---
title: Free-Body Diagrams
module: Dynamics
moduleNumber: 2
lessonNumber: 2
order: 202
summary: >
  Once several forces act on a body at once, the reliable way to predict its motion is
  to isolate that one body and draw every external push and pull on it — nothing more,
  nothing less. The free-body diagram is that discipline. We fix a system boundary,
  resolve $\sum\vec F=m\vec a$ into components along axes chosen to fit the geometry,
  and solve for the unknowns a problem hands us — normal forces, tensions, friction,
  and the acceleration a constraint permits — seeing why internal forces drop out only
  when the boundary encloses both bodies that share them.
topics: [Dynamics]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 4 — Newton's Laws; §§4-5–4-8"
---

## System boundaries and force inventory

A free-body diagram isolates one selected object or system. Every arrow represents
an external interaction acting on that selected boundary. Motion arrows, coordinate
axes, and forces exerted by the selected object on its surroundings are excluded.
Fix the system boundary before judging the diagram complete. The same rope force is
external to a block but internal to a block-rope system.

> **Definition (constraint).** A constraint restricts the relative positions or
> motions of bodies. Contact, a taut string, a rigid rod, and a prescribed track
> are constraints; their forces enforce the permitted motion.

Field interactions belong in the inventory even when no object touches the selected
body. Gravity, electric fields, and magnetic fields have sources outside the drawn
boundary and contribute force arrows only when the stated model predicts a coupling.
Labeling the source prevents a field force from being confused with a contact reaction
or with a coordinate-direction arrow.

The resulting component equations are

$$
\sum F_x=ma_x,
\qquad
\sum F_y=ma_y,
$$

with axes selected to match the geometry. A negative solved force means the
assumed arrow direction is opposite the physical direction; it does not invalidate
the equation.

$$
% caption: A block isolated from its surroundings. Only external interactions are drawn:
% the upward normal force N, the downward weight mg, and a horizontal applied pull.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (-0.3,0) -- (4.4,0);
  \draw[thick] (1.3,0) rectangle (2.6,1.0);
  \draw[->, very thick] (1.95,1.0) -- (1.95,2.25) node[above] {N};
  \draw[->, black, very thick] (1.95,0.5) -- (1.95,-1.25) node[below] {mg};
  \draw[->, very thick] (2.6,0.5) -- (4.0,0.5) node[right] {pull};
\end{tikzpicture}
$$

**Contact and incline constraints.**

The normal force is perpendicular to the local contact surface. Its magnitude is
not universally $mg$; it follows from the normal component equation. For a block on
an incline at angle $\theta$ with no acceleration normal to the surface,

$$
N-mg\cos\theta=0,
\qquad N=mg\cos\theta.
$$

Along the incline, gravity contributes $mg\sin\theta$ downhill. Static or kinetic
friction, when present, is tangent to the surface. These axes avoid resolving the
normal force itself and place the constraint directly in the normal equation.

$$
% caption: Incline-aligned axes for a block on a slope. The normal force N is
% perpendicular to the surface, friction f acts along it, and the weight mg is vertical.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0,0) -- (6.0,0) -- (6.0,3.0) -- cycle;
  \draw[black] (1.4,0) arc[start angle=0, end angle=27, radius=1.4];
  \begin{scope}[shift={(3.6,1.83)}, rotate=27]
    \draw[thick] (-0.6,0) rectangle (0.6,0.7);
    \draw[->, very thick] (0,0.7) -- (0,1.9) node[above] {N};
    \draw[->, thick] (0,0.35) -- (-1.35,0.35) node[left] {f};
  \end{scope}
  \draw[->, black, very thick] (3.33,2.15) -- (3.33,0.55) node[below] {mg};
\end{tikzpicture}
$$

**Strings, pulleys, and connected bodies.**

An ideal massless taut string has one tension magnitude along a straight segment.
A frictionless massless pulley changes tension direction but not magnitude. String
length constrains the accelerations of connected objects. For a simple two-mass
pulley system, the accelerations have equal magnitudes and opposite directions.

$$
% caption: A massless string transmits one tension around an ideal pulley, so the two
% hanging masses share a single tension magnitude and equal, opposite accelerations.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (2.9,3.15) circle (0.42);
  \draw[black, thick] (2.48,3.15) -- (2.48,1.5);
  \draw[black, thick] (3.32,3.15) -- (3.32,2.65);
  \draw[fill=acc!12, draw=acc, thick] (2.0,0.75) rectangle (2.96,1.5);
  \node at (2.48,1.12) {$m_1$};
  \draw[fill=black!10, draw=black, thick] (2.84,1.9) rectangle (3.8,2.65);
  \node at (3.32,2.27) {$m_2$};
  \draw[->, acc, very thick] (2.48,0.6) -- (2.48,-0.15) node[below] {a};
  \draw[->, black, very thick] (4.05,1.95) -- (4.05,2.7) node[above] {a};
\end{tikzpicture}
$$

For $m_1$ descending and $m_2$ rising,

$$
m_1g-T=m_1a,
\qquad T-m_2g=m_2a.
$$

Addition removes tension and produces

$$
a=\frac{(m_1-m_2)g}{m_1+m_2}.
$$

This holds in the ideal model: massless string, frictionless massless pulley,
and a taut string throughout, so the two blocks share one acceleration magnitude
and one tension.

## Diagram construction and coordinate choices

Write an interaction inventory before drawing a free-body diagram. State the selected
body, list every object that touches it or acts on it at a distance, and translate
each interaction into one force arrow on the selected body. A table can exert a
normal force and perhaps friction; Earth exerts
weight; a string exerts tension along its own direction; a hand, spring, or motor
can exert an applied contact force. Each item should have an identifiable source.

The inventory prevents double counting of contact and gravitational interactions.
“Normal force” and “support force” are usually two names for the same contact
interaction, not separate arrows. A block
does not receive both $mg$ and a second “gravity force.” The reaction force to the
block's weight acts on Earth and belongs on Earth's diagram, not on the block's.
Likewise, an acceleration arrow may be placed nearby as analysis information, but it
is not one of the forces in the sum.

The number of arrows is set by interactions, not by the number of coordinate axes.
A two-dimensional diagram can have three, four, or more external forces. Conversely,
a body moving in a curved path can have only one external force. Coordinates are
chosen after the interactions are identified, so an inconvenient axis never causes a
real force to disappear from the inventory.

Applied-force labels need a physical source as well. “Pull” may denote a hand, a
motorized cable, a spring, or a fluid actuator; those models can impose different
directions, time dependences, and reaction forces. State the source whenever later
parts of the calculation require a system boundary or a third-law partner.

A lamp hangs motionless from a vertical cable. The selected body is the lamp. Its
inventory contains cable tension upward and Earth's gravitational force downward.
The ceiling exerts a force on the cable, not directly on the lamp. Adding a ceiling
force to the lamp diagram would introduce a nonexistent interaction and produce an
incorrect three-force balance.

Choose axes to match the constraint geometry. A block on a straight incline is often
simplest with one axis tangent to the surface and one normal to it. The contact
constraint then gives a direct normal-acceleration condition, usually
$a_\perp=0$. A horizontal-vertical basis remains valid, but it forces the normal
force and any friction force to be resolved into components, producing more algebra
without adding information.

The direction of a constraint force follows the geometry. A smooth surface can push
perpendicular to itself but cannot pull an unattached object toward itself. A taut
string pulls along its length and has no compressive response. A negative solved
tension under the selected sign convention means the string has gone slack, so the
fixed-length constraint no longer applies. A rigid rod may push or pull, which is why
a rod model and a string model cannot be exchanged casually.

Normal force is determined by the normal component equation. It equals $mg$ only
when the surface is horizontal and normal acceleration is zero with no other normal
forces. An elevator floor, a curved track, an inclined plane, or a hand pressing on
the body changes the result. The normal arrow should therefore be drawn first from
the contact geometry and assigned a magnitude only after the component equation is
written.

> **Worked example.** A $4.0\ \mathrm{kg}$ crate rests on a frictionless $30^\circ$
> incline while a rope pulls it parallel to the surface ($g=9.8\ \mathrm{m\,s^{-2}}$).
> With axes along and normal to the surface, the rope has no normal component and the
> normal acceleration is zero, so the normal equation is
> $$
> N-mg\cos30^\circ=0.
> $$
> The normal force is
> $$
> N=(4.0\ \mathrm{kg})(9.8\ \mathrm{m\,s^{-2}})\cos30^\circ
> =(39.2\ \mathrm N)(0.866)=34\ \mathrm N,
> $$
> smaller than the crate's weight $mg=39\ \mathrm N$ because the incline is tilted,
> not because gravity has weakened. The normal force follows from the normal
> component equation, never from a memorized $N=mg$.

**Tension models and string-length constraints.**

The statement “the tension is the same everywhere” belongs to an ideal model with a
massless taut string and frictionless massless pulleys. It is not a universal
property of rope. A massive rope can have different tension at different locations
because each segment must accelerate its own mass. A pulley with rotational inertia
or axle friction can support unequal tensions on its two sides. A slack string has
zero tension and imposes no length constraint until it becomes taut again.

An ideal straight string connecting two bodies has fixed length that links their
displacements. Select one positive coordinate for each body and express total string
length in those coordinates. Differentiating once gives the velocity relation and
twice gives the acceleration relation. Signs follow the coordinate definitions. The
familiar “equal magnitudes, opposite directions” result is only one special choice
of coordinates for a simple one-pulley arrangement.

