题目
Problem
6.
The uniform lamina PQRSTUV shown in Figure 2 is formed from two identical rectangles, PQUV and QRSTU.
The rectangles have sides PQ=RS=2a and PV=QR=ka.
(a) Show that the centre of mass of the lamina is (46+k)a from PV
(3)
The lamina is freely suspended from P and hangs in equilibrium with PR at an angle of α to the downward vertical.
Given that tanα=157
(b) find the value of k.
(6)
题目中文翻译
图 2 所示的均匀薄片 PQRSTUV 由两个全等长方形 PQUV、QRSTU 组成。
长方形的边满足 PQ=RS=2a,PV=QR=ka。
(a) 证明薄片的质心到 PV 的距离为 (46+k)a。
(3)
薄片从 P 点自由悬挂并处于平衡状态,PR 与竖直向下方向成 α 角。
已知 tanα=157,
(b) 求 k 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 PV 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 PV towards R. The centroids of PQUV and QRSTU are at distances
a,2a+2ka
from PV. The rectangles are identical and the lamina is uniform, so their masses are equal. Let each have mass M.
Taking moments about PV,
2Mxˉ=M(a)+M(2a+2ka).
Thus
xˉ=21(3a+2ka)=46+ka.
(b)
解法一
思路
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还要求质心到原图中水平线 PR 的向下距离。自由悬挂时 PG 竖直向下,PR 与竖直的夹角就是原图中 PR 与 PG 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 G be the combined centre of mass and let yˉ be its distance below PR in Figure 2. The centroids of PQUV and QRSTU are at distances ka/2 and a below PR.
Taking moments about PR,
2Myˉ=M(2ka)+M(a),
so
yˉ=4k+2a.
When suspended from P, the line PG is vertically downwards. Since PR is horizontal in the original figure, the angle between PR and PG satisfies
tanα=xˉyˉ.
Therefore
157=k+6k+2.
Cross-multiplying,
7k+42=15k+30,
and hence
k=23.