Gravitational torque on pivoted sign Steps:

• Center of mass is midway down the sign, so distance d from hinge = 0.8 m / 2 = 0.4 m.
• Torque magnitude τ = weight × d × sinθ = 40 N × 0.4 m × sinθ = 16 sinθ Nm, with θ from vertical.
• Given maximum torque of 6.0 Nm, set 16 sinθ = 6.
• sinθ = 6/16 = 0.375, so θ = arcsin(0.375) ≈ 22°.

Why B is correct:

• At 22°, sinθ = 0.375 exactly, so τ = 16 × 0.375 = 6.0 Nm per the torque formula τ = mgd sinθ.

Why the others are wrong:

• A. sin11° ≈ 0.191, τ ≈ 3.1 Nm (too small).
• C. sin68° ≈ 0.927, τ ≈ 14.8 Nm (exceeds 6.0 Nm).
• D. sin79° ≈ 0.982, τ ≈ 15.7 Nm (exceeds 6.0 Nm).

Final answer: B