---
title: Static Equilibrium
module: Gravitation and Matter
moduleNumber: 6
lessonNumber: 3
order: 603
summary: >
  What does it take for a loaded structure to stay put? A body at rest needs its
  forces to cancel and its turning effects to cancel — $\sum\vec F=0$ and
  $\sum\vec\tau=0$ about any point — and almost all of statics is the craft of
  turning a physical setup into those equations. We build free-body diagrams,
  replace supports, cables, friction, couples, and distributed loads with their
  idealized reactions, and locate the centre of gravity that decides whether a body
  tips. We also count equations against unknowns to separate a determinate problem
  from one that needs the material's deformation to resolve, and read every negative
  or inconsistent reaction as a sign that a contact or a boundary was chosen wrong.
topics: [Gravitation and Matter]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 12 — Static Equilibrium and Elasticity; §§12-1–12-6"
---

## Rigid-body balance and torque

Static equilibrium means that a body remains at rest and keeps the same orientation
in an inertial reference frame. A rigid-body model separates translation of the
centre of mass from rotation about it. The external forces must have zero resultant,
and the external torques must have zero resultant:

$$
\sum\vec F=\vec 0,
\qquad
\sum\vec\tau_O=\vec 0.
$$

The point $O$ may be selected anywhere. The first equation prevents linear
acceleration. The second prevents angular acceleration. A force balance alone can
hold the centre of mass stationary while a couple rotates the body. A torque balance
alone can hold a body against rotation while a nonzero resultant force accelerates
it. Both conditions are required for a rigid body at rest.

The word rigid is a modelling assumption. Under the loads considered here, distances
between marked points are treated as constant, so force locations and moment arms
remain known. Actual supports, cables, and bodies deform. Their deformation becomes
important in an indeterminate system, a precision measurement, or a large-deflection
problem. The equilibrium equations still govern external balance, but additional
constitutive or compatibility information becomes necessary to resolve every
reaction force.

In a planar problem, all forces lie in the $xy$ plane and all torque vectors are
parallel to the $z$ axis. The balance conditions reduce to three scalar equations:

$$
\sum F_x=0,
\qquad
\sum F_y=0,
\qquad
\sum\tau_z=0.
$$

Declare one sign convention and retain it. Counterclockwise-positive rotation gives
a signed scalar torque. A force whose line of action passes through
the selected point has zero moment arm and zero torque about that point. This
geometric fact removes one unknown reaction when it lies on the torque axis.

In three dimensions, force balance gives three scalar equations and torque balance
gives three more:

$$
\sum F_x=\sum F_y=\sum F_z=0,
\qquad
\sum\tau_x=\sum\tau_y=\sum\tau_z=0.
$$

The six equations are independent only when the geometry permits independent force
and moment components. Parallel forces, concurrent force lines, or constraints that
remove a direction can reduce the number of independent equations. Counting equations
without considering their directions can conceal a singular model.

**Torque, moment arms, and axis selection.**

Torque about a point is the vector product

$$
\vec\tau_O=\vec r\times\vec F,
\qquad
|\vec\tau_O|=rF\sin\alpha=F\ell.
$$

The vector $\vec r$ runs from the selected point to any point on the force line.
The angle $\alpha$ is between $\vec r$ and $\vec F$. The perpendicular
distance $\ell$ from the selected point to the force line is the moment arm. Drawing
$\ell$ avoids ambiguity in a slanted sketch. A force applied
far from an axis can have little torque when its line of action nearly passes through
the axis.

$$
% caption: Moment-arm geometry for one force. The torque magnitude equals the force
% times the perpendicular distance $\ell$ from the pivot to the line of action, not
% the force times the distance $r$ to the point where the arrow is drawn; the dashed
% moment arm meets the force line at a right angle.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[fill=white,draw=black] (1.0,0.8) circle (2.4pt);
\node[left] at (0.9,0.8) {pivot};
\draw[black,dashed] (4.6,0.8)--(4.6,3.5);
\draw[acc,very thick] (4.6,1.6)--(4.6,3.0);
\draw[->,acc,very thick] (4.6,3.0)--(4.6,3.6);
\node[right] at (4.6,3.3) {$F$};
\draw[->,black] (1.0,0.8)--(4.6,2.4);
\node[above left] at (2.9,1.7) {$r$};
\draw[black,dashed] (1.0,0.8)--(4.6,0.8);
\node[below] at (2.8,0.74) {moment arm};
\draw[black] (4.4,0.8)--(4.4,1.0)--(4.6,1.0);
\end{tikzpicture}
$$

Planar scalar torque can be calculated from component forces. For a force
$\vec F=(F_x,F_y)$ applied at $\vec r=(x,y)$ relative to $O$,

$$
\tau_z=xF_y-yF_x.
$$

This expression handles oblique forces without a separate trigonometric moment-arm
construction. It also gives an algebraic check on a diagram method. Coordinates
must be measured from the same origin used for every force in the torque sum.
Changing origin changes each individual torque, but the total torque of a system
with zero resultant force remains the same.

An efficient torque axis passes through a force line whose magnitude or direction is
unknown. That force then disappears from the torque equation. The choice reduces
algebra while preserving the physical model. After finding another reaction from
the torque equation, use force balance to recover the force components at the axis.
A torque axis through a point of contact is often convenient for a beam on a hinge,
a ladder at its foot, or a wheel on the edge of a step.

## Free-body diagrams and reaction models

An equilibrium calculation begins by isolating one body or one chosen group of
bodies. Replace each physical interaction crossing the system boundary with an
external force or applied couple. Internal forces between objects retained inside
the same system are omitted because they occur in equal-and-opposite pairs. Weight,
contact forces, cable tensions, spring forces, pressure resultants, and applied
moments are external only when their source lies outside the selected boundary.

