B Simple mass/scale puzzle

[Ibix]

What I don't seem to see is a formal solution of the initial scenario that does not involve an altered scenario along with a long intuitive explanation.

It isn't an intuitive thing. It's that the question of how the scale is labelled is arbitrary (edit: or perhaps "conventional" is a better word than arbitrary). It could be labelled in anything, or it could be labelled incorrectly.

So you can show analytically (as has been done several times) that the scale has a 100N force on each end. But the only way you can say that "therefore the scale should read 100N" is by arguing that a 100N force on each end of the scale is exactly the circumstance that you want the scale to read 100N (or 10kg) for its usual purpose of weighing stuff.

 

[kuruman]

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

Consider the drawing shown on the right.

• The tension in the left string is equal to the weight, T = 100 N because the mass is at rest.
• Likewise, the tension in the right string is T = 100 N.
• The pulleys are ideal meaning that they change the direction of the tension without affecting its magnitude.
• Therefore the tension on each side of the spring scale is 100 N as shown.

The gist of your question is what does the spring scale read when it is at rest and a matched pair of equal and opposite forces F act on each end?

Well, if no forces act on the spring it does not stretch. If a matched pair of forces $F$ act on the spring and it stretches by amount $x$, the force $kx$ exerted by the spring on the strings attached to either of its ends must be equal to that force F otherwise the end will accelerate. This allows calibration markings to be put on the scale so that spring displacements in units of length can be read directly as units of force.

The answer then to your question is, the spring scale, when at rest, reads the value of the force F exerted on one of its ends. A spring bathroom scale works the same way except that, here, the spring is compressed instead of stretched. There are two forces acting in opposite directions on the scale, the person's weight $W$ and the normal force $N$ from the floor.

I don't know if this is an "elegant" explanation, but it's simple.

 

[Herman Trivilino]

I just find it hard to believe the original diagram must be altered to a different scenario in order for the problem to be answered.

Why? Almost every bit of knowledge we gain involves a process of comparison to something else that we already understand.

Anyway, the only time we do these alterations you speak of is when a person doesn't understand the following essential feature of how force-measuring devices work.

All that need be done to answer the original question is to possess the understanding that in any static situation only when two forces, each of magnitude $F$, are exerted in opposite directions on a force-measuring device it reads $F$.

The comparisons to other situations are purely heuristic devices designed to prompt people, using more familiar scenarios, who don't appear to understand how force-measuring devices work that they really do understand how they work.

Your objection is rather like a scenario where two walls are separated by a distance of 3 m, and a tape measure is placed on the floor spanning the gap between the two walls. And the question asked is what does the tape measure read?

And of course the answer is 3 m because that's the way the device is designed to work.

 

[sophiecentaur]

Just because you have found two equal but opposite forces somewhere, doesn't mean they are a 3rd Law pair.

I thought my wording was clear that the 'Equilibrium condition' is not N3. Each pair of the chain of objects (including adjacent fibres in the string) has a third law relationship with its neighbour. The values of those third law pairs may not all be the same when there's no equilibrium.

But the thread does bring up the fact that there's a choice of definition between 100N and 200N. But 200N would be needlessly confusing and would make fluid pressure difficult (amongst other interacting forces).

 

[Herman Trivilino]

What I don't seem to see is a formal solution of the initial scenario that does not involve an altered scenario along with a long intuitive explanation.

I'm assuming a properly labeled diagram would do.

(Maybe it's gotten lost somewhere in the last 99 posts).

Post #46.

 

[jbriggs444]

So, if I blew out the centre wall and tied the two strings to each other, they'd still read 10kg each.
And then if I bridged one of them with a taut piece of string, then cut out the scale, the remaining one would still read 10kg.

Which is fine.


I just find it hard to believe the original diagram must be altered to a different scenario in order for the problem to be answered. If this diagram were presented on an exam, I doubt any of these long "look at it this other way" scenarios would cut it.

Can you please elaborate for us the broken intuition that you need fixed? Why does your intuition hold that the forces on the two ends of a scale should add to produce the scale reading?

Why do you not have the same intuition when one end is simply fixed in place instead of being pulled by a more complicated arrangement.

Does your intuition think of a force as requiring some sort of action? So that a restraint does not qualify?

 

[Herman Trivilino]

But the thread does bring up the fact that there's a choice of definition between 100N and 200N. But 200N would be needlessly confusing and would make fluid pressure difficult (amongst other interacting forces).

