Working Content> MacroModels > Fluids
5.2.2
Prerequisites
In our initial discussion of the concept of pressure we explicitly ignored gravity. In this webpage, we'll consider the effects and see that gravity has a significant impact on the behavior of fluids and of objects immersed in them.
Most of you have had lots of personal experiences immersing yourself in a fluid. We're almost always immersed in air, but its density is so low -- about 1 kg/m3 -- that we hardly notice it (unless we and it are moving at a significant relative velocity). When we immerse ourselves in water, which has a density of about 1000 kg/m3 -- comparable to our own -- then we personally can see effects of gravity on fluids.
Think about going swimming in a pool or lake. Do you float? Sink? Most people float, at least somewhat. What if you take a beachball filled with air and try to hold it under water? Does the ball sink? Float? Fly up out of the water? Get pushed down to the bottom? If you don't know what happens, find a ping-pong or tennis ball and hold it under water to see.
Although most of us know what happens, it's a bit difficult to reconcile this with the physics we are learning. We know that gravity pulls everything down. If I hold the beachball under water there is water on top of it. Gravity is pulling that water down. Why doesn't that push the ball down to add to the downward pull of gravity on the ball? Why does the ball fly up in the air?
Warning! Notational variation -- In chemistry and biology, density is often represented by the symbol "d". In physics, the standard notation for density is "ρ" (Greek letter "rho"). (Though this is also often used for charge density in physics as well as for mass density.) We will use that notation here since what matters to this analysis is the depth -- and we want to use "d" to represent that.
Pressure under gravity
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The way to understand what's happening to objects that live in water is to use Newton's theoretical framework to analyze the forces on a block of water in a container of still water. Let's assume that the water is open on the top to the air. The cylinder has a cross sectional area A.
From the Newtonian framework we know we can isolate a part of an object and consider the forces acting on it. Let's do that for the top disk of water, a thickness d (the depth), shown in lighter blue. (This is really no different from any of the other water -- this is just the part we are considering.)
There are three forces acting on the disk:
- the force of the air pushing down on top, p0A, where p0 is the air pressure;
- the weight of the disk due to gravity, W = ρgV = ρgdA; density x volume x g.
- the force of the water beneath the disc pushing up on it, pA, where p is the pressure of the water.
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Balancing the up forces against the down forces, we get
pA = p0A + ρgdA;
In words: the upward force from the water underneath the disk balances the weight of the disk plus the force of the air pushing down on the top. We see that there is an A in each term so we can cancel it to get
p = p0 + ρgd.
The pressure in the water is equal to the pressure at the top and it increases with the depth d.
This makes good sense since the farther down you go, the more water that's above you has to be held up. The water "trying" to fall squeezes the water beneath it and increases the pressure.
Archimedes' Principle
Well that seems interesting, but what does it have to do with the beach ball? The implication is actually quite striking.
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Suppose we immerse a small cylinder of height h and area A in the water at a depth d as shown in the figure at the right. The water on the top exerts a pressure downwards on the object, and the water on the bottom exerts a pressure upwards on the object. But since the bottom is deeper than the top, the pressure at the bottom is greater than the pressure at the top. (The sideways forces all cancel.)
The result is a net upward force! The result can be expressed in an interesting way. Taking up to be positive, if we write the balance of forces, we get
There's a lot of cancellation, but the result is the density of the water times the volume of the object times the gravitational field, g. Since the density of water times the volume of the object is the mass of the water that would have been in the place where the object is, the result is:
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Archimedes' Principle: When an object is imbedded in a fluid, it feels an upward (Buoyant) force that is equal to the weight of the displaced fluid.
This seems a bit strange. Why should the displaced fluid matter? It isn't there anymore? Did it just come out of the math this way as an accident? Or can we make sense of it?
Consider an arbitrary object (a rock) that we are going to place underwater as shown in the figure at the right below.
If we consider the shape of the object in the water before the water was there -- like an imaginary bag of water being held up by the swimmer at the left -- that water would just sit there, despite having weight. That tells us that the forces being exerted on that "imaginary bag" by the water outside is exactly equal to the weight of the water inside. The pressure must grow with depth in order to stabilize the weights of the water above it. Since the water outside the rock doesn't "know" what's inside the surface, all it can do is exert the same forces it would exert on the water that used to be there. The result is an upward force equal to "the weight of the water displaced"!
Workout: Archimedes principle
Joe Redish 10/23/11
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