Kamis, 05 Mei 2011

Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas

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Today we're going to change topics.

I'm going to talk to you about fluids, hydrostatic pressure and barometric pressure.

If, for now, we forget gravity and I would have a compartment closed off and filled with a fluid--

could be either a gas or it could be a liquid--

this has area A, here--

and I apply a force on it in this direction, then I apply a pressure.

Pressure is defined as the force divided by area--

has units newtons per square meter which is also called pascal.

One newton per square meter is one pascal.

Now, in the absence of gravity, the pressure is, everywhere in this vessel, the same.

And that is what's called Pascal's principle.

Pascal's principle says that the pressure applied to an enclosed fluid is transmitted undiminished to every point in the fluid and to the walls of the container.

Keep in mind, pressure is a scalar, it has no direction.

Force has a direction and the force exerted by the fluid on anything--

therefore also on the wall--

must be everywhere perpendicular to the wall, because if there were any tangential component, then the fluid would start to move.

Action equals minus reaction, so it starts to move and we are talking here about a static fluid.

So if I take any element--

I take one here at the surface, little element delta A and the force must be perpendicular to that surface, delta F, and so delta F divided by delta A--

in the limiting case for delta A goes to zero--

is, then, that pressure P.

This has some truly amazing consequences which are by no means so intuitive.

This is the idea of an hydraulic jack.

I have here a vessel which has a very peculiar shape.

Ooh, ooh, an opening here.

And let there be here a piston on it with area A1 and here one with area A2.

It's filled with liquid everywhere and I apply here a force F1 and here a force F2.

So the pressure that I apply here is F1 divided by A1.

So according to Pascal, everywhere in the fluid, that pressure must be the same.

For now, I just assume that the effect of gravity, which I will discuss shortly, doesn't change the situation very significantly.

But I will address the gravity very shortly.

So the pressure, then, will be the same everywhere, but the pressure due to this side is F2 divided by A2...

and so the two must be the same, if the liquid is not moving.

So what that means is that if A2 over A1 were 100, it means that this force could be a hundred times less than that one.

In other words, I could put on here a weight, a mass of ten kilograms, and here I could put 1,000 kilograms and it would be completely in equilibrium.

That's not so intuitive.

This is used in all garages.

What they do is, they put on top of this--

if I blow that up here, so this is this platform, there's a rod here and on top of it is a car.

And someone pushes here and then this goes up.

The car goes up.

If I push here with a force a little bit more than ten kilograms--

so that would be 100 newtons--

this level would go up.

And so your first thought may be, "Gee, isn't that a violation of the conservation of energy? Am I not getting something for nothing?" Well, not really.

Suppose I push this down over a distance d1, then the amount of fluid that I displace--

that is, the volume, is A1 times d1.

That fluid ends up here.

So this one will go up over a distance d2.

But the same amount of fluid that leaves here adds there.

In other words, Al d1 must be A2 times d2.

Now, if the force here is a hundred times less than the force there, the work that I am doing on the left side is F1 times d1.

If the force here is a hundred times less than that, the distance that I move is a hundred times larger than d2, because A2 over A1 is 100.

And so F1 d1 will be the same as F2 d2--

100 times lower force but over 100 times larger distance, and so the product is the same.

So the work that I do when I push this down I get back in terms of gravitational potential energy by lifting the car.

So if I wanted to move the car up by one meter and if the ratio is 100 to one I would have to move this down by 100 meters.

That's a little bit impractical so these hydraulic presses are designed in such a way that you can just jack it like that and every time that you bring it up, that liquid flows back in again into this side of the hydraulic jack.

But, indeed, you will have to go effectively 100 meters, then, for that car to go up by one meter if the ratio is 100 to one.

Now, gravity, of course, has an effect on the pressure in the fluid.

If you go down into the oceans, we know that the pressure will go up, and that is the result of gravity.

And I would like to derive the pressure increase.

