B Simple mass/scale puzzle

[DaveC426913]

TL;DR Summary

What does the scale read in this diagram?

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


What does the scale read?

I say the scale reads 10kg.

My logic is this:

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

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

 

[russ_watters]

That's fine, but also: The rope can only have one tension. It can't be 10kg by each weight and 20 in the middle (a common error).

 

[berkeman]

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

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

I think you meant to describe it in a different way. If you fix the left pulley and cut the cable at the left weight, the left cable fragment slips over the fixed pulley and the scale is pulled over the right pulley by the right weight. Fixing the left pulley does not constrain the left cable (modulo the low friction between the left cable and the left fixed pulley).

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

 

[DaveC426913]

I think you meant to describe it in a different way. If you fix the left pulley and cut the cable at the left weight, the left cable fragment slips over the fixed pulley and the scale is pulled over the right pulley by the right weight.

Sorry, you're right, what I wrote was ambiguous.

I meant a giant glob of glue that affixes both pulley and cable in place.

 

[DaveC426913]

That's fine, but also: The rope can only have one tension. It can't be 10kg by each weight and 20 in the middle (a common error).

I don't follow. Who says it would be 10kg by each weight?

 

[renormalize]

I don't follow. Who says it would be 10kg by each weight?

Since the weight is stationary (i.e., not accelerating), the tension pulling up on the weight must exactly balance the gravitational force pulling down on it.

 

[Herman Trivilino]

I don't follow. Who says it would be 10kg by each weight?

Suppose you eliminate the pulley on the left and tie the cable to a fixed block resting on the table top, as shown in the figure below. What will the scale read?

Next, hang the scale and a 10 kg block from the ceiling, as shown in the figure below. What will the scale read?

Hopefully, you can see that in all cases the tension in the cable is the same.

 

[russ_watters]

I don't follow. Who says it would be 10kg by each weight?

I mean the weight is 10kg, so where the cable attaches to the weight the upward force on the weight is 10kg.

 

[PeroK]

Since the weight is stationary (i.e., not accelerating), the tension pulling up on the weight must exactly balance the gravitational force pulling down on it.

Just to note that is the definition of tension. It's the force transmitted; not twice the force transmitted!

 

[DaveC426913]

Suppose you eliminate the pulley on the left and tie the cable to a fixed block resting on the table top, as shown in the figure below. What will the scale read?
View attachment 363602

Next, hang the scale and a 10 kg block from the ceiling, as shown in the figure below. What will the scale read?
View attachment 363603

Hopefully, you can see that in all cases the tension in the cable is the same.

Yes. That was exactly my logic in the opening post. (Ijust glued the left pulley and cable solid.)

What I'm not grasping is Russ' alternate explanation in post 2.

 

[PeroK]

Yes. That was exactly my logic in the opening post. (Ijust glued the left pulley and cable solid.)

What I'm not grasping is Russ' alternate explanation in post 2.

Newton's laws demand that a rope in equilibrium can only have one tension throughout. Unless it has intermediate contact points.

There's no logic as such. The definition of tension is the force. Not twice the force transmitted.

 

[Herman Trivilino]

What I'm not grasping is Russ' alternate explanation in post 2.

I don't see it as an alternate explanation.

Once you go through the sequence of scenarios associated with each situation shown in the figures, you conclude that the tension is the same in all three scenarios. It's therefore an error to think the tension doubles in the first scenario. In all cases you are attaching the ends of a cable to a scale in a place where you could have just tied the two ends together instead. In other words, cut a cable anywhere and insert a scale between the cut ends, the scale by definition measures the tension in the cable.

 

[DaveC426913]

I don't see it as an alternate explanation.

Once you go through the sequence of scenarios associated with each situation shown in the figures, you conclude that the tension is the same in all three scenarios. It's therefore an error to think the tension doubles in the first scenario.

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

In all cases you are attaching the ends of a cable to a scale in a place where you could have just tied the two ends together instead. In other words, cut a cable anywhere and insert a scale between the cut ends, the scale by definition measures the tension in the cable.

I'm not sure that helps me.

I see what you're saying:

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

 

[DaveC426913]

Let me rephrase:

Can you show - formally, and without using an alternate scenario for contrast - why the scale shows only 10kg, when it appears to the naive eye, to be under 20 kg of tension.

 

[PeroK]

Let me rephrase:

Can you show - formally, and without using an alternate scenario for contrast - why the scale shows only 10kg, when it appears to the naive eye, to be under 20 kg of tension.
View attachment 363617

If the spring is stretched to the point where it shows $10g$ on the display, then it provides a $10g$ force at either end. That's how its calibrated. You could calibrate it so that it shows $20g$ at that point. But, all you've done is redefined the measure of tension.

You can't prove a definition.

 

[Herman Trivilino]

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'm not sure that helps me.

It appears in Paul G. Hewitt's famous "Next Time Questions". In the instructor's guide, which accompanies his Conceptual Physics textbook, that is the way he explains the answer.

