\subsection{Normal Force, Tension, and Constrained Motion}

This subsection treats normal force and tension as contact forces set by constraints. In AP mechanics, a surface constrains motion perpendicular to itself, and an ideal string constrains connected objects to have linked motions.

\dfn{Normal force, tension, and mechanical constraints}{Let a body be in contact with a surface whose outward unit normal is $\hat{n}$. The \emph{normal force} on the body is the contact force exerted by the surface perpendicular to the surface, so
\[
\vec{N}=N\hat{n},
\]
where $N\ge 0$ is its magnitude.

Let a body be attached to a taut string, rope, or cable whose direction at the body is given by the unit vector $\hat{t}$. The \emph{tension force} exerted by that connector acts along the connector, so
\[
\vec{T}=T\hat{t},
\]
where $T\ge 0$ is its magnitude. In the ideal AP model, the string is massless and inextensible, and any pulley is massless and frictionless.

A \emph{constraint} is a geometric restriction on motion. Contact with a rigid surface constrains motion perpendicular to the surface, and an inextensible string constrains connected bodies to move so that their accelerations along the string are related. Therefore normal force and tension are found from the constraint together with Newton's second law, not chosen independently.}

\nt{The symbols $N$ and $T$ do not automatically mean $mg$. The equality $N=mg$ holds only in special cases such as a body on a horizontal surface with no vertical acceleration and no other vertical forces. Likewise, $T=mg$ holds only for special cases such as a hanging mass in equilibrium or moving with constant velocity. On an incline, $N$ is often less than $mg$. In an accelerating elevator, $N$ can be greater than or less than $mg$. In a connected multi-body system, $T$ is usually set by the common acceleration of the system, so it is generally not equal to the weight of either mass.}

\ex{Short example: normal force in an accelerating elevator}{Choose the positive $y$-axis upward. A student of mass $m=60.0\,\mathrm{kg}$ stands on a scale in an elevator that accelerates upward with magnitude $a=2.0\,\mathrm{m/s^2}$. Let $N$ denote the magnitude of the scale's normal force on the student, and let $g=9.8\,\mathrm{m/s^2}$.

Newton's second law in the vertical direction gives
\[
N-mg=ma.
\]
Therefore,
\[
N=m(g+a)=(60.0\,\mathrm{kg})(11.8\,\mathrm{m/s^2})=708\,\mathrm{N}.
\]
So here $N>mg$. The normal force is determined by the acceleration constraint, not by a rule that it must equal the weight.}

\mprop{Operational rules for frictionless contact and ideal strings}{Let $\vec{a}$ denote the acceleration of a body in contact with a frictionless surface, let $\hat{n}$ denote a unit vector perpendicular to the surface, and let $a_{\perp}=\vec{a}\cdot \hat{n}$ denote the acceleration component perpendicular to the surface. Then Newton's second law in the perpendicular direction is
\[
\sum F_{\perp}=ma_{\perp}.
\]
If the contact surface is a fixed plane and the body remains in contact without leaving that plane, then $a_{\perp}=0$, so the perpendicular force equation determines $N$. For a plane at angle $\theta$ to the horizontal, if the only force with a perpendicular component besides $\vec{N}$ is the weight, then
\[
N=mg\cos\theta.
\]
But if another force has a perpendicular component, it must also be included, so $N$ is not automatically $mg\cos\theta$.

Let two or more bodies be connected by one ideal string over massless, frictionless pulleys. Then the tension magnitude is the same everywhere in that string:
\[
T_1=T_2=\cdots=T.
\]
If $a_1$ and $a_2$ denote acceleration components of two connected bodies measured along their allowed directions of motion, then the inextensible-string constraint gives equal magnitudes:
\[
|a_1|=|a_2|.
\]
With a consistent choice of positive directions, this often becomes a signed relation such as
\[
a_1=a_2
\qquad\text{or}\qquad
a_1=-a_2.
\]
Then write one Newton's second law equation for each body and solve those equations together with the constraint relation.}

\qs{Worked example}{A block of mass $m_1=3.0\,\mathrm{kg}$ rests on a frictionless incline that makes an angle $\theta=30^\circ$ with the horizontal. The block is connected by a light inextensible string that passes over a massless frictionless pulley to a hanging block of mass $m_2=2.0\,\mathrm{kg}$. Let $g=9.8\,\mathrm{m/s^2}$.

