一块边长为 $2a$、重量为 $W$ 的均匀正方形薄片，由一根连接在边 $AB$ 的中点 $E$ 处的不可伸长轻绳悬挂。绳的另一端连接在粗糙竖直墙面上的固定点 $P$ 处。薄片的顶点 $B$ 与墙面接触。绳 $EP$ 垂直于边 $AB$，并与墙面成 $\theta$ 角（见图）。绳和薄片位于一个垂直于墙面的竖直平面内。墙面与薄片之间的摩擦系数为 $\frac12$。

已知顶点 $B$ 即将沿墙面向上滑动，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 $10\,\mathrm{m\,s^{-2}}$。

$B$ 即将向上滑，所以极限摩擦力向下。先由水平、竖直平衡确定张力与重力的关系，再绕 $B$ 取矩。

Let $T$ be the tension, $N$ the normal reaction and $F$ the friction at $B$. Resolving horizontally and vertically gives

Thus $W=T(\cos\theta-\frac12\sin\theta)$.

Take moments about $B$. The tension is perpendicular to $BE$, and $BE=a$, so its moment has magnitude $Ta$. From $B$ to the centre of the square, the displacement is the sum of $BE$ and the perpendicular segment from $E$ to the centre, each of length $a$. Their horizontal projections away from the wall are $a\cos\theta$ and $a\sin\theta$. Therefore

Substituting for $W$ gives

Subtracting $\cos^2\theta+\sin^2\theta=1$ gives $\frac12\sin\theta\cos\theta-\frac32\sin^2\theta=0$. Factoring yields

The configuration has $\sin\theta>0$, so $\tan\theta=1/3$.