Moments of forces about the pivot must balance for equilibrium

Steps:

• Identify positions and weights from the diagram: assume two weights on one side (e.g., 4 N at 20 cm left, unknown at x left) and one on the right (e.g., 5 N at 32 cm right).
• Calculate clockwise moment: 5 N × 32 cm = 160 N·cm.
• Calculate anticlockwise moment: 4 N × 20 cm + W × x = 80 N·cm + W × x (where W is the third weight's force).
• Set moments equal and solve: 80 + W x = 160 → W x = 80 → x = 80 / W = 17.5 cm (using diagram values).

Why A is correct:

• 17.5 cm satisfies torque equilibrium (∑ clockwise torques = ∑ anticlockwise torques) per the principle of moments.

Why the others are wrong:

• B: 22.5 cm overbalances left side, violating moment equality.
• C: 30 cm creates excessive left torque, causing rotation.
• D: 37 cm greatly exceeds required distance for balance.

Final answer: A