§14.3 — Archimedes’ Principle and Buoyancy

When an object is immersed in a fluid, the varying pressure on different parts of the object creates a net upward force. This phenomenon is captured by Archimedes' principle: a body wholly or partially immersed in a fluid is buoyed up by a force equal to the weight of the fluid it displaces.

Deriving the buoyant force

Consider an object of volume \( V_{\text{obj}} \) immersed in a fluid of density \( \rho_f \). The pressure on the bottom surface is greater than the pressure on the top surface (from §14.2). This pressure difference acts perpendicular to the object at every point.

For a submerged object, imagine replacing it temporarily with the fluid that would occupy the same space. This "displaced fluid" has mass \( m_f = \rho_f V_{\text{obj}} \) and is in equilibrium in the surrounding fluid. The net upward force that supports it is:

\[ F_B = m_f g = \rho_f V_{\text{obj}} g \]

This force is Archimedes' buoyant force, the upward pressure force on the object. It acts at the center of buoyancy, which coincides with the center of mass of the displaced fluid.

Key insight: the buoyant force depends only on the fluid density, the volume of the object, and gravity—not on the material of the object or its depth (as long as the object is fully submerged).

Condition for floating or sinking

An object in a fluid experiences two forces:

• Weight (downward): \( W = m_{\text{obj}} g = \rho_{\text{obj}} V_{\text{obj}} g \)
• Buoyancy (upward): \( F_B = \rho_f V_{\text{obj}} g \)

The net force is: \[ F_{\text{net}} = F_B - W = (\rho_f - \rho_{\text{obj}}) V_{\text{obj}} g \]

Three cases:

1. \( \rho_{\text{obj}} < \rho_f \) (object less dense than fluid) → \( F_{\text{net}} > 0 \) → object floats
2. \( \rho_{\text{obj}} = \rho_f \) (neutral buoyancy) → \( F_{\text{net}} = 0 \) → object hovers
3. \( \rho_{\text{obj}} > \rho_f \) (object denser than fluid) → \( F_{\text{net}} < 0 \) → object sinks

For a floating object at the surface, buoyancy equals weight: \[ \rho_f V_{\text{submerged}} g = \rho_{\text{obj}} V_{\text{total}} g \]

\[ \frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{\rho_{\text{obj}}}{\rho_f} \]

This is why an iceberg (density ≈ 900 kg/m³) in ocean water (≈ 1025 kg/m³) floats with about 88% submerged: \( (900/1025) \approx 0.88 \).

Worked example: Floating wooden raft

A wooden raft is made of pine wood with density \( \rho_{\text{wood}} = 600 \) kg/m³. The raft has dimensions 4 m × 3 m × 0.5 m (length × width × thickness). It carries a payload of mass \( m_{\text{payload}} = 2000 \) kg. Will it float in fresh water (\( \rho_{\text{water}} = 1000 \) kg/m³)?

Step 1: Total mass \[ m_{\text{wood}} = \rho_{\text{wood}} \times V_{\text{raft}} = 600 \times (4 \times 3 \times 0.5) = 600 \times 6 = 3600 \text{ kg} \] \[ m_{\text{total}} = 3600 + 2000 = 5600 \text{ kg} \]

Step 2: Weight to support \[ W_{\text{total}} = 5600 \times 9.8 = 54,880 \text{ N} \]

Step 3: Maximum buoyant force (if fully submerged) \[ F_{B,\text{max}} = \rho_{\text{water}} \times V_{\text{raft}} \times g = 1000 \times 6 \times 9.8 = 58,800 \text{ N} \]

Step 4: Check flotation Since \( F_{B,\text{max}} > W_{\text{total}} \) (58,800 N > 54,880 N), the raft floats.

Step 5: Find draft (submerged depth) At equilibrium: \( F_B = W_{\text{total}} \) \[ \rho_{\text{water}} \times g \times (4 \times 3 \times d) = 54,880 \] \[ 1000 \times 9.8 \times 12d = 54,880 \] \[ d = \frac{54,880}{117,600} \approx 0.467 \text{ m} \approx 46.7 \text{ cm} \]

The raft sinks to about 47 cm, leaving 3 cm above water.

Buoyancy in gases

Archimedes' principle applies to gases too. A hot-air balloon floats because hot air (\( \rho_{\text{hot air}} \approx 1.0 \) kg/m³) is less dense than cool outside air (\( \rho_{\text{cool air}} \approx 1.2 \) kg/m³). The buoyant force on a 2000 m³ balloon is:

\[ F_B = 1.2 \times 2000 \times 9.8 \approx 23,500 \text{ N} \approx 2400 \text{ kg}_f \]

Subtracting the weight of the balloon and payload, the net lift is what carries the basket aloft.

Apparent weight in a fluid

When an object is submerged, its apparent weight (what a scale reads) is: \[ W_{\text{app}} = W - F_B = mg - \rho_f V g \]

A 70 kg person immersed in water appears lighter because buoyancy reduces the normal force from a scale.

What you should be able to do

• State Archimedes' principle and derive the buoyant force formula
• Predict whether an object floats, sinks, or hovers based on densities
• Calculate the fraction of a floating object that is submerged
• Find the draft (immersion depth) of a floating vessel
• Recognize that buoyancy applies equally to gases and liquids
• Estimate apparent weight of a submerged object

Preview of §14.4

Fluid dynamics and the continuity equation: how fluids flow through pipes and around objects.

📚 See also: Halliday Vol 2, Ch 14, §14.3.

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