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concepts/mechanics/u2/m2-5-friction.tex
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concepts/mechanics/u2/m2-5-friction.tex
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\subsection{Static and Kinetic Friction}
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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.
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\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.
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The friction force acts parallel to the contact surface and opposes the relative motion or the tendency of relative motion between the surfaces.
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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:
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\[
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f_s\le \mu_s N,
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\]
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where $\mu_s$ is the coefficient of static friction.
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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,
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\[
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f_k=\mu_k N,
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\]
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where $\mu_k$ is the coefficient of kinetic friction.
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The coefficients $\mu_s$ and $\mu_k$ are dimensionless constants for the pair of surfaces in the chosen model.}
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\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 $f_s=\mu_s N$.}
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\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 $\mu_s N=5.0\,\mathrm{N}$.
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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
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\[
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\vec{f}_s=-(3.0\,\mathrm{N})\hat{\imath},
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\]
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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.}
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\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.
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\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
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\item If there is no slipping at the contact, then the friction is static. Its magnitude must satisfy
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\[
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f_s\le \mu_s N.
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\]
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Its direction is opposite the direction the body would move \emph{relative to the surface} if friction were absent.
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\item If the body slides relative to the surface, then the friction is kinetic. Its magnitude is
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\[
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f_k=\mu_k N,
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\]
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and its direction is opposite the relative velocity of the sliding surfaces.
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\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.
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\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
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\[
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N=mg\cos\theta.
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\]
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Thus the friction magnitudes are often written as
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\[
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f_{s,\max}=\mu_s mg\cos\theta,
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\qquad
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f_k=\mu_k mg\cos\theta.
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\]
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\end{enumerate}}
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\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 $\mu_k=0.30$.
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Determine whether the block remains at rest or starts to slide. If it slides, find the magnitude and direction of its acceleration.}
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\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.
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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.
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Because the block remains on the surface, there is no acceleration perpendicular to the incline, so
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\[
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a_y=0.
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\]
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Resolve the weight into components relative to the chosen axes. The component parallel to the incline has magnitude
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\[
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W_x=mg\sin\theta,
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\]
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and the component perpendicular to the incline has magnitude
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\[
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W_y=mg\cos\theta.
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\]
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Apply Newton's second law perpendicular to the incline:
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\[
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\sum F_y=ma_y.
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\]
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Since the positive $y$-direction is away from the surface,
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\[
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N-mg\cos\theta=0.
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\]
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Therefore,
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\[
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N=mg\cos\theta.
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\]
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Substitute the given values:
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\[
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N=(6.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})\cos 35^\circ\approx 48.2\,\mathrm{N}.
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\]
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Now check whether static friction can prevent motion. The component of the weight down the incline is
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\[
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mg\sin\theta=(6.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})\sin 35^\circ\approx 33.7\,\mathrm{N}.
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\]
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The maximum possible static-friction magnitude is
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\[
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f_{s,\max}=\mu_s N=(0.40)(48.2\,\mathrm{N})\approx 19.3\,\mathrm{N}.
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\]
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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
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\[
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33.7\,\mathrm{N}>19.3\,\mathrm{N}.
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\]
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So static friction is not large enough to hold the block at rest. The block starts to slide down the incline.
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Once the block is sliding, the friction is kinetic and points up the incline. Its magnitude is
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\[
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f_k=\mu_k N=(0.30)(48.2\,\mathrm{N})\approx 14.5\,\mathrm{N}.
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\]
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Now apply Newton's second law parallel to the incline:
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\[
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\sum F_x=ma_x.
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\]
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Taking down the incline as positive, the component of the weight is positive and the kinetic friction is negative, so
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\[
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mg\sin\theta-f_k=ma_x.
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\]
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Substitute the values:
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\[
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33.7\,\mathrm{N}-14.5\,\mathrm{N}=(6.0\,\mathrm{kg})a_x.
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\]
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Thus,
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\[
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a_x=\frac{19.2\,\mathrm{N}}{6.0\,\mathrm{kg}}\approx 3.2\,\mathrm{m/s^2}.
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\]
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Therefore, the block does not remain at rest. It slides down the incline with acceleration
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\[
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3.2\,\mathrm{m/s^2}
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\]
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down the incline.
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