That's not quite correct Brig. lowering an object into water adds downward force due to the displacement of the fluid. The mass of the object lowered in doesn't matter - it's purely a function of the density of the fluid and the volume of the object. So the ping pong ball and steel ball of equal volume both displace equal amount of fluid and therefore add equal buoyant force. The confusing bit is the 'floating' of ping pong ball and the 'sinking' of the steel ball which appear to be opposite forces. But in fact the steel ball adds no force at all due to mass because it's suspended outside of the system. Whereas the ping pong ball and string are within the system so add a small mass to the left.
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- Thread starter AronSearle
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Yes John, but only when the system has "settled down". My guess is that, while the steel ball is in the process of being lowered into the water, it is conferring some force to that beaker, which will indeed (as per the youtube video) go down.
Only when the ball is not moving downwards will the true (LHS slightly heavier) state be shown.
Which begs the question, does the act of withdrawing the steel ball actually cause the RHS beaker to rise? (at least until the ball stops again, or is fully out of the water)...
I just typed in "steel ball ping pong ball" into the youtube search bar.
Sorry, can't link as on works computer.
Sorry, can't link as on works computer.
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E
EddieM
The video of you tube does not match the described scenario.
Not sure if I got the same one. The only one I can find is an Asian guy in a college or school who says the ping pong ball doesn't need to be tied down. He then does the experiment with it floating on the surface. But that's not correct. Bouyant force is equal to the weight of displaced water and a floating ping pong ball displaces much less water than a tied down one.
E
EddieM
Sounds like the same one. It doesnt replicate the described scenario. As I think I said before my understanding (which could well be flawed!!) is that only the left hand side has added mass (the ping pong ball, plus the string) the right hand side has no additional mass. Any buoyancy forces positive or negative are cancelled out by the string applying an opposite and equal force?
I think I may have almost repeated verbatim what jeds said !!
I think I may have almost repeated verbatim what jeds said !!
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Anchor 4cm diam table tennis ball weighing 2.7gms, to the inside base of a flask using length of cotton and glue such that the uppermost part of ball is 1/3 the internal height of the flask, fill the flask with water until the ball is submerged by a small amount, note the precise amount of water used.
Place a lid of known weight upon the flask, Weigh the flask, lid and contents, note the weight with lid weight subtracted.
Upend the flask such that the ball is hanging above the water - not touching.
Weigh the flask and contents, note the weight, with lid weight subtracted.
The weights noted should be identical. = wt of flask + water + cotton and glue + table tennis ball.
Fill an identical flask with an identical amount of water hang a steel ball 4cm diameter, carefully, using cotton thread hanging from a point outside of the apparatus into the water until fully immersed without touching the insides of flask.
This ball will displace around 33.5cc approx 33.5gms of water, the flask plus inserted ball and cotton thread will gain 33.5gms in weight, whereas the first flask can only have the weight of the table tennis ball and it's associated thread - say 2.8 gms total in addition to the weight of water added.
I say the flask containing the steel ball immersed in water will outweigh the flask containing the tethered t/t ball and therefore sink or fall or drop as in the you tube vid.
On the other hand ... If the two weighings of the t/t ball flask differ, then I am scuppered !
Place a lid of known weight upon the flask, Weigh the flask, lid and contents, note the weight with lid weight subtracted.
Upend the flask such that the ball is hanging above the water - not touching.
Weigh the flask and contents, note the weight, with lid weight subtracted.
The weights noted should be identical. = wt of flask + water + cotton and glue + table tennis ball.
Fill an identical flask with an identical amount of water hang a steel ball 4cm diameter, carefully, using cotton thread hanging from a point outside of the apparatus into the water until fully immersed without touching the insides of flask.
This ball will displace around 33.5cc approx 33.5gms of water, the flask plus inserted ball and cotton thread will gain 33.5gms in weight, whereas the first flask can only have the weight of the table tennis ball and it's associated thread - say 2.8 gms total in addition to the weight of water added.
I say the flask containing the steel ball immersed in water will outweigh the flask containing the tethered t/t ball and therefore sink or fall or drop as in the you tube vid.
On the other hand ... If the two weighings of the t/t ball flask differ, then I am scuppered !
With respect, I think this is where you've gone wrong. The displaced water will still be there, even though displaced by the metal ball. No more water is added, merely displaced. The metal ball will add absolutely no weight to the container, as it is suspended externally.
No, I'm sure the balance will remain static, with the possible very slight tip to the left due to the additonal weight of the ping pong ball and string - if the balance is sensitive enough.
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Ah but, the tether is pulling the t/t ball down, the ball is pulling upward on the tether.
The ball would float by displacing 2.7gms (cc) of water.
I reckon the residual weight is the tension in the tether with this tension at 30.8 gms cancelling out between forced upthrust and tethered pull down, the difference will be the weight of the ball.
