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M1 2022 Oct Q3

A Level / Edexcel / M1

IAL 2022 Oct Paper · Question 3

题目

Problem

3.

A rough plane is inclined to the horizontal at an angle α\alpha, where tan⁡α=34\tan\alpha=\frac{3}{4}.

A particle PP of mass 2 kg2\,\mathrm{kg} is held in equilibrium on the plane by a horizontal force of magnitude XX newtons, as shown in Figure 2. The force acts in a vertical plane which contains a line of greatest slope of the inclined plane.

(a) Show that when X=14.7X=14.7 there is no frictional force acting on PP.

(3)

The coefficient of friction between PP and the plane is 0.50.5.

(b) Find the smallest possible value of XX.

(8)
题目中文翻译

一个粗糙斜面与水平面成 α\alpha 角，其中 tan⁡α=34\tan\alpha=\frac{3}{4}。

如图 2 所示，一个大小为 XX 牛顿的水平力使质量为 2 kg2\,\mathrm{kg} 的质点 PP 在斜面上保持平衡。该力位于包含斜面一条最大坡度线的竖直平面内。

(a) 证明：当 X=14.7X=14.7 时，PP 不受摩擦力。

(3)

PP 与斜面之间的摩擦系数为 0.50.5。

(b) 求 XX 的最小可能值。

(8)

解答

(a)

解法一

思路

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沿斜面分解即可消去支持力。先保留未知摩擦力，代入水平力后计算它是否为零，不能预先假定摩擦已达到极限。

答题过程

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Since tan⁡α=3/4\tan\alpha=3/4, sin⁡α=3/5\sin\alpha=3/5 and cos⁡α=4/5\cos\alpha=4/5.

Let FF be the frictional force taken as positive down the plane. Resolving parallel to the plane when X=14.7X=14.7,

14.7cos⁡α= 2gsin⁡α+F,F= 14.7(45)−19.6(35),= 11.76−11.76=0.\begin{align*} 14.7\cos\alpha=&\,2g\sin\alpha+F,\\ F=&\,14.7\left(\frac45\right)-19.6\left(\frac35\right),\\ =&\,11.76-11.76=0. \end{align*}

Thus there is no frictional force acting on PP, as required.

(b)

解法一

思路

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水平推力最小时，粒子将要沿斜面下滑，极限摩擦力沿斜面向上。水平推力还会把粒子压向斜面，因此支持力包含它的垂直斜面分量。

答题过程

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At the smallest possible XX, the particle is on the point of sliding down the plane. Friction acts up the plane and has magnitude F1=0.5SF_1=0.5S, where SS is the normal reaction.

Resolving perpendicular and parallel to the plane respectively,

S=Xsin⁡α+2gcos⁡α,Xcos⁡α+0.5S=2gsin⁡α.S=X\sin\alpha+2g\cos\alpha,\qquad X\cos\alpha+0.5S=2g\sin\alpha.

Substituting for SS and the trigonometric values,

45X+12(35X+85g)= 65g,1110X= 25g,X= 4g11=3.56 N(3 s.f.).\begin{align*} \frac45X+\frac12\left(\frac35X+\frac85g\right)=&\,\frac65g,\\ \frac{11}{10}X=&\,\frac25g,\\ X=&\,\frac{4g}{11}=\boxed{3.56\ \mathrm N}\quad\text{(3 s.f.)}. \end{align*}

解法二

思路

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也可在临界下滑状态下，沿水平和竖直方向分解，结合 F1=0.5SF_1=0.5S 消去支持力，作为独立的分解方法。

答题过程

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At impending downward motion, let the upward frictional force be F1=0.5SF_1=0.5S.

Resolving vertically,

Scos⁡α+F1sin⁡α= 2g,45S+12S(35)= 2g,S= 20g11.\begin{align*} S\cos\alpha+F_1\sin\alpha=&\,2g,\\ \frac45S+\frac12S\left(\frac35\right)=&\,2g,\\ S=&\,\frac{20g}{11}. \end{align*}

Resolving horizontally,

X+F1cos⁡α= Ssin⁡α,X= S(35−12⋅45)=S5,X= 4g11=3.56 N(3 s.f.).\begin{align*} X+F_1\cos\alpha=&\,S\sin\alpha,\\ X=&\,S\left(\frac35-\frac12\cdot\frac45\right)=\frac S5,\\ X=&\,\frac{4g}{11}=\boxed{3.56\ \mathrm N}\quad\text{(3 s.f.)}. \end{align*}