Torque equilibrium about the pivot

Steps:

• Model boom as horizontal rod of length L, pivoted at mast end.
• Weight of 200 N acts at midpoint, producing clockwise torque of 200 × (L/2) = 100L Nm.
• Tension T acts at far end along rope at 40° to horizontal, so perpendicular component T sin 40° produces counterclockwise torque of T sin 40° × L.
• For rotational equilibrium, torques balance: T sin 40° × L = 100L, so T = 100 / sin 40°.
• Using sin 40° ≈ 0.625, T = 100 / 0.625 = 160 N.

Why B is correct:

• Matches torque balance formula where T provides the exact moment to counter the boom's weight torque.

Why the others are wrong:

• A: Underestimates by using cos 40° (≈0.766) in place of sin 40°, giving ≈130 N.
• C: Equals total weight, ignoring lever arm and angle effects.
• D: Overestimates by considering full vertical support without torque reduction by sin 40°.

Final answer: B