Chapter 4: Buoyant Forces & Archimedes’ Principle

Course: Prep4Uni Fluid Mechanics 1

Chapter 1: Pressure
Chapter 2: Variation of Pressure with Depth
Chapter 3: Pressure Measurement
Chapter 4: Buoyant Forces & Archimedes’ Principle
Chapter 5: Fluid Dynamics
Chapter 6: Bernoulli’s Equation
Chapter 7: Applications of Fluid Dynamics

🚁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.

📖Contents

Archimedes’ Principle
Buoyant Force Calculations
Flotation & Stability

🎯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:
F_b = ρ_fluid × g × V_disp.

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
(M_air − M_water) is the buoyant force exerted by the fluid.
Archimedes’ Principle tells us that this buoyant force equals the weight of the fluid displaced.
Since the crown is fully submerged, the volume of water displaced equals the crown’s volume.

According to legend, King Hiero II asked Archimedes to determine whether his crown was pure gold.
By measuring its density, Archimedes could uncover any alloying material.

Example Calculation

- M_air = 7.84 N (weight in air)
- M_water = 6.84 N (apparent weight in water)

Buoyant force:
F_b = M_air − M_water = 7.84 − 6.84 = 1.00 N

Mass of the crown:
m_crown = M_air / g = 7.84 N / 9.81 m/s² ≈ 0.800 kg

Mass of displaced water:
m_water = F_b / g = 1.00 N / 9.81 m/s² ≈ 0.102 kg

Volume of water displaced (crown’s volume):
V = m_water / ρ_water = 0.102 kg / 1000 kg/m³ ≈ 1.02×10⁻⁴ m³

Density of the crown:
ρ_crown = m_crown / V ≈ 0.800 kg / 1.02×10⁻⁴ m³ ≈ 7 840 kg/m³

Gold has a density of about 19 320 kg/m³, so this result shows the crown cannot be pure gold.

4.2 Buoyant Force Calculations

Fully submerged:
F_b = ρ_f × g × V_obj,
where V_obj is the object’s volume.
Partially submerged (floating):
The displaced volume V_disp satisfies
ρ_f g V_disp = m_obj g,
so the buoyant force equals the object’s weight.
Net force:
F_net = F_b − W = ρ_f g V_disp − ρ_obj g V_obj.

4.3 Flotation & Equilibrium

Condition for flotation:
A floating body displaces just enough fluid so that
ρ_f V_disp = ρ_obj V_obj.
Equilibrium:
The buoyant force line of action through the center of buoyancy B must pass through the center of gravity G to avoid a net moment.

4.4 Stability of Floating Bodies

When a floating object is tilted by a small angle, its center of buoyancy shifts and the buoyant force line of action meets the vertical through G at the metacenter M. The distance GM is the metacentric height:

BM = I / V_disp, where I is the second moment of area of the waterplane about the tilt axis.
GM = BM − BG, where BG is the distance between B and G.
Stable equilibrium: GM > 0 (metacenter above G) → righting moment restores upright position.
Unstable: GM < 0 (metacenter below G) → overturning moment capsizes the body.

Example: A rectangular barge of beam b and draft d has waterplane moment of inertia I = b³L/12; one can compute BM and compare to BG to check stability.

4.5 Partially Submerged Objects

How Much of an Iceberg Is Hidden?

It’s often asked: “How much of an iceberg lies hidden beneath the water?” Let’s find out.

Given Data

Density of ice: ρ_ice = 917 kg/m³
Density of seawater: ρ_water = 1025 kg/m³

Derivation

Let the iceberg’s mass be M (kg). Then its volume is

V_ice = M / ρ_ice = M / 917 m³.

At equilibrium the buoyant force equals the iceberg’s weight:

F_b = M g ⇒ displaced water mass = M.

Thus the volume of displaced seawater is

V_disp = M / ρ_water = M / 1025 m³.

Fraction Submerged

Fraction = V_disp / V_ice = (M/1025) / (M/917) = 917/1025 ≈ 0.8946 = 89.46 %

Conclusion: Approximately 89.5 % of an iceberg is submerged, so only about 10.5 % remains visible above the surface.

4.6 Example Worked Examples

Example 1: A metal cube of side 0.1 m (ρ_obj = 7800 kg/m³) is fully submerged in water (ρ_f = 1000 kg/m³). Find the buoyant and net forces.Solution:
Volume: V = (0.1 m)³ = 1×10⁻³ m³.
Buoyant force: F_b = 1000×9.81×1×10⁻³ ≈ 9.81 N.
Weight: W = 7800×9.81×1×10⁻³ ≈ 76.5 N.
Net: F_net = 9.81 − 76.5 = −66.7 N (downward).
Example 2: A wooden block (ρ_obj = 600 kg/m³, V = 0.02 m³) floats in water. What fraction of its volume is submerged?Solution:
For flotation: ρ_f V_disp = ρ_obj V_obj
⇒ V_disp/V_obj = 600/1000 = 0.6.
So 60% is submerged.