A movable pulley introduces a factor of two. If the moving pulley is supported by
two segments of the same string, raising the pulley by a distance $y$ shortens both
segments by $y$. The free end must provide $2y$ of string. Its speed and acceleration
therefore have twice the magnitude of the pulley's values, with direction set by the
chosen coordinates. Draw the full string path before differentiating the length
constraint; the two supporting segments set the factor of two.

$$
% caption: A movable pulley is held by two string segments. One end is fixed to the ceiling;
% the string passes under the movable pulley that carries the load, up over a fixed pulley, and
% down to the free end. A unit rise of the load shortens each supporting segment, so the free
% end must be pulled down through twice that length.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0.2,3.4) -- (5.4,3.4);
  \draw[black, thick] (1.45,3.4) -- (1.45,1.7);
  \fill[black] (1.45,3.4) circle (1.5pt);
  \draw[black, thick] (1.9,1.7) circle (0.45);
  \draw[black, thick] (2.35,1.7) -- (2.35,3.05);
  \draw[black, thick] (2.7,3.05) circle (0.35);
  \draw[black, thick] (3.05,3.05) -- (3.05,1.8);
  \draw[->, acc, very thick] (3.05,1.8) -- (3.05,0.3) node[below] {pull};
  \draw[black, thick] (1.9,1.25) -- (1.9,0.9);
  \draw[thick] (1.4,0.3) rectangle (2.4,0.9);
  \node at (1.9,0.6) {load};
  \draw[->, acc, thick] (0.8,0.72) -- (0.8,1.47) node[above] {rise};
\end{tikzpicture}
$$

Let downward displacement of the free end be $x$ and upward displacement of the
movable pulley be $y$. Apart from fixed portions, string length is $x-2y$. Since
the string is taut, $x-2y$ is constant. Thus $v_x=2v_y$ and $a_x=2a_y$. If the free
end is pulled down at $0.80\ \mathrm{m\,s^{-1}}$, the pulley and attached load rise
at $0.40\ \mathrm{m\,s^{-1}}$. The factor is geometric; it does not depend on the
mass of the load.

## Connected systems and force models

Connected-body problems usually need more than one diagram. An individual-body
diagram retains tension and contact forces because they act externally on that one
body. A combined-system diagram removes forces between bodies inside the selected
boundary. The combined-system equation can be short because internal forces cancel.
An internal tension or contact force still requires an individual-body diagram after
the common acceleration has been found.

The system boundary should be selected to match the question. For two blocks joined
by a string on a smooth table, choosing both blocks as the system removes tension
from the horizontal external-force sum. Choosing one block retains tension and is
necessary to find it. Neither diagram is more correct; each answers a different
part of the problem. Mixing an internal tension cancellation from the two-block
system with a one-block mass on the right side is an inconsistent boundary change.

$$
% caption: One connected pair, two boundaries. Tension is external on either isolated
% block but internal to the two-block system, where it cancels from the momentum balance.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (-0.2,0) -- (6.2,0);
  \draw[thick] (0.8,0) rectangle (1.9,0.9);
  \node at (1.35,0.45) {A};
  \draw[thick] (3.6,0) rectangle (4.7,0.9);
  \node at (4.15,0.45) {B};
  \draw[black, thick] (1.9,0.45) -- (3.6,0.45);
  \draw[->, acc, thick] (2.35,0.75) -- (1.95,0.75) node[left] {T};
  \draw[->, acc, thick] (3.15,0.75) -- (3.55,0.75) node[right] {T};
  \draw[black, dashed] (0.55,-0.35) rectangle (4.95,1.25);
  \node[black, below right] at (0.55,1.25) {system};
\end{tikzpicture}
$$

> **Worked example.** Blocks of mass $m_1=2.0\ \mathrm{kg}$ and
> $m_2=3.0\ \mathrm{kg}$ are joined by a light string on a frictionless horizontal
> surface; a $20\ \mathrm N$ pull acts on $m_2$. Taking the two-block system (string
> tension internal, so it cancels) gives the common acceleration:
> $$
> a=\frac{F}{m_1+m_2}=\frac{20\ \mathrm N}{2.0\ \mathrm{kg}+3.0\ \mathrm{kg}}
> =4.0\ \mathrm{m\,s^{-2}}.
> $$
> The tension then follows from the isolated $2.0\ \mathrm{kg}$ block, whose only
> horizontal force is the string:
> $$
> T=m_1a=(2.0\ \mathrm{kg})(4.0\ \mathrm{m\,s^{-2}})=8.0\ \mathrm N.
> $$
> Setting $T=20\ \mathrm N$ would confuse the external pull on the whole system with
> the internal force transmitted through the string.

**Contact models beyond a level surface.**

Contact constraints are local. The normal direction is perpendicular to the surface
at the point of contact, even when the surface is curved. A bead inside a circular
track has a normal force directed toward the centre if it remains pressed against the
inside surface; a bead on the outside receives a normal force away from the surface.
The force direction follows the side of contact, whereas the required normal
acceleration follows the trajectory geometry. These are separate facts and should
not be inferred from a memorized sign alone.

A surface can lose contact. The normal force cannot become negative for a simple
unattached contact, because the surface cannot pull the body toward itself. A solved
value $N<0$ means the assumed contact has already failed; the replacement model has
$N=0$ and a new unconstrained trajectory. This interpretation is essential for
objects cresting a hill, a block on a track, or a vehicle crossing a curved bridge.

