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Archimedes Principle and Buoyancy

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics
0 formulas
10 exam-style examples
~45 min
All Fluid Mechanics lectures

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

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 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=2.0fth = 2.0 ft
  • SG=12.60SG = 12.60
  • γwater=62.4lb/ft3\gamma_water = 62.4 lb/ft^{3}

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.
h = 2.0 ft

Figure 1 — schematic for Hydrostatic pressure at depth — Archimedes Principle and Buoyancy

Step-by-step solution

  1. Specific weight

    γ=SG×γw=12.60(62.4)=786.2lb/ft3\gamma = SG \times \gamma_w = 12.60(62.4) = 786.2 lb/ft^{3}

  2. Hydrostatic — p = γh

  3. Substituting

    p=786.2(2.0)=1,572psfp = 786.2(2.0) = 1,572 psf

  4. Convert

    p=1,572/144=10.92psip = 1,572/144 = 10.92 psi
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 2
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=6.0m,B=3.0mL = 6.0 m, B = 3.0 m
  • Draft=1.28mDraft = 1.28 m
  • KG=1.42mKG = 1.42 m
  • ρ=1000kg/m3\rho = 1000 kg/m^{3}

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

  1. Displaced volume

    ∀=LBd=6.0(3.0)(1.28)=22.950m3\forall = L B d = 6.0(3.0)(1.28) = 22.950 m^{3}

  2. Formula

    FB=ρg∀F_B = \rho g \forall

  3. Substituting

    FB=1000(9.81)(22.950)=225.1kNF_B = 1000(9.81)(22.950) = 225.1 kN

  4. Waterplane inertia

    I=LB3/12=6.0(3.0)3/12=13.500m4I = L B^{3}/12 = 6.0(3.0)^{3}/12 = 13.500 m^{4}

  5. Formula

    BM=I/∀BM = I/\forall

  6. Substituting

    BM=13.500/22.950=0.588mBM = 13.500/22.950 = 0.588 m

  7. Formula

    GM=KB+BM−KGGM = KB + BM - KG

  8. Substituting

    GM=0.638+0.588−1.42=−0.199m→unstableGM = 0.638 + 0.588 - 1.42 = -0.199 m \to unstable
Answer:
FB=225.1kN,GM=−0.20m(unstable)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 3
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=12.5fth = 12.5 ft
  • SG=4.60SG = 4.60
  • γwater=62.4lb/ft3\gamma_water = 62.4 lb/ft^{3}

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.
h = 12.5 ft

Figure 3 — schematic for Hydrostatic pressure at depth — Archimedes Principle and Buoyancy (2)

Step-by-step solution

  1. Specific weight

    γ=SG×γw=4.60(62.4)=287.0lb/ft3\gamma = SG \times \gamma_w = 4.60(62.4) = 287.0 lb/ft^{3}

  2. Hydrostatic — p = γh

  3. Substituting

    p=287.0(12.5)=3,588psfp = 287.0(12.5) = 3,588 psf

  4. Convert

    p=3,588/144=24.92psip = 3,588/144 = 24.92 psi
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 4
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=8.5m,B=6.0mL = 8.5 m, B = 6.0 m
  • Draft=1.37mDraft = 1.37 m
  • KG=1.58mKG = 1.58 m
  • ρ=1000kg/m3\rho = 1000 kg/m^{3}

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

  1. Displaced volume

    ∀=LBd=8.5(6.0)(1.37)=69.615m3\forall = L B d = 8.5(6.0)(1.37) = 69.615 m^{3}

  2. Formula

    FB=ρg∀F_B = \rho g \forall

  3. Substituting

    FB=1000(9.81)(69.615)=682.9kNF_B = 1000(9.81)(69.615) = 682.9 kN

  4. Waterplane inertia

    I=LB3/12=8.5(6.0)3/12=153.0m4I = L B^{3}/12 = 8.5(6.0)^{3}/12 = 153.0 m^{4}

  5. Formula

    BM=I/∀BM = I/\forall

  6. Substituting

    BM=153.0/69.615=2.198mBM = 153.0/69.615 = 2.198 m

  7. Formula

    GM=KB+BM−KGGM = KB + BM - KG

  8. Substituting

    GM=0.683+2.198−1.58=1.305m→stableGM = 0.683 + 2.198 - 1.58 = 1.305 m \to stable
Answer:
FB=682.9kN,GM=1.31m(stable)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 5
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=5.5fth = 5.5 ft
  • SG=8.60SG = 8.60
  • γwater=62.4lb/ft3\gamma_water = 62.4 lb/ft^{3}

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.
h = 5.5 ft

Figure 5 — schematic for Hydrostatic pressure at depth — Archimedes Principle and Buoyancy (3)

Step-by-step solution

  1. Specific weight

    γ=SG×γw=8.60(62.4)=536.6lb/ft3\gamma = SG \times \gamma_w = 8.60(62.4) = 536.6 lb/ft^{3}

  2. Hydrostatic — p = γh

  3. Substituting

    p=536.6(5.5)=2,952psfp = 536.6(5.5) = 2,952 psf

  4. Convert

    p=2,952/144=20.50psip = 2,952/144 = 20.50 psi
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 6
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=3.5m,B=3.0mL = 3.5 m, B = 3.0 m
  • Draft=1.26mDraft = 1.26 m
  • KG=1.89mKG = 1.89 m
  • ρ=1000kg/m3\rho = 1000 kg/m^{3}

