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

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

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Bernoulli's Equation |

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Concept #1: Pascal's Law and Hydraulic Lift

**Transcript**

Hey guys. For this video we're going to talk about the Pascal's law and its most important application in physics which is the hydraulic lift, let's check out. Alright, so Pascal's law states, and you should probably memorize this, that pressure in a confined fluid is transmitted equally throughout the fluid. So, what does that mean? That means that, let's say you have a contraption like this, and I apply a little bit of pressure here, p? Well, that pressure is going to be transmitted equally throughout the fluid. Now, what does that mean in terms of problem solving? Not much. So, let's not get too hung up on that definition, I just want you to memorize this in case you need it, I have other definitions that I think are very more useful in problem solving, a second related rule, it isn't like a law named after anyone but you should know that pressure in the liquids. So, do you think pressure to liquid does depend on the shape of the container or do you think it does not depend on the shape of the container. So, for example, is the pressure here, at this point, the same as the pressure here or is it different? And the answer is that the pressure is actually the same, the pressure, you should know, does not depend on the shape of the container it only depends on some other stuff, if you remember P bottom equals P top plus Rho, g, h and h is depth. So, pressure at the bottom over here, it only depends on the pressure on the top, it only depends on the density of the liquid and gravity and on depth, right? So, if you had air up here, then air is pushing down on all of these, just the same with the same amount of pressure. So, pressure at the top therefore is the same, if you have the same liquid everywhere density is the same, gravity is going to be the same for all of them because you're all sitting next to each other and if the height. So, with all these things being the same, if the depth of them is the same then the pressure will be the same. So, I can actually say that the pressure at all of these points here is the same. So, all other things being equal pressure depends only on depth. So, I can say all things being equal pressure is proportional to depth, okay? Now, remember the depth is top down, depth is top down, okay? It does not depend on height, it's not measured from here, because then these two guys would have different amounts of height see how the two blue lines have different lengths, right? So, this is higher point than this point, it has to do with depth. So, you measure from the top down and if the depth is the same and all the other things are the same then the pressure in this line is going to be the same, so all of these, all four pressures at these four points are the same, cool? So, that's what point two was referring to, point three is referring to this image over here and it says, in connected column, so if you have something like this some weird contraption that has connected columns with liquid inside, the liquid height, this level up here is going to be the same as long as the pressure at the top is the same. So, let's say, for example, you have air up here, right? Then that means that air is going to equally push on all of these but it could also be a different gas that's not air, it's going to equally push in all these and it's going to basically balance out those heights, if I were to add a little bit of liquid, let's say, we add liquid here, add liquid, then the additional liquid is going to be spread out in such a way that the heights are the same that does mean that most of the liquid is going to end up on that fat column but that's because they have to balance itself out, one sort of way they remember this is, you might have heard this phrase water seeks its own level, it just means it's going to balance itself out. So, you should know these three rules, they will be helpful throughout the rest. So, now let's talk about the Pascal, I'm sorry, the hydraulic lift, which is the most important application of Pascal's law, and the hydraulic lift has this rule here, that I will talk about in just a bit but let me first explain how it works. So, on the first thing is the Pascal lift, hydraulic lift is going to be built out of two columns of liquid that have different sizes, right? So, you can imagine these usually going to be cylindrical columns. So, they kind of look like this, right? It's like a tube cylindrical tube on that side. So, maybe you draw a little thing here to show that this is cylindrical and the first thing, we're going to do is we are going to put some liquid in here and we're going to have no pistons, so just liquid and because of this rule over here, that we just discussed, if you pour some liquid the liquid is going to level out, and it doesn't matter, what the liquid is, it makes no difference, okay? So, you just have liquid, the liquid levels up and it's important to also note that this pressure at the top here is the same as the pressure at the top here, why? Because they're both touching air or whatever is up here, right? Air is pushing down on both of these things so the pressure is the same, also another way to think about why the pressure is the same, is because pressure depends on height, as we just discussed over here in this picture, pressure depends on the height these two guys have the same height because this thing says they have the same height, so the pressures will then be the same, however you want to remember it but that's not enough. So, first, we put liquids and then we're going to put pistons and what pistons are, is just a little cap that goes on top of it. So, we're going to put a little cap here and you put that cap as like a platform. So, you can put object on top of it otherwise, if you were to put it, let's say a car hydraulic lifter, use to lift cars, then the car would just go into the liquid and that wouldn't be good. So, you put a little cap here, that floats on top of it and the key thing that the cap must have the same thickness. So, this little thickness here, has to be the same as this thickness here, otherwise this thing is going to be unbalanced. So, for example, if you have the same thickness they're going to just stay like this but it's this one here's thicker then it's going to be heavier over all, right? Or it's going to be, it's going to be heavier per area. So, it's going to do this, okay? So, they have to have the same thickness so that they have the same pressure, okay? So, this one here, remember, pressure is force over area and what's causing, what's causing the force here, the force that's applying the pressure is the weight of the piston, so this is going to be m, g over A. Now, the piston on the right side is heavier but it also has a bigger area and those two things will cancel itself out so that you have the same pressure on both sides by the two Pistons, okay? So, one is heavier but it's sitting on a larger area. cancels itself out, because the two pistons have exactly the same amount of pressure that they're applying, they basically, they cancel each other out and effectively it's as, if they weren't there, okay? So, it's just a little platform just, to talk about that real quick. Now, let's get to the real thing here, which is you're going to push down on one of the pistons. So, let's say this stuff is originally here at this height and we're now going to push down on a piston until it's here, I'm no longer going to draw the piston because we know the pistons are going to have the same thickness and they don't affect, they don't make any, they don't have any impact on the problem itself, if you push this down with a force F1, if you remember Pascal's law. that additional pressure, we just talked about that, right? Up here, this pressure is going to be transmitted throughout the fluid. So, if you push down the pressure is going to be transmitted this way and push the other side up, another way to think about this is, this is a contained liquid, contained, it's a closed container. So, if you push down here, the water has nowhere else to go or the liquid has nowhere else to go so it has to go up, right? So, it's going to go up here somewhere, another thing to note is that this extra amount of water here or liquid that got displaced, okay? Over here, it's not going to go up as much, because the volume has to be the same but this is wider. So, it's not going to be as tall, okay? So, that's actually the first thing I want to talk about, which is this change in volume over here, change in volume on the left, has to equal the change in volume on the right, so the right side isn't going to go up as much, okay?

Now, this is where my little rule comes in, which is pressure within the same liquid on a hydraulic lift is the same at the same height, okay? So, what does this mean? This is the most important part. So, look at this point right here, and we have liquid over here, everywhere, okay? And, if you go across to this side. Notice that you have the same liquid, this is the same liquid and the pressure within the same liquid is equal at the same height, these two points right here, this red dot and this red dot are the same liquid the red liquid and they are at the same height therefore I can write that P1 equals P2, okay? And those, those two statements are the most important statements on a hydraulic lift, the fact that the pressure is the same at the same height and the fact that the change in volumes are the same and there's some interesting consequences to this but rather than keep talking I'm just going to do this in an example and show you how that works, cool? So, let's do an example that uses the hydraulic lift and we're going to use these two very, very important rules over here. So, hidraulic lift is designed with columns having areas 1 and 4. Alright, so I'm going to draw, it's always going to look like that. So, something like this and we're going to say that the area of this column here is 1. So, let's call this area 1 and in this area here, area two is going to be 4, 1 square meter, 4 square meter, by the way, this is the surface area, it's the area of the sort of top of the piston, if you will, okay? And it says, pistons of equal thickness are placed on top, by the way, if the problem doesn't say pistons of same thickness, right? If it doesn't say of equal thickness you can assume that. Alright, and then it says, you push down with 10 on the smaller column. So, you're going to push here, this is going to be, we're going to call this F1 equals 10 and it causes it to lower by 20 centimeters. So, when you push it drops the height. So, let's call this height1 or change in height1 equals 20 centimeters, which is 0.2 meters, okay? And we want to know some stuff, we want to know first, how much force acts on the right piston? So, remember, when you're push here this goes up. So, if you apply a force here,you're applying a pressure, this pressure gets transmitted which means the force gets transmitted as well. So, we want to know what is this force here F2, that goes up on the other side, cool? And then the second question is we want to know, how much does the right piston rise by. So, this one here goes down by 0.2. So, this one here has to go up by certain amount. Remember, it's getting less than 0.2 but we're going to figure out how to calculate that. So, we want to know what is the change in height on the right side, cool? So, how do we do this? So, we're going to do, we're going to solve both of these using these two equations over here. So, first remember, force is related to pressure, right? So, I'm going to write that pressure1, pressure1 equals pressure2 on the two sides, that's how you're always going to start these questions and I'm going to rewrite pressure because remember, pressure is force over area. So, I'm going to rewrite P1 as F1 over A1 and P2 as F2 over A2, we're looking for F2. So, all I got to do is move some stuff around. So, F2 is going to be F1 times, I'm going to move the A up here. So, it's going to be A2 over A1, okay? And we are ready to plug in numbers, the force, force is 10, A2 is 4, and A1 is 1. So, this becomes 40, okay? So, this is really important. Notice what happened here, which is the original force, 10, became 4 times greater because the second column is 4 times the area of the first column, so this is super important, what happens here is the hydraulic lift, hydraulic lift will multiply the force by a factor of area 2 over area 1, the ratio of the area's serve as a multiply of a force, this is awesome because you can put a really heavy object like a car and have a normal person just push down this thing and if the, if the second column is much bigger than the first column you get a much bigger force kind magic you can multiply your force by doing this, cool? So, the second one is, by the way, this is called mechanical advantage, gives you an advantage in that you can multiply your force, mechanical advantage, a few times where you'll see mechanical advantage in physics, this is one of them, so the second question is, how much does the piston rise by. So, it's, what is the change in height? And remember, height is related not to pressure but height is related to volume, height is related to volume because the volume, volume is always area times height, and by the way, this is the volume, this is the general volume equation, meaning this works for every single shape, volume is always the cross-sectional area times the height of it. So, now instead of starting with P1 equals P2, I'm going to start with the volume equation up here, this one, okay? So, I'm going to start with, let me start with Delta V1 equals Delta V2, okay? So, let's do that and then similarly, we're going to expand using this equation here. So, we're going to write A1 and Delta h1 equals A2 Delta h2 and we're looking for Delta h2, how much does it rise by. So, Delta h2 equals A1, Delta h1 divided by Delta, I'm sorry, by up, let me, let me go back a little bit here, it's going to Delta h1, A1 over A2, that is the final equation, we're ready to plug in numbers, the first one change, the height change in the first one is 20 centimeters or 0.2 meters and area1 is 1, and area2 is 4, so this is 0.2 divided by 4, which is 5 centimeters or 0.05 meters and I want to point out something, again, really important, the original was 10 but now instead of getting multiply 4 as it was here, it got divided by 4, okay? And now that you know this, by the way, you don't necessarily have to do this whole thing you can just know that it changes by the ratio, cool? So, remember I said here, that hydraulic lift multiplies a force, it's like magic? Well, it's not really magic, you gain more force, which is awesome, but you lose, you lose height change, you lose height change, meaning, yeah you're four times stronger but now this thing is only going to rise 1/4 of the way, okay? So, that means that you're going to have to push a very long distance here to get this thing to move a little bit but it's probably worth it depending on your case, okay? So, it reduces by the same factor, the factor here, in this example was a factor of 4, right? So, it grew by a factor of 4 and here it got reduced by a factor 4. So, it's not magic, it's a trade off, more force but less height gained as a result, okay? And in the last point I want to make here, I mentioned this briefly earlier multi drop lift problems will have cylindrical columns something like this and the area the area of a cylindrical column, the surface area of a cylindrical column is the area of a circle which is pi, R squared and I mentioned that volume is always area times height but then the volume of a cylinder, therefore, it is the area, which is this, pi, R squared, times the height. So, you should know this, you should know this, and you should know this when you are solving these questions, okay? That's it for this one, let's keep going.

