流体中的密度与压强

AP 物理1· AP 物理1 课程与考试说明 — 流体· 14 分钟阅读

1. 密度★★☆☆☆⏱ 4 min

密度描述了给定体积的流体或固体中填充的质量多少。它是一种强度性质，意味着其数值与你拥有的材料总量无关：一小块铝片的密度和一大块铝块的密度完全相同。对于空心球、混合材料这类复合物体，我们将其平均密度定义为物体总质量除以物体总体积（包含空心部分的体积）。

📘 定义

密度

$ρ$ (Greek rho)

材料或物体的质量与体积的比值，国际单位为 $kg/m^{3}$ 。

例:

Pure water has a standard density of $1000 kg/m^{3}$ .

\rho = \frac{m}{V}

AP 物理1 经常考察密度相关的比例推理：若两个物体质量相等，则密度与体积成反比；若两个物体体积相等，则密度与质量成正比。

📐 例题

一个空心铝球的外半径为 $0.10 m$ ，总质量为 $2.7 kg$ 。已知实心铝的密度为 $2700 kg/m^{3}$ ，求球内部空心腔的半径是多少？

1
首先通过变形密度公式，求出制作该球所用铝材料的体积：
$V_{\text{Al}} = \frac{m}{\rho_{\text{Al}}} = \frac{2.7 \ \text{kg}}{2700 \ \text{kg/m}^3} = 0.001 \ \text{m}^3$
2
计算该球的外部总体积：
$V_{\text{outer}} = \frac{4}{3}\pi R_{\text{outer}}^3 = \frac{4}{3}\pi (0.10 \ \text{m})^3 \approx 0.00419 \ \text{m}^3$
3
空心腔的体积等于球的外部总体积减去铝材料的体积：
$V_{\text{cavity}} = V_{\text{outer}} - V_{\text{Al}} = 0.00419 - 0.001 = 0.00319 \ \text{m}^3$
4
求解空心腔的半径 $r$ ：
$r = \left(\frac{3V_{\text{cavity}}}{4\pi}\right)^{1/3} = \left(\frac{3(0.00319)}{4\pi}\right)^{1/3} \approx 0.092 \ \text{m}$

Exam tip:

在比例推理类选择题中，代入数值前先约去所有常量，既可以节省时间，也能降低计算错误的概率。

2. 流体静压强★★★☆☆⏱ 5 min

流体静压强是静止流体中某一深度处的压强，由该点上方的流体自重产生。一个重要结论是：流体静压强仅与深度、流体密度和重力加速度 $g$ 有关，与容器的形状无关，这一结论也被称为「流体静力学佯谬」。

🔬 推导

目标:

推导流体静压强随深度变化的公式

起点:

压强的定义：单位面积受到的力

1
取流体表面下深度为 $h$ 处的一个水平面积 $A$ ，该面积上方的流体质量为：
2
$m = \rho V = \rho A h$
3
作用在该面积上的力等于上方流体的重力：
4
$F = mg = \rho A h g$
5
压强是单位面积受到的力，因此从液面开始的压强变化量为：
6
$\Delta P = \frac{F}{A} = \rho g h$

结果:

压强随深度的变化量仅与 $ρ$ 、 $g$ 、 $h$ 有关，与容器形状无关。

我们将压强分为两类：表压是相对于大气压的压强，数值等于 $ρ g h$ 。绝对压强 是包含液面处大气压在内的总压强，因此满足 $P_{abs} = P_{atm} + P_{gauge}$ 。静止流体中同一水平深度处的压强处处相等，这是求解压力计问题的核心原理。

📐 例题

一个U型管压力计的一端开口与大气连通，另一端连接装有压缩气体的密封储罐。密度为 $13600 kg/m^{3}$ 的汞柱在开口端的液面比储气罐端的液面高 $0.25 m$ ，已知大气压为 $1.013 \times 1 0^{5} Pa$ ，求储罐内气体的绝对压强是多少？

1
对于静止流体，同一水平高度处的压强相等，我们选取储气罐一侧汞液面所在的水平面作为参考面。
2
该参考面处来自储气罐侧的压强等于气体压强 $P_{gas}$ ，来自开口侧的压强等于大气压加上汞柱产生的压强，因此：
$P_{\text{gas}} = P_{\text{atm}} + \rho g h$
3
计算汞柱产生的表压：
$\rho g h = (13600)(9.8)(0.25) \approx 3.33 \times 10^4 \ \text{Pa}$
4
将表压与大气压相加，得到绝对压强：
$P_{\text{gas}} = 1.013 \times 10^5 + 3.33 \times 10^4 \approx 1.35 \times 10^5 \ \text{Pa}$

Exam tip:

做题时务必确认题目要求的是表压还是绝对压强，出题人经常在此设置陷阱，错误的压强类型往往是高频错误选项。

3. 帕斯卡原理★★★☆☆⏱ 3 min

帕斯卡原理指出，对密闭不可压缩流体施加的压强变化会毫无衰减地传递到流体的每一个点以及容器的壁面上。该原理是汽车升降机、制动管路等液压系统的基础，这类场景在AP考试中十分常见。

📘 定义

帕斯卡原理

对于密闭不可压缩流体，压强变化会均匀传递至整个流体，由此可推导出双活塞液压系统的如下关系。

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

该关系意味着，作用在小活塞上的较小输入力可以在大活塞上产生很大的输出力——它起到力的放大作用，与杠杆类似。功依然守恒：小活塞的移动距离远大于大活塞，因此输入功等于输出功（忽略摩擦）。帕斯卡原理仅适用于不可压缩流体（通常为液体）。

