图 2 阴影所示的均匀薄片 $L$，由边长为 $8a$ 的均匀正方形薄片 $ABCD$ 挖去边长为 $2a$ 的正方形 $PQRV$ 和边长为 $4a$ 的正方形 $RSTU$ 得到。$Q$、$R$、$U$ 三点共线，$V$、$R$、$S$ 三点共线。边 $AD$ 平行于 $PV$，边 $AB$ 平行于 $PQ$。$AD$ 与 $PV$ 之间的距离为 $a$，$AB$ 与 $PQ$ 之间的距离也为 $a$。$L$ 的质心为 $G$。

(a) 证明 $G$ 到边 $AD$ 的距离为 $\frac{42}{11}a$。

$L$ 的质量为 $M$。将一个质量为 $kM$ 的质点连接于 $L$ 的 $C$ 点。

带有该质点的薄片从 $B$ 自由悬挂，平衡时 $BC$ 与水平面成 $45^\circ$ 角。

(b) 求 $k$ 的值。

均匀薄片质量与面积成正比。将挖去的两个正方形视为负质量，分别列它们到 $AD$ 的重心距离。

The areas of the original square and the two removed squares are in the ratio

Their centres are respectively $4a$, $2a$ and $5a$ from $AD$. The remaining area is $44a^2$. If $G$ is distance $d$ from $AD$, taking first moments about $AD$ gives

Thus

as required.

整个图形关于对角线 $AC$ 对称，加在 $C$ 的质点也在此线上，所以组合重心仍在 $AC$ 上。悬挂平衡时重心在 $B$ 的正下方；结合 $45^\circ$ 条件可确定它为大正方形中心，再取矩。

The lamina is symmetric about $AC$, so $G$ lies on $AC$. Adding a particle at $C$ leaves the combined centre of mass on $AC$.

When suspended from $B$, the combined centre of mass is vertically below $B$. Since $BC$ makes $45^\circ$ with the horizontal, the diagonal $BD$ is vertical in the hanging position. Therefore the combined centre of mass is the intersection $H$ of the diagonals, the midpoint of $AC$.

In the original square coordinates, $G$ has equal distances $42a/11$ from $AD$ and $AB$, whereas $H$ has both distances $4a$. Hence

while $HC=4\sqrt2a$. The diagonal $AC$ is horizontal in the hanging position. Taking moments about $H$,

Cancelling the common factors gives

另一方法直接计算组合重心坐标。以 $B$ 为参考，左、下两个分量相等才会在悬挂后位于其正下方。

Use coordinates measured rightwards from $AD$ and downwards from $AB$. By symmetry and part (a),

Let the combined centre of mass be $(\bar x,\bar y)$. Then

Since the vertical line through the suspension point $B=(8a,0)$ makes equal angles with the two adjacent sides, it satisfies

Substituting the expressions for the centre of mass,