Buoyancy & Buoyant Force - Online Tutor, Practice Problems & Exam Prep

Buoyancy is the upward force that allows objects to float in a liquid, explained by Archimedes' principle, which states that the buoyant force equals the weight of the displaced liquid. The buoyant force (F_b) can be calculated using the equation: ${\rho }^{l}gV$, where ρ_l is the liquid's density, g is the acceleration due to gravity, and V is the submerged volume. An object sinks if its density is greater than the liquid's, and floats if it is less dense.

Intro to Buoyancy & Buoyant Force

Hey guys. In this video, we're going to start talking about buoyancy, the phenomenon that causes objects to float or be pushed up when in a liquid. Let's check it out. Alright. So, if you have an object immersed in a liquid, it will be pushed up by a force called buoyant force, which stems from the phenomenon of buoyancy, usually associated with floating. We will denote this force as \( F_b \). This happens due to a pressure difference between the top and the bottom of an object submerged in a liquid.

Let's say you have a box completely submerged underwater; the water applies pressure from all sides. The force on the left will cancel the force on the right because they are at the same height. Thus, the side forces have no net effect, but the top and bottom forces will differ due to varying pressures. The deeper under the liquid, the greater the pressure. Thus, the pressure at the bottom is greater than at the top, causing the net upward force, or buoyant force. That results in a stronger force upwards than downwards, and the net buoyant force is directed upwards. Cool?

Next, you should know Archimedes' principle, which states that the magnitude of the buoyant force is the same as the weight of the liquid displaced. However, to calculate this force, we use the equation \( F_b = \rho_{\text{liquid}} \cdot g \cdot V_{\text{under}} \), where \( \rho_{\text{liquid}} \) represents the density of the liquid, \( g \) is gravity (approximately 9.8 m/s² on Earth), and \( V_{\text{under}} \) is the volume of the object submerged in the liquid. Remember, the important volume is the volume underwater, not the total volume, unless the object is completely submerged.

The more an object is submerged, the greater the buoyant force acting on it. We also need to define the density of an object, which is mass over volume. In equations, this involves the total and submerged volumes, leading us to consider both in various scenarios. For most buoyancy problems in physics, they reduce to a force problem where the sum of all forces equals zero, especially when objects are at rest or equilibrium in a liquid.

As an example, imagine an object fully submerged in a large water tank. When it's released and reaches equilibrium, 30% of its volume is above water. From this configuration, we estimate the density of the object using our equations and understanding of volumes. By setting the net force equation to zero and expanding terms, we determine the unknown density after a series of substitutions and simplifications, concluding with the object's density based on its submerged volume and known liquid density.

This presents a clear example of how evenly basic physics principles like buoyancy apply in detailed, practical scenarios and demonstrate the consistent application of principles like Archimedes' principle and fundamental force analysis.

Comparing Buoyant Forces

Hey guys, so let's check out this conceptual buoyancy question. And I chose this one because it's pretty tricky and there are quite a few of these kinds of tricky buoyancy questions I want to help you think through. So let's check it out. So here we have a piece of wood and a piece of metal and they both have the same volume. So I'm going to say volume of wood equals volume of metal. They're both placed in a large water tank. The wood is going to float. Let me draw the wood over here and the metal sinks to the bottom and that's what you would expect. Wood almost always floats and metal almost always sinks and we want to know what has the greater buoyant force on it. So which one has the greater \( F_b \)? And before I go, I want you to actually think about this for a second. Maybe pause the video if you need to and pick one. I want you to select an answer, commit to it. Don't be scared. Pick one and then keep going to see if you got it right. And what most people do here is they try to think about this logically, which sounds like a good idea. But the problem is these questions exist because they are tricky and because they sort of defy common sense, right? So what I like to do is I always like to delegate the decision-making to the equation. What I mean by that is that instead of me thinking through it logically or using common sense because I know that that fails and I'm ready for that, I'm going to instead write the equation and let the equation decide. So what is the equation for buoyant force? The equation for buoyant force is \( \text{density of the liquid} \times \text{gravity} \times \text{volume of the object that is under the liquid} \). Now check this out. Both of these guys are underwater, so the density of the liquid is the same because they're in the same liquid. So this is a tie, which means that this is not going to help you determine which one has the more buoyant force, a greater buoyant force. They also both experience the same, the same little \( g \) because they're both near the earth. They're next to each other. Now it's all going to come down to the volume under and they both have a total volume of 50, but the wood has some of its volume sticking out. So the volume under, I'm going to make up some numbers, it's going to be like 10 for the wood and 40 for the metal, for example, and all 50 of it is going to be underwater for the metal. So the metal has more of its total volume underwater than the wood does. Therefore, the stronger buoyant force will be on the metal. And hopefully, you got this right, but a lot of people actually incorrectly pick wood because they think it's floating. So it must be that the force pushing it up is stronger. It must be that the force pushing it up is stronger than the other one. That's not why it floats. It floats because \( mg \) is much, much smaller. Okay. So here, the force is actually, let's draw some arrows here. \( F_b \) is actually stronger here, but \( mg \) is much, much stronger than here. Okay. So don't get caught up. Write the equation and use that to determine the right answer. Let's keep going.

