This is a very well explained. Better than any other video explaining this phenomenon which I have seen.

for those who's figuring about spheres, it's possible to do surface integration around the curvature of the sphere. Remember that the flux is going upwards only Assume we can negate the side influx from the sides. But if you need the proof, youc an just take the integral for inward flux and it will return 0. Btw, this works for any type of object, just do the integration.

@briancreech9990

Жыл бұрын

Couldn't you just find the volume of the sphere? Would that cut the difficulty down some?

Wow, that's beautiful. Thanks for sharing!

@JamesCharbonneau

Жыл бұрын

Thanks! I’m glad you liked it.

Thank you very much for this video. Only in my adulthood do I finally understand how this works (and not just by memorizing some formula). I intuitively thought that it must be because of the difference in pressure and had to look it up. Your video is on a perfect level of abstraction needed to fully understand this phenomenon.

@JamesCharbonneau

4 ай бұрын

Thanks for the kids words! I’m glad it was useful for you.

Excelente! Gracias!

Thank you, it helped me fully understand it

thank you for this!

God bless you

thanks so much!!

Thanks a lot sir 👍

THANK YOU

Can you show how it works for an arbitrarily-shaped object rather than just a cylinder? I’m having trouble finding examples of the general case. I assume it involves some integrals, but I haven’t figured out how it works for myself

@JamesCharbonneau

3 жыл бұрын

Good question. You’re right that it would be an integral, but I think it’s a pretty straight forward extension of the idea in the video. First, if you understand the cylinder derivation, then it would be pretty easy to imagine the same argument, but with a square on the top and bottom rather than a circle. This would make the volume shaped like a long rectangular prism. Then, instead of a finite square being the top and bottom, imagine it being an infinitely small area element dxdy, which would make a thin column. You could then imagine the buoyancy of any arbitrary volume as being the sum of all these infinitely thin columns times the density of the fluid, which is an integral. The integral of all these infinitely thin columns over the area dxdy gives the volume of the object. For shapes where a vertical line running parallel to the z axis would go in and out of the surface multiple times forming multiple columns you would need to do the integral piece-wise. edit: For an arbitrary object it has to be an abstract argument like this, because we don’t know the shape, so we can’t get a formula out.

@friiq0

3 жыл бұрын

@@JamesCharbonneau I considered integrating using vertical rectangular prisms, but I was wondering if that would give an error when calculating the total area on top and bottom. When using the prisms to approximate a sphere, for instance, the limit of the area of the bottom surface will approach the largest cross-sectional area of the sphere rather than half the surface area of the sphere. Will that affect our final answer? I found this paper that’s pretty over my head, but on page 15 it has a diagram that seems to suggest that you have to keep track of the normal vector to the surface in the process of integrating: arxiv.org/pdf/1110.5264.pdf

@JamesCharbonneau

3 жыл бұрын

@@friiq0 Those choices can’t affect the final answer simply because things are infinitesimal. Worrying about the corner cases of infinitesimals is definitely interesting in math, but physics never chooses those corner cases. Unless there’s some good reason to think something is up, you keep things simple. I think the paper you linked unnecessarily complicates things. They’re talking about the details of the math that aren’t really pertinent. One thing they mention is deviations from the basic principle, like when the object is touching the walls. This is expected even with the arguments I’ve given, but they go through a long mathematical treatment to comes at the same conclusion. But, there could be physically interesting ways in which Archimedes principle deviates from the law, such as when a viscous fluid might form bubbles or gaps in creases of objects, that aren’t addressed at all. I think these would be interesting places to look for deviations. I think as a general rule, there’s no need to complicate things unnecessarily. If the mathematical arguments work, and agree with experiments, then it’s good. The places that need attention should be more obvious and not happen in the minutia. As I heard in a great talk once “new physics was never found in the 6th term of a Taylor expansion.”

@friiq0

3 жыл бұрын

@@JamesCharbonneau yeah, I think that’s fair. The paper I linked to was pretty over my head, to be honest

@JamesCharbonneau

3 жыл бұрын

@@friiq0 Yeah, I don’t mean to discount the paper entirely, but it’s more an exercise in math than physics.

Basically net upward force is more5 不 B o (Net The bressure exerted by the liquid on the phase AD and BC is equal Pressure exerted by the liquid on the top is downwards and on the bottom is upwards. Totol downward Force F-Pressure X Area F= Pa [F₁ = (h,dgia] (dis density) (P= h.dg also) 0 Upward force exerted by liquid Ep Pressure x Area Fp= (h₂dg) a,

mr james one doubt when we take a solid material like this cylinder there should not be any PGH2 pressure right? becuase there is no water in the h2 lenght so how does it actually included in the equation thats the only thing confusing me

@JamesCharbonneau

3 жыл бұрын

Sorry, I don’t quite understand your question. h_2 is the top portion, which contains the water above the container. Just like the atmosphere is a fluid pushing down on the to of the water, the water above the cylinder pushes down on it. Do you mean the h_1 length. The h_1 length is included because it is the different western the pressure of the water pushing down on the top and the pressure of water pushing up on the bottom.

