一个质量为 $m\,\mathrm{kg}$ 的质点 $P$ 连接在一根长 $2.6\,\mathrm{m}$ 的轻质不可伸长细绳的一端。细绳的另一端连接在水平天花板上的一个固定点，且细绳绷紧。一个大小为 $35\,\mathrm{N}$ 的力使质点保持平衡。该力位于一个垂直于天花板且包含细绳的竖直平面内，方向垂直于细绳（见图）。细绳中的张力为 $T\,\mathrm{N}$，$P$ 到天花板的竖直距离为 $2.4\,\mathrm{m}$。

按任意顺序求 $m$ 和 $T$ 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 $10\,\mathrm{m\,s^{-2}}$。

绳与竖直方向夹角的cos为12/13，sin为5/13。沿绳垂直方向平衡先求重力，再沿绳求张力。

Let $\phi$ be the angle between the string and the vertical. From the given lengths,

The applied force is perpendicular to the string. Resolving perpendicular to the string,

Resolving along the string,

也可用三力平衡的正弦关系，所用角度由绳与竖直的夹角确定。

Let $\phi$ be the string’s angle to the vertical, so $\cos\phi=12/13$ and $\sin\phi=5/13$.

The angle between the tension and the $35\,\mathrm{N}$ force is $90^\circ$. The angle between tension and weight is $180^\circ-\phi$, and the angle between the applied force and weight is $90^\circ+\phi$. Thus the sine rule for the triangle of forces gives

Therefore