An ideal cable transmits tension along its own direction. It cannot push. A frictionless
pin joint in a planar model supplies two force components and no reaction couple.
A roller on a smooth surface supplies one force normal to that surface. A rough
surface may supply a normal force plus a tangential friction force. A built-in
support can supply two force components and a reaction couple in planar statics.
Each reaction is an unknown only if the physical connection permits it.

$$
% caption: Common planar support idealizations and the reactions they supply. A
% cable exerts one tension along its length, a smooth roller one normal reaction, a
% pin two force components, and a built-in end two components plus a reaction couple.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% cable
\draw[fill=black!30] (1.05,2.5) rectangle (1.35,2.62);
\draw[->,acc,very thick] (1.2,1.3)--(1.2,2.45);
\node[below] at (1.2,1.2) {cable};
% roller
\draw[black,thick] (2.7,1.05)--(3.7,1.05);
\draw[fill=white,draw=black] (3.2,1.27) circle (0.2);
\draw[->,acc,very thick] (3.2,1.5)--(3.2,2.5);
\node[below] at (3.2,0.95) {roller};
% pin
\draw[black,thick] (4.7,1.05)--(5.7,1.05);
\draw[fill=black!12,draw=black] (5.2,1.35)--(5.0,1.05)--(5.4,1.05)--cycle;
\draw[fill=white,draw=black] (5.2,1.35) circle (2pt);
\draw[->,acc,very thick] (5.2,1.35)--(6.0,1.35);
\draw[->,acc,very thick] (5.2,1.35)--(5.2,2.35);
\node[below] at (5.2,0.95) {pin};
% built-in end
\draw[black,very thick] (6.9,0.95)--(6.9,2.55);
\draw[acc,very thick] (6.9,1.75)--(7.8,1.75);
\draw[->,acc,very thick] (6.9,1.75)--(6.3,1.75);
\draw[->,acc,very thick] (6.9,1.75)--(6.9,2.5);
\draw[->,acc,thick] (7.35,2.35) arc (55:-215:0.3);
\node[below] at (7.2,0.95) {built-in end};
\end{tikzpicture}
$$

The assumed direction of an unknown reaction is arbitrary. A negative solution
means the actual reaction points opposite the arrow drawn in the free-body diagram.
A cable tension requires a separate contact-state check: if the
solution for a tension magnitude is negative after an assumed pulling direction,
the cable would be slack and the proposed contact geometry has changed. A normal
contact reaction that calculates as negative similarly indicates separation from an
ideal one-sided contact.

$$
% caption: Free-body diagram of a ladder against a smooth wall on a rough floor. The
% wall supplies only a horizontal normal force; the floor supplies an upward normal
% force and friction, whose uphill direction opposes the impending slide of the foot.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black,thick] (0.6,0.7)--(6.4,0.7);
\draw[black,thick] (5.9,0.7)--(5.9,3.6);
\draw[acc,very thick] (1.4,0.7)--(5.9,3.3);
% wall normal
\draw[->,acc,thick] (5.9,3.3)--(5.0,3.3);
\node[above left] at (5.4,3.32) {wall normal};
% weight at centre
\draw[->,black,thick] (3.65,2.0)--(3.65,1.1);
\node[right] at (3.65,1.55) {weight};
% floor normal
\draw[->,acc,thick] (1.4,0.7)--(1.4,1.7);
\node[left] at (1.4,1.3) {ground normal};
% friction
\draw[->,acc,thick] (1.4,0.7)--(2.4,0.7);
\node[below] at (2.0,0.62) {friction};
\end{tikzpicture}
$$

Reaction models must be compatible with the object under study. A pin in a drawing
may be a welded connection in hardware; a nominal roller may develop friction; a
cable may sag under its own weight. The idealization is selected from the intended
load path, then checked against measured motion or known construction. Adding
unsupported reaction components can turn a determinate problem into an artificial
indeterminate one. Omitting a permitted reaction can make a physically viable
configuration appear impossible.

## Cables, couples, and distributed loads

A taut massless cable has one scalar unknown tension magnitude. Its force on an
attached body points away from that body along the cable. A cable at angle $\theta$
to the horizontal has components

$$
T_x=T\cos\theta,
\qquad
T_y=T\sin\theta.
$$

The geometry of the cable controls its mechanical advantage. A nearly horizontal
cable provides a small vertical component for a large tension, which can produce
large support reactions even when the supported load is modest. A statics diagram
should show the cable line before resolving components; reversing a component by
eye is a frequent source of sign error.

A planar body acted on by three nonparallel forces in static equilibrium has three
lines of action that meet at one point. The condition follows from torque balance.
The intersection of any two force lines has zero moment arm for both forces. The
third force must also have zero torque about that point, so its line must pass
through it. This construction can determine the direction of an unknown contact
reaction without first resolving components.

The condition applies only to three external forces without an applied couple. If
two forces are parallel, their intersection lies at infinity and the simple drawing
test becomes a limiting case. If a fourth external force appears, concurrent lines
can still occur but are not required by the three-force argument.

$$
% caption: A planar body in equilibrium under three nonparallel forces. Extending the
% two known load lines locates their common point; the unknown support reaction must
% pass through it, otherwise it leaves an uncompensated moment about that point.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[fill=acc!10,draw=acc,thick] (1.6,0.8)--(5.4,1.0)--(4.7,2.8)--(1.2,2.4)--cycle;
% concurrency lines (dashed)
\draw[black,dashed] (3.5,1.9)--(1.0,3.4);
\draw[black,dashed] (3.5,1.9)--(6.2,0.6);
\draw[black,dashed] (3.5,1.9)--(3.5,3.6);
% forces along those lines
\draw[->,acc,very thick] (2.0,2.8)--(1.0,3.4);
\draw[->,acc,very thick] (5.12,1.12)--(6.2,0.6);
\draw[->,acc,very thick] (3.5,2.92)--(3.5,3.6);
\draw[fill=white,draw=black] (3.5,1.9) circle (1.8pt);
\node[left] at (1.0,3.4) {load one};
\node[right] at (6.2,0.6) {load two};
\node[above] at (3.5,3.62) {support reaction};
\node[below left] at (3.3,1.78) {common point};
\end{tikzpicture}
$$

Two equal, opposite, parallel forces whose lines of action are separated form a
couple. Their net force vanishes, while their torque has magnitude

$$
\tau_{\rm couple}=FD,
$$

where $D$ is the perpendicular distance between the two lines. The torque of a
couple has the same value about every point. Moving the torque axis changes the
moments of the individual forces, but the additional terms cancel because the net
force is zero. A free couple is therefore represented in a free-body diagram by a
curved arrow and its signed magnitude.

