Archimedes Principle and Buoyancy

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

0 formulas

10 exam-style examples

~45 min

All Fluid Mechanics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Archimedes Principle and Buoyancy within Fluid Mechanics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what archimedes principle and buoyancy describes physically and when it applies.
State every one of the 0 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early.

Lecture

Why this section exists. Archimedes Principle and Buoyancy is the part of Fluid Mechanics that lets you connect a pipeline, jet or submerged surface to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as continuity plus energy, with one head-loss or force term. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Crane lowering a steel plate girder onto bridge bearings while ironworkers guide it. — Photo 1. Where this shows up in practice: archimedes principle and buoyancy.
Capstone Studio instructional photograph

Fluid Mechanics — Archimedes Principle and Buoyancy: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a pipeline, jet or submerged surface. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 0 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

1. The buoyant force exerted on a submerged or floating body is equal to the weight of the fluid displaced by the body.
2. A floating body displaces a weight of fluid equal to its own weight; i.e., a floating body is in equilibrium.
The center of buoyancy is located at the centroid of the displaced fluid volume.
In the case of a body lying at the interface of two immiscible fluids, the buoyant force equals the sum of the weights of the fluids
displaced by the body.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Hydrostatic force on a vertical gate

A vertical rectangular gate 2.0 m wide and 3.0 m tall has its top at the water surface. Find the resultant force and its depth of application.

Given

b = 2.0 m
h = 3.0 m
Top at the free surface
γ = 9.81 kN/m³

Find

F and its line of action

Start with the thinking

Resultant equals pressure at the centroid times the area.
For a surface-piercing rectangle the resultant acts at 2h/3.

Step-by-step solution

Centroid depth
$h ˉ = 3.0/2 = 1.50 m$
Area
$A = 2.0 (3.0) = 6.00 m^{2}$
Resultant — F = γh̄A = 9.81(1.50)(6.00)
Evaluate
$F = 88.3 k N$
Line of action
$y_{p} = 2 h /3 = 2.00 mb e l o w t h es u r f a ce$

Answer: F = 88.3 kN acting 2.00 m below the surface

Why the other options are there

177 kN (bottom pressure used over the whole area)
y_p = 1.50 m (centroid taken as the pressure centre)

Reference: FE Reference Handbook — Fluid Mechanics — Hydrostatic forces

Example 2

Hydrostatic pressure at depth — Archimedes Principle and Buoyancy

Find the gauge pressure 2.0 ft below the surface of a fluid with specific gravity 12.60.

Given

h = 2.0 ft
SG = 12.60
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Archimedes Principle and Buoyancy

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 12.60 (62.4) = 786.2 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 786.2 (2.0) = 1, 572 p s f$
Convert
$p = 1, 572/144 = 10.92 p s i$

Answer: p ≈ 1,572 psf (10.92 psi)

Why the other options are there

124.8 psf (specific gravity ignored)
226,437 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Archimedes Principle and Buoyancy

Example 3

Buoyant force and metacentric height of a floating barge — Archimedes Principle and Buoyancy

A rectangular barge 6.0 m long by 3.0 m wide floats at a draft of 1.28 m in fresh water, with its centre of gravity 1.42 m above the keel. Compute the buoyant force (and hence its weight) and the metacentric height, and judge its stability.

Given

L = 6.0 m, B = 3.0 m
Draft = 1.28 m
KG = 1.42 m
ρ = 1000 kg/m³

Find

Buoyant force and metacentric height GM

Start with the thinking

Archimedes: the buoyant force equals the weight of displaced fluid, so a floating body's weight equals that force.
Positive GM means the metacentre lies above the centre of gravity and the vessel is stable.

