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AP 物理 1 · 知识点 8

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8:06

流体

一根钢钉投入水中会直沉底部。现在看这艘船。它由同样的钢材建造,重达数万吨,但它……

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8.1

内部结构与密度

教学大纲

学习目标 8.1.A: 描述流体的性质。

  • 8.1.A.1 固体、液体和气体性质的差异源于原子和分子间相互作用的强弱不同。
  • 8.1.A.2 流体是指没有固定形状的物质。
  • 8.1.A.3 流体可以通过其密度来表征。密度定义为质量与体积的比值。
    • 公式: $\rho = \dfrac{m}{V}$
  • 8.1.A.4 理想流体是不可压缩的,且没有黏性(粘度)。

来源:College Board AP 课程与考试说明

English

A steel ship weighing many thousands of tonnes floats, while a steel bolt sinks at once. A mechanic lifts a car with one hand on a pump handle. Both follow from one idea: a fluid pushes back with pressure, and pressure depends on depth and area in ways this unit makes exact.

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.

探索

通过密度决定漂浮或下沉

如果物体的密度小于流体,它会漂浮。改变密度,观察它浮得更高或下沉,并排开与其自身重量相等的流体。

词汇 训练
English 中文 拼音
fluid/ˈfluːɪd/ 流体 liú tǐ
density/ˈdensɪti/ 密度 mì dù
ideal fluid/aɪˈdɪəl ˈfluːɪd/ 理想流体 lǐ xiǎng liú tǐ
incompressible/ˌɪnkəmˈpresəbl/ 不可压缩 bù kě yā suō
viscosity/vɪˈskɒsɪti/ 黏度 nián dù
8.2

压强

教学大纲

学习目标 8.2.A: 描述给定力在表面产生的压强。

  • 8.2.A.1 压强定义为单位面积上垂直作用力的分量大小,如以下公式所示:
    • 公式: $P = \dfrac{F_\perp}{A}$
  • 8.2.A.2 压强是一个标量。
  • 8.2.A.3 对于一定质量的不可压缩流体,其体积和密度保持恒定,不随所施加的压强而改变。

学习目标 8.2.B: 描述流体产生的压强。

  • 8.2.B.1 流体产生的压强是流体组成粒子与接触表面之间所有相互作用的总体结果。
  • 8.2.B.2 流体中某一点的绝对压强等于参考压强 $P_0$(例如大气压 $P_{\text{atm}}$)与表压 $P_{\text{gauge}}$ 之和。
    • 公式: $P = P_0 + \rho g h$
  • 8.2.B.3 竖直液柱的表压由以下公式描述:
    • 公式: $P_{\text{gauge}} = \rho g h$

来源:College Board AP 课程与考试说明

English

Pressure 压强 is the perpendicular 垂直 force per unit area, a scalar 标量 measured in pascals 帕斯卡 (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.

Two hydraulic bottle jacks standing side by side, each with a narrow pump lever and a wide lifting ram
Hydraulic jacks: pressure applied to a small piston is transmitted undiminished through the fluid to a large one, so a small force lifts a car

Distinguish the two pressures AP asks about: the gauge pressure 表压 is pressure above the surrounding atmospheric pressure 大气压强. For a stationary liquid with its surface open to the atmosphere, $P_{\text{gauge}}=\rho g h$. The absolute pressure 绝对压强 is measured from vacuum, $P=P_{\text{atm}}+P_{\text{gauge}}$. 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

Worked example (a hydraulic lift). Pressure applied to an enclosed fluid is passed on equally throughout it (Pascal's principle 帕斯卡原理). A hydraulic lift has a small piston of area $0.0010\ \text{m}^2$ and a large piston of area $0.10\ \text{m}^2$. What force on the small piston holds up a $1500\ \text{kg}$ car?

$$\frac{F_1}{A_1}=\frac{F_2}{A_2}\;\Rightarrow\;F_1=\frac{A_1}{A_2}F_2=\frac{0.0010}{0.10}(1500)(9.8)=147\ \text{N}.$$
The force is 100 times smaller, but nothing is free: to raise the car $0.10\ \text{m}$, the small piston must move $10\ \text{m}$. The work done on each piston is the same, $1470\ \text{J}$.

