In figure, the bar is uniform and weighing 500 N. How large must W be if T1 and T2 are to be equal ?
Answer & explanation
Correct answer: option 4
Taking torque about the attachment point for W, we get :
\(-T_1 (0.4 L) + T_2 (0.3 L) + 500 (0.2 L) = 0\)
\(T = 1000 N \text{ ; where }T_1 = T_2 = T\)
\(\Sigma F_y = 0\)
\(\Rightarrow 2 T - W - 500 = 0\)
\(\Rightarrow W = 1500 \text{ N}\)