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If you wanted to continue the problem in the last video to find D₂, what would have been the easiest way to go about doing this? Sum moments about BD Sum forces in the z direction Sum forces in the x direction Sum moments about AB Sum moments about z axis Sum forces in the y direction 0 = - \ (490.5) + 1 TBC TBC = 491 N C EMO=0 B 1 m 0.5 m 0.5 m E Z. A D EMD=O ΣMoy=0 EM D₂ = 0 Dz x 100 kg

If you wanted to continue the problem in the last video to find D₂, what would have been the easiest way to go about doing this? Sum moments about BD Sum forces in the z direction Sum forces in the x direction Sum moments about AB Sum moments about z axis Sum forces in the y direction 0 = - \ (490.5) + 1 TBC TBC = 491 N C EMO=0 B 1 m 0.5 m 0.5 m E Z. A D EMD=O ΣMoy=0 EM D₂ = 0 Dz x 100 kg

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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If you wanted to continue the problem in the last video to find D₂, what would have been the easiest way to go about doing this? Sum moments about BD Sum forces in the z direction Sum forces in the x direction Sum moments about AB Sum moments about z axis Sum forces in the y direction
Transcribed Image Text:If you wanted to continue the problem in the last video to find D₂, what would have been the easiest way to go about doing this? Sum moments about BD Sum forces in the z direction Sum forces in the x direction Sum moments about AB Sum moments about z axis Sum forces in the y direction
0 = - \ (490.5) + 1 TBC TBC = 491 N C EMO=0 B 1 m 0.5 m 0.5 m E Z. A D EMD=O ΣMoy=0 EM D₂ = 0 Dz x 100 kg
Transcribed Image Text:0 = - \ (490.5) + 1 TBC TBC = 491 N C EMO=0 B 1 m 0.5 m 0.5 m E Z. A D EMD=O ΣMoy=0 EM D₂ = 0 Dz x 100 kg
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