14. A thin wooden rod is suspended from one end and can freely rotate relative to the point of suspension. The lower end of the rod is lowered into water. At a certain depth of immersion, the equilibrium position of the rod becomes unstable, and it begins to deviate from the vertical position. Find this depth and the angle of deviation of the rod from the vertical depending on the depth of immersion, if the length of the rod is 1 m, the density of its material is $0.6 \mathrm{r} / \mathrm{cm}^{3}$, and the density of water is $1 \mathrm{r} / \mathrm{cm}^{3}$.
14. Let the distance from the point of suspension of the rod to the water surface be \( h \). The length of the rod is denoted by \( l \). Suppose the rod is deflected from the vertical by an angle \( \boldsymbol{\alpha} \). Since the rod is in equilibrium, the algebraic sum of the moments of the forces of gravity and the buoyant force acting on it relative to the point of equilibrium must be zero. Since the force of gravity is applied to the center of the rod, its moment relative to the point of suspension of the rod is

\[
M_{1}=m g \frac{l}{2} \sin \alpha
\]

where \( \mathrm{m} \) is the mass of the rod. The buoyant force is directed vertically upwards and is applied to the midpoint of the part of the rod submerged in water. Its value is found using the formula

\[
F=m \frac{\rho_{\epsilon}}{\rho_{\partial}} g \frac{l-d}{l}
\]

The moment of this force relative to the point of suspension of the rod is

\[
M_{2}=m \frac{\rho_{\varepsilon}}{\rho_{\delta}} g \frac{l-d l+d}{l} \frac{l}{2} \sin \alpha
\]

Here,

\[
d=\frac{h}{\cos \alpha}-
\]

is the length of the part of the rod remaining above the water surface; \( \rho_{\varepsilon} \) is the density of water; \( \rho_{\varnothing} \) is the density of the material of the rod.

Since the moments of the force of gravity and the buoyant force rotate the rod in opposite directions, their algebraic sum will be zero if

\[
M_{1}=M_{2}
\]

From this relation, we find

\[
1-\frac{h^{2}}{l^{2} \cos ^{2} \alpha}=\frac{\rho_{\delta}}{\rho_{\xi}}
\]

From this, for the cosine of the angle of deflection of the rod from the vertical, we obtain

\[
\cos \alpha=\sqrt{\frac{\rho_{s}}{\rho_{s}-\rho_{\partial}}} \frac{h}{l}
\]

Since the value of the cosine cannot exceed one, the inequality must be satisfied

\[
\sqrt{\frac{\rho_{\varepsilon}}{\rho_{\varepsilon}-\rho_{\delta}}} \frac{h}{l} \leq 1
\]

which will be valid if

\[
\frac{h}{l} \leq \sqrt{\frac{\rho_{s}-\rho_{j}}{\rho_{\mathrm{s}}}}=\sqrt{0.4}=0.63.
\]

Thus, the rod will begin to deviate from the vertical if the length of the part of the rod submerged in water is greater than 37 cm. The angle of its deflection can be calculated using formula (1).