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Buoyant Forces and Archimedes’ Principle

🚁 Overview

Any object submerged in (or floating on) a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces. In this chapter you will:

  • State and apply Archimedes’ principle.
  • Calculate buoyant forces on fully and partially submerged bodies.
  • Determine the conditions for flotation and assess stability of floating objects.

Course Hub

Prep4Uni Fluid Mechanics 1
Main syllabus portal organizing all sub-modules, problem sets, and fluid statics roadmap milestones.

Chapter 1: Pressure Fundamentals

Chapter 1: Pressure
Establishes basic definitions of pressure, normal forces, unit systems, and absolute versus gauge pressure scales.

Chapter 3: Pressure Measurement

Chapter 3: Pressure Measurement
Explores manometers, Bourdon tube gauges, and transducers for measuring fluid static pressure levels.

Chapter 5: Fluid Dynamics

Chapter 5: Fluid Dynamics
Transitions from static buoyancy considerations into moving fluids, continuity equations, and momentum balance.

Chapter 6: Bernoulli Equation

Chapter 6: Bernoulli’s Equation
Connects hydrostatic elevation potential and static pressure head along streamlines in ideal dynamic flows.

📖 Contents

  1. Archimedes’ Principle
  2. Buoyant Force Calculations
  3. Flotation & Equilibrium
  4. Stability of Floating Bodies
  5. Partially Submerged Objects
  6. Worked Examples
  7. Example Problems & Solutions
  8. Key Formula Recap
  9. Comprehensive Practice Exercises

🎯 Learning Outcomes

In this chapter you will:

  • State and apply Archimedes’ principle.
  • Calculate buoyant forces on fully and partially submerged bodies.
  • Determine the conditions for flotation and assess stability of floating objects.

4.1 Archimedes’ Principle

Archimedes’ principle: A body immersed in a fluid is buoyed up by a force equal to the weight of the fluid displaced:

Fb = ρfluid · g · Vdisp
Crown in Air and Submerged
Crown in Air and Submerged

Archimedes’ Principle: Verifying a Crown’s Purity

When an object is weighed in air and then again while fully submerged in water, the apparent loss of weight Mair − Mwater is the buoyant force. Since the crown is fully submerged, the displaced-water volume equals the crown’s volume.

Given: Mair = 7.84 N, Mwater = 6.84 N.

Buoyant force:
Fb = Mair − Mwater = 1.00 N.

Mass of the crown:
mcrown = Mair / g = 7.84 / 9.81 kg ≈ 0.800 kg.

Mass of displaced water:
mwater = Fb / g = 1.00 / 9.81 kg ≈ 0.102 kg.

Volume of displaced water (crown’s volume):
V = mwater / ρwater = 0.102 / 1000 m3 ≈ 1.02 × 10-4 m3.

Density of the crown:
ρcrown = mcrown / V ≈ 0.800 / (1.02 × 10-4) kg/m3 ≈ 7.84 × 103 kg/m3.

Pure gold has ρ ≈ 1.932 × 104 kg/m3, so the crown is not pure gold.

4.2 Buoyant Force Calculations

  • Fully submerged: Fb = ρf · g · Vobj.
  • Floating (partially submerged):
    ρf · g · Vdisp = mobj · g ⇒ Fb = Wobj.
  • Net force:
    Fnet = Fb − W = ρf · g · Vdisp − ρobj · g · Vobj.

4.3 Flotation & Equilibrium

  • Condition for flotation:
    ρf · Vdisp = ρobj · Vobj.
  • Equilibrium (no net moment): The line of action of Fb through the center of buoyancy B should pass vertically through the center of gravity G.

4.4 Stability of Floating Bodies

For a small heel angle, the line of action of the shifted buoyant force intersects the vertical centerline through G at the metacenter M. The metacentric height is GM:

  • BM = I / Vdisp (waterplane second moment of area I about the heel axis).
  • GM = BM − BG.
  • Stable: GM > 0 → creates a righting moment. Unstable: GM < 0 → creates an overturning moment.

