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A boat of volume 200 m³ floats on sea water of density 1030 kg m⁻³. Calculate the upthrust of the sea water on the boat. [g = 10 m s⁻²]

Physics

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6.d)

ii) A boat of volume 200 m3200 \text{ m}^3 floats on sea water of density 1030 kg m−31030 \text{ kg m}^{-3}. Calculate the upthrust of the sea water on the boat. [g=10 m s−2g = 10 \text{ m s}^{-2}]

  1. Identify the given values.

    Volume of the boat (VboatV_{boat}) = 200 m3200 \text{ m}^3
    Density of sea water (ρwater\rho_{water}) = 1030 kg m−31030 \text{ kg m}^{-3}
    Acceleration due to gravity (gg) = 10 m s−210 \text{ m s}^{-2}

  2. Apply Archimedes' principle.

    Upthrust (FBF_B) is equal to the weight of the fluid displaced. The weight of the fluid displaced is calculated as Vdisplaced×ρwater×gV_{displaced} \times \rho_{water} \times g.

  3. Make an assumption to solve the problem.

    Since the mass of the boat is not provided, and the question asks for the upthrust, we assume that the given volume of the boat (200 m3200 \text{ m}^3) refers to the volume of water displaced for the purpose of this calculation. This implies the boat is fully submerged or the question is asking for the upthrust if it were fully submerged.

  4. Substitute the values and calculate the upthrust.

    FB=Vdisplaced×ρwater×gF_B = V_{displaced} \times \rho_{water} \times g FB=200 m3×1030 kg m−3×10 m s−2F_B = 200 \text{ m}^3 \times 1030 \text{ kg m}^{-3} \times 10 \text{ m s}^{-2} FB=2060000 NF_B = 2060000 \text{ N}

    The upthrust of the sea water on the boat is 2060000 N\boxed{\text{2060000 N}}.

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