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PROBLEM 3: Consider the following cable system: cables AE, BE, CE, and DE intersect at point E. The coordinates of A, B, C and E are given: A (-2, 3, 2) m B (0, -2,5) m C (-3, 3, -1) m E (4, Y, 1) m Projection angles of D are 0 = 24°, and a = 10°, and distance L = 6 m. The length of cable AE is 4.8 m. Note the at D is on plane x-z. The forces in each cable are: EA = 118 KN EB = 98 KN EC = 95 KN ED = 102 KN

PROBLEM 3: Consider the following cable system: cables AE, BE, CE, and DE intersect at point E. The coordinates of A, B, C and E are given: A (-2, 3, 2) m B (0, -2,5) m C (-3, 3, -1) m E (4, Y, 1) m Projection angles of D are 0 = 24°, and a = 10°, and distance L = 6 m. The length of cable AE is 4.8 m. Note the at D is on plane x-z. The forces in each cable are: EA = 118 KN EB = 98 KN EC = 95 KN ED = 102 KN

Elements Of Electromagnetics
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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Calculate the x-component of force EA and the y-component of the resultant

PROBLEM 3: Consider the following cable system: cables AE, BE, CE, and DE intersect at point E. The coordinates of A, B, C and E are given: A (-2, 3, 2) m B (0, -2,5) m C (-3, 3, -1) m E (4, Y, 1) m Projection angles of D are 0 = 24°, and a = 10°, and distance L = 6 m. The length of cable AE is 4.8 m. Note the at D is on plane x-z. The forces in each cable are: EA = 118 KN EB = 98 KN EC = 95 KN ED = 102 KN
Transcribed Image Text:PROBLEM 3: Consider the following cable system: cables AE, BE, CE, and DE intersect at point E. The coordinates of A, B, C and E are given: A (-2, 3, 2) m B (0, -2,5) m C (-3, 3, -1) m E (4, Y, 1) m Projection angles of D are 0 = 24°, and a = 10°, and distance L = 6 m. The length of cable AE is 4.8 m. Note the at D is on plane x-z. The forces in each cable are: EA = 118 KN EB = 98 KN EC = 95 KN ED = 102 KN
A(-2,3,2)m B(0,-2,5)m C(-3,3,-1) m € (4, Y₁1) m lution! Projection Angle 0 = 24° a = 10° Distance, L = com Length cable AE=4.8m y EBC EC= ED=
Transcribed Image Text:A(-2,3,2)m B(0,-2,5)m C(-3,3,-1) m € (4, Y₁1) m lution! Projection Angle 0 = 24° a = 10° Distance, L = com Length cable AE=4.8m y EBC EC= ED=
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