Friction belongs in the tangential direction and depends on the contact model. Static
friction is an adjustable response up to its limiting magnitude; it is not
automatically equal to $\mu_sN$. Kinetic friction is used only while sliding and
points opposite the relative sliding direction. If the direction of impending motion
is uncertain, choose a trial friction direction, solve the equations, and interpret
a negative answer as reversal of the assumed direction.

$$
% caption: A local normal-tangent basis at a point on a curved guide. The normal force
% follows the surface and can vanish where contact is lost; the weight stays vertical.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0,2.6) .. controls (2.0,0.2) and (4.0,0.2) .. (6.0,2.6);
  \draw[thick] (3.0,0.72) circle (0.3);
  \draw[->, acc, very thick] (3.0,0.72) -- (3.0,2.15) node[above] {N};
  \draw[->, black, thick] (3.0,0.72) -- (4.4,0.72) node[right] {tangent};
  \draw[->, black, very thick] (3.0,0.72) -- (3.0,-0.75) node[below] {mg};
\end{tikzpicture}
$$

> **Worked example.** A cart of mass $m$ passes over the crest of a circular hill of
> radius $R$ at speed $v$. The centre of curvature lies below the crest, so the
> required centripetal acceleration $v^2/R$ points downward. Weight $mg$ acts
> downward (toward the centre); the track's normal force $N$ acts upward (away from
> the centre). Taking the inward (downward) direction positive,
> $$
> mg-N=\frac{mv^2}{R}
> \quad\Longrightarrow\quad
> N=m\!\left(g-\frac{v^2}{R}\right).
> $$
> Contact is lost when $N=0$, at
> $$
> v^2=gR.
> $$
> Above that speed the track would have to pull the cart inward, which a push-only
> contact cannot do, so the cart leaves the surface. The same threshold governs a
> vehicle cresting a curved bridge and a block leaving a rounded track.

A connected system requires a separate free-body diagram for every body whose force
balance is needed. Mark a positive coordinate for each diagram and write the
constraint relation in the same coordinate language. Only then write component
equations. This order prevents signs from being carried informally from one diagram
to another, especially when one body moves vertically while another moves along an
incline or a table.

Unknowns usually include acceleration, tension, normal force, and perhaps friction.
The count of independent equations must match the unknowns after the constraint
relations are included. A string-length relation imposes an equation even though it
is not a force equation. Conversely, writing Newton's second law for a combined
system and for all members can create redundant equations; select independent
equations that determine the requested quantities without subtracting the same
information twice.

After solving, test each force against its model. Tension must be nonnegative for a
taut string, normal force must be nonnegative for simple contact, and static
friction must not exceed its maximum magnitude. A violation does not signal failed
algebra. It signals that the assumed constraint state—taut string, contact, or
sticking—was physically inconsistent and must be replaced by a different case.

$$
% caption: A coupled pair gets one free-body diagram each plus a shared constraint. The
% string transmits equal tensions; the linking equation ties the two accelerations.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0.2,1.9) -- (2.8,1.9);
  \draw[thick] (1.0,1.9) rectangle (2.0,2.6);
  \node at (1.5,2.25) {A};
  \draw[->, black, thick] (1.5,2.6) -- (1.5,3.35) node[above] {N};
  \draw[->, acc, thick] (2.0,2.25) -- (2.95,2.25) node[right] {T};
  \draw[->, black, very thick] (1.5,1.9) -- (1.5,1.15) node[below] {$m_A g$};
  \draw[thick] (4.05,0.9) rectangle (4.75,1.7);
  \node at (4.4,1.3) {B};
  \draw[->, acc, thick] (4.4,1.7) -- (4.4,2.45) node[above] {T};
  \draw[->, black, very thick] (4.4,0.9) -- (4.4,0.15) node[below] {$m_B g$};
\end{tikzpicture}
$$

> **Worked example.** A $3.0\ \mathrm{kg}$ block on a frictionless table is connected
> over an ideal (massless, frictionless) pulley to a $2.0\ \mathrm{kg}$ hanging mass
> ($g=9.8\ \mathrm{m\,s^{-2}}$). Take the block positive toward the pulley and the
> hanging mass positive downward; the taut string gives both bodies one acceleration
> magnitude $a$ and one tension $T$. The two component equations are
> $$
> T=(3.0\ \mathrm{kg})a,
> \qquad
> (2.0\ \mathrm{kg})g-T=(2.0\ \mathrm{kg})a.
> $$
> Adding eliminates $T$:
> $$
> a=\frac{(2.0\ \mathrm{kg})g}{3.0\ \mathrm{kg}+2.0\ \mathrm{kg}}
> =\frac{(2.0)(9.8)}{5.0}\ \mathrm{m\,s^{-2}}=3.9\ \mathrm{m\,s^{-2}},
> $$
> and back-substitution gives
> $$
> T=(3.0\ \mathrm{kg})(3.9\ \mathrm{m\,s^{-2}})=12\ \mathrm N.
> $$
> The tension is less than the hanging weight $mg=20\ \mathrm N$ because the hanging
> mass accelerates downward rather than resting in equilibrium.

Audit a completed free-body diagram before calculation. Every arrow must name an
interaction source rather than a direction such as “up.” Every force must act on the
selected body rather than on a neighboring object. Contact forces should match the
local surface direction, and string forces should follow the string. An unattached
surface supplies no normal force, and a slack rope supplies no tension. Apply these
checks instead of relying on a standard arrow pattern.

