---
title: Friction and Curved Motion
module: Dynamics
moduleNumber: 2
lessonNumber: 3
order: 203
summary: >
  Real surfaces grip before they slip, fluids push back harder the faster you move
  through them, and anything rounding a bend must be pulled toward the inside of the
  curve by something. This lesson supplies the force laws for those three cases. We
  bound static friction by $|f_s|\leq\mu_sN$ and switch to kinetic friction $\mu_kN$
  once sliding starts, model drag as a speed-dependent resistance that levels off at a
  terminal speed, and show that circular motion demands an inward net force $mv^2/r$
  furnished by real interactions — friction, a banked normal force, tension — never by
  an invented outward one.
topics: [Dynamics]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 5 — Additional Applications of Newton's Laws; §§5-1–5-3"
---

Static friction adjusts to prevent relative sliding, up to a maximum:

$$
|f_s|\leq \mu_sN.
$$

Once surfaces slide, the usual model is $|f_k|=\mu_kN$, directed opposite the
relative motion. The equality $f_s=\mu_sN$ applies only at impending slip.
Rolling without slipping is different: the contact point is instantaneously at
rest relative to the surface, so static friction may be zero or nonzero.

Determine a static-friction direction from the relative motion that would occur in
the absence of tangential contact response. On a level surface, a horizontal pull
tends to move the block with respect to the surface; on an incline, gravity and any
applied force set the tendency. A trial arrow is sufficient for the force equation.
Its solved sign fixes the physical direction, and its magnitude must then lie within
the static bound.

Fluid drag opposes motion through the fluid. At low Reynolds number it is often
proportional to $v$; for many fast macroscopic motions it is approximately
$D=cv^2$. A falling object reaches terminal speed when its upward drag balances
its weight. With quadratic drag, $mg=cv_t^2$, so

$$
v_t=\sqrt{\frac{mg}{c}}.
$$

$$
% caption: Free-body diagram of an object falling at speed v. Drag acts upward and
% grows with speed; at terminal speed it balances the weight and the acceleration is zero.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[thick] (0,0) circle (0.42);
  \draw[->, acc, very thick] (0,0.42) -- (0,2.15) node[above] {drag};
  \draw[->, black, very thick] (0,-0.42) -- (0,-2.15) node[below] {mg};
  \draw[->, black, thick] (2.0,0.95) -- (2.0,-0.95) node[midway, right] {v};
\end{tikzpicture}
$$

On a curve of radius $r$, resolve forces in radial and tangential directions.
The radial equation is $\sum F_r=mv^2/r$, with inward chosen positive. A car on
a level curve has static friction as its inward force, so the largest speed before
skidding is $v_{\max}=\sqrt{\mu_sgr}$. Banking can supply the inward component
through the normal force and reduce the reliance on friction.

## Friction, drag, and curvature

Static friction is a constraint force. It takes the value required to prevent
sliding, within the bound

$$
|f_s|\leq\mu_sN.
$$

At the threshold of sliding, the equality holds. During sliding, the elementary
model is $|f_k|=\mu_kN$. The coefficients depend on the material pair and surface
condition; neither is a universal constant. A friction force opposes relative
motion or its tendency, not necessarily the selected object's coordinate
velocity.

Fluid resistance is not a contact-friction law. At sufficiently low speed for
small objects, drag is often proportional to speed, $D=bv$. At higher Reynolds
number, it is commonly approximated by $D=cv^2$. A falling body reaches terminal
speed when its acceleration vanishes. With quadratic drag,

$$
mg=cv_t^2,\qquad v_t=\sqrt{\frac{mg}{c}}.
$$

At terminal speed, drag and weight balance while the body continues moving.

A curved path has a local tangent and inward normal, and the acceleration decomposes
along them:

$$
\vec a=\frac{\d v}{\d t}\hat t
+\frac{v^2}{r}\hat n.
$$

The tangential component changes speed; the normal component changes direction.
A level road curve uses static friction as the inward force, so
$\mu_smg\geq mv^2/r$. The no-skid upper speed is
$v_{\max}=\sqrt{\mu_sgr}$. This bound assumes a level road and neglects aerodynamic
downforce; bank angle changes the normal-force geometry.

Static friction is set by force balance, not assigned its maximum at the outset.

> **Worked example.** A $10\ \mathrm{kg}$ crate rests on a level floor with
> $\mu_s=0.25$, pulled horizontally. The static-friction bound is
> $$
> f_{s,\max}=\mu_smg=(0.25)(10\ \mathrm{kg})(9.8\ \mathrm{m\,s^{-2}})=24.5\ \mathrm N.
> $$
> Under an $18\ \mathrm N$ pull the crate stays at rest and static friction adjusts to
> match the pull exactly: $f_s=18\ \mathrm N<24.5\ \mathrm N$, so $a=0$. The friction
> equals what equilibrium demands, not $\mu_sN$. A $30\ \mathrm N$ pull exceeds the
> bound, the crate breaks free, and the kinetic model takes over.

