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

[russ_watters]

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I know it gets more silly the further you take it, but I feel like it could even be better to leave the original scale in place and just add two scales to either side of it. The scales on each side show the forces you're applying to the scale in the center, and the one in the middle shows the "total".

 

[DaveC426913]

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

 

[Herman Trivilino]

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

Yes.

And for those interested about that normal force, it's an upward force exerted on the book by the table, so its 3rd Law partner is the downward force exerted on the table by the book.

 

[Herman Trivilino]

Can you expand on this by stating the correct answer?

See Post #93.

 

[Herman Trivilino]

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

10 kg.

 

[A.T.]

Fer cryin out loud...

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.

 

[A.T.]

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.

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.

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.

 

[DaveC426913]

10 kg.

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.

 

[PAllen]

All the scenarios are the same, including the original. A string or spring under tension T means each element has a pair of Newtonian opposing forces on each end - right end of one element pulls with force T on left end of next, and left end of next pulls on right element of first. This state exists all along the string/spring under tension. To avoid snap back, each final end element must have force T applied to it - else it will snap back. Any method of providing that force T on each end is equivalent: holding with your hand, glueing to massive body, rocket, counterweight, etc.

There is no changing of scenario with respect to the string/spring.

 

[DaveC426913]

All the scenarios are the same, including the original.

Yes.

There is no changing of scenario with respect to the string/spring.

Agree.

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).

 

[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: