Part 2 · Chapter 12

Equilibrium and Elasticity

What keeps a structure standing still — the balance of forces and torques, and how real materials stretch, shear, and squeeze under load

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 45 min

i What you'll learn

The two requirements for equilibrium — constant linear momentum and constant angular momentum — and the extra condition that makes it static.
How to apply the balance of forces (\(F_{\text{net}} = 0\)) and the balance of torques (\(\tau _{\text{net}} = 0\)), and why a wise choice of rotation axis eliminates unknowns.
The idea of the center of gravity and when it coincides with the center of mass.
Why some real structures are indeterminate — more unknowns than equations — and how elasticity rescues us.
The three elastic moduli (Young's \(E\), shear \(G\), bulk \(B\)) and the master relation \(\text{stress} = \text{modulus} \times \text{strain}\).

Section 12-1

What Is Physics?

Human constructions are supposed to stay put. A building must stand despite gravity and the wind; a bridge must hold despite its own weight and the relentless jolting of traffic. One central concern of physics is therefore deceptively simple: what allows an object to remain stable under the forces acting on it? This chapter treats the two faces of that question — the equilibrium of forces and torques on rigid bodies, and the elasticity of real (non-rigid) bodies that governs how they deform. Done well, this physics fills engineering journals; done badly, it fills newspapers and courtrooms.

Section 12-2

Equilibrium

Consider four objects: a book on a table, a puck gliding at constant velocity on frictionless ice, the steadily turning blades of a ceiling fan, and a bicycle wheel rolling straight at constant speed. Each shares two features — its center-of-mass linear momentum is constant, and its angular momentum about any point is constant. We call such objects in equilibrium.

The two requirements for equilibrium

\[ \vec{P} = \text{a constant} \qquad \vec{L} = \text{a constant} \]

Both the linear momentum of the center of mass and the angular momentum about any point must be unchanging. Our concern in this chapter is the special case where both constants are zero.

When those constants are zero — the body is neither translating nor rotating in our frame — the object is in static equilibrium. Of the four objects above, only the book qualifies. The balancing rock of a desert formation, cathedrals, filing cabinets, and taco stands all share this quiet property of standing still over time.

⊕ Stable equilibrium

If a small displacing force is removed and the body returns to its original state, the equilibrium is stable — like a marble at the bottom of a bowl. Restoring torques push it back.

↡ Unstable equilibrium

A domino balanced on one edge sits in equilibrium (the weight's line of action passes through the edge, so its torque is zero), but the slightest nudge tips it: the torque then grows the rotation.

Stability is geometry. A domino on its narrow side is steadier than one balanced on an edge, and a child's square block is steadier still — to topple any of them, the center of mass must be carried up and over a supporting edge. The farther that journey, the more stable the object. The construction worker on a beam knows this in his body: wide stance along the beam, stable; narrow stance across it, at the mercy of the wind.

Section 12-3

The Requirements of Equilibrium

Translation is governed by \(\vec{F}_{\text{net}} = d\vec{P}/dt\). If \(\vec{P}\) is constant, the derivative vanishes and the net force is zero. Rotation is governed by the angular form \(\vec{\tau}_{\text{net}} = d\vec{L}/dt\). If \(\vec{L}\) is constant, the net torque is zero. These give the two vector requirements for equilibrium:

Balance of forces and balance of torques

\[ \vec{F}_{\text{net}} = 0 \qquad \vec{\tau}_{\text{net}} = 0 \]

The vector sum of all external forces is zero, and the vector sum of all external torques about any point is zero. The torque condition holds about every choice of point — which we will exploit shamelessly.

If we restrict the forces to the \(xy\) plane — as we will throughout this chapter — every torque points along \(z\), and the six component equations collapse to just three useful ones:

⚖

The three working equations (planar problems)

F_net,x = 0 · F_net,y = 0 · τ_net,z = 0

Two force equations and one torque equation. A puck gliding at constant velocity satisfies all three yet is not in static equilibrium — so there is a third requirement for the static case: the linear momentum \(\vec{P}\) must itself be zero (the body must be at rest).

