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M2 2022 Oct Q6

A Level / Edexcel / M2

IAL 2022 Oct Paper · Question 6

题目

Problem

6.

The uniform lamina PQRSTUVPQRSTUV shown in Figure 2 is formed from two identical rectangles, PQUVPQUV and QRSTUQRSTU. The rectangles have sides PQ=RS=2aPQ=RS=2a and PV=QR=kaPV=QR=ka.

(a) Show that the centre of mass of the lamina is (6+k4)a\big(\frac{6+k}{4}\big)a from PVPV

(3)

The lamina is freely suspended from PP and hangs in equilibrium with PRPR at an angle of α\alpha to the downward vertical.

Given that tan⁡α=715\tan\alpha=\frac7{15}

(b) find the value of kk.

(6)
题目中文翻译

图 2 所示的均匀薄片 PQRSTUVPQRSTUV 由两个全等长方形 PQUVPQUV、QRSTUQRSTU 组成。 长方形的边满足 PQ=RS=2aPQ=RS=2a,PV=QR=kaPV=QR=ka。

(a) 证明薄片的质心到 PVPV 的距离为 (6+k4)a\big(\frac{6+k}{4}\big)a。

(3)

薄片从 PP 点自由悬挂并处于平衡状态,PRPR 与竖直向下方向成 α\alpha 角。

已知 tan⁡α=715\tan\alpha=\frac7{15},

(b) 求 kk 的值。

(6)

解答

(a)

解法一

思路

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两个全等长方形质量相同,所以组合质心坐标就是两个长方形中心坐标的平均值。先以 PVPV 为竖直参考轴,求到它的水平距离。

答题过程

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Measure horizontal distances from PVPV towards RR. The centroids of PQUVPQUV and QRSTUQRSTU are at distances

a,2a+ka2a,\qquad2a+\frac{ka}{2}

from PVPV. The rectangles are identical and the lamina is uniform, so their masses are equal. Let each have mass MM.

Taking moments about PVPV,

2Mxˉ=M(a)+M(2a+ka2).2M\bar x=M(a)+M\left(2a+\frac{ka}{2}\right).

Thus

xˉ=12(3a+ka2)=6+k4a.\bar x=\frac12\left(3a+\frac{ka}{2}\right)=\boxed{\frac{6+k}{4}a}.

(b)

解法一

思路

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还要求质心到原图中水平线 PRPR 的向下距离。自由悬挂时 PGPG 竖直向下,PRPR 与竖直的夹角就是原图中 PRPR 与 PGPG 的夹角,因此用向下距离除以向右距离。

答题过程

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Let GG be the combined centre of mass and let yˉ\bar y be its distance below PRPR in Figure 2. The centroids of PQUVPQUV and QRSTUQRSTU are at distances ka/2ka/2 and aa below PRPR.

Taking moments about PRPR,

2Myˉ=M(ka2)+M(a),2M\bar y=M\left(\frac{ka}{2}\right)+M(a),

so

yˉ=k+24a.\bar y=\frac{k+2}{4}a.

When suspended from PP, the line PGPG is vertically downwards. Since PRPR is horizontal in the original figure, the angle between PRPR and PGPG satisfies

tan⁡α=yˉxˉ.\tan\alpha=\frac{\bar y}{\bar x}.

Therefore

715=k+2k+6.\frac7{15}=\frac{k+2}{k+6}.

Cross-multiplying,

7k+42=15k+30,7k+42=15k+30,

and hence

k=32.\boxed{k=\frac32}.