4.2.11. With what force a heavy stick presses on the bottom of the reservoir, if a hollow ball of radius $r$, rigidly connected to the stick, is half immersed in the liquid? The density of the liquid is $\rho$, the length of the stick is $l$.
Solution
Let's consider the following figure
Force analysis
The system is on equilibrium in above position, so total torque is zero ($\sum \vec{\tau} = \vec{0}$). Considering as a pivot point the center of mass of the stick (located at stick's half),
Let's suppose the pivot point is at the contact of the stick with the bottom. Applying Newton's Second Law for rotation ($\sum \vec{\tau} = 0$) and considering the massless ball ($m_b = 0$): $$F_A (l + r) \cos\alpha - m_s g \frac{l}{2} \cos\alpha = 0$$ where $m_s$ is the mass of the stick. Hence, the mass of the stick is: $$m_s = \frac{2 F_A (l + r)}{g l} = \frac{2 \left(\rho g \frac{V}{2}\right) (l + r)}{g l} = \rho V \left(1 + \frac{r}{l}\right) \quad (1)$$
Applying Newton's Second Law for translation (vertical axis): $$F_A + N = m_s g$$ $$\rho g \frac{V}{2} + N = m_s g \quad (2)$$
From $(1)$ and $(2)$, and taking $V = \frac{4}{3}\pi r^3$, the above answer is obtained again: $$N = \rho V g \left(1 + \frac{r}{l}\right) - \rho g \frac{V}{2} = \rho g V \left(\frac{1}{2} + \frac{r}{l}\right) = \frac{\rho g V}{2} \left(1 + 2\frac{r}{l}\right)$$ $$N = \frac{2}{3}\pi r^3 \rho g \left(1 + 2\frac{r}{l}\right)$$
Note: By Newton's Third Law, the force $F$ with which the stick presses on the bottom is equal in magnitude to the normal reaction force $N$.
Answer
$$F = \frac{2}{3}\pi r^3 \rho g \left(1 + \frac{2r}{l}\right)$$