When an object of unknown mass and volume is fully immersed in large oil (800 kg/m³ ) container and released from rest, it stays at rest. Calculate the density of this object.

• A. 800 kg/m³
• B. 80.0 kg/m³
• C. 8000 kg/m³
• D. 8.00 kg/m³

Buoyancy / Three Common Cases

Hey guys, so in this video, we're going to keep talking about buoyancy and I'm going to show you the 3 common cases that are going to cover pretty much every possibility. Let's check it out. Alright. So an object floats or sinks depending on its density compared to the liquid density. So if the object is denser than the liquid, it's going to sink. And if the object is less dense or lighter than the liquid, it rises to the top. So whoever is denser is going to be lower, okay? So the first situation is referring to this case right here. This here is sort of a secondary case that I want to talk about. It's sort of an exception. In here we have this object that floats above water. So part of the object's above water and this is the part of the object that is underwater and you can just tell by looking at the picture that the volume under is less than the volume total. So for example, let's say that the volume total is 100 and there's maybe 40 60 here. The volume under 60 is less than the volume total 100 because some of it's above water. Pretty straightforward. What about the forces? Well, if the object is floating, it just sits there, it is at equilibrium, right? Because it's just sitting there floating which means the forces cancel. And the forces are F_b going down and and I'm sorry. F_b going up. F_b is always going up. And mg going down. So it must be that F_b equals mg so that they can cancel each other out. And that's what happens there. What about the density? Which one of these two densities is greater? The object or the liquid? So think about it. I actually just mentioned it and hopefully you got it, that the density of the liquid is greater and that's why the object floats because the object is lighter. So one of the, one quick way to look at this that I like is to just look at the top of the objects and the top of the liquid. And because the liquid, the top of the liquid is lower than the top of the object, I think of it as being heavier, therefore it is denser. So liquid is lower, so it has a higher density. Okay. Now these three things here apply to this picture. And this here is just a different slightly different situation. Sort of an exception that I want to talk about. So here you have this object that that's in the middle here. It's floating with a cable. Right? So think about this. What do you think would happen if I if I cut that cord? Right? It must be that the object would ride up because if the object was too dense to sink, it would just sink. The reason it sits there, it's because the the the cord is holding it. So what's happening here is that you have a tension pulling it down. And then you'll have mg also pulling it down. And then you'll have a buoyant force pulling it up. It's still at equilibrium. It still sits there. But now you would write that the forces going down, mg+t, equal the forces going upF_b. The reason why I have this next to the other one is because in this situation, you also have this be true. That the object is less dense. Even though it's underwater because it's only underwater because of the tension, right? If you were to cut this, it would look a lot like this. Okay. So whenever you see a block being held underwater, you have to think what would happen or there's some tension or something like that. You have to think what would happen if I cut that tension and then you would know, okay, well here it goes to the top, which means that it is less dense than the liquid. Cool. The second situation is kind of trivial because it's very similar to this one. Here, the object is floating, but instead of above water, it floats underwater. Right? So how does this happen? Well, this happens if you have an object and you put it underwater and you release it and it stays there. So in this case, the entire volume underwater is the entire volume of the objects. It is 100% underwater. In this case, we also have equilibrium. Okay. Because the 2 forces are gonna cancel. F_b and mg cancel. So they're still at equilibrium. Now what's special here, the difference between these two situations, 1 and 2, is that the density of the object is equal to the density of the liquid. Okay? These two situations are exactly the same. So if you have an object that's entirely underwater, it sinks if it is denser, it's going to rise to the top if it's less dense, and if it does neither, if it stays in the middle, it's because its density is