QUESTION 5 The uniform lamina shown in the diagram is formed by removing a square DEFG of side length a from the equilateral triangle ABC of side length Sa. The centre of mass of DEFG lies on the perpendicular bisector of AC Given that AC and DG are parallel and a distance a apart, work out the dist of the centre of mass of the whole lamina from B. O 0.898 O 1.89a O 289a O 3.89a W Sa
QUESTION 5 The uniform lamina shown in the diagram is formed by removing a square DEFG of side length a from the equilateral triangle ABC of side length Sa. The centre of mass of DEFG lies on the perpendicular bisector of AC Given that AC and DG are parallel and a distance a apart, work out the dist of the centre of mass of the whole lamina from B. O 0.898 O 1.89a O 289a O 3.89a W Sa
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Transcribed Image Text:3:28
QUESTION 5
The uniform lamina shown in the diagram is formed by
removing a square DEFG of side length a from the equilateral
triangle ABC of side length Sa. The centre of mass of DEFG
lies on the perpendicular bisector of AC Given that AC and
DG are parallel and a distance a apart, work out the dist :
of the centre of mass of the whole lamina from B.
O 0.89a
O 1.89a
O 2.893
O 3.89a
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