Fluid mechanics questions on the AP Physics B Exam cover four major topics. Two of these deal with static fluids, two with flowing fluids:
- Pressure in a static column
- Archimedes' principle
- Bernoulli's equation
I spend just shy of two weeks covering these topics, usually in the period between the Thanksgiving and winter holiday breaks.
General Course Philosophy
I use the "less is more" approach, as detailed in "Less Is More: Teaching Strategies for Limited Class Time." (Follow the link below in "See also.") Briefly, this means that:
- Most "example problems" done in class are quantitative predictions of the measurable results of demonstrations.
- I assign only two problems per night.
- Most classes begin with a brief quiz.
- This section of the course does not culminate with a "unit test." Rather, I set test dates at the beginning of the year, and fluids questions become fair game for inclusion on the next scheduled cumulative test.
- Class time focuses on quantitative demonstrations, concept development, and brief, simple derivations of some of the principles involved. We do not spend time going over homework.
Pressure in a Static Column
A large (approximately 30 cm tall) graduated cylinder filled with water sits on the demonstration table. A pressure sensor (connected to Logger Pro software) is available.
The question: What will the pressure sensor read if placed at the bottom of the cylinder?
To answer this, we must derive an equation for the pressure in a static column of fluid.
Derivation: We have previously defined pressure P as force per unit area. So consider the force applied by the water on the bottom of the tank. That force is merely the weight of the column of water, mg.
In fluids, though, we don't like to talk about the mass of a fluid. Instead, we usually refer to the fluid's density,
, mass is equal to
. Get used to replacing mass with
. Back to the column of water: the weight of the water in the cylinder is
. The pressure on the bottom of the tank is force divided by the area of the bottom,
Now consider the V/A portion of the equation. In this case, the volume of the water divided by the area of the cylinder's base is the height of the fluid, h, so the pressure due to the column of water is
There's only one more issue: the pressure at the top of the fluid column is nonzero. The total pressure exerted on the cylinder's base is this pressure at the top (call it
) plus the pressure due to the fluid itself. Thus we get the equation on the equation sheet:
Prediction: The column of water on the table is 30 cm high. The pressure at the top of the water is atmospheric pressure; the pressure sensor indicates this pressure to be 100 kPa.1
Write in your notebook: What is the density of water?
Many students will reflexively answer, "One." Briefly remind them that we must use consistent units throughout this calculation. The density of water is 1 g/cm3, or 1,000 kg/m3. Note that all the other relevant values are rendered in consistent units, without going through any laborious conversions:
Demonstration: Let the pressure probe record as a function of time. Lower the probe into the column of water. The reading at the bottom will be 103 kPa, confirming the prediction.
Write in your notebook: What will the pressure be at half the depth?
The reading should be 101 to 102 kPa, halfway between the surface pressure and the bottom pressure. Verify this reading with the probe.
Definitions: The equation for pressure due to a static column,
, includes two terms. The
term is the pressure at the top of the column. The
term, called the gauge pressure, is the pressure caused by the fluid only. Absolute pressure refers to the sum of the two terms.
Check your neighbor: Remember when you were little, you used to put your finger on top of the straw to lift some of your drink out? Consider the air pressure inside the straw, above the fluid but below your finger. Is this pressure:
A. Greater than atmospheric pressure?
B. Less than atmospheric pressure?
C. Equal to atmospheric pressure?
Write the answer in your notebook. After a minute, argue with your neighbor about the answer.
Answer: (B), the pressure is less than atmospheric. This is clearly a static column of fluid, so the equation
does not represent atmospheric pressure in this case; rather,
represents the pressure at the top of the fluid column, which is what we're looking for. P is the pressure at the bottom of the column, which in this case is atmospheric pressure. [Any time a liquid is exposed to the atmosphere, you can consider it to be at atmospheric pressure.] Solving for
, you get
. So the pressure at the top is atmospheric pressure minus something.
Clever alternate answer: Occasionally a student will reason from the ideal gas law. You might have noticed that when you withdraw the straw from your drink, the liquid drops down a little bit before settling. This means that the air trapped under your finger expanded a little bit. The temperature of the air was unlikely to change, and none escaped; so by PV = nRT, increased volume means reduced pressure.
A toy submarine floats on top of a vat of water, while a 100 g brass mass sinks to the bottom. How can you tell whether an object should float or sink?
The answer has nothing to do with whether an object is "heavier" than water, as many students will suggest. Such a comparison makes no sense. The vat of water is much, much heavier than the brass mass; in fact, the toy submarine has a mass of 210 g, heavier than the brass mass, and yet the submarine floats!
The class will come to the determination that the density of the object is important. Guide them to the testable suggestion that objects that are denser than water sink, and those that are less dense float.
The question: This toy submarine2 has an empty mass of 210 g and a volume of 280 cm3. How much water must you add to make it sink?
Prediction: Set the density of the sub equal to the density of water, 1 g/cm3. To make this submarine sink, its mass, equal to
, should be (1 g/cm3)(280 cm3) = 280 g. Its empty mass is 210 g, so add 70 g (= 70 cm3, or 70 mL) of water.
Demonstration: This particular toy allows you to add water using an included modified syringe. When the sub sinks, you can read the syringe's graduations to see that you added about 70 mL of water.
Definition: The specific gravity of a substance is defined as its density relative to the density of water. Something with a specific gravity of 0.5 is half as dense as water and thus will float with half of the object submerged. Something with a specific gravity greater than 1 will sink in water.
