Practice: When an object of unknown mass and volume is fully immersed in large oil (800 kg/m^{3} ) container and released from rest, it stays at rest. Calculate the density of this object.

Subjects

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Density | 34 mins | 0 completed | Learn |

Intro to Pressure | 76 mins | 0 completed | Learn |

Pascal's Law & Hydraulic Lift | 28 mins | 0 completed | Learn |

Pressure Gauge: Barometer | 14 mins | 0 completed | Learn |

Pressure Gauge: Manometer | 15 mins | 0 completed | Learn |

Pressure Gauge: U-shaped Tube | 23 mins | 0 completed | Learn |

Buoyancy & Buoyant Force | 64 mins | 0 completed | Learn |

Ideal vs Real Fluids | 5 mins | 0 completed | Learn |

Fluid Flow & Continuity Equation | 22 mins | 0 completed | Learn |

Additional Practice |
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Bernoulli's Equation |

Viscous Flow & Poiseuille's Law |

Concept #1: Intro to Buoyancy & Buoyant Force

**Transcript**

Hey guys. In this video we're going to start talking about buoyancy which is the thing 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 that is immersed in a liquid the object will be pushed up by a force called buoyant force and buoyant force comes from the fact that the physical phenomenon responsible for this is called buoyancy, right? Which usually is associated with floating and because, this is buoyant force with the B, we're going to call this FB. Now, the reason this happens is because of a pressure difference between the top and the bottom of an object that's under a liquid. So, let's say you have a box that is completely submerged, completely underwater and we know the water is going to push on all sides, right? The water is going to apply pressure on all sides and pressure is associated with with force. So, if there's a pressure there's a force. Now, the force on the left will cancel the force on the right because they are at the same height for every force on the left I can draw an equivalent force on the other side on the right that will exactly cancel it. So, the side forces don't do anything but the top and bottom forces will be different because the top and bottom pressures are different and we know that the deeper you go under liquid the greater the pressure. So, the pressure at the bottom over here at this point is greater than the pressure at the top therefore the force with which the liquid pushes up in this box is greater at the bottom than it is at the top and the difference between these two forces is the buoyant force, okay? So, you're always going to have a stronger force up than a force down. So, the net buoyant force is going to be up, cool? So that's the number one, thing about you need to know and you probably should memorize this is Archimedes principle, he said he figured out that the magnitude of this buoyant force is the same as the weight of the liquid displaced, okay? So, you should memorize that quote because that could show up in some way and it's important to know that, it's a big principle in physics. Now, the problem is that this quote here doesn't really help you calculate this force. So I'm just going to give you the equation, if you were to sort of translate this quote into a physics equation you would get that the buoyant force is the density of the liquid, density of the liquid, gravity 9.8 or whatever planet you're on and the volume of the object that is under water. So, I'm going to call it rho liquid, g and v under. So, there's two really important things here, distinctions to make, in this equation the density that matters, the density that you should be looking into is the density of the liquid even though the object is inside and this is the force on the object the density that matters the density of the liquid, don't get that messed up, okay? And the volume that matters is the volume of the object but only the amounts of volume that is under water. So, it's the volume under water or under the liquid, okay? Or the volume submerged of the object that is submmersed or immersed in the liquid, okay? I'm just going to call it under to keep these words simple and it's not necessarily the entire volume, unless of course you are entirely under. So, let's think about this real quick, the idea here is that the more of the of an object is under water the stronger the force and you might play this out in your head and just sort of imagine that if you're getting a piece of styrofoam as you push it down under water there's more resistance there's more force pulling it up and that's because the more volume you have the more buoyant force you have, okay? So, that's that and then one last thing I want to remind you because you're going to be using this a lot is the density of an object the definition of density is mass over volume. So, the density of the object will be the mass of the object and the total volume of the object so I'm going to write v, I'm going to write actually mass total, it's the entire mass, it won't go away, it's the entire mass of the object divided by the entire volume of the object and the reason I'm being careful that distiction is because you have volume under which isn't necessarily the entire thing and you have volume total which is the entire thing and we're going to be using these two equations all the time to solve these questions, okay? The last point I want to make and we're gonna do an example here, is that almost every single buoyancy force, a problem, is just a force problem. So, it's just a good old F equals m, a problem, we could have done this right after you learned f equals m, a in some ways and also that most of the time it's going to be even simpler than F equals m, a because most of the time the object will be at rest, meaning the velocity 0 and they will be in equilibrium which means the acceleration is 0 and the next force is 0. So, it's really an f sum of all forces equals 0 problem most of the time, it's just equilibrium, forces are just going to cancel, cool? So, let's do a quick example here, So, it says, when an object of unknown mass , so we don't know the mass and volume, unknown volume, we don't know the volume, great, it's fully immersed, fully immersed, let me draw that fully immersed means entirely underwater. Remember I told you there's two kinds of volumes, right? There's volume under and there's volume total because you are fully immersed this means that volume under equals volume total, okay? In a large water tank and released from rest it accelerates up. So, when you release the object that's going to accelerate up and then it says, once the object reaches equilibrium 30% of its volume is above water. So, here's the object it's going to ,eventually it stops up there so when you release it accelerates up eventually it stops and 30% of the object is above water, okay? Now, if you remember on this equation here, the only two volumes that matter are the total volume and the volume under. So, 30% above doesn't do anything for us, what you really care about is the volume under. So, you're going to change that into seventy percent under, that's one of the things that you have to do these questions is you always want to know the volume under, that's what matters, okay? Now, with just that information we're going to calculate the density of this object. So, what is Rho object, cool? Now, again all these questions are going to start with F equals ma equals 0 because you are at equilibrium. So, I'm going to write that the sum of all forces equals 0 and remember the way to solve force problems has always been to draw Free-body diagram and then write F equals ma. So, let's draw a Free-body diagram, what's going on here with forces, well, this object is here. So, there has to be an mg pulling down and the new thing here is that because it's under a liquid it's going to be pushed up by a buoyant force and fb is always going to be up, okay? Let me write that here, just so you have it always up, because as, because the density down here, is stronger as you go lower, cool? So, if the this system is at equilibrium the object is at equilibrium it means that the forces cancel. So, you can simply write that f,b equals m,g. The next step is always going to be or almost always going to be to expand this f,b, what I mean by that, we're going to rewrite f,b based on its equation which is this. So, instead of writing f,b memory write as Rho g v. So, let's do that density, remember of the liquid gravity and volume under equals to m,g, right away g cancels and this is going to happen quite a bit, not always but quite a bit and we want to know the density of the object. Now, if you stop for a second and look here, you will see that density Rho object isn't anywhere in this equation, right? It's not here, you have density of the liquid but that's not what we want. So, it actually is here, but you have to look a little bit more carefully, it is going to be inside, hiding inside of one of these things here, right? Are there any of these things that you can rewrite, that you can rewrite to make this variable show up and there is and I warned you about this that we were going to be using this equation all the time. So, very often you're gonna rewrite things to make variables show up, okay? So, the way we're going to do this is, density of the object is mass total divided by volume total therefore mass total is density of the object times volume total. Alright, so we're going to do is we're going to replace mass with density volume and now we're going to have our variable there, okay? So, the density of the liquid times volume under equals density of the object which is what we're looking for, that's good, times volume total, cool? Now, realize also that we know, we know the volume under, okay? We know that we are 70% under water. So I can rewrite volume under as 70 of or 0.7 times volume total, okay? Screen writing this here. So, if you do that you can cancel these two out and then we are basically done, the density of the object is 0.7 the density of the liquid, we are in water, we're in water so this is just 1,000. So, the density of the object is 700 kilograms divided by or per cubic meter, okay? It makes sense, it should make sense that the object is floating above water because it is less dense than water therefore it rises to the top, I want to talk about one last thing here, notice here that I had said that the the volume under the water is the same as the total volume but that was before, that was before you released and the object moves this way here. Now, the volume under is less than the volume total because some of it is above, cool? That's it for this one, let's keep going.

