Two forces of 8.0 N form a couple. They act in opposite directions at the two ends of a rod 0.30 m long, each at 50° to the rod. What is the torque of the couple? (sin 50° = 0.766)
- The forces are not at right angles to the rod, so find the perpendicular distance between their lines of action.
- Perpendicular distance = 0.30 × sin 50° = 0.30 × 0.766 = 0.230 m.
- Torque = one force × perpendicular distance = 8.0 × 0.230 = 1.84 N m.
Answer: About 1.8 N m.