A lever balances when force times distance from the pivot is the same on both sides. Torque is the general version of that rule. It measures how strongly a force turns something about an axis, and it depends on more than the size of the force. To find it you will use force components.
Why it works
Push on a door. Push near the hinges and it barely moves; push at the edge and it swings easily. Push straight toward the hinges and nothing happens, however hard you push. Only the part of the force at right angles to the door, $F_\perp = F\sin\theta$, turns it, and its effect grows with the distance $r$ from the hinge. So the turning effect is $\tau = rF\sin\theta$. You can read the same product another way: $r\sin\theta$ is the lever arm $r_\perp$, the perpendicular distance from the axis to the line along which the force acts.
Key idea
$$\tau = rF\sin\theta = rF_\perp = r_\perp F,$$ where $r$ is the distance from the axis to where the force acts and $\theta$ is the angle between $\vec{r}$ and $\vec{F}$. Unit: $\text{N·m}$. Give torques a sign: counterclockwise positive, clockwise negative, and add them for the net torque.
Worked examples
Worked example: A wrench at right angles
You pull with $80\ \text{N}$ at right angles to a wrench, $0.25\ \text{m}$ from the bolt. Find the torque.
- $\theta = 90^\circ$, so $\sin\theta = 1$: $\tau = (0.25\ \text{m})(80\ \text{N}) = 20\ \text{N·m}$.
Answer: $20\ \text{N·m}$
Worked example: A door pushed at an angle
You push a door $0.80\ \text{m}$ from its hinges with $40\ \text{N}$, at $30^\circ$ to the surface of the door. Find the torque.
- The angle between $\vec{r}$ (along the door) and $\vec{F}$ is $30^\circ$.
- $\tau = (0.80\ \text{m})(40\ \text{N})\sin 30^\circ = 16\ \text{N·m}$, half of what a perpendicular push would give.
Answer: $16\ \text{N·m}$
Worked example: Net torque on a beam
A beam pivots at its middle. $30\ \text{N}$ pushes down $1.5\ \text{m}$ left of the pivot, and $40\ \text{N}$ pushes down $0.5\ \text{m}$ right of it. Find the net torque (counterclockwise positive).
- Left force turns the beam counterclockwise: $+(1.5)(30) = +45\ \text{N·m}$.
- Right force turns it clockwise: $-(0.5)(40) = -20\ \text{N·m}$.
- $\Sigma\tau = +25\ \text{N·m}$: it starts turning counterclockwise even though the smaller force is on that side.
Answer: $+25\ \text{N·m}$
Common mistake
Thinking a bigger force always makes a bigger torque. Torque depends on three things: the force, where it acts and its direction. A $150\ \text{N}$ push straight toward the hinges makes no torque, while $40\ \text{N}$ at the edge, at right angles, turns the door easily. Check each force's lever arm before comparing.
Torque and work share the unit $\text{N·m}$, but they are different things: torque is a turning effect, work is energy. Physicists write work in joules and torque in $\text{N·m}$ to keep them apart.