题目

Problem

Three coplanar forces of magnitudes $F\,\mathrm{N}$ , $65\,\mathrm{N}$ and $50\,\mathrm{N}$ act at a point in the directions shown in the diagram.

(a) It is initially given that $F=40$ .

Find the magnitude and direction of the resultant force.

(5)

(b) It is given instead that the value of $F$ is such that the component of the resultant force in the $y$ -direction is zero.

Find the magnitude and direction of the resultant force.

(3)

题目中文翻译

三个大小分别为 $F\,\mathrm{N}$ 、 $65\,\mathrm{N}$ 和 $50\,\mathrm{N}$ 的共面力沿图示方向作用于同一点。

(a) 最初给定 $F=40$ 。

求合力的大小和方向。

(5)

(b) 改为给定 $F$ 的值使合力在 $y$ 方向的分量为零。

求合力的大小和方向。

(3)

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

(a)

解法一

思路

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按图的x、y正方向分解，两个分量均负，方向在负x轴下方。

答题过程

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Using the axes shown, with $F=40$ ,

\begin{align*} R_x=&\,40\cos30^\circ-50\cos40^\circ-65\cos60^\circ\\[4mm] =&\,-36.161206\ldots\text{ N}, \end{align*}

\begin{align*} R_y=&\,40\sin30^\circ+50\sin40^\circ-65\sin60^\circ\\[4mm] =&\,-4.152271\ldots\text{ N}. \end{align*}

The magnitude is

R=\sqrt{R_x^2+R_y^2}=36.398821\ldots\approx\boxed{36.4\text{ N}}.

Since both components are negative, the resultant is directed to the left and downwards. Its angle below the negative $x$ -axis is

\phi=\tan^{-1}\biggl(\frac{|R_y|}{|R_x|}\biggr) =6.550395\ldots^\circ\approx\boxed{6.6^\circ}.

(b)

解法一

思路

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重新由竖直合分量为零求F，不能沿用40；再看水平分量的符号确定方向。

答题过程

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For the resultant to have zero $y$ -component,

F\sin30^\circ+50\sin40^\circ-65\sin60^\circ=0.

Hence

F=\frac{65\sin60^\circ-50\sin40^\circ}{\sin30^\circ} =48.304542\ldots.

The horizontal component is therefore

\begin{align*} R_x=&\,F\cos30^\circ-50\cos40^\circ-65\cos60^\circ\\[4mm] =&\,-28.969262\ldots\text{ N}. \end{align*}

Thus the resultant has magnitude $\boxed{29.0\text{ N}}$ and acts along the negative $x$ -axis.

CIE 9709 2026 June Paper 43 Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程