How did they get the Fab and Fbc?

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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How did they get the Fab and Fbc?
= 0 + FAB cos (45.91) + FBC cos (37.75) =0
=0.6958 FAB + 0.7907FBC =0
ŹFy = 0
ZFx = 0 - FAB sin (45.91) + FBC sin (37.75) + P = 0
= 0.7182 FAB
0.6122 FBC = P
we solve equations
we get
FAR= 0.7956P and FBC = -0.7P
D
142
Transcribed Image Text:= 0 + FAB cos (45.91) + FBC cos (37.75) =0 =0.6958 FAB + 0.7907FBC =0 ŹFy = 0 ZFx = 0 - FAB sin (45.91) + FBC sin (37.75) + P = 0 = 0.7182 FAB 0.6122 FBC = P we solve equations we get FAR= 0.7956P and FBC = -0.7P D 142
Messenger
ne/Downloads/SAPLE-Q-AND-ANSWER.pdf
4,3
3
E
equilibrium conditions
$
A"
2Fy=0
<-40
SAPLE-Q-AND-ANSWER.pdf X +
1=0 + FAB cos (45.91) + FBC Cos (37.75)=0
= 0.6958 FAB + 0.7907 FBC =0
R
=
ZFx=0-FAB sin (45.91) + FBC sin (37.75) +.P =0
0.6122 FBC = P.
=P-
(2)
Calculation
LL
we solve equations
we get
FAB 0.7956P and FBC = -0.7P
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Transcribed Image Text:Messenger ne/Downloads/SAPLE-Q-AND-ANSWER.pdf 4,3 3 E equilibrium conditions $ A" 2Fy=0 <-40 SAPLE-Q-AND-ANSWER.pdf X + 1=0 + FAB cos (45.91) + FBC Cos (37.75)=0 = 0.6958 FAB + 0.7907 FBC =0 R = ZFx=0-FAB sin (45.91) + FBC sin (37.75) +.P =0 0.6122 FBC = P. =P- (2) Calculation LL we solve equations we get FAB 0.7956P and FBC = -0.7P % 5 F S 0.7182 FAB F4 6/24 - | For A T G 6 P, - B 1 & 2 & Y 148% + a-can 7 H FB A U 8 3.5) - 31.10 F> RO 1 U 9 K D > L FO 0/8 P : C { 2
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