What force is required to lift a metal part with a mass of 270 g and a volume of 30 cm3 under water.

The Archimedes force and the force of gravity act on a metal part in water. To lift the part, you need to apply a force in the direction of the Archimedes force in order to compensate for the effect of gravity.
F = Fт – Fа = m * g – ρ * g * V, where m is the mass of the part (m = 270 g = 0.27 kg), g is the acceleration of gravity (g = 10 m / s ^ 2), ρ is the density of water (ρ = 1000 kg / m ^ 3), V is the volume of the part (V = 30 cm ^ 3 = 30 * 10 ^ 6 m ^ 3).
F = m * g – ρ * g * V = 0.27 * 10 – 1000 * 10 * 30 * 10 ^ 6 = 2.7 – 0.3 = 2.4 N.
Answer: It is necessary to apply a force of 2.4 N.