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Buoyancy and Archimedes' Principle

Buoyant force, floating and sinking, apparent weight in fluids

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⬆️ 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.

FB=ρfluid⋅g⋅VdisplacedF_B = \rho_{fluid} \cdot g \cdot V_{displaced}

where:

  • FBF_B = buoyant force (N)
  • ρfluid\rho_{fluid} = density of the fluid (kg/m³)
  • gg = 9.8 m/s²
  • VdisplacedV_{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: Fg=mg=ρobjectgVobjectF_g = mg = \rho_{object} g V_{object} ⬇️
  • Buoyant force: FB=ρfluidgVdisplacedF_B = \rho_{fluid} g V_{displaced} ⬆️

Three Cases:

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

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

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

  • Partially submerged
  • At equilibrium: FB=FgF_B = F_g
  • Fraction submerged: VsubVtotal=ρobjectρfluid\frac{V_{sub}}{V_{total}} = \frac{\rho_{object}}{\rho_{fluid}}

3. Neutral Buoyancy (ρobject=ρfluid\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:

VsubmergedVtotal=ρobjectρfluid\frac{V_{submerged}}{V_{total}} = \frac{\rho_{object}}{\rho_{fluid}}

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

VsubVtotal=9171000=0.917=91.7%\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:

Wapparent=Wactual−FBW_{apparent} = W_{actual} - F_B

For a fully submerged object:

Wapparent=ρobjectgV−ρfluidgV=(ρobject−ρfluid)gVW_{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 hh, area AA: FB=(Pbottom−Ptop)⋅A=ρgh⋅A=ρgVF_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': FB=ρfluidgVdisplacedF_B = \rho_{fluid} g V_{displaced}
  4. For equilibrium (floating): FB=FgF_B = F_g
  5. For apparent weight: Wapp=W−FBW_{app} = W - F_B
  6. Watch volumes: fully or partially submerged?

Common Mistakes

❌ Using object density for FBF_B (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: L=10L = 10 cm =0.10= 0.10 m
  • Water density: ρ=1000\rho = 1000 kg/m³

Find: Buoyant force FBF_B

Solution:

Step 1: Calculate volume. V=L3=(0.10)3=1.0×10−3 m3=1.0 LV = L^3 = (0.10)^3 = 1.0 \times 10^{-3} \text{ m}^3 = 1.0 \text{ L}

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

Since fully submerged: Vdisplaced=VV_{displaced} = V

FB=(1000)(9.8)(1.0×10−3)=9.8 NF_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! ✓

2Problem 2easy

❓ 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: L=10L = 10 cm =0.10= 0.10 m
  • Water density: ρ=1000\rho = 1000 kg/m³

Find: Buoyant force FBF_B

Solution:

Step 1: Calculate volume. V=L3=(0.10)3=1.0×10−3 m3=1.0 LV = L^3 = (0.10)^3 = 1.0 \times 10^{-3} \text{ m}^3 = 1.0 \text{ L}

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

Since fully submerged: Vdisplaced=VV_{displaced} = V

FB=(1000)(9.8)(1.0×10−3)=9.8 NF_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! ✓

3Problem 3medium

❓ 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: ρwood=600\rho_{wood} = 600 kg/m³
  • Water: ρwater=1000\rho_{water} = 1000 kg/m³
  • Mass: m=2.0m = 2.0 kg

Solution:

Part (a): Fraction submerged

VsubVtotal=ρobjectρfluid=6001000=0.60=60%\frac{V_{sub}}{V_{total}} = \frac{\rho_{object}}{\rho_{fluid}} = \frac{600}{1000} = 0.60 = 60\%

Part (b): Volume displaced

Step 1: Find total volume. Vtotal=mρwood=2.0600=3.33×10−3 m3V_{total} = \frac{m}{\rho_{wood}} = \frac{2.0}{600} = 3.33 \times 10^{-3} \text{ m}^3

