P1-VL：连接物理奥林匹克竞赛中的视觉感知与科学推理

To analyze the situation for case b in Cox's timepiece, we consider the vertical tube of length $H$, cross-sectional area $S$, and mass $m$, lifted so that its bottom end is at the bath surface ($z = 0$) and its top is at altitude $h$. In this configuration, the water level inside the tube is at the same height as the top of the tube, i.e., $z_\ell = h$, indicating that the tube is completely filled with water.

(1) Pressure at the Top of the Tube

The tube is sealed at the top and open at the bottom, and the water inside is in hydrostatic equilibrium. The pressure at the bottom of the tube (at $z = 0$) must equal the atmospheric pressure $P_0$, as the tube is open to the bath at this level.

Using the hydrostatic pressure law, the pressure at any height $z$ inside the water column is given by:

$$P(z) = P_0 - \rho g z$$

At the top of the tube, where $z = h$, the pressure in the water is:

$$P_{\rm w} = P_0 - \rho g h$$

This result reflects that the pressure at the top of the water column is reduced by the weight of the water column of height $h$.

(2) Force Required to Maintain the Tube in Equilibrium

To determine the force $\vec{F}$ necessary to hold the tube in place, we perform a force balance on the tube alone.

Forces on the Tube:

• Gravitational force: The tube has a mass $m$, so its weight is $-mg\vec{u_z}$.
• Pressure forces:
  • The atmospheric pressure $P_0$ acts downward on the top cap of the tube, exerting a force $-P_0 S \vec{u_z}$.
  • The water pressure $P_{\rm w}$ at the top of the tube acts upward on the inner surface of the cap, exerting a force $+P_{\rm w} S \vec{u_z}$.

Thus, the net pressure force on the tube is:

$$(P_{\rm w} - P_0) S \vec{u_z} = -\rho g h S \vec{u_z}$$

This is a downward force due to the pressure difference between the outside atmosphere and the water column inside the tube.

• Applied force: $\vec{F}$ is applied upward to maintain equilibrium.

Equilibrium Condition:

The net force on the tube must be zero:

$$\vec{F} + (P_{\rm w} - P_0) S \vec{u_z} - m g \vec{u_z} = \vec{0}$$

Substituting $P_{\rm w} = P_0 - \rho g h$:

$$\vec{F} - \rho g h S \vec{u_z} - m g \vec{u_z} = \vec{0}$$

$$\Rightarrow \vec{F} = (m g + \rho g h S) \vec{u_z}$$

This result is physically intuitive: the force required to hold the tube in place must counteract both the tube's own weight and the net downward pressure force due to the difference between atmospheric pressure and the reduced pressure at the top of the water column.

Final Answers

(1) The pressure in the water at the top of the tube is:

$$\boxed{P_{\mathrm{w}} = P_0 - \rho g h}$$

(2) The force required to maintain the tube in equilibrium is:

$$\boxed{\vec{F} = \left( m g + \rho g h S \right) \vec{u_{z}}}$$

(1) Since the tube is slowly lifted and contains no gas, the water fills the tube up to the top inner surface as long as the pressure remains above the vapor pressure (which is the case here given the dimensions and atmospheric pressure). Thus, the water level inside the tube is $z_{\ell} = h$.

The pressure in the stationary water column varies with height according to the hydrostatic law. Taking the pressure at the surface of the bath ($z=0$) as the atmospheric pressure $P_{0}$, the pressure $P_{\mathrm{w}}$ at the altitude $h$ inside the water column is given by:

$$P_{\mathrm{w}} = P_{0} - \rho g h$$

$\boxed{P_{\mathrm{w}} = P_{0} - \rho g h}$

(2) To find the force $\vec{F}$ necessary to maintain the tube in equilibrium, we consider the balance of forces acting on the tube. The tube has mass $m$, so the gravitational force is $-mg\vec{u_{z}}$. The external lifting force is $\vec{F}$.

The pressure forces acting on the tube are due to the atmosphere on the outer top surface and the water on the inner top surface. The thickness of the walls is neglected, so the area for both surfaces is $S$. The side walls are vertical, so pressure forces on them cancel horizontally and have no vertical component.

• Force from atmospheric pressure on the top (downward): $\vec{F}_{\text{air}} = -P_{0} S \vec{u_{z}}$.
• Force from water pressure on the top inside (upward): $\vec{F}_{\text{water}} = P_{\mathrm{w}} S \vec{u_{z}} = (P_{0} - \rho g h) S \vec{u_{z}}$.