It would require that the spring constant be defined as $k=2F/x$. A convention that I have never seen in the hundreds of textbooks I've either used or reviewed. In fact, any textbook author who adopted this convention would have it labelled as an error by the reviewers and editors, insisting that it be changed to $k=F/x$ before the book could be published.

Introducing unconventional conventions is a very confusing, and in my opinion a silly, thing to do, unless there is some compelling reason to do so. Or, as you say, needless.

 

[Herman Trivilino]

How would you explain it to a 5 year old?

That in every situation where a scale is used, that's the way scales work. And proceed to show, by use of other examples, that that is indeed the case. The 5 year old, if smart enough, would then reach this as a generalization from observation.

Which is in fact the very thing you did for yourself.

I just don't understand why you continue to insist that it doesn't make sense. In every case forces of magnitude $F$ are applied in opposite directions to a force-measuring device that then reads $F$. That's the way spring scales and all other force-measuring devices are designed to work. They are engineered to work that way, by choice.

 

[kuruman]

I just don't understand why you continue to insist that it doesn't make sense. In every case forces of magnitude $F$ are applied in opposite directions to a force-measuring device that then reads $F$. That's the way spring scales and all other force-measuring devices are designed to work. They are engineered to work that way, by choice.

I completely agree. If you have a force measuring device with two strings attached to it on either end, it's only useful if the tensions in the strings were equal else it will accelerate. Given then that it's not accelerating, the obvious reading for it to display is the value of the tension on either end so that you can hang it from the ceiling and find the weight of a bunch of bananas.

 

[jbriggs444]

It would require that the spring constant be defined as k=2F/x.

That depends on how you define the spring constant.

You could define it as $\frac{F}{x}$ or you could define it as $\frac{T}{x}$. If you make the latter choice then, as you suggest, you would have $k=\frac{2F}{x}$ If you make the former choice then you would have $k=\frac{T}{2x}$

I do not see anyone here advocating for defining $T=2F$. We are simply pointing out that it is a possible definition. Silly, but possible.

Until we have a definition for $T$ in hand, no mathematical proof of the value of $T$ in this scenario is possible.

 

[Nugatory]

I just find it hard to believe the original diagram must be altered to a different scenario in order for the problem to be answered.

It doesn't. Several posts have already done that: Demystifier's #71, multiple posts from multiple people suggesting that you draw a free-body diagram for the two pieces of the scale (hook and case) that are exactly what we're looking at when we ask "what does the scale read", probably some other that I have missed.

 

[Herman Trivilino]

That's all well and good, but if I understand the request by @DaveC426913 correctly, he doesn't want comparisons to other scenarios, but a direct derivation of the scale reading for his scenario.

Well, it is a direct derivation of the scale reading for the given scenario. An elegant one.

Your result for that scale reading still contains the spring constant k, which by convention is based on the magnitude F, for two opposite forces of magnitude F each, acting on the ends of the spring, and the elongation they cause. But that's just a convention and we could just as well define k based on 2F.

True, but it doesn't make your point. It's still a direct derivation of the scale reading for the given scenario.

Yes, it depends on the convention that $k=F/x$, but I don't think the convention is what confuses people who don't understand the answer. Those people are confused about the way force-measuring devices measure force. I don't think they're questioning the definition $k=F/x$.

 

[Herman Trivilino]

What I don't seem to see is a formal solution of the initial scenario that does not involve an altered scenario along with a long intuitive explanation.

Alright, here's a better answer.

Referring to the figure in your OP, there are two hanging blocks, each of mass 10 kg. The block on the left exerts a force $F=mg=(10\ \mathrm{kg})(9.8\ \mathrm{N/kg}) \approx 100\ \mathrm{N}$ on the left side of the scale. The block on the right exerts a force $F=mg=(10\ \mathrm{kg})(9.8\ \mathrm{N/kg}) \approx 100\ \mathrm{N}$ on the right side of the scale.

When you have these two forces acting on the scale in this way, the scale reads 100 N, by definition, design, engineering, and convention. Or in this case, in the no-longer-officially-sanctioned unit of kilogram-force, 10 kg.

 

[Herman Trivilino]

(which is peculiar in that it is magically capable of measuring fish in both mass and weight)

If you tried to use that device in a place where the free fall acceleration were significantly different, you would find that it does indeed not measure mass. What it measures is force.

