How much force must be applied to a 17 kg brass rod to lift it in water?

m = 17 kg.

g = 10 m / s2.

ρl = 8500 kg / m3.

ρw = 1000 kg / m3.

F -?

Any body, including a brass rod, in a liquid, in addition to gravity m * g, is acted upon by the buoyancy force of Archimedes Farkh, directed vertically upwards. Therefore, the resultant of these forces will be the force F with which it is necessary to keep it under water.

Since the forces are directed oppositely, then F = m * g – Farkh.

The force of Archimedes Farch is expressed by the formula: Farch = V * ρв * g.

We express the volume of the rod V by the formula: V = m / ρl, where ρl is the density of brass, we will take it from the table of the density of substances.

Farch = m * ρw * g / ρl.

F = m * g – m * ρw * g / ρl = m * g * (1 – ρw / ρl).

F = 17 kg * 10 m / s2 * (1 – 1000 kg / m3 / 8500 kg / m3) = 150 N.

Answer: to hold the brass rod under water, it is necessary to apply a force F = 150 N vertically upward.