题目

Problem

Coplanar forces of magnitudes $16\,\mathrm{N}$ , $12\,\mathrm{N}$ , $24\,\mathrm{N}$ and $8\,\mathrm{N}$ act at a point in the directions shown in the diagram.

Find the magnitude and direction of the single additional force acting at the same point which will produce equilibrium.

(6)

题目中文翻译

大小分别为 $16\,\mathrm{N}$ 、 $12\,\mathrm{N}$ 、 $24\,\mathrm{N}$ 和 $8\,\mathrm{N}$ 的共面力沿图示方向作用于同一点。

求为使系统达到平衡，需要在该点施加的一个附加力的大小和方向。

(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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先求四个已知力的合力分量，再整体反向得到平衡力。方向必须说明所在象限，不能只写反正切的锐角。

答题过程

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Take the positive $x$ -axis to the right and the positive $y$ -axis upwards. The resultant of the four given forces has components

\begin{align*} R_x=&\,12\sin30^\circ+24+8\cos30^\circ\\[4mm] &\,\hspace{2pt}-16\cos30^\circ\\[4mm] =&\,30-4\sqrt3,\\[4mm] R_y=&\,12\cos30^\circ-8\sin30^\circ-16\sin30^\circ\\[4mm] =&\,6\sqrt3-12. \end{align*}

The additional force is opposite to this resultant, so its components are

\begin{align*} F_x=&\,4\sqrt3-30<0,\\[4mm] F_y=&\,12-6\sqrt3>0. \end{align*}

Its magnitude is

\begin{align*} F=&\,\sqrt{(30-4\sqrt3)^2+(12-6\sqrt3)^2}\\[4mm] \approx&\,\boxed{23.1\,\mathrm{N}}. \end{align*}

If $\phi$ is the angle above the negative $x$ -axis,

\tan\phi=\frac{12-6\sqrt3}{30-4\sqrt3}.

Thus the direction is

\boxed{4.0^\circ\text{ above the negative }x\text{-axis}}.

CIE 9709 2024 November Paper 43 Q2

题目

解答

解法一

思路

答题过程