Torque as a vector

A force does more than push a body, it can twist it. Torque measures that twisting effect about an axis. In Physics C torque is a vector: $\vec{\tau} = \vec{r} \times \vec{F}$, the cross product of the position vector $\vec{r}$, from the axis to where the force acts, and the force $\vec{F}$.

Its size is $\tau = rF\sin\theta$, where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$. The $\sin\theta$ keeps only the part of the force perpendicular to $\vec{r}$, the part that actually turns the body. Its direction follows the right-hand rule, out of or into the page along $\pm\hat{k}$.

For students arriving from Physics 1, the size $rF\sin\theta$ and the lever-arm idea look familiar. The Physics C work is the vector form: the cross product, its direction, and keeping $\vec{r} \times \vec{F}$ in the right order.

Torque is the cross product $\vec{\tau} = \vec{r} \times \vec{F}$. Its magnitude is $\tau = rF\sin\theta$: the distance $r$ times the force $F$ times the sine of the angle between them.

The $\sin\theta$ matters. Only the part of $\vec{F}$ perpendicular to $\vec{r}$, which is $F\sin\theta$, turns the body. The part along $\vec{r}$ only pushes the point toward or away from the axis, doing nothing. So a force aimed straight along $\vec{r}$ gives zero torque ($\sin 0 = 0$), and a force perpendicular to $\vec{r}$ gives the most ($\sin 90^\circ = 1$).

The common slip is to drop the angle and write $\tau = rF$. That is right only when the force is perpendicular. At any other angle, $rF$ overcounts, because part of the force is wasted along $\vec{r}$.

Torque is $\vec{\tau} = \vec{r} \times \vec{F}$ with magnitude $\tau = rF\sin\theta$. Keep the $\sin\theta$: only the perpendicular part of the force turns the body.

There is a second way to read $rF\sin\theta$. Group it as $F$ times $(r\sin\theta)$. That $r\sin\theta$ is the lever arm: the perpendicular distance from the axis to the force's line of action, the straight line the force points along.

So torque is the force times the lever arm. This makes two things clear. Slide the force along its own line of action and the torque does not change: the perpendicular distance to that line stays the same. And a force whose line of action passes through the axis has a lever arm of zero, so it makes zero torque no matter how far out it acts.

The slip here is to use the full distance $r$ to where the force acts as if it were the lever arm. Unless the force is perpendicular, the lever arm is shorter: $r\sin\theta$, not $r$.

Torque is $F$ times the lever arm $r\sin\theta$, the perpendicular distance to the line of action. A line of action through the axis gives zero torque.

The direction of $\vec{\tau} = \vec{r} \times \vec{F}$ comes from the right-hand rule. Point your fingers along $\vec{r}$, curl them toward $\vec{F}$, and your thumb points along $\vec{\tau}$. The result is always perpendicular to the plane that holds $\vec{r}$ and $\vec{F}$, never in that plane.

For vectors in the page, $\vec{\tau}$ points out of the page ($+\hat{k}$) or into it ($-\hat{k}$). A force that would spin the body counterclockwise gives $+\hat{k}$; clockwise gives $-\hat{k}$.

Order matters. Swap the two vectors and the sign flips: $\vec{F} \times \vec{r} = -(\vec{r} \times \vec{F})$. The size $rF\sin\theta$ stays the same, but the direction reverses. Torque is $\vec{r} \times \vec{F}$, position first.

$\vec{\tau} = \vec{r} \times \vec{F}$ is perpendicular to the plane of $\vec{r}$ and $\vec{F}$, out of or into the page by the right-hand rule. Reversing to $\vec{F} \times \vec{r}$ flips the direction.

A force $F = 20$ N acts at a point $r = 0.30$ m from an axle, at $\theta = 40^\circ$ to the position vector. The magnitude of the torque is $\tau = rF\sin\theta = 0.30 \times 20 \times \sin 40^\circ \approx 3.9$ N$\cdot$m.

Read the same number as a lever arm: the perpendicular distance to the line of action is $r\sin\theta = 0.30 \times \sin 40^\circ \approx 0.19$ m, and $F$ times that arm is $20 \times 0.19 \approx 3.9$ N$\cdot$m, the same answer.

Two traps to avoid: writing $\tau = rF = 6.0$ N$\cdot$m drops the $\sin\theta$ and overcounts; using the full $r = 0.30$ m as the lever arm makes the same error the other way. If the force pointed straight along $\vec{r}$, then $\theta = 0$ and the torque would be zero.

Use $\tau = rF\sin\theta$, or equally $F$ times the lever arm $r\sin\theta$. Both give the same value; the bare product $rF$ does not.

Torque problems collapse in recognizable ways—each trap below matches a misconception code on the diagnostic.