Find the resultant of the force vectors in the following images. In each case, finding the Force vector in Cartesian form (x, y, and z components) is totally fine. 1. . A FB = 840 N 6 m Fc = 420 N 3 m B 2 m 3 m 2 m 2. . 2 ft Fc = 490 lb FB = 600 lb 6 ft 4 ft 3 ft B 4 ft 2 ft 4 ft
Find the resultant of the force vectors in the following images. In each case, finding the Force vector in Cartesian form (x, y, and z components) is totally fine. 1. . A FB = 840 N 6 m Fc = 420 N 3 m B 2 m 3 m 2 m 2. . 2 ft Fc = 490 lb FB = 600 lb 6 ft 4 ft 3 ft B 4 ft 2 ft 4 ft
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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![= 360i-240j-720k+ 1201'+180oJ-360k
Position d points
A C0,0,6)m , B(3,-2,0)m cC2,3,0)m
民= (480t- 60] ー1080k) N
AB = e-0) i+(-2-j+ Co-6)K
=(31-25-6K) m
(岡
%3D
The coordehatos o A(2,0,6) ft, B(0,4,22ft, CE4,314)ft
R=(0-2)「+(4-のJ+(2-6)k
ミ 7m
ニ
盛,3-2j-6K
1砲
MAB =
=(-21+4)-4k)ft
ニ
FAB UAB =
840N (31-2)
UAB =
-2i+43-4K
(3601- 240) -720k) N
- FAB UAB = 600 Ub/
AC = 2-01+C3 -0)+(0-6)K
=2i+3j -6k)m
=E2001'+400)–40ok) lb
AC = -4-2)i+e-oj+C4OK
= -6l+3)-2k.
応- J-6)+4 -2 =7
= 7m
ニ
%3D
UAC =
IRI
2+3j-6K
%3D
礼
= FAC UAC = 420N/
UAC =
bl+3j-2k
- (120i +1 80j –360k) N
R- 品+成
盛= Fhe Uac=
490lb (-6143)-2k)
屋-ヒ42+210) ー4ok) 6
= -2001'+ 400J-400k- 4201+210}-140k
屋--620j+6o) -540k) b](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa658c71c-1a2c-47f6-b23a-f861466702f5%2F0654fc12-c75f-4d13-b825-cc0cf63161b9%2Fdx4y48l_processed.png&w=3840&q=75)
Transcribed Image Text:= 360i-240j-720k+ 1201'+180oJ-360k
Position d points
A C0,0,6)m , B(3,-2,0)m cC2,3,0)m
民= (480t- 60] ー1080k) N
AB = e-0) i+(-2-j+ Co-6)K
=(31-25-6K) m
(岡
%3D
The coordehatos o A(2,0,6) ft, B(0,4,22ft, CE4,314)ft
R=(0-2)「+(4-のJ+(2-6)k
ミ 7m
ニ
盛,3-2j-6K
1砲
MAB =
=(-21+4)-4k)ft
ニ
FAB UAB =
840N (31-2)
UAB =
-2i+43-4K
(3601- 240) -720k) N
- FAB UAB = 600 Ub/
AC = 2-01+C3 -0)+(0-6)K
=2i+3j -6k)m
=E2001'+400)–40ok) lb
AC = -4-2)i+e-oj+C4OK
= -6l+3)-2k.
応- J-6)+4 -2 =7
= 7m
ニ
%3D
UAC =
IRI
2+3j-6K
%3D
礼
= FAC UAC = 420N/
UAC =
bl+3j-2k
- (120i +1 80j –360k) N
R- 品+成
盛= Fhe Uac=
490lb (-6143)-2k)
屋-ヒ42+210) ー4ok) 6
= -2001'+ 400J-400k- 4201+210}-140k
屋--620j+6o) -540k) b

Transcribed Image Text:Find the resultant of the force vectors in the following images. In each case, finding the Force vector in
Cartesian form (x, y, and z components) is totally fine.
1.
A
FB = 840 N
6 m
Fc = 420 N
3 m
B
2 m.
3 m
- 2 m
C
2.
2 ft
Fc=
= 490 lb
Fв 3D 600 Ib
6 ft
3 ft
В
4 ft 2 ft
4 ft
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