Example #1: Hydraulic Lift / Proportional Reasoning

**Transcript**

Hey guys. So, let's check out this hydraulic lift example and this is a very straightforward example, I really want to nail this point that you can solve some of these very easily if you just know the ratio between the areas, let's check it out. So, it says you have a hydraulic lift is designed with cylindrical columns, one having double the radius of the other. So, let's draw this real quick. So, I've got little cylindrical columns there and one is double the radius, so I'm going to call this R1 is just R and then R2 will be 2R because it's double the radius, it says, both columns are capped with pistons of the same density and thickness, that's your standard language so that you know that the pistons basically don't have an impact on anything, they cancel each other out. So, looks like that and then it says, if you push on the thinner column with the force F. So, if you push with F. So, I'm going to say that if your force F1 has a magnitude of F, how much force will act on the other piston, so, how much force do you get here, cool? And I hope you remember that the way to start this is to say, hey the pressure on both sides is the same. So, to say P1 equals P2 therefore F1 equals, F1 over A1 is F2 over A2, this is because of course pressure is force over area. So, then I can write that F2 must be F1 times A2 over A1. Now, the areas here, the areas here are the areas of a circle because the surface area of a cylindrical column is going to be the area of the circle, pi, R squared, because it is cylindrical, cool. So, this means are going to rewrite this as pi, R2 squared, the radius of the second one, divided by pi, R1 squared, the radius of the first one, and I can cancel out the pi's and if you want you can even rewrite, you can factor out the square and it's going to look like this. So, it's proportional, your new force is proportional to the square of the ratio of the radii, that's sounds like of a mouthful, but if you have F here, which is the original force and then the second radius is double the first so it looks like this, second radius is double at first, so the R's cancel and you're left with 2 square, which is 4, and we're done, the answer is 4F. Now, there's a lot of math here, you could have solved this more simply by knowing that hey, if the radius is double and the area is pi, R squared, if you double the radius and the radius squared that means that the area is going to be four times greater, if the radius is double the area is quadruple, and if the area is quadruple that means that the new force is also going to be quadruple or four times greater the original force, okay? Double the radius means quadruple the area, which means you quadruple the force, you could have just done that as well, hopefully this served as a little bit of a review how to do the full solution but you could have been that quick also and if you remember from a previous video, I told you that if you, if the force becomes four times larger then the height difference or the height gain on the right side is going to become four times smaller, it's just the opposite of what happens. So, if the force becomes four times bigger then the height becomes four times smaller, that's it, okay? That's all you need to know now I'm going to show you how to solve this but at this point you could have already known this by just knowing that the amount that you're going to gain on this side here is going to be smaller by the same factor that the force gets multiplied by, but let's solve this real quick, and we solve this by knowing that the change in volume on the left, the amount of volume that goes down here is the same amount of volume that goes up here, I didn't draw that properly because I'm trying to move this quickly but the volume on the left is the same as the volume on the right and volume. Remember, is, volume if you remember is area times height, right? So, I can write A1, h1 or Delta h1 because the change in height equals A2, Delta h2 and we're looking for Delta h2. So, Delta h 2 is Delta h1, A1 over A2, all I've done is move the A2 to the other side of the equation and I now know that this area here, this area here is four times larger, we know this from over here, right? So, I can just say this is Delta h1, which we call just big H and I have an area and then I have an area that is 4 times that size. So, these cancel and you see how you end up with h over 4, okay? That's it for this one, let's keep going.

