Solution: b) ΣΜ -0 Forx AF = 0 DF FOr = 0 Answer: ΣFy - 0 T sin 300- 4 = 0 T = 8 kN 30 a) FG and AG 2 m Σ Fx-0 T cos 300 + FAR + 8 (cos 300) + FAB+ 0 = 0 DF Foe 3 kN 4 kN Answer: 4 kN 2 m 2 m 2 m FAB= -6.928 kN IDENTIFY THE ZERO FORCE MEMBERS AT THE TRUSS. B C ) Σ Fx 0 FBcsin 45 - FAB= 0 FBcsin 45 - (-6.928) = 0 DETERMINE THE FORCE AT MEMBER AB USING FAB METHOD OF SECTION AND THE FORCE AT MEMBER BC USING METHOD OF JOINTS. Required: F BC= - 6.928 450 sin 45 a) Zero Force members b) Force at AB c) Force at BC Activate Windows Go to Setigs to activate Windows. Answer: FBc= -9.797 kN FBD BC
Solution: b) ΣΜ -0 Forx AF = 0 DF FOr = 0 Answer: ΣFy - 0 T sin 300- 4 = 0 T = 8 kN 30 a) FG and AG 2 m Σ Fx-0 T cos 300 + FAR + 8 (cos 300) + FAB+ 0 = 0 DF Foe 3 kN 4 kN Answer: 4 kN 2 m 2 m 2 m FAB= -6.928 kN IDENTIFY THE ZERO FORCE MEMBERS AT THE TRUSS. B C ) Σ Fx 0 FBcsin 45 - FAB= 0 FBcsin 45 - (-6.928) = 0 DETERMINE THE FORCE AT MEMBER AB USING FAB METHOD OF SECTION AND THE FORCE AT MEMBER BC USING METHOD OF JOINTS. Required: F BC= - 6.928 450 sin 45 a) Zero Force members b) Force at AB c) Force at BC Activate Windows Go to Setigs to activate Windows. Answer: FBc= -9.797 kN FBD BC
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
Related questions
Question
Can you please explain the solution and how the answer is obtained?
![Solution:
b) ΣΜ 0
AF = 0
FOr
DF
300
%3D
Answer:
DE = 0
ΣFy -0
T sin 30°- 4 = 0
T = 8 kN
30
a) FG and AG
2 m
Σ Fx-0
T cos 300 + FAR + F=0
8 (cos 300) + FAB + 0 = 0
АВ
DF
FOF E
F
3 kN
4 kN
Answer:
4 kN
2 m
2 m
2 m
FAB= -6.928 kN
IDENTIFY THE ZERO FORCE MEMBERS AT THE TRUSS.
В
C ) ΣFx =0
FBcsin 45 - FAB= 0
FBcsin 45 - (-6.928) = 0
F BC= - 6.928
DETERMINE THE FORCE AT MEMBER AB USING
FAB
METHOD OF SECTION AND THE FORCE AT MEMBER
BC USING METHOD OF JOINTS.
Required:
45°
sin 45
a) Zero Force members
b) Force at AB
c) Force at BC
Activate Windows
Go to Settings to activate Windows.
Answer:
FBc= -9.797 kN
FBD
BC](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F137f3595-e04a-40b2-9581-70171dc25cc1%2Fa6d7188a-e8d1-4028-a319-6c367138e3bc%2Fveymly_processed.png&w=3840&q=75)
Transcribed Image Text:Solution:
b) ΣΜ 0
AF = 0
FOr
DF
300
%3D
Answer:
DE = 0
ΣFy -0
T sin 30°- 4 = 0
T = 8 kN
30
a) FG and AG
2 m
Σ Fx-0
T cos 300 + FAR + F=0
8 (cos 300) + FAB + 0 = 0
АВ
DF
FOF E
F
3 kN
4 kN
Answer:
4 kN
2 m
2 m
2 m
FAB= -6.928 kN
IDENTIFY THE ZERO FORCE MEMBERS AT THE TRUSS.
В
C ) ΣFx =0
FBcsin 45 - FAB= 0
FBcsin 45 - (-6.928) = 0
F BC= - 6.928
DETERMINE THE FORCE AT MEMBER AB USING
FAB
METHOD OF SECTION AND THE FORCE AT MEMBER
BC USING METHOD OF JOINTS.
Required:
45°
sin 45
a) Zero Force members
b) Force at AB
c) Force at BC
Activate Windows
Go to Settings to activate Windows.
Answer:
FBc= -9.797 kN
FBD
BC
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