题目

Problem

A block of mass $12\,\mathrm{kg}$ is placed on a plane which is inclined at an angle of $24^\circ$ to the horizontal. A light string, making an angle of $36^\circ$ above a line of greatest slope, is attached to the block. The tension in the string is $65\,\mathrm{N}$ (see diagram). The coefficient of friction between the block and plane is $\mu$ . The block is in limiting equilibrium and is on the point of sliding up the plane.

Find $\mu$ .

(6)

题目中文翻译

一个质量为 $12\,\mathrm{kg}$ 的物块放在与水平面成 $24^\circ$ 角的斜面上。一根轻绳连接着物块，其方向位于一条最大坡度线上方，并与该线成 $36^\circ$ 角。绳中的张力为 $65\,\mathrm{N}$ （见图）。物块与斜面之间的摩擦系数为 $\mu$ 。物块处于极限平衡状态，恰要沿斜面向上滑动。

求 $\mu$ 。

(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

展开

恰要上滑，摩擦力沿斜面向下。先垂直斜面求支持力，再沿斜面求极限摩擦力。

答题过程

展开

Let the normal reaction be $R$ and the limiting friction be $F$ . Since the block is on the point of sliding upwards, friction acts down the plane.

Resolving perpendicular to the plane,

\begin{align*} R+65\sin36^\circ=&\,120\cos24^\circ, \\[4mm] R=&\,71.41941352\ldots\,\mathrm{N}. \end{align*}

Resolving up the plane,

\begin{align*} 65\cos36^\circ=&\,120\sin24^\circ+F, \\[4mm] F=&\,3.777707465\ldots\,\mathrm{N}. \end{align*}

Using $F=\mu R$ gives

\begin{align*} \mu=&\,\frac{3.777707465\ldots}{71.41941352\ldots}\\[4mm] =&\,0.05289468618\ldots\\[4mm] \approx&\,\boxed{0.0529}. \end{align*}

CIE 9709 2022 June Paper 42 Q5

题目

解答

解法一

思路

答题过程