Static Equilibrium and Levers

Beginner

A bridge, a shelf and a balanced seesaw have one thing in common: they stay perfectly still. Physicists call this static equilibrium. Here is what it takes, step by step.

Step 1. The pushes and pulls must cancel. A book on a table is pulled down by its weight and pushed up by the table just as hard, so the net force is zero and it does not start to move.

Step 2. Cancelling forces is not enough. Push the two ends of a ruler lying on a table equally hard in opposite directions and it spins on the spot. So the turning effects must cancel too. The turning effect of a force, its torque, grows with the force and with its distance from the pivot.

Step 3. On a seesaw the heavier side wins unless it sits closer to the middle. Someone twice as heavy must sit half as far out, so that weight times distance matches on both sides. This is the law of the lever, known since Archimedes; it is how a long crowbar shifts a rock you could never lift. In the animation the heavy blue box slides toward the pivot, and at 2 marks against the light box's 4 the beam balances level.

Step 4. Every object behaves as if its whole weight acted at one point, its centre of mass. A bookcase, or a person leaning over, stays up while that point is above the base it stands on; lean further and the weight starts to turn it over.

Check yourself: a 20 kg child sits 3 m from the pivot of a seesaw. Where must a 60 kg parent sit to balance? Answer: 1 m from the pivot on the other side, three times as heavy and a third as far.

Intermediate

Two conditions for equilibrium

Push both ends of a ruler equally hard in opposite directions, and it spins although the forces cancel. A rigid body, one that keeps its shape, is in static equilibrium when it stays at rest: it neither starts to slide nor starts to turn. The forces must cancel, or the centre of mass would accelerate, and so must their turning effects, the torques, or the body would start to rotate:
$\sum F=0,\sum \tau =0$

In a plane the first is two equations, for horizontal and vertical components. Neither condition implies the other: two equal and opposite forces acting along different lines cancel as forces but form a couple, which is how your hands turn a steering wheel. Start every problem with a free-body diagram: the body on its own, with every external force drawn where it acts, each weight at its centre of mass and each support force at its contact.

Torque: force times lever arm

Torque measures how strongly a force turns a body about a chosen point. Take the force's line of action, the straight line along which it acts, and the perpendicular distance $r$ from the point to that line: the lever arm. Then
$\tau =rF\text{(}r\text{: lever arm, perpendicular to the line of action)}$

The unit is the newton metre (N·m), never written as joules, because a torque is not an energy. Counter-clockwise torques count as positive. A force whose line passes through the point has no torque, however large it is: pushing a door toward its hinges does nothing. If the force acts at a distance $d$ from the point, at an angle $\varphi$ to the line joining them, the lever arm is $d\sin \varphi$.

The law of the lever

Put a load $m_1$ at a distance $r_1$ on one side of a pivot and a load $m_2$ at $r_2$ on the other. Each weight $mg$ acts straight down, so its lever arm is its horizontal distance from the pivot. The two torques have opposite signs, and $\sum \tau =0$ gives
$m_{1}g{r}_1=m_{2}g{r}_2\text{(law of the lever)}$

The $g$ cancels, so only masses and distances matter: twice the mass needs half the lever arm. The pivot's push has no torque about the pivot and drops out; $\sum F=0$ then fixes its size, the total weight it carries, the beam's own included. The law also multiplies force: 300 N pushing on a crowbar 1.2 m from its fulcrum presses with 3600 N on a stone 0.10 m from it.

Worked example: the balance in the animation

The light box, $m_2=1.0\mathrm{kg}$, sits $r_2=0.40\mathrm{m}$ right of the pin, so with the beam level its clockwise torque is
${\tau }_2=r_2{m}_{2}g=0.40\cdot 1.0\cdot 9.81=3.92\mathrm{N}\cdot \mathrm{m}$
The heavy box, $m_1=2.0\mathrm{kg}$, balances it at
$r_1=\frac{m_2{r}_2}{m_1}=\frac{1.0\cdot 0.40}{2.0}=0.20\mathrm{m}$

That is 2 of the beam's 10 cm marks against 4. There both readouts show 3.92 N·m and the beam is level: the arrow $F_1=19.6\mathrm{N}$ is twice as long as $F_2=9.81\mathrm{N}$, and the lever arms $r_1$ and $r_2$ marked along the bottom are in the ratio 1 : 2.