State the chosen axes. Resolve only forces that are not already aligned with an
equation axis. A component equation must include every force
component along that axis, including components that oppose the assumed positive
direction. The acceleration component belongs on the right side and is determined by
the motion constraint; it is not selected to make the algebra convenient.

Compare the solved signs and magnitudes with the model. A negative normal
or tension identifies a changed contact or string state. A static-friction value
larger than its limit identifies impending slip. These outcomes are part of the
physical solution: they state that the original free-body diagram described a trial
case, not the realized constraint configuration.

Clear arrows make later algebra auditable. Ambiguous arrows create ambiguous
components, while an isolated body exposes the model assumptions before numerical
substitution.

## Distributed loads and string constraints

Some contacts act over an extended region rather than at a single geometric point.
A floor supports a box through a distributed pressure over its base; a beam support
may apply forces across a bracket; a fluid exerts pressure over a surface. A free-
body diagram at introductory scale often replaces this distribution by one resultant
force. The resultant has the same total force and the same net turning effect about
the points relevant to the model. Its line of action is therefore part of the
replacement, not a decorative placement of an arrow.

Uniform load density on a symmetric contact region places the resultant through its
geometric centre. A nonuniform load shifts it toward the
more heavily loaded side. A block resting level on a broad table can be modelled by
one normal resultant through the base centre when no tipping tendency is present.
If an applied force shifts the load distribution toward an edge, the resultant shifts
with it. Once its line of action reaches the edge, the contact model may change to
impending tip; beyond the edge, the original full-base contact is impossible.

The distributed contact is still one interaction category, but its resultant can
have more modelling content than a point-contact normal force. A diagram should
state whether the arrow represents an ideal point contact or an equivalent resultant
of a spread load. This distinction matters whenever torque, stability, or partial
loss of contact is being analysed.

A crate base is $0.80\ \mathrm m$ wide. A horizontal push at the top shifts the
normal resultant $0.25\ \mathrm m$ from the base centre toward the leading edge.
The edge lies $0.40\ \mathrm m$ from the centre, so the crate remains in full
contact under the point-resultant model. A calculated shift of $0.46\ \mathrm m$
would not mean a larger normal force; it would mean that the full-base contact
assumption has failed and a tipping or new-contact model is required.

Multi-segment strings are handled by writing total variable length explicitly. Each
straight segment contributes its signed coordinate length. Portions wrapped around
fixed pulleys and distances between fixed supports are constants and disappear on
differentiation. The resulting length equation is often simpler than the drawing
first suggests, but only after every variable segment has been counted once.

A load supported by $n$ vertical segments of the same ideal string shortens all $n$
segments during upward displacement. The free end must therefore move $n$
times as far in the opposite sense. Differentiating gives the corresponding velocity
and acceleration factors. This relation is purely kinematic. The force advantage
comes from the same geometry through the multiple tension segments, but the two
arguments should not be merged into one unexplained rule.

Coordinate signs matter. If both the free-end coordinate and the load coordinate
are defined downward, a typical relation is $x+ny=\mathrm{constant}$, giving
$a_x=-na_y$. If one coordinate is upward and the other downward, the same physical
constraint may appear with the opposite sign. State the coordinates beside the
diagram before differentiating so that a later negative acceleration has a clear
meaning.

With downward free-end coordinate $x$ and upward load coordinate $y$, the variable
string length is $x-3y$. Tautness gives $x-3y=\mathrm{constant}$ and therefore
$a_x=3a_y$. Pulling the free end downward with acceleration
$1.2\ \mathrm{m\,s^{-2}}$ raises the load with acceleration
$0.40\ \mathrm{m\,s^{-2}}$. The relation would be invalid if one segment went
slack, because the string length would no longer constrain all parts of the system.

**Inclined contact with applied forces.**

Inclines become more subtle when an applied force is not parallel to the surface.
The normal force then depends on both gravity and the applied force's normal
component. A pull angled away from the surface can reduce contact pressure; a push
into the surface can increase it. The free-body diagram should show the actual pull
direction before it is resolved. Replacing it prematurely with an along-slope arrow
erases the component responsible for changing the normal force and, consequently,
any friction limit.

A block on a fixed incline uses an incline-aligned basis that separates tangent and
normal components. The no-penetration constraint determines the normal
acceleration condition while contact persists. The tangential equation determines
whether the block speeds uphill, slows while moving uphill, remains at rest, or
slides downhill. Static friction has whatever tangent direction is needed to oppose
the anticipated relative motion, subject to its magnitude limit. It should not be
assigned from the current direction of a block that may still be at rest.