Find the magnitude and direction of the acceleration of the system, the tension $T$ in the string, and the magnitude of the normal force $N$ exerted on $m_1$ by the incline.}

\sol Choose the system to be both blocks, but write Newton's second-law equations separately for each block.

For block $m_1$, choose the positive $x$-axis up the incline and the positive $y$-axis perpendicular to the incline, away from the surface. Let $a$ denote the common acceleration magnitude. If $m_2$ moves downward, then $m_1$ moves up the incline with the same acceleration magnitude because the string is ideal and inextensible.

For block $m_2$, choose the positive direction downward so that both bodies have acceleration component $+a$ along their chosen directions.

Now identify the forces.

On $m_1$, the forces are the weight $\vec{W}_1$, the normal force $\vec{N}$, and the tension $\vec{T}$. Resolve the weight into components relative to the incline:
\[
W_{1,\parallel}=m_1g\sin\theta,
\qquad
W_{1,\perp}=m_1g\cos\theta.
\]

On $m_2$, the forces are its weight $\vec{W}_2$ downward and the tension $\vec{T}$ upward.

First check the likely direction of motion. The hanging weight has magnitude
\[
m_2g=(2.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})=19.6\,\mathrm{N},
\]
while the component of $m_1$'s weight down the incline has magnitude
\[
m_1g\sin\theta=(3.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})\sin 30^\circ=14.7\,\mathrm{N}.
\]
Since $19.6\,\mathrm{N}>14.7\,\mathrm{N}$, the system accelerates with $m_2$ downward and $m_1$ up the incline, consistent with the chosen positive directions.

Apply Newton's second law to $m_1$ along the incline:
\[
\sum F_{\parallel}=m_1a.
\]
The positive direction is up the incline, so
\[
T-m_1g\sin\theta=m_1a.
\]

Apply Newton's second law to $m_1$ perpendicular to the incline:
\[
\sum F_{\perp}=m_1a_{\perp}.
\]
Because the block stays in contact with the incline, $a_{\perp}=0$. Therefore,
\[
N-m_1g\cos\theta=0,
\]
so
\[
N=m_1g\cos\theta.
\]

Apply Newton's second law to $m_2$ in the downward positive direction:
\[
\sum F=m_2a.
\]
Thus,
\[
m_2g-T=m_2a.
\]

Now solve the two equations containing $a$ and $T$:
\[
T-m_1g\sin\theta=m_1a,
\]
\[
m_2g-T=m_2a.
\]
Add them to eliminate $T$:
\[
m_2g-m_1g\sin\theta=(m_1+m_2)a.
\]
Hence
\[
a=\frac{m_2g-m_1g\sin\theta}{m_1+m_2}.
\]
Substitute the values:
\[
a=\frac{19.6-14.7}{3.0+2.0}\,\mathrm{m/s^2}
=\frac{4.9}{5.0}\,\mathrm{m/s^2}
=0.98\,\mathrm{m/s^2}.
\]

Now find the tension from either block's equation. Using block $m_2$,
\[
m_2g-T=m_2a,
\]
so
\[
T=m_2g-m_2a.
\]
Substitute:
\[
T=(2.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})-(2.0\,\mathrm{kg})(0.98\,\mathrm{m/s^2})
=19.6\,\mathrm{N}-1.96\,\mathrm{N}
=17.64\,\mathrm{N}.
\]
Thus
\[
T\approx 17.6\,\mathrm{N}.
\]

Finally, compute the normal force:
\[
N=m_1g\cos\theta=(3.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})\cos 30^\circ.
\]
Using $\cos 30^\circ\approx 0.866$ gives
\[
N\approx (29.4\,\mathrm{N})(0.866)=25.5\,\mathrm{N}.
\]

Therefore the system accelerates with magnitude
\[
0.98\,\mathrm{m/s^2},
\]
with $m_2$ moving downward and $m_1$ moving up the incline, the string tension is
\[
T\approx 17.6\,\mathrm{N},
\]
and the normal force on the incline block is
\[
N\approx 25.5\,\mathrm{N}.
\]
