Sections | |||
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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: Pressure and Atmospheric Pressure

**Transcript**

Hey guys. So, in this video we're going to start talking about pressure and atmospheric pressure both of which are huge topics in this chapter, let's check out. Alright, so pressure is defined as force divided by area, force divided by area. So, it's a measurement of how much a particular force is spread out over a surface area and it has units of Newton per square meter and that's because force is measured in Newtons an area is measured in square meters. Now, they got tired of writing Newtons per square meters over and over again. So, they decided to call this something, and this is called Pascal, Pascal named after mr. Pascal, abbreviated Pa, and it just means that if you have 1 Pascal, you have 1 Newton per square meter, let's look at a quick example here. So, two identical wood blocks, these two guys here, one and two, and it specifies here, that they are, they have 800 kilograms per cubic meter, hopefully right away you identify that this is density because of the unit's, it doesn't say that, but you need to know that, so the density, which is Rho, Greek letter Rho, is 800 kilograms per cubic meter, this is why we cover density earlier and these are the dimensions of the box. So, 0.2 by 0.2 by 1. So, if you notice, this is the long side here. So, this must be 1 meter and these guys here will be the 0.2s meters, and here it's just oriented in a different direction, this is the long side so this is going to be 1 meter and this must be 0.2 height and the depth here must be 0.2 as well, okay? So, they're placed outdoors, meaning that there's a bunch of air around them and horizontal surface, on horizontal surfaces, so the idea is that it's placed on sort of a surface here, the floor, something like that, we want to know the pressure of each block on the surfaces that they sit on, so the idea is that if you have a block and it sits on a surface, it is pushing against the surface and it's applying a pressure, why? Because there is a force over an area and then whenever you have a force over an area you have a pressure. So, I want to know how much pressure is this block over here, applying on the surface, right underneath, underneath it. So, you might imagine that it looks kind of like that, right? If you draw sort of the 3d version here and you might imagine that there's the bottom here of this guy, is also pushing against the surface, against the floor, and I want to know the pressure. So, we're calculating pressure, pressure against the floor, let's call that Pf, how do we find pressure? Well, the equation for pressure is force over area. So, let's write that, it's the amount of force that the block applies on the floor divided by the area, the area that they're touching, how much area is it is there between the two of them, which is just this area down here, the area of interaction, okay? So, what is the force? This block pushes against the floor because it has a weight, because of gravity, right? So, gravity pulls on a block down, the earth pulls down the block, the block pulls on the table or on the surface or on the floor, so the force that's causing the block to push against the surface is m, g. So, I'm just going to rename this to m, g, and this happens a lot by the way, that the force on a pressure problem is the weight force, divided by the area and I can just sort of start plugging in that the area here is maybe 0.2 times 0.2, okay? Obviously, we know gravity is 9.8 for the sake of this problem to keep it simple, we're going to use that gravity is approximately 10 meters per second squared, to make our life easier, but I still have to find the mass and when I find a mass I plug it in and we're done, how do we find mass? You may remember that if you have density, which you do, and if you can have volume, which you do, you can find mass because density is mass over volume therefore mass is density times volume. Now, please don't get the little p, the little curvy Greek p, which is Rho confused, that's density, don't get that confused with big P, pressure, okay? Those are different things it's unfortunate that they look so similar. So, do I have pressure and volume? Yes so that I can find, I'm sorry, do I have density see I just did it, do I have density and volume? We do. So, we're going to be able to just plug all the stuff in and figure out the mass. So, let's do that real quick, mass is going to be density, which is 800 kilograms per cubic meter. Remember to always put units like this, it's easier to cancel times the volume, the volume is just the three sides multiplied. So, 0.2 times 0.2 times 1.0 and because this is meter, meter, meter this is cubic meter and this is nice because cubic meter cancels here and we end up with the mass, the mass will be, the mass will be, I have it here, 32 kilograms. Now that I have the mass I can plug it in here, 32 gravity's 10, this is 0.04 and if you do this entire thing you get that the pressure is 8,000. Now, the question is what are the units here? Well, because I'm using the standard units, this is just going to be Pascal. Now, if you don't, if you don't see that just keep in mind that m, g is, because this is in kilograms and this is in meters per second squared, this m, g here is in Newtons and this was meter, meter so this is meter squared. So, Newtons per meter squared gives you a Pascal, so that is the answer to this one, okay?

