Appearance
1.10 — Statics and Elasticity
A ladder leans against a wall with a person on it. Nothing moves. That sounds like the easiest possible mechanics problem — everything is zero — and it is where most students first discover that "not moving" is two separate conditions, not one, and that satisfying only the first one gives you a ladder that spins on the spot.
Then there is the other half. Every structure in the previous chapters was treated as perfectly rigid, which no real object is. Push hard enough on anything and it stretches, bends, and eventually breaks, and the second half of this chapter is about the numbers that decide when.
Two conditions, not one
A rigid body is in equilibrium when it has no linear acceleration and no angular acceleration. Those are two independent statements and each gives its own equation:
\boxed{\sum \vec{F} = 0} \qquad\text{and}\qquad \boxed{\sum \vec{\tau} = 0}
The first is Newton's second law with a = 0. The second is Chapter 1.8's rotational law with \alpha = 0.
You need both because they are genuinely independent. Take a rod and push its two ends in opposite directions with equal force. The forces sum to zero, so the centre of mass does not move — but the torques add rather than cancel, and the rod spins. That arrangement is called a couple, and it is the standard proof that force balance alone is not equilibrium.
In two dimensions this gives three scalar equations:
\sum F_x = 0, \qquad \sum F_y = 0, \qquad \sum \tau = 0
Three equations means you can solve for at most three unknowns. If a problem has more unknown forces than that, it is statically indeterminate — a four-legged table is the everyday example, and you genuinely cannot find the four leg forces from statics alone. You need to know how much the legs and floor deform, which is the second half of this chapter.
The torque axis is yours to choose, and choosing well is the whole skill
Here is the piece that makes these problems easy. If a body is in equilibrium, the net torque is zero about every point, not just about its actual pivot. So you may take torques about any point you like.
Why that is true: suppose the net torque is zero about point O. Take any other point P, displaced from O by \vec{d}. The torque about P is
\sum(\vec{r}_i - \vec{d})\times\vec{F}_i = \sum\vec{r}_i\times\vec{F}_i - \vec{d}\times\sum\vec{F}_i
The first term is the torque about O, which is zero. The second contains \sum\vec{F}_i, which is zero by the first equilibrium condition. So the torque about P is zero too.
The practical consequence: choose the axis so that an unknown force you do not care about passes straight through it. A force whose line of action goes through the axis has zero moment arm and therefore zero torque, so it drops out of the equation entirely. One well-chosen axis routinely turns a three-unknown simultaneous system into a single equation with one unknown.
Worked example: the ladder
A uniform ladder of mass 20 kg and length 5.0 m leans against a smooth vertical wall at 60° to the ground. A 70 kg person stands 3.5 m up the ladder. The ground is rough, the wall is frictionless. Find all three forces on the ladder, and the minimum coefficient of friction needed to stop it slipping.
Set up. Forces on the ladder:
- Its weight, W_L = 20\times9.81 = 196\ \text{N}, acting at its centre, 2.5 m along, because it is uniform.
- The person's weight, W_P = 70\times9.81 = 687\ \text{N}, acting 3.5 m along.
- The wall's normal force N_W, horizontal (perpendicular to the wall), pushing the ladder away from the wall. Frictionless wall means no vertical force there.
- The ground's normal force N_G, vertical.
- The ground's friction f, horizontal, pointing towards the wall (opposing the ladder's tendency to slide out).
Vertical force balance. The wall contributes nothing vertical:
N_G = W_L + W_P = 196 + 687 = 883\ \text{N}
Horizontal force balance:
f = N_W
Two equations, three unknowns. The third comes from torque.
Torque, taken about the foot of the ladder — chosen precisely because both N_G and f act there and therefore contribute nothing.
The horizontal distance from the foot to a point d along the ladder is d\cos 60° = 0.5d; the vertical height is d\sin 60° = 0.866d.
Clockwise torques (trying to rotate the ladder down and out), taken as negative:
- Ladder's weight: 196 \times (2.5\times0.5) = 196\times1.25 = 245\ \text{N·m}
- Person's weight: 687 \times (3.5\times0.5) = 687\times1.75 = 1202\ \text{N·m}
Anticlockwise torque, from the wall pushing at the top of the ladder, whose height is 5.0\times0.866 = 4.33 m:
- N_W \times 4.33
Setting the sum to zero:
N_W \times 4.33 = 245 + 1202 = 1447
N_W = \frac{1447}{4.33} = 334\ \text{N}
And therefore f = 334\ \text{N}.
The minimum coefficient of friction. Slipping is avoided as long as f \le \mu_s N_G:
\mu_s \ge \frac{334}{883} = 0.378
Now change one thing and see the physics. Move the person to the very top, d = 5.0 m:
N_W = \frac{245 + 687\times2.5}{4.33} = \frac{245+1718}{4.33} = 453\ \text{N} \quad\Longrightarrow\quad \mu_s \ge \frac{453}{883} = 0.513
Climbing higher makes slipping more likely, sharply. And steepening the ladder helps: at 75° the horizontal lever arms shrink and the vertical one grows, and the required \mu_s drops to about 0.2. That is why the standard safety rule is one unit out for every four units up — it puts the ladder at about 76°.
