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CIE 9709 2022 March Paper 42 Q5

A Level / CIE / M1

CIE 9709 2022 March Paper 42 Paper · Question 5

题目

Problem

Four coplanar forces act at a point. The magnitudes of the forces are 10 N10\,\mathrm{N}, F NF\,\mathrm{N}, G NG\,\mathrm{N} and 2F N2F\,\mathrm{N}. The directions of the forces are as shown in the diagram.

(a) Given that the forces are in equilibrium, find the values of FF and GG.

(5)

(b) Given instead that F=3F=3, find the value of GG for which the resultant of the forces is perpendicular to the 10 N10\,\mathrm{N} force.

(2)

题目中文翻译

四个共面力作用于同一点，大小分别为 10 N10\,\mathrm{N}、F NF\,\mathrm{N}、G NG\,\mathrm{N} 和 2F N2F\,\mathrm{N}。各力的方向如图所示。

(a) 已知这些力处于平衡状态，求 FF 和 GG 的值。

(5)

(b) 现改为已知 F=3F=3，求使这些力的合力垂直于 10 N10\,\mathrm{N} 的力的 GG 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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水平方向与竖直方向分别合力为零，利用水平方程把G表示成F，再代入竖直方程。

答题过程

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Taking rightwards and upwards as positive, equilibrium gives

F+Gcos⁡60∘−2Fcos⁡40∘= 0,Gsin⁡60∘+2Fsin⁡40∘−10= 0.\begin{align*} F+G\cos60^\circ-2F\cos40^\circ=&\,0, \\[4mm] G\sin60^\circ+2F\sin40^\circ-10=&\,0. \end{align*}

From the horizontal equation,

G=(4cos⁡40∘−2)F.G=(4\cos40^\circ-2)F.

Substituting into the vertical equation,

 (4cos⁡40∘−2)Fsin⁡60∘ +2Fsin⁡40∘=10.\begin{align*} &\,(4\cos40^\circ-2)F\sin60^\circ\\[4mm] &\,+2F\sin40^\circ=10. \end{align*}

Hence F=4.530667673…F=4.530667673\ldots and G=4.821435832…G=4.821435832\ldots, so

F≈4.53,G≈4.82.\boxed{F\approx4.53},\qquad\boxed{G\approx4.82}.

(b)

解法一

思路

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10N的力竖直，因此合力垂直于它意味着竖直合力为零；水平方向不必平衡。

答题过程

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The 10 N10\,\mathrm{N} force is vertical, so the resultant must be horizontal. With F=3F=3, its vertical component is zero when

Gsin⁡60∘+6sin⁡40∘−10=0.G\sin60^\circ+6\sin40^\circ-10=0.

Thus

G= 10−6sin⁡40∘sin⁡60∘= 7.093642190…≈ 7.09.\begin{align*} G=&\,\frac{10-6\sin40^\circ}{\sin60^\circ}\\[4mm] =&\,7.093642190\ldots\\[4mm] \approx&\,\boxed{7.09}. \end{align*}

The horizontal component is then 3+Gcos⁡60∘−6cos⁡40∘=1.950554436… N3+G\cos60^\circ-6\cos40^\circ=1.950554436\ldots\,\mathrm{N}, so the resultant is non-zero and horizontal.