Question

A crate is supported by three cables as shown. If tension in cable AC (force C) is 500+ xx N, where xx is the last two digits of your ID. Determine the tension developed in cable AB (force B) and tension in cable AD (force D) and the weight of the crate W at point A (force W). If your student ID is 60-xxxxxx-xxx1-2, C = 500 + 12 = 512 N. В. 36 cm 40 cm 32 cm 27 cm 60 cm

Answer 1

AY Solution B (-3600, -27) D(40,0-27) c(0,0,32) 7x Z A(0,-60, 0) V are as follow The position vectors AB = (-36-0) Î + (0 +60) ĵ + (-27-0) Ê AB = -36 i +60 § 27 Ř (cm) Hence, ABx = - 36 ; ABy = 60 ; A B ₂ = -27 Now, Ac = (0-0) | +60+60) Î +(12-0) AC = o i +60 7 +² + 32 (cm) Hence, ACx=0 , ACY = 60 ; Acz - 32 wow. AD = (40-0) î +(0+60) Î +(-27-0) Ê AD 40 î +60 9 - 21 Â (cm) Hence, ADR = 40 ; Ady = 60 ; AD₂ = -27

grow, Cable Length of is given by AB AB = -36)'+ (60)+(-27,2 - 75 cm Length of Cable Ac is given by AC = 02 + (60)?+ (32)2 68 cm Length of Cable AD is given by AD - 40? + 60?+ (-27)2 Il 77 cm 60 i + 32 Now, FAC 512. FAC AC 11 AC (68) Face 451.8 ģ 4.51.8 ſ + 240.96 Â AB -36 i +60 î - 27 FAB Also, FAB il 1 FAB AB 75 . 11 -0.48 FAB ARE + 0.8 FAB ŷ - 0.36 FAIS 40 i +60 7-21 AD 40 Also FAD FAR FAD 1 АР 77 0.519 AD ? + 0.779 FAD # 0.355k - 'AD. Tue 13 Also, Ñ =-W

E Fx = 0 -0.48 FAB +0.519 AD o { Fy = 0 45108 + 0.8 FAB + 0.779 FAD - W = Fz-O 240.96 -0.36 FAB -0-35 FAD = O 3 on solving Eq" o @f 3 we get. FAB = 352 .4 N 325.9 987.66 N