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M1 2024 June Q4

A Level / Edexcel / M1

IAL 2024 June Paper · Question 4

题目

Problem

4.

A non-uniform rod ABAB has length 6.5 m6.5\,\mathrm{m} and mass 1.2 kg1.2\,\mathrm{kg}. The centre of mass of the rod is 3 m3\,\mathrm{m} from AA. The rod rests on a horizontal step and overhangs the end of the step CC by 1.5 m1.5\,\mathrm{m}, as shown in Figure 2.

The rod is perpendicular to the edge of the step.

A particle of mass 4 kg4\,\mathrm{kg} is placed on the rod at BB and another particle, whose mass is M kgM\,\mathrm{kg}, is placed on the rod at DD, where AD=0.5 mAD=0.5\,\mathrm{m}.

The rod remains in equilibrium in a horizontal position.

(a) Find the smallest possible value of MM.

(3)

The particle at BB and the particle at DD are now removed.

A new particle is placed on the rod at the point EE, where EB=0.9 mEB=0.9\,\mathrm{m}.

The rod remains in equilibrium in a horizontal position but is on the point of tilting about CC.

(b) Find the magnitude of the force acting on the rod at CC.

(3)
题目中文翻译

一根非均匀杆 ABAB 长 6.5 m6.5\,\mathrm{m},质量为 1.2 kg1.2\,\mathrm{kg},其质心距 AA 点 3 m3\,\mathrm{m}。杆放在水平台阶上,伸出台阶边缘 CC 的长度为 1.5 m1.5\,\mathrm{m},如图 2 所示。

杆与台阶边缘垂直。

一个质量为 4 kg4\,\mathrm{kg} 的质点放在杆上的 BB 点,另一个质量为 M kgM\,\mathrm{kg} 的质点放在杆上的 DD 点,其中 AD=0.5 mAD=0.5\,\mathrm{m}。

杆保持水平平衡。

(a) 求 MM 的最小可能值。

(3)

现移去 BB 点和 DD 点的质点。

在杆上的 EE 点放置一个新质点,其中 EB=0.9 mEB=0.9\,\mathrm{m}。

杆保持水平平衡,但恰好要绕 CC 倾覆。

(b) 求 CC 点处作用在杆上的力的大小。

(3)

解答

(a)

解法一

思路

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MM 最小时,杆恰好绕台阶边缘 CC 倾覆,支持力作用于 CC。杆非均匀,重心必须用题给距 AA 为 3 m,而不是杆的中点。

答题过程

展开

For the smallest possible MM, the rod is on the point of tilting about CC.

Since AC=6.5−1.5=5AC=6.5-1.5=5 m, the rod’s centre of mass is 2 m to the left of CC, and DD is 5−0.5=4.55-0.5=4.5 m to the left. Taking moments about CC,

Mg(4.5)+1.2g(2)=4g(1.5).Mg(4.5)+1.2g(2)=4g(1.5).

Cancelling gg,

4.5M=6−2.4=3.6,M=0.8.4.5M=6-2.4=3.6,\qquad \boxed{M=0.8}.

(b)

解法一

思路

展开

原来的两个质点已经移去。新质点位于 CC 右侧 0.6 m;绕新质点处 EE 取矩,可直接消去未知质点重力并求 CC 处支持力。

答题过程

展开

At impending tilt, the support force acts at CC. The new particle is at EE, where

CE=CB−EB=1.5−0.9=0.6 m.CE=CB-EB=1.5-0.9=0.6\ \mathrm m.

The rod’s centre of mass is 2+0.6=2.62+0.6=2.6 m to the left of EE. Taking moments about EE eliminates the new particle’s weight:

RC(0.6)=1.2g(2.6).R_C(0.6)=1.2g(2.6).

Hence

RC=5.2g=50.96 N=51.0 N(3 s.f.).R_C=5.2g=50.96\ \mathrm N =\boxed{51.0\ \mathrm N}\quad\text{(3 s.f.)}.