Question

The tree surgeon attempts to pull down the partially sawn-through tree trunk. He creates a tension TA 209 N in the rope, which has a mass of 0.69 kg per meter of its length. Determine the angle θΑ at which he pulls, the length L of rope between points A and B, and the tension Tg at point B. 11.7 m 5.5 ml Answers: 25.17 87.5 419.95

Answer 1

Given that T_A = 209 N Rope has a mass of 0.69 kg/m theta_A = Tan^-1 (5.5/11.7) theta_A = tan^-1 (0.4700) = 25.17 degree = theta_A Assume Rope AB is straight the length of Rope AB AB^2 = (AC)^2 + CB^2 = (5.5)^2 + (1.7)^2 \r\n AB^2 = 30.25 + 136.89 AB^2 = 167014 AB = squareroot 167.14 AB = 12.9283 m omega = 12.928 times 0.69 times 9.81 = 87.50 N sigma_1 v = 0 209 sin (25.17) + w - T_B sin (25.17) 209 times 0.42530 + 87.50 = T_B times 0.42530 T_B 176.3888/0.42530 = T_B T_B = 414.7398 N