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 $\alpha, \beta$ are both acute angles. If there is a small wooden block of mass $m$ on the board that has not slipped off, find the frictional force acting on the small wooden block.
Construct the 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 wooden 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 between the final plane of the wooden board and the horizontal plane be $\gamma$.
By the theorem of area projection, we have
$$
\begin{array}{c}
\cos \gamma=\frac{S_{\triangle A B D}}{S_{\triangle A B_{1} D_{1}}}=\frac{\frac{1}{2} A D \cdot A B}{\frac{1}{2} A D_{1} \cdot Q B_{1}} \\
=\frac{A D}{A D_{1}} \cdot \frac{P Q}{Q B_{1}}=\cos \alpha \cdot \cos \beta .
\end{array}
$$

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