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CIE 9709 2023 June Paper 42 Q3

A Level / CIE / M1

CIE 9709 2023 June Paper 42 Paper · Question 3

题目

Problem

Coplanar forces of magnitudes 30 N30\,\mathrm{N}, 15 N15\,\mathrm{N}, 33 N33\,\mathrm{N} and P NP\,\mathrm{N} act at a point in the directions shown in the diagram, where tan⁡α=43\tan\alpha=\dfrac43. The system is in equilibrium.

(a) Show that

(14.430−P)2+(28.8P+30)2=1.\biggl(\frac{14.4}{30-P}\biggr)^2+\biggl(\frac{28.8}{P+30}\biggr)^2=1.
(4)

(b) Verify that P=6P=6 satisfies this equation and find the value of θ\theta.

(2)

题目中文翻译

大小分别为 30 N30\,\mathrm{N}、15 N15\,\mathrm{N}、33 N33\,\mathrm{N} 和 P NP\,\mathrm{N} 的共面力作用于同一点,方向如图所示,其中 tan⁡α=43\tan\alpha=\dfrac43。系统处于平衡状态。

(a) 证明:

(14.430−P)2+(28.8P+30)2=1.\biggl(\frac{14.4}{30-P}\biggr)^2+\biggl(\frac{28.8}{P+30}\biggr)^2=1.
(4)

(b) 验证 P=6P=6 满足此方程,并求 θ\theta 的值。

(2)

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

(a)

解法一

思路

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先把已知正切转换为精确正弦、余弦,不用舍入角度。分别平衡两方向后,用正弦平方加余弦平方恒等式消去θ。

答题过程

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Since tan⁡α=4/3\tan\alpha=4/3 and α\alpha is acute,

sin⁡α=45,cos⁡α=35.\sin\alpha=\frac45,\qquad\cos\alpha=\frac35.

Horizontal equilibrium gives

(33+15)(35)=(P+30)cos⁡θ,(33+15)\bigg(\frac35\bigg)=(P+30)\cos\theta,

so (P+30)cos⁡θ=28.8(P+30)\cos\theta=28.8. Vertical equilibrium gives

15(45)+30sin⁡θ=33(45)+Psin⁡θ,15\bigg(\frac45\bigg)+30\sin\theta =33\bigg(\frac45\bigg)+P\sin\theta,

so (30−P)sin⁡θ=14.4(30-P)\sin\theta=14.4. Thus

cos⁡θ=28.8P+30,sin⁡θ=14.430−P.\cos\theta=\frac{28.8}{P+30},\qquad \sin\theta=\frac{14.4}{30-P}.

Using sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 yields

(14.430−P)2+(28.8P+30)2=1.\boxed{\bigg(\frac{14.4}{30-P}\bigg)^2 +\bigg(\frac{28.8}{P+30}\bigg)^2=1}.

(b)

解法一

思路

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明确代入P等于6验证两个平方和为1,再从上一问正弦、余弦分量求锐角。

答题过程

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For P=6P=6, the left-hand side becomes

 (14.424)2+(28.836)2= (35)2+(45)2= 925+1625=1.\begin{align*} &\,\bigg(\frac{14.4}{24}\bigg)^2+\bigg(\frac{28.8}{36}\bigg)^2\\[4mm] =&\,\bigg(\frac35\bigg)^2+\bigg(\frac45\bigg)^2\\[4mm] =&\,\frac9{25}+\frac{16}{25}=1. \end{align*}

Thus P=6P=6 satisfies the equation. Also sin⁡θ=3/5\sin\theta=3/5 and cos⁡θ=4/5\cos\theta=4/5, consistent with an acute angle. Therefore

θ≈36.9∘.\theta\approx\boxed{36.9^\circ}.