Question

(10%) Problem 6: After a mishap, a m-85 kg circus performer clings to a trapeze which is being pulled to the side by another circus artist as displayed in the figure. The left rope (with tension T1) is at an angle θ,-15° to the vertical. The right rope (with tension T2) is an at angle θ,-11° from the horizontal. Assume the person is momentarily motionless. 25% Part (a) Write an equation for the net horizontal force on the performer Fr in terms of the tensions T1 and T2 as well as the angles θ! and θ. Let to the right be the positive x direction. Grade Summary Deductions Potential 100% cos(α) cos(81) sin(φ) sin(92) | | | cos(q) cos(92) sin(θ) cos(9) sin(a) sin(81) Submissions Attempts remaining:19 (00.0 per attempt) detailed view | 090 Submit Feedbackgup! Hints: 0 for a 0% deduction. Hints remaining: 0 Feedback: 500 deduction per feedback. 25% Part b) What is the sum of the forces n the y direction F, based on the tensions in ropes TI and T the mass m of the circus performer, the acceleration due to gravity g, and the angles θ 1 and θ.? Let up be the positive y direction. 25% Part (c) what is the value of the tension in rope l (T) in newtons? 2500 Part (d) What is the value of the tension in rope 2 (12) in newtons?

Answer 1

a) Sigma F_x = T_2 cos theta_2 - T_1 sin theta_1 = T_2 cos 11 degree - T_1 sin 15 degree = 0 b) Sigma F_y = T_1 cos theta_1 + T_2 sin theta_2 - w = T_1 cos 15 degree + T_2 sin 11 degree - m g = 0 c) From Sigma F_x = 0 Therefore T_2 cos 11 - T_1 sin 15 degree = 0 Therefore T_2 cos 11 = T_2 sin 15 Therefore T_2 = T_1 sin 15/cos 11 from Sigma F_y = 0 rightarrow T_1 cos 15 + T_2 sin 11 = mg rightarrow T_1 cos 15 + T_1 sin 15 sin 11/cos 11 = mg = T_1 [cos 15 + sin 11 sin 75] = mg m = 85 Kg Therefore T_1 = 85 times 9.8/sin 15 + sin 11 sin 15 = 819.692 N Therefore T_1 = 819.692 N \r\n d) T_2 = T_1 times sin 15/cos 11 Therefore T_2 = 216.123 N