At the start the heavy box sits 0.42 m out, and its 8.30 N·m beats the light box's 3.79 N·m. The beam turns counter-clockwise by about 10°, until the weighted pointer under the pin, swung toward the light side, supplies the missing 4.51 N·m. As the box slides in, ${\tau }_1$ falls; inside 0.20 m the light side wins, and at 0.10 m, with ${\tau }_1\approx 1.87\mathrm{N}\cdot \mathrm{m}$, the beam leans about 4.6° the other way.

Notice that ${\tau }_2$ drifts between 3.79 and 3.95 N·m although its box never moves: tilting the beam changes the horizontal distance from the pin to the dashed line of its weight. The lever arm is measured perpendicular to the line of action, not along the beam.

Worked example: a painter on a plank

A uniform 20 kg plank 3.0 m long rests on trestles under its ends, and a 70 kg painter stands 1.0 m from the left end. The plank's weight, $196.2\mathrm{N}$, acts at its middle, and the painter's, $686.7\mathrm{N}$, at 1.0 m. Torques about the left trestle remove its unknown force:
$F_R\cdot 3.0=686.7\cdot 1.0+196.2\cdot 1.5=981\mathrm{N}\cdot \mathrm{m}$
so $F_R=327\mathrm{N}$, and $\sum F=0$ gives $F_L=686.7+196.2-327=556\mathrm{N}$: the trestle nearer the painter carries more.

Now move the right trestle 1.0 m inward, leaving a 1.0 m overhang, and let the painter walk out. The plank tips about that trestle when the left push falls to zero, that is, when the painter's torque about it equals the plank's, whose centre lies 0.50 m on the other side: $70d=20\cdot 0.50$, so $d=0.14\mathrm{m}$. One short step past the trestle tips the plank, which is why scaffold boards are fixed down.

Centre of mass, stability and tipping

For torques, a body's whole weight acts at one point, its centre of mass, the mass-weighted average position:
$x_{\mathrm{CM}}=\frac{\sum m_i{x}_i}{\sum m_i}$
For the boxes at balance, $x_{\mathrm{CM}}=(2.0\cdot (-0.20)+1.0\cdot 0.40)/3.0=0$: straight above the pin, which is another way of saying that the beam is level.

An equilibrium is stable if a small tilt raises the centre of mass, so that the weight turns the body back, as in the balance of the animation, whose heavy pointer keeps the centre of mass below the pin. It is unstable if a tilt lowers the centre of mass, like a pencil on its tip or a seesaw with two riders.

A body on a base topples once the vertical through its centre of mass passes outside the base. A uniform fridge 0.70 m wide and 1.80 m tall, tilted onto one edge, rocks back if released before 21.3°, where $\tan \theta =0.70/1.80$, and falls beyond it. On a ramp a uniform box tips where $\tan \theta =w/h$.

Common mistakes

• Thinking that balance needs equal weights. It needs equal torques: 2.0 kg at 0.20 m balances 1.0 kg at 0.40 m.
• Measuring the lever arm along the beam or the door instead of perpendicular to the line of action. A 50 N push at 30° to a door, 0.80 m from the hinges, gives $0.80\cdot 50\cdot \sin 30^{\circ }=20\mathrm{N}\cdot \mathrm{m}$, not 40 N·m.
• Leaving out the beam's own weight. Without the plank's 20 kg, the right trestle would seem to push 229 N instead of 327 N.
• Believing that torques must be taken about the real pivot. $\sum \tau =0$ holds about every point, so choose one where an unknown force acts.
• Expecting a heavier object to tip at a steeper angle. Only the shape matters: an evenly loaded crate 0.40 m wide and 1.00 m tall tips at 21.8° whether it holds 20 kg or 200 kg.

