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CIE 9709 2024 March Paper 42 Q4

A Level / CIE / M1

CIE 9709 2024 March Paper 42 Paper · Question 4

题目

Problem

Four coplanar forces act at a point. The magnitudes of the forces are F NF\,\mathrm{N}, 2F N2F\,\mathrm{N}, 3F N3F\,\mathrm{N} and 30 N30\,\mathrm{N}. The directions of the forces are as shown in the diagram.

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

(6)

题目中文翻译

四个共面力作用于同一点，大小分别为 F NF\,\mathrm{N}、2F N2F\,\mathrm{N}、3F N3F\,\mathrm{N} 和 30 N30\,\mathrm{N}。各力的方向如图所示。

已知这些力处于平衡状态，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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按实际图沿水平、竖直方向列平衡。相除消去F求锐角θ，再用平方和求F。

答题过程

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Resolving horizontally and vertically according to the diagram gives

30cos⁡θ=3F+2Fsin⁡45∘,30\cos\theta=3F+2F\sin45^\circ, 30sin⁡θ=F+2Fcos⁡45∘.30\sin\theta=F+2F\cos45^\circ.

Dividing,

tan⁡θ=1+23+2.\tan\theta=\frac{1+\sqrt2}{3+\sqrt2}.

The diagram gives an acute angle, so

θ≈28.7∘.\theta\approx\boxed{28.7^\circ}.

Squaring and adding the component equations,

900=F2[(3+2)2+(1+2)2].900=F^2\big[(3+\sqrt2)^2+(1+\sqrt2)^2\big].

Thus

F=3014+82≈5.96.\boxed{F=\frac{30}{\sqrt{14+8\sqrt2}} \approx5.96}.