Question

Course Contents»..» Unit 6: Torque » Practice exam 6 Inf Print Evaluate Communicate Notes Bookmark A 5-m long, 200-kg bar is supported by 3 ropes: On the far right end, the Rope A make an angle of 15° to the right of vertical. On the far left end, Rope B makes an angle of 40° to the left of vertical, and Rope C (also on the far left end) is vertical. For the purpose of calculating torque, place the pivot point on the far left end of the bar. What is the magnitude of the weight of the bar? Submit Answer Tries 0/99 What is the magnitude of the torque that the weight of the bar exerts on the bar? Submit Answer Tries 0/99 What is the tension in Rope A? Tries 0/99 Submit Answer What is the tension in Rope B? Submit Answer Tries 0/99 What is the tension in Rope C? Submit Answer Tries 0/99

Answer 1

a] Weight of the bar = mg = 200*9.8 = 1960 N

b] Torque exerted = mgL/2 = 1960*5/2 = 4900 Nm

c] Let tension in rope A be TA,

torque about left has to be zero,

TA*cos 15 degree * L - 1960*L/2 = 0

TA = (1960/2)/cos 15degree = 1015 N

d] Net Force in horizontal = 0

TB*sin 40 degree - TA*sin 15 degree

TB = TA*sin 15 degree/sin 40 degree = 1015*sin 15 degree/sin 40 degree = 408.7 N

e] Net force in vertical = 0

TC + TB cos 40 degree + TA*cos 15 degree - mg = 0

TC + 408.7*cos 40 degree + 1015*cos 15 degree - 1960 = 0

TC = 1960 -408.7*cos 40 degree -1015*cos 15 degree

= 666.5 N