Can you please show me the full solution on how the tensions were taken in detail. Thank you. Substituting into EF = 0, factoring i, j, k, and setting each coefficient equal to gives:4812i: P=- T +53j:k:12AB TAC +240 N5319533T. +AB TAC=180 N1313TAB413TAC = 55 NSolving the system of linear equations using conventional algorithms gives:TAR=446.71 NTAC = 341.71 N

Given equations, P=4853TAB+1213TAC+240 .......... (1)1253TAB+313TAC=180…

Answered: Substituting into 2F = 0, factoring i,…

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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Substituting into 2F = 0, factoring i, j, k, and setting each coefficient equal to gives: 48. i: P= TAB + +TAC 53 13 3 +TAC=180 N 13 j: k: 53 19. 53 TAC +240 N AR TAC=55 N 13 Solving the system of linear equations using conventional algorithms gives: TAR=446.71 N TAC = 341.71 N