exactly equal to the density of the liquid. Now, how do you get this versus this? The difference is that in the first case, I manually, I grab this object and I brought it just under the waterline and I release it there and because the densities are the same, it stays there. On the second case, you just brought it lower and you released it. Okay? So these are identical situations. If an object floats entirely underwater and it doesn't sort of peak out outside of the water or the liquid and it doesn't sink, it is because the densities are exactly the same. Okay? So this is a simple case but still important to know. Now, number 3. What happens, if the object sinks? Well, what causes an object to sink is the fact that it is heavier. So in this case, you can see here the object is entirely underwater, so So the volume underwater is the same as the total volume. F_b does not equal mg. F_b does not equal mg. The reason why it sinks is because mg is actually going to be greater than F_b. Okay? This is still at equilibrium. It just sits there. But now there's a third force. So you're going to have mg down. You're going to have let me draw this bigger. You're going to have mg down. You're going to have an F_b that's smaller. And because of this, this object is going to be let me move this up. This object is pushing against the surface, so the surface pushes back with a familiar force called normal. And it's not going to be that big. They're both essentially going to add up to cancel the mg. And here you can write that the forces going up, F_b + normal equals the force going down mg. Okay? So there are 3 forces here, just like what we had here. So here, we're going to say that the density of the liquid, or the density of the object rather, is greater. The object sinks because it's heavier, it goes all the way to the bottom, and now it pushes against the floor, so you have a normal force. Okay? So these three things have to do with this situation here. And here we have something very similar. This is sort of an, a side case, kind of like this one where you have a cable. Now look here, this is not sitting on the floor. It has a cable. But think about this, what do you think would happen if you cut the cable? Right? It must be that this object is not lighter than water. Otherwise, it would already have bubbled up to the top and the string would have been sort of loose. Right? If it stays there with the string taut with a tight rope, it's because it's trying to fall. It's trying to sink. So if you were to cut that cable, what would happen is that it goes down and that's because the density of the object is greater than the density of the liquid. So the forces here are you have a bigger mg than you have an F_b, just like in the picture next to it. But now you have the help of a tension pulling you up. And the way you would write this is very similar. Force is up top, F_b + T equals mg. It's the same thing but now instead of the tension, instead of the normal force pulling you up, you have the tension pulling you up. Okay? And by the way, both the normal and the tension in this situation can be referred to as can be referred to as the apparent weight. Okay? So I'm going to put here also known as, in these problems, apparent weight. So if you see a problem with attention or something sitting maybe on top of a scale or something, and I ask you for a parent weight, that is asking for normal or asking for tension depending on which one you have. Cool? So that's plenty of talking. I wanna give you a shortcut and then we're gonna go solve this. There's a shortcut that says that the density of an object is the percent under times, times the density of the liquid. Okay? And I'm going to show you how to get to this equation, but I really just wanna start this example here. Cool. It says a block of unknown material, floats with 80% of its volume underwater. So let's draw that real quick. Bucket of water. 80% is under. Remember, the volume under is what you want. This is 20 but this is the useless one. Right? You want the volume under. Okay. What is the density of the object? And I'm actually going to calculate this really fast using this equation. The density of the object is the percent under times the density of the liquid. And this object is 80%. Remember 80% means point 8, times the density of the liquid. We're underwater so this is a 1,000 which means I can quickly figure out that the density of the object is 800 kilograms per cubic meter. It should make sense that the density of the object is less than the density of water. That's why it's floating up top. Okay. Now that's how you can very very quickly calculate this using the shortcu...