It may be calibrated in units of mass, for purposes of measuring mass, but such devices, especially in cases where sufficient precision is required, have to be calibrated for the location in which they are intended to be used. Use them in a different location and they may no longer accurately measure mass.

 

[PAllen]

If you tried to use that device in a place where the free fall acceleration were significantly different, you would find that it does indeed not measure mass. What it measures is force.

It may be calibrated in units of mass, for purposes of measuring mass, but such devices, especially in cases where sufficient precision is required, have to be calibrated for the location in which they are intended to be used. Use them in a different location and they may no longer accurately measure mass.

The advantages of the old fashioned agate knife edge balance with weights. Good to one milligram easily. Tarnish on the weights was the enemy…

 

[Herman Trivilino]

The advantages of the old fashioned agate knife edge balance with weights. Good to one milligram easily. Tarnish on the weights was the enemy…

Comparison by means of a balance is still the most precise way to measure mass.

 

[DaveC426913]

What's the matter? This whole thread is about the fact that a standard spring scale needs two forces of magnitude F pulling on each side, in order to indicate F. So the fact that you omitted one of those forces in your 'correct' diagram is quite telling.

Oh fer cryin out loud the scales are hanging from the ceiling.

 

[DaveC426913]

Post #46.

Absolutely not.

As I said, you've shown no correlation between the single equation you posted and its relevance to the diagram.

 

[Herman Trivilino]

Absolutely not.

As I said, you've shown no correlation between the single equation you posted and its relevance to the diagram.

See Post #113 for a revised version that hopefully addresses your concerns.

 

[PAllen]

I don't know if this will help or not, and I don't have any drawing software, so here is a free hand diagram:

 

[OmCheeto]

If you tried to use that device in a place where the free fall acceleration were significantly different, you would find that it does indeed not measure mass. What it measures is force.

It may be calibrated in units of mass, for purposes of measuring mass, but such devices, especially in cases where sufficient precision is required, have to be calibrated for the location in which they are intended to be used. Use them in a different location and they may no longer accurately measure mass.

Please get back to me when our planet has moved from 0.001 on the Kardashev scale, to 0.01.
Thanks!

 

[ergospherical]

The scale itself is an elastic coil. You can imagine replacing it with a segment of rope with a different elastic modulus to the rest of the rope. Now you just have a rope. The tension everywhere in the rope is the same: it's precisely the size of the force that each infinitesimal bit of rope applies on the next bit, and vice versa.

 

[Herman Trivilino]

Please get back to me when our planet has moved from 0.001 on the Kardashev scale, to 0.01.
Thanks!

Free fall acceleration varies by 0.5% across Earth's surface at sea level. Probably not enough to bother with for fish, but it is for other things.

Regardless, your fish-weighing device measures force in units of mass.

 

[.Scott]

I didn't see this. So what do the scales read?

Presuming they are calibrated for local gravity, they will read 10Kg.
When I put that graphic together, I was running out of time.
Had I more time, I would have put little hooks where the (massless) rope attached to that new wall, just to make the intent more clear.
As everyone has explained, those scales are basically measuring the tension on the rope.

 

[OmCheeto]

The scale itself is an elastic coil. You can imagine replacing it with a segment of rope with a different elastic modulus to the rest of the rope. Now you just have a rope. The tension everywhere in the rope is the same: it's precisely the size of the force that each infinitesimal bit of rope applies on the next bit, and vice versa.

As someone who doesn't seem to be able to solve problems verbally, I generally resort to building models to solve problems. Your 'scale is an elastic coil' is very similar to what I was considering building yesterday. Rather than a bulky fish scale, I was going to replace it with a rubber band.

Would anyone consider it a faux pas if I replaced the two pulleys with one for this lab experiment?

 

[kuruman]

I think that the allure of this problem is based on the tendency of people to draw an incorrect inference from a very familiar observation as they lead themselves down the garden path.

Let me explain. Assume that a person is shown what is presented below and asked questions about it.

The observation

A scale hanging from the ceiling with nothing in its tray reads zero.
When 6 lbs of bananas are placed in the tray, the scale reads 6 lbs.
This works the same way with any goods of any weight in the tray. (I will stick with bananas because I like them.)

What the person will say if asked
The weight of bananas on the tray is the reading on the scale. This weight is a force in the direction away from the scale (down) and acts on the "banana side" of the scale.

