CivilSolve

Fluid Mechanics & Hydraulics

Buoyancy & Archimedes' Principle

Archimedes: the buoyant force equals the weight of displaced fluid. Handles the three classic setups — (1) does it float, and how much sits below the surface (submerged fraction = SG_object/SG_fluid); (2) apparent weight of a fully submerged object; (3) the crown problem: weight in air + weight in water gives the volume and the object's specific gravity. Use for: floating blocks and drafts, 'weighs X in air and Y in water', anchor/apparent-weight questions. Not for: stability/metacentric height.

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Inputs

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Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

wood block SG 0.6 floats 60% submerged

Given

object volume
0.1 m^3
object specific gravity
0.6

Assumptions

  • Uniform (fully mixed) fluid; the object is rigid and non-absorbent; SG referenced to water at 1000 kg/m³.

Solution steps

  1. Compare weight with full-submersion buoyancy

    If the fluid displaced by the WHOLE object outweighs the object, it floats; otherwise it sinks.

    FB(full) = γf·V vs W

    FB(full) = 980.7 N vs W = 588.4 N — it FLOATS

  2. Submerged fraction

    Floating, it sinks just enough for the displaced fluid to match its weight.

    Vsub/V = W/(γf·V) = SG_object/SG_fluid

    fraction = 0.6 (60 percent below the surface); Vsub = 0.06 m^3

    = 0.6

Results

Object weight W

0.5884kN

Floats? (1 = yes, 0 = no)

1

Fraction of volume below the surface

0.6

Submerged volume

0.06m^3

Buoyant force (equals W when floating)

0.5884kN

Where this answer was checked
source
Submerged fraction = SG_object/SG_fluid — 0.1 m³ block in water
verified by
hand-recomputed
derivation
W = 0.6·1000·9.80665·0.1 = 588.4 N < FB(full) = 980.7 N → floats at fraction 0.6.

Example 2

steel sinks: apparent weight of 0.01 m³ at SG 7.85

Given

object volume
0.01 m^3
object specific gravity
7.85

Assumptions

  • Uniform (fully mixed) fluid; the object is rigid and non-absorbent; SG referenced to water at 1000 kg/m³.

Solution steps

  1. Compare weight with full-submersion buoyancy

    If the fluid displaced by the WHOLE object outweighs the object, it floats; otherwise it sinks.

    FB(full) = γf·V vs W

    FB(full) = 98.07 N vs W = 769.8 N — it SINKS

  2. Apparent weight when submerged

    Fully underwater, the scale reads the true weight minus the buoyant force.

    W_apparent = W − γf·V

    W_apparent = 671.8 N

    = 671.8 N

Results

Object weight W

0.7698kN

Floats? (1 = yes, 0 = no)

0

Buoyant force FB

0.09807kN

Apparent (submerged) weight

0.6718kN

Where this answer was checked
source
W_apparent = W − γ·V — submerged apparent weight
verified by
hand-recomputed
derivation
W = 7850·9.80665·0.01 = 769.8 N; FB = 98.07 N; apparent = 671.7 N.

Example 3

the crown problem: 100 lbf in air, 60 lbf in water

Given

object weight
100 lbf
apparent weight submerged
60 lbf

Assumptions

  • Uniform (fully mixed) fluid; the object is rigid and non-absorbent; SG referenced to water at 1000 kg/m³.

Solution steps

  1. Buoyant force from the two weighings

    The scale reads less underwater by exactly the buoyant force.

    FB = W_air − W_sub

    FB = 444.8 N − 266.9 N = 177.9 N

  2. Volume and specific gravity

    The buoyant force equals the displaced fluid's weight, revealing the volume — and the SG follows from weight over buoyancy.

    V = FB/γf; SG = (W_air/FB)·SG_fluid

    V = 0.01814 m^3; SG = 2.5

    = 2.5

Results

Buoyant force FB

0.1779kN

Object volume V

0.01814m^3

Object specific gravity

2.5

Where this answer was checked
source
Archimedes' classic: V = FB/γw; SG = W/FB — volume and SG from two weighings
verified by
hand-recomputed
derivation
FB = 40 lbf; V = 40/62.4 = 0.6410 ft³; SG = 100/40 = 2.5.