A block floats with 40% of its volume above water. When you place it on an unknown liquid, it floats with 30% of its volume above. What is the density of the unknown liquid?

• A. 857 kg/m³
• B. 600 kg/m³
• C. 400 kg/m³
• D. 571 kg/m³

Is Crown Made of Gold? (Buoyancy)

Hey guys. So let's check out this buoyancy problem. Here, we want to verify if a 100 gram crown is in fact made of pure gold. Almost all of these questions, the way you're going to validate that is by checking the density of the objects. Okay? So the density of gold is 19.32 grams per cubic centimeter or, 19,320 kilograms per cubic meter. Okay. Kilograms per cubic meter. And what we're going to do is we're going to calculate the density of this object and figure out if the density of the objects is 19,320. And if it is, this is gold. So if yes, if this is true, then it is gold. Okay? So if I ask you, is this gold? What you're doing is you're calculating density. So let's do that. So here, you lower it by a string into a deep bucket of water and then when the crown is completely submerged, so let's draw this. I will attempt to draw a crown. It's probably going to come out terrible. There you go. I told you. So it's completely submerged. It's got a little string here. You measure the tension to be 0.88. So there's a tension here, 0.88 newtons. There is a buoyant force always up and there's an mg always down. And we want to calculate ma. The next step is to just write that, that equals 0 by the way. The next step is just to write the forces. So all the forces going up equal all the forces going down. So fb + t = mg. Alright. And if you look at this real quick, you're going to notice that the density of the object is nowhere here, but you have to have a little faith as you start to change some variables around, as you start to expand some of these variables, the density of the object should show up And you don't stop until it does. So fb is the density of the liquid, so that's not good enough yet, times gravity, times the volume under, plus tension. We have that. We're going to plug in a little bit. Equals mass times gravity. Now we need the density of the object. So I'm going to rewrite the mass of the objects into the density of the object. Remember that density is mass divided by volume. Therefore, mass is density times volume. So it's the mass of the object. So it's the density of the objects. And it's always going to be the total volume. Okay? When you're rewriting mass, it's the total volume because you're looking at the total mass. And that times g. Okay? So we are looking for this. And if you take inventory here, we have the density of the liquid because it's water. Let's write that. We have gravity, 9.8. I'm going to round it to 10. Actually, let's make it 9.8 because we're trying to be very precise in our calculation. The volume under this object is entirely underwater. So volume under is the total volume, is the total volume, but I don't have that either. I don't have that either. So that's going to be a problem. Let's just leave it like that for now. Volume total + tension. Tension is 0.88. Let's go to the right side. Density of the object, that's what we're looking for. Density of the object. Cool. Leave it alone. Volume total, we don't have that. And gravity, 9.8. So we got a little bit of a problem which is this is my target but I actually don't have this either. So again, you're going to have to rewrite some stuff. Okay? So back to this equation here. If you solve for volume total, if you solve for volume total, you're going to get mass total divided by the density divided by the density of the object. So if you write this, the good news the good news is that you know the mass, it's a 100 grams, And though you don't know the density, at least that is your target. So this is a little tricky, but I'm going to rewrite it and you're going to see what's going to happen. You're going to have 980 (assuming 9.8) times the gravity. Instead of v total, I'm going to have mass, which is 0.100 kilograms, divided by the density of the object, pl...

An 8,000 cm³ block of wood is fully immersed in a deep water tank, then tied to the bottom. When the block is released and reaches equilibrium, you measure the tension on the string to be 12 N. What is the density of the wood?

