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17 Comments
kulpimssays...*promote
siftbotsays...Promoting this video and sending it back into the queue for one more try; last queued Wednesday, October 28th, 2015 8:10am PDT - promote requested by kulpims.
Stormsingersays...Problem 1: The scale will tilt towards the lead block. It's the same principle as Archimedes, except using air instead of water. When there is air, there is a buoyant force exerted on any object immersed in it. Remove the air, and the weight of the object goes up, by the weight of the same volume of air.
Problem 2: 20*pi meters. I'm not sure how extreme physics is involved in this one at all. It's trivially derived from the definition of circumference.
Barbarsays...Wouldn't it tilt towards the wooden block (meaning indicate the wooden block was heavier) since bouyancy would have contributed more to the larger wooden block?
On the other hand she states they have identical mass (not weight), so in a vacuum they should register even.
Problem 1: The scale will tilt towards the lead block. It's the same principle as Archimedes, except using air instead of water. When there is air, there is a buoyant force exerted on any object immersed in it. Remove the air, and the weight of the object goes up, by the weight of the same volume of air.
Problem 2: 20*pi meters. I'm not sure how extreme physics is involved in this one at all. It's trivially derived from the definition of circumference.
Stormsingersays...You're absolutely correct. Especially the last part (this is a sloppy video). If they balance on the scales -with- an atmosphere, they won't balance without one. And in atmosphere, the wood will have have more buoyancy than the lead (my screwup).
Wouldn't it tilt towards the wooden block (meaning indicate the wooden block was heavier) since bouyancy would have contributed more to the larger wooden block?
On the other hand she states they have identical mass (not weight), so in a vacuum they should register even.
rkonesays...I'm not sure where air comes into play for problem one. There's no "air" factor in the gravity equation. F=GM1M2/(r^2)
But yeah, by definition, you need to use the equation to get the 20pi answer...
newtboysays...In the opening question she blew it. What if the rock is lava rock, which is LIGHTER than water? That means you can't figure out the answer without knowing the density of the rock.
Archimedes equation is only useful in figuring out weight for things that are buoyant. Anything more dense than water (or whatever medium you're in) will only displace it's own volume in water, not it's mass.
That's why I think the wood block should weigh more in a vacuum. It displaced more air, so was more buoyant, and so had more buoyancy to lose. It seems to me she set it up poorly again, because if they weigh the same in air, but are different densities, they would seem to need to have different masses to achieve balance, but she said they have the same mass, but I think she should have said 'they weigh the same'....just as @Barbar and @Stormsinger indicated above.
ChaosEnginesays...@Stormsinger @Barbar
what is the "extreme case" for the scales problem?
Stormsingersays...Beats the hell out of me.
Just to noodle around a bit, the only extreme I can think of about the scales would be to substitute an extremely low density object for the wood. Say, a helium filled balloon? But that assumes that she did in fact mean equal mass for the two objects, and wouldn't actually give valid readings on a scale in atmosphere anyway.
Extreme cases are a rather specialized approach, as I remember...its not really a common, or easy way to get answers. I got the feeling this was kind of a "wannabee" presentation. Like she wanted to do "Smarter every day" stuff but isn't quite able to find and explain interesting non-intuitive problems well.
@Stormsinger @Barbar
what is the "extreme case" for the scales problem?
Stormsingersays...If the rock sinks, you can float it on something buoyant, and measure the water levels with and without the rock. It's an extra step, but still doable. Your other points are definite flaws in her logic/presentation.
Archimedes equation is only useful in figuring out weight for things that are buoyant. Anything more dense than water (or whatever medium you're in) will only displace it's own volume in water, not it's mass.
robbersdog49says...Problem 1: Tip toward the wood, as the wood will lose more buoyancy from the air than the lead.
As for the extreme case here, let's use the helium balloon. You tie the helium balloon to the right hand side of the scale. Now, to get the bar on the scale horizontal (balanced) you need to hang the lead weight closer to the fulcrum but on the right hand side of the scale too.