Example #1: Comparing Buoyant Forces

**Transcript**

Hey guys. So, let's check out this conceptual buoyancy question and I chose this one because it's pretty tricky and there's quite a few of these kinds of tricky buoyancy questions that want to help you think through. So, let's it check 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 wood equals volume metal, they're both placed on 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 you want to know what is the greater buoyant force on it. So, which one has the greater FB, 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 a buoyant force, the equation for buoyant force is density of the liquid, gravity and the volume of the object that is under the liquid, okay? 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 tye, which means that this is not going to help you determine who, which one has the more buoyant, more buoyant force, a greater buoyant force, they also both experienced 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 is, has some of its volume sticking out, so the volume under, I'm going to make up some numbers, is going to be like 10 and 40, for example, and this is going to be all 50 of it, are going to be underwater. So, the metal has more, the metal has more of its total volume underwater than the wood does, therefore the stronger blowing force will be metal and hopefully you got this right, what a lot of people actually incorrectly pick wood because they think it's floating, so it must be, if it's floating, 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 the m, g is much, much smaller, okay? So, here the force is actually, let's draw some arrows here, the FB is actually stronger than here, but the m, g is much, much stronger than here, okay? So, don't get caught up, write the equation and use that to determine the right answer. So, let's keep going.

Practice: When an object of unknown mass and volume is fully immersed in large oil (800 kg/m^{3} ) container and released from rest, it stays at rest. Calculate the density of this object.

