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CIE 9709 2023 June Paper 42 Q5

A Level / CIE / M1

CIE 9709 2023 June Paper 42 Paper · Question 5

题目

Problem

A particle of mass 0.6 kg0.6\,\mathrm{kg} is placed on a rough plane which is inclined at an angle of 35∘35^\circ to the horizontal. The particle is kept in equilibrium by a horizontal force of magnitude P NP\,\mathrm{N} acting in a vertical plane containing a line of greatest slope (see diagram). The coefficient of friction between the particle and plane is 0.40.4.

Find the least possible value of PP.

(6)

题目中文翻译

一个质量为 0.6 kg0.6\,\mathrm{kg} 的质点放在与水平面成 35∘35^\circ 角的粗糙斜面上。一个大小为 P NP\,\mathrm{N} 的水平力使质点保持平衡,该力位于包含一条最大坡度线的竖直平面内(见图)。质点与斜面之间的摩擦系数为 0.40.4。

求 PP 的最小可能值。

(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 (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

解法一

思路

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最小水平推力对应即将下滑,摩擦向上且达到极限。水平推力同时压紧斜面,支持力也含P的分量。

答题过程

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For the least PP, the particle is on the point of sliding down, so limiting friction acts up the plane. Resolving perpendicular to the plane,

R=Psin⁡35∘+6cos⁡35∘.R=P\sin35^\circ+6\cos35^\circ.

Resolving up the plane with F=0.4RF=0.4R,

Pcos⁡35∘+0.4R=6sin⁡35∘.P\cos35^\circ+0.4R=6\sin35^\circ.

Substituting for RR gives

P(cos⁡35∘+0.4sin⁡35∘)=6(sin⁡35∘−0.4cos⁡35∘).P(\cos35^\circ+0.4\sin35^\circ) =6(\sin35^\circ-0.4\cos35^\circ).

Hence

Pmin⁡=6(sin⁡35∘−0.4cos⁡35∘)cos⁡35∘+0.4sin⁡35∘≈1.41.P_{\min}=\frac{6(\sin35^\circ-0.4\cos35^\circ)} {\cos35^\circ+0.4\sin35^\circ} \approx\boxed{1.41}.

解法二

思路

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也可沿水平、竖直方向分解支持力与摩擦,先解出支持力,再求水平推力。

答题过程

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At the least PP, limiting friction acts up the slope, with F=0.4RF=0.4R. Vertical and horizontal equilibrium give

Rcos⁡35∘+Fsin⁡35∘=6,R\cos35^\circ+F\sin35^\circ=6, P+Fcos⁡35∘=Rsin⁡35∘.P+F\cos35^\circ=R\sin35^\circ.

Thus

R=6cos⁡35∘+0.4sin⁡35∘=5.722009… N.R=\frac{6}{\cos35^\circ+0.4\sin35^\circ} =5.722009\ldots\,\mathrm{N}.

Therefore

Pmin⁡= R(sin⁡35∘−0.4cos⁡35∘)≈ 1.41.\begin{align*} P_{\min}=&\,R(\sin35^\circ-0.4\cos35^\circ)\\[4mm] \approx&\,\boxed{1.41}. \end{align*}