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M1 2021 June Q7

A Level / Edexcel / M1

IAL 2021 June Paper · Question 7

题目

Problem

7.

A non-uniform beam ABAB, of mass 60 kg60\,\mathrm{kg} and length 8a8a metres, rests in equilibrium in a horizontal position on two vertical supports. One support is at CC, where AC=aAC=a metres and the other support is at DD, where DB=2aDB=2a metres, as shown in Figure 2.

The magnitude of the normal reaction between the beam and the support at DD is three times the magnitude of the normal reaction between the beam and the support at CC.

By modelling the beam as a non-uniform rod whose centre of mass is at a distance xx metres from AA,

(a) find an expression for xx in terms of aa.

(5)

A box of mass M kgM\,\mathrm{kg} is placed on the beam at EE, where AE=2aAE=2a metres.

The beam remains in equilibrium in a horizontal position.

The magnitude of the normal reaction between the beam and the support at CC is now equal to the magnitude of the normal reaction between the beam and the support at DD.

By modelling the box as a particle,

(b) find the value of MM.

(5)
题目中文翻译

一根非均匀横梁 ABAB 的质量为 60 kg60\,\mathrm{kg}，长 8a8a 米，水平放置在两个竖直支座上并保持平衡。如图 2 所示，一个支座位于 CC 点，其中 AC=aAC=a 米；另一个位于 DD 点，其中 DB=2aDB=2a 米。

梁与 DD 处支座之间法向反作用力的大小，是梁与 CC 处支座之间法向反作用力大小的三倍。

将梁视为非均匀杆，其质心距 AA 点 xx 米，

(a) 求用 aa 表示 xx 的表达式。

(5)

现将一个质量为 M kgM\,\mathrm{kg} 的箱子放在梁上的 EE 点，其中 AE=2aAE=2a 米。

横梁仍保持水平平衡。

此时，梁与 CC 处支座之间法向反作用力的大小，等于梁与 DD 处支座之间法向反作用力的大小。

将箱子视为质点，

(b) 求 MM 的值。

(5)

解答

(a)

解法一

思路

展开

支点 DD 到 AA 的距离为 6a6a。设两支持力分别为 RR 和 3R3R，先用竖直平衡求 RR，再绕 AA 取矩求重心位置。

答题过程

展开

Let the reactions at CC and DD be RR and 3R3R respectively. Vertical equilibrium gives

R+3R=60g,R=15g.R+3R=60g,\qquad R=15g.

Since AD=8a−2a=6aAD=8a-2a=6a, taking moments about AA gives

60gx= R(a)+3R(6a),60gx= 19Ra=285ga,x= 19a4.\begin{align*} 60gx=&\,R(a)+3R(6a),\\ 60gx=&\,19Ra=285ga,\\ x=&\,\boxed{\frac{19a}{4}}. \end{align*}

(b)

解法一

思路

展开

加箱子后两支持力相等，但不再是之前的 RR 和 3R3R。用新的未知量 SS 表示每个支持力，联立竖直平衡与力矩方程；梁本身的重心不变。

答题过程

展开

Let each new support reaction be SS. Vertical equilibrium gives

2S=(60+M)g.(1)2S=(60+M)g.\tag{1}

Taking moments about AA, using x=19a/4x=19a/4,

60g(19a4)+Mg(2a)= Sa+S(6a),285+2M= 7Sg.\begin{align*} 60g\left(\frac{19a}{4}\right)+Mg(2a)=&\,Sa+S(6a),\\ 285+2M=&\,\frac{7S}{g}.\tag{2} \end{align*}

Substituting S/g=(60+M)/2S/g=(60+M)/2 from (1) into (2),

285+2M= 72(60+M),570+4M= 420+7M,M= 50.\begin{align*} 285+2M=&\,\frac72(60+M),\\ 570+4M=&\,420+7M,\\ M=&\,\boxed{50}. \end{align*}