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M2 2025 Oct A Q5

A Level / Edexcel / M2

IAL 2025 Oct A Paper · Question 5

题目

Problem

5.

A uniform rod ABAB has mass 6 kg6\,\mathrm{kg} and length 2 m2\,\mathrm{m}. The end AA of the rod rests against a rough vertical wall.

One end of a light string is attached to the rod at BB. The other end of the string is attached to the wall at CC, which is vertically above AA.

The angle between the rod and the string is 30∘30^\circ and the angle between the rod and the wall is 70∘70^\circ, as shown in Figure 2.

The rod is in a vertical plane perpendicular to the wall and rests in limiting equilibrium.

Find

(a) the tension in the string,

(4)

(b) the coefficient of friction between the rod and the wall,

(5)

(c) the direction of the force exerted on the rod by the wall at AA.

(2)
题目中文翻译

均匀杆 ABAB 的质量为 6 kg6\,\mathrm{kg}，长为 2 m2\,\mathrm{m}。杆的 AA 端靠在粗糙竖直墙上。

轻绳的一端连接于杆的 BB 点，另一端连接于墙上的 CC 点，CC 位于 AA 的正上方。

如图 2 所示，杆与绳之间的夹角为 30∘30^\circ，杆与墙之间的夹角为 70∘70^\circ。

杆位于垂直于墙面的竖直平面内，并处于极限平衡。

求

(a) 绳中的张力；

(4)

(b) 杆与墙之间的摩擦系数；

(5)

(c) 墙在 AA 点对杆施加的力的方向。

(2)

解答

(a)

解法一

思路

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绕 A 取矩消去墙的支持力和摩擦力。杆重作用于中点，其水平力臂是 sin70°；张力与杆夹角为 30°。

答题过程

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The rod’s centre of mass is 11 m from AA. Taking moments about AA,

6g(1)sin⁡70∘=T(2)sin⁡30∘.6g(1)\sin70^\circ=T(2)\sin30^\circ.

Hence

T=6gsin⁡70∘=55.3 N(3 s.f.).T=6g\sin70^\circ=\boxed{55.3\text{ N}}\quad(3\text{ s.f.}).

(b)

解法一

思路

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杆向右下方，与水平成 20°；绳子向左上方，与水平成 50°。分解张力，先确定摩擦力向上，再用极限摩擦条件。

答题过程

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The rod is 20∘20^\circ below the horizontal, and the string from BB to CC is 50∘50^\circ above the horizontal towards the wall. Let RR be the normal reaction at AA, acting horizontally away from the wall. Horizontal equilibrium gives

R=Tcos⁡50∘.R=T\cos50^\circ.

Let FF be the signed frictional force upwards. Vertical equilibrium gives

F+Tsin⁡50∘=6g,F+T\sin50^\circ=6g, F=6g−Tsin⁡50∘>0.F=6g-T\sin50^\circ>0.

Thus friction is upwards. Since equilibrium is limiting, F=μRF=\mu R, so

μ=6g−Tsin⁡50∘Tcos⁡50∘.\mu=\frac{6g-T\sin50^\circ}{T\cos50^\circ}.

Using the unrounded value T=6gsin⁡70∘T=6g\sin70^\circ,

μ=0.464(3 s.f.).\boxed{\mu=0.464}\quad(3\text{ s.f.}).

(c)

解法一

思路

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墙的合力由向右的支持力和向上的摩擦力组成。必须把角的参考方向写清楚。

答题过程

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The force exerted by the wall has components RR horizontally away from the wall and FF vertically upwards. If its angle above the horizontal is ψ\psi,

tan⁡ψ=FR=μ.\tan\psi=\frac FR=\mu.

Thus

ψ=tan⁡−1(0.463813…)=24.882…∘.\psi=\tan^{-1}(0.463813\ldots)=24.882\ldots^\circ.

The force acts 24.9∘ above the horizontal, away from the wall\boxed{24.9^\circ\text{ above the horizontal, away from the wall}}, equivalently 65.1∘65.1^\circ to the upward vertical.