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AP Physics C: Fluids Unit

1. Introduction to Fluids

A fluid is any substance that can flow and continuously deform under shear stress. Liquids and gases are both fluids. The key idea is that fluids do not maintain a fixed shape the way solids do.

Core Properties

  • Fluids exert pressure perpendicular to surfaces.
  • Static fluids cannot support shear stress.
  • Fluid behavior depends on density, pressure, velocity, and viscosity.

Physics C Lens

Most fluid problems reduce to one of three big tools: force balance, conservation of mass, or conservation of energy.

2. Density and Pressure

Density describes how much mass exists per unit volume. Pressure describes how force is distributed over an area.

Density

\[\rho = \frac{m}{V}\]

where \(\rho\) is density, \(m\) is mass, and \(V\) is volume.

MaterialDensity \((\text{kg/m}^3)\)
Water1000
Ice917
Air1.2
Mercury13,600
Steel7800

Pressure

\[P = \frac{F}{A}\]

Pressure is a scalar, but the force caused by pressure acts perpendicular to a surface.

High heels exert more pressure than flat shoes because the same force is applied over a smaller area.

Sample Problem 2.1 — Pressure from a Force

A student stands on one foot with an area of contact of \(0.015\,\text{m}^2\). If the student has a mass of \(65\,\text{kg}\), determine the pressure exerted on the floor.

Solution

First find the student's weight:

\[F_g = mg = (65)(9.8)=637\,\text{N}\]

Then divide by contact area:

\[P=\frac{F}{A}=\frac{637}{0.015}=4.25\times10^4\,\text{Pa}\]

Final answer: \(\boxed{4.3\times10^4\,\text{Pa}}\)

Practice Problems

  1. A cube of aluminum has side length \(5.0\,\text{cm}\). Find its mass.
  2. A force of \(120\,\text{N}\) acts on an area of \(0.020\,\text{m}^2\). Determine the pressure.
  3. Explain why high heels exert more pressure on a floor than flat shoes.
  4. A diver experiences a pressure of \(2.5\times10^5\,\text{Pa}\). If the diver's effective body area is \(1.8\,\text{m}^2\), what force acts on the diver?

3. Pressure Variation with Depth

Pressure increases with depth because the fluid above a point has weight. This is hydrostatic pressure.

\[P=P_0+\rho gh\]
Important: In many problems, \(\rho gh\) is gauge pressure. Absolute pressure includes atmospheric pressure.

Sample Problem 3.1 — Pressure Underwater

How much gauge pressure exists \(15\,\text{m}\) below the surface of fresh water?

Solution
\[P=\rho gh=(1000)(9.8)(15)=1.47\times10^5\,\text{Pa}\]

Final answer: \(\boxed{1.47\times10^5\,\text{Pa}}\)

Practice Problems

  1. Find the gauge pressure \(40\,\text{m}\) below the ocean surface. Use \(\rho=1025\,\text{kg/m}^3\).
  2. How deep must a diver descend in fresh water to experience \(3.0\) atmospheres of absolute pressure?
  3. Explain why a dam is thicker at the bottom than at the top.
  4. Compare the pressure increase over \(5.0\,\text{m}\) in water versus mercury.

4. Pascal’s Principle and Hydraulic Systems

A pressure change applied to an enclosed fluid is transmitted equally throughout the fluid.

\[\frac{F_1}{A_1}=\frac{F_2}{A_2}\]

Sample Problem 4.1 — Hydraulic Lift

A mechanic applies \(200\,\text{N}\) to a small piston of area \(0.010\,\text{m}^2\). The large piston has area \(0.50\,\text{m}^2\). Determine the output force.

Solution
\[F_2=F_1\frac{A_2}{A_1}=200\left(\frac{0.50}{0.010}\right)=1.0\times10^4\,\text{N}\]

Final answer: \(\boxed{1.0\times10^4\,\text{N}}\)

5. Buoyancy and Archimedes’ Principle

A fluid exerts an upward force on submerged objects because pressure is greater at larger depths.

\[F_B=\rho_f V_{\text{disp}}g\]

Sample Problem 5.1 — Buoyant Force

A \(0.020\,\text{m}^3\) block is completely submerged in water. Determine the buoyant force.

Solution
\[F_B=\rho Vg=(1000)(0.020)(9.8)=196\,\text{N}\]

Final answer: \(\boxed{196\,\text{N}}\)

Sample Problem 5.2 — Fraction Submerged

A block of wood with density \(650\,\text{kg/m}^3\) floats in water. What fraction of the block is submerged?

Solution

For floating equilibrium, \(F_B=mg\).

\[\rho_f V_{sub}g=\rho_o Vg\]
\[\frac{V_{sub}}{V}=\frac{\rho_o}{\rho_f}=\frac{650}{1000}=0.65\]

Final answer: \(\boxed{65\%}\)

Practice Problems

  1. A metal sphere displaces \(0.0030\,\text{m}^3\) of water. Find the buoyant force.
  2. Why do steel ships float while solid steel blocks sink?
  3. A floating object has \(80\%\) of its volume submerged. Determine its density.
  4. A submerged object experiences a buoyant force of \(500\,\text{N}\). Determine the volume displaced in water.

6. Fluid Flow and the Equation of Continuity

For an incompressible fluid, the volume flow rate must stay constant along a pipe.

\[Q=Av\]
\[A_1v_1=A_2v_2\]

Sample Problem 6.1 — Continuity

Water moves through a pipe with initial radius \(4.0\,\text{cm}\) at speed \(2.0\,\text{m/s}\). The pipe narrows to radius \(2.0\,\text{cm}\). Determine the new speed.