A steering wheel, screwdriver, and two hands turning a valve can be represented by
couples. In a rigid-body equilibrium calculation, an applied couple may be placed
at any chosen point without changing the external effect. An ordinary force cannot
be moved in this way. Shifting a force to another point requires adding a couple
equal to the force times the offset, otherwise the torque balance changes.

**Distributed loads and equivalent resultants.**

Loads are often distributed over a length, area, or volume. A beam-like member
carrying a load intensity $w(x)$ has a differential vertical force

$$
\d F=w(x)\,\d x.
$$

The total load and its line of action follow from force and moment equivalence:

$$
F_R=\int_a^b w(x)\,\d x,
\qquad
x_R=\frac{\int_a^b xw(x)\,\d x}{\int_a^b w(x)\,\d x}.
$$

The resultant has the same total force and the same torque about every point as the
original distribution. It replaces the distribution only for external equilibrium.
It does not preserve the detailed internal force or deformation inside the loaded
member, which require a separate structural model outside the present scope.

$$
% caption: A uniform line load has constant intensity across the span. Its equivalent
% resultant equals the area of the shaded rectangle and acts through the midpoint,
% where the first moment of the symmetric distribution vanishes.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[acc,very thick] (0.7,0.8)--(6.5,0.8);
\fill[acc!10] (0.9,0.8) rectangle (6.3,2.0);
\draw[acc,thick] (0.9,2.0)--(6.3,2.0);
\draw[acc,thick] (0.9,0.8)--(0.9,2.0);
\draw[acc,thick] (6.3,0.8)--(6.3,2.0);
\foreach \x in {1.2,1.8,2.4,3.0,3.6,4.2,4.8,5.4,6.0}{
  \draw[->,black] (\x,2.0)--(\x,1.0);
}
\draw[->,acc,very thick] (3.6,3.1)--(3.6,0.95);
\draw[black,dashed] (3.6,0.55)--(3.6,3.1);
\node[above] at (3.6,3.15) {resultant};
\node[below] at (3.6,0.5) {midpoint};
\node[right] at (6.3,1.4) {load intensity};
\end{tikzpicture}
$$

With a uniform load $w_0$ on a span of length $L$,

$$
F_R=w_0L,
\qquad
x_R=\frac{L}{2}.
$$

The resultant is the area under the load-intensity graph. Units verify the result:
$w_0$ has units $\mathrm{N\,m^{-1}}$, so $w_0L$ has units of newtons. The
first-moment numerator has units $\mathrm{N\,m}$, so division by total force gives
a distance.

Piecewise distributions are handled by summing simpler shapes or by integrating
each interval. A point load is included directly as a force. A uniformly distributed
load has a rectangular area. A linearly varying load has a triangular or
trapezoidal area. The combined resultant location follows from total moment divided
by total force, not from an unweighted average of separate centroid locations.

Resultant replacement reduces pressure loads to force and moment equivalents. A hydrostatic pressure on
a vertical surface varies with depth, giving a triangular pressure diagram. The
resultant force is the diagram area times the surface width, while its line of action
lies below the geometric centroid because the greater pressure occurs at larger
depth. The pressure force may be treated as a distributed load in a static
equilibrium calculation, provided its direction and loaded area are specified.

## Centers of gravity and stability

Gravity acts on every mass element of an extended body. In a uniform gravitational
field, the distributed gravitational forces have the same net force and net torque
as one weight $M\vec g$ acting at the centre of mass:

$$
\vec r_{\rm cm}=\frac{1}{M}\sum_i m_i\vec r_i,
\qquad
M=\sum_i m_i.
$$

For continuous distributions, replace the sum with an integral:

$$
\vec r_{\rm cm}=\frac{1}{M}\int\vec r\,\d m.
$$

The centre of gravity is the point through which the resultant gravitational force
acts. It coincides with the centre of mass when $\vec g$ is effectively uniform
across the body. That approximation is excellent for ordinary laboratory and
engineering dimensions near Earth. It becomes less accurate for an extended
astronomical system or a body spanning a strong gravity gradient.

$$
% caption: When gravity is uniform, the distributed weights of the mass elements
% reduce to a single resultant at the centre of gravity. That resultant carries the
% same total force and the same total moment about any point as the distribution.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[fill=acc!10,draw=acc,thick]
  (1.1,1.0) .. controls (0.9,2.4) and (2.1,2.9) .. (3.3,2.7)
  .. controls (4.7,2.5) and (5.9,2.8) .. (6.0,1.6)
  .. controls (6.1,0.7) and (4.6,0.6) .. (3.3,0.6)
  .. controls (2.0,0.6) and (1.3,0.5) .. (1.1,1.0);
\foreach \x/\y in {1.8/2.5,2.6/2.6,3.4/2.55,4.3/2.5,5.1/2.45}{
  \draw[->,black] (\x,\y)--(\x,\y-0.45);
}
\draw[fill=acc!10,draw=acc] (3.55,1.5) circle (2pt);
\draw[->,acc,very thick] (3.55,1.5)--(3.55,0.3);
\node[right] at (3.65,1.55) {centre of gravity};
\node[right] at (3.65,0.4) {total weight};
\end{tikzpicture}
$$

For point masses on one coordinate axis,

$$
x_{\rm cm}=\frac{\sum_i m_ix_i}{\sum_i m_i}.
$$

The numerator is a mass-weighted first moment. A centre of mass must lie inside the
smallest interval containing all the masses, which is a fast check on a numerical
result. A large mass placed near an origin can outweigh several smaller masses at a
larger distance. Coordinate signs matter: masses on opposite sides of the origin
contribute moments with opposite signs.

