Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2024 November Paper 41 Q3

A Level / CIE / M1

CIE 9709 2024 Nov Paper 41 Paper · Question 3

题目

Problem

Coplanar forces of magnitudes 52 N52\,\mathrm{N}, 39 N39\,\mathrm{N} and P NP\,\mathrm{N} act at a point in the directions shown in the diagram. The system is in equilibrium.

Find the values of PP and θ\theta.

(4)
题目中文翻译

大小分别为 52 N52\,\mathrm{N}、39 N39\,\mathrm{N} 和 P NP\,\mathrm{N} 的共面力,沿图示方向作用于同一点。系统处于平衡状态。

求 PP 和 θ\theta 的值。

(4)
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

分别沿水平和竖直方向平衡,求未知力的两个分量,再由平方和与比值求力及锐角。

答题过程

展开

Resolving horizontally and vertically gives

Pcos⁡θ=39,Psin⁡θ=52.P\cos\theta=39,\qquad P\sin\theta=52.

Hence

P=392+522=65 N.P=\sqrt{39^2+52^2}=\boxed{65\,\mathrm{N}}.

Since the diagram gives an acute angle,

θ=tan⁡−1(5239)≈53.1∘.\theta=\tan^{-1}\bigg(\frac{52}{39}\bigg) \approx\boxed{53.1^\circ}.

解法二

思路

展开

力平衡也可以转为闭合直角力三角形,直接求斜边与角。

答题过程

展开

The three forces form a closed right-angled triangle with perpendicular sides 3939 and 5252. Thus

P2=392+522,P=65 N.P^2=39^2+52^2,\qquad \boxed{P=65\,\mathrm{N}}.

The angle corresponding to θ\theta satisfies

tan⁡θ= 5239,θ= tan⁡−1(5239)≈53.1∘.\begin{align*} \tan\theta=&\,\frac{52}{39},\\[4mm] \theta=&\,\tan^{-1}\biggl(\frac{52}{39}\biggr)\approx\boxed{53.1^\circ}. \end{align*}