Example 7 A rectangular wooden board (neglecting thickness) is placed on a horizontal desktop, denoted as rectangle $A B C D$. Now the board is rotated upward around side $A B$ by an angle $\alpha$, and then rotated upward around side $A D$ by an angle $\beta$ and fixed, where $\alpha \backslash \beta$ are both acute angles. If at this moment there is a small wooden block of mass $m$ on the board that has not slipped, find the frictional force acting on the small wooden block.
【Analysis】The core of this physics problem is actually a solid geometry problem: To decompose gravity orthogonally, we need to know the angle formed between the board and the horizontal plane after two rotations. Note that, the problem only involves the size of the dihedral angle, and is unrelated to the shape of the plane (do not be limited by the "rectangular" board).

Solution Construct a rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, and connect $B D, A D_{1}$, as shown in Figure 9.

Assume the plane of the rectangular board after the first rotation is the plane $A B C_{1} D_{1}$, and the plane after the second rotation is the plane $A B_{1} D_{1}$.

In the plane $B C C_{1} B_{1}$, draw $B_{1} P \perp B C_{1}$ at point $P$, and through $P$, draw $P Q / / A B$, intersecting $A D_{1}$ at point $Q$, and connect $B_{1} Q$.
It is easy to prove that $\angle C B C_{1}=\alpha, \angle P Q B_{1}=\beta$.
Let the angle formed between the final plane of the board and the horizontal plane be $\gamma$.
By the theorem of area projection, we have
$$
\begin{array}{l}
\cos \gamma=\frac{S_{\triangle A B D}}{S_{\triangle A B_{1} D_{1}}}=\frac{\frac{1}{2} A D \cdot A B}{\frac{1}{2} A D_{1} \cdot Q B_{1}} \\
=\frac{A D}{A D_{1}} \cdot \frac{P Q}{Q B_{1}}=\cos \alpha \cdot \cos \beta .
\end{array}
$$

Thus, the friction force acting on the small block is
$$
m g \sin \gamma=m g \sqrt{1-\cos ^{2} \alpha \cdot \cos ^{2} \beta} .
$$