Now, I'm going to do the second one in a different way. So, you can see another way that you could have done this that is going to be a little bit easier and it's going to be helpful later on but this is sort of this the most straightforward way you could have done it without anything fancy, okay? So, let's do this a little bit different and the first thing you might be wondering is isn't it just the same thing because it's the same block? Well, pressure is force over area and while the force is the same because the mass is the same, because is the same block, right? The area is different, the floor is touching, is interacting with the surface underneath it via a much larger area, so the area that they are touching against each other is much bigger and if the area is bigger, you might imagine that the pressure will be smaller, okay? The pressure will be smaller but we're going to calculate this a little bit different so the pressure with the floor it's still going to be the force against the floor divided by the area and the force against the floor by the way it's still m, g divided by the area but I'm going to show you something a little bit different now. So, what is the area? The area is the, are these two dimensions here, right? These two dimensions here. Now, all three of them but just two, which is the width times the depth, okay? So, let's leave it there and in mass, remember we just did this here, mass is right here, mass is density times volume but what is volume, volume is width times depth times height, okay? Width, depth and height. So, I can plug in this stuff in here and I can say the m is going to become Rho, W, D, H don't forget the g over here, divided by W times D, okay? And, this is the only time I'm going to do this just to show you this is actually very helpful for you, W cancels, D cancels and you're left with, you're left with that the pressure against the floor is going to be Rho, Rho not P, this is Rho, be careful, and I'm going to just move the letters a little bit here, Rho, g, H, Rho g, H, so this is interesting because the pressure actually does not depend on the area, it only depends on how high this thing is, why doesn't the pressure depend on the area? Well, as you have a bigger base you have a bigger object but that force is being distributed over a bigger area, so it doesn't really matter, it only matters what the height of the object is, and that's good news because if you know this, this question is much simpler to solve because you can just plug a bunch of stuff in here, let me move this up a little bit, so the density is 800, gravity is 10 and the height is 0.2 and if you do this, if you do this you get that this is 8,000 divided by 5, this is going to be 1600 Pascal, 1600 Pascal, and notice that this is a smaller number than the other number over here and that's because even though it's the same mass therefore the same weight its distributed over a bigger surface area. So, there's less pressure, okay? So, that's a quick example of how pressure works and two different ways you can calculate it and I even sort of showed you or derived this nice equation that you can use, and this equation is going to come back later. So, that's good news, okay? So, now let's talk about something else, this is the last point I'm gonna make here. So, and just like how you can have an object that is applying pressure against a surface you can also have air molecules around objects applying pressure on them as well, so that is called atmospheric pressure, atmospheric pressure is the pressure due to air molecules around you that are pushing against you or against an object, okay? So, the idea is that a box will apply pressure on the floor but then air molecules directly above will apply pressure on the object. Alright, and that pressure has a standard value at sea level. So, what does that mean? That means that the amount of pressure that the air exerts on you actually changes if you go, let's say, up a mountain but if you are sitting, hanging out next to the ocean you know for a fact that, that pressure has to be 1.1 times 10 to the 5 Pascal, so, it's the standard value that we're always going to assume, if they don't give us a pressure, we're going to assume that the pressure of air around us is 1.01 times 10 to the 5 Pascal and this is the pressure of air, sometimes I'll refer to this as P air. Sound funny, some french dude. So, pressure of air, p air, and that's the amount, sometimes it gets simplified as 10 to the 5 Pascal but typically you do have to remember the 1.01, kind of annoying. Now, they got tired of writing this over and over because it's the bigger number and they decided to invent this thing called in 1 atm. So, 1 atm just standards for 1.01 times 10 to the 5, it's sort of a shortcut because they got lazy, you can also have pressure in British units, so instead of Pascal, which remember, Pascal was Newton per square meter, you can have it in pounds per square inch and you can also have it in terms of millimeters of, millimeters of mercury, right? 760 millimeters of mercury you see that in lab, when you're doing chemistry, okay? So, you should know all these. So, you can convert between them, but remember, that this is the standard unit one so this is the one that's going to go into all your equations. So, if I give you 760 millimeters of mercury you have to, you have to convert this into Pascal or if I give you any millimeters of mercury you have to convert that into Pascal.

