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## Physics library

### Course: Physics library > Unit 9

Lesson 1: Density and Pressure# 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.

## Want to join the conversation?

- 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)

- 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(5 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.(5 votes)

- Can some water be magnetic? I'm just wondering.(4 votes)
- conservation of energy tells us that the work we put into the system is equal to the work that we get out of the system.and we also know that gas is compressible.so if we put work into gas how does it come out?(3 votes)
- You've misstated conservation of energy. It doesn't say anything at all about work in and work out. If you do work on a gas, the gas can just get hotter.(7 votes)

- 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

F1=hose force

A1=hose area

F2=nozzle force

A2=nozzle area

A2< A1

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)

- At around3:25, he establishes that P1V1=P2V2. I was just wondering if this had anything to do with Boyle's gas law?(3 votes)
- PV = nRT. If T is constant, then the ideal gas law becomes PV = constant, which we can say another way as P1V1 = P2V2. In other words, Boyle's Law and Charles Law are both special cases of the ideal gas law.(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)

## Video transcript

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.