\subsection{Torque and Lever Arm}

This subsection introduces torque as the rotational effect of a force about a chosen point or axis. In AP mechanics, the key computational idea is that only the part of the force with a nonzero lever arm contributes to the turning effect.

\dfn{Torque vector, lever arm, and fixed-axis sign convention}{Let $O$ denote the pivot or reference point. Let $\vec{r}$ denote the position vector from $O$ to the point where a force $\vec{F}$ is applied. The \emph{torque vector} of $\vec{F}$ about $O$ is the vector quantity that measures the tendency of the force to cause rotation about $O$.

Let $\phi$ denote the angle between $\vec{r}$ and $\vec{F}$. The \emph{lever arm} $\ell$ is the perpendicular distance from the pivot to the line of action of the force, so
\[
\ell=r\sin\phi,
\]
where $r=|\vec{r}|$.

For a fixed-axis problem, choose an axis with unit vector $\hat{k}$ and declare a positive sense of rotation by the right-hand rule. The corresponding signed scalar torque is
\[
\tau=(\vec{r}\times \vec{F})\cdot \hat{k}.
\]
If the force tends to rotate the object in the chosen positive sense, then $\tau>0$; if it tends to rotate the object in the opposite sense, then $\tau<0$.}

\thm{Torque formulas}{Let $\vec{r}$ denote the position vector from the pivot to the point of application of a force $\vec{F}$. Let $\phi$ denote the angle between $\vec{r}$ and $\vec{F}$, let $r=|\vec{r}|$, let $F=|\vec{F}|$, and let $\ell$ denote the lever arm. Then the torque vector is
\[
\vec{\tau}=\vec{r}\times \vec{F},
\]
and its magnitude is
\[
|\vec{\tau}|=rF\sin\phi=F\ell.
\]
For rotation about a chosen fixed axis with unit vector $\hat{k}$,
\[
\tau=(\vec{r}\times \vec{F})\cdot \hat{k},
\]
so the scalar $\tau$ is positive or negative according to the declared sign convention.}

Torque about a fixed axis connects directly to angular acceleration through Newton's second law for rotation (Section 5.6), just as net force connects to linear acceleration through Newton's second law (Section 2.1).

\pf{Why $rF\sin\phi$ equals $F\ell$}{From the cross-product magnitude formula,
\[
|\vec{\tau}|=|\vec{r}\times \vec{F}|=rF\sin\phi.
\]
But by geometry, the lever arm is the perpendicular distance from the pivot to the line of action of the force, so
\[
\ell=r\sin\phi.
\]
Substituting this into the previous expression gives
\[
|\vec{\tau}|=rF\sin\phi=F\ell.
\]
Thus torque can be found either from the perpendicular component of the force or from the full force multiplied by the lever arm.}

\cor{If the line of action passes through the pivot, the torque is zero}{If the line of action of $\vec{F}$ passes through the pivot, then the perpendicular distance from the pivot to that line is $\ell=0$. Therefore
\[
|\vec{\tau}|=F\ell=0.
\]
Equivalently, in this case $\phi=0$ or $\phi=\pi$, so $rF\sin\phi=0$. Thus a force directed exactly through the pivot can change the net force on an object without producing any torque about that pivot.}

\qs{Worked example}{A door rotates about a vertical hinge through its left edge. View the door from above and choose counterclockwise rotation as positive. Let $r=0.90\,\mathrm{m}$ be the distance from the hinge to the point where the force is applied at the outer edge. A student pushes with force magnitude $F=40\,\mathrm{N}$. The force lies in the horizontal plane and makes an angle $\phi=30^\circ$ with the door, so it tends to rotate the door counterclockwise.

Find:
\begin{enumerate}[label=(\alph*)]
\item the lever arm $\ell$,
\item the signed torque $\tau$ about the hinge, and
\item the magnitude of a force applied perpendicular to the door at the same point that would produce the same torque.
\end{enumerate}}

\sol Because the position vector from the hinge to the point of application lies along the door, the given angle $\phi=30^\circ$ is the angle between $\vec{r}$ and $\vec{F}$.

For part (a), the lever arm is
\[
\ell=r\sin\phi=(0.90\,\mathrm{m})\sin 30^\circ.
\]
Since $\sin 30^\circ=0.50$,
\[
\ell=(0.90)(0.50)=0.45\,\mathrm{m}.
\]
So the lever arm is
\[
\ell=0.45\,\mathrm{m}.
\]

For part (b), the torque magnitude is
\[
|\tau|=rF\sin\phi=F\ell.
\]
Using either form,
\[
|\tau|=(0.90\,\mathrm{m})(40\,\mathrm{N})\sin 30^\circ=(40\,\mathrm{N})(0.45\,\mathrm{m}).
\]
Thus
\[
|\tau|=18\,\mathrm{N\cdot m}.
\]
Because the force tends to rotate the door counterclockwise and counterclockwise was chosen as positive,
\[
\tau=+18\,\mathrm{N\cdot m}.
\]

For part (c), let $F_{\perp}$ denote the force magnitude that would act perpendicular to the door at the same point. A perpendicular force has lever arm equal to the full distance $r=0.90\,\mathrm{m}$, so its torque magnitude is
\[
|\tau|=rF_{\perp}.
\]
Set this equal to the required $18\,\mathrm{N\cdot m}$:
\[
18\,\mathrm{N\cdot m}=(0.90\,\mathrm{m})F_{\perp}.
\]
Therefore,
\[
F_{\perp}=\frac{18}{0.90}=20\,\mathrm{N}.
\]

The results are
\[
\ell=0.45\,\mathrm{m},
\qquad
\tau=+18\,\mathrm{N\cdot m},
\qquad
F_{\perp}=20\,\mathrm{N}.
\]
This example shows that the same torque can be produced either by a larger force with a shorter lever arm or by a smaller force applied more effectively.
