[ 半径为 $r$ 的半圆弧的质心到圆心的距离为 $\frac{2r}{\pi}$。 ]

用均匀细金属丝制成图 2 所示的框架。

在框架中：

• $ABC$ 为直线段，长为 $25a$；
• $ADE$ 为直线段，长为 $24a$；
• $ABD$ 是半径为 $7a$ 的半圆弧；
• $EC=7a$；
• $\angle AEC=90^\circ$；
• $A$、$B$、$C$、$D$、$E$ 各点均位于同一平面内。

框架的质心到 $AE$ 的距离为 $d$。

(a) 证明 $d=\frac{53}{2(7+\pi)}a$。

框架从 $A$ 点自由悬挂并处于平衡状态，$AC$ 与竖直向下方向成 $\alpha^\circ$ 角。

(b) 求 $\alpha$ 的值。

这是均匀金属丝框架，质量与长度成正比。只有两条直线段和半圆弧；$EC$ 是距离标记，不是另一根杆。按各段长度与质心到 $AE$ 的距离取矩。

Use the orientation of Figure 2 and measure horizontal distances to the right from $AE$. Since the wire is uniform, masses are proportional to lengths.

The three pieces have the following lengths and centroid distances from $AE$:

• $AE$: length $24a$, distance $0$;
• $AC$: length $25a$, distance $7a/2$;
• semicircular arc $ABD$: length $7\pi a$, distance $14a/\pi$.

The total length is $24a+25a+7\pi a=7a(7+\pi)$. Taking moments about $AE$,

Thus

and hence

还须求质心沿 $AE$ 从 $A$ 向下的投影距离。悬挂后 $AG$ 竖直向下，所求是 $AC$ 与 $AG$ 的夹角；它是两条线各自相对于 $AE$ 的角度之差。

Let $G$ be the centre of mass and let $y$ be its distance below the horizontal line through $A$ in the original orientation.

The centroids of straight pieces $AE$ and $AC$ are both $12a$ below $A$. The arc has vertical diameter $AD=14a$, so its centroid is $7a$ below $A$.

Taking moments about the horizontal line through $A$,

Therefore

The angle $\phi$ between $AG$ and $AE$ satisfies

The angle $\gamma$ between $AC$ and $AE$ satisfies

When suspended from $A$, the line $AG$ is vertically downwards.

Since $AG$ lies between $AE$ and $AC$, the required angle is

Hence