On an incline, $N=mg\cos\theta$ only when no other force has a normal component
and normal acceleration is zero. The downhill gravitational component is
$mg\sin\theta$. The condition for rest is

$$
mg\sin\theta\leq\mu_smg\cos\theta,
\qquad \tan\theta\leq\mu_s.
$$

The equality defines the angle of repose. It is a threshold relation, not the
friction law for every block on every incline.

$$
% caption: A block resting on an incline. The weight acts vertically down; the normal
% force N is perpendicular to the surface and static friction f_s acts up the slope.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0,0) -- (6.0,0) -- (6.0,2.8) -- cycle;
  \draw[black] (1.4,0) arc[start angle=0, end angle=25, radius=1.4];
  \begin{scope}[shift={(3.7,1.73)}, rotate=25]
    \draw[thick] (-0.6,0) rectangle (0.6,0.7);
    \draw[->, very thick] (0,0.7) -- (0,1.85) node[above] {N};
    \draw[->, thick] (0,0.35) -- (-1.35,0.35) node[left] {$f_s$};
  \end{scope}
  \draw[->, black, very thick] (3.44,2.04) -- (3.44,0.55) node[below] {mg};
\end{tikzpicture}
$$

Drag depends on velocity relative to the fluid. In the linear model $D=bv$, a
falling body with downward positive obeys

$$
m\frac{\d v}{\d t}=mg-bv.
$$

Its terminal speed is $v_t=mg/b$. Integration for release from rest yields

$$
v(t)=v_t\left(1-e^{-t/\tau}\right),
\qquad \tau=\frac{m}{b}.
$$

The velocity approaches $v_t$ asymptotically. Quadratic drag has a different time
dependence, but the terminal condition remains $mg-cv_t^2=0$. At terminal speed,
drag and weight balance while the body remains in motion.

Banking replaces friction with geometry. On a frictionless banked road the normal
force tilts inward, and its components carry both weight and the turning force:

$$
N\cos\beta=mg,
\qquad N\sin\beta=\frac{mv^2}{r},
\qquad v=\sqrt{rg\tan\beta}.
$$

The design speed $\sqrt{rg\tan\beta}$ needs no friction. At other speeds static
friction acts up or down the bank, following the impending slip. In a vertical
circle the inward direction changes with position: at the top $T+mg=mv^2/r$, at the
bottom $T-mg=mv^2/r$. A string cannot push, so a taut string at the top requires
$v^2\geq gr$.

> **Worked example.** A curve of radius $r=50\ \mathrm m$ is banked for a design
> speed of $v=20\ \mathrm{m\,s^{-1}}$ with no friction. The bank angle satisfies
> $\tan\beta=v^2/(rg)$:
> $$
> \tan\beta=\frac{(20\ \mathrm{m\,s^{-1}})^2}{(50\ \mathrm m)(9.8\ \mathrm{m\,s^{-2}})}
> =0.816,\qquad \beta=39.2^\circ.
> $$
> At this speed the horizontal component $N\sin\beta$ supplies exactly $mv^2/r$ while
> $N\cos\beta$ balances $mg$; a car needs no sideways friction. Below $20\ \mathrm{m\,s^{-1}}$
> it tends to slide down the bank and friction acts up-slope; above it, friction acts
> down-slope.

Kinetic friction converts a force balance into an acceleration prediction.

> **Worked example.** A $5.0\ \mathrm{kg}$ block slides on a level floor with
> $\mu_k=0.20$ under a $20\ \mathrm N$ horizontal pull. Only gravity and the normal
> force act vertically, so $N=mg$ and
> $$
> f_k=\mu_kmg=(0.20)(5.0\ \mathrm{kg})(9.8\ \mathrm{m\,s^{-2}})=9.8\ \mathrm N.
> $$
> The horizontal net force is $20-9.8=10.2\ \mathrm N$, giving
> $$
> a=\frac{10.2\ \mathrm N}{5.0\ \mathrm{kg}}=2.04\ \mathrm{m\,s^{-2}}.
> $$
> Kinetic friction opposes the block's velocity relative to the floor. Remove the
> pull while it still slides and friction keeps that direction, decelerating the block
> at $\mu_kg=1.96\ \mathrm{m\,s^{-2}}$ until it stops.

For two blocks in contact, friction between them is internal to the two-block
system but external to either separate block. A no-slip condition requires that the
static friction demanded by the acceleration not exceed $\mu_sN$. Compare the
required friction with that inequality to determine whether the objects remain
together.

## Resistance models and circular motion

The linear drag model $D=bv$ is appropriate in regimes where viscous effects
dominate. The quadratic model $D=cv^2$ commonly describes large objects at higher
speeds in air or water. The coefficients have different units: $b$ has units
$\mathrm{kg\,s^{-1}}$ and $c$ has units $\mathrm{kg\,m^{-1}}$. Consequently,
coefficients cannot be compared numerically without their associated model and
units. The actual drag law may cross over between these regimes.

With quadratic drag and downward motion, the net force is $mg-cv^2$. It is positive
below terminal speed and negative above terminal speed. Thus a body moving faster
than terminal speed slows down while still moving downward. Terminal velocity is a
stable equilibrium of the velocity equation.

> **Worked example.** A car rounds a level curve of radius $r=80\ \mathrm m$ with
> $\mu_s=0.60$. Static friction supplies the entire inward force, so the no-skid
> condition is $\mu_smg\geq mv^2/r$. The maximum speed is
> $$
> v_{\max}=\sqrt{\mu_sgr}
> =\sqrt{(0.60)(9.8\ \mathrm{m\,s^{-2}})(80\ \mathrm m)}
> =21.7\ \mathrm{m\,s^{-1}}\approx78\ \mathrm{km\,h^{-1}}.
> $$
> Mass cancels because both the friction bound and the required radial force scale
> with mass. Rain or ice that halves $\mu_s$ drops the safe speed only by
> $\sqrt{2}$, to $15.3\ \mathrm{m\,s^{-1}}$.

When speed changes around a path, the radial and tangential force equations are
separate:

$$
\sum F_r=m\frac{v^2}{r},
\qquad
\sum F_t=m\frac{\d v}{\d t}.
$$

The same tire-contact force can have both components. Its required magnitude is
$\sqrt{F_r^2+F_t^2}$ when the components are perpendicular. A vehicle braking while
turning can exceed the available static-friction limit even when either braking or
turning alone would be possible.

Off the design speed the normal force alone cannot turn the vehicle: below
$\sqrt{rg\tan\beta}$ it tends to slide down the bank and static friction acts
up-slope; above it, friction acts down-slope. The inequality $|f_s|\leq\mu_sN$ then
sets a range of permitted speeds around the design speed rather than one value.

A body on the inside of a circular track requires inward net force at every point.
At the top of the track, downward is inward, so

$$
N+mg=\frac{mv^2}{r}.
$$

Contact persists only while $N\geq0$. The limiting condition is $v^2=gr$ at the
top. Below that speed, the required normal force would be negative, which a rigid
surface cannot provide; the body leaves the track and subsequently follows a
projectile path. At the bottom of the circle,

$$
N-mg=\frac{mv^2}{r}.
$$

The normal force can therefore greatly exceed weight at the bottom, felt as the
apparent heaviness on a fast vertical loop.

> **Worked example.** A $2.0\ \mathrm{kg}$ object falls under linear drag $D=bv$ with
> $b=0.80\ \mathrm{kg\,s^{-1}}$. Balancing $mg$ against $bv$ gives the terminal speed
> and time constant
> $$
> v_t=\frac{mg}{b}=\frac{(2.0\ \mathrm{kg})(9.8\ \mathrm{m\,s^{-2}})}{0.80\ \mathrm{kg\,s^{-1}}}
> =24.5\ \mathrm{m\,s^{-1}},
> \qquad \tau=\frac{m}{b}=2.50\ \mathrm s.
> $$
> Released from rest, $v(t)=v_t(1-e^{-t/\tau})$. After one time constant
> $t=\tau=2.50\ \mathrm s$,
> $$
> v=v_t(1-e^{-1})=(24.5\ \mathrm{m\,s^{-1}})(0.632)=15.5\ \mathrm{m\,s^{-1}}.
> $$
> The speed is not yet terminal: one time constant closes the gap to $e^{-1}\approx37\%$
> of its initial size, not to zero. Approach to $v_t$ is asymptotic.

## Assumptions, energy, and model choice

The elementary friction model treats $\mu_s$ and $\mu_k$ as constants and ignores
the area of apparent contact and speed dependence. Real surfaces can depart from
that model through heating, lubrication, wear, deformation, and vibration. Drag
models similarly omit turbulence details and changing fluid density. These limits
do not invalidate the equations; they identify the regime in which a coefficient
measured under one condition can be used in another.

Radial and tangential axes are local. As a body moves along a curve, their directions
rotate. A fixed Cartesian analysis gives the same result but may require resolving
the changing directions at each position. The radial-tangential form is efficient
when the path geometry is known.

**Energy and force viewpoints.**

Friction work over a sliding distance $d$ on a level surface is

$$
W_f=-f_kd=-\mu_kmgd.
$$

The force equation determines acceleration at an instant; the work expression
determines the associated kinetic-energy change over a displacement. They are
consistent descriptions of the same interaction. For constant kinetic friction,
$v_f^2=v_i^2-2\mu_kgd$ follows from either approach. Static friction may do zero
work at a fixed contact, but it can transfer energy when the contact point moves,
as in an accelerating conveyor or rolling system.

Drag also removes mechanical energy. For a body moving through still air, drag
power is $P_D=\vec D\cdot\vec v=-Dv$. Quadratic drag therefore dissipates
power proportional to $v^3$. This steep speed dependence explains why aerodynamic
losses become prominent at high road and flight speeds. The lost mechanical energy
appears primarily as internal energy of the fluid and object.

Two scaling checks guard the algebra. The radial force $mv^2/r$ depends on speed
squared, so doubling speed on the same curve requires four times the inward force
and four times the tire friction; this holds independent of vehicle mass. Terminal
speed must rise with mass and fall with the drag coefficient, and $\sqrt{\mu_sgr}$
has units $\mathrm{m\,s^{-1}}$ since $gr$ carries $\mathrm{m^2\,s^{-2}}$. A result
with the opposite trend signals an inverted balance. A coefficient measured on dry,
rigid surfaces in still air need not represent wet contact or compressible flow.

## Friction and contact thresholds

Static friction is best viewed as a response range. Before slipping begins, the
surface supplies whatever tangential contact force is needed to satisfy the stated
motion constraint, from zero up to a limiting magnitude. A graph of static-friction
magnitude against a slowly increased applied force follows the applied demand while
the object remains at rest. At the threshold, the response reaches
$\mu_sN$. Beyond that point, the contact state changes to sliding and the elementary
kinetic-friction model applies. The graph is not a material law giving one friction
value for every situation; it is a record of two different contact states.