Where you meet it

Engineers size beams and foundations with these two equations, and a tower crane balances its load with a concrete counterweight on its short arm. Your forearm is a lever: the biceps pulls about 5 cm from the elbow, so holding a 5.0 kg dumbbell 35 cm from the elbow takes about $49\cdot 0.35/0.05\approx 343\mathrm{N}$, seven times its weight. These ideas continue in Simple Machines: Levers, Pulleys and Gears, where levers trade force for distance, and in Elasticity: Hooke's Law, Stress and Strain, where real beams bend under the loads found here.

Advanced

Equilibrium as zero virtual work

Two conditions keep a rigid body at rest: the forces on it add up to zero, and so do their torques. Both follow from a single statement about work, and the same statement later decides whether an equilibrium is stable, so it is worth deriving them that way once.

Imagine moving a body at rest by a tiny amount that its supports allow: a virtual displacement, made only on paper. Add up the work that every force on the body would do during it; the total is the virtual work $\delta W$. If $\delta W$ were positive for some allowed motion, the forces themselves would supply kinetic energy along it, and the body would set off that way on its own. Reversing a small displacement reverses the sign of $\delta W$, so it cannot be negative either. In equilibrium, therefore, the virtual work vanishes, to first order in the displacement, for every displacement the supports allow.

A rigid body moving in a plane has three independent small motions: a shift in each of two directions, and a turn. In a shift $\delta \vec{s}$ every point of the body moves by the same $\delta \vec{s}$, so $\delta W=(\sum \vec{F})\cdot \delta \vec{s}$, which vanishes for shifts in every direction only if $\sum \vec{F}=0$. In a small turn $\delta \theta$ about any point, a force whose torque about that point is $\tau$ does the work $\tau \delta \theta$: torque is exactly the work per radian of turn. So $\delta W=(\sum \tau )\delta \theta$, which vanishes only if $\sum \tau =0$.

Internal forces, the pushes and pulls between parts of the body, never enter. They come in equal and opposite pairs along the line joining the two parts, and a rigid body keeps that distance fixed, so the two forces of each pair do equal and opposite work.

The law of the lever, from work alone

The simplest test of the principle is a lever. Virtual work gives its law straight from weights and heights, without defining torque at all.

Legend

$m_{1}g$	Weight of load 1, whose mass is $m_1$.	$\mathrm{N}$
$m_{2}g$	Weight of load 2, whose mass is $m_2$.	$\mathrm{N}$
$r_1$	Distance from the pin to load 1, along the beam.	$\mathrm{m}$
$r_2$	Distance from the pin to load 2, along the beam; here twice $r_1$.	$\mathrm{m}$
$\delta \theta$	Small virtual turn of the beam about the pin, exaggerated in the drawing.	$\mathrm{rad}$
$\delta h_1$	Drop of load 1 during the turn: $r_1\delta \theta$.	$\mathrm{m}$
$\delta h_2$	Rise of load 2 during the turn: $r_2\delta \theta$.	$\mathrm{m}$

Figure 1 shows a beam on a pin, with load 1, of mass $m_1$, at the distance $r_1$ on one side and load 2, of mass $m_2$, at the distance $r_2$ on the other. The beam is balanced, $m_1{r}_1=m_2{r}_2$, and we give it a small virtual turn $\delta \theta$ about the pin, drawn much larger than it really is. Look at the two short vertical brackets. A point of the beam at the distance $r$ from the pin moves along an arc of length $r\delta \theta$, and for a small angle that arc is practically vertical. So load 1 drops by $\delta h_1=r_1\delta \theta$ and load 2 rises by $\delta h_2=r_2\delta \theta$.

