$ is its length when no external forces act on it. Since $L=const-z_1-z_2$, the total potential energy is the sum of all three potential energies
$$V=m_1gz_1 + m_2gz_2 + \frac{k}{2}(-z_1-z_2+c)^2$$
where $c$ is a constant. The forces on the first and the second weight are $-\partial V/\partial z_1$ and $-\partial V/\partial z_2$, respectively. In the static case they vanish, so
$$ \frac{\partial V}{\partial z_1} = m_1g -k(-z_1-z_2+c)=0$$
$$ \frac{\partial V}{\partial z_2} = m_2g -k(-z_1-z_2+c)=0$$
Both equations can be simultaneously satisfied only if $m_1=m_2$. In that case the two equations are equivalent, leading to
$$c-z_1-z_2=m_1g/k=m_2g/k$$
We see that the total elongation $c-z_1-z_2$ of the spring is as if only one of the weights with mass $m_1$ or $m_2$ was stretching it.

 

[DaveC426913]

_{Well I can't say I didn't ask for that...}


.

 

[Demystifier]

_{Well I can't say I' didn't ask for that...}

The formalization is not only for its own sake. It's very useful when one has to solve a more complicated problem of the same type. When things get complicated, the intuitive reasoning becomes almost impossible, but formalism comes to the rescue.

 

[hutchphd]

I don't want it explained. I want a formal solution.

Here is the formal solution: All is static
Draw a free body diagram for the hook. There are two forces on the hook, the Tension in the string and the spring force ( as indicated on the scale.....that's why it is called a scale)
Draw a free body diagram for the connected mass. There are two forces on the mass F=mg and Tension.
Because all is static the forces on each body sum to zero: mg=T and F_scale=T Therefore F_scale=mg


Formal dress required RSVP

 

[.Scott]

 

[sophiecentaur]

Which is, I guess, the central point. Just because you can add two numbers together does not mean that you should.

There are dozens of such spoof statements about forces, money etc., where you get an apparent paradox because you are fooled into adding the wrong things together. If the string can only just support a 100N force and would break with a 101N force then the string would break when 20N is applied. But we know it won't.

Newtons Third Law insists that there is always a pair of forces involved, whatever happens. (Action and reaction are equal and opposite and there's absolutely no way to deny it.) The two forces at the ends of the string are equal and opposite and so there is no resultant force anywhere along the string; it's a difference of zero.

 

[A.T.]

Newtons Third Law insists that there is always a pair of forces involved, whatever happens. (Action and reaction are equal and opposite and there's absolutely no way to deny it.) The two forces at the ends of the string are equal and opposite and so there is no resultant force anywhere along the string; it's a difference of zero.

The two forces acting on the ends of the string are not a Newtons Third Law pair.

 

[sophiecentaur]

The two forces acting on the ends of the string are not a Newtons Third Law pair.

The forces between the strings and the two end connections of the scale are two third law pairs. What's the difference? Look at any point along the strings, there are third law pairs acting at that point or the string would be moving (accelerating). All those forces are equal magnitude. (Massless measuring instrument, of course)

 

[A.T.]

I think we can agree both diagrams are correct.

Are they? With the scales attached to nothing on top?

They would be correct, if you had the same but opposite forces F at both ends of the scale that also shows F. The same your horizontal scale with pulleys.

How would you "formalize" the result for this simple vertical scale hanging from the ceiling? The same applies to your horizontal scale with pulleys.

 

[A.T.]

there are third law pairs acting at that point or the string would be moving (accelerating).

You are confusing the 3rd Law with the 2nd Law. Two forces acting on the same piece of string are not a 3rd Law pair. 3rd Law pair forces act on different objects each. Just because you have found two equal but opposite forces somewhere, doesn't mean they are a 3rd Law pair.

 

[Herman Trivilino]

The forces between the strings and the two end connections of the scale are two third law pairs. What's the difference?

Third Law Pairs never act on the same object. String pulls on scale, scale pulls on string. That's a Third Law Pair.

The forces acting on opposite ends of a string may happen to be equal but opposite, but they are not a Third Law Pair. In fact, if the string is massive and accelerating, one force will have a greater magnitude than the other.

