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M2 2025 June A Q6

A Level / Edexcel / M2

IAL 2025 June A Paper · Question 6

题目

Problem

6.

A uniform rod ABAB has mass mm and length 2d2d.

The rod is held in equilibrium in a horizontal position, with the end AA against a rough vertical wall, by a light inextensible string.

One end of the string is attached to the rod at BB.

The other end of the string is attached to the wall at the point CC, where AC=0.8dAC=0.8d and CC is vertically above AA, as shown in Figure 2.

A particle of mass kmkm is attached at BB and the rod ABAB remains in equilibrium in a horizontal position.

The particle, rod and string are in a vertical plane perpendicular to the wall.

(a) Show that the tension in the string is

mg294(2k+1)\frac{mg\sqrt{29}}4(2k+1)
(4)

The resultant force acting on the rod at AA makes an angle θ\theta with the horizontal.

(b) Given that tan⁡θ=18\tan\theta=\frac18, find the exact value of kk.

(6)
题目中文翻译

均匀杆 ABAB 的质量为 mm，长为 2d2d。

一根不可伸长的轻绳使杆保持水平平衡，AA 端靠在粗糙竖直墙上。

绳的一端连接于杆的 BB 点。

如图 2 所示，绳的另一端连接于墙上的 CC 点，其中 AC=0.8dAC=0.8d，且 CC 位于 AA 的正上方。

一个质量为 kmkm 的质点连接于 BB 点，杆 ABAB 仍保持水平平衡。

质点、杆与绳均位于垂直于墙面的竖直平面内。

(a) 证明绳中的张力为

mg294(2k+1)\frac{mg\sqrt{29}}4(2k+1)
(4)

AA 点处作用于杆的合力与水平面成 θ\theta 角。

(b) 已知 tan⁡θ=18\tan\theta=\frac18，求 kk 的精确值。

(6)

解答

(a)

解法一

思路

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由直角三角形求绳子与水平线的夹角。绕 A 取矩，墙的作用力被消去；注意杆重作用于中点，而附重作用于 B。

答题过程

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Let α\alpha be the angle between the string and the horizontal. From the geometry,

tan⁡α=0.8d2d=25,sin⁡α=229,cos⁡α=529.\tan\alpha=\frac{0.8d}{2d}=\frac25,\qquad \sin\alpha=\frac2{\sqrt{29}},\qquad \cos\alpha=\frac5{\sqrt{29}}.

Taking moments about AA,

Tsin⁡α(2d)=mg(d)+kmg(2d).T\sin\alpha(2d)=mg(d)+kmg(2d).

Therefore

T229=mg(1+2k)2,T\frac2{\sqrt{29}}=\frac{mg(1+2k)}2, T=mg294(2k+1).\boxed{T=\frac{mg\sqrt{29}}4(2k+1)}.

(b)

解法一

思路

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将 A 点合力分成水平支持力和竖直摩擦力。先用竖直平衡确定摩擦力确实向上，再以竖直分量除水平分量得到角的正切。

答题过程

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Let the force at AA have horizontal component XX to the right and signed vertical component YY upwards. Horizontal equilibrium gives

X=Tcos⁡α=5mg(1+2k)4.X=T\cos\alpha=\frac{5mg(1+2k)}4.

Vertical equilibrium gives

Y+Tsin⁡α=(1+k)mg,Y+T\sin\alpha=(1+k)mg, Y=(1+k)mg−mg(1+2k)2=mg2>0.Y=(1+k)mg-\frac{mg(1+2k)}2=\frac{mg}2>0.

Thus the resultant acts above the horizontal, and

tan⁡θ=YX=25(1+2k)=18.\tan\theta=\frac YX=\frac2{5(1+2k)}=\frac18.

Hence

16=5+10k,k=1110.16=5+10k,\qquad \boxed{k=\frac{11}{10}}.