Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2023 November Paper 41 Q2

A Level / CIE / M1

CIE 9709 2023 Nov Paper 41 Paper · Question 2

题目

Problem

A particle of mass 2.4 kg2.4\,\mathrm{kg} is held in equilibrium by two light inextensible strings, one of which is attached to point AA and the other attached to point BB. The strings make angles of 35∘35^\circ and 40∘40^\circ with the horizontal (see diagram).

Find the tension in each of the two strings.

(5)

题目中文翻译

一个质量为 2.4 kg2.4\,\mathrm{kg} 的质点由两根不可伸长的轻绳保持平衡,其中一根连接到点 AA,另一根连接到点 BB。两根绳与水平方向的夹角分别为 35∘35^\circ 和 40∘40^\circ(见图)。

求两根绳各自的张力。

(5)

原卷通用作答要求

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 (gg) is needed, use 10 m s−210\,\mathrm{m\,s^{-2}}.

需要使用重力加速度 gg 的数值时,取 10 m s−210\,\mathrm{m\,s^{-2}}。

解答

解法一

思路

展开

两根不同绳的张力不同。用A、B端命名张力,分别平衡水平与竖直分量,防止数值和绳端对应错位。

答题过程

展开

Let TAT_A and TBT_B be the tensions in the strings attached to AA and BB, respectively. Resolving horizontally and vertically,

TAcos⁡35∘=TBcos⁡40∘,T_A\cos35^\circ=T_B\cos40^\circ, TAsin⁡35∘+TBsin⁡40∘=24.T_A\sin35^\circ+T_B\sin40^\circ=24.

From the first equation, TA=TBcos⁡40∘/cos⁡35∘T_A=T_B\cos40^\circ/\cos35^\circ. Substituting,

 TB(cos⁡40∘sin⁡35∘cos⁡35∘+sin⁡40∘)= 24.\begin{align*} &\,T_B\bigg(\frac{\cos40^\circ\sin35^\circ}{\cos35^\circ} +\sin40^\circ\bigg)\\[4mm] =&\,24. \end{align*}

Hence

TB=24cos⁡35∘sin⁡75∘≈20.4 N,T_B=\frac{24\cos35^\circ}{\sin75^\circ} \approx\boxed{20.4\,\mathrm{N}}, TA=24cos⁡40∘sin⁡75∘≈19.0 N.T_A=\frac{24\cos40^\circ}{\sin75^\circ} \approx\boxed{19.0\,\mathrm{N}}.