Ejercicio 4.116 Resuélvase el problema 4.115 suponiendo que el

Ejercicio 4.116 Resuélvase el problema 4.115 suponiendo que el

Ejercicio 4.116
Resuélvase el problema 4.115 suponiendo que el cable EF es
reemplazado or un cable unido a los puntos E y H.
EF = (−30) 2 + (12) 2 + (−20) 2 = 38in
FEHx = -30 in * EH = -0.79 EH
38 in
FEHy = 12 in * EH = 0.315 EH
38 in
FEHz = -20 in * EH = -0,526 EH
38 in
∑ MA = 0 = M WA + M AB + M AE + M AA
∑ MA =
[(15” i + 10” k) x (-75 lb j)] + [(30” i) x (RBYj + RBz k)] +
[(26” i + 20” k) x (-0,79 EH i + 0.315 EH j – 0.526 EH k)] = 0
∑ MA =
0 = (-1125 lb” k + 750 lb” i)+(30” By k – 30 Bz j) + (8,19” EH
k)
+ (13,67” EH j – 15,8” EH j – 6.3” EH i)
∑ MAx =
EH =
∑ MA i = 0 = 750 lb” – 6,3” EH
119 lbs
FEHx =
FEHy =
FEhz =
-0,79 (119 lb) → FEHx = -94 lb
0,315 (119 lb) → FEhy = 37.5 lb
-0,526 (119 lb) →
FEhz = -62,6 lb
∑ MAY =
-30 Bk + 13,67” EH – 15,8 EH = 0
-30” Bk + 1626,73” – 1880,20” = 0
Bk = -8,45 lb
∑ MAZ =
∑ MA k = -1125 lb” + 30” By + 8.19 EH = 0
-1125 lb” + 30” By + 974.61 = 0
By = 5 lb
B = (5 lb) j – (8.45 lb) k
∑ Fx = 0 = Ax – 94 lb Ax = 94 lb
∑ Fy = 0 = Ay + 37.5 lb – 76 lb + 5 lbs
Ay = 32,5 lb
∑ F2 = Az – 8,45 lb – 62,6 lb = 0
Az = 71 lbs
A = (94 lb) i + (32,5 lb) j + (71 lb) k
A = (94 lb) i + (32,5 lb) j + (71 lb) k
A = 122,20 lb
B = (5 lb) i – (8,45 lb) k
B = 9.82 lb