Concept #2: Buoyancy / Three Common Cases

**Transcript**

Hey guys. So, in this video we're going to keep talking about buoyancy and I'm going to show the three 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 go, 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, first situation is referring to this case right here, this here is sort of a secondary case that I want to talk about sort of exception, in here we have this object that floats above water. So, part of the object is above the water and this is the part of the object that is under water 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 the volume total is 100 and there's maybe 40 and 60 here, the volume under 60 is less than the volume total 100 because some of its above water pretty straightforward, what about the forces? Well, if the object is floating just sits there it is at equilibrium, right? Because it's just sitting there floating which means the force is canceled and the forces are FB going down and, I'm sorry, FB going up FB is always going up in m, g going down. So, it must be the FB equals m, g so that they can't cancel each other out and that's what happens there, what about the density? Which one of these two then sities 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 quick way to look at this that I like is to just look at the top of the object 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 slightly different situation sort of an exception that I want to talk about. So, here you have this object that, that's in the metal here, it's floating with a cable, right? So, think about this, what do you think would happen if I cut, 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 cord is holding it. So, what's happening here is that you have a tension pulling it down and then you'll have m, g 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 force is going down m, g plus T equal the force is going up FB, the reason why I have this next the other one is because in this situation you also have this be true, that's the object is less dense even though it's under water because it's only under water 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 under water you have to think, what would happen or there's some tension reason 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 on but instead of above water it floats under water, right? So, how does this happen? Well, this happens if you have an object and you put it under water and you release it and it stays there. So, in this case the entire volume under water is the entire volume of the object it is 100% under water in this case we also have equilibrium, okay? Because the two forces are going to cancel FB + m, g cancel. So, they're still at equilibrium. Now, what's special here, the difference between these two situations one & two 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 is it's going to rise to the topic 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 grabbed this object that I brought it just under the water line and I release it there and because the density are the same it stayed 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 peek out outside of the water 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 three, 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, the volume underwater is the same as a total volume, FB does not equal m, g, FB does not equal m, g, the reason why it sinks is because m, g is actually going to be greater than FB, okay? This is still at equilibrium, it just sits there but now there's a third force. So, you're going to have m, g down, you're going to have, let me draw this bigger you're going to have m, g down, you're going to have an FB that's smaller and because of this this object is going to be, let me move this up, this objects pushing against the surface so the surface pushes back with the familiar force called normal and it's not going to be that big, they're both, essentially they're gonna add up to cancel the m, g and you can write that the force is going up FB plus normal equals the force going down m, g, okay? So, there are three 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 pushes it 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 a side case kind of like this one where we 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 m, g then you have an FB just like in the picture next to it but now you have the help of a tension pulling you up and the way you'd write this is very similar, forces up top FB plus T equals m, g, 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 the tension or something sitting maybe on top of a scale or something and I ask you for apparent weight that is asking for normal or asking for tension depending which one you have, cool? So, that's plenty of talking I want to give you a shortcut and then we're going to go solve this, there's a shortcut that says that the density of an object is the percent under times the density of the liquid, okay? And, I'm going to show you how to get to this equation but I really just want to start this example here, cool? it says, It's a block of unknown material, floats with 80% of its volume under water. 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 18%. Remember, 80% means 0.8 times the density of the liquid, we're underwater so this is 1,000 which means I can quickly figure out 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 shortcut, where does this shortcut come from? Let me show you really quickly, in case your professor doesn't like you using shortcuts and he wants to see the full solution. Remember, that all these questions start with F equals m, a, sum of all forces equals m, a and because we are at equilibrium that equals 0, the next step is to write the forces and then you have that FB cancels out with m, g the next step is to expand FB which is going to be density of the liquid always gravity and then volume under water and that's going to equal to m, g, I can cancel the g's, okay? Now, what I'm going to do is I'm going to rewrite M and remember, density is mass over volume. So, mass is density times volume, this is the density of the object and the total volume of the object. So, you're going to have density liquid, volume under equals density of the object, volume total and we're solving for density of the object. So, I'm going to move the volume total to the other side density object equals volume under divided by volume total times density of the liquid. Now, check this out, let's say volume under is 800 and volume total is 1000, if you divide the two you end up with 0.8 and that's the percentage under, that means that this is 80%. So, that's why this ratio here of partial over total is the percentage. So, we can say that density of the object is the percent under, which is a decimal, times density of the liquid, that's how we arrived at that, okay? Last thing I want to show you with this shortcut before we go into example two is, what happens if the object floats thus by the way only works if the object floats, okay? Now, what happens with the object floats with 100%, 100% underwater which is situation number two over here. So, if you have 100% underwater 100% is 1 this would be a 1 which means you have density of the object equals density of the liquid and we already knew that, if you float entirely underwater it's because de densities are the same. So, this equation actually works for cases 1 & 2 but it doesn't work for case 3.

