The h 120 ft tall communication tower is supported by three cables at A and a ball-and-socket support atO. Cable AB is pre-tensioned with a magnitude TAB= 6 kips. The coordinates of points B, C, and D in theground (x, y) plane are:B = (25, 30) ftC=(-15, 25) ftD = (10, −20) ftDetermine the magnitude of the tension in cables AC and AD and the normal force in tower AO.••TACTADPAO=011.7301kipskips → xoxkipsC

To find: The tension in cable AC and AD along with the force on the tower AO. Given: The tension in the cable AB, TAB =6 kips. The coordinates of the cable where it is mounted: B = (25, 20) ftC = (-15, 25) ftD = (10, -20) ft The height of the tower, h = 120 ft. Formula used: The Lami's theorem is: F1sin(a) =F2sin(b) = F3sin(c) The resultant force due to force on three different planes: R = F12+F22+F32 where, F1 is the force acting on x-y plane, F2 is the force acting y-z plane and F3 is the force acting on the z-x plane.The shortest distance between the cable mounting point and the point O in the tower: From ∆ OLB, OB = 252+302 = 43.82 ft. Similarly from ∆OKC and ∆OMD, OC = 252+152 = 29.15 ft.OD = 102+202 = 22.36 ft. The respective angle with which the points B, C and D are mounted with respect to point A are as follows: From ∆AOB, cos(θ) = 1201202+43.822θ = 20.06o Similarly from ∆AOD and ∆AOC cos(α) = 1201202+22.362α = 10.55o cos(∅) = 1201202+29.152∅ = 13.65o The angle at which the tension due to the cable are oriented with respect to A: tan(P) = 1020P = 26.56otan(R) = 2530R = 39.86otan(Q) = 1525Q = 30.96oThe tension in the cable AC and AD are as follows: AD×sin(10.55)sin(90-30.96+39.86) = AB×sin(20.06)sin(90+26.56+30.96)=AD×sin(13.65)sin(180-26.56-39.86)AD×0.183sin(98.9) = 6×0.343sin(156.36)=AD×0.236sin(98.9) From this the tension in cables are as follows: TAB =6 kips.TAD =27.70 kips.TAc =19.99 kips.The resultant value of the above forces will give the force acting on the communication tower: R = 62 + 27.7072 + 19.992 = 34.68 kips