Alright, so let's do a quick example to drive this point home, so it says, for the blocks above calculate the force applied by the air above them to their top surfaces. So, we have these blocks, I'm going to draw them both real quick and I want to know, I want to know the force applied by the air above them. So, there's a bunch of air molecules everywhere and I want to know, how much force is the air applying to their top surface, how much force is air applying here. Now, to be clear, air is applying force to all sides of this block, the back as well, except the bottom because the bottom is clutching the surface, right? But the air is applying everywhere but I just want to know how much force is applied to the top so it is a force on the top. For Part B it's the same thing except that it's laid out a little bit differently, so it looks sort of like this and I want to know how much force is applied to the top surface over here, F top, cool? And. So, how are you doing this? We're going to do this using the fact that we know what the pressure of air around you is most of the time, we can assume that pressure is going to be, the pressure of air is the atmospheric pressure, which is 1.01 times 10 to the 15 and if we know that and we know the area we can find the force, let me show you, that's because pressure is force over area and if I'm looking for force and I have the other two I can rewrite that force is pressure times area, okay? So, the pressure will be the pressure of air up here, which is 1.01 times 10 to the 5 Pascal and the area is, the area, if you remember the dimensions here, we're 0.2, 0.2 and the height was 1 but if I want the top surface I'm going to use this in this measurement, these two measurements here. So, 0.2 times 0.2, it's meters times meters so it's square meters and if I multiply this I get 4040 Newtons. Now, notice how I use the standard unit for pressure, I use the standard unit for area therefore when you combine those two you're going to get the standard unit Newtons for force, I don't have to sort of combine those two and figure out, do dimension analysis, figure out what unit gets left out at the end here, because if I'm using standard units as an input, I'm going to get the standard unit for the output. So, 4,000 Newtons you might thinking that's a lot of force and it is a lot of force, it's the equivalent of having something that is about 400 kilograms on top of you or roughly 880 pounds, right? And 20 by 20 is a square of this big and if you have 800 pounds on top of this very heavy, and does that make sense? Air is very light, how come it's going to be so heavy on top of you? Well, the reason it's so heavy on top of you is because there's a huge column of air that starts from right over on top of you all the way to the atmosphere, right? So, you know thousand of or many, many miles above you. So, it's a lot of air. So, it's pretty heavy, adds up to a lot, you're just used to it so it doesn't bother you at all. Alright, so for the second one it's going to be very similar, but we're just going to use different numbers, so the force, remember, we can just start from here, force is pressure times area, the pressure is 1.01 times 10 to the 5 and the area now is going to be the lone one, the long dimension here, which is 1 and the one of the smaller dimensions, which is 0.2, so this is going to be one times 0.2 square meters, okay? And this means that the pressure, this means that the pressure will be, the pressure will be, I have it here, I'm sorry, the force will be 20,200 Newtons. Now, notice that the force here turned out to be much greater than the force here, why? Well, because there's more air on top of you, right? That top base, the top area of this of this block is supporting a lot more air on top of it therefore it's more force, and you can also think of this as more weight, essentially what we're doing here is calculating the weight of the air on top of you, alright? So, that's how this works, let's keep going.

Practice: A large warehouse is 100 m wide, 100 m deep, 10 m high:

a) What is the total weight of the air inside the warehouse?

b) How much pressure does the weight of the air apply on the floor?