Example: For a rectangular barge with beam b and length L: I = b3L / 12. Use this to compute BM and compare with BG.

4.5 Partially Submerged Objects

Ice Berg
Iceberg

How Much of an Iceberg Is Hidden?

  • ρice = 917 kg/m3
  • ρwater = 1025 kg/m3

Floating equilibrium gives ρwater · Vdisp = ρice · Vice. Hence the fraction submerged is:

Vdisp / Vice = ρice / ρwater = 917 / 1025 ≈ 0.8946 (89.46%)

So about 10.54% of the iceberg remains above the water surface.

4.6 Worked Examples

  1. Example 1: Metal cube a = 0.1 m, ρobj = 7800 kg/m3, water ρf = 1000 kg/m3.
    Solution:
    V = a3 = 1.0 × 10-3 m3.
    Fb = ρf · g · V ≈ 1000 × 9.81 × 1.0 × 10-3 = 9.81 N.
    W = ρobj · g · V ≈ 7800 × 9.81 × 1.0 × 10-3 = 76.5 N.
    Fnet = Fb − W ≈ 9.81 − 76.5 = -66.7 N (downward).
  2. Example 2: Wooden block ρobj = 600 kg/m3, V = 0.02 m3 in water.
    Solution:
    Flotation ratio: Vdisp / Vobj = ρobj / ρf = 600 / 1000 = 0.6 ⇒ 60% submerged.

4.7 Example Problems & Solutions

  1. Problem 1: Sphere r = 0.2 m in oil ρ = 850 kg/m3. Find Fb.
    Solution:
    V = (4/3) · π · r3 = (4/3) · π · (0.2)3 ≈ 3.351 × 10-2 m3.
    Fb = ρ · g · V = 850 × 9.81 × 3.351 × 10-2 ≈ 279.4 N.
  2. Problem 2: Wooden log r = 0.05 m, L = 1 m, ρ = 700 kg/m3 in freshwater.
    Solution:
    Vobj = π · r2 · L ≈ π · (0.05)2 · (1) ≈ 7.854 × 10-3 m3.
    Fraction submerged = ρobj / ρf = 700 / 1000 = 0.7.
    Vdisp = 0.7 · Vobj ≈ 5.498 × 10-3 m3.
    Submerged draft height equivalent hsub = 0.70 · L = 0.70 m.
  3. Problem 3: Explain the effect of metacentric height GM on vessel flotation stability.
    Solution:
    GM > 0 ⇒ buoyant force creates a righting moment restoring upright posture (stable). GM < 0 ⇒ buoyant force creates an overturning moment driving capsizing (unstable).
  4. Problem 4: Barge with waterplane moment of inertia I = 3 m4, displaced volume V = 20 m3, and BG = 0.3 m. Find GM.
    Solution:
    BM = I / V = 3 / 20 = 0.15 m.
    GM = BM − BG = 0.15 − 0.30 = -0.15 m < 0 ⇒ vessel is unstable.

4.8 Key Formula Recap

FormulaDescription
Fb = ρf · g · VdispBuoyant force = weight of displaced fluid
Vdisp / Vobj = ρobj / ρfFraction submerged for floating bodies
Fnet = Fb − ρobj · g · VobjNet force on a fully submerged object
GM = BM − BGMetacentric height for flotation stability analysis

Interactive Buoyancy & Flotation Simulator

Adjust object density (ρobj), fluid density (ρf), and center of gravity position (BG) to observe submerged fraction and metacentric height stability across timeline steps.

Submerged Fraction: 60.0%
Metacentric Height (GM): +0.15 m (STABLE)

Interactive Buoyancy Concept Toggles

Click each section below to toggle detailed explanations regarding buoyancy principles and metacentric mechanics.

