Question

7.5 For the structure loaded as shown below, calculate the tension in cable AB and the reaction forces at pin C for the case where L = 8 ft. (Ans. TAB-6,212.90 lb, Cx 5,840.13 lb, Cy 875.0 lb) 20° 2L 2,500 lb 500 lb 7.6 Calculate the reaction forces at point A and the tension in cable BC for the given structure. (Ans. T-2, 121.32 N, Ax = 1,500 N, A,-3,500 N) 1.5 m B 2,000 N : 2 m 45°

Answer 1

We need to resolve vector in x and y direction .

Then apply the static equilibrium condition .. All the net forces in x and y direction is zero .

As this body is in static equilibrium sigma F_x = 0 implies T Cos(20) - C-x = 0 sigma F_y = 0 C_y + T Sin (20) - 2500 mu_0 - 500 mu_0 = 0 C_y + T Sin (20) = 3000 lb sigma T_c = 0 T Sin(20 degrees) times 4L - (2500 times 3L + 500 times 2L) = 0 T Sin 2 times 4 = 7500 + 1000 T = 5800/4 times Sin(20 degrees) T = 6212.90 lb C_x = T Cos 20 C_x = 1 Cos 20 degrees C_x = 5840 lb C_y = 3000 - T Sin 20 C_y = 875 lb \r\n sigma F_x = 0 implies T Sin(45) - A_x = 0 sigma F_y = 0 implies - T Cos(45) + A_y - 2000 = 0 A_y - T Cos(45) = 2000 sigma T_A = 0 implies T Sin (45) degrees times 2 - 2000 times 1.5 = 0 T Sin(45 degrees)