Solution
\[v_2=v_1\frac{A_1}{A_2}=v_1\frac{r_1^2}{r_2^2}=2.0\left(\frac{4.0^2}{2.0^2}\right)=8.0\,\text{m/s}\]

Final answer: \(\boxed{8.0\,\text{m/s}}\)

7. Bernoulli’s Equation

Bernoulli’s equation is conservation of mechanical energy per unit volume for an ideal fluid moving steadily along a streamline.

\[P+\frac{1}{2}\rho v^2+\rho gh=\text{constant}\]
Teaching warning: Students often memorize “faster means lower pressure” without checking height changes. Make them identify which Bernoulli terms increase, decrease, or stay constant.

Interactive Demonstration — Bernoulli in a Pipe

Adjust the pipe geometry and fluid conditions. The demo uses continuity to find the outlet speed, then Bernoulli’s equation to predict the outlet pressure.

Outlet speed

5.56 m/s

Outlet pressure

187 kPa

Flow rate

0.0157 m³/s

Energy check

Balanced

When the pipe narrows, continuity requires the fluid speed to increase. If height is unchanged, the pressure usually drops.

Sample Problem 7.1 — Horizontal Pipe

Water moves through a horizontal pipe. At a wide section the speed is \(2.0\,\text{m/s}\) and pressure is \(180\,\text{kPa}\). At a narrow section the speed is \(6.0\,\text{m/s}\). Determine the pressure in the narrow section.

Solution

Because the pipe is horizontal, the gravitational terms cancel.

\[P_1+\frac12\rho v_1^2=P_2+\frac12\rho v_2^2\]
\[P_2=P_1+\frac12\rho(v_1^2-v_2^2)\]
\[P_2=180000+500(4-36)=164000\,\text{Pa}\]

Final answer: \(\boxed{1.64\times10^5\,\text{Pa}}\)

Sample Problem 7.2 — Torricelli’s Law

Water exits a hole \(5.0\,\text{m}\) below the surface of a large tank. Determine the exit speed.

Solution

Pressure is atmospheric at the surface and hole. The tank is large, so surface speed is negligible.

\[\rho gh=\frac12\rho v^2\]
\[v=\sqrt{2gh}=\sqrt{2(9.8)(5.0)}=9.9\,\text{m/s}\]

Final answer: \(\boxed{9.9\,\text{m/s}}\)

Practice Problems

  1. Water exits a tank hole \(10\,\text{m}\) below the surface. Determine the exit speed.
  2. A pipe rises upward while narrowing. Describe how speed and pressure change.
  3. Derive Torricelli’s law from Bernoulli’s equation.
  4. Explain why a shower curtain moves inward when water flows.

8. Viscosity and Real Fluids

Real fluids are not ideal. Viscosity causes internal friction and energy loss.

Laminar Flow

Smooth, orderly flow in layers. Neighboring layers slide past each other predictably.

Turbulent Flow

Chaotic flow with eddies and mixing. Energy is dissipated more strongly.

Poiseuille’s Law

\[Q=\frac{\pi r^4\Delta P}{8\eta L}\]
The fourth-power dependence is the key result: a small change in radius creates a large change in flow rate.

Sample Problem 8.1 — Radius Dependence

If a pipe radius doubles, by what factor does flow rate change, assuming all other variables remain constant?

Solution
\[Q\propto r^4\Rightarrow Q_{new}=2^4Q=16Q\]

Final answer: \(\boxed{16}\)

Narrative Inquiry Activity

Can the Field Station Fill Its Emergency Water Tank?

A multi-step AP Physics C free-response investigation combining pressure, density, continuity, volume flow rate, and Bernoulli's equation in one real-world engineering decision. Includes automated scoring and answer checks.

Open Narrative Inquiry →

9. AP-Style Unit Review Problems

These problems combine multiple ideas and are better suited for quizzes, review days, or AP-style free response practice.

Review Problem 1 — Floating Cylinder

A wooden cylinder of density \(700\,\text{kg/m}^3\) floats vertically in water. The cylinder has radius \(0.20\,\text{m}\) and height \(1.0\,\text{m}\).

  1. Determine the fraction submerged.
  2. Determine the submerged depth.
  3. Determine the buoyant force.
Solution
\[\frac{V_{sub}}{V}=\frac{700}{1000}=0.70\]

The submerged depth is \(0.70(1.0)=0.70\,\text{m}\).

\[V=\pi r^2h=\pi(0.20)^2(1.0)=0.126\,\text{m}^3\]
\[V_{sub}=0.70V=0.0882\,\text{m}^3\]
\[F_B=(1000)(0.0882)(9.8)=864\,\text{N}\]

Review Problem 2 — Continuity and Bernoulli

Water flows through a pipe that narrows from radius \(5.0\,\text{cm}\) to radius \(2.0\,\text{cm}\). The initial pressure is \(220\,\text{kPa}\) and initial speed is \(1.5\,\text{m/s}\).

  1. Determine the speed in the narrow section.
  2. Determine the pressure in the narrow section.
Solution
\[v_2=v_1\frac{r_1^2}{r_2^2}=1.5\left(\frac{5.0^2}{2.0^2}\right)=9.38\,\text{m/s}\]
\[P_2=P_1+\frac12\rho(v_1^2-v_2^2)\]
\[P_2=220000+500(2.25-87.98)=1.77\times10^5\,\text{Pa}\]

Formula Summary

\[\rho=\frac{m}{V}\]
\[P=\frac{F}{A}\]
\[P=P_0+\rho gh\]
\[F_B=\rho_f V_{disp}g\]
\[A_1v_1=A_2v_2\]
\[P+\frac12\rho v^2+\rho gh=\text{constant}\]