Concept #2: Pressure In Air and In Liquids

**Transcript**

Hey guys. So in this video we're going to look into how pressure works if you have an object that is surrounded by air as opposed to if you have an object surrounded by a liquid such as water, let's check it out. Alright, so imagine that you are next to the ocean, okay? So, you're next next to the ocean and if you're next to the ocean. Remember that the pressure of the air molecules around you, let's make the air molecules green, the pressure that the air is going to exert on you at this level next to the ocean is going to be 1 atm which is the standard atmospheric pressure at sea level, okay? So, if you're out in the open you have air molecules around you and that's what happens. Now, as you go up in height, as you go up in height the pressure will change and the easiest way I think to remember what happens to the air pressure whether it increases or decreases is just to think that if you go up in air there's going to be less air above you. So, if you are here, you can imagine there's a column of air molecules on top of you but if you're here, there's a smaller column of air molecules on top of you. So, because there's less air on top of you the air pressure will be lower, it will decrease, okay? Remember, air pressure, air pressure decreases, air pressure comes from the weight of air molecules on top of you. So, if there's less air on top of you there's less of a weight so the air pressure decreases, okay? Because there's less weight pushing down on you. Now, the air density, the air density is going to decrease as well, and number 2 follows from number 1. So, here you have a ton of air molecules on top of you and the air molecules up here push down against the air molecules down here so the air molecules down here are more squished together because all this weight on top of them, okay? So, you can think that this is, this is low-density and this is higher density of molecules, okay? The molecules are more spread out up there because they're not being squished by the weight of the molecules on top of them. So, if you just remember that, I think if you think about in terms of what's on top of you, you don't even have to memorize that as you go up in height your p air goes down and then your rho goes down as well, as you go up in Heights, okay? Now, here's what's even more important for you to remember is that the density of air is very low as it is. So, both of these changes are very insignificant, in fact most of the time we're going to ignore changes in pressure and density of air, okay? So, this next example deals with that, which of the following is the best approximation for the atmospheric pressure, p air, that's 100 metres above sea level. So remember, changes are only significant over large distances and I should say for very, very large distances such as how high an airplane is flying. So, 100 metres is not a very large distance even though it'd be pretty tall but it's not significant therefore the answer is that the atmospheric pressure here is basically going to be the same as it is at sea level, okay? So, it's the same because there's very little different it's approximately the same, cool? So, if you're not sure which pressure to use you should be using 1 atm which of course is this number right here, it's 1.01, I made 1.00 just because I was rounding and if they don't tell you, you can use that number, cool? So, it's a little bit different if you have liquids however. So, if you were an object or under a liquid submerged in a liquid the pressure differences will be much more pronounced they're going to be much bigger differences in pressure even for a little bit of a distance because liquids have much higher density than air, okay? So, but now in air we moved up and our pressure change but if you are in water you're going to move down, okay? So, here the pressure depends on your height and here it depends on your depth, okay? Now, we just used h for both of those but the idea is that the pressure will increase as you go down here and everyone knows that, if you start swimming, if you start going underwater and the deeper you go your ears start to feel a lot of pressure and that's because the water pressure increases as you go down in height or depth, okay? It increases and it increases because there's more liquid above you. So, before, if you went up it would go down because there's less air. Now, if you go down the pressure will go up because there's more stuff on top of you there's more liquid on top of you. So, there's more weight pushing down it's the same logic as before, the difference here is that changes are significant even for small distances, right? And, if you're swimming and you just go a little bit lower underwater you can tell those differences are pretty significant, water density does not change much. So, we're always going to assume that water density is constant because the changes are very insignificant even for very large distances. So, you can pretty much assume, you could even assume that I never even mentioned this and just pretended water density is always the same always, cool? And then the last point here is that the pressure in the liquid such as water but really any liquid depends on this equation r can be calculated according to this equation so this is a very important equation and it tells us that the pressure at the bottom of a column so let's draw a little beaker here and let's say we have, let's say we have some liquid and our two lines that are important here, the highest point here, okay? And the lowest point of the liquid here, so the pressure at the bottom right here, pressure at the bottom, is going to be equal to the pressure at the top, which is this, plus Rho, this is density of the liquid, g gravity and h which is the height difference between these two or the depth of the liquid, okay? So, I can calculate the pressure at the bottom if I know the pressure at the top and if I know the h, okay? We're going to use this equation quite a bit. Now, you should know that the pressure at the bottom is called the absolute pressure, the pressure at the top is called. the relative pressure and the pressure difference between these two is called the gauge pressure, okay? Gauge pressure is the difference between the two pressures, how much greater one is than the other and the idea that this pressure here is relative to the top pressure, the pressure at the bottom depends on the pressure at the top, that's why this one's called relative. So, sometimes you see questions that will throw this terms at you. So, you should know what they are.

Let's do an example real quick and then we'll be done with it, so it says, suppose you are 1.8 meters tall and your heart is located 1.4 meters from your feet. So, I'm going to draw, I'm going to draw a person here, pretty big, so that we can do this and your heart, let's say over here and your total height is 1.8 meters and your heart is 1.4 meters away from your feet so it follows that your, this is 1.4, this is 1.8, this gap here, hearts to top of your head must be the difference between those two which is 0.4 meters. So, that's you, it says, the blood pressure near your heart is 1.3 so right here the pressure at your heart is going to be it's going to be 1.3 times 10 to the fourth pascal and you want to know the, we want to calculate the blood pressure at the top of your head. So, we want to know the blood pressure here, pressure of head in 100, the pressure at the bottom of your feet, pressure feet and guess what we're going to use this equation highlighted in green right here to figure this out one at a time so the first one, we want to know what is the pressure of your head, okay? Blood pressure of your head, this here by the way is the density of blood some of the right here that rho blood is 1060 and we're going to that number. Alright, so check this out, we know this here, this is our known and these are our unknowns. So, for both of these questions were going to the same thing, we're going to set up an equation with a known pressure in an unknown pressure and if you look at this known pressure and this unknown pressure, we know, we know this distance right here, which means we can set up an equation between these two guys. So, if you set up an equation between these two guys it's always going to be that p bottom equals p top plus Rho g, h and the h is the gap between them which is 0.4. So, I know p bottom, I'm looking for p top and this is just because this is at the bottom, this is at the top, it's that simple, right? This is the guy at the bottom, that's at the top. So, I know, I want p top I know the density 1060, I know gravity, we're going to use 10 and actually for gravity we're going to use 9.8 because I want to be more accurate since we're dealing with the human body here and h is going to be the distance between top and bottom, so this is very important h is the distance between top and bottom which in this case is 0.4. So, let's set this up I can write that p bottom is 1.3 times 10 to the 4th equals p top which is what we want plus Rho 1060, gravity 9.8, h 0.4 for the sake of time I'm not including units here, but all the units are standard which means my pressure will have standard units at the end pascal. So, if I move this around you end up with p top equals 1.3 times 10 to the fourth minus, right? This goes to the other side minus and I have it here, 1060, 9.8, 0.4 and this is, I'm rounding here, 8800 pa. So, let me write this here, this went from 1.3 times 10 to the fourth, if I want to rewrite this with a 10 to the fourth, I'm going to do this kind of quickly but it would look like this .088 times 10 to the fourth, okay? You can validate that, if you would like, let me get out of the way. Alright, and I want to do that so that we can write all of these answers with a power of 4.

