B Simple mass/scale puzzle

[Herman Trivilino]

That's just a misunderstanding of how springs work.

There are ways to measure force that don't make use of springs. They all share this property, though, that if you apply forces of equal magnitude $F$ in opposite directions they report a value of $F$ for the force. That's by design.

So I would say that this is a misunderstanding of how forces are measured. And the misunderstanding is weird because it occurs in some scenarios and not in others.

It appears to be a stubborn refusal to form a consistent worldview.

 

[ergospherical]

Right, sure. It'll probably be some sort of strain-based sensor in any case. Whether it's the extension of a spring, the refractive index of the material, resistance, etc., due to the strain that gives rise to the reading internally is not important for this question.

I don't think "tension" is very clearly explained, at least not when I was in school. In the sense that it's a "thing" (read: tensor, hence why it's not explained in school) that exists everywhere in the body. Every part of the elastic body is applying an internal tensile force on every other bit. So until you get to that stage, you just have to accept that elastic materials apply a force equal to their internal tension on anything attached to them. (Again, at least in a linear 1d geometry.)

 

[tech99]

Aftyer all that, as we measure weight in Newtons, the question is meaningless.

 

[PAllen]

I should point out that most laypeople, including myself, don't usually think of weights and scales in terms of Newtons.

I see a diagram with two 10kg weights being pulled down by gravity on a single scale, I intuit the scale is going to read 20kg. It is very hard to break that conclusion without resorting to an alteration of the scenario.

Well if you look at the crude diagram I posted some poste ago, you will see what balances the 20 kg.

 

[DaveC426913]

Well if you look at the crude diagram I posted some poste ago, you will see what balances the 20 kg.

I did look at it and tried to follow it. I'll have to spend more time on it.

 

[Herman Trivilino]

Aftyer all that, as we measure weight in Newtons, the question is meaningless.

The $\mathrm{SI}$ unit of force is the newton. Weight is defined in physics and engineering differently than in international treaty and law. In the latter case the $\mathrm{SI}$ unit of weight is the kilogram.

 

[OmCheeto]

Ok. I'm out. I thought about it in terms of a tug of war, couldn't figure out the answer, found the following video, and the replacement of the one group of tuggers with the tree answered the question.

 

[kuruman]

So I would say that this is a misunderstanding of how forces are measured. And the misunderstanding is weird because it occurs in some scenarios and not in others.

I think it is also due to partial understanding of Newton's Third law. Here is a diagnostic question related to @OmCheeto's video that probes one's understanding of the 3rd law and what a spring scale displays.

The figure shows a team of four men engaged in a tug o wall. Nothing is moving. Which of the forces on the list has a magnitude that matches the reading on the scale? Circle all that apply.

(A) The force exerted by the left side of the scale on the rope.
(B) The force exerted by the right side of the rope on the scale.
(C) The force exerted by the rope on the wall.
(D) The total horizontal force exerted by the men's hands on the rope.
(E) The total static friction force exerted by the men's feet on the ground.

Spoiler: Answer

All five choices (A)-(E) must be circled.

 

[A.T.]

I should point out that most laypeople, including myself, don't usually think of weights and scales in terms of Newtons.

I see a diagram with two 10kg weights being pulled down by gravity on a single scale, I intuit the scale is going to read 20kg. It is very hard to break that conclusion without resorting to an alteration of the scenario.

You talk about masses not forces. Thus you ignore the force that is always needed at the other side of the scale, even in the simple vertical setup, because you see no bannanas there.

 

[sophiecentaur]

I should point out that most laypeople, including myself, don't usually think of weights and scales in terms of Newtons.

Weights are appropriate when talking to the handyman or the DIY store but you can't use weight in any context except when the Earth's gravity is producing it. A member of PF, even if he/she is a 'layperson', should make the effort and be an honorary Physicist / Engineer / Mathematician. Mass v weight is a standing joke with many comedians and can be funny as the context is usually a cranky Science teacher; fair do's. But you can't join a club if you don't use the right terms. That's even down to the direction the referee points after an infringement in rugby and 'soccer'.

