\subsection{Static and Kinetic Friction}

This subsection gives the AP dry-friction model for a body in contact with a surface. Friction is a contact force parallel to the surface, tied to the tendency for slipping or to actual slipping, while the normal force is perpendicular to the surface.

\dfn{Static friction, kinetic friction, and coefficients of friction}{Let a body be in contact with a surface. Let $\vec{N}$ denote the normal force exerted by the surface on the body, let $N=|\vec{N}|$ denote its magnitude, and let $\vec{f}$ denote the friction force exerted by the surface on the body.

The friction force acts parallel to the contact surface and opposes the relative motion or the tendency of relative motion between the surfaces.

If the surfaces are not slipping relative to each other, the friction is called \emph{static friction}. Let $\vec{f}_s$ denote the static-friction force and let $f_s=|\vec{f}_s|$ denote its magnitude. Then static friction can adjust in magnitude up to a maximum value:
\[
f_s\le \mu_s N,
\]
where $\mu_s$ is the coefficient of static friction.

If the surfaces are sliding relative to each other, the friction is called \emph{kinetic friction}. Let $\vec{f}_k$ denote the kinetic-friction force and let $f_k=|\vec{f}_k|$ denote its magnitude. In the idealized AP dry-friction model,
\[
f_k=\mu_k N,
\]
where $\mu_k$ is the coefficient of kinetic friction.

The coefficients $\mu_s$ and $\mu_k$ are dimensionless constants for the pair of surfaces in the chosen model.}

\nt{Static friction is \emph{not} always equal to $\mu_s N$. The quantity $\mu_s N$ is the \emph{maximum possible} static-friction magnitude. The actual static-friction magnitude is whatever value is required by Newton's second law to prevent slipping, as long as that required value does not exceed $\mu_s N$. Only at the threshold of slipping does $f_s=\mu_s N$.}

\ex{Illustrative example}{Let a block rest on a horizontal table. Let $\vec{P}=(3.0\,\mathrm{N})\hat{\imath}$ denote a horizontal applied force on the block, and let the maximum possible static-friction magnitude be $\mu_s N=5.0\,\mathrm{N}$.

Because the required friction to prevent motion is only $3.0\,\mathrm{N}$, the block remains at rest and the actual static-friction force is
\[
\vec{f}_s=-(3.0\,\mathrm{N})\hat{\imath},
\]
not $-(5.0\,\mathrm{N})\hat{\imath}$. If the applied-force magnitude were increased beyond $5.0\,\mathrm{N}$, static friction could no longer hold the block at rest and slipping would begin.}

\mprop{Operational laws and direction rules}{Let $\vec{N}$ denote the normal force on a body, let $N=|\vec{N}|$, and let a tangent axis be chosen along the contact surface.

\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
\item If there is no slipping at the contact, then the friction is static. Its magnitude must satisfy
\[
f_s\le \mu_s N.
\]
Its direction is opposite the direction the body would move \emph{relative to the surface} if friction were absent.

\item If the body slides relative to the surface, then the friction is kinetic. Its magnitude is
\[
f_k=\mu_k N,
\]
and its direction is opposite the relative velocity of the sliding surfaces.

\item Friction and the normal force are different parts of the same contact interaction: $\vec{N}$ is perpendicular to the surface, while $\vec{f}$ is parallel to the surface.

\item On an incline that makes an angle $\theta$ with the horizontal, if the only forces perpendicular to the surface are the normal force and the perpendicular component of the weight, then
\[
N=mg\cos\theta.
\]
Thus the friction magnitudes are often written as
\[
f_{s,\max}=\mu_s mg\cos\theta,
\qquad
f_k=\mu_k mg\cos\theta.
\]
\end{enumerate}}

\qs{Worked example}{A block of mass $m=6.0\,\mathrm{kg}$ is released from rest on a rough incline that makes an angle $\theta=35^\circ$ with the horizontal. Let $g=9.8\,\mathrm{m/s^2}$ denote the magnitude of the gravitational field. Let the coefficient of static friction be $\mu_s=0.40$ and the coefficient of kinetic friction be $\mu_k=0.30$.

Determine whether the block remains at rest or starts to slide. If it slides, find the magnitude and direction of its acceleration.}

\sol Choose axes so that the $x$-axis is parallel to the incline and positive down the incline, and the $y$-axis is perpendicular to the incline and positive away from the surface. Let $a_x$ and $a_y$ denote the acceleration components in these directions.

The forces on the block are the weight $\vec{W}$, the normal force $\vec{N}$ exerted by the incline, and a friction force $\vec{f}$ exerted by the incline.

Because the block remains on the surface, there is no acceleration perpendicular to the incline, so
\[
a_y=0.
\]
Resolve the weight into components relative to the chosen axes. The component parallel to the incline has magnitude
\[
W_x=mg\sin\theta,
\]
and the component perpendicular to the incline has magnitude
\[
W_y=mg\cos\theta.
\]

Apply Newton's second law perpendicular to the incline:
\[
\sum F_y=ma_y.
\]
Since the positive $y$-direction is away from the surface,
\[
N-mg\cos\theta=0.
\]
Therefore,
\[
N=mg\cos\theta.
\]
Substitute the given values:
\[
N=(6.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})\cos 35^\circ\approx 48.2\,\mathrm{N}.
\]

Now check whether static friction can prevent motion. The component of the weight down the incline is
\[
mg\sin\theta=(6.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})\sin 35^\circ\approx 33.7\,\mathrm{N}.
\]
The maximum possible static-friction magnitude is
\[
f_{s,\max}=\mu_s N=(0.40)(48.2\,\mathrm{N})\approx 19.3\,\mathrm{N}.
\]

To keep the block at rest, static friction would need to balance the $33.7\,\mathrm{N}$ downslope component of the weight by acting upslope with magnitude $33.7\,\mathrm{N}$. But
\[
33.7\,\mathrm{N}>19.3\,\mathrm{N}.
\]
So static friction is not large enough to hold the block at rest. The block starts to slide down the incline.

Once the block is sliding, the friction is kinetic and points up the incline. Its magnitude is
\[
f_k=\mu_k N=(0.30)(48.2\,\mathrm{N})\approx 14.5\,\mathrm{N}.
\]

Now apply Newton's second law parallel to the incline:
\[
\sum F_x=ma_x.
\]
Taking down the incline as positive, the component of the weight is positive and the kinetic friction is negative, so
\[
mg\sin\theta-f_k=ma_x.
\]
Substitute the values:
\[
33.7\,\mathrm{N}-14.5\,\mathrm{N}=(6.0\,\mathrm{kg})a_x.
\]
Thus,
\[
a_x=\frac{19.2\,\mathrm{N}}{6.0\,\mathrm{kg}}\approx 3.2\,\mathrm{m/s^2}.
\]

Therefore, the block does not remain at rest. It slides down the incline with acceleration
\[
3.2\,\mathrm{m/s^2}
\]
down the incline.
