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

1500 N

Taking torque about the attachment point for W, we get :

\(-T_1 (0.4 L) + T_2 (0.3 L) + 500 (0.2 L) = 0\)

\(T = 1000 N \text{ ; where }T_1 = T_2 = T\) 

\(\Sigma F_y = 0\)

\(\Rightarrow 2 T - W - 500 = 0\)

\(\Rightarrow W = 1500 \text{ N}\)