B F C A z 5 3 4 Solution Let Tension developed in cables AB

Solution

>> Let Tension developed in cables AB, AC & AD are Tab, Tac & Tad

Resolving each force in vector components

>> Considering AB

Tab = Tab i

>> Considering AC

Tac = - Tac*3/5 i + Tac*4/5 j

=> Tac = Tac(- 0.6 i + 0.8 j)

>> Considering AD

Tad = - Tad*3/5 j + Tad*4/5 k

=> Tad = Tad(- 0.6 j + 0.8 k)

and, F is in vertical downward direction

=> F = - 860 k

>> As, whole system is under equilibrium,

=> F + Tab + Tac + Tad = 0

=> -860 k + Tab i + Tac(- 0.6 i + 0.8 j) + Tad(- 0.6 j + 0.8 k) = 0

>> Comparing coefficients both sides,

=> Tab - 0.6*Tac = 0

, 0.8*Tac -0.6 Tad = 0

& 0.8*Tad - 860 = 0

>> Solving above three equations,

=> Tab = 483.75 N

Tac = 806.25 N

and, Tad = 1075 N ...................ANSWER...........