题目

Problem

A uniform rod, of length $8a$ and mass $M$ , has one end freely hinged to a fixed point $A$ on a vertical wall. One end of a light inextensible string is attached to the rod at the point $B$ , where $AB=5a$ . The other end of the string is attached to the wall at the point $C$ , where $AC=5a$ and $C$ is vertically above $A$ . The rod rests in equilibrium in a vertical plane perpendicular to the wall with angle $BAC=70^\circ$ , as shown in Figure 3.

(a) Find, in terms of $M$ and $g$ , the tension in the string.

(3)

The magnitude of the force acting on the rod at $A$ is $\lambda Mg$ , where $\lambda$ is a constant.

(b) Find, to 2 significant figures, the value of $\lambda$ .

(6)

题目中文翻译

一根长为 $8a$ 、质量为 $M$ 的均匀杆，其一端通过自由铰链连接于竖直墙上的固定点 $A$ 。一根不可伸长的轻绳的一端连接于杆上的 $B$ 点，其中 $AB=5a$ 。绳的另一端连接于墙上的 $C$ 点，其中 $AC=5a$ ， $C$ 位于 $A$ 的正上方。如图 3 所示，杆在垂直于墙面的竖直平面内静止平衡，且 $\angle BAC=70^\circ$ 。

(a) 用 $M$ 和 $g$ 表示绳中的张力。

(3)

$A$ 点处作用在杆上的力的大小为 $\lambda Mg$ ，其中 $\lambda$ 为常数。

(b) 求 $\lambda$ 的值，保留 2 位有效数字。

(6)

解答

(a)

解法一

思路

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$AB=AC$ ，所以先由等腰三角形确定绳与杆的夹角。均匀杆的重心在中点；对铰链取矩消去铰链反力。

答题过程

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Since $AB=AC=5a$ , triangle $ABC$ is isosceles. Therefore

\angle ABC=\angle BCA=\frac{180^\circ-70^\circ}{2}=55^\circ.

The rod is inclined at $20^\circ$ to the horizontal. Its weight acts at its midpoint, a distance $4a$ from $A$ .

Let the tension be $T$ . Taking moments about $A$ ,

T(5a)\sin55^\circ=Mg(4a)\cos20^\circ.

Hence

\boxed{T=\frac{4\cos20^\circ}{5\sin55^\circ}Mg}.

(b)

解法一

思路

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铰链反力通常既不沿杆也不水平。分别求其水平、竖直分量，再用勾股定理求合力。绳与竖直方向的夹角为 $55^\circ$ 。

答题过程

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Let the force at $A$ have horizontal component $H$ to the left and vertical component $V$ upwards. Resolving horizontally and vertically,

H=T\sin55^\circ,\qquad V+T\cos55^\circ=Mg.

Using the unrounded expression for $T$ ,

\frac H{Mg}=\frac45\cos20^\circ,

\frac V{Mg}=1-\frac{4\cos20^\circ\cos55^\circ}{5\sin55^\circ}.

Since the magnitude of this force is $\lambda Mg$ ,

\lambda=\sqrt{\left(\frac H{Mg}\right)^2+\left(\frac V{Mg}\right)^2}.

Substitution gives, to 2 significant figures,

\boxed{\lambda=0.89}.

解法二

思路

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也可以不使用张力：对 $C$ 取矩消去张力和竖直铰链分量，对 $B$ 取矩再求竖直分量。注意杆重心在 $A$ 与 $B$ 之间，距 $B$ 为 $a$ 。

答题过程

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Let the hinge-force components be $H$ to the left and $V$ upwards. Taking moments about $C$ , both the tension and $V$ have zero moment, so

5aH=4aMg\cos20^\circ.

Thus

H=\frac45Mg\cos20^\circ.

Taking moments about $B$ ,

aMg\cos20^\circ+5aH\cos70^\circ=5aV\cos20^\circ.

Consequently

V=\frac{Mg}{5}+\frac{H\cos70^\circ}{\cos20^\circ}.

Therefore

\lambda=\frac{\sqrt{H^2+V^2}}{Mg}=0.89\quad\text{(2 s.f.)}.

M2 2021 June Q5

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

解法二

思路

答题过程