where:

• $F_B$ = buoyant force (N)
• $\rho_{fluid}$ = density of the fluid (kg/m³)
• $g$ = 9.8 m/s²
• $V_{displaced}$ = volume of fluid displaced (m³)

Key Insights:

1. Buoyant force depends on fluid density, not object density
2. Buoyant force depends on displaced volume
3. Direction is always upward (opposes gravity)

Floating, Sinking, and Equilibrium

Two forces act on a submerged object:

• Weight: $F_g = mg = \rho_{object} g V_{object}$ ⬇️
• Buoyant force: $F_B = \rho_{fluid} g V_{displaced}$ ⬆️

Three Cases:

1. Object Sinks ($\rho_{object} > \rho_{fluid}$)

• Weight > Buoyant force
• Net force downward
• Accelerates to bottom

2. Object Floats ($\rho_{object} < \rho_{fluid}$)

• Partially submerged
• At equilibrium: $F_B = F_g$
• Fraction submerged: $\frac{V_{sub}}{V_{total}} = \frac{\rho_{object}}{\rho_{fluid}}$

3. Neutral Buoyancy ($\rho_{object} = \rho_{fluid}$)

• Weight = Buoyant force
• Remains at any depth (like submarine)
• Fully submerged, no net force

Fraction Submerged

For a floating object in equilibrium:

$\frac{V_{submerged}}{V_{total}} = \frac{\rho_{object}}{\rho_{fluid}}$

Example: Ice ($\rho = 917$ kg/m³) in water ($\rho = 1000$ kg/m³):

$\frac{V_{sub}}{V_{total}} = \frac{917}{1000} = 0.917 = 91.7\%$

This is why icebergs show only ~10% above water!

Apparent Weight

When submerged in a fluid, an object's apparent weight is reduced:

$W_{apparent} = W_{actual} - F_B$

For a fully submerged object:

$W_{apparent} = \rho_{object} gV - \rho_{fluid} gV = (\rho_{object} - \rho_{fluid}) gV$

💡 This is why you feel lighter in a pool - buoyant force reduces apparent weight!

Why Buoyancy Occurs (Pressure Explanation)

Buoyancy arises from pressure differences in the fluid:

• Bottom of object: higher pressure (greater depth)
• Top of object: lower pressure
• Net upward force from pressure difference

For a cube of height $h$, area $A$: $F_B = (P_{bottom} - P_{top}) \cdot A = \rho g h \cdot A = \rho g V$

Matches Archimedes' Principle! ✓

Applications

Ships and Boats

• Average density (including air spaces) < water density
• Displace enough water so buoyant force = weight

Hot Air Balloons

• Hot air less dense than cool air
• Buoyant force from cool air > weight → rises

Submarines

Control depth by adjusting average density:

• Dive: Fill ballast tanks with water → ↑ density → sink
• Surface: Blow water out → ↓ density → rise
• Cruise: Balance water/air → neutral buoyancy

Problem-Solving Strategy

1. Draw free body diagram (weight down, buoyancy up)
2. Identify situation: floating, sinking, or neutral?
3. Apply Archimedes': $F_B = \rho_{fluid} g V_{displaced}$
4. For equilibrium (floating): $F_B = F_g$
5. For apparent weight: $W_{app} = W - F_B$
6. Watch volumes: fully or partially submerged?

Common Mistakes

❌ Using object density for $F_B$ (use fluid density!) ❌ Forgetting partial submersion (floating objects) ❌ Sign errors with apparent weight ❌ Assuming floating = fully submerged ❌ Unit inconsistency (keep kg/m³, m³)

Find: Buoyant force $F_B$

Solution:

Step 1: Calculate volume. $V = L^3 = (0.10)^3 = 1.0 \times 10^{-3} \text{ m}^3 = 1.0 \text{ L}$

Step 2: Apply Archimedes' Principle. $F_B = \rho_{fluid} g V_{displaced}$

Since fully submerged: $V_{displaced} = V$

$F_B = (1000)(9.8)(1.0 \times 10^{-3}) = 9.8 \text{ N}$

Answer: 9.8 N upward

This equals the weight of 1.0 kg (1 liter) of water displaced! ✓