 

[Herman Trivilino]

I see a diagram with two 10kg weights being pulled down by gravity on a single scale, I intuit the scale is going to read 20kg. It is very hard to break that conclusion without resorting to an alteration of the scscenario.

But that is the process of altering an intuition. It won't change by itself, and leaving it unchanged means a discord in your own worldview. So if you're not happy with either choice, for no other reason than it prevents you from advancing your knowledge, and resorting to alternate scenarios helps you change your intuition, then why not do that?

 

[Herman Trivilino]

Weights are appropriate when talking to the handyman or the DIY store but you can't use weight in any context except when the Earth's gravity is producing it.

True in the world of physics and engineering, which is indeed where we are right now as we participate in this forum.

But by international treaty and its concomitant laws, the definition of weight is identical to the physicist's definition of mass. That is of course the reason you see it being used that way in the DIY store or by the handyman or in countless other contexts.

It's a definition, so contrary to what appears in nearly every physics textbook, it can't be wrong. It's a definition!

 

[sophiecentaur]

That is of course the reason

No it's not. The reason people use weight for mass is because all the others in their everyday lives use it. In the same way they confuse kW with kWh and use the verbs for current and volts the wrong way round? Everyday language just doesn't cut it for discussing any serious activity.

You seem to be holding a torch for this sort of error. Why?

 

[jbriggs444]

You seem to be holding a torch for this sort of error. Why?

I share the same sentiment as @Mister T. Because it is not always an error. We (as scientists and physics students) need to avoid the self righteous attitude that makes us look down at the merchants and regulatory bodies who use the term "weight" in a way that goes counter to our textbooks.

The merchants have thousands of years of experience and good reason to use the term "weight" to refer to mass. For them, "weight" is what a balance scale measures. Using the word to refer to a force would be erroneous in that context. (Force is used as a proxy for mass under the reasonable assumption that the local gravitational acceleration does not vary greatly over the diameter of the apparatus)

"Weight" is also what a spring scale measures. A spring scale that is properly calibrated for use in commerce will measure mass (using force as a proxy to compare the unknown mass with the known reference weights used in the calibration under the reasonable assumption that neither the spring constant nor the local gravitational acceleration vary greatly over time).

 

[A.T.]

The reason people use weight for mass is because all the others in their everyday lives use it.

Wasn't the word 'weight' around long before physists started using it as a technical term, and telling everyone how to use it properly? Just like power, force, work, current? If you want to avoid this confusion, then you have to make up completely new words for new physical concepts.

 

[Herman Trivilino]

You seem to be holding a torch for this sort of error. Why?

Because it's an error made in virtually every introductory physics textbook and classroom. The so-called Treaty of the Meter was signed in Paris in 1875 by, among other nations, the United States. It established the International Bureau of Weights and Measures (BIPM) as the governing body over what is now called the SI. It requires a meeting every 4 years IIRC of the General Conference on Weights and Measures (GCPM) to keep the system updated.

It established the meter as the unit of length and the kilogram as the unit of weight. Member nations are free to use other units, but they must be defined in terms of SI units. Thus the foot is exactly 0.3048 m and the pound Avoirdupois is exactly 0.453 592 37 kg.

The newton was later added as the unit of force.

Weight is a term with a definition. It's defined by law one way and in physics another way. Neither can possibly be wrong because they are definitions.

The error made in physics books and classrooms is the insistence that the only correct definition of weight is one where it's a force.

 

[AlexB23]

TL;DR Summary: What does the scale read in this diagram?

My learning is a bit rusty. I can intuit the answer but I can't formalize it.


What does the scale read?
View attachment 363598


I say the scale reads 10kg.

My logic is this:

1. Fix the left cable and pulley in place with a giant glob of glue.
2. Cut the cable holding the left weight.
3. The scale is holding up a 10kg weight. Nothing has changed as far as the scale is concerned.

