一个表面粗糙、半径为 $a$ 的均匀圆柱体被固定，其轴水平。两根相同的均匀杆 $AB$ 和 $BC$ 在 $B$ 点刚性连接，且 $AB$ 垂直于 $BC$；每根杆的重量均为 $W$，长度均为 $2a$。两杆靠在圆柱体上，位于垂直于圆柱体轴的竖直平面内，且 $AB$ 与水平方向成 $\theta$ 角。$D$ 和 $E$ 分别为 $AB$ 和 $BC$ 的中点，同时也是两杆与圆柱体的接触点（见图）。两杆即将沿顺时针方向滑动。每根杆与圆柱体之间的摩擦系数均为 $\mu$。

$AB$ 与圆柱体之间的法向反作用力为 $R$，$BC$ 与圆柱体之间的法向反作用力为 $N$。

(a) 求比值 $R:N$，用 $\mu$ 表示。

(b) 已知 $\mu=\frac13$，求 $\tan\theta$ 的值。

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 $10\,\mathrm{m\,s^{-2}}$。

在两个接触点，摩擦力都反抗图示的顺时针滑动。沿 $AB$ 及其垂直方向列平衡方程，再绕接头 $B$ 取矩。

Take one axis along $AB$ from $A$ to $B$, and the other perpendicular to $AB$ towards the upper side of the diagram. The reaction $N$ is along the first positive axis and $R$ along the second. The limiting friction at $D$ is opposite the first axis with magnitude $\mu R$, while the limiting friction at $E$ is along the second axis with magnitude $\mu N$. Force balance gives

Taking moments about $B$, the friction forces have zero moment since they lie along the rods. Balancing the moments of the reactions and the weights gives

Substituting the force balances gives

Hence $R:N=(1+\mu):(1-\mu)$.

也可分别绕两个接触点取矩。在每个方程中，该接触点的反力与摩擦力，以及该杆的重力，均不产生力矩。

Let the two perpendicular axes be as in the first solution. Relative to $D$, point $E$ is a distance $a$ along the first axis and $a$ opposite the second. The reaction and friction at $E$ therefore produce moments in the same sense with total magnitude $a(N+\mu N)$. The weight at $E$ produces the opposite moment of magnitude $Wa(\cos\theta+\sin\theta)$. Hence

Taking moments about $E$, the reaction $R$ at $D$ has moment $aR$, opposed by the moment $a\mu R$ of its friction and the moment $Wa(\cos\theta+\sin\theta)$ of the weight there. Thus

Eliminating the common right-hand side gives $R:N=(1+\mu):(1-\mu)$.

代入 $\mu=1/3$ 得到两个支持力的比例，再将沿杆和垂直于杆的平衡式相除。

At $\mu=1/3$, part (a) gives $R=2N$. The force balances then give