题目

Problem

Coplanar forces of magnitudes $55\,\mathrm{N}$ , $23\,\mathrm{N}$ , $20\,\mathrm{N}$ and $16\,\mathrm{N}$ 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 $55\,\mathrm{N}$ 、 $23\,\mathrm{N}$ 、 $20\,\mathrm{N}$ 和 $16\,\mathrm{N}$ 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 ( $g$ ) is needed, use $10\,\mathrm{m\,s^{-2}}$ .

需要使用重力加速度 $g$ 的数值时，取 $10\,\mathrm{m\,s^{-2}}$ 。

解答

解法一

思路

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取右为水平正方向、上为竖直正方向，先分解合成，再用两个正分量确定第一象限方向。

答题过程

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Take rightwards and upwards as positive. The resultant components are

\begin{align*} R_x=&\,55\sin60^\circ+23\cos15^\circ\\[4mm] &\,\hspace{2pt}-20\cos40^\circ-16\\[4mm] =&\,38.526802\ldots\text{ N}, \end{align*}

\begin{align*} R_y=&\,55\cos60^\circ-23\sin15^\circ-20\sin40^\circ\\[4mm] =&\,8.691410\ldots\text{ N}. \end{align*}

Thus the magnitude is

R=\sqrt{R_x^2+R_y^2}=39.495\ldots \approx\boxed{39.5\text{ N}}.

Both components are positive, so the resultant acts to the right and upwards. Its angle above the rightward horizontal is

\theta=\tan^{-1}\biggl(\frac{R_y}{R_x}\biggr) =12.712\ldots^\circ\approx\boxed{12.7^\circ}.

CIE 9709 2026 June Paper 41 Q1

题目

解答

解法一

思路

答题过程