题目

Problem

A non-uniform rod $AB$ of length $3a$ rests in equilibrium on a horizontal ledge and overhangs the edge of the ledge at $C$ . The point $C$ is such that $AC=2a$ and $CB=a$ , as shown in Figure 3.

The rod has weight $W$ .

The distance of the centre of mass of the rod from $A$ is $d$ .

The rod is perpendicular to the edge of the ledge.

When a force of magnitude $P$ , acting vertically upwards, is applied to the rod at $B$ , the rod is on the point of tilting about $A$ .

When the force applied at $B$ is replaced by a force of magnitude $1.25P$ , acting vertically downwards at $B$ , the rod is on the point of tilting about $C$ .

Find $d$ in terms of $a$ .

(6)

题目中文翻译

一根长为 $3a$ 的非均匀杆 $AB$ 平衡地静止在水平凸台上，并伸出凸台边缘 $C$ 。如图 3 所示， $C$ 点满足 $AC=2a$ ， $CB=a$ 。

杆的重量为 $W$ 。

杆的质心距 $A$ 点的距离为 $d$ 。

杆与凸台边缘垂直。

当在杆的 $B$ 点施加大小为 $P$ 、竖直向上的力时，杆恰好要绕 $A$ 倾覆。

当把 $B$ 点的力替换为大小为 $1.25P$ 、在 $B$ 点竖直向下作用的力时，杆恰好要绕 $C$ 倾覆。

用 $a$ 表示 $d$ 。

(6)

解答

解法一

思路

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两次临界倾覆的支点不同：向上拉时支点为 $A$ ，向下压时支点为 $C$ 。分别绕实际支点取矩，再消去 $P/W$ 求重心位置。

答题过程

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When the upward force $P$ makes the rod just tilt about $A$ , the ledge’s resultant reaction acts at $A$ . Taking moments about $A$ ,

3aP=Wd. \tag{1}

When the downward force $1.25P$ makes the rod just tilt about $C$ , the reaction acts at $C$ . The centre of mass is to the left of $C$ , giving

1.25Pa=W(2a-d). \tag{2}

From (1), $P=Wd/(3a)$ . Substituting into (2),

\begin{align*} \frac54\frac{Wd}{3a}a=&\,W(2a-d),\\ \frac{5d}{12}=&\,2a-d,\\ \frac{17d}{12}=&\,2a,\\ d=&\,\boxed{\frac{24a}{17}}. \end{align*}

This satisfies $0<d<2a$ , consistent with the centre of mass lying on the supported side of $C$ .

M1 2025 June Q5

题目

解答

解法一

思路

答题过程