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CIE 9709 2025 March Paper 42 Q1

A Level / CIE / M1

CIE 9709 2025 March Paper 42 Paper · Question 1

题目

Problem

Three coplanar forces of magnitudes 40 N40\,\mathrm{N}, 30 N30\,\mathrm{N} and X NX\,\mathrm{N} act at a point in the directions shown in the diagram.

Given that the forces are in equilibrium, find the values of θ\theta and XX.

(4)
题目中文翻译

三个大小分别为 40 N40\,\mathrm{N}、30 N30\,\mathrm{N} 和 X NX\,\mathrm{N} 的共面力,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 XX 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

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把未知力沿水平、竖直方向分解,分别平衡另外两个力。由图可知角为锐角。

答题过程

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Resolving horizontally and vertically gives

Xcos⁡θ=40,Xsin⁡θ=30.X\cos\theta=40,\qquad X\sin\theta=30.

Squaring and adding,

X2=402+302=2500,X=50 N.X^2=40^2+30^2=2500, \qquad \boxed{X=50\,\mathrm{N}}.

Since 0<θ<90∘0<\theta<90^\circ,

tan⁡θ= 3040,θ= tan⁡−1(34)≈36.9∘.\begin{align*} \tan\theta=&\,\frac{30}{40},\\[4mm] \theta=&\,\tan^{-1}\biggl(\frac34\biggr)\approx\boxed{36.9^\circ}. \end{align*}

解法二

思路

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平衡的三个力首尾相接形成闭合直角三角形,直接用勾股定理及三角比。

答题过程

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The three forces form a closed triangle. The 40 N40\,\mathrm{N} and 30 N30\,\mathrm{N} sides are perpendicular, so

X=402+302=50 N.X=\sqrt{40^2+30^2} =\boxed{50\,\mathrm{N}}.

The acute angle corresponding to the diagram satisfies

tan⁡θ= 3040,θ= tan⁡−1(34)≈36.9∘.\begin{align*} \tan\theta=&\,\frac{30}{40},\\[4mm] \theta=&\,\tan^{-1}\biggl(\frac34\biggr)\approx\boxed{36.9^\circ}. \end{align*}

解法三

思路

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用 Lami 定理,各力除以其余两力夹角的正弦相等。先求角,再求力。

答题过程

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The angles between the other two forces are 90∘90^\circ, 180∘−θ180^\circ-\theta and 90∘+θ90^\circ+\theta, respectively. By Lami’s theorem,

Xsin⁡90∘= 30sin⁡(180∘−θ)= 40sin⁡(90∘+θ).\begin{align*} \frac{X}{\sin90^\circ} =&\,\frac{30}{\sin(180^\circ-\theta)}\\[4mm] =&\,\frac{40}{\sin(90^\circ+\theta)}. \end{align*}

Thus

30sin⁡θ=40cos⁡θ⟹tan⁡θ=34.\frac{30}{\sin\theta}=\frac{40}{\cos\theta} \quad\Longrightarrow\quad \tan\theta=\frac34.

Therefore

θ=tan⁡−1(34)≈36.9∘.\theta=\tan^{-1}\biggl(\frac34\biggr)\approx\boxed{36.9^\circ}.

Since θ\theta is acute and tan⁡θ=3/4\tan\theta=3/4, sin⁡θ=3/5\sin\theta=3/5. Thus

X=303/5=50 N.X=\frac{30}{3/5}=\boxed{50\,\mathrm{N}}.