What the person will take away from all this
Bananas in the tray cause a reading in the scale. No bananas in the tray cause no reading in the scale.

This is correct but incomplete. At this point it is necessary to acknowledge the importance of the tension in the upper string that attaches the scale to the ceiling. It has to be there regardless of whether the scale reads zero or not because the scale does not accelerate.

Asking at this point "Does the tension in the upper string contribute to the reading? Why or why not?" will probably prevent the person from falling into the trap associated with this problem.

Springing the trap

Now you present the person with the double-banana assembly as shown on the right and you pop the question What does the scale read?

The person will most likely say 12 lbs and if you ask why, the response would be something like "We have seen that bananas in the tray cause a reading in the scale. No bananas in the tray cause no reading in the scale. Since we now have two bunches of bananas pulling on each side of the spring scale, it follows that the reading should be twice as much."

This response is based on the faulty conclusion that the force at each end of the scale contributes to the reading of the scale independently of what what force acts at the other end. Of course, if that were true the spring scale would accelerate. For that reason the forces must be equal and opposite and that is why it is important to ask about the contribution of the upper string to the reading of the scale.

 

[Herman Trivilino]

At this point it is necessary to acknowledge the importance of the tension in the upper string that attaches the scale to the ceiling. It has to be there regardless of whether the scale reads zero or not because the scale does not accelerate

The ceiling exerts a passive force on the vertical scale, unlike the active force exerted by the other bananas on the horizontal scale. That force is there because gravity pulls down on the bananas, and yes it must be there because the scale is in equilibrium. Those are two reasons. The notion of gravity pulling is familiar, but the notion of equilibrium requiring balanced forces is further removed from experience.

In the case of the vertical scale you have given only one reason why the support cable must pull upward on the scale, and that is because the scale is in equilibrium. But there is another reason. The ceiling is strong enough to withstand the flexing caused by the downward pull, and so is the supporting cable. This point must be made to complete an understanding. The cable doesn't pull up because it has to, it pulls up because its being stretched. Likewise, the ceiling doesn't pull up because it has to, it pulls up because it is being deformed. Both the ceiling and the supporting cable are being deformed, and it is their ability to withstand the deformation that is the cause of this passive force.

People have trouble with this notion of a passive force. How can a cable exert a force? It has no p-p-p-power!

 

[ergospherical]

Split the scale (i.e. the coil) into 100 little imaginary pieces. The tension T in the coil is the size of the force that each piece of coil exerts on the one next to it. Consider the piece right at one end of the coil, where it connects to the rope. It’s pulled to one side by a force T by the piece of coil next to it. So it must also be pulled by a force T by the rope. This is how elastic materials work and why the tension is uniform.

 

[A.T.]

but I don't think the convention is what confuses people who don't understand the answer.

And I think this is exactly what confuses some of them. Especially those who demand a formal derivation from some fundamental principles, of something that is a mere convention.

 

[A.T.]

People have trouble with this notion of a passive force.

Because quantitatively there is no such distinction, so the whole causation prose doesn't affect the quantitative result.

 

[Herman Trivilino]

And I think this is exactly what confuses some of them. Especially those who demand a formal derivation from some fundamental principles, of something that is a mere convention.

Perhaps I've interacted with a lower caliber of student. But in my experience they don't demand or even ask for a formal solution, and they readily accept the conventional definitions that lead to $k=F/x=T/x$. What stumps them I think is that both hanging blocks exert active forces (due to gravity) on the scale. Replace one of the hanging blocks with a rigid structure and their confusion seems to disappear because the passive force (due to the rigid block) is not seen as equivalent to the active force it replaced?

It's difficult to figure out what these students are thinking. We tend to attribute their errors to misconceptions, but perhaps they don't even have well-formed preconceptions. Instead they have what's been called protoconcepts.

The demand for a formal derivation may just be a knee-jerk reaction to having their misconception exposed. A defense mechanism.

Sometimes it seems my teaching devolves into psychoanalysis.

 

[kuruman]

People have trouble with this notion of a passive force. How can a cable exert a force? It has no p-p-p-power!

Precisely. The departure from reality starts when they develop the misconception that since the passive force has no "power", it is doing nothing therefore it might as well not be there.

 

[Herman Trivilino]

Precisely. The departure from reality starts when they develop the misconception that since the passive force has no "power", it is doing nothing therefore it might as well not be there.

