题目

Problem

A non-uniform beam $AB$ has length $6\,\mathrm{m}$ and weight $W$ newtons. The beam is supported in equilibrium in a horizontal position by two vertical ropes, one attached to the beam at $A$ and the other attached to the beam at $C$ , where $CB=1.5\,\mathrm{m}$ , as shown in Figure 1.

The centre of mass of the beam is $2.625\,\mathrm{m}$ from $A$ .

The ropes are modelled as light strings. The beam is modelled as a non-uniform rod.

Given that the tension in the rope attached at $C$ is $20\,\mathrm{N}$ greater than the tension in the rope attached at $A$ ,

(a) find the value of $W$ .

(6)

(b) State how you have used the fact that the beam is modelled as a rod.

(1)

题目中文翻译

一根非均匀横梁 $AB$ 长 $6\,\mathrm{m}$ ，重 $W$ 牛顿。如图 1 所示，两根竖直绳分别连接在梁上的 $A$ 点和 $C$ 点，使梁保持水平平衡，其中 $CB=1.5\,\mathrm{m}$ 。

横梁的质心距 $A$ 点 $2.625\,\mathrm{m}$ 。

将绳视为轻绳，将横梁视为非均匀杆。

已知连接在 $C$ 点的绳中张力比连接在 $A$ 点的绳中张力大 $20\,\mathrm{N}$ ，

(a) 求 $W$ 的值。

(6)

(b) 说明如何使用了“将横梁视为杆”这一假设。

(1)

解答

(a)

解法一

思路

展开

两张力的差是 $20$ 牛顿，不是张力的比。设 $A$ 端张力为 $T$ ，则 $C$ 端为 $T+20$ ；用竖直平衡与力矩平衡联立。

答题过程

展开

Let the tension at $A$ be $T\,\mathrm{N}$ ; the tension at $C$ is then $(T+20)\,\mathrm{N}$ . Since $AC=6-1.5=4.5\,\mathrm{m}$ , vertical equilibrium and moments about $A$ give

T+(T+20)=W,

4.5(T+20)=2.625W.

Substituting $W=2T+20$ into the moments equation,

\begin{align*} 4.5T+90=&\,5.25T+52.5,\\[4mm] 37.5=&\,0.75T,\\[4mm] T=&\,50. \end{align*}

Hence $\boxed{W=2(50)+20=120\,\mathrm{N}}$ .

(b)

解法一

思路

展开

杆模型使各力的作用点保持在同一直线上，不会因横梁弯曲而改变力臂。

答题过程

展开

The beam is assumed to remain straight and not bend under the forces acting on it.

M1 2020 Jan Q2

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程