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Density | 33 mins | 0 completed | Learn |

Intro to Pressure | 71 mins | 0 completed | Learn |

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

Pressure Gauge: Barometer | 13 mins | 0 completed | Learn |

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

Pressure Gauge: U-shaped Tube | 22 mins | 0 completed | Learn |

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Concept #1: Pascal's Law and Hydraulic Lift

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

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