But what is a more elegant way of getting the answer?

Is this like the same type of thing where you have a car crash into a stationary car at 10 m/s, then have both cars travel at 10 m/s towards each other, and the damage is the same as in what Mythbusters did many years ago?

Source:
https://www.wired.com/2010/05/mythbusters-energy-explanation/

 

[tech99]

Is this like the same type of thing where you have a car crash into a stationary car at 10 m/s, then have both cars travel at 10 m/s towards each other, and the damage is the same as in what Mythbusters did many years ago?

Source:
https://www.wired.com/2010/05/mythbusters-energy-explanation/

Or the supposed impossibility of moving an object because every applied force has an equal and opposing reaction.

 

[AlexB23]

Or the supposed impossibility of moving an object because every applied force has an equal and opposing reaction.

Yep, Newton's third law. So this scale problem is the same exact thing.

 

[hutchphd]

That's just a misunderstanding of how springs work.

Perhaps but who cares. For the purposes of this problem a scale is a black box with a hook and a calibrated readout of the static force on that hook. It could be filled with microprocessors and elves and it wouyld not be relevant to the question. Have humams here (and everywhere!) lost all ability to concentrate.???.......this approaches the absurd

 

[PAllen]

Is this like the same type of thing where you have a car crash into a stationary car at 10 m/s, then have both cars travel at 10 m/s towards each other, and the damage is the same as in what Mythbusters did many years ago?

Source:
https://www.wired.com/2010/05/mythbusters-energy-explanation/

Wait a minute - depends on whether the stationary car is able to roll, or how slippery the surface is. Extreme case: stationary car on ice starting at collision point. To whatever extent both cars move after collision, there is less damage in the stationary car case.

 

[ergospherical]

Perhaps but who cares. For the purposes of this problem a scale is a black box with a hook and a calibrated readout of the static force on that hook. It could be filled with microprocessors and elves and it wouyld not be relevant to the question. Have humams here (and everywhere!) lost all ability to concentrate.???.......this approaches the absurd

Right. But I gathered the whole confusion of the OP was why the scale reads T if you pull on both sides with a force of size T. It’s somewhat helpful to know how a spring works (!).

 

[kuruman]

It established the meter as the unit of length and the kilogram as the unit of weight.

I don't know what "it" established in 1875, but the NIST site today says that the kilogram is the unit of mass. I go by that.

 

[AlexB23]

Wait a minute - depends on whether the stationary car is able to roll, or how slippery the surface is. Extreme case: stationary car on ice starting at collision point. To whatever extent both cars move after collision, there is less damage in the stationary car case.

Assuming it does not roll.

 

[Herman Trivilino]

I don't know what "it" established in 1875, but the NIST site today says that the kilogram is the unit of mass. I go by that.

Me, too. But "it" established the BIPM, and as part of the treaty agreement each member nation must establish its own organization to insure compliance. In the US that organization is the NIST.

What you're saying is not incompatible with what I said.

 

[hutchphd]

Right. But I gathered the whole confusion of the OP was why the scale reads T if you pull on both sides with a force of size T. It’s somewhat helpful to know how a spring works (!).

You need to pull equally on any "scale" to keep it in place while in use. and in general it is very useful to understand how a spring works. I don't think the two are equally necessary here. This post headed for the weeds faster than I would have deemed possible. I was just saying K. I .S.S. I guess.

 

[DaveC426913]

Right. But I gathered the whole confusion of the OP was why the scale reads T if you pull on both sides with a force of size T. It’s somewhat helpful to know how a spring works (!).

That is an illuminating way of phrasing it...

Even though it's been alluded to many times by many people in this thread, you've phrased it in a way that is so simple, it is immediately intuitive.

The scale does not know it is suspended from the ceiling. All it knows is that there is a force pulling on each end (equally - as they must - or the scale would shoot off screen.)

