图 2 所示的均匀薄片 $ABCD$ 为等腰梯形。

• $BC$ 平行于 $AD$，且 $\angle BAD=\angle ADC$；
• $BC=5a$，$AD=7a$；
• $BC$ 与 $AD$ 之间的垂直距离为 $3a$；
• $ABCD$ 的质心到 $AD$ 的距离为 $d$。

(a) 证明 $d=\frac{17}{12}a$。

均匀薄片 $PQRS$ 为长方形，$PQ=5a$，$QR=9a$。

从 $PQRS$ 中挖去图 2 的薄片 $ABCD$，形成图 3 阴影所示的模板。

• $PS$ 平行于 $AD$；
• $PS$ 与 $AD$ 之间的垂直距离为 $a$；
• $A$ 到 $PQ$ 的垂直距离为 $a$。

模板从 $P$ 点自由悬挂并处于平衡状态，$PS$ 与竖直向下方向成 $\theta^\circ$ 角。

(b) 求 $\theta$ 的值。

把等腰梯形分成中间长方形和两侧全等直角三角形。三角形底边各为 $a$，质心距 $AD$ 为高度的三分之一。列面积矩并化简证明给定结果。

Split the trapezium into a central rectangle of width $5a$ and height $3a$, and two congruent side triangles.

Each triangle has base

so their combined area is

Their centroids are each $3a/3=a$ from $AD$. The rectangle has area $15a^2$ and centroid distance $3a/2$ from $AD$.

The total area is $18a^2$. Taking moments of areas about $AD$,

Hence

Thus

先用对称性确定质心到 $PQ$ 的距离，再以负面积矩求其到 $PS$ 的距离。悬挂时 $PG$ 竖直，角度正切必须按相对于 $PS$ 的两分量来取。

Let $G$ be the template’s centre of mass. In the orientation of Figure 3, measure horizontal distances from $PQ$ and vertical distances above $PS$.

Since $A$ is $a$ from $PQ$ and $AD=7a$, the trapezium’s symmetry axis is

from $PQ$. This is also the symmetry axis of the rectangle, so $G$ has horizontal coordinate

The original rectangle has area $45a^2$ and centroid height $5a/2$. The removed trapezium has area $18a^2$ and centroid height

The remaining area is $45a^2-18a^2=27a^2$. Taking moments about $PS$,

Thus

so

When freely suspended from $P$, $PG$ is vertically downwards. Therefore the angle between $PS$ and $PG$ satisfies

Hence