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

  1. Displaced volume

    ∀=LBd=3.5(3.0)(1.26)=13.230m3\forall = L B d = 3.5(3.0)(1.26) = 13.230 m^{3}

  2. Formula

    FB=ρg∀F_B = \rho g \forall

  3. Substituting

    FB=1000(9.81)(13.230)=129.8kNF_B = 1000(9.81)(13.230) = 129.8 kN

  4. Waterplane inertia

    I=LB3/12=3.5(3.0)3/12=7.875m4I = L B^{3}/12 = 3.5(3.0)^{3}/12 = 7.875 m^{4}

  5. Formula

    BM=I/∀BM = I/\forall

  6. Substituting

    BM=7.875/13.230=0.595mBM = 7.875/13.230 = 0.595 m

  7. Formula

    GM=KB+BM−KGGM = KB + BM - KG

  8. Substituting

    GM=0.630+0.595−1.89=−0.665m→unstableGM = 0.630 + 0.595 - 1.89 = -0.665 m \to unstable
Answer:
FB=129.8kN,GM=−0.66m(unstable)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 7
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=13.0fth = 13.0 ft
  • SG=2.10SG = 2.10
  • γwater=62.4lb/ft3\gamma_water = 62.4 lb/ft^{3}

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.
h = 13.0 ft

Figure 7 — schematic for Hydrostatic pressure at depth — Archimedes Principle and Buoyancy (4)

Step-by-step solution

  1. Specific weight

    γ=SG×γw=2.10(62.4)=131.0lb/ft3\gamma = SG \times \gamma_w = 2.10(62.4) = 131.0 lb/ft^{3}

  2. Hydrostatic — p = γh

  3. Substituting

    p=131.0(13.0)=1,704psfp = 131.0(13.0) = 1,704 psf

  4. Convert

    p=1,704/144=11.83psip = 1,704/144 = 11.83 psi
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 8
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=8.5m,B=2.0mL = 8.5 m, B = 2.0 m
  • Draft=1.19mDraft = 1.19 m
  • KG=1.95mKG = 1.95 m
  • ρ=1000kg/m3\rho = 1000 kg/m^{3}

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

  1. Displaced volume

    ∀=LBd=8.5(2.0)(1.19)=20.230m3\forall = L B d = 8.5(2.0)(1.19) = 20.230 m^{3}

  2. Formula

    FB=ρg∀F_B = \rho g \forall

  3. Substituting

    FB=1000(9.81)(20.230)=198.5kNF_B = 1000(9.81)(20.230) = 198.5 kN

  4. Waterplane inertia

    I=LB3/12=8.5(2.0)3/12=5.667m4I = L B^{3}/12 = 8.5(2.0)^{3}/12 = 5.667 m^{4}

  5. Formula

    BM=I/∀BM = I/\forall

  6. Substituting

    BM=5.667/20.230=0.280mBM = 5.667/20.230 = 0.280 m

  7. Formula

    GM=KB+BM−KGGM = KB + BM - KG

  8. Substituting

    GM=0.595+0.280−1.95=−1.080m→unstableGM = 0.595 + 0.280 - 1.95 = -1.080 m \to unstable
Answer:
FB=198.5kN,GM=−1.08m(unstable)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 9
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=7.0fth = 7.0 ft
  • SG=2.10SG = 2.10
  • γwater=62.4lb/ft3\gamma_water = 62.4 lb/ft^{3}

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.
h = 7.0 ft

Figure 9 — schematic for Hydrostatic pressure at depth — Archimedes Principle and Buoyancy (5)

Step-by-step solution

  1. Specific weight

    γ=SG×γw=2.10(62.4)=131.0lb/ft3\gamma = SG \times \gamma_w = 2.10(62.4) = 131.0 lb/ft^{3}

  2. Hydrostatic — p = γh

  3. Substituting

    p=131.0(7.0)=917.3psfp = 131.0(7.0) = 917.3 psf

  4. Convert

    p=917.3/144=6.37psip = 917.3/144 = 6.37 psi
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

Example 10
Buoyant force and metacentric height of a floating barge — Archimedes Principle and Buoyancy (5)

A rectangular barge 7.5 m long by 2.0 m wide floats at a draft of 1.72 m in fresh water, with its centre of gravity 2.88 m above the keel. Compute the buoyant force (and hence its weight) and the metacentric height, and judge its stability.

Given

  • L=7.5m,B=2.0mL = 7.5 m, B = 2.0 m
  • Draft=1.72mDraft = 1.72 m
  • KG=2.88mKG = 2.88 m
  • ρ=1000kg/m3\rho = 1000 kg/m^{3}

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

  1. Displaced volume

    ∀=LBd=7.5(2.0)(1.72)=25.875m3\forall = L B d = 7.5(2.0)(1.72) = 25.875 m^{3}

  2. Formula

    FB=ρg∀F_B = \rho g \forall

  3. Substituting

    FB=1000(9.81)(25.875)=253.8kNF_B = 1000(9.81)(25.875) = 253.8 kN

  4. Waterplane inertia

    I=LB3/12=7.5(2.0)3/12=5.000m4I = L B^{3}/12 = 7.5(2.0)^{3}/12 = 5.000 m^{4}

  5. Formula

    BM=I/∀BM = I/\forall

  6. Substituting

    BM=5.000/25.875=0.193mBM = 5.000/25.875 = 0.193 m

  7. Formula

    GM=KB+BM−KGGM = KB + BM - KG

  8. Substituting

    GM=0.862+0.193−2.88=−1.819m→unstableGM = 0.862 + 0.193 - 2.88 = -1.819 m \to unstable
Answer:
FB=253.8kN,GM=−1.82m(unstable)F_B = 253.8 kN, GM = -1.82 m (unstable)

Why the other options are there

  • 338.4 kN (full depth used instead of draft)
  • GM = -2.68 m (KB omitted)

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

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