Alright, let's check out example 2 real quick and then we're going to move on. So, here, we have an aluminum block. So, the density of this object is going to be 2700 and it has these dimensions here, if I know all three dimensions by the way I can calculate the volume, the total volume of this object is going to be 1 times 1 times 1 which is 1 cubic meter and it sits on a scale at the bottom of a 2 by 2 by 2 tank. So, you have a tank like this that is 2 by 2 by 2, let's make it 3d here, 2 by 2 by 2, okay? And you have this box sitting here, the box a little bit smaller, cool? 1 by 1 by 1 and it is filled with water to the very top over here, how much does the scale read? Now, this this diagram is a little bit too complicated, you can just look at it as a block that is one on every side and it sits on top of the scale, the 2 by 2 by 2 just tells you that there's enough water here for this thing to be completely submerged, okay? How much does the scale read, I hope you remember that the reading of the scale is the same thing as apparent weight and it's the same thing as the normal force in this case. So, asking how much the scale read is the same thing as asking, what is the normal force? and because normal is a force we're going to start this question like to start every first question with F equals m, a. So, the sum of all forces on the object equals to m, a and because it sits there at equilibrium that is 0, what are the forces on the object? So, the object gets pulled down with m, g, it gets pushed up with FB, it's always like that, but because the object is pushing against the scale, the scale pushes back against the object. So, we have a normal force going up. So, the sum of all forces is that you have the force going up against the force going down. So, I can rewrite this to be normal plus FB equals m, g because the forces are just cancelling up equals down, remember that stuff, and we're looking for the reading of the scale. So, we're looking for normal and to calculate normal I just have to now move FB to the other side. So, it's going to go to a a negative you're going to have that normal is m, g minus FB and instead of writing FB I'm actually going to already go ahead and expand FB, FB is density of the liquid. Remember, it's always liquid, gravity and the volume under, volume under, okay? Now, before, we start plugging in numbers here, let's take stock, let's take inventory of what we have and don't have. So, let's say, let's see, I have gravity, that's easy, I have density of the liquid, that's easy, we're in water. So, the density of the liquid is just 1,000 and the volume under. Now, this block is entirely submerged so the volume under is the same as the volume total which we calculate it to be 1 cubic metre. So, we have everything except mass and that happens quite a bit, we're going to use the fact that the density of the object is mass total divided by volume total, okay? And we're going to use that to our advantage to be able to solve for mass. So, mass equals mass of the object equals density of the object times volume total and I can plug that in here or I can just go ahead and calculate this. So, the density of the object, we have it here, it's aluminum so, it's 2700, the volume is 1 so the mass is 2700. So, over here, I'm going to put 2700, gravity, I'm going to write it as a 10 just to make the multiplication little easier here, density of the liquid is 1,000, gravity 10, volume under 1, cool? So that's it, this is, actually already got the result here, this is 2700, sorry about that, I hope you caught that, 2700, it's going to be 27,000, when you multiply by 10. So this is gonna be 27 thousand minus 10 thousand which is 17,000, 17,000, we're looking for normal which is a force. So, this has units of Newton's, units of Newton's 17,000 and that's it, we're done, let's keep get going.

Practice: 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?

Example #2: Is Crown Made of Gold? (Buoyancy)

**Transcript**

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 these questions the way you're going to validate that is by checking the density of the object, 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 gold, of this object and figure out if the density of the object is the density of the object 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 is a tension here, 0.88 Newtons there is a buoyant force always up and there's an m, g always down and we want to calculate the density of the object, all of these questions are going to start with F equals m, so the sum of all forces equals m, a 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 plus T equals m, g. 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 density of the liquid. So, that's not good enough yet, times gravity times volume under plus tension, we have that, we're going to plug in a little bit, equals mass times gravity, we need density of the object so I'm going to rewrite mass of the object into 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 object 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 a 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 plus 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've 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 density divided by the density of the object. So, if you write this, the good news, the good news is that you know mass, it's 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 1,000 times 9.8, instead of V total I'm going to have mass which is 100 grams so 0.1 hundred kilograms divided by the density of the object plus 0.88 equals the density of the object times volume which we're rewriting is mass. 100 divided by the density of the object times 9.8. Now, at this point you might be freaking out just a bit but notice that this cancels with this and then you end up having just one unknown out of this entire thing. So, it's a little bit messy because we expanded, right? We rewrote mass so that the density of the object shows up and it turns out that it actually cancelled here. So, you could have known ahead of time, not to do that because you have to undo it anyway but the chances are that you wouldn't know that, right? You wouldn't know that that was coming. So, I try to solve problems in a way that most people would do which is sort of systematic and not already knowing things in advance. So, you have to be able to be good at manipulating these things which is why I wanted to show you this question as an example, you have to be back and forth so you have to be very fluent if you will with this little equation so you can move some stuff around and just keep going and keep changing some stuff until you're left with one target so this is just good solid physics hustle to get to the target variable. Now, we're going to move a bunch of stuff around.

So, if you multiply all of this, let's see, this is going to be, this is going to be 9, this is only 980 over here divided by Rho of the object, density of the object plus 0.88 and on this side, this is going to be 0.98. So, when you move this over here you're going to get 98, 0.98 minus 0.88 which is 0.1. So, 980, density of the object equals 0.1. Now, I'm going to move the density of the object up here. So, 980 equals 0.1 density of the object, finally I can move the. One over here. So, 980 divided by 0.1 is the density of the object therefore the density of the object must be 9800 kilograms per cubic meter and this is a problem because this is nowhere near gold, gold is 19,320 kilograms per cubic meter, this is like less than half or a little bit more than half of this guy's so it's way off from gold, by the way, if you get something that was very close, if you got something that was like 19,200, if it's that close then whoever wrote the question meant for it to have been gold. So, even, if it's not exactly the same, if it's really close, that's gold, if they didn't mean it for it to be gold then they'll make a number that is very, very different, that's the case here, this is clearly not gold, okay? So, the answer here would be not, let's get out of the way, the answer here is not gold, okay? Now, I want to quickly talk about something else, there's another way you could have solved this question, I don't like it but it works which is once you get to the big equation right here, you could've, I mean, this is kind of a hack but what you could have done is you could have plugged in the density right here, you could have plugged in the density of gold 19320 and what would have happened is that the left side of the equation and the right side of the equation would not equal to each other, right? You would end up with something, I'm just going to make it up, you'd end up with something like 20 equals 40 and then you would say, 20 is not equal 40 because this didn't turn out to be true it must be that the density of this thing is actually not the density of gold which means that this thing is not gold, okay? Long story short, this is not gold, we are done here, let's keep going.