Now consider a free-body diagram of the submarine floating. The sub's weight acts down. Since it's in equilibrium, some force must act upward to cancel mg. What is that force? This is not what we would term a "normal force," which is a contact force acting perpendicular to a solid surface.
Rather, we call this upward force a buoyant force. Any time an object is wholly or partially submerged in a fluid, the fluid applies a buoyant force.
The question: You probably have noticed that objects seem to weigh less underwater. We want to figure out how much less an object might weigh.
On the demonstration table is a 200 g mass hanging from a force probe. The force probe now reads, of course, mg = 2.0 N. What is the effective weight of the 200 g mass when it's completely submerged in the water? Or, to phrase it another way, what will the force probe read when we completely submerge the mass in this beaker of water?
To answer this question, we must derive an equation for the buoyant force on a submerged object.
Derivation: Consider a cylinder of base area A submerged in a fluid. The fluid has density
. The top of the cylinder is at a depth h1 below the fluid's surface; the bottom of the cylinder is at depth h2.
The water applies pressure to the cylinder. This pressure applies a force inward on the cylinder in all directions. So the water pushes down on the top face of the cylinder and pushes up on the bottom face. Force equals pressure times area; thus the total force applied by the water is the difference between the force on the top and bottom surfaces.
What is the force on the top of the cylinder? The pressure at the top of the cylinder is due to the static column of fluid.
The force on the bottom of the cylinder is due to a slightly deeper static column of fluid:
The total force applied by the water is the buoyant force
, which is what we're looking for.
Well, (h2 - h1) is just the height of the cylinder itself. And A(h2 – h1) is the area of the base times the height, also known as the volume of the cylinder that is submerged. Thus we get the equation on the equation sheet:
Prediction: A free-body diagram of the fully submerged mass looks like this:
The mass is in equilibrium, so
is the tension in the rope, which is also the force probe reading.)
We can substitute for the buoyant force using the equation we derived and solve for the tension.
The density term
represents the fluid density, 1,000 kg/m3. V represents the submerged volume; this we have to find in a clever way. Ask for suggestions from the class.
Since the masses I use are approximately cylindrical, some folks suggest measuring the height and diameter of the mass; others suggest displacing fluid in the graduated cylinder that I conveniently have left set up from the day before. Either way, we measure the mass to have a volume of about 50 cm3.3
Now we can just plug and chug:
Demonstration: Lift the beaker of water to envelop the mass, and the reading on the force probe drops from 2.0 N to 1.5 N, as predicted.
Definition: In words, we can state the buoyant force equation above as: "The buoyant force on an object is equal to the weight of the fluid the object displaces."
Check your neighbor: A student stands on the bottom of a deep swimming pool. He drops a bowling ball with specific gravity 5.0 from a height of 1.0 m above the bottom of the pool. How much time does it take for the bowling ball to hit the bottom?
A. 0.005 s
B. 0.02 s
C. 0.5 s
D. 2.0 s
E. 5.0 s
Write the answer in your notebook. After a minute, argue with your neighbor.
Answer: (I go through this really, really fast, because, well, you'll see.) A free-body diagram of the ball has the buoyant force upward and mg downward. We can use Newton's second law to find the acceleration:
To make problem solving clearer, it's helpful to make generous use of subscripts to differentiate between quantities referring to the ball and to the water. The buoyant force is
, since the ball is entirely submerged. We can replace the mass of the ball by
The volume terms cancel. Solving for a,
Now using kinematics, the ball was released from rest, so the time of the drop becomes
We can easily evaluate this without a calculator -- the square root of 1/4 is 1/2. The answer is 0.5 s, choice (C).
That's an awful lot of work for one measly multiple-choice question, isn't it? Way too much. I smell a rat.
Look at the answer choices, and think about dropping a bowling ball in a swimming pool. We can be pretty darn sure that the ball will drop from about half your height in much less than 2 seconds, but since we can measure the time with a stopwatch, it must be more than two-hundredths of a second.
Or, we could use kinematics to show that in a vacuum, the ball would drop in a bit less than a third of a second (
s). So with a buoyant force acting, the ball should take just a bit longer to fall. With a ball that heavy, we wouldn't expect the time to increase by almost a factor of 10!
1. Don't just use the book value of 101 kPa. Depending on the weather, the altitude, and the calibration of your sensor, the sensor might read all kinds of things for the ambient pressure.
2. A similar sub is available from Frey Scientific, catalog number 05529080.
3. I make a point to never dwell on unit conversions. In fact, for conversions I can do in my head, like cm to m or Pa to kPa, I don't even mention that I'm converting -- unit conversions are a fifth-grade, not an eleventh-grade, skill. However, students botch the conversion from cm3 to m3 so often that I make a brief point about how to get this one right, reminding everyone that 50 cm3 is not 0.50 m3.
Greg Jacobs teaches AP Physics B and C at Woodberry Forest School in central Virginia. He is a graduate of Haverford College, and has a master's degree in engineering from Northwestern University. When he is not teaching, Greg broadcasts Woodberry Forest varsity baseball games over the Internet; he is a reporter for STATS, Inc., covering baseball, basketball, and football; and he is a Reader and consultant for the College Board's AP Physics program. Greg lives on campus at Woodberry with his wife Shari and their son Milo Cebu.