A cube floats in a vessel with water so that its upper edge is parallel to the water surface.Half of the cube is immersed in the water.What layer of oil must be added to make the cube float completely immersed in the liquid, if the density of oil is half that of water and the length of the cube's edge is $a$? Oil and water do not mix.
Decision:
First, let us determine the density of the substance of the cube $\rho$.Let us denote the density of water by $\rho_{w}$. First, the cube is subjected to the downward force of gravity
$F=\rho ga^{3}$ (1)
and the water pressure force $F_{A}$ directed upward and equal to the weight of the water displaced by the cube:
$F_{A}=\frac{\rho_{w}ga^{3}}{2}$. (2)
Since the cube was in equilibrium, then
$F = F_{A}$. (3)
Substituting expressions (1) and (2) into formula (3), we obtain:
$\rho=\frac{1}{2} \rho_{w}$, (4)
i.e., the density of the substance of the cube is equal to the density of the oil.
Suppose that the vessel is filled with such a layer of oil that the cube is completely submerged and its upper and lower edges are horizontal. Let $a_{1}$ be the height of the part of the cube in the oil; then $a-a_{1}$ is the height of its part immersed in water. The force of gravity $F$ and the expulsive force $F_{A}^{\prime}$ act on the cube, equal to
$F_{A}=\rho_{w} g(a-a_{1})a^{2}+\rho ga_{1}a^{2}$.
Since even now the cube is in equilibrium, we can equate $F = F^{\prime}_{A}$ and get an expression for finding $a$:
$\rho ga=\rho ga_{1}+\rho_{c}g(a-a_{1})$
Given equality (4), we obtain that $a_{1} = a$, i.e. in the equilibrium state the cube is completely immersed in oil. Since the densities of the oil and the cube are equal, the thickness h of the oil layer does not matter, as long as the cube can be completely immersed in it. Thus, it is necessary to add a layer of oil $h \geq a$.
Decision:
First, let us determine the density of the substance of the cube $\rho$.Let us denote the density of water by $\rho_{w}$. First, the cube is subjected to the downward force of gravity
$F=\rho ga^{3}$ (1)
and the water pressure force $F_{A}$ directed upward and equal to the weight of the water displaced by the cube:
$F_{A}=\frac{\rho_{w}ga^{3}}{2}$. (2)
Since the cube was in equilibrium, then
$F = F_{A}$. (3)
Substituting expressions (1) and (2) into formula (3), we obtain:
$\rho=\frac{1}{2} \rho_{w}$, (4)
i.e., the density of the substance of the cube is equal to the density of the oil.
Suppose that the vessel is filled with such a layer of oil that the cube is completely submerged and its upper and lower edges are horizontal. Let $a_{1}$ be the height of the part of the cube in the oil; then $a-a_{1}$ is the height of its part immersed in water. The force of gravity $F$ and the expulsive force $F_{A}^{\prime}$ act on the cube, equal to
$F_{A}=\rho_{w} g(a-a_{1})a^{2}+\rho ga_{1}a^{2}$.
Since even now the cube is in equilibrium, we can equate $F = F^{\prime}_{A}$ and get an expression for finding $a$:
$\rho ga=\rho ga_{1}+\rho_{c}g(a-a_{1})$
Given equality (4), we obtain that $a_{1} = a$, i.e. in the equilibrium state the cube is completely immersed in oil. Since the densities of the oil and the cube are equal, the thickness h of the oil layer does not matter, as long as the cube can be completely immersed in it. Thus, it is necessary to add a layer of oil $h \geq a$.