Buoyancy and Archimedes' Principle
Buoyant force, floating and sinking, apparent weight in fluids
Try the Interactive Version!
Learn step-by-step with practice exercises built right in.
⬆️ Buoyancy and Archimedes' Principle
Introduction
When objects are submerged in fluids, they experience an upward force called buoyancy. This explains why ships float, balloons rise, and you feel lighter in water.
Archimedes' Principle
Archimedes' Principle: An object fully or partially submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced.
where:
- = buoyant force (N)
- = density of the fluid (kg/m³)
- = 9.8 m/s²
- = volume of fluid displaced (m³)
Key Insights:
- Buoyant force depends on fluid density, not object density
- Buoyant force depends on displaced volume
- Direction is always upward (opposes gravity)
Floating, Sinking, and Equilibrium
Two forces act on a submerged object:
- Weight: ⬇️
- Buoyant force: ⬆️
Three Cases:
1. Object Sinks ()
- Weight > Buoyant force
- Net force downward
- Accelerates to bottom
2. Object Floats ()
- Partially submerged
- At equilibrium:
- Fraction submerged:
3. Neutral Buoyancy ()
- Weight = Buoyant force
- Remains at any depth (like submarine)
- Fully submerged, no net force
Fraction Submerged
For a floating object in equilibrium:
Example: Ice ( kg/m³) in water ( kg/m³):
This is why icebergs show only ~10% above water!
Apparent Weight
When submerged in a fluid, an object's apparent weight is reduced:
For a fully submerged object:
💡 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 , area :
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
- Draw free body diagram (weight down, buoyancy up)
- Identify situation: floating, sinking, or neutral?
- Apply Archimedes':
- For equilibrium (floating):
- For apparent weight:
- Watch volumes: fully or partially submerged?
Common Mistakes
❌ Using object density for (use fluid density!) ❌ Forgetting partial submersion (floating objects) ❌ Sign errors with apparent weight ❌ Assuming floating = fully submerged ❌ Unit inconsistency (keep kg/m³, m³)
📚 Practice Problems
1Problem 1easy
❓ Question:
A solid cube with sides of 10 cm is fully submerged in water. What is the buoyant force on the cube? (ρ_water = 1000 kg/m³)
💡 Show Solution
Given:
- Side length: cm m
- Water density: kg/m³
Find: Buoyant force
Solution:
Step 1: Calculate volume.
Step 2: Apply Archimedes' Principle.
Since fully submerged:
Answer: 9.8 N upward
This equals the weight of 1.0 kg (1 liter) of water displaced! ✓
2Problem 2medium
❓ Question:
A wooden block (ρ = 600 kg/m³) floats in water (ρ = 1000 kg/m³). What fraction is submerged? If the block has mass 2.0 kg, what volume of water is displaced?
💡 Show Solution
Given:
- Wood: kg/m³
- Water: kg/m³
- Mass: kg
Solution:
Part (a): Fraction submerged
Part (b): Volume displaced
Step 1: Find total volume.
Step 2: Find displaced volume.
Verification: ✓
Answer:
- (a) 60% submerged
- (b) 2.0 L displaced
3Problem 3hard
❓ Question:
A brass weight (m = 5.0 kg, ρ = 8500 kg/m³) is held underwater by a string. (a) What is its weight in air? (b) What is its apparent weight in water? (c) What is the string tension?
💡 Show Solution
Given:
- Mass: kg
- Brass: kg/m³
- Water: kg/m³
Solution:
Part (a): Weight in air
Part (b): Apparent weight
Step 1: Find volume.
Step 2: Calculate buoyant force.
Step 3: Find apparent weight.
Part (c): String tension
The string provides the apparent weight:
Alternative method: ✓
Answer:
- (a) 49 N in air
- (b) 43.2 N apparent (12% lighter!)
- (c) 43.2 N tension