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Archimedes' buoyancy — submerged object

An iron cube (ρ_iron = 7874 kg/m³) with side L = 0.1 m is completely submerged in water (ρ_H₂O = 1000 kg/m³).\nCalculate: (a) the volume, (b) the buoyant force, (c) whether it sinks or floats.
Given data
L = 0.1 mρ_Fe = 7874 kg/m³ρ_H₂O = 1000 kg/m³g = 9.81 m/s²
Review the theory: Fluidi
Steps
0 / 4
  1. What is the volume of the cube with side L = 0.1 m?
  2. What is the mass of the iron cube?
  3. What is the buoyant force exerted by the water on the cube?
  4. What is the weight of the iron cube?
Full worked solution
  1. What is the volume of the cube with side L = 0.1 m?
    V=L3=0.13V = L^3 = 0.1^3
    V=L3=(0.1)3=0.001 m3=1 LV = L^3 = (0.1)^3 = \mathbf{0.001\,m^3} = \mathbf{1\,L}. The volume of a cube is the side length cubed.
  2. What is the mass of the iron cube?
    m=ρFe⋅V=7874⋅0.001m = \rho_{Fe} \cdot V = 7874 \cdot 0.001
    m=ρFeV=7874×0.001=7.874 kgm = \rho_{\mathrm{Fe}} V = 7874 \times 0.001 = \mathbf{7.874\,kg}. Mass is the product of density and volume.
  3. What is the buoyant force exerted by the water on the cube?
    FA=ρH2O⋅V⋅g=1000⋅0.001⋅9.81F_A = \rho_{H_2O} \cdot V \cdot g = 1000 \cdot 0.001 \cdot 9.81
    FA=ρH2OVg=1000×0.001×9.81=9.81 NF_A = \rho_{\mathrm{H_2O}} V g = 1000 \times 0.001 \times 9.81 = \mathbf{9.81\,N}. The buoyant force equals the weight of the displaced water (equivalent to 1 kg of water).
  4. What is the weight of the iron cube?
    P=m⋅g=7.874⋅9.81P = m \cdot g = 7.874 \cdot 9.81
    P=mg=7.874×9.81=77.24 N≫FA=9.81 NP = mg = 7.874 \times 9.81 = \mathbf{77.24\,N} \gg F_A = 9.81\,\mathrm{N} → the cube sinks. Since ρFe≫ρH2O\rho_{\mathrm{Fe}} \gg \rho_{\mathrm{H_2O}}, the weight far exceeds the buoyant force.
Result:Buoyant force = 9.81 N, Weight = 77.24 N. Iron sinks because ρ_Fe ≫ ρ_H₂O. To float, ρ_object ≤ ρ_fluid is required.
TheoryFluidiFormula sheetPhysics 1

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