Symmetry can determine components without integration. A uniform rectangle has its
centre at the intersection of diagonal symmetry lines. A uniform disk lies at its
geometric centre. A uniform semicircular wire and a uniform semicircular plate have
different centres because one distributes mass along arc length and the other over
area. The outline alone does not identify the mass distribution; thickness and
density must be included when symmetry arguments are applied.

The torque of weight about an arbitrary point $O$ is

$$
\vec\tau_{g,O}
=\vec r_{\rm cg}\times M\vec g.
$$

The coordinate $\vec r_{\rm cg}$ must be measured from the same point used for
the torque equation. A beam's own weight acts at its centre of mass if its density
and cross-section are uniform. A mounted instrument, counterweight, or liquid
container shifts the total centre of mass. Treat each item as a point mass only
when its physical size is small compared with the relevant lever arms.

A suspension method locates the centre of gravity of an irregular flat object.
Suspend the object from one hole and draw the vertical line passing through the
suspension point with a plumb line. Repeat from a second hole. In static
equilibrium, the centre of gravity lies directly below the suspension point, because
its weight has zero torque about that point. The two plumb lines intersect at the
centre of gravity, subject to line thickness and measurement uncertainty.

Use three or more suspension points, fit their best common intersection, and report
the spread of the inferred centre of mass. A calibrated level can establish the
vertical direction in place of a plumb line. A broad suspension hole, a bent plate,
or a nearby air current broadens the intersection region. A knife-edge balance gives
an independent coordinate: move the support until the object balances, then measure
its position relative to a reference mark.

Centre-of-gravity calculations give a direct stability test. A body resting
on a base can remain statically supported while the vertical projection of its
centre of gravity lies inside the support region. The boundary of that region marks
a tipping threshold. This test accounts for gravity and normal support reactions;
friction, acceleration, deformation, and a nonhorizontal surface can add further
constraints.

**Rotational stability, support regions, and tipping thresholds.**

An equilibrium orientation can be stable, unstable, or neutral under a small
allowed rotation. In stable equilibrium, a small displacement produces a restoring
torque and raises gravitational potential energy. In unstable equilibrium, a small
displacement produces a torque in the same direction and lowers gravitational
potential energy. Neutral equilibrium has no first restoring or overturning torque
along the allowed motion. The classification depends on the constraint: a body can
be stable with respect to one rotation and free to move in another direction.

$$
U_g=Mgy_{\rm cg}.
$$

With one rotation coordinate $q$, equilibrium occurs when $\frac{\d U_g}{\d q}=0$.
Positive curvature $\frac{\d^2U_g}{\d q^2}$ indicates local stable equilibrium, negative
curvature indicates local unstable equilibrium, and zero curvature calls for
higher-order or constraint analysis. Potential-energy language and torque language
are equivalent because generalized torque is $-\frac{\d U_g}{\d q}$.

$$
% caption: The three rotational-equilibrium types by how the centre of gravity moves
% under a small displacement. In a valley it rises and a restoring moment returns the
% body; on a crest it falls and the moment overturns; on a flat surface it is unchanged.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% stable: valley
\draw[black,thick] (0.4,2.2) .. controls (1.2,0.6) and (1.8,0.6) .. (2.6,2.2);
\draw[fill=acc!10,draw=acc,thick] (1.5,1.22) circle (0.22);
\draw[->,black] (1.5,1.22)--(1.5,0.6);
\node[below] at (1.5,0.2) {stable};
% unstable: crest
\draw[black,thick] (3.2,0.7) .. controls (4.0,2.3) and (4.6,2.3) .. (5.4,0.7);
\draw[fill=acc!10,draw=acc,thick] (4.3,2.12) circle (0.22);
\draw[->,black] (4.3,2.12)--(4.3,1.5);
\node[below] at (4.3,0.2) {unstable};
% neutral: flat
\draw[black,thick] (6.0,1.0)--(8.0,1.0);
\draw[fill=acc!10,draw=acc,thick] (7.0,1.22) circle (0.22);
\draw[->,black] (7.0,1.22)--(7.0,0.6);
\node[below] at (7.0,0.2) {neutral};
\end{tikzpicture}
$$

On a flat support, tipping begins when the resultant normal reaction
reaches an edge of the support region. At that threshold, the vertical line through
the centre of gravity passes through the edge for gravity-only loading. Beyond it,
the required normal reaction would lie outside the physical contact patch, so the
body rotates about the edge. The prior normal-pressure distribution can shift across
the base; representing it by one effective resultant is valid for external force and
torque balance.

$$
% caption: The vertical projection of the centre of gravity sets the support status.
% It lies within the base when supported, reaches the pivot edge at the tipping
% threshold, and passes beyond the edge once the body must rotate about that corner.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black,thick] (0.4,0.6)--(7.6,0.6);
% supported: upright, projection well inside
\draw[acc,thick,fill=acc!10] (0.9,0.6) rectangle (2.3,2.4);
\draw[fill=acc!10,draw=acc] (1.6,1.5) circle (1.8pt);
\draw[black,dashed] (1.6,1.5)--(1.6,0.62);
\node[below] at (1.6,0.2) {supported};
% edge threshold: projection at pivot corner
\draw[acc,thick,fill=acc!10] (3.3,0.6)--(4.7,0.6)--(4.95,2.35)--(3.55,2.35)--cycle;
\draw[fill=acc!10,draw=acc] (4.7,1.5) circle (1.8pt);
\draw[black,dashed] (4.7,1.5)--(4.7,0.62);
\node[below] at (4.1,0.2) {edge threshold};
% overturning: projection beyond corner
\draw[acc,thick,fill=acc!10] (5.7,0.6)--(6.8,0.6)--(7.2,2.2)--(6.0,2.35)--cycle;
\draw[fill=acc!10,draw=acc] (6.95,1.55) circle (1.8pt);
\draw[black,dashed] (6.95,1.55)--(6.95,0.62);
\node[below] at (6.6,0.2) {overturning};
\end{tikzpicture}
$$

The same condition can be written as a torque balance about the impending pivot.
Consider a rectangular block of base width $b$, centre-of-gravity height $h_{\rm cg}$,
and applied horizontal force $P$ at height $h_P$. At the threshold for tipping
about the lower edge,

$$
P h_P=Mg\frac{b}{2},
\qquad
P_{\rm tip}=\frac{Mgb}{2h_P}.
$$

The result assumes sufficient friction to avoid sliding before tipping. A friction
check is separate: horizontal equilibrium requires friction magnitude $P$, and the
available static friction depends on the contact law. A low applied point requires
larger force because it has a shorter moment arm; increasing the base width raises
the resisting gravitational moment.