Let this be the direction of increasing y, and I choose a liquid element, so this is in the liquid itself.

I can choose it in any shape that I want to.

I just take a nice horizontal slab.

And this is area A so the bottom is also A.

And let this be at height y plus delta y and this is at height y.

And the pressure here is P y plus delta y and the pressure here is P of y.

And this object has a mass, delta m, and the liquid has a density, rho, which could be a function of y--

we will leave that open for now.

And so this mass--

the mass that I have here--

is the volume times the density.

And the volume is A--

this area--

times delta y, and then times the density, which may be a function of y.

So if now I put in all the forces at work here, there is gravity, which is delta m times g in this direction.

Then I have a force upwards due to the pressure of the fluid.

That's what we want to evaluate.

It's always perpendicular to the surfaces.

We talked about that earlier.

So in this side, it comes in like this and here it comes in like this, the force.

From the bottom, it comes in like this and from the top, I call this F2.

I only consider the vertical direction, because all forces in the horizontal plane will cancel, for obvious reasons.

So, now, there has to be equilibrium.

This fluid element is not going anywhere--

it's just sitting still in the fluid.

And so I now have that F1--

which is in this direction--

minus F2 minus delta mg must be zero.

Only then is the fluid element in static equilibrium.

But F1 is this pressure times the area--

so that is P at level y times the area, and F2 is P at level y plus delta y times the area, minus delta m is A delta y times rho, so I get minus A times delta y, which could be a function...

rho could be a function of y times g, and that equals zero.

Notice I lose my area.

I'm going to rearrange this slightly and divide by delta y.

And so I get that P at the level y plus delta y minus the pressure at level y divided by delta y equals--

if I switch that around, so I bring this to the other side--

equals minus rho y times g.

And if I take the limiting case of this--

for delta y goes to zero--

then we would call this dP/dy.

And this tells you that when you go to increasing values of y, that the pressure will go down, it's a minus sign.

Very natural.

If you go with decreasing values of y, then the pressure will go up.

And we call this hydrostatic pressure.

So it's due to the fact that there is gravity; without gravity, there is no hydrostatic pressure.

Now, most fluids, most liquids are practically incompressible.

In other words, the density of the liquid cannot really change.

And so therefore, you could remove this and simply always use the same density.

It's exceedingly difficult.

It takes horrendous forces and pressures to change the density of a liquid, unlike that of a gas.

A gas is compressible and you can very easily change the density of gas.

So liquid is incompressible.

If I have here a piston and I have here a liquid and I put a force on here, it would be impossible for me to make that volume smaller--

even by a fraction of a percent, it would be impossible.

If this, however, were gas, then it would be very easy for me to push that in and to change the volume, make the volume smaller and thereby make the density of the gas go up.

If I took a sledgehammer and I would hit a plastic pillow, just bang, and the pillow was filled with air, it acts like a cushion and I could squeeze it.

If I hit the sledgehammer on the marble floor, I could not squeeze it and the force on the marble floor and on the hammer would be way higher, because I don't have this cushion action.

If I take a paint can--

and we have one here, we have two--

and this paint can is filled to the brim with water and another one is filled with air and I hit it with a sledgehammer, then this acts like a cushion.

This one, however, doesn't want the volume to be decreased, so the force, like on the marble floor, would be way higher.

But remember that force divided by area is pressure, and according to Pascal, that pressure propagates undiminished in the whole fluid.

And so if I would shoot a bullet through here, then I get a huge force--

extremely small area of the bullet.

And so the pressure inside the liquid would go up enormously, and the can might explode, provided that it's really filled to the brim with water, because if there is air left, then you have this cushion action.

Now, I don't remember whether there's air in here or whether there's air in there.

I'll leave you to decide.

So we'll fire a bullet from this side, and then we'll see which can explodes and which does not.

And the one that doesn't is the one that has air in it, and the one that explodes...

has the water in it, provided that we really filled it to the brim.

Oh, boy, there's still something in there.



Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

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