The strategist's move. Because the torque balance holds about any axis, choose the axis to pass through an unknown force you would rather not deal with. That force then has zero moment arm and drops out of the torque equation entirely — turning a tangle of unknowns into one clean equation.

Section 12-4

The Center of Gravity

Gravity pulls on every atom of a body. Rather than track all those tiny forces, we replace them by a single equivalent force \(\vec{F}_{g}\) acting at one point — the center of gravity (cog). "Equivalent" means: if you switched off all the individual pulls and switched on \(\vec{F}_{g}\) at the cog, neither the net force nor the net torque about any point would change.

Summing the torques of the element-by-element gravitational forces about the origin and matching them to the torque of the single force \(\vec{F}_{g}\) gives \(x_{\text{cog}}\sum m_{i}g_{i} = \sum x_{i}m_{i}g_{i}\). If \(g\) is the same at every element, it cancels:

When g is uniform, the cog sits at the com

\[ x_{\text{cog}} = \frac{1}{M}\sum x_{i}m_{i} = x_{\text{com}} \]

For everyday objects g varies only slightly over their extent, so the center of gravity coincides with the center of mass — which is why we have been free to let gravity "act at the center of mass" all along.

Section 12-5

Some Examples of Static Equilibrium

The recipe never changes: (1) choose a system, (2) draw a free-body diagram showing every external force, (3) write \(F_{\text{net},x} = 0\) and \(F_{\text{net},y} = 0\), and (4) write \(\tau_{\text{net},z} = 0\) about a cleverly chosen axis. The skill lies entirely in two choices — the system and the axis. The worked examples below put the recipe to work; here it is worth fixing the sign convention.

↺ Torque signs

A torque that would turn a stationary body counterclockwise about the chosen axis is positive; one that would turn it clockwise is negative. Write each torque as \(\pm r_{\perp}F\).

→ Sliding the weight

A force may be slid along its own line of action without changing its torque about any axis. This lets you put a block's weight vector wherever the diagram is cleanest.

Section 12-6

Indeterminate Structures

Planar statics hands us only three equations. So if a problem carries more than three unknowns, we are stuck. The four (all-different) forces on a car's four tires, or the four legs of a table, cannot be found from statics alone — these are indeterminate problems.

Where did the missing equation go? We quietly assumed the body is perfectly rigid. No real body is. The tires flatten, the table legs compress, and each deforms until the loads sort themselves out into definite values. Rest a car's tires on four scales and each reads something specific. The extra physics we need is elasticity — a relation between how hard you push and how much the material gives.

Section 12-7

Elasticity

Atoms in a metallic solid sit on a stiff three-dimensional lattice, bound by interatomic "springs." That stiffness is why a steel ladder feels rigid — yet every real body is at least slightly elastic. Hang a subcompact car from a 1 m steel rod 1 cm across and it stretches about half a millimetre, then springs back. Hang three cars and it snaps. Across the useful range, the response is linear, captured by one master relation:

The master relation of elasticity

\[ \text{stress} = \text{modulus} \times \text{strain} \]

Stress is a deforming force per unit area; strain is the resulting unit (fractional) deformation. The constant of proportionality, a modulus of elasticity, has the same units as stress because strain is dimensionless.

Three kinds of deformation give three moduli. Tension or compression (force perpendicular to a face, length changes by \(\Delta L\)) uses Young's modulus \(E\). Shearing (force in the plane of a face, like skewing a deck of cards by \(\Delta x\)) uses the shear modulus \(G\). Hydraulic compression (fluid pressure \(p\) squeezing volume by \(\Delta V\)) uses the bulk modulus \(B\).

The three elastic moduli

\[ \frac{F}{A} = E\,\frac{\Delta L}{L} \qquad \frac{F}{A} = G\,\frac{\Delta x}{L} \qquad p = B\,\frac{\Delta V}{V} \]

Same form, three contexts. Beyond a material's yield strength S_y the deformation becomes permanent; beyond its ultimate strength S_u it ruptures. Concrete is strong in compression but weak in tension — which is why it is reinforced.