Step 2: Find displaced volume. Vdisplaced=0.60×3.33×10−3=2.0×10−3 m3=2.0 LV_{displaced} = 0.60 \times 3.33 \times 10^{-3} = 2.0 \times 10^{-3} \text{ m}^3 = 2.0 \text{ L}

Verification: FB=(1000)(9.8)(2.0×10−3)=19.6 NF_B = (1000)(9.8)(2.0 \times 10^{-3}) = 19.6 \text{ N} Fg=(2.0)(9.8)=19.6 NF_g = (2.0)(9.8) = 19.6 \text{ N} ✓

Answer:

  • (a) 60% submerged
  • (b) 2.0 L displaced

4Problem 4medium

❓ 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: ρwood=600\rho_{wood} = 600 kg/m³
  • Water: ρwater=1000\rho_{water} = 1000 kg/m³
  • Mass: m=2.0m = 2.0 kg

Solution:

Part (a): Fraction submerged

VsubVtotal=ρobjectρfluid=6001000=0.60=60%\frac{V_{sub}}{V_{total}} = \frac{\rho_{object}}{\rho_{fluid}} = \frac{600}{1000} = 0.60 = 60\%

Part (b): Volume displaced

Step 1: Find total volume. Vtotal=mρwood=2.0600=3.33×10−3 m3V_{total} = \frac{m}{\rho_{wood}} = \frac{2.0}{600} = 3.33 \times 10^{-3} \text{ m}^3

Step 2: Find displaced volume. Vdisplaced=0.60×3.33×10−3=2.0×10−3 m3=2.0 LV_{displaced} = 0.60 \times 3.33 \times 10^{-3} = 2.0 \times 10^{-3} \text{ m}^3 = 2.0 \text{ L}

Verification: FB=(1000)(9.8)(2.0×10−3)=19.6 NF_B = (1000)(9.8)(2.0 \times 10^{-3}) = 19.6 \text{ N} Fg=(2.0)(9.8)=19.6 NF_g = (2.0)(9.8) = 19.6 \text{ N} ✓

Answer:

  • (a) 60% submerged
  • (b) 2.0 L displaced

5Problem 5medium

❓ Question:

A 5.0 kg block with density 800 kg/m³ is submerged in water (ρ = 1000 kg/m³). (a) What is the buoyant force on the block? (b) What is the block's apparent weight? (c) What is the block's acceleration if released from rest?

💡 Show Solution

Solution:

Given: m = 5.0 kg, ρ_block = 800 kg/m³, ρ_water = 1000 kg/m³, g = 10 m/s²

(a) Buoyant force: Volume of block: V = m/ρ = 5.0/800 = 6.25 × 10⁻³ m³ F_B = ρ_water × V × g = (1000)(6.25 × 10⁻³)(10) F_B = 62.5 N or 63 N

(b) Apparent weight: Weight: W = mg = 5.0 × 10 = 50 N W_apparent = W - F_B = 50 - 62.5 = -12.5 N

Negative means net upward force! Block will rise.

(c) Acceleration: F_net = F_B - W = 62.5 - 50 = 12.5 N (upward) a = F_net/m = 12.5/5.0 = 2.5 m/s² (upward)

6Problem 6hard

❓ 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: m=5.0m = 5.0 kg
  • Brass: ρbrass=8500\rho_{brass} = 8500 kg/m³
  • Water: ρwater=1000\rho_{water} = 1000 kg/m³

Solution:

Part (a): Weight in air W=mg=(5.0)(9.8)=49 NW = mg = (5.0)(9.8) = 49 \text{ N}

Part (b): Apparent weight

Step 1: Find volume. V=mρbrass=5.08500=5.88×10−4 m3V = \frac{m}{\rho_{brass}} = \frac{5.0}{8500} = 5.88 \times 10^{-4} \text{ m}^3