@akshinbarathi8914

3 жыл бұрын

@@JamesCharbonneau i mean when we normly calculae pgh or gauge pressure we take it as the mass of water above the level but here since we are having an object actualy inside the water some water has escaped so for the particular height there is no water above but only that object exists so how do we include the pgh2 when there is no water above that level which is replaced by an object?

@JamesCharbonneau

3 жыл бұрын

@@akshinbarathi8914 Ah, I see. Very good question. There are perhaps a couple of ways to think about. The first is to remember that’s fluid is made of particles, and the reasons they exert pressure is because these particles with themselves to transfer the pressure and collide with surfaces of objects and transfer momentum. This transfer of momentum of many particle collisions over a surface is called pressure. This is essentially a microscopic way of thinking about Pascal’s principle. Pascal’s principle is the idea that a pressure exerted on one part of a fluid is transferred to everywhere in that fluid. If you think of the weight of the fluid above the height of the bottom of the object exerting pressure on the fluid below it, then every part of the fluid below must have that pressure, even the part under the container. The result of this is that in hydrostatic equilibrium, all fluid at the same depth underwater has the same pressure. Another way to think of this is to realize that a fluid by its nature can’t support a shear force. Another thing to remember is that because the fluid pressure is caused by the collisions of particles, the collisions exert a pressure upwards on the bottom of the container with the same pressure as the rest of the fluid at that depth. The most important argument for why this is true, however, is to remember that physics is simply an attempt to model phenomena we see in the real world. Our models can get very sophisticated, but we only accept them because they model the real world. The model that I describe above to for the pressure on the bottom of the surface works time and time again when we test it. So, even if it seems weird and counterintuitive (which fluids often is) we must accept it because it accurately describes our world.

@akshinbarathi8914

3 жыл бұрын

@@JamesCharbonneau ok !!!!!!!!! i get it!! so if i take a cylinder in a bucket according to my question i woul have said that there is no pgh2 pressure at height h2. but at the bottom level of the cyclinder the pressure level should be same, so if i take a cylinder, and imagine the bottom height is h2 as you said, then the whole level of water at h2 height will have pgh2 and due to pascals law and microscopic transer of pressure even the underpart of the cylider will face the same pressure sinc ethe whole fluid and cylinder should be in equilibrium. in a way from the sides there will be pressre that will act. WOW~!!! james truley amazing i got my doubt cleared, i had a hard timke aceptin g how it is true but now surely iam clear. thanks amillion sir. i love it thanks for spending yout ime to help me out!!🙏👍

@JamesCharbonneau

3 жыл бұрын

@@akshinbarathi8914 Yes, exactly! You’ve got it. I’m glad I could help out.

sir, can you make us understand about e=mc^2

does pwg equal specific gravity

@JamesCharbonneau

2 жыл бұрын

No, the specific density is a ratio of a density to a reference density, usually water. Because it is a ratio it is a unitless number. ho_w g is the boundary force per volume the object has. It’s not something that is really used, but is an interesting idea.

Why the reaction made by the water caused by the weight of the object doesn't participate in the F up force, but instead we pretend that the object is a portion of water? Doesn't the weight produces a force downwards in the bottom?

@JamesCharbonneau

Жыл бұрын

This is a derivation of just the buoyancy force, which is the force from the water on the object. The mass comes into things when we consider the force of gravity on the object. To find the Net Force of the object we add together the buoyant force and the for due to gravity F_N = F_b - F_G.

@rede_neural

Жыл бұрын

@@JamesCharbonneau But that indeed would be a force from the water on the object, the reaction that the water does because of the weight of the object... Or maybe the water doesn't exert any reaction? Or maybe the object doesn't exert a force on the bottom of the liquid? Which one?

@JamesCharbonneau

Жыл бұрын

@@rede_neural I think I misunderstood your original question. By bottom, I thought you meant the bottom of the object, which I was somewhat confused by. You mean the bottom of the container. Yes, the weight of the object certainly pushes on the water, which then pushes down on the bottom of the container. Newton’s law says that every force has an opposite and equal reaction. So, the buoyancy force from the water on the object has an equal an opposite force acting from the object to the water. If the object is floating, this is equal to the weight of the object. If it’s sinking then the object touches the bottom and exerts though force though both buoyancy and a contact force. A common question that deals with this kind of thing is when you have a container of water on a scale. If you place an object in the container, the scale reading will increase by the weight of that object regardless if it’s floating or not. However, if you suspend an object denser than water in water, then the weight only increases by the buoyancy force. These are very subtle problems. I have some other videos on my channel where I work out solutions to buoyancy problems. Perhaps one on those would help.

