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CIE 9709 2025 November Paper 43 Q7

A Level / CIE / M1

CIE 9709 2025 Nov Paper 43 Paper · Question 7

题目

Problem

A particle PP of mass 6 kg6\,\mathrm{kg} lies on a rough plane inclined at an acute angle θ\theta to the horizontal. A horizontal force of magnitude 12 N12\,\mathrm{N} acts on PP as shown in the diagram. The line of action of this horizontal force lies in a vertical plane which contains the line of greatest slope of the plane that passes through PP. The coefficient of friction between PP and the plane is μ\mu.

PP is in equilibrium and on the point of sliding down the plane.

(a) Show that μ=5tan⁡θ−15+tan⁡θ\mu=\frac{5\tan\theta-1}{5+\tan\theta}.

(5)

(b) Hence find the set of possible values of θ\theta.

(2)
题目中文翻译

一个质量为 6 kg6\,\mathrm{kg} 的质点 PP 位于粗糙斜面上,斜面与水平面成锐角 θ\theta。一个大小为 12 N12\,\mathrm{N} 的水平力作用于 PP,如图所示。该水平力的作用线位于一个竖直平面内,该竖直平面包含斜面上经过 PP 的最大坡度线。PP 与斜面之间的摩擦系数为 μ\mu。

PP 处于平衡状态,且恰好即将沿斜面向下滑动。

(a) 证明 μ=5tan⁡θ−15+tan⁡θ\mu=\frac{5\tan\theta-1}{5+\tan\theta}。

(5)

(b) 由此求 θ\theta 的可能取值集合。

(2)
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 (gg) is needed, use 10 m s−210\,\mathrm{m\,s^{-2}}.

需要使用重力加速度 gg 的数值时,取 10 m s−210\,\mathrm{m\,s^{-2}}。

解答

(a)

解法一

思路

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即将下滑,摩擦沿坡向上。水平力既帮助上坡也压向斜面,分别列平行、垂直平衡。

答题过程

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Let FF be the friction and RR the normal contact force. Friction acts up the plane, and at limiting equilibrium F=μRF=\mu R.

Resolving perpendicular to the plane,

R=60cos⁡θ+12sin⁡θ.R=60\cos\theta+12\sin\theta.

Resolving parallel to the plane,

F+12cos⁡θ=60sin⁡θ,F=60sin⁡θ−12cos⁡θ.F+12\cos\theta=60\sin\theta, \qquad F=60\sin\theta-12\cos\theta.

Therefore

μ= 60sin⁡θ−12cos⁡θ60cos⁡θ+12sin⁡θ= 5sin⁡θ−cos⁡θ5cos⁡θ+sin⁡θ.\begin{align*} \mu=&\,\frac{60\sin\theta-12\cos\theta}{60\cos\theta+12\sin\theta}\\[4mm] =&\,\frac{5\sin\theta-\cos\theta}{5\cos\theta+\sin\theta}. \end{align*}

Since θ\theta is acute, cos⁡θ>0\cos\theta>0. Dividing numerator and denominator by cos⁡θ\cos\theta gives

μ=5tan⁡θ−15+tan⁡θ,\boxed{\mu=\frac{5\tan\theta-1}{5+\tan\theta}},

as required.

(b)

解法一

思路

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粗糙接触摩擦系数为正;锐角保证分母正,故只需分子正,再合并锐角范围。

答题过程

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For a rough plane, μ>0\mu>0. Since 0<θ<90∘0<\theta<90^\circ, the denominator 5+tan⁡θ5+\tan\theta is positive. Hence

5tan⁡θ−1>0,θ>tan⁡−1(15).5\tan\theta-1>0, \qquad \theta>\tan^{-1}\biggl(\frac15\biggr).

Thus the possible angles satisfy

tan⁡−1(1/5)<θ<90∘,\boxed{\tan^{-1}(1/5)<\theta<90^\circ},

with the lower boundary approximately 11.3∘11.3^\circ.