4.7 Example Problems & Solutions

Problem 1: Compute the buoyant force on a sphere (r = 0.2 m) fully submerged in oil (ρ = 850 kg/m³).
Solution:
Sphere volume: V = (4/3)πr³ = (4/3)π(0.2)³ ≈ 0.03351 m³.
Buoyant force: F_b = ρ g V = 850×9.81×0.03351 ≈ 2.79×10² N.
Problem 2: Find the submerged depth of a log (r = 0.05 m, L = 1 m, ρ = 700 kg/m³) floating in freshwater.
Solution:
Object volume: V_obj = π r²L = π×(0.05)²×1 ≈ 7.85×10⁻³ m³.
Fraction submerged: V_disp/V_obj = ρ_obj/ρ_f = 700/1000 = 0.7.
Displaced volume: V_disp = 0.7×V_obj ≈ 5.50×10⁻³ m³.
Submerged depth: h = V_disp / (π r²) = 0.00550 / (π×0.0025) ≈ 0.70 m.
Problem 3: Explain how metacentric height GM affects the rolling stability of a boat.
Solution:
If GM > 0, the metacenter M lies above the center of gravity G, so when the boat tilts the buoyant force produces a righting moment, restoring upright equilibrium.
If GM < 0, M lies below G, so tilting produces an overturning moment and the boat becomes unstable.
Problem: A barge has waterplane inertia I = 3 m⁴, displaced volume V = 20 m³, and BG = 0.3 m. Compute GM and state whether it is stable.
Solution:
BM = I/V = 3/20 = 0.15 m → GM = 0.15−0.3 = −0.15 m < 0: unstable.

4.8 Key Formula Recap

Formula	Description
`F_b = ρ_f g V_disp`	Buoyant force = weight of displaced fluid
`V_disp/V_obj = ρ_obj/ρ_f`	Fraction submerged for floating bodies
`F_net = F_b − ρ_obj g V_obj`	Net force on a submerged object
`GM = BM − BG`	Metacentric height for stability analysis

Proceed to: Chapter 5: Fluid Dynamics

Return to: Prep4Uni Fluid Mechanics 1

📝EXERCISES

20 Questions & Answers

1. What is the buoyant force?
The buoyant force is the upward force exerted by a fluid on an immersed body, equal to the weight of fluid displaced.
2. State Archimedes’ principle.
Archimedes’ principle: A body immersed in a fluid experiences a buoyant force equal in magnitude to the weight of the fluid it displaces.
3. Write the formula for buoyant force.
F_b = ρ_fluid g V_disp, where V_disp is displaced volume.
4. What is the unit of buoyant force?
The SI unit of buoyant force is the newton (N).
5. How do you compute the displaced volume of a floating object?
For a floating object, set buoyant force equal to weight: ρ_f g V_disp = m_obj g, solve for V_disp.
6. What fraction of a floating body is submerged?
Fraction submerged = V_disp / V_obj = ρ_obj / ρ_fluid.
7. Under what condition does an object sink?
If its weight exceeds the maximum buoyant force (ρ_fluid g V_obj), it sinks.
8. Define center of buoyancy.
The center of buoyancy is the centroid of the displaced volume of fluid, where the buoyant force acts.
9. What is metacentric height?
Metacentric height (GM) is the distance between the center of gravity G and the metacenter M, indicating stability.
10. What is the condition for stable flotation?
Stable flotation requires GM > 0, so the metacenter lies above the center of gravity.
11. How does density affect floatation?
Lower object density relative to fluid density increases the fraction of volume displaced for flotation.
12. Why does an object experience no net buoyant force if fully submerged and weightless?
If weightless, buoyant force equals displaced fluid weight but there is no gravity acting on the object, so net force is zero.
13. How do you find the submerged depth of a floating cylinder?
Use V_disp = πr²h_sub and set ρ_fluid g V_disp = m_obj g to solve for h_sub.
14. What happens to buoyant force if fluid density doubles?
Buoyant force doubles, since F_b = ρ_fluid g V_disp.
15. Can Archimedes’ principle be applied to gases?
Yes, for gases with significant density, e.g. hot-air balloon buoyancy in air.
16. In what direction does the buoyant force act?
Upward, opposite to gravity.
17. How do you compute net force on a submerged object?
F_net = F_b − W = ρ_f g V_obj − ρ_obj g V_obj.
18. What is the weight of displaced fluid for a floating wood log?
Equal to the weight of the log, m_log g, by Archimedes’ principle.
19. Why does a steel ship float?
Because its overall density (including air spaces) is lower than water, so it displaces enough water for buoyancy.
20. How is the displaced volume measured in a laboratory?
By measuring the rise in fluid level in a graduated container when the object is submerged.

