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
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf17c1c9-1319-4132-9da8-63907ba88f8e%2F58d1c9a9-a0af-47fe-9413-02de9cb29072%2F5ynkj5r_processed.png&w=3840&q=75)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf17c1c9-1319-4132-9da8-63907ba88f8e%2F58d1c9a9-a0af-47fe-9413-02de9cb29072%2Fmgxnhw3_processed.png&w=3840&q=75)
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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