Archimedes Principle

Overview

Engineering reference data for Archimedes Principle in fluid mechanics.

Key Formulas

Reynolds Number

$Re = \frac{\rho v D}{\mu}$

Ratio of inertial to viscous forces — determines flow regime.

Bernoulli's Equation

$P + \frac{1}{2}\rho v^2 + \rho g h = \text{const}$

Conservation of energy for steady, inviscid, incompressible flow.

Continuity Equation

$A_1 v_1 = A_2 v_2$

Conservation of mass for incompressible flow.

Darcy-Weisbach

$\Delta P = f \frac{L}{D} \frac{\rho v^2}{2}$

Pressure drop due to friction in a pipe.

Variables

Symbol	Description	Unit
$Re$	Reynolds number	—
$\rho$	Fluid density	kg/m³
$v$	Flow velocity	m/s
$D$	Characteristic dimension	m
$\mu$	Dynamic viscosity	Pa·s
$P$	Pressure	Pa
$f$	Darcy friction factor	—

References

• Original Source

Advanced Worked Examples

Example: Calculating Density of a Floating Body

A floating body is 95% submerged in water with density 1000 kg/m³. For a floating body, the buoyant force equals the weight of the displaced liquid.

Given:

• Submerged fraction: 95%
• Water density ($\rho_w$): 1000 kg/m³

Derivation: For equilibrium: $F_B = W \quad \Rightarrow \quad V_b \rho_b g = V_w \rho_w g$

Where:

• $V_b$ = volume of the body (m³)
• $V_w$ = volume of displaced water (m³)

Solving for body density: $\rho_b = \frac{V_w}{V_b} \rho_w$

Since 95% is submerged: $V_w = 0.95 V_b$

Therefore: $\rho_b = 0.95 \times 1000 \, \text{kg/m}^3 = 950 \, \text{kg/m}^3$

Example: Buoyant and Net Force on a Submerged Brick

A standard brick (dimensions 3⅝" × 2¼" × 8") is submerged in water ($\rho = 1.940 \, \text{slugs/ft}^3$).

Step 1: Calculate Brick Volume $V_{\text{brick}} = \left(3\frac{5}{8} \, \text{in}\right) \times \left(2\frac{1}{4} \, \text{in}\right) \times (8 \, \text{in}) = 65.25 \, \text{in}^3$

Convert to ft³: $V_{\text{brick}} = \frac{65.25 \, \text{in}^3}{1728 \, \text{in}^3/\text{ft}^3} = 0.03776 \, \text{ft}^3$

Step 2: Buoyant Force Calculation $F_B = V \rho g = (0.03776 \, \text{ft}^3)(1.940 \, \text{slugs/ft}^3)(32.174 \, \text{ft/s}^2) = 2.36 \, \text{lbf}$

Step 3: Weight of the Brick Using specific gravity (SG = 1.75): $W_B = F_B \times \text{SG} = 2.36 \, \text{lbf} \times 1.75 = 4.12 \, \text{lbf}$

Step 4: Net Force $F_{\text{net}} = W_B - F_B = 4.12 \, \text{lbf} - 2.36 \, \text{lbf} = 1.76 \, \text{lbf} \, \text{(downward)}$

Floating Body Density Relation

For a floating body，the fraction submerged equals the ratio of the body's density to the fluid's density.

$$\frac{V_{submerged}}{V_{body}} = \frac{\rho_{body}}{\rho_{fluid}}$$

This relation is derived from the equilibrium condition where the buoyant force equals the body's weight.

Practical Calculation Examples

Example 1：Determining Density from Submersion

A body floats with 95% of its volume submerged in water ($\rho_w = 1000 \, \text{kg/m}^3$). The body's density is calculated as：

$$\rho_b = \frac{V_w}{V_b} \cdot \rho_w = 0.95 \times 1000 \, \text{kg/m}^3 = 950 \, \text{kg/m}^3$$

Example 2：Net Force on a Submerged Brick

For a standard brick ($3\frac{5}{8} \times 2\frac{1}{4} \times 8 \, \text{inches}$) submerged in water ($\rho = 1.940 \, \text{slugs/ft}^3$)：

1. Brick Volume：

$$V_{brick} = \left(3\frac{5}{8} \, \text{in}\right) \times \left(2\frac{1}{4} \, \text{in}\right) \times \left(8 \, \text{in}\right) = 65.25 \, \text{in}^3$$

2. Buoyant Force：

$$F_B = \left(\frac{65.25 \, \text{in}^3}{1728 \, \text{in}^3/\text{ft}^3}\right) \times (1.940 \, \text{slugs/ft}^3) \times (32.174 \, \text{ft/s}^2) = 2.36 \, \text{lbf}$$