Why did you not considered weight of object i.e, mg

@JamesCharbonneau

3 жыл бұрын

This is a derivation of the formula for the buoyancy force. I’m trying to give a feel for why there is an upward force on the object from the fluid. If you were to solve a statics or dynamics problem you would consider all the forces acting on the object, including weight.

@JamesCharbonneau

3 жыл бұрын

@TOP GAMERS ARENA The weight of the object would be considered when looking at the net force on the object to determine whether it sinks or floats. In that case the net for would be F_net = F_buoyancy - F_gravity = ma. In this video we are just looking at the buoyancy force, which is the force on an object due to a displaced fluid in gravity.

@manansharma6348

3 жыл бұрын

@@JamesCharbonneau hi i wanted to knoow i placed a cd on top of water it floated but not when placed in middle whyy buoyant force is same

@JamesCharbonneau

3 жыл бұрын

@@manansharma6348 Good question. You might have to experiment more, but I can think of two reasons why the buoyant force wouldn’t be the same. If the can was close to floating when on top, then putting it under water might have compressed the can enough such that the volume of displaced water was not enough to provide a buoyant force to make it float. This actually happens with divers as they get deeper under water. Their lungs get compressed and take up less volume, so they sink after a certain point. Another explanation is that perhaps the can cooled down in the water. The air in the hot can takes up more volume than the air in the cooled can. However, this would mean that the can sinks all time after it cools, so you might have to experiment. In a honestly, it might be a combination of the two. You get the can to just the right temperature such that is barely floats, then putting it under water compresses it to make it sink. The equation of interest here is the Ideal gas law PV=nRT. You see that rearanging gives V = NRT/P, which shows that decreasing the temp and increasing the pressure give a smaller volume of air in the container, thus decreasing the volume of the actual container and decreasing the buoyant force.

@manansharma6348

3 жыл бұрын

@@JamesCharbonneau thanks i believe surface tension could also be the reason for same

You don't explain why the pressure is bigger when going deeper.

@JamesCharbonneau

Жыл бұрын

Here is a video where I derive the hydrostatic equation and explain why pressure gets higher as your go deeper: kzread.info/dash/bejne/eXlhxMqgcrm0iMY.html.

@veronicanoordzee6440

Жыл бұрын

@@JamesCharbonneau Thanks a lot. I removed my disappointment ;-)

@JamesCharbonneau

Жыл бұрын

@@veronicanoordzee6440 Haha, I’m glad I could help.

@HassanKhan-sc7id

8 ай бұрын

Beacuse pressure increases with the increased in depth so p=rho×g×h which equal to yh and elevation decreases in this medium

@veronicanoordzee6440

8 ай бұрын

@@HassanKhan-sc7id Thanks!

The thing is that objects don't just stay submerged inside a volume of water. They either float or sink. This proof of buoyancy doesn't really prove anything, it doesn't describe at all a real life objet in the water. When an object floats there isnt any water above it so how whould we proove the buoyancy force there?

@JamesCharbonneau

Жыл бұрын

There are a couple things to discuss here I suppose. First, this describes the boundary force on a fully submerged object. The derivation of the buoyancy force on a partially submerged object is very similar to this one. The difference is there in no pressure from the water of the top of the object, and the pressure on the bottom is given by the submerged height of the object, F_buoyancy = F_bottom - F_top = P_atm A - (P_atmA + ho g hA) = ho g h A, where now h is the distance below water rather than the full height of the cylinder. We then recognize that hA = V_submerged, the submerged lump of the solid. This submerged volume is the amount of fluid displaced by the object. We get that the buoyancy force is given by the weight of displaced fluid F_buoyancy = ho g V_submerged. This is how Archimedes’ principle was originally formulated. You are right that this does not describe an object floating in water. This is simply a derivation of the upward force on a submerged object. To model the situation you are talking about you need to include a force to balance that upward force, which is the force of gravity. The net force on the object, which described it’s acceleration, F_net = ma, is F_net = F_b - F_g, where F_b is the boundary force pointing upwards and F_g is the force due to gravity, pointing downwards. Combining these we get ma = F_b-F_g. So, to get to your question, when F_b - F_g > 0, the acceleration upwards is positive, and the object will float to the surface. When F_b - F_g What’s important to remember when modelling a real system like you are discussing is to remember to include all forces acting on the object. This is simply a derivation of one of those forces.

Hjj