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M1 2025 Jan Q4

A Level / Edexcel / M1

IAL 2025 Jan Paper · Question 4

题目

Problem

4.

A uniform rod ABAB has length 3 m3\,\mathrm{m} and weight WW newtons. The rod is suspended by two light vertical ropes. One rope is attached to the rod at AA and the other rope is attached to the rod at CC, where AC=2.1 mAC=2.1\,\mathrm{m}.

The rod is in equilibrium in a horizontal position, as shown in Figure 2.

The tension in the rope at CC is 350 N350\,\mathrm{N}.

(a) Show that W=490W=490

(3)

A particle PP of weight 210 N210\,\mathrm{N} is attached to the rod at a distance dd metres from AA. The tension in the rope at CC is now 600 N600\,\mathrm{N}. The rod remains in equilibrium in a horizontal position.

(b) Find the value of dd.

(3)

Particle PP is removed from the rod. A particle QQ of weight XX newtons is now attached at BB. The rod remains in equilibrium in a horizontal position and is now on the point of tilting.

(c) Find the value of XX.

(4)
题目中文翻译

一根均匀杆 ABAB 长 3 m3\,\mathrm{m},重 WW 牛顿。 杆由两根竖直轻绳悬挂,一根绳连接于杆上的 AA 点,另一根连接于 CC 点,其中 AC=2.1 mAC=2.1\,\mathrm{m}。

如图 2 所示,杆处于水平平衡状态。

CC 点处绳中的张力为 350 N350\,\mathrm{N}。

(a) 证明 W=490W=490。

(3)

一个重 210 N210\,\mathrm{N} 的质点 PP 连接在杆上距 AA 点 dd 米的位置。 此时 CC 点处绳中的张力为 600 N600\,\mathrm{N}。 杆仍保持水平平衡。

(b) 求 dd 的值。

(3)

从杆上移去质点 PP。 现将重 XX 牛顿的质点 QQ 连接在 BB 点。 杆仍保持水平平衡,且恰好要倾覆。

(c) 求 XX 的值。

(4)

解答

(a)

解法一

思路

展开

均匀杆的重力作用在距 AA 为 1.5 m 的中点。绕 AA 取矩消去该处张力;题中已经给重力,不应再乘 gg。

答题过程

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The weight of the uniform rod acts at its midpoint, 1.5 m from AA. Taking moments about AA,

350(2.1)=W(1.5).350(2.1)=W(1.5).

Therefore

W=7351.5=490,W=\frac{735}{1.5}=\boxed{490},

as required.

(b)

解法一

思路

展开

加入质点后重新绕 AA 取矩,两项重力矩由 CC 处新张力矩平衡。

答题过程

展开

Taking moments about AA in the new equilibrium,

600(2.1)=490(1.5)+210d.600(2.1)=490(1.5)+210d.

Thus

210d=1260−735=525,d=2.5.210d=1260-735=525,\qquad \boxed{d=2.5}.

(c)

解法一

思路

展开

原质点已移去;在 BB 加重物使 AA 处绳将松弛。临界时 AA 处张力为零,绕 CC 取矩求 XX。

答题过程

展开

At the point of tilting about CC, the tension at AA is zero. The rod’s midpoint is 2.1−1.5=0.62.1-1.5=0.6 m to the left of CC, and BB is 3−2.1=0.93-2.1=0.9 m to the right.

Taking moments about CC,

490(0.6)=X(0.9).490(0.6)=X(0.9).

Hence X=9803\boxed{X=\dfrac{980}{3}} (newtons).