Torque
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 4 practice questions.
Torque is the turning effect of a force: τ = r⊥F, where the lever arm r⊥ is the perpendicular distance from the axis to the force’s line of action. Equivalently, τ = rF sin θ. A force aimed straight through the axis makes no torque.
1. Lever arm
2. Balancing torques
Push a door at the edge, perpendicular to it, and it swings easily. Push near the hinges, or at a slant, and it barely moves: the lever arm is smaller.
Push straight toward the hinges and the line of the force passes through the axis: r⊥ = 0, so τ = 0.
An object is in rotational equilibrium when the torques balance: Στ = 0. Clockwise torques equal counterclockwise torques. That’s how a seesaw balances a heavy child near the middle against a light child near the end.
3. Example, then try it
A 30 kg child sits 1.5 m from a seesaw’s pivot. Where must a 45 kg child sit on the other side to balance it?
Balance the torques from their weights: (30)(10)(1.5) = (45)(10)x.
x = 450 ÷ 450 = 1.0 m from the pivot. The heavier child sits closer.
A door is 0.80 m wide. Find the torque about the hinges for a 20 N push (a) at the edge, perpendicular to the door; (b) at the edge, at 30° to the door; (c) at the middle, perpendicular; (d) at the edge, aimed at the hinges.
(a) τ = rF = (0.80)(20) = 16 N·m.
(b) τ = rF sin 30° = (0.80)(20)(0.50) = 8.0 N·m.
(c) τ = (0.40)(20) = 8.0 N·m.
(d) The force points through the axis: 0 N·m.
Answer: 16, 8.0, 8.0, and 0 N·m. Same force, very different turning effects.
4. Common mistakes
You push on a door at its outer edge, aiming straight toward the hinges. What torque does your push produce about the hinges?
The force’s line of action passes through the hinges, so the lever arm is zero and τ = r⊥F = 0. The door doesn’t turn.
Ready to practice?
4 questions on Torque, each with a hint and an explanation for every wrong answer.