A few representative numbers (note the enormous span — bone to steel):

Table 12-1 · Some elastic properties of selected materials
Material	Density ρ (kg/m³)	Young's modulus E (10⁹ N/m²)	Ultimate strength S_u (10⁶ N/m²)	Yield strength S_y (10⁶ N/m²)
Steel	7860	200	400	250
Aluminum	2710	70	110	95
Glass	2190	65	50	—
Concrete	2320	30	40	—
Wood (Douglas fir)	525	13	50	—
Bone	1900	9	170	—
Polystyrene	1050	3	48	—

Solids resist squeezing. The bulk modulus is \(2.2 \times 10^{9}\,\mathrm{N/m^{2}}\) for water but \(1.6 \times 10^{11}\,\mathrm{N/m^{2}}\) for steel. At the Pacific's average depth (≈4000 m, pressure ≈\(4.0 \times 10^{7}\,\mathrm{N/m^{2}}\)), water compresses about 1.8% but steel only ≈0.025%. Rigid atomic lattices simply give less ground than loosely coupled liquids.

Worked Examples

Putting It to Work

1 Beam on two scales — reading the supports

Problem. A uniform beam of length \(L\) and mass \(m = 1.8\,\mathrm{kg}\) rests on two scales. A block of mass \(M = 2.7\,\mathrm{kg}\) sits with its center a distance \(L/4\) from the left end. What do the scales read?

Solution. Take the beam-plus-block as the system. Put the rotation axis at the left end to find the right-hand force \(F_{r}\), then use vertical balance for \(F_{l}\).

Torque about the left end, then F_net,y = 0

\[\begin{gathered} F_{r}L - Mg\tfrac{L}{4} - mg\tfrac{L}{2} = 0 \;\Rightarrow\; F_{r} = \tfrac{1}{4}Mg + \tfrac{1}{2}mg \approx 15.4\,\mathrm{N} \\ F_{l} + F_{r} = (M+m)g \;\Rightarrow\; F_{l} = (4.5)(9.8) - 15.4 \approx 28.7\,\mathrm{N} \end{gathered}\]

The heavier-loaded left scale carries more, as it should. Notice \(L\) cancelled completely — only the load fractions matter.

2 The leaning ladder — a firefighter climbs

Problem. A ladder of length \(L = 12\,\mathrm{m}\), mass \(m = 45\,\mathrm{kg}\), leans against a frictionless wall with its top at height \(h = 9.3\,\mathrm{m}\). A firefighter (\(M = 72\,\mathrm{kg}\)) stands halfway up; the ladder's com is one-third up. Find the forces from the wall and from the pavement.

Solution. The wall (frictionless) gives only a horizontal push \(F_{w}\); the pavement gives both a normal force \(F_{py}\) and friction \(F_{px}\). Put the axis at the foot (\(O\)) so the pavement forces vanish from the torque equation. The horizontal run is \(a = \sqrt{L^{2}-h^{2}} = 7.58\,\mathrm{m}\).

Torque about the foot, then the two force balances

\[\begin{gathered} (h)F_{w} - \tfrac{a}{2}Mg - \tfrac{a}{3}mg = 0 \;\Rightarrow\; F_{w} = \frac{ga\left(\tfrac{M}{2} + \tfrac{m}{3}\right)}{h} \approx 410\,\mathrm{N} \\ F_{px} = F_{w} \approx 410\,\mathrm{N}, \qquad F_{py} = (M+m)g \approx 1100\,\mathrm{N} \end{gathered}\]

The wall push and the floor friction are equal and opposite (\(F_{px} = F_{w}\)); the floor also bears the full weight vertically. Placing the axis at \(O\) turned three unknowns into one.

