Fluid Mechanics for AP Physics B
Practice problems for these concepts can be found at:
Pressure and Density
A fluid, just like a gas, exerts a pressure outward in all directions on the sides of the container holding it. The formula in the box tells us that the pressure exerted by a fluid is equal to the amount of force with which it pushes on the walls of its container divided by the surface area of those walls. Recall that the standard units of pressure are newtons per square meter, which are also called pascals.
Besides pressure, the other important characteristic of a fluid is its density, abbreviated as the Greek letter ρ(rho).
As you could guess, the standard units of density are kg/m3. You probably learned about density in sixth grade science, so this should be a familiar topic for you. But make sure you've memorized the definition of density, because it's critical for studying fluids.
It is useful to know the density of water, just because water is such a common fluid. So let us ask you, "What is the density of water?"
You said, "1," right? And being the brilliant physics ace that you are, you quickly put some units on your answer: "1 g/cm3." But look back at our definition of density!
The standard units of density are kg/m3, so your answer would be more useful in these units. The best answer, then, is that water has a density of 1,000 kg/m3.
Because water is so ubiquitous, the density of a substance is often compared to water; for instance, alcohol is eight tenths as dense as water; lead is eleven times as dense. We call this comparison the specific gravity of a substance—lead's specific gravity is 11 because it is 11 times more dense than water. So its density in standard units is 11,000 kg/m3.
Density and Mass
When dealing with fluids, we don't generally talk about mass—we prefer to talk about density and volume. But, of course, mass is related to these two measurements: mass = density · volume.
So the mass of any fluid equals ρV. And because weight equals mg, the weight of a fluid can be written as ρVg.
Pressure in a Static Fluid Column
Consider a column of water in a container, such as the one shown in Figure 19.1.
The water has significant weight, all pushing down on the bottom of the container (as well as out on the sides). So we expect water pressure to increase as the water gets deeper. Sure enough, the pressure due to the column of fluid of density ρ is given by this formula:
In this formula, P0 is the pressure at the top of the column, and h is the vertical height of the column. This formula actually makes intuitive sense—the pressure at the bottom of the column should equal the pressure at the top plus the pressure due to all the water in the column. And the pressure of the water in the column equals force divided by area; the force is the weight of the water (ρVg), and the area is the surface area of the base of the column. If you divide the volume of a column of water by its crosssectional area, you're just left with height. So there we have it—pressure due to a column of fluid equals the pressure at the top of the column, P0, plus the weight of the fluid divided by area, ρgh.
Note that h is a vertical height, and it does not depend on the width of the container, or even the shape of the container. A town's water tower is usually located at the highest available elevation, because the water pressure in someone's sink depends only on the altitude difference between the sink and the tower. The horizontal distance to the tower is irrelevant.
P0 represents the pressure at the top of the column, which is often atmospheric pressure. But not always! For example, if you put your finger on top of a straw and lift up some water, the pressure at the bottom of the straw is atmospheric (because that part of the water is open to the atmosphere). So, the pressure at the top of the column P0 must be less than atmospheric pressure by P = P0 + ρgh.
The pressure P = P0 + ρgh is called the absolute pressure of a fluid. Gauge pressure refers to just the ρgh part of the absolute pressure. These are terms to know because they might show up on the AP exam.
Let's start by calculating the water pressure at the bottom of the container: P = P0 + ρgh. P0 is the pressure at the top of the container, which we'll assume is atmospheric pressure (atmospheric pressure is available on the constant sheet). And h is the depth of the cooler, 70 cm, which in standard units is 0.70 m.
- P = (1 × 105 N/m2) + (1,000 kg/m3)(10 N/kg)(0.70 m)
- P = 107,000 N/m2
This pressure acts outward on the container everywhere at the bottom. This means the pressure pushes down on the container's bottom, and out on the container's sides near the bottom. So this pressure pushes on the plug.
How do we get the force on the plug? Since pressure is force/area, force is just pressure times area (and the area of the plug is πr2, where r is the radius of the plug, which is 1 cm):
- F = 107,000 N/m2• π(0.01 m)2 = 34 N.
That's about 6 pounds of force—that plug had better be wedged in well! This is why you've usually got to push hard on the button at the bottom of a full water cooler to get anything out—you're opposing several pounds of force.
Practice problems for these concepts can be found at:
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