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M1 2024 Oct Q2

A Level / Edexcel / M1

IAL 2024 Oct Paper · Question 2

题目

Problem

2.

A non-uniform beam ABAB has length 6 m6\,\mathrm{m} and mass 50 kg50\,\mathrm{kg}. The beam rests horizontally on two supports at CC and DD, where AC=0.9 mAC=0.9\,\mathrm{m} and DB=1.8 mDB=1.8\,\mathrm{m}.

A child of mass 25 kg25\,\mathrm{kg} stands on the beam at EE, where AE=EB=3 mAE=EB=3\,\mathrm{m}, as shown in Figure 1.

The beam is in equilibrium.

The magnitude of the normal reaction between the beam and the support at CC is RCR_C newtons.

The magnitude of the normal reaction between the beam and the support at DD is RDR_D newtons.

The beam is modelled as a rod and the child is modelled as a particle.

The centre of mass of the beam is between CC and DD and is a distance xx metres from DD.

Given that 2RD=3RC2R_D=3R_C

(a) show that x=1.38x=1.38

(6)

The child remains at EE and a block of mass M kgM\,\mathrm{kg} is placed on the beam at BB.

The block is modelled as a particle.

Given that the beam is on the point of tilting,

(b) find the value of MM.

(3)
题目中文翻译

一根非均匀梁 ABAB 长 6 m6\,\mathrm{m}，质量为 50 kg50\,\mathrm{kg}。梁水平放在 CC、DD 两点的支座上，其中 AC=0.9 mAC=0.9\,\mathrm{m}，DB=1.8 mDB=1.8\,\mathrm{m}。

如图 1 所示，一个质量为 25 kg25\,\mathrm{kg} 的孩子站在梁上的 EE 点，其中 AE=EB=3 mAE=EB=3\,\mathrm{m}。

梁处于平衡状态。

梁与 CC 点支座之间的法向支持力大小为 RCR_C 牛顿。

梁与 DD 点支座之间的法向支持力大小为 RDR_D 牛顿。

将梁视为杆，将孩子视为质点。

梁的质心位于 CC 和 DD 之间，距 DD 点 xx 米。

已知 2RD=3RC2R_D=3R_C，

(a) 证明 x=1.38x=1.38。

(6)

孩子仍站在 EE 点，一个质量为 M kgM\,\mathrm{kg} 的物块被放在梁上的 BB 点。

将物块视为质点。

已知梁恰好要倾覆，

(b) 求 MM 的值。

(3)

解答

(a)

解法一

思路

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先把支持力比例与竖直平衡联立，注意 2RD=3RC2R_D=3R_C 表示右侧支持力更大。再绕 DD 取矩求非均匀梁重心的位置。

答题过程

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Vertical equilibrium gives

RC+RD=(50+25)g=75g.R_C+R_D=(50+25)g=75g.

Since 2RD=3RC2R_D=3R_C, RD=32RCR_D=\tfrac32R_C, so

52RC=75g,RC=30g.\frac52R_C=75g,\qquad R_C=30g.

The distances from DD are CD=6−0.9−1.8=3.3CD=6-0.9-1.8=3.3 m and DE=4.2−3=1.2DE=4.2-3=1.2 m. Taking moments about DD,

50gx+25g(1.2)= 3.3RC,50x+30= 3.3(30)=99,x= 6950=1.38,\begin{align*} 50gx+25g(1.2)=&\,3.3R_C,\\ 50x+30=&\,3.3(30)=99,\\ x=&\,\frac{69}{50}=\boxed{1.38}, \end{align*}

as required.

(b)

解法一

思路

展开

在右端加物块会使梁绕 DD 倾覆，临界时 CC 处支持力为零。孩子仍留在梁上，必须计入其力矩。

答题过程

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At the point of tilting, the beam is about to rotate about DD and the reaction at CC is zero. Taking moments about DD,

25g(1.2)+50g(1.38)=Mg(1.8).25g(1.2)+50g(1.38)=Mg(1.8).

Cancelling gg gives

30+69=1.8M,M=55.30+69=1.8M,\qquad \boxed{M=55}.