The equilibrium condition $\sum \vec{F} = \vec{0}$ gives:

$$\vec{F} - mg\vec{u_{z}} - P_{0} S \vec{u_{z}} + (P_{0} - \rho g h) S \vec{u_{z}} = \vec{0}$$

$$\vec{F} = \left( mg + P_{0} S - P_{0} S + \rho g h S \right) \vec{u_{z}}$$

$$\vec{F} = (mg + \rho S h g) \vec{u_{z}}$$

Alternatively, one can consider that the external force must support the weight of the tube plus the weight of the suspended water column of height $h$ (volume $Sh$) pulled up above the bath level.

Weight of tube: $mg$.

Weight of lifted water: $\rho (Sh) g$.

Total magnitude: $F = mg + \rho S h g$.

Direction is upward ($\vec{u_{z}}$).

$\boxed{\vec{F} = (m + \rho S h)g \vec{u_{z}}}$

Problem Solution: Tube Lifted from Water Bath

Part (1): Finding the pressure $P_w$ at the top of the tube

Step 1: Analyze the physical situation

The problem describes a cylindrical tube (length $H=1 \text{m}$, closed top, open bottom) initially filled with water and submerged such that its top is at the water surface ($h=0$). No gas is inside the tube. We study the configuration (Fig. 2b) where the tube's top is at height $h$ above the external water level.

Step 2: Determine the water level inside the tube, $z_l$

The space above the water column in the tube is a vacuum (no initial gas, neglect water vapor pressure). Atmospheric pressure $P_0$ supports a water column of maximum (barometric) height:

$$H_{\text{baro}} = \frac{P_0}{\rho g}$$

Using $P_0 \approx 1.013 \times 10^5 \text{Pa}$, $\rho \approx 1000 \text{kg/m}^3$, we get $H_{\text{baro}} \approx 10.3 \text{m}$. Since the tube length $H=1 \text{m} < 10.3 \text{m}$, atmospheric pressure keeps the tube fully filled with water. Thus:

$$z_l = h$$

Step 3: Calculate $P_w$ using hydrostatics

The hydrostatic principle for a static fluid: $P_2 = P_1 - \rho g (z_2 - z_1)$.

• Point 1: External water surface ($z_1=0$, $P_1=P_0$)
• Point 2: Top of the water column inside the tube ($z_2=h$, $P_2=P_w$)

Substituting gives:

$$P_w = P_0 - \rho g h$$

Part (2): Finding the force $\vec{F}$

Step 1: Define the system and identify external forces

Treat the tube + water inside as the system (static equilibrium, net force = 0). External forces:

1. Pulling force: $\vec{F} = F \vec{u_z}$ (upward)
2. Tube's weight: $\vec{W}_{\text{tube}} = -mg \vec{u_z}$ (downward)
3. Water's weight (volume $V_w = S \cdot H$): $\vec{W}_{\text{water}} = -\rho S H g \vec{u_z}$ (downward)
4. Atmospheric force on tube top: $\vec{F}_{\text{air}} = -P_0 S \vec{u_z}$ (downward)
5. Water pressure force on tube bottom (height $z_{\text{bottom}} = h - H$, pressure $P(z_{\text{bottom}}) = P_0 - \rho g (h-H)$): $\vec{F}_{\text{water}} = (P_0 - \rho g (h-H)) S \vec{u_z}$ (upward)

Step 2: Apply static equilibrium (vertical direction)

$$\sum F_z = F - mg - \rho S H g - P_0 S + (P_0 - \rho g (h-H)) S = 0$$

Step 3: Solve for the magnitude of the force $F$

Let's simplify the equation:

$$F - mg - \rho g h S = 0 \implies F = mg + \rho g h S$$

Step 4: Express as a vector

The force acts upward (along $\vec{u_z}$):

$$\vec{F} = (mg + \rho g h S) \vec{u_z}$$

Final Check

• $P_w$ decreases linearly with $h$; at $h=0$, $P_w = P_0$ (reasonable).
• $F$ increases linearly with $h$; at $h=0$, $F=mg$ (supports only the tube's weight). The term $\rho g h S$ is the weight of the lifted water column.

Answers

(1) Pressure at the top of the tube:

$$\boxed{P_0 - \rho g h}$$

(2) Force to maintain equilibrium:

$$\boxed{(mg + \rho g h S)\vec{u_z}}$$

To determine the behavior of the tube during lifting and the corresponding values of $F_{\max}$ and $h^*$, we analyze the pressure equilibrium and the conditions under which vaporization (cavitation) occurs.

