Swimming ice floe. An ice floe is floating on the surface of the lake, in the form of a rectangular parallelepiped

Swimming ice floe. An ice floe is floating on the surface of the lake, in the form of a rectangular parallelepiped with a square base. An ice floe is put out of the water at h = 5 cm. A load weighing m = 30 kg is placed in the center of the ice floe. At what area of the base of an ice floe with a load will not completely submerge in water? What is the height of the ice floe? Water density ρw = 1000 kg / m3, ice density ρl = 900 kg / m3.

h = 5 cm = 0.05 m.

m = 30 kg.

g = 10 N / kg.

ρw = 1000 kg / m3.

ρl = 900 kg / m3.

S -?

h1 -?

When a load is immersed on an ice floe, the gravity of the load m * g should be equal to the force of Archimedes Farch, which will act on the part of the ice floe that was above the surface of the water: m * g = Farch.

Farch = ρw * g * V, where ρw is the density of the water, V is the volume of the ice floe that was above the water.

V = S * h.

m * g = ρw * g * S * h.

S = m / ρw * h.

S = 30 kg / 1000 kg / m3 * 0.05 m = 0.6 m2.

Let us write down the conditions for the balance of the load on the ice floe: (m + ml) * g = Farch1, Farch1 is the Archimedes force that acts on the entire ice floe.

Farch1 = ρw * g * V1, where V1 is the volume of the entire ice floe.

V1 = S * h1.

ml = V1 * ρl = S * h1 * ρl.

(m + S * h1 * ρl) * g = ρw * g * S * h1.

m + S * h1 * ρl = ρw * S * h1.

h1 = m / S * (ρv – ρl).

h1 = 30 kg / 0.6 m2 * (1000 kg / m3 – 900 kg / m3) = 0.5 m.

Answer: with an area larger than S = 0.6 m2, the ice floe will not completely submerge into the water, the height of the ice floe is h1 = 0.5 m.