Let's do Part B. So, for Part B we want to know what is the pressure of blood or blood pressure on your feet. So, again, we're going to set up an equation, p bottom equals p top plus Rho g, h but now, we are talking about this interval here, okay? So, this green height was this height right here, but now the blue height has to do with this height right here, and perhaps obviously this is and now for this equation bottom is the feet and top is the heart, okay? Just to be very careful here, when I did this the bottom was the hearts and the top was the head but this is all sort of relative, right? So, now that I'm writing another equation for a different interval for this height here, bottom and top change, okay? So, be careful there and we are looking for p bottom whereas before we were looking for p cop, okay? Be careful, if you're careful it's going to be easy. So, let me write this over here, p bottom, p top is the heart so, 1.3 times 10 to the fourth plus rho 10 60, gravity 9.8 and h is 1.4 meters, all the units are standard so I'm going to get the answer in Pascal and if you do all of this you get 27,500 or if you want to write it in terms of a power of 10 to the fourth, right? You can write this as 2.75 times 10 to the fourth pascal, okay? 2.75 10 to the fourth pascal, let me get out of the way and last I'm going to do is put it over here, that this is 2.75 times 10 to the fourth pascal. So, I want to quickly show you these answers, this is 0.8, this is 1.3 and this is 2.75 and the important point to make here is that as you go down you have more and more pressure and that's what you should have because at the bottom of your feet you have all the way to the blood and your entire body pushing down on you so the lowest blood pressure should be all the way to top and then the highest blood pressure should be all the way at the bottom. Now this is a little bit simplistic, the human body is a little more complicated than that but this is good enough for physics approximations, the last, last point I'm going to make here is that this equation actually technically works for air pressure but we're not going to use it because most of the time we're just going to ignore changes in air pressure, okay? So, if I tell you that you are on top of a building, that's 100 meters tall you're not going to calculate the pressure up there because it's not going to be very different from atmospheric pressure, you can try it and and you'll see that it's a very small difference. So, what we tend to think of this equation as an equation for pressure in liquids even though it would work for air it's just that air pressure changes are very, very subtle over small distances, cool? That's it for this one, let's keep rolling.

