Torque equilibrium for horizontal rod Steps:

• Identify pivot at attachment point Q, 25 cm from X (lever arm XQ = 0.25 m) and 50 cm from Y (lever arm QY = 0.5 m).
• Weight W = 11 × 9.8 = 107.8 N acts downward at Y.
• For rotational equilibrium about Q, upward force F at X produces counterclockwise torque F × 0.25 = clockwise torque W × 0.5.
• Solve: F = (0.5 / 0.25) × 107.8 = 215.6 N; vertical force balance T + F = W gives T negative, impossible for string tension. Not enough information to resolve configuration without diagram showing string orientations or mass attachment details.

Final answer: Not enough information.