Support-region reasoning generalizes to a three-dimensional base. The vertical
projection of the centre of gravity must lie inside the convex polygon formed by
active contact points for gravity-only equilibrium. A chair with four feet has a
support polygon joining the foot contacts. A load shifted near one side transfers
normal force toward that side; at a boundary, one or more opposite-foot reactions
fall to zero. The model needs contact geometry, because a broad pad and a point foot
provide different possible reaction distributions.

Stability has a scale. A broad low object requires a larger rotation before its
centre crosses the support boundary than a tall narrow object. The energy barrier
to tipping is the rise in centre-of-gravity height needed to reach the edge. A body
may be locally stable under a small disturbance but still tip after a larger
disturbance. Static calculations identify threshold configurations; the subsequent
motion depends on rotational inertia, applied impulse, damping, and contact impacts.

Static stability should be separated from material strength. A block can remain
inside its support region while a support connection yields, and a strong component
can still tip through lack of geometric support. Similarly, friction may prevent
sliding at one applied force while the same force creates a tipping moment. Compare
sliding and tipping thresholds whenever a horizontal or inclined load is present.

## Accelerated frames and reaction consistency

An object stationary relative to a translating, nonrotating frame has the frame's
acceleration in an inertial frame. Force balance is therefore

$$
\sum\vec F=M\vec a_{\rm frame}.
$$

Rotation is most simply balanced about the centre of mass:

$$
\sum\vec\tau_{\rm cm}=\vec 0.
$$

The second equation follows because a body that keeps the same orientation has zero
angular acceleration. Taking torques about an accelerating point other than the
centre of mass requires an additional translation term. The centre-of-mass torque
equation avoids that term and is usually the clearest choice.

In the accelerating frame, introduce the inertial force

$$
\vec F_{\rm in}=-M\vec a_{\rm frame}.
$$

Then the body can be treated with static-style equations:

$$
\sum\vec F+\vec F_{\rm in}=\vec 0,
\qquad
\sum\vec\tau_O+\vec\tau_{{\rm in},O}=\vec 0.
$$

The inertial force is a reference-frame bookkeeping term. It acts at the centre of
mass for rigid-body translational balance and has no torque about that point. Its
use is valid for a frame with uniform translational acceleration; a rotating frame
requires additional terms and a separate kinematic treatment.

$$
% caption: A block on the floor of a vehicle accelerating to the right. In the
% vehicle frame the inertial force points rearward, so weight and inertial force
% combine into an effective gravity that leans toward the back of the vehicle.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black,thick] (0.5,0.6)--(6.8,0.6);
\draw[acc,thick,fill=acc!10] (2.4,0.6) rectangle (4.5,2.6);
\draw[fill=acc!10,draw=acc] (3.45,1.6) circle (1.8pt);
\draw[->,black,thick] (3.45,1.6)--(3.45,0.72);
\node[right] at (3.5,1.05) {weight};
\draw[->,acc,thick] (3.45,1.6)--(2.3,1.6);
\node[above] at (2.75,1.62) {inertial force};
\draw[->,acc,very thick] (3.45,1.6)--(2.3,0.72);
\node[below] at (2.15,0.7) {ef{}fective gravity};
\draw[->,black,thick] (5.1,0.3)--(6.4,0.3);
\node[above] at (5.75,0.32) {vehicle acceleration};
\end{tikzpicture}
$$

Consider a block of base width $b$ and centre-of-mass height $h_{\rm cg}$ on a
vehicle accelerating horizontally with magnitude $a$. In the inertial frame, floor
friction supplies horizontal force $Ma$ and the normal reaction supplies $Mg$:

$$
F_{\rm friction}=Ma,
\qquad
N=Mg.
$$

Balancing torques about the centre of mass places the effective normal reaction a
distance $d$ from the centreline:

$$
Ma h_{\rm cg}=Nd.
$$

Thus

$$
d=\frac{a h_{\rm cg}}{g}.
$$

At the tipping threshold $d=b/2$, so

$$
a_{\rm tip}=\frac{b}{2h_{\rm cg}}g.
$$

The result assumes that static friction can provide $Ma$ before the contact shifts
to the edge. The sliding threshold is $a_{\rm slide}=\mu_s g$ for a horizontal
floor. The lower of these two thresholds controls the first loss of the assumed
static state.

An elevator or laboratory platform accelerating vertically changes support force
without changing torque geometry. A mass $m$ standing on a scale has

$$
N-mg=ma_y,
\qquad
N=m(g+a_y).
$$

Upward acceleration increases the scale reading; downward acceleration decreases
it. The apparent weight becomes zero in free fall when $a_y=-g$. A hanging
instrument in the same cabin aligns with the effective gravity direction, which is
vertical for pure vertical acceleration and tilted for horizontal acceleration.

The effective-gravity construction gives a compact stability test. In the
accelerated frame, combine gravitational acceleration and inertial acceleration:

$$
\vec g_{\rm eff}=\vec g-\vec a_{\rm frame}.
$$

The line through the centre of mass parallel to $\vec g_{\rm eff}$ must intersect
the support region. This construction avoids resolving each force separately for a
rigid body at rest in the accelerated frame. It still assumes uniform acceleration
and a contact model capable of supplying the required resultant reaction.

Acceleration-frame analysis needs an explicit frame statement. A passenger seated
in a turning vehicle experiences additional effects from rotation; treating that
case as a uniformly translating frame omits centripetal and rotational inertial
terms. A body on a ship can have time-varying translation and rotation. Static
equilibrium methods then apply only over a specified instant or after a suitable
quasistatic approximation is justified.

