In Fig., the bar is uniform and weighing 500 N. How large must W be if T1 and T2 are to be equal?

Taking torque about the attachment point for W, we get-T1(0.4L) + T2 (0.3L) + 500 (0.2L) = 0 T = 1000 N, where T1 = T2 = TSFy = 0 Þ 2T - W - 500 = 0  ÞW = 1500 N