A single incline geometry permits two trial friction directions. One trial places
friction uphill because gravity tends to draw the block down. Another places friction
downhill because a strong applied pull tends to drag the block up. The solved sign
selects the physically consistent direction. This procedure is cleaner
than memorizing a fixed friction arrow for every incline.

$$
% caption: A pull applied at an angle to an incline has both an along-surface component
% and a normal component; the latter changes N and the available friction.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0,0) -- (6.0,0) -- (6.0,3.0) -- cycle;
  \begin{scope}[shift={(3.4,1.72)}, rotate=27]
    \draw[thick] (-0.6,0) rectangle (0.6,0.7);
    \draw[->, acc, very thick] (0.6,0.35) -- (2.1,0.95) node[right] {pull};
    \draw[->, very thick] (0,0.7) -- (0,1.8) node[above] {N};
  \end{scope}
  \draw[->, black, very thick] (3.13,2.04) -- (3.13,0.5) node[below] {mg};
\end{tikzpicture}
$$

> **Worked example.** A $10\ \mathrm{kg}$ crate rests on a $20^\circ$ incline while a
> rope pulls at $15^\circ$ above the surface with magnitude $60\ \mathrm N$
> ($g=9.8\ \mathrm{m\,s^{-2}}$). In the incline-aligned basis, the rope's outward
> normal component is
> $$
> T_\perp=(60\ \mathrm N)\sin15^\circ=15.5\ \mathrm N,
> $$
> and gravity's inward normal component is
> $$
> mg\cos20^\circ=(10\ \mathrm{kg})(9.8\ \mathrm{m\,s^{-2}})\cos20^\circ=92.1\ \mathrm N.
> $$
> With zero normal acceleration, the normal equation $N+T_\perp-mg\cos20^\circ=0$
> gives
> $$
> N=mg\cos20^\circ-T_\perp=92.1\ \mathrm N-15.5\ \mathrm N=77\ \mathrm N.
> $$
> Using $mg\cos20^\circ$ alone would overstate $N$ and, through $\mu_sN$, the
> available friction. This step only sets the contact model; whether the crate slides
> is decided later by the tangential equation.

## Curved paths and frame selection

On a curved guide, the normal and tangent directions rotate from point to point.
The diagram should be drawn at the specific location under analysis, not with a
single global “up” and “right” copied around the curve. The normal direction points
toward the local centre of curvature for the usual inward-positive choice, while the
tangent direction follows the instantaneous path. Weight, contact force, and any
friction must be projected onto these local directions.

The normal component of acceleration is set by speed and curvature, whereas the
tangent component is set by change in speed. This separation is a kinematic
constraint attached to the path. A body may have zero tangential acceleration while
still requiring a nonzero normal force, or it may change speed on a straight guide
with no curvature at all. Conflating the two creates the false rule that every
normal force equals weight or that every force along a curve changes speed.

Contact side again matters. An object inside a loop receives an inward normal from
the track; an object on the outside receives an outward normal. Before writing a
component equation, mark the contact side and draw the force in the direction the
surface can actually push. The solved magnitude then tests whether that contact can
be maintained.

At the side of a vertical circular track, the inward normal is horizontal while
weight is vertical. Weight therefore has no normal component at that instant, even
though it may have a full tangent component. A diagram that draws both normal force
and weight horizontally at the side has imported the geometry from the top or bottom
of the circle and cannot produce a consistent component equation.

The diagram and the coordinate frame must be declared together. In an inertial
ground frame, draw only interactions exerted on the selected body and write the
acceleration measured in that frame. A frame moving at constant velocity changes
coordinate values but does not require a new force inventory. An accelerating frame
can be convenient for describing relative positions, but its diagram requires a
clearly labelled inertial term if that frame is used for force equations. Mixing a
ground-frame acceleration with forces interpreted in a vehicle frame is not a
change of notation; it is an inconsistent model.

Use one inertial frame for the free-body diagrams and express constraints relative to
the guides or supports in that same frame.
For example, a block held against an accelerating wall has a normal force due to the
wall contact, while its no-separation condition states that the block shares the
wall's normal acceleration. The wall acceleration is not a force arrow. It appears
in the kinematic constraint that connects the diagram to the motion.

Equation consistency can be checked before solving. Each isolated body has one
component equation per selected axis. Each string or rigid linkage imposes a
geometric relation. Each contact state imposes a condition such as zero normal
relative acceleration, nonnegative normal force, or a friction bound. List unknowns
beside these relations. If a tension, normal force, and common acceleration are
unknown, the set must contain enough independent statements to determine them, but
duplicating a system equation that is merely the sum of two individual equations
does not add information.

After a solution, substitute the values back into every diagram separately. Internal
forces must appear with equal magnitude and opposite direction on the two bodies they
connect. Apply constraint checks after the algebra: a computed negative tension,
negative normal, or excessive static-friction magnitude identifies the diagram that
requires revision. The check ties each solved value to its physical constraint state.

If a block is pressed against a vertical guide that accelerates horizontally, the
normal direction is horizontal and the no-separation condition concerns horizontal
acceleration. Weight remains a separate vertical interaction. A solution that puts
the guide acceleration into the vertical component equation, or that draws it as a
third contact force, has mixed the geometry of the constraint with the force
inventory. The corrected diagram makes each role distinct: wall force, weight, and
shared horizontal acceleration.

Before numerical substitution, label every equation by its source: an isolated-body
component balance, a geometric string relation, a contact condition, or a declared
model limit. Those labels expose missing constraints and duplicate equations. If
contact is lost, replace the contact condition and its normal-force arrow without
rebuilding an unrelated string or system equation.