Concept #3: Calculating Pressure in Liquids

**Transcript**

Hey guys. So in this video I'm going to show how to calculate pressure at different points within a liquid using two common examples, let's check out, cool? So, first remember that the pressure in the liquid changes with depth, as you go down in the liquid the pressure is going to increase and the exact calculation is used, is done using this equation, the pressure equation, you should also know that in between two liquids, whenever two liquids touch or two materials, they don't have to be liquids, whenever the two materials touch we have something called an interface. So, for example right here in this little cup beaker, this line right here where these two things touch, is called an interface, there's another interface here, and I want you to see if you can figure out where it goes and hopefully you're thinking that the other interface is over here, and that's true so this is the interface between air and a yellow liquid, those are two different materials and then there's an interface between yellow and blue liquid. Now, technically there's an interface between legs down here, between blue liquid in glass those are two different materials but you're never going to use that. So, just forget about it, what's important about interface is that at these points the pressure of both materials is the same. So, what does that mean? So, imagine you have a molecule all the way at the bottom of the liquid here, blue molecule and the molecule all the way at the top, these two molecules you're gonna have different pressures because they're in different heights and the molecule at the bottom is going to have a greater pressure than the molecule at the top because it has more stuff on top of it. Now, what this, what this over here is saying, is that the top most molecule the one it's like all the way at the top has the same pressure as the bottom most molecule in the yellow liquid, I drew it right here, but imagine this is a yellow molecule, okay? So, this has some consequences but I want you to remember that the pressure is the same, the first consequence is that everywhere a liquid touches air the pressure of the liquid equals the pressure of the air. So, what does that mean? So, if you look at the molecule all the way at the top of the yellow column, this is all the way to pop, is a topmost molecule it is right under air therefore the pressure at the top of the yellow is the same as the pressure of air and this is going to be very useful. So, let's do example one to see how this might work. So, it says, suppose the density of a particular beer is 1050. So, density rho is 1050, this is using the standard units, I don't have to do anything weird and we want to know, what is the gauge pressure at the bottom of a solo cup if it is filled with beer to it's very top. So, I have a solo cup looks something like this, it's not a perfect straight cylinder but here it doesn't make a difference and it just has beer, I'm going to make beer blue. has beer all the way to the top. So, it's got 12 centimeters height 0.12 meters and it's got beer everywhere and we want to know what is the gauge pressure, gauge pressure, okay? Now, almost every time you're asked for a pressure and you have a height you're going to use the pressure equation right here, which combines lengths, pressures with Heights, okay? So, let's write this real quick . So we're going to say because we have a height and I want a pressure, I'm going to write that p bottom equals p top plus Rho g, h. okay? And p bottom is down here, this is the p bottom and p top is the top of the liquid p top. Now, because the top of the liquid touches air because it interfaces with air the pressure at the top here will be the pressure of air for more precisely the atmospheric pressure to standard atmospheric pressure which is 1 atm. Now, if you remember, we can't really use one each here, we have to use the Pascal version. So, the pressure at the top is 1.01 times 10 to the fifth pascals, it's just a known number, you're supposed to memorize that, plus Rho g, h. Now, before, we get further here, I wanted to talk about this but you may have noticed that the question says that we want the gage pressure and this is a little tricky and I did this on purpose, I hope you remember that what gage pressure means is this right here, okay? So, let me make that yellow as well, meaning, we don't want the pressure at the bottom, that's not, what we want, we want the pressure difference. So, I don't want p bottom, I just want, I just want Rho g, h, okay? So, I was almost going the wrong way, I wanted to talk about how the pressure at the top was going to be the pressure of air so it made that connection but this question actually wants just the gauge pressure which means you don't even use p top, you just use Rho g, h, okay? Rho is density which we have it's 1050, gravity, I'm going to use 10 instead of 9.8 to make things a little simpler and the height is 0.12 meters and all the units are standard so the pressure will come out in the pascal and if you multiply everything the answer is 1260 pascal, this is the gauge pressure here, cool? Let's do another example and on the second example we have two liquids on top of each other which is kind of common. So, you're going to get something like this with two liquids and the first liquid here, it says, it's blue. So I've got some blue liquid here, and the second liquid is yellow. So, I'm just going to sort of do this and I'm going to put a little line here, cool? And it says, the blue liquid, you pour six centimeter column of blue liquid. So, you pour liquid essentially until it reaches six centimeters of height, that's what it means by 6 centimeter column. So, this height here is 0.06 meters and the density of blue liquid is 1200 standard units and then you add a 4 centimeter column. So, of yellow liquid, so this is four centimeter column and the density of the yellow liquid is 800 also standard units and this is about a 12 centimeter beaker. So, this entire thing is 12 centimeters. So, 0.12 meters by the way, if you look at this quickly you got 6, 10, 6, 4, so that's 10 which means there's sort of a two centimeter gap here, I don't know if we're going to need this, let's write here, just in case, cool? So, the liquids don't mix, this is very important, if they look liquids mix this would be a mess and you wouldn't be able to solve it like this and we want to calculate the absolute pressure at the blue yellow interface. So, there is blue here, let's make this blue and I want to know, what is the pressure here. So, Part A is asking for this, what is the pressure in the blue yellow interface? how do we do this? well, there's two ways we can set this up, we can set up an equation that goes from here to here, and then we would use the 6 centimeter height or we can set up an equation from here to here and then we would use the 4 centimeter height, okay? So, we have a choice there but we don't really have a choice because if you have an unknown pressure, this is my unknown pressure, I have to couple it with a known pressure, I don't know the pressure here, I also don't know the pressure here, but I do know the pressure at this point, the pressure at this point, because the top of the yellow liquid touches air, the pressure at the top of the yellow is the same as the pressure of air or the atmospheric pressure which is 1.01 times 10 to the fifth. So, therefore I have to use the top interval to calculate this, okay? So, let's do that, we're going to say that now we're using this piece here, the top. So, I'm going to write that p bottom the equation is always the same equals p top plus Rho g, h. Now, when we do this, this is the density because we're talking about this interval here, this interval, this height of liquid, the density here is the density of yellow and it's the height of the yellow column and p bottom means blue yellow interface and p top is air, I hope you see this I'm just doing this slowly so we don't screw this up, bottom is over here, so this is the interface between the two which by the way is what we want and top is where we touch air, okay? So, let's calculate, it's just a matter of plugging stuff in, we already know that the pressure of air is 1.01 times 10 to the fifth, the density of yellow is 800 gravity, I'm going to round it to 10 just to make life easier and the height here is 0.04 meters, okay? And, if you do all of this and I have it here, you know, this piece here is, this entire thing here is just 320, not much, okay? And of you do this whole thing you end up with 101,320 which if you were to round to two significant figures or three significant figures you basically end up with this number, in other words there's negligible difference here and that's because it's only 4 centimeters which is a tiny column of air or liquid but that is the pressure in that middle point, this blue, this thick blue line over here, cool? For Part B, Part B asks us, we are out of the way, Part B asks us for the gauge pressure. Remember, absolute pressure means p bottom, right? Right there between those two, gauge pressure is just a difference. So, if I'm asking for the gauge pressure from the blue and yellow interface, it's just, I hope you see it, it's just this part here, it's how much higher is the pressure of blue relative to the pressure directly on top of it, okay? So, when I say pressure gauge, we're just saying is the pressure gauge between these two things here, So, it's between blue and the gap between the air and the blue over here so it's the pressure gauge across the yellow, is going to be rho yellow, g, height yellow and we already did this, the entire thing here, if you multiply is just 320 p, h, just asking, how much higher is the pressure of the blue line relative to the water on top of it, okay? So, this is related to this and C is about a new point, we want the absolute pressure. So we want the absolute pressure, you can also think of this as p bottom, right? At the bottom of the blue liquid. So, all the way at the bottom of blue, cool? And again, we're going to use the p bottom equation. So, p bottom equals p top plus Rho g, h, but now we're going to set up an interval from here to here, right? Across the 6 centimeters, why? Because I know this pressure, I now know this pressure over here and to find this pressure, I'm going to write an equation from here to here, I hope that makes sense. So, when I write p bottom, this is all the way at the bottom of blue, the p top is going to be the blue yellow interface, this is going to be, we're going to go through, think of it as you start at the top of the liquid and you're going to go through the liquid, you're going to go through the blue liquid and the height will be the height of the blue liquid which is 6, I'm doing this very slowly so that hopefully everyone can follow that. So, p bottoms is what we are looking for, p top we just calculated it, it's this number right here, Okay? So, 101,320 plus the density of blue which is 1200, gravity which I'm going to round as 10 and the height of blue right here, 6, 0.06 and if you do all of this you get that the answer is 102,040 pascal, very similar not a big difference but a little bit higher pressure than before, okay? So, if I want the pressure at the bottom of the second liquid I can find the pressure in the middle, right? So, I start at the top, I know this one, I can find the next one then I can find it next one. Now, if you didn't want to do this one at a time, let's say this question asked you directly for C which sometimes they'll do that, we're just going to say, hey find the bottom, the pressure all the way at the bottom, can you do this in one step and the answer is yes. So, or you could have done this in one shot, you could have said, the pressure, let me rewrite this just to make this a little bit cleaner, let's say you have a liquid here and you have another liquid here, and let's call this pressure A, pressure B, and pressure C and by the way if this is air pressure a is the pressure of air because it's touching air but if you have a situation like this which is what we have, we can say that pressure C is the pressure a plus Rho g, h for the first liquid, let's call that Rho1 g, h1. So we're adding that tiny pressure plus the next column of air or a column of liquid Rho g, h but now it's going to be the density of the second liquid plus the height of the second liquid here, okay? So this is 1, this is 2, if you were to do this, you would get something like this, pressure of air which is 101,000 plus the first density which up here was 800, gravity 10 and the first height 0.04 then we do the same thing for the second column. So, 1200, let me get out of the way 1210 and the height is 0.06, if you do this you would end up with the exact same answer and you can try if you want, you end up with the same exact answer here, So, point that I want to make here is it's very important, if you have two columns or three columns or four columns with liquid and you want to find the pressure at the very bottom you can just keep stacking Rho g, hs for every column you have and that is a faster way of solving this, if you have to do it in one step, cool? That's it for this one, let's keep going.