You must agree with inverting the flask and finding the same weight? Therefore there can be no addition or reduction due to water displacement by the ball.
Stick a measuring jug on your kitchen scales loer something, a hand will do, into the water - note the weight change.
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The ball would float by displacing 2.7gms (cc) of water.
I reckon the residual weight is the tension in the tether with this tension at 30.8 gms cancelling out between forced upthrust and tethered pull down, the difference will be the weight of the ball.
You must agree with inverting the flask and finding the same weight? Therefore there can be no addition or reduction due to water displacement by the ball.
Stick a measuring jug on your kitchen scales loer something, a hand will do, into the water - note the weight change.
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Good thinking
But use two items, first one that floats, and then one that sinks but is suspended as per the question.
But you wouldn't find the same weight. In one the ball would be immersed and in the other it wouldn't.
When the steel ball is immersed in water it displaces some water. If it was a hollow ball that floated then the weight of water displaced would be equal to the weight of the ball and the ball would be "weighless" as measured by the pull on the string.
If the ball is too heavy to float it will still displace water but not enough weight of water to support the ball so it will sink until the string is taut. But the weigh of the ball on the string will be less than the weigh out of water due to the floatation force the water is exerting upwards onto the ball.
The equal and opposite reaction to that force is acting on the sides and bottom of the flask so the flask will be heavier, the extra weight being equal to the the weight "lost" by the ball.
If the ball is too heavy to float it will still displace water but not enough weight of water to support the ball so it will sink until the string is taut. But the weigh of the ball on the string will be less than the weigh out of water due to the floatation force the water is exerting upwards onto the ball.
The equal and opposite reaction to that force is acting on the sides and bottom of the flask so the flask will be heavier, the extra weight being equal to the the weight "lost" by the ball.
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Somewhat done...
Apparatus :- 1 Measuring jug, 1 small 'Acura' (don't ask !) ladies face cream jar - empty and cleansed with decent sealing screw top. 1 length of cotton, 2 stainless steel nuts, 1 Salter electronic scales. Approx 800 cc water.
Trap a few cm cotton in jar, lid on tightly, weight = 115 gms
Weigh jug and water = 1816 gms leave jug on scales.
Gently hang jar into water - result ! it only just sank, hold steady not touching insides of jug weight total 1926 gms.
Increase = 110 gms suggesting volume of jar is around 110 cc displacing 110 gms of water.
Jar shape more or less frustrum of cone having base diam. 3.1cm minor diam 2.7cm with a height of 4.3cm... Theoretical volume 113.8 cc
http://tinyurl.com/n2ymgg2
Error to volume actual measure (113.8 / 110) - 1 = 0.035 or + 3.5% - reasonable.
Just as a checker, placed two 30 gm s/s nuts in jar, closed lid tight dangled into water - steady state weight increase, as before 110 gms.
Additional weight of nuts had no effect (as expected) only volume displacement weight of water adding to overall weight.
Haven't tethered, but absolutely 100 % certain there can be no weight difference between t/tennis ball + flask + water with flask upright or inverted - cannot get something for nothing.
Here is the crux - how can, as depicted, one inversion weigh more than the other without adding or subtracting mass ?
When one realises they cannot - then one realises why the steel ball insertion causes its flask to descend or gain weight.
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Apparatus :- 1 Measuring jug, 1 small 'Acura' (don't ask !) ladies face cream jar - empty and cleansed with decent sealing screw top. 1 length of cotton, 2 stainless steel nuts, 1 Salter electronic scales. Approx 800 cc water.
Trap a few cm cotton in jar, lid on tightly, weight = 115 gms
Weigh jug and water = 1816 gms leave jug on scales.
Gently hang jar into water - result ! it only just sank, hold steady not touching insides of jug weight total 1926 gms.
Increase = 110 gms suggesting volume of jar is around 110 cc displacing 110 gms of water.
Jar shape more or less frustrum of cone having base diam. 3.1cm minor diam 2.7cm with a height of 4.3cm... Theoretical volume 113.8 cc
http://tinyurl.com/n2ymgg2
Error to volume actual measure (113.8 / 110) - 1 = 0.035 or + 3.5% - reasonable.
Just as a checker, placed two 30 gm s/s nuts in jar, closed lid tight dangled into water - steady state weight increase, as before 110 gms.
Additional weight of nuts had no effect (as expected) only volume displacement weight of water adding to overall weight.
Haven't tethered, but absolutely 100 % certain there can be no weight difference between t/tennis ball + flask + water with flask upright or inverted - cannot get something for nothing.
Here is the crux - how can, as depicted, one inversion weigh more than the other without adding or subtracting mass ?
When one realises they cannot - then one realises why the steel ball insertion causes its flask to descend or gain weight.
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