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CIE 9709 2026 June Paper 45 Q3

A Level / CIE / M1

CIE 9709 2026 June Paper 45 Paper · Question 3

题目

Problem

Four coplanar forces of magnitudes 20 N20\,\mathrm{N}, 15 N15\,\mathrm{N}, 25 N25\,\mathrm{N} and F NF\,\mathrm{N} act at a point in the directions shown in the diagram.

Given that the four forces are in equilibrium, find the value of FF and the value of θ\theta.

(6)
题目中文翻译

四个大小分别为 20 N20\,\mathrm{N}、15 N15\,\mathrm{N}、25 N25\,\mathrm{N} 和 F NF\,\mathrm{N} 的共面力沿图示方向作用于同一点。

已知这四个力处于平衡状态,求 FF 和 θ\theta 的值。

(6)
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

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分别水平、竖直平衡,得到F的两个分量,取比值求角、平方和求力大小。

答题过程

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Resolving horizontally,

15+25cos⁡30∘=20cos⁡70∘+Fcos⁡θ,15+25\cos30^\circ=20\cos70^\circ+F\cos\theta,

so

Fcos⁡θ=29.810232….(1)F\cos\theta=29.810232\ldots.\tag{1}

Resolving vertically,

20sin⁡70∘=25sin⁡30∘+Fsin⁡θ,20\sin70^\circ=25\sin30^\circ+F\sin\theta,

so

Fsin⁡θ=6.293852….(2)F\sin\theta=6.293852\ldots.\tag{2}

Dividing (2) by (1),

tan⁡θ= 6.293852…29.810232…,θ≈ 11.9∘.\begin{align*} \tan\theta=&\,\frac{6.293852\ldots}{29.810232\ldots},\\[4mm] \theta\approx&\,\boxed{11.9^\circ}. \end{align*}

Also,

F= (29.810232…)2+(6.293852…)2= 30.4674…,\begin{align*} F=&\,\sqrt{(29.810232\ldots)^2+(6.293852\ldots)^2}\\[4mm] =&\,30.4674\ldots, \end{align*}

so F=30.5\boxed{F=30.5}.