题目

Problem

Three blocks $P$ , $Q$ and $R$ , of masses $25\,\mathrm{kg}$ , $20\,\mathrm{kg}$ and $m\,\mathrm{kg}$ respectively, are held in equilibrium by three light inextensible strings $OP$ , $OQ$ and $OR$ . The strings $OP$ and $OR$ both pass over small fixed smooth pulleys $A$ and $B$ respectively, with $P$ and $R$ hanging vertically below the pulleys. The block $Q$ hangs vertically below the point $O$ . The angle between $OA$ and the vertical is $30^\circ$ and the angle $BOQ=\alpha^\circ$ (see diagram).

Find the value of $m$ and the value of $\alpha$ .

(6)

题目中文翻译

质量分别为 $25\,\mathrm{kg}$ 、 $20\,\mathrm{kg}$ 和 $m\,\mathrm{kg}$ 的三个物块 $P$ 、 $Q$ 和 $R$ ，由三根轻质不可伸长细绳 $OP$ 、 $OQ$ 和 $OR$ 保持平衡。细绳 $OP$ 和 $OR$ 分别跨过小光滑定滑轮 $A$ 和 $B$ ， $P$ 和 $R$ 竖直悬挂在滑轮下方。物块 $Q$ 竖直悬挂在点 $O$ 下方。 $OA$ 与竖直方向的夹角为 $30^\circ$ ，且 $\angle BOQ=\alpha^\circ$ （见图）。

求 $m$ 和 $\alpha$ 的值。

(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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悬挂物平衡使三绳张力分别25g、20g、mg；在结点O列水平、竖直平衡，角α从向下竖直量起。

答题过程

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The smooth pulleys and equilibrium give tensions $25g$ , $20g$ and $mg$ in $OP$ , $OQ$ and $OR$ respectively.

At $O$ , resolving horizontally,

mg\sin\alpha=25g\sin30^\circ=125\text{ N}.

Resolving vertically,

25g\cos30^\circ=20g+mg\cos\alpha,

mg\cos\alpha=250\cos30^\circ-200=16.506350946\ldots\text{ N}.

Both components of the $OR$ tension are positive in the rightward and downward directions. Therefore

\begin{align*} \tan\alpha=&\,\frac{125}{250\cos30^\circ-200},\\[4mm] \alpha\approx&\,\boxed{82.5^\circ}. \end{align*}

Also,

mg=\sqrt{125^2+(250\cos30^\circ-200)^2},

m=12.608512847\ldots\approx\boxed{12.6}.

CIE 9709 2025 June Paper 43 Q4

题目

解答

解法一

思路

答题过程