A door of width 0.90 m is opened with a horizontal force of 15 N applied at the edge. What torque is exerted about the hinge?

• A. 6.8 N·m
• B. 13.5 N·m
• C. 27 N·m
• D. 19 N·m

Answer: (B)

The question tests how torque arises from a force about a pivot, using τ = r × F, whose magnitude is τ = r F sin(θ). Here the lever arm is the distance from the hinge to where the force is applied, which is the door’s width: 0.90 m. The force is directed to open the door in a way that makes the force perpendicular to that lever arm, so the angle between r and F is 90°, and sin(90°) = 1. Therefore the torque is τ = 0.90 m × 15 N = 13.5 N·m. This torque acts to rotate the door about the hinge in the opening direction. The other numbers would require a different lever arm or a non-90° angle, which isn’t the given setup.