F6-20. Determine the reactions at D. 4 m B 10 kN 3 m3 m3 -3 m m3 m3 m -3m- 15 kN Prob. F6-20 C D
F6-20. Determine the reactions at D. 4 m B 10 kN 3 m3 m3 -3 m m3 m3 m -3m- 15 kN Prob. F6-20 C D
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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![F6-20. Determine the reactions at D.
4 m
B
10 kN
3 m3 m3
-3
m
m3 m3 m
-3m-
15 kN
Prob. F6-20
C
D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F652e5396-12b6-4e5e-865e-eecd6b9e3fee%2F3a3ca46a-913c-4c08-b31c-3401f9ab4909%2Fqzxcktd_processed.png&w=3840&q=75)
Transcribed Image Text:F6-20. Determine the reactions at D.
4 m
B
10 kN
3 m3 m3
-3
m
m3 m3 m
-3m-
15 kN
Prob. F6-20
C
D
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Follow-up Question
In my textbook these are the correct answer but all are positive, why is that I thought we can assume the the forces.
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Follow-up Question
This is how I solved for reaction D, is this also correct?
![Problem (6-20)
LOKN
FAB (35)
FAB
3m
√ ↑
FAB (5)
2Fx= FAB (2) + (x = 0
4m
Dx
3m
15KN
12Fy= FAB (4/5)-10 KN-15KN + Cy=0
(Mc = (3m 15KN) + (6m 10 KN) - (9m. FAB (45)) -0
.
FAB= 14.583 KN
13.333
Dy
3m
ум
Cx=-(14.583 HN) (²) -- 8.749 KN
(y=-14.583 (4) KN + 10 + 15 KN = 13.333 KN
F37
→8.749 KN
cy
Ax
ZMD= M-(4m 8.749 KN) = 0
ZFx=Dx+8.749 KN = 0
Fy=-13.333 + Dy = ²
DX = -8.749 KN
Dy = 13.333 KN
M= 34.996 KN.m](https://content.bartleby.com/qna-images/question/652e5396-12b6-4e5e-865e-eecd6b9e3fee/e5dfe3fa-55c8-48bf-aa47-44feaee62f05/8toi3yo_thumbnail.png)
Transcribed Image Text:Problem (6-20)
LOKN
FAB (35)
FAB
3m
√ ↑
FAB (5)
2Fx= FAB (2) + (x = 0
4m
Dx
3m
15KN
12Fy= FAB (4/5)-10 KN-15KN + Cy=0
(Mc = (3m 15KN) + (6m 10 KN) - (9m. FAB (45)) -0
.
FAB= 14.583 KN
13.333
Dy
3m
ум
Cx=-(14.583 HN) (²) -- 8.749 KN
(y=-14.583 (4) KN + 10 + 15 KN = 13.333 KN
F37
→8.749 KN
cy
Ax
ZMD= M-(4m 8.749 KN) = 0
ZFx=Dx+8.749 KN = 0
Fy=-13.333 + Dy = ²
DX = -8.749 KN
Dy = 13.333 KN
M= 34.996 KN.m
Solution
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