5. A sphere of radius $R$, made of a material with a density equal to that of water, floats in water, touching its surface (see figure). Find the force with which the water acts on the lower half of the sphere. The density of water $\rho$ is known.

![](https://cdn.mathpix.com/cropped/2024_05_06_151e295255b2d1aa1930g-1.jpg?height=231&width=300&top_left_y=1666&top_left_x=153)
5. Obviously, the forces acting on different parts of the surface of the lower hemisphere from the water are different and are also directed differently. Therefore, calculating this force by summing the forces acting on small elements of the hemisphere's surface is quite complex. However, we have another method of summation. The fact is that similar sums of forces over the surface of a body are included in Archimedes' law. Therefore, if we take an auxiliary body such that the lower surface of the sphere is its surface, and use Archimedes' law, we can calculate the required force.

Thus, let's mentally cut and remove the upper hemisphere, i.e., consider a body in the form of only the lower hemisphere. On one hand, the Archimedes' force acting on this hemisphere from the water is known to us. This is

$$
F_{A}=\rho g V_{\text {n.u. }}=\frac{2}{3} \pi \rho g R^{3}
$$

where $\rho$ is the density of water, $g$ is the acceleration due to gravity, and $V_{\text {n.u. }}=2 \pi R^{3} / 3$ is the volume of the hemisphere. On the other hand, this force is the resultant of the forces acting on the hemisphere from the liquid. Therefore, it can be represented as the sum of the forces, one of which acts on the lower surface of the hemisphere

![](https://cdn.mathpix.com/cropped/2024_05_06_0aa2685a2c7968513d04g-4.jpg?height=257&width=322&top_left_y=1216&top_left_x=1615)
$\vec{F}_{n . \text {.u. }}$. (which we need to find), and the second acts on its flat circular surface closing the hemisphere $\vec{F}_{k p}$. Since these forces are directed oppositely, in the expression for the Archimedes' force, they are subtracted

$$
F_{A}=F_{n . u .}-F_{k p}
$$

The last force is easy to calculate. Since the circular surface closing the hemisphere is at a depth of $R$, the hydrostatic pressure of the liquid at this depth is $p=\rho g R$. Therefore,

$$
F_{k p}=p \pi R^{2}=\pi \rho g R^{3}
$$

From here and formulas $(*),(* *)$ we find

$$
F_{n . u .}=F_{A}+F_{k p}=\frac{5}{3} \pi \rho g R^{3}
$$

Criteria for evaluating the problem (maximum score for the problem - 2 points)