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

If you pull on a cable with a force of magnitude $F$ the cable pulls back on you with a force of magnitude $F$. You can measure this magnitude by inserting a scale between you and the cable. Orient it so that it reads the force exerted by you. Flip it around so that it reads the force exerted by the cable. In both cases the scale reads $F$, verifying the validity of Newton's Third Law. The value of $F$ is by definition the tension in the cable.

I see what you're saying:
View attachment 363616
I just don't follow why it's still only 10kg except by using the "look at this alternate scenario" method.

It's not the only method. It's just a heuristic method used to help clarify or reinforce. Replacing the 10 kg block hanging on the left with the fixed block attached to the table top shows that they both perform the same function. They each exert the same magnitude force on the cable, therefore the hanging block doesn't add anything to the magnitude of the force exerted on the scale.

 

[Bandersnatch]

Are you guys pulling my leg? How is that not 20kg in tension?

 

[PeroK]

Are you guys pulling my leg? How is that not 20kg in tension?

You could define it as 20kg if you want, but then the force would be given by $F = T/2$.

 

[jbriggs444]

Are you guys pulling my leg? How is that not 20kg in tension?

So the [not fully baked] idea is that you sum the magnitude of the force exerted out the one way and the magnitude of the force exerted out the other way? These forces being exerted by a slice of negligible thickness?

That idea is not ideal since tension can be a signed quantity. A negative tension is a pressure. For example in a stiff rod.

One can generalize the notions of tension and pressure and shear forces with the Cauchy Stress Tensor. By convention (or, equivalently, by definition), the components of that tensor are not doubled. Now instead of looking at magnitudes, one is looking at force per unit area (tensor) multiplied by an incremental directed area (vector) yielding an incremental force (vector). And the signs come out perfectly.

 

[DaveC426913]

If the spring is stretched to the point where it shows $10g$ on the display, then it provides a $10g$ force at either end. That's how its calibrated. You could calibrate it so that it shows $20g$ at that point. But, all you've done is redefined the measure of tension.

You can't prove a definition.

But ... there's 20kg pulling on it!

 

[PeroK]

But ... there's 20kg pulling on it!

Define it as 20kg then! It won't change the physics. As above, you'll just need to use $F = T/2$ in your free-body diagrams.

 

[jbriggs444]

But ... there's 20kg pulling on it!

But forces are vectors. That's -100 N on the left and +100 N on the right for a total of 0 N. No surprise there. The rope is not accelerating.

Which is, I guess, the central point. Just because you can add two numbers together does not mean that you should.

[Yes, I know that you are only trying to help pull our collective legs!]

 

[DaveC426913]

Are you guys pulling my leg? How is that not 20kg in tension?

I know, right?


I mean, I am satisfied that it is 10kg, but only because of the way I went about it:

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.

 

[DaveC426913]

But forces are vectors. That's -200 N on the left and +200 N on the right for a total of 0 N. No surprise there. The rope is not accelerating.

OK. I see that explains why it's not moving, but that still doesn't really explain why it's 10 and not 20.

Which is, I guess, the central point. Just because you can add two numbers together does not mean that you should.

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.

[Yes, I know that you are only trying to help pull our collective legs!]

I'm really not.

It's a conundrum.

I know the correct answer. I even have my own way to demonstrate it intuitively.
But I also know it's not the way they would teach the solution to it in a physics class.

 

[Ibix]

OK. I see that explains why it's not moving, but that still doesn't really explain why it's 10 and not 20.

Use the Newton meter in its usual configuration hanging from a ceiling, hang a 10kg mass from it and let it settle. In that configuration it has a 100N force pulling it down from the mass. It must also have a 100N force pulling it up from the ceiling attachment, otherwise it would be accelerating by Newton's second law. You want this to read 100N.

In the horizontal case in the OP you have 100N forces on each end, just as in the experiment I described above. So it should read 100N (or 10kg depending how it's calibrated).

 

[Herman Trivilino]

Are you guys pulling my leg? How is that not 20kg in tension?

Did you not read Post #7?

 

[PeroK]

In the horizontal case in the OP you have 100N forces on each end, just as in the experiment I described above. So it should read 100N (or 10kg depending how it's calibrated).

Or, 200N if you want to calibrate it that way!

 

[Herman Trivilino]

But ... there's 20kg pulling on it!

Is that not also true of the two alternate scenarios I described in Post #7? Do you honestly believe that when you hang 3 lb of bananas from a spring scale the scale will read 6 lb? Remember you have 3 lb of force pulling down on the scale and 3 lb of force pulling up on the scale. The scale is in a state of static equilibrium, the net force on it is zero.

 

[DaveC426913]

Is that not also true of the two alternate scenarios I described in Post #7? Do you honestly believe that when you hang 3 lb of bananas from a spring scale the scale will read 6 lb?

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

 

[DaveC426913]

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?

 

[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).