posted by Anonymous on .
A wooden cube with a specific gravity of 0.90 and side length 0.120 m is placed into a
bucket of water and floats upright with its sides in a horizontal or vertical orientation.
What is the mass of the cube, what is the buoyancy force acting on the cube and how
much of the cube projects above the surface?
Let the length above the surface of the water is «x”,
a =0.120 m,
ρ1/ρ2 = 0.9. Since the density of water is ρ2 = 1000 kg/m³,
the density of wood is ρ1 = 900 kg/m³.
m•g = F(buoyancy force),
ρ1•V•g = ρ2•V1 •g
ρ1•a³•g = ρ2•a²•(a-x)•g,
ρ1/ ρ2 = (a-x)/a =0.9.
x = a – 0.9•a = 0.1•a =0.012 m.
The mass of the cube is
m = ρ1•a³ = 900•(0.12)³=1.56 kg.
The buoyancy force is
F = ρ2•a²•(a-x)•g = 1000•(0.12)² •(0.12 -0.012) •9.8 = 15.24 N.