134 lines
6.1 KiB
TeX
134 lines
6.1 KiB
TeX
\subsection{System Choice, Free-Body Diagrams, and Newton's Laws}
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This subsection gives the standard AP mechanics workflow: choose a system, identify the interactions on that system, draw a free-body diagram, choose axes, and then apply Newton's laws in an inertial frame.
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\dfn{System choice, interactions, free-body diagrams, and net external force}{Let $S$ denote a chosen \textbf{system}. In this subsection, $S$ will usually be a single body treated as a particle or rigid object.
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An \textbf{interaction} is a physical influence between the system $S$ and something in the environment, such as gravity from Earth or contact with a surface, rope, or another object. Each interaction can exert a force on $S$.
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A \textbf{free-body diagram} for $S$ is a diagram that shows only the forces acting \emph{on} $S$. If a force is exerted by the environment on $S$, that force belongs on the free-body diagram. Forces exerted by $S$ on other objects do not belong on the free-body diagram of $S$.
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Let $\vec{F}_{\mathrm{ext}}$ denote the net external force on $S$. If the external forces acting on $S$ are $\vec{F}_1$, $\vec{F}_2$, $\dots$, and $\vec{F}_n$, then
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\[
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\vec{F}_{\mathrm{ext}}=\sum_{i=1}^n \vec{F}_i=\vec{F}_1+\vec{F}_2+\cdots+\vec{F}_n.
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\]
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This is a vector sum. After axes are chosen, one may resolve a force into components for calculation, but those components are not additional physical forces to be added to the free-body diagram.}
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\thm{Newton's laws in AP-usable form}{Work in an inertial reference frame. Let $m$ denote the mass of the chosen body, let $\vec{v}$ denote its velocity, let $\vec{a}$ denote its acceleration, and let $\vec{F}_{\mathrm{ext}}$ denote the vector sum of all external forces on that body.
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\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
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\item \textbf{Newton I.} If $\vec{F}_{\mathrm{ext}}=\vec{0}$, then $\vec{a}=\vec{0}$. Thus the body is either at rest or moves with constant velocity.
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\item \textbf{Newton II.} For the chosen body,
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\[
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\vec{F}_{\mathrm{ext}}=m\vec{a}.
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\]
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This is the main working law for AP mechanics.
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\item \textbf{Newton III.} If body $A$ exerts a force $\vec{F}_{A\to B}$ on body $B$, then body $B$ exerts a force $\vec{F}_{B\to A}$ on body $A$ such that
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\[
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\vec{F}_{B\to A}=-\vec{F}_{A\to B}.
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\]
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These two forces act on different bodies, so they do not cancel on a single free-body diagram.
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\end{enumerate}}
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\pf{Why the vector law becomes component equations}{Choose Cartesian axes with unit vectors $\hat{\imath}$ and $\hat{\jmath}$. Let the acceleration be
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\[
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\vec{a}=a_x\hat{\imath}+a_y\hat{\jmath},
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\]
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where $a_x$ and $a_y$ are scalar components. Let the net external force be
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\[
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\vec{F}_{\mathrm{ext}}=(\sum F_x)\hat{\imath}+(\sum F_y)\hat{\jmath},
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\]
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where $\sum F_x$ and $\sum F_y$ are the scalar sums of force components along the chosen axes.
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Substitute these into Newton II:
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\[
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(\sum F_x)\hat{\imath}+(\sum F_y)\hat{\jmath}=m a_x\hat{\imath}+m a_y\hat{\jmath}.
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\]
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Because the unit vectors $\hat{\imath}$ and $\hat{\jmath}$ are independent, the corresponding scalar components must match. Therefore,
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\[
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\sum F_x=ma_x,
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\qquad
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\sum F_y=ma_y.
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\]
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In practice, one first draws only the actual forces on the free-body diagram, then chooses axes, and only then resolves forces into components if that makes the equations simpler.}
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\cor{Zero net force and equilibrium}{If $\vec{F}_{\mathrm{ext}}=\vec{0}$, then Newton II gives $\vec{a}=\vec{0}$. This does \emph{not} mean the velocity must be zero. A body can have zero net force while moving with a nonzero constant velocity. A special case is equilibrium: if a body is initially at rest and $\vec{F}_{\mathrm{ext}}=\vec{0}$, then it remains at rest.}
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\qs{Worked example}{A block of mass $m=5.0\,\mathrm{kg}$ rests on a frictionless incline that makes an angle $\theta=30^\circ$ with the horizontal. Let $g=9.8\,\mathrm{m/s^2}$ denote the magnitude of the gravitational field near Earth. Choose the system to be the block.
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Find the acceleration of the block and the magnitude of the normal force exerted by the incline on the block.}
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\sol The system is the block. The interactions on the block are the gravitational interaction with Earth and the contact interaction with the incline. Therefore the free-body diagram contains only two forces: the weight $\vec{W}$ exerted by Earth on the block and the normal force $\vec{N}$ exerted by the incline on the block. Let $N$ denote the magnitude of $\vec{N}$.
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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 block stays in contact with the plane, so there is no acceleration perpendicular to the surface. Thus
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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 of $\vec{W}$ 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 of $\vec{W}$ 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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These are components of the single force $\vec{W}$; they are not extra forces on the free-body diagram.
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Now apply Newton II by components.
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Along the incline,
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\[
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\sum F_x=ma_x.
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\]
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The only force component along the incline is $mg\sin\theta$ in the positive $x$-direction, so
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\[
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mg\sin\theta=ma_x.
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\]
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Cancel $m$:
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\[
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a_x=g\sin\theta.
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\]
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Substitute the stated values:
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\[
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a_x=(9.8\,\mathrm{m/s^2})\sin 30^\circ=(9.8\,\mathrm{m/s^2})(0.50)=4.9\,\mathrm{m/s^2}.
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\]
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So the block accelerates at
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\[
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4.9\,\mathrm{m/s^2}
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\]
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down the incline.
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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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The positive $y$-direction is away from the surface, so the normal force is positive and the perpendicular component of the weight is negative. Since $a_y=0$,
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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 stated values:
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\[
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N=(5.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})\cos 30^\circ.
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\]
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Using $\cos 30^\circ\approx 0.866$ gives
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\[
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N\approx (49.0\,\mathrm{N})(0.866)=42.4\,\mathrm{N}.
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\]
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Therefore, the block's acceleration is
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\[
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4.9\,\mathrm{m/s^2}
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\]
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down the incline, and the magnitude of the normal force is
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\[
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42.4\,\mathrm{N}.
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\]
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