On the slope $A B$, a weight $q$ is balanced by three forces, each of magnitude $q / 3$. One is in the direction opposite to the gravitational force, the second is parallel to the base of the slope, and the third acts upward along the slope. What is the angle between the slope and the horizontal direction?
Due to the effect of the upward force, the body exerts only $2 q / 3$ force on the slope. The component of this force in the direction of the slope, if we denote the sought angle by $\alpha$, is $2 q \sin \alpha / 3$. The sum of the components of the balancing force in the direction of the slope is $(1 + \cos \alpha) q / 3$. These two must be equal, from which

$$
1 + \cos \alpha = 2 \sin \alpha
$$

The left side cannot be negative, so $\sin \alpha$ cannot be negative either. Squaring both sides,

$$
\begin{gathered}
(1 + \cos \alpha)^{2} = 4 \sin ^{2} \alpha = 4(1 - \cos ^{2} \alpha) = 4(1 + \cos \alpha)(1 - \cos \alpha) \\
(1 + \cos \alpha)(5 \cos \alpha - 4) = 0
\end{gathered}
$$

The first factor is zero when $\alpha = 180^{\circ}$. This is not a solution to the original problem. The second factor vanishes when $\cos \alpha = 4 / 5$, and $\alpha$ is an acute angle because $\sin \alpha$ is positive. The angle of inclination of the slope is $\alpha = 36^{\circ} 58^{\prime}$.