Static-friction capacity and actual response are distinct. A crate can have a large
maximum capacity while its actual static friction is zero, as when no tangential
force tends to move it. A crate at rest on a slope can instead have nonzero static
friction even with zero coordinate velocity. Friction responds to relative sliding or
to a stated tendency toward relative sliding at the contact interface, not to a label
such as “moving object.”

The normal force must be established independently before a friction limit is used.
A downward push, an upward rope component, an accelerating support, or curved-path
contact can alter $N$ and hence alter both the static bound and the kinetic-friction
prediction. Substituting $N=mg$ without checking can conceal the only force component
that matters to the contact model.

$$
% caption: Friction versus applied force. Static friction rises to match the pull up
% to the limit f_s max; once sliding starts the force drops to the kinetic value f_k.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.0,0) node[right] {applied force};
  \draw[->, black] (0,0) -- (0,3.4) node[above] {friction};
  \draw[acc, very thick] (0,0) -- (3.0,2.7);
  \draw[acc, very thick] (3.0,2.1) -- (5.6,2.1);
  \draw[fill=acc, draw=acc] (3.0,2.7) circle (1.6pt);
  \draw[black, dashed] (0,2.7) -- (3.0,2.7);
  \draw[black, dashed] (0,2.1) -- (3.0,2.1);
  \node[black, left] at (0,2.7) {$f_s$ max};
  \node[black, left] at (0,2.1) {$f_k$};
  \node[black, below] at (3.0,0) {slip};
\end{tikzpicture}
$$

> **Worked example.** A $12\ \mathrm{kg}$ crate is pulled horizontally while a second
> rope lifts it with $30\ \mathrm N$ straight up. The vertical balance sets the normal
> force below $mg$:
> $$
> N=mg-30\ \mathrm N=(12\ \mathrm{kg})(9.8\ \mathrm{m\,s^{-2}})-30\ \mathrm N=87.6\ \mathrm N.
> $$
> With $\mu_s=0.40$ the static bound is $f_{s,\max}=\mu_sN=35.0\ \mathrm N$. A
> horizontal pull of $20\ \mathrm N$ produces exactly $20\ \mathrm N$ of static
> friction and no motion; a pull of $38\ \mathrm N$ exceeds the bound and the crate
> slips. The upward rope lowers the slip threshold without supplying any horizontal
> force, so an unchecked $N=mg$ would overstate the friction the crate can hold.

The reference object for friction is the other surface at the contact. A luggage
case on a moving conveyor can have zero ground-frame speed while the belt slides
beneath it; kinetic friction then acts in the belt's direction on the case. A tire
rolling without slipping has an instantaneous contact point at rest relative to the
road, so static friction is possible even though the vehicle centre moves rapidly.
“Static” classifies the interface, whereas the object may translate rapidly.

Rolling constraints couple translation and rotation. For a wheel of radius $R$
rolling without slip on a fixed surface, the centre speed and angular speed satisfy
$v=R\omega$ with a sign determined by the rotation convention. Static friction
provides the tangential contact interaction required by the other forces and torques;
its value may be zero in uniform rolling on a level surface. Introducing kinetic
friction merely because a wheel is rolling would contradict the no-slip assumption.

A belt moving right at $1.5\ \mathrm{m\,s^{-1}}$ beneath a case initially at rest
illustrates the reference-frame rule. Relative to the belt the case slides left, so
kinetic friction on the case points right and accelerates it until its speed matches
the belt. At that instant the sliding source vanishes; drawing kinetic friction past
speed matching invents a force with no relative motion behind it.

## Drag and terminal motion

Drag is set by velocity relative to the surrounding fluid, not by velocity relative
to the ground. A cyclist riding at $8\ \mathrm{m\,s^{-1}}$ into an
$4\ \mathrm{m\,s^{-1}}$ headwind has air-relative speed
$12\ \mathrm{m\,s^{-1}}$, whereas the same ground speed with a tailwind has
air-relative speed $4\ \mathrm{m\,s^{-1}}$. For quadratic drag, the force magnitude
changes with the square of those values, so the wind effect is far larger than a
simple ground-speed comparison suggests. The drag arrow is always opposite the
relative-fluid velocity vector.

Linear and quadratic drag laws are approximations for different flow regimes. The
linear form applies when viscous effects dominate and the flow remains orderly
around small or slowly moving bodies. The quadratic form better describes
rapid motion of larger bodies through air or water, where inertial flow effects
dominate. Neither coefficient can be transferred between the models: their units,
physical interpretation, and fitted range differ. A graph of measured drag against
speed is more informative than a coefficient quoted without its model.

Drag can also have components. In a crosswind, the air-relative velocity is not
collinear with a vehicle's ground path, so the aerodynamic force can have both
longitudinal and lateral parts. A scalar “drag force” may be adequate for a straight
fall or one-dimensional coast, but a vector model is needed when heading and track
separate.

$$
% caption: Air-relative velocity sets aerodynamic drag. In a headwind the wind velocity
% adds to the ground velocity in the air frame; in a tailwind it subtracts.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[black, left] at (-0.05,2.0) {headwind};
  \draw[->, black] (0.2,2.0) -- (3.0,2.0) node[midway, above] {ground v};
  \draw[->, black] (3.0,2.0) -- (4.5,2.0) node[midway, above] {wind};
  \draw[->, acc, very thick] (0.2,1.25) -- (4.5,1.25) node[midway, below] {relative};
  \node[black, left] at (-0.05,0.1) {tailwind};
  \draw[->, black] (0.2,0.1) -- (3.0,0.1) node[midway, above] {ground v};
  \draw[->, black] (3.0,0.1) -- (1.85,0.1);
  \draw[->, acc, very thick] (0.2,-0.65) -- (1.85,-0.65) node[midway, below] {relative};
\end{tikzpicture}
$$