探索

压强与深度

在流体中,压强随表面以下的深度增加而稳定增大。

词汇 训练
English 中文 拼音
pressure/ˈpreʃə/ 压强 yā qiáng
perpendicular/ˌpɜːpənˈdɪkjʊlə/ 垂直 chuí zhí
scalar/ˈskeɪlə/ 标量 biāo liàng
pascals 帕斯卡 pà sī kǎ
depth/depθ/ 深度 shēn dù
gauge pressure/ɡeɪdʒ ˈpreʃə/ 表压 biǎo yā
atmospheric pressure/ˌætməsˈferɪk ˈpreʃə/ 大气压强 dà qì yā qiáng
absolute pressure/ˈæbsəluːt ˈpreʃə/ 绝对压强 jué duì yā qiáng
Pascal's principle 帕斯卡原理 pà sī kǎ yuán lǐ
8.3

流体与牛顿定律

教学大纲

学习目标 8.3.A: 描述流体速度发生变化的条件。

  • 8.3.A.1 牛顿定律可用于描述流体内粒子的运动。
  • 8.3.A.2 流体的宏观行为是流体组成粒子之间的内部相互作用以及施加在流体上的外力共同作用的结果。

学习目标 8.3.B: 描述物体与流体相互作用时所受的浮力。

  • 8.3.B.1 浮力是流体对物体施加的向上的净力。
  • 8.3.B.2 流体对物体施加的浮力是构成流体的粒子对物体施加的合力(集体力)的结果。
  • 8.3.B.3 流体对物体施加的浮力大小等于物体排开流体的重量。
    • 公式: $F_b = \rho V g$

来源:College Board AP 课程与考试说明

English
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.

Worked example (apparent weight). A $2.0\ \text{kg}$ rock with volume $8.0\times10^{-4}\ \text{m}^3$ hangs from a spring scale, fully under water. Its weight is $19.6\ \text{N}$. The buoyant force is $(1000)(9.8)(8.0\times10^{-4})=7.8\ \text{N}$. The rock is at rest, so the scale reads $19.6-7.8=11.8\ \text{N}$: this is its apparent weight 视重. The rock seems lighter, though its weight has not changed.

探索

它会漂浮吗?

比较物体密度与流体密度,观察浮力如何决定沉浮。

词汇 训练
English 中文 拼音
buoyant force/ˈbɔɪənt fɔːs/ 浮力 fú lì
Archimedes' principle/ˌɑːkɪˈmiːdiːz ˈprɪnsɪpl/ 阿基米德原理 ā jī mǐ dé yuán lǐ
apparent weight 视重 shì zhòng
8.4

流体与守恒定律

教学大纲

学习目标 8.4.A: 利用质量守恒描述不可压缩流体通过横截面的流动。

  • 8.4.A.1 两个位置之间的压强差会导致流体流动。
    • 8.4.A.1.i 进入两端开口充满流体的管子的物质速率必须等于从管子排出的物质速率。
    • 8.4.A.1.ii 流入某位置的物质速率与流动的横截面积和流体流速成正比。
      • 导出方程: $\dfrac{V}{t} = A v$
  • 8.4.A.2 流体流动的连续性方程描述了不可压缩流体中质量流量率的守恒。
    • 方程: $A_1 v_1 = A_2 v_2$

来源:College Board AP 课程与考试说明

English

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 质量守恒): for steady incompressible flow, the volume flow rate 体积流量 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 能量守恒 per volume): for steady, incompressible, non-viscous flow along a streamline 流线,
    $$P+\tfrac{1}{2}\rho v^2+\rho g y = \text{constant}.$$

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.

Continue the example with Bernoulli's equation. The pipe is level, so the height terms cancel:

$$P_1-P_2=\tfrac12\rho\left(v_2^2-v_1^2\right)=\tfrac12(1000)\left(5.0^2-2.0^2\right)=1.05\times10^4\ \text{Pa}.$$
The pressure in the narrow section is about $10\ \text{kPa}$ lower, roughly a tenth of an atmosphere.

