Determine the density of a body floating on the surface of the water if 2/3 of its volume is submerged in water.

According to Archimedes’ law:
A buoyant force equal to the weight of the displaced fluid acts on a body immersed in a liquid, and is called the Archimedes force:
Fa = ρ * g * V, where ρ is the density of the liquid, g is the acceleration of gravity of a body raised above the Earth g = 9.8 m / s², V is the volume of the body.
In our case:
Fа = ρж * g * V * (2/3)
The body, under the influence of gravity, presses on the water with the force of gravity:
Fт = m * g
Let’s write down the formula for determining the mass, through the density:
m = ρ * V
we substitute it in the formula for gravity:
Fт = m * g = ρ * V * g
We have a body in a state of equilibrium, which means Fа = Fт:
ρzh * g * V * (2/3) = ρ * V * g
Let us express the density of the body from this expression, after dividing both sides of the equality by V * g:
ρzh * (2/3) = ρ
Let’s substitute the numerical values, taking ρzh according to the reference book ρzh = 1000 kg / m³
ρ = 1000 * (2/3) = 667 kg / m³
Answer: body density 667 kg / m³.