**Determinacy, reaction consistency, and static measurement.**

A planar rigid body has three independent equilibrium equations. Three independent
reaction components can therefore be determined from

$$
\sum F_x=0,\qquad \sum F_y=0,\qquad \sum\tau_z=0.
$$

For example, a beam with a pin at one end and a smooth roller at the other has two
pin components and one roller reaction. The support types and their orientations
matter. A roller on an inclined guide has a reaction normal to the guide, so its
horizontal and vertical components are linked by that known direction and count as
one unknown magnitude.

Static indeterminacy occurs when equilibrium equations alone leave reaction
components unresolved. A beam with a pin and two vertical rollers has four apparent
reaction components in a planar model. Force and torque balance supply three
equations. The remaining relationship depends on deformation of the beam and
supports, thermal changes, assembly clearance, or load sharing. A numerical answer
requires those extra physical conditions; an arbitrary division of load among
supports has no general basis.

Equation counting supplements a free-body diagram. Two
unknown force components can be dependent if their directions are constrained to be
parallel. A body acted on only by vertical forces has one independent force equation
and two independent torque equations in three dimensions; horizontal force equations add
identities rather than information. A mechanism with an unrestrained rotation can
have the correct number of reaction components on paper while remaining unable to
support a general load.

An isolated three-dimensional rigid body has up to six independent equations when
available. A ball-and-socket joint supplies three force components. A hinge can
supply force components and moments about axes that it prevents, while allowing
rotation about its hinge axis. Connection hardware must be translated into permitted
motions and reaction components before counting. A generic label such as \"joint\"
does not define a reaction model.

Load-cell measurements test an equilibrium model. Support readings give
reaction forces. A force plate measures the location of a normal resultant from its
moments. A plumb line or level establishes gravity direction. Position measurements
locate lever arms. Before comparing data with a model, state the coordinate origin,
sign convention, calibration range, and whether the measured load is a time average
or an instantaneous value.

Residual force and moment quantify consistency:

$$
\vec R_F=\sum\vec F,
\qquad
\vec R_\tau=\sum\vec\tau_O.
$$

Their units differ. A small force residual can coexist with a large torque residual
when forces have long lever arms. Report both relative to a meaningful force scale
and moment scale. Changing the torque origin is also a strong check: for a correct
force balance, the calculated net torque remains zero about every origin within
measurement uncertainty.

Practical checks include unit consistency, reaction directions, contact admissibility,
and limiting cases. A cable must carry nonnegative tension. A smooth one-sided
support must carry a compressive normal reaction or separate. If a load is moved
toward one support, that support's reaction should increase in a simple two-support
model. A reported centre of mass should lie within the measured mass envelope.
These checks catch setup errors before small numerical rounding differences receive
attention.

## Measurement protocols and force-system diagnostics

> **Worked example.** A horizontal uniform rod of length $L=2.40\ \mathrm m$ and mass
> $M=6.00\ \mathrm{kg}$ is pinned to a wall at its left end, and an $18.0\ \mathrm{kg}$
> sign hangs from its right end. A cable runs from that right end to a wall point
> $1.80\ \mathrm m$ above the pin. Treating the rod as rigid and the cable as massless,
> find the cable tension and the wall reaction on the rod.
>
> The cable spans $1.80\ \mathrm m$ vertically and $2.40\ \mathrm m$ horizontally, a
> $3.00\ \mathrm m$ hypotenuse, so
>
> $$
> \sin\theta=\tfrac{1.80}{3.00}=0.600,\qquad
> \cos\theta=\tfrac{2.40}{3.00}=0.800.
> $$
>
> Take the rod as the system. Moments about the pin remove both unknown pin
> components. With the rod weight at $L/2$ and the sign weight $mg$ at $L$,
>
> $$
> (T\sin\theta)L-(Mg)\tfrac{L}{2}-(mg)L=0,
> $$
>
> so
>
> $$
> T\sin\theta=\left(\tfrac{M}{2}+m\right)g=(3.00+18.0)(9.81)=206\ \mathrm N,
> \qquad
> T=\frac{206\ \mathrm N}{0.600}=343\ \mathrm N.
> $$

$$
% caption: Free-body diagram for the cable-supported sign. Taking moments about the
% wall pin gives both pin components a zero moment arm, so the cable tension, rod
% weight at midspan, and sign weight at the end fix the tension before force balance.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black,thick] (0.8,0.5)--(0.8,3.4);
\draw[acc,very thick] (0.8,1.0)--(6.0,1.0);
% cable tension (arrow along cable, toward wall)
\draw[->,acc,very thick] (6.0,1.0)--(0.9,3.05);
\node[above] at (3.6,2.15) {cable tension};
% weights
\draw[->,black,thick] (3.4,1.0)--(3.4,0.3);
\node[below] at (3.4,0.25) {rod weight};
\draw[->,black,thick] (6.0,1.0)--(6.0,0.3);
\node[below] at (6.0,0.25) {sign weight};
% pin and its reaction components
\draw[fill=white,draw=black] (0.8,1.0) circle (2.3pt);
\draw[->,acc,thick] (0.8,1.0)--(1.5,1.0);
\draw[->,acc,thick] (0.8,1.0)--(0.8,1.7);
\node[left] at (0.75,1.55) {pin reaction};
\end{tikzpicture}
$$

Force balance gives the pin reaction. Horizontal equilibrium gives

$$
R_x-T\cos\theta=0
\quad\Longrightarrow\quad
R_x=(343)(0.800)=275\ \mathrm N.
$$

The positive direction is to the right, so the wall pushes the rod rightward. Vertical
equilibrium gives

$$
R_y+T\sin\theta-(M+m)g=0,
$$

$$
R_y=(24.0)(9.81)-206=29.4\ \mathrm N.
$$

The small upward pin component follows from the cable supplying nearly all the
vertical support. The rod exerts the equal-and-opposite force on the wall. Compare
the cable rating with the calculated tension; its angle controls the tension required
to support the vertical component, so the sign weight alone is insufficient.