Example #2: Force to Lift a Car

**Transcript**

Hey guys. So, in this hydraulic lift example we're being asked to find the minimum force needed to lift a car, let's check it out. Alright, so it says, a hydraulic lift is designed with cylindrical columns having these radii right here, so I'm going to draw a standard hydraulic lift, looks something like this, the radius of the first one here, we're going to call this R1 is 20 centimeters or 0.3 meters and then this one is radius 2 is 2 meters and notice that radius 2 is 10 times larger than radius 1, cool? And you are going to push down here with a force F1 and that's what we want to know. So that you lift a car, I'm going to draw our ugly car here. So, that you can lift the car, lift car with it, okay? And we want to know how much force that is. Now remember, every hydraulic lift question or almost all of them are going to start with either the fact that the pressure on the left equals the pressure on the right or it's going to start with the fact that the change in volume on the left has to equal the changing volume on the right side, okay? And because we're looking for force and force has to do with pressure, this is the equation we're going to use. Remember, that pressure is force over area. So, we're going to immediately as soon as you write this the next step is going to write that P1 becomes F1 over A1 and p2 becomes F2 over A2 and here we're looking for F1, the input force, I can calculate the areas because I know the radii, right? Remember, area, when you have a cylindrical column it's going to be pi, R squared. So, if I have R, I have A, I can calculate that, what about F2? Well, F2 is the force required to lift the car, how much force you need to lift something? The amount of force they need to lift something, you should remember this, the amount of force to lift something is the weight of that something, the weight of the object. Now, you might be thinking, if I, if m, g is 100 and you push with 100 isn't that just going to cancel itself out? It is, technically you should push with 100.00001, right? You have to be just barely enough but this is kind of silly. So, we just set them equal to each other, okay? But don't get confused, don't think that just because they're equal to each other it's not going to move, the idea is that it is slightly more than that, but we use that amount, so hopefully that makes sense F lift will be replaced with m, g. So, this right here, will be m, g, okay? So, now I can solve for F1 by moving A1 to the other side. So, it's going to be F1 equals F2, which is m, g times A1, A1 goes to the top divided by A2, okay? And the mass of the car is 800, gravity, I'm going to use 10, just to make this easier, the first area, let's calculate the first area, it's going to be pi, R squared, I'm going to do this slowly here, and then here, pi, R squared R1, R2, we're going to cancel these two, 800 times 10 is 8,000, the first radius is 0.2, and the second radius is 2.0. Alright, so if you do this, if you do this, the best way to do 0.2 divided by 20, by the way, by 2, is to multiply both sides by 10. So, this becomes 2 over 20 and now you can easily see that this is just 1 over 10 okay, hopefully you get that or you can just do the calculator but this is going to be 8,000, 1 divided by 10 squared, if you square the top and the bottom you get 8000 divided by 100. So, now two zeros are going to cancel and you're left with 80, so this is the final answer, this means that you need a force of 80 Newtons. Now, check out what's happening here, if you apply a force of 80 Newtons, you're going to be able to lift this car even though it takes 800, I'm sorry, 8,000 Newton's to lift this car. Notice that this force here is 100 times greater than your input force and that should make sense because your radius was 10 times greater, let me write this out, the second radius was 10 times greater than the first radius. Remember, area is pi, R squared. So, if the radius is 10 times larger then the area is 10 squared times larger so the area is going to be 100 times larger and therefore the force will be magnified by a factor of 100, coo? That's it for this one, classic example in hydraulic lift, let's keep going.

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Concept #1: Pascal's Law and Hydraulic Lift

Example #1: Hydraulic Lift / Proportional Reasoning

Example #2: Force to Lift a Car

A hydrolic lift system is power by lowering a weight onto an "input pad" with an area of 100 cm2, transmitting the pressure through hydrolic fluid, which is then used to lift a 10 m 2 platform that a car rests on. If a 2000 kg car is on top of the platform, how much weight must be placed on the "input pad" to raise the car?

The hydraulic lift (the figure below) has a small piston with the surface A1= 0.039 m2 and a large piston with the surface A2 = 5.0 m2. F1 = 21 N. What force will the large piston provide if the small piston in a hydraulic lift is moved down as shown in the figure? (Express your answer to two significant figures.)

The hydraulic lift (the figure below) has a small piston with the surface A1= 0.039 m2 and a large piston with the surface A2 = 5.0 m2. F1 = 21 N. If the small piston is depressed a distance of Δy1, by how much will the large piston rise? (Express your answer to two significant figures.)

The hydraulic lift (the figure below) has a small piston with the surface A1= 0.039 m2 and a large piston with the surface A2 = 5.0 m2. F1 = 21 N. How much work is done in slowly pushing down the small piston if Δy1= 0.25 m? (Express your answer to two significant figures.)

A hydraulic lift is designed to raise a 900-kg car. The "large" piston has a radius of 36.0 cm and the "small" piston has a radius of 4.00 cm. Determine the minimum force exerted on the small piston to accomplish the task. (Express your answer to three significant figures.)

A crass host pours the remnants of several bottles of wine into a jug after a party. He then inserts a cork with a 2.00-cm diameter into the bottle, placing it in direct contact with the wine. He is amazed when he pounds the cork into place and the bottom of the jug (with a 14.0-cm diameter) breaks away. Calculate the extra force exerted against the bottom if he pounded the cork with a 120-N force.

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