Pascal's Law & Hydraulic Lift

Hey, guys. So in this video, we're gonna talk about Moscow's law and its most important application in physics, which is the hydraulic lift. Let's check it out. Alright, so Pasko's law states and you 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 contraction like this and I apply a little bit of pressure here. P well, that pressure is going to be 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. Um, I have other definitions that I think I would be more useful in problem solving a second related rule. Um, it isn't like a law named after anyone, but you should know that pressure in the liquid. So do you think pressure the 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, Um, 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 row. G H and H is depth. So pressure at the bottom over here on Lee depends on the pressure at the top. It only depends on the density of the liquid on gravity and in depth. Right. So if you have a Rupp 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. Um, if you have the same liquid everywhere, density is the same. Um, gravity is going to be the same for all of them because they're all sitting next, 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 on Lee on on depth. So I could 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 heights. 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 points than this point. It has to do with depth. So you measured from the top down. And if the depth is the same and all of 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 0.2 was referring to. 0.3 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 liquids 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 Arab here right then. That means that there is going to equally push on all of these. But it could also be a difference. Gas. That's not air. It's going to equally pushing all of these, and it's gonna 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 gonna end up on that fat column. Um, but that's because they have to balance itself out. One sort of way to remember this is you might have heard this phrase water seeks its own level. It just means it's gonna balance itself out. So you should know these three rules. Um, they will be helpful throughout the rest. So now let's talk about the past. I'm sorry. The hydraulic lift, which is the most important application of Moscow'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, um, the first thing is the paschal lift. A hydraulic lift is going to be built out of two columns of liquid that have different sizes. Right. So you can imagine these air 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 on. But the first thing we're gonna do is we're gonna put some liquid in here, and we're gonna have no pistons. So So just liquid. And because of this rule over here that we just discussed, if you pour some liquid, the liquid is going toe level out, and it doesn't matter what the liquid is, it makes no difference. Okay, so you just have liquid liquid levels out. 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 the same. Also another way to think about why the pressure is the same is because pressure depends on heights, 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 heights. Eso the pressures would then be the same. However you want to remember it, but that's not enough. So first we put liquids and then we're gonna put pistons. And what pistons are is just a little cap that goes on top of it. So we're gonna put a little cap here, and you put that cap is like a platform so you can put objects on top of it. Otherwise, if you were to put it, let's say a car hydraulic lifter used to lift cars. Then the car was 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 things 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 gonna be unbalanced. So, for example, if you have the same thickness, they're gonna just stay like this. But if this one here is thicker, then it's gonna be heavier overall, right, or it's gonna be It's gonna be heavier per area. So it's gonna do this so they have to have the same thickness so that they have the same pressure. So the this one here, remember, pressure is forced over em 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 gonna be mg over a now. The piston on the right side is heavier, but it also has a bigger area. And those two things would cancel itself out so that you have the same pressure on both sides by the two pistons. Okay, so once heavier, but it's sitting on a larger area, cancels itself out because of two pistons have exactly the same amount of pressure that they're applying. They basically 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 the 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 heights 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 that pistons, they're gonna have the same thickness and they don't affect they don't make any any any. They don't have any impact on the problem itself. If you push this down with a Force F one, if you remember, pass cows 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 pushed the other side up. Another way to think about this is this is a contained liquid contained, um, 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 gonna 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 gonna go up a much because the volume has to be the same. But this is wider, so it's not gonna be a stall. Okay, so that's actually the first thing I want to talk about, Which is this change in volume over here? Changing volume on the left has to equal to change in volume on the rights, so the right side isn't gonna 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 heights. 