 

[DaveC426913]

I'm starting to understand how to intuit the 10kg reading in the scale without resorting to altering the scenario.

We only need to look at one tiny section of the diagram:

A 10kg mass is supported on a string. It is static, meaning something is holding it up else it would fall.

We (i.e. a layperson) can immediately see that any scale above this would read 10kg. The string is under 10kg of tension (100 Newtons, if you wish.)


We expand our scope:

Nothin has changed. Everything is static. String is still under 10kg tension.

We expand again:

Literally nothing has changed. All I've done is show more of the setup.

It must still be under 10kg of tension. And indeed, our spring scale says so.

Again, nothing has changed.



Finally, we reveal what has ensured the string and mass stays static:

Even the layperson will have to acknowledge that merely revealing what is hidden cannot suddenly change the reading on the scale from 10kg to 20kg.


That wasn't quite what I had been appealing for; I'd been appealing for it a single diagram, labeled in such away as to show why it works. But at least the above doesn't require altering the scenario.

 

[DaveC426913]

Weights are appropriate when talking to the handyman or the DIY store

Yes. I am trying to intuit the diagram from a handyman's (layperson's) point of view.

They don't know from Newtons; forces are hard to see. What they know is that a 10kg weight will read 10kg on a spring scale.

but you can't use weight in any context except when the Earth's gravity is producing it

I'd hoped it went without saying that the diagram was in the context of Earth.

If you were in the presence of laypeople, and started insisting "this only works here on Earth" and "you need to be an honorary physicist/engineer/mathematicin" you would get a lot of eyerolling - including from me.

i.e. it is an unnecessary complication to a very simple scenario.

. A member of PF, even if he/she is a 'layperson', should make the effort and be an honorary Physicist / Engineer / Mathematician.

But the diagram in post 1 is not aimed at PF members. Everyone here got it right away, of course.

Laypeople do not. But could, if the diagram were labeled in such a way as to make it obvious and inescapable.

 

[Herman Trivilino]

We only need to look at one tiny section of the diagram:

A 10kg mass is supported on a string. It is static, meaning something is holding it up else it would fall.

We (i.e. a layperson) can immediately see that any scale above this would read 10kg. The string is under 10kg of tension (100 Newtons, if you wish.)

That's what I was trying to explain to you in Post #7.

 

[renormalize]

That's what I was trying to explain to you in Post #7.

Yes, and as I tried to do in post #34 and @vela in post #60.

 

[DaveC426913]

That's what I was trying to explain to you in Post #7.

Yes, and as I tried to do in post #34 and @vela in post #60.

Yes. This is the culmination. I would not have been able to get here without the explanations y'all have been providing.


The only difference here is that I haven't resorted to an alternate diagram/scenario to explain the principle; I wanted to remain entirely with the original diagram/scenario.

 

[PAllen]

I'm starting to understand how to inuit the 10kg reading in the scale without resorting to altering the scenario.

We only need to look at one tiny section of the diagram:

A 10kg mass is supported on a string. It is static, meaning something is holding it up else it would fall.
View attachment 363781
We (i.e. a layperson) can immediately see that any scale above this would read 10kg. The string is under 10kg of tension (100 Newtons, if you wish.)


We expand our scope:
View attachment 363782
Nothin has changed. Everything is static. String is still under 10kg tension.

We expand again:
View attachment 363783
Literally nothing has changed. All I've done is show more of the setup.

It must still be under 10kg of tension. And indeed, our spring scale says so.

View attachment 363784
Again, nothing has changed.



Finally, we reveal what has ensured the string and mass stays static:
View attachment 363785
Even the layperson will have to acknowledge that merely revealing what is hidden cannot suddenly change the reading on the scale from 10kg to 20kg.


That wasn't quite what I had been appealing for; I'd been appealing for it a single diagram, labeled in such away as to show why it works. But at least the above doesn't require altering the scenario.

I gave you a single diagram showing that, including where there is 20 kg force.

