For part B.
T_A = T_C*(sind(37.1))
729N = T_C*(sind(37.1))
T_C = 1250N
In a repair shop a truck engine that has mass 429 kg is held in place by four light cables (Figure 1) . Cable A is horizontal, cables B and D are vertical, and cable C makes an angle of 37.1∘ with a vertical wall. Tension in cable A is 729 N .
a.What is the tension in cable B?
b.What is the tension in cable C?
2 answers
For Part A
T_B = mg + costheta*T_C
T_B = 429*9.8 + cos(37.1)*1250
T_B = 5200N
T_B = mg + costheta*T_C
T_B = 429*9.8 + cos(37.1)*1250
T_B = 5200N