> **Worked example.** A $4.00\ \mathrm m$ horizontal beam is pinned at its left end and
> rests on a roller at its right end. It carries a uniform downward load
> $w=1.20\ \mathrm{kN\,m^{-1}}$ over the full span plus a $3.00\ \mathrm{kN}$ point load
> $1.00\ \mathrm m$ from the left support. Find the vertical support reactions.
>
> Replace the uniform load by its resultant, acting at midspan:
>
> $$
> W=wL=(1.20)(4.00)=4.80\ \mathrm{kN}\quad\text{at }2.00\ \mathrm m.
> $$
>
> With $R_A$ and $R_B$ the upward reactions at left and right, moments about the left
> pin isolate $R_B$:
>
> $$
> (4.00)R_B-(4.80)(2.00)-(3.00)(1.00)=0
> \quad\Longrightarrow\quad
> R_B=3.15\ \mathrm{kN},
> $$
>
> and vertical force balance gives
>
> $$
> R_A+R_B-4.80-3.00=0,
> \qquad
> R_A=4.65\ \mathrm{kN}.
> $$
>
> The reactions sum to $7.80\ \mathrm{kN}$, the total downward load, and their moments
> about the left support cancel:
>
> $$
> (3.15)(4.00)-(4.80)(2.00)-(3.00)(1.00)=0\quad\mathrm{kN\,m}.
> $$
>
> The two checks probe different parts of the calculation: a misplaced resultant can
> satisfy force balance while breaking torque balance.

$$
% caption: Free-body diagram of the mixed-load beam. The uniform load and the point
% load act downward on the span; taking moments about the left support isolates the
% right reaction, then vertical force balance gives the left reaction.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[acc,very thick] (0.7,1.4)--(6.5,1.4);
% uniform load
\draw[acc,thick] (1.0,2.3)--(6.2,2.3);
\foreach \x in {1.3,1.9,2.5,3.1,3.7,4.3,4.9,5.5}{
  \draw[->,black] (\x,2.3)--(\x,1.5);
}
\node[above] at (4.5,2.35) {uniform load};
% point load at 1 m from left support
\draw[->,acc,very thick] (2.45,3.1)--(2.45,1.5);
\node[above] at (2.45,3.15) {point load};
% reactions
\draw[->,acc,very thick] (1.0,0.4)--(1.0,1.3);
\draw[->,acc,very thick] (6.1,0.4)--(6.1,1.3);
\node[below] at (1.0,0.35) {left reaction};
\node[below] at (6.1,0.35) {right reaction};
\end{tikzpicture}
$$

> **Worked example.** A block of base width $b=0.800\ \mathrm m$ and centre-of-mass
> height $h_{\rm cg}=0.600\ \mathrm m$ sits on a horizontal vehicle floor with static
> friction coefficient $\mu_s=0.500$. As the vehicle accelerates, does the block slide
> or tip first?
>
> The tipping threshold takes moments about the trailing edge:
>
> $$
> a_{\rm tip}=\frac{b}{2h_{\rm cg}}\,g=\frac{0.800}{2(0.600)}(9.81)
> =6.54\ \mathrm{m\,s^{-2}}.
> $$
>
> The sliding threshold is where friction saturates, $a_{\rm slide}=\mu_s g$:
>
> $$
> a_{\rm slide}=\mu_s g=(0.500)(9.81)=4.91\ \mathrm{m\,s^{-2}}.
> $$
>
> Since $a_{\rm slide}<a_{\rm tip}$, the block slides first: the assumed static state
> is lost to relative sliding before the tipping threshold is reached. A rougher floor
> or a restraining stop would raise $a_{\rm slide}$ and could reverse the order.

**Centre-of-gravity measurement.** For an irregular plate, suspend it from three
holes and record the three plumb lines in a common coordinate system. Fit their
intersection rather than selecting one pixel by eye. If line $i$ has equation
$a_ix+b_iy+c_i=0$, estimate the centre $(x_c,y_c)$ by minimizing

$$
S(x_c,y_c)=\sum_i\left(a_ix_c+b_iy_c+c_i\right)^2,
$$

after normalizing each line so $a_i^2+b_i^2=1$. The residual distances provide an
uncertainty scale for the measured centre. A systematic offset in the vertical
reference affects all lines similarly and requires a calibration check rather than
more repeated suspensions.

The measurement protocol should record plate orientation, hole positions, plumb-line
width, image scale, and uncertainty in the vertical direction. Verify the inferred
centre with a knife-edge balance along one or two axes. Agreement between suspension
and balance methods tests both the geometry record and the assumption that the
plate's mass distribution stayed unchanged between measurements.

Static analysis ends with a physical report: state the isolated system, all support
models, coordinate axes, torque point, sign convention, load values, resultant
locations, solved reactions, residual force and moment, and admissibility checks.
This record permits a reader to distinguish an equilibrium result from an unstated
assumption about a cable, contact, or support.

**Equivalent force systems and balance diagnostics.**

Forces may be combined only when both their resultant force and resultant torque are
preserved. Several parallel vertical forces $F_i$ at coordinates $x_i$ have

$$
F_R=\sum_i F_i,
\qquad
x_R=\frac{\sum_i x_iF_i}{F_R},
$$

provided $F_R\ne0$. The result is the discrete form of the distributed-load
calculation. If $F_R=0$ while the force moments do not cancel, the system reduces
to a pure couple instead of a single force with a finite line of action.

Moving a force from point $A$ to point $B$ changes its torque about every reference
point. An equivalent system at $B$ contains the same force plus a couple:

$$
\vec M_B=(\vec r_A-\vec r_B)\times\vec F.
$$

The added couple restores the missing moment. This reduction applies when a
force acts through a bracket, a distributed contact patch is represented by a
resultant, or several loads are transferred to a convenient calculation point. It
also explains why a force cannot be placed arbitrarily on a free-body diagram.

