题目

Problem

Figure 3 shows a shaded uniform lamina $OAB$ which is in the shape of the region enclosed by the $x$ -axis and the curve with equation $y=\dfrac1a(ax-x^2)$ , where $a$ is a positive constant. The centre of mass of the lamina is at $G$ .

(a) Find, in terms of $a$ , the $y$ coordinate of $G$ .

(6)

The lamina is now freely pivoted in a vertical plane about a smooth horizontal axis which passes through $O$ and is perpendicular to the plane of the lamina.

The lamina is held in equilibrium by a horizontal force which is applied to the lamina at $A$ . The line $OA$ makes an angle $\theta$ with the downward vertical through $O$ , where $\tan\theta=\dfrac34$ , as shown in Figure 4.

The line of action of the force lies in the plane of the lamina.

The mass of the lamina is $M$ and the magnitude of the applied force is $F$ .

(b) Find $F$ in terms of $M$ and $g$ .

(6)

题目中文翻译

图 3 阴影所示的均匀薄片 $OAB$ 的形状为 $x$ 轴与曲线 $y=\dfrac1a(ax-x^2)$ 围成的区域，其中 $a$ 为正常数。薄片的质心为 $G$ 。

(a) 用 $a$ 表示 $G$ 的 $y$ 坐标。

(6)

现在薄片可绕过 $O$ 且垂直于薄片平面的光滑水平轴，在竖直平面内自由转动。

如图 4 所示，在 $A$ 点对薄片施加水平力，使其保持平衡，直线 $OA$ 与过 $O$ 的竖直向下方向成 $\theta$ 角，其中 $\tan\theta=\dfrac34$ 。

力的作用线位于薄片平面内。

薄片的质量为 $M$ ，所施加的力的大小为 $F$ 。

(b) 用 $M$ 和 $g$ 表示 $F$ 。

(6)

解答

(a)

解法一

思路

展开

竖条的质心高度是条高的一半。先求薄片面积，再将 $y^2/2$ 展开并积分求对 $x$ 轴的面积矩。

答题过程

展开

The curve meets the $x$ -axis where $x(a-x)=0$ , so the lamina extends from $x=0$ to $x=a$ . Its area is

\begin{align*} \mathcal A=&\,\int_0^a\frac{ax-x^2}{a}\,\mathrm{d}x\\ =&\,\frac1a\left[\frac{ax^2}{2}-\frac{x^3}{3}\right]_0^a\\ =&\,\frac1a\left(\frac{a^3}{2}-\frac{a^3}{3}\right) =\frac{a^2}{6}. \end{align*}

A thin vertical strip has its centre at height $y/2$ . As the lamina is uniform, its first-moment equation is

\mathcal A\bar y=\frac12\int_0^a y^2\,\mathrm{d}x.

Expanding and integrating,

\begin{align*} \mathcal A\bar y =&\,\frac1{2a^2}\int_0^a(a^2x^2-2ax^3+x^4)\,\mathrm{d}x\\ =&\,\frac1{2a^2}\left[\frac{a^2x^3}{3}-\frac{ax^4}{2}+\frac{x^5}{5}\right]_0^a\\ =&\,\frac{a^3}{2}\left(\frac13-\frac12+\frac15\right) =\frac{a^3}{60}. \end{align*}

Hence

\boxed{\bar y=\frac{a^3/60}{a^2/6}=\frac a{10}.}

(b)

解法一

思路

展开

薄片关于 $x=a/2$ 对称，因此旋转前质心的坐标为 $(a/2,a/10)$ 。旋转后重力的力臂是这两个分量在水平方向的投影之和；水平外力的力臂是 $A$ 到 $O$ 的竖直距离。

答题过程

展开

The original lamina is symmetric about $x=a/2$ , so

\bar x=\frac a2,\qquad \bar y=\frac a{10}.

In the pivoted position, let $H$ be the foot of the perpendicular from $G$ to $OA$ . Then $OH=a/2$ and $HG=a/10$ . Both contribute positively to the horizontal distance $d$ of $G$ from the vertical through $O$ :

d=\frac a2\sin\theta+\frac a{10}\cos\theta.

Since $\tan\theta=3/4$ , $\sin\theta=3/5$ and $\cos\theta=4/5$ . Hence

d=\frac a2\frac35+\frac a{10}\frac45 =\frac{3a}{10}+\frac{2a}{25} =\frac{19a}{50}.

The horizontal force at $A$ has perpendicular distance $a\cos\theta=4a/5$ from $O$ . Taking moments about $O$ eliminates the pivot reaction:

F(a\cos\theta)=Mg\,d, \qquad F\frac{4a}{5}=Mg\frac{19a}{50}.

Therefore

\boxed{F=\frac{19}{40}Mg.}

M3 2025 Jan Q4

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程