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 thing is the same liquid, and the pressure within the same liquid is equal at the same heights. These two points right here, this red dot in this red dot are the same liquid, the red liquid, and they are at the same heights. Therefore, I can write that p one equals p chew. 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 gonna 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 gonna use these two very, very important rules over here. So hydraulic lift is designed with columns having areas one and four. All right, so I'm gonna draw. It's always gonna look like that. So it's something like this, and we're going to say that the area of this column here is one. So let's call this area one. And in this area here, area to is going to be 41 square meter, four square meter. By the way, this is the surface area. It's the area off the sort of top of the piston, if you will. Okay, and it says pistons of equal thickness or placed on top. By the way, if the problem doesn't say pistons of same thickness, right, if he doesn't say of equal thickness, you can assume that Alright on. Bennett says you push down with 10 on the smaller column. So you're gonna push here? This is gonna be we're gonna call. This F one equals 10 and it causes it to lower by 20 centimeters. So when you push it drops the heights. Let's call this heights one or change in height, one equals 20 centimeters, which is 200.2 m. Okay. And we wanna know some stuff we wanna know first. How much force acts on the right piston. So remember, when you 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? F two that goes up on the other side. Cool. And then the second question is we want to know how much does the right pissed and rise by? So this this one here goes down by points to so this one here has to go up by some amounts. Remember, it's gonna be less than point to, but we're gonna figure out how to calculate that. So we wanna know what is the change in height on the right side. Cool. So how do we do this? So we're gonna do We're gonna solve both of these using these two equations over here. So first, remember, force is related to pressure. Right? So I'm gonna write that pressure one pressure one equals pressure to on the two sides. That's how you're always gonna start these questions. And I'm gonna rewrite pressure, because, remember, pressure is force over area. So I'm gonna rewrite P one as f one over a one and pick us F two over a two. We're looking for F two, So all I gotta do is move some stuff around, so f two is gonna be f one times. I'm gonna move the A up here so it's gonna be a two over a one, okay? And we are ready to plug in numbers. The force forces 10. A two is four and a one is one, so this becomes 40. Okay, so this is really important. Notice what happened here, which is the original Force 10 became four times greater because the second column is four times the area of the first column. So this is super important. What happens here is the hydraulic lift. Hydraulic lifts will multiply the force by a factor of area to over area one. The ratio of the areas serves as a multiplier 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 on this thing. And if the if the second column is much bigger than the first column, you get a much bigger force. It's kind of like magic. You could 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. There's a few a few times where you 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 heights and by the way, this is the volume. This is the general volume equation meeting. This works for every single shape. Volume is always the cross sectional area times the height of it. So now, instead of starting with P one equals P two. I'm going to start with the volume equation up here. This one. Okay, so I'm gonna start with going to start with Delta View One equals Delta V two. Okay, so let's do that. And then similarly, we're going to expand using this equation here. So we're gonna write a one. Delta H one equals a two Delta H two and we're looking for Delta H two. How much does it rise by? So Delta H two equals a one delta H one divided by Delta. I'm sorry by Oh, let me let me go back a little bit. Here. It's gonna delta H one a one over A to that is the final equation. We're ready to plug in numbers. The first one changed the height changing. The first one was 20 centimeters, 1.2 m and area one is one in area two is four. So this is 0.2 divided by four, which is five centimeters or 0.5 m. And I want to point out something again. Really important. The original was 10. But now, instead of getting multiplied by four as it was here, it got divided by four. Okay, And now that you know this, by the way, you don't necessarily have to do this whole thing. You could just know that it changes by the ratio. Cool. So remember I said here that hydraulic lifts multiply 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 heights, change. Meaning? Yeah, you're four times stronger. But now this thing is only going to rise one quarter of the way. Okay, So that means that you're gonna 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 four, right? So it grew by a factor for and here it got reduced by a factor of four. So it's not magic. It's a trade off. More force, but less heights gained as a result. Okay. And in the last point I wanna make here. I mentioned this briefly earlier. Most hydraulic lift problems will have cylindrical columns. Something like this in 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 square. And I mentioned that volume is always area times heights, 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're solving these questions. Okay, That's it for this one. Let's keep going.

