$6 \cdot 25$ A uniform circular disk is suspended from a fine thread tied at the center $O$ of the disk and is in a horizontal position. Weights $p_{1}, p_{2}, p_{3}$ are placed at three different points $A, B, C$ on the edge of the disk, and the balance of the disk is not disturbed. Calculate $\angle A O B, \angle B O C, \angle C O A$.
[Solution] Forces $p_{1}$ and $p_{2}$, parallel to the same direction, have their resultant passing through a point $M$ on the line segment $AB$ (as shown in the figure). Since the disk remains balanced after placing the weight, the resultant of the three forces $p_{1}, p_{2}, p_{3}$ passes through point $O$. Therefore, point $M$ does not coincide with the suspension point $O$, and the point of application of force $p_{3}$ (i.e., point $C$) coincides with the intersection of ray $MO$ and the edge of the disk. The sum of the angles $\angle AOB, \angle BOC, \angle COA$ is $360^{\circ}$.

From statics, the position of point $M$ is determined by the following equation:
$$\frac{AM}{MB}=\frac{p_{2}}{p_{1}}$$

In $\triangle AOM$ and $\triangle BOM$, the following relationships hold:
$$\begin{array}{l}
\frac{AM}{\sin \angle AOM}=\frac{OM}{\sin \angle A} \\
\frac{MB}{\sin \angle MOB}=\frac{OM}{\sin \angle B}
\end{array}$$

Since $\angle A = \angle B$, we have
$$\frac{AM}{\sin \angle AOM}=\frac{MB}{\sin \angle MOB}$$

From equations (1) and (2), we get
$$\frac{p_{2}}{\sin \angle AOM}=\frac{p_{1}}{\sin \angle MOB}$$

But since $\angle AOM = 180^{\circ} - \angle COA$ and $\angle MOB = 180^{\circ} - \angle BOC$, we have
$$\frac{p_{1}}{\sin \angle BOC}=\frac{p_{2}}{\sin \angle COA}$$

A similar equation holds for the other pair of angles. Combining them, we get
$$\frac{p_{1}}{\sin \angle BOC}=\frac{p_{2}}{\sin \angle COA}=\frac{p_{3}}{\sin \angle AOB}.$$

Equation (3) allows us to calculate $\angle BOC, \angle COA, \angle AOB$. In fact, let $\alpha = 180^{\circ} - \angle BOC$, $\beta = 180^{\circ} - \angle COA$, $\gamma = 180^{\circ} - \angle AOB$. Since $\angle BOC + \angle COA + \angle AOB = 360^{\circ}$, we have $\alpha + \beta + \gamma = 180^{\circ}$. Thus, equation (3) becomes
$$\frac{p_{1}}{\sin \alpha}=\frac{p_{2}}{\sin \beta}=\frac{p_{3}}{\sin \gamma}$$

Here, $\alpha, \beta, \gamma$ are the interior angles of a triangle with sides $p_{1}, p_{2}, p_{3}$. Applying the cosine rule to this triangle, we can find $\alpha, \beta, \gamma$, and $\angle BOC, \angle COA, \angle AOB$ are the supplementary angles of these angles. The actual calculations yield
$$\left.\begin{array}{l}
\cos \angle BOC=\frac{p_{1}^{2}-p_{2}^{2}-p_{3}^{2}}{2 p_{2} p_{3}} \\
\cos \angle COA=\frac{p_{2}^{2}-p_{3}^{2}-p_{1}^{2}}{2 p_{3} p_{1}} \\
\cos \angle AOB=\frac{p_{3}^{2}-p_{1}^{2}-p_{2}^{2}}{2 p_{1} p_{2}}
\end{array}\right\}$$

Since $\angle BOC, \angle COA, \angle AOB$ are all between $0^{\circ}$ and $180^{\circ}$, equation (4) uniquely determines the required angles.