3 Stress, strain, and stretch of a steel rod

Problem. A steel rod (radius \(R = 9.5\,\mathrm{mm}\), length \(L = 0.81\,\mathrm{m}\)) is clamped at one end. A force \(F = 62\,\mathrm{kN}\) pulls perpendicularly across the free face. Find the stress, the elongation, and the strain.

Solution. Stress is \(F/A\); elongation follows from \(F/A = E\,\Delta L/L\) with \(E = 2.0 \times 10^{11}\,\mathrm{N/m^{2}}\); strain is \(\Delta L/L\).

Stress → elongation → strain

\[\begin{gathered} \frac{F}{A} = \frac{62000}{\pi(9.5\times 10^{-3})^{2}} \approx 2.2 \times 10^{8}\,\mathrm{N/m^{2}} \\ \Delta L = \frac{(F/A)L}{E} = \frac{(2.2\times 10^{8})(0.81)}{2.0\times 10^{11}} \approx 0.89\,\mathrm{mm} \\ \frac{\Delta L}{L} \approx 1.1 \times 10^{-3} = 0.11\% \end{gathered}\]

That stress sits dangerously close to steel's yield strength (≈\(2.5 \times 10^{8}\,\mathrm{N/m^{2}}\)) — the rod is near permanent deformation.

4 The wobbly table — statics meets elasticity

Problem. A table has three legs of length \(L = 1.00\,\mathrm{m}\) and one longer by \(d = 0.50\,\mathrm{mm}\), so it wobbles. A steel cylinder (\(M = 290\,\mathrm{kg}\)) is set on top; all four wooden legs (\(A = 1.0\,\mathrm{cm^{2}}\), \(E = 1.3 \times 10^{10}\,\mathrm{N/m^{2}}\)) compress until the top is level. Find the leg forces.

Solution. Statics alone leaves two unknowns (\(F_{3}\) on each short leg, \(F_{4}\) on the long one). The missing relation is elasticity: the long leg must compress by \(d\) more, so \(\Delta L_{4} - \Delta L_{3} = d\) with \(\Delta L = FL/AE\). Combine with vertical balance \(3F_{3} + F_{4} = Mg\).

Elastic compatibility + force balance

\[\begin{gathered} F_{3} = \frac{Mg}{4} - \frac{dAE}{4L} = \frac{(290)(9.8)}{4} - \frac{(5.0\times10^{-4})(10^{-4})(1.3\times10^{10})}{4(1.00)} \approx 548\,\mathrm{N} \\ F_{4} = Mg - 3F_{3} = (290)(9.8) - 3(548) \approx 1.2\,\mathrm{kN} \end{gathered}\]

The single long leg carries roughly twice the load of each short one — and reaching the level configuration compresses the short legs ≈0.42 mm and the long leg ≈0.92 mm.

5 Squeezing a copper cube — bulk modulus

Problem. A solid copper cube of edge \(85.5\,\mathrm{cm}\) is uniformly compressed until each edge becomes \(85.0\,\mathrm{cm}\). The bulk modulus of copper is \(B = 1.4 \times 10^{11}\,\mathrm{N/m^{2}}\). What hydraulic stress is required?

Solution. Use \(p = B\,\Delta V/V\) with the volume ratio computed from the edge ratio.

Volume strain, then pressure

\[\begin{gathered} \frac{\Delta V}{V} = 1 - \left(\frac{85.0}{85.5}\right)^{3} \approx 0.0174 \\ p = B\,\frac{\Delta V}{V} = (1.4\times 10^{11})(0.0174) \approx 2.4 \times 10^{9}\,\mathrm{N/m^{2}} \end{gathered}\]

A volume change of under 2% demands gigapascal pressures — a vivid measure of how stoutly a metal lattice resists being squeezed.

Review

Chapter Summary

✓ Equilibrium

Constant \(\vec{P}\) and constant \(\vec{L}\). Static equilibrium adds \(\vec{P} = 0\) — the body is at rest. Stability is set by how far the com must travel over a supporting edge.