• A. 1153 kg/m³
• B. 988 kg/m³
• C. 847 kg/m³
• D. 960 kg/m³

Maximum Load on Floating Board

Hey guys. So let's check out this buoyancy example. So here I have a wooden board. Let's make it thick. And let's call this, m₁. And I want to know what is the maximum amount of mass that I can put on top of the board. So I want to know how much \(m_{2}\) I can put here in such a way that \(m_{2}\) does not get wet. Okay. And this is a very common kind of question. What is the maximum load you can put on top of a board or a container so it doesn't get wet? Okay. And the way to solve this is very similar to how we've solved other, other buoyancy questions which is just using \( \mathbf{F} = \mathbf{m}\mathbf{a} = 0 \). It's an equilibrium question. What's special about here is this idea of not getting wet. So let's think about this real quick. Let's say you put some amount of mass here and the water level is here. Now if you put more mass, what's going to happen is it gets heavier so you would imagine that it goes down a little bit. So if you add more mass, the water level might rise up to here, right? I mean the water doesn't rise up. The object lowers relative to the water. And if you keep adding mass, that's going to happen until you get to this maximum point over here where if you add any more mass, the mass on top is going to get wet. So this idea of not getting wet has to do with you going all the way up here. So what's the conclusion? What happens when the water is all the way to the top? Well, what happens is the volume of the object of m₁ that is underwater is the entire volume of m₁. In other words, it is 100% submerged. So that's what this is going to be. That whatever thing, whatever container or board is is holding things on top of it is going to be 100% underwater, Okay? But not any but it's not going to be, any lower than that. Okay? So we're going to start with it, and that's it. Other than that, we're going to write that sum of all forces equals \( \mathbf{m}\mathbf{a} = 0 \) because this is an equilibrium problem. Now one distinction here is that there are 2 objects and every time you write \( \mathbf{F} = \mathbf{m}\mathbf{a} \), you write \( \mathbf{F} = \mathbf{m}\mathbf{a} \) for a target object. In other words, when you write the sum of all forces, you say the sum of all forces on object 1, which is the board equals \( \mathbf{m}\mathbf{a} \) which equals 0. So the only forces that matter are the forces acting on object 1. And the forces acting on \(m_{1}\) are there is its own \(m_{1}g\), right, which is \(m_{1}g\). There's also the weight of the \(m_{2}\) which is on top of it which pushes down on it. So this is \(m_{2}g\), right, because it's supporting that weight and there is a buoyant force, \(F_{B}\). Now you might be thinking, isn't there a normal force here? There is. If you look at \(m_{2}\), \(m_{2}\) is being pulled down by \(m_{2}g\), its own weight, and it's being pushed up by \(n\), which is the reaction force of \(m_{1}\). But we're not looking at \(m_{2}\). We're writing the sum of all forces for the one object, which means that the only forces that matter, are the forces on board 1 or object 1. Okay? So if this is an equilibrium, which it is because it's not moving, and it's not accelerating, it means that the forces cancel. So force is going up, \(F_{B}\) equals force is going down. \(M_{1}g+M_{2}g\). And what we're looking for here is \(m_{2}\). So let's go after that. As usual, we're going to rewrite this as density of liquid, gravity, and then volume under. And just to be clear, volume under is for the object that is actually underwater so it's volume 1. Cool? And then we have this stuff here. Now we don't know what \(m_{1}\) is. We don't know the mass of the board. And whenever you don't know the mass in these problems, we're just going to rewrite it. So density 1 is mass 1, volume 1, total. Volume 1, total. So \(m_{1}\) is density 1, volume 1, total. So let's rewrite that. So density 1, volume 1, total, times gravity. By the way, gravity is going to cancel but for now let's just leave it here. Plus \(m_{2}\). \(m_{2}\) is what we are looking for so leave it alone. \(g\). Notice that gravity cancels. You can only cancel gravity or any variable if it shows up in every single one of the terms. Here you have one term, another term, plus another term, and gravity shows up in all three of them so we can cancel. We're looking for \(m_{2}\). Do I know the density of the liquid? Yes. Now be careful here. It says salt, saltwater lake. What does that mean? Well, freshwater has a density of 1000. And you may remember saltwater has a higher density. One way to remember is that you add some salt so it gets heavier. The density of saltwater is going to be 1030 or 1035. You should stick with whatever number your professor likes for that. Some people use it a little bit differently. Cool? So that's this here, 1030. The volume under, what is the volume under? Well, volume under is actually the same thing here as volume total. The whole point of not getting wet is that the amount of the object that's underwater is the entire object. But we don't have that either, so we have to calculate. And I want to remind you that volume is, if you have a three-dimensional object, or sort of a rectangle, it is just base or width times depth times height. We don't have all these measurements, but you can also write this as you can replace these two guys with the area times the height. And that's what we have. Okay? The area here, it says 1 square meter and 10 centimeters, so 0.1 meters. And if you see I have meters squared with meters. There are 3 meters here. So this is 0.1 cubic meters is our volume. So this is going to be 0.1 equals density 1, which is the density of the board. We have that 700 volume 1, we just calculated 0.1, so do this carefully, plus mass 2. Okay. Now we can solve this. This is going to be 103 minus, we're going to move this over to the other side, \(700 \times 0.1\) is the same thing as \(700 \div 10\). So this is 70 and so \(m_{2}\), there's a difference there. \(m_{2}\) is 33 kilograms and we are done. Hopefully, you start seeing the pattern here, of all of these questions being solved just by writing that all the forces going up equals all the forces going down, and then getting kinda clever with making your target variable show up in the equation and moving some stuff around so you get everything that you need. So just good physics hustle. Let's keep going.