Practice: An 8,000 cm^{3} 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?

Example #3: Maximum Load on Floating Board

**Transcript**

Hey guys. So, in this video we're going to start talking about fluids in motion or fluid flow and the first thing we're going to cover is the distinction between real fluids and ideal fluids, let's check it out. Alright, so the motion of fluids or the flow of fluids can get pretty complicated it's actually an area, that's still under active research in physics but we're going to simplify things by using a model called ideal fluid. So, the idea that fluids are pretty complicated but if we eliminate a ton of the complexities, the more complicated complexities of real fluids you end up with something that's called ideal fluids, they don't exist but they are a way to make things more manageable, cool? And there are some things you should know first you should know that these fluids will always be, ideal fluids will always be incompressible, that's a simplification, incompressible means constant density. Remember, density has to do with how tightly packed molecules are and in ideal fluids that's going to be constant, in real fluids molecules could get tighter or less dense, less tightly packed depending what's going on, okay? So, incompressible constant density, that simplifies things a bunch and the second thing you should know is that ideal fluids are always going to have what's called laminar flow, and laminar flow just means steady flow. So, if you look at water going to a pipe, if it's let's say see-through pipe or something like that you would see that there's just a constant stream of water that looks very clean and neat as opposed to real fluids that could have what's called turbulent flow, turbulent flow or turbulence, you could have turbulence if the liquid is moving or the fluid more generally is moving too fast, okay? So, imagine, if water is going way too fast that it goes through a little a little clot type thing here, some sort of constriction, then it could be that the water starts going all over the place and this is turbulence and this is generally bad news, lucky for you, you're probably not going to see any turbulence questions, you may just have to know this conceptually, cool? And the third distinction between them and I'm going to start over here is whether this motion is going to have viscosity, whether the fluid is going to have viscosity or not, okay? So, the defining characteristic, this is the most important of the three here, the defining characteristic of real fluids is that they have what's called viscous flow, viscous flow, in other words, the liquid has viscosity and viscosity has to do with the thickness, thickness of the fluid. So, for example, honey, right? If you get a cup full of honey and you turn it like this it's going to move very, very slowly it's because there's a lot of viscosity, ideal fluids have no viscosity at all they have no resistance, viscosity is essentially fluid friction, it's essentially fluid friction and it works very similar to air resistance or a kinetic friction and that it slows it down, okay? So, real fluids could have viscosity and they could have a viscous friction, ideal fluids are always going to have what's called non viscous, not very creative name, non viscous flow, in other words, no friction. So, that's the big difference, if you have an ideal fluid it's going to flow, it's going to flow smoothly with no friction, a real fluid could have turbulence and it has viscosity, okay? Now, lucky for you, most problems you see and maybe even all of the problems you see will be about ideal fluids, in fact a lot of professors don't even get into real fluids. So, if yours doesn't your life is simple, we're going to also assume ideal fluids unless something says explicitly that this is a real fluid or if they refer to viscosity which is fluid resistance, right? So, if they say that there's some sort of viscosity then that means there's resistance which mean it's a real, which now it means that you have a real fluid.

In most real fluid problems you're going to have viscosity because it's the defining characteristic but you're not going to have turbulence and you're also not going to have compression of the fluid, okay? So, then you're really not going to see this most of the time, you're not going to see this most of the time but you are going to see this quite a bit. So, you're going to have no turbulence and no compression of the fluid, okay? So, really it's going to come down to whether or not it has viscosity. So it is a quick intro so you can know some of the terminology and that's it, let's keep going.

Practice: 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?

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Concept #1: Intro to Buoyancy & Buoyant Force

Example #1: Comparing Buoyant Forces

Practice #1: Density of Object Floating Underwater

Concept #2: Buoyancy / Three Common Cases

Practice #2: Density of Liquid with Unknown Block

Example #2: Is Crown Made of Gold? (Buoyancy)

Practice #3: Density of Submerged Block on String

Example #3: Maximum Load on Floating Board

Practice #4: Height of Sides on Floating Container

A rectangular block of wood, 12 cm x 18 cm x 42 cm, has a specific gravity of 0.60. Determine the buoyant force that acts on the block when it is placed in a pool of fresh water.

Two wooden boxes of equal mass but different density are held beneath the surface of a large container of water. Box A has smaller average density than box B. When the boxes are released, they accelerate upward to the surface. Which box has the greater acceleration?
a. Box A
b. Box B
c. They are the same.
d. We need to know the actual densities of the boxes in order to answer the question.
e. It depends on the contents of the boxes.

