弹性轻绳 $AB$ 的原长为 $3a$，弹性模量为 $\dfrac{20mg}7$。

绳未被拉伸时，将两个质量均为 $m$ 的质点连接于绳上，一个在 $P$ 点，$AP=a$，另一个在 $Q$ 点，$AQ=2a$。

随后将绳的 $A$ 端连接于水平天花板上的 $X$ 点，$B$ 端连接于天花板上的另一个点 $Y$，其中 $XY>3a$。

如图 1 所示，两个质点静止悬挂并处于平衡，绳的 $XP$、$YQ$ 两段均与水平面成 $\theta$ 角。

已知 $\tan\theta=\dfrac34$，用 $a$ 表示 $XY$。

先证明两侧绳段的张力相同、中间绳段水平，再对一个质点列平衡方程。每一段的原长都是 $a$，但中段与外侧段张力不同，伸长也不同。

Let the outer tensions be $T_X$ in $XP$ and $T_Y$ in $QY$. Considering both particles together and resolving horizontally,

so $T_X=T_Y=T_1$. The two outer portions have the same natural length $a$ and modulus $20mg/7$. By Hooke’s law they therefore have the same extension and the same length. Since $X,Y$ are at the same height and the outer portions make the same angle with the horizontal, $P,Q$ are at the same height. Thus $PQ$ is horizontal.

Let $T_2$ be the tension in $PQ$. At $P$, resolving vertically and horizontally gives

As $\tan\theta=3/4$ and $\theta$ is acute, $\sin\theta=3/5$ and $\cos\theta=4/5$. Therefore

Write $e_1$ for the extension of each outer portion and $e_2$ for that of $PQ$. Each portion has natural length $a$, so Hooke’s law gives

Hence

Adding the horizontal projections,