Torque
Turning effect depends on how hard, where, and in which direction you push.
Lesson at a glanceShow the key points
- Torque of a force about a point O: τ = r × F, where r runs from O to the point where the force acts.
- Torque about an axis = component of r × F along the axis; it does not depend on where on the axis O is chosen.
- A force parallel to the axis, or one whose line of action passes through the axis, has zero torque about it.
- For a force perpendicular to the axis: τ = F × (lever arm), where the lever arm is the perpendicular distance from the axis to the line of action; τ = rF sin θ.
- Anticlockwise torque is usually taken positive. Net torque is the sum of the individual torques; it can be non-zero even when the net force is zero (a couple).
A wheelchair doesn’t start spinning just because you wish it to; you have to push. But a ceiling fan spins up while its centre stays still, so the net force on it is zero. Something other than net force decides whether a body starts to turn. That something is torque.
Torque about a point
Being a cross product, is perpendicular to both and , and its size is . Its unit is N·m (not written as joule, though dimensionally the same).
Torque about an axis
For a body turning about a fixed axis, what matters is the torque about that axis: pick any point O on the axis and take the component of along the axis. Shifting O along the axis adds a vector perpendicular to the axis, so the answer doesn’t depend on which point you pick. Four situations cover everything:
Case 1: force parallel to the axis
is perpendicular to , hence to the axis. Zero torque about the axis. Pushing a door upward along its hinges doesn’t open it.
Case 2: force whose line passes through the axis
Choose O where the line meets the axis; then r and F are parallel. Zero torque. Pushing a door straight at its hinges does nothing.
Case 3: force perpendicular to the axis but not meeting it
This is the everyday case. The torque about the axis is the force times the length of the common perpendicular between the axis and the force’s line of action: the lever arm (or moment arm).
Case 4: anything else
Split the force into a part parallel to the axis (no torque, case 1) and a part perpendicular to it (case 3).
By convention a torque that would turn the body anticlockwise, as seen looking back along the axis toward you, is positive.
A door is 0.80 m wide. A 30 N push is applied at the handle edge. Find the torque about the hinges if the push is (a) perpendicular to the door, (b) at 60° to the door’s surface, (c) straight toward the hinges.
A pulley is clamped at its centre, and a rope wound on it hangs down from the rim with tension T. Which forces act on the pulley, and what torque does each have about its axle?
Adding torques
The total torque on a body is the sum of the individual torques:
A driver pushes up with 15 N on one side of a 36 cm diameter steering wheel and pulls down with 15 N on the other. Find the net force and net torque.
- 1; about an axis, take its component along the axis.
- 2Forces parallel to the axis, or passing through it, have no torque about it.
- 3For a force perpendicular to the axis: lever arm .
- 4Add torques, not forces: a couple has zero net force but a net torque.
Check yourself
Keep exploring
Hand-picked videos, simulations and readings that explain this idea from a different angle.