> **Worked example.** Two identical bicycles travel at $10\ \mathrm{m\,s^{-1}}$ over
> the ground, one into a $3\ \mathrm{m\,s^{-1}}$ headwind, the other with a
> $3\ \mathrm{m\,s^{-1}}$ tailwind. Their air-relative speeds are
> $$
> v_{\rm rel}=10+3=13\ \mathrm{m\,s^{-1}}\quad\text{and}\quad 10-3=7\ \mathrm{m\,s^{-1}}.
> $$
> Quadratic drag $D=cv_{\rm rel}^2$ then gives a drag ratio
> $$
> \frac{D_{\rm head}}{D_{\rm tail}}=\frac{13^2}{7^2}=\frac{169}{49}=3.45.
> $$
> The headwind rider fights nearly $3.5$ times the drag at the same ground speed. The
> ratio comes from relative velocity alone, before any coefficient is fixed.

Terminal speed is the steady solution of the velocity equation. For downward motion
with quadratic drag, the net downward force $mg-cv^2$ is positive below terminal
speed and negative above it, so small deviations decay back toward $v_t$: the
terminal state is stable in a still fluid. The force balance fixes the limiting
value; the velocity-dependent law fixes the approach rate. Linear drag approaches
$v_t$ with a single time constant, quadratic drag along a different curve, and a
sub-terminal body keeps a nonzero acceleration despite substantial drag.

Changes in fluid density, projected area, or orientation change the drag coefficient
and hence the terminal state. A skydiver who changes body posture changes effective
area and shifts terminal speed while still falling. A terminal-speed calculation
should state which orientation and fluid conditions were used; otherwise the
coefficient has no stable physical meaning.

$$
% caption: Net downward force for quadratic drag. Weight is constant; drag grows with
% speed and cancels it at the terminal speed v_t, where the net force crosses zero.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.0,0) node[right] {speed};
  \draw[->, black] (0,-1.6) -- (0,2.6) node[above] {net force};
  \draw[black, thick] (0,1.9) -- (5.6,1.9) node[right] {weight};
  \draw[acc, very thick, domain=0:5.2, samples=60] plot (\x,{1.9-0.09*\x*\x});
  \node[acc, right] at (5.2,{1.9-0.09*5.2*5.2}) {net};
  \draw[fill=acc, draw=acc] (4.6,0) circle (1.6pt);
  \node[black, below] at (4.6,0) {$v_t$};
\end{tikzpicture}
$$

Because $v_t=\sqrt{mg/c}$, doubling mass at fixed geometry raises terminal speed by
$\sqrt2$, not by two, and doubling the effective area (raising $c$) lowers it by
$1/\sqrt2$. These square-root trends confirm the balance was solved for speed rather
than for its square.

## Curved-path force analysis

Curved-path force analysis begins by selecting a local inward normal direction and a
local tangent direction at the position of interest. The inward component of net
force is constrained by the required change in velocity direction. The tangential
component controls the change in speed. A single force can contribute to both, and
several forces can combine to supply either component. The diagram should therefore
show the local axes before labels such as “centripetal force” are used.

Use “centripetal force” for the net inward component of real forces such as friction,
tension, normal force, gravity, or aerodynamic lift. Drawing both tire friction and a
second inward centripetal arrow double-counts that component. Draw the real contact
and field forces, then project their sum onto the inward direction.

When a vehicle both turns and changes speed, the tire-force vector must supply a
radial component and a tangential component. Its required magnitude is the vector
combination of those demands, which is why braking on a curve can skid even when
braking on a straight road and turning at constant speed are each safe. The available
static friction bounds the combined contact demand, not each direction separately.

$$
% caption: Top view of a car braking through a level curve. The inward force mv^2/r and
% the tangential braking force are perpendicular components of one tire contact force.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, dashed] (0,0) circle (2.3);
  \draw[thick] (0.03,1.95) rectangle (0.93,2.55);
  \coordinate (c) at (0.48,2.25);
  \draw[->, very thick] (c) -- (0.086,0.405) node[midway, right] {inward};
  \draw[->, black, very thick] (c) -- (-0.92,2.55) node[above left] {braking};
  \draw[->, acc, thick] (c) -- (-1.31,0.70);
  \node[acc] at (-1.18,1.55) {total};
  \draw[fill=white, draw=black] (0,0) circle (1.6pt) node[below right] {center};
\end{tikzpicture}
$$

> **Worked example.** A $1200\ \mathrm{kg}$ car takes a $60\ \mathrm m$ level curve at
> $15\ \mathrm{m\,s^{-1}}$ while braking at $2.0\ \mathrm{m\,s^{-2}}$. The inward and
> tangential demands are perpendicular:
> $$
> F_r=\frac{mv^2}{r}=\frac{(1200)(15)^2}{60}=4500\ \mathrm N,
> \qquad F_t=ma_t=(1200)(2.0)=2400\ \mathrm N.
> $$
> The tire must supply their vector sum,
> $$
> F=\sqrt{F_r^2+F_t^2}=\sqrt{4500^2+2400^2}=5100\ \mathrm N.
> $$
> With $\mu_s=0.50$ the available bound is $\mu_smg=(0.50)(1200)(9.8)=5880\ \mathrm N$,
> so the contact can meet the combined demand. Checking only $F_r=4500\ \mathrm N$
> against $5880\ \mathrm N$ would ignore the braking term and understate the risk of
> skidding.