 

[DaveC426913]

I gave you a single diagram showing that, including where there is 20 kg force.

Yeah. I could not have gotten here without you.

I should clarify: I didn't mean to suggest I succeeded where yall failed - I'm saying I'm parroting back what I learned from yall.

 

[kuruman]

The only difference here is that I haven't resorted to an alternate diagram/scenario to explain the principle; I wanted to remain entirely with the original diagram/scenario.

Refer to the original diagram in post #1.
Fact: The hanging mass on the left is and remains at rest. This means that the tension in the string to which it is attached is equal the weight.
Fact: The hanging mass on the right is and remains at rest. This means that the tension in the string to which it is attached is equal the weight.
Fact: The pulleys are assumed to be ideal. This means they change the direction of the tension at each side but do not affect its magnitude.
Conclusion from the facts above: The tension at each side of the scale is equal to the weight of a mass.

Answer to original question: Spring masses are designed to display one of the equal and opposite tensions applied at their ends^*. Therefore, the reading on the scale is the weight of one of the masses.

---

^*How do we know that? Simple logic. In order to measure a force with a gadget, we need to exert the unknown force on the gadget, but we don't want the gadget to accelerate. Therefore, we need to apply an equal and opposite "stopper" force to keep the gadget at rest. However, that second force does not count. Only the unknown force counts and that's what the gadget displays. In your original diagram one of the weights is being measured and the other serves as the "stopper" force. Obviously, it doesn't matter which is which.

 

[DaveC426913]

^*How do we know that? Simple logic. In order to measure a force with a gadget, we need to exert the unknown force on the gadget, but we don't want the gadget to accelerate. Therefore, we need to apply an equal and opposite "stopper" force to keep the gadget at rest. However, that second force does not count. Only the unknown force counts and that's what the gadget displays. In your original diagram one of the weights is being measured and the other serves as the "stopper" force. Obviously, it doesn't matter which is which.

Yes. This is a very difficult concept for the layperson to grasp in one go.

I am not a layperson; I have been a very active PF member for decades, and I still struggle with this.

 

[kuruman]

Yes. This is a very difficult concept for the layperson to grasp in one go.

I am not a layperson; I have been a very active PF member for decades, and I still struggle with this.

I think it would be easy to explain to a layperson how force scales work. I outlined the gist of my explanation post #185 footnote. Laypersons do not consciously appreciate the importance of the equal and opposite "stopper" force although they make sure that it's there before they use a spring scale. If Alice, a layperson, wants to find the weight of a bunch of bananas, she would first hang the scale from a fixed support and then place the bananas in the tray. She already knows that if she placed the bananas in the tray and then let go of the scale without anchoring the other end, the bananas and tray would be in free fall.

So if I were to explain this to Alice, I would first ask her to explain to me, in her own words, why it is necessary to hang the scale from the ceiling before placing the bananas in the tray and then lead her to the goal of what a spring scale reads. I would start by exploring her answer to "what, do you think, the scale would read if there were bananas in the tray and the tray were in free fall?"

 

[DaveC426913]

I think it would be easy to explain to a layperson how force scales work. I outlined the gist of my explanation post #185 footnote. Laypersons do not consciously appreciate the importance of the equal and opposite "stopper" force although they make sure that it's there before they use a spring scale. If Alice, a layperson, wants to find the weight of a bunch of bananas, she would first hang the scale from a fixed support and then place the bananas in the tray. She already knows that if she placed the bananas in the tray and then let go of the scale without anchoring the other end, the bananas and tray would be in free fall.

So if I were to explain this to Alice, I would first ask her to explain to me, in her own words, why it is necessary to hang the scale from the ceiling before placing the bananas in the tray and then lead her to the goal of what a spring scale reads. I would start by exploring her answer to "what, do you think, the scale would read if there were bananas in the tray and the tray were in free fall?"

Yes. We take for granted that things hang, and that the ceiling exerts its own force.