 

[Meir Achuz]

How could there be 81 posts on this?

 

[Herman Trivilino]

A question asked of prospective physics instructors: A book rests on a level table top. There's a downward gravitational force $\vec{w}$ exerted on the book. Newton's Third Law requires that this force have an equal-but-opposite partner. Identify that force.

The answer would almost always be the upward normal force $\vec{n}$ exerted on the book.

That answer is of course wrong. All of the responders had an advanced degree in physics.

 

[DaveC426913]

Are they? With the scales attached to nothing on top?

Fer cryin out loud...

 

[OmCheeto]

How could there be 81 posts on this?

One guess is that people are answering the question as if they know the answer.
Another guess is that Dave34567whatever is a visual learner like myself and all the blah blah blah in the world isn't going to solve this.
A third guess is that people often use the old 'well newtons laws of course' answer, which is not an answer at all. It annoys the heck out of me. It's like saying; "Well, the Encyclopædia Britannica says that's the way it is. Joila! I've answered your question!"

Like Dave456etc, I didn't understand the WHY, after having read ALL the explanations in this thread. So I went into my kitchen and found my fish weighing scale (which is peculiar in that it is magically capable of measuring fish in both mass and weight), got out a 0.84 kg lead fish weight, and visualized transitioning from vertical to multiple pulleys in the mix, and I saw that Dave's scale would always read 10 kg, as I, holding the fish scale, would have to be replaced by a 10 kg mass, when I reached the point of opposite verticality.

ps. I failed 'Statics' in university at least twice. I blame it on the subject nearly boring me to death. Aced 'Dynamics' on the first try. Pat on back. Pat on back.

As in my olden days: OK to delete, infract, and ban.

 

[OmCheeto]

Hmmmm.... Actually, I don't think I understand my explanation. This still doesn't make any sense.

 

[OmCheeto]

How could there be 81 posts on this?

How would you explain it to a 5 year old?

Not saying Dave and I are 5 year olds, but as far as I can tell, Dave is 20 times smarter than I am, and I had a guesstimated IQ of 162 about 20 years ago. It's around 15 right now, as far as I can tell.

 

[russ_watters]

A question asked of prospective physics instructors: A book rests on a level table top. There's a downward gravitational force $\vec{w}$ exerted on the book. Newton's Third Law requires that this force have an equal-but-opposite partner. Identify that force.

The answer would almost always be the upward normal force $\vec{n}$ exerted on the book.

That answer is of course wrong. All of the responders had an advanced degree in physics.

I'm not a physics professor: is the answer the upwards gravitational force the book exerts on the Earth?

 

[renormalize]

The answer would almost always be the upward normal force $\vec{n}$ exerted on the book.

That answer is of course wrong. All of the responders had an advanced degree in physics.

Can you expand on this by stating the correct answer?

 

[russ_watters]

One guess is that people are answering the question as if they know the answer.
Another guess is that Dave34567whatever is a visual learner like myself and all the blah blah blah in the world isn't going to solve this.
A third guess is that people often use the old 'well newtons laws of course' answer, which is not an answer at all. It annoys the heck out of me. It's like saying; "Well, the Encyclopædia Britannica says that's the way it is. Joila! I've answered your question!"

My choice is we don't know what sort of answer would satisfy him, and the answer may be "none". There's some things that just don't "click" for some people and in those cases you just have to memorize and apply the definitions/rules.

Like Dave456etc, I didn't understand the WHY, after having read ALL the explanations in this thread. So I went into my kitchen and found my fish weighing scale (which is peculiar in that it is magically capable of measuring fish in both mass and weight), got out a 0.84 kg lead fish weight, and visualized transitioning from vertical to multiple pulleys in the mix, and I saw that Dave's scale would always read 10 kg, as I, holding the fish scale, would have to be replaced by a 10 kg mass, when I reached the point of opposite verticality.

I think when I took Physics 1 the prof did a demonstration where they instructed two people to each pull on a spring scale in opposite directions with different forces. It looked like they were air-sawing. Thus proving there is/can be only one force/tension.