MasterinqEngineering MastenngComputerScience: HW 2:CH·3-Google Chrome 0 HW 2: CH
ID: 1578027 • Letter: M
Question
MasterinqEngineering MastenngComputerScience: HW 2:CH·3-Google Chrome 0 HW 2: CH. 3 Problem 3.64 Part A If cable ADis tightened by a tumbuckle and develops a tension of 1500 lb determine the tension developed in cable AßFigure 1) Expreee your answer to two aignificant figurea and include the appropriate units. FAB-1310 lb Submit Pravious Answors Raquest Answar Incorrect; Try Again; 26 attempts remaining Part B Figure 1 of 1 If cable AD is tightened by a tumbuckle and develops tension of 1600 lb , determine the tension developed in cable AC Express your answer to two significant igures and include the appropriate units. 30 ft FAC 994 lb 10 ft Submit Previous Answers Request Answer 15 ft × Incorrect: Ty Again; 28 attempts remaining 12.5 :43 PM 29/2018 3Explanation / Answer
Ans:-
Given data :-
A = 0i + 0j + 30k uEA= 0i + 0j + 1k
B = 10i -15j + 0k
C= -15i -10j + 0k
D = 0i + 12.5j + 0k
E = 0i + 0j + 0k
AB = B-A = 10i -15j -30k
|AB|= sqrt (x^2 + y^2 +z^2) = 35
uAB = AB/|AB|= 0.285i-0.429j-0.357k
AC= -15i +0j -30k
|AC|= 35
uAC = -0.429i -0.215j -0.857k
AD= 0i + 12.5j -30k
|AD|=32.5
uAD = 0i + 0.385j- 0.923k
TAB = |TAB|.uAB = 0.285ABi-0.429ABj-0.357ABk
TAC = |TAC|.uAC = -0.429ACi -0.285ACj -0.857ACk
TAD = |TAD|.uAD = 0i + 0.385ADj – 0.923ADk
FAE = |FAE|.uAE = 0i + 0j + AEk
= 0.285AB - 0.429AC=0………1
-0.429AB-0.285AC + 0.385AD=0…….2
-0.357AB-0.857AC – 0.923AD = 0……….3
AD=1600lb
-0.429AB-0.285AC + 616=0
-0.357AB-0.857AC – 1476.8 =0
Eq 1 multiply by 0.429 and divided by 0.285
0.429AB - 0646AC = 0
+
-0.429AB- 0.285AC = -616
-0.931 AC = -616
FAC= 661.65lb
FAB = 996.34lb