Banking changes the direction of the normal contact force. At the design speed on a
frictionless bank, its horizontal inward component supplies the turning requirement
while its vertical component balances weight. Off design speed the friction direction
follows the impending slip, not the bank angle: fix it from a trial tendency or from
the sign of the solved friction force. A slower vehicle slides down-slope, so friction
acts up-slope; a faster one climbs, reversing the arrow. The normal-force geometry is
the same in both cases.

Contact limits appear whenever a normal force would have to reverse direction. On
the inside of a vertical loop, the track can push inward; when the computed normal
becomes zero, the body is at the contact-loss boundary. On the outside of a loop or
over a hill, the surface geometry reverses the allowed normal direction. A diagram
must identify which side is contacted before the inward component equation is
interpreted. The word “normal” does not automatically mean inward.

$$
% caption: Cross-section of a banked curve. The normal force tilts inward; its vertical
% component supports the weight and its horizontal component supplies the turning force.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0,0) -- (5.4,0) -- (5.4,2.3) -- cycle;
  \draw[black] (0.9,0) arc[start angle=0, end angle=23, radius=0.9];
  \begin{scope}[shift={(3.0,1.278)}, rotate=23]
    \draw[thick] (-0.5,0) rectangle (0.5,0.5);
  \end{scope}
  \draw[->, acc, very thick] (3.0,1.278) -- (2.374,2.752) node[above left] {N};
  \draw[->, black, thick] (3.0,1.278) -- (3.0,2.752) node[right] {N cos};
  \draw[->, black, thick] (3.0,1.278) -- (2.374,1.278) node[left] {N sin};
  \draw[->, black, very thick] (3.0,1.278) -- (3.0,-0.322) node[below] {mg};
  \fill[black!70] (3.0,1.278) circle (1.4pt);
\end{tikzpicture}
$$

Friction, drag, and curved-path relations each carry their own required state, and
mixing them is a common error source. Friction coefficients relate tangential to
normal contact force within a stated sticking or sliding state; drag coefficients
relate fluid force to relative-fluid velocity within a flow regime; radial equations
constrain the inward net force for a known path geometry. Three quick checks keep them
separate: friction vanishes when contact is removed, drag reverses when relative-fluid
velocity reverses, and radial demand grows with speed squared at fixed radius.

| Model | Required state | Relation or test |
| --- | --- | --- |
| Static friction | contact remains stuck | $\lvert f_s\rvert\leq\mu_sN$ |
| Kinetic friction | relative surface motion | $\lvert f_k\rvert=\mu_kN$ in the elementary model |
| Drag | velocity relative to the fluid | direction opposite $\vec v-\vec v_{\rm fluid}$ |
| Curved path | local radius and inward axis | $\sum F_{\rm in}=mv^2/r$ |

When a calculation predicts a negative normal force, friction based on that normal
force is no longer meaningful because the assumed contact has been lost. When a
static-friction demand exceeds its bound, replace the sticking condition with a
sliding model. When a drag law is fitted outside its speed range, obtain new data or
state the extrapolation. Each correction changes the model rather than merely
changing a numerical sign. A final result should name the active model and its
source: a coefficient for a stated pair of surfaces, a drag law for a particular flow
regime, or a local radius for a prescribed path.

## Coupled contacts and transient drag

Find static friction in a multi-body system from the acceleration required by the
shared constraint, then compare it with the available bound. The coefficient alone
does not fix the force. For a small block resting on an accelerating platform,
friction must accelerate the block with the platform. Its required magnitude is
$m|a|$. The maximum static friction is $\mu_sN$. The no-slip condition is therefore
$m|a|\leq\mu_sN$, which becomes a limit on platform acceleration.

The platform acceleration may arise from a motor, a pull on another attached body,
or gravity along an incline. Static friction transmits the required tangential
acceleration to the block. Once the demand exceeds the bound, the bodies have
different tangential accelerations and a kinetic-friction model replaces the
no-slip constraint.

The normal force must still be obtained from the normal geometry. A block inside an
accelerating elevator or pressed by an angled force can have a friction capacity very
different from $\mu_smg$. In a vertical support problem, friction may hold a block
against a wall while normal force is supplied by a horizontal push. Weight then
creates the tangential tendency, and the required static friction is set by that
weight rather than by horizontal motion.

$$
% caption: A package rides an accelerating platform. Static friction is the only
% horizontal force on the package and must supply m a; beyond its limit the package slips.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[fill=black!8, draw=black, thick] (0,0) rectangle (5.2,0.5);
  \draw[thick] (1.6,0.5) rectangle (2.8,1.4);
  \draw[->, acc, very thick] (2.2,0.95) -- (3.6,0.95) node[right] {$f_s$};
  \draw[->, black, very thick] (4.4,0.95) -- (5.7,0.95) node[right] {a};
  \node[black, below] at (2.6,0) {platform};
\end{tikzpicture}
$$

> **Worked example.** A $1.5\ \mathrm{kg}$ package rides a horizontal platform with
> $\mu_s=0.35$. Static friction alone accelerates it, so the no-slip demand $m|a|$ must
> stay under $\mu_sN$. With $N=mg=14.7\ \mathrm N$ the bound is $\mu_sN=5.15\ \mathrm N$,
> giving a top platform acceleration
> $$
> a_{\max}=\frac{\mu_sN}{m}=\frac{5.15\ \mathrm N}{1.5\ \mathrm{kg}}=3.43\ \mathrm{m\,s^{-2}}.
> $$
> At $a=2.0\ \mathrm{m\,s^{-2}}$ the actual friction is only $m a=3.0\ \mathrm N$, well
> below the bound. At $a=4.0\ \mathrm{m\,s^{-2}}$ the demand $6.0\ \mathrm N$ exceeds
> $5.15\ \mathrm N$, and the package slides backward relative to the platform.