✓ The three equations

\(F_{\text{net},x} = 0\), \(F_{\text{net},y} = 0\), \(\tau_{\text{net},z} = 0\). Torque balances about any axis — choose it through an unknown force to delete it.

✓ Center of gravity

Gravity acts effectively at the cog. If \(g\) is uniform over the body, \(x_{\text{cog}} = x_{\text{com}}\).

✓ Indeterminate structures

More unknowns than equations (e.g. a four-legged table). Statics alone fails; the resolution is elasticity — real bodies deform until loads are determined.

✓ Elastic moduli

\(\text{stress} = \text{modulus} \times \text{strain}\). Tension/compression: \(F/A = E\,\Delta L/L\). Shear: \(F/A = G\,\Delta x/L\). Hydraulic: \(p = B\,\Delta V/V\).

✓ Yield & rupture

Linear (recoverable) response below \(S_{y}\); permanent deformation beyond it; rupture at the ultimate strength \(S_{u}\).

Practice

Problems

Equilibrium problems reward a careful free-body diagram and a shrewd axis: write the two force balances and one torque balance, then solve. Elasticity problems reduce to the master relation \(\text{stress} = \text{modulus} \times \text{strain}\). Take \(g = 9.8\,\mathrm{m/s^{2}}\).

A uniform horizontal bar of weight 10 N hangs from a ceiling by two vertical wires. If one wire is attached at the bar's center and the other at one end, can the system be in static equilibrium? Explain using the balance of torques.
A uniform beam of weight 500 N and length 3.0 m is hinged to a wall at its left end and held horizontal by a cable bolted to the wall a distance D above the right end. The cable snaps at 1200 N. (a) Find the value of D that puts the cable exactly at that tension. (b) Should D be larger or smaller to keep the cable safe?
A 5.0 kg uniform rod 1.0 m long is held against a wall by a rope and by friction. If you must find the rope tension with a single equation, where should the rotation axis be placed, and why?
A uniform ladder of weight 400 N and length 5.0 m leans against a frictionless vertical wall; the coefficient of static friction at the floor is 0.46. What is the greatest distance the foot of the ladder can stand from the wall without slipping?
In the firefighter-ladder example, let the coefficient of static friction between ladder and pavement be 0.53. How far (as a percentage of the ladder's length) can the firefighter climb before the ladder is on the verge of sliding?
A door 2.1 m tall and 0.91 m wide (mass 27 kg) hangs on two hinges, one 0.30 m from the top and one 0.30 m from the bottom, each carrying half the weight. In unit-vector notation, find the force on the door at each hinge.
A horizontal aluminum rod 4.8 cm in diameter projects 5.3 cm from a wall; a 1200 kg object hangs from its end. With shear modulus \(G = 3.0 \times 10^{10}\,\mathrm{N/m^{2}}\), find (a) the shear stress and (b) the vertical deflection of the rod's end.
A vertical steel rod 1.0 m long and 1.0 cm in diameter (\(E = 2.0 \times 10^{11}\,\mathrm{N/m^{2}}\)) supports a 250 kg load hung from its lower end. Find (a) the stress, (b) the strain, and (c) the elongation.
A solid copper cube of edge 85.5 cm is to have its edge reduced to 85.0 cm by uniform pressure. With \(B = 1.4 \times 10^{11}\,\mathrm{N/m^{2}}\), what hydraulic stress is required?
A cubical crate (edge 1.2 m) has its center of mass 0.30 m above the geometric center. It rests on a ramp whose angle is slowly increased. With \(\mu_{s} = 0.60\) between crate and ramp, does the crate tip or slide first, and at what angle? (Hint: at the onset of tipping, where does the normal force act?)

Tip: for every equilibrium problem the same two decisions decide the difficulty — which body you isolate and which axis you balance torques about. Pick the axis so an unwanted unknown force has zero moment arm, and a three-unknown tangle usually collapses to a one-line equation. For elasticity, always ask first which of the three deformations you are looking at, then reach for the matching modulus.