Step 2: Calculate buoyant force. FB=ρwatergV=(1000)(9.8)(5.88×10−4)=5.76 NF_B = \rho_{water} g V = (1000)(9.8)(5.88 \times 10^{-4}) = 5.76 \text{ N}

Step 3: Find apparent weight. Wapparent=W−FB=49−5.76=43.2 NW_{apparent} = W - F_B = 49 - 5.76 = 43.2 \text{ N}

Part (c): String tension

The string provides the apparent weight: T=Wapparent=43.2 NT = W_{apparent} = 43.2 \text{ N}

Alternative method: Wapp=W(1−ρfluidρobject)=49(1−10008500)=43.2 NW_{app} = W\left(1 - \frac{\rho_{fluid}}{\rho_{object}}\right) = 49\left(1 - \frac{1000}{8500}\right) = 43.2 \text{ N} ✓

Answer:

  • (a) 49 N in air
  • (b) 43.2 N apparent (12% lighter!)
  • (c) 43.2 N tension

7Problem 7hard

❓ Question:

A wooden block (ρ = 600 kg/m³) floats in water with 40% of its volume above the surface. (a) Verify this using Archimedes' principle. (b) If the block has volume 0.010 m³, what is its mass? (c) What additional mass must be added to just submerge it?

💡 Show Solution

Solution:

Given: ρ_wood = 600 kg/m³, ρ_water = 1000 kg/m³, 40% above → 60% submerged

(a) Verify floating fraction: At equilibrium: F_B = Weight ρ_water × V_submerged × g = ρ_wood × V_total × g V_submerged/V_total = ρ_wood/ρ_water = 600/1000 = 0.60 = 60% submerged ✓

(b) Mass of block: V = 0.010 m³ m = ρV = 600 × 0.010 = 6.0 kg

(c) Additional mass to submerge: When fully submerged (V_sub = V_total = 0.010 m³): F_B = ρ_water × V × g = 1000 × 0.010 × 10 = 100 N

Currently: Weight = 6.0 × 10 = 60 N Additional weight needed: 100 - 60 = 40 N Additional mass: m_add = 40/10 = 4.0 kg

8Problem 8hard

❓ 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: m=5.0m = 5.0 kg
  • Brass: ρbrass=8500\rho_{brass} = 8500 kg/m³
  • Water: ρwater=1000\rho_{water} = 1000 kg/m³

Solution:

Part (a): Weight in air W=mg=(5.0)(9.8)=49 NW = mg = (5.0)(9.8) = 49 \text{ N}

Part (b): Apparent weight

Step 1: Find volume. V=mρbrass=5.08500=5.88×10−4 m3V = \frac{m}{\rho_{brass}} = \frac{5.0}{8500} = 5.88 \times 10^{-4} \text{ m}^3

Step 2: Calculate buoyant force. FB=ρwatergV=(1000)(9.8)(5.88×10−4)=5.76 NF_B = \rho_{water} g V = (1000)(9.8)(5.88 \times 10^{-4}) = 5.76 \text{ N}

Step 3: Find apparent weight. Wapparent=W−FB=49−5.76=43.2 NW_{apparent} = W - F_B = 49 - 5.76 = 43.2 \text{ N}

Part (c): String tension

The string provides the apparent weight: T=Wapparent=43.2 NT = W_{apparent} = 43.2 \text{ N}

Alternative method: Wapp=W(1−ρfluidρobject)=49(1−10008500)=43.2 NW_{app} = W\left(1 - \frac{\rho_{fluid}}{\rho_{object}}\right) = 49\left(1 - \frac{1000}{8500}\right) = 43.2 \text{ N} ✓

Answer:

  • (a) 49 N in air
  • (b) 43.2 N apparent (12% lighter!)
  • (c) 43.2 N tension
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❓ Frequently Asked Questions

What is Buoyancy and Archimedes' Principle?▾
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Yes, this page includes 8 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.