Now add up the work. The pin does not move, so its force does no work. Gravity does work only on vertical motion: on load 1, which falls, it does $+m_{1}g\delta h_1$; on load 2, which rises, it does $-m_{2}g\delta h_2$. Setting the total to zero gives $m_{1}g{r}_1\delta \theta =m_{2}g{r}_2\delta \theta$, and dividing by $g\delta \theta$ leaves $m_1{r}_1=m_2{r}_2$. That is the law of the lever, obtained from energy bookkeeping alone.

In the animation the balanced beam carries a 2.0 kg box at $r_1=0.20\mathrm{m}$ and a 1.0 kg box at $r_2=0.40\mathrm{m}$. Tilting it by 0.01 rad lowers the heavy box by 2 mm and raises the light one by 4 mm, and $2.0\cdot 2=1.0\cdot 4$: what gravity gains on one side, it pays back on the other.

Why could the rule be nothing else? A balanced, frictionless lever can be turned slowly by a push as small as you like. If it balanced at any other ratio of masses and distances, turning it one way would raise the total potential energy of the loads with nothing paying for it, and the trick could be repeated without limit: a perpetual-motion machine. So the law of the lever follows from the impossibility of perpetual motion. Simon Stevin argued in this way in 1586 about the inclined plane: a closed loop of chain hung over a wedge never starts to circulate by itself.

Three equations, and any point as pivot

In practice one writes the equilibrium conditions as equations and picks the pivot that makes them easiest. This section shows why any pivot is allowed, and how many unknowns the equations can fix.

In a plane, equilibrium is three equations: $\sum F_x=0$, $\sum F_y=0$ and $\sum {\tau }_z=0$. The torque about a point $B$ of a force ${\vec{F}}_i$ applied at ${\vec{r}}_i$ is $({\vec{r}}_i-{\vec{r}}_B)\times {\vec{F}}_i$. Split the arm as $({\vec{r}}_i-{\vec{r}}_A)+({\vec{r}}_A-{\vec{r}}_B)$, where $A$ is any other point, and add over all the forces:
${\vec{\tau }}_B={\vec{\tau }}_A+({\vec{r}}_A-{\vec{r}}_B)\times \sum \vec{F}$
When the net force is zero, the second term vanishes and the torque is the same about every point. So the pivot can go wherever is convenient, usually at a support: that support's unknown force has no arm there and drops out of the torque equation.

Three equations fix at most three unknowns. A beam on two supports, or a ladder against a smooth wall, has no more than that and is called statically determinate. A four-legged table has four unknown leg forces but still only three useful equations, so it is statically indeterminate: statics alone cannot say how the load is shared, and only the slight give of the legs and the floor decides it. On an uneven floor one leg carries nothing, and the table rocks.

Three forces meet in one point

When exactly three forces hold a body still, a geometric rule fixes the direction of one of them before any equation is written. A ladder leaning on a wall is the standard example.

Legend

$W$	Weight of the ladder, acting at its midpoint.	$\mathrm{N}$
$P$	Push of the wall. A smooth wall pushes only at right angles to itself, so $P$ is horizontal.	$\mathrm{N}$
$R$	Reaction of the floor at the foot of the ladder: the normal force and the friction added together. Its line passes through $T$.	$\mathrm{N}$
$T$	Point where the lines of action of $W$ and $P$ cross.
$\theta$	Angle between the ladder and the floor.	$\mathrm{rad}$
$\phi$	Angle between $R$ and the vertical; $\tan \phi =a/h$, the friction divided by the normal force.	$\mathrm{rad}$
$h$	Height of the top of the ladder, which is also the height of $T$.	$\mathrm{m}$
$a$	Horizontal distance from the foot of the ladder to the vertical through its midpoint, which also passes through $T$.	$\mathrm{m}$

Figure 2 shows a uniform ladder of length $L$ and weight $W$, standing on the floor at the angle $\theta$ and leaning on a smooth wall. Three forces act on it. The weight $W$ acts straight down through the midpoint. A smooth wall can push only at right angles to itself, so its push $P$ at the top is horizontal. The floor's reaction $R$ at the foot has, so far, an unknown direction. Follow the dashed lines of action of $W$ and $P$: they cross at $T$, level with the top of the ladder and straight above its midpoint.