Centre of gravity and stability
The centre of gravity is the point where the whole weight can be taken to act. In a uniform gravitational field it coincides exactly with the centre of mass from Chapter 1.7, and for everyday purposes the two words are interchangeable.
An object standing on a base tips over when its centre of gravity passes outside the base of support — the area enclosed by its contact points. Up to that moment, gravity's torque about the tipping edge pushes it back; past it, gravity's torque pushes it over.
That single rule explains a great deal:
- A double-decker bus is tested by tilting it on a platform. It must not tip below 28°, which is what limits how high the upper deck can be.
- A rally car takes corners with the inside wheels lifting because the cornering force is shifting the effective line of gravity-plus-inertia towards the outer wheels.
- Standing up from a chair without leaning forward is impossible. Your centre of gravity starts behind your feet, outside the base of support, and no amount of leg strength can fix a torque problem. Try it with your back against a wall.
- A forklift carries its load low and has a heavy counterweight at the back, because the moment the combined centre of gravity crosses the front axle it tips forwards.
Structures: why a triangle is the only shape that holds
Take four rods pinned at the corners into a square. Push sideways and it collapses into a rhombus without any rod changing length. Now take three rods pinned into a triangle. To deform it at all, at least one rod must get longer or shorter — and rods are enormously stiff along their length. The triangle is rigid by geometry, not by the strength of its joints.
That is why every truss bridge, roof, pylon, crane and bicycle frame is made of triangles. It is also why a square gate needs a diagonal brace, and why the brace must run from the bottom hinge corner upwards: in that orientation the brace is in compression, which timber handles well, rather than tension, which its fixings do not.
Members in a structure carry two kinds of load:
- Tension — being pulled apart. Cables, ropes and chains carry only this, and it is why a suspension bridge's main cables can be thin.
- Compression — being squeezed. Stone, brick and concrete are excellent in compression and nearly useless in tension, which is the entire reason for the arch: an arch converts a downward load into compression all the way to the ground.
Reinforced concrete exists because concrete is strong in compression and weak in tension, while steel is strong in both. Casting steel bars into the parts of a beam that will be in tension gives you a material that is strong in both directions, for the price of concrete. Which parts are in tension is a torque question: in a simply supported beam, the bottom stretches and the top squeezes, so the steel goes near the bottom. In a cantilever it is the other way round, and getting that backwards has collapsed real buildings.
Elasticity: what happens when the body is not rigid
Everything above assumed nothing deforms. Now drop that.
Stress and strain
Stress is force per unit area:
\sigma = \frac{F}{A}
Units are pascals, the same as pressure, and for the same reason. Stress is what a material actually feels — a 1000 N load on a thick cable is nothing and on a thread is fatal, and stress is the quantity that knows the difference.
Strain is the fractional change in length:
\varepsilon = \frac{\Delta L}{L_0}
It is dimensionless. A 2 m rod stretched by 1 mm has a strain of 0.0005.
Young's modulus
For small deformations, stress and strain are proportional. That proportionality is Hooke's law in its material form, and the constant is Young's modulus:
\boxed{E = \frac{\sigma}{\varepsilon} = \frac{F/A}{\Delta L/L_0}}
E is a property of the material, not of the object. Steel has E = 200 GPa whatever shape it is in. Rearranged to give the stretch of a particular object:
\Delta L = \frac{FL_0}{AE}
Which reads: longer things stretch more, thicker things stretch less, stiffer materials stretch less. All three of those are what you would guess, which is a good sign.
Comparing this with the spring law F = kx from Chapter 1.6 shows what a spring constant really is:
k = \frac{AE}{L_0}
A spring constant is not fundamental. It is a material property mixed with a shape.
| Material | Young's modulus E (GPa) |
|---|---|
| Rubber | 0.01–0.1 |
| Wood (along grain) | 10 |
| Bone | 15 |
| Concrete | 30 |
| Aluminium | 70 |
| Steel | 200 |
| Diamond | 1200 |
| Carbon nanotube | ~1000 |
Steel is about 20 000 times stiffer than rubber. Note that stiffness and strength are different properties: cast iron is stiffer than steel in some respects and far more likely to shatter, because stiffness is about how much it moves under load and strength is about when it fails.
Reading the stress–strain curve
Pull a metal bar steadily and plot stress against strain. Five features matter, and every one of them is an engineering decision somewhere.
The elastic region. A straight line. Stress is proportional to strain, its slope is Young's modulus, and if you release the load the bar returns exactly to its original length. Everything in normal service lives here.
The yield point. Where the line stops being straight. Beyond this, deformation is permanent — release the load and the bar stays longer than it started. This is the number that structural design is actually built around, because a bridge that has yielded is ruined even if it has not broken.
