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.
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?
1 answer
1 answer