题目

Problem

Two particles $P$ and $Q$ , of masses $0.3\,\mathrm{kg}$ and $0.2\,\mathrm{kg}$ respectively, are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley at $B$ which is attached to two inclined planes. $P$ lies on a smooth plane $AB$ which is inclined at $60^\circ$ to the horizontal. $Q$ lies on a plane $BC$ which is inclined at $30^\circ$ to the horizontal. The string is taut and the particles can move on lines of greatest slope of the two planes (see diagram).

(a) It is given that the plane $BC$ is smooth and that the particles are released from rest.

Find the tension in the string and the magnitude of the acceleration of the particles.

(5)

(b) It is given instead that the plane $BC$ is rough. A force of magnitude $3\,\mathrm{N}$ is applied to $Q$ directly up the plane along a line of greatest slope of the plane.

Find the least value of the coefficient of friction between $Q$ and the plane $BC$ for which the particles remain at rest.

(5)

题目中文翻译

质量分别为 $0.3\,\mathrm{kg}$ 和 $0.2\,\mathrm{kg}$ 的两个质点 $P$ 和 $Q$ 连接在一根轻质不可伸长绳子的两端。绳子绕过位于 $B$ 点、固定在两个斜面上的光滑滑轮。 $P$ 位于与水平面成 $60^\circ$ 角的光滑斜面 $AB$ 上。 $Q$ 位于与水平面成 $30^\circ$ 角的斜面 $BC$ 上。绳子绷紧，两质点可以沿各斜面的最大坡度线运动（见图）。

(a) 已知斜面 $BC$ 光滑，两质点从静止释放。

求绳中的张力及两质点的加速度大小。

(5)

(b) 现改为斜面 $BC$ 粗糙。一个大小为 $3\,\mathrm{N}$ 的力直接作用于 $Q$ ，方向沿该斜面的一条最大坡度线向上。

求使两质点保持静止的 $Q$ 与斜面 $BC$ 之间的摩擦系数的最小值。

(5)

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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图中P远离滑轮向下，Q朝滑轮也向下，两个重力分量都驱动系统。张力在P的方程中阻碍运动，在Q的方程中促进运动。

答题过程

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The string constraint means that $P$ moves down $AB$ away from $B$ , while $Q$ moves down $CB$ towards $B$ , with equal acceleration magnitude $a$ . Let the tension be $T$ .

Taking each particle’s direction of motion as positive,

\begin{align*} 0.3(10)\sin60^\circ-T=&\,0.3a, \\[4mm] T+0.2(10)\sin30^\circ=&\,0.2a. \end{align*}

Adding gives

\begin{align*} a=&\,\frac{3\sin60^\circ+2\sin30^\circ}{0.5}\\[4mm] =&\,3\sqrt3+2. \end{align*}

Hence the acceleration magnitude is $\boxed{7.20\,\mathrm{m\,s^{-2}}}$ .

Substitution into the equation for $Q$ gives

\begin{align*} T=&\,0.2(3\sqrt3+2)-1\\[4mm] =&\,0.6\sqrt3-0.6\\[4mm] \approx&\,\boxed{0.439\,\mathrm{N}}. \end{align*}

(b)

解法一

思路

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先由P平衡求张力。Q所受向下的张力与重力分量大于向上外力3N，因此摩擦力必须向上。最小摩擦系数对应极限平衡。

答题过程

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For equilibrium of $P$ ,

T=0.3(10)\sin60^\circ=\frac{3\sqrt3}{2}\,\mathrm{N}.

For $Q$ , the normal reaction is

R=0.2(10)\cos30^\circ=\sqrt3\,\mathrm{N}.

The total downslope force excluding friction is $T+1-3=3\sqrt3/2-2>0$ . Thus friction acts up the plane and must have magnitude

f=T+1-3=\frac{3\sqrt3}{2}-2.

Equilibrium is possible if $f\le\mu R$ . The least coefficient occurs at equality:

\begin{align*} \mu_{\min}=&\,\frac{3\sqrt3/2-2}{\sqrt3}\\[4mm] =&\,\frac32-\frac2{\sqrt3}\\[4mm] \approx&\,\boxed{0.345}\quad\text{(3 s.f.)}. \end{align*}

CIE 9709 2022 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