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CIE 9709 2024 June Paper 42 Q5

A Level / CIE / M1

CIE 9709 2024 June Paper 42 Paper · Question 5

题目

Problem

A particle of mass 0.8 kg0.8\,\mathrm{kg} lies on a rough plane which is inclined at an angle of 28∘28^\circ to the horizontal. The particle is kept in equilibrium by a force of magnitude T NT\,\mathrm{N}. This force acts at an angle of 35∘35^\circ above a line of greatest slope of the plane (see diagram). The coefficient of friction between the particle and the plane is 0.20.2.

Find the least and greatest possible values of TT.

(8)

题目中文翻译

一个质量为 0.8 kg0.8\,\mathrm{kg} 的质点位于与水平面成 28∘28^\circ 角的粗糙斜面上。一个大小为 T NT\,\mathrm{N} 的力使质点保持平衡。该力的方向在斜面最大坡度线上方,与该线成 35∘35^\circ 角(见图)。质点与斜面之间的摩擦系数为 0.20.2。

求 TT 可能的最小值和最大值。

(8)

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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张力向上分量会减小支持力。最小张力对应将要下滑、摩擦向上;最大张力对应将要上滑、摩擦向下。两种状态都用各自的支持力。

答题过程

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Resolving perpendicular to the plane gives

R=8cos⁡28∘−Tsin⁡35∘.R=8\cos28^\circ-T\sin35^\circ.

For the least TT, limiting friction acts up the plane. Hence

Tcos⁡35∘+0.2R=8sin⁡28∘.T\cos35^\circ+0.2R=8\sin28^\circ.

Substituting for RR,

T(cos⁡35∘−0.2sin⁡35∘)=8(sin⁡28∘−0.2cos⁡28∘).T(\cos35^\circ-0.2\sin35^\circ) =8(\sin28^\circ-0.2\cos28^\circ).

Therefore

Tmin⁡=8(sin⁡28∘−0.2cos⁡28∘)cos⁡35∘−0.2sin⁡35∘≈3.33 N.T_{\min}=\frac{8(\sin28^\circ-0.2\cos28^\circ)} {\cos35^\circ-0.2\sin35^\circ} \approx\boxed{3.33\,\mathrm{N}}.

For the greatest TT, limiting friction acts down the plane, giving

Tcos⁡35∘=8sin⁡28∘+0.2R.T\cos35^\circ=8\sin28^\circ+0.2R.

Thus

T(cos⁡35∘+0.2sin⁡35∘)=8(sin⁡28∘+0.2cos⁡28∘),T(\cos35^\circ+0.2\sin35^\circ) =8(\sin28^\circ+0.2\cos28^\circ),

so

Tmax⁡=8(sin⁡28∘+0.2cos⁡28∘)cos⁡35∘+0.2sin⁡35∘≈5.53 N.T_{\max}=\frac{8(\sin28^\circ+0.2\cos28^\circ)} {\cos35^\circ+0.2\sin35^\circ} \approx\boxed{5.53\,\mathrm{N}}.

The corresponding reactions are approximately 5.16 N5.16\,\mathrm{N} and 3.89 N3.89\,\mathrm{N}, both positive, so contact with the plane is maintained.