Now remove the air.
The balloon was only pulling up because of the air. Without the air it will hang down. So, we now have two things hanging down on the same side of the scale, so it's very obvious which way the scale will swing...
Problem 2: pi*20m Circumference = pi*diameter. Poles increase diameter by 20m. Really not sure where the 'extreme case' comes into this though?
Beats the hell out of me.
Just to noodle around a bit, the only extreme I can think of about the scales would be to substitute an extremely low density object for the wood. Say, a helium filled balloon? But that assumes that she did in fact mean equal mass for the two objects, and wouldn't actually give valid readings on a scale in atmosphere anyway.
Extreme cases are a rather specialized approach, as I remember...its not really a common, or easy way to get answers. I got the feeling this was kind of a "wannabee" presentation. Like she wanted to do "Smarter every day" stuff but isn't quite able to find and explain interesting non-intuitive problems well.
Stormsingersays...@robbersdog49 I think you've nailed it.
newtboyjokingly says...That's cheating. ;-)
If it's not buoyant, adding something more buoyant so the two together are buoyant then subtracting the original displacement will work, but you have to make the thing you measure 'buoyant' to use Archimedes equation. For things that aren't buoyant, (at all) it doesn't work, you have to change the property of the thing you measure to measure it.
If the rock sinks, you can float it on something buoyant, and measure the water levels with and without the rock. It's an extra step, but still doable. Your other points are definite flaws in her logic/presentation.
rkonesays...Good insight, and good use of the extreme example. I guess the question is sloppy as newtboy said. I got hung up on "they have the same mass" which would definitely mean they would be balanced with no air.
Some excellent points
Stormsingersays...I know! That's what makes it cool.
That's cheating. ;-)
L0ckysays...The extreme case I thought of here wasn't about the balloon, it was about the air. Imagine the air was a lot denser... like water, or jelly.
I wasn't familiar with the term 'extreme cases' but use this all the time when making logical arguments, and think of it as exaggeration to help illustrate a point.
Unfortunately, while I think it makes it easier for me to visualize the point, people are often blinded by the comical nature of the exaggeration
"What do you mean if there were 10 zillion Chinese people on the island of Malta and 3 Brits had the Americas to themselves? The world should still learn English!"
(Someone trying to argue that even if Chinese was predominant, English would still be used in a larger geographical area, and me attempting to use exaggeration to illustrate that it's besides the point).
Problem 1: Tip toward the wood, as the wood will lose more buoyancy from the air than the lead.
As for the extreme case here, let's use the helium balloon. You tie the helium balloon to the right hand side of the scale. Now, to get the bar on the scale horizontal (balanced) you need to hang the lead weight closer to the fulcrum but on the right hand side of the scale too.
Now remove the air.
The balloon was only pulling up because of the air. Without the air it will hang down. So, we now have two things hanging down on the same side of the scale, so it's very obvious which way the scale will swing...
Problem 2: pi*20m Circumference = pi*diameter. Poles increase diameter by 20m. Really not sure where the 'extreme case' comes into this though?
robbersdog49says...Genius! Make it water and the wood floats. (Same effect as mine but feels a lot more relevant to the question).
The extreme case I thought of here wasn't about the balloon, it was about the air. Imagine the air was a lot denser... like water, or jelly.
I wasn't familiar with the term 'extreme cases' but use this all the time when making logical arguments, and think of it as exaggeration to help illustrate a point.
Unfortunately, while I think it makes it easier for me to visualize the point, people are often blinded by the comical nature of the exaggeration
"What do you mean if there were 10 zillion Chinese people on the island of Malta and 3 Brits had the Americas to themselves? The world should still learn English!"
(Someone trying to argue that even if Chinese was predominant, English would still be used in a larger geographical area, and me attempting to use exaggeration to illustrate that it's besides the point).
Discuss...
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