By the way, I did run this past a number of laypeople, as well as the smartest person I know - an engineer. They all got it wrong.

After I walked him through it, the smartest person I know was silent for several minutes, you could hear his brain cogs squeaking.

 

[Herman Trivilino]

I think it would be easy to explain to a layperson how force scales work

It's difficult to predict, without experimenting, how a person will react to an explanation. Ask someone who thinks there's no gravity on the moon what holds the astronauts down. Some will then concede that there must be gravity but a very common response is "their boots"!

 

[kuruman]

Yes. We take for granted that things hang, and that the ceiling exerts its own force.

We do take it for granted. Everybody should agree that there is a stopping force behind the screen the prevents the weight on the left from falling down.

A stopping force is a stopping force. It makes no difference if the other end of the string is (a) attached to a rigid wall, (b) is held by Bob hiding behind the screen, (c) goes over another pulley and is tied to the floor or (d) goes over another pulley to another 10 kg hanging mass.

That's what should be pointed out to anyone who is unsure. If they are still unsure, it is worth asking them whether they believe that, when any of (a) - (c) above is surreptitiously replaced by (d) behind the screen, they will see a jump in the reading of scale. Those who do are laboring under a faulty logic based on their everyday observation of spring scales. It's probably something like this.

If I place 6 lbs of bananas in the tray of a vertical spring scale attached to the ceiling, the reading is 6 lbs.
Therefore, the scale reads the weight of whatever is hanging at the free end.

I am now asked what the scale will read in the double hanging bananas picture as shown on the right.

Probable faulty reasoning

1. I have established that the spring scale reads the hanging weight.
2. Here we have have gravity pulling at twice the weight.
3. Therefore the scale reads twice the weight.

The fault lies in not recognizing that both ends of the scale need to be included in what has been established.

Correct reasoning that a layperson should be able to understand (if not construct)

1. I have established that the spring scale reads weight hanging from one end when the other end is supported by the ceiling.
2. Here we have gravity pulling at twice the weight, but there is no ceiling. Nevertheless, the hanging masses are at rest. If there is no ceiling, what keeps the masses at rest?
3. (Aha moment) This picture would be the same as the single hanging mass picture if one of the hanging weights assumes the function of the missing ceiling and supports the other hanging mass.
4. Therefore, the scale reads the weight of either one of the hanging masses.

A physicist would call step 3 "modeling the double-pulley, double-hanging-mass system as an equivalent single mass hanging from the ceiling."

 

[DaveC426913]

We do take it for granted

'We' being laypeople.

. Everybody should agree that there is a stopping force behind the screen the prevents the weight on the left from falling down.

A stopping force is a stopping force. It makes no difference if the other end of the string is (a) attached to a rigid wall, (b) is held by Bob hiding behind the screen, (c) goes over another pulley and is tied to the floor or (d) goes over another pulley to another 10 kg hanging mass.

This is where, having had it explained, laypeople begin to cease taking it for granted.

That's what should be pointed out to anyone who is unsure. If they are still unsure, it is worth asking them whether they believe that, when any of (a) - (c) above is surreptitiously replaced by (d) behind the screen, they will see a jump in the reading of scale. Those who do are laboring under a faulty logic based on their everyday observation of spring scales. It's probably something like this.

Yes.

 

[Andrew Mason]

 

[kuruman]

View attachment 363807

I know what you're trying to say but the figure doesn't look right. I assume that the vertical scale is attached to the horizontal scale. The point of attachment is in equilibrium. I can see the horizontal forces cancelling at the point of attachment but what cancels the vertical tension at that point? The string going from pulley to pulley should be drawn as two equal sides of an isosceles triangle, not a straight line.

