Buoyant Cork
The diagram shows a bucket filled with water hanging on a string, with a cork tied to the base of the bucket by another string.
When the cork's string is cut, since the cork is less dense than water, the cork will rise up due to buoyancy .
When the string holding the bucket is cut, the bucket and its contents will fall due to gravity.
What happens if both strings are cut simultaneously ? How would the cork move relative to the bucket?
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8 solutions
The gravitational force always exists, even during free fall. I think you should rather write that, in the frame of reference of the bucket + water in free fall, the net volumic force acting on water is zero (gravity is compensated by inertial force). Since net volumic force is zero, water pressure is constant, thus there is no buoyant force.
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Perfect judging !
"gravity is compensated by inertial force" it's a naive conscious. When you view it from the outside perspective, there is no inertial force here. Inertial force only appears due to the choice of reference, and it's not real. I agree with you that during the fall, gravity still exists (where could it go?!), but there is no pressure on the water. That's the point. And I like the way Arjen Vreugdenhil observed this, very subtle.
You could also use the sum of forces in the y direction=0. The cork is part of the falling system so the all fall at the same rate.
I agree that the straightforward answer is that in free fall the bucket, water and cork will all fall at the same rate, so there will be no relative movement of the cork to the bucket.
However, there are other considerations!
Firstly, you need a vacuum to be in free fall. If you're in air your acceleration (towards earth) will be gradually reduced, reaching zero when you're at terminal velocity. In this scenario the cork will rise, but more slowly than when the bucket wasn't falling.
Secondly, even if you're in free fall in a vacuum, the constituents of the bucket, water and cork still have gravitational attraction between themselves. Think of the bucket as like a little planet in space. If you had a planet that was just water, with a cork inside it somewhere, which way would the cork move? It would move away from the center of the planet, and pop up on the surface. In our scenario, the cork would move away from the center of mass of the bucket/water/cork system. Since it's near the bottom of the bucket, the cork would move downward. This would be a small effect. :-)
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Both are small effects. Estimating the size of the bucket as 30 cm across and 50 cm tall, the forces are F gravity = ρ w g h π r 2 = 1 0 0 0 ⋅ 1 0 ⋅ 0 . 5 0 ⋅ 3 . 1 4 ⋅ 0 . 1 5 2 ∼ 3 0 0 N , F drag = 2 1 ρ a C D π r 2 v 2 = 2 1 ⋅ 1 , 2 ⋅ 1 ⋅ 3 . 1 4 ⋅ 0 . 1 5 2 ⋅ v 2 ∼ 0 . 0 4 N / ( m / s ) 2 ⋅ v 2 . The bucket must fall for more than a second (about 15 ft) to make the drag force 1% of the bucket's weight; therefore the buoyancy will remain less than 1% of what it would be normally.
Glad to see deeper thought. Don't forget Coriolis effect, which will minutely twist the falling system horizontally about its N-S axis of mass, and the 'flotation' of the cork on its arc toward the top surface as the system approaches terminal velocity. The missing info about the altitude & latitude constrain our more detailed exploration. Some of us enjoy exploring the shape of the sometimes transparent thinking-box.
This neglects air resistance on the bucket. The cork would fall, albeit very slowly.
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Ahhh, but this neglects water resistance on the cork. It would still fall, but not quite as fast if this didn't exist. Ahhhhhhhhhhh, but this neglects the air resistance on the bucket due to the cork not falling quite as fast as it otherwise would. Ahhhhhhhhhhhhhhhhhhhhhh, but this neglects the resistance of the water on the cork which is not falling as fast as it should, while the bucket is falling slightly faster than it should while the ... oh, man, my head hurts.
Arjen Vreugdenhil is correct for the initial moment after the strings were cut: The cork would initially fall at the same rate as the water, so it would be at rest relative to the bucket.
As Justin Case correctly points out, the rate of acceleration of the bucket decreases over time due to air friction. So, relative to the bucket, the cork will start to rise.
Think what happens to Arjen's analysis in the limit where the bucket has reached its terminal velocity---in that case, the buoyancy analysis would be exactly the same as for a bucket at rest.
I think the author meant "at the initial moment after the string was cut". Unfortunately this was not stated, so really none of the answers listed was correct. (This is why I despise multiple choice answers for physics problems, except for definitions.)
1 free fall like in space they should float 0 gravity. 2. frame of reference of bucket + water, gravitational force relative upwards and actual same magnitude downwards net g=0, let atmospheric pressure be 0, mass of cork miniscule but weight =0, pressure from water buoyancy not counted due to g=0; stable position. Is it correct?
I dissent before you cut the string is an existing force that keep the string taut so at the cutting the cork will move upward.
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This force disappears at the very moment the bucket starts falling.
There is no water pressure on the cork during the free fall. Therefore the water above the cork will not try to replace the cork's position.
Of course there will be the water pressure. But it is almost uniform. Molecules of the water will still move due to temperature, thus creating pressure on the surfaces in contact.
That's not totally true, but due to different causes... the tensional attraction on the surface of the water makes a membrane that has some tension on it, which is the same experimented uniformly inside the water, so there's actually pressure (and of course the pressure of atmosphere in which the system is embedded) What is not there is a gradient of pressure which is what makes the cork to recibe a force that forces it to float.
Awesome simple solution, with much appreciated addendums explaining further.
By applying Einstein's equivalence principle, the cork will not experience is own weight because it is falling together with the bucket. Therefore in the cork's reference freme the "perceived" gravity acceleration will be 0.
The instant both strings are cut the bucket and cork are in free fall, hence no gravity on the water, no head of water pressure, hence no buoyancy effect. No buoyancy effect = no relative motion.
Since both cords are cut simultaneously, all three objects (the bucket, the water, and the cork) are in free fall. During free fall, all objects fall at the same rate since the pull of gravity is greater for more massive objects, but the inertia of the more massive objects exactly cancels this out. Thus the cork and the water, and even the bucket, accelerate towards the ground at the exact same rate.
Of course, once the bucket hits the ground, the water cannot move down further, so the cork will then rise to the top of the water.
I think this is about the best answer that I've seen. You only need to consider the initial moment after the cord is cut. The answer changes over time, due to air friction. Once the bucket is falling near its terminal speed, the analysis of the motion of the cork would be the same as for a bucket at rest.
Cool! For several decades I have used this (inertia v mass) and the 'thread v weld' explanations for intuitive fall speed confusion.
Technically the correct answer is not listed. The cork does not remain at rest. It falls at the same rate as the bucket and water, staying in the same place relative to the water and the bucket.
The question does say 'relative to the bucket'.
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Unfortunately it doesn't use the word "initially". The cork starts moving upwards relative to the bucket once air friction stops being negligible. Think about what happens once the bucket reaches its terminal speed...
Simple explained ,Newton 2 says that within an isolated system the sum of forces is zero. So , no relative motion .
Even after looking through some solutions I still don't really understand the multiple choice options. I understand relativity to be about spatial coordinates as well as time.
I didn't really know, which approximations I was supposed to make:
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Do I really have to understand the concept of buoyancy? (I decided that it was more about relativity)
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Can I work with linear approximations? (I haven't studied the equations of general relativity yet, therefore couldn't come up with a clear answer. Decided to skip it.)
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Which time frame does the questioner have in mind? (Simultaneity is relative. Me as an imagined physicist looking at the experiment? Or me as a the bucket itself? [Then I would have used "bucket language" in the answers provided. " I am at rest." Although this isn't a usual format for sciences, I suppose.])
I'm more experienced in linguistics than in some formal calculus. Often, I don't know what the problem stated really is about. I can only answer a question, if I know, which concepts I (don't) have to apply. In this case: Do I have to read more about buoyancy or not? (Probably, I should have invested the time.)
Actually here everywhere inside the bucket pressure remains same equal to atmospheric pressure . As we know that pressure static at any point in a liquid is given by :- P(atm) +hdg(eff) Where d :- density of liquid ,h:- depth from the free surface of liquid g(eff):- effective accleration with respect to bucket Since here g(eff) is zero(g-g), Therfore the second part of the above equation becomes equal to zero ,and as we know that buoyancy arises due to the differences in pressure at different points in a liquid, due to no pressure difference ,here buoyant force equals zero Now draw fbd of cork with respect to bucket You will notice by applying psuedo force that net force on cork in the frame of bucket becomes equal to zero .( as mg-mg = 0)
From the frame of reference of the bucket + water in free fall, there is no gravitational force and therefore also no buoyant force. In this frame of reference, the net force on the cork is zero. Therefore it will remain at rest (relative to the bucket).
From an outside perspective, the cork will only experience gravitational force and no buoyant force. It will therefore be in free fall, just like the bucket and the water.
There is no buoyant force because the water pressure is constant. Normally, the water pressure increases with depth. The reason for this hydrostatic pressure is as follows: Consider a layer of water of area d A and thickness d y . There are three forces acting on it: the gravitational force ρ g d A d y ; upward force due to pressure from the layer underneath P ↓ d A ; and downward force due to pressure from the layer above P ↑ d A . Together, they keep the layer of water in equilibrium, so ρ g d A d y + ( P ↑ − P ↓ ) d A = 0 . Thus the pressure gradient is d P = P ↓ − P ↑ = ρ g d y .
In the case of the falling bucket, the "0" in this derivation must be replaced by ρ g d A d y , since the layer of water is in free fall, accelerating at rate g . This leaves d P = 0 .