The vertical cylinder, sealed at the top and open at the bottom, is lifted from the bath (where $h = 0$) to the position where its bottom end reaches the surface (i.e., $h = H = 1 \, \mathrm{m}$). The pressure at the top of the tube is governed by:

$$P_{\mathrm{top}} = P_0 - \rho g h$$

Cavitation (formation of vapor) occurs when $P_{\mathrm{top}} = P_{\mathrm{sat}}$, i.e., when:

$$h^* = \frac{P_0 - P_{\mathrm{sat}}}{\rho g}$$

• Behavior A occurs when $h^* > H$, meaning the tube remains completely filled with liquid throughout the lifting process ($z_\ell = h$).
• Behavior B occurs when $h^* < H$, meaning the water column inside the tube stops rising at $h^*$, and the top of the tube continues to rise while the water level inside remains at $z_\ell = h^*$.

The force $F$ required to hold the tube in equilibrium is given by:

$$F = mg + (P_0 - P_{\mathrm{top}})S$$

• For Behavior A, $P_{\mathrm{top}} = P_0 - \rho g h$, so: $$F = mg + \rho g S h$$ At $h = H$, $$F_{\max} = mg + \rho g S H$$
• For Behavior B, once $h > h^*$, $P_{\mathrm{top}} = P_{\mathrm{sat}}$, so: $$F_{\max} = mg + (P_0 - P_{\mathrm{sat}})S$$

Data Recap:

• Cross-sectional area: $S = 10 \, \mathrm{cm^2} = 1 \times 10^{-3} \, \mathrm{m^2}$
• Mass of the tube: $m = 0.5 \, \mathrm{kg}$
• Gravitational acceleration: $g = 9.8 \, \mathrm{m/s^2}$
• Atmospheric pressure: $P_0 = 1.000 \times 10^5 \, \mathrm{Pa}$

Experiment 1: Water at $20^\circ \mathrm{C}$

• $\rho = 1.00 \times 10^3 \, \mathrm{kg/m^3}$
• $P_{\mathrm{sat}} = 2.34 \times 10^3 \, \mathrm{Pa}$

$$h^* = \frac{1.000 \times 10^5 - 2.34 \times 10^3}{1.00 \times 10^3 \times 9.8} \approx 9.96 \, \mathrm{m} > H$$

$\rightarrow$ Behavior A

$$F_{\max} = mg + \rho g S H = 0.5 \times 9.8 + 1.00 \times 10^3 \times 9.8 \times 1 \times 10^{-3} = 4.9 + 9.8 = 14.7 \, \mathrm{N}$$

Experiment 2: Water at $80^\circ \mathrm{C}$

• $\rho = 0.97 \times 10^3 \, \mathrm{kg/m^3}$
• $P_{\mathrm{sat}} = 47.4 \times 10^3 \, \mathrm{Pa}$

$$h^* = \frac{1.000 \times 10^5 - 47.4 \times 10^3}{0.97 \times 10^3 \times 9.8} \approx 5.53 \, \mathrm{m} > H$$

$\rightarrow$ Behavior A

$$F_{\max} = mg + \rho g S H = 4.9 + 0.97 \times 10^3 \times 9.8 \times 1 \times 10^{-3} \approx 4.9 + 9.506 = 14.4 \, \mathrm{N}$$

Experiment 3: Water at $99^\circ \mathrm{C}$

• $\rho = 0.96 \times 10^3 \, \mathrm{kg/m^3}$
• $P_{\mathrm{sat}} = 99.8 \times 10^3 \, \mathrm{Pa}$

$$h^* = \frac{1.000 \times 10^5 - 99.8 \times 10^3}{0.96 \times 10^3 \times 9.8} \approx 0.0213 \, \mathrm{m} = 2.1 \, \mathrm{cm} < H$$

$\rightarrow$ Behavior B

$$F_{\max} = mg + (P_0 - P_{\mathrm{sat}})S = 4.9 + (200)(1 \times 10^{-3}) = 4.9 + 0.2 = 5.1 \, \mathrm{N}$$

Answer:

Experiment	Behaviour (A or B?)	$h^*$ (cm)	$F_{\max}$ (N)
1	A	None	14.7
2	A	None	14.4
3	B	2.1	5.1

To solve this problem, we need to analyze the forces acting on the tube as it is lifted from the bath.

Part A Analysis:

Let $h$ be the height of the top of the tube above the bath surface ($z=0$). The tube has length $H$, mass $m$, and cross-sectional area $S$.

The forces acting on the tube are gravity ($mg$ downwards), the external pulling force ($F$ upwards), the atmospheric pressure force on the top outer surface ($P_0 S$ downwards), and the fluid pressure force on the top inner surface ($P_{\mathrm{w}} S$ upwards).

