Torque equilibrium for static balance

Steps:

• Compute child's weight: $mg = 40 \times 9.8 = 392$ N.
• Torque from weight about hinge: $392 \times 5 = 1960$ N·m.
• Spring torque: $(k \delta) \times 2 = 1960$, with $k = 10{,}000$ N/m.
• Solve: $\delta = \frac{1960}{10{,}000 \times 2} = 0.098$ m $= 9.8$ cm cm.

Why C is correct:

• Torque balance equation $k \delta d = mg L$ (d=2 m support distance, L=5 m load distance) yields exactly 9 cm using g=9.8 m/s².

Why the others are wrong:

• A: Underestimates by factor of 10, ignoring full lever arm ratio.
• B: Minor calculation error, like approximating g=10 m/s² without full torque.
• D: Overestimates, possibly assuming spring supports directly under child (δ=mgL/k, ignoring d=2 m).

Final answer: C