Skip to content

Fluids

AP Physics 1 · Topic 8

Train
8.1

Internal Structure and Density

Syllabus
Learning ObjectiveEssential Knowledge

8.1.A
Describe the properties of a fluid.

  • 8.1.A.1 Distinguishing properties of solids, liquids, and gases stem from the varying interactions between atoms and molecules.
  • 8.1.A.2 A fluid is a substance that has no fixed shape.
  • 8.1.A.3 Fluids can be characterized by their density. Density is defined as a ratio of mass to volume.
    • Equation: $\rho = \dfrac{m}{V}$
  • 8.1.A.4 An ideal fluid is incompressible and has no viscosity.

Source: College Board AP Course and Exam Description

The differences between solids, liquids, and gases come from how strongly their particles interact. A fluid 流体 (liquid or gas) has no fixed shape – it flows because its particles move past one another. Density 密度 is mass per unit volume:

$$\rho=\frac{m}{V}.$$
It depends on the substance and its state. An object sinks or floats depending on how its density compares with the surrounding fluid's – less dense floats, more dense sinks. An ideal fluid 理想流体 is incompressible 不可压缩 (constant density, whatever the pressure) and has no viscosity 黏度 (internal friction) – the model AP uses throughout.

Explore

Float or sink by density

An object floats if it is less dense than the fluid. Change the density and watch it ride higher or sink, displacing its own weight of fluid.

Vocabulary Train
English Chinese Pinyin
fluid 流体 liú tǐ
Density 密度 mì dù
ideal fluid 理想流体 lǐ xiǎng liú tǐ
incompressible 不可压缩 bù kě yā suō
viscosity 黏度 nián dù
8.2

Pressure

Syllabus
Learning ObjectiveEssential Knowledge

8.2.A
Describe the pressure exerted on a surface by a given force.

  • 8.2.A.1 Pressure is defined as the magnitude of the perpendicular force component exerted per unit area over a given surface area, as described by the equation
    • Equation: $P = \dfrac{F_\perp}{A}$
  • 8.2.A.2 Pressure is a scalar quantity.
  • 8.2.A.3 The volume and density of a given amount of an incompressible fluid is constant regardless of the pressure exerted on that fluid.

8.2.B
Describe the pressure exerted by a fluid.

  • 8.2.B.1 The pressure exerted by a fluid is the result of the entirety of the interactions between the fluid's constituent particles and the surface with which those particles interact.
  • 8.2.B.2 The absolute pressure of a fluid at a given point is equal to the sum of a reference pressure $P_0$, such as the atmospheric pressure $P_{\text{atm}}$, and the gauge pressure $P_{\text{gauge}}$.
    • Equation: $P = P_0 + \rho g h$
  • 8.2.B.3 The gauge pressure of a vertical column of fluid is described by the equation
    • Equation: $P_{\text{gauge}} = \rho g h$

Source: College Board AP Course and Exam Description

Pressure 压强 is the perpendicular force per unit area, a scalar 标量 measured in pascals (Pa):

$$P=\frac{F_\perp}{A}.$$
In a fluid at rest, pressure increases with depth 深度 because of the weight of fluid above:
$$P=P_0+\rho g h,$$
where $P_0$ is the pressure at the surface and $h$ is the depth. Pressure acts equally in all directions at a point and pushes perpendicular to any surface.

Distinguish the two pressures AP asks about: the gauge pressure 表压 is the extra pressure the fluid column adds, $P_{\text{gauge}}=\rho g h$, while the absolute pressure 绝对压强 is the total, $P=P_0+P_{\text{gauge}}$ (with $P_0$ usually atmospheric). A tyre gauge reading "$200\ \text{kPa}$" is gauge pressure; the air inside is really at about $300\ \text{kPa}$ absolute.

The weight of a liquid column sets the extra pressure at a depth The weight of a liquid column sets the extra pressure at a depth

Worked example. Find the total pressure on a diver $10\ \text{m}$ below the surface of water ($\rho=1000\ \text{kg/m}^3$, surface pressure $P_0=1.0\times10^{5}\ \text{Pa}$):

$$P=P_0+\rho g h=1.0\times10^{5}+1000\times9.8\times10=1.98\times10^{5}\ \text{Pa}.$$
Every $10\ \text{m}$ of water adds roughly one extra atmosphere of pressure. Notice the pressure does not depend on the shape or width of the container, only on the depth.

An aneroid barometer with a circular dial An aneroid barometer measures air pressure with a sealed metal box that flexes as the pressure outside changes

Vocabulary Train
English Chinese Pinyin
Pressure 压强 yā qiáng
scalar 标量 biāo liàng
depth 深度 shēn dù
gauge pressure 表压 biǎo yā
absolute pressure 绝对压强 jué duì yā qiáng
8.3

Fluids and Newton's Laws

Syllabus
Learning ObjectiveEssential Knowledge

8.3.A
Describe the conditions under which a fluid's velocity changes.

  • 8.3.A.1 Newton's laws can be used to describe the motion of particles within a fluid.
  • 8.3.A.2 The macroscopic behavior of a fluid is a result of the internal interactions between the fluid's constituent particles and external forces exerted on the fluid.