A block of density ρ1 and volume V1 is submerged in a liquid of density ρL. A second block of density ρ2 and volume V2 is placed on top of the first block. The two blocks are floating in the manner shown in the figure below.
Find V2 such that the two blocks are just submerged, as shown in figure above.

Suppose that a volleyball A and a bowling ball B are completely submerged in water and have the same volume, as in the figure. (Of course, you would have to hold the volleyball beneath the water to keep it from popping up to the surface.)
Which feels a greater buoyant force?
A. bowling ball B
B. Unable to determine
C. They feel the same buoyant force.
D. volleyball A

Consider a rectangular block of ice floating in water in an open vessel. Let a denote the height of the ice block showing above the water surface, and b the height of the submerged part of the ice block.
Given a, b, the density ρwater of liquid water and the density ρice of the ice, which relationship is correct?

A slab of ice floats on a freshwater lake. What minimum volume must the slab have for a 45 kg woman to be able to stand on it without getting her feet wet?

A diver wishes to recover a treasure chest she found at the bottom of the sea, 60 m below the surface. To do this, she inflates a plastic balloon to a radius of 40 cm with the air from her compressed air tanks. The mass of the treasure chest is 200 kg and its dimensions are 20 cm x 40 cm x 10 cm. Take the density of sea water as 1025 kg/m3. How does the acceleration change (qualitatively) as the balloon and chest float to the top? Does it get larger, smaller, or stay the same? Justify your answer with equations.

A boat's hull displaces 1600 L of water when floating. What is the weight of the boat?

An object floats on the surface of water. If the object's specific gravity is 0.7, what percentage of the object's volume is above the water-line?

A boat is floating on the surface of the water when the captain throws the anchor overboard. When the anchor is thrown out of the boat, which of the following is true?
(a) The weight of the boat decreases, so the buoyant force increases
(b) The weight of the boat decreases, so the buoyant force decreases
(c) The weight of the boat increases, so the buoyant force increases
(d) The weight of the boat increases, so the buoyant force decreases

A barge is 10.0 m wide and 60.0 m long and has vertical sides. The bottom of the hull is 1.20 m below the water surface. What is the weight of the barge and its cargo, if it is floating in fresh water?
(A) 3.53 MN
(B) 6.39 MN
(C) 7.06 MN
(D) 6.82 MN
(E) 9.54 MN

A rectangular block of wood, 12 cm x 18 cm x 42 cm, has a specific gravity of 0.60. What fraction of the block is submerged?(Express your answer to two significant figures.)

A cube of side s is completely submerged in a pool of fresh water. After drawing a free-body diagram, derive an algebraic expression for the net force on the cube.a. ΣFy = Pfluidgs - Mcubegb. ΣFy = Pfluidgs3 - Mcubegc.. ΣFy = Pcubegs3 - Mfluidgd. ΣFy = Pfluidgs3

A cube of side s is completely submerged in a pool of fresh water. What is the weight of the displaced water when the cube is submerged?a. wwater = ρfluidsgb. wwater = ρcubes3gc. wwater = ρfluids3gd. wwater = mcubeg

A woman floats in a region of the Great Salt Lake where the water is about four times saltier than the ocean and has a density of about 1130 kg/m3. The woman has a mass of 56 kg, and her density is 971 kg/m3 after exhaling as much air as possible from her lungs. Determine the percentage of her volume that will be above the waterline of the Great Salt Lake. (Express your answer to three significant figures.)

A beaker of mass 1 kg containing 2.5 kg of water rests on a scale. A 3.1 kg block of a metallic alloy of density 4900 kg/m3 is suspended from a spring scale and is submerged in the water of density 1000 kg/m3 as shown in the figure.
What does the hanging scale read? The acceleration due to gravity is 9.8 m/s 2.
A. 21.3715
B. 16.043
C. 23.0577
D. 28.3774
E. 14.3733
F. 31.6615
G. 27.8826
H. 24.18
I. 23.8994
J. 27.6033

A spar buoy consists of a circular cylinder, which floats with its axis oriented vertically. One such buoy has a radius of 1.00 m, a height of 2.00 m and weighs 40.0 kN. What portion of it is submerged when it is floating in fresh water?
[A] 1.35 m
[B] 1.30 m
[C] 1.25 m
[D] 1.20 m
[E] 1.50 m

Suppose the vessel containing the water and the ice is full: The water level is at the vessel’s rim. What happens once the ice melts?
A. The water overflows.
B. The level of the water remains at the rim.
C. There is not enough information given, the outcome is not definite.
D. The level of the water drops below the rim.

A solid aluminum cylinder of length L, radius r and density pal is suspended vertically from a thin wire and slowly lowered into a still lake. Let z denote the depth of the bottom of the cylinder. (When z < 0 no portion of the cylinder is submerged in water. Note that z is assumed here to increase downwards.) Assume that both the mass of the wire and the density of air are negligibly small. Let pω denote the density of water. What is the tension T in the wire
a) for z < 0?
b) for 0 < z < L?
c) for z > L?
d) Plot T(z).

