AP®︎/College Physics 2
Pressure and Pascal's principle (part 2)
Sal finishes the calculation of work to determine the mechanical advantage in a U-shaped tube. He also explains pressure and Pascal's Principle. Created by Sal Khan.
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- I don't get something at10:00. It doesn't intuitively make sense to me that on one side the force is lesser and on the other its more, shouldn't force be conserved? I know if I look at the mathematical equations, it makes sense there, but how could the force be larger on one side, even though its being applied on the same amount of volume, and just a different surface area? Shouldn't it be smaller since its on a larger surface area?(109 votes)
- In this system, and for that matter all systems, it's the work that is conserved. work being force x distance means that the larger piston will travel a shorter distance than the smaller piston with a greater amount of force. It took me several times reviewing this video to understand the conclusion that he was coming to. he probably should have put that in the end of the video but we are all only human.(178 votes)
- Paradox: Take two infinitely powerful pistons and place each piston onto the ends of an indefinitely resistant cylindrical container full of liquid. What would happen to the system? Would the liquid turn to gas from all the pressure, or would the system be at a deadlock?(21 votes)
- that wouldn't really work, they would be at a deadlock( assuming the pistons have the same force and area). Think about it, a solid or liquid becomes a gas after it has enough energy to break the bonds between molecules. Compressing a gas may make it a solid OR liquid eventually, but compressing a solid, well the solid stays a solid, and same for liquid.(21 votes)
- Is this how car brakes work? One presses on the brake pedal and the pressure of the pressing is translated to all the brake pads?(26 votes)
- ya!when you press the break you are actually pushing the small piston and when this happens the pressure is transmitted equally and undiminished in all directions and this will lead the liquid level to rise leading to the uplift of the other piston leading to the transmit of the pressure to all the break pads(5 votes)
- at 6.05 it is told that balloon expands uniformly but in the real situation sometime wen we blow a balloon...just one part of it expands first i.e the right side or the left side of it and then the whole balloon starts expanding uniformly. why does this happen?(9 votes)
- I'm not sure, but it may be because the thickness (and thus the k-constant in N/m) of the elastic surface of the balloon is not uniform. This would make it so that it takes less effort to stretch one side than the other, making that one side expand first.(17 votes)
- At around9:58the denominator is 4m when it should be 4 m^2, but I suppose it doesn't matter since we all get it anyway.(12 votes)
- You are right. Thank you for pointing out this mistake because it might help other confused people(5 votes)
- Can some water be magnetic? I'm just wondering.(4 votes)
- This Veritasium video might help youi.
- This concept seems to make sense to me, except when i think about a hose. Specifically a hose, with a nozzle, or tapered end where the water comes out. The area is smaller than the input side. So if the pressure going in (larger area) is equal to the pressure coming out (smaller area) then the force of the water coming out is less than the force of the water pushing in. This seems extremely counterintuitive to me. Can someone please clarify this for me? Thanks for the help!(4 votes)
- P1= P2=F1/A1=F2/A2
Therefore, F2< F1. The force at the nozzle is less than the force in the hose.
Momentum is what is increasing at the nozzle. Momentum is mass x velocity. Momentum can change by varying the mass or velocity, or both. The nozzle example increases the momentum by increasing the velocity of the fluid. Consider boxing: a heavyweight has more mass than a lightweight and can transfer more momentum than a lightweight, given the same force and distance of the punch.(5 votes)
- Is this how car brakes work? One presses on the brake pedal and the pressure of the pressing is translated to all the brake pads?(3 votes)
- It's similar. Your foot presses on the pedal, which then activates a hydraulic mechanism, that makes pads apply pressure on the wheels. Friction creates heat and reduces the kinetic energy of the wheel - car slows down.(6 votes)
- what if we had two output holes? Will the force be equally distributed among the two openings or the force will be greater in one and lesser in other?(3 votes)
- if they are at the same level, the pressure would be the same. The force = P x A. as long as the pressure does not change then the Force will be the same from both holes(3 votes)
- This just blowed my mind
What if we had a machine like this and in place of the piston2 u would have an elevator(or some kind of box in which people would fit in ,)
So if you apply a force on piston1 you would get double the force you applied
Wouldn't that save energy(2 votes)
- Good question, but the answer is NO.
Because work is force times distance. You need to push the piston twice the distance which the second one is moving upward (consider the volumes of liquid!).