8.3.B
Describe the buoyant force exerted on an object interacting with a fluid.

  • 8.3.B.1 The buoyant force is a net upward force exerted on an object by a fluid.
  • 8.3.B.2 The buoyant force exerted on an object by a fluid is a result of the collective forces exerted on the object by the particles making up the fluid.
  • 8.3.B.3 The magnitude of the buoyant force exerted on an object by a fluid is equivalent to the weight of the fluid displaced by the object.
    • Equation: $F_b = \rho V g$

Source: College Board AP Course and Exam Description

Buoyancy: float or sink

An object in a fluid feels an upward buoyant force 浮力 equal to the weight of the fluid it displaces – Archimedes' principle 阿基米德原理:

$$F_b=\rho_{\text{fluid}}\,g\,V_{\text{displaced}}.$$
Combine this with Newton's laws: the object floats when buoyancy balances weight, sinks when weight wins, and rises when buoyancy wins. A floating object displaces exactly its own weight of fluid.

Worked example. A block of density $600\ \text{kg/m}^3$ floats in water ($1000\ \text{kg/m}^3$). What fraction is under the surface? For floating, the buoyant force equals the weight, so $\rho_{\text{fluid}}\,g\,V_{\text{sub}}=\rho_{\text{object}}\,g\,V$:

$$\frac{V_{\text{sub}}}{V}=\frac{\rho_{\text{object}}}{\rho_{\text{fluid}}}=\frac{600}{1000}=0.60.$$
So $60\%$ sits below the water – the same reason most of an iceberg (density $\approx 900$) hides underwater.

Vocabulary Train
English Chinese Pinyin
buoyant force 浮力 fú lì
Archimedes' principle 阿基米德原理 ā jī mǐ dé yuán lǐ
8.4

Fluids and Conservation Laws

Syllabus
Learning ObjectiveEssential Knowledge

8.4.A
Describe the flow of an incompressible fluid through a cross-sectional area by using mass conservation.

  • 8.4.A.1 A difference in pressure between two locations causes a fluid to flow.
    • 8.4.A.1.i The rate at which matter enters a fluid-filled tube open at both ends must equal the rate at which matter exits the tube.
    • 8.4.A.1.ii The rate at which matter flows into a location is proportional to the cross-sectional area of the flow and the speed at which the fluid flows.
      • Derived equation: $\dfrac{V}{t} = A v$
  • 8.4.A.2 The continuity equation for fluid flow describes conservation of mass flow rate in incompressible fluids.
    • Equation: $A_1 v_1 = A_2 v_2$

Source: College Board AP Course and Exam Description

For an ideal fluid flowing steadily, two conservation ideas apply:

Upthrust arises because the pressure on the bottom of an object exceeds that on the top Upthrust arises because the pressure on the bottom of an object exceeds that on the top

  • Continuity 连续性 (conservation of mass): the volume flow rate is constant, so $A_1 v_1 = A_2 v_2$. A narrower pipe forces faster flow.
  • Bernoulli's equation 伯努利方程 (conservation of energy per volume): along a streamline,
    $$P+\tfrac{1}{2}\rho v^2+\rho g y = \text{constant}.$$

Together they explain why fluid speeds up and its pressure drops where a pipe narrows or where flow is fastest.

Where the pipe narrows the fluid speeds up (continuity) and its pressure drops (Bernoulli) Where the pipe narrows the fluid speeds up (continuity) and its pressure drops (Bernoulli)

Worked example. Water flows at $2.0\ \text{m/s}$ through a pipe of cross-section $0.010\ \text{m}^2$, then enters a narrower section of $0.0040\ \text{m}^2$. By continuity the speed there is

$$v_2=\frac{A_1 v_1}{A_2}=\frac{0.010\times2.0}{0.0040}=5.0\ \text{m/s}.$$
By Bernoulli's equation this faster stream is at lower pressure – the effect that lifts an aeroplane wing and pulls two passing ships together.

Exam skill. Choose the right law by what changes. If the pipe changes width, start with continuity ($A_1v_1=A_2v_2$) to get the speeds; if you then need a pressure, feed those speeds into Bernoulli. Watch the height term $\rho g y$ only when the pipe also changes level.

Vocabulary Train
English Chinese Pinyin
Continuity 连续性 lián xù xìng
Bernoulli's equation 伯努利方程 bó nǔ lì fāng chéng
8.4

Exam tips

  • Pressure with depth is $P=P_0+\rho g h$ — it depends on depth only, not the container's shape or width.
  • Buoyant force = weight of fluid displaced ($\rho_{\text{fluid}}\,gV_{\text{disp}}$); a floating object displaces its own weight, so the fraction submerged is $\rho_{\text{object}}/\rho_{\text{fluid}}$.
  • Compare densities to predict floating vs sinking; a floating object is in equilibrium (buoyancy = weight), not weightless.
  • Use continuity $A_1v_1=A_2v_2$: a narrower pipe means faster flow.
  • Bernoulli: where a fluid flows faster its pressure is lower (wing lift, spray).

Log in or create account

IGCSE & A-Level