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CIE 9709 2022 June Paper 42 Q2

A Level / CIE / M1

CIE 9709 2022 June Paper 42 Paper · Question 2

题目

Problem

Coplanar forces of magnitudes 60 N60\,\mathrm{N}, 20 N20\,\mathrm{N}, 16 N16\,\mathrm{N} and 14 N14\,\mathrm{N} act at a point in the directions shown in the diagram.

Find the magnitude and direction of the resultant force.

(6)

题目中文翻译

大小分别为 60 N60\,\mathrm{N}、20 N20\,\mathrm{N}、16 N16\,\mathrm{N} 和 14 N14\,\mathrm{N} 的共面力沿图示方向作用于同一点。

求合力的大小和方向。

(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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取右和上为正方向,分别合成水平、竖直分量。合力向左上方,必须结合象限说明角度。

答题过程

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Taking rightwards and upwards as positive, the components of the resultant are

Rx= 20cos⁡60∘−14−16cos⁡50∘= −14.28460175… N,Ry= 60−20sin⁡60∘−16sin⁡50∘= 30.42278083… N.\begin{align*} R_x=&\,20\cos60^\circ-14-16\cos50^\circ \\[4mm] =&\,-14.28460175\ldots\,\mathrm{N}, \\[4mm] R_y=&\,60-20\sin60^\circ-16\sin50^\circ \\[4mm] =&\,30.42278083\ldots\,\mathrm{N}. \end{align*}

Thus the magnitude is

R= Rx2+Ry2= 33.60945464…≈ 33.6 N.\begin{align*} R=&\,\sqrt{R_x^2+R_y^2}\\[4mm] =&\,33.60945464\ldots\\[4mm] \approx&\,\boxed{33.6\,\mathrm{N}}. \end{align*}

Since Rx<0R_x<0 and Ry>0R_y>0, the resultant acts upwards and to the left. Its angle θ\theta above the leftward horizontal satisfies

tan⁡θ=30.42278083…14.28460175….\tan\theta=\frac{30.42278083\ldots}{14.28460175\ldots}.

Hence its direction is 64.8∘ above the 14 N force\boxed{64.8^\circ\text{ above the }14\,\mathrm{N}\text{ force}}, equivalently 25.2∘25.2^\circ to the left of the upward vertical.