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

[PeroK]

I get that, if you hung a 10kg mass from the ceiling, there is 10kg pulling down and 10 kg pulling up.

I think that's the crux. All the pulleys do is change one of them from 'up' to 'down'. No, that would cancel to zero.

Is there a FBD diagram that would demonstrate the solution?

Maybe show why I think there's a missing 10kg? Will it show 5kg pulling on each end of the scale?

How would you solve this problem - with formulae, say, on an test?

I don't know many many times and how many different ways it needs to be explained.

 

[DaveC426913]

I don't know many many times and how many different ways it needs to be explained.

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

 

[Andy Resnick]

I know, right?


I mean, I am satisfied that it is 10kg, but only because of the way I went about it:
View attachment 363623
It obviously shows 10kg in diagram 2. And diagram 1 is the sam, as far as the scale goes so it must also be 10kg.

I just think that's a pretty sloppy way to demonstrating it.

How about this explanation: cut one of the strings (no glue, no anchoring, just cut the string) so the scale reads 0 N and accelerates (that should be uncontroversial). In order to keep the scale in position (or to allow it to move with constant velocity!), how much force do you need to apply to the other end of the scale?

But this puzzle is just a warm-up to something more interesting: predict the scale reading if the weights have different masses.

 

[renormalize]

It's a conundrum.

Is it? Consider the two configurations 1 and 2 below:

Anthropomorphizing the scale, it can only "see" (respond to) what's inside the red ellipses. From the scale's local perspective (neglecting the gravitational sag of the horizontal cable in 1) there is no difference in the two configurations and so it responds identically to both. How could it be otherwise? It doesn't even require a calculation!

 

[PeroK]

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

The definition of tension is the force exerted by the rope or string. It can't exert a force at one end and not the other, so you must have the same force at both ends. I've said some or all of that in posts #9, #11, #15, #18, #21. So, this is the sixth time I've told you that. One tension (T) throughout, the same magnitude of force (F) at either end, $T = F$ by definition.

 

[DaveC426913]

The definition of tension is the force exerted by the rope or string. It can't exert a force at one end and not the other, so you must have the same force at both ends. I've said some or all of that in posts #9, #11, #15, #18, #21. So, this is the sixth time I've told you that.

Yes you have. It's not resonating.
Because you're still not explicitly saying why it's 10 and not 20.

One tension (T) throughout, the same magnitude of force (F) at either end, $T = F$ by definition.

Sure. But why not 20? (Do you see what I mean? I agree it's "the same magnitude" throughout. But thats doesn't answer why it's 10 and not 20.)


What are you guys taking for granted that I don't know? None of these explanations end with "... which is only 10kg, not 20kg".

 

[DaveC426913]

How about this explanation:

I don't want explanations. I have explanations. I even have my own. I know it is, in fact, 10 and not 20. The answer is not in doubt.

I think what I'm looking for is a FBD, or equivalent.

 

[PeroK]

Sure. But why not 20? (Do you see what I mean? I agree it's "the same magnitude" throughout. But thats doesn't answer why it's 10 and not 20.)

If the string pulls a mass of 10kg with a force of 200N, then the mass will accelerate against gravity.

Not only that, but the force from the string is a reaction to being pulled by the mass of 10kg with a force of 100N (approx).

What are you guys taking for granted that I don't know? None of these explanations end with "... which is only 10kg, not 20kg".

Newton's laws. Several posts in this thread have indicated that.

 

[PeroK]

I don't want explanations. I have explanations. I even have my own. I know it is, in fact, 10 and not 20. The answer is not in doubt.

I think what I'm looking for is a FBD, or equivalent.

I think you've lost your reason in this thread and should maybe take a break and come back when your head is a bit clearer. How many diagrams do you want to see?

 

[jbriggs444]

Well, if I added 10 kg to each side, that would certainly add to the tension put on the cable. So it makes sense that I should be adding the weights up.

I am trying to wrap my head around the intuition you have here.

You start with a scale with no weights on either side. It has zero tension across it. Surely we agree on this.

Now you add a tension on one side... But you cannot. The rope moves under the unbalanced force and the tensionometer in the middle reads near zero. This part of the tension test does not really work. Can we agree on this?

Then you add a weight on the other side. Now the left hand rope is now pulling with 100 N the one way on the tensionometer and the right hand rope is pulling with 100 N the other way on the tensionometer.

You want a reading of 200 N to reflect the sum of the pulling forces on the two sides.

If we now nail one side of the tensionometer to the wall do you want a reading of 100 N since only one side is exerting an "active" force while the other side is "passive"?

In my view, there is no distinction to be made between an "active" force and a "passive" force. So I do not think that nailing one side to the wall will change anything.


I am 100% with @PeroK on this. If you want to define the tension in a small section of rope as the sum of the forces pulling in each direction (arranging for a positive sign on both forces since both are pulling) then that is just fine and dandy. But if you then want the force supplied to the object on one side then you must divide this "tension" by two: $F_\text{right} = \frac{T}{2}$ and $F_\text{left} = \frac{T}{2}$.

 

[DaveC426913]

How many diagrams do you want to see?

