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CIE 9709 2024 June Paper 42 Q3

A Level / CIE / M1

CIE 9709 2024 June Paper 42 Paper · Question 3

题目

Problem

Four coplanar forces of magnitude P NP\,\mathrm{N}, 10 N10\,\mathrm{N}, 16 N16\,\mathrm{N} and 2 N2\,\mathrm{N} act at a point in the directions shown in the diagram. It is given that the forces are in equilibrium.

Find the values of θ\theta and PP.

(6)

题目中文翻译

大小分别为 P NP\,\mathrm{N}、10 N10\,\mathrm{N}、16 N16\,\mathrm{N} 和 2 N2\,\mathrm{N} 的四个共面力沿图示方向作用于同一点。已知这些力处于平衡状态。

求 θ\theta 和 PP 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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先水平平衡求θ,借图选择锐角;再竖直平衡求向上的P。

答题过程

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Resolving horizontally,

10cos⁡25∘=2cos⁡40∘+16sin⁡θ.10\cos25^\circ=2\cos40^\circ+16\sin\theta.

Thus

sin⁡θ=10cos⁡25∘−2cos⁡40∘16=0.470686811….\sin\theta=\frac{10\cos25^\circ-2\cos40^\circ}{16} =0.470686811\ldots.

The diagram specifies an acute angle, so

θ≈28.1∘.\theta\approx\boxed{28.1^\circ}.

Resolving vertically and retaining the unrounded angle,

P= 10sin⁡25∘+16cos⁡θ+2sin⁡40∘≈ 19.6.\begin{align*} P=&\,10\sin25^\circ+16\cos\theta+2\sin40^\circ\\[4mm] \approx&\,\boxed{19.6}. \end{align*}