The equation of motion (equilibrium) is:

$$F - mg - P_0 S + P_{\mathrm{w}} S = 0 \implies F = mg + (P_0 - P_{\mathrm{w}}) S$$

The pressure $P_{\mathrm{w}}$ inside the tube at the top depends on the height of the water column. As long as the tube is full of water, the pressure at height $h$ is related to the pressure at the bath surface ($P_0$) by hydrostatic equilibrium:

$$P_{\mathrm{w}} = P_0 - \rho g h$$

Substituting this into the force equation gives:

$$F(h) = mg + (P_0 - (P_0 - \rho g h)) S = mg + \rho g S h$$

This represents a linear increase in force with height $h$. This corresponds to the initial part of both behaviours.

However, the pressure $P_{\mathrm{w}}$ cannot drop below the saturated vapour pressure $P_{\text{sat}}$. If $P_{\mathrm{w}}$ reaches $P_{\text{sat}}$, the water column detaches from the top of the tube (vapor forms), and the pressure $P_{\mathrm{w}}$ remains constant at $P_{\text{sat}}$.

The height $h^{\star}$ at which this transition occurs is found by setting $P_{\mathrm{w}} = P_{\text{sat}}$:

$$P_{\text{sat}} = P_0 - \rho g h^{\star} \implies h^{\star} = \frac{P_0 - P_{\text{sat}}}{\rho g}$$

If $h^{\star} \ge H$, the pressure never reaches $P_{\text{sat}}$ before the tube is fully lifted ($h=H$). The force increases linearly throughout. This is Behaviour A.

The maximum force is at $h=H$:

$$F_{\max} = mg + \rho g S H$$

If $h^{\star} < H$, the pressure reaches $P_{\text{sat}}$ at $h = h^{\star}$. For $h > h^{\star}$, the pressure $P_{\mathrm{w}}$ is fixed at $P_{\text{sat}}$, and the force becomes constant:

$$F(h) = mg + (P_0 - P_{\text{sat}}) S$$

This plateau corresponds to Behaviour B.

The maximum force is:

$$F_{\max} = mg + (P_0 - P_{\text{sat}}) S$$

Calculations for the Experiments:

Given: $H = 1 \mathrm{m}$, $S = 10 \mathrm{cm}^2 = 10^{-3} \mathrm{m}^2$, $m = 0.5 \mathrm{kg}$, $g = 9.8 \mathrm{m/s^2}$, $P_0 = 1.000 \times 10^5 \mathrm{Pa}$.

Experiment 1:

$\rho = 1.00 \times 10^3 \mathrm{kg/m^3}$, $P_{\text{sat}} = 2.34 \times 10^3 \mathrm{Pa}$.

$$h^{\star} = \frac{100000 - 2340}{1000 \times 9.8} = \frac{97660}{9800} \approx 9.97 \mathrm{m}$$

Since $h^{\star} > H$, it is Behaviour A.

$h^{\star}$ is not pertinent (marked as 'None').

$$F_{\max} = mg + \rho g S H = 0.5(9.8) + 1000(9.8)(10^{-3})(1) = 4.9 + 9.8 = 14.7 \mathrm{N}$$

Experiment 2:

$\rho = 0.97 \times 10^3 \mathrm{kg/m^3}$, $P_{\text{sat}} = 47.4 \times 10^3 \mathrm{Pa}$.

$$h^{\star} = \frac{100000 - 47400}{970 \times 9.8} = \frac{52600}{9506} \approx 5.53 \mathrm{m}$$

Since $h^{\star} > H$, it is Behaviour A.

$h^{\star}$ is 'None'.

$$F_{\max} = mg + \rho g S H = 0.5(9.8) + 970(9.8)(10^{-3})(1) = 4.9 + 9.506 = 14.406 \approx 14.4 \mathrm{N}$$

Experiment 3:

$\rho = 0.96 \times 10^3 \mathrm{kg/m^3}$, $P_{\text{sat}} = 99.8 \times 10^3 \mathrm{Pa}$.

$$h^{\star} = \frac{100000 - 99800}{960 \times 9.8} = \frac{200}{9408} \approx 0.02126 \mathrm{m} = 2.13 \mathrm{cm}$$

Since $h^{\star} < H$, it is Behaviour B.

