In the $ABCD$ rhombus, whose sides can move freely around the vertices; at points $A$ and $C$, a force $P$ acts in the direction of the line $AC$, which are of equal magnitude but opposite in sign; what equal magnitude but opposite sign forces $Q$ need to be applied at points $B$ and $D$ in the direction of the line $BD$ so that the rhombus is in equilibrium?
Suppose that under the influence of force $P$, the distance $A C$ has changed to $A^{\prime} C^{\prime}B D$. We can now assume that the side $A B$ of the rhombus has moved to the position $A^{\prime} B^{\prime}$ by rotating around point $E$, which is the intersection of the lines $A B$ and $A^{\prime} B^{\prime}$.

This assumption will deviate less from reality the smaller we choose the distance $A A^{\prime}=C C^{\prime}$.

Let $A B=a, \quad B A C \angle=\alpha, \quad B^{\prime} A^{\prime} C^{\prime} \angle=\alpha^{\prime}$. Let the distances of point $E$ from the lines $B D$ and $A C$ be denoted by $x$ and $y$. Then we can write the following relations:

$$
\frac{x}{\cos \alpha}+\frac{y}{\sin \alpha}=\frac{x}{\cos \alpha^{\prime}}+\frac{y}{\sin \alpha^{\prime}}=a
$$

From these equations, the values of $x$ and $y$ are as follows:

$$
\begin{gathered}
x=\frac{a \cos \alpha \cos \alpha^{\prime}\left(\sin \alpha-\sin \alpha^{\prime}\right)}{\sin \alpha \cos \alpha^{\prime}-\cos \alpha \sin \alpha^{\prime}} \\
y=\frac{a \sin \alpha \sin \alpha^{\prime}\left(\cos \alpha-\cos \alpha^{\prime}\right)}{\cos \alpha \sin \alpha^{\prime}-\sin \alpha \cos \alpha^{\prime}} \\
x=\frac{2 a \cos \alpha \cos \alpha^{\prime} \cos \frac{\alpha+\alpha^{\prime}}{2} \sin \frac{\alpha-\alpha^{\prime}}{2}}{2 \sin \frac{\alpha-\alpha^{\prime}}{2} \cos \frac{\alpha-\alpha^{\prime}}{2}} \\
y=\frac{2 a \sin \alpha \sin \alpha^{\prime} \sin \frac{\alpha+\alpha^{\prime}}{2} \sin \frac{\alpha-\alpha^{\prime}}{2}}{2 \sin \frac{\alpha-\alpha^{\prime}}{2} \cos \frac{\alpha-\alpha^{\prime}}{2}} \\
x=\frac{a \cos \alpha \cos \alpha^{\prime} \cos \frac{1}{2}\left(\alpha+\alpha^{\prime}\right)}{\cos \frac{1}{2}\left(\alpha-\alpha^{\prime}\right)} \\
y=\frac{a \sin \alpha \sin \alpha^{\prime} \sin \frac{1}{2}\left(\alpha+\alpha^{\prime}\right)}{\cos \frac{1}{2}\left(\alpha-\alpha^{\prime}\right)}
\end{gathered}
$$

These last two values will more accurately determine the position of the rotation point $E$, the less $\alpha$ differs from $\alpha^{\prime}$; if therefore $\alpha=\alpha^{\prime}$ or $\alpha-\alpha^{\prime}=0$, then $x$ and $y$ determine the position of the point around which $A B$ rotates to the nearest, infinitesimally different position $A^{\prime} B^{\prime}$.

But if $\alpha=\alpha^{\prime}, \cos \alpha=\cos \alpha^{\prime}=\cos \frac{1}{2}\left(\alpha+\alpha^{\prime}\right), \sin \alpha=\sin \alpha^{\prime}=\sin \frac{1}{2}\left(\alpha+\alpha^{\prime}\right)$ and $\cos \frac{1}{2}\left(\alpha-\alpha^{\prime}\right)=1$; and thus

$$
\begin{aligned}
& x=a \cos ^{3} \alpha \\
& y=a \sin ^{3} \alpha
\end{aligned}
$$

In this case, we can consider $A B$ as a lever, whose arms are $A E$ and $B E$, and on whose endpoints the forces $P$ and $Q$ act in the directions of the lines $A C$ and $B D$. For the forces $P$ and $Q$ to balance each other, the following equation must hold:

$$
P a \sin ^{3} \alpha=Q a \cos ^{3} \alpha
$$

From which the value of the sought $Q$ is

$$
Q=P \cdot \operatorname{tg}^{3} \alpha
$$