Torque equilibrium about the hinge

Steps:

• Consider torques about hinge X for rotational equilibrium; net torque is zero.
• Weight of rod (10 N) acts at midpoint (L/2 from X), producing clockwise torque of 10 × (L/2) = 5L N·m.
• Force F at Y (distance L from X) has vertical component F sin 30° = F/2 upward, producing counterclockwise torque of (F/2) × L = (F L)/2 N·m.
• Set torques equal: (F L)/2 = 5L; cancel L to get F/2 = 5, so F = 10 N.

Why C is correct:

• Torque balance requires F sin 30° × L = weight × (L/2), simplifying to F = 10 N via the lever arm and angle component.

Why the others are wrong:

• A: 5 N ignores the 30° angle, treating F as fully vertical at the end.
• B: 8.7 N (≈5√3) mistakenly uses cos 30° for vertical lift instead of sin 30°.
• D: 20 N doubles the weight, neglecting the midpoint acting distance (L/2).

Final answer: C