Hey, guys. So let's check out this hydraulic lift example, and this is a very straightforward example. I really wanna 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. Eso it says here hydraulically to design with cylindrical columns, one having double the radius of the other. So let's draw this real quick. So we got a little cylindrical columns there and one is double the radius. I'm gonna call this. Our one is just our and then our two will be to our because it's double the radius, says both columns, Air capital, pistons of the same density and thickness. That's just standard language so that you know that the pistons basically don't have an impact. Um, on anything they cancel each other out. So it 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 your force F one 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 P one equals p two. Therefore, F one equals F one over a one is F two over a To this is because, of course, pressure is force over area. So then I can write that f two must be f one times a two over a one. Now the areas here, the areas here are the areas of a of a circle because the surface area of a cylindrical, um, column is going to be the area of a circle pie are square because it is cylindrical quote. So this means you're gonna rewrite this as pi r two square the radius of the second one divided by pi r one square the radius of the first one and I can cancel out the pies and if you want, you can even rewrites. You can factor out the square, and it's gonna look like this. So it's proportional. Your your new force is proportional to the square of the ratio of the radio. I That sounds like 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 rate This is double the first. So the ours cancel and you left the two square, which is four. And we're done the answers for F Now there's a lot of Matthew you could have solved this were simply by knowing that. Hey, if the radius is double in in the area is pi r squared. If you double the radius and the radius is 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 quadrupled, that means that the new forces also going to be quadruple or four times greater the original force. Okay, double the radius means you 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, um if the force becomes four times larger than 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 gonna show how to solve this, but at this point, you could have already known this. But you by just knowing that the amount of that you're gonna get on the side here is gonna 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 of 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 a one h one or Delta H one because it's the change in height equals H two Delta H two and we're looking for a Delta H two. So Delta H two is Delta H one A one over a two. All I've done is movie A to 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 H one, which we call this Big H and I have an area, and then I have an area that is four times that size. So these can so and you see how you end up with HPE over four. Okay, that's it for this one. Let's keep going.

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. All right. So it says hydraulic lift is designed with cylindrical columns having these radio here, So I'm gonna draw a standard hydraulic lift. Looks something like this. Um, the radius of the first one here. We're gonna call this. Our one is 20 centimeters or 200.2 m. And then this one is radius two is 2 m and notice that radius two is 10 times larger than radius one. Cool. And you are going to push down here with a Force F one, and that's what we want to know. So that's you Lift a car. I'm gonna draw ugly car here so that you can lift the car lift car with it, okay? And we wanna know how much force that ISS 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 and left has to equal the change in 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 gonna immediately As soon as we write this, the next step is gonna be the right. That p one becomes F one over a one and p two becomes F two over a two. And here we're looking for F one. The input force. I can calculate the areas because I know the radio. I right. Remember area. When you have a cylindrical columns, it's going to be pi r Square. So if I have are I have a I can calculate that. What about F two? Well, F two is the force required to lift a car? How much force do you need to lift something? The amount of force that you need to lift something. You should remember this theme Mount 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 you push with 100 isn't that just gonna cancel itself out. It is. Technically, you should push with 100.1 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 each other, it's not going to move. The idea is that it's the slightly more then that's what 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 F one by moving a one to the other side. So it's gonna be f one equals, um F two, which is mg times a one. A one goes to the top, divided by a to 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 is gonna be pi r squared. I'm gonna do this slowly here and then here. Pi r square are one or two. We're gonna cancel these two 800 times. 10 is 8000. The first radius is 0. and the second radius is 2.0. All right, So if you do this, if you do this, the best way to do 0.2 divided by 20 by the way, by true is to multiply both sides by 10. So this becomes to over 20. And now you can easily see that this is just 1/10. Okay, hopefully you get that, or you could just doing a calculator. But this is gonna be 8000 one divided by 10 square. If you square the top in the bottom, you get 8000 divided by 100. So now 20 they're gonna 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 gonna be able to lift this car even though it takes 800. I'm sorry. 8000 Newtons toe, lift this car. Notice that this force here is 100 times greater then your input force. And that should make sense because your radius was times greater let me write this south. The second radius was 10 times greater than the first radius. Remember, area is Pi R Square, so if the radius is 10 times larger than the area is square times larger, so the area is going to be 100 times larger and therefore the force will be magnified by a factor of 100. Cool. That's it for this one. Classic example in hydraulic lift, let's keep going.