Practice: The deepest known point on Earth is called the Mariana Trench, at ~11,000 m (~36,000 ft). If the surface area of the average human ear is 20 cm^{2} , how much average force would be exerted on your ear at that depth?

Practice: A tall cylindrical beaker 10 cm in radius is placed on a picnic table outside. You pour 5 L of an 8,000 kg/m^{3} liquid and 10 L of a 6,000 kg/m^{3} liquid into. Calculate the total pressure at the bottom of the beaker.

Practice: A wooden cube, 1 m on all sides and having density 800 kg/m^{3} , is held under water in a large container by a string, as shown below. The top of the cube is exactly 2 m below the water line. Calculate the difference between the force applied by water to the top and to the bottom faces of the cube (Hint: calculate the two forces, then subtract).

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Concept #1: Pressure and Atmospheric Pressure

Practice #1: Weight and Pressure of Air

Concept #2: Pressure In Air and In Liquids

Concept #3: Calculating Pressure in Liquids

Practice #2: Force on Ear (Mariana Trench)

Practice #3: Pressure on Two Liquid Beaker

Practice #4: Net Force Underwater

A window in a submersible needs to be able to withstand the incredible forces that water will exert at depth. If the window is 30 cm x 15 cm, what would the force exerted by the water be at a depth of 1 km? Assume that the density of sea water is 1029 kg/m3.