One would suffice.

Not of some other scenario (as almost every diagram here has contained) but of the indicated scenario.


Anyway, clearly the problem is me, if only by consensus.

 

[jbriggs444]

Yes you have. It's not resonating.
Because you're still not explicitly saying why it's 10 and not 20.

It is a definition. There is no physical "why". There is only mathematical convenience.

It is easier to write $F_\text{right} = T$ instead of $F_\text{right} = \frac{T}{2}$.

Do you have a proposal for a physical test to see what the tension in a rope "really" is other than by measuring the forces that the rope exerts?

The one way that comes to mind is Hooke's law. That yields a result that does not have the factor of two.

 

[DaveC426913]

I am trying to wrap my head around the intuition you have here.

You start with a scale with no weights on either side. It has zero tension across it. Surely we agree on this.

Now you add a tension on one side... But you cannot. The rope moves under the unbalanced force and the tensionometer in the middle reads near zero. This part of the tension test does not really work. Can we agree on this?

Then you add a weight on the other side. Now the left hand rope is now pulling with 100 N the one way on the tensionometer and the right hand rope is pulling with 100 N the other way on the tensionometer.

You want a reading of 200 N to reflect the sum of the pulling forces on the two sides.

If we now nail one side of the tensionometer to the wall do you want a reading of 100 N since only one side is exerting an "active" force while the other side is "passive"?

In my view, there is no distinction to be made between an "active" force and a "passive" force. So I do not think that nailing one side to the wall will change anything.

I don't know how many times I can say that "I figured this out in the opening post".

1. Glue the left pulley and its cable together so it can't move.
2. Cut left weight loose.
3. Nothing happens. The scale stays where it is. And it it now obvious at a glance that the answer must be 10kg now, which means it must have been 10kg before.
I get that.

What I don't get is how one can solve this without a long explanation that requires changing the scenario to prove the point. Surely basic physics problems aren't solved by saying "Here is a long explanation".

Surely the diagram, as-given, can be labelled.


But it seems not. I give up. It's not worth aggravating everyone and tarnishing my immaculate reputation.

 

[Herman Trivilino]

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

Nope. I've done this demonstration for students many many times.

What difference does it make what pulls with 3 lb of force? It could be a ceiling, it could be a brick fastened to the table top, it could be another 3 lb of bananas.

If the scale hangs from the ceiling by a cable, and the scale reads 3 lb because 3 lb of bananas hang from the scale, what is the tension in that cable that hangs from the ceiling?

 

[Andy Resnick]

I don't want explanations. I have explanations. I even have my own. I know it is, in fact, 10 and not 20. The answer is not in doubt.

I think what I'm looking for is a FBD, or equivalent.

Have you tried drawing some? You said your learning is a bit rusty, this seems like a good 'homework assignment' :)

Edit: when you do, remember to 'explode' the scale to make the spring mechanism explicit.

 

[Herman Trivilino]

What I don't get is how one can solve this without a long explanation that requires changing the scenario to prove the point. Surely basic physics problems aren't solved by saying "Here is a long explanation".

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

 

[russ_watters]

Right. That's the same as my opening analysis. But it's not elegant because it requires saying "OK, let's look at this alternate layout and now you can see it's the same as the given layout." I would be that is not how it is solved in physics classes.

I can't speak for physics classes, but that is in fact how it's done in engineering classes: cut the problem up literally and figuratively into separate parts, solve them separately and then recombine them to get the final result.

...At least in the beginning until the important points become intuitive (one tension, same everywhere) and override the instinct (10+10=20). But it can always be brought out again when needed.

Also can't speak for "elegant".

 

[russ_watters]

Yes you have. It's not resonating.
Because you're still not explicitly saying why it's 10 and not 20.


Sure. But why not 20? (Do you see what I mean? I agree it's "the same magnitude" throughout. But thats doesn't answer why it's 10 and not 20.)


What are you guys taking for granted that I don't know? None of these explanations end with "... which is only 10kg, not 20kg".

Because you can't pull on a 10kg weight with 20kg of force and have it remain stationary.

 

[DaveC426913]

Nope. I've done this demonstration for students many many times.

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

What difference does it make what pulls with 3 lb of force? It could be a ceiling, it could be a brick fastened to the table top, it could be another 3 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.

If the scale hangs from the ceiling by a cable, and the scale reads 3 lb because 3 lb of bananas hang from the scale, what is the tension in that cable that hangs from the ceiling?

Er. 3 lb?

But this is a different scenario.

 

[DaveC426913]

The definition of tension is the force exerted by the rope or string. It can't exert a force at one end and not the other, so you must have the same force at both ends. I've said some or all of that in posts #9, #11, #15, #18, #21. So, this is the sixth time I've told you that. One tension (T) throughout, the same magnitude of force (F) at either end, $T = F$ by definition.

And about sixth time I've said:

"Sure. It's the same everywhere. Never doubted you."

That's not what's being asked.

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



Why are you not understanding that the conclusion (it's 10kg) does not immediately follow from your stated premise (it must be the same force at both ends).

 

[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.