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Equilibrium & Elasticity

Source lecture(s): SC133 Lec 15

Intuition

Bridges, cranes, bookshelves, ladders against walls: most of engineering is things that don't move. Statics is Newton's laws in the special case \(\vec a = 0\) and \(\alpha = 0\) — but with a twist: for extended bodies you must balance not just forces but torques, or your bookshelf rotates off the wall. And because nothing is perfectly rigid, "not moving" really means "deforming slightly and pushing back" — elasticity.

The two conditions of static equilibrium

\[\boxed{\;\sum \vec F = 0 \qquad\text{and}\qquad \sum \vec\tau = 0\;\text{(about any axis)}\;}\]

The freedom to choose the torque axis is the problem-solver's best weapon: put the axis where the most annoying unknown force acts, and that force vanishes from the torque equation.

Worked example: the leaning ladder

A uniform ladder (mass \(m\), length \(L\)) leans at angle \(\theta\) against a frictionless wall; floor friction coefficient \(\mu_s\). When does it slip?

Forces: weight \(mg\) at the center, wall normal \(N_w\) (horizontal), floor normal \(N_f\) and friction \(f\).

  • \(\sum F_y = 0\): \(N_f = mg\)
  • \(\sum F_x = 0\): \(f = N_w\)
  • Torque about the floor contact (killing \(N_f\) and \(f\)): \(N_w L\sin\theta = mg\frac{L}{2}\cos\theta \Rightarrow N_w = \frac{mg}{2\tan\theta}\)

Slipping when \(f > \mu_s N_f\):

\[\tan\theta_\text{min} = \frac{1}{2\mu_s}\]

Steeper is safer; a slick floor (\(\mu_s \to 0\)) makes every angle unsafe — exactly as intuition (and ladder-safety posters) insist.

Elasticity: stress and strain

Real "rigid" bodies are stiff springs. Define

\[\text{stress} = \frac{F}{A}\ \text{(Pa)} \qquad \text{strain} = \frac{\Delta L}{L}\ \text{(dimensionless)}\]

For modest loads they are proportional — Hooke's law in grown-up units:

\[\frac{F}{A} = E\,\frac{\Delta L}{L}\]

with Young's modulus \(E\) a material property (steel \(\approx 200\) GPa, bone \(\approx 15\) GPa, rubber \(\approx 0.01\) GPa). Analogous moduli handle shear (shear modulus \(G\)) and squeezing (bulk modulus \(B\)). Push past the elastic limit and deformation becomes permanent; past ultimate strength, it becomes news.

Worked example: stretching a cable

A 2 m steel cable, cross-section \(1\,\text{cm}^2\), hangs a 500 kg load. Stretch?

\[\Delta L = \frac{FL}{AE} = \frac{4900 \times 2}{10^{-4}\times 2\times10^{11}} \approx 0.5\,\text{mm}\]

Stiff — but not zero. Skyscrapers, bridges and bones all live on this small-but-finite give; the stress–strain language also marks the exact boundary between solids (sustain shear) and fluids (don't).

Stability of equilibrium

Balanced is not the same as stable: displace slightly and ask what gravity does. Center of mass above the support polygon = balanced; CM falling when tipped = stable. Lower CM and wider base ⇒ harder to topple — why racing cars are low and wide, and why you widen your stance on a bus. (The energy view — valleys vs hilltops of \(U\) — is on the potential energy page, and the full dynamical treatment awaits in hydrodynamic stability.)

Common mistakes

  • Balancing forces but not torques. Both conditions are mandatory; a couple (equal, opposite, offset forces) has zero net force yet spins the body.
  • Choosing a torque axis and then forgetting forces at that axis still exist — they're absent from the torque equation only.
  • Confusing stress with force and strain with stretch — both are normalized, so material properties don't depend on specimen size.
  • Treating the elastic limit as failure. Permanent bending comes first; fracture later — different design margins.

Knowledge graph position

Prerequisites: Newton's laws, Torque. Leads to: Fluids (the no-shear limit), structural mechanics beyond this course.

Quiz

Q1 (computational). A 60 kg person stands 1.5 m from the left end of a uniform 4 m, 20 kg plank supported at both ends. Force on each support?

Answer

Torque about the left support: \(N_R(4) = 20g(2) + 60g(1.5) \Rightarrow N_R = \frac{40g + 90g}{4} = 32.5g \approx 319\,\text{N}\). Then \(N_L = 80g - N_R = 47.5g \approx 466\,\text{N}\) — the nearer support carries more, as it must.

Q2 (conceptual). Why does a crane need a massive counterweight?

Answer

The load creates a huge torque about the tower; the counterweight supplies an opposing torque so the net torque at the base stays small. Force balance alone would be satisfied by the tower pushing up — it's the torque condition that demands the counterweight.

Q3 (multiple choice). Two wires of the same material and length; wire B has twice the diameter. Under the same load, B stretches: (a) half as much (b) a quarter as much (c) the same

Answer

(b). \(\Delta L \propto 1/A \propto 1/d^2\) — doubling diameter quadruples the area.