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FA 400kN Fe 1-5 {Fy=0 FA+ Fc + FB = 400 X 10²³ 2FA +Fc = 400×10 (-SA) + (SA) T For exa ZM₂ = 0 FAX +-5 = √₂x1-5 FA = Fg p → [= {a) + (Sal+ = (-6) + (60) - (SA)T - • (-5B)₂ + (√₂³)₁ = (-Sc)p+ (de)s - Fala + LAATlA = -Felo + Leste le RE स्मृ id AT 0

FA 400kN Fe 1-5 {Fy=0 FA+ Fc + FB = 400 X 10²³ 2FA +Fc = 400×10 (-SA) + (SA) T For exa ZM₂ = 0 FAX +-5 = √₂x1-5 FA = Fg p → [= {a) + (Sal+ = (-6) + (60) - (SA)T - • (-5B)₂ + (√₂³)₁ = (-Sc)p+ (de)s - Fala + LAATlA = -Felo + Leste le RE स्मृ id AT 0

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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what are the contents of the circled formula and what is it called?

and what is the relationship between δp and δt, why are the solutions different in the following line? (arrows)

FA 400KN Fe 1.5 Ебу=0 FA+ F₁+F₂ = 400 X 10²³ 2 FA +Fc 400×10 = For equ еди ZM₂ = 0 - Fala + LA ATALA = RE FAX1.5 = ₂x1.5 FA = FB (-SA) ₁ + (SA) = (-SB)₂ + (Se ³) = (-Sc)p+ (Sc) => [= √₂)₁ + (Sa) = (-6) + (60) - p 2 - -Fel + Leste le AEE & AT
Transcribed Image Text:FA 400KN Fe 1.5 Ебу=0 FA+ F₁+F₂ = 400 X 10²³ 2 FA +Fc 400×10 = For equ еди ZM₂ = 0 - Fala + LA ATALA = RE FAX1.5 = ₂x1.5 FA = FB (-SA) ₁ + (SA) = (-SB)₂ + (Se ³) = (-Sc)p+ (Sc) => [= √₂)₁ + (Sa) = (-6) + (60) - p 2 - -Fel + Leste le AEE & AT
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