It's essential that they understand that it's always due to a deformation of an elastic material, whether it be a stretched cord or a deformed surface.

One thing you can do, or even just describe, is a laser beam reflecting at a low angle off the surface of a table top and making a spot on the wall. The slightest disturbance of the table top makes the spot on the wall move.

Another useful exercise is to have the students measure the deformation of a rubber band when something with weight is hung from it. Then repeat with a stiffer rubber band. Progress to a string and finally a chain. It's essential to understand that even if the deformation is immeasurably small, it still must be there. Otherwise there's no way the weight can be supported.

 

[Ibix]

Instead they have what's been called protoconcepts.

I remember being confused by it the first time I came across it. The experiment we did at school was to measure how much force was needed to open a can of Coke, which we did by putting the can on a desk, tying a string to the ring pull, passing it over a pulley, and hanging weights from the free end of the string. At some point we had to add weights on top of the can because we needed more force than the weight of the can, and it occurred to me that we needed the force on both ends to open the can so (I argued) the force needed was twice what we were applying at one end.

It isn't an unreasonable definition (which I suspect is the problem students have), but it isn't consistent with how we define the behaviour of a force meter in its role as a weighing scale.

 

[erobz]

To me the misconception is a simple one rooted in common experience with weight measurement. We have a rope with two weights hanging off of it vs. one weight when we hypothetically tie it to the fixed structure to "explain it". To me it's natural to feel like we have halved the "force experience" of the rope when we do this... It's not until you look individually at free body diagrams of the three parts (left mass, scale, right mass) applying Newtons Laws ( in static equilibrium ) that you find yourself in a theoretical pickle with that "common sense" scenario.

 

[kuruman]

This thread is the gift that keeps on giving. Here is a related question that came to me last night that could form the basis for a class discussion.

Two students are presented with the picture on the right. It shows a 100 N weight hanging at the end of two identical (massless) spring scales. The students are asked what the reading on each scale is and to explain their reasoning.

Alice says, "A spring scale reads the weight that is attached to it. The bottom scale reads 100 N because it is attached to the 100 N weight. The top scale also reads 100 N because it can't tell whether there is weight pulling down or another spring."

Bob says, "Alice you are wrong. At the top string (point P in the drawing) the tension is clearly 100 N. The total force that supports the 100 N weight is provided by the intervening springs that stretch under the load. Since the springs are identical, they stretch by the same amount so that together they provide the necessary 100 N. Therefore each scale reads 50 N."

Question for the class: Who is right and why is the other person wrong?

Spoiler

Alice is right. Bob is wrong because when you remove a piece of string under tension at both ends and replace it with a spring scale under the same tension at both ends, you are adding a zero net force, not a force F=kx. Bob bamboozled himself by incorrectly invoking the correct result that the sum of two identical vectors is twice one of the vectors.

 

[DaveC426913]

What stumps them I think is that both hanging blocks exert active forces (due to gravity) on the scale. Replace one of the hanging blocks with a rigid structure and their confusion seems to disappear because the passive force (due to the rigid block) is not seen as equivalent to the active force it replaced?

Yup. I would say this is the crux.

I can get to the right answer by altering the scenario and then altering it back, but I cannot look at the initial scenario and say "Yup. 10kg on each side means the scale shows 10kg. "

 

[DaveC426913]

Alice says, "A spring scale reads the weight that is attached to it. The bottom scale reads 100 N because it is attached to the 100 N weight. The top scale also reads 100 N because it can't tell whether there is weight pulling down or another spring."

If you replaced that string with 10 scales, they would all read 100N?

The vertical column of things would be so stretched it would reach the floor?

It must be so.

The only alternative is to imagine the 10 springs sharing the load, at 10N each, which is preposterous.

 

[hutchphd]

Static means static. This is an idealization but well defined and not a "fuzzy" force of some weird character. So what is an "active force" (other than an unfortunate complication?). I am mystified by much of this, but not without reason.

ADDENDUM: the "many stacked scale" problem is in fact likely support much more complicated dynamics (e.g. like dropping the slinky) but the original problem is by definition (and fiat) static.

 

[Herman Trivilino]

So what is an "active force" (other than an unfortunate complication?). I am mystified by much of this, but not without reason.

Well, gravity doesn't fit this scheme. But otherwise, an active force is a force that arises from the expenditure of fuel. A person pushes on a wall. The person can do this because a fuel source keeps him alive and gives him the needed strength. Novices have a hard time understanding how the wall is able to push on the person because the wall requires no fuel source and is thus a passive force.