Treat the diagram as the record of the selected model. Update it before reusing
equations when the system boundary, contact state, string tautness, coordinate frame,
or applied-force direction changes.

**Three-dimensional support contacts.**

In three dimensions, a support may restrict motion in one direction while allowing
motion in two others, or it may restrict several translations and rotations at once.
The free-body diagram should represent only the reactions permitted by the idealized
support. A smooth collar on a straight rod can exert force perpendicular to the rod
but not along it. A frictionless ball-and-socket joint can exert three force
components but no couple moment. A fixed support can exert force components and a
reaction couple because it prevents both translation and rotation of the attached
body.

These models are not interchangeable. Drawing three force components at a smooth
contact that can only push normal to one surface over-constrains the body. Drawing a
single normal force at a socket that prevents motion in every direction under-
constrains it. The geometry of the support and the allowed motions determine which
unknown reactions belong in the diagram. A short note such as “smooth collar” or
“pinned joint” carries physical content that an unlabeled point cannot.

Coordinate axes are particularly valuable in three dimensions. Resolve a reaction
only into the directions needed by the subsequent equations; an arrow in an oblique
direction may be left as a vector until the geometry is clear. Do not add an extra
reaction merely because a third axis exists. The number of components follows the
support constraint, not the coordinate-system dimension.

$$
% caption: Support models fix the allowed reaction components. A smooth collar reacts
% perpendicular to its guide only, while a pin can react along two independent directions.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0.4,0.4) -- (2.6,0.4);
  \draw[thick] (1.2,0.4) rectangle (1.8,1.0);
  \draw[->, acc, very thick] (1.5,1.0) -- (1.5,2.1) node[above] {reaction};
  \node[black, below] at (1.5,0.4) {collar};
  \draw[fill=black!10, draw=black, thick] (5.0,0.4) circle (0.28);
  \draw[->, acc, very thick] (5.0,0.4) -- (6.1,0.9) node[right] {$R_x$};
  \draw[->, acc, very thick] (5.0,0.4) -- (5.0,1.7) node[above] {$R_y$};
  \node[black, below] at (5.0,0.1) {pin};
\end{tikzpicture}
$$

A bead slides without friction on a vertical circular wire. At a given point, the
wire can push the bead in directions perpendicular to the local wire tangent, but
it cannot exert a tangential friction force. A free-body diagram with a single
vertical “support force” is incomplete because the contact direction rotates around
the wire. A diagram with a tangential reaction would contradict the stated smooth
contact model. The correct reaction representation follows the local three-
dimensional guide geometry.

## Constraint checks and noninertial frames

Every body in a coupled system may use a different positive coordinate. The
constraint equation is the place where these choices are reconciled. For a string
joining a block on a table to a hanging mass, defining the block positive toward the
pulley and the hanging mass positive downward makes their accelerations equal. If
both coordinates are defined away from the pulley, the same physical motion gives
opposite signed accelerations. Neither convention is preferred; an unstated switch
between them is the error.

The sign check starts from the variable string length. Mark each segment that grows
when its associated coordinate grows with a plus sign and each segment that shrinks
with a minus sign. Differentiate only after the length expression is complete. This
method works for pulley systems, rods, rolling constraints, and linked sliders. It
also tests whether a supposed constraint remains active: a slack string does not
provide a fixed-length equation, and a lost contact no longer links normal motion.

Internal force arrows provide a second sign audit. The tension exerted by a string
on each attached body points toward the string segment. Contact forces on two bodies
are opposite in direction but need not share a convenient global positive sign. Draw
them separately on their own diagrams before applying any cancellation in a combined
system. A force cannot be cancelled merely because two arrows look similar; it is
internal only when both interacting bodies lie inside the same selected boundary.

$$
% caption: A sign audit for a table block joined over a pulley to a hanging mass. Named
% axes and a string-length relation give the two acceleration signs before any numbers.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (-0.2,1.7) -- (3.9,1.7);
  \draw[thick] (0.9,1.7) rectangle (1.9,2.4);
  \node at (1.4,2.05) {block};
  \draw[->, acc, thick] (1.9,2.05) -- (3.5,2.05) node[right] {T};
  \draw[black, thick] (3.9,2.4) circle (0.3);
  \draw[fill=black!10, draw=black, thick] (3.6,0.2) rectangle (4.2,1.0);
  \node at (3.9,0.6) {load};
  \draw[->, acc, thick] (3.9,1.0) -- (3.9,1.75) node[right] {T};
  \draw[->, black, very thick] (3.9,0.2) -- (3.9,-0.6) node[below] {mg};
  \draw[->, black] (5.0,1.6) -- (5.0,0.6) node[midway,right] {$+$};
\end{tikzpicture}
$$

Let a table block coordinate increase away from the pulley and let a hanging-mass
coordinate increase downward. If the string has one horizontal and one vertical
variable segment, its length is $x+y+\mathrm{constant}$. Tautness gives
$a_x+a_y=0$. The hanging mass moving downward therefore corresponds to the table
block moving toward the pulley, which is negative in this coordinate convention.
Writing both accelerations as a positive symbol without this relation would impose a
different physical motion from the one drawn.

