The tension in cable AD is 570N From the perspective of poin

The tension in cable AD is 570N From the perspective of point A provide the force vector in CVF.

Also give the scalar value listed

360 mm 450 mn P. s : 500 min 320 mm \" . 600 mm

>> Writing all the coordinates :

A = (0, - 600, 0) mm = (0, -0.6, 0) m

B = (450, 0, 0) mm = (0.45, 0, 0) m

C = (0, 0, -320) mm = (0, 0, -0.32) m

D = (- 500, 0, 360) mm = (- 0.5, 0, 0.36) m

>> Now, Let Tensions in cables are : Tab, Tac and Tad

>> Considering Tab

It is acting along AB

and, AB = (0.45 i) - (- 0.6 j) = 0.45 i + 0.6 j

Magnitude = [0.45² + 0.6²]^1/2 = 0.5625

So, Unit Vector along AB = (0.45 i + 0.6 j)/0.5625 = 0.8 i + 1.0667 j

=> Tab = Tab(0.8 i + 1.0667 j)

>> Considering Tac

It is acting along AC

and, AC = (- 0.32 k) - (- 0.6 j) = - 0.32 k + 0.6 j

Magnitude = [0.32² + 0.6²]^1/2 = 0.68

So, Unit Vector along AC = ( - 0.2 k + 0.6 j)/0.68 = 0.882 j - 0.471 k

=> Tac = Tac(0.882 j - 0.471 k)

>> Considering Tad

It is acting along AD

and, AD = (- 0.5 i + 0.36 k) - (- 0.6 j) = - 0.5 i + 0.6 j + 0.36 k

Magnitude = [0.5² + 0.6² + 0.36²]^1/2 = 0.86

So, Unit Vector along AD = ( - 0.5 i + 0.6 j + 0.36 k)/0.86 = - 0.58 i + 0.698 j +0.419 k

=> Tad = Tad(- 0.58 i + 0.698 j +0.419 k)

>> Now, under Equilibrium, Net Forces = 0

=> Tab + Tac + Tad = 0

=> Tab(0.8 i + 1.0667 j) + Tac(0.882 j - 0.471 k) + Tad(- 0.58 i + 0.698 j +0.419 k) = 0

As, Tad = 570 N given

>> Now, Comparing Coefficients,

=> 0.8*Tab - 0.58*Tad = 0 , => Tab = 413.25 N

>> - 0.471*Tac + 0.419*Tad = 0 , => Tac = 507.07 N