A crate is hung by ropes attached to a steel ring Assume tha

A crate is hung by ropes attached to a steel ring. Assume that the center of gravity (CG) of the crate shown lies at its geometric center, and that ring A lies directly above the crate\'s CG. The top surface of the crate is horizontal, and ring A lies 50 inches above it. Other dimensions are given as: a = 48 in b = 33 in c = 12 in A load cell measures the tension in rope AC to be TAC = 180 lb. Find the weight of the crate and the tension in the other two ropes.

Solution

>> First, writing all coordinates

B = (0,0,0)

C = (33,12,0)

D = (0,48,0)

G = Center of Gravity of Crate = (16.5,24,0)

A = (16.5,24,50)

>> Considering AB

Vector AB = - 16.5 i - 24 j - 50 k

Unit Vector = ( - 16.5 i - 24 j - 50 k )/[16.5² + 24² + 50²]^1/2 = - 0.285 i - 0.415 j - 0.864 k

=> Tension in AB, Tab = Tab (-0.285 i - 0.415 j - 0.864 k)

>> Considering AC

Vector AC = 16.5 i - 12 j - 50 k

Unit Vector = ( 16.5 i - 12 j - 50 k )/[16.5² + 12² + 50²]^1/2 = 0.306 i - 0.222 j - 0.926 k

=> Tension in AB, Tac = Tac ( 0.306 i - 0.222 j - 0.926 k)

>> Considering AD

Vector AD = - 16.5 i + 24 j - 50 k

Unit Vector = ( - 16.5 i + 24 j - 50 k )/[16.5² + 24² + 50²]^1/2 = - 0.285 i + 0.415 j - 0.864 k

=> Tension in AD, Tad = Tad (-0.285 i + 0.415 j - 0.864 k)

>> Weight of Crate is acting vertically

=> W = - w k

>> Now, as the system is in equilibrium.

=> Tab + Tac + Tad + W = 0

>> Tab (-0.285 i - 0.415 j - 0.864 k) + Tac ( 0.306 i - 0.222 j - 0.926 k) + Tad (-0.285 i + 0.415 j - 0.864 k) - w k = 0

>> Equating Coefficients on both sides,

- 0.285*Tab + 0.306*Tac - 0.285*Tad = 0

- 0.415*Tab - 0.222*Tac + 0.415*Tad = 0

- 0.864*Tab - 0.926*Tac - 0.864*Tad - w = 0

>> Now putting Tac = 180 lb and solving equations (1), (2) & (3)

Tab = 48.49,

Tad = 144.78

w= 333.66 ....REQUIRED WEIGHT.....