Skip to content
CalcGospel 國際數學圖譜
返回

M1 2022 Oct Q2

A Level / Edexcel / M1

IAL 2022 Oct Paper · Question 2

题目

Problem

2.

A uniform rod ABAB has length 2a2a and mass MM. The rod is held in equilibrium in a horizontal position by two vertical light strings which are attached to the rod at CC and DD, where AC=25aAC=\frac{2}{5}a and DB=35aDB=\frac{3}{5}a, as shown in Figure 1.

A particle PP is placed on the rod at BB.

The rod remains horizontal and in equilibrium.

(a) Find, in terms of MM, the largest possible mass of the particle PP.

(3)

Given that the mass of PP is 12M\frac{1}{2}M,

(b) find, in terms of MM and gg, the tension in the string that is attached to the rod at CC.

(3)
题目中文翻译

一根均匀杆 ABAB 长 2a2a,质量为 MM。如图 1 所示,两根竖直轻绳分别连接在杆上的 CC 点和 DD 点,使杆保持水平平衡,其中 AC=25aAC=\frac{2}{5}a,DB=35aDB=\frac{3}{5}a。

在杆的 BB 点放置一个质点 PP。

杆仍保持水平且处于平衡状态。

(a) 用 MM 表示质点 PP 的最大可能质量。

(3)

已知 PP 的质量为 12M\frac{1}{2}M,

(b) 用 MM 和 gg 表示连接在 CC 点的绳中的张力。

(3)

解答

(a)

解法一

思路

展开

增加 BB 端质量会使 CC 处绳子的张力减小。最大质量对应 CC 处张力恰好为零,此时绕 DD 取矩求临界质量。

答题过程

展开

Let GG be the midpoint of the rod and let the mass at BB be XX. The relevant distances are

DG=2a5,DB=3a5,CD=a.DG=\frac{2a}{5},\qquad DB=\frac{3a}{5},\qquad CD=a.

If the tension at CC is TCT_C, taking moments about DD gives

TCa+Xg(3a5)=Mg(2a5).T_Ca+Xg\left(\frac{3a}{5}\right)=Mg\left(\frac{2a}{5}\right).

Since TC≥0T_C\geq0, it follows that X≤2M/3X\leq2M/3. At the largest possible mass, TC=0T_C=0, so

Xmax⁡=2M3.\boxed{X_{\max}=\frac{2M}{3}}.

(b)

解法一

思路

展开

这一问的粒子质量已改为 M/2M/2,不能继续使用上一问的临界质量。仍绕 DD 取矩,可直接求 CC 处张力。

答题过程

展开

With the particle mass equal to M/2M/2, take moments about DD:

TCa+12Mg(3a5)= Mg(2a5),TCa= Mga10,TC= Mg10.\begin{align*} T_Ca+\frac12Mg\left(\frac{3a}{5}\right)=&\,Mg\left(\frac{2a}{5}\right),\\ T_Ca=&\,\frac{Mga}{10},\\ T_C=&\,\boxed{\frac{Mg}{10}}. \end{align*}