Problem 2.173 — Surface Tension: Measurement by Capillary Rise

Problem Statement

Solve the fluid mechanics problem: Solve the Newton’s Laws / mechanics problem: Water rises to $h=10.5\ \text{cm}$ in a capillary of $r=0.14\ \text{mm}$. Find $\sigma$ for water. ($\rho=1000\ \text{kg/m}^3$, $\theta=0°$) $$h = \frac{2\sigma\cos\theta}{\rho g r} \implies \sigma = \frac{h\rho g r}{2\cos\theta}$$ $$\sigma = \frac{0.105\

Given Information

• See problem statement for all given quantities.

Physical Concepts & Formulas

Newton’s second law $\mathbf{F}_\text{net} = m\mathbf{a}$ is the fundamental relation between net force and acceleration. For systems of connected objects (Atwood machine, blocks on inclines), each body is treated separately with a free-body diagram, and the constraint equations (same rope length, etc.) link the accelerations.

• $\mathbf{F}_{\text{net}} = m\mathbf{a}$ — Newton’s second law
• Atwood: $a = (m_1-m_2)g/(m_1+m_2)$, $T = 2m_1m_2g/(m_1+m_2)$
• $f_k = \mu_k N$ — kinetic friction

Step-by-Step Solution

Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Worked Calculation

$$h = \frac{2\sigma\cos\theta}{\rho g r} \implies \sigma = \frac{h\rho g r}{2\cos\theta}$$

$$\sigma = \frac{0.105\

Given Information

• Fluid density $\rho$, velocities, cross-sections, and heights as given
• Atmospheric pressure $P_0 = 1.013\times10^5\,\text{Pa}$
• $g = 9.8\,\text{m/s}^2$

Physical Concepts & Formulas

Fluid statics is governed by Pascal’s Law and the hydrostatic pressure formula $P = P_0 + \rho g h$. Archimedes’ principle states that the buoyant force equals the weight of fluid displaced: $F_b = \rho_f V g$. Fluid dynamics for ideal (incompressible, non-viscous, steady) flow uses two key results: the continuity equation $A_1 v_1 = A_2 v_2$ (mass conservation) and Bernoulli’s equation $P + \frac{1}{2}\rho v^2 + \rho g h = \text{const}$ (energy conservation per unit volume).

• $P = P_0 + \rho g h$ — hydrostatic pressure
• $F_b = \rho_f V g$ — Archimedes buoyancy
• $A_1 v_1 = A_2 v_2$ — continuity equation
• $P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{const}$ — Bernoulli’s equation
• $v_{\text{efflux}} = \sqrt{2gh}$ — Torricelli’s theorem

Step-by-Step Solution

Step 1 — Identify the situation: Static (use $P = P_0+\rho gh$ and Archimedes) or dynamic (use continuity + Bernoulli).

Step 2 — Apply Bernoulli between two points at the same streamline:

$$

$$

Step 3 — Use continuity to relate $v_1$ and $v_2$: $v_2 = v_1 A_1/A_2$.

Step 4 — Solve for the unknown (pressure, velocity, or flow rate).

Worked Calculation

Substituting all values with units:

Torricelli: Tank depth $h = 2\,\text{m}$, hole at bottom:

$$

Answer

$$\boxed{v_{\text{efflux}} = \sqrt{2gh}}$$

Physical Interpretation

Capillary action allows plants to draw water from roots to leaves against gravity. The thinner the tube, the higher the rise — but also the smaller the volume transported.