What force must be applied to hold a stone whose mass is 7.2 kg under water?

What force must be applied to hold a stone whose mass is 7.2 kg under water? (Density of a stone 2400 kg / m3, density of water 1000 kg / m3)

A stone in water is acted upon by the force of Archimedes and the force of gravity. These forces are exactly the opposite.
P = Ft – Fa = m * g – ρw * g * V, where m is the mass of the stone (m = 7.2 kg), g is the acceleration of gravity (we take g = 10 m / s ^ 2), ρ is the density water (ρ = 1000 kg / m ^ 3), V is the volume of the stone.
Stone volume: V = m / ρк, where ρк is the density of the stone (ρк = 2400 kg / m ^ 3).
P = m * g – ρw * g * V = mg – ρw * g * m / ρk = 7.2 * 10 – 1000 * 10 * 7.2 / 2400 = 72 – 30 = 42 N.
Answer. To hold the stone in the water, a force of 42 N must be applied.