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

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter9: Moments And Products Of Inertia Of Areas

Section: Chapter Questions

Problem 9.60P: Determine Iu for the inverted T-section shown. Note that the section is symmetric about the y-axis.

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