The trunk of a tree floats in water, sinking to 4/5 of its volume. If a buoyant force equal to 980N

The trunk of a tree floats in water, sinking to 4/5 of its volume. If a buoyant force equal to 980N acts on it, then the volume of the barrel?

Vp = 4 * V / 5.

Fв = 980 N.

g = 9.8 m / s2.

ρ = 1000 kg / m3.

V -?

m -?

Since the trunk of a tree floats in water, it is in a state of equilibrium. This means that the effect of forces on him is compensated. The force of gravity m * g is balanced by the buoyancy force of Archimedes Fw: Fw = m * g.

The buoyancy force of Archimedes Fв is determined by the formula: Fв = ρ * g * Vп. Where ρ is the density of the fluid in which the body is immersed, g is the acceleration of gravity, Vp is the volume of the immersed part of the body in the fluid.

We express the mass of the tree trunk m by the formula: m = V * ρd, where ρd is the density of the tree.

ρ * g * Vп = V * ρд.

ρд = ρ * g * Vп / V = ​​ρ * g * 4 * V / 5 * V = ρ * g * 4 * / 5.

ρd = 1000 kg / m3 * 9.8 m / s2 * 4/5 = 7840 kg / m3.

Vп = Fв / ρ * g.

Vp = 980 N / 9.8 m / s2 * 1000 kg / m3 = 0.1 m3.

V = 5 * Vp / 4.

V = 5 * 0.1 m3 / 4 = 0.125 m3.

m = 0.125 m3 * 7840 kg / m3 = 980 kg.

Answer: the log has a mass m = 980 kg and a volume V = 0.125 m3.