📐 例题

一台用于举升汽车的液压升降机，其小输入活塞的横截面积为 $0.005 m^{2}$ ，大输出活塞的横截面积为 $0.4 m^{2}$ 。输出活塞支撑着一辆质量为 $2000 kg$ 的汽车。若要保持汽车静止，所需的最小输入力是多少？

1
帕斯卡原理给出两个活塞处的压强相等：
$\frac{F_1}{A_1} = \frac{F_2}{A_2}$
2
输出力 $F_{2}$ 等于汽车的重力：
$F_2 = mg = (2000)(9.8) = 19600 \ \text{N}$
3
整理公式求解输入力：
$F_1 = F_2 \cdot \frac{A_1}{A_2} = 19600 \cdot \frac{0.005}{0.4} = 245 \ \text{N} \approx 250 \ \text{N}$
4
该结果是合理的：一个很小的输入力（相当于25kg物体的重力）就可以举升2000kg的汽车，这与液压升降机的设计原理相符。

Exam tip:

如果你得到的是直径（或半径）而非面积，请记住面积与直径的平方成正比，因此面积比为 $(d_{2} / d_{1})^{2}$ 。切勿直接使用直径比。

4. AP风格概念检测★★★☆☆⏱ 2 min

✓ 快速检测

请完成这道AP风格的选择题，检验你对知识点的理解：

Two vertical cylinders are filled with water to the same depth $h$ . Cylinder A has twice the diameter of Cylinder B. What is the ratio of the total hydrostatic force on the bottom of Cylinder A to the total hydrostatic force on the bottom of Cylinder B?
- $1 : 1$
- $2 : 1$
- $4 : 1$
- $8 : 1$
显示答案
$4:1$ —
Hydrostatic pressure at the bottom is equal for both cylinders, but total force equals pressure times area. Area scales with the square of diameter, so A has 4x the area and 4x the total force.

5. 常见陷阱

错误做法:

Using the diameter ratio directly instead of squaring it to get the area ratio for Pascal's principle.

原因:

Students confuse linear and area proportionality, and forget that area depends on the square of linear dimensions.

正确做法:

Always confirm that $A_{2} / A_{1} = (d_{2} / d_{1})^{2}$ before solving; the $π$ and $1/4$ terms cancel out when taking the ratio.

错误做法:

Forgetting to add atmospheric pressure to gauge pressure when asked for absolute pressure.

原因:

Most problems default to gauge pressure for fluid columns, so students develop a habit of only calculating $P = ρ g h$ without checking the question.

正确做法:

Circle the words "gauge" or "absolute" in the question, and write $P = P_{atm} + ρ g h$ explicitly for absolute pressure before starting calculations.

错误做法:

Using the height of a submerged object instead of the depth of the point of interest below the surface when calculating hydrostatic pressure.

原因:

Students mix up object height and depth, and assume the object's own dimension is the $h$ in $P = ρ g h$ .

正确做法:

Label $h$ on your diagram as the distance from the fluid surface to the point you are analyzing, and confirm it is depth, not an object dimension.

错误做法:

Calculating the density of a hollow object's material by dividing total mass by total outer volume including the hollow cavity.

原因:

Students confuse average density of the whole object with density of the material it is made from.

正确做法:

Check whether the question asks for material density or average density. For material density, only use the volume of the material itself, not the total outer volume of the object.

错误做法:

Assuming pressure at the same depth is higher in a wider container because it holds more total mass of fluid.

原因:

Intuition about total weight overrides the definition of pressure as force per unit area.

正确做法:

Remember hydrostatic pressure only depends on depth, density, and $g$ — container shape has no effect, so always use $P = ρ g h$ regardless of container width.

错误做法:

Using $g = 10 m/s^{2}$ when the problem expects $9.8 m/s^{2}$ , leading to an answer that does not match MCQ options.

原因:

Students memorize one value for $g$ and do not check the problem's convention.

正确做法:

Always use $g = 9.8 m/s^{2}$ unless the problem explicitly states to use $10 m/s^{2}$ for approximation.

6. 速查表

Category	Formula	Notes
Density	$ρ = \frac{m}{V}$	Units: $kg/m^{3}$ . Average density = total mass / total volume for composites.
Gauge Pressure	$P_{gauge} = ρ g h$	Pressure relative to atmospheric pressure, $h$ = depth below surface.
Absolute Pressure	$P_{abs} = P_{atm} + P_{gauge}$	Use this when asked for total, absolute pressure.
Pressure Change with Depth	$Δ P = ρ g Δ h$	Pressure increases with depth, decreases with altitude.
Pascal's Principle	$\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}}$	For enclosed incompressible fluids. Area ratio = (diameter ratio) $^{2}$ .
Hydrostatic Equilibrium	$P_{1} = P_{2}$ at same depth	Core principle for solving U-tube manometer problems.
Force from Uniform Pressure	$F = P A$	Total force on a surface equals pressure times surface area.

下一步

This topic lays the foundation for all other fluid concepts in AP Physics 1 Unit 8. Next, you will apply density and pressure concepts to understand buoyancy and Archimedes' principle, which is one of the most heavily tested topics in the fluid unit. Without a solid grasp of hydrostatic pressure and density, you cannot correctly derive the buoyant force or solve force balance problems for submerged or floating objects, which frequently appear in both multiple-choice and free-response sections. It also reinforces force and Newton's laws from earlier in the course, as most static fluid problems rely on force balance analysis.