题目

Problem

The number “4”, shown in Figure 2, is a rigid framework made from three uniform rods, $AB$ , $BC$ and $CD$ , where

AB=6a,\quad BC=5a\quad\text{and}\quad CD=4a

The point $E$ is on $AB$ and $CD$ , where $BE=4a$ , $CE=3a$ and angle $CEB=90^\circ$

The three rods are all made from the same material and they all lie in the same plane.

The framework is suspended from $B$ and hangs in equilibrium with $BA$ at an angle $\theta$ to the downward vertical.

Find $\theta$ to the nearest degree.

(9)

题目中文翻译

图 2 所示的数字“4”是由三根均匀杆 $AB$ 、 $BC$ 和 $CD$ 组成的刚性框架，其中

AB=6a,\quad BC=5a,\quad CD=4a

$E$ 点位于 $AB$ 和 $CD$ 上，其中 $BE=4a$ ， $CE=3a$ ， $\angle CEB=90^\circ$ 。

三根杆均由同一种材料制成，且位于同一个平面内。

框架从 $B$ 悬挂，平衡时 $BA$ 与竖直向下方向成 $\theta$ 角。

求 $\theta$ ，精确到最接近的整数度。

(9)

解答

解法一

思路

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三根均匀杆的质量比等于长度比。以悬挂点 $B$ 为原点，在原图中向左为 $x$ 、沿 $BA$ 向下为 $y$ ；分别求组合重心的两个坐标，悬挂后 $BG$ 为竖直线。

答题过程

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Let the masses of rods $AB$ , $BC$ and $CD$ be $6m$ , $5m$ and $4m$ respectively. Measure $x$ leftwards from $AB$ and $y$ downwards from $B$ in the original diagram.

The rods’ midpoints have coordinates

AB:(0,3a),\qquad BC:\left(\frac{3a}{2},2a\right),

CD:(a,4a).

For the centre of mass $G=(\bar x,\bar y)$ , taking first moments gives

\begin{align*} 15m\bar x=&\,6m(0)+5m\frac{3a}{2}+4m(a),\\ \bar x=&\,\frac{23a}{30}, \end{align*}

\begin{align*} 15m\bar y=&\,6m(3a)+5m(2a)+4m(4a),\\ \bar y=&\,\frac{44a}{15}. \end{align*}

When the framework hangs in equilibrium, $G$ is vertically below $B$ . Thus the angle between $BA$ and the downward vertical is the angle between $BA$ and $BG$ in the original diagram. Hence

\tan\theta=\frac{\bar x}{\bar y} =\frac{23a/30}{44a/15}=\frac{23}{88}.

Therefore

\theta=14.647\ldots^\circ =\boxed{15^\circ}

to the nearest degree.

M2 2021 Jan Q4

题目

解答

解法一

思路

答题过程