In the loaded system shown find the tension in cable EG Assu

In the loaded system shown, find the tension in cable EG. Assume that the coordinates of point G are x=0, y=15 ft, and z=0 (G is on xy plane). BC is parallel to x-axis, DC is parallel to a-axis, an AB is parallel to y-axis.

Solution

>> Writing coordinates of point E

E = (6, 12, 0) ft

as, G = (0, 15, 0)

>> and, C = (12, 12, 0)

>> Let Tension in cable EG =T and it is directed along EG

As, EG = - 6 i + 3 j

So, Unit Vector along EG = [ -6 i + 3 j ]/[6² + 3²]^1/2 = - 0.894 i + 0.447 j

=> Teg = T( - 0.894 i + 0.447 j)

As, W = -450*32.2 j = - 14490 j

Now, as system is in equilibrium

So, Moment about A, Ma = 0

=> (AE) X (Teg) + (AC) X (W) = 0

=> [ (6 i + 12 j) X T( - 0.894 i + 0.447 j) ] + [ (12 i + 12 j ) X (- 14490 j) ] = 0

=> 2.682*T k + 10.728*T k - 12*14490 k = 0

=> T = 12966.443 lbf .....ANSWER......