一个质量为 $4\,\mathrm{kg}$ 的环穿在以 $C$ 为圆心的光滑圆形刚性金属丝上。金属丝固定在竖直平面内，一根连接至圆的最高点 $A$ 的轻绳使环保持静止。绳子与竖直方向成 $25^\circ$ 角（见图）。

求绳中的张力及金属丝对环的法向反作用力的大小。

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 $10\,\mathrm{m\,s^{-2}}$。

先用两条半径相等推出圆心角130度，因此环所在半径与向下竖直夹50度。光滑线的反力沿半径向外，沿水平、竖直平衡。

Let the ring be at $D$. Since $CA=CD$, triangle $ACD$ is isosceles. The angles at $A$ and $D$ are $25^\circ$, so $\angle ACD=130^\circ$. Thus $CD$ makes $50^\circ$ with the downward vertical.

The normal reaction $R$ is radial, outwards. Horizontal and vertical equilibrium give

Using $\sin50^\circ=2\sin25^\circ\cos25^\circ$, the first equation gives $T=2R\cos25^\circ$. Substituting into the second,

Since $2\cos^225^\circ=1+\cos50^\circ$, this gives $R=40$. Hence

沿圆的切线分解可消去反力，再沿半径求反力。所需角度来自等腰三角形。

The radii from $C$ to $A$ and to the ring are equal. Thus the radius to the ring makes $50^\circ$ with the downward vertical, while the string makes $25^\circ$ with the inward radial direction.

Resolving tangentially,

so

Resolving radially inwards gives

Hence

也可沿绳及垂直绳分解；反力与重力在垂绳方向的分量大小相等，直接得R。

The isosceles triangle formed by the two radii and the string shows that the angle between the string and the inward radius is $25^\circ$.

Resolving perpendicular to the string,

Resolving parallel to the string,

Therefore