\subsection{System Choice, Free-Body Diagrams, and Newton's Laws}

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.

\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.

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 $S$.

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 $S$.

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
\[
\vec{F}_{\mathrm{ext}}=\sum_{i=1}^n \vec{F}_i=\vec{F}_1+\vec{F}_2+\cdots+\vec{F}_n.
\]
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.}

\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.

\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
\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.

\item \textbf{Newton II.} For the chosen body,
\[
\vec{F}_{\mathrm{ext}}=m\vec{a}.
\]
This is the main working law for AP mechanics.

\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
\[
\vec{F}_{B\to A}=-\vec{F}_{A\to B}.
\]
These two forces act on different bodies, so they do not cancel on a single free-body diagram.
\end{enumerate}}

\wc{Net force causes acceleration, not motion}{A body can move at constant velocity with \emph{zero} net force (Newton's law I). Net force determines \emph{how} motion changes, not \emph{if} motion exists. An object with zero net force does not stop --- it continues at whatever velocity it already had.}

\pf{Why the vector law becomes component equations}{Choose Cartesian axes with unit vectors $\hat{\imath}$ and $\hat{\jmath}$. Let the acceleration be
\[
\vec{a}=a_x\hat{\imath}+a_y\hat{\jmath},
\]
where $a_x$ and $a_y$ are scalar components. Let the net external force be
\[
\vec{F}_{\mathrm{ext}}=(\sum F_x)\hat{\imath}+(\sum F_y)\hat{\jmath},
\]
where $\sum F_x$ and $\sum F_y$ are the scalar sums of force components along the chosen axes.

Substitute these into Newton II:
\[
(\sum F_x)\hat{\imath}+(\sum F_y)\hat{\jmath}=m a_x\hat{\imath}+m a_y\hat{\jmath}.
\]
Because the unit vectors $\hat{\imath}$ and $\hat{\jmath}$ are independent, the corresponding scalar components must match. Therefore,
\[
\sum F_x=ma_x,
\qquad
\sum F_y=ma_y.
\]
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.}

\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.}

See also the kinematics in Unit 1 for the constant-velocity case, and Unit 3 for the work-energy connection to zero net force.

\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.

Find the acceleration of the block and the magnitude of the normal force exerted by the incline on the block.}

\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 $\vec{N}$.

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 block stays in contact with the plane, so there is no acceleration perpendicular to the surface. Thus
\[
a_y=0.
\]
Resolve the weight into components relative to the chosen axes. The component of $\vec{W}$ parallel to the incline has magnitude
\[
W_x=mg\sin\theta,
\]
and the component of $\vec{W}$ perpendicular to the incline has magnitude
\[
W_y=mg\cos\theta.
\]
These are components of the single force $\vec{W}$; they are not extra forces on the free-body diagram.

Now apply Newton II by components.

Along the incline,
\[
\sum F_x=ma_x.
\]
The only force component along the incline is $mg\sin\theta$ in the positive $x$-direction, so
\[
mg\sin\theta=ma_x.
\]
Cancel $m$:
\[
a_x=g\sin\theta.
\]
Substitute the stated values:
\[
a_x=(9.8\,\mathrm{m/s^2})\sin 30^\circ=(9.8\,\mathrm{m/s^2})(0.50)=4.9\,\mathrm{m/s^2}.
\]
So the block accelerates at
\[
4.9\,\mathrm{m/s^2}
\]
down the incline.

Perpendicular to the incline,
\[
\sum F_y=ma_y.
\]
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 $a_y=0$,
\[
N-mg\cos\theta=0.
\]
Therefore,
\[
N=mg\cos\theta.
\]

\wc{Normal force is not always equal to weight}{On a horizontal surface with only gravity and normal force, $N=mg$. But on an incline $N=mg\cos\theta$, and if another force pushes on the object, $N$ adjusts accordingly. The normal force is \emph{whatever value is required} to prevent the object from penetrating the surface. Never assume $N=mg$.}

Substitute the stated values:
\[
N=(5.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})\cos 30^\circ.
\]
Using $\cos 30^\circ\approx 0.866$ gives
\[
N\approx (49.0\,\mathrm{N})(0.866)=42.4\,\mathrm{N}.
\]

Therefore, the block's acceleration is
\[
4.9\,\mathrm{m/s^2}
\]
down the incline, and the magnitude of the normal force is
\[
42.4\,\mathrm{N}.
\]