**Noninertial-frame diagram handling.**

An accelerating vehicle frame can simplify a constraint because stationary features
of the vehicle remain fixed in that coordinate description. The diagram must then
make the frame choice explicit. One option is to work in an inertial ground frame:
draw only real interactions and write the acceleration required by the vehicle
constraint. The other option is to work in the accelerating vehicle frame:
draw the same real interactions and include a labelled inertial term associated with
the frame acceleration. Either approach can be consistent; combining the two is not.

The two descriptions use different equation forms for the same physical contact.
Keeping the frame choice in the record prevents a real wall force from being counted
again as a frame term.

| Description | Arrows on the block | Acceleration statement |
| --- | --- | --- |
| Inertial ground frame | real contact and field interactions only | block acceleration equals the vehicle constraint |
| Accelerating vehicle frame | same real interactions plus labelled inertial term | block may be stationary in vehicle coordinates |
| Mixed description | incompatible arrow set | reject before component summation |

The inertial term is not a new contact source. It represents the use of an
accelerating coordinate frame and acts on every mass in the frame analysis opposite
the frame's acceleration. It should be visually distinguished from a normal force,
tension, or applied push. The selected system boundary remains important: an inertial
term belongs on each mass included in the noninertial analysis, while internal
interactions still cancel only for a combined system.

Rotating frames require additional care because their axes change direction. A
simple one-dimensional vehicle diagram can often use an accelerating translation
model, but a turntable or rotating arm needs rotation-dependent terms and a local
axis convention. When the course has not introduced those terms, the safer choice is
an inertial frame with time-dependent contact directions rather than an incomplete
rotating-frame free-body diagram.

A block remains against the rear wall of a truck accelerating forward. In the ground
frame, the wall's normal force accelerates the block forward; the no-separation
condition sets the block's forward acceleration equal to the truck's. In the truck
frame, the block may be at rest, but the diagram must include a backward inertial
term in addition to the forward wall force. Drawing the wall force and setting the
acceleration to zero without the inertial term would describe equilibrium in the
wrong frame.

## Force-inventory diagnostics

When a solution fails a sign or magnitude check, return to the interaction inventory
rather than adding arrows until the equations balance. A missing force usually has a
missing source: an overlooked rope, a second contact surface, a magnetic field, or
an applied support. A double-counted force usually has two names for one interaction,
such as both “normal” and “table support,” or it places the action-reaction partner
on the same body's diagram. The latter is especially common with strings and blocks
in contact.

Use a one-to-one test. For every arrow on the selected body, name the other object
or field that exerts it. For every listed interaction, locate exactly one arrow on
the selected body. Then inspect the arrow direction against the model: a string
pulls, a smooth surface pushes normal to itself, friction lies tangent to contact,
and weight points toward the local gravitational source. This test separates a
physical omission from a mere component-resolution mistake.

Finally, distinguish forces from constraints and from outcomes. “The block stays on
the surface” states a normal-motion condition, not an extra arrow. “The block
accelerates with the cart” states a kinematic relation, not a force. “The normal
force is zero” can be a solved outcome that changes the contact model. Keeping these
three roles separate makes an overloaded diagram easier to repair than an equation
list alone.

The inventory can be recorded before component equations are written. The distinction
between interaction, constraint, and solved outcome prevents a diagram from acquiring
arrows that have no physical source.

| Entry | Diagram treatment | Equation role |
| --- | --- | --- |
| Contact, string, field, or applied interaction | one arrow on the selected body | force component in $\sum\vec F$ |
| Geometric or no-separation condition | no additional arrow | relation among positions, velocities, or accelerations |
| Contact loss or slip state | no assumed arrow magnitude | result that may change the active model |

Consider a suitcase pulled across a floor by a strap angled upward. An initial diagram
contains weight, a vertical normal force, and the strap tension. The horizontal
equation appears to predict the observed acceleration, but the calculated normal
force has been set equal to weight by habit. The inventory is not missing an arrow;
it is missing a component. The upward component of the strap reduces the normal
force, so the normal equation must include both weight and that component. If kinetic
friction is present, its magnitude changes as a consequence because it depends on
the corrected normal force.

Now suppose an extra “reaction to the pull” arrow is drawn backward on the suitcase.
That arrow is the suitcase's force on the person or strap, not a second force on the
suitcase. It is a double count, and its inclusion would incorrectly cancel part of
the genuine strap tension. The corrected diagram has one strap force acting on the
suitcase, with its source and direction explicit, followed by component resolution
in the selected axes.

A finished solution should state the active contact, string, and frame assumptions
in words. “The string remains taut, the block stays in contact with the guide, and
the ground frame is used” explains why the selected arrows and constraint equations
apply. If later data contradict one of those assumptions, the diagram can be revised
at its source instead of patched by adding an unexplained force term.
The assumptions should be stated explicitly alongside the final numerical result.

[^tipler46]: Tipler and Mosca, _Physics for Scientists and Engineers_, 6th ed., §§4-5–4-8.