Step-by-step solution

Displaced volume
$\forall = L B d = 6.0 (3.0) (1.28) = 22.950 m^{3}$
Formula
$F_{B} = ρ g \forall$
Substituting
$F_{B} = 1000 (9.81) (22.950) = 225.1 k N$
Waterplane inertia
$I = L B^{3} /12 = 6.0 (3.0)^{3} /12 = 13.500 m^{4}$
Formula
$B M = I /\forall$
Substituting
$B M = 13.500/22.950 = 0.588 m$
Formula
$GM = K B + B M - K G$
Substituting
$GM = 0.638 + 0.588 - 1.42 = - 0.199 m \to u n s t ab l e$

Answer: F_B = 225.1 kN, GM = -0.20 m (unstable)

Why the other options are there

264.9 kN (full depth used instead of draft)
GM = -0.84 m (KB omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Archimedes Principle and Buoyancy

Example 4

Hydrostatic pressure at depth — Archimedes Principle and Buoyancy (2)

Find the gauge pressure 12.5 ft below the surface of a fluid with specific gravity 4.60.

Given

h = 12.5 ft
SG = 4.60
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Archimedes Principle and Buoyancy (2)

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 4.60 (62.4) = 287.0 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 287.0 (12.5) = 3, 588 p s f$
Convert
$p = 3, 588/144 = 24.92 p s i$

Answer: p ≈ 3,588 psf (24.92 psi)

Why the other options are there

780.0 psf (specific gravity ignored)
516,672 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Archimedes Principle and Buoyancy

Example 5

Buoyant force and metacentric height of a floating barge — Archimedes Principle and Buoyancy (2)

A rectangular barge 8.5 m long by 6.0 m wide floats at a draft of 1.37 m in fresh water, with its centre of gravity 1.58 m above the keel. Compute the buoyant force (and hence its weight) and the metacentric height, and judge its stability.

Given

L = 8.5 m, B = 6.0 m
Draft = 1.37 m
KG = 1.58 m
ρ = 1000 kg/m³

Find

Buoyant force and metacentric height GM

Start with the thinking

Archimedes: the buoyant force equals the weight of displaced fluid, so a floating body's weight equals that force.
Positive GM means the metacentre lies above the centre of gravity and the vessel is stable.

Step-by-step solution

Displaced volume
$\forall = L B d = 8.5 (6.0) (1.37) = 69.615 m^{3}$
Formula
$F_{B} = ρ g \forall$
Substituting
$F_{B} = 1000 (9.81) (69.615) = 682.9 k N$
Waterplane inertia
$I = L B^{3} /12 = 8.5 (6.0)^{3} /12 = 153.0 m^{4}$
Formula
$B M = I /\forall$
Substituting
$B M = 153.0/69.615 = 2.198 m$
Formula
$GM = K B + B M - K G$
Substituting
$GM = 0.683 + 2.198 - 1.58 = 1.305 m \to s t ab l e$

Answer: F_B = 682.9 kN, GM = 1.31 m (stable)

Why the other options are there

1,051 kN (full depth used instead of draft)
GM = 0.62 m (KB omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Archimedes Principle and Buoyancy

Example 6

Hydrostatic pressure at depth — Archimedes Principle and Buoyancy (3)

Find the gauge pressure 5.5 ft below the surface of a fluid with specific gravity 8.60.

Given

h = 5.5 ft
SG = 8.60
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Archimedes Principle and Buoyancy (3)

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 8.60 (62.4) = 536.6 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 536.6 (5.5) = 2, 952 p s f$
Convert
$p = 2, 952/144 = 20.50 p s i$

Answer: p ≈ 2,952 psf (20.50 psi)

Why the other options are there

343.2 psf (specific gravity ignored)
425,019 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Archimedes Principle and Buoyancy

Example 7

Buoyant force and metacentric height of a floating barge — Archimedes Principle and Buoyancy (3)

A rectangular barge 3.5 m long by 3.0 m wide floats at a draft of 1.26 m in fresh water, with its centre of gravity 1.89 m above the keel. Compute the buoyant force (and hence its weight) and the metacentric height, and judge its stability.

Given

L = 3.5 m, B = 3.0 m
Draft = 1.26 m
KG = 1.89 m
ρ = 1000 kg/m³

Find

Buoyant force and metacentric height GM

Start with the thinking

Archimedes: the buoyant force equals the weight of displaced fluid, so a floating body's weight equals that force.
Positive GM means the metacentre lies above the centre of gravity and the vessel is stable.