A table-tennis ball has a diameter of 3.80 cm and average density of 0.0840 g/cm 3. What force is required to hold it completely submerged under water?

A 1.00-kg beaker containing 2.00 kg of oil (density = 916 kg/m 3) rests on a scale. A 2.00-kg block of iron is suspended from a spring scale and is completely submerged in the oil (Fig. P9.43). Find the equilibrium readings of both scales.

A geologist finds that a Moon rock whose mass is 9.30 kg has an apparent mass of 6.30 kg when submerged in water. What is the density of the rock?

Archimedes principle can be used not only to determine the specific gravity of a solid using a known liquid; the reverse can be done as well. As an example, a 3.80-kg aluminum ball has an apparent mass of 2.30 kg when submerged in a particular liquid: calculate the density of the liquid. Derive a formula for determining the density of a liquid using this procedure.

A cube of side length 13.0 cm and made of unknown material floats at the surface between water and oil. The oil has a density of 810 kg/m3.a) If the cube floats so that it is 73 % in the water and 27 % in the oil, what is the mass of the cube?b) What is the buoyant force on the cube?

A 3.60-kg piece of wood (exttip{SG}{SG}SG = 0.51) floats on water. What minimum mass of lead, hung from the wood by a string, will cause it to sink?

Oil having a density of 930 kg/m3 floats on water. A rectangular block of wood 4.00 cm high and with a density of 960 kg/m3 floats partly in the oil and partly in the water. The oil completely covers the block. How far below the interface between the two liquids is the bottom of the block?

A 7.4 kg solid sphere made of metal whose density is 2300 kg/m3, is suspended by a cord. When the sphere is immersed in a liquid of unknown density, the tension in the cord is 18 N. The density of the liquid is closest to:
A. 1700 kg/m3
B. 1600 kg/m3
C. 1500 kg/m3
D. 1400 kg/m3
E. 1300 kg/m3

A diver wishes to recover a treasure chest she found at the bottom of the sea, 60 m below the surface. To do this, she inflates a plastic balloon to a radius of 40 cm with the air from her compressed air tanks. The mass of the treasure chest is 200 kg and its dimensions are 20 cm x 40 cm x 10 cm. Take the density of sea water as 1025 kg/m3. The diver attaches the inflated balloon to the treasure chest using a rope. Calculate the initial acceleration of the balloon and the chest. Ignore the mass of the plastic of the balloon and of the rope connecting the balloon and the chest.

A rock with density 2300 kg/m3 is suspended from the lower end of a light string. When the rock is in air, the tension in the string is 43.0 N. What is the tension in the string when the
rock is totally immersed in a liquid with density 750 kg/m3?

A hydrometer consists of a spherical bulb and a cylindrical stem with a cross-section area of 0.400 cm2. The total volume of bulb and stem is 13.2 cm3. When immersed in water, the hydrometer floats with 8.00 cm of the stem above the water surface. When immersed in an organic fluid, 3.20 cm of the stem is above the surface. (Note: This illustrates the precision of such a hydrometer. Relatively small density differences give rise to relatively large differences in hydrometer readings.) Find the density of the organic fluid.

A toy floats in a swimming pool. The buoyant force exerted on the toy depends on the volume ofa. water in the pool.b. the pool.c. the toy under water.d. the toy above water.e. none of the above choices.

An object floats in water with 5/8 of its volume submerged. The ratio of the density of the object to that of water isa. 8/5.b. 5/8.c. 1/2.d. 2/1.e. 3/8.

An ice cube floats in a glass of water. As the ice melts, what happens to the water level?a. It rises.b. It remains the samec. It falls by an amount that cannot be determined from the information given.d. It falls by an amount proportional to the volume of the ice cube.e. It falls by an amount proportional to the volume of the ice cube that was initially above the water line.

An iron anchor with mass 35.5 kg and density 7860 kg/m3 lies on the deck of a small barge that has vertical sides and floats in a freshwater river. The area of the bottom of the barge is 8.05 m2 . The anchor is thrown overboard but is suspended above the bottom of the river by a rope; the mass and volume of the rope are small enough to ignore. After the anchor is overboard and the barge has finally stopped bobbing up and down, has the barge risen or sunk down in the water? By what vertical distance?

A slab of ice floats on a freshwater lake. What minimum volume must the slab have for a 72.0 kg woman to be able to stand on it without getting her feet wet?

A hollow, plastic sphere is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.660 m3 and the tension in the cord is 2120 N.a) Calculate the buoyant force exerted by the water on the sphere.b) What is the mass of the sphere?c) The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged?

Suppose your body has a density of 992 kg/m3.(Part A) What fraction of you will be submerged when floating gently in freshwater (with density 1000 kg/m3)?(Part B) What fraction of you will be submerged when floating in salt water (with a density of 1027 kg/m3)?

A rectangular wooden block of weight W floats with exactly one-half of its volume below the waterline.A. What is the buoyant force acting on the block?B. The density of water is 1.00g/cm3 . What is the density of the block?