$h^{\star} = 2.13 \mathrm{cm}$.

$$F_{\max} = mg + (P_0 - P_{\text{sat}}) S = 0.5(9.8) + (200)(10^{-3}) = 4.9 + 0.2 = 5.1 \mathrm{N}$$

Answers:

Experiment	Behaviour (A or B?)	$h^{\star}$ (cm)	$F_{\max}$ (N)
1	A	None	14.7
2	A	None	14.4
3	B	2.13	5.1

An elegant and often robust method to solve this kind of problem is to use a free body diagram on a cleverly chosen system. Let's define our system as the tube plus all its contents (the water and the gas/vapor inside). We will assume the tube wall thickness is negligible, so its inner and outer cross-sectional areas are both equal to $S$.

Let's analyze the forces acting on this system in the vertical direction ($\vec{u_z}$). The system is in equilibrium, so the net force is zero.

The forces are:

1. The pulling force $\vec{F} = F\vec{u_z}$.
2. The total weight of the system $\vec{W}_{\text{tot}}$. This is the weight of the tube, $m\vec{g}$, plus the weight of the water inside it, $m_w\vec{g}$. The weight of the gas/vapor is negligible.
3. The force exerted by the surrounding fluid (air and water). This force can be split into two parts:
   • The buoyant force, $\vec{F}_B$, due to the pressure gradient in the surrounding fluid.
   • The pressure force on the top horizontal surface of the system, $\vec{F}_P$, which is exposed to the uniform atmospheric pressure $P_0$.

The equilibrium equation is: $\vec{F} + \vec{W}_{\text{tot}} + \vec{F}_B + \vec{F}_P = \vec{0}$.

For a given tube top altitude $h$:

• Total Weight: $W_{\text{tot}} = (m + \rho S (z_{\ell} - h + H))g$
• Buoyant Force: $F_B = \rho g S (H-h)$
• Top Pressure Force: $F_P = -P_0 S$

This gives: $F = mg + P_0 S + \rho g S z_{\ell}$

Phase 1: Tube is full of water

When $z_{\ell} = h$ and $P_w = P_0 - \rho g h \ge P_{\text{sat}}$, the force is:

$$F(h) = mg + P_0 S + \rho g S h$$

Cavitation occurs at $h_c = \frac{P_0 - P_{\text{sat}}}{\rho g}$.

Phase 2: Vapor bubble has formed

For $h > h_c$, $P_w = P_{\text{sat}}$ and $z_{\ell} = h_c$, so:

$$F(h) = mg + P_0 S + \rho g S h_c$$

Numerical Calculations:

Given: $m=0.5 \mathrm{kg}$, $S=10^{-3} \mathrm{m}^2$, $g=9.8 \mathrm{m/s}^2$, $P_0=1.000 \times 10^5 \mathrm{Pa}$, $H=1 \mathrm{m}$.

Experiment 1: Water at $20^{\circ}\mathrm{C}$

$\rho = 1.00 \times 10^3 \mathrm{kg/m^3}$, $P_{\text{sat}} = 2.34 \times 10^3 \mathrm{Pa}$.

$h_c = \frac{1.000 \times 10^5 - 2.34 \times 10^3}{1.00 \times 10^3 \times 9.8} \approx 9.965 \mathrm{m} > H$

Behavior: A, $h^*$: None

$F_{\max} = mg + P_0 S + \rho g S H = 4.9 + 100 + 9.8 = 114.7 \mathrm{N}$

Experiment 2: Water at $80^{\circ}\mathrm{C}$

$\rho = 0.97 \times 10^3 \mathrm{kg/m^3}$, $P_{\text{sat}} = 47.4 \times 10^3 \mathrm{Pa}$.

$h_c = \frac{1.000 \times 10^5 - 47.4 \times 10^3}{0.97 \times 10^3 \times 9.8} \approx 5.533 \mathrm{m} > H$

Behavior: A, $h^*$: None

$F_{\max} = mg + P_0 S + \rho g S H = 4.9 + 100 + 9.506 = 114.406 \approx 114.4 \mathrm{N}$

Experiment 3: Water at $99^{\circ}\mathrm{C}$

$\rho = 0.96 \times 10^3 \mathrm{kg/m^3}$, $P_{\text{sat}} = 99.8 \times 10^3 \mathrm{Pa}$.

$h_c = \frac{1.000 \times 10^5 - 99.8 \times 10^3}{0.96 \times 10^3 \times 9.8} \approx 0.02126 \mathrm{m} = 2.126 \mathrm{cm}$

Behavior: A, $h^*$: 2.13 cm

$F_{\max} = mg + P_0 S + \rho g S h_c = 4.9 + 100 + 0.1999 \approx 105.1 \mathrm{N}$

Final Table:

Experiment	Behaviour	$h^{\star}$ (cm)	$F_{\max} (N)$
1	A	None	114.7
2	A	None	114.4
3	A	2.13	105.1