Take torques about $T$. Neither $W$ nor $P$ has a torque there, because both lines pass through $T$. The total torque is zero, so $R$ must have no torque about $T$ either, which is possible only if its line of action also passes through $T$. This fixes the direction of $R$ before friction has even been mentioned. The rule is general: when exactly three non-parallel forces hold a body in equilibrium, their lines of action meet in one point.

Now split $R$ into the normal force $N$, straight up, and the friction $f$, along the floor towards the wall. Balancing vertical forces gives $N=W$, and horizontal ones $f=P$. In Figure 2, $R$ leans from the vertical by the angle $\phi$, so $\tan \phi =f/N$. The same angle sits at the foot of the right triangle whose vertical side is the height of the top (and of $T$), $h=L\sin \theta$, and whose horizontal side is the distance from the foot to the vertical through the midpoint, $a=\frac{1}{2}L\cos \theta$. Hence $\tan \phi =a/h$, and
$\frac{f}{N}=\frac{a}{h}=\frac{1}{2\tan \theta }$
For a uniform ladder at $60^{\circ }$ to the floor this is $1/(2\tan 60^{\circ })\approx 0.29$, whatever the ladder weighs and however long it is. The ladder therefore stays up only if the coefficient of static friction at the floor is at least 0.29. A steeper ladder needs less.

Gravity acts at the centre of mass

Almost every statics problem replaces a body's weight by one force at its centre of mass. In a uniform field that replacement is exact, and it also predicts how the balance in the animation behaves.

Each part $i$ of a body, of mass $m_i$ at the position ${\vec{r}}_i$ from a pivot, feels the weight $m_i\vec{g}$. Because $\vec{g}$ is the same for every part, it comes out of the sum: $\sum {\vec{r}}_i\times m_i\vec{g}=(\sum m_i{\vec{r}}_i)\times \vec{g}=M{\vec{r}}_{\mathrm{CM}}\times \vec{g}$, where $M$ is the total mass and ${\vec{r}}_{\mathrm{CM}}=\sum m_i{\vec{r}}_i/M$. The whole weight acts at the centre of mass. With $x$ measured horizontally from the pivot and counter-clockwise torques positive,
$x_{\mathrm{CM}}=\frac{\sum m_i{x}_i}{\sum m_i}$
$\sum {\tau }_{\mathrm{gravity}}=-Mg{x}_{\mathrm{CM}}$
The minus sign says that a centre of mass to the right of the pivot turns the body clockwise.

Two consequences follow. A body hung from a pin settles with its centre of mass straight below the pin, where gravity has no torque. And the plumb lines from two different hanging points cross at the centre of mass, which is how it is found in practice.

Apply this to the two boxes alone. Write ${\tau }_1$ for the counter-clockwise torque of the heavy box, ${\tau }_2$ for the clockwise torque of the light one, and $x_{\mathrm{CM}}$ for the boxes' centre of mass; then ${\tau }_1-{\tau }_2=-(m_1+m_2)g{x}_{\mathrm{CM}}$. The expert graph of the animation plots it against the heavy box's position $x_1$: 4.51 N·m at $x_1=-0.42\mathrm{m}$, zero at $x_1=-m_2{r}_2/m_1=-0.20\mathrm{m}$, and $-2.07\mathrm{N}\cdot \mathrm{m}$ at $x_1=-0.10\mathrm{m}$.