A flat slab of styrofoam, with a density of 32kg/m3, floats on a lake.What minimum volume must the slab have so that a 33kg boy can sit on the slab without it sinking?

Question 1: Why does an ocean liner float?A. It is made of steel, which floats.B. It is held up in the water by large Styrofoam compartments.C. The average density of the ocean liner is less than that of seawater.D. Its very big size changes the way water supports it.E. Remember the Titanic -ocean liners do not float.

A 10 cm × 10 cm × 10 cm wood block with a density of 700 kg/m3 floats in water. What is the distance from the top of the block to the water in cm...(a) if the water is fresh?(b) What if it is sea water?

A flask of water rests on a scale that reads 100N. Then, a small block of unknown material is held completely submerged in the water. The block does not touch any part of the flask, and the person holding the block will not tell you whether the block is being pulled up (keeping it from falling further) or pushed down (keeping it from bobbing back up).The experiment is repeated with the six different blocks listed below. In each case, the blocks are held completely submerged in the water.Mass (g)Volume (cm3)A10050B100200C20050D50100E200100F40050Rank these blocks on the basis of the scale reading when the blocks are completely submerged.Rank from largest to smallest. To rank items as equivalent, overlap them.

A flask of water rests on a scale that reads 100N. Then, a small block of unknown material is held completely submerged in the water. The block does not touch any part of the flask, and the person holding the block will not tell you whether the block is being pulled up (keeping it from falling further) or pushed down (keeping it from bobbing back up).The experiment is repeated with the six different blocks listed below. In each case, the blocks are held completely submerged in the water.Mass (g)Volume (cm3)A10050B100200C20050D50100E200100F40050If the blocks were released while submerged, which, if any, would sink to the bottom of the flask?

A fat slab of styrofoam, with a density of 32 kg/m3 floats on a lake. What minimum volume must the slab have so that a 40 kg boy can sit on the slab without sinking?

Three balls with the same radius 11 cm are in water. Ball 1 floats, with half of it exposed above the water level. Ball 2, with a density 815 kg/m3 is held below the surface by a cord anchored to the bottom of the container, so that it is fully submerged. Ball 3, of density 1342 kg/m3 is suspended from a rope so that it is fully submerged. Assume the density of water is 1000 kg/m3 in this problem.What is the tension on the rope holding the 2nd ball in Newtons?

The buoyant force on an object submerged in a liquid depends on:a. the object's mass. b. the object's volume. c. the density of the liquid. d. the object's mass and volume. e. the object's volume and density of the liquid. f. the object's mass and volume and the density of the liquid.

A fishbowl contains a single goldfish and is filled with water to the level indicated.At each of the designated points, rotate the given vector to indicate the direction of the force exerted by the water on either the inside of the fishbowl (for points A and B)or the outside of the goldfish (for points C, D,and E).

Hot air balloons float in the air because of the difference in density between cold and hot air. Consider a balloon in which the mass of the pilot basket together with the mass of the balloon fabric and other equipment is mb. The volume of the hot air inside the balloon is V1 and the volume of the basket, fabric, and other equipment is V2. The absolute temperature of the hot air at the bottom of the balloon is Th (where Th > Tc). The balloon is open at the bottom, so that the pressure inside and outside the balloon is the same. Assume that we can treat air as an ideal gas. Use g for the magnitude of the acceleration due to gravity. For the balloon to float, what is the minimum temperature Tmin of the hot air inside in terms of the variables given?

A 6.00cm -diameter sphere with a mass of 89.3g is neutrally buoyant in a liquid. Identify the liquid.

A flask of water rests on a scale that reads 100N. Then, a small block of unknown material is held completely submerged in the water. The block does not touch any part of the flask, and the person holding the block will not tell you whether the block is being pulled up (keeping it from falling further) or pushed down (keeping it from bobbing back up).The experiment is repeated with the six different blocks listed below. In each case, the blocks are held completely submerged in the water.Mass (g)Volume (cm3)A10050B100200C20050D50100E200100F40050What is the new reading on the scale?

A cylindrical beaker of height 0.100 m and negligible weight is filled to the brim with a fluid of density ρ = 890 kg/m3. When the beaker is placed on a scale, its weight is measured to be 1.00 N. A ball of density ρb = 5000 kg/m3 and volume V = 60.0 cm3 is then submerged in the fluid so that some of the fluid spills over the side of the beaker. The ball is held in place by a stiff rod of negligible volume and weight. Throughout the problem, assume the acceleration due to gravity is g = 9.81 m/s2A. What is the weight Wb of the ball? Express your answer numerically in newtons.B. What is the reading W2 of the scale when the ball is held in this submerged position? Assume that none of the water that spills over stays on the scale.C. What is the force Fr applied to the ball by the rod? Take upward forces to be positive (e.g., if the force on the ball is downward, your answer should be negative). Calculate your answer from the quantities given in the problem and express it numerically in newtons.D. The rod is now shortened and attached to the bottom of the beaker. The beaker is again filled with fluid, the ball is submerged and attached to the rod, and the beaker with fluid and the submerged ball is placed on the scale. What weight W3 does the scale now show? Express your answer numerically in newtons.

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