图 1 所示的均匀平面薄片 $ABCDEF$ 由两个相同的菱形组成。

每个菱形的边长为 $a$，且 $\angle BAD=\angle DAF=\theta$。

薄片的质心到 $A$ 的距离为 $0.9a$。

(a) 证明 $\cos\theta=0.8$。

薄片的重量为 $W$。

一个重为 $kW$ 的质点固定于薄片的 $A$ 点。

薄片从 $B$ 点自由悬挂并处于平衡，且 $DA$ 水平。

(b) 求 $k$ 的值。

菱形是平行四边形，质心在对角线中点。两个菱形质量相等，因此整体质心是这两个质心的中点；利用对称性求它到 A 的距离。

Use an axis from $A$ along $AD$. In the orientation shown, $B$ has component $a\cos\theta$ along this axis, while $BC$ is parallel and equal to $AD$ and has length $a$. Thus $C$ has component

along $AD$. The centre of mass of rhombus $ABCD$ is the midpoint of its diagonal $AC$, so its component from $A$ along $AD$ is

By symmetry, the other rhombus has the same component and the opposite perpendicular component. The rhombuses have equal mass, so the combined centre of mass $G$ lies on $AD$ with

Since $AG=0.9a$,

也可使用菱形对角线平分顶角及对角线中点性质。先求单个菱形质心到 A 的距离，再投影到对称轴，并使用半角公式。

Let $X$ be the centre of rhombus $ABCD$. Since its diagonals bisect each other, $X$ is the midpoint of $AC$. The diagonal $AC$ bisects angle $BAD$, so

The length of $AC$ is $2a\cos(\theta/2)$, hence

The component of $AX$ along $AD$ is therefore $a\cos^2(\theta/2)$. The other rhombus has the same mass and the reflected centre, so the combined centre lies on $AD$ with

Using $\cos\theta=2\cos^2(\theta/2)-1$,

DA 水平时，沿 AD 的投影就是重力的水平力臂。绕悬点 B 取矩，附重在一侧，原薄片质心在另一侧。

When $AD$ is horizontal, the horizontal distance from $A$ to $B$ is

The original centre of mass $G$ is on $AD$, at distance $0.9a$ from $A$. Thus $A$ and $G$ lie on opposite sides of the vertical line through $B$, with horizontal moment arms $0.8a$ and $0.1a$ respectively. Taking moments about $B$,

Hence