The line is nearly straight, with a slope close to $-m_{1}g=-19.6\mathrm{N}\cdot \mathrm{m}$ per metre: each centimetre the box slides inward removes about 0.20 N·m. The plotted slope is about 5 % steeper. The boxes' centres ride some 5 cm above the pin line, so the change of tilt that comes with each shift of the box also moves them sideways a little.

This is not the total torque on the beam. The 14 kg bob on the pointer supplies an opposite torque and keeps the total at zero, which is why the beam tilts only part of the way (worked out below). The gold plumb line in the animation shows the same thing geometrically: the beam leans towards the side on which the boxes' centre of mass lies.

Stability and potential energy

An equilibrium can be like the bottom of a bowl, which the body returns to, or like the top of a hill, which it falls from. The potential energy tells the two apart.

For a body turning about a pin, the potential energy is $U(\theta )=Mg{y}_{\mathrm{CM}}$, where $y_{\mathrm{CM}}$ is the height of the centre of mass. When the body turns by $d\theta$, gravity's torque does the work $\tau d\theta$, which is taken from $U$, so $dU=-\tau d\theta$. Equilibrium, zero torque, therefore means $dU/d\theta =0$. It is stable if $d^{2}U/d{\theta }^2>0$: a tilt then raises the centre of mass, and the torque that appears pushes the body back. It is unstable if $d^{2}U/d{\theta }^2<0$, and neutral if $U$ stays constant, as for a uniform wheel on its axle.

Put the centre of mass at the distance $\ell$ from the pin and at the angle $\psi$ from straight below it. Its height relative to the pin is $-\ell \cos \psi$, so $U=-Mg\ell \cos \psi$ and $d^{2}U/d{\psi }^2=Mg\ell \cos \psi$. This is positive at $\psi =0$, so hanging below the pin is stable, and negative at $\psi =180^{\circ }$, so balancing above it is unstable. A seesaw with two riders has its centre of mass above the pivot: it does not stay level but comes to rest on one end, held there by the ground. The balance in the animation hangs a 14 kg bob 0.19 m below the pin precisely to avoid this.

Worked problem: how far the balance tilts

With the box off-centre, the balance in the animation does not fall to one end: it settles at a definite tilt. That angle follows first from geometry, then from energy.

Legend

$O$	The pin about which the beam turns.
$G$	Centre of mass of the whole balance: boxes, beam, pointer and bob.
$Mg$	Weight of the whole balance, of total mass $M$, acting at $G$.	$\mathrm{N}$
$u_G$	Position of $G$ along the beam, measured from the pin and positive to the right; negative here.	$\mathrm{m}$
$w_G$	Position of $G$ across the beam, positive upward; negative here, because $G$ is below the beam.	$\mathrm{m}$
$\theta$	Tilt of the beam from the horizontal, counter-clockwise positive. The same angle sits at $G$ in the right triangle.	$\mathrm{rad}$

Figure 3 sketches the balance with its proportions exaggerated. Use coordinates fixed to the beam and measured from the pin $O$: $u_i$ along the beam (positive to the right) and $w_i$ across it (positive upward). Together the parts, of masses $m_i$, weigh like one body of mass $M=\sum m_i$ at their centre of mass $G$, whose coordinates are $u_G=A/M$ and $w_G=B/M$, with
$A=\sum m_i{u}_i,B=\sum m_i{w}_i$
The tilt $\theta$ of the beam from the horizontal is counted counter-clockwise.

The balance is at rest when $G$ hangs straight below $O$, because then the weight $Mg$ passes through the pin and has no torque. Straight above $O$ is also an equilibrium, but an unstable one.

