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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