Step-by-step solution

Displaced volume
$\forall = L B d = 3.5 (3.0) (1.26) = 13.230 m^{3}$
Formula
$F_{B} = ρ g \forall$
Substituting
$F_{B} = 1000 (9.81) (13.230) = 129.8 k N$
Waterplane inertia
$I = L B^{3} /12 = 3.5 (3.0)^{3} /12 = 7.875 m^{4}$
Formula
$B M = I /\forall$
Substituting
$B M = 7.875/13.230 = 0.595 m$
Formula
$GM = K B + B M - K G$
Substituting
$GM = 0.630 + 0.595 - 1.89 = - 0.665 m \to u n s t ab l e$

Answer: F_B = 129.8 kN, GM = -0.66 m (unstable)

Why the other options are there

216.3 kN (full depth used instead of draft)
GM = -1.29 m (KB omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Archimedes Principle and Buoyancy

Example 8

Hydrostatic pressure at depth — Archimedes Principle and Buoyancy (4)

Find the gauge pressure 13.0 ft below the surface of a fluid with specific gravity 2.10.

Given

h = 13.0 ft
SG = 2.10
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Archimedes Principle and Buoyancy (4)

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 2.10 (62.4) = 131.0 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 131.0 (13.0) = 1, 704 p s f$
Convert
$p = 1, 704/144 = 11.83 p s i$

Answer: p ≈ 1,704 psf (11.83 psi)

Why the other options are there

811.2 psf (specific gravity ignored)
245,307 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Archimedes Principle and Buoyancy

Example 9

Buoyant force and metacentric height of a floating barge — Archimedes Principle and Buoyancy (4)

A rectangular barge 8.5 m long by 2.0 m wide floats at a draft of 1.19 m in fresh water, with its centre of gravity 1.95 m above the keel. Compute the buoyant force (and hence its weight) and the metacentric height, and judge its stability.

Given

L = 8.5 m, B = 2.0 m
Draft = 1.19 m
KG = 1.95 m
ρ = 1000 kg/m³

Find

Buoyant force and metacentric height GM

Start with the thinking

Archimedes: the buoyant force equals the weight of displaced fluid, so a floating body's weight equals that force.
Positive GM means the metacentre lies above the centre of gravity and the vessel is stable.

Step-by-step solution

Displaced volume
$\forall = L B d = 8.5 (2.0) (1.19) = 20.230 m^{3}$
Formula
$F_{B} = ρ g \forall$
Substituting
$F_{B} = 1000 (9.81) (20.230) = 198.5 k N$
Waterplane inertia
$I = L B^{3} /12 = 8.5 (2.0)^{3} /12 = 5.667 m^{4}$
Formula
$B M = I /\forall$
Substituting
$B M = 5.667/20.230 = 0.280 m$
Formula
$GM = K B + B M - K G$
Substituting
$GM = 0.595 + 0.280 - 1.95 = - 1.080 m \to u n s t ab l e$

Answer: F_B = 198.5 kN, GM = -1.08 m (unstable)

Why the other options are there

283.5 kN (full depth used instead of draft)
GM = -1.67 m (KB omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Archimedes Principle and Buoyancy

Example 10

Hydrostatic pressure at depth — Archimedes Principle and Buoyancy (5)

Find the gauge pressure 7.0 ft below the surface of a fluid with specific gravity 2.10.

Given

h = 7.0 ft
SG = 2.10
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Archimedes Principle and Buoyancy (5)

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 2.10 (62.4) = 131.0 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 131.0 (7.0) = 917.3 p s f$
Convert
$p = 917.3/144 = 6.37 p s i$

Answer: p ≈ 917.3 psf (6.37 psi)

Why the other options are there

436.8 psf (specific gravity ignored)
132,088 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Archimedes Principle and Buoyancy

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a pipeline, jet or submerged surface, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Archimedes Principle and Buoyancy contains 0 relations; you must be able to find this page in under 15 seconds.
Exam style: continuity plus energy, with one head-loss or force term.
Unit rule: γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.