20 Problems & Solutions

Problem 1: A cube of side 0.2 m and density 600 kg/m³ floats in water. What fraction of its height is submerged?
Solution:
Fraction submerged = ρ_obj/ρ_water = 600/1000 = 0.6, so 60% of its height (0.12 m) is submerged.
Problem 2: A sphere of radius 0.1 m is fully submerged in oil (ρ = 850 kg/m³). Calculate the buoyant force.
Solution:
V = (4/3)π(0.1)³ ≈ 4.19×10⁻³ m³.
F_b = 850×9.81×4.19×10⁻³ ≈ 34.9 N.
Problem 3: A solid cylinder (r = 0.05 m, L = 1 m, ρ = 700 kg/m³) floats in freshwater. Find the submerged length.
Solution:
V_obj = πr²L = π(0.05)²×1 ≈ 7.85×10⁻³ m³.
V_disp/V_obj = 700/1000 = 0.7 ⇒ h_sub = 0.7×1 = 0.7 m.
Problem 4: A block of wood (0.5×0.3×0.2 m, ρ = 600 kg/m³) is placed in water. Determine the buoyant force.
Solution:
Submerged volume = ρ_obj/ρ_water × V_obj = 0.6 × (0.5×0.3×0.2) = 0.018 m³.
F_b = 1000×9.81×0.018 ≈ 176.6 N.
Problem 5: A steel ship displaces 2000 m³ of water. Find its buoyant force.
Solution:
F_b = ρ_water g V = 1000×9.81×2000 ≈ 1.962×10⁷ N.
Problem 6: A metal cube (ρ = 7800 kg/m³, side = 0.1 m) is fully submerged. Calculate net force.
Solution:
V = 1×10⁻³ m³; F_b = 1000×9.81×10⁻³ = 9.81 N; W = 7800×9.81×10⁻³ ≈ 76.5 N; F_net = 9.81−76.5 = −66.7 N.
Problem 7: A submerged object weighs 50 N in air and 30 N in water. Find its buoyant force and volume.
Solution:
F_b = 50−30 = 20 N; V = F_b/(ρg) = 20/(1000×9.81) ≈ 2.04×10⁻³ m³.
Problem 8: A hot-air balloon displaces 500 m³ of air (ρ = 1.2 kg/m³). What is its lifting buoyant force?
Solution:
F_b = 1.2×9.81×500 ≈ 5886 N.
Problem 9: A boat has G above B. What does this imply for stability?
Solution:
G above B gives a positive righting moment, so the boat is stable if GM > 0.
Problem 10: Derive the condition for neutral buoyancy.
Solution:
Neutral buoyancy: F_b = W ⇒ ρ_f V = ρ_obj V ⇒ ρ_obj = ρ_f.
Problem 11: A wooden plank (ρ = 600 kg/m³) floats with half its volume submerged. Verify.
Solution:
Fraction = ρ_obj/ρ_water = 600/1000 = 0.6, so 60% submerged—not half. Statement is false.
Problem 12: A cube floats with 0.4 m submerged; side 0.5 m. Find its density.
Solution:
Fraction = 0.4/0.5 = 0.8 ⇒ ρ_obj = 0.8×1000 = 800 kg/m³.
Problem 13: A cylinder sinks with net force 100 N downward. What is F_b if W = 300 N?
Solution:
F_net = W − F_b ⇒ 100 = 300 − F_b ⇒ F_b = 200 N.
Problem 14: A floating object displaces 0.02 m³ of water. What mass does it support?
Solution:
Supported mass = ρ_water V = 1000×0.02 = 20 kg.
Problem 15: Show that two objects with same volume but different densities have different submerged depths.
Solution:
h_sub/H = ρ_obj/ρ_f, so denser object has greater fraction submerged.
Problem 16: A submerged cube experiences F_b = 50 N. What is ρ_f if V = 5×10⁻³ m³?
Solution:
50 = ρ_f×9.81×5×10⁻³ ⇒ ρ_f ≈ 1019 kg/m³.
Problem 17: A submarine adjusts its buoyancy by pumping water. Explain principle.
Solution:
Changing displaced volume changes buoyant force; adding water increases density ⇒ submerges, removing water floats.
Problem 18: Why is Archimedes’ principle valid for irregular shapes?
Solution:
Because buoyant force depends only on displaced volume, not object shape.
Problem 19: Calculate GM for a barge if B shifts by 0.1 m when tilted by 5°.
Solution:
BM ≈ Δx/θ = 0.1/(5°×π/180) ≈ 1.15 m; GM = BM − BG (given BG) as needed.
Problem 20: A material of density 500 kg/m³ is placed in liquid of ρ = 1200 kg/m³. Will it float?
Solution:
Since ρ_obj < ρ_fluid, it will float and 41.7% of volume will be submerged.

Prepare for University Studies & Career Advancement

Buoyant Forces and Archimedes’ Principle