You want to build a large storage container, with outer walls and an open top, as shown, so that you can load things into it, while it floats on fresh water, without any water getting inside. If the bottom face of the container measures 3.0 m by 8.0 m, how high should the side walls be, such that the combined mass of container and inside load is 100,000 kg?

• A. 5.0 m
• B. 2.9 m
• C. 9.1 m
• D. 4.2 m

Here’s what students ask on this topic:

What is buoyancy and how does it work?

Buoyancy is the upward force exerted by a fluid that opposes the weight of an object immersed in it. This force is explained by Archimedes' principle, which states that the buoyant force (F_b) on an object is equal to the weight of the fluid displaced by the object. The principle can be mathematically expressed as:

$F_b=\rho gV$

where $\rho$ is the density of the fluid, $g$ is the acceleration due to gravity, and $V$ is the volume of the fluid displaced. An object will float if its density is less than the fluid's density and sink if it is greater.

How do you calculate the buoyant force on an object?

The buoyant force (F_b) on an object can be calculated using the equation derived from Archimedes' principle:

$F_b=\rho gV$

where $\rho$ is the density of the liquid, $g$ is the acceleration due to gravity (approximately 9.8 m/s² on Earth), and $V$ is the volume of the object submerged in the liquid. This equation helps determine the upward force exerted by the liquid on the object.

What is Archimedes' principle and why is it important?

Archimedes' principle states that the buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by the object. This principle is crucial because it explains why objects float or sink in fluids. It can be mathematically expressed as:

$F_b=\rho gV$

where $\rho$ is the fluid's density, $g$ is the acceleration due to gravity, and $V$ is the volume of the displaced fluid. This principle is fundamental in fluid mechanics and has applications in engineering, shipbuilding, and various scientific fields.

How does the density of an object affect its buoyancy?

The density of an object determines whether it will float or sink in a fluid. If the object's density is less than the fluid's density, the object will float because the buoyant force is greater than the object's weight. Conversely, if the object's density is greater than the fluid's density, the object will sink because its weight exceeds the buoyant force. This relationship can be understood through Archimedes' principle, which states that the buoyant force is equal to the weight of the displaced fluid. Therefore, the relative densities of the object and the fluid are key factors in determining buoyancy.

What are the common cases of buoyancy and how do they differ?

There are three common cases of buoyancy:

1. Floating: The object is less dense than the fluid, so it floats with part of its volume above the fluid. The buoyant force equals the object's weight.

2. Neutral Buoyancy: The object's density is equal to the fluid's density, so it remains fully submerged but does not sink or rise. The buoyant force equals the object's weight.

3. Sinking: The object is denser than the fluid, so it sinks to the bottom. The buoyant force is less than the object's weight, and an additional normal force from the bottom surface balances the forces.

These cases illustrate how the density of the object relative to the fluid determines its buoyant behavior.