Why does the buoyant force act through the centroid of displaced volume (Center of Buoyancy B)?
The buoyant force represents the resultant upward force of all hydrostatic surface pressure distributions acting on the submerged object boundary. Integrating hydrostatic forces over an arbitrary shape produces a single equivalent force passing precisely through the geometric centroid B of the displaced fluid volume.
How does a vessel maintain flotation stability when its Center of Gravity G lies above its Center of Buoyancy B?
When a ship heels by a small angle, its submerged profile reshapes, causing the center of buoyancy to shift laterally to B'. If the vertical line through B' intersects the central symmetry axis at metacenter M above G (GM > 0), the buoyant force generates a restoring righting moment that corrects the tilt.
Why does adding ballast water to lower compartments increase vessel stability?
Adding heavy ballast mass low in the hull shifts the overall center of gravity G downward. This reduces distance BG, directly increasing metacentric height GM = BM − BG, which strengthens the vessel's righting moment against roll over.

Frequently Asked Questions

Does Archimedes' principle apply to objects resting firmly against container bottoms?

If an object rests tightly against a smooth container bottom with zero fluid film underneath, hydrostatic pressure acts only downward on top, generating no upward buoyant force. True buoyancy requires fluid contact along bottom boundaries.

Why do hollow steel cargo ships float while solid steel spheres sink?

Hollow ships enclose large internal air volumes, yielding an average bulk density (mtotal / Vhull) far below water density (≈ 1000 kg/m3), enabling flotation equilibrium.

Can Archimedes' principle be applied to gases like air?

Yes. Hot air and helium balloons rise because surrounding atmospheric air exerts an upward buoyant force equal to the weight of displaced air (Fb = ρair · g · Vballoon).

📝 Comprehensive Practice Module

Section 1: Foundational Core Assessment

  1. What is the buoyant force?
    Answer: The net upward force exerted by a fluid on an immersed body equal to the weight of displaced fluid: Fb = ρfluid · g · Vdisp.
  2. State Archimedes’ principle.
    Answer: Any body wholly or partially submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced.
  3. Write the fundamental mathematical formula for buoyant force.
    Answer: Fb = ρfluid · g · Vdisp.
  4. What is the SI unit of buoyant force?
    Answer: Newton (N = kg · m / s2).
  5. How is displaced volume calculated for a partially submerged floating body in static equilibrium?
    Answer: Equating weight to buoyancy gives Vdisp = mobj / ρf = (ρobj / ρf) · Vobj.
  6. What fraction of a floating body's volume is submerged below the free surface?
    Answer: Vdisp / Vobj = ρobj / ρfluid.
  7. Under what density condition will a solid object sink completely in a fluid?
    Answer: When average object density exceeds fluid density (ρobj > ρfluid).
  8. Define the center of buoyancy B.
    Answer: The geometric centroid of the displaced fluid volume through which the resultant buoyant force acts.
  9. What is the metacentric height GM?
    Answer: The distance between a floating body's center of gravity G and its metacenter M.
  10. What is the critical criterion for stable vessel flotation?
    Answer: Positive metacentric height (GM > 0), producing a righting moment.

Section 2: Analytical & Conceptual Analysis

  1. How does fluid density shift the submerged draft of a floating vessel?
    Answer: Higher fluid density reduces required displaced volume (Vdisp = m / ρf), raising the vessel higher in the water.
  2. Why does buoyant force vanish in a zero-gravity environment?
    Answer: Without gravity (g = 0), hydrostatic depth pressure gradients vanish, leaving no net upward force.
  3. How is submerged depth calculated for a uniform vertical cylinder of height H floating in water?
    Answer: hsub = H · (ρobj / ρf).
  4. What happens to buoyant force if fluid density doubles while submerged volume stays constant?
    Answer: Buoyant force doubles because Fb is directly proportional to fluid density ρf.
  5. Explain how Archimedes' principle governs hot-air balloon atmospheric lift.
    Answer: Heating air inside the envelope lowers its internal density relative to ambient air, producing upward buoyancy (Fb = ρambient · g · V) that exceeds total balloon weight.
  6. In what direction does buoyant force act relative to local gravity?
    Answer: Vertically upward, directly opposing the gravitational acceleration vector.
  7. How is net force computed for a fully submerged sinking object?
    Answer: Fnet = Fb − W = (ρf − ρobj) · g · Vobj.
  8. Why can a dense steel ship float safely on ocean water?
    Answer: The hollow geometry encloses air, making bulk ship density lower than seawater density.
  9. How is displaced volume experimentally measured in a laboratory?
    Answer: Submerging the object in a graduated cylinder and recording liquid displacement volume change ΔV.

