一根长为 $6a$、重为 $W$ 的均匀杆 $AB$，其端点 $A$ 与粗糙竖直墙壁接触。一根长为 $3a$ 的不可伸长轻绳的一端连接在杆的中点 $C$，另一端连接在墙上位于 $A$ 正上方的点 $D$。当杆与墙壁的夹角为 $\theta$ 时，杆处于平衡状态，其中 $\tan\theta=\frac32$。一个重为 $W$ 的质点连接在杆上的点 $E$ 处，距离 $AE$ 等于 $ka$ $(3<k<6)$（见图）。杆和绳位于垂直于墙壁的同一竖直平面内。杆与墙壁之间的摩擦系数为 $\frac13$。杆即将沿墙壁向下滑动。

(a) 求 $k$ 的值。

(b) 用 $W$ 表示杆与墙壁之间摩擦力的大小。

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 $10\,\mathrm{m\,s^{-2}}$。

$AC=CD$，所以杆与绳分别在墙上形成相等的角。先用力的平衡和极限摩擦求张力的分量，再绕 $A$ 取矩。

Let $T$ be the tension, $R$ the normal reaction and $F$ the friction at $A$. Triangle $ACD$ is isosceles since $AC=CD=3a$, so the string also makes angle $\theta$ with the vertical. Friction acts upwards and $F=R/3$. The horizontal and vertical equilibrium equations are

Since $\tan\theta=3/2$, we have $F=\frac13T\sin\theta=\frac12T\cos\theta$. Thus $\frac32T\cos\theta=2W$, so $T\cos\theta=4W/3$ and $T\sin\theta=2W$. Taking moments about $A$ gives

The two tension terms are equal. Dividing by $aW\sin\theta$ gives $3+k=6(T\cos\theta)/W=8$, hence $k=5$.

也可绕中点 $C$ 取矩，直接消去作用于该点的张力和杆的重力。

The equal sides $AC$ and $CD$ give the same angle $\theta$ between the string and the vertical as between the rod and the vertical. Horizontal equilibrium gives $R=T\sin\theta$ and hence $T\cos\theta=R\cot\theta=2R/3$. Since $F=R/3$, vertical equilibrium gives $2R/3+R/3=2W$, so $R=2W$ and $F=2W/3$.

Taking moments about $C$, the reaction at $A$ has anticlockwise moment $R(3a\cos\theta)$. This balances the clockwise moments of friction at $A$ and the particle’s weight at $E$:

Dividing by $aW\sin\theta$ gives $6\cot\theta=2+k-3$, so $4=k-1$ and $k=5$.

极限摩擦力等于摩擦系数乘支持力，支持力由水平平衡得到。

From part (a), $R=T\sin\theta=2W$. Thus the magnitude of the limiting friction is $R/3=2W/3$.