Posted by **Anonymous** on Tuesday, May 1, 2012 at 10:35am.

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?

- physics -
**Elena**, Tuesday, May 1, 2012 at 4:09pm
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.

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