On an incline the sticking condition is a two-sided bound on the applied force, not a
single case. A small uphill pull leaves a downhill tendency, so friction points
uphill; a large uphill pull reverses the tendency and friction points downhill;
between them one pull needs no friction. Take uphill as positive: the required static
friction is the negative of the non-friction tangential sum, and $|f_s|\leq\mu_sN$
sets a lowest and highest allowed pull.

$$
% caption: A block held on an incline by an up-slope pull P. Static friction reverses
% direction as P passes the downslope gravity component, giving a two-sided sticking range.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0,0) -- (6.0,0) -- (6.0,2.8) -- cycle;
  \begin{scope}[shift={(3.7,1.73)}, rotate=25]
    \draw[thick] (-0.6,0) rectangle (0.6,0.7);
    \draw[->, very thick] (0,0.35) -- (1.55,0.35) node[right] {P};
    \draw[<->, acc, thick] (-0.05,0.9) -- (-1.35,0.9) node[left] {$f_s$ range};
  \end{scope}
  \draw[->, black, very thick] (3.44,2.04) -- (3.44,0.55) node[below] {mg};
\end{tikzpicture}
$$

> **Worked example.** A $5.0\ \mathrm{kg}$ block sits on a $25^\circ$ incline with
> $\mu_s=0.30$. The normal force and static bound are
> $$
> N=mg\cos25^\circ=44.4\ \mathrm N,\qquad
> f_{s,\max}=\mu_sN=13.3\ \mathrm N,
> $$
> and the downslope gravity component is $mg\sin25^\circ=20.7\ \mathrm N$. An uphill
> pull $P$ leaves a required friction $20.7-P$ (positive means uphill). The rest
> condition $|20.7-P|\leq13.3$ gives
> $$
> 7.4\ \mathrm N\leq P\leq34.0\ \mathrm N.
> $$
> At $P=20.7\ \mathrm N$ friction is zero; below it friction acts uphill, above it
> downhill. Outside the interval the block slides and the kinetic model applies.

A velocity-time record also carries information about the drag law before terminal
speed is reached. Under linear drag the gap to terminal speed decays by the same
fraction over equal time intervals; under quadratic drag the approach has a different
shape. Plotting acceleration against speed distinguishes a linear trend from a
parabolic one when mass and gravity are known.

The initial acceleration of a released falling object is close to $g$ only when its
initial air-relative speed is small. A body thrown downward faster than its terminal
speed has an upward drag force larger than its weight and therefore has upward
acceleration in a downward-positive coordinate system. It continues downward while
slowing. The signs of velocity and acceleration are independent; the force model
must be evaluated at the actual relative-fluid velocity rather than at an assumed
direction of motion.

Position measurements introduce a second integration step. A drag force law predicts
acceleration from velocity, velocity from elapsed time, and displacement from the
area under velocity. Using a terminal speed as though it applied throughout a fall
overestimates distance during the accelerating portion. Conversely, using the
initial speed throughout a long descent ignores the later drag-limited regime.

> **Worked example.** A falling sensor under linear drag has terminal speed
> $v_t=18\ \mathrm{m\,s^{-1}}$ and time constant $\tau=3.0\ \mathrm s$. After one time
> constant its speed is $v_t(1-e^{-1})=11.4\ \mathrm{m\,s^{-1}}$, but the distance is
> not $11.4(3.0)$ because the speed was lower earlier. Integrating
> $v(t)=v_t(1-e^{-t/\tau})$ over $0$ to $\tau$,
> $$
> x=v_t\bigl[\tau-\tau(1-e^{-1})\bigr]=18\bigl[3.0-3.0(1-e^{-1})\bigr]=19.9\ \mathrm m.
> $$
> Assuming terminal speed throughout would give $v_t\tau=54\ \mathrm m$, nearly triple
> the true drop, so the terminal approximation cannot be used during the transient.

In two dimensions drag is antiparallel to the full velocity relative to the fluid, not
to the ground track. Build the relative-fluid velocity $\vec v-\vec v_{\rm fluid}$
before applying a linear or quadratic law: its magnitude sets the drag magnitude and
its unit vector sets the direction. Reducing ground, wind, and current speeds to
unsigned magnitudes before subtracting loses the lateral component that drives
cross-track acceleration. The same construction serves a rising balloon, a boat in a
current, and a vehicle in a gust. If a crosswind blows east
and an object has no eastward ground velocity, its air-relative velocity points west;
drag on the object points east. As the object acquires eastward speed, the horizontal
drag component decreases and can change sign only after the object exceeds the wind
speed eastward. A scalar drag magnitude cannot capture that reversal.