Worked example (a draining tank). Water leaves a small hole $1.8\ \text{m}$ below the surface of an open tank. The surface and the jet are both at atmospheric pressure, and the surface barely moves, so Bernoulli's equation reduces to $\rho g h=\tfrac12\rho v^2$:

$$v=\sqrt{2gh}=\sqrt{2(9.8)(1.8)}=5.9\ \text{m/s}.$$
This is Torricelli's result 托里拆利定理: the water leaves as fast as if it had fallen freely from the surface.

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.

探索

追踪管道中的流动

逐步展示变窄的管道,观察流体在狭窄处流速加快。

词汇 训练
English 中文 拼音
continuity/kɒntɪˈnjuːɪti/ 连续性 lián xù xìng
conservation of mass/ˌkɒnsəˈveɪʃn ɒv mæs/ 质量守恒 zhì liàng shǒu héng
volume flow rate/ˈvɒljuːm fləʊ reɪt/ 体积流量 tǐ jī liú liàng
Bernoulli's equation/ˈbɜːnaʊliz ɪˈkweɪʒn/ 伯努利方程 bó nǔ lì fāng chéng
conservation of energy/ˌkɒnsəˈveɪʃn ɒv ˈenədʒi/ 能量守恒 néng liàng shǒu héng
streamline 流线 liú xiàn
Torricelli's result 托里拆利定理 tuō lǐ chāi lì dìng lǐ
8.4

Practice

English

Answer each before reading the answers below. Take $g=9.8\ \text{m/s}^2$ and $P_0=1.0\times10^5\ \text{Pa}$.

  1. Find the gauge pressure and the absolute pressure $25\ \text{m}$ below the surface of sea water, density $1030\ \text{kg/m}^3$.
  2. Ice of density $920\ \text{kg/m}^3$ floats in sea water of density $1030\ \text{kg/m}^3$. What fraction of the ice is under the surface?
  3. A hydraulic press has pistons of area $2.0\ \text{cm}^2$ and $400\ \text{cm}^2$. What force can a $50\ \text{N}$ push on the small piston produce?
  4. Water flows through a pipe whose radius halves at a narrowing. By what factor does the speed change?
  5. A student says: "Steel is denser than water, so a steel ship should sink." Correct the student.
8.4

答案

English
  1. Gauge: $(1030)(9.8)(25)=2.5\times10^5\ \text{Pa}$. Absolute: $1.0\times10^5+2.5\times10^5=3.5\times10^5\ \text{Pa}$.
  2. $920/1030=0.89$, so about $89\%$ is under the surface.
  3. $F_2=F_1\dfrac{A_2}{A_1}=50\times\dfrac{400}{2.0}=1.0\times10^4\ \text{N}$.
  4. Halving the radius divides the area by 4, so by continuity the speed is 4 times larger.
  5. What matters is the average density of the whole ship, including the air inside the hull. The hull pushes aside a large volume of water; the ship floats when the weight of that displaced water equals its own weight. A solid steel bolt cannot displace enough water, so it sinks.
8.4

考试技巧

English
  • 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 平衡 (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).

Common mistakes

  • Using the volume of the whole object in $F_b = \rho g V$ for a floating object. Use the submerged volume, and the density of the fluid, not of the object.
  • Mixing gauge and absolute pressure. $\rho g h$ alone is the gauge pressure; add $P_0$ for the absolute pressure.
  • Making the pressure at a depth depend on the width or shape of the container. Only the depth below the surface (and the fluid's density) matter.
  • Writing $A_1 v_1 = A_2 v_2$ with radii instead of areas. Area goes as $r^2$, so halving the radius makes the flow 4 times faster.
  • Saying that faster-moving fluid has higher pressure. In Bernoulli's equation, where the speed rises (same height), the pressure falls.
  • Saying a hydraulic lift multiplies energy. It multiplies force; the small piston moves further, so the work in equals the work out.
词汇 训练
English 中文 拼音
equilibrium/ˌiːkwɪˈlɪbrɪəm/ 平衡 píng héng

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