An open container is filled with water, to a depth of 5 cm. Above that, a layer of oil 2 cm thick is poured on top of the water. If the density of the oil is 700 kg/m3, what the pressure at the top of the oil? At the water-oil boundary? At the bottom of the container? Take atmospheric pressure to be 1x105 Pa.

In a lecture demonstration, a professor pulls apart two hemispherical, steel shells (diameter D) with ease using their attached handles. She then places them together, pumps out the air to an absolute pressure of p, and hands them to a bodybuilder in the back row to pull apart. If atmospheric pressure is p0, how much force must the bodybuilder exert on each shell? Evaluate your answer for the case p=0.025 , D=10.0 .

A cube of side s is completely submerged in a pool of fresh water. Derive an expression for the pressure difference between the bottom and top of the cube.a. Pbottom - Ptop = Pfluidgsb. Pbottom - Ptop = Pfluidsc. Pbottom - Ptop = Pcubegsd. Pbottom - Ptop = Patm + Pfluidgs

The Aswan High Dam is 111 m high. What is the absolute pressure at the foot of the dam?
[A] 1.09 × 106 Pa
[B] 1.09 × 103 Pa
[C] 1.19 × 106 Pa
[D] 1.11 × 102 Pa

The four tires of an automobile are inflated to a gauge pressure of 2.0 x 10 5 Pa. Each tire has an area of 0.024 m2 in contact with the ground. Determine the weight of the automobile.

A container is filled to a depth of 20.0 cm with water. On top of the water floats a 30.0-cm-thick layer of oil with specific gravity 0.700. What is the absolute pressure at the bottom of the container?

a) Estimate the pressure exerted on a floor by one pointed chair leg (66 kg on all four legs) of area = 2.6×10−2 cm2. b) Estimate the pressure exerted on a floor by a 1200-kg elephant standing on one foot (area = 860 cm2 ).

A patient is to be given a blood transfusion. The blood is to flow through a tube from a raised bottle to a needle inserted in the vein. The inside diameter of the 26-mm-long needle is 0.81 mm, and the required flow rate is 2.3 cm3 of blood per minute. How high h should the bottle be placed above the needle? Assume the blood pressure is 78 torr m torr above atmospheric pressure.

Intravenous infusions are often made under gravity, as shown in the figure.a) Assuming the fluid has a density of 1.00 g/cm3, at what height h1 should the bottle be placed so the liquid pressure is 58 mm-Hg ?b) At what height h2 should the bottle be placed so the liquid pressure is 660 mm-H2O ?c) If the blood pressure is 78 mm-Hg above atmospheric pressure, how high should the bottle be placed so that the fluid just barely enters the vein?

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. What is the pressure at this depth?

The lower end of a long plastic straw is immersed below the surface of the water in a plastic cup. An average person sucking on the upper end of the straw can pull water into the straw to a vertical height of 1.1 m above the surface of the water in the cup. What is the lowest gauge pressure that the average person
can achieve inside his lungs?

If the gauge pressure is doubled, the absolute pressure willa. be halved.b. be doubled.c. be unchanged.d. be increased, but not necessarily doubled.e. be decreased, but not necessarily halved.

A swimming pool 6.0 m wide by 13 m long is filled to a depth of 13 m. What is the pressure on the bottom of the pool? (Express your answer to two significant figures.)

At 25°C the density of ether is 72.7 kg/m3 and the density of iodine is 4930 kg/m3. A cylinder is filled with iodine to a depth of 1.6 m. How tall would a cylinder filled with ether need to be so that the pressure at the bottom is the same as the pressure at the bottom of the cylinder filled with iodine? (Express your answer to two significant figures.)

Convert the following units of pressure to the SI unit of pascals (Pa), where 1 Pa = 1 N/m2. 2300 kPa = _____ Pa (Express your answer to two significant figures.)

Convert the following units of pressure to the SI unit of pascals (Pa), where 1 Pa = 1 N/m2. 871 torr = _____ Pa (Express your answer to three significant figures.)

A rectangular swimming pool is 8.0 m × 30 m in area. The depth varies uniformly from 1.0 m in the shallow end to 3.0 m in the deep end.Determine the pressure at the bottom of the deep end of the pool. (Express your answer to two significant figures.)

Calculate the difference in blood pressure between the feet and top of the head for a person who is 1.80 m tall. Consider a cylindrical segment of a blood vessel 2.70 cm long and 2.40 mm in diameter. What additional outward force would such a vessel need to withstand in the persons feet compared to a similar vessel in her head?

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