Section 3: Computational & Numerical Exercises

  1. A wooden cube of side 0.2 m and density 600 kg/m3 floats in pure water (ρ = 1000 kg/m3). Calculate the submerged height.
    Solution:
    Submerged fraction = 600 / 1000 = 0.6.
    Submerged height = 0.6 × 0.2 m = 0.12 m (12 cm).
  2. A solid sphere of radius 0.1 m is fully submerged in oil (ρ = 850 kg/m3). Calculate the buoyant force (g = 9.81 m/s2).
    Solution:
    V = (4/3) · π · (0.1)3 ≈ 4.189 × 10-3 m3.
    Fb = ρ · g · V = 850 × 9.81 × 4.189 × 10-3 ≈ 34.93 N.
  3. A solid log (radius r = 0.05 m, length L = 1 m, density ρ = 700 kg/m3) floats in freshwater. Find its submerged length.
    Solution:
    Submerged fraction = ρobj / ρf = 700 / 1000 = 0.7.
    Submerged length = 0.7 × 1.0 m = 0.70 m.
  4. A block of wood measuring 0.5 m × 0.3 m × 0.2 m with density 600 kg/m3 floats in water. Determine total buoyant force.
    Solution:
    Total Volume V = 0.5 × 0.3 × 0.2 = 0.03 m3.
    Displaced Volume Vdisp = (600 / 1000) × 0.03 = 0.018 m3.
    Fb = 1000 × 9.81 × 0.018 = 176.58 N.
  5. A steel cargo ship displaces 2000 m3 of seawater (ρ = 1025 kg/m3). Calculate the total buoyant force (g = 9.81 m/s2).
    Solution:
    Fb = ρ · g · Vdisp = 1025 × 9.81 × 2000 = 20 110 500 N ≈ 20.11 MN.
  6. A metal cube (density 7800 kg/m3, side 0.1 m) is fully submerged in water. Calculate the net downward force.
    Solution:
    V = (0.1)3 = 0.001 m3.
    Fb = 1000 × 9.81 × 0.001 = 9.81 N.
    W = 7800 × 9.81 × 0.001 = 76.518 N.
    Fnet = Fb − W = 9.81 − 76.518 = -66.708 N (66.71 N downward).
  7. An object weighs 50 N in air and 30 N when submerged in water. Calculate buoyant force and object volume.
    Solution:
    Fb = Wair − Wwater = 50 − 30 = 20 N.
    V = Fb / (ρ · g) = 20 / (1000 × 9.81) ≈ 2.039 × 10-3 m3.
  8. A hot-air balloon displaces 500 m3 of ambient air (ρ = 1.2 kg/m3). Calculate its buoyant lift force.
    Solution:
    Fb = ρ · g · V = 1.2 × 9.81 × 500 = 5886 N (5.886 kN).
  9. A floating object displaces 0.02 m3 of water. Determine the total mass of the object supported at equilibrium.
    Solution:
    mobj = ρwater · Vdisp = 1000 kg/m3 × 0.02 m3 = 20 kg.
  10. A submerged object experiences a buoyant force of 50 N in a liquid. Calculate fluid density if object volume is 5 × 10-3 m3.
    Solution:
    ρf = Fb / (g · V) = 50 / (9.81 × 5 × 10-3) ≈ 1019.37 kg/m3.
  11. Estimate BM for a flat-bottom barge if its waterplane moment of inertia is I = 12 m4 and displaced volume is V = 15 m3.
    Solution:
    BM = I / V = 12 / 15 = 0.80 m.
  12. A solid material of density 500 kg/m3 is placed into a chemical fluid of density 1200 kg/m3. Calculate the percentage volume submerged.
    Solution:
    Submerged ratio = ρobj / ρf = 500 / 1200 ≈ 0.4167 ⇒ 41.67% submerged.
Last updated: 05 Aug 2026