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M2 2020 Oct Q3

A Level / Edexcel / M2

IAL 2020 Oct Paper · Question 3

题目

Problem

3.

A uniform rod ABAB, of mass 25 kg25\,\mathrm{kg} and length 3 m3\,\mathrm{m}, has end AA resting on rough horizontal ground. The end BB rests against a rough vertical wall.

The rod is in a vertical plane perpendicular to the wall.

The coefficient of friction between the rod and the ground is 45\frac45

The coefficient of friction between the rod and the wall is 35\frac35

The rod rests in limiting equilibrium.

The rod is at an angle of θ\theta to the ground, as shown in Figure 1.

Find the exact value of tan⁡θ\tan\theta.

(9)
题目中文翻译

一根质量为 25 kg25\,\mathrm{kg}、长为 3 m3\,\mathrm{m} 的均匀杆 ABAB，其 AA 端放在粗糙水平地面上，BB 端靠在粗糙竖直墙上。

杆位于垂直于墙面的竖直平面内。

杆与地面之间的摩擦系数为 45\frac45。

杆与墙之间的摩擦系数为 35\frac35。

杆静止并处于极限平衡状态。

如图 1 所示，杆与地面成 θ\theta 角。

求 tan⁡θ\tan\theta 的精确值。

(9)

解答

解法一

思路

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杆有下滑趋势，墙端摩擦力向上、地面摩擦力向墙。两处均处于极限摩擦状态。先用水平和竖直平衡求支持力，再绕地面接触点取矩。

答题过程

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Let the normal reactions at the ground and wall be RR and SS respectively. At limiting equilibrium, the friction forces have magnitudes

FA=45Rtowards the wall,F_A=\frac45R\quad\text{towards the wall}, FB=35Sup the wall.F_B=\frac35S\quad\text{up the wall}.

Horizontal and vertical equilibrium give

S=45R,R+35S=25g.S=\frac45R,\qquad R+\frac35S=25g.

Using R=5S/4R=5S/4,

(54+35)S=25g⟹S=500g37.\left(\frac54+\frac35\right)S=25g \quad\Longrightarrow\quad S=\frac{500g}{37}.

Taking moments about AA, the rod’s weight acts at its midpoint:

25g(1.5cos⁡θ)=S(3sin⁡θ)+35S(3cos⁡θ).25g(1.5\cos\theta) =S(3\sin\theta)+\frac35S(3\cos\theta).

Dividing by 1.5cos⁡θ1.5\cos\theta, which is positive,

25g=2Stan⁡θ+65S.25g=2S\tan\theta+\frac65S.

Hence

tan⁡θ= 25g−65S2S= 25−600/371000/37= 1340.\begin{align*} \tan\theta=&\,\frac{25g-\frac65S}{2S}\\ =&\,\frac{25-600/37}{1000/37}\\ =&\,\boxed{\frac{13}{40}}. \end{align*}