The plastic region. The bar stretches a great deal for very little extra stress. Metals can extend by tens of percent here, and this is what makes them formable: every pressed car body panel and every drawn wire is a metal that has been deliberately taken past its yield point.
Ultimate tensile strength. The peak of the curve, the highest stress the material ever carries. After this the bar starts to neck — thin locally at one spot. Because stress is force over area and the area is now shrinking, that spot fails faster, which is why necking runs away once it starts.
Fracture. It breaks.
The area under the curve is the energy absorbed per unit volume before fracture, and that quantity is toughness. It is not the same as strength, and the difference is why glass is a bad structural material: glass is genuinely strong, with a high peak, but its curve is a straight line that stops dead with no plastic region at all, so the area under it is tiny. It absorbs almost no energy before failing. Mild steel is weaker than glass in some measures and vastly tougher, because it deforms enormously before letting go — and that deformation is a warning, which glass never gives.
That difference has a name: ductile materials (steel, copper, aluminium) stretch and warn before failing; brittle ones (glass, cast iron, concrete, ceramic) fail suddenly. Every safety-critical structure is designed to fail ductilely if it fails at all, so that it sags visibly before it collapses.
Safety factors, and why they are large
Engineers do not design to the yield stress. They divide it by a safety factor:
\sigma_{\text{allowed}} = \frac{\sigma_{\text{yield}}}{n}
Typical values: n \approx 1.5 for aircraft, where every kilogram costs fuel and the material is tested to death; n \approx 2 to 3 for buildings; n \approx 8 or more for lift cables, where a failure kills people instantly and inspection is imperfect.
The factor is not covering ignorance of the physics. It is covering variation in the material, corrosion, manufacturing flaws, loads nobody predicted, and fatigue — the fact that a metal cycled repeatedly at a stress far below its yield point will eventually crack anyway. Fatigue is what brought down the de Havilland Comet, the first jet airliner, in 1954: repeated pressurisation cycles grew cracks from the corners of its square windows. Aircraft windows have been round ever since, because a rounded corner spreads the stress that a sharp corner concentrates.
The other two moduli
Young's modulus is for stretching. Two companions cover the other ways to deform something.
Shear modulus G describes resistance to a sideways sliding deformation — the top face pushed one way while the bottom stays put. It is what a bolt resists, and what fails when a bolt is sheared off. For most metals G \approx 0.4E.
Bulk modulus K describes resistance to being squeezed from all sides at once:
K = -V\frac{dP}{dV}
The minus sign is because increasing pressure decreases volume. Water has K = 2.2 GPa, which is why hydraulic systems work: at the bottom of the Mariana Trench, under 1100 atmospheres, seawater is compressed by only about 5%. Liquids are, for practical purposes, incompressible, and the next chapter leans on that constantly.
Beams: why an I-beam is that shape
Bend a beam and the material on one side stretches while the other side compresses. Somewhere in the middle is a surface that does neither, called the neutral axis, and material sitting on it is doing nothing at all.
The stiffness of a beam in bending depends on the second moment of area, which is \int y^2\,dA — the same r^2 weighting as the moment of inertia in Chapter 1.8, and for the same reason. Material far from the neutral axis contributes as the square of its distance.
So the efficient shape puts as much material as possible far from the middle and as little as possible near it, joined by just enough web to stop the two halves sliding relative to each other. That is an I-beam, and it is why an I-beam is stiffer than a solid rectangular bar of the same weight.
The same principle explains a hollow bicycle frame, a bird's bone, a bamboo stem and a drinking straw. A tube is a beam with all its material at maximum distance from the axis, and it is stiffer per kilogram than any solid rod. It is also why an I-beam turned on its side is dramatically weaker — the material is now near the neutral axis and its y^2 has collapsed.
Where this shows up in your life
Bone is a living stress-optimised structure. It remodels continuously, adding material where stress is high and removing it where stress is low, which is called Wolff's law. This is why weight-bearing exercise builds bone density and why astronauts lose 1–2% of their bone mass per month in orbit, where there is no stress signal telling the bone it is needed. It is also why a broken bone heals thicker than it was.
Every crane has a load chart rather than a load limit, because the limit depends on the radius. Torque is force times distance, so a crane rated at 20 tonnes at 5 m might manage only 5 tonnes at 20 m. Crane accidents are almost always someone using the tonnage without reading the radius.
And the reason you tighten wheel nuts to a specified torque rather than as hard as you can is that the bolt is a spring. The torque figure is really a specification for how far to stretch the bolt into its elastic region, so it clamps hard and stays clamped. Under-tighten and the joint works loose; over-tighten and you take the bolt past its yield point, after which it stays stretched, loses its clamping force, and fails on the road.
What the next chapter fixes
Solids hold their shape and can be pushed on at a point. Liquids and gases cannot: they flow, they push on every surface they touch, and the force they exert depends on depth and on how fast they are moving. Chapter 1.11 derives the pressure-depth law, proves Archimedes' principle rather than quoting it, gets Bernoulli's equation from energy conservation, and settles the question of why an aeroplane wing actually generates lift — which is not the reason most books give.