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concepts/mechanics/u5/m5-3-torque.tex

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+\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.}
+
+\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.