> **Worked example.** A weather balloon rises at $4.0\ \mathrm{m\,s^{-1}}$ and drifts
> east at $2.0\ \mathrm{m\,s^{-1}}$ over the ground while a $6.0\ \mathrm{m\,s^{-1}}$
> east wind blows. Its air-relative velocity subtracts the wind:
> $$
> \vec v-\vec v_{\rm fluid}=(2.0-6.0)\,\hat\imath+4.0\,\hat\jmath
> =-4.0\,\hat\imath+4.0\,\hat\jmath\ \mathrm{m\,s^{-1}}.
> $$
> Drag opposes this vector, so it points east and down: an eastward component
> $+4.0$ and a downward component from the $+4.0$ rise. A diagram drawing drag purely
> downward would drop the eastward component that changes the balloon's lateral drift.

## Contact loss and variable curvature

Vertical circles provide a precise test of local force directions because the inward
axis changes from one point to another. At the top of an inside loop, both weight
and a taut string's tension point inward. At the bottom, tension points inward while
weight points outward relative to the centre. The force arrows remain tied to their
interaction sources; only their signs in the local inward equation change. Reusing
the top equation at the bottom reverses the weight contribution and produces a
normal-force error of twice the weight.

The limiting top condition for a string or an inside track is contact retention.
At the boundary, tension or normal force becomes zero, not negative. The required
inward force is then supplied by weight alone. A speed below this boundary means the
object cannot follow the assumed circle at the top; it leaves the guide. A speed
above it increases tension or normal force and maintains contact. This is a contact
model check, not a separate conservation-law condition.

$$
% caption: Inside a vertical circle the inward axis points to the centre at both top and
% bottom, but the weight reverses relative to it, so the normal force differs top to bottom.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, dashed] (0,0) circle (1.9);
  \draw[thick] (0,1.9) circle (0.3);
  \draw[thick] (0,-1.9) circle (0.3);
  \draw[->, acc, very thick] (0,1.6) -- (0,0.55) node[midway, left] {N};
  \draw[->, black, very thick] (0,1.6) -- (0,0.9) node[midway, right] {mg};
  \draw[->, acc, very thick] (0,-1.6) -- (0,-0.55) node[midway, left] {N};
  \draw[->, black, very thick] (0,-2.2) -- (0,-2.9) node[below] {mg};
  \node[black, right] at (0.35,1.9) {top};
  \node[black, right] at (0.35,-1.9) {bottom};
\end{tikzpicture}
$$

> **Worked example.** A $0.50\ \mathrm{kg}$ mass runs on the inside of a vertical track
> of radius $r=2.0\ \mathrm m$ at $v=8.0\ \mathrm{m\,s^{-1}}$. The inward force needed is
> $$
> \frac{mv^2}{r}=\frac{(0.50)(8.0)^2}{2.0}=16\ \mathrm N,
> \qquad mg=(0.50)(9.8)=4.9\ \mathrm N.
> $$
> At the top, weight points inward, so $N+mg=mv^2/r$ gives $N=16-4.9=11.1\ \mathrm N$.
> At the bottom, weight points outward, so $N-mg=mv^2/r$ gives $N=16+4.9=20.9\ \mathrm N$.
> The same speed presses the track almost twice as hard at the bottom, and contact at
> the top would fail below $v^2=gr$, i.e. $v=\sqrt{(9.8)(2.0)}=4.4\ \mathrm{m\,s^{-1}}$.

A banked curve uses local vertical and inward-horizontal axes distinct from the tilted
surface normal. Resolve the normal force into those two axes: at the design speed
$N\cos\beta=mg$ and $N\sin\beta=mv^2/r$ with no surface friction. Setting the full
normal force equal to both weight and the inward turning force uses one vector twice
without resolving it. Away from design speed, static friction adds a signed component
along the surface, its direction fixed by the predicted slip.

Two feasibility checks are distinct: contact needs $N\geq0$, sticking needs
$|f_s|\leq\mu_sN$. A vehicle can stay pressed into the road yet skid when the required
friction exceeds the bound; conversely a curved guide can lose contact with no friction
model involved.

The bank calculation holds two simultaneous component conditions; keeping them
separate shows whether friction is absent, required, or unable to maintain contact.

| Condition | Vertical component | Inward component |
| --- | --- | --- |
| Frictionless design speed | $N\cos\beta=mg$ | $N\sin\beta=mv^2/r$ |
| Lower or higher speed | include signed friction component | include the same friction component |
| Contact boundary | $N=0$ | path can no longer impose the assumed curvature |

The radial equation uses the local radius of curvature, not the radius of a circle
drawn elsewhere in the problem. Where a track straightens the radius is effectively
infinite and the inward demand is zero; where it tightens the demand rises at fixed
speed, so a vehicle can meet the tire-friction limit on a broad section and exceed it
on a tighter one without speeding up. Draw the force diagram at the location of
interest, inward axis aimed at that point's centre of curvature.

Path constraints have a hierarchy: a guide first requires contact through a physically
allowed normal force, and only while contact holds can a rough guide add a bounded
tangential friction. When $N$ falls to zero the friction model based on that contact
vanishes with it. Loop problems often err here, correctly finding $N=0$ at the top but
then continuing to apply a surface coefficient. Aerodynamic downforce works the other
way: it raises the normal loading and so the available tire friction without changing
mass, which is why the level-road formula $N=mg$ is a special case, not a universal
law of turning.