Archimedes’ Principle
Archimedes’ principle states: a body wholly or partially submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces.
Core Statement and Notation
The buoyant force magnitude for an incompressible fluid is
\[ F_B \;=\; \rho_f \, g \, V_{\text{disp}} \]- \(\rho_f\): fluid density
- \(g\): gravitational acceleration
- \(V_{\text{disp}}\): volume of displaced fluid (volume of the submerged part of the body)
The buoyant force acts vertically upward through the center of buoyancy, the centroid of the displaced fluid volume.
Hydrostatics Background
In a static fluid of uniform density, pressure varies with depth as
\[ p(h) \;=\; p_0 \;+\; \rho_f \, g \, h, \]where \(p_0\) is a reference pressure and \(h\) is the vertical depth below the free surface (measured positive downward).
Derivation by Pressure Integration
The pressure field exerts a normal force \(\mathrm{d}\mathbf{F} = -p\,\mathbf{n}\,\mathrm{d}A\) on each surface element. Integrating over a closed surface \(S\) of the submerged body:
\[ \mathbf{F}_B \;=\; \oint_S \! (-p\,\mathbf{n})\,\mathrm{d}A \;=\; -\oint_S \! (p_0 + \rho_f g h)\,\mathbf{n}\,\mathrm{d}A. \]The \(p_0\) term integrates to zero over a closed surface. The remaining term is equivalent to the weight of the displaced fluid:
\[ \mathbf{F}_B \;=\; \rho_f g \, \hat{\mathbf{z}} \, V_{\text{disp}}, \]pointing upward. Thus, the net hydrostatic force equals the weight of the displaced fluid, independent of the object’s shape (hydrostatic paradox).
Float, Sink, and Equilibrium
Weight vs Buoyancy
\[ W \;=\; m g,\qquad W_{\text{app}} \;=\; W - F_B. \]- Float: \(F_B \ge W\) at equilibrium (for a floating body at rest, \(F_B = W\)).
- Sink: \(F_B < W\).
- Neutral buoyancy: \(F_B = W\) with full submergence (e.g., tuned submarines, fish with swim bladders).
Fraction Submerged (Floating Bodies)
\[ \rho_f \, g \, V_{\text{sub}} \;=\; m g \;\;\Rightarrow\;\; \frac{V_{\text{sub}}}{V_{\text{obj}}} \;=\; \frac{\rho_{\text{obj}}}{\rho_f}. \]A body floats with a submerged fraction equal to its average density divided by the fluid density.
Fully Submerged, Tethered Objects
For a submerged object held by a vertical tether of tension \(T\):
\[ T \;=\; W - F_B \;=\; m g - \rho_f g V_{\text{obj}}. \]If \(T < 0\), the object tends to rise (tether would go slack).
Center of Buoyancy and Stability
The buoyant force acts through the center of buoyancy \(B\) (centroid of displaced volume). Stability of a floating body depends on the relative positions of the center of gravity \(G\), center of buoyancy \(B\), and the metacenter \(M\).
- Metacentric height: \(GM = BM - BG\), with \[ BM \;=\; \frac{I_w}{V_{\text{disp}}}, \] where \(I_w\) is the second moment of area of the waterplane about the roll axis.
- Static stability: small-angle stability requires \(GM > 0\) (the righting moment restores the upright position).
Practical Computations
Determining Density via Weighing in a Fluid
Let \(W_a\) be weight in air and \(W_f\) apparent weight when immersed in a fluid of density \(\rho_f\). Then
\[ V_{\text{obj}} \;=\; \frac{W_a - W_f}{\rho_f g}, \qquad \rho_{\text{obj}} \;=\; \frac{m}{V_{\text{obj}}} \;=\; \rho_f \,\frac{W_a}{W_a - W_f}. \]Hydrometers
A hydrometer floats deeper in a less dense fluid so that the displaced fluid weight equals the hydrometer’s weight. The stem scale maps immersion depth to fluid density via \( \rho_f = \rho_{\text{hyd}}\,V_{\text{hyd}}/V_{\text{sub}} \).