 

[OmCheeto]

My outline for my thesis of my analysis of this thread:

1. My 4th grade sister taught me 4rd grade maths when I was preschool.
Sis; "Take 4 pies away from 3 pies and you get negative 1 pie."
Me; "You can't take away more pies than you have. Maths is stupid."
1.5 Subsequently self studied maths, well in advance of schooling level.
2. 7th grade teacher giving me an F on an exam where I had every answer right
Me; "Why?"
Teacher; "You didn't show your work, so you cheated."
Me; "But I knew the answers"
Teacher; "Pffft! You still get an F"
Me, internally; "No point arguing."
Result: Never took an unrequired maths class for the rest of my life. What's the point?
2.5 Scored 204 out of 402 upon graduating from high school. I was the epitome of average!
3. Weeks later, scored so high on the Navy entrance exam, it qualified me for Mensa.
4. Missing that ONE answer out of 80 after Navy Nuclear School.
Most missed question I've run across since.
And now, the 2nd most missed question.
4.5 Ran across people who were smarter than I was on the internet who claimed to know their IQs.
5. Watched the Ramanujan movie and seeing my #2 'reminded me of'.
He, like myself, was self trained, and hence couldn't speak 'proper' maths.
6. This thread, where instructors seem to be at battle with students.
7. I'm 66 year old, and am tired of infering from what others say, that I'm stupid, in spite of the evidence.
8. #7 reminds me of that genius stupid cowgirl lady. Can't remember her name.
Senility Now!

 

[DaveC426913]

I know what you're trying to say but the figure doesn't look right. I assume that the vertical scale is attached to the horizontal scale. The point of attachment is in equilibrium. I can see the horizontal forces cancelling at the point of attachment but what cancels the vertical tension at that point? The string going from pulley to pulley should be drawn as two equal sides of an isosceles triangle, not a straight line.

That's a detail. Pretend the string is in a rigid horizontal tube so it is not deflected.

 

[DaveC426913]

2. 7th grade teacher giving me an F on an exam where I had every answer right
Me; "Why?"
Teacher; "You didn't show your work, so you cheated."
Me; "But I knew the answers"
Teacher; "Pffft! You still get an F"

In defense of math, the same thing occurs in English.

OmCheeto's exam essay: "Shylock was definitely a nasty dude."
Teacher: "You get an F."
OmCheeto: "But I got the right answer!"
Teacher: "You didn't show your work."

 

[Lnewqban]

Yes. We take for granted that things hang, and that the ceiling exerts its own force.

By the way, I did run this past a number of laypeople, as well as the smartest person I know - an engineer. They all got it wrong.

After I walked him through it, the smartest person I know was silent for several minutes, you could hear his brain cogs squeaking.

Strain on a section under tension that finally yields, is acting on both axial directions, increasing the length, and trying to separate the bonds of two layers of molecules.
Only one interior force was acting on each layer, which ends up on a different section of the broken bar.
The bond was the opposite force, which resulted overwhelmed by the pulling machine.

 

[OmCheeto]

4.5 Ran across people who were smarter than I was on the internet who claimed to know their IQs.

This was on Michio Kaku's forum. Which I was later to learn was hosted by a high school kid who was ""failing"" so he tried something and it turned into THIS.

Curious how his wife, an educator, deals with children who have IQ's higher than hers.

 

[OmCheeto]

In defense of math, the same thing occurs in English.

OmCheeto's exam essay: "Shylock was definitely a nasty dude."
Teacher: "You get an F."
OmCheeto: "But I got the right answer!"
Teacher: "You didn't show your work."

I have never read that story, and hence, have zero knowledge of this 'Shylock' of whom you speak.
And therefore don't understand any of your post.

 

[Andrew Mason]

I know what you're trying to say but the figure doesn't look right. I assume that the vertical scale is attached to the horizontal scale. The point of attachment is in equilibrium. I can see the horizontal forces cancelling at the point of attachment but what cancels the vertical tension at that point? The string going from pulley to pulley should be drawn as two equal sides of an isosceles triangle, not a straight line.

Ok. Suppose the top vertical scale was directly connected to the rigid (but massless) pulleys and frame like this: