The three cables are used to support the 40kg flowerpot Dete

The three cables are used to support the 40-kg flowerpot. Determine the force developed in each cable for equilibrium.

AC is along +ve x axis.

Fac = Tac i

Ab is along +ve y axis.

Fab = Tab j

AD is in the direction of -1.5i - 2j + 1.5k

unit vector = (-1.5i - 2j + 1.5k ) / sqrt(1.5^2 + 2^2 + 1.5^2)

= - 0.514i - 0.686j + 0.514k

Fad = Tad ( - 0.514i - 0.686j + 0.514k )

and gravitation force, Fg = 40g (-k)

Fg = - 392k N

balancing force,

Fnet = Fab + Fad +Fac + mg = 0

Tac i + Tab j + Tad ( - 0.514i - 0.686j + 0.514k ) - 392k = 0

Tac - 0.514Tad = 0

Tab - 0.686Tad = 0

0.514Tad - 392 = 0

Tad = 762.6 N ........Tension in AD cable.

Tab = 0.686 x 762.6 = 523.17 N ...Tension in AB

Tac = 0.514 x 762.6 = 392 N ---- Tension in AC