In the Arctic ice in the center of a small flat ice floe with an area of 70m ^ 3 there is a polar bear weighing 700 kg

In the Arctic ice in the center of a small flat ice floe with an area of 70m ^ 3 there is a polar bear weighing 700 kg. In this case, the surface of the ice floe protrudes 10 cm above the water surface. At what depth under water is the lower surface of the ice floe.

S = 70 m2.

mm = 700 kg.

g = 10 m / s2.

h1 = 10 cm = 0.1 m.

ρw = 1000 kg / m3.

ρl = 900 kg / m3.

h2 -?

The force of gravity Ft, which acts on the ice floe with the bear, is balanced by the buoyancy force of Archimedes Farch: Ft = Farch.

Ft = (mm + ml) * g.

The buoyancy force of Archimedes Farch is determined by the formula: Farch = ρw * g * V2. Where ρw is the density of the fluid in which the body is immersed, g is the acceleration of gravity, V2 is the volume of the immersed part of the body in the fluid.

ml = ρl * V = ρl * S * h, where h is the height of the entire ice floe.

V2 = S * h2, where h2 is the underwater thickness of the ice floe.

h = h1 + h2, where h1 is the surface thickness of the ice floe.

ml = ρl * S * (h1 + h2).

(mm + ρl * S * (h1 + h2)) * g = ρw * g * S * h2.

mm + ρl * S * (h1 + h2) = ρw * S * h2.

mm + ρl * S * h1 + ρl * S * h2 = ρw * S * h2.

mm + ρl * S * h1 = ρw * S * h2 – ρl * S * h2.

h2 = (mm + ρl * S * h1) / (ρv – ρl) * S.

h2 = (700 kg + 900 kg / m3 * 70 m2 * 0.1 m) / (1000 kg / m3 – 900 kg / m3) * 70 m2 = 1 m.

Answer: the underwater thickness of the ice floe is h2 = 1 m.