Help please with this problem of statics! thanks a lot Determine the force gener

Question

Help please with this problem of statics! thanks a lot

Determine the force generated on the AB, AC, and AD cables. The tower is undergoing a force F representing the force exerted by an earthquake in two directions along the x-axis and y-axis. The components Es and of the force F are shown in Fig. 3 and are located at the base of the tower. F_x equals 500 kN and F_y is 400 kN. The weight of the tower is equal to 700 kN. The point O has coordinates (0, 0, 0). The height of the tower is equal to 50 m. (Figure 3). Perform the free-body diagram.

Explanation / Answer

given coordinates of point A (xa,ya,za) = (0,0,50)
given coordinates of point B (xb,yb,zb) = (-13,-17,0)
given coordinates of point C (xc,yc,zc) = (0,19,0)
given coordinates of point D (xd,yd,zd) = (18,-35,0)

let the magnitude of tension in wire AB = |T1|, in AC = |T2| , in AD = |T3|
these forces act away from point A along the given directions of wire
so a unit vector along AB = ((xb-xa)i + (yb-ya)j + (zb-za)k)/sqroot((xb-xa)^2 + (yb-ya)^2 + (zb-za)^2) = 0.01838(-13i -17j - 50k)
so a unit vector along AC = ((xc-xa)i + (yc-ya)j + (zc-za)k)/sqroot((xc-xa)^2 + (yc-ya)^2 + (zc-za)^2) = 0.01869(19j - 50k)
so a unit vector along AD = ((xd-xa)i + (yd-ya)j + (zd-za)k)/sqroot((xd-xa)^2 + (yd-ya)^2 + (zd-za)^2) = 0.01571(18i -35j - 50k)

so, T1 = |T1|0.01838(-13i -17j - 50k)
T2 = |T2| 0.01869(19j - 50k)
T3 = |T3|0.01571(18i -35j - 50k)

and given, Fy = 400,000 j
Fx = 500,000 i
let compression in the tower be F = |F|k
so from force balance we get

|T1|0.01838(-13i -17j - 50k) + |T2| 0.01869(19j - 50k) + |T3|0.01571(18i -35j - 50k) + 400,000 j + 500,000 i + |F|k = 0
i(-0.23894|T1| + 0.28278|T3| + 500,000) + j(-0.31246|T1| + 0.35511|T2| - 0.54985|T3| + 400,000) + k(-0.919|T1| - 0.9345|T2| - 0.7855|T3| + |F|) = 0

also, from moment balance about O
T1 x (-50k) + T2 x (-50k) + T3 x (-50 k ) = 0
(T1 + T2 + T3)x (-50k) = 0
(|T1|0.01838(-13i -17j - 50k) + |T2| 0.01869(19j - 50k) +|T3|0.01571(18i -35j - 50k) ) x (-50k) = 0
(-0.23894|T1|i - 0.31246|T1|j - 0.919|T1|k + 0.35511|T2|j - 0.9345|T2|k + 0.28278|T3|i - 0.54985|T3|j - 0.7855|T3|k) x (-50 k) = 0
(i(-0.23894|T1| + 0.28278|T3|) + j(- 0.31246|T1|+ 0.35511|T2| - 0.54985|T3|) + k(- 0.919|T1| - 0.9345|T2| - 0.7855|T3| )) x (-50 k) = 0
(-0.23894|T1| + 0.28278|T3|)*50j + (- 0.31246|T1|+ 0.35511|T2| - 0.54985|T3|)*(-50)i = 0
comparing we get

-0.23894|T1| + 0.28278|T3| = 0
- 0.31246|T1|+ 0.35511|T2| - 0.54985|T3| = 0
-0.23894|T1| + 0.28278|T3| + 500,000 = 0
-0.31246|T1| + 0.35511|T2| - 0.54985|T3| + 400,000 = 0
-0.919|T1| - 0.9345|T2| - 0.7855|T3| + |F| = 0

solving we get
|T1| = 152339.499 N
|T2| = -793056.2986 N
|T3| = 128721.97467 N

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