Torque comparison about two origins gives the identity. If $\vec r_{O'O}$
points from $O'$ to $O$, then

$$
\sum\vec\tau_{O'}
=\sum\vec\tau_O
+\vec r_{O'O}\times\sum\vec F.
$$

When the net force is zero, a zero net torque about one point is a zero net torque
about every point. When the net force is nonzero, net torque changes with origin as
expected for an accelerating system. This identity explains why a complete
equilibrium solution can use any convenient torque point and why a partial torque
calculation should never be moved to a new origin without carrying the force
resultant along.

Numerical statics benefits from a declared tolerance. Suppose force readings are
reported to $1\ \mathrm N$ and lever arms to $1\ \mathrm{mm}$. A reaction residual
of a few newtons and a moment residual comparable with the summed force times
position uncertainty may be consistent with measurement resolution. A residual
orders of magnitude larger than those scales indicates a sign, unit, geometry, or
model error. Rounding all intermediate quantities before moment calculations can
create an apparent imbalance; retain guard digits until the final report.

An equilibrium model is complete when its external forces and couples are identified,
their lines of action are defined, the number of independent unknowns is justified,
and the solved reactions satisfy all relevant equations and contact conditions.
The algebra occupies only the final stage. The physical information enters through
system boundary, geometry, support model, load direction, and admissible contact
states.

## Inclined contacts and reaction-direction checks

A smooth contact force is normal to the contact surface. For a body on an incline
at angle $\beta$, resolve weight into axes perpendicular and parallel to the plane:

$$
W_\perp=Mg\cos\beta,
\qquad
W_\parallel=Mg\sin\beta.
$$

With no other forces, the normal reaction has magnitude $N=Mg\cos\beta$. Static
friction must supply an uphill force of magnitude $Mg\sin\beta$ to prevent sliding.
The contact condition is

$$
|f_s|\le\mu_sN.
$$

At the threshold of downward sliding,

$$
Mg\sin\beta=\mu_sMg\cos\beta,
\qquad
\tan\beta=\mu_s.
$$

The equality applies only at impending relative motion. Below that threshold,
static friction takes the value required by force balance and can be much smaller
than $\mu_sN$.

$$
% caption: Free-body diagram of a block at rest on an incline. The normal reaction is
% perpendicular to the surface and static friction points uphill, opposing the
% downhill tendency of the weight, which acts vertically through the centre of mass.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black,thick] (0.6,0.6)--(6.4,3.0);
\draw[black,thick] (0.6,0.6)--(6.4,0.6);
\draw[acc,thick,fill=acc!10] (2.6,1.43)--(3.75,1.9)--(3.37,2.82)--(2.22,2.35)--cycle;
\draw[fill=acc!10,draw=acc] (2.98,2.12) circle (1.8pt);
% weight
\draw[->,black,thick] (2.98,2.12)--(2.98,0.9);
\node[right] at (2.98,1.35) {weight};
% normal
\draw[->,acc,thick] (2.98,2.12)--(2.6,3.04);
\node[above] at (2.55,3.06) {normal};
% friction
\draw[->,acc,thick] (2.98,2.12)--(3.9,2.5);
\node[right] at (3.9,2.5) {friction};
\end{tikzpicture}
$$

An inclined cable, guide, or contact surface should be represented by a unit
direction vector before equations are written. A reaction of unknown magnitude $R$
normal to a surface with outward normal
$\hat n=(n_x,n_y)$ has components

$$
R_x=Rn_x,
\qquad
R_y=Rn_y.
$$

This one-magnitude representation prevents a smooth roller reaction from being
mistakenly counted as two independent Cartesian unknowns. It also keeps the
reaction direction tied to the observed guide geometry when force balance is solved.

Contact admissibility remains part of the result. A normal reaction is compressive
for a one-sided smooth surface. A calculated reaction pointing into the surface
with the wrong sign indicates that the assumed contact is inactive. A friction
force must remain within its available bound. A cable tension must pull along the
cable. These inequalities distinguish a formal solution of linear equations from a
physically available static configuration.

A contact model can change during a load sequence. A wheel pressed against a step
can lose its lower-floor reaction as the applied force rises, leaving the step edge
as the rotation point. A block with two support pads can unload one pad when its
normal resultant reaches the other pad. Treat each contact set as a candidate static
state, solve the balance equations, and retain only states satisfying the appropriate
normal and friction inequalities. The transition between admissible contact sets is
often the event of interest in tipping and support-reaction problems.

The same force can have different signs in different equations without any
contradiction. A downward load is negative in a vertical force sum if upward is
positive. Its torque can be positive or negative according to whether its line lies
to the left or right of the selected torque point. A friction force directed uphill
on an incline can have a positive component in a global horizontal axis and a
negative component in a local down-slope axis. State the axes beside the free-body
diagram, then project each vector onto those axes before inserting signs.

Moment arms should be measured to force lines, not to support symbols or arrow
tails. A reaction force drawn beside a pin acts through the pin centre even when the
arrow is displaced visually for readability. A distributed-load resultant acts at
its centroid, not at the tallest point of a load diagram. A cable tension acts
through the attachment point along the cable direction. These geometric
specifications prevent a correct force magnitude from being paired with an
incorrect torque.

Repeated calculations can be organized as a compact data record:

- **System boundary:** list the bodies retained and the interactions crossing out of
  the selected system.
- **Geometry:** record coordinates, load locations, cable slopes, support normals,
  and centre-of-gravity locations in one unit system.
- **Equations:** write force components and moments with their signs before
  substituting numerical values.
- **Admissibility:** test cable tension, normal-contact sign, friction bound, support
  reactions, and any stated tipping condition.
- **Residuals:** recompute force and moment sums from rounded reported values, then
  compare them with measurement precision.

This format exposes a missing load or a changed contact state early. It also keeps
static calculations reproducible when the same apparatus is tested with another
load position, support spacing, or acceleration history.

Signed reaction results carry physical information. A negative horizontal pin
component merely reverses the assumed arrow. A negative cable magnitude changes the
admissible force system because the cable has no compressive branch. A negative
normal reaction changes the active contact set because a one-sided support cannot
pull a body toward its surface. Record those distinctions in the final calculation
instead of replacing a negative number with its magnitude. The sign identifies
whether the originally drawn interaction remains available under the stated load.