Gravity, I suppose, is considered an active force because of its familiarity as a force.

I agree that the distinction is not well-formed. It's more of a heuristic, or a tool of pedagogy. It took a fuel source to make that wall!

 

[DaveC426913]

ADDENDUM: the "many stacked scale" problem is in fact likely support much more complicated dynamics (e.g. like dropping the slinky) but the original problem is by definition (and fiat) static.

The many stacked scales scenario can be perfectly static too. That is what I was describing.

 

[Herman Trivilino]

It isn't an unreasonable definition (which I suspect is the problem students have), but it isn't consistent with how we define the behaviour of a force meter in its role as a weighing scale.

It wouldn't be a valid measure of the force needed to open the can. Instead of putting weights on the can you could have just held the can.

 

[hutchphd]

It wouldn't be a valid measure of the force needed to open the can. Instead of putting weights on the can you could have just held the can.

An+d this makes things less complicated somehow? The issue of extended bodies and forces where things are "overdetermined" gives most students difficulty. But more rigor and crisper definition is the solution.

 

[kuruman]

I can get to the right answer by altering the scenario and then altering it back, but I cannot look at the initial scenario and say "Yup. 10kg on each side means the scale shows 10kg. "

Suppose I asked you to answer the following binary choice question.

The figure on the right shows a 10 kg hanging mass connected to a spring scale by a string going over an ideal pulley. The system is at equilibrium. A screen hides what's beyond the scale to the right.
What does the scale read?
(A) 10 kg
(B) It depends on my knowing what's behind the screen.

What would your answer be? You have no other choices; only one is correct and makes sense.

 

[Frabjous]

There are so many posts in this thread, I apologize if I am repeating things.

There are two forces in this problem. There is also 1 tension.
The forces are vectors. The tension is a tensor.
At the ends of the rope, the forces are calculated using the normals and the tension tensor. So adding things is not the way to go.

 

[jbriggs444]

Bob says, "Alice you are wrong. At the top string (point P in the drawing) the tension is clearly 100 N. The total force that supports the 100 N weight is provided by the intervening springs that stretch under the load. Since the springs are identical, they stretch by the same amount so that together they provide the necessary 100 N. Therefore each scale reads 50 N."

I've seen that misconception in the wild. It is the "transitive property of forces". If you pull on an object with a force of 50 N then that object passes the 50 N force along to whatever it is attached to. This is in addition to any tension force supplied by the body.

Of course, as an incoherent preconception (love that word), it is difficult to attack coherently.

 

[DaveC426913]

View attachment 363732Suppose I asked you to answer the following binary choice question.

The figure on the right shows a 10 kg hanging mass connected to a spring scale by a string going over an ideal pulley. The system is at equilibrium. A screen hides what's beyond the scale to the right.
What does the scale read?
(A) 10 kg
(B) It depends on my knowing what's behind the screen.

What would your answer be? You have no other choices; only one is correct and makes sense.

Yup. I know exactly what you mean. I often use the "opaque screen" technique to help people understand things too.

Theres a million ways to explain it. But that's still not really a solution like you might see on a test.

 

[DaveC426913]

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.

 

[ergospherical]

That's just a misunderstanding of how springs work. If the tension in a spring is T, the spring exerts a force of size T on anything attached to it. It pulls whatever is on one side with a force T, and whatever is on the other side with a force T. And a scale is a spring: the number you see is the tension (divided by $g$, if it's calibrated to display units of mass).

 

[kuruman]

But that's still not really a solution like you might see on a test.

The solution to what problem? This question probes whether one knows and understands the context of a spring scale's reading. Here it is: If I cut a piece of string under tension and attach a spring scale end at each cut end of the string, the spring scale will read the local tension in the string.

If I asked the original question in post #1 on a test, I would give full credit to any student who first establishes that the tension in the string without the scale in line is 100 N and then argues that since the spring scale measures the local tension in the string, the reading would be 100 N or 10 kg depending on the scale calibration. This answer shows an understanding of statics and what the reading of the spring scale means.

What more is there to say in a solution?

BTW, the context of an ammeter reading is similar to that of a spring scale: If I cut a piece of wire in which there is a current and attach an ideal ammeter terminal at each cut end of the wire, the ammeter will read the local current in the wire.