By the way: The same phenomenon applies to a lever or a lifting block.(2 votes)
Welcome back. To just review what I was doing on the last video before I ran out of time, I said that conservation of energy tells us that the work I've put into the system or the energy that I've put into the system-- because they're really the same thing-- is equal to the work that I get out of the system, or the energy that I get out of the system. That means that the input work is equal to the output work, or that the input force times the input distance is equal to the output force times the output distance-- that's just the definition of work. Let me just rewrite this equation here. If I could just rewrite this exact equation, I could say-- the input force, and let me just divide it by this area. The input here-- I'm pressing down this piston that's pressing down on this area of water. So this input force-- times the input area. Let's call the input 1, and call the output 2 for simplicity. Let's say I have a piston on the top here. Let me do this in a good color-- brown is good color. I have another piston here, and there's going to be some outward force F2. The general notion is that I'm pushing on this water, the water can't be compressed, so the water's going to push up on this end. The input force times the input distance is going to be equal to the output force times the output distance right-- this is just the law of conservation of energy and everything we did with work, et cetera. I'm rewriting this equation, so if I take the input force and divide by the input area-- let me switch back to green-- then I multiply by the area, and then I just multiply times D1. You see what I did here-- I just multiplied and divided by A1, which you can do. You can multiply and divide by any number, and these two cancel out. It's equal to the same thing on the other side, which is F2-- I'm not good at managing my space on my whiteboard-- over A2 times A2 times D2. Hopefully that makes sense. What's this quantity right here, this F1 divided by A1? Force divided by area, if you haven't been familiar with it already, and if you're just watching my videos there's no reason for you to be, is defined as pressure. Pressure is force in a given area, so this is pressure-- we'll call this the pressure that I'm inputting into the system. What's area 1 times distance 1? That's the area of the tube at this point, the cross-sectional area, times this distance. That's equal to this volume that I calculated in the previous video-- we could say that's the input volume, or V1. Pressure times V1 is equal to the output pressure-- force 2 divided by area 2 is the output pressure that the water is exerting on this piston. So that's the output pressure, P2. And what's area 2 times D2? The cross sectional area, times the height at which how much the water's being displaced upward, that is equal to volume 2. But what do we know about these two volumes? I went over it probably redundantly in the previous video-- those two volumes are equal, V1 is equal to V2, so we could just divide both sides by that equation. You get the pressure input is equal to the pressure output, so P1 is equal to P2. I did all of that just to show you that this isn't a new concept: this is just the conservation of energy. The only new thing I did is I divided-- we have this notion of the cross-sectional area, and we have this notion of pressure-- so where does that help us? This actually tells us-- and you can do this example in multiple situations, but I like to think of if we didn't have gravity first, because gravity tends to confuse things, but we'll introduce gravity in a video or two-- is that when you have any external pressure onto a liquid, onto an incompressible fluid, that pressure is distributed evenly throughout the fluid. That's what we essentially just proved just using the law of conservation of energy, and everything we know about work. What I just said is called Pascal's principle: if any external pressure is applied to a fluid, that pressure is distributed throughout the fluid equally. Another way to think about it-- we proved it with this little drawing here-- is, let's say that I have a tube, and at the end of the tube is a balloon. Let's say I'm doing this on the Space Shuttle. It's saying that if I increase-- say I have some piston here. This is stable, and I have water throughout this whole thing. Let me see if I can use that field function again-- oh no, there must have been a hole in my drawing. Let me just draw the water. I have water throughout this whole thing, and all Pascal's principle is telling us that if I were to apply some pressure here, that that net pressure, that extra pressure I'm applying, is going to compress this little bit. That extra compression is going to be distributed through the whole balloon. Let's say that this right here is rigid-- it's some kind of middle structure. The rest of the balloon is going to expand uniformly, so that increased pressure I'm doing is going through the whole thing. It's not like the balloon will get longer, or that the pressure is just translated down here, or that just up here the balloon's going to get wider and it's just going to stay the same length there. Hopefully, that gives you a little bit of intuition. Going back to what I had drawn before, that's actually interesting, because that's actually another simple or maybe not so simple machine that we've constructed. I almost defined it as a simple machine when I initially drew it. Let's draw that weird thing again, where it looks like this, where I have water in it. Let's make sure I fill it, so that when I do the fill, it will completely fill, and doesn't fill other things. This is cool, because this is now another simple machine. We know that the pressure in is equal to the pressure out. And pressure is force divided by area, so the force in, divided by the area in, is equal to the force out divided by the area out. Let me give you an example: let's say that I were to apply with a pressure in equal to 10 pascals. That's a new word, and it's named after Pascal's principle, for Blaise Pascal. What is a pascal? That is just equal to 10 newtons per meter squared. That's all a pascal is-- it's a newton per meter squared, it's a very natural unit. Let's say my pressure in is 10 pascals, and let's say that my input area is 2 square meters. If I looked the surface of the water there it would be 2 square meters, and let's say that my output area is equal to 4 meters squared. What I'm saying is that I can push on a piston here, and that the water's going to push up with some piston here. First of all, I told you what my input pressure is-- what's my input force? Input pressure is equal to input force divided by input area, so 10 pascals is equal to my input force divided by my area, so I multiply both sides by 2. I get input force is equal to 20 newtons. My question to you is what is the output force? How much force is the system going to push upwards at this end? We know that must if my input pressure was 10 pascals, my output pressure would also be 10 pascals. So I also have 10 pascals is equal to my out force over my out cross-sectional area. So I'll have a piston here, and it goes up like that. That's 4 meters, so I do 4 times 10, and so I get 40 newtons is equal to my output force. So what just happened here? I inputted-- so my input force is equal to 20 newtons, and my output force is equal to 40 newtons, so I just doubled my force, or essentially I had a mechanical advantage of 2. This is an example of a simple machine, and it's a hydraulic machine. Anyway, I've just run out of time. I'll see you in the next video.