3. Weight of Brick (specific gravity $SG = 1.75$ for common red brick):

$$W_B = F_B \times SG = 2.36 \, \text{lbf} \times 1.75 = 4.12 \, \text{lbf}$$

4. Net Force (downward, since brick sinks):

$$F_{net} = W_B - F_B = 4.12 \, \text{lbf} - 2.36 \, \text{lbf} = 1.76 \, \text{lbf}$$

Buoyant Force Formula

The fundamental formula for buoyant force is expressed as:

$F_B = V \cdot \gamma = V \cdot \rho \cdot g$

Where:

• $F_B$ is the buoyant force (N, lbf)
• $V$ is the volume of the submerged portion of the body or displaced fluid (m³, ft³)
• $\gamma = \rho g$ is the specific weight of the fluid (N/m³, lbf/ft³)
• $\rho$ is the density of the fluid (kg/m³, slugs/ft³)
• $g$ is the acceleration due to gravity (9.81 m/s², 32.174 ft/s²)

Additional Practical Examples

Density of a Floating Body

For a body floating in equilibrium, the buoyant force equals the body's weight:

$F_B = W \quad \Rightarrow \quad V_w \cdot \rho_w \cdot g = V_b \cdot \rho_b \cdot g$

This simplifies to:

$\rho_b = \rho_w \cdot \frac{V_w}{V_b}$

If a body is 95% submerged in water ($\rho_w = 1000\ \text{kg/m}^3$), its density is:

$\rho_b = 1000\ \text{kg/m}^3 \times 0.95 = 950\ \text{kg/m}^3$

Buoyant Force on a Submerged Brick

Using imperial units for a standard brick (dimensions: $3\frac{5}{8} \times 2\frac{1}{4} \times 8\ \text{in}$):

1. Calculate brick volume: $V_{\text{brick}} = (3.625\ \text{in}) \times (2.25\ \text{in}) \times (8\ \text{in}) = 65.25\ \text{in}^3$
2. Convert volume to cubic feet: $V_{\text{brick}} = \frac{65.25\ \text{in}^3}{1728\ \text{in}^3/\text{ft}^3} \approx 0.03776\ \text{ft}^3$
3. Calculate buoyant force (water density $\rho_w = 1.940\ \text{slugs/ft}^3$): $F_B = V_{\text{brick}} \cdot \rho_w \cdot g = (0.03776\ \text{ft}^3) \cdot (1.940\ \text{slugs/ft}^3) \cdot (32.174\ \text{ft/s}^2) \approx 2.36\ \text{lbf}$
4. Calculate brick's weight (specific gravity SG = 1.75): $W_{\text{brick}} = F_B \cdot \text{SG} = (2.36\ \text{lbf}) \cdot 1.75 \approx 4.12\ \text{lbf}$
5. Net downward force: $F_{\text{net}} = W_{\text{brick}} - F_B = 4.12\ \text{lbf} - 2.36\ \text{lbf} = 1.76\ \text{lbf}$

Variables Reference Table

Additional Example: Brick Submersion Analysis

This detailed example demonstrates force analysis on a submerged object with known specific gravity.

Given:

• A standard red brick with dimensions: 3 5/8 in × 2 1/4 in × 8 in
• Water density, $ρ_w = 1.940 \text{ slugs/ft}^3$
• Acceleration of gravity, $g = 32.174 \text{ ft/s}^2$
• Specific gravity of common red brick, $SG = 1.75$

Step 1: Calculate the brick's volume. $V_{\text{brick}} = \left(3\frac{5}{8}\right) \times \left(2\frac{1}{4}\right) \times 8 = 65.25 \text{ in}^3$

Step 2: Calculate the buoyant force. The buoyant force equals the weight of the water displaced by the brick. $F_B = \left( \frac{V_{\text{brick}}}{1728 \text{ in}^3/\text{ft}^3} \right) \rho_w g$ $F_B = \left( \frac{65.25}{1728} \right) \times 1.940 \times 32.174 \approx 2.36 \text{ lbf}$

Step 3: Calculate the brick's weight in water. Using the specific gravity, the weight in air ($W_{\text{air}}$) is related to the buoyant force. $W_{\text{air}} = F_B \times SG = 2.36 \text{ lbf} \times 1.75 \approx 4.13 \text{ lbf}$

Step 4: Calculate the resulting (net) force. The net force is the difference between the brick's weight in air and the buoyant force. $F_{\text{net}} = W_{\text{air}} - F_B = 4.13 - 2.36 \approx 1.77 \text{ lbf (downward)}$

This net force determines that the brick will sink in water.