SME
Fluid Mechanics · Problem type
Buoyancy of a floating body
Use this for “0.06 m³ of wood of 600 kg/m³ floats in water. How much is under, and what is FB?”
Diagram
Floating:
Vsub
V
=
ρbody
ρ
=
600
1,000
=
0.6.
FB = ρgVsub = 1,000 × 9.81 × 0.036 = 353.2 N, the block’s weight.
This is an example. Type your own values to start a new problem.
Vsub (m³)
V (m³)
=
ρbody (kg/m³)
ρ (kg/m³)
FB (N)
=
ρ (kg/m³)
×
9.81
×
Vsub (m³)
Formulas
Vsub
V
=
ρbody
ρ
0.036
0.06
=
600
1,000
FB
=
ρ
×
9.81
×
Vsub
353.2 = 1,000 × 9.81 × 0.036
Assumptions
•
The fluid is at rest and has the same density at every depth (incompressible).
•
g = 9.81 m/s².
•
The body floats at rest, so its weight equals the buoyant force (Archimedes).
Step-by-step
WE KNOW
V = 0.06 m³, ρbody = 600 kg/m³, ρ = 1,000 kg/m³
FIND
volume under the surface (Vsub), buoyant force (FB)
Step 1 · Find volume under the surface (Vsub)
Use Vsub ÷ V = ρbody ÷ ρ
Floating,
the
weight
ρbody
gV
equals
the
buoyant
force
ρgVsub,
so
Vsub
=
V
×
ρbody
ρ
.
Vsub
=
V
×
ρbody
ρ
Vsub
=
0.06
×
600
1,000
Vsub
=
36
1,000
Vsub = 0.036 m³
Step 2 · Find buoyant force (FB)
Use FB = ρ × 9.81 × Vsub
The buoyant force is the weight of the fluid the body pushes aside: ρg times Vsub.
FB = 1,000 × 9.81 × 0.036
FB = 9,810 × 0.036
FB = 353.2 N
Check
Put the numbers back in. Both sides should match.
0.036
0.06
=
600
1,000
✓
353.2 = 1,000 × 9.81 × 0.036 ✓
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