Balloons and Buoyancy in Gases
For a balloon of volume \(V\), filled with gas of density \(\rho_g\) in air of density \(\rho_{\text{air}}\), the buoyant force is \(F_B = \rho_{\text{air}} g V\). The gas inside has weight \(\rho_g g V\). Neglecting small skin volume, the ideal net lift is
\[ L_{\text{ideal}} \;=\; (\rho_{\text{air}} - \rho_g)\, g \, V. \]Accounting for envelope and payload mass \(m_s\), the useful lift is \(L_{\text{useful}} = L_{\text{ideal}} - m_s g\).
Worked Examples
Example 1: Submerged Fraction of a Floating Block
A wooden block has average density \(\rho_{\text{obj}} = 600\ \text{kg/m}^3\) and floats in water \(\rho_f = 1000\ \text{kg/m}^3\). The submerged fraction is
\[ \frac{V_{\text{sub}}}{V_{\text{obj}}} \;=\; \frac{\rho_{\text{obj}}}{\rho_f} \;=\; \frac{600}{1000} \;=\; 0.60. \]Therefore, 60% of its volume lies below the waterline.
Example 2: Apparent Weight of a Submerged Metal Object
A solid object has mass \(m = 2.0\ \text{kg}\) and volume \(V = 2.2 \times 10^{-4}\ \text{m}^3\). In water (\(\rho_f = 1000\ \text{kg/m}^3\)), its apparent weight is
\[ W = m g,\quad F_B = \rho_f g V \;\Rightarrow\; W_{\text{app}} = (m - \rho_f V)\, g. \]Numerically, \(\rho_f V = 0.22\ \text{kg}\), so \(W_{\text{app}} = (2.0 - 0.22)g = 1.78\,g \approx 17.5\ \text{N}\).
Example 3: Hot-Air Balloon Payload
A balloon of volume \(V = 2800\ \text{m}^3\) operates where \(\rho_{\text{air}} = 1.20\ \text{kg/m}^3\). The heated air inside averages \(\rho_g = 0.95\ \text{kg/m}^3\). Envelope plus basket mass is \(m_s = 300\ \text{kg}\). The useful lift is
\[ L_{\text{useful}} = \big[(1.20 - 0.95)\,g\,2800\big] - 300\,g \;=\; (0.25 \times 2800 - 300)\, g \;=\; (700 - 300)\, g \;=\; 400\, g. \]The maximum additional payload is about \(400\ \text{kg}\) (neglecting small corrections).
Common Pitfalls and Edge Cases
- Confusing mass and weight: compare forces; use \(W = m g\).
- Wrong fluid density: \(\rho_f\) varies with temperature, salinity (water), or altitude (air).
- Compressibility: in gases, \(\rho_f\) changes with height; \(F_B\) can vary with position for large vertical extents.
- Shape vs. material: buoyancy depends on displaced volume, not material; hollow steel ships float, solid blocks sink.
- Stability ≠ floating: a body can float yet be unstable if \(GM \le 0\).
Quick Calculator
Enter any three values to evaluate buoyant force, float/sink, submerged fraction (if floating), and apparent weight. Use SI units.
Results
Key Formulas Summary
- Buoyancy (incompressible): \[ F_B = \rho_f g V_{\text{disp}} \]
- Apparent weight: \[ W_{\text{app}} = W - F_B \]
- Floating equilibrium: \[ \rho_f g V_{\text{sub}} = m g \;\Rightarrow\; \dfrac{V_{\text{sub}}}{V_{\text{obj}}} = \dfrac{\rho_{\text{obj}}}{\rho_f} \]
- Tether tension (fully submerged): \[ T = m g - \rho_f g V_{\text{obj}} \]
- Metacentric height: \[ GM = BM - BG,\quad BM = \dfrac{I_w}{V_{\text{disp}}} \]