An ice floe floats in the water, a part of which with a volume of 200m3 is above the water.

An ice floe floats in the water, a part of which with a volume of 200m3 is above the water. What is the volume of the entire ice floe if the density of the ice is 900 kg / m3, and the density of water is 1000 kg / m3?

Vн = 200 m3.

g = 9.8 m / s2.

ρl = 900 kg / m3.

ρw = 1000 kg / m3.

V -?

Since the ice floe is at rest, then, according to 1 Newton’s law, the forces that act on it are compensated.

Two forces act on the ice floe in the water: the force of gravity Ft directed vertically downward, and the buoyancy force of Archimedes Farkh directed vertically upward.

Ft = Farch.

The force of gravity is determined by the formula: Ft = m * g, where m is the mass of the ice floe, g is the acceleration of gravity.

m = V * ρl.

The buoyancy force of Archimedes is determined by the formula: Farch = ρw * g * Vpog. Where ρw is the density of water, g is the acceleration of gravity, Vpog is the volume of the submerged part of the body in the liquid.

Vpog = V – Vn.

Farch = ρw * g * (V – Vn).

V * ρl * g = ρw * g * (V – Vn).

V * ρl = ρw * V – ρw * Vn.

ρw * Vn = ρw * V – V * ρl.

V = ρw * Vn / (ρw – ρl).

V = 1000 kg / m3 * 200 m3 / (1000 kg / m3 – 900 kg / m3) = 2000 m3.

Answer: the volume of the entire ice floe is V = 2000 m3.