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CIE 9709 2023 June Paper 41 Q5

A Level / CIE / M1

CIE 9709 2023 June Paper 41 Paper · Question 5

题目

Problem

Four coplanar forces act at a point. The magnitudes of the forces are F NF\,\mathrm{N}, 10 N10\,\mathrm{N}, 50 N50\,\mathrm{N} and 40 N40\,\mathrm{N}. The directions of the forces are as shown in the diagram.

(a) Given that the forces are in equilibrium, find the value of FF and the value of θ\theta.

(6)

(b) Given instead that F=102F=10\sqrt2 and θ=45\theta=45, find the direction and the exact magnitude the resultant force.

(3)

题目中文翻译

四个共面力作用于同一点。各力的大小分别为 F NF\,\mathrm{N}、10 N10\,\mathrm{N}、50 N50\,\mathrm{N} 和 40 N40\,\mathrm{N}，方向如图所示。

(a) 已知各力处于平衡状态，求 FF 和 θ\theta 的值。

(6)

(b) 若改为已知 F=102F=10\sqrt2 且 θ=45\theta=45，求合力的方向及大小的精确值。

(3)

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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水平、竖直方向分别平衡，求F的两个分量，再由比值与平方和求θ与F。

答题过程

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Resolving horizontally and vertically gives

Fcos⁡θ+10−40cos⁡60∘=0,F\cos\theta+10-40\cos60^\circ=0, Fsin⁡θ+40sin⁡60∘−50=0.F\sin\theta+40\sin60^\circ-50=0.

Hence

Fcos⁡θ=10,Fsin⁡θ=50−203.F\cos\theta=10,\qquad F\sin\theta=50-20\sqrt3.

Both components are positive, consistent with the acute angle in the diagram. Thus

θ=tan⁡−1(5−23)≈56.9∘,\theta=\tan^{-1}(5-2\sqrt3)\approx\boxed{56.9^\circ}, F=102+(50−203)2≈18.3(3 s.f.).F=\sqrt{10^2+(50-20\sqrt3)^2} \approx\boxed{18.3} \quad\text{(3 s.f.)}.

(b)

解法一

思路

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换用指定F与角，水平分量相消，竖直合力为负。最终给出正的大小和向下方向。

答题过程

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The resultant components are

Rx= 102cos⁡45∘+10−40cos⁡60∘= 0.\begin{align*} R_x=&\,10\sqrt2\cos45^\circ+10-40\cos60^\circ\\[4mm] =&\,0. \end{align*} Ry= 102sin⁡45∘+40sin⁡60∘−50= 203−40<0.\begin{align*} R_y=&\,10\sqrt2\sin45^\circ+40\sin60^\circ-50\\[4mm] =&\,20\sqrt3-40<0. \end{align*}

Therefore the exact magnitude is

40−203 N,\boxed{40-20\sqrt3\,\mathrm{N}},

and the direction is vertically downwards\boxed{\text{vertically downwards}}, the same direction as the 50 N50\,\mathrm{N} force.