题目

Problem

A uniform lamina $AEFGD$ is formed by removing two triangles from a square $ABCD$ of side length $a$ . Triangle $FCG$ is isosceles and the two equal sides each have length $\frac12a$ . The angle between $EF$ and $BF$ is $\alpha$ , where $\tan\alpha=\frac12$ (see diagram).

The centre of mass of the lamina is at a distance $\overline{x}$ from $AD$ and at a distance $\overline{y}$ from $AE$ .

(a) Show that $\overline{x}=\frac5{12}a$ and find a corresponding expression for $\overline{y}$ in terms of $a$ .

(5)

(b) The lamina is freely suspended from the point $G$ and hangs in equilibrium. The angle between $DA$ and the downward vertical through $G$ is $\theta$ .

Find the value of $\theta$ .

(2)

题目中文翻译

从一个边长为 $a$ 的正方形 $ABCD$ 中去掉两个三角形，得到均匀薄片 $AEFGD$ 。三角形 $FCG$ 是等腰三角形，两条相等的边的长度均为 $\frac12a$ 。 $EF$ 与 $BF$ 之间的夹角为 $\alpha$ ，其中 $\tan\alpha=\frac12$ （见图）。

薄片的质心到 $AD$ 的距离为 $\overline{x}$ ，到 $AE$ 的距离为 $\overline{y}$ 。

(a) 证明 $\overline{x}=\frac5{12}a$ ，并求 $\overline{y}$ 的相应表达式，用 $a$ 表示。

(5)

(b) 薄片从点 $G$ 自由悬挂，处于平衡状态。 $DA$ 与经过 $G$ 的竖直向下方向之间的夹角为 $\theta$ 。

求 $\theta$ 的值。

(2)

原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

Where a numerical value for the acceleration due to gravity ( $g$ ) is needed, use $10\,\mathrm{m\,s^{-2}}$ .

需要使用重力加速度 $g$ 的数值时，取 $10\,\mathrm{m\,s^{-2}}$ 。

解答

(a)

解法一

思路

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以 $A$ 为原点，用完整正方形的面积矩减去两块挖去的三角形。先从图中读出两条等边，再用 $\tan\alpha$ 求 $BE$ 。

答题过程

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Take coordinates measured from $AD$ to the right and from $AE$ upwards. From the diagram, $BF=FC=CG=a/2$ and

BE=BF\tan\alpha=\frac a4.

The centres of mass of the removed triangles are the averages of their three vertex coordinates. Their areas and centres are

\begin{align*} A_{EBF}=&\,\frac12\cdot\frac a4\cdot\frac a2=\frac{a^2}{16},\\[4mm] C_{EBF}:&\,\bigg(\frac{11a}{12},\frac a6\bigg),\\[4mm] A_{FCG}=&\,\frac{a^2}{8},\\[4mm] C_{FCG}:&\,\bigg(\frac{5a}{6},\frac{5a}{6}\bigg). \end{align*}

The remaining area is $13a^2/16$ . The uniform density cancels from the moments, so

\begin{align*} \frac{\overline{x}}a =&\,\frac{\frac12-\frac1{16}\frac{11}{12}-\frac18\frac56}{13/16} =\frac5{12},\\ \frac{\overline{y}}a =&\,\frac{\frac12-\frac1{16}\frac16-\frac18\frac56}{13/16} =\frac{37}{78}. \end{align*}

Thus $\overline{x}=5a/12$ and $\overline{y}=37a/78$ .

解法二

思路

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也可把剩下的薄片直接分成左侧矩形和右侧两个三角形，再把各部分的面积矩相加。

答题过程

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In units of $a$ , draw the vertical segment from $G=(1/2,1)$ to $H=(1/2,0)$ and join $G$ to $E=(3/4,0)$ . The rectangle on the left has area $a^2/2$ and centre $(a/4,a/2)$ . Triangle $GHE$ has area $a^2/8$ and centre $(7a/12,a/3)$ . Triangle $GFE$ has area $3a^2/16$ and centre $(3a/4,a/2)$ . Therefore

\begin{align*} \frac{\overline{x}}a =&\,\frac{\frac12\frac14+\frac18\frac7{12}+\frac3{16}\frac34}{13/16} =\frac5{12},\\ \frac{\overline{y}}a =&\,\frac{\frac12\frac12+\frac18\frac13+\frac3{16}\frac12}{13/16} =\frac{37}{78}. \end{align*}

(b)

解法一

思路

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悬挂平衡时，质心在 $G$ 的正下方。用薄片原坐标系中 $G$ 到质心的水平、竖直位移求转角。

答题过程

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In the lamina’s coordinates, the horizontal and vertical distances from $G$ to the centre of mass are

\begin{align*} \frac a2-\frac{5a}{12}=&\,\frac a{12},\\[4mm] a-\frac{37a}{78}=&\,\frac{41a}{78}. \end{align*}

The segment from $G$ to the centre of mass becomes vertical in equilibrium, while $DA$ is parallel to the downward coordinate direction in the lamina. Hence

\begin{align*} \tan\theta=&\,\frac{a/12}{41a/78}=\frac{13}{82},\\[4mm] \theta=&\,9.0^\circ. \end{align*}

CIE 9231 2026 June Paper 34 Q2

题目

解答

(a)

解法一

思路

答题过程

解法二

思路

答题过程

(b)

解法一

思路

答题过程