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CIE 9231 2025 November Paper 34 Q3

A Level / CIE / FM

CIE 9231 2025 Nov Paper 34 Paper · Question 3

题目

Problem

The lamina BFDEBFDE is obtained by removing triangles AEDAED and BCFBCF from a uniform square lamina ABCDABCD of side 2a2a. The length of side AEAE is aa and the length of side FCFC is hh (see diagram). The centre of mass of BFDEBFDE is at a distance xˉ\bar{x} from ADAD, and at a distance yˉ\bar{y} from ABAB.

(a) Show that

xˉ=h26ah+11a23(3ah)\bar{x}=\frac{h^2-6ah+11a^2}{3(3a-h)}

and find a corresponding expression for yˉ\bar{y}.

(5)

(b) The lamina BFDEBFDE is placed vertically on its edge EBEB on a smooth horizontal surface.

Find, in terms of aa, the set of possible values of hh for which the lamina remains in equilibrium.

(2)
题目中文翻译

薄片 BFDEBFDE 由边长为 2a2a 的均匀正方形薄片 ABCDABCD 去除三角形 AEDAEDBCFBCF 得到。边 AEAE 的长度为 aa,边 FCFC 的长度为 hh(见图)。BFDEBFDE 的质心距 ADAD 的距离为 xˉ\bar{x},距 ABAB 的距离为 yˉ\bar{y}

(a) 证明

xˉ=h26ah+11a23(3ah)\bar{x}=\frac{h^2-6ah+11a^2}{3(3a-h)}

并求 yˉ\bar{y} 的相应表达式。

(b) 将薄片 BFDEBFDE 竖直放置在光滑水平面上的边 EBEB 上。

aa 表示使薄片保持平衡的 hh 的可能取值集合。

解答