In Figure 3, go from $O$ along the beam by $u_G$, then across it by $w_G$, and you reach $G$; the plumb line from $G$ back up to $O$ closes a right triangle, with the right angle where $u_G$ meets $w_G$. The angle at $G$ equals the tilt $\theta$. Two lines drawn at right angles to two others cross at the same angle as those do; the beam and the horizontal cross at $\theta$, so the side $w_G$, at right angles to the beam, and the plumb line, at right angles to the horizontal, meet at $\theta$ as well. Seen from $G$, $u_G$ is the opposite side and $w_G$ the adjacent one, so
$\tan \theta =\frac{u_G}{w_G}=\frac{A}{B}$
In the figure both $u_G$ and $w_G$ are negative, so $\tan \theta >0$: the beam has turned counter-clockwise, left side down.

The potential energy gives the same result, and it also settles stability. Turned counter-clockwise by $\theta$, part $i$ sits at the height $u_i\sin \theta +w_i\cos \theta$ above the pin, so
$U(\theta )=g(A\sin \theta +B\cos \theta )$
Setting $dU/d\theta =g(A\cos \theta -B\sin \theta )$ to zero gives $\tan \theta =A/B$ once more.

Now the numbers. The boxes' centres sit 0.053 m and 0.043 m above the pin line and the bob 0.19 m below it, so $B=2.0\cdot 0.053+1.0\cdot 0.043-14\cdot 0.19\approx -2.51\mathrm{kg}\cdot \mathrm{m}$; the beam's own weight acts at the pin and adds nothing.

At the start the heavy box is at $x_1=-0.42\mathrm{m}$ along the beam and the light one at 0.40 m, so $A=2.0\cdot (-0.42)+1.0\cdot 0.40=-0.44\mathrm{kg}\cdot \mathrm{m}$. Then $\tan \theta =0.175$ and $\theta =9.9^{\circ }$, heavy side down, as in Figure 3. At $x_1=-0.10\mathrm{m}$, $A=+0.20\mathrm{kg}\cdot \mathrm{m}$ and $\theta =-4.6^{\circ }$, the other side down; at $x_1=-0.20\mathrm{m}$, $A=0$ and the beam is level. As a check, the bob's clockwise torque at the start, $14\cdot 9.81\cdot 0.19\sin \theta \approx 4.51\mathrm{N}\cdot \mathrm{m}$, cancels the boxes' torque at the start of the graph.

Is this rest position stable? The second derivative is $d^{2}U/d{\theta }^2=-g(A\sin \theta +B\cos \theta )$, and at the equilibrium with $G$ below $O$ (here $B<0$) it works out to $g\sqrt{A^2+B^2}$. That is never less than $g|B|=9.81\cdot 2.51=24.6\mathrm{N}\cdot \mathrm{m}$ per radian, so the balance is stable wherever the box sits.

Near level $\theta \approx A/B$: sliding the box 1 cm changes $A$ by 0.020 kg·m and tilts the beam by $0.020/2.51\approx 0.008\mathrm{rad}$, about 0.46°. That is the balance's sensitivity. A smaller $|B|$, from a lighter bob or a shorter pointer, would raise it, at the price of a weaker restoring torque, $g|B|$ per radian near level. Without the bob, $B=+0.15\mathrm{kg}\cdot \mathrm{m}$: $G$ lies above the pin, the second derivative at the level position is negative, and the slightest imbalance would tip the beam right over, like a seesaw.

Tipping and sliding

A body standing on a slope can fail in two ways: it slides down, or it tips over its lower edge. Which comes first depends only on its shape and on the friction.

Legend

$w$	Width of the crate's base, along the ramp.	$\mathrm{m}$
$h$	Height of the crate.	$\mathrm{m}$
$\theta$	Angle of the ramp. The same angle lies between $R$ and the normal to the ramp.	$\mathrm{rad}$
$mg$	Weight of the crate, acting at its centre.	$\mathrm{N}$
$R$	Contact force of the ramp on the crate (normal force plus friction): equal and opposite to the weight, so vertical.	$\mathrm{N}$
$d$	Distance from the lower edge of the base to the point where $R$ acts: $d=w/2-(h/2)\tan \theta$.	$\mathrm{m}$

Figure 4 shows a uniform crate, of base $w$ and height $h$, on a ramp at the angle $\theta$. Only two forces act on it: its weight $mg$, at the centre, and the ramp's contact force $R$, the normal force and the friction over the whole base added together into one force. Two forces balance only if they are equal, opposite and on one line, so $R$ points straight up along the vertical through the centre, and acts where that vertical meets the base.

To find that point, start from the centre, which is $w/2$ along the base from the lower edge and $h/2$ above the base. The vertical through the centre leans from the normal to the ramp by $\theta$, so on its way down through the height $h/2$ it drifts $(h/2)\tan \theta$ downhill. It therefore meets the base $(h/2)\tan \theta$ downhill of the middle of the base, at the distance
$d=\frac{w}{2}-\frac{h}{2}\tan \theta$
from the lower edge, marked $d$ in Figure 4. As the ramp steepens, $d$ shrinks. The contact force cannot act outside the base, so when $d$ reaches zero it can no longer stand under the centre, and the crate tips about its lower edge:
$\tan \theta =\frac{w}{h}\text{(tipping of a uniform crate on a ramp)}$

Sliding is read off the same picture. $R$ is vertical and the normal leans at $\theta$ from the vertical, so $R$ leans at $\theta$ from the normal: the angle marked between $R$ and the dashed normal in Figure 4. Its friction part divided by its normal part is therefore $\tan \theta$, and static friction can supply that only while $\mu \ge \tan \theta$, where $\mu$ is the coefficient of static friction. Comparing the two limits, the crate tips first if $w/h<\mu$ and slides first if $\mu <w/h$.

A filing cabinet 0.45 m wide and 1.30 m tall has $w/h\approx 0.35$. On rubber, with $\mu \approx 0.8$, it tips at 19.1°; on a wet plastic ramp with $\mu =0.20$, it slides first, at 11.3°. For a body that is not uniform, use the real distances of its centre of mass from the lower edge and from the ramp in place of $w/2$ and $h/2$. A low centre of mass makes $d$ shrink more slowly, which is why heavy goods are loaded low.

Limits of validity

Everything here assumes rigid bodies. Real beams bend, and in a statically indeterminate structure it is the deformations, through Hooke's law, that decide the forces. Static friction is an inequality, $f\le \mu N$: the equations give the friction a body needs, and the inequality only says whether the surfaces can supply it.

Placing the weight at the centre of mass also needs a uniform field. Gravity weakens with the height $y$ by a fraction of about $2y/R_{\mathrm{E}}$, where $R_{\mathrm{E}}$ is the Earth's radius. In a uniform column $H=800\mathrm{m}$ tall, the centre of gravity, the point where the weight effectively acts, therefore sits $H^2/(6R_{\mathrm{E}})\approx 1.7\mathrm{cm}$ below the centre of mass: negligible for buildings. In orbit the same gradient turns a long satellite to point at the Earth, and tidal torques have turned the Moon to keep one face towards us.

Finally, statics is the limit of slow change. Move the box suddenly and the beam overshoots and swings; then $\tau =I\alpha$ from Rotational Dynamics and Torque takes over.

Where this leads

Virtual work runs every ideal machine in Simple Machines: Levers, Pulleys and Gears, and sagging beams lead to Elasticity: Hooke's Law, Stress and Strain. A stable equilibrium is where Simple Harmonic Motion begins: displaced slightly, a body oscillates about it with ${\omega }^2=(d^{2}U/d{\theta }^2)/I$, where $I$ is its moment of inertia about the pin. Fluids obey statics too: in Buoyancy and Archimedes' Principle a ship stays upright when, as it heels, its weight and the buoyant force form a righting couple. When the torques do not cancel, the body starts to turn, the subject of Angular Momentum and Its Conservation. Charges have their own statics, in Electric Field and Potential, with a twist: electrostatic forces alone can never hold a charge in stable equilibrium (Earnshaw's theorem).