Net torque about pivot P is 6 N m clockwise

Steps:

• Calculate torque from 16 N force at 0.3 m from P: 16 × 0.3 = 4.8 N m clockwise.
• Calculate torque from 20 N force at 0.9 m from P: 20 × 0.9 = 18 N m counterclockwise.
• Net torque without F: 4.8 - 18 = -13.2 N m (13.2 N m counterclockwise).
• For net +6 N m clockwise, torque from F at 1.2 m: 1.2F = 13.2 + 6 = 19.2; F = 19.2 / 1.2 = 16 N.

Why C is correct:

• 16 N produces exactly 19.2 N m clockwise torque, balancing to net 6 N m clockwise per torque equation ∑M_P = 6.

Why the others are wrong:

• A: 9.2 N gives 11.0 N m torque, net -2.2 N m (counterclockwise).
• B: 11 N gives 13.2 N m torque, net 0 N m.
• D: 24 N gives 28.8 N m torque, net 15.6 N m (excess clockwise).

Final answer: C