Q. 8.) ZV=0 RA+RI = L0+20 +20+20 20+20 +20+20十 ZMA=0 20x3 +2OX6+20x9+20x12. +20 X 1S +20x18+20x21 +10X 24 = RqX2Y RI=80RNし) RA = 80RN(↑) 20RN 20RN
Q. 8.) ZV=0 RA+RI = L0+20 +20+20 20+20 +20+20十 ZMA=0 20x3 +2OX6+20x9+20x12. +20 X 1S +20x18+20x21 +10X 24 = RqX2Y RI=80RNし) RA = 80RN(↑) 20RN 20RN
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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What are you guys doing I gave you just one simple task, you just have to copy paste the given photos onto the white page and upload ( only copy paste) pls do that I'll UPVOTE for sure pls diagram clear pls do it man, u dont have to find anything it's not against honor code violation u can do pls
Transcribed Image Text:Joint A
FAB sin o=70
FAB = 218.75 KN (C)
Joint B
e20RN
fBc = 218.45-6.4
218.719
FBC
80
= 212.35 KN(C)
FABCOs
> FAJ
FAJ =207.8 kN (T)
FBJ
19RN (C)
Joint J
1990-0
90-e
Joint C
by substitution method
fje=30.08 RNCT)
FJK =17.66 RN(T)
2078
Joint K
17.16
FKLXO.6
→FKLXO.8=22-84
6.4 19 , K
17766
2288
177-64
212. 85E
FDE =170.95RN(C)
FKL X0.6 = 17.16
FKL= 28-6 RN
28.5 30.08
FKN = 131.9 kN
fcp =177.35RNCC)
%3D
FCK
= 28.6 KN ((C)
FCL = B0.08 RNCT)
Transcribed Image Text:Q. 8.)
ZH=0 =DHA=0
ZV=0
Re+RI = 10+20 +20+20+20 +20 +20+20 t
ZMA=0
20x3 +2OX6 +20x9 +20x12. +20 X 1s +20x18+20x21 +10X 24 = RqX2Y
RI=8ORN()
RA = 80RN(↑)
20RN
20RN
20KN
20RN
ER
20kN
20PN
Vo 1705
EO
17 구.35
20kN
1ORN
55.7
LORN
G 212.35CT) H 218.7SIT)
3008
2213
28
286 C)
207.8
194-4 K
+319(T)
R
ST).
LT)
tane = W12
RA
e - tan- (0.33) = 18. 43°
sin 9 = 032
cosO = O.95
7.66
207.8LT)
4EQ = AE
AE COsO= AS
AP = 3m
sin 20= 0.6
coo 29 =0.8
EQ = 3-16m
24m
9-434
80-08
28.6
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