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CIE 9709 2024 June Paper 41 Q2

A Level / CIE / M1

CIE 9709 2024 June Paper 41 Paper · Question 2

题目

Problem

Two forces of magnitudes 20 N20\,\mathrm{N} and F NF\,\mathrm{N} act at a point PP in the directions shown in the diagram.

(a) Given that the resultant force has no component in the yy-direction, calculate the value of FF.

(2)

(b) Given instead that F=10F=10, find the magnitude and direction of the resultant force.

(5)

题目中文翻译

大小分别为 20 N20\,\mathrm{N} 和 F NF\,\mathrm{N} 的两个力沿图示方向作用于点 PP。

(a) 已知合力在 yy 方向上没有分量,计算 FF 的值。

(2)

(b) 改为已知 F=10F=10,求合力的大小和方向。

(5)

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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合力没有竖直分量,向上的20牛力竖直分量必须恰好抵消向下的F。

答题过程

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Resolving in the yy-direction,

20sin⁡60∘−F=0.20\sin60^\circ-F=0.

Therefore

F=103≈17.3.\boxed{F=10\sqrt3\approx17.3}.

(b)

解法一

思路

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重新使用F等于10;求合力分量后由平方和与反正切求大小和方向。

答题过程

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The resultant has components

Rx= 20cos⁡60∘=10,Ry= 20sin⁡60∘−10=103−10.\begin{align*} R_x=&\,20\cos60^\circ=10,\\[4mm] R_y=&\,20\sin60^\circ-10=10\sqrt3-10. \end{align*}

Thus its magnitude is

R=102+(103−10)2≈12.4 N.R=\sqrt{10^2+(10\sqrt3-10)^2} \approx\boxed{12.4\,\mathrm{N}}.

Both components are positive. If ϕ\phi is its angle above the positive xx-axis,

tan⁡ϕ=103−1010=3−1.\tan\phi=\frac{10\sqrt3-10}{10}=\sqrt3-1.

Hence the direction is

36.2∘ above the positive x-axis.\boxed{36.2^\circ\text{ above the positive }x\text{-axis}}.