B Simple mass/scale puzzle

[DaveC426913]

$F=mg=(10 \mathrm{kg})(9.8 \mathrm{N/kg}) \approx 100 \mathrm{N}$

How does this relate to the diagram? I see only one instance of "10kg" in that formula, yet the digram contains two.

 

[Herman Trivilino]

No. It really would require 6lb of bananas for the a scale to read 6lb. I think we are talking past each other.

This is what you said:

No, it would read 6lb if I had 3lb of bananas hanging off one side and 3lb off the other side.

I don't see that as talking past each other. It's you contradicting yourself. The scale either reads 3 lb or it doesn't. And I'm telling you that if you hang 3 lb of bananas on each side, the scale reads 3 lb, even though you have a total of 6 lb of bananas.

Those 3 lb of bananas hanging off the other side of the scale are pulling down on the scale, not up. The scale will read 6 lbs in the scenario you describe in post 28.

So you're saying that hanging 3 lb of bananas from a scale (which is what I describe in Post #28) makes the scale read 6 lb?! I suggest a trip to the produce department of your local grocery store.

Er. 3 lb?

The tension in the cable pulling down on the scale is 3 lb, and the tension in the cable pulling up on the scale is 3 lb, and the scale reads 3 lb.

But this is a different scenario.

In the different scenario the tension in the cable pulling to the right on the scale is 100 N, the tension in the cable pulling to the left on the scale is 100 N, and the scale reads 10 kg.

Note: Edited for clarity.

 

[Herman Trivilino]

How does this relate to the diagram? I see only one instance of "10kg" in that formula, yet the digram contains two.

Because the formula applies to each instance.

 

[Ibix]

What's being asked is "why isn't that same force everywhere equal to 20?"

Think about one mass. It has a downwards force of 100N due to gravity, so the upward force from the string must also be 100N or it would move.

Now think of the string attached to that mass. It has a 100N force on one end, so it must have a 100N force on the other end or it would move.

So the spring must be exerting a 100N force.

Repeat the process from the other end.

So the spring is exerting a 100N force at each end. Its scale is calibrated so that it reads 10kg under those circumstances (because otherwise it wouldn't do its job in normal vertical operation). The end.

 

[Nugatory]

I just don't follow why it's still only 10kg except by using the "look at this alternate scenario" method.

You can try a free body diagram that includes as separate components the hook on the scale and the case of the scale. The hook is subject to the force from the string and from the spring; it is not accelerating so these forces must cancel. The case is subject to the force from the spring and from the string on the other side; again these must cancel as there is no acceleration. Then you can add the third law partners of all four of these forces to satisfy yourself that the force pulling the hook end of the spring away from the case (which is what the scale actually reads) is equal to the tension in the string is equal to the force required to hold either of the weights stationary against gravity.

 

[jbriggs444]

What's being asked is "why isn't that same force everywhere equal to 20?"

Units, darn it. 200 N.

What would be the effects of this mystical "force" that is everywhere equal to 200 N?

The only effects I see are 100 N on one end and 100 N on the other. Well, that and 141 N on each pulley.

Let me try explaining things another way...

A force is a momentum flow from one object to another without the exchange of mass. If we have a 10 kg set of books sitting on a table we have 100 kg m/s of downward momentum flowing from book to table every second. By Newton's third law we also have 100 kg m/s of upward momentum flowing from table to book every second.

You propose adding these two momentum flows together and saying that we have 200 kg m/s² total.

But that is double dipping. You would be counting the same momentum flow twice. It is the same with a rope. If you tried to add the 100 N of rightward momentum flowing left and the 100 N of leftward momentum flowing right you would be counting the same momentum flow twice.

 

[DaveC426913]

I don't see that as talking past each other. It's you contradicting yourself. The scale either reads 3 lb or it doesn't. And I'm telling you that if you hang 3 lb of bananas on each side, the scale reads 3 lb, even though you have a total of 6 lb of bananas.

I was trying to give you the benefit of the doubt.

You appear to be doubling-down on the assertion that a scale suspended from the ceiling that is holding 3lb of bananas and another 3lb of bananas will show only 3lb, not 6lb.

Mister T: " Do you honestly believe that when you hang 3 lb of bananas from a spring scale the scale will read 6 lb?
DaveC: "No, it would read 6lb if I had 3lb of bananas hanging off one side and 3lb off the other side."


Oh. I see whats happening.

A spring scale for weighing bananas is vertical; it is not like the initial scenario. It doesn't have pulleys, and its spring is not between two weighed items.

Diagram 1 is what I pictured when you raised the idea of hanging bananas from a spring scale.

Diagram 2 is what I meant in my response when I said "you'd have to hang another 3 lb. off the other side to get 6 lb.".

I think we can agree both diagrams are correct. But probably not what you were picturing.

So you're saying that hanging 3 lb of bananas from a scale (which is what I describe in Post #28) makes the scale read 6 lb?! I suggest a trip to the produce department of your local grocery store.

No.

We were talking past each other. We had two different ideas of the banana scenario in our head.

I thought you were changing more of the scenario than you may have meant - as I explain above. (In my defense, if you follow the thread, you will see why. Several contributors have proposed scenarios with scales hanging from the ceiling, so it's not a big leap to make.)

The tension in the cable pulling down on the scale is 3 lb, and the tension in the cable pulling up on the scale is 3 lb, and the scale reads 3 lb.

Right, and the only way it would read 6 lb. is if there is a second 3 lb. bunch of bananas in the mix.

 

[DaveC426913]

Units, darn it. 200 N.

I have been talking in kg since post 1.

What would be the effects of this mystical "force" that is everywhere equal to 200 N?

The only effects I see are 100 N on one end and 100 N on the other. Well, that and 141 N on each pulley.

I didn't know I didn't introduce Newtons.

Let me try explaining things another way...

Another explanation.

A force is a momentum flow from one object to another without the exchange of mass. If we have a 10 kg set of books sitting on a table we have 100 kg m/s of downward momentum flowing from book to table every second. By Newton's third law we also have 100 kg m/s of upward momentum flowing from table to book every second.

You propose adding these two momentum flows together and saying that we have 200 kg m/s² total.

I do not, no. Because this analogy is too far from the original. I made no proposals about your scenario at all.




I would really like all participants to read and understand this so you don't keep misinterpreting me:

1. The spring scale reads 10kg. As I state in post 1. There is no argument there.

2. The explanation I posted in post one explains why it must be 10kg. Like almost every other explanaton, it does so by altering the scenario (with glue and scissors, in my case), and then showing that nothing changes. We do not need to repeat that explanation, or any other.

So we're all simpatico, so far...

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.


I'm not yelling, but dang a lot people seem to be not reading the assignment. I just want to head that off once and for all by recapping where we are.

 

[hutchphd]

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

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

I think you meant something more like "attach the left cable to a wall instead of the pulley and weight...

I have to assume there's more direct explanation, one that simply analyzes the given diagram irecxtly.

I have to ask pointedly what could be more elegant and direct than " take a big old wad of glue and splooge one of the pulleys to the